Science, Irving K. Barber Faculty of (Okanagan)
Computer Science, Mathematics, Physics and Statistics, Department of (Okanagan)
DSpace
UBCO
Alwadani, Salihah Thabet
Doctor of Philosophy - PhD
University of British Columbia
Finding a zero of the sum of two maximally monotone operators is of fundamental importance in variational analysis and optimization. There are various algorithms for solving this problem such as the Douglas–Rachford splitting algorithm (DR) and Arag ́on Artacho-Campoy algorithm (AACA), which is a new method in this area. Moreover, it is well known that there is a connection between the set of zeros of maximally monotone operators and the fixed point set of composition of nonexpansive operators. In this thesis we will explore the connection between algorithms and fixed point sets. This thesis consists of two parts. In the first part of this thesis, we advance the analysis of AACA. We use the proximal and resolvent average in our analysis. We re-derive the central convergence result on AACA and demonstrate that the underlying curve converges to the nearest zero of the sum of the two operators. Using the Attouch-Th ́era duality, we study the cycles, gap vectors and fixed point sets of compositions of general nonexpansive mappings in the second part. We also use the framework of monotone operator theory to resolve the geometry conjecture completely, which states that the fixed point set of the composition of projectors onto nonempty closed convex sets in Hilbert space is actually equal to the intersection of certain translations of the underlying sets. Then we extend the results from projectors to proximal mappings.
On the behaviour of algorithmsfeaturing compositions of projectorsand proximal mappings with nosolutionsbySalihah Thabet AlwadaniB.Sc. Hons., Umm Al-Qura University, 2013M.Sc., Acadia University, 2017A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinTHE COLLEGE OF GRADUATE STUDIES(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Okanagan)November 2021c© Salihah Thabet Alwadani, 2021AbstractFinding a zero of the sum of two maximally monotone operators is offundamental importance in variational analysis and optimization. There arevarious algorithms for solving this problem such as the Douglas–Rachfordsplitting algorithm (DR) and Arago´n Artacho-Campoy algorithm (AACA),which is a new method in this area. Moreover, it is well known that thereis a connection between the set of zeros of maximally monotone operatorsand the fixed point set of composition of nonexpansive operators. In thisthesis we will explore the connection between algorithms and fixed pointsets. This thesis consists of two parts.In the first part of this thesis, we advance the analysis of AACA. Weuse the proximal and resolvent average in our analysis. We re-derive thecentral convergence result on AACA and demonstrate that the underlyingcurve converges to the nearest zero of the sum of the two operators.Using the Attouch-The´ra duality, we study the cycles, gap vectors andfixed point sets of compositions of general nonexpansive mappings in thesecond part. We also use the framework of monotone operator theory toresolve the geometry conjecture completely, which states that the fixed pointset of the composition of projectors onto nonempty closed convex sets inHilbert space is actually equal to the intersection of certain translations ofthe underlying sets. Then we extend the results from projectors to proximalmappings.iiLay SummaryOptimization and feasibility problems are important in mathematics andthe natural sciences. These are solved by algorithms featuring compositionsof projection operators and proximal mappings. In this thesis, I will providean analysis of the behaviour of some of these algorithms in the case whenthere is no solution.iiiPrefaceThis thesis is based on the following papers:Refereed Journal PublicationsJ1. Salihah Alwadani, Heinz Bauschke, Walaa Moursi, and Xianfu Wang,“On the asymmetric behaviour of the Arago´n Artacho-Campoy algo-rithm,” Operation Research Letters, vol. 46, no. 6, pp. 585–587, 2018.https://doi.org/10.1016/j.orl.2018.10.003. This paper formsthe basis of Chapter 4.J2. Salihah Alwadani, Heinz Bauschke, and Xianfu Wang, “Fixed pointsof compositions of nonexpansive mappings: Finitely many linear re-flectors, to appear in Pure and Applied Functional Analysis. http://arXiv.org/abs/2004.12582. This paper forms the basis of Chap-ter 6.J3. Salihah Alwadani, Heinz Bauschke, Julian Revalski, and Xianfu Wang,“Resolvents and Yosida Approximations of Displacement Mappings ofIsometries,” Set-Valued and Variational Analysis, vol. 29, pp. 721–733, 2021. https://doi.org/10.1007/s11228-021-00584-2. Thispaper forms the basis of Chapter 7.J4. Salihah Alwadani, Heinz Bauschke, Julian Revalski, and Xianfu Wang,“The difference vectors for convex sets and a resolution of the ge-ometry conjecture,” Open Journal of Mathematical Optimization andSet-Valued and Variational Analysis, vol. 2, pp. 1–18, 2021. https://doi.org/10.5802/ojmo.7. It forms the basis of Chapter 5, Chap-ter 8, Chapter 9 and Chapter 10.J5. Salihah Alwadani, Heinz Bauschke, and Xianfu Wang, “Attouch-The´raDuality, Generalized Cycles and Gap Vectors,” SIAM Journal on Op-timization, vol. 31, no. 3, pp. 1926–1946, 2021. https://doi.org/10.1137/21M1392085. This paper form the basis of Chapter 8, Chap-ter 11, and Chapter 12.ivTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . vList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiGlossary of Notation and Symbols . . . . . . . . . . . . . . . . xiiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . xvDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xviChapter 1: Introduction . . . . . . . . . . . . . . . . . . . . . . . 11.1 The goal of the thesis . . . . . . . . . . . . . . . . . . . . . . 11.2 Contributions in this thesis . . . . . . . . . . . . . . . . . . . 2Chapter 2: Preliminary Details . . . . . . . . . . . . . . . . . . 52.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3.1 Single-valued operators . . . . . . . . . . . . . . . . . 62.3.2 Set-valued operators . . . . . . . . . . . . . . . . . . . 72.4 Convex analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 72.5 Convex functions . . . . . . . . . . . . . . . . . . . . . . . . . 102.6 Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15vTABLE OF CONTENTSChapter 3: Monotonicity and Firm Nonexpansiveness . . . . 173.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Firmly nonexpansiveness operators . . . . . . . . . . . . . . . 173.3 Monotone operator theory . . . . . . . . . . . . . . . . . . . . 203.4 Attouch-The´ra duality . . . . . . . . . . . . . . . . . . . . . . 263.5 Douglas-Rachford and Arago´n Artacho-Campoy algorithms . 27Chapter 4: On the asymptotic behaviour of AACA . . . . . . 334.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.2 Resolvent and the reflected resolvent of Cγ . . . . . . . . . . 344.3 Properties of the operator Cγ . . . . . . . . . . . . . . . . . . 374.4 Properties of the operator Aγ +Bγ . . . . . . . . . . . . . . . 424.5 The Arago´n Artacho-Campoy algorithm (AACA) . . . . . . . 464.6 Original AACA with fixed γ . . . . . . . . . . . . . . . . . . . 494.7 The behaviour of AACA when γ → 1− . . . . . . . . . . . . . 504.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Chapter 5: Fixed Point Sets of Compositions of Projectorsand the Magic Difference Vectors . . . . . . . . . . 665.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 665.2 Reviewing the case when S 6= ∅ . . . . . . . . . . . . . . . . 675.3 Cycles and the fixed point sets of the compositions of projec-tions mappings . . . . . . . . . . . . . . . . . . . . . . . . . . 675.4 When m = 2 with empty intersection . . . . . . . . . . . . . . 705.5 When m ≥ 3 with empty intersection . . . . . . . . . . . . . . 735.6 Cycles and difference vectors . . . . . . . . . . . . . . . . . . 74Chapter 6: The Composition of Reflection Operators . . . . . 776.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 776.2 The composition of two reflectors and its fixed point set . . . 786.3 The composition of (three or more) reflectors and its fixedpoint set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806.4 The Euclidean plane R2 . . . . . . . . . . . . . . . . . . . . . 866.4.1 The composition of reflectors and rotators . . . . . . . 866.4.2 The fixed point sets of compositions of classical reflec-tors on R2 . . . . . . . . . . . . . . . . . . . . . . . . . 876.5 R3 and beyond . . . . . . . . . . . . . . . . . . . . . . . . . . 92Chapter 7: Resolvents and Yosida Approximation of Displace-ment Mappings of Isometries . . . . . . . . . . . . 93viTABLE OF CONTENTS7.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 937.2 Properties of operator M and γM . . . . . . . . . . . . . . . . 947.3 The resolvent of γM and its asymptotic behaviour . . . . . . 987.4 Yosida approximations . . . . . . . . . . . . . . . . . . . . . . 1047.5 Properties of T operator . . . . . . . . . . . . . . . . . . . . . 1067.6 The set valued inverse M−1 . . . . . . . . . . . . . . . . . . . 1107.7 Moore-Penrose inverse M † . . . . . . . . . . . . . . . . . . . . 1137.8 Rational rotation Example . . . . . . . . . . . . . . . . . . . . 1147.9 The right shift operator R . . . . . . . . . . . . . . . . . . . . 1157.10 Cyclic right-shift operator R and 12Id + T operator . . . . . . 1157.11 Cyclic right-shift operator example . . . . . . . . . . . . . . . 118Chapter 8: More Convex Analysis . . . . . . . . . . . . . . . . 1218.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1218.2 Properties of σC+∆ . . . . . . . . . . . . . . . . . . . . . . . . 1228.3 Properties of ιC and its subdifferential . . . . . . . . . . . . . 1248.4 Properties of ιCι∆,(ιCι∆)∗, (ι∗Cι∗∆)∗ and their subdif-ferentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1258.5 Visiting well-known results for the separable sum f . . . . . . 1288.6 Properties of f∗ + ι∗∆, fι∆, cl(fι∆), ∂(fι∆), as well as∂[cl(fι∆)]. . . . . . . . . . . . . . . . . . . . . . . . . . . 1298.7 Properties of subdifferentials . . . . . . . . . . . . . . . . . . 1328.8 When is fι∆ lower semicontinuous and exact? . . . . . . . . 136Chapter 9: Attouch-The´ra Duality, Fixed Point Sets of Com-position of Projectors, and the Full GeometryConjecture . . . . . . . . . . . . . . . . . . . . . . . . 1429.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1429.2 The fixed point sets of the compositions of projectors and thecycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1449.3 The relationship among Z and Fi’s . . . . . . . . . . . . . . . 1469.4 Enlarged dual problem and the general gap vector . . . . . . 1499.5 A characterization of y, e and the sets of the cycles Z . . . . 1509.6 The full geometry conjecture is true . . . . . . . . . . . . . . 1549.7 Exploring the m = 2 case and visiting known results . . . . . 1569.8 Exploring the m = 3 case . . . . . . . . . . . . . . . . . . . . 1619.8.1 Three lines . . . . . . . . . . . . . . . . . . . . . . . . 1629.8.2 The epigraph of the exponential function and two hor-izontal lines . . . . . . . . . . . . . . . . . . . . . . . . 167viiTABLE OF CONTENTS9.9 Finding the difference vectors for m ≤ 5 using Banach con-traction principle . . . . . . . . . . . . . . . . . . . . . . . . . 1739.10 Finding the difference vectors by forward-backward . . . . . . 174Chapter 10:Projection mappings, Generalized Cycles, and GapVectors . . . . . . . . . . . . . . . . . . . . . . . . . . 17810.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17810.2 Extending the primal approach . . . . . . . . . . . . . . . . . 17910.3 The Relationship between the generalized cycles and the gen-eralized gap vectors . . . . . . . . . . . . . . . . . . . . . . . . 18010.4 The Relationship among the classical cycles, generalized cy-cles and the generalized gap vectors . . . . . . . . . . . . . . 18110.5 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183Chapter 11:Attouch-The´ra Duality and Fixed Point Sets ofCompositions of Proximal Mappings . . . . . . . . 18811.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18811.2 Important setup! . . . . . . . . . . . . . . . . . . . . . . . . . 18911.3 The fixed point sets of the compositions of the proximal map-pings and the set of cycles . . . . . . . . . . . . . . . . . . . . 19111.4 Classical cycles, gap vectors, and the set of all the cycles . . . 19111.5 The fixed point sets of cyclic proximal mappings . . . . . . . 19311.6 The generalized gap vectors . . . . . . . . . . . . . . . . . . . 197Chapter 12:Proximal Mappings, Generalized Cycles, and GapVectors . . . . . . . . . . . . . . . . . . . . . . . . . . 19912.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19912.2 The generalized cycles . . . . . . . . . . . . . . . . . . . . . . 20012.3 The primal-dual approach . . . . . . . . . . . . . . . . . . . . 20112.4 The relationship between classical cycles and generalized cycles20212.5 The relationship among the classical cycles, generalized cyclesand the generalized gap vectors . . . . . . . . . . . . . . . . . 20512.6 More on generalized cycles . . . . . . . . . . . . . . . . . . . . 20812.7 Finding the difference vectors by forward-backward . . . . . . 20912.8 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212Chapter 13:On the Approximation Curve . . . . . . . . . . . . 21513.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21513.2 Useful properties and the composition JγM−1JγN−1C . . . . . . 21613.3 Finding the approximation curve for m = 2 . . . . . . . . . . 220viiiTABLE OF CONTENTS13.4 Examples on m = 2 case . . . . . . . . . . . . . . . . . . . . . 22213.4.1 Two intervals . . . . . . . . . . . . . . . . . . . . . . . 22213.4.2 Linear subspace and single set . . . . . . . . . . . . . 24113.5 Finding the approximation curve for m = 3 . . . . . . . . . . 24313.6 Examples on m = 3 case . . . . . . . . . . . . . . . . . . . . 24513.6.1 One interval . . . . . . . . . . . . . . . . . . . . . . . . 245Chapter 14:Conclusion . . . . . . . . . . . . . . . . . . . . . . . . 25414.1 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25414.2 Further results . . . . . . . . . . . . . . . . . . . . . . . . . . 25514.3 Future work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261Appendix A: Matlab code for Example 5.9 . . . . . . . . . . . . . . 261Appendix B: Mathematica code . . . . . . . . . . . . . . . . . . . . 267B.1 Code to verify Example 7.12 . . . . . . . . . . . . . . . . . . . 267B.2 Code to verify Example 7.14 . . . . . . . . . . . . . . . . . . . 269B.3 Code to verify Section 9.9 . . . . . . . . . . . . . . . . . . . . 271Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274ixList of TablesTable 7.1 T operator with the right shift operator and Rationalrotation operator (see Example 7.12 and Example 7.14for more details). . . . . . . . . . . . . . . . . . . . . . 110Table 9.1 Computing the operator norm of 12Id−T. . . . . . . . 174xList of FiguresFigure 5.1 When S = C1 ∩ C2 ∩ · · · ∩ Cm 6= ∅, then all fixedpoint sets Fi must coincide with S. . . . . . . . . . . 68Figure 5.2 An illustration for the case when m = 2 and δ isattained. . . . . . . . . . . . . . . . . . . . . . . . . . 71Figure 5.3 An illustration for the case when the infimum is notattained. . . . . . . . . . . . . . . . . . . . . . . . . . 72Figure 5.4 An illustration for the sets F1 and F2. . . . . . . . . . 72Figure 5.5 An illustration for Example 5.8 with x0 =(− 4, 1). . 73Figure 5.6 An illustration for Example 5.9 with x0 =(− 3, 5, 2.9). 75Figure 6.1 The fixed point sets for Example 6.15 . . . . . . . . . 89Figure 9.1 Visualization the cycle and the difference vectors forthree lines in R3 that are parallel. See Proposition 9.17.165Figure 9.2 Visualization of the cycle and the difference vectorsfor three lines in R3 that are not parallel. See Propo-sition 9.18. . . . . . . . . . . . . . . . . . . . . . . . . 167Figure 9.3 Visualization of the cycle and the difference vectorsfor three lines in plane that are not parallel. SeeProposition 9.18. . . . . . . . . . . . . . . . . . . . . . 168Figure 9.4 Visualization of the cycle and the difference vectorsfor three lines in plane that are not parallel. See Sub-section 9.8.2. . . . . . . . . . . . . . . . . . . . . . . . 171xiGlossary of Notation andSymbolsN p. 5 The set of natural numbers {0, 1, 2, 3, . . .}.Z p. 5 The set of integers {. . . ,−2,−1, 0, 1, 2, . . .}.R p. 5 The set of real numbers.R+ p. 5 The set of positive real numbers [0,+∞[.R++ p. 5 The set of strictly positive real numbers]0,+∞[.R− p. 5 The set of negative real numbers ]−∞, 0].R−− p. 5 The set of strictly negative real numbers]−∞, 0[.H p. 5 A real Hilbert space.Hm p. 5 A Hilbert product space.xn → x p. 5 Strong convergence in a Hilbert space.xn ⇀ x p. 5 Weak convergence in a Hilbert space.Pµ(f ,λ)p. 15 Proximal average.Rµ(A,λ)p. 25 Resolvent average.‖x‖ p. 5 The norm of x〈·, ·〉 p. 5 Inner productId p. 6 Identity mappingSets:int C p. 8 Interior of a set CB(H) p. 6 The set of continuous linear mappings from Hto H.Γ0(H) p. 10 The set of all proper convex lower semicontin-uous functions defined from H to ]−∞,+∞].C p. 8 Closure of a set Cint C p. 8 Interior of a set Cint C p. 8 Interior of a set Cri C p. 8 Relative interior of a set CxiiGlossary of Notation and Symbolssri C p. 8 Strong relative interior of a set Cconv C p. 7 Convex hull of a set Caff C p. 8 Affine hull of a set Ccone C p. 8 Conic hull of a set Cspan C p. 8 Linear span of a set CC⊥ p. 8 Orthogonal complement of a set CPC x p. 8 The projection of x onto the set CFunctions:dom f p. 10 Domain of a function f .epif p. 10 Epigraph of a function f .ιC p. 10 Indicator function of a set C.fg p. 10 Infimal convolution of the functions f and g.∂f p. 11 Subdifferential of a function f .f∗ p. 11 Conjugate of a function f .f∗∗ p. 11 Biconjugate of a function f .Proxf p. 12 Proximal mapping of a function of f .σC p. 12 Support function of a set C.Set-valued operators:A : H⇒ H p. 7 A is a set-valued operator from H to H.A−1 p. 7 Inverse of an operator A.domA p. 7 Domain of an operator A.ranA p. 7 Range of an operator A.graA p. 7 Graph of an operator A.zerA p. 7 Set of zeros of an operator A.JA p. 22 Resolvent of an operator A.RA p. 22 Reflected resolvent of an operator A.A> p. 26 The reversal of an operator A.A−> p. 26 The reversal of the inverse of an operator A.λA p. 24 Yosida λ-regularization of A.A† p.9 Moore-Penrose inverse of A.Single-valued operators:T : C→ H p. 6 T is a single-valued operator on C defined ev-erywhereranT p. 6 Range of an operator TxiiiGlossary of Notation and SymbolskerT p. 6 Kernel o an operator TT ∗ p. 6 Adjoint of a bounded operator TFix T p. 17 The set of fixed points of an operator TT |C The restriction of an operator T to a set C.xivAcknowledgementsWords cannot express how grateful I am to Dr. Heinz Bauschke and Dr. Xi-anfu Wang who became chapters in my story. I cannot thank you enough forall your help, constant encouragement, continuous support, immense knowl-edge, and providing me with an amazing atmosphere to do my research in.Your excellent guidance and patience allowed for the successful completionof this thesis.I deeply thank Dr. Julian Cheng, Dr. Yves Lucet, and Dr. Boris Mor-dukhovich for serving on my thesis committee and for their constructivecomments. I also would like to extend my thanks to all the faculty membersin the Mathematics Department at UBC (Okanagan campus) for adding tomy knowledge of Mathematics.Special thanks to my family Ahmed, Ranse, and my parents who sup-ported and helped me through many challenges on my journey to completemy PhD degree. Thanks to my colleagues Hui, Manish, Shambavi, andZiyuan for being supportive over the last three years.Finally, I would like to thank the ministry of higher education in SaudiArabia and the Saudi Arabia Cultural Bureau for allowing me to have thisopportunity to study abroad, and supporting me on my journey to attaingreater knowledge.xvDedicationTo Ahmed, Ranse and my parents.xviChapter 1IntroductionThe connection between firmly nonexpansive mappings and monotone op-erators helps to study a large number of problems. The general problem isthe convex feasibility problem which asks tofind x ∈ C := C1 ∩ C2 ∩ · · · ∩ Cm,where m ∈ {2, 3, . . .} and C1,C2, . . . ,Cm are nonempty closed convex sub-sets of a real Hilbert space H (see Definition 2.1). In such formulations,each set can be specified in various forms, e.g., as the fixed point set of anonexpansive operator or the set of zeros of maximally monotone opera-tors. Finding a zero of a maximally monotone operator is of fundamentalimportance in optimization and variational analysis. This problem can bemodelled asfind x ∈ H such that 0 ∈ E(x), (1.1)where A, B, and E := A + B are maximally monotone operators. (Fordetailed discussions on problem (1.1) and the connection to optimizationproblems we refer the reader to [10], [24], [26], [27], [28], [31], [43], [44], [45],[54], [51], [52], [53], and the references therein).1.1 The goal of the thesisLet T1, T2, . . . , Tm be firmly nonexpansive operators defined on H. The fixedpoint sets of the cyclic compositions are given byF1 := Fix(T1Tm · · ·T2), (1.2)F2 := Fix(T2T1Tm · · ·T3), (1.3)... (1.4)Fm := Fix(TmTm−1 · · ·T1). (1.5)Our main goal is to understand the fixed point sets F1, F2, . . . , Fm. If for all1 ≤ i ≤ m, (Ti)1≤i≤m=(PCi )1≤i≤m, where PCi is the projection operator11.2. Contributions in this thesisonto Ci and C 6= ∅, then solving (1.2)-(1.5) is equivalent to solving a convexfeasibility problem. Moreover, a point in C can be found as the weak limitof the sequence(xn)n∈N by using the method of cyclic projections. (Formore details we refer the reader to [11], [12], [13], [48], [39], [33], and thereferences therein.) The central questions are:Q1: What happens when C = ∅ ?Q2: What can we say about the fixed point sets F1, F2, . . . , Fm for anyfirmly nonexpansive operator, e.g., projection operators, proximal op-erators, and the resolvent of maximally monotone operators?Q3: What can we say about the fixed point sets F1, F2, . . . , Fm for nonex-pansive operator, e.g., reflection operators?Moreover, if the goal is not just finding a point in the intersection of nonemptyclosed convex sets but also finding the closest point in that intersection toa given point w ∈ H, then the problem is a best approximation problem,namelyfind p ∈ C1 ∩ C2 such that ‖p− w‖ = infx∈C1∩C2‖x− w‖ . (1.6)There are numerous algorithms have been invented to solve this type ofproblems such as the classical Dykstra algorithm DR and Arago´n Artacho-Campoy algorithm AACA. Part of this thesis will focus on the analysis andthe convergence results of AACA algorithm. More details are given in thenext section.1.2 Contributions in this thesisIn this section, we summarize the main contributions of the research thatconstitutes this thesis, in the same order they appear in the sequel.Chapter 2 and Chapter 3 present standard material and basic facts fromconvex analysis and monotone operator theory.Our new results start in Chapter 4. In this chapter, we develop theanalysis of the Arago´n Artacho-Campoy algorithm using the proximal andresolvent average. We advance the analysis of [7] by demonstrating thatthe underlying curve converges to the nearest zero of the sum of the twooperators. We also provide a new interpretation of the underlying operatorsin terms of the resolvent and the proximal average. We illustrate our resultsby considering the projection operator to a closed linear subspace and theprojection operator to affine subspace.21.2. Contributions in this thesisIn Chapter 5, we start by reviewing some well known results about thefixed point sets of composition of projection mappings. We sketch the con-nection between the cycles and the fixed point sets of composition of projec-tion mappings and we also present a new proof for [13, Remark 5.1.7], whichindicates that half of the geometry conjecture, which concerns the structureof fixed point of composition, is true when there are cycles.Chapter 6 presents a comprehensive study to the fixed point set of thecomposition of finitely many reflectors associated with closed linear sub-spaces. In particular, we show the effect of cyclically shifting a compositionof m reflectors on the fixed point set of the composition (see Lemma 6.4).We also show the existence of a nice permutation of the reflectors whichyields the same fixed point set of the composition of reflectors (see Theo-rem 6.8). Theorem 6.6 and Theorem 6.9 show that the location of the fixedpoint set of the composition of an odd number of reflectors can be narroweddown. Moreover, the upper bound in Theorem 6.9 is sometimes sharp (seeExample 6.11) but not always attained (see Example 6.12).Chapter 7 is dedicated to study in detail the displacement mappings ofisometries of finite order. We derive explicit formulas for resolvents, Yosidaapproximations, and (set-valued and Moore-Penrose) inverses of displace-ment mappings (see Theorem 7.2, Lemma 7.5, Theorem 7.7, and Theo-rem 7.11 respectively). We also study the asymptotic behaviour of its resol-vents (see Lemma 7.4).Chapter 8 focuses on some conditions under which infimal convolutionsare closed and exact (see Section 8.8). It also provides additional convexanalysis results which turn to be useful for our study.In Chapter 9, by using the framework of monotone operator theory, weresolve the geometry conjecture completely (see Theorem 9.10). We studythe relationship among the set of cycles and fixed point sets for compositionsof projectors (see Lemma 9.4). We obtain alternative descriptions of the setof cycles (see Theorem 9.7). We find the difference vectors using the forward-backward method (see Section 9.10) and by using the Banach principle (seeSection 9.9). We also sketch numerical approches for the computation ofthe difference vector by using Seeger’s algorithm (see Fact 9.19).Chapter 10 then introduces the generalized cycle which is a variant ofthe classical one that is defined by projection mappings to tackle the casewhen the classical ones do not exist. We prove that this generalized cycle isunique and we derive a formula for it (see Theorem 10.2). We study the rela-tionship between the generalized cycles and the generalized gap vectors (seeTheorem 10.2). We also study the relationship among the classical cycles,generalized cycles and the generalized gap vectors (see Theorem 10.3).31.2. Contributions in this thesisChapter 11 shows that the set of the classical cycles that is defined byusing the proximal mappings is at most singleton and possibly empty (seeProposition 11.6). We study the set of the classical cycles that is definedby using the proximal mappings and the set of classical gap vectors (seeProposition 11.7). We also study the relationship among the set of cyclesand fixed point sets for compositions of proximal mapping (see Lemma 11.8and Lemma 11.9). We prove the uniqueness of the solution to the enlargeddual problem and we show that if the enlarged dual solution exists, then itis exactly the solution to the classical dual problem (see Theorem 11.12).We derive a formula for the gap vector of the enlarged dual problem (seeTheorem 11.12(i)).In Chapter 12, we extend results of Chapter 10 from the projectionmappings to the proximal mappings.In Chapter 13, we state formulas for the resolvents of the normal cone onthe products of nonempty closed convex sets, the resolvents of the normalcone inverse on the products of nonempty closed convex sets, the Yosidaapproximation of the normal cone, and the the Yosida approximation of thenormal cone inverse (see Theorem 13.2). We prove that the composition ofthe resolvents of the displacement mappings inverse and the resolvents ofthe normal cones inverse is a Banach contraction (see Lemma 13.3). Wederive a formula for the fixed point xγ of JγM−1JγN−1C (see Section 13.3).We also prove that the curve (xγ) belongs to ∆⊥ (see Lemma 13.4). Wealso prove that every weak cluster point of the curve (xγ) coincides withPC1−C2(0) (see Theorem 13.5(ii)).Finally, in Chapter 14, we give a brief summary of our results and providea list of possible avenues of future research.4Chapter 2Preliminary Details2.1 OverviewIn this chapter, I pick my (mathematical) tool box with various resultsmostly from Functional Analysis, Fixed Point Theory, and Convex Analysis.The results are drawn from [34], [10], [32], and [46].2.2 SetupThroughout this thesis, we denote the nonnegative integers {0, 1, 2, . . .}by N. The real numbers, nonnegative real numbers, and strictly positivereal numbers are indicated by R, R+, and R++, respectively. Euclidean andHilbert spaces are the fundamental spaces we work in.Definition 2.1. A Hilbert space, H, is a complete inner product space.We assume thatH is a real Hilbert space (2.1)with inner product 〈·, ·〉 : H × H → R and induced norm denoted by ‖·‖.Let I be an index set with I = {1, 2, . . . ,m} for some integer positive m.ThenHm = {x = (xi)i∈I | (∀i ∈ I) xi ∈ H},denotes the Hilbert product space with inner product 〈x,y〉 = ∑i∈I xiyi.Definition 2.2. A sequence (xn)n∈N in H converges strongly to a point x,written xn → x, iflimn→∞ ‖xn − x‖ = 0.A sequence (xn)n∈N in H converges weakly to a point x, written xn ⇀ x, iffor every u ∈ H,limn→∞ 〈xn, u〉 = 〈x, u〉 .52.3. Operators2.3 Operators2.3.1 Single-valued operatorsLet C ⊆ H and T : C → H denote an operator (or mapping) T thatmaps every point x ∈ C to a point Tx ∈ H. The range of T isranT = {y ∈ H | ∃x ∈ H with Tx = y}.SetB(H) = {T : H → H | T is linear and continuous}.The kernel of T iskerT = {x ∈ H | Tx = 0}.Fact 2.3. [34, 8.25] For T ∈ B(H), the adjoint of T is the unique operatorT ∗ that satisfies (∀x ∈ H)(∀y ∈ H) 〈Tx, y〉 = 〈x, T ∗y〉 .Example 2.4. Let R be the cyclic right-shift operator,R : Hm → Hm : (x1, x2, . . . , xm) 7→ (xm, x1, . . . , xm−1).Then R∗ is the cyclic left-shift operator,R∗ : Hm → Hm : (x1, x2, . . . , xm) 7→ (x2, x3, . . . , xm, x1).Definition 2.5. Let T : H → H. Then T is Lipschitz continuous withconstant β ≥ 0 if(∀x ∈ H)(∀y ∈ H) ‖Tx− Ty‖ ≤ β ‖x− y‖ .If β ∈ [0, 1), then T is called a Banach contraction.Definition 2.6. A mapping T : H → H is an isometry if(∀x ∈ H)(∀y ∈ H) ‖Tx− Ty‖ = ‖x− y‖ .The identity mapping is the operator Id : H → H : x 7→ x.62.4. Convex analysis2.3.2 Set-valued operatorsLet A : H ⇒ H be an arbitrary set valued operator, i.e., Ax ⊆ H (∀x ∈H). The graph of A, denoted by graA, is defined asgraA = {(x, y) ∈ H ×H | y ∈ Ax}.The domain of A, denoted by domA, is defined asdomA = {x ∈ H | Ax 6= ∅}.The range of A, denoted by ranA, is defined asranA = {y ∈ H | ∃x ∈ H such that y ∈ Ax}.The set-valued inverse A−1 of A is defined by(y, x) ∈ graA−1 ⇔ (x, y) ∈ graA.The set of zeros of A is written aszerA = A−1(0) = {x ∈ H | 0 ∈ Ax}.Fact 2.7. [10, Proposition 20.17] Let A ∈ B(H) be monotone. Then kerA =kerA∗ and ranA = ranA∗.Fact 2.8. [10, Corollary 15.34] Let A ∈ B(H). Then ranA is closed if andonly if ranA∗ is closed.Definition 2.9. Let A : H ⇒ H. Then A is a linear relation if graA is alinear subspace of H×H.Fact 2.10. [32, I.2.3] Let A be a linear relation on H. Then A−1 is a linearrelation.2.4 Convex analysisA subset C of H is convex if for all x, y ∈ C and λ ∈ (0, 1),λx+ (1− λ)y ∈ C. (2.2)The intersection of all the convex sets containing C is called the convex hullof C, i.e., the smallest convex subset of H containing C, and is denoted byconv C. The intersection of all affine subspaces containing C is called the72.4. Convex analysisaffine hull of C and is denoted by aff C. The closure of C is denoted by C.The smallest linear subspace of H containing C is span C. The interior ofC is the largest open set contained in C,int C = {x | ∃  > 0, B(x, ) ⊆ C}.A subset C of H is a cone if C = R++C. That is, x ∈ C and λ > 0 impliesλx ∈ C. A convex cone is a set that is both convex and a cone. The conicalhull of C, cone C, is the intersection of all the cones in H containing C.Moreover, the relative interior of C isri C = {x ∈ aff C | ∃  > 0, B(x, ) ∩ aff C ⊆ C}.Definition 2.11. The strong relative interior of a set C issri C = {x ∈ C | cone(C− x) = span(C− x)}.Definition 2.12. Let C be a nonempty closed convex subset of H and letz ∈ H. The projection operator (this is also known as the closest pointmapping) of z onto C is the unique point in C denoted by PC z that satisfies‖z − x‖ = infc∈C‖c− z‖ , where x = PC(z). (2.3)Fact 2.13. [10, Corollary 3.22] Let C be a closed convex subset of H. Forevery x and p in H,p = PC x⇐⇒ p ∈ C and(∀c ∈ C) 〈c− p, x− p〉 ≤ 0. (2.4)Fact 2.14. [10, Proposition 3.17] Let C be a nonempty subset of H, and lety ∈ H. Then (∀x ∈ H) Py+C x = y + PC (x− y). (2.5)Fact 2.15. [10, Corollary 3.20 & Corollary 3.22] Let C be a closed affinesubspace of H and let x ∈ H. Then the following hold:(i) PC is affine.If C is a closed linear subspace of H, then we additionally have:(ii) PC is linear.(iii) PC⊥ = Id−PC.82.4. Convex analysisExample 2.16. [10, Proposition 25.4(iii) and (iv)] Suppose ∆ and ∆⊥are subsets of Hm, where Hm be the product space of Hilbert space, ∆ ={(x, x, . . . , x) ∈ Hm | x ∈ H} and ∆⊥ = {(x1, x2, . . . , xm) | m∑i=1xi = 0}.Then(i) P∆(x1, . . . , xm) =(1mm∑k=1xk,1mm∑k=1xk, . . . ,1mm∑k=1xk).(ii) P∆⊥(x1, . . . , xm) =(x1− 1mm∑k=1xk, x2− 1mm∑k=1xk, . . . , xm− 1mm∑k=1xk).Fact 2.17. [47, Proposition 2.1] Let C1, C2, and C3 be subset ofH. Supposethat C2 is nonempty and bounded, that C3 is closed and convex, and thatC1 + C2 ⊆ C3 + C2. Then C1 ⊆ C3.Definition 2.18. Let C be a nonempty convex subset of H and x ∈ H.Then the normal cone operator to C at x isNC : H⇒ H : x 7→{{y ∈ H | sup 〈C− x, y〉 ≤ 0}, if x ∈ C;∅, otherwise.(2.6)Example 2.19. [10, Proposition 25.4(iii) and (iv)] Let ∆⊥ ⊆ Hm, where∆⊥ = {(x1, x2, . . . , xm) | m∑i=1xi = 0}. ThenN∆⊥ : Hm ⇒ Hm : x 7→{∆, if x ∈∆⊥;∅, otherwise.(2.7)Definition 2.20 (Moore-Penrose inverse). Let T ∈ B(H) be such thatranT is closed and for every y ∈ H, set Cy ={x ∈ H∣∣ T ∗Tx = T ∗y}. TheMoore-Penrose inverse of T isT † : H → H : y 7→ PCy 0.For more information about Moore-Penrose inverses see [38].Fact 2.21. [19, Proposition 2.1] Let Y be a real Hilbert space, and let B bea continuous linear operator from H to Y with adjoint B∗ such that ranBis closed. Then the Moore-Penrose inverse of B satisfiesB† = PranB∗ ◦B−1 ◦ PranB .92.5. Convex functions2.5 Convex functionsA function f : H → ]−∞,+∞] = R ∪ {+∞} is said to be convex if itsdomain, dom f = {x ∈ H | f(x) < +∞}, is a convex set and ∀x, y ∈ H, and0 < λ < 1, we havef(λx+(1− λ)y) ≤ λf(x)+ (1− λ)f(y). (2.8)A function f : H → [−∞,+∞] is proper if(∀x ∈ H) f(x) > −∞ and (∃x0 ∈ H) such that f(x0) < +∞. (2.9)A function f is lower semicontinuous if its epigraph,epi f ={(x, r) ∈ H × R | f(x) ≤ r}, (2.10)is a closed set; equivalently for every sequence(xn)n∈N in H,xn → x⇒ f(x) ≤ lim infn→∞ f(xn). (2.11)Example 2.22. [46, Example 5.15] Let C be a convex and closed subset ofH. Then the indicator function of C at x isιC(x)={0, if x ∈ C;+∞, otherwise. (2.12)It is convex and lower semicontinuous.Definition 2.23. Let f : H → ]−∞,+∞] be proper and let β ∈ R++. Thenf is strongly convex with constant β if(∀x ∈ dom f), (∀y ∈ dom f), and(∀λ ∈ ]0, 1[ ),f(λx+ (1− λ)y)+ λ(1− λ)β2 ‖x− y‖2 ≤ λf(x) + (1− λ)f(y).A function f : H → [−∞,+∞] is coercive iflim‖x‖→+∞f(x)= +∞. (2.13)Let Γ0(H) denote the class of proper lower semicontinuous convex functionsfrom H → ]−∞,+∞]. For f, g ∈ Γ0(H), then fg stands for the infimalconvolution of f with g, which is given by(∀x ∈ H) (fg)(x) = inf{f(x1)+ g(x2) | x1 + x2 = x}. (2.14)102.5. Convex functionsFact 2.24. [10, Proposition 13.12(ii)] Let f : H → ]−∞,+∞]. Then f∗ 6≡+∞ if and only if f possesses a continuous affine minorant.Let f : H → ]−∞,+∞] be proper. Then ∂f is the convex subdifferentialof f is the set-valued operator∂f : H⇒ H : x¯ 7→{y ∈ H∣∣∣〈x− x¯, y〉+ f(x¯) ≤ f(x) (∀x ∈ H)}, (2.15)and dom(∂f)= {x ∈ H∣∣∂f(x) 6= ∅}. The next fact will provide someimportant properties of subdifferential of the sum.Fact 2.25. [46, Theorem 5.38 and Theorem 5.39] Let f1, f2 be properconvex functions on H and let x ∈ H. Then(i) ∂f1(x)+ ∂f2(x) ⊆ ∂(f1 + f2)(x).(ii) If dom(f1) ∩ dom (f2) 6= ∅ and f1 is continuous, then∂f1(x)+ ∂f2(x)= ∂(f1 + f2)(x).(iii) If ri(dom(f1)) ∩ ri ( dom (f2)) 6= ∅, then∂f1(x)+ ∂f2(x)= ∂(f1 + f2)(x).Let f : H → [−∞,+∞]. The Fenchel conjugate of f isf∗ : H → [−∞,+∞] : u→ supx∈H(〈x, u〉 − f(x)), (2.16)and the biconjugate of f is f∗∗ =(f∗)∗.Fact 2.26. [10, Theorem 16.23] Let f ∈ Γ0(H), let x ∈ H and let u ∈ H.Then the following statements are equivalent:(i) u ∈ ∂f(x).(ii) f(x)+ f∗(u)= 〈x, u〉 .(iii) x ∈ ∂f∗(u).Fact 2.27. [10, Theorem 23.5] Let f ∈ Γ0(H). Then∂f∗ =(∂f)−1.Fact 2.28. [10, Example 13.43(ii)] Let C be a nonempty subset of H. Thenι∗∗C = ι∗∗conv C = ιconv C.112.5. Convex functionsFact 2.29. [10, Theorem 15.3] Let f and g be functions in Γ0(H) such thatdom f − dom g is a closed linear subspace, i.e.,0 ∈ sri ( dom f − dom g).Then(f + g)∗ = f∗g∗ ∈ Γ0(H).Fact 2.30. [10, Fact 6.14] Let C be a nonempty convex subset of H, whereH is finite-dimensional. Then riC is the interior of C relative to aff C andriC 6= ∅. Moreover,riC = sriC.Fact 2.31. [10, Corollary 6.15] Let C and D be nonempty convex subsetsof H, where H is finite-dimensional. Thenri(C −D) = riC − riD.Definition 2.32. [10, Definition 12.23] Let f ∈ Γ0(H) and let x ∈ H. ThenProxf is the unique point in H that satisfiesminy∈H(f(y)+ 12 ‖x− y‖2)= f(Proxf x)+ 12 ‖x− Proxf x‖2 . (2.17)The operator Proxf : H → H is the proximity operator or the proximalmapping of f .Fact 2.33. [10, Proposition 16.34] Let f ∈ Γ0(H) and x, p ∈ H. Thenp = Proxf(x)⇔ x− p ∈ ∂f(p) and Proxf = ( Id +∂f)−1. (2.18)For numerous examples and properties of proximal mappings, we referthe reader to [10, Chapter 24] and [22, Chapter 6].Definition 2.34. Let C ⊂ H. Then the support function of C given byσC(y): H → ]−∞,+∞] : y 7→ sup 〈C, y〉 . (2.19)Remark 2.35. [10, Example 13.3] Let C ⊂ H. Then there is a strong rela-tionship between the support function and the indicator function becauseι∗C = σC.Fact 2.36. [10, Example 16.12] Let C be a nonempty closed convex subsetof H. Then ∂ιC = NC.122.5. Convex functionsExample 2.37. Given ∆,∆⊥ ⊂ Hm, both sets defined in Example 2.16.Then the following hold:N∆⊥ = ∂ι∆⊥ = ∂ι∗∆. (2.20)Proof. Assume x0 ∈∆⊥ and let u ∈ Hm. From Definition 2.18u ∈ N∆⊥(x0)⇔ (∀x ∈∆⊥) 〈u,x− x0〉 ≤ 0 (∀x ∈∆⊥)⇔ (∀x ∈∆⊥) 〈u,x− x0〉+ ι∆⊥(x0) ≤ ι∆⊥(x)⇔ u ∈ ∂ι∆⊥(x0).The last equation is obtained by using (2.15). Hence,N∆⊥ = ∂ι∆⊥ . (2.21)Moreover,u ∈ ∂ι∆⊥(x0)⇔ (∀x ∈∆⊥) 〈u,x− x0〉+ ι∆⊥(x0) ≤ ι∆⊥(x)⇔ (∀x ∈∆⊥) 〈u,x− x0〉 ≤ 0⇔ (∀x ∈∆⊥) 〈u, x− x0〉+ sup{ 〈x0,u〉 | u ∈∆}≤ sup{〈x,u〉 | u ∈∆}⇔ (∀x ∈∆⊥)(∀u ∈∆) 〈u,x− x0〉+ σ∆(x0) ≤ σ∆(x)⇔ (∀x ∈∆⊥)(∀u ∈∆) 〈u,x− x0〉+ ι∗∆(x0) ≤ ι∗∆(x)⇔ u ∈ ∂ι∗∆.Hence,∂ι∆⊥ = ∂ι∗∆. (2.22)Combining (2.21) and (2.22) givesN∆⊥ = ∂ι∆⊥ = ∂ι∗∆.Proposition 2.38. Let C1, C2 be nonempty closed and convex subset of Hand let σC1 , σC2 : H → ]−∞,+∞] and let x ∈ H. Then the following hold:(i) σC1(x)+ σC2(x)= σC1+C2(x).(ii) ∂σC1(x)+ ∂σC2(x) ⊆ ∂(σC1 + σC2)(x) = ∂(σC1+C2)(x).132.5. Convex functionsProof. (i): We haveσC1(x)+ σC2(x)= supy1∈C1〈x, y1〉+ supy2∈C2〈x, y2〉= supy1∈C1, y2∈C2( 〈x, y1〉+ 〈x, y2〉 )= supy1∈C1, y2∈C2〈x, y1 + y2〉= supy∈C1+C2〈x, y〉= σC1+C2(x).(ii): Let u ∈ ∂σC1(x)and let v ∈ ∂σC2(x). Then(∀y ∈ H) σC1(x) +〈u, y − x〉 ≤ σC1(y)and σC2(x)+ 〈v, y − x〉 ≤ σC2(y), which upon addingresults in(σC1 + σC2)(x)+ 〈u+ v, y − x〉 ≤ (σC1 + σC2)(y) and by (i), weobtainσC1+C2(x)+ 〈u+ v, y − x〉 ≤ σC1+C2(y).Thus, u+ v ∈ ∂σC1+C2(y).Definition 2.39. [43, Section 19] A polyhedral convex set in H is by defini-tion a set which can be expressed as the intersection of some finite collectionof closed half-spaces, i.e. as the set of solutions to some finite system of in-equalities of the form〈x, bi〉 ≤ βi, i = 1, . . . ,m,with bi ∈ H and βi ∈ R.Fact 2.40. [43, Corollary 19.3.4] If f1 and f2 are proper polyhedral convexfunctions on Rn, then f1f2 is a polyhedral convex function too. More-over, if f1f2 is proper, then the infimum in the definition of(f1f2)(x)isattained for each x.Fact 2.41. [43, Theorem 19.4] If f1 and f2 are proper polyhedral convexfunctions, then f1 + f2 is polyhedral.Fact 2.42. [10, Theorem 16.58] Let f ∈ Γ0(H), let (y, v) ∈ dom(f ⊕ f∗),and λ and µ in R+ satisfy f(y) + f∗(v) ≤ 〈y, v〉 + λµ. Then there exists(z, w) ∈ ∂f such that ‖z − y‖ ≤ λ and ‖w − v‖ ≤ µ.142.6. Averages2.6 AveragesThere are many methods of averaging; this section gathers the definitionsof some methods that will be of interest.Definition 2.43 (proximal average). Let(∀i ∈ {1, . . . , n}), fi ∈ Γ0(H) andλi be n strictly positive real numbers with∑ni=1 λi = 1. The λ-weightedproximal average of f =(f1, . . . , fn)with parameter µ > 0 is defined byPµ(f ,λ):=(λ1(f1 +1µq)∗ + λ2(f2 + 1µq)∗ + · · ·+ λn(fn + 1µq)∗)∗ − 1µq.The function Pµ(f ,λ)is a proper lower semi-continuous convex function onH, and it inherits many desirable properties from the underlying functionsfi. Moreover,(∀x ∈ H) Pµ(f ,λ)(x) = 1µ(− 12 ‖x‖2+ inf∑i∈I xi=x∑i∈I(µfi(xi/λi)+ 12 ‖xi/λi‖2)).It is also written as P(f ,λ) when µ = 1. For more information about theproximal average, see [17] and [18].Example 2.44. Let w ∈ H, f ∈ Γ0(H), and g = ι{w} ∈ Γ0(H). Letγ ∈ R++ and λ1, λ2 ∈ R such that λ1 = γ and λ2 = 1− γ. ThenP(f, ι{w}, λ1, λ2) = γf(γ−1(x− (1− γ)w))+ 1− γ2γ ‖x− w‖2 .Proof. By Definition 2.43, we obtainP(f, ι{w}, λ1, λ2) = infx1+x2=x(λ1f(x1λ1)+ 12∥∥∥∥x1λ1∥∥∥∥2+ λ2ι{w}(x2λ2)+ 12∥∥∥∥x2λ2∥∥∥∥2 )− 12 ‖x‖2 . (2.23)By changing variables:y1 =x1λ1⇔ x1 = λ1y1,y2 =x2λ2⇔ x2 = λ2y2,and x = λ1y1 + λ2y2.152.6. AveragesSubstituting this back into (2.23) givesP(f, ι{w}, λ1, λ2) = inf {λ1f(y1)+ 12 ‖y1‖2 + λ2ι{w}(y2)+ 12 ‖y2‖2| λ1y1 + λ2y2 = x}− 12 ‖λ1y1 + λ2y2‖= inf{λ1f(y1)+ λ2ι{w}(y2)+ λ1λ22 ‖λ1y1 + λ2y2‖2| λ1y1 + λ2y2 = x}= inf{λ1f(y1)+ λ1λ22 ‖λ1y1 + λ2w‖2 | λ1y1 + λ2w = x}.Hence,P(f, ι{w}, λ1, λ2) = inf {λ1f(y1)+ λ1λ22 ‖y1 − w‖2 | λ1y1 + λ2w = x}. (2.24)Let λ1 = γ and λ2 = 1− γ. Solving λ1y1 + λ2w for y1 givesy1 = γ−1(x− (1− γ)w).Substituting this back into (2.24) givesP(f, ι{w}, γ, (1− γ)) = γf(γ−1(x− (1− γ)w))+γ(1− γ)2∥∥γ−1(x− (1− γ)w)− w∥∥2= γf(γ−1(x− (1− γ)w))+ 1− γ2γ ‖x− w‖2 .The resolvent average will be given in the next chapter (see Defini-tion 3.47).16Chapter 3Monotonicity and FirmNonexpansiveness3.1 OverviewIn this chapter, we collect mathematical results from Functional Analy-sis, Fixed Point Theory, Convex Analysis, and Monotone Operator Theory.Most of the results in this chapter are drawn from [10], [42], [23], [29], [16],[35], [37], and [43].3.2 Firmly nonexpansiveness operatorsDefinition 3.1. [10, Definition 4.1] Let C be a nonempty subset H. Amapping T : C→ H is(i) nonexpansive, or Lipschitz continuous with constant 1, if(∀x ∈ C)(∀y ∈ C) ‖Tx− Ty‖ ≤ ‖x− y‖ ; (3.1)(ii) firmly nonexpansive if(∀x ∈ C)(∀y ∈ C) ‖Tx− Ty‖2+∥∥( Id−T )x− ( Id−T )y∥∥2 ≤ ‖x− y‖2 ;(3.2)(iii) quasinonexpansive if(∀x ∈ C)(∀y ∈ Fix T ) ‖Tx− y‖ ≤ ‖x− y‖ , (3.3)where Fix T is the set of fixed point of T defines asFix T = {x ∈ H | x = Tx}.Example 3.2. [10, Proposition 4.8] Let C be a nonempty closed convexsubset H. Then the projector PC is firmly nonexpansive.173.2. Firmly nonexpansiveness operatorsExample 3.3. Let C be a nonempty closed convex subset H and let RC =2 PC− Id be a nonexpansive operator on C. ThenFixRC = C.Proof. Let x ∈ H. Then x = RC(x) ⇔ x = 2 PC(x) − x ⇔ PC(x) = x ⇔x ∈ C.Fact 3.4. [42, Page 408 in Section X.144] Let T : H → H be a nonexpansivelinear operator. ThenFix T = Fix T ∗.Definition 3.5. Let C be a nonempty subset of H, let T : C→ H, and letβ ∈ R++. Then T is β-cocoercive (or β-inverse strongly monotone) if βT isfirmly nonexpansive, i.e.,(∀x ∈ C)(∀y ∈ C) 〈x− y, Tx− Ty〉 ≥ β ‖Tx− Ty‖2 .Fact 3.6. [10, Proposition 14.16] Let f be a function in Γ0(H). Then thefollowing are equivalent:(i) f is coercive.(ii) 0 ∈ int dom f∗.Fact 3.7. [10, Proposition 4.11] Let C be a nonempty subset of H and letT : C→ H. Then T is nonexpansive if and only if Id−T is (1/2)-cocoercive.Fact 3.8. [23, Theorem VI.5.1] Let R : H → H be a linear isometry of finiteorder m ∈ {2, 3, . . . }:Rm = Id .Moreover, R is a Lipschitz continuous with constant 1. ThenR∗ = R−1 = Rm−1.Fact 3.9. [10, Corollary 4.24] Let C be a nonempty closed convex subset ofH and let T : C→ H be nonexpansive. Then Fix T is closed and convex.Fact 3.10. [10, Proposition 4.2] Let C be a nonempty closed convex subsetof H and let T : C→ H . Then the following are equivalent:(i) T is firmly nonexpansive.(ii) Id−T is firmly nonexpansive.183.2. Firmly nonexpansiveness operators(iii) 2T − Id is nonexpansive.(iv)(∀x ∈ C)(∀y ∈ C) ‖Tx− Ty‖2 ≤ 〈x− y, Tx− Ty〉.(v)(∀x ∈ C)(∀y ∈ C) 0 ≤ 〈Tx− Ty, ( Id−T )x− ( Id−T )y〉.Fact 3.11. [29, Lemma 4.1.12(ii)] Let Si : H → H, i ∈ I := {1, 2, . . . ,m},be nonexpansive. Then the composition SmSm−1 · · ·S2S1 is nonexpansive.Fact 3.12. [16, Fact 2.2(iii)] Let T and(Ti)1≤i≤m be operators fromH toH.Suppose that T is strongly nonexpansive and let x0 ∈ H. If Fix T 6= ∅, thenthe sequence(Tnx0)n∈N converges weakly to some point in Fix T ; otherwise,‖Tnx0‖ → +∞.Definition 3.13. Let C ⊆ H with C 6= ∅ and T : C → H be nonexpan-sive. Let α ∈ ]0, 1[. Then T is averaged with constant α if there exists anonexpansive operator R : C→ H such thatT =(1− α) Id +αR.Fact 3.14. [10, Proposition 4.35] T : H → H is α-averaged if and only if itsatisfies((1− α)α−1) ∥∥( Id−T )x− ( Id−T )y∥∥2 + ‖Tx− Ty‖2 ≤ ‖x− y‖2 (3.4)for all x, y ∈ H.Fact 3.15. [10, Corollary 4.51] Let C be a nonempty subset of H, let m bea strictly positive integer, set I = {1, 2, . . . ,m}. Let (Ti)i∈I be a family ofaveraged nonexpansive operators from C to C such that ⋂i∈I Fix Ti 6= ∅,and set T = T1 · · ·Tm. ThenFix T =⋂i∈IFix Ti.Definition 3.16. An operator T : H → H is θ-negatively averaged withθ ∈ (0, 1) if −T is θ-averaged.It follows directly from Fact 3.14 and Definition 3.16 that T is θ-negativelyaveraged if and only if((1− θ)θ−1) ∥∥( Id +T )x− ( Id +T )y∥∥2 + ‖Tx− Ty‖2 ≤ ‖x− y‖2 (3.5)since −T is θ-averaged.193.3. Monotone operator theory3.3 Monotone operator theoryDefinition 3.17. [10, Definition 20.1 and Definition 22.1] A set-valued op-erator A : H⇒ H is(i) monotone if(∀(x, u) ∈ graA) (∀(y, v) ∈ graA) 〈x− y, u− v〉 ≥ 0. (3.6)(ii) paramonotone if it is monotone and (∀(x, u) ∈ graA) (∀(y, v) ∈ graA)〈x− y, u− v〉 = 0⇒ (x, v) ∈ graA and (y, u) ∈ graA. (3.7)(iii) strictly monotone if (∀(x, u) ∈ graA) (∀(y, v) ∈ graA)x 6= y ⇒ 〈x− y, u− v〉 > 0. (3.8)(iv) strongly monotone with constant β ∈ R++ if A − β Id is monotone,i.e.,(∀(x, u) ∈ graA) (∀(y, v) ∈ graA) 〈x− y, u− v〉 ≥ β ‖x− y‖2 .(3.9)Example 3.18. [10, Example 22.3(iv)] Let f : H → ]−∞,+∞] be properand strongly convex with constant β ∈ R++. Then ∂f is strongly monotonewith constant β.A monotone operator A is maximally monotone if there exists no monotoneoperator B such that graA ⊂ graB. That is, for every (x, u) ∈ H ×H,(x, u) ∈ graA⇔ (∀(y, v) ∈ graA) 〈x− y, u− v〉 ≥ 0. (3.10)Example 3.19. Let A : H ⇒ H : x 7→ −v. Then A is monotone. By using(3.6) we get that(∀x ∈ H) and (∀y ∈ H) 〈x− y,−v − (−v)〉 = 0.Example 3.20. Let A : H ⇒ H : x 7→ Id. Then A is monotone. By using(3.6) we get that(∀x ∈ H) and (∀y ∈ H) 〈x− y, x− y〉 = ‖x− y‖2 ≥ 0.Hence, A is strongly monotone with constant 1.Lemma 3.21. Let A,B : H ⇒ H be monotone operators. Then A + B ismonotone.203.3. Monotone operator theoryProof. Take(x1, x∗1)and(x2, x∗2)in gra(A + B). Then x∗1 = a∗1 + b∗1 andx∗2 = a∗2 + b∗2, where a∗i ∈ Axi and b∗i ∈ Bxi. Then〈x1 − x2, x∗1 − x∗2〉 =〈x1 − x2,(a∗1 − a∗2)+(b∗1 − b∗2)〉= 〈x1 − x2, a∗1 − a∗2〉+ 〈x1 − x2, b∗1 − b∗2〉≥ 0 + 0 = 0.Fact 3.22. [10, Corollary 20.28] Let A : H → H be monotone and continu-ous. Then A is maximally monotone.Fact 3.23. [10, Example 20.26] Let T : H → H be nonexpansive and letα ∈ [−1, 1]. Then Id +αT is maximally monotone.Fact 3.24. [10, Example 20.30] Let A ∈ B(H) be such that A∗ = −A. ThenA is maximally monotone.Example 3.25. [10, Theorem 20.25] Let f ∈ Γ0(H). Then ∂f is maximallymonotone.Example 3.26. Let M : H ⇒ H be maximally monotone and λ ∈ R++.Then γM is maximally monotone.Proof. We have(x, u) ∈ gra γM⇔ u ∈ γM(x)⇔ (1/γ)u ∈ Mx⇔ (x, (1/γ)u)∈ gra M. Since M is maximally monotone, (3.10) gives(x, (1/γ)u) ∈ gra M⇔ (∀(y, (1/γ)v) ∈ gra M) 〈x− y, (1/γ)u− (1/γ)v〉 ≥ 0.Therefore, ∀(y, v) ∈ gra γM, and we obtain〈x− y, u− v〉 = γ 〈x− y, (1/γ)u− (1/γ)v〉 ≥ 0.To prove the other direction, let(x, u) ∈ H ×H such that(∀(y, v) ∈ gra γM) 〈x− y, u− v〉 ≥ 0;therefore,〈x− y, u− v〉 = γ 〈x− y, (1/γ)u− (1/γ)v〉 ≥ 0⇒ 〈x− y, (1/γ)u− (1/γ)v〉 ≥ 0.We obtain, for every(x, (1/γ)u) ∈ H×H and for every (y, (1/γ)v) ∈ gra M,〈x− y, (1/γ)u− (1/γ)v〉 ≥ 0.Hence, by (3.10),(x, (1/γ)u) ∈ gra M and therefore, (x, u) ∈ gra γM.213.3. Monotone operator theoryLemma 3.27. Let σA > 0 and σB > 0. Suppose A : H⇒ H is σA-monotoneand B : H⇒ H is σB-monotone. Then A+B is σA + σB-monotone.Proof. Note that A is σA-monotone ⇔ A − σA Id is monotone and B isσB-monotone ⇔ B − σB Id is monotone. From Lemma 3.21, we obtain(A− σA)+(B − σB)is monotone ⇔ (A + B)− (σA + σB) Id is monotone⇔ A+B is (σA + σB)-monotone.Definition 3.28. Let A be a monotone operator from H ⇒ H and denotethe associated resolvent byJA = (Id +A)−1. (3.11)The reflected resolvent of A is denoted by RA and defined byRA = 2JA − Id . (3.12)Definition 3.29. [10, Definition] Let A : H⇒ H be an operator. Given anyγ ∈ ]0, 1], the operator 2γJA − Id is sometimes called a modified reflectedresolvent of A.The next theorem characterizes maximal monotonicity.Fact 3.30. [10, Theorem 20.1] Let A : H ⇒ H be monotone. Then A ismaximally monotone if and only if ran(Id +A)= H.The following is the very useful Minty parameterization.Fact 3.31. [40, Theorem 1] Let A : H⇒ H be maximally monotone. ThengraA → H : (a, a∗) 7→ a + a∗ is a continuous bijection with continuousinverse x 7→ (JAx, x− JAx); thusgraA ={(JAx, x− JAx) | x ∈ H}. (3.13)Example 3.32. Let A = Id. Then JA = 12 Id, RA = 0, and the modifiedreflected resolvent is 0.Proof. Let y ∈ H and set x = JAy. Then y ∈ (Id +A)x⇔ y = x+ x⇔ y =2x⇔ x = 12y ⇔ JA = 12 Id. Using (3.12), we see thatRA = 2(12) Id− Id = 0.Finally, the modified reflected resolvent is equal to 0 by Definition 3.29.223.3. Monotone operator theoryExample 3.33. Let A = −v. Then JA = Id +v, RA = Id +2v, and themodified reflected resolvent is(2γ − 1) Id +2γv.Proof. Let y ∈ H and set x = JAy. Then y ∈ (Id +A)x ⇔ y = x − v ⇔x = y + v ⇒ JA = Id +v. From (3.12), we obtain RA = 2(Id +v) − Id ⇔RA = Id +2v and the modified reflected resolvent is(2γ − 1) Id +2γv byDefinition 3.29.Fact 3.34. [10, Example 20.41 & Example 23.4] Let C be a nonempty closedconvex subset of H. Then(i) NC is maximally monotone.(ii) JNC =(Id +NC)−1 = PC. Moreover, RNC = 2 PC− Id.Fact 3.35. [10, Corollary 16.29 & Proposition 16.34] Let f ∈ Γ0(H). Thenthe following hold:(i) dom ∂f is a dense subset of dom f .(ii) Proxf =(Id +∂f)−1 = J∂f .Fact 3.36. [35] Let A : H⇒ H and T : H → H. Then the following hold:(i) If T is firmly nonexpansive, then B := T−1−Id is maximally monotoneand JB = T .(ii) If A is maximally monotone, then JA has full domain, is single valued,firmly nonexpansive, and A = J−1A − Id.Fact 3.37. [25, Corollary 23.10] Let A : H⇒ H be a maximally monotoneoperator. Let γ ∈ R++. Then(i) JγA : H → H is firmly nonexpansive, i.e.,∥∥JγA(x)− JγA(y)∥∥2 + ∥∥( Id−JγA(x)− ( Id−JγA(y))∥∥ ≤ ‖x− y‖2 .Moreover, JγA is maximally monotone.(ii) RγA : H → H is nonexpansive, i.e.,∥∥RγA(x)−RγA(y)∥∥ ≤ ‖x− y‖ (∀x ∈ H)(∀y ∈ H).233.3. Monotone operator theoryFact 3.38. [37, Proposition 5.4] The reflected resolvent of ζA-strongly mono-tone and maximally monotone operator A : H⇒ H is11 + ζA-negatively averaged.Definition 3.39. Let A : H ⇒ H and w ∈ H. The corresponding innerw-perturbation of A is the operator Aw : H⇒ H defined byAw(x):= A(x− w), for all x ∈ H.For λ > 0, the Yosida λ-regularization of A is,λA = 1λ(Id−JλA). (3.14)The resolvent satisfies the useful resolvent identity,JA = Id−JA−1 . (3.15)Fact 3.40. [10, Proposition 23.11] Let A : H ⇒ H be monotone and letβ ∈ R++. Then A is strongly monotone with constant β if and only if JA is(β + 1)-cocoercive, in which case JA is Lipschitz continuous with constant1/(β + 1) ∈ ]0, 1[.Fact 3.41. [10, Proposition 23.35] Let A : H ⇒ H be strictly monotone.Then zerA is at most a singleton.Fact 3.42. [10, Corollay 23.37(ii)] Suppose A : H⇒ H is maximally mono-tone operator and strongly monotone. Then zerA is nonempty and single-ton.If A and B are maximally monotone operators, then the sum of themcan fail to be maximally monotone.Example 3.43. Let H = R2 and set C be the closed unit ball centered at(0,−1) and D be the closed unit ball centered at (0, 1). Let f = ιC , g = ιD,A = ∂f = NC , and B = ∂g = ND. Then A and B are maximally monotoneby Fact 3.34(i) and we havedomA ∩ domB = domNC ∩ domND = {0} 6= ∅.Note that ∂f(x)+ ∂g(x)= ∅ if x 6= 0. Then∂f(0)= NC(0)= R+ ·(0, 1)243.3. Monotone operator theoryand∂g(0)= ND(0)= R− ·(1, 0),Combining these together, we haveNC(0)+ND(0)= R · (1, 0) 6= R2.Therefore, NC +ND is not maximally monotone becauseran(Id +(NC +ND))= R · (1, 0) 6= R2.The next fact will discuss some properties that will ensure that the sumof two maximally monotone operators is maximally monotone.Fact 3.44. [10, Corollary 25.5] Let A,B : H⇒ H be maximally monotoneoperators. If domB = H, then A+B is maximally monotone.Suppose A and B are monotone operators. Then finding the zeros of thesum of two monotone operators can be written asFind x ∈ H such that 0 ∈ Ax+Bx⇔ zer(A+B)−1(0). (3.16)Fact 3.45. [10, Proposition 25.1(ii)] Let A,B : H⇒ H be monotone oper-ator. Let γ ∈ R++. Thenzer(A+B)= JγB(FixRγARγB).Example 3.46. Let A = PU , where U is a closed linear subspace of H. LetB = −v. Thenzer(A+B)= (A+B)−1(0) = {v + U⊥}. (3.17)Proof. Let z ∈ H, where zer (A+B) = {z}. Our goal is to find {z}. Then(A+B)z = PU(z)− v. (3.18)If (3.18) equals zero, then we must have PU(z)= v ⇒ v ∈ U . Since twovectors are equal if and only if their components are equal, we must havePU(z)= PU(v). Therefore, v ∈ U and PU(z)= PU(v) ⇒ v ∈ U andPU(z − v) = 0⇒ v ∈ U and z − v ∈ U⊥ ⇒ v ∈ U and z ∈ v + U⊥.Definition 3.47 (Resolvent average). Let Ai, i = 1, 2, . . . , n be monotoneoperators, λi > 0 with∑ni=1 λi = 1, and µ > 0. For A =(A1, A2, . . . , An)and λ =(λ1, λ2, . . . , λn)the resolvent average of A is,Rµ(A,λ):=[λ1(A1 + µ−1 Id)−1 + · · ·+ λn(An + Id )−1]− µ−1 Id .253.4. Attouch-The´ra dualityWhen µ = 1, the resolvent average is(R1(A,λ)+ Id )−1 = λ1(A1 + Id )−1 + · · ·+ λn(An + Id )−1,which indicates that the resolvent of R1(A,λ)is the arithmetic average ofthe resolvents of Ai, with weight λ =(λ1, λ2, . . . , λn). For more informationabout the resolvent average, see [20] and [9].3.4 Attouch-The´ra dualityFor convenience, setA> = (− Id ) ◦A ◦ (− Id ),andA−> := (A−1)> = (A>)−1.Fact 3.48 (primal problem). [14, Definition 2.1] Suppose that A and B aremaximally monotone operators on H. The primal problem, for the orderpair(A,B), is to find the zero of A + B. We define the primal problemwith respect to the (ordered) pair(A,B). The reason we must do this is toassociate a unique dual problem.Since A and B form a pair of maximally monotone operators, so do A−1and B−>: We thus define the dual pair(A,B)∗ := (A−1, B−>). (3.19)Moreover,(A−1)−1=A, (B>)>=B and (B>)−1=(B−1)>. We have ”bidual-ity”: (A,B)∗∗ = (A,B). (3.20)Fact 3.49 ((Attouch-The´ra) dual problem). [14, Definition 2.2] Supposethat A and B are maximally monotone operators on H. The Attouch-The´radual problem, for the order pair(A,B), is to find the zero of A−1 + B−>.We define the primal problem with respect to the (ordered) pair(A,B)tofind a zero of A+B.Fact 3.50 (primal and dual solutions). [14, Definition 2.3] The primal so-lutions are the solutions to the primal problem and analogously for the dualsolutions. DefineZ :=(A+B)−1(0) (3.21)263.5. Douglas-Rachford and Arago´n Artacho-Campoy algorithmsandK :=(A−1 +B−>)−1(0). (3.22)Then:(i) Z = {x ∈ H | (∃x∗ ∈ K) x∗ ∈ Ax and − x∗ ∈ Bx}.(ii) K = {x∗ ∈ H | (∃x ∈ Z) x ∈ A−1x∗ and − x ∈ B−>x∗}.Let us make this simple but important equivalence a little more precise. Inorder to do so, we define(∀z ∈ H) Kz = (Az) ∩ (−Bz), (3.23)and (∀k ∈ H) Zk = (A−1k) ∩ (−B−>k) = (A−1k) ∩ (B−1(−k)). (3.24)Fact 3.51. [14, Corollary 3.2 and Proposition 2.4 (iii) and (vi)] Let z ∈ Hand k ∈ H. Recall (3.23) and (3.24). Then the following hold:(i) Z 6= ∅⇔ K 6= ∅.(ii) z ∈ Z ⇔ Kz 6= ∅.(iii) k ∈ K ⇔ Zk 6= ∅.3.5 Douglas-Rachford and Arago´nArtacho-Campoy algorithmsLet C1 and C2 be nonempty, closed and convex subsets of H, whereC1 ∩ C2 6= ∅,Find x ∈ C1 ∩ C2. (3.25)Equation (3.25) is referred to as the convex feasibility problem. In order tosolve these types of problems many algorithms have been created. One ofthose is the Douglas-Rachford (DR) (see [10, 14, 21]). For any initial pointx0 ∈ H, the DR algorithm is generates a sequence that iteratively definedbyxn+1 :=(1−α)xn+α(2 PC2 − Id )(2 PC1 − Id )(xn), n = 0, 1, 2, . . . , (3.26)where α ∈ ]0, 1[, and PC and Id denote the projector onto the set C and theidentity mapping respectively. It is known the DR algorithm can be applied273.5. Douglas-Rachford and Arago´n Artacho-Campoy algorithmsto monotone operators, specifically, it can be used to solve problem of theformFind x ∈ zer (A+B) = {x ∈ H | 0 ∈ Ax+Bx}, (3.27)where A : H⇒ H and B : H⇒ H are maximally monotone operators. Thestructure of the iteration is the same as (3.26), but the projectors onto thesets are replaced by the resolvents JA and JB, i.e.,xn+1 :=(1− α)xn + α(2JB − Id )(2JA − Id )(xn), n = 0, 1, 2, . . . . (3.28)The resolvent of a normal cone to a convex set coincides with the projectoronto the set, see Fact 3.34(ii). Hence, the feasibility problem (3.25) can bewritten in the form (3.27) by taking A = NC1 and B = NC2 and (3.28)becomes the DR iteration for solving the feasibility problem (3.25).If the goal is not just finding a point in the intersection of nonempty closedconvex sets but also finding the closest point in that intersection to a givenpoint w ∈ H, then the problem will be characterized as the best approxima-tion problem, which isFind p ∈ C1 ∩ C2 such that ‖p− w‖ = infx∈C1∩C2‖x− w‖ . (3.29)The Arago´n Artacho-Campoy algorithm (AACA) has been invented to solvethis type of problem. For any initial point x0 ∈ H, the AACA algorithmgenerates a sequence that is iteratively defined byxn+1 :=(1− α)xn + α(2γ PC2−w− Id )(2γ PC1−w− Id )(xn), (3.30)where n = 0, 1, 2, . . . and α, γ ∈ ]0, 1[ . The authors assume the constraintqualificationw ∈ ( Id +NC1 +NC2)(PC1∩C2 (w)). (3.31)In 2018, the AACA was generalized from normal cone operators to maxi-mally monotone operators (see [6]). Given a point w ∈ dom (JA+B), thegeneralized version of best approximation problem (3.29) can be written asFind p = JA+B(w), (3.32)whereA,B are maximally monotone operators. If (3.31) holds andA = NC1 ,B = NC2 , then we havePC1∩C2(w)= JNC1+NC2(w),and therefore (3.32) becomes (3.29). For more details about AACA we referthe reader to [6] and [7].283.5. Douglas-Rachford and Arago´n Artacho-Campoy algorithmsFact 3.52 (Douglas-Rachford splitting algorithm with flexible parameters).[10, Theorem 26.11]. Let A,B : H ⇒ H be maximally monotone operatorssuch that zer(A + B) 6= ∅, let β > 0 and (λn)∞n=0 be a sequence in [0, 1]such that ∑n≥0 λn(1− λn) = +∞. Given any x0 ∈ H, setxn+1 =(1− λn)xn + λnRβBRβA(xn), for n = 0, 1, 2, . . .Then there exists x¯ ∈ Fix (RβBRβA) such that the following hold:(i)(xn+1 − xn)∞n=0 converges strongly to 0.(ii)(xn)∞n=0 converges weakly to x¯, and JβA(x¯) ∈ zer (A+B).(iii)(JβA(xn))∞n=0 converges weakly to JβA(x¯).(iv) Suppose that A and B is µ-strongly monotone for some constant µ >0. Then the sequence(JβA(xn))∞n=0 converges strongly to the uniquepoint in zer(A+B).Fact 3.53. [10, Theorem 26.11 and Proposition 26.13] Let A and B bemaximally monotone on H. Suppose that zer (A+B) = (A+B)−1(0) 6= ∅.Let λ ∈ ]0, 1] and setT =(1− λ) Id +λRBRA, (3.33)Let x0 ∈ H and define (∀n ∈ N) xn+1 = Txn. (3.34)Then there exists x¯ ∈ Fix T such that z¯ = JAx¯ ∈ zer(A + B)and thefollowing hold:(i) If A or B is strongly monotone, then zer(A+B)= {z¯}.(ii) If λ < 1, then xn ⇀ x¯ and JAxn ⇀ z¯.(iii) If λ < 1 and A or B is strongly monotone, then JAxn → z¯.(iv) If λ = 1 and A is strongly monotone, then JAxn → z¯.Fact 3.54. [10, Theorem 23.44] Let C : H ⇒ H be maximally monotoneand let x ∈ H. Then the inclusions(∀δ ∈ ]0, 1[) 0 ∈ Cδ + δ(xδ − x)define a unique curve(xδ)δ∈]0,1[. Moreover, exactly one of the followingholds:293.5. Douglas-Rachford and Arago´n Artacho-Campoy algorithms(i) zerC 6= ∅ and xδ → PzerC x as δ → 0.(ii) zerC = ∅ and ‖xδ‖ → +∞ as δ → 0.Proposition 3.55. Suppose A,B : H⇒ H be maximally monotone opera-tors. Let ζA > 0 and ζB > 0. Suppose that A is ζA-strongly monotone andB is ζB-strongly monotone. Then the following hold:(i) RA is θA-negatively averaged and RB is θB-negatively averaged, whereθA = 11+ζA and θB =11+ζB .(ii) The composition RBRA is α :=(k(k + 1)−1)-averaged wherek = θB(1− θB)−1 + θA(1− θA)−1 =(ζB + ζA)(ζAζB)−1.Proof. (i): Set θA :=(1 + ζA)−1 and θB := (1 + ζB)−1. It follows fromFact 3.38 that RA and RB are θA-negatively averaged and θB-negativelyaveraged respectively.(ii): Set S := Id−RBRA and letkθA = θA(1− θA)−1 = (ζA)−1andkθB = θB(1− θB)−1 = (ζB)−1.Then‖Sx− Sy‖2 = ∥∥( Id−RBRA)x− ( Id−RBRA)y∥∥2=∥∥(x− y)− (RAx−RAy)+ (RAx−RAy)− (RBRAx−RBRAx)∥∥2.Hence,‖Sx− Sy‖2 = ∥∥( Id−RA)x− ( Id−RA)y + ( Id−RB)RAx− ( Id−RB)RAy∥∥2. (3.35)Lett : =(kθA(kθB + kθA)−1)= ζB(ζB + ζA)−1 ∈ ]0, 1[,which implies1− t=(kθB (kθB + kθA)−1) = ζA(ζB + ζA)−1 ∈ ]0, 1[.Moreover, for simplicity, letx˜ =(Id−RA)x− ( Id−RA)y and y˜ = ( Id−RB)RAx− ( Id−RB)RAy.303.5. Douglas-Rachford and Arago´n Artacho-Campoy algorithmsThen the convexity of ‖·‖2 and (3.35) give‖Sx− Sy‖2 = ‖x˜+ y˜‖2 = ∥∥t(1/t)x˜+ (1− t)(1/(1− t))y˜∥∥2≤ t ‖(1/t)x˜‖2 + (1− t) ∥∥(1/(1− t))y˜∥∥2= (1/t) ‖x˜‖2 + (1/(1− t)) ‖y˜‖2 .Hence,‖x˜+ y˜‖2 ≤ (ζB + ζA)ζ−1B ‖x˜‖2 + (ζB + ζA)ζ−1A ‖y˜‖2 . (3.36)It follows from Fact 3.14 that(ζB + ζA)ζ−1B ‖x˜‖2 ≤(ζB + ζA)(ζAζB)−1( ‖x− y‖2 − ‖RAx−RAy‖2 ),and (ζB + ζA)ζ−1A ‖y˜‖2 ≤(ζB + ζA)(ζAζB)−1( ‖RAx−RAy‖2− ‖RBRAx−RBRAy‖2).Substituting this back into (3.36) gives‖x˜+ y˜‖2 ≤ (ζB + ζA)(ζAζB)−1( ‖x− y‖2 − ‖RBRAx−RBRAy‖2 )= k(‖x− y‖2 − ‖RBRAx−RBRAy‖2).Hence, we have‖Sx− Sy‖2 ≤ k(‖x− y‖2 − ‖RBRAx−RBRAy‖2). (3.37)The relation in (3.37) coincides with the definition of the averagedness inFact 3.14. Thus, RBRA is α-averaged with α satisfying 1−αα =1k . This givesα = kk+1 and the proof is complete.Fact 3.56 (Krasnosel´ski˘ı-Mann algorithm). [10, Proposition 5.15] Let C bea nonempty closed convex subset of H, let T : C → C be a nonexpansiveoperator such that Fix T 6= ∅, let (λn)∞n=0 be a sequence in [0, 1] such that∑n∈N λn(1− λn)= +∞, and let x0 ∈ C. Set(∀n ∈ N) xn+1 = xn + λn(Txn − xn).Then the following hold:(i)(xn+1 − xn)∞n=0 converges strongly to 0.313.5. Douglas-Rachford and Arago´n Artacho-Campoy algorithms(ii)(xn)∞n=0 converges weakly to a point in Fix T .Fact 3.57. [10, Proposition 5.15] Let α ∈ ]0, 1[, let T : H → H be α-averaged operator such that Fix T 6= ∅, let (λn)∞n=0 be a sequence in [0, 1/α]such that ∑n∈N λn(1− αλn) = +∞, and let x0 ∈ H. Set(∀n ∈ N) xn+1 = xn + λn(Txn − xn).Then the following hold:(i)(xn+1 − xn)∞n=0 converges strongly to 0.(ii)(xn)∞n=0 converges weakly to a point in Fix T .Fact 3.58. [7, Theorem 3.1 ] Let A and B be maximally monotone on H,let α > 0 and suppose that zer(A+ B)=(A+ B)−1(0) 6= ∅. Let (λn)∞n=0be a sequence in[0, 1]such that ∑n≥0 λn(1−λn) = +∞. Let γ ∈ ]0, 1[ andsuppose that w ∈ ran(Id + α2(1−γ)(A + B)). Given any x0 ∈ H, for everyn = 0, 1, . . . , setxn+1 =(1− λn)xn + λn(2γJ(αB−w) − Id)(2γJ(αA−w) − Id)(xn). (3.38)Then there exists x¯ ∈ Fix(2γJ(αB−w)− Id)(2γJ(αA−w)− Id)such that thefollowing hold:(i)(xn+1 − xn)∞n=0 converges strongly to 0.(ii)(xn)∞n=0 converges weakly to x¯, and JαA(w + x¯)= J α2(1−γ) (A+B)(w).(iii)(JαA(w + xn))∞n=0 converges strongly to J α2(1−γ) (A+B)(w).Fact 3.59. [10, Theorem 26.14] Let A : H⇒ H, let β ∈ R++, let B : H → Hbe β-cocoercive, let γ ∈ ]0, 2β[, and set δ = 2 − γ/(2β). Furthermore, let(λn)∞n=0 be a sequence in [0, δ] such that∑n≥0 λn(1 − λn)= +∞, and letx0 ∈ H. Suppose that zer(A+B) 6= ∅ and setxn+1 = xn + λn(JγA(xn − γBxn)− xn).Then the following hold:(i)(xn)∞n=0 converges weakly to a point in zer(A+B).(ii) Let x ∈ zer (A+B). Then (Bxn)∞n=0 converges strongly to the uniquedual solution of Bx.32Chapter 4On the AsymptoticBehaviour of the Arago´nArtacho-Campoy AlgorithmFrom now on, suppose that C : H⇒ H, w ∈ H, γ ∈ ]0, 1[. SetCγ : H⇒ H : x 7→ C(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w). (4.1)4.1 OverviewThis chapter is based on our paper [1]. In this chapter I advanced theanalysis of the Arago´n Artacho-Campoy algorithm (AACA) using the prox-imal and resolvent average point of view. We complete the analysis of [7]by demonstrating that the underlying curve converges to the nearest zero ofthe sum of the two operators. We also provide a new interpretation of theunderlying operators in terms of the resolvent and the proximal average.We summarize our main results in this chapter as follows:• We derive explicit formulas for resolvent and the reflected resolvent ofCγ (see Theorem 4.1(i)&(ii)). We illustrate our results by consideringlinear and affine operators (see Lemma 4.3 and Lemma 4.5).• We collect several useful properties of the operator Cγ , JCγ and RCγ(see Theorem 4.10, Theorem 4.1(iv), Lemma 4.6, Lemma 4.7, Lemma4.8, and Lemma 4.9 ). We also provide a new interpretation of Cγ interms of the resolvent and the proximal average (see Theorem 4.10(i)and Proposition 4.11).• Theorem 4.15, Lemma 4.12, Lemma 4.13, and Lemma 4.14 give im-portant properties of the sum of two operators Aγ and Bγ , whereAγ : H⇒ H : x 7→ A(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w).334.2. Resolvent and the reflected resolvent of CγBγ : H⇒ H : x 7→ B(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w),w ∈ H, and γ ∈ ]0, 1[.• A new variant of a convergence result for AACA (see Theorem 4.17).• We rewrite the operator of the original AACA with fixed γ (see [7])using the new operator 4.16. We also study the convergence result ofthe redefine one (see Proposition 4.20).• The announced asymptotic behaviour of the curve is given in Theo-rem 4.21.• We end the chapter by illustrating our results by two Examples (seeExample 4.23 and Example 4.25).4.2 Resolvent and the reflected resolvent of CγIn this section we derive explicit formulas for resolvent and the reflectedresolvent of Cγ . We illustrate our results by considering linear and affineoperators. We also collect some useful properties of the resolvent and thereflected resolvent of Cγ .Theorem 4.1. Let ζCγ ≥ 0 and recallCγ : H⇒ H : x→ C(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w),where x ∈ H, w ∈ H, and γ ∈ ]0, 1[. Then the following hold:(i) JCγ = γJC +(1− γ)w.(ii) RCγ = 2γJC + 2(1− γ)w − Id .(iii) JCγ is a convex combination of the resolvents JC and P{w}.(iv) If Cγ is ζCγ -strongly monotone and maximally monotone, then RCγ is(γ/(1 + ζCγ))-negatively averaged.Proof. (i): For any x ∈ H, we have (1− γ)w + γJCx = JCγx⇔ (1− γ)w +γJCx = (Id +Cγ)−1x ⇔ x ∈ (1 − γ)w + γJCx + Cγ((1 − γ)w + γJCx) ⇔x − (1 − γ)w − γJCx ∈ Cγ((1 − γ)w + γJCx) ⇔ x − (1 − γ)w − γJCx ∈C(JCx)+γ−1(1−γ)((1−γ)w+γJCx)−γ−1(1−γ)w ⇔ x−(1−γ)w−γJCx ∈344.2. Resolvent and the reflected resolvent of CγC(JCx)+(1−γ)JCx−(1−γ)w ⇔ x ∈(Id +C)JCx⇔ JCx =(Id +C)−1(x),which is true. Hence,JCγ = γJC +(1− γ)w.(ii): This follows from (3.12) and (i).(iii): Note that P{w} = w, γ > 0,(1 − γ) > 0, and γ + (1 − γ) = 1. Thenfrom (i), we haveJCγ = γJC +(1− γ)w = γJC + (1− γ)P{w},i.e., JCγ is a convex combinations of JC and P{w}.(iv): It follows from Fact 3.38 that RCγ is(γ/(1+ζCγ))-negatively averaged.Example 4.2. Suppose C = −v. Let x,w ∈ H and γ ∈ ]0, 1[. Then byExample 3.33, we have JC = Id +v and RC = Id +2v. By using Theo-rem 4.1(i)&(ii), we obtain JCγ = γ(Id +v)+(1 − γ)w and RCγ = (2γ −1)Id +2γv + 2(1− γ)w.Lemma 4.3. Let C = a+ PU where U is a linear closed subspace of H anda ∈ H. Let Cγ : H ⇒ H : x→ C(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w),where x ∈ H, w ∈ H and γ ∈ ]0, 1[. Then(i) JC = (Id−12 PU ) + (12 PU − Id) a.(ii) RC = (Id−PU ) + (PU −2 Id) a.(iii) JCγ = γ((Id−12 PU ) + (12 PU − Id)a)+ (1− γ)w.(iv) RCγ = (2γ − 1) Id−γ PU +γ(PU −2 Id)a+ 2(1− γ)w.Proof. (i): Let y ∈ H and set x = JCy. Our goal is to find x. Using (3.28)givesy ∈ (Id +C)x⇔ y = x+a+PU x⇔ y = x+a+x∗, where x∗ = PU (x) (4.2)Applying PU to (4.2) givesPU y = PU (x+ a+ x∗) = 2x∗ + PU a⇔ x∗ = 12 PU y −12 PU a. (4.3)Substituting (4.3) back into (4.2) givesy = x+ a+ 12 PU y −12 PU a⇔ x = y − a−12 PU y +12 PU a.354.2. Resolvent and the reflected resolvent of CγHence,JC =(Id−12 PU)+(12 PU − Id)a.(ii): From (3.12) and (i), we haveRC =(Id−PU)+(PU −2 Id)a.(iii): This follows from Theorem 4.1(i) and (i).(iv): This follows from Theorem 4.1(ii) and (ii).Remark 4.4. If C = PU where U is closed linear subspace, then by Lemma 4.3,we obtain JC = Id−12 PU , RC = Id−PU , JγC = γ (Id−12 PU ) + (1 − γ)w,and RCγ = (2γ − 1) Id−γ PU +2(1− γ)w.Lemma 4.5. Let C = Pa+U , where U is a closed linear subspace and a ∈ H.Let Cγ : H ⇒ H : x → C(γ−1(x − (1 − γ)w)) + (1 − γ)γ−1(x − w), wherex ∈ H, w ∈ H and γ ∈ ]0, 1[. Then(i) JC = (Id−12 PU )− PU⊥ a.(ii) RC = (Id−PU )− 2 PU⊥ a.(iii) JCγ = γ((Id−12 PU )− PU⊥ a)+ (1− γ)w.(iv) RCγ = (2γ − 1) Id−γ PU −2γ PU⊥ a+ 2(1− γ)w.Proof. (i): Let y ∈ H and set x = JCy. Our goal is to find x. By using(3.28) and Fact 2.14, we obtainy ∈ (Id + Pa+U )x⇔ y = x+ a+ PU (x− a)⇔ y = x+ (Id−PU )a+ PU x⇔ y = x+ PU⊥ a+ PU x.Hence,y = x+ PU⊥ a+ x∗, where x∗ = PU x. (4.4)Applying PU to (4.4) givesPU y = PU x+ PU PU⊥ a+ x∗ ⇔ PU y = 2x∗ ⇔ x∗ =12 PU y. (4.5)Substituting (4.5) back into (4.4) givesy = x+ PU⊥ a+12 PU y ⇔ x =(Id−12 PU)y − PU⊥ a.364.3. Properties of the operator CγTherefore,JC = (Id−12 PU )− PU⊥ a. (4.6)(ii): By using (4.6) and (3.12), we obtainRC = 2(Id−12 PU)− 2 PU⊥ a− Id = (Id−PU )− 2 PU⊥ a. (4.7)(iii): This follows from (i) and Theorem 4.1(i).4.3 Properties of the operator CγIn this section, we collect several useful properties of the operator Cγ .We also provide a new interpretation of Cγ in terms of the resolvent and theproximal average.Lemma 4.6. Let C : H⇒ H and let (x, u) ∈ H ×H. Then(x, u) ∈ graCγ ⇔ (γ−1(x− (1− γ)w), u− (1− γ)γ−1(x− w)) ∈ graC.Proof. Let(x, u) ∈ graCγ ⇔ u ∈ Cγ(x)⇔ u ∈ C(γ−1(x− (1−γ)w))+ (1−γ)γ−1(x−w) by (4.1) ⇔ u− (1− γ)γ−1(x−w) ∈ C(γ−1(x− (1− γ)w))⇔(γ−1(x− (1− γ)w), u− (1− γ)γ−1(x− w)) ∈ graC.Lemma 4.7. Let C : H⇒ H and let (x˜, u˜) ∈ H ×H. Then(x˜, u˜) ∈ graC ⇔ (γx˜+ (1− γ)w, u˜+ (1− γ)(x˜− w)) ∈ graCγ .Proof. Let(γx˜+ (1− γ)w, u˜+ (1− γ)(x˜−w)) ∈ graCγ ⇔ (γ−1(γx˜+ (1−γ)w− (1−γ)w), u˜+(1−γ)(x˜−w)− (1−γ)γ−1(γx˜+(1−γ)w−w)) ∈ graCby Lemma 4.6 ⇔ (x˜, u˜ + (1 − γ)(x˜ − w) − (1 − γ)γ−1(γx˜ − γw)) graC ⇔(x˜, u˜+ (1− γ)(x˜− w)− γ−1γ(1− γ)(x˜− w)) graC ⇔ (x˜, u˜) ∈ graC.Lemma 4.8. Let C : H⇒ H, (x¯, u¯) ∈ H ×H, and ζC ≥ 0. Then(x¯, u¯) ∈ gra (Cγ − γ−1(ζC + 1− γ) Id )m(γ−1(x¯− (1− γ)w), u¯+ γ−1(1 + ζC)(1− γ)w) ∈ gra (C − ζC Id ).374.3. Properties of the operator CγProof. Let(x¯, u¯) ∈ gra (Cγ − γ−1(ζC + 1− γ) Id )⇔ u¯+ γ−1(ζC + 1− γ)x¯ ∈Cγ(x¯). This is equivalent by (4.1) to⇔ u¯+ γ−1(ζC + 1− γ)x¯− (1− γ)γ−1(x¯− w) ∈ C(γ−1(x¯− (1− γ)w))⇔ u¯+ γ−1(ζC x¯+ (1− γ)w) ∈ C(γ−1(x¯− (1− γ)w))⇔ u¯+ γ−1(ζC x¯+ (1− γ)w)− ζC(γ−1(x¯− (1− γ)w))∈ C(γ−1(x¯− (1− γ)w))− ζC(γ−1(x¯− (1− γ)w))⇔ u¯+ γ−1(1 + ζC)(1− γ)w ∈ (C − ζC Id )(γ−1(x¯− (1− γ)w)),which is equivalent to((γ−1(x¯− (1− γ)w)), u¯+ γ−1(1 + ζC)(1− γ)w) ∈ gra (C − ζC Id ).Lemma 4.9. Let C : H⇒ H, (x´, u´) ∈ H ×H, and ζC ≥ 0. Then(x´, u´) ∈ gra (C − ζC Id )m(γx´+ (1− γ)w, u´− γ−1(1 + ζC)(1− γ)w) ∈ gra (Cγ − γ−1(ζC + 1− γ) Id ).Proof. Let(γx´+ (1− γ)w, u´− γ−1(1 + ζC)(1− γ)w) ∈ gra (Cγ − γ−1(ζC +1 − γ) Id ) ⇔ (γ−1(γx´ + (1 − γ)w − (1 − γ)w), u´ − γ−1(1 + ζC)(1 − γ)w +γ−1(1 + ζC)(1 − γ)w) ∈ gra (C − ζC Id ) by Lemma 4.8. Hence, we obtain(x´, u´) ∈ gra (C − ζC Id ).Theorem 4.10. Let C : H⇒ H, γ ∈ ]0, 1[, and ζC ≥ 0. Recall thatCγ : H⇒ H : x 7→ C(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w).Then the following hold:(i) Cγ is a resolvent averages of C and N{w}.(ii) If C is monotone, then Cγ is (1− γ)γ−1-monotone.(iii) If C is ζC-strongly monotone, i.e., C − ζC Id is monotone. Then Cγ isγ−1(ζC + 1− γ)-strongly monotone.(iv) If C is maximally monotone, then D := Cγ−(1−γ)γ−1 Id is maximallymonotone.384.3. Properties of the operator Cγ(v) If C is maximally monotone, then Cγ is also maximally monotone.(vi) If C is ζC-strongly monotone, then Cγ is γ−1(ζC + 1 − γ)-stronglymonotone.(vii) If C is ζC-strongly monotone and maximally monotone, then zerCγ isa nonempty and singleton.Proof. (i): It follows from Theorem 4.1(i)&(iii), (3.11), and Fact 3.34(ii).(ii): Take(x, u)and(y, v)in graCγ . Then u ∈ Cγ(x)and v ∈ Cγ(y). ByLemma 4.6, we obtain(γ−1(x− (1− γ)w), u− (1− γ)γ−1(x− w)) ∈ graC,and (γ−1(y − (1− γ)w), v − (1− γ)γ−1(y − w)) ∈ graC.Since C is monotone operator on H, (3.6) gives0 ≤〈γ−1(x− (1− γ)w)− γ−1(y − (1− γ)w),(u− (1− γ)γ−1(x− w))− (v − (1− γ)γ−1(y − w))〉= γ−1〈x− y, (u− v)− (1− γ)γ−1(x− y)〉 .Multiplying by γ because 0 < γ < 1 gives0 ≤〈x− y, (u− v)− (1− γ)γ−1(x− y)〉⇔(1− γ)γ−1 ‖x− y‖2≤ 〈x− y, u− v〉 .Hence, Cγ is (1− γ)γ−1-monotone.(iii): Let (x, u) ∈ graCγ . By (??), u ∈ Cγ(x) gives u− γ−1(1− γ)(x−w) ∈C(γ−1(x− (1− γ)w))⇔ u− ζC(γ−1(x− (1− γ)w))+ γ−1(1− γ)(x−w) ∈(C − ζC Id)(γ−1(x− (1− γ)w)). Hence,(γ−1(x−(1−γ)w), u−ζC(γ−1(x−(1−γ)w))+γ−1(1−γ)(x−w)) ∈ graC−ζC Id .Similarly, (y, v) ∈ graCγ , we have(γ−1(y−(1−γ)w), v−ζC(γ−1(y−(1−γ)w))+γ−1(1−γ)(y−w)) ∈ graC−ζC Id .(iv): Let(x, u) ∈ H ×H and assume(∀(y, v) ∈ graD) 〈x− y, u− v〉 ≥ 0. (4.8)394.3. Properties of the operator CγWe must show that(x, u) ∈ graD ⇔ u ∈ D(x), which is equivalent tou ∈ Cγ(x)− (1− γ)γ−1x⇔ u+ (1− γ)γ−1x ∈ Cγ(x)⇔ (x, u+ (1− γ)γ−1x) ∈ graCγ⇔ (γ−1(x− (1− γ)w), u+ (1− γ)γ−1x− (1− γ)γ−1(x− w)) ∈ graC,from Lemma 4.6. Hence, we must prove that(γ−1(x− (1− γ)w), u+ (1− γ)γ−1w) ∈ graC (4.9)Set x˜ := γ−1(x − (1 − γ)w) and u˜ := u + (1 − γ)γ−1w. Note that x =γx˜ + (1 − γ)w and u = u˜ − (1 − γ)γ−1w. Now let (y˜, v˜) ∈ graC and sety := γy˜ +(1− γ)w and v := v˜ − (1− γ)γ−1w. Then let (y, v) ∈ graD. By(4.8), we obtain0 ≤ 〈x− y, u− v〉=〈(γx˜+(1− γ)w)− (γy˜ + (1− γ)w),(u˜− (1− γ)γ−1w)− (v˜ − (1− γ)γ−1w)〉= γ 〈x˜− y˜, u˜− v˜〉 γ>0==⇒ 0 ≤ 〈x˜− y˜, u˜− v˜〉 .Since C is maximally monotone on H, we have (x˜, u˜) ∈ graC. Hence, (4.9)does hold, as desired. This implies that(x, u) ∈ graD. Therefore, D ismaximally monotone operator on H.(v): Clearly,(1 − γ)γ−1 Id is maximally monotone and full domain. Also,Cγ −(1− γ)γ−1γ−1 Id is maximally monotone by (iv). Hence,Cγ = Cγ −(1− γ)γ−1 Id +(1− γ)γ−1 Idis maximally monotone by Fact 3.44.(vi): Note that C is ζC-strongly monotone implies that C − ζC Id is mono-tone. Our goal is to show that Cγ − γ−1(ζC + 1− γ)Id is monotone. Take(x, u) ∈ gra (Cγ − γ−1(ζC + 1− γ) Id ) and (y, v) ∈ gra (Cγ − γ−1(ζC + 1−γ)Id). By Lemma 4.8, we obtain(γ−1(x− (1− γ)w), u+ γ−1(1 + ζC)(1− γ)w) ∈ gra (C − ζC Id ),and (γ−1(y − (1− γ)w), v + γ−1(1 + ζC)(1− γ)w) ∈ gra (C − ζC Id ).404.3. Properties of the operator CγNow C − ζC Id is monotone, we have0 ≤〈γ−1(x− (1− γ)w)− γ−1(y − (1− γ)w),u+ γ−1(1 + ζC)(1− γ)w)− (v + γ−1(1 + ζC)(1− γ)w))〉= γ−1 〈x− y, u− v〉 .Hence,γ−1 〈x− y, u− v〉 ≥ 0 γ>0==⇒ 〈x− y, u− v〉 ≥ 0.Therefore, Cγ − γ−1(ζC + 1 − γ)Id is monotone ⇔ Cγ is γ−1(ζC + 1 − γ)-strongly monotone.(vii): Items (v) and (vi) imply that Cγ is maximally monotone and stronglymonotone and hence zerCγ is a nonempty and singleton by Fact 3.42.Proposition 4.11. Let f : H → ]−∞,+∞] is proper, lower semicontinuousconvex function on H. Let γ ∈ ]0, 1[ and setfγ : H → ]−∞,+∞]: x 7→ inf{γf(y1)+(1− γ)ι{w}(y2)+ γ(1− γ)2 ‖y1 − y2‖∣∣∣γy1 + (1− γ)y2 = x}= γf(γ−1(x− (1− γ)w))+ 1− γ2γ ‖x− w‖2 ,which is the proximal average of f and ι{w}. Suppose that C = ∂f . Then(i) Cγ = ∂fγ .(ii) JCγ = Proxfγ .Proof. (i): Since C = ∂f , then by (4.1), we obtain∂fγ = ∂f(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w)= Cγ(x)(ii): It is clear from (i) and Fact 3.35(ii).414.4. Properties of the operator Aγ +Bγ4.4 Properties of the operator Aγ +BγIn this section, we collect some important properties of the sum of twooperators Aγ and Bγ . The following results will be effective. Let A : H⇒ Hand B : H⇒ H. Let w ∈ H and γ ∈ ]0, 1[. DefineAγ : H⇒ H : x 7→ A(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w). (4.10)Bγ : H⇒ H : x 7→ B(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w). (4.11)Lemma 4.12. Let A,B : H⇒ H and let (x, u) ∈ H ×H. Then(x, u) ∈ gra (Aγ +Bγ)m(γ−1(x− (1− γ)w), u− 2(1− γ)γ−1(x− w)) ∈ gra (A+B).Proof. Let(x, u) ∈ gra (Aγ + Bγ) ⇔ u ∈ (Aγ + Bγ)(x) ⇔ u ∈ (A +B)(γ−1(x − (1 − γ)w)) + 2(1 − γ)γ−1(x − w) by (4.10) & (4.11). This isequivalent tou− 2(1− γ)γ−1(x− w) ∈ (A+B)(γ−1(x− (1− γ)w))⇔ (γ−1(x− (1− γ)w), u− 2(1− γ)γ−1(x− w)) ∈ gra (A+B).Lemma 4.13. Let A,B : H⇒ H and let (x˜, u˜) ∈ H ×H. Then(x˜, u˜) ∈ gra (A+B)⇔ (γx˜+(1−γ)w, u˜+2(1−γ)(x˜−w)) ∈ gra (Aγ +Bγ).Proof. From Lemma 4.12, we have(γx˜ + (1 − γ)w, u˜ + 2(1 − γ)(x˜ − w)) ∈gra(Aγ + Bγ) ⇔ (γ−1(γx˜ + (1 − γ)w − (1 − γ)w), u˜ + 2(1 − γ)(x˜ − w) −2(1− γ)γ−1(γx˜+ (1− γ)w−w)) ∈ gra (A+B)⇔ (x˜, u˜+ 2(1− γ)(x˜−w)−2(1− γ)γ−1(γx˜− γw)) gra (A+B)⇔ (x˜, u˜) ∈ gra (A+B)Lemma 4.14. Let A,B : H⇒ H, (x˜, u˜) ∈ H×H. Let ζA ≥ 0 and ζB ≥ 0.Then (x˜, u˜) ∈ gra (Aγ +Bγ − γ−1(ζA + ζB + 2− 2γ) Id )m(γ−1(x˜− (1− γ)w), u˜+ (2 + ζA + ζB)(1− γ)γ−1w)∈ gra (A+B − (ζA + ζB) Id ).424.4. Properties of the operator Aγ +BγProof. Let(x˜, u˜) ∈ gra (Aγ+Bγ−γ−1(ζA+ζB+2−2γ) Id )⇔ u˜+γ−1(ζA+ζB + 2− 2γ)x˜ ∈ (Aγ +Bγ)(x˜). From (4.10) and (4.11), this is equivalent tou˜+γ−1(ζA+ζB+2−2γ)x˜ ∈ (A+B)(γ−1(x˜−(1−γ)w))+2(1−γ)γ−1(x˜−w)⇔u˜+(ζA+ζB)γ−1x˜+2(1−γ)γ−1w ∈ (A+B)(γ−1(x˜−(1−γ)w))⇔ u˜+2(1−γ)γ−1w+(ζA+ζB)(1−γ)γ−1w ∈ (A+B−(ζB+ζB) Id )(γ−1(x˜−(1−γ)w))⇔u˜+(2 + ζA + ζB)(1− γ)γ−1w ∈ (A+B − (ζA + ζB) Id )(γ−1(x˜− (1− γ)w).Hence,(γ−1(x˜−(1−γ)w, u˜+(2+ζA+ζB)(1−γ)γ−1w) ∈ gra (A+B−(ζA+ζB) Id ).Theorem 4.15. Let A,B : H ⇒ H, γ ∈ ]0, 1[, ζA ≥ 0, and ζB ≥ 0. RecallthatAγ : H⇒ H : x 7→ A(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w),andBγ : H⇒ H : x 7→ B(γ−1(x− (1− γ)w))+ (1− γ)γ−1(x− w).Then the following hold:(i) Suppose A and B are monotone. ThenAγ +Bγ is 2(1− γ)γ−1-monotone.(ii) Suppose that A + B is maximally monotone. Then E := Aγ + Bγ −2(1− γ)γ−1 Id is maximally monotone.(iii) Suppose A + B is maximally monotone. Then Aγ + Bγ is maximallymonotone.(iv) Suppose A and B are strongly monotone with constants ζA and ζBrespectively. ThenAγ +Bγ is γ−1(ζA + ζB + 2− 2γ)-strongly monotone.(v) Suppose that Aγ+Bγ is maximally monotone and γ−1(ζA+ζB+2−2γ)-strongly monotone. Then zer(Aγ +Bγ)is a singleton. Moreover,zer(Aγ +Bγ)={γJ(2(1−γ))−1(A+B)(w)+(1− γ)w}.434.4. Properties of the operator Aγ +BγProof. (i): It follows from (4.10), (4.11), and Theorem 4.10(ii) that Aγ is(1−γ)γ−1-monotone and Bγ is(1 − γ)γ−1-monotone. Now apply Lemma 3.27gives that Aγ +Bγ is 2(1− γ)γ−1-monotone.(ii): Let(x, u) ∈ H ×H and assume that∀(y, v) ∈ graE 〈x− y, u− v〉 ≥ 0. (4.12)We must show that(x, u) ∈ graE ⇔ u ∈ E(x), which is equivalent tou ∈ (Aγ +Bγ)(x)− 2(1− γ)γ−1x⇔ u+ 2(1− γ)γ−1x ∈ gra (Aγ +Bγ)(x)⇔ (x, u+ 2(1− γ)γ−1x) ∈ gra (Aγ +Bγ).By Lemma 4.12, we have(γ−1(x− (1− γ)w), u+ 2(1− γ)γ−1x− 2(1− γ)γ−1(x−w)) ∈ gra (A+B).Hence, we need to prove that(γ−1(x− (1− γ)w), u+ 2(1− γ)γ−1w) ∈ gra (A+B). (4.13)Set x˜ := γ−1(x − (1 − γ)w) and u˜ := u + 2(1 − γ)γ−1w. Note that x =γx˜+(1− γ)w and u = u˜− 2(1− γ)γ−1w. Now let (y˜, v˜) ∈ gra (A+B) andset y := γy˜ +(1− γ)w and v := v˜ − 2(1− γ)γ−1w. Then let (y, v) ∈ graE.By (4.12), we obtain0 ≤ 〈x− y, u− v〉=〈(γx˜+ (1− γ)w)− (γy˜ + (1− γ)w),u˜− 2(1− γ)γ−1w − (v˜ − 2(1− γ)γ−1w)〉= γ 〈x˜− y˜, u˜− v˜〉 .This implies that0 ≤ 〈x˜− y˜, u˜− v˜〉 .Since A + B is maximally monotone on H, we have (x˜, u˜) ∈ gra (A + B).Hence, (4.13) does hold, as desired. This implies that(x, u) ∈ graE. There-fore, E is maximally monotone on H.(iii): Clearly, 2(1−γ)γ−1 Id is maximally monotone and full domain. More-over, Aγ +Bγ − 2(1− γ)γ−1 Id is maximally monotone by (ii). Hence,Aγ +Bγ − 2(1− γ)γ−1 Id +2(1− γ)γ−1 Id = Aγ +Bγ ,444.4. Properties of the operator Aγ +Bγis maximally monotone by Fact 3.44.(iv): Note that A is ζA-strongly monotone⇔ A is ζA-monotone and B is ζB-strongly monotone⇔ B is ζB-monotone. Hence, A+B is ζA+ζB-monotoneby Lemma 3.27. Hence, A + B − (ζA + ζB) Id is monotone. Our goal is toshow that E := Aγ + Bγ − γ−1(ζA + ζB + 2 − 2γ)Id is monotone. Take(x, u) ∈ graE and (y, v) ∈ graE. Then by Lemma 4.14, we obtain(γ−1(x−(1−γ)w), u+γ−1(2+ζA+ζB)(1−γ)w) ∈ gra (A+B−(ζA+ζB) Id ),and(γ−1(y−(1−γ)w), v+γ−1(2+ζA+ζB)(1−γ)w) ∈ gra (A+B−(ζA+ζB) Id ).Now A+B − (ζA + ζB) Id is monotone, we have0 ≤〈γ−1(x− (1− γ)w)− γ−1(y − (1− γ)w),u+ γ−1(2 + ζA + ζB)(1− γ)w)− (v + γ−1(2 + ζA + ζB)(1− γ)w))〉= γ−1 〈x− y, u− v〉 .Hence,γ−1 〈x− y, u− v〉 ≥ 0 γ>0==⇒ 〈x− y, u− v〉 ≥ 0.Therefore, Aγ + Bγ − γ−1(ζA + ζB + 2− 2γ)Id is monotone ⇔ Aγ + Bγ isγ−1(ζA + ζB + 2− 2γ)-strongly monotone.(v): This is follows from Fact 3.42. Now let x ∈ H. Then x ∈ zer (Aγ +Bγ)⇔0 ∈ (Aγ +Bγ)(x) (By (4.10) & (4.11))⇔ 0 ∈ A(γ−1(x− (1− γ)w))+B(γ−1(x− (1− γ)w))+ 2γ−1(1− γ)(x− w)⇔ 0 ∈ (A+B)(γ−1(x− (1− γ)w))+ 2γ−1(1− γ)(x− w)⇔ 0 ∈ (2(1− γ))−1(A+B)(γ−1(x− (1− γ)w))+ γ−1(x− w)⇔ 0 ∈ (2(1− γ))−1(A+B)(γ−1(x− (1− γ)w))+ γ−1(x− (1− γ)w − γw)⇔ w ∈ (2(1− γ))−1(A+B)(γ−1(x− (1− γ)w))+ (γ−1(x− (1− γ)w))⇔ w ∈(Id +(2(1− γ))−1(A+B))(γ−1(x− (1− γ)w))⇔ γ−1(x− (1− γ)w) = J(2(1−γ))−1(A+B)(w) (By (3.11))⇔ x = γJ(2(1−γ))−1(A+B)(w)+(1− γ)w.454.5. The Arago´n Artacho-Campoy algorithm (AACA)In view of Theorem 4.15(v), we will denote the unique point in zer(Aγ +Bγ)by zγ :zer(Aγ +Bγ)= {zγ}. (4.14)4.5 The Arago´n Artacho-Campoy algorithm(AACA)Proposition 4.16. Let γ ∈ ]0, 1[, ζA ≥ 0 and ζB ≥ 0. Let Aγ and Bγbe as in (4.10) and (4.11) respectively. Suppose that A and B are stronglymonotone with constants ζA and ζB respectively. Moreover, suppose thatboth A and B are maximally monotone. Recall thatRAγ = 2γJA + 2(1− γ)w − Id and RBγ = 2γJB + 2(1− γ)w − Id .Then the following hold:(i) RAγ and RBγ are θAγ -negatively averaged and θBγ -negatively averagedrespectively, where θAγ = γ(ζA + 1)−1 and θBγ = γ(ζB + 1)−1.(ii) RBγRAγ is(k/(k + 1))-averaged operator wherek = γ(ζB + 1− γ)−1 + γ(ζA + 1− γ)−1.Proof. (i): Since A is ζA-strongly monotone and B is ζB-strongly mono-tone, then it follows from Theorem 4.10(vi) that Aγ and Bγ are stronglymonotone with constants γ−1(ζA + 1− γ)and γ−1(ζB + 1− γ)respectively.Moreover, A and B are maximally monotone operators implies that Aγ andBγ are maximally monotone operators by Theorem 4.10(v). Next applyingFact 3.38 gives that RAγ and RBγ are negatively averaged with constants(1 + (1/γ)(ζA + 1− γ))−1 = γ/(ζA + 1) := θAγ ,and (1 + (1/γ)(ζB + 1− γ))−1 = γ/(ζB + 1) := θBγrespectively.(ii): This follows from (i) and Proposition 3.55(ii) applied to RBγRAγ insteadof RBRA with θAγ = γ(ζA + 1)−1, θBγ = γ(ζB + 1)−1, andk = θBγ(1− θBγ)−1 + θAγ (1− θAγ )−1= γ(ζB + 1− γ)−1 + γ(ζA + 1− γ).464.5. The Arago´n Artacho-Campoy algorithm (AACA)Blanket assumptions. For the rest of this chapter, we suppose the fol-lowing(i) A and B are maximally monotone operators on H, w ∈ H, and γ ∈]0, 1[.(ii) Let ζA ≥ 0 and ζB ≥ 0 be such that A − ζA Id and B − ζB Id aremonotone.(iii) A+B is maximally monotone.In the following Theorem, we essentially rederive the central convergenceresult of Arago´n Artacho-Campoy algorithm Fact 3.58.Theorem 4.17 (AACA for fixed γ). Recall thatAγ : H⇒ H : x 7→ A(γ−1(x− (1− γ)w))+ γ−1(1− γ)(x− w),Bγ : H⇒ H : x 7→ B(γ−1(x− (1− γ)w))+ γ−1(1− γ)(x− w),andJAγ = γJA +(1− γ)w,where γ ∈ ]0, 1[. Let ζA ≥ 0, ζB ≥ 0, and λ ∈]0, 1]. Let x0 ∈ H and definethe sequence(xn)n∈N via (∀n ∈ N) xn+1 = Tnγ xn, (4.15)whereTγ =(1−λ) Id +λ(2γJB+2(1−γ)w−Id )◦(2γJA+2(1−γ)w−Id ). (4.16)Then the following hold:(i) There exists x¯ ∈ Fix (RBγRAγ ) = Fix Tγ such that xn ⇀ x¯ andγJAxn + (1− γ)w → zγ .(ii) JAxn → J(2(1−γ))−1(A+B)(w).Proof. From Theorem 4.10(ii), we haveRAγ = 2γJA + 2(1− γ)w − Id,RBγ = 2γJB + 2(1− γ)w − Id,474.5. The Arago´n Artacho-Campoy algorithm (AACA)andRBγRAγ =(2γJB + 2(1− γ)w − Id ) ◦ (2γJA + 2(1− γ)w − Id ).From Theorem 4.10(vi) both Aγ and Bγ are strongly monotone with con-stants γ−1(ζA+1−γ)and γ−1(ζB+1−γ)respectively. Therefore, zer(Aγ+Bγ) 6= ∅ by Theorem 4.15(v). Now, there are two cases depending on thevalue of λ:Case 1. Let 0 < λ < 1. We have RAγ and RBγ are nonexpansive byFact 3.37(ii). By using Fact 3.11, we have RBγRAγ is nonexpansive. Next,by using Fact 3.45, we obtain JAγ(Fix Tγ)= zer(Aγ + Bγ) 6= ∅. Then theresult follow from Fact 3.56 and Fact 3.53 applied to(Aγ , Bγ)instead of(A,B).Case 2. Let λ = 1. We obtain RAγ is γ(ζA + 1)−1-negatively averaged andRBγ is γ(ζB + 1)−1-negatively averaged by Proposition 4.16(i). Moreover,RBγRAγ is α-averaged by Proposition 4.16(ii). By using Fact 3.45, we obtainJAγ(Fix Tγ)= zer(Aγ +Bγ) 6= ∅. Then the result follow from Fact 3.57(ii)and Fact 3.53 applied to(Aγ , Bγ)instead of(A,B).(ii): It follows from (i) that JAxn →(zγ − (1− γ)w)/γ and from Theo-rem 4.15(v) thatzγ − (1− γ)w = γJ(2(1−γ))−1(A+B)(w)⇔ (zγ − (1− γ)w)/γ = J(2(1−γ))−1(A+B)(w).Hence,JAxn → J(2(1−γ))−1(A+B)(w).Remark 4.18. Several comments regarding regarding Theorem 4.17:(i) We focus on an easier-to-use version of AACA where the effect of w isrecorded.(ii) While one could make λ depending on n as [7], we decided instead tostress the new case λ = 1, corresponding to the Peaceman-Rachfordversion and notably absent in [7]. This case deserves interest becauseit turned out to be the best parameter choice in [15].(iii) Our assumption of maximal monotonicity makes for a tidy theory. Itis used chiefly to guarantee the existence of each zγ ; in [7], this isreplaced by some condition regarding the existence of zγ which seemsto be not so easy to check in practice.484.6. Original AACA with fixed γ4.6 Original AACA with fixed γLemma 4.19. [7, Lemma 2.2] Let A : H⇒ H and let w ∈ H. ThenJA−w = JA(·+ w)− w.Proof. For any x ∈ H, we havep ∈ JA−w(x)⇔ x ∈ p+A−w(p)⇔ x ∈ p+A(p+ w) (From Definition 3.39)x+ w ∈ p+ w +A(p+ w)⇔ p+ w ∈ JA(x+ w).Proposition 4.20. Let A,B : H ⇒ H be maximally monotone operators.Let w ∈ H and recall thatJA−w = JA( ·+w)− w.Let y0 ∈ H, γ ∈ ]0, 1[, and λ ∈]0, 1]. Define the sequence(yn)n∈N via(∀n ∈ N) yn+1 = T˜nγ yn, (4.17)whereT˜γ =(1− λ) Id +λ((2γJB−w − Id ) ◦ (2γJA−w − Id )).Let Tγ be as in (4.16). Then the following hold:(i) T˜γy =(1−λ)y+λ(2γJB(2γJA(y+w)−2γw−y+w)−2γJA(y+w)+y).(ii) T˜γ(y)= −w + Tγ(y + w).(iii) Let x0 ∈ H, set y0 = x0−w, set(∀n ∈ N) xn = Tnγ x0, and yn = T˜nγ y0.Then (yn)n∈N =(− w + xn)n∈N.(iv) JA(w + yn)→ J(2(1−γ))−1(A+B)(w).Proof. (i): Let y ∈ H. ThenT˜γy =(1− λ)y + λ((2γJB−w − Id ) ◦ (2γJA−w − Id ))y=(1− λ)y + λ((2γ(JB(·+ w)− w)− Id ) ◦ (2γ(JA(·+ w)− w)− Id ))y=(1− λ)y + λ((2γ(JB(·+ w)− w)− Id ) ◦ (2γJA(y + w)− 2γw − y))=(1− λ)y + λ(2γJB(2γJA(y + w)− 2γw − y + w)− 2γJA(y + w) + y).494.7. The behaviour of AACA when γ → 1−(ii): From Theorem 4.1(ii), we haveRAγ = 2γJA + 2(1− γ)w − Id and RBγ = 2γJB + 2(1− γ)w − Id .Note that (4.16) can be written asTγ(x)=(1− λ)x+ λRBγ (RAγ (x)), (4.18)and 2γJA = RAγ − 2(1− γ)w + Id. From (i), we haveT˜γy =(1− λ)y + λ(2γJB(2γJA(y + w)− 2γw − y + w)− 2γJA(y + w) + y)=(1− λ)y + λ(2γJB(2γJA(y + w) + 2w − 2γw − (y + w))− 2γJA(y + w)+ y)=(1− λ)y + λ(2γJB(RAγ (y + w))− 2γJA(y + w)+ y)=(1− λ)y + λ(2γJB(RAγ (y + w))− (RAγ (y + w)− 2(1− γ)w + (y + w))+ y)=(1− λ)y + λ(2γJB(RAγ (y + w))−RAγ (y + w) + 2(1− γ)w − (y + w) + y)=(1− λ)y + λ(2γJB(RAγ (y + w))+ 2(1− γ)w −RAγ (y + w)− w)=(1− λ)y + λ(RBγ(RAγ (y + w))− w)=(1− λ)y + λRBγ(RAγ (y + w))− λw=(1− λ)y + λRBγ(RAγ (y + w))+ (1− λ)w − w=(1− λ)(y + w)+ λRBγ(RAγ (y + w))− w.Using (4.18), we can see that this is precisely Tγ(y + w)− w.(iii): Let xn+1 = Tγxn and yn+1 = T˜γyn. Let x0 ∈ H and set y0 = x0 − w.Now, assume that yn = xn − w for some n ∈ N, and consider yn+1:yn+1 = T˜γ(yn)= T˜γ(xn − w)(by induction)= Tγ((xn − w)+ w)− w (by (ii))= Tγ(xn)− w= xn+1 − w.(iv): Combine (iii), Theorem 4.15(v), and Theorem 4.17(ii).4.7 The behaviour of AACA when γ → 1−Theorem 4.21 (dichotomy for AACA when γ → 1−1). Let γ ∈ ]0, 1[ andx,w ∈ H. Suppose A,B : H⇒ H are maximally monotone operators on H.504.7. The behaviour of AACA when γ → 1−RecallAγ : H⇒ H : x 7→ A(γ−1(x− (1− γ)w))+ γ−1(1− γ)(x− w),andBγ : H⇒ H : x 7→ B(γ−1(x− (1− γ)w))+ γ−1(1− γ)(x− w).Let zγ be as in (4.14). Then exactly one of the following holds:(i) zer(A+B) 6= ∅ and zγ → Pzer(A+B)(w) as γ → 1−.(ii) zer(A+B)= ∅ and ‖zγ‖ → ∞ as γ → 1−.Proof. (i): Let zγ be the unique zero of(Aγ +Bγ). Then0 ∈ Aγzγ +Bγzγ= A(γ−1(zγ − (1− γ)w))+B(γ−1(zγ − (1− γ)w))+ γ−12(1− γ)(zγ − w)=(A+B)(γ−1(zγ − (1− γ)w))+ γ−12(1− γ)(zγ − w)=(A+B)(γ−1(zγ − (1− γ)w))+ 2(1− γ)(γ−1(zγ − w + γw − γw))=(A+B)(γ−1(zγ − (1− γ)w))+ 2(1− γ)(γ−1(zγ − (1− γ)w)− w)Set yδ := γ−1(zγ − (1− γ)w)and δ := 2(1− γ). Then we have0 ∈ (A+B)(yδ)+ δ(yδ − w),where(A + B)is maximally monotone by assumption and δ → 0+ ⇔ γ →1−. By using Fact 3.54 we have two cases:(i) If zer(A+B) 6= ∅, thenlimδ→0+yδ = Pzer(A+B)w, (4.19)or equivalently,limγ→1−zγ = Pzer(A+B)w.(ii) If zer(A+B)= ∅, thenlimδ→0+‖yδ‖ = +∞, (4.20)or equivalently,limγ→1−‖zγ‖ = +∞.514.8. ExamplesRemark 4.22. Here are some comments on Theorem 4.21:(i) The information presented in Theorem 4.21(ii) is new even when Aand B are normal cone operators as in [7].(ii) Computing Pzer(A+B)w via Theorem 4.21 is doubly iterative: one mustfirst employ an algorithm to find zγ and then let γ tend 1−. However,we can avoid this double iteration and solve the problem via a singleiteration instead. See the discussion in [14, Section 8].4.8 ExamplesExample 4.23. Let λ ∈ ]0, 1] and γ ∈ ]0, 1[. Suppose that A = a + PU ,where a ∈ H and U is a closed linear subspace of H. Let B ≡ −v andT =(1− λ) Id +λRBRA.Recall from (4.16)Tγ =(1− λ) Id +λ(2γJB + 2(1− γ)w − Id ) ◦ (2γJA + 2(1− γ)w − Id ).Then the following hold:(i) zer(A+B) 6= ∅⇔ (v − a) ∈ U , in which casezer(A+B)=(v − a)+ U⊥.(ii) T(x)= x− λPU x+ λ(PU −2 Id)a+ 2λv.(iii) Fix T ={2(v − a)+ PU a+ U⊥, if (v − a) ∈ U∅, if(v − a) /∈ U(iv) JA(Fix T)= v +(PU −2 Id)a+ U⊥.(v) We have(a) If v − a ∈ U , then zer (Aγ +Bγ) = {(3− 2γ)−1(3(1− γ) PU w +γ(v − a))+ PU⊥ w}.524.8. Examples(b) If v − a ∈ U⊥, then zer (Aγ +Bγ) = {(3− 2γ)−13(1− γ) PU w+γ(2(1− γ))−1(v − a)+ PU⊥ w}.(c) If v − a /∈ U and v − a /∈ U⊥, then zer (Aγ + Bγ) = {(3 −2γ)−1(3(1−γ) PU w+γ PU (v−a))+γ(2(1−γ))−1 PU⊥ (v−a)+PU⊥ w}.(vi) Tγ(x)=(1− 4λγ(1− γ))x+ λγ(1− 2γ) PU x+ k, where k := λγ((1−2γ)(2 Id−PU)a+ 4(1− γ)w + 2v).(vii) Fix Tγ ={(3−2γ)−1(2 PU v+ (1−2γ) PU a+ 4(1−γ) PU w)+ (2(1−γ))−1((1− 2γ) PU⊥ a+ PU⊥ v)+ PU⊥ w}.(viii) JAγ(Fix Tγ)= Φ1(a)+ Φ2(w)+ Φ3(v), whereΦ1 := γ(2−1(1− 2γ)(3− 2γ)−1 PU − Id +(1− 2γ)(2(1− γ))−1 PU⊥ ),Φ2 :=(3− 2γ)−12γ(1− γ) PU +γ PU⊥ +(1− γ) Id,andΦ3 := γ((3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ ).(ix) Let w = 0. Then there are two cases:Case 1: If v − a ∈ U , thenlimγ→1−zγ = v − a = Pzer(A+B)(0).Case 2: If v − a /∈ U , thenlimγ→1−‖zγ‖ → ∞.Proof. (i): ”⇒”: Let z ∈ H, where zer (A+B) = {z}. Then(A+B)z = a+ PU z − v (4.21)If this equals zero, then we must have PU z = v − a ⇒ (v − a) ∈ U .Furthermore, PU (z) = PU (v)− PU (a)⇒ PU (z) = PU (v − a). Then(v − a) ∈ U and PU z = PU (v − a)⇔ (v − a) ∈ U and PU(z − (v − a)) = 0⇔ (v − a) ∈ U and z − (v − a) ∈ U⊥⇔ (v − a) ∈ U and z ∈ (v − a) + U⊥.534.8. ExamplesHence,zer(A+B) ⊆ (v − a) + U⊥. (4.22)”⇐”: Suppose x ∈ (v − a) + U⊥, where u := v − a ∈ U . Then there existsu⊥ ∈ U⊥ such that x = u+ u⊥. Then(A+B)x = a+ PU x− v = a+ PU(u+ u⊥)− v= a− v + PU x = a− v + u = −u+ u = 0.Hence, (v − a)+ U⊥ ⊆ zer (A+B). (4.23)Combining (4.22) and (4.23) giveszer(A+B)=(v − a)+ U⊥.(ii): Using Lemma 4.3(ii) and Example 4.2 givesT(x)=(1− λ)x+ λ( Id +2v)(x− PU x+ (PU −2 Id )a)=(1− λ)x+ λ(x− PU x+ (PU −2 Id )a)+ 2λv= x− λPU x+ λ(PU −2 Id)a+ 2λv.(iii): It follows from (ii) thatx = T(x)= x− λPU x+ λ(PU −2 Id)a+ 2λv,and,x− λPU x+ λ(PU −2 Id)a+ 2λv − x = 0.Thus, PU x = PU a+ 2(v − a). Therefore, we obtainFix T ={PU a+ 2(v − a)+ U⊥, if (v − a) ∈ U∅, if(v − a) /∈ U.(iv): Set k := PU a+ 2(v − a) ∈ U . Using (iii) and 4.3(i) giveJA Fix T =((Id−2−1 PU)+(2−1 PU − Id)a)(k + y), where y ∈ U⊥= k + y − 2−1k + (2−1 PU − Id )a=(PU −2 Id)a+ 2v + y − 2−1((PU −2 Id )a+ 2v)+ (2−1 PU − Id )a= v +(PU −2 Id)a+ y.544.8. Examples(v): Let zγ ∈ H, where(Aγ + Bγ)= {zγ}. By using (4.10) and (4.11), wehave(Aγ +Bγ)zγ = a− v + γ−1(PU +2(1− γ) Id)zγ − γ−1(1− γ)(PU +2 Id)w.(4.24)Applying PU and PU⊥ to (4.24) givesPU zγ = 3(1− γ)(3− 2γ)−1 PU w + γ(3− 2γ)−1(PU v − PU a),andPU⊥ zγ = PU⊥ w + γ(2(1− γ))−1(PU⊥ v − PU⊥ a).Thenzγ = PU zγ + PU⊥ zγ= 3(1− γ)(3− 2γ)−1 PU w + γ(3− 2γ)−1(PU v − PU a)+ PU⊥ w + γ(2(1− γ))−1(PU⊥ v − PU⊥ a).If(v − a) ∈ U , thenzγ =(3− 2γ)−1(3(1− γ) PU w + γ(v − a))+ PU⊥ w,and if(v − a) ∈ U⊥, thenzγ =(3− 2γ)−13(1− γ) PU w + γ(2(1− γ))−1(v − a)+ PU⊥ w.(vi): Using (4.16), Example 4.2, and Lemma 4.3(iv) giveTγ(x)=(1− λ)x+ λ[((2γ − 1) Id +2γv + 2(1− γ)w)((2γ − 1)x− γ PU x+ γ(PU −2 Id)a+ 2(1− γ)w)]=(1− λ)x+ λ((2γ − 1)2x− γ(2γ − 1) PU x+ γ(2γ − 1)(PU −2 Id )a+ 2(1− γ)(2γ − 1)w + 2γv + 2(1− γ)w)=((1 + 4λγ(γ − 1)) Id−λγ(2γ − 1) PU )x+ λγ((2γ − 1)(PU −2 Id )a+ 4(1− γ)w + 2v).(vii): It follows from (vi) thatx = g(x)=(1− 4λγ(1− γ))x+ λγ(1− 2γ) PU x+ λγ((1− 2γ)(2 Id−PU )a+ 4(1− γ)w + 2v),554.8. Examplesand−4(1−γ)x+(1−2γ) PU x+(1−2γ)(2 Id−PU)a+4(1−γ)w+2v = 0. (4.25)Applying PU and PU⊥ to (4.25) givesPU x =(3− 2γ)−1(2 PU v + (1− 2γ) PU a+ 4(1− γ) PU w),andPU⊥ x =(2(1− γ))−1((1− 2γ) PU⊥ a+ PU⊥ v)+ PU⊥ w.Thenx =(3− 2γ)−1(2 PU v + (1− 2γ) PU a+ 4(1− γ) PU w)+(2(1− γ))−1((1− 2γ) PU⊥ a+ PU⊥ v)+ PU⊥ w.(viii): For simplicity, setΨ1(a):=(1− 2γ)((3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ )(a),Ψ2(w):=(4(1− γ)(3− 2γ)−1 PU + PU⊥ )(w),andΨ3(v):=(2(3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ )(v).From (vii), we can see that Fix Tγ = {Ψ1(a)+ Ψ2(w)+ Ψ3(v)}. Now using(vii) and Lemma 4.3(iii) givesJAγ(Fix Tγ)= γ(Ψ1(a)+ Ψ2(w)+ Ψ3(v))− γ2−1 PU (Ψ1(a)+ Ψ2(w)+ Ψ3(v))+ γ(2−1 PU − Id)a+ (1− γ)w. (4.26)We haveγΨ1(a)= γ(1− 2γ)((3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ )(a), (4.27)γΨ2(w)= γ(4(1− γ)(3− 2γ)−1 PU + PU⊥ )(w), (4.28)andγΨ3(v)= γ(2(3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ )(v). (4.29)Moreover,2−1γ PU Ψ1(a)= 2−1γ(1− 2γ)(3− 2γ)−1 PU (a), (4.30)564.8. Examples2−1γ PU Ψ2(w)= 2γ(1− γ)(3− 2γ)−1 PU (w), (4.31)and2−1γ PU Ψ3(v)= γ(3− 2γ)−1 PU (v). (4.32)Substituting (4.27), (4.28), (4.29), (4.30), (4.31), (4.32) back into (4.26)givesJAγ(Fix Tγ)= γ(2−1(1− 2γ)(3− 2γ)−1 PU − Id+ (1− 2γ)(2(1− γ))−1 PU⊥ )(a)+((3− 2γ)−12γ(1− γ) PU +γ PU⊥ +(1− γ) Id )(w)+ γ((3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ )(v).(ix): Let w = 0. Then there are three cases:Case 1: If v− a ∈ U , then zγ =(3− 2γ)−1γ(v− a), by (v). Set u := v− a.Then (i) gives zer(A+B)= u+ U⊥, where u ∈ U . We havePzer(A+B) x = Pu+U⊥ x= u+ PU⊥(x− u) ( From Fact 2.14)= u+ PU⊥ x.This implies thatPzer(A+B)(0)= u = v − a.Hence, zγ → v − a = Pzer(A+B)(0).Case 2: If v − a /∈ U , thenzγ =(3− 2γ)−1γ PU (v − a)+ γ(2(1− γ))−1 PU⊥ (v − a),by (v). Moreover,‖zγ‖ =∥∥∥(3− 2γ)−1γ PU (v − a)− (− γ(2(1− γ))−1 PU⊥ (v − a))∥∥∥≥∥∥∥(3− 2γ)−1γ PU (v − a)∥∥∥− ∥∥∥−γ(2(1− γ))−1 PU⊥ (v − a)∥∥∥=∥∥∥(3− 2γ)−1γ PU (v − a)∥∥∥− ∥∥∥γ(2(1− γ))−1 PU⊥ (v − a)∥∥∥→ ∥∥PU (v − a)∥∥+∞ =∞, when γ → 1−.Hence,‖zγ‖ → ∞ as γ → 1−.So both cases illustrate Theorem 4.21.574.8. ExamplesRemark 4.24. Considering Example 4.23 with a = 0. Then the followinghold:(i) zer(A+B) 6= ∅⇔ v ∈ U , in which casezer(A+B)= v + U⊥.(ii) T(x)= x− λPU x+ 2λv.(iii) Fix T ={2v + U⊥, if v ∈ U∅, if v /∈ U(iv) JA(Fix T)= v + U⊥.(v) We have(a) If v ∈ U , then zer (Aγ+Bγ) = {(3−2γ)−1(3(1−γ) PU w+γv)+PU⊥ w}.(b) If v ∈ U⊥, then zer (Aγ + Bγ) = {(3 − 2γ)−13(1 − γ) PU w +γ(2(1− γ))−1v + PU⊥ w}.(c) If v /∈ U and v /∈ U⊥, then zer (Aγ + Bγ) = {(3 − 2γ)−1(3(1 −γ) PU w + γ PU v)+ γ(2(1− γ))−1 PU⊥ v + PU⊥ w}.(vi) Tγ(x)=(1−4λγ(1−γ))x+λγ(1−2γ) PU x+k, where k := 2λγ(2(1−γ)w + v).(vii) Fix Tγ ={(3− 2γ)−1(2 PU v+ 4(1− γ) PU w)+ (2(1− γ))−1 PU⊥ v+PU⊥ w}.(viii) JAγ(Fix Tγ)= Φ1(w)+ Φ2(v), whereΦ1 :=(3− 2γ)−12γ(1− γ) PU +γ PU⊥ +(1− γ) Id,andΦ2 := γ((3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ ).584.8. Examples(ix) Let w = 0. Then there are two cases:Case 1: If v ∈ U , thenlimγ→1−zγ = v = Pzer(A+B)(0).Case 2: If v /∈ U , thenlimγ→1−‖zγ‖ → ∞.So both cases illustrate Theorem 4.21.Example 4.25. Let λ ∈ ]0, 1] and γ ∈ ]0, 1[. Suppose that A = Pa+U ,where a ∈ H and U is a closed linear subspace of H. Let B = −v andT =(1− λ) Id +λRBRA.Recall from (4.16)Tγ =(1− λ) Id +λ(2γJB + 2(1− γ)w − Id ) ◦ (2γJA + 2(1− γ)w − Id ).Then the following hold:(i) zer(A+B)(x) = ∅⇔ a− v ∈ U , in which casezer(A+B)= PU v + U⊥.(ii) T(x)= x− λPU x− 2λPU⊥ a+ 2λv.(iii) Fix T 6= ∅⇔ v − a ∈ U , in which caseFix T = 2 PU v + U⊥.(iv) JA(Fix T)= PU v + U⊥.(v) We have(a) If a−v ∈ U , then zer (Aγ+Bγ) = {PU w+PU⊥ w+(3−2γ)−1γv}.(b) If a − v ∈ U⊥, then zer (Aγ + Bγ) = {PU w + PU⊥ w − (2(1 −γ))−1γ(a− v)}.594.8. Examples(c) If a− v /∈ U and a− v /∈ U⊥, then zer (Aγ +Bγ) = {PU w+ (3−2γ)−1γ PU v + PU⊥ w −(2(1− γ))−1γ(PU⊥ a− PU⊥ v)}.(vi) Tγ(x)=(1− 4λγ(1− γ))x+ λγ(1− 2γ) PU x+ k, where k = 2λγ(v +2(1− γ)w + (1− 2γ) PU⊥ a).(vii) Fix Tγ ={(3 − 2γ)−1(4(1 − γ) PU w + 2 PU v) + PU⊥ w + (2(1 −γ))−1((1− 2γ) PU⊥ a+ PU⊥ v)}.(viii) JAγ Fix Tγ = Ψ1(a)+ Ψ2(w)+ Ψ3(v), whereΨ1 := −(2(1− γ))−1γ PU⊥ ,Ψ2 :=(3− 2γ)−13(1− γ) PU + PU⊥ ,andΨ3 := γ((3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ ).(ix) Let v ∈ U⊥ and w = 0. Then there are two cases:Case 1: If a− v ∈ U , thenlimγ→1−zγ → PU v = Pzer(A+B)(0).Case 2: If a− v /∈ U , thenlimγ→1−‖zγ‖ → ∞.Proof. (i): Let z ∈ H. Then Fact 2.14 givesA(z)= Pa+U z = a+ PU (z − a) = PU⊥ a+ PU z.Therefore,(A+B)z = PU⊥ a+ PU z − v.Note thatz ∈ zer (A+B)⇔ (A+B)z = PU⊥ a+ PU z − v = 0604.8. Examples⇔ (PU z − PU v)+ (PU⊥ a− PU⊥ v) = 0⇔ PU(z − v)+ PU⊥ (a− v) = PU 0 + PU⊥ 0⇔ PU(z − v) = 0 and PU⊥ (a− v) = 0⇔ z − v ∈ U⊥ and a− v ∈ U⇔ z ∈ v + U⊥ and a− v ∈ U.Hence, zer(A+B)(z) = ∅⇔ a− v ∈ U , in which casezer(A+B)= v + U⊥= PU v + PU⊥ v + U⊥= PU v + U⊥.(ii): Using Lemma 4.5(ii) and Example 4.2 givesT(x)=(1− λ)x+ λ(( Id +2v)((Id−PU )− 2 PU⊥ a))(x)=(1− λ)x+ λ(( Id +2v)(x− PU x− 2 PU⊥ a))= x− λPU x− 2λPU⊥ a+ 2λv.(iii): Let x ∈ H such that x ∈ Fix T . Then it follows from (ii) thatx = T(x)= x− λPU x− 2λPU⊥ a+ 2λv,and,x− x+ λPU x+ 2λPU⊥ a− 2λv = 0.Thus,PU x = 2v − 2 PU⊥ a= 2 PU v + 2 PU⊥ v − 2 PU⊥ a= 2 PU v − 2 PU⊥(v − a)and⇔ PU x− 2 PU v = PU(x− 2v) = PU⊥ (2(v − a))⇔ PU(x− 2v) = PU⊥ (2(v − a)) (Because U ∩ U⊥ = {0})⇔ x− 2v ∈ U⊥ and 2(v − a) ∈ U⇔ x− 2 PU v ∈ U⊥ and v − a ∈ U (Because U and U⊥ are subspaces).614.8. ExamplesHence, Fix T 6= ∅⇔ v − a ∈ U, in which caseFix T = 2 PU v + U⊥.(iv): By using Lemma 4.5(i) and (ii), we obtainJA(Fix T)= JA(2 PU v + y), where y ∈ U⊥=(Id−2−1 PU)(2 PU v + y)− PU⊥ a= 2 PU v + y − 2−1(2 PU v)− PU⊥ a= PU v + y − PU⊥ a= PU v + U⊥.(v): Let zγ ∈ H, where {zγ} =(Aγ + Bγ)−1(0). Using (4.10), (4.11), andFact 2.14 gives(Aγ +Bγ)zγ = Pa+U(γ−1(zγ − (1− γ)w))+ 2γ−1((1− γ)(zγ − w))− v= a+ PU(γ−1(zγ − (1− γ)w)− a)+ 2γ−1((1− γ)(zγ − w))− v=(Id−PU)a+ γ−1 PU zγ − γ−1(1− γ) PU w+ 2γ−1((1− γ)(zγ − w))− v.Hence,(Aγ +Bγ)zγ = PU⊥ a+ γ−1 PU zγ − γ−1(1− γ) PU w+ 2γ−1((1− γ)(zγ −w))− v.(4.33)Applying PU and PU⊥ to (4.33) givesPU zγ = PU w +(3− 2γ)−1γ PU v,andPU⊥ zγ = PU⊥ w +(2(1− γ))−1γ(PU⊥ v − PU⊥ a).Hence,zγ = PU zγ + PU⊥ zγ= PU w +(3− 2γ)−1γ PU v + PU⊥ w − (2(1− γ))−1γ(PU⊥ a− PU⊥ v).If a− v ∈ U , thenzer(Aγ +Bγ)={PU w + PU⊥ w +(3− 2γ)−1γv}.624.8. ExamplesIf a− v ∈ U⊥, thenzer(Aγ +Bγ)={PU w + PU⊥ w −(2(1− γ))−1γ(a− v)}.If a− v /∈ U and a− v /∈ U⊥, thenzer(Aγ+Bγ)={PU w+(3−2γ)−1γ PU v+PU⊥ w−(2(1−γ))−1γ(PU⊥ a−PU⊥ v)}.(vi): By using Lemma 4.5 (iv) and Example 4.2, we obtainTγ(x)=(1− λ)x+ λ(((2γ − 1) Id +2γv + 2(1− γ)w) ◦ ((2γ − 1) Id−γ PU− 2γ PU⊥ a+ 2(1− γ)w))(x)=(1− λ)x+ λ(((2γ − 1) Id +2γv + 2(1− γ)w) ◦ ((2γ − 1)x− γ PU x− 2γ PU⊥ a+ 2(1− γ)w))(x))=(1− λ)x+ λ((2γ − 1)2x− γ(2γ − 1) PU x− 2γ(2γ − 1) PU⊥ a+ 4γ(1− γ)w + 2γv)=(1 + 4λγ2 − 4λγ)x− λγ(2γ − 1) PU x− 2λγ(2γ − 1) PU⊥ a+ 4λγ(1− γ)w + 2λγv.Hence,Tγ(x)=(1− 4λγ(1− γ))x+ λγ(1− 2γ)PU x+ k,where k = 2λγ((1− 2γ)PU⊥ a+ 2(1− γ)w + v) .(vii): It follows from (vi) thatx = Tγ(x)=(1− 4λγ(1− γ))x+ λγ(1− 2γ)PU x+ 2λγ((1− 2γ)PU⊥ a+ 2(1− γ)w + v),and4λγ(1−γ)x−λγ(1−2γ)PU x−2λγ((1−2γ) PU⊥ a+ 2(1−γ)w+ v) = 0. (4.34)Applying PU and PU⊥ to (4.34) givesPU x =(3− 2γ)−1(4(1− γ) PU w + 2 PU v),andPU⊥ x = PU⊥ w +(2(1− γ))−1((1− 2γ) PU⊥ a+ PU⊥ v).634.8. ExamplesHence,x = PU x+ PU⊥ x=(3− 2γ)−1(4(1− γ) PU w + 2 PU v)+ PU⊥ w+(2(1− γ))−1((1− 2γ) PU⊥ a+ PU⊥ v).(viii): For simplicity setΦ1(a):=(2(1− γ))−1(1− 2γ)PU⊥ (a),Φ2(w):=(3− 2γ)−14(1− γ)PU w + PU⊥ w,andΦ3(v):=(3− 2γ)−12 PU v + (2(1− γ))−1 PU⊥ v.From (vii), we can see that Fix Tγ = {Φ1(a)+ Φ2(w)+ Φ3(v)}. Now usingLemma 4.5(iii) and (vii) givesJAγ Fix Tγ = γ(Φ1(a)+ Φ2(w)+ Φ3(v))− 2−1γ(Φ1(a)+ Φ2(w)+ Φ3(v))− γ PU⊥ a+(1− γ)w. (4.35)We haveγΦ1(a)=(2(1− γ))−1γ(1− 2γ)PU⊥ (a), (4.36)γΦ2(w)=(3− 2γ)−14γ(1− γ)PU w + γ PU⊥ w, (4.37)andγΦ3(v)=(3− 2γ)−12γ PU v + (2(1− γ))−1γ PU⊥ v. (4.38)Moreover,γ2−1 PU Φ1(a)= 0, (4.39)γ2−1 PU Φ2(w)=(3− 2γ)−12γ(1− γ)PU w, (4.40)andγ2−1 PU Φ3(v)=(3− 2γ)−1γ PU v. (4.41)Substituting (4.36), (4.37), (4.38), (4.39), (4.40), and (4.41) back into (4.35)givesJAγ Fix Tγ = −(2(1− γ))−1γ PU⊥ (a)+((3− 2γ)−13(1− γ) PU + PU⊥ )(w)+ γ((3− 2γ)−1 PU +(2(1− γ))−1 PU⊥ )(v).644.8. Examples(ix): Let w = 0. If a− v ∈ U , then zγ = {(3− 2γ)−1γ PU v}, by (v). From(i), we obtain zer(A+B)= PU v + U⊥. ThenPPU v+U⊥ x = PU v + PU⊥(x− PU v)= PU v + PU⊥ x.This implies that Pzer(A+B)(0)= PU v. Hence,limγ→1−zγ → PU v = Pzer(A+B)(0).If a− v /∈ U , thenzγ =(3− 2γ)−1γ PU v + γ(2(1− γ))−1 PU⊥ (v − a),by (v). Moreover,‖zγ‖ =∥∥∥(3− 2γ)−1γ PU v − (− γ(2(1− γ))−1 PU⊥ (v − a))∥∥∥≥∥∥∥(3− 2γ)−1γ PU v∥∥∥− ∥∥∥−γ(2(1− γ))−1 PU⊥ (v − a)∥∥∥=∥∥∥(3− 2γ)−1γ PU v∥∥∥− ∥∥∥γ(2(1− γ))−1 PU⊥ (v − a)∥∥∥→ ‖PU v‖+∞ =∞, when γ → 1−.Hence,‖zγ‖ → ∞ as γ → 1−.So both cases illustrate Theorem 4.21.65Chapter 5Fixed Point Sets ofCompositions of Projectorsand the Magic DifferenceVectors5.1 OverviewThis chapter is based on our paper [3]. Let m ∈ {2, 3, . . . } and letC1,C2, . . . ,Cm be nonempty closed convex subsets with corresponding pro-jections P1,P2, . . . ,Pm in Hilbert space H. The fixed point sets of the cycliccompositions are given byFm := Fix(Pm Pm−1 · · ·P1),Fm−1 := Fix(Pm−1 Pm−2 · · ·Pm),...F1 := Fix(P1 Pm · · ·P2).(5.1)(5.2)(5.3)(5.4)Our goal is to understand the fixed point sets F1, F2, . . . , Fm. The motiva-tion behind this is a significant number of convex optimization problem areconvex feasibility problem of the formFind x ∈ C1 ∩ C2 ∩ · · · ∩ Cm := S.The central questions are:What happens when S = ∅ ? What are F1, F2, . . . , Fm?We summarize our main results in this chapter as follows:• We collect well-known facts and mathematical tools, which help us tostudy the fixed point sets of composition of projection mappings. See665.2. Reviewing the case when S 6= ∅Section 5.4 for m = 2 and Section 5.5 for m ≥ 3. The well knownresults in these two sections are drawn from [13], [12], [11], [30], and[8].• We sketch the connection between the cycles and the fixed point setsof the compositions of projections mappings (see Theorem 5.2).• We present a new proof for [13, Remark 5.1.7] which states that halfof the geometry conjecture is true when there are cycles (see Theo-rem 5.10). The full geometry conjecture is proven in Chapter 9.Except for facts with explicit references, this chapter is based on results thatappear in our paper [3].5.2 Reviewing the case when S 6= ∅When S 6= ∅, it is well known thatS = Fm = Fm−1 = · · · = F2 = F1.Moreover, a point in S can be found as the weak limit of the sequence(xn)n∈N viax0PC17−−→ x1PC27−−→ x2 · · ·xm−1 PCm7−−−→ xmPC17−−→ xm+1PC27−−→ xm+2 · · ·generated by the method of cyclic projections, which is governed by theiterationsx0,P1 x0,P2 P1 x0, . . . ,Pm · · ·P1 x0,P1 Pm · · ·P1 x0, . . . ,(Pm · · ·P1)2x0,P1(Pm · · ·P1)2x0, . . . ,(Pm · · ·P1)3x0, . . .(For more detail we refer the reader to [11], [12], [13], [48], [39], [33], andthe references therein).5.3 Cycles and the fixed point sets of thecompositions of projections mappingsDefinition 5.1. [13, Definition 5.1.1] Let z1 ∈ F1, and set z2 := P2 z1, z3 :=P3 z2, . . . , zm−1 := Pm−1 zm−2, and zm := Pm zm−1. We call the tuple z =(z1, z2, . . . , zm−1, zm)a cycle, and the difference vectors are defined byv1 := z2 − z1, v2 := z3 − z2, . . . , vm−1 := zm − zm−1, vm := z1 − zm. (5.5)675.3. Cycles and the fixed point sets of the compositions of projections mappingsFigure 5.1: When S = C1 ∩ C2 ∩ · · · ∩ Cm 6= ∅, then all fixed point sets Fimust coincide with S.In the next theorem, we sketch the connection between the cycles andthe fixed point sets of the compositions of projections mappingsTheorem 5.2. The following are equivalent(i) Cycle exists.(ii) For all 1 ≤ i ≤ m, Fi 6= ∅.Proof. ”(i)⇒(ii)”: Let z = (z1, z2, . . . , zm−1, zm) be a cycle. Then z1 =P1 zm, z2 = P2 z1, . . . , zm−1 = Pm−1 zm−2, and zm = Pm zm−1 by Defini-tion 5.1. This implies thatz1 = P1 Pm · · ·P2 z1...zi = Pi Pi−1 · · ·P1 Pm · · ·Pi+1 zi...zm = Pm Pm−1 · · ·P1 zmHence, From (5.1)-(5.4), we have z1 ∈ F1 ⇒ F1 6= ∅, z2 ∈ F2 ⇒ F2 6= ∅,. . . , zm−1 ∈ Fm−1 ⇒ Fm−1 6= ∅ and finally, zm ∈ Fm ⇒ Fm 6= ∅.”(ii)⇒(i)”: Suppose Fm 6= ∅. Let zm ∈ Fm. From (5.1), we havezm ∈ Fix(Pm Pm−1 · · ·P3 P2 P1)zm.Thus,zm = Pm Pm−1 · · ·P3 P2 P1 zm.685.3. Cycles and the fixed point sets of the compositions of projections mappingsThenP1 zm = P1(Pm Pm−1 · · ·P3 P2)(P1 zm),⇔ P1 zm ∈ Fix(P1 Pm Pm−1 · · ·P3 P2) ⇔ P1 zm ∈ F1 6= ∅. Next, letz1 ∈ F1 such that z1 = P1 zm andz1 = P1 Pm Pm−1 · · ·P3 P2 z1.ThenP2 z1 = P2(P1 Pm Pm−1 · · ·P3)(P2 z1),⇔ P2 z1 ∈ Fix(P2 P1 Pm Pm−1 · · ·P3) ⇔ P2 z1 ∈ F2 6= ∅. Now let z2 ∈ F2such that z2 = P2 z1. Keep doing this, we obtainzm−2 = Pm−2 Pm−3 · · ·P1 Pm Pm−1 zm−2.ThenPm−1 zm−2 = Pm−1(Pm−2 Pm−3 . . .P1 Pm)(Pm−1 zm−2),andPm−1 zm−2 ∈ Fix(Pm−1 Pm−2 Pm−3 · · ·P1 Pm)⇔ Pm−1 Pm−2 ∈ Fm−1 6= ∅.Let zm−1 ∈ Fm−1 such that zm−1 = Pm−1 zm−2. Finally, we havezm−1 = Pm−1 Pm−2 Pm−3 · · ·P1 Pm zm−1.ThenPm zm−1 = Pm(Pm−1 Pm−2 Pm−3 · · ·P1)(Pm zm−1),and hence, we obtainPm zm−1 ∈ Fix(Pm Pm−1 Pm−2 Pm−3 · · ·P1)⇔ Pm zm−1 ∈ Fm 6= ∅,Combining all these together we have(z1, z2, . . . , zm−1, zm)in Hm satisfyingthat (z1, z2, . . . , zm−1, zm)=(P1 zm,P2 z1, . . . ,Pm−1 zm−2,Pm zm−1).695.4. When m = 2 with empty intersectionDefinition 5.3 (geometry conjecture). The geometry conjecture formulatedin 1997 (see [13]), states that there exists a list of vectors v1, v2, . . . , vm inH such thatv1 + v2 + · · ·+ vm = 0, (5.6)andFm = Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (C1 + v1 + · · ·+ vm−1),Fm−1 = Cm−1 ∩(Cm−2 + vm−2) ∩ · · · ∩ (Cm + vm + · · ·+ vm−2),...F1 = C1 ∩(Cm + vm) ∩ · · · (C2 + v2 + · · ·+ vm).5.4 When m = 2 with empty intersectionDefine the gap vector between the two sets C1 and C2 byv := PC2−C1(0), (5.7)the generalized solutions sets byE1 := C1 ∩(C2 − v)and E2 :=(C1 + v) ∩ C2, (5.8)andδ := inf ‖C2 − C1‖ ={ ‖c1 − c2‖ | c1 ∈ C1, c2 ∈ C2}. (5.9)Fact 5.4. [12, Lemma 2.2] and [11, Fact 1.1] The following hold:(i) F1 = Fix(PC1 PC2)= E1 and F2 = Fix(PC2 PC1)= E2. Moreover,both F1 and F2 are closed and convex.If δ is actually attend, i.e., the infimum is a minimum, then the fol-lowing hold:(ii) F1 and F2 are nonempty.(iii) ‖z1 − z2‖ = δ ⇔ z2 = PC2 z1 and z1 = PC1 z2.(iv) ‖z1 − z2‖ = δ ⇒ z2 ∈ F2 and z1 ∈ F1.Proposition 5.5. [12, Fact 5.1(i)] If C2 is bounded, then δ is attained.705.4. When m = 2 with empty intersectionFigure 5.2: An illustration for the case when m = 2 and δ is attained.Proof. DefinePC2 PC1 : C2 → C2,is a nonexpansive (Lipschitz-1) self-map of C2. Then by the Browder-Go¨hde-Kirk Fixed point theorem, this map has a fixed point x∗ ∈ C2 that satisfiesx∗ = PC2 PC1 x∗. Then we get a best approximation pair(P1 x∗, x∗)andattains the infimum.It also could happen that the infimum δ is not attained.Example 5.6. Let H = R2, C1 = R × {0} and C2 = epi(1 + exp)={(x, λ) ∈ R2 | 1 + exp(x) ≤ λ}. Then v = PC2−C1 (0) = (0, 1), and δ =‖v‖ = 1. However, the infimum is not attained, F1 = C1 ∩(C2 − v)= ∅and F2 = C2 ∩(C1 + v)= ∅. See fig. 5.3 for an illustration, created withGeoGebra [36].Hence, when m = 2 whether or not δ is attained, we can always definedifference vectorsv1 := PC2−C1(0), and v2 := PC1−C2(0)= −v1, (5.10)such that ‖v1‖ = ‖v2‖ = δ and v1 + v2 = 0. Moreover, the difference vectorssatisfyF2 = C2 ∩(v1 + C1), and F1 = C1 ∩(v2 + C2),andx2n+2 − x2n+1 → v1, x2n+1 − x2n → v2.See [11, Fact 1.2] for more details.715.4. When m = 2 with empty intersectionFigure 5.3: An illustration for the case when the infimum is not attained.Figure 5.4: An illustration for the sets F1 and F2.Fact 5.7. [30, Theorem 2] We have z =(z1, z2) ∈ H2 is a cycle if and onlyif z is a minimizer of the functionx =(x1, x2) 7→ ιC1(x1)+ ιC2(x2)+ ‖x1 − x2‖ . (5.11)Example 5.8. Let H = R2. Let C1 = R × {−1} and C2 = {x ∈ R2 |‖x− 3‖ ≤ 3}. Then ∩2i=1Ci = ∅. For starting point x0 ∈ R2, the method ofalternating projections produces a cycle z =(z1, z2)where z1 =(0,−1) ∈ C1and z2 =(0, 0) ∈ C2. Moreover,ιC1(z1)+ ιC2(z2)+ ‖z1 − z2‖ = 1,which is the minimum distance between two sets. Hence, the cycle is thesolution to the variational problem (5.11).725.5. When m ≥ 3 with empty intersectionFigure 5.5: An illustration for Example 5.8 with x0 =(− 4, 1).5.5 When m ≥ 3 with empty intersectionIf m ≥ 3, then Fact 5.7 suggests the conjecture that z = (z1, z2, . . . , zm−1, zm)is a cycle if it minimizes a function of the formx =(x1, x2, . . . , xm−1, xm) 7→ m∑i=1ιCi(xi)+ ϕ(x1, x2, . . . , xm−1, xm). (5.12)However, in 2012 Baillon, Combettes, and Cominetti proved in [8] that thereis no function ϕ : Hm → R such that for every ordered family of nonempty,closed and convex subset C1,C2, . . . ,Cm of H, the set of cycles is the set ofminimizers over C1 × C2 × · · · × Cm.Example 5.9. Let H = R2. Let C1 = (−2.5,−2.5) + R(0.7071, 0.7071),C2 = R × {0}, and C3 = (2.5, 2.5) + R( − 0.7071, 0.7071). Moreover, let(x1, x2, x3) ∈ C1 × C2 × C3,ϕ(x1, x2, x3)= ‖x1 − x2‖+ ‖x2 − x3‖+ ‖x3 − x1‖ ,735.6. Cycles and difference vectorsandϑ(x1, x2, x3)= 12 ‖x1 − x2‖2 + 12 ‖x2 − x3‖2 + 12 ‖x3 − x1‖2 .Suppose a = (−1.32, 3.7). Thenb := PC2(a) = (−1.32, 0)andc := PC3(b) = (1.84, 3.16).Then we obtainϕ(a, b, c)= ‖a− b‖+ ‖b− c‖+ ‖c− a‖ = 11.3519,andϑ(a, b, c)= 12 ‖a− b‖2 + 12 ‖b− c‖2 + 12 ‖c− a‖2 = 21.8822.Additionally, the method of cyclic projections initialized at x0 ∈ R2 producesthe cycle z =(z1, z2, z3)where z1 =(0, 5), z2 =(0, 0), and z3 =(2.5, 2.5).Moreover, we haveϕ(z1, z2, z3)= ‖z1 − z2‖+ ‖z2 − z3‖+ ‖z3 − z1‖ = 12.0711 > 11.3519,andϑ(z1, z2, z3)= 12 ‖z1 − z2‖2 + 12 ‖z2 − z3‖2 + 12 ‖z3 − z1‖2 = 25 > 21.8822.Therefore, the cycle (z1, z2, z3) obtained by the method of cyclic projectionsfails to minimize the functions ϕ and ϑ. See Figure 5.6 for visualization.The code that we used to verify this example can be found in Appendix A.5.6 Cycles and difference vectorsTheorem 5.10. If cycles exist, then difference vectors exist,Fm ⊆ Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (C1 + v1 + · · ·+ vm−1),Fm−1 ⊆ Cm−1 ∩(Cm−2 + vm−2) ∩ · · · ∩ (Cm + vm + · · ·+ vm−2),...and,F1 ⊆ C1 ∩(Cm + vm) ∩ · · · (C2 + v2 + · · ·+ vm).745.6. Cycles and difference vectorsFigure 5.6: An illustration for Example 5.9 with x0 =(− 3, 5, 2.9).Proof. Let z :=(z1, z2, . . . , zm−1, zm)and z˜ :=(z˜1, z˜2, . . . , z˜m−1, z˜m)be twocycles. Then by Definition 5.1, we havez1 = P1 zm, z2 = P2 z1, . . . , zm−1 = Pm−1 zm−2, zm = Pm zm−1,andz˜1 = P1 z˜m, z˜2 = P2 z˜1, . . . , z˜m−1 = Pm−1 z˜m−2, z˜m = Pm z˜m−1.By using Example 3.2, we obtain‖z˜m − zm‖2 = ‖Pm z˜m−1 − Pm zm−1‖2≤ ‖z˜m−1 − zm−1‖2 −∥∥( Id−Pm )z˜m−1 − ( Id−Pm )zm−1∥∥2≤ ‖z˜m−2 − zm−2‖2 −∥∥( Id−Pm−1 )z˜m−2 − ( Id−Pm−1 )zm−2∥∥2− ∥∥( Id−Pm )z˜m−1 − ( Id−Pm )zm−1∥∥2≤ ...≤ ‖z˜1 − z1‖2 −∥∥( Id−P2 )z˜1 − ( Id−P2 )z1∥∥2− ∥∥( Id−P3 )z˜2 − ( Id−P3 )z2∥∥2 − · · ·− · · · − ∥∥( Id−Pm−1 )z˜m−2 − ( Id−Pm−1 )zm−2∥∥2− ∥∥( Id−Pm )z˜m−1 − ( Id−Pm )zm−1∥∥2≤ ‖z˜m − zm‖2 −∥∥( Id−P1 )z˜m − ( Id−P1 )zm∥∥2755.6. Cycles and difference vectors− ∥∥( Id−P2 )z˜1 − ( Id−P2 )z1∥∥2 − · · ·−− ∥∥( Id−Pm )z˜m−1 − ( Id−Pm )zm−1∥∥2 .Hence, equality holds throughout which implies that all non positive termsare equal to zero and thusz˜1 − P2 z˜1 = z1 − P2 z1 ⇔ z˜1 − z˜2 = z1 − z2z˜2 − P3 z˜2 = z2 − P3 z2 ⇔ z˜2 − z˜3 = z2 − z3...z˜m−1 − Pm z˜m−1 = zm−1 − Pm zm−1 ⇔ z˜m−1 − z˜m = zm−1 − zmz˜m − P1 z˜m = zm − Pm zm ⇔ z˜m − z˜1 = zm − z1.Thus, definev1 := z2 − z1, v2 := z3 − z2, . . . , vm−1 := zm − zm−1, vm := z1 − zm,where z =(z1, z2, . . . , zm−1, zm)is any cycle. Thenzm ∈ Cmzm = zm−1 + vm−1 ⇔ zm ∈ Cm−1 + vm−1zm = zm−2 + vm−1 + vm−2 ⇔ zm ∈ Cm−2 + vm−1 + vm−2...zm = v1 + v2 + · · ·+ vm−2 + vm−1 + z1 ⇔ zm ∈ v1 + · · ·+ vm−1 + C1.Thus,Fm ⊆ Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (v1 + v2 + · · ·+ vm−2 + vm−1 + C1).The other fixed point sets Fm−1, . . . , F1 are treated similarly.Remark 5.11. Regarding to Definition 5.3, we know that the geometry con-jecture is true when C1∩C2∩ · · ·∩Cm 6= ∅ with v1 = v2 = · · · = vm = 0. Itis also true when m = 2 from Fact 5.4(i) and (5.8). Moreover, in general thehalf of the geometry conjecture is known to be true when there are cyclesby Theorem 5.10.Chapter 9 will provide new results that will be useful in resolving thegeometry conjecture.76Chapter 6The Composition ofReflection Operators6.1 OverviewNonexpansive mappings are essential to modern optimization and mono-tone operator theory because their fixed points can describe solutions to op-timization problems. This chapter is based on our paper [4]. We study thefixed point set of the composition of finitely many reflectors associated withclosed linear subspaces. A central question is: Given nonexpansive mapsR1, . . . , Rm on H with ⋂mi=1 FixRi 6= ∅,what can we say about Fix(RmRm−1 · · ·R1)?We summarize our main results in this chapter as follows:• Important properties about the fixed point set of the composition oftwo reflectors are given in Proposition 6.1 and Lemma 6.2. Theseproperties will be useful for studying the general case.• We show that the order of the reflection operators does influence thefixed point set of the composition (see Proposition 6.3 and Proposi-tion 6.5). These propositions provide knowledge about the fixed pointset of the composition of three reflectors.• We obtain a general result which shows the effect of cyclically shiftinga composition of m reflectors (see Lemma 6.4).• We also prove the existence of a nice permutation of the reflectorswhich yields the same fixed point set of the composition of reflectors(see Theorem 6.8).• We show that the location of the fixed point set of the composition ofan odd number of reflectors can be narrowed down (see Theorem 6.6and Theorem 6.9).776.2. The composition of two reflectors and its fixed point set• We illustrate that the upper bound in Theorem 6.9 is sometimes sharp(see Example 6.11) but not always attained (see Example 6.12).• We illustrate our results by focussing on classical reflectors in the Eu-clidean plane (see Section 6.4).6.2 The composition of two reflectors and itsfixed point setIn this section we will collect important properties about the compositionof two reflectors and the fixed point set of the composition. These propertieswill be useful for our study.Proposition 6.1. Let V be a closed linear subspace of H. Then the fol-lowing hold:(i) −RV = RV ⊥ .(ii) RV ◦(− Id ) = −RV = RV ⊥ .(iii) Fix(RV(− Id )) = Fix (−RV ) = V ⊥.Proof. (i): From (3.12), we have−RV = −(2 PV − Id)= −(PV −(Id−PV ))= −(PV −PV ⊥ )= −(PV −PV ⊥ + PV ⊥ −PV ⊥ )= −(PV + PV ⊥ −2 PV ⊥ ) (from Fact 2.15(iii))= −( Id−2 PV ⊥ )= 2 PV ⊥ − Id = RV ⊥ ,by (3.12).(ii): Let x ∈ H. By using (3.12), we have(RV ◦ (− Id))(x) = RV (−x)= 2 PV (−x)− (−x)= −2 PV (x) + (x)= −(2 PV (x)− (x))= −RV= RV ⊥ ,786.2. The composition of two reflectors and its fixed point setby (i).(iii): It follows from Fact 3.3, (ii), and (iii) thatFix(RV(− Id )) = Fix (−RV ) = Fix (RV ⊥) = V ⊥.Lemma 6.2. Let U be a closed linear subspace of H. Then the followinghold:(i) RU⊥RU = RURU⊥ = − Id.(ii) −RU = RU ◦ (− Id) = RU⊥ .(iii) −RU⊥ = RU⊥ ◦ (− Id) = RU .(iv) Fix(−RU ) = FixRU⊥ = U⊥.Proof. (i): By using (3.12), we obtainRU⊥RU =(2 PU⊥ − Id)RU=(PU⊥ −(Id−PU⊥))RU=(PU⊥ −PU)RU (from Fact 2.15(iii))=(PU⊥ −PU)(2 PU − Id)(from (3.12))=(PU⊥ −PU)(PU −(Id−PU))=(PU⊥ −PU)(PU −PU⊥)= PU⊥ PU −PU⊥ −PU + PU PU⊥= 0− PU⊥ −PU +0= −(PU⊥ + PU ) = − Id .Moreover,RURU⊥ =(2 PU − Id)(2 PU⊥ − Id)=(PU −PU⊥)(PU⊥ −PU)= PU PU⊥ −PU −PU⊥ + PU⊥ PU= 0− PU −PU⊥ +0= −(PU + PU⊥ ) = − Id .796.3. The composition of (three or more) reflectors and its fixed point set(ii): By using (3.12), we have−RU = −(2 PU − Id)= −(PU −( Id−PU⊥ )) = −(PU −PU⊥ )=(PU⊥ −PU)=(PU⊥ −PU⊥⊥)(because U⊥⊥ = U)= PU⊥ −(Id−PU⊥)= 2 PU⊥ − Id = RU⊥ .(iii): Apply (ii) to U⊥.(iv): It follows from (ii) that −RU = RU⊥ and by using Fact 3.3, we obtainFix(−RU) = Fix (RU⊥) = U⊥.Similarly, it follows from (iii) that −RU⊥ = RU and by using Fact 3.3, weobtainFix(−RU⊥) = Fix (RU) = U.6.3 The composition of (three or more) reflectorsand its fixed point setThe following result, which is a consequence of Lemma 6.2, provides acase when we have precise knowledge of the fixed point set of the compositionof three reflectors:Proposition 6.3. Let U and V be closed linear subspaces of H. Then thefollowing hold:(i) RVRU⊥RU = RV ⊥ .(ii) RVRURU⊥ = RV ⊥ .(iii) RU⊥RURV = RV ⊥ .(iv) RURU⊥RV = RV ⊥ .(v) Fix(RVRU⊥RU)= V ⊥.(vi) Fix(RVRURU⊥)= V ⊥.(vii) Fix(RU⊥RURV)= V ⊥.806.3. The composition of (three or more) reflectors and its fixed point set(viii) Fix(RURU⊥RV)= V ⊥.Proof. (i): We haveRVRU⊥RU = RV(− Id ) (from Lemma 6.2(i))= −RV (from Lemma 6.2(ii))= RV ⊥ (from Lemma 6.2(ii)).(ii), (iii), and (iv): The proof is similar to (i).(v): Combining (i) and Fact 3.3 givesFix(RVRU⊥RU)= FixRV ⊥ = V ⊥.(vi), (vii), and (viii): The proof is similar to (v).Lemma 6.4. Let R1, . . . , Rm be arbitrary maps from H to H. ThenFix(RmRm−1 · · ·R2R1) = (RmRm−1 · · ·R3R2)(Fix(R1RmRm−1 · · ·R3R2))= (RmRm−1 · · ·R3)(Fix(R2R1RmRm−1 · · ·R4R3))...= RmRm−1(Fix(Rm−2Rm−3 · · ·R2R1RmRm−1))= Rm(Fix(Rm−1Rm−2 · · ·R2R1Rm)).Proof. Let x ∈ Fix(RmRm−1 · · ·R2R1). ThenR1x = R1(RmRm−1 · · ·R2R1)x = (R1RmRm−1 · · ·R3R2)(R1x)and so R1x ∈ Fix(R1RmRm−1 · · ·R3R2). It follows thatR1(Fix(RmRm−1 · · ·R2R1)) ⊆ Fix(R1RmRm−1 · · ·R3R2).The same reasoning gives(R2R1)(Fix(RmRm−1 · · ·R2R1)) ⊆ R2(Fix(R1RmRm−1 · · ·R3R2))⊆ Fix(R2R1RmRm−1 · · ·R4R3),hence(R3R2R1)(Fix(RmRm−1 · · ·R2R1)) ⊆ (R3R2)(Fix(R1RmRm−1 · · ·R3R2))⊆ R3(Fix(R2R1RmRm−1 · · ·R4R3))⊆ Fix(R3R2R1RmRm−1 · · ·R5R4)816.3. The composition of (three or more) reflectors and its fixed point setuntil finallyFix(RmRm−1 · · ·R2R1) = (RmRm−1 · · ·R2R1)(Fix(RmRm−1 · · ·R2R1))⊆ (RmRm−1 · · ·R2)(Fix(R1RmRm−1 · · ·R3R2))...⊆ Rm(Fix(Rm−1Rm−2 · · ·R2R1Rm))⊆ Fix(RmRm−1 · · ·R2R1).Hence, equality holds throughout and we are done.Proposition 6.5. Let U and V be closed linear subspaces of H. Then thefollowing hold:(i) Fix(RURVRU⊥) = RU (V ⊥).(ii) Fix(RU⊥RVRU ) = RU (V ⊥).Proof. (i): From Lemma 6.4, we haveFix(RURVRU⊥) = RU(Fix(RVRU⊥RU ))= RU(Fix(RV (− Id)))(from Lemma 6.2(i))= RU(Fix(RV ⊥))(from Proposition 6.1(ii))= RU(V ⊥).(ii): From Lemma 6.4, we haveFix(RU⊥RVRU ) = RU⊥(Fix(RVRURU⊥))= RU⊥(RV (− Id)))(from Lemma 6.2(i))= RU⊥(Fix(RV ⊥))(from Proposition 6.1(ii))= RU ◦ (− Id)(V ⊥) (from Lemma 6.2(ii))= RU(− V ⊥) = RU(V ⊥) (because V ⊥ is a subspace)= RU(V ⊥).Theorem 6.6. Let U, V,W be closed linear subspaces of H. ThenFix(RWRVRU ) = Fix(4 PW PV PU −2(PW PV + PW PU + PV PU )+ PW + PV + PU)(6.5)andFix(RWRVRU ) ⊆ U + V +W. (6.6)826.3. The composition of (three or more) reflectors and its fixed point setProof. From (3.12), we have RU = 2 PU − Id, which implies thatRVRU =(2 PV − Id)(2 PU − Id)= 4 PV PU −2 PV −2 PU + Id .Moreover,RWRVRU = 2 PW(4 PV PU −2 PV −2 PU + Id)− (4 PV PU −2 PV −2 PU+ Id)= 8 PW PV PU −4 PW PV −4 PW PU +2 PW −4 PV PU+ 2 PV +2 PU − Id .Hence,RWRVRU = 8 PW PV PU −4 PW PV −4 PW PU +2 PW −4 PV PU+ 2 PV +2 PU − Id . (6.7)Let x ∈ Fix (RWRVRU). By using (6.7), we havex = RWRVRU (x)⇔ 2x = 8 PW PV PU x− 4 PW PV x− 4 PW PU x+ 2 PW x− 4 PV PU x+ 2 PV x+ 2 PU x⇔ x = 4 PW PV PU x− 2 PW PV x− 2 PW PU x+ PW x− 2 PV PU x+ PV x+ PU x.Hence,Fix(RWRVRU ) = Fix(4 PW PV PU −2(PW PV + PW PU + PV PU )+ PW + PV + PU)(6.8)Therefore,Fix(RWRVRU ) ⊆ U + V +W.Example 6.7. Suppose U , V , W are pairwise orthogonal. Then by (6.5),we have: x = PU x+ PV x+ PW x. Hence, Fix(RWRVRU ) = U + V +W .Theorem 6.8. Let U1, . . . , Um be closed linear subspaces of H. ThenFix(RUmRUm−1 · · ·RU2RU1) = Fix(RU1RU2 · · ·RUm−1RUm). (6.9)836.3. The composition of (three or more) reflectors and its fixed point setProof. Since U1, U2, . . ., Um−1, Um are closed linear subspaces of H, thenthe projectors PU1 , PU2 , . . ., PUm−1 , PUm are linear. Because(RUmRUm−1 . . . RU2RU1)∗ = R∗U1R∗U2 . . . R∗Um−1R∗Um= RU1RU2 . . . RUm−1RUm .The the reflectors RU1 , RU2 , . . ., RUm−1 , RUm are self-adjoint.Hence, by Fact 3.11 and Fact 3.4, we obtainFix(RUmRUm−1 . . . RU2RU1)= Fix(RUmRUm−1 . . . RU2RU1)∗= Fix(RU1RU2 . . . RUm−1RUm).Theorem 6.9. Let U1, U2, . . . , Um−1, Um be an odd number of closed linearsubspaces of H. ThenFix(RUmRUm−1 · · ·RU2RU1) ⊆ U1 + U2 + · · ·+ Um−1 + Um. (6.10)Proof. Let x ∈ H. Then x ∈ Fix(RUmRUm−1 · · ·RU2RU1). This equivalentto x = RUmRUm−1 · · ·RU2RU1x, which is equivalent tox = (2 PUm − Id)(2 PUm−1 − Id) · · · (2 PU2 − Id)(2 PU1 − Id)x.This implies x ∈ (−1)mx+ ran(∑i PUi) ⇒ 2x ∈ ran(∑i PUi) ⊆∑i Ui.Remark 6.10. Theorem 6.9 is false when m is assumed to be even: indeed,assume that U is a proper closed linear subspace of H, and set U1 := U2 :=U . Then RU2RU1 = Id and henceFix(RU2RU1) = H % U = U1 + U2. (6.11)The next example shows that the upper bound provided in Theorem 6.9is sometimes sharp:Example 6.11. Let U1, U2, . . . , Um−1, Um be closed linear subspaces of Hwhich are assumed to be pairwise orthogonal: Ui ⊥ Uj whenever i 6= j.Then the following holds:(i) m is odd andFix(RUmRUm−1 · · ·RU2RU1) = U1 + U2 + · · ·+ Um−1 + Um (6.12)846.3. The composition of (three or more) reflectors and its fixed point set(ii) m is even andFix(−RUmRUm−1 · · ·RU2RU1) = U1 + U2 + · · ·+ Um−1 + Um. (6.13)Proof. (i): Assume first that m is odd. Let (x1, . . . , xm) ∈ U1 × U2 × · · · ×Um−1×Um and set x = x1 +x2 + · · ·+xm−1 +xm. Write RU1 = PU1 −PU⊥1 ,RU2 = PU2 −PU⊥2 , . . ., RUm−1 = PUm−1 −PU⊥m−1 , RUm = PUm −PU⊥m . ThenRU1x = (PU1 −PU⊥1 )(x1 + x2 + · · ·+ xm−1 + xm) (6.14)= x1 − x2 − · · · − xm−1 − xm. (6.15)HenceRU2RU1x = (PU2 −PU⊥2 )(x1 − x2 · · · − xm−1 − xm) (6.16)= −x1 − x2 + x3 · · ·+ xm−1 + xm. (6.17)and furtherRU3RU2RU1x = (PU3 −PU⊥3 )(−x1 − x2 + x3 · · ·+ xm−1 + xm) (6.18)= x1 + x2 + x3 − x4 − · · · − xm−1 − xm. (6.19)In general, for 1 ≤ k ≤ m, we have(−1)k−1RUkRUk−1 · · ·RU2RU1x = (x1+x2+· · ·+xk)−(xk+1+· · ·+xm−1+xm).(6.20)In particular, because m− 1 is even, we obtain RUmRUm−1 · · ·RU2RU1 = x.This completes the proof of (6.12).(ii): Now assume that m is even. Set Um+1 := {0}. Then RUm+1 = − Idand m + 1 is odd. Therefore, we obtain (6.13) from the odd case we justproved.Example 6.12. Assume that U is a closed linear subspace of H such that{0} $ U . Let m be an odd positive integer, and let i ∈ {1, . . . ,m}. Then setUi := {0} and Uj := U for every j 6= i. Then m−1 is even and R{0} = − Id.HenceRUmRUm−1 · · ·RU2RU1 = −Rm−1U = − Idand thereforeFix(RUmRUm−1 · · ·RU2RU1) = Fix(− Id) = {0} $ U = U1 + U2 + · · ·+ Um.856.4. The Euclidean plane R26.4 The Euclidean plane R2Let us now specialize the general result of the last section to the Eu-clidean plane and classical reflectors. We start with some well known resultswhose statements can be found, e.g., in [50]. SetRefl: R→ R2×2 : α 7→(cos(2α) sin(2α)sin(2α) − cos(2α)). (6.21)It is clear that Refl is periodic, with minimal period pi. The importanceof Refl stems from the fact that it describes all classical reflectors on R2;indeed,RR·(cos(α),sin(α)) = Refl(α) (6.22)for every α ∈ R. It is also convenient to defineRot: R→ R2×2 : α 7→(cos(α) − sin(α)sin(α) cos(α)). (6.23)Note that for every α ∈ R, Rot(α) describes the counterclockwise rotationby α; the operator Rot is periodic with minimal period 2pi.6.4.1 The composition of reflectors and rotatorsFact 6.13. Let α and β be in R. Then the following hold:(i) Refl(β)Refl(α) = Rot(2(β − α)).(ii) Rot(β)Rot(α) = Rot(α+ β).(iii) Rot(β)Refl(α) = Refl(α+ 12β).(iv) Refl(β)Rot(α) = Refl(β − 12α).Proof. (i): Leta := cos(2β) cos(2α) + sin(2β) sin(2α) = cos(2(β − α)),b := cos(2β) sin(2α)− sin(2β) cos(2α) = − sin (2(β − α)),c := sin(2β) cos(2α)− cos(2β) sin(2α) = sin (2(β − α)),andd := sin(2β) sin(2α) + cos(2β) cos(2α) = cos(2(β − α)).866.4. The Euclidean plane R2By using (6.21), we obtainRefl(β)Refl(α) =(cos(2β) sin(2β)sin(2β) − cos(2β))·(cos(2α) sin(2α)sin(2α) − cos(2α))=(a bc d)=(cos(2(β − α)) − sin(2(β − α))sin(2(β − α)) cos(2(β − α)))= Rot(2(β − α)),by (6.23).(ii), (iii), and (iv): The proof is similar to (i).6.4.2 The fixed point sets of compositions of classicalreflectors on R2In this section we provide results about the fixed point sets of composi-tions of classical reflectors on R2:Theorem 6.14. Let α1, α2, . . . , αm−1, αm be in R. Consider the composi-tion of m classical reflectors,Sm := Refl(αm) · · ·Refl(α1)Refl(α1), (6.24)and set βm := αm−αm−1 + · · ·−(−1)mα1. Then exactly one of the followingholds:(i) m is odd, Sm = Refl(βm), andFixSm = R(cos(βm), sin(βm)).(ii) m is even, Sm = Rot(2βm), andFixSm ={R2, if βm ∈ Zpi;{0}, otherwise.Proof. We will prove the two cases using induction on m.First, assume that m = 1. Then β1 = α1 and S1 = Refl(β1) = Refl(α1).Using (6.22) givesS1 = Refl(α1) = RR·(cos(α1),sin(α1)).876.4. The Euclidean plane R2Then by Fact 3.3 and (6.22), we obtainFix(S1) = Fix(Refl(α1))= Fix(RR·(cos(α1),sin(α1)))= R · (cos(α1), sin(α1))Second, assume that m = 2. ThenS2 = Refl(α2)Refl(α1),and β2 = α2 − α1. Then by Fact 6.13(i), we haveS2 = Refl(α2)Refl(α1)= Rot(2(α2 − α1))= Rot(2β2).Hence,FixS2 = Fix(Rot(2β2)) ={R2, if βm ∈ Zpi;{0}, otherwise. .Hypothesis: We assume that the result is true for some integer m ≥ 2. ThenSm+1 = Refl(αm+1)Refl(αm) · · ·Refl(α2)Refl(α1) = Refl(αm+1)Sm (6.25)andβm+1 = αm+1 − βm. (6.26)(i): Supposem+1 is odd. Thenm is even. Using the hypothesis, Fact 6.13(iv),and (6.26), we obtainSm+1 = Refl(αm+1)Sm = Refl(αm+1)Rot(2βm)= Refl(αm+1 − 12(2βm)) = Refl(βm+1)and the result follows.(ii): Supposem+1 is even. Thenm is odd. Using the hypothesis, Fact 6.13(i),and (6.26), we obtainSm+1 = Refl(αm+1)Sm = Refl(αm+1)Refl(βm)= Rot(2(αm+1 − βm)) = Rot(2βm+1)and the result follows.886.4. The Euclidean plane R2Figure 6.1: The fixed point sets for Example 6.15The next example shows that the order of the reflectors influences thefixed point set. See also Figure 6.1 for a visualization.Example 6.15. Set U := R(1, 0) so that U⊥ = R(0, 1), and V := R(1, 1).ThenRU = RR(1,0) = Refl(0),RU⊥ = RR(0,1) = Refl(pi/2),andRV = RR(1,1) = Refl(pi/4).Moreover, the following hold:(i) Fix(RU⊥RVRU ) = Fix(Refl(pi/2)Refl(pi/4)Refl(0)) = R(cos(pi/4),sin(pi/4)) = R(1, 1) = V = RU (V ⊥).(ii) Fix(RURVRU⊥) = Fix(Refl(0)Refl(pi/4)Refl(pi/2)) = R(cos(pi/4),sin(pi/4)) = R(1, 1) = V = RU (V ⊥).(iii) Fix(RVRU⊥RU ) = Fix(Refl(pi/4)Refl(pi/2)Refl(0)) = R(cos(−pi/4),sin(−pi/4)) = R(1,−1) = V ⊥.(iv) Fix(RVRURU⊥) = Fix(Refl(pi/4)Refl(0)Refl(pi/2)) = R(cos(3pi/4),sin(3pi/4)) = R(−1, 1) = V ⊥.896.4. The Euclidean plane R2(v) Fix(RU⊥RURV ) = Fix(Refl(pi/2)Refl(0)Refl(pi/4)) = R(cos(3pi/4),sin(3pi/4)) = R(−1, 1) = V ⊥.(vi) Fix(RURU⊥RV ) = Fix(Refl(0)Refl(pi/2)Refl(pi/4)) = R(cos(−pi/4),sin(−pi/4)) = R(1,−1) = V ⊥.Proof. (i): We haveFix(RU⊥RVRU ) = Fix(Refl(pi/2)Refl(pi/4)Refl(0))= Fix(Refl(pi/2)Rot(2((pi/4)− 0))) (from Fact 6.13(i))= Fix(Refl(pi/2)Rot(pi/2))= Fix(Refl((pi/2)− (1/2)(pi/2))) (from Fact 6.13(iv))= Fix(Refl(pi/4)).By (6.22) and Fact 3.3, we obtainFix(RU⊥RVRU ) = Fix(Refl(pi/4))= Fix(RR·(cos(pi/4),sin(pi/4)))= R · ( cos(pi/4), sin(pi/4))= R · (1, 1) = V = RU (V ⊥).(ii): The proof is similar to (i).(iii): We haveFix(RVRU⊥RU ) = Fix(Refl(pi/4)Refl(pi/2)Refl(0))= Fix(Refl(pi/4)Rot(2((pi/2)− 0))) (from Fact 6.13(i))= Fix(Refl(pi/4)Rot(pi))= Fix(Refl((pi/4)− (1/2)(pi))) (from Fact 6.13(iv))= Fix(Refl(−pi/4)).By (6.22) and Fact 3.3, we obtainFix(RVRU⊥RU ) = Fix(Refl(−pi/4))= Fix(RR·(cos(−pi/4),sin(−pi/4)))= R · ( cos(−pi/4), sin(−pi/4))= R · (1,−1) = V ⊥.(iv), (v), (vi): The proof is similar to (iii).906.4. The Euclidean plane R2Example 6.16. Let γ ∈ R, and let ε1, ε2, ε3 all be small in absolute value.Set α1 := γ+pi/6+ε1, α2 := γ+ε2, α3 := γ−pi/6+ε3, and ε := ε1−ε2 +ε3,and suppose thatU1 = R(cos(α1), sin(α1)),U2 = R(cos(α2), sin(α2)),andU3 = R(cos(α3), sin(α3)).Then it follows from (6.22) thatRU1 = RR(cos(α1),sin(α1)) = Refl(α1),RU2 = RR(cos(α2),sin(α2)) = Refl(α2),andRU3 = RR(cos(α3),sin(α3)) = Refl(α3).Then we haveRU3RU2RU1 = Refl(α3)Refl(α2)Refl(α1)= Refl(α3)Rot(2(α2 − β1)) (from Fact 6.13(i))= Refl(α3 − 12(2(α2 − β1))) (from Fact 6.13(iv))= Refl(α3 − α2 + α1)= Refl(γ − pi6 + ε3 − γ − ε2 + γ +pi6 + ε1)= Refl(γ + ε1 − ε2 + ε3)= Refl(γ + ε),and by using (6.22), we obtainRefl(γ + ε) = RR·(cos(γ+ε),sin(γ+ε)).From Fact 3.3 and (6.22), we haveFix(RU3RU2RU1) = Fix(Refl(γ + ε))= Fix(RR·(cos(γ+ε),sin(γ+ε)))= R(cos(γ + ε), sin(γ + ε)).However, U1 ∩ U2 ∩ U3 = {0}.Remark 6.17. Consider the setting of Example 6.16.916.5. R3 and beyond(i) No matter which of the operators in {PU1 ,PU2 ,PU3 ,PU⊥1 ,PU⊥2 ,PU⊥3 }we apply to Fix(RU3RU2RU1), we always obtain a line and never thesingleton U1 ∩ U2 ∩ U3 = {0}.(ii) If each εi = 0, then ε = 0 and Fix(RU3RU2RU1) = U2. If additionallyγ = pi/6, then U2 = R(cos(pi/6), sin(pi/6)) = R(√3/2, 1/2) and werecover precisely Example 6.16.6.5 R3 and beyondConsidering reflectors and rotations in R3 (see, e.g., [49]) or even Rnis more complicated because there is no “easy” counterpart of Fact 6.13.However, using the fact that eigenvalues of isometries are always drawn from±1 or from nonreal complex conjugate pairs of magnitude 1, one obtains atleast the parity result thatm+ n ≡ dim(Fix(RmRm−1 · · ·R1)) mod 2 (6.29)for m classical reflectors R1, . . . , Rm on Rn.92Chapter 7Resolvents and YosidaApproximation ofDisplacement Mappings ofIsometries7.1 OverviewThis chapter is based on our papers [2] and [3]. We study in detail an-other nice class of monotone operators: displacement mappings of isometriesof finite order. Throughout this chapter, we assume thatR : H → H is a linear isometry of finite order m: Rm = Id. (7.1)Note that R is surjective and ‖R‖ = 1, and hence R is nonexpansive. Forconvenience, setM := Id−R (7.2)to be the displacement mapping of R. We summarize our main results inthis chapter as follows:• Important properties of the displacement mapping M are given in The-orem 7.1.• We give an explicit formula for the resolvent of M (see Theorem 7.2).We also collect some useful properties of the resolvent of γM whereγ ∈ R++ (see Proposition 7.3).• We study the asymptotic behaviour of JγM as γ tends either to 0+ orto +∞ (see Lemma 7.4).• Lemma 7.5 gives formulas for the Yosida approximations of the dis-placement mapping γM, the Yosida approximations of the inverse of937.2. Properties of operator M and γMdisplacement mapping γ(M−1), and J(1/γ)M where γ ∈ R++. It alsoprovides nice properties of γM.• Theorem 7.6 gives a formula for the skew linear operator T. It alsogives some of its properties.• We derive a formula for the inverse of the displacement mapping(see Theorem 7.7). We also collect important properties of M−1 (seeLemma 7.8).• We prove that the resolvent of γM−1 is a Banach contraction withconstant 2/(2 + γ) < 1 (see Lemma 7.9 and Remark 7.10).• A formula for the Moore-Penrose inverse of the displacement mappingand its properties are given in Theorem 7.11.• We illustrate the results by giving two examples. The first one is whenR is rational rotation (see Example 7.12) and the second one is whenR is the right shift operator (see Example 7.14).• Lemma 7.13 gives formulas for 12Id + T operator and its inverse. Italso gives important properties for both operators.Except for facts with explicit references, this chapter is based on results thatappear in our papers [2] and [3].7.2 Properties of operator M and γMThe next Theorem will show important results and some useful proper-ties of M and γM, where γ ∈ R++. SetD := Fix R ={x ∈ H∣∣x = Rx}. (7.3)Theorem 7.1. Let γ > 0 and setL := 1mm−2∑k=0(m− 1− k)Rk. (7.4)Then the following hold:(i) M is a singled-valued, continuous, linear, and maximally monotoneoperator with dom M = H. Moreover, γM is maximally monotone.947.2. Properties of operator M and γM(ii) ker M = D = ker M∗.(iii) D is a closed linear subspace of H. Moreover,PD =1mm−1∑k=0Rk, (7.5)and,PD⊥ = Id−PD =m− 1mId− 1mm−1∑k=1Rk. (7.6)(iv) M ◦ L = Id |D⊥ .(v) ran M = ran M∗ = D⊥ is closed.Proof. (i): It follows from (7.2) that M is a singled-valued, that dom M = Hand that R is nonexpansive. Hence, it follows from Fact 3.23 that M ismaximally monotone. Since γ > 0 and M is maximally monotone, then γMis maximally monotone by Example 3.26.(ii): Clearly,ker M = {x ∈ H | Mx = 0} = {x ∈ H | x− Rx = 0} = Fix R = D.By Fact 2.7, ker M∗ = D.(iii): Let x ∈ H and set p := 1m∑m−1k=0 Rkx. By using the fact Rm = R0 = Id,we havep− Rp = 1mm−1∑k=0Rkx− 1mm−1∑k=0Rk+1x = 1mm−1∑k=0Rkx− 1mm∑k=1Rkx = 0.Hence, p ∈ Fix R and by (7.3), we obtain p ∈ D. Moreover, R is isometrythen it preserve inner products, so for all x, y ∈ H, we have〈x, y〉 = 〈Rx,Ry〉 .Thus, for all d ∈ D〈d− p,p− x〉 = 〈R(d− p),R(p− x)〉= 〈Rd− Rp,Rp− Rx〉= 〈d− p,p− Rx〉 .Applying R repeatedly, we obtain〈d− p, p− x〉 =〈d− p, p− Rkx〉,957.2. Properties of operator M and γMfor all k ∈ N. Thenm 〈d− p,p− x〉 =m−1∑k=0〈d− p, p− Rkx〉=〈d− p,m−1∑k=0(p− Rkx)〉=〈d− p,mp−m−1∑k=0Rkx〉= 〈d− p,mp−mp〉 = 0.Therefore,PD =1mm−1∑k=0Rk.Next, by (7.5) we havePD⊥ = Id−PD = Id−1mm−1∑k=0Rk = m− 1mId− 1mm−1∑k=1Rk.(iv): Using (7.2) and (7.4), we obtainM ◦ L = ( Id−R)( 1mm−2∑k=0(m− 1− k)Rk) (7.7)= 1mm−2∑k=0(m− 1− k)Rk − 1mm−2∑k=0(m− 1− k)Rk+1 (7.8)= 1mm−2∑k=0(m− 1− k)Rk − 1mm−1∑k=1(m− k)Rk. (7.9)Moreover,1mm−2∑k=0(m− 1− k)Rk = 1mm−1∑k=0(m− 1− k)Rk (7.10)=m−1∑k=0Rk − 1mm−1∑k=0Rk − 1mm−1∑k=0k Rk (7.11)and− 1mm−1∑k=1(m− k)Rk = −m−1∑k=0Rk + R0 + 1mm−1∑k=0k Rk. (7.12)967.2. Properties of operator M and γMSubstituting (7.11) and (7.12) back into (7.9) with the fact that R0 = Id,we obtainM ◦ L = Id− 1mm−1∑k=0Rk. (7.13)Hence, combining (7.13) with (ii) gives(M ◦ L)|D⊥ = ( Id−PD )|D⊥ = Id |D⊥ .(v):”⇒”: From (iii), we have ker M = D = ker M∗. So for all x ∈ D, we haveMx = 0 and M∗x = 0. Fix x ∈ D and y ∈ H.〈x,My〉 = 〈M∗x, y〉 = 〈0, y〉 = 0.Therefore,ran M ⊆ D⊥,and by Fact 2.7, we obtainran M = ran M∗ ⊆ D⊥. (7.14)”⇐”: Assume that y ∈ D⊥ and setx = 1mm−2∑k=0(m− 1− k)Rky. (7.15)From the definitions of M and x, we haveMx = x− Rx (7.16)= 1mm−2∑k=0(m− 1− k)Rky − 1mm−2∑k=0(m− 1− k)Rk+1y. (7.17)Clearly,1mm−2∑k=0(m− 1− k)Rk+1y = 1mm−1∑k=1(m− 1− (k − 1))Rky (7.18)= 1mm−1∑k=1(m− k)Rky. (7.19)977.3. The resolvent of γM and its asymptotic behaviourSubstituting (7.19) in (7.17), we haveMx = 1mm−2∑k=0(m− 1− k)Rky − 1mm−1∑k=1(m− k)Rky= m− 1mR0y + 1mm−2∑k=1(m− 1− k)Rky − 1mm−1∑k=1(m− k)Rky= m− 1my + 1mm−1∑k=1(m− 1− k)Rky − 1mm−1∑k=1(m− k)Rky= m− 1my − 1mm−1∑k=1Rky= y − 1my − 1mm−1∑k=1Rky=(Id− 1mm−1∑k=0Rk)y = PD⊥ y = y.Therefore, D⊥ ⊆ ran M. By Fact 2.7, we obtainD⊥ ⊆ ran M = ran M∗. (7.20)Combining (7.14) and (7.20) yieldsran M = ran M∗ = D⊥,which is closed by Fact 2.8.7.3 The resolvent of γM and its asymptoticbehaviourIn the next we derive a formula of the resolvents of γM and show usefulproperties and JγM. Moreover, the relationship between γM and JγM isgiven too.Theorem 7.2. Let γ > 0. Then the following hold:(i) JγM = 11+γ∞∑r=0( γ1+γ)rRr.(ii) ‖ γ1+γR‖ < 1.987.3. The resolvent of γM and its asymptotic behaviour(iii) We haveJγM =1(1 + γ)m − γmm−1∑k=0(1 + γ)m−1−kγkRk. (7.21)(iv) JγM : H → H and Id−JγM : H → H are firmly nonexpansive andmaximally monotone.Proof. (i): SetS := 11 + γ∞∑r=0( γ1 + γ)rRr. (7.22)From (3.11), we haveJγM =(Id +γM)−1.Since M and S are both defined using only powers of R and scalar multipli-cation, then everything commutes. To prove S =(Id +γM)inverse, it thensuffices to show thatS ◦ ( Id +γM) = Id .Note thatS ◦ R = 11 + γ∞∑r=0( γ1 + γ)rRr+1 (7.23)=(1γ)( γ1 + γ) ∞∑r=0( γ1 + γ)rRr+1 (7.24)= 1γ∞∑r=0( γ1 + γ)r+1Rr+1 (7.25)= 1γ∞∑r=1( γ1 + γ)rRr (7.26)= −1γId + 1γ∞∑r=0( γ1 + γ)rRr (7.27)= −1γId +(1 + γγ)S (7.28)= 1γ((1 + γ)S − Id). (7.29)Moreover,Id +γM = Id +γ(Id−R) = (1 + γ) Id−γR. (7.30)997.3. The resolvent of γM and its asymptotic behaviourBy using (7.30) and (7.29), we obtainS ◦ ( Id +γM) = S ◦ ((1 + γ) Id−γR)=(1 + γ)S − γS ◦ R=(1 + γ)S − γ(1γ((1 + γ)S − Id))=(1 + γ)S − (1 + γ)S + Id= Id .(ii): By the assumption ‖R‖ = 1, we obtain∥∥∥∥ γ1 + γR∥∥∥∥ = γ1 + γ ‖R‖ = γ1 + γ < 1. (7.31)The last inequality in (7.31) holds since γ > 0.(iii): Let r = mq + k, where 0 ≤ k ≤ m − 1 and 0 ≤ q < ∞. From (i) andthe fact that Rm = Id, we obtainJγM =11 + γ∞∑r=0( γ1 + γ)rRr= 11 + γm−1∑k=0∞∑q=0( γ1 + γ)mq+kRmq+k= 11 + γ(m−1∑k=0( γ1 + γ)kRk ∞∑q=0( γ1 + γ)mqRmq)= 11 + γ(m−1∑k=0( γ1 + γ)kRk(1 + ( γ1 + γ )m + ( γ1 + γ )2m + · · ·)Id)= 11 + γ(m−1∑k=0( γ1 + γ)kRk( 11− (γ/(1 + γ))m)Id)= 11 + γ(m−1∑k=0( γ1 + γ)kRk( (1 + γ)m(1 + γ)m − γm)Id)= 1(1 + γ)m − γmm−1∑k=0(1 + γ)m−1−kγkRk.Proposition 7.3. We have1007.3. The resolvent of γM and its asymptotic behaviour(i) Let m be fixed. Then for all k ∈ {0, 1, . . . ,m− 1},(1 + γ)m−1(1 + γ)m − γm( γ1 + γ)k> 0 and(1 + γ)m−1(1 + γ)m − γmm−1∑k=0( γ1 + γ)k= 1.(ii) JγM ∈ conv {Id,R, . . . ,Rm−1}.(iii) If D 6= {0}, then JγM cannot be a Banach contraction.Proof. (i): Clearly, (1 + γ)m−1(1 + γ)m − γm( γ1 + γ)k> 0.Sets :=m−1∑k=0( γ1 + γ)k= 1 +( γ1 + γ)+( γ1 + γ)2+ · · ·+( γ1 + γ)m−1(7.32)Then( γ1 + γ)s =( γ1 + γ)+( γ1 + γ)2+ · · ·+( γ1 + γ)m−1+( γ1 + γ)m. (7.33)Using (7.32) and (7.33), we haves1 + γ = s−( γ1 + γ)s = 1−( γ1 + γ)m ⇒ s = (1 + γ)m − γm(1 + γ)m−1 . (7.34)Using (7.34), we obtain(1 + γ)m−1(1 + γ)m − γmm−1∑k=0( γ1 + γ)k=(1 + γ)m−1(1 + γ)m − γm(1 + γ)m − γm(1 + γ)m−1 = 1.(ii): Clear from (i) and (7.21).(iii): If D 6= {0}, then D = ker M = ker (γM) = Fix JγM contains infinitelymany points. Therefore, JγM cannot be a Banach contraction.In the next lemma, we will describe the asymptotic behaviour of JγM asγ tends either to 0+ or to +∞.Lemma 7.4. Let γ > 0. We have1017.3. The resolvent of γM and its asymptotic behaviour(i) limγ→0+JγM = Id.(ii) limγ→+∞ JγM =1mm−1∑k=0Rk = PD.Proof. From Theorem 7.2(iii), we haveJγM =1(1 + γ)m − γmm−1∑k=0(1 + γ)m−1−kγkRk=(1 + γ)m−1(1 + γ)m − γmm−1∑k=0( γ1 + γ)kRk.Note that γ appears in the fraction, so we can first take the limits of thetwo fractions, and then use those limits to evaluate the overall sum.(i): We havelimγ→0+(1 + γ)m−1(1 + γ)m − γm =(1 + 0)m−1(1 + 0)m − 0m = 1. (7.35)For k > 0, we havelimγ→0+( γ1 + γ)k=( 01 + 0)k= 0. (7.36)for k = 0, on the other hand, we havelimγ→0+( γ1 + γ)0= limγ→0+1 = 1. (7.37)Using (7.35), (7.36) and (7.37), we obtain the overall limit aslimγ→0+JγM = limγ→0+(1 + γ)m−1(1 + γ)m − γmm−1∑k=0( γ1 + γ)kRk=(limγ→0+(1 + γ)m−1(1 + γ)m − γm)m−1∑k=0(limγ→0+( γ1 + γ)k)Rk= 1((limγ→0+( γ1 + γ)0)R0 +m−1∑k=1(limγ→0+( γ1 + γ)k)Rk)= 1 · R0 +m−1∑k=10 · Rk= Id .1027.3. The resolvent of γM and its asymptotic behaviour(ii): We evaluate the limit by comparing the leading terms. For the insidefraction, we havelimγ→+∞( γ1 + γ)k= limγ→+∞( 11γ + 1)k= limγ→∞ 1k = 1. (7.38)For the outside fraction, we first need to use binomial theorem to simplifythe denominator. Therefore,(1 + γ)m = 1 +mγ + · · ·+mγm−1 + γm,(1 + γ)m − γm = 1 +mγ + · · ·+mγm−1,and (1 + γ)m−1 = 1 + (m− 1)γ + · · ·+ (m− 1)γm−2 + γm−1.Taking the leading terms again, we havelimγ→+∞(1 + γ)m−1(1 + γ)m − γm = limγ→+∞1 + · · ·+ (m− 1)γm−2 + γm−11 +mγ + · · ·+mγm−1= limγ→+∞(1/γm−1) + · · ·+ (m− 1)(γm−2/γm−1)+ 1(1/γm−1) + · · ·+ (m− 1)(γm−2/γm−1)+m.Hence,limγ→+∞(1 + γ)m−1(1 + γ)m − γm =1m. (7.39)Using (7.38) and (7.39), we obtain the overall limitlimγ→∞ JγM = limγ→∞(1 + γ)m−1(1 + γ)m − γmm−1∑k=0( γ1 + γ)kRk=(limγ→∞(1 + γ)m−1(1 + γ)m − γm)m−1∑k=0(limγ→∞( γ1 + γ)k)Rk= 1mm−1∑k=0Rk.Now apply (7.5).1037.4. Yosida approximations7.4 Yosida approximationsRecall that the Yosida approximation of A of index γ > 0 is defined byγA := 1γ(Id−JγA). Yosida approximations can be viewed as regularizationand approximation of A because γA is a single-valued Lipschitz continuousoperator on H and it approximates A because γA → PAx(0) ∈ Ax asγ → 0+, for this and more information see [10, Chapter 23]. In this sectionwe will derive formulas for γM, γ(M−1). We also present formulas for J(1/γ)Mand JγM−1 .Lemma 7.5. Let γ > 0 and recall thatJγM =1(1 + γ)m − γmm−1∑k=0(1 + γ)m−1−kγkRk.Then the following hold:(i)γ(M−1) = 1γ(Id−JγM−1)=(γ Id +M)−1 = J(1/γ)M◦(γ−1 Id ) (7.40)andγM = 1γ(Id−JγM)=(γ Id +M−1)−1 = (J 1γM−1) ◦ (γ−1 Id ). (7.41)(ii) γM is maximally monotone.(iii) γM : H → H is Lipchitz continuous.(iv) We haveJγM−1 = Id−J(1/γ)M. (7.42)(v) We haveJ(1/γ) M =γ(γ + 1)m − 1m−1∑k=0(γ + 1)m−1−kRk. (7.43)(vi) We haveγ(M−1) = 1(1 + γ)m − 1m−1∑k=0(1 + γ)m−1−kRk. (7.44)1047.4. Yosida approximationsProof. (i): It follows from (3.14) that γ(M−1)= 1γ(Id−JγM−1). Let(x, u) ∈H × H. Then we have u ∈ γ(M−1)x ⇔ γu = x − JγM−1x ⇔ x − γu =JγM−1x ⇔ x ∈(x − γu) + γM−1(x − γu) ⇔ u ∈ M−1(x − γu) ⇔ x − γu ∈Mu⇔ u ∈ (γ Id +M)−1x. Hence,γM−1 =(γ Id +M)−1.Moreover, we have (1/γ)x ∈ ( Id +(1/γ)M)u⇔ u ∈ ( Id +(1/γ)M)−1((1/γ)x).Hence,γ(M−1) = J(1/γ)M ◦ (γ−1 Id ).Similar approach shows thatγM = 1γ(Id−JγM)=(γ Id +M−1)−1 = (J(1/γ)M−1) ◦ (γ−1 Id ).(ii): By Theorem 7.2(iv) we obtain that γ(Id−JγM)/γ := γ(γM) is firmlynonexpansive and maxiamlly monotone. Since γ > 0, then γM is still maxi-mally monotone.(iii): Take x and y be in H. From (3.14), Theorem 7.2(iv), and Cauchy-Schwarz inequality we obtain∥∥γ(M)x−γ (M)y∥∥2 = ∥∥∥∥1γ ( Id−JγM)x− 1γ ( Id−JγM)y∥∥∥∥2= 1γ2∥∥( Id−JγM)x− ( Id−JγM)y∥∥2≤ 1γ2〈x− y, ( Id−JγM)x− ( Id−JγM)y〉= 1γ〈x− y, 1γ(Id−JγM)x− 1γ(Id−JγM)y〉= 1γ‖x− y‖∥∥∥∥1γ ( Id−JγM)x− 1γ ( Id−JγM)x∥∥∥∥= 1γ‖x− y‖ ∥∥γ(M)x−γ (M)y∥∥ .Hence, for all x, y ∈ H:∥∥γ(M)x−γ (M)y∥∥ ≤ 1γ‖x− y‖ .1057.5. Properties of T operator(iv): From (i) and using the linearity of M, we obtainId = JγM−1 + γ(γ(M−1))= JγM−1 + γJ(1/γ)M ◦(γ−1 Id)= JγM−1 + J(1/γ)M.Therefore,JγM−1 = Id−J(1/γ)M.(v): By (7.21), we haveJ(1/γ)M =1(1 + (1/γ))m − (1/γ)mm−1∑k=0(1 + (1/γ))m−1−k(1/γ)kRk= 1((γ + 1)/γ)m − (1/γ)mm−1∑k=0((1 + γ)/γ)m−1−k(1/γ)kRk= γm(γ + 1)m − 1m−1∑k=0(γ + 1)m−1−k(1/γ)m−1Rk= γ(γ + 1)m − 1m−1∑k=0(γ + 1)m−1−k Rk.(vi): By using (3.14) and (iv), we obtainγ(M−1) = 1γ(Id−JγM−1)= 1γ(Id−( Id−J(1/γ)M)) = γ−1J(1/γ)M. (7.45)On the other hand, using (v) gives1γJ(1/γ)M =1(γ + 1)m − 1m−1∑k=0(γ + 1)m−1−kRk. (7.46)Altogether, with (7.45) and (7.46), the result follows.7.5 Properties of T operatorTheorem 7.6. SetT : H → H : x 7→ 12mm−1∑k=1(m− 2k)Rk. (7.47)Then the following hold:1067.5. Properties of T operator(i) T is a continuous linear operator.(ii) T = 12mbm2 c∑k=1(m− 2k)(Rk − Rm−k).(iii) T∗ = −T, i.e., T is skew-linear.(iv) T is maximally monotone.(v) ran T ⊆ D⊥, where D = ker M.(vi) PD⊥ T = T PD⊥ = T.Proof. (i): This is clear because T is defined using R and R is a continuouslinear. (ii): We split into odd and even cases.Case 1: m is even. We have⌊m2⌋= m2 . Let l = m− k. Since k ranges from1 to m2 , then l = m − k ranges from m2 to m − 1, and m − 2k = 2l − m.Hence, we have12mm2∑k=1(m− 2k)(Rk − Rm−k)= 12m( m2∑k=1(m− 2k)Rk − m2∑k=1(m− 2k)Rm−k)= 12m( m2∑k=1(m− 2k)Rk − m−1∑l=m2(2l −m)Rl)= 12m( m2∑k=1(m− 2k)Rk + m−1∑l=m2(m− 2l)Rl).Since the label on the index is arbitrary, we have12m( m2∑k=1(m− 2k) Rk + m−1∑l=m2(m− 2l) Rl) (7.48)= 12m( m2∑k=1(m− 2k) Rk + m−1∑k=m2(m− 2k) Rk). (7.49)Note that the term with k = m2 is counted twice. Luckily, when k =m2 ,m − 2k = m −m = 0, and 0 + 0 = 0. Thus, the doubled term is then just1077.5. Properties of T operator0. Therefore (7.49) gives12mm2∑k=1(m− 2k)(Rk − Rm−k) = 12m( m2∑k=1(m− 2k)Rk + m−1∑k=m2(m− 2k)Rk)= 12mm−1∑k=1(m− 2k)Rk= T.Case 2: m is odd, we have⌊m2⌋= m−12 . Let l = m − k Then l will rangefrom m−12 + 1 to m − 1. Hence by using the same arguments as before, wehave12mm−12∑k=1(m−2k)(Rk−Rm−k) = 12m( m−12∑k=1(m−2k)Rk+ m−1∑k=m−12 +1(m−2k)Rk).(7.50)Therefore,12m( m−12∑k=1(m− 2k)Rk + m−1∑k=m−12 +1(m− 2k)Rk) = 12mm−1∑k=1(m− 2k)Rk= T.(iii): By Fact 3.8, we obtainT∗ =( 12mm−1∑k=1(m− 2k)Rk)∗= 12mm−1∑k=1(m− 2k)(Rk)∗= 12mm−1∑k=1(m− 2k)Rm−k (let l = m− k.)= 12mm−1∑l=1(2l −m)Rl= 12mm−1∑l=1−(m− 2l)Rl= − 12mm−1∑k=1(m− 2l)Rl1087.5. Properties of T operator= −T,(iv): From (iii) we have(∀y ∈ H) 〈x− y,Tx− Ty〉 = 〈x− y,T(x− y)〉 = 0.Hence, T is monotone and it is also a continuous by (i). Then T is maximallymonotone by Fact 3.22.(v): Fix x ∈ D = Fix R. Then Rx = x andTx = 12mm−1∑k=1(m− 2k)Rk(x)= 12mm−1∑k=1(m− 2k)x= 12mm−1∑k=1(mx− 2kx)= 12m(m−1∑k=1mx− 2m−1∑k=1kx)= 12m((m− 1)mx− 2 (m(m− 1)2)x)= 12m((m− 1)mx−m(m− 1)x)= 0.Hence, Tx = 0 for all x ∈ D. In other words, D ⊆ ker T. Next, for all x ∈ Dand y ∈ H, we have〈x,Ty〉 = 〈T∗x, y〉 = 〈−Tx, y〉 = 〈0, y〉 = 0.Therefore, ran T ⊆ D⊥.(vi): Since ran T ⊆ D⊥ by using (v), we obtainPD⊥ T = T.Moreover, both T and Id−PD = PD⊥ are polynomials in R by (7.47) and(7.6). Hence, T and PD⊥ commute and soT PD⊥ = PD⊥ T = T.1097.6. The set valued inverse M−1m T with the right shift operator T with Rational rotation operator2(0 00 0) (0 00 0)3 16 0 −1 11 0 −1−1 1 0 12(0 −1/√31/√3 0)4 140 −1 0 11 0 −1 00 1 0 −1−1 0 1 0 12(0 −11 0)Table 7.1: T operator with the right shift operator and Rational rotationoperator (see Example 7.12 and Example 7.14 for more details).7.6 The set valued inverse M−1In this section we derive a formula for the inverse of the operator M andwe collect some useful properties of M−1. These will be useful in the nextchapters.Theorem 7.7. We haveM−1 − 12 Id = T +ND⊥ .andM−1 = 12 Id +T +ND⊥ .Proof. Let y ∈ D⊥ and x ∈ M−1y. Then Mx = y and M−1y = x+ ker M =x+D. Hence we must show thatx+D − 12y?= Ty +D. (7.51)Using (7.2) and y = Mx, we have (7.51) asx+D − 12(x−Rx) ?= T (x−Rx)+D ⇔ D + (x+Rx) ?= 2T (x−Rx)+D.Note that PD⊥(x+ Rx)= PD⊥(2T(x− Rx)) ⇔ 2T (x− Rx)− (x+ Rx) ∈D = ker M⇔ M(2T (x−Rx)) = M(x+Rx) = ( Id−R)(x+Rx). Hence wemust prove thatM(2T (x−Rx)) ?= x−R2x. (7.52)1107.6. The set valued inverse M−1By using (7.47), we have2mT(x−Rx) = m−1∑k=1(m− 2k)Rk(x−Rx)=m−1∑k=1(m− 2k)Rkx− m∑k=2(m+ 2− 2k)Rkx=(m− 2)x+ (m− 2)Rx− 2m−1∑k=2Rkx.The last equation implies that2T(x−Rx) = m− 2mx+ m− 2mRx− 2mm−1∑k=2Rkx.Hence, we haveM(2T(x−Rx)) = ( Id−R)(2T (x−Rx))=(2T(x−Rx))−R(2T (x−Rx))= m− 2mx+ m− 2mRx− 2mm−1∑k=2Rkx−R(m− 2mx+ m− 2mRx− 2mm−1∑k=2Rkx)= m− 2mx− 2mm−1∑k=2Rkx− m− 2mR2x+ 2mm−1∑k=2Rk+1x= m− 2mx− 2mR2x− m− 2mR2x+ 2mx= x−R2x,which verifies (7.52).Lemma 7.8. Let γ > 0. Then the following hold:(i) dom M−1 = D⊥.(ii) M−1 − 12 Id is skew and hence monotone.(iii) γM−1 − (γ/2) Id is skew and hence monotone.(iv) γM−1 is(γ/2)-strongly monotone.1117.6. The set valued inverse M−1Proof. (i): Clearly, dom(T+ND⊥)= D⊥ and from Theorem 7.1(v), we haveD⊥ = ran M = dom M−1 = dom(M−1 − 12 Id).(ii): From Theorem 7.7, we obtain M−1 = 12 Id +T +ND⊥ . ThenM−1 − 12 Id = T +ND⊥ .Clearly, ND⊥ is skew and T is a skew from Theorem 7.6(iii). Hence, T+ND⊥is a skew. Therefore, M−1 − 12 Id is a skew and monotone.(iii): It follows from (ii) that M−1 − 12 Id is monotone and a skew. ThenγM−1 − γ2 Id is skew and monotone since γ > 0.(iv): This follows from Definition 3.17(iv) and (iii).Lemma 7.9. Recalling from Lemma 7.8 (iv) that γM−1 is strongly mono-tone with constant β = γ/2. Then JγM−1 is((2 + γ)/2)-cocoercive andJγM−1 is Lipschitz continuous with constant 1/(1 + β)= 2/(2 + γ) ∈ ]0, 1[.Proof. Set u = JγM−1x and v = JγM−1y. Then x ∈ u + γM−1u ⇔ x −u ∈ γM−1u and y − v ∈ γM−1v. Since γM−1 is strongly monotone withconstant β = γ/2, then we obtain〈(x− u)− (y − v), u− v〉 ≥ γ2 ‖u− v‖2 ⇔〈x− y, u− v〉 − ‖u− v‖2 ≥ γ2 ‖u− v‖2 ⇔ 〈x− y, u− v〉 ≥ 2+γ2 ‖u− v‖2.This shows that JγM−1 is((2 + γ)/2)-cocoercive. By using Cauchy-Schwarzinequality we have‖x− y‖ ‖u− v‖ ≥ 〈x− y, u− v〉 ≥ 2 + γ2 ‖u− v‖2⇒ ‖x− y‖ ≥ 2 + γ2 ‖u− v‖ .Hence, we obtain ∥∥∥JγM−1x− JγM−1y∥∥∥ ≤ 22 + γ ‖x− y‖ . (7.53)This shows that JγM−1 is Lipschitz continuous with constant 2/(2 + γ) ∈]0, 1[.Remark 7.10. We show in Lemma 7.9 that JγM−1 is Banach contraction withconstant 2/(2 + γ)< 1.1127.7. Moore-Penrose inverse M †7.7 Moore-Penrose inverse M †Theorem 7.11. The following hold:(i) M † =(Id−R)† = 12 PD⊥ ( Id +2T) = ∑m−1k=0 m−1−2k2m Rk.(ii) For all y ∈ ran M = D, we haveM−1y = M † + D = D +m−1∑k=0m− 1− 2k2m Rky.(iii) M † = 0(M−1).Proof. (i): By using (7.6), (7.47) and Fact 2.21 we obtainM† = Pran M∗ ◦M−1 ◦ Pran M = PD⊥ ◦(12 Id +T +ND⊥) ◦ PD⊥= 12 PD⊥ + PD⊥ T + PD⊥(D)= 12 PD⊥ + PD⊥ T + 0= 12 PD⊥ + PD⊥ T= 12 PD⊥ +T= 12(Id− 1mm−1∑k=0Rk)+ 12mm−1∑k=1(m− 2k)Rk= 12(m− 1mId + 1mm−1∑k=1(m− 2k − 1)Rk)= 12mm−1∑k=0(m− 1− 2k)Rk.(ii): Suppose y ∈ ran M = D⊥. ThenM−1y = 12y + Ty +ND⊥y= 12y +12mm−1∑k=1(m− 2k)Rky + D= 12mm−1∑k=0(m− 1− 2k)Rky + D + 1mm−1∑k=0Rky= M† + D + 1mm−1∑k=0Rky.1137.8. Rational rotation ExampleSince PD = 1m∑m−1k=0 Rk and y ∈ D⊥, then 1m∑m−1k=0 Rky = 0. Hence, weobtainM−1y = M†y + D.(iii): It follows from Definition 2.20 that M†y = PM−1y(0).7.8 Rational rotation ExampleWe illustrate the results by giving two examples. The first one when Ris rational rotation. See example bellow:Example 7.12. Let m ∈ {1, 2, . . . } and suppose H = R2 andR =cos(2pim) − sin (2pim )sin(2pim)cos(2pim) .Then Rm = Id and D = Fix R = {(0, 0)} ⊂ H. Using (7.21), (7.42),(3.14) and (7.44), we record the following formulas for JγM, JγM−1 , γM, andγ(M)−1:Let m = 2. ThenJγM =11 + 2γ(1 00 1), JγM−1 =22 + γ(1 00 1),γM = 21 + 2γ(1 00 1), and γ(M−1)= 12 + γ(1 00 1).Let m = 3. ThenJγM =16γ2 + 6γ + 2(2 + 3γ −√3γ√3γ 2 + 3γ),JγM−1 =12γ2 + 6γ + 6(6 + 3γ√3γ−√3γ 6 + 3γ),γM = 16γ2 + 6γ + 2(3 + 6γ√3−√3 3 + 6γ),andγ(M−1) = 12γ2 + 6γ + 6(3 + 2γ −√3√3 3 + 2γ).1147.9. The right shift operator RLet m = 4. ThenJγM =12γ2 + 2γ + 1(1 + γ −γγ 1 + γ),JγM−1 =1γ2 + 2γ + 2(2 + γ γ−γ 2 + γ),γM = 12γ2 + 2γ + 1(1 + 2γ 1−1 1 + 2γ),andγ(M−1) = 1γ2 + 2γ + 2(1 + γ −11 1 + γ).Appendix B.1 contains the code that can be used to verify the above exam-ple.7.9 The right shift operator RLet Hm be the Hilbert product space. Define the cyclic right-shift oper-atorR : Hm → Hm : (x1, x2, . . . , xm) 7→ (xm, x1, . . . , xm−1).Recall from (7.2) thatM := Id−R.7.10 Cyclic right-shift operator R and 12Id + ToperatorLemma 7.13. SetQ := 1mm−1∑k=1kRk, (7.54)and recall from (7.47) thatT = 12mm−1∑k=1(m− 2k)Rk.Then the following hold:(i) We have P∆ R = R P∆ = P∆ and P∆⊥ R = R − P∆.1157.10. Cyclic right-shift operator R and 12Id + T operator(ii) 2Q P∆ =(m− 1)P∆.(iii) −Q(Id−R) = P∆⊥ .(iv) 12Id + T =m2 P∆−Q.(v)(12Id + T)−1 = Id−R + 2 P∆ .(vi)(12Id + T)−1|∆⊥ = (Id−R)|∆⊥ = M|∆⊥ .(vii) 12Id + T is12 -strongly monotone.(viii)(12Id + T)−1(x) is 12 -cocoercive.Proof. (i): Note that P∆ R = R P∆ from (7.5). ThenR P∆ =1mm−1∑k=0Rk+1= 1mm∑k=1Rk= 1mm−1∑k=0Rk (Since R0 = Rm = Id )= P∆ ( From (7.5)).Hence,P∆⊥ R =(Id− P∆)R= R − P∆ R= R −R P∆= R − P∆ .(ii): By using (7.54) and (i), we obtainQ P∆ =1mm−1∑k=1kRk P∆= 1mm−1∑k=1kP∆= 1m(m(m− 1)2)P∆ =m− 12 P∆ .1167.10. Cyclic right-shift operator R and 12Id + T operator(iii): By using (7.54) and (7.5), we obtain−mQ(Id−R) = (R − Id)(mQ) = (R − Id)m−1∑k=1kRk=m−1∑k=1kRk+1 −m−1∑k=1kRk =m∑k=2(k − 1)Rk − m−1∑k=1kRk=(m− 1)Rm + (m−1∑k=2((k − 1)− k)Rk)−R=(m− 1)Id− (m−1∑k=2Rk)−R (Because Rm = Id)= m Id−m−1∑k=0Rk = m Id−mP∆ = mP∆⊥ .(iv): From (7.47), we haveT = 12mm−1∑k=1(m− 2k)Rk = 12m−1∑k=1Rk − 1mm−1∑k=1kRk= −12 Id +m2mm−1∑k=0Rk − 1mm−1∑k=1kRk.Hence,12Id + T =m2 P∆−1mm−1∑k=1kRk = m2 P∆−Q,by (7.54).(v): By using (iv), (i), (ii) , and (iii), we obtain(12Id + T)(Id−R + 2 P∆)=(m2 P∆−Q)(Id−R + 2 P∆)= m2(P∆−P∆ R)+mP∆−Q(Id−R)− 2Q P∆= mP∆ + P∆⊥ −(m− 1)P∆ = P∆ + P∆⊥= Id.(vi): Combining (v) and (7.2), we have(12Id + T)−1|∆⊥ = (Id−R + 2 P∆ )|∆⊥ = (Id−R)|∆⊥ = M|∆⊥ .1177.11. Cyclic right-shift operator example(vii): Our goal is to show that 12Id + T is12 -strongly monotone ⇔ 12Id +T− 12Id = T is monotone. By Theorem 7.6(iv), T is maximally monotone⇒ T is monotone.Hence, 12Id + T is 12 -strongly monotone.(viii): For simplicity, set A := 12Id + T and B :=(12Id + T)−1. Take(x,u) ∈ gra A and (y,v) ∈ gra A. Then u ∈ Ax⇔ u = (12Id+T)x⇔ x =(12Id + T)−1u by (v). Similarly, y = (12Id + T)−1v. Since A is 12 -stronglymonotone by (vii), then〈u− v,x− y〉 ≥ 12 ‖x− y‖2⇒〈u− v,(12Id + T)−1u−(12Id + T)−1v〉≥ 12∥∥∥∥(12Id + T)−1u−(12Id + T)−1v∥∥∥∥2= 〈u− v,Bu−Bv〉 ≥ 12 ‖Bu−Bv‖2 .Hence,(12Id + T)−1is 12 -cocoercive.7.11 Cyclic right-shift operator exampleExample 7.14. Let R be the cyclic right-shift operator. Then Rm = Id,D = Fix R = {(x, x, . . . , x) ∈ Hm | x ∈ H}, andD⊥ = {(x1, x2, . . . , xm) ∈ Hm | x1 + x2 + · · ·+ xm = 0}.Using (7.21), (7.42), (3.14) and (7.44), we record the following formulas forJγM, JγM−1 , γM, and γ(M)−1:Let m = 2. ThenJγM =11 + 2γ(1 + γ γγ 1 + γ), JγM−1 =12 + γ(1 −1−1 1),γM = 11 + 2γ(1 −1−1 1), γ(M−1)= 1(2 + γ)γ(1 + γ 11 1 + γ).Let m = 3. ThenJγM =13γ2 + 3γ + 1(1 + γ)2 γ2 (1 + γ)γ(1 + γ)γ (1 + γ)2 γ2γ2 (1 + γ)γ (1 + γ)2 ;1187.11. Cyclic right-shift operator exampleJγM−1 =1γ2 + 3γ + 3 2 + γ −1 −(1 + γ)−(1 + γ) 2 + γ −1−1 −(1 + γ) 2 + γ ;γM = 13γ2 + 3γ + 1 1 + 2γ −γ −(1 + γ)−(1 + γ) 1 + 2γ −γ−γ −(1 + γ) 1 + 2γ ;andγ(M−1) = 1(γ2 + 3γ + 3)γ(1 + 2γ)2 1 1 + γ1 + γ (1 + 2γ)2 11 1 + γ (1 + γ)2 .Letaγ :=1(1 + γ)4 − γ4 ,bγ :=14 + 6γ + 4γ2 + γ3 ,cγ :=11 + 4γ + 6γ2 + 4γ3 ,anddγ :=1−1 + (1 + γ)4 .Let m = 4. ThenJγM = aγ(1 + γ)3 γ3 γ2(1 + γ) γ(1 + γ)2γ(1 + γ)2 (1 + γ)3 γ3 γ2(1 + γ)γ2(1 + γ) γ(1 + γ)2 (1 + γ)3 γ3γ3 γ2(1 + γ) γ(1 + γ)2 (1 + γ)3 ;JγM−1 = bγ3 + 3γ + γ2 −1 −(1 + γ) −(1 + γ)2−(1 + γ)2 3 + 3γ + γ2 −1 −(1 + γ)−(1 + γ) −(1 + γ)2 3 + 3γ + γ2 −1−1 −(1 + γ) −(1 + γ)2 3 + 3γ + γ2 ;γM = cγ1 + 3γ + 3γ2 −γ2 −γ(1 + γ) −(1 + γ)2−(1 + γ)2 1 + 3γ + 3γ2 −γ2 −γ(1 + γ)−γ(1 + γ) −(1 + γ)2 1 + 3γ + 3γ2 −γ2−γ2 −γ(1 + γ) −(1 + γ)2 1 + 3γ + 3γ2 ;1197.11. Cyclic right-shift operator exampleandγ(M−1) = dγ(1 + 2γ)3 1 1 + γ (1 + γ)2(1 + γ)2 (1 + γ)3 1 1 + γ1 + γ (1 + γ)2 (1 + γ)3 11 1 + γ (1 + γ)2 (1 + γ)3 .Appendix B.2 contains the code that can be used to verify the aboveexample. The results from this chapter will be used to resolve the geometryconjecture when m ≥ 3.120Chapter 8More Convex Analysis8.1 OverviewThis chapter is based on our papers [3] and [5]. Even though the mainresults are formulated in the given Hilbert space H, it will turn out that thekey space to work in is the product space,Hm = {x = (xi)i∈I | (∀i ∈ I) xi ∈ H}, (8.1)where m ∈ {2, 3, 4, . . . } and I = {1, 2, . . . ,m}. The product space containsan embedding of the original space H via the diagonal subspace∆ ={x =(x)i∈I | x ∈ H}. (8.2)Let C1,C2, . . . ,Cm be nonempty closed convex subsets of H with projectorsPC1 , PC2 , . . . , PCm which could be written more simply as P1,P2, . . . ,Pm.SetC := C1 × C2 × · · · × Cm. (8.3)Let us make Hm into the Hilbert product spaceH = Hm, with 〈x,y〉 =∑i∈I〈xi, yi〉 . (8.4)An important operator acts on H is the cyclic right-shift operatorR :H→H : (x1, x2, . . . , xm) 7→ (xm, x1, . . . , xm−1). (8.5)We summarize our main results in this chapter as follows:• We collect useful properties of σC+∆ (see Lemma 8.3).• Important properties of ιC, ιCι∆,(ιCι∆)∗, and (ι∗Cι∗∆)∗ are givenin Proposition 8.4 and Proposition 8.5.• We show that the range of ∂ιC+∆ is a subset of ∆⊥ (for more detailssee Lemma 8.6).1218.2. Properties of σC+∆• Properties of f∗+ι∗∆, fι∆, cl(fι∆), ∂(fι∆) as well as ∂[cl(fι∆)]are given in Proposition 8.8, Lemma 8.10, and Lemma 8.12.• We show that the range of ∂[cl(fι∆)] is a subset of ∆⊥ (for moredetails see Lemma 8.9).• We collect important properties of subdifferentials (see Lemma 8.11and Lemma 8.12).• We define some conditions that ensure that fιC is lower semicontin-uous and exact (see Proposition 8.13, Proposition 8.14, Corollary 8.15,Proposition 8.16, and Corollary 8.17).Except for facts with explicit references, this chapter is based on results thatappear in our paper [3] and [5].Fact 8.1. [10, Proposition 29.4] Set C := C1 × C2 × · · · × Cm and letx =(x1, x2, . . . , xm) ∈H. ThenPC :H→H :(x1, x2, . . . , xm) 7→ (P1 x1,P2 x2, . . . ,Pm xm). (8.6)Fact 8.2. [10, Proposition 26.4(i) and (iii)] Set ∆ = {x = (xi)i∈I | x ∈ H}and let x =(x1, x2, . . . , xm) ∈H. ThenP∆ :H→H :(x1, x2, . . . , xm) 7→ ( 1mm∑i=1xi, . . . ,1mm∑i=1xi). (8.7)Moreover, the orthogonal complement of ∆ with respect to this standardinner product is∆⊥ ={x =(xi)i∈I |∑i∈Ixi = 0}. (8.8)8.2 Properties of σC+∆Lemma 8.3. Let σC+∆ :H→ ]−∞,+∞] and T be as (7.47). Let x,y ∈Hand λ ∈ [0, 1]. Then the following hold:(i) σC+∆ is a convex.(ii) σC+∆ is lower semicontinuous.1228.2. Properties of σC+∆Proof. (i): Let x,y ∈ dom σC+∆. Then for all λ ∈ [0, 1] and(∀(c,d) ∈C×∆) we have〈λx, c + d〉 ≤ sup(c,d)∈C×∆〈λx, c + d〉 = λ sup(c,d)∈C×∆〈x, c + d〉and〈(1− λ)y, c + d〉 ≤ sup(c,d)∈C×∆〈λy, c + d〉 = (1− λ) sup(c,d)∈C×∆〈y, c + d〉 .Hence, by adding both equations we obtain for all(c,d) ∈ C×∆〈λx + (1− λ)y, c + d〉 ≤ λ sup(c,d)∈C×∆〈x, c + d〉+ (1− λ) sup(c,d)∈C×∆〈y, c + d〉 .Taking the supremum on the left-hand side, we havesup(c,d)∈C×∆〈λx + (1− λ)y, c + d〉 ≤ λ sup(c,d)∈C×∆〈x, c + d〉+ (1− λ) sup(c,d)∈C×∆〈y, c + d〉 .By Definition 2.34, we obtainσC+∆(λx + (1−λy)) ≤ λσC+∆(x)+ (1−λ)σC+∆(y) (∀(c,d) ∈ C×∆).Hence, σC+∆ is convex.(ii): It is sufficient to show epi(σC+∆)is closed. Let(xn, tn) ∈ epi(σC+∆)that converge to(x, t) ∈ H. We want to show that (x, t) ∈ epi(σC+∆).From (2.10) and Definition 2.34, we have for all w ∈ C + ∆σC+∆(xn) ≤ tn ⇒ (∀w ∈ C + ∆) supw∈C+∆〈xn,w〉 ≤ tn ⇒ 〈xn,w〉 ≤ tn.Hence,(∀w ∈ C + ∆) limn→∞ 〈xn,w〉 ≤ limn→∞ tn ⇔〈limn→∞xn,w〉≤ limn→∞ tn⇔ 〈x, t〉 ≤ t (∀w ∈ C + ∆) and by taking the supremum on the left-handside we getsupw∈C+∆〈x,w〉 ≤ t (∀w)⇔ (x, t) ∈ epi(σC+∆).Hence, epi(σC+∆)is closed. Therefore, σC+∆ is lower semicontinuous.1238.3. Properties of ιC and its subdifferential8.3 Properties of ιC and its subdifferentialProposition 8.4. The following are hold(i) ιC is lower semicontinuous.(ii) ∆⊥ = ∂ι∆(d), where d ∈∆.Proof. (i): It is sufficient to show that epi(ιC)is closed. Let λ ∈ R. If λ ≥ 0,then{x ∈H | ιC ≤ λ} = {x ∈H | ιC(x)= 0} = C,which is closed because C1,C2, . . . ,Cm are closed. On the other hand, ifλ < 0, then{x ∈H | ιC ≤ λ} = ∅,which is also closed set. Altogether epi(ιC)is closed. Hence, ιC is lowersemicontinuous.(ii): ”⊇”: Let d ∈∆ and u ∈ ∂ι∆(d). Then using (2.15), we have〈u,x− d〉+ ι∆(d) ≤ ι∆(x) ⇒ 〈u,x− d〉 ≤ ι∆(x)(∀x ∈H).Therefore,〈u,x− d〉 ≤ 0, (∀x ∈ dom ι∆). (8.9)Since ∆ is subspace, then −(x− d) ∈∆. Hence,(∀x ∈ dom ι∆) 〈u,−(x− d)〉 ≤ 0 ⇒ 〈u,x− d〉 ≥ 0. (8.10)Combining (8.9) and (8.10) gives〈u,x− d〉 = 0 ⇔ u ∈∆⊥.Hence,∂ι∆(d) ⊆∆⊥. (8.11)”⊆”: Let u ∈∆⊥ and fix d ∈∆. Then for all x ∈∆, we have〈u,x− d〉 = 0,〈u,x− d〉+ ι∆(d) = ι∆(x),(∀x ∈ dom ι∆).This implies that u ∈ ∂ι∆(d). Therefore,∆⊥ ⊆ ∂ι∆(d). (8.12)Combining (8.11) and (8.12), we have∆⊥ = ∂ι∆(d).1248.4. Properties of ιCι∆,(ιCι∆)∗, (ι∗Cι∗∆)∗ and their subdifferentials8.4 Properties of ιCι∆,(ιCι∆)∗, (ι∗Cι∗∆)∗ andtheir subdifferentialsProposition 8.5. Let x ∈ C + ∆, y ∈ C, and x − y ∈ ∆. Then thefollowing hold:(i) ιCι∆ = ιC+∆.(ii) dom(ι∗C + ι∗∆)= dom ι∗C ∩∆⊥.(iii) ι∗C + ι∗∆ =(ιCι∆)∗.(iv)(ι∗C + ι∗∆)∗ = (ιCι∆)∗∗.(v) cl(ιCι∆)= ιC+∆.(vi) ∂(ιCι∆)= ∂ιC+∆.(vii) ∂cl(ιCι∆)= ∂ιC+∆.(viii) ∂ιC+∆(x)= ∂ιC(y) ∩ ∂ι∆(x− y).(ix) ∂ιC(y) ∩ ∂ι∆(x− y) = ∂ιC+∆(x) ⊆ ∂ιC+∆(x).Proof. (i): From (2.14), we have(ιCι∆)(x)= infy+d=x{ιC(y)+ ι∆(d)}.Now,ιC ={0, if y ∈ C+∞, if y /∈ C.Hence,infy∈Cd∈∆y+d=x={0, if x ∈ C + ∆+∞, if x /∈ C + ∆. = ιC+∆(x).(ii): Clearly,dom(ι∗C + ι∗∆)= dom ι∗C ∩ dom ι∗∆= dom ι∗C ∩ dom ι∆⊥= dom ι∗C ∩∆⊥.1258.4. Properties of ιCι∆,(ιCι∆)∗, (ι∗Cι∗∆)∗ and their subdifferentials(iii): Let w ∈ dom (ι∗C + ι∗∆). Then(ι∗C + ι∗∆)w = supx1∈ dom ι∗C{〈w,x1〉 − ιC(x1)}+ supx2∈ dom ι∗∆{〈w,x2〉 − ι∆(x2)}= supx1∈ dom ι∗C, x2 ∈∆⊥{ 〈w,x1 + x2〉 − (ιC(x1) + ι∆(x2))}= supx∈ dom ι∗C∩∆⊥{ 〈w,x〉 − infx{ιC(x1) + ι∆(x2) | x = x1 + x2}}= supx∈ dom ι∗C∩∆⊥{ 〈w,x〉 − (ιCι∆)(x)}=(ιCι∆)∗w.(iv): Let u ∈ dom (ι∗C + ι∗∆)∗(u) and from (iii), we obtain(ι∗C + ι∗∆)∗(u) = supx∈ dom ι∗C∩∆⊥{ 〈u,x〉 − (ι∗C + ι∗∆)(x)}= supx∈ dom ι∗C∩∆⊥{ 〈u,x〉 − (ιCι∆)∗(x)}=(ιCι∆)∗∗u.(v): Because ι∗C + ι∗∆ =(ιCι∆)∗, from (iii). Hence,(ι∗C + ι∗∆)∗ = (ιCι∆)∗∗= ι∗∗conv (C+∆) ( From Fact 2.28)= ιC+∆.(vi): It follows from (i) and (2.15).(vii): It follows from (v) and (2.15).(ix): ”⊆”: Let x∗ ∈ ∂(ιCι∆)(x). Then by (2.15), we have(ιCι∆)(x)+ 〈z− x,x∗〉 ≤ (ιCι∆)(z) (∀z ∈H)⇒ ιC(y)+ ι∆(x− y)+ 〈z− x,x∗〉 ≤ (ιCι∆)(z) (∀z ∈H).Note that(∀u ∈H), (∀v ∈H) satisfy that u + v = z, we have(ιCι∆)(z)= infu∈Hv∈Hu+v=z{ιC(u)+ ι∆(v)} ≤ ιC(u)+ ι∆(v).Hence,(∀u ∈H), (∀v ∈H) such that u + v = z, we obtainιC(y)+ ι∆(x− y)+ 〈z− x,x∗〉 ≤ ιC(u)+ ι∆(v). (8.13)1268.4. Properties of ιCι∆,(ιCι∆)∗, (ι∗Cι∗∆)∗ and their subdifferentialsSet u = y. Then z = y + v and (8.13) givesιC(y)+ ι∆(x− y)+ 〈(y + v)− x,x∗〉 ≤ ιC(y)+ ι∆(v)⇒ ι∆(x− y)+ 〈v− (x− y),x∗〉 ≤ ι∆(v)This implies thatx∗ ∈ ∂ι∆(x− y). (8.14)Now, set v = x− y. Then z = u + (x− y) and (8.13) givesιC(y)+ ι∆(v)+〈(u +(x− y))− x,x∗〉 ≤ ιC(u)+ ι∆(v)⇒ ιC(y)+ 〈u− y,x∗〉 ≤ ιC(u)This implies thatx∗ ∈ ∂ιC(y). (8.15)Combining (8.14) and (8.15), we obtainx∗ ∈ ∂ιC(y) ∩ ∂ι∆(x− y).Hence,∂(ιCι∆) ⊆ ∂ιC(y) ∩ ∂ι∆(x− y). (8.16)”⊇”: Let x∗ ∈ ∂ιC(y) ∩ ∂ιC(x− y). Then by (2.15) we obtain〈x∗,u− y〉+ ιC(y) ≤ ιC(u) (∀u ∈H), (8.17)and 〈x∗,v− (x− y)〉+ ι∆(x− y) ≤ ι∆(v) (∀v ∈H). (8.18)Summing (8.17) and (8.18) gives(∀u ∈H), (v ∈H), we haveιC(y)+ ι∆(x− y)+ 〈(u + v)− x,x∗〉 ≤ ιC(u)+ ι∆(v).Hence, if z ∈H satisfies z = u + v, then we have(ιCι∆)(x)+ 〈x∗, z− x〉 ≤ ιC(u)+ ι∆(v).Taking the infimum for the right hand side overall z such that z = u + v,we obtain (ιCι∆)(x)+ 〈x∗, z− x〉 ≤ (ιCι∆)(z)1278.5. Visiting well-known results for the separable sum fThis implies thatx∗ ∈ ∂(ιCι∆)(x).Hence,∂ιC(y) ∩ ∂ιC(x− y) ⊆ ∂(ιCι∆). (8.19)Combining (8.16) and (8.19), we obtain∂ιC(y) ∩ ∂ιC(x− y) = ∂(ιCι∆).(ix): It follows from the fact that ιC+∆ ⊆ ιC+∆ and (viii).Lemma 8.6. We haveran ∂ιC+∆(x)= ranNC+∆(x) ⊆∆⊥.Proof. Let w ∈ ∂(ιC+∆)(x). Then from (2.15), we obtain〈w, z− x〉+ ιC+∆(x) ≤ ιC+∆(z), (∀z ∈H).Let v ∈∆, u ∈ C, and set z = u + v. From Proposition 8.4(i)〈w,(u + v)− x〉 ≤ ιC+∆(z) = (ιCι∆)(u + v)〈w,u− x〉+ 〈w,v〉 ≤ ιC(u)+ ι∆(v)= 0.Hence,〈w,v〉 ≤ − 〈w,u− x〉 . (8.20)Moreover, ∆ is subspace, therefore −v ∈∆. This gives that〈w,v〉 ≥ 〈w,u− x〉 . (8.21)Then summing (8.20) and (8.21), gives〈w,v〉 = 0 ⇔ w ∈∆⊥.8.5 Visiting well-known results for the separablesum fSuppose(fi)mi=1 be a finite family in Γ0(H) and define its separable sum:f := f1 ⊕ f2 ⊕ · · · ⊕ fm :H→]−∞,+∞] : x 7→ m∑i=1fi(xi), (8.22)where x =(x1, x2, . . . , xm) ∈H.1288.6. Properties of f∗ + ι∗∆, fι∆, cl(fι∆), ∂(fι∆), as well as ∂[cl(fι∆)]Fact 8.7. [22, Lemma 2.4] The following hold:(i) ∂f = ∂f1 × ∂f2 × · · · × ∂fm.(ii) f∗ = f∗1 ⊕ f∗2 ⊕ · · · ⊕ f∗m :(x1, x2, . . . , xm) 7→∑mi=1 f∗i (xi).(iii) Proxf =(Proxf1 ,Proxf2 , . . . ,Proxfm).8.6 Properties of f ∗ + ι∗∆, fι∆, cl(fι∆), ∂(fι∆),as well as ∂[cl(fι∆)]We recall from (2.14) that(fι∆)(x)= infy∈H{f(y)+ ι∆(x− y)}.Proposition 8.8. The following are hold:(i) dom(f∗ + ι∗∆)= dom f∗ ∩∆⊥.(ii) f∗ + ι∗∆ =(fι∆)∗.(iii)(f∗ + ι∗∆)∗ = (fι∆)∗∗. If dom(f∗ + ι∗∆) 6= ∅, then(f∗ + ι∗∆)∗ = cl(fι∆).Proof. (i): Clearly, dom(f∗ + ι∗∆)= dom f∗ ∩ dom ι∗∆ = dom f∗ ∩ ι∆⊥ =dom f∗ ∩∆⊥.(ii): Let w ∈ dom (f∗ + ι∗∆). Then(f∗ + ι∗∆)(w)= supx1∈ dom f∗{〈w,x1〉 − f(x1)}+ supx2∈ ∆⊥{〈w,x2〉 − ι∆(x2)}= supx1∈ dom f∗, x2 ∈∆⊥{ 〈w,x1 + x2〉 − (f(x1) + ι∆(x2))}= supx∈ dom f∗∩∆⊥{ 〈w,x〉 − infx{f(x1) + ι∆(x2) | x = x1 + x2}}= supx∈ dom f∗∩∆⊥{ 〈w,x〉 − (fι∆)(x)} (from (2.14))=(fι∆)∗(w). (from (2.16))(iii): Let w ∈ dom (f∗ + ι∗∆)∗ and from (ii), we obtain(f∗ + ι∗∆)∗(w) = supx∈ dom f∗∩∆⊥{ 〈w,x〉 − (f∗ + ι∗∆)(x)}= supx∈ dom f∗∩∆⊥{ 〈w,x〉 − (fι∆)∗(x)} (From (ii))=(fι∆)∗∗(w) (from (2.16) ).1298.6. Properties of f∗ + ι∗∆, fι∆, cl(fι∆), ∂(fι∆), as well as ∂[cl(fι∆)]Finally,(fι∆)∗∗ = cl(fι∆) because under the assumption, we have (f∗+ι∗∆)∗ is proper, so (f∗ + ι∗∆)∗ = cl(fι∆).Lemma 8.9. We haveran ∂[cl(fι∆)](x) ⊆∆⊥.Proof. Let w ∈ ∂[cl(fι∆)](x). Then from Definition 2.15, we obtain〈w, z− x〉+ cl(fι∆)(x) ≤ cl(fι∆)(z), (∀z ∈H).Set z = u + v with u ∈ dom f and v ∈∆. We have〈w,(u + v)− x〉+ cl(fι∆)(x) ≤ f(u)+ ι∆(v)〈w,u− x〉+ 〈w,v〉+ cl(fι∆)(x) ≤ f(u),because ι∆(v)= 0. Hence,〈w,v〉 ≤ f(u)− 〈w,u− x〉 − cl(fι∆)(x). (8.23)Note that x ∈ dom cl(fι∆) so cl(fι∆)(x) ∈ R. Moreover, f(u) < +∞.Finally, the fact that ∆ is subspace gives two cases:Case 1: For all n ∈ N, we have nv ∈∆. Then+∞ > f(u) ≥ cl(fι∆)(x)+ 〈w,u− x〉+ 〈w, nv〉⇒ +∞ > f(u) ≥ cl(fι∆)(x)+ 〈w,u− x〉+ n 〈w,v〉 .Taking the limit for both side when n→ +∞ gives+∞ > f(u) ≥ cl(fι∆)(x)+ 〈w,u− x〉+∞ = +∞,which is contradiction.Case 2: For all n ∈ N, we have −nv ∈∆. Then+∞ > f(u) ≥ cl(fι∆)(x)+ 〈w,u− x〉+ 〈w,−nv〉⇒ +∞ > f(u) ≥ cl(fι∆)(x)+ 〈w,u− x〉 − n 〈w,v〉 .Taking the limit for both side when n→ +∞ gives+∞ > f(u) ≥ cl(fι∆)(x)+ 〈w,u− x〉 −∞ = −∞,which is a contradiction since f is proper. Hence,〈w,v〉 = 0 ⇒ w ∈∆⊥.1308.6. Properties of f∗ + ι∗∆, fι∆, cl(fι∆), ∂(fι∆), as well as ∂[cl(fι∆)]Lemma 8.10. For every d ∈∆ and x ∈H. Then the following hold:(i)(fι∆)(x + d)=(fι∆)(x).(ii) cl(fι∆)(x + d)= cl(fι∆)(x).(iii) ∂(fι∆)(x + d)= ∂(fι∆)(x).(iv) ∂[cl(fι∆)](x + d)= ∂[cl(fι∆)](x).Proof. (i): Note that ∆ is subspace, then d −∆ = ∆ −∆ = ∆. Then itfollows from Definition 2.14 that(fι∆)(x + d)= infw∈∆{f((x + d)−w)+ ι∆(w)}= infw∈∆{f(x + d−w)} = infw∈d−∆{f(x + w)}= infw∈∆{f(x + w)}= inf−w∈∆{f(x− (−w))}= inf−w∈∆{f(x + 0− (−w))+ ι∆(−w)}=(fι∆)(x + 0)=(fι∆)(x)(from 2.14).(ii): It follows from (i) and Proposition 8.8(iii).(iii): From (2.15), we obtainw ∈ ∂(fι∆)(x + d)is equivalent to(∀y ∈H) 〈w,y− (x + d)〉+ (fι∆)(x + d) ≤ (fι∆)(y)⇔ (∀y ∈H) 〈w,y− (x + d)〉+ (fι∆)(x) ≤ (fι∆)(y) (from (i))⇔ (∀y ∈H) 〈w,y− x〉 − 〈w,d〉+ (fι∆)(x) ≤ (fι∆)(y)⇔ (∀y ∈H) 〈w,y− x〉 − 0 + (fι∆)(x) ≤ (fι∆)(y),by From Lemma 8.9. This is equivalent to w ∈ ∂(fι∆)(x) by (2.15).Hence,∂(fι∆)(x + d)= ∂(fι∆)(x).(iv): It follows from (ii) and (iii).1318.7. Properties of subdifferentials8.7 Properties of subdifferentialsIn this section we will collect some useful properties that will be usefulfor our studyLemma 8.11. Let f : H → ]−∞,+∞] be proper and convex. Let x ∈ H,then the following hold:(i) If ∂f(x) 6= ∅, then f is lower semicontinous at x.(ii) If f(x)= clf(x), f is lower semicontinous at x, then∂f(x)= ∂(clf)(x).(iii) In general, ∂f ⊆ ∂(clf).Proof. (i): y ∈ ∂f(x). Then by (2.15), we have〈y, z − x〉+ f(x) ≤ f(z), (∀z ∈ H)⇔ 〈y, z〉 − 〈y, x〉 ≤ f(z)− f(x), (∀z ∈ H)⇔ −〈y, x〉 ≤ f(z)− f(x)− 〈y, z〉 , (∀z ∈ H)⇔ 〈y, x〉 − f(x) ≥ 〈y, z〉 − f(z) (∀z ∈ H)⇔ 〈y, x〉 − f(x) ≥ supz∈H( 〈y, z〉 − f(z)).Hence,〈y, x〉 ≥ f∗(y)+ f(x). (8.24)From Young inequality, we have〈y, x〉 ≤ f∗(y)+ f(x). (8.25)Therefore, using (8.24) and (8.25) gives〈y, x〉 = f∗(y)+ f(x) ⇔ 〈y, x〉 − f∗(y) = f(x),⇒ supy∈H{〈y, x〉 − f∗(y)} = f(x)f∗∗(x)= f(x).Hence, f is lower semicontinous at x.(ii): ”k”: We have clf(x)= f(x)). From (2.15), we havey ∈ ∂(clf)(x)⇒ 〈y, z − x〉+ clf(x) ≤ clf(z) (∀z ∈ H)⇒ 〈y, z − x〉+ f(x) ≤ clf(z), (∀z ∈ H)⇒ 〈y, z − x〉+ f(x) ≤ clf(z) ≤ f(z) (∀z ∈ H).1328.7. Properties of subdifferentialsHence,〈y, z − x〉+ f(x) ≤ f(z) (∀z ∈ H)⇒ y ∈ ∂f(x).Therefore,∂(clf) ⊆ ∂f. (8.26)”j”: From (2.15), we havey ∈ ∂f(x)⇒ 〈y, z − x〉+ f(x) ≤ f(z), (∀z ∈ H).By taking the limit as z → w on both sides gives〈y, w − x〉+ f(x) ≤ clf(w),because limz→w f(z)=(clf)(w). Since f(x)= clf(x), then we obtain〈y, w − x〉+ clf(x) ≤ clf(w).This implies that, y ∈ ∂(clf)(x). Hence,∂f ⊆ ∂(clf). (8.27)Combining (8.26) and (8.27), we have∂f(x)= ∂(clf)(x).(iii): If ∂f(x) 6= ∅, then by (i) f is is lower semicontinous at x. Hence,f(x)=(clf)(x),and by (ii) we obtain ∂f(x)= ∂(clf)(x). However, if ∂f(x)= ∅, then∂f(x) ⊆ ∂(clf)(x).Lemma 8.12. Let x ∈ dom (fg), y ∈ H, and f, g ∈ Γ0(H). Then thefollowing hold:(i) If(fg)(x)= f(y)+ g(x− y), then ∂(fg)(x) = ∂f(y) ∩ ∂g(x− y).(ii) If ∂f(y) ∩ ∂g(x− y) 6= ∅, then(fg)(x)= f(y)+ g(x− y), and ∂(fg)(x) = ∂f(y) ∩ ∂g(x− y).(iii) In general, ∂f(y) ∩ ∂g(x− y) ⊆ ∂(fg)(x).1338.7. Properties of subdifferentialsProof. (i): ”⊆”: Let x∗ ∈ ∂(fg)(x). Then by (2.15), we have(fg)(x)+ 〈z − x, x∗〉 ≤ (fg)(z) (∀z ∈ H)⇒ f(y)+ g(x− y)+ 〈z − x, x∗〉 ≤ (fg)(z) (∀z ∈ H).Note that(∀u ∈ H), (∀v ∈ H) satisfy that u+ v = z, we have(fg)(z)= infu∈Hv∈Hu+v=z{f(u)+ g(v)} ≤ f(u)+ g(v).Hence,(∀u ∈ H), (∀v ∈ H) such that u+ v = z, we obtainf(y)+ g(x− y)+ 〈z − x, x∗〉 ≤ f(u)+ g(v). (8.28)Set u = y. Then z = y + v and (8.28) givesf(y)+ g(x− y)+ 〈(y + v)− x, x∗〉 ≤ f(y)+ g(v)⇒ g(x− y)+ 〈v − (x− y), x∗〉 ≤ g(v)This implies thatx∗ ∈ ∂g(x− y). (8.29)Now, set v = x− y. Then z = u+ (x− y) and (8.28) givesf(y)+ g(v)+〈(u+(x− y))− x, x∗〉 ≤ f(u)+ g(v)⇒ f(y)+ 〈u− y, x∗〉 ≤ f(u).This implies thatx∗ ∈ ∂f(y). (8.30)Combining (8.29) and (8.30), we obtainx∗ ∈ ∂f(y) ∩ ∂g(x− y).Hence,∂(fg) ⊆ ∂f(y) ∩ ∂g(x− y). (8.31)”⊇”: Let x∗ ∈ ∂f(y) ∩ ∂g(x− y). Then by (2.15) we obtain〈x∗, u− y〉+ f(y) ≤ f(u) (∀u ∈ H), (8.32)1348.7. Properties of subdifferentialsand 〈x∗, v − (x− y)〉+ g(x− y) ≤ g(v) (∀v ∈ H). (8.33)Summing (8.32) and (8.33) gives(∀u ∈ H), (v ∈ H), we havef(y)+ g(x− y)+ 〈(u+ v)− x, x∗〉 ≤ f(u)+ g(v).Hence, ∀z ∈ H satisfies that z = u+ v, we have(fg)(x)+ 〈x∗, z − x〉 ≤ f(u)+ g(v).Taking the infimum for the right hand side overall z such that z = u + v,we obtain (fg)(x)+ 〈x∗, z − x〉 ≤ (fg)(z).This implies thatx∗ ∈ ∂(fg)(x).Hence,∂g(y) ∩ ∂f(x− y) ⊆ ∂(fg)(x). (8.34)Combining (8.31) and (8.34), we obtain∂f(y) ∩ ∂g(x− y) = ∂(fg)(x).(ii):”⇒”: Let y∗ ∈ ∂f(y) ∩ ∂g(x− y). This implies thatf(y)+ 〈y∗, u− y〉 ≤ f(u) (∀u ∈ H), (8.35)andg(x− y)+ 〈y∗, v − (x− y)〉 ≤ g(v) (∀v ∈ H). (8.36)Summing (8.35) and (8.36), we obtainf(y)+ g(x− y)+ 〈y∗, (u+ v)− x〉 ≤ f(u)+ g(v) (∀u ∈ H)(∀v ∈ H).By taking the infimum for all u ∈ H, v ∈ H satisfies that x = u + v. Thenwe havef(y)+ g(x− y)+ 〈y∗, x− x〉 ≤ infu∈Hv∈Hu+v=x{f(u)+ g(v)}⇒ f(y)+ g(x− y)+ 0 ≤ (fg)(x),by (2.14). Hence,f(y)+ g(x− y) ≤ (fg)(x) (8.37)1358.8. When is fι∆ lower semicontinuous and exact?”⇐”: From (2.14) we have(fg)(x)= infy∈H{f(y)+ g(x− y)} ≤ f(y)+ g(x− y).Therefore, (fg)(x) ≤ f(y)+ g(x− y). (8.38)Combing (8.37) and (8.38) gives(fg)(x)= f(y)+ g(x− y).Then apply (i) to obtain∂f(y) ∩ ∂g(x− y) = ∂(fg)(x).(iii): If ∂f(y) ∩ ∂g(x− y) 6= ∅, then apply (ii) gives ∂(fg)(x) = ∂f(y) ∩∂g(x−y). Moreover, if ∂f(y)∩∂g(x−y) = ∅, then ∅ ⊂ ∂(fg)(x). Hence,∂f(y) ∩ ∂g(x− y) ⊆ ∂(fg)(x).8.8 When is fι∆ lower semicontinuous andexact?Proposition 8.13. Suppose that one of the following holds:(i) The conical hull of (dom f∗1 ×· · ·×dom f∗m)−∆⊥ is a closed subspace.(ii) The function d 7→ (f1(d) + · · · + fm(d)) is coercive and the canonicalhull of(dom f1 × · · · × dom fm)−∆ is a closed subspace.(iii) dom f∗1 × · · · × dom f∗m =H.(iv) H is finite-dimensional and (ri dom f∗1 × · · · × ri dom f∗m) ∩∆⊥ 6= ∅.Then fι∆ is proper, lower semicontinuous and convex, and exact on H.Proof. (i): We have(dom f∗1 × . . .×dom f∗m)−∆⊥ ⊂ cone (( dom f∗1 × . . .×dom f∗m)−∆⊥) ⊂ span (( dom f∗1 × . . .×dom f∗m)−∆⊥) ⊂ span (( dom f∗1 ×. . .× dom f∗m)−∆⊥). Then our assumption implies that(dom f∗1 × . . .× dom f∗m)−∆⊥ = span (( dom f∗1 × . . .× dom f∗m)−∆⊥).1368.8. When is fι∆ lower semicontinuous and exact?Hence, we obtainedcone((dom f∗1×. . .×dom f∗m)−∆⊥) = span (( dom f∗1×. . .×dom f∗m)−∆⊥).Therefore,0 ∈ sri (( dom f∗1 × . . .× dom f∗m)−∆⊥).Hence, from Fact 2.29, we have(f∗ + ι∆⊥)∗ = fι∆ ∈ Γ0(H).(ii): Since(dom f1×dom f2×× . . .×dom fm)−∆ is closed subspace, thenit is easy show that0 ∈ sri (( dom f1 × dom f2 × . . .× dom fm)−∆).Then Fact 2.29 gives(f + ι∆)∗ = f∗ι∗∆ = f∗ι∆⊥ ∈ Γ0(H).Next, let d =(d, d, . . . , d) ∈∆ and d ∈ H. Thenf(d)+ ι∆(d)= f(d)= f1(d) + f2(d) + . . .+ fm(d),which is coercive by assumption. Therefore, by Fact 3.6, we obtain0 ∈ int dom (f + ι∆)∗ = int dom (f∗ι∗∆) = int dom (f∗ι∆⊥)= int((dom f∗1 × dom f∗2 × . . .× dom f∗m)+ ∆⊥)= int((dom f∗1 × dom f∗2 × . . .× dom f∗m)−∆⊥)⊂ sri (( dom f∗1 × dom f∗2 × . . .× dom f∗m)−∆⊥).We have obtained0 ∈ sri ((( dom f∗1 × dom f∗2 × . . .× dom f∗m)−∆⊥).Hence, from Fact 2.29, we have(f∗ + ι∆⊥)∗ = fι∆ ∈ Γ0(H).(iii): If dom f∗1 ×· · ·×dom f∗m =H, then 0 ∈ sri(((dom f∗1 ×dom f∗2 × . . .×dom f∗m)−∆⊥) and by Fact 2.29. we obtain(f∗ + ι∆⊥)∗ = fι∆ ∈ Γ0(H).1378.8. When is fι∆ lower semicontinuous and exact?(iv): It follows from Fact 2.30 thatri((dom f∗1 × . . .× dom f∗m)−∆⊥) = sri (( dom f∗1 × . . .× dom f∗m)−∆⊥).(8.39)By using Fact 2.31, we have((ri dom f∗1×ri dom f∗2×. . .×ri dom f∗m)−∆⊥) 6=∅ ⇔ 0 ∈ ( ri dom f∗1 × ri dom f∗2 × . . . × ri dom f∗m) −∆⊥ = sri (( dom f∗1 ×dom f∗2 × . . .× dom f∗m)−∆⊥), by (8.39). Hence, we obtain0 ∈ sri ((( dom f∗1 × dom f∗2 × . . .× dom f∗m)−∆⊥).Hence, from Fact 2.29, we have(f∗ + ι∆⊥)∗ = fι∆ ∈ Γ0(H).Proposition 8.14. Let inf fi > −∞ for i = 1, . . . ,m, and suppose that atleast one of fi is coercive. Then fι∆ is proper, lower semicontinuous andconvex, and exact on H.Proof. Because f and ι∆ are proper, then there exists u and v such thatf(u)< +∞ and ι∆(v)< +∞. If x = v + u, then x− u = v and thereforeF(x,u)= f(u)+ ι∆(x− u) (8.40)= f(u)+ ι∆(v)<∞+∞ =∞. (8.41)From (2.14), we have (fι∆)(x)= inf F(x,H). (8.42)Using (8.42) and (8.41), we obtain(fι∆)(x) ≤ F(x,u) <∞.Moreover, dom(fι∆)= dom f + ∆ and we assumed inf fi > −∞ for alli = 1, 2 . . . ,m. Therefore, fι∆ is proper. Next, take a :=(x,y) ∈H×H,b :=(u,v) ∈H×H, and λ ∈ ]0, 1[. We will show that F(λa + (1− λ)b ≤λF(a)+(1 − λ)F(b) ⇔ F(λ(x,y) + (1 − λ)(u,v)) ≤ λF(x,y) + (1 −λ)F(u,v)⇔F(λx +(1− λ)u, λy + (1− λ)v) ≤ λF(x,y)+(1− λ)F(u,v). (8.43)1388.8. When is fι∆ lower semicontinuous and exact?Let z = λx + (1− λ)u, λy + (1− λ)v. Then by using the convexity of f andι∆, we obtainF(z)= F(λx + (1− λ)u, λy + (1− λ)v)≤ f(λy +(1− λ)v)+ ι∆(λx + (1− λ)u− λy− (1− λ)v)= f(λy + (1− λ)v)+ ι∆(λ(x− y)+ (1− λ)(u− v))≤ λf(y)+ (1− λ)f(v)+ λι∆(x− y)+ (1− λ)ι∆(u− v)= λ(f(y) + ι∆(x− y))+ (1− λ)(f(v)+ ι∆(u− v))= λF(x,y)+ (1− λ)F(u,v).Hence,F(λx +(1− λ)u, λy + (1− λ)v) ≤ λF(x,y)+ (1− λ)F(u,v). (8.44)Next, we will show that fι∆ is convex. Let x ∈ dom(fι∆)and u ∈dom(fι∆). Moreover, let ζ1 ∈](fι∆)(x),+∞[ and ζ2 ∈ ](fι∆)(u),+∞[.Then there exists y ∈H and v ∈H such thatF(x,y)< ζ1 and F(u,v)< ζ2. (8.45)Hence, from this, (8.42), and (8.44), we obtain(fι∆)(λx +(1− λ)u) ≤ F(λx + (1− λ)u, λy + (1− λ)v) (From (8.42))≤ λF(x,y)+ (1− λ)F(u,v) (From (8.44))< λζ1 +(1− λ)ζ2 (From (8.45)).Letting ζ1 →(fι∆)(x)and ζ2 →(fι∆)(u), we obtain(fι∆)(λx +(1− λ)u) ≤ λ(fι∆)(x)+ (1− λ)(fι∆)(u).Now, we will show that fι∆ is exact on H. Let x =(x1, x2, . . . , xm) ∈Hand d =(d, d, . . . , d) ∈∆ and d ∈ H. Setg(d):= f1(x1 − d)+ f2(x2 − d)+ · · ·+ fm(xm − d).Assume that f1 is coercive. From (2.14), we have(fι∆)(x)= infd∈∆(f(x− d)+ ι∆(d))= infd∈∆(f(x− d))= infd∈H(f1(x1 − d) + f2(x2 − d) + . . .+ fm(xm − d))= infd∈Hg(d).1398.8. When is fι∆ lower semicontinuous and exact?Since f1 is coercive, andg(d) ≥ f1(x1 − d) +m∑i=2inf fiwe have that g is coercive, so g has a minimizer overH. Hence fι∆ is exact.Next, we show that fι∆ is lower semicontinuous. Let x = (x1, . . . , xm) ∈H and xn → x with xn = (x(n)1 , . . . , x(n)m ) ∈H. We show that(fι∆)(x) ≤ lim infn→∞ (fι∆)(xn). (8.46)Set µ = lim infn→∞(fι∆)(xn). When µ = +∞, then (8.46) clearly holds.We thus assume that µ < +∞. After passing to a subsequence and rela-belling, we can assume that (fι∆)(xn) → µ ∈ [−∞,+∞[. Let (dn)n∈N bea sequence in H such that(fι∆)(xn) = f1(x(n)1 − dn) + · · ·+ fm(x(n)m − dn).We show that (dn)n∈N is bounded. Suppose this is not the case. Taking asubsequence if necessary, we can assume that ‖dn‖ → ∞. Then+∞ > µ← (fι∆)(xn) = f1(x(n)1 − dn) + · · ·+ fm(x(n)m − dn) (8.47)≥ f1(x(n)1 − dn) +m∑i=2inf fi → +∞, (8.48)because (‖xn1‖)n∈N is bounded, ‖dn‖ → ∞, and f1 is coercive. This contra-diction shows that (dn)n∈N has to be bounded. After passing to a subse-quence and relabelling, we can assume that dn ⇀ d ∈ H, i.e., dn convergesto d weakly. Then x(n)i −dn ⇀ xi−d, and using weakly lower semicontinuityof fi we haveµ = limn→∞(fι∆)(xn) = limn→∞(f1(x(n)1 − dn) + · · ·+ fm(x(n)m − dn)) (8.49)≥m∑i=1lim infn→∞ fi(x(n)i − dn) ≥m∑i=1fi(xi − d) ≥ (fι∆)(x), (8.50)which gives (8.46).Corollary 8.15. Suppose that x is coercive. Then fι∆ is proper, lowersemicontinuous and convex, and exact.1408.8. When is fι∆ lower semicontinuous and exact?Proof. Because f = f1⊕· · ·⊕fm onH, we have that f is coercive if and onlyif each fi is coercive on H; in particular, 0 ∈ int dom f∗1 × · · · × int dom f∗m,see Proposition 8.14.Without the coercivity of fi, one can exploit the structure of fi. Recallthat when H is finite-dimensional, we call f : H → ]−∞,+∞] a polyhedralconvex function if f(x) = h(x) + ιC(x) where for every x ∈ H,h(x) = max{〈x, bi〉 − βi | i = 1, . . . , k},C = {x ∈ H | 〈x, bi〉 ≤ βi, i = k + 1, . . . ,m}with fixed bi ∈ H, βi ∈ R.Proposition 8.16. Suppose that H is finite-dimensional, and each fi :H → ]−∞,+∞] is a proper polyhedral convex function for i = 1, . . . ,m.If fι∆ is proper, then fι∆ is a polyhedral convex function (so lowersemicontinuous) and fι∆ is exact.Proof. f = f1 ⊕ · · · ⊕ fm is a proper polyhedral convex function on H byFact 2.41 Also ι∆ is a proper polyhedral convex function. Because fι∆ isproper, then the result is follow from Fact 2.40.Corollary 8.17. Suppose that H is finite-dimensional, and fi = ιCi witheach Ci ⊆ H being a polyhedral convex set for i = 1, . . . ,m. Then fι∆ =ιC+∆ is lower semicontinuous and exact; equivalently, C+∆ is a closed set.Proof. From Proposition 8.5(i), we obtain fι∆ = ιC+∆ is proper, thenapply Proposition 8.16.141Chapter 9Attouch-The´ra Duality,Fixed Point Sets ofComposition of Projectors,and the Full GeometryConjecture9.1 OverviewThis chapter is based on our paper [3]. Recall from (8.1) the Hilbertproduct space,Hm = {x = (xi)i∈I | (∀i ∈ I) xi ∈ H},where m ∈ {2, 3, 4, . . . } and I = {1, 2, . . . ,m}. Let C1,C2, . . . ,Cm arenonempty closed convex subsets of H with projectors PC1 , PC2 , . . . , PCmwhich could be written more simply as P1,P2, . . . ,Pm. SetC := C1 × C2 × · · · × Cm.Recall from (8.5) the cyclic right-shift operatorR :H→H : (x1, x2, . . . , xm) 7→ (xm, x1, . . . , xm−1).Recall from Chapter 5 the fixed point sets of the cyclic compositions ofprojectorsF1 = Fix(P1 Pm · · ·P2),F2 = Fix(P2 P1 Pm · · ·P3),...Fm = Fix(Pm Pm−1 · · ·P1).We summarize our main results in this chapter as follows:1429.1. Overview• We prove that the cycles can be formulated in the Hilbert productspace as solutions to a fixed point equation (see Proposition 9.1).• We show that the set of the classical cycles that is defined by usingthe projection operators is at most singleton and possibly empty (seeProposition 9.2).• We study the set of the classical cycles and the set of classical gapvectors (see Proposition 9.3).• The relationship among the set of cycles and fixed point sets for com-positions of projectors is given in Lemma 9.4.• The enlargement of the dual problem and the general gap vectors aregiven in Section 9.4.• We prove the uniqueness of the solution to the enlarged dual problemand we show that if the enlarged dual solution exists, then it is exactlythe solution to the classical dual problem (see Theorem 9.5).• We derive a formula for the gap vector of the enlarged dual problem(see Theorem 9.5(i)).• A characterization of y, e, v, and the sets of the cycles Z is given inLemma 9.6 and Proposition 9.8.• We derive a new formula for the set of all the cycles (see Theorem 9.7).• We prove that the full geometric conjecture is true by using the ”magic”difference vectors (see Theorem 9.10).• We explore the case when m = 2, derive a new formula for the gapvector, and visit well known results (see Lemma 9.11).• We also explore the case when m = 3 (see Theorem 9.15).• We illustrate our results by various examples. See Proposition 9.13and Proposition 9.14 for m = 2 Proposition 9.17 and Proposition 9.18for m = 3.• When m = 3, we show that the order of the sets will influence theexistence of the cycles and the fixed point set of the composition ofprojectors. However, the difference vectors always exist (see Subsec-tion 9.8.2).1439.2. The fixed point sets of the compositions of projectors and the cycles• We find the difference vectors for m ≤ 5 using the Banach contractionprinciple (see Section 9.9).• We find the difference vectors by using the forward-backward algo-rithm (see Theorem 9.23).9.2 The fixed point sets of the compositions ofprojectors and the cyclesRecall the fixed point sets of the cyclic compositions of projectorsF1 = Fix(P1 Pm · · ·P2),F2 = Fix(P2 P1 Pm · · ·P3),...Fm = Fix(Pm Pm−1 · · ·P1).Let z1 ∈ F1, and set z2 := P2 z1, z3 := P3 z2, . . . , zm−1 := Pm−1 zm−2, andzm := Pm zm−1. We call the tuple z =(z1, z2, . . . , zm−1, z)as a cycle.Proposition 9.1. Let z ∈H is a cycle. Thenz = PC(Rz). (9.1)Moreover, solving (9.1) is equivalent to solve the following monotone inclu-sion problem0 ∈ NC(z)+(Id−R)(z). (9.2)Proof. Since z =(z1, z2, . . . , zm) ∈H is a cycle, then z1 = P1 zm, z2 = P2 z1,. . . , zm−1 = Pm−1 zm−2, and zm = Pm zm−1. This implies thatz =(z1, z2, . . . , zm−1, zm)=(P1 zm,P2 z1, . . . ,Pm−1 zm−2,Pm zm−1)= PC(R(z1, z2, . . . , zm−1, zm))(From (8.5))= PC(Rz).Moreover, z = PC(Rz) ⇔ z = (Id + NC)−1(Rz) by Fact 3.34(ii). Hence,Rz ∈ z +NC(z)and 0 ∈ NC(z)+(Id−R)(z).Define the set of all cycles byZ := Fix(PC R). (9.3)1449.2. The fixed point sets of the compositions of projectors and the cyclesBy fact 3.48 we view (9.2) as primal (Attouch-The´ra) problem for the pair(NC, Id−R). Since R is linear, then the dual pair:(NC, Id−R)∗=(N−1C , (Id−R)−>)=(N−1C , (Id−R)−1).Hence, by Fact 3.49 the Attouch-The´ra dual is0 ∈ N−1C(y)+ (Id−R)−1(y). (9.4)Proposition 9.2. The following hold:12Id + T + N∆⊥ is (1/2)- strongly monotone,and12Id +(N∆⊥ + T +N−1C)is strongly monotone.Moreover, the solution set of (9.4) is at most a singleton (possibly empty).Proof. From Theorem 7.7, we have(Id −R)−1 = 12Id + T + N∆⊥ . SinceR is nonexpansive, then Id−R is (1/2)-cocoercive by Fact 3.7. Moreover,(Id−R)−1 = 12Id+T+N∆⊥ is (1/2)- strongly monotone by Lemma 7.8(iv)where γ = 1 andN−1C +(Id−R)−1 = 12Id + (N∆⊥ + T +N−1C )is strongly monotone. Hence, by Fact 3.41, we havezer(N−1C +(Id−R)−1) = [N−1C + (Id−R)−1]−1(0)is at most a singleton.Proposition 9.3 (cycle and gap vector). Consider the sets of classical cyclesand classical gap vectors defined respectively byZ :={z ∈H | 0 ∈ NC(z)+(Id−R)(z)}, (9.5)K :={y ∈H | 0 ∈ N−1C(y)+(Id−R)−1(y)}. (9.6)We have(i) Z = ∪y∈K(Id−R)−1(− y) ∩N−1C (y).1459.3. The relationship among Z and Fi’s(ii) K = ∪z∈Z{Rz− z | z ∈ Z}. If K 6= ∅, thenK is a singleton, y ∈∆⊥, and y = Rz− z for every z ∈ Z.Proof. (i): Let z ∈ ∪y∈K(Id−R)−1(− y) ∩N−1C (y). Then we havez ∈ (Id−R)−1(− y) ∩N−1C (y) (∀y ∈ K)⇔ z ∈ (Id−R)−1(− y) and z ∈ N−1C (y)⇔ −y = (Id−R)(z) and y ∈ NC(z)⇔ 0 ∈ (NC + (Id−R))(z)⇔ z ∈ (NC + (Id−R))−1(0)⇔ z ∈ Z.(ii): Let y ∈ K. Then (9.6) gives0 ∈ N−1C(y)+(Id−R)−1(y), (∀z ∈ Z)⇔ z ∈ N−1C(y)and − z ∈ (Id−R)−1(y) (∀z ∈ Z)⇔ y ∈ NC(z)and y = Rz− z (∀z ∈ Z),hence,y = Rz− z (∀z ∈ Z),andK = ∪z∈Z{Rz− z | z ∈ Z}.If K 6= ∅, then K is singleton by Proposition 9.2. Moreover, dom (Id −R)−1 = ∆⊥, which implies that y ∈∆⊥.9.3 The relationship among the set of cycles andfixed point sets for compositions of projectorsSome basic properties and relationship among the Fi’s, and Z are givenbelow. DefineQi :H→ H : z 7→ zi.Lemma 9.4. For every 1 ≤ i, j ≤ m, the following hold:1469.3. The relationship among Z and Fi’s(i) Fi are closed and convex. Moreover,Fm =(Pm Pm−1 . . .P3 P2)(F1)= · · · = Pm Pm−1(Fm−2)= Pm(Fm−1).Fm−1 =(Pm−1 . . .P3 P2 P1)(Fm)= · · · = Pm−1(Fm−2)....F2 =(P2 P1 Pm . . .P4)(F3)= · · · = P2 P1(Fm)= P2(F1).F1 =(P1 Pm . . .P4 P3)(F2)= · · · = (P1 Pm )(Fm−1) = P1 (Fm).(ii) ∩mi=1 Fix Pi ⊆ ∩mi=1Fi and if Fi = ∅, then ∩mi=1 Fix Pi = ∅.(iii) For 1 ≤ i ≤ m− 1, Pi+1(Fi)=(Fi+1), and P1(Fm)= F1. Therefore,PCR(F1 × F2 × · · · × Fm)= F1 × F2 × · · · × Fm. (9.7)(iv) Fi 6= ∅ if and only if Fj 6= ∅ if and only if Z = ∅.(v) Z is closed and convex, and Z ⊆ F1 × F2 × · · · × Fm.(vi) The mapping Qi|Z : Z→ Fi is bijective and Qi(Z)= Fi.Proof. (i): For each 1 ≤ i ≤ m the projector operators Pi is firmly nonex-pansive by Example 3.2 and therefore it is nonexpansive. Hence, by Fact 3.11the mappingU := Pi · · ·P1 Pm · · ·Pi+1,is nonexpansive and Fix(U)is closed and convex by Fact 3.9. Furthermore,Let x ∈ Fm ⇔ x ∈ Fix(Pm Pm−1 · · ·P3 P2 P1)and hencex = Pm Pm−1 · · ·P3 P2 P1 x.ThenP1 x = P1(Pm Pm−1 . . .P3 P2 P1)x =(P1 Pm Pm−1 . . .P3 P2)(P1 x)and hence, P1 x ∈ Fix(P1 Pm Pm−1 . . .P3 P2)⇔ P1 x ∈ F1. It follows thatP1(Fm) ⊆ F1 (9.8)Moreover,(P2 P1)(Fm)=(P2 P1)(Fix(Pm Pm−1 . . .P2 P1))and(P2 P1)(Fix(Pm Pm−1 . . .P2 P1)) ⊆ P2 (Fix (P1 Pm Pm−1 . . .P3 P2 ))⊆ Fix (P2 P1 Pm Pm−1 . . .P4 P3 ) = F2,1479.3. The relationship among Z and Fi’sand(P3 P2 P1)(Fm)=(P3 P2 P1)(Fix(Pm Pm−1 . . .P2 P1)). Hence,(P3 P2 P1)(Fix(Pm Pm−1 . . .P2 P1)) ⊆ (P3 P2 )(Fix (P1 Pm Pm−1 . . .P3 P2 ))⊆ P3(Fix(P2 P1 Pm Pm−1 . . .P4 P3))⊆ Fix (P3 P2 P1 Pm Pm−1 . . .P5 P4 ) = F3,until finally Fm = Fix(Pm Pm−1 . . .P2 P1)andFm =(Pm Pm−1 . . .P2 P1)(Fix(Pm Pm−1 . . .P2 P1))(9.9)⊆ (Pm Pm−1 . . .P2 )(Fix (P1 Pm . . .P3 P2 )) (9.10)... (9.11)⊆ Pm(Fix(Pm−1 Pm−2 . . .P2 P1 Pm))= Pm(Fm−1)(9.12)⊆ Fix (Pm Pm−1 . . .P2 P1 ) = Fm. (9.13)Hence, equality holds throughout (9.9)-(9.13) and we are done. The sameapproach will verify the rest.(ii): Clearly, ∩mi=1 Fix Pi ⊆ F1, ∩mi=1 Fix Pi ⊆ F2, . . . , ∩mi=1 Fix Pi ⊆ Fm.Hencem⋂i=1Fix Pi ⊆m⋂i=1Fi.This also implies that ∩mi=1 Fix Pi = ∅ if Fi = ∅.(iii): From (i), we haveP1(Fm)= F1, P2(F1)= F2, . . . , Pm(Fm−1)= Fm. (9.14)Using (8.5), (8.6), and (9.14), we obtainPCR(F1 × F2 × · · · × Fm−1 × Fm)= F1 × F2 × · · · × Fm−1 × Fm.(iv): This is obvious from the definition of Fi, Fj and from (9.3).(v): Since R is nonexpansive and PC is firmly nonexpansive by Example 3.2,then PCR is nonexpansive by Fact 3.11 and Z = Fix(PCR)is closed andconvex by Fact 3.9. Next, let z =(z1, z2, . . . , zm) ∈ Fix (PCR) ⇔ z =PCRz. This impliesz1 = P1 Pm . . .P2 z1,...zi = Pi Pi−1 . . .P1 Pm . . .Pi+1 zi,...zm = Pm Pm−1 . . .P1 zm.1489.4. Enlarged dual problem and the general gap vectorHence, z =(z1, z2, . . . , zm) ∈ F1 × F2 × · · · × Fm−1 × Fm. Since this is truefor all z ∈ Z, therefore,Z ⊆ F1 × F2 × · · · × Fm−1 × Fm.(vi): It is clear from (i) that Qi : Z → Fi is surjective. To show Qi isinjective, suppose z =(z1, z2, . . . , zm), z˜ =(z˜1, z˜2, . . . , z˜m) ∈ Z and Qi(z) =Qi(z˜)⇒ zi = z˜i. Because z and z˜ are cycles then we obtainzi+1 = Pi+1 zi = Pi+1 z˜i = z˜i+1 (9.15)... (9.16)zm = Pm zm−1 = Pm z˜m−1 = z˜m (9.17)z1 = P1 zm = P1 z˜m = z˜1 (9.18)z2 = P2 z1 = P2 z˜1 = z˜2 (9.19)... (9.20)zi−1 = Pi−1 zi−2 = Pi−1 z˜i−2 = z˜i−1. (9.21)Hence from (9.15)-(9.21), we havez = z˜.9.4 Enlarged dual problem and the general gapvectorTo deal with the empty-set possibility, we will enlarge (9.4) so it has asolution as follows:N−1C +(Id−R)−1 = N−1C + 12Id + T +N∆⊥ (Theorem 7.7)= 12Id + T +N−1C +N−1∆= 12Id + T + ∂σC + ∂σ∆ ( Fact 2.35 & Example 2.37)⊆ 12Id + T + ∂σC+∆. (Proposition 2.38(ii))Hence, instead of working with (9.4), which has no solution if there are nocycles, we will work with the enlarged dual problem:0 ∈ 12y + Ty + ∂σC+∆(y). (9.22)1499.5. A characterization of y, e and the sets of the cycles ZTheorem 9.5. The following hold:(i) The enlarged dual problem (9.22) always has a unique solution y sat-isfies thaty = J2T+2∂σC+∆(0)(ii) If (9.4) has a solution, then it is exactly the solution of (9.22).Proof. (i): σC+∆ is proper, convex and lower-semicontinous by Lemma 8.3(i)and (ii). Hence, ∂σC+∆ is maximally monotone by Example 3.25. T is fulldomain and maximally monotone by Theorem 7.6(iv). Then(∂σC+∆ + T)is maximally monotone by Fact 3.44. Therefore,ran(Id +(∂σC+∆ + T))=H.Moreover, from (9.22) we have0 ∈ y + 2Ty + 2∂σC+∆(y)⇔ 0 ∈ (Id + (2T + 2∂σC+∆))(y)hence,y =(Id +(2T + 2∂σC+∆))−1(0)⇔ y = J2T+2∂σC+∆(0).(ii): This is clear sinceN−1C +(Id−R)−1 = 12Id + T + ∂σC + ∂σ∆⊆ 12Id + T + ∂σC+∆.9.5 A characterization of y, e and the sets of thecycles ZLemma 9.6 (a characterization of y and e). Let y be the unique solutionto (9.22) and sete := −12y−Ty. (9.23)Then the following hold:(i) y ∈∆⊥.1509.5. A characterization of y, e and the sets of the cycles Z(ii) e ∈∆⊥.(iii) y ∈ ∂ιC+∆(e)= NC+∆(e). Consequently, e ∈ C + ∆.(iv) y = Re− e.(v)(∀c ∈ C) 〈c,y〉 ≤ −12 ‖y‖2.Proof. (i): From (9.22), we havey ∈ dom ∂σC+∆ ⊆ dom σC+∆ = dom(σC + σ∆)= dom σC ∩ dom σ∆⊆ (recC) ∩ (rec∆) = (recC) ∩∆⊥.Therefore, y ∈∆⊥.(ii): Combining (i) and Theorem 7.6(v) gives e ∈∆⊥.(iii): From (9.22) and (9.23), we obtain −12y − Ty ∈ ∂σC+∆(y) ⇔ e ∈∂σC+∆(y) ⇔ e ∈ ∂ι∗C+∆(y) ⇔ e ∈ (∂ιC+∆)−1(y) ⇔ y ∈ ∂ιC+∆(e).Hence,e ∈ dom ∂ιC+∆ = dom ιC+∆ = C + ∆.(iv): From (9.23), we have y ∈ (12Id + T)−1( − e) and by Lemma 7.13(vi)and (ii), we havey =(Id−R + 2 P∆)(− e) = (Id−R)(− e) = Re− e.(v): By (iii), we have((∀(c,d)) ∈ C×∆) 〈y, c + d− e〉 = 〈y, c + d + 12y + Ty〉≤ 0= 〈y, c〉+ 〈y,d〉+ 12 ‖y‖2 + 〈y,Ty〉= 〈y, c〉+ 0 + 12 ‖y‖2 + 0.By Theorem 7.6(iii) & (v), we obtain〈y, c〉+ 12 ‖y‖2 ≤ 0.Hence, (∀c ∈ C) 〈y, c〉 ≤ −12 ‖y‖2 .1519.5. A characterization of y, e and the sets of the cycles ZTheorem 9.7. Let y and e be as in (9.22) and (9.23) respectively. Thenthe set of cycles is given byZ = N−1C(y) ∩ (e + ∆) (9.24)= e +(∆ ∩ (C− e)). (9.25)Proof. From Theorem 9.5(ii), we haveZ = N−1C(y) ∩ (Id−R)−1(− y).Let x ∈ Z ⇔ x ∈ N−1C(y) ∩ (Id − R)−1( − y) ⇔ x ∈ N−1C (y) ∩ −(Id −R)−1(y) ⇔ x ∈ N−1C (y) and x ∈ −(Id − R)−1(y) ⇔ x ∈ N−1C (y) andx ∈ (− 12y−Ty−N∆⊥(y)) by Theorem 7.7. This equivalent to x ∈ N−1C (y)and x ∈ (e + ∆) by (9.23) and Lemma 9.6(i). Hence,x ∈ N−1C(y) ∩ (e + ∆). Hence, Z = N−1C (y) ∩ (e + ∆).Next, let x ∈H and d := x− e. Then from (9.24) we obtain x ∈ Z⇔ x ∈N−1C (y) ∩(e + ∆) ⇔ x ∈ N−1C (y) and x ∈ (e + ∆) ⇔ x ∈ N−1C (y) andx = e+d, where d ∈∆⇔ y ∈ NC(x) and d = x−e ∈∆⇔ y ∈ NC(e+d)and d = x − e ∈ ∆⇔ d ∈ ∆, e + d ∈ C, and (∀c ∈ C) 〈y, c− (e + d)〉 ≤0 ⇔ d ∈ ∆,d ∈ C − e, and (∀c ∈ C) 〈y, c− e〉 ≤ 0 ⇔ d ∈ ∆ ∩ (C − e),and(∀c ∈ C) 〈y, c〉 ≤ 〈y, e〉 ⇔ d ∈ ∆ ∩ (C − e), and σC(y) ≤ 〈y, e〉. ByLemma 9.6(v), we obtainx ∈∆ ∩ (C− e) + e.Hence,Z = ∆ ∩ (C− e) + e.Proposition 9.8. Setv := R∗e− e, (9.26)where R∗ is the circular left shift operator. Then the following hold:(i) v ∈∆⊥.(ii) v = −R∗y and y = −Rv.(iii) P∆⊥(Z) ⊆ {e} and P∆ (Z) = ∆ ∩ (C− e).(iv) Z 6= ∅⇔ e ∈ C + ∆.1529.5. A characterization of y, e and the sets of the cycles Z(v) If e ∈ C + ∆, say e = c + d, where d ∈∆ and c ∈ C, then c ∈ Z.(vi) If z ∈ Z, then e = P∆⊥ z ∈(C + ∆) ∩∆⊥.Proof. (i): From Theorem 7.1(v), we obtainran M∗ = ran(Id−R)∗ = ran (Id−R∗) = ∆⊥.Hence, v ∈∆⊥.(ii): v = R∗e− e = −R∗(Re− e) = −R∗y by Lemma (9.6)(iv). Moreover,y = Re− e = −R(R∗e− e) = −Rv by (9.26).(iii): From (9.25) we obtain Z = e +(∆ ∩ (C− e)). Note that e ∈ ∆⊥ byLemma 9.6(ii) and ∆ ∩ (C− e) ⊆∆. ThenP∆(Z)= ∆ ∩ (C− e),andP∆⊥(Z)= P∆⊥(e +(∆ ∩ (C− e))) ⊆ {e}.(iv): From (9.25) we obtain Z = e +(∆ ∩ (C − e)). Then Z 6= ∅ ⇔e +(∆ ∩ (C − e)) 6= ∅ ⇔ ∆ ∩ (C − e) 6= ∅ ⇔ (∃c ∈ C) such thatc− e ∈∆⇔ e ∈ C + ∆.(v): Let e = c + d, where c ∈ C and d ∈∆. Note that ∆ is a subspace so−d ∈∆ and ∆ ∩ (C− e) ⊆∆. Hence−d = c− e ∈∆⇔ c− e ∈∆ ∩ (C− e)⇔ c ∈ e + (∆ ∩ (C− e)).Therefore, c ∈ Z by (9.25).(vi): From (iii), we obtain P∆⊥(Z) ⊆ {e} and from Lemma 9.6(ii), we havee ∈∆⊥. Hence,e = P∆⊥ z= z− P∆ z∈ (Z−∆) ∩∆⊥ (From Lemma 9.6(ii))⊆ (C + ∆) ∩∆⊥. (Since ∆ is subspace, then ∆ = −∆)Remark 9.9. Let e =(e1, e2, . . . , em)and y =(y1, y2, . . . , ym). Let R∗ isthe circular left shift operator. Then (9.26) and Lemma 9.6(iv) yield tov =(e2 − e1, e3 − e2, . . . , em − em−1, e1 − em). (9.27)1539.6. The full geometry conjecture is trueandy =(em − e1, e1 − e2, . . . , em−2 − em−1, em−1 − em). (9.28)Then (9.27) and (9.28) givev =(− y2,−y3, . . . ,−ym,−y1). (9.29)andy =(− vm,−v1, . . . ,−vm−2,−vm−1). (9.30)9.6 The full geometry conjecture is trueTheorem 9.10 (The geometry conjecture is true). Let v =(v1, v2, . . . ,vm−1, vm)be the vector defined in (9.26). ThenFm = Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (C1 + v1 + v2 + · · ·+ vm−1). (9.31)and analogously for Fm−1, Fm−2, . . . , F2, F1.Proof. ”j”: Let e =(e1, e2, . . . , em−1, em)be the vector defined in (9.23)and Z = e +(∆ ∩ (C − e)) from (9.25). Fm 6= ∅ ⇔ Z 6= ∅, fromLemma 9.4(iv). Let zm ∈ Fm. Then there exists x =(x, x, . . . , x, x) ∈ ∆such that z = e + x. Then(z1, z2, . . . , zm−1, zm)=(e1, e2, . . . , em−1, em)+(x, x, . . . , x, x).Hence,z1 = e1 + x, z2 = e2 + x, . . . , zm−1 = em−1 + x, zm = em + x. (9.32)Since zm is the m-th component of a cycle z, thenzm ∈ Cm. (9.33)From (9.27) and (9.32) we have the following:zm = em + x = (em− em−1) + (em−1 + x) = vm−1 + zm−1 ∈ vm−1 + Cm−1. (9.34)Also,zm =(x+ em−2)+(em−1− em−2)+(em− em−1) ∈ Cm−2 + vm−2 + vm−1. (9.35)Keep doing that we getzm =(x+ e1)+(e2− e1)+ · · ·+ (em− em−1) ∈ C1 + v1 + v2 + · · ·+ vm−1. (9.36)1549.6. The full geometry conjecture is trueCombing these equations together, we obtainzm ∈ Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (C1 + v1 + v2 + · · ·+ vm−1).Hence,Fm ⊆ Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (C1 + v1 + v2 + · · ·+ vm−1). (9.37)”k”: Letcm ∈ Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (C1 + v1 + v2 + · · ·+ vm−1). (9.38)Hence, there exists c1 ∈ C1, c2 ∈ C2, . . . , and cm−1 ∈ Cm−1 such thatcm ∈ Cm (9.39)cm = cm−1 + vm−1 ∈ Cm−1 + vm−1 (9.40)cm = cm−2 + vm−1 + vm−2 ∈ Cm−2 + vm−1 + vm−2 (9.41)cm = cm−3 + vm−1 + vm−2 + vm−3 ∈ Cm−3 + vm−1 + vm−2 + vm−3 (9.42)... (9.43)cm = c2 + v2 + v3 + · · ·+ vm−1 ∈ C2 + v2 + v3 + · · ·+ vm−1 (9.44)cm = c1 + v1 + v2 + · · ·+ vm−1 ∈ C1 + v1 + v2 + · · ·+ vm−1 (9.45)solving the above system for cm−1, cm−2, . . . , c2, c1, we obtain0 = cm−1 − cm−2 − vm−2 ⇔ cm−1 = cm−2 + vm−2,0 = cm−2 − cm−3 − vm−3 ⇔ cm−2 = cm−3 + vm−3,...0 = c2 − v1 ⇔ c2 = c1 + v10 = c1 + v1 + v2 + · · ·+ vm−1 − cm ⇔ c1 = cm + vm.Thus, we obtainc =(c1, c2, . . . , cm−1, cm) ∈ C=(cm + vm, c1 + v1, . . . , cm−2 + vm−2, cm−1 + vm−1)= Rc + Rv= Rc + R(R∗e− e) = Rc + e−Re,by (9.26). Therefore, we havec−Rc− e + Re = 0 ⇔ (Id−R)(c− e) = 0⇔M(c− e) = 0,1559.7. Exploring the m = 2 case and visiting known resultsand from Theorem 7.1(ii), we obtain ker M = ∆. Hence,(c − e) ∈ ∆ and(c−e) ∈ C−e. Therefore, (c−e) ∈∆∩(C−e) ⇔ c ∈ e+(∆∩(C−e)) = Zby (9.25). Finally, from Lemma 9.4(v) and (9.38) we have(c1, c2, . . . , cm−1, cm) ∈ Z ⊆ F1 × F2 × · · · × Fm−1 × Fm ⇔ cm ∈ Fm.Hence,Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (C1 + v1 + v2 + · · ·+ vm−1) ⊆ Fm. (9.46)Combining (9.37) and (9.46) givesFm = Cm ∩(Cm−1 + vm−1) ∩ · · · ∩ (C1 + v1 + v2 + · · ·+ vm−1).The same approach can be used to prove Fm−1, Fm−2, . . . , F2, F1.9.7 Exploring the m = 2 case and visiting knownresultsLemma 9.11. Let m = 2 and y is the unique solution of (9.22). Let e andv be as in (9.23) and (9.26) respectively. Then the following hold:(i) y = P2(∆−C)(0).(ii) e = P(∆+C)(0).(iii) y =(PC2−C1(0),PC1−C2(0)).(iv) e = 12(PC1−C2(0),PC2−C1(0)).(v) v =(PC2−C1(0),PC1−C2(0)).(vi) v = P2(∆−C)(0).Proof. Using (7.47) and (8.5), we haveT = 14(2− 2)R = 0 and R = R∗.(i): From (9.22) we obtain 0 ∈ 12y+Ty+∂σC+∆(y)⇔ 0 ∈ 12y+∂σC+∆(y)⇔−12y ∈ ∂σC+∆(y)⇔ −12y ∈ ∂ι∗C+∆(y) by Fact 2.35 ⇔ −12y ∈ (∂ιC+∆)−1(y)by Fact 2.27 ⇔ y ∈ ∂ιC+∆( − 12y) ⇔ y ∈ NC+∆( − 12y) by Fact 2.361569.7. Exploring the m = 2 case and visiting known results⇔ y − 12y ∈(Id + NC+∆)( − 12y) ⇔ −12y ∈ (Id + NC+∆)−1(12y). Hence,by Fact 3.34(ii), we obtain−12y = PC+∆(12y). Then − 12y ∈ C + ∆.Therefore, we have〈(c + d)− (− 12y), 12y− (− 12y)〉≤ 0(∀(c,d) ∈ C×∆).Hence,〈2(d− c)− y, 0− y〉 ≤ 0 (∀(c,d) ∈ C×∆). Therefore,y = P2(∆−C)(0). (9.47)(ii): By using (9.23) and (i), we obtain e = −12y −Ty ⇔ e = −12y ⇔ e =−12 P2(∆−C)(0) ⇔ e = P−(∆−C) (0) by using the dilation formula of theprojection and because −∆ = ∆, we havee = P(∆+C)(0). (9.48)(iii): Sety = PC2−C1(0), (9.49)and let(c1,n, c2,n)n∈N be a sequence in C = C1×C2 such that c1,n−c2,n → y,where(c1,n)n∈N and(c2,n)n∈N be a sequences in C1 and C2 respectively.Then for every n ∈ N, we obtainc1,n − c2,n = 2(12(c1,n + c2,n)− c2,n),andc2,n − c1,n = 2(12(c1,n + c2,n)− c1,n).Hence, (c2,n − c1,n, c1,n − c2,n) ∈ 2(∆− C1 × C2) = 2(∆−C). (9.50)We claim thaty =(PC2−C1(0),PC1−C2(0))=(PC2−C1(0),−PC2−C1(0))=(y,−y).1579.7. Exploring the m = 2 case and visiting known results(y,−y) ∈ 2(∆ − C) by (9.50). Moreover, let c = (c1, c2) ∈ C1 × C2 andd =(x, x) ∈∆, where x ∈ H. Then〈2(d− c)− (y,−y), (0, 0)− (y,−y)〉 = 〈2(x− c1, x− c2)− (y,−y), (− y, y)〉=〈2(x− c1)− y, 2(x− c2)+ y, (− y, y)〉= 〈2x− 2c1 − y,−y〉+ 〈2x− 2c2 + y, y〉= 〈2c2 − 2c1 − 2y,−y〉= 2〈(c2 − c1)− y, 0− y〉 ≤ 0,from (9.49). Therefore, by Fact 2.13 we obtainy =(PC2−C1(0),PC1−C2(0)).(iv): From (9.23), we havee = −12y = −12(PC2−C1(0),PC1−C2(0))= 12(PC1−C2(0),PC2−C1(0)).(v): By (iii) and Proposition 9.8(ii) we obtain v = −R∗y = −Ry =−R(PC2−C1 (0),PC1−C2 (0)) = −(PC1−C2 (0),PC2−C1 (0)). Therefore,v =(PC2−C1(0),PC1−C2(0)).(vi): Combining (i), (iii) and (v) givesv = P2(∆−C)(0)= 2 P(∆−C)(0).Lemma 9.12 (two lines). Let c1, c2, u1, and u2 ∈ H. Let C1 = c1 + Ru1and C2 = c2 + Ru2 be two lines in H, where ‖u1‖ = ‖u2‖ = 1, c1 ⊥ u1 andc2 ⊥ u2. Then for all i = 1, 2 and for all x ∈ HPi(x)= ci + 〈ui, x〉ui. (9.51)Let z =(z1, z2) ∈ C1×C2, where z1 = c1+ζ1u1 ∈ C1 and z2 = c2+ζ2u2 ∈ C2.Assume that z is a cycle. Thenζ1(1− 〈u1, u2〉2)= 〈u1, c2〉+ 〈u1, u2〉 〈u2, c1〉 , (9.52)andζ2(1− 〈u1, u2〉2)= 〈u2, c1〉+ 〈u1, u2〉 〈u1, c2〉 . (9.53)1589.7. Exploring the m = 2 case and visiting known resultsMoreover, if 〈u1, u2〉 6= 1 or 〈u1, u2〉 6= −1, thenζ1 =〈u1, c2〉+ 〈u1, u2〉 〈u2, c1〉1− 〈u1, u2〉2, (9.54)andζ2 =〈u2, c1〉+ 〈u1, u2〉 〈u1, c2〉1− 〈u1, u2〉2. (9.55)Proof. Let z be a cycle. Thenz2 = P2 z1 and z1 = P1 z2, (9.56)by Definition 5.1. By using (9.51) and (9.56), we havec1 + ζ1u1 = z1= P1 z2= P1 P2 z1= c1 + 〈u1,P2 z1〉u1= c1 + 〈u1, c2 + 〈u2, z1〉u2〉u1= c1 +( 〈u1, c2〉+ 〈u1, u2〉 〈u2, z1〉 )u1= c1 +( 〈u1, c2〉+ 〈u1, u2〉 〈u2, c1 + ζ1u1〉 )u1= c1 +( 〈u1, c2〉+ 〈u1, u2〉 〈u2, c1〉+ ζ1 〈u1, u2〉 〈u2, u1〉 )u1.Hence, (1− 〈u1, u2〉2)ζ1 = 〈u1, c2〉+ 〈u1, u2〉 〈u2, c1〉 .if 〈u1, u2〉 6= 1 or 〈u1, u2〉 6= −1, then we could solve for ζ1:ζ1 =〈u1, c2〉+ 〈u1, u2〉 〈u2, c1〉1− 〈u1, u2〉2.Doing the same approach for z2 = P2 P1 z2 givesζ2 =〈u2, c1〉+ 〈u1, u2〉 〈u1, c2〉1− 〈u1, u2〉2.Proposition 9.13. Recall that C1 = c1 + Ru1 and C2 = c2 + Ru2 be twoparallel lines. Recall from (9.52) and (9.53) :ζ1(1− 〈u1, u2〉2)= 〈u1, c2〉+ 〈u1, u2〉 〈u2, c1〉 ,andζ2(1− 〈u1, u2〉2)= 〈u2, c1〉+ 〈u1, u2〉 〈u1, c2〉 ,respectively. Then the following hold:1599.7. Exploring the m = 2 case and visiting known results(i) Z =(c1, c2)+ R(u, u).(ii) v =(c2 − c1, c1 − c2).(iii) y =(c2 − c1, c1 − c2)= v.(iv) e = 12(c1 − c2, c2 − c1).Proof. (i): Since the lines are parallel, then 〈u1, u2〉2 = 1. Assume thatu := u1 = u2. Then every ζ1, ζ2 ∈ R solves (9.52) and (9.53). Hence,Z =(c1, c2)+ R(u, u).(ii): By Using Definition 5.1, we have c1 = P1 c2 and c2 = P2 c1. Therefore,v =(c2 − c1, c1 − c2)by (5.5). (iii): From Lemma 9.11(iii)&(v) and (ii), wehave y =(c2 − c1, c1 − c2)= v. (iv): From (9.23), we havee = −12(c2 − c1, c1 − c2)−T(c2 − c1, c1 − c2) = 12(c1 − c2, c2 − c1).Proposition 9.14. Recall that C1 = c1 + Ru1 and C2 = c2 + Ru2 be twonot parallel lines. Recall from (9.52) and (9.53):ζ1(1− 〈u1, u2〉2)= 〈u1, c2〉+ 〈u1, u2〉 〈u2, c1〉 ,andζ2(1− 〈u1, u2〉2)= 〈u2, c1〉+ 〈u1, u2〉 〈u1, c2〉 ,respectively. Then the following hold:(i) The set of cycles Z has only one element, which isZ =(c1 + ζ1u1, c2 + ζ2u2).(ii) v =((c2 − c1) + (ζ2u2 − ζ1u1), (c1 − c2) + (ζ1u1 − ζ2u2)).(iii) y = v.(iv) e = 12((c1 − c2) + (ζ1u1 − ζ2u2), (c2 − c1) + (ζ2u2 − ζ1u1)).1609.8. Exploring the m = 3 caseProof. (i): Since C1 and C2 are not parallel lines, then 〈u1, u2〉2 < 1. By(9.54) and (9.55) the set of cycles Z has only one elementz =(z1, z2)=(c1 + ζ1u1, c2 + ζ2u2).(ii): Using Definition 5.1 gives that z1 = P1 z2, z2 = P2 z1, andv =(z2 − z1, z1 − z2)=((c2 − c1) + (ζ2u2 − ζ1u1), (c1 − c2) + (ζ1u1 − ζ2u2)).(iii): From Lemma 9.11(iii)&(v) and (ii), we have y =((c2 − c1) + (ζ2u2 −ζ1u1), (c1 − c2) + (ζ1u1 − ζ2u2))= v. (iv): From (9.23) and (iii), we obtaine = −12(z2 − z1, z1 − z2)= −12((c2 − c1) + (ζ2u2 − ζ1u1), (c1 − c2) + (ζ1u1 − ζ2u2))= 12((c1 − c2) + (ζ1u1 − ζ2u2), (c2 − c1) + (ζ2u2 − ζ1u1)).9.8 Exploring the m = 3 caseLet m = 3. Then (7.47) givesT = 160 0 11 0 00 1 0− 160 1 00 0 11 0 0 = 16 0 −1 11 0 −1−1 1 0 . (9.57)Theorem 9.15. Let y =(y1, y2, y3)satisfy that y1 + y2 + y3 = 0. Assumethat there exists sequences(c1,n)n∈N ∈ C1,(c2,n)n∈N ∈ C2,(c3,n)n∈N ∈ C3,and xn ∈ H such thatc1,n + xn → 16(− 3y1 + y2 − y3), (9.58)c2,n + xn → 16(− y1 − 3y2 + y3), (9.59)c3,n + xn → 16(y1 − y2 − 3y3). (9.60)Moreover, for all(c1, c2, c3) ∈ C1 × C2 × C3, we have〈y1, c1〉+ 〈y2, c2〉+ 〈y3, c3〉 ≤ −12( ‖y1‖2 + ‖y2‖2 + ‖y3‖2 ). (9.61)Then the following hold:1619.8. Exploring the m = 3 case(i) e = 16(− 3y1 + y2 − y3,−y1 − 3y2 + y3, y1 − y2 − 3y3).(ii) v = 13(y1 − 2y2 + y3, y1 + y2 − 2y3,−2y1 + y2 + y3).Proof. (i): We have y =(y1, y2, y3) ∈ H3 that satisfies all the given condi-tions. Then by (9.23) and (9.57) we havee1e2e3 = −12y1 0 00 y2 10 0 y3− 16 0 −y2 y3y1 0 −y3−y1 y2 0 = 16−3y1 y2 −y3−y1 −3y2 y3y1 −y2 −3y3 .Hence,e = 16(− 3y1 + y2 − y3,−y1 − 3y2 + y3, y1 − y2 − 3y3).(ii): From (i) and (9.26), we havev = R∗e− e= 16(2y1 − 4y2 + 2y3, 2y1 + 2y2 − 4y3,−4y1 + 2y2 + 2y3)= 13(y1 − 2y2 + y3, y1 + y2 − 2y3,−2y1 + y2 + y3).9.8.1 Three linesLemma 9.16 (Three lines). Let c1, c2, c3, u1, u2, u3 ∈ H. Let C1 = c1+Ru1,C2 = c2 + Ru2 and C3 = c3 + Ru3 be 3 lines of H, where ‖u1‖ = ‖u2‖ =‖u3‖ = 1, c1 ⊥ u1, c2 ⊥ u2 and c3 ⊥ u3. Let z =(z1, z2, z3) ∈ C1 × C2 × C3be a cycle, where z1 = c1 + ζ1u1, z2 = c2 + ζ2u2 and z3 = c3 + ζ3u3. Thenζ1(1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉)= 〈u1, c3〉+ 〈u1, u3〉 〈u3, c2〉+ 〈u1, u3〉 〈u3, u2〉 〈u2, c1〉 , (9.62)ζ2(1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉)= 〈u2, c1〉+ 〈u2, u1〉 〈u1, c3〉+ 〈u2, u1〉 〈u1, u3〉 〈u3, c2〉 , (9.63)andζ3(1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉)= 〈u3, c2〉+ 〈u3, u2〉 〈u2, c1〉+ 〈u3, u2〉 〈u2, u1〉 〈u1, c3〉 . (9.64)1629.8. Exploring the m = 3 caseMoreover, if 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉 6= 1, thenζ1 =〈u1, c3〉+ 〈u1, u3〉 〈u3, c2〉+ 〈u1, u3〉 〈u3, u2〉 〈u2, c1〉1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉 , (9.65)ζ2 =〈u2, c1〉+ 〈u2, u1〉 〈u1, c3〉+ 〈u2, u1〉 〈u1, u3〉 〈u3, c2〉1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉 , (9.66)andζ3 =〈u3, c2〉+ 〈u3, u2〉 〈u2, c1〉+ 〈u3, u2〉 〈u2, u1〉 〈u1, c3〉1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉 . (9.67)Proof. Let z =(z1, z2, z3) ∈ C1 ×C2 ×C3 be a cycle. By Definition 5.1, wehave z1 = P1 z3, z2 = P2 z1 and z3 = P3 z2. Thenc1 + ζ1u1 = z1= P1 z3= c1 + 〈u1, z3〉u1= c1 + 〈u1,P3 z2〉u1= c1 + 〈u1, c3 + 〈u3, z2〉u3〉u1= c1 + 〈u1, c3〉u1 + 〈u3, z2〉 〈u1, u3〉u1= c1 + 〈u1, c3〉u1 + 〈u3,P2 z1〉 〈u1, u3〉u1= c1 + 〈u1, c3〉u1 + 〈u3, c2 + 〈u2, z1〉u2〉 〈u1, u3〉u1= c1 + 〈u1, c3〉u1 + 〈u3, c2 + 〈u2, c1 + ζ1u1〉u2〉 〈u1, u3〉u1.Hence,ζ1(1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉)= 〈u1, c3〉+ 〈u1, u3〉 〈u3, c2〉+ 〈u1, u3〉 〈u3, u2〉 〈u2, c1〉 . (9.68)If 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉 6= 1, thenζ1 =〈u1, c3〉+ 〈u1, u3〉 〈u3, c2〉+ 〈u1, u3〉 〈u3, u2〉 〈u2, c1〉1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉 .The same approach can be used to prove the value of ζ2 and ζ3.Proposition 9.17. Let C1 = c1 + Ru1, C2 = c2 + Ru2 and C3 = c3 + Ru3be three parallel lines. Recall from (9.62), (9.63), and (9.64) thatζ1(1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉)= 〈u1, c3〉+ 〈u1, u3〉 〈u3, c2〉+ 〈u1, u3〉 〈u3, u2〉 〈u2, c1〉 ,1639.8. Exploring the m = 3 caseζ2(1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉)= 〈u2, c1〉+ 〈u2, u1〉 〈u1, c3〉+ 〈u2, u1〉 〈u1, u3〉 〈u3, c2〉 ,andζ3(1− 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉)= 〈u3, c2〉+ 〈u3, u2〉 〈u2, c1〉+ 〈u3, u2〉 〈u2, u1〉 〈u1, c3〉 ,respectively. Then the following hold:(i) Z =(c1, c2, c3)+ R(u, u, u).(ii) v =(c2 − c1, c3 − c2, c1 − c3).(iii) y =(c3 − c1, c1 − c2, c2 − c3).(iv) e = 13(2c1 − c2 − c3,−c1 + 2c2 − c3,−c1 − c2 + 2c3).Proof. (i): Since the lines are parallel, then 〈u3, u2〉 〈u2, u1〉 〈u1, u3〉 = 1.Assume that u := u1 = u2 = u3. Then every ζ1, ζ2, ζ3 ∈ R solves (9.62),(9.63), and (9.64). Thus, the set of all cycles Z =(c1, c2, c3)+ R(u, u, u).(ii): By Definition 5.1, we have c1 = P1 c3, c2 = P2 c1, c3 = P3 c2, andv =(c2 − c1, c3 − c2, c1 − c3).(iii): From (ii) and Proposition 9.8(ii), we have y = −Rv = (c3 − c1, c1 −c2, c2 − c3). (iv): From (iii) and (9.23), we have y1 = c3 − c1, y2 = c1 − c2,y3 = c2 − c3, ande1e2e3 = −12y1 0 00 y2 10 0 y3− 16 0 −y2 y3y1 0 −y3−y1 y2 0= 16−3y1 y2 −y3−y1 −3y2 y3y1 −y2 −3y3 .= 163(c1 − c3) c1 − c2 c3 − c2c1 − c3 3(c2 − c1) c2 − c3c3 − c1 c2 − c1 3(c3 − c2) .Hence,e1 =16(3c1 − 3c3 + c1 − c2 + c3 − c2)= 13(2c1 − c2 − c3),e2 =16(c1 − c3 + 3c2 − 3c1 + c2 − c3)= 13(− c1 + 2c2 − c3),e3 =16(c3 − c1 + c2 − c1 + 3c3 − 3c2)= 13(− c1 − c2 + 2c3).1649.8. Exploring the m = 3 caseFigure 9.1: Visualization the cycle and the difference vectors for three linesin R3 that are parallel. See Proposition 9.17.Proposition 9.18. Let C1 = c1 + Ru1, C2 = c2 + Ru2 and C3 = c3 + Ru3be three not parallel lines. Let ζ1, ζ2 and ζ3 are given in (9.62),(9.63), and(9.64) respectively, then the following hold:(i) The set of cycles Z has only one element, which isZ =(c1 + ζ1u1, c2 + ζ2u2, c3 + ζ3u3).(ii) v =(c2− c1 + ζ2u2− ζ1u1, c3− c2 + ζ3u3− ζ2u2, c1− c3 + ζ1u1− ζ3u3).(iii) y =(c3− c1 + ζ3u3− ζ1u1, c1− c2 + ζ1u1− ζ2u2, c2− c3 + ζ2u2− ζ3u3).(iv) e = 13(2c1− c2− c3 + 2ζ1u1− ζ2u2− ζ3u3, 2c2− c1− c3 + 2ζ2u2− ζ1u1−ζ3u3, 2c3 − c1 − c2 + 2ζ3u3 − ζ1u1 − ζ2u2).Proof. (i): Since the lines C1, C2 and C3 are not parallel, then〈u3, u2〉 〈u2, u1〉 〈u1, u3〉 < 1.1659.8. Exploring the m = 3 caseBy (9.65), (9.66) and (9.67), the set of cycles Z has only one elementz =(z1, z2, z3)=(c1 + ζ1u1, c2 + ζ2u2, c3 + ζ3u3),where ζ1, ζ2 and ζ3 are given in (9.65), (9.66) and (9.67) respectively.(ii): By using Definition 5.1, we have z1 = P1 z3, z2 = P2 z1, z3 = P3 z2, andv =(z2 − z1, z3 − z2, z1 − z3)=(c2 − c1 + ζ2u2 − ζ1u1, c3 − c2 + ζ3u3 − ζ2u2, c1 − c3 + ζ1u1 − ζ3u3).(iii): From (ii) and Proposition 9.8(ii), we havey = −Rv=(c3 − c1 + ζ3u3 − ζ1u1, c1 − c2 + ζ1u1 − ζ2u2, c2 − c3 + ζ2u2 − ζ3u3).(iv): From (iii) and (9.23), we have y1 = c3 − c1 + ζ3u3 − ζ1u1, y2 = c1 −c2 + ζ1u1 − ζ2u2, y3 = c2 − c3 + ζ2u2 − ζ3u3, ande1e2e3 = −12y1 0 00 y2 10 0 y3− 16 0 −y2 y3y1 0 −y3−y1 y2 0= 16−3y1 y2 −y3−y1 −3y2 y3y1 −y2 −3y3 .Hence,e1 =16(− 3y1 + y2 − y3)= 16(3(c1 − c3 + ζ1u1 − ζ3u3) + c1 − c2 + ζ1u1 − ζ2u2 + c3 − c2+ ζ3u3 − ζ2u2)= 13(2c1 − c2 − c3 + 2ζ1u1 − ζ2u2 − ζ3u3),e2 =16(− y1 − 3y2 + y3)= 16(c1 − c3 + ζ1u1 − ζ3u3 + 3(c2 − c1 + ζ2u2 − ζ1u1) + c2 − c3+ ζ2u2 − ζ3u3)= 13(2c2 − c1 − c3 + 2ζ2u2 − ζ1u1 − ζ3u3),1669.8. Exploring the m = 3 caseande3 =16(y1 − y2 − 3y3)= 16(c3 − c1 + ζ3u3 − ζ1u1 + c2 − c1 + ζ2u2 − ζ1u1+ 3(c3 − c2 + ζ3u3 − ζ2u2))= 13(2c3 − c1 − c2 + 2ζ3u3 − ζ1u1 − ζ2u2).Figure 9.2: Visualization of the cycle and the difference vectors for threelines in R3 that are not parallel. See Proposition 9.18.9.8.2 The epigraph of the exponential function and twohorizontal linesIn this section, we assume thatH = R2. (9.69)1679.8. Exploring the m = 3 caseFigure 9.3: Visualization of the cycle and the difference vectors for threelines in plane that are not parallel. See Proposition 9.18.We present three sets in the Euclidean plane and consider two differentorderings. The sets are the epigraph of the exponential function,epi(exp)= {(ξ, η) ∈ R2 | exp (ξ) ≤ η} = gra ( exp )+ ({0} × R), (9.70)two horizontal linesR× {0} and R× {1}.We work on two different ordering. One of the order lead to the existenceof the cycle and the other order lead to their absence.An ordering with cyclesAssume thatC1 = R× {0}, C2 = R× {1}, C3 = epi(exp). (9.71)1689.8. Exploring the m = 3 casey =(y1, y2, y3)=((0, 1),(0,−1), (0, 0)). (9.72)Then the following are hold:(i) (9.72) satisfies the characterization provided by Theorem 9.15.(ii) e =(e1, e2, e3)=((0,−(2/3)), (0, (1/3)), (0, (1/3))).(iii) v =(v1, v2, v3)=((0, 1),(0, 0),(0,−1)).(iv) F1 = R− × {0}, F2 = R− × {1}, and F3 = R− × {1}.Proof. (i): It is clear that y1 + y2 + y3 =(0 + 0 + 0, 1 − 1 + 0) = (0, 0).Hence, (9.72) holds. Set c1,n ≡(0, 0) ∈ C1, c2,n ≡ (0, 1) ∈ C2, c3,n ≡(0, 1)=(0, exp(0)) ∈ C3, and xn ≡ (0,−(2/3)) ∈ R2. Thenc1,n + xn ≡(0,−(2/3)) = (1/6)(− 3(0, 1) + (0,−1)− (0, 0))= (1/6)(− 3y1 + y2 − y3),c2,n + xn ≡(0, (1/3))= (1/6)(− (0, 1)− 3(0,−1) + (0, 0))= (1/6)(− y1 − 3y2 + y3),andc3,n + xn ≡(0, (1/3))= (1/6)((0, 1)− (0,−1)− 3(0, 0))= (1/6)(y1 − y2 − 3y3).Hence, (9.58), (9.59), and (9.60) hold. Next, let c1 =(γ1, 0) ∈ C1, c2 =(γ2, 1) ∈ C2, and c3 = (γ3, exp(γ3) + δ3) ∈ C3, where {γ1, γ2, γ3} ⊆ R, andδ3 ∈ R+. Then〈y1, c1〉+ 〈y2, c2〉+ 〈y3, c3〉=〈(0, 1),(γ1, 0)〉+〈(0,−1), (γ2, 1)〉+ 〈(0, 0), (γ3, exp(γ3) + δ3)〉= −1= −(1/2)(1 + 1 + 0)= −(1/2)( ‖(0, 1)‖2 + ‖(0,−1)‖2 + ‖(0, 0)‖2 )= −(1/2)( ‖y1‖2 + ‖y2‖2 + ‖y3‖2 ).Hence, (9.61) holds. (ii): By using Theorem 9.15(i), we obtaine =(e1, e2, e3)= (1/6)(− 3y1 + y2 − y3,−y1 − 3y2 + y3, y1 − y2 − 3y3)=((0,−(2/3)), (0, (1/3)), (0, (1/3))).1699.8. Exploring the m = 3 case(iii): By using Theorem 9.15(ii), we havev =(v1, v2, v3)= −(y2, y3, y1)=((0, 1),(0, 0),(0,−1).(iv): From (9.71), (iii) and Definition 5.3, we haveF1 = C1 ∩(C3 + v3) ∩ (C2 + v2 + v3)=(R× {0}) ∩ (epi(exp) + (0,−1)) ∩ (R× {1}+ (0, 0)+ (0,−1))=(R× {0}) ∩ (epi(exp) + (0,−1)) ∩ (R× {0})=(R× {0}) ∩ (epi(exp) + (0,−1))= R− × {0}.Next,F2 = C2 ∩(C1 + v1) ∩ (C3 + v3 + v1)=(R× {1}) ∩ (R× {0}+ (0, 1)) ∩ (epi( exp )+ (0,−1)+ (0, 1))=(R× {1}) ∩ (R× {1}) ∩ epi( exp )= R− × {1}.Finally,F3 = C3 ∩(C2 + v2) ∩ (C1 + v1 + v2)= epi(exp) ∩ (R× {1}+ (0, 0)) ∩ (R× {0}+ (0, 1))= epi(exp) ∩ (R× {1}) ∩ (R× {1})= R− × {1}.An ordering without cyclesAssume thatC1 = R× {1}, C2 = R× {0}, C3 = epi(exp). (9.73)y =(y1, y2, y3)=((0,−1), (0, 1), (0, 0)). (9.74)Then the following are hold:1709.8. Exploring the m = 3 caseFigure 9.4: Visualization of the cycle and the difference vectors for threelines in plane that are not parallel. See Subsection 9.8.2.(i) (9.74) satisfies the characterization provided by Theorem 9.15.(ii) e =(e1, e2, e3)=((0, (2/3)),(0,−(1/3)), (0,−(1/3))).(iii) v =(v1, v2, v3)=((0,−1), (0, 0), (0, 1)).(iv) F1 = ∅, F2 = ∅, and F3 = ∅.Proof. (i): It is clear that y1+y2+y3 =(0+0+0,−1+1+0) = (0, 0). Hence,(9.72) holds. Set for all n ∈ N, c1,n ≡( − n, 1) ∈ C1, c2,n ≡ ( − n, 0) ∈ C2,c3,n ≡(− n, exp(−n)) ∈ C3, and xn ≡ (n,−(1/3)) ∈ R2. Thenc1,n + xn ≡(0, (2/3))= (1/6)(− 3(0,−1) + (0, 1)− (0, 0))= (1/6)(− 3y1 + y2 − y3),c2,n + xn ≡(0,−(1/3)) = (1/6)(− (0,−1)− 3(0, 1) + (0, 0))= (1/6)(− y1 − 3y2 + y3),andc3,n + xn ≡(0, exp(−n)− (1/3))→ (0,−(1/3))= (1/6)((0,−1)− (0, 1)− 3(0, 0)) = (1/6)(y1 − y2 − 3y3).1719.8. Exploring the m = 3 caseHence, (9.58), (9.59), and (9.60) hold. Next, let c1 =(γ1, 1) ∈ C1, c2 =(γ2, 0) ∈ C2, and c3 = (γ3, exp(γ3) + δ3) ∈ C3, where {γ1, γ2, γ3} ⊆ R, andδ3 ∈ R+. Then〈y1, c1〉+ 〈y2, c2〉+ 〈y3, c3〉=〈(0,−1), (γ1, 1)〉+ 〈(0, 1), (γ2, 0)〉+ 〈(0, 0), (γ3, exp(γ3) + δ3)〉= −1= −(1/2)(1 + 1 + 0)= −(1/2)( ‖(0,−1)‖2 + ‖(0, 1)‖2 + ‖(0, 0)‖2 )= −(1/2)( ‖y1‖2 + ‖y2‖2 + ‖y3‖2 ).Hence, (9.61) holds.(ii): By using Theorem 9.15(i), we obtaine =(e1, e2, e3)= (1/6)(− 3y1 + y2 − y3,−y1 − 3y2 + y3, y1 − y2 − 3y3)=((0, (2/3)),(0,−(1/3)), (0,−(1/3))).(iii): By Theorem 9.15(ii), we havev =(v1, v2, v3)= −(y2, y3, y1)=((0,−1), (0, 0), (0, 1).(iv): From (9.73), (iii) and Definition 5.3, we haveF1 = C1 ∩(C3 + v3) ∩ (C2 + v2 + v3)=(R× {1}) ∩ (epi(exp) + (0, 1)) ∩ (R× {0}+ (0, 0)+ (0, 1))=(R× {1}) ∩ (epi(exp) + (0, 1)) ∩ (R× {1})=(R× {1}) ∩ (epi(exp) + (0, 1))= ∅.Next,F2 = C2 ∩(C1 + v1) ∩ (C3 + v3 + v1)=(R× {0}) ∩ (R× {1}+ (0,−1)) ∩ (epi( exp )+ (0, 1)+ (0,−1))=(R× {0}) ∩ (R× {0}) ∩ epi( exp )=(R× {0}) ∩ epi( exp )= ∅.1729.9. Finding the difference vectors for m ≤ 5 using Banach contraction principleFinally,F3 = C3 ∩(C2 + v2) ∩ (C1 + v1 + v2)= epi(exp) ∩ (R× {0}+ (0, 0)) ∩ (R× {1}+ (0,−1)+ (0, 0))= epi(exp) ∩ (R× {0}) ∩ (R× {0})= ∅.Fact 9.19 (Seeger´s algorithm). Given x ∈ H and d0 ∈ H, generate thesequences (cn)n≥1 and (dn)n≥1 iteratively viacn := PC(x− dn−1), dn := P∆(x− cn).Thencn + dn → PC+∆(x).9.9 Finding the difference vectors for m ≤ 5 usingBanach contraction principleLemma 9.20. Let y be the unique solution to (9.22). Then the followinghold:(i) PC+∆ is firmly nonexpansive.(ii) Id− PC+∆ is nonexpansive (1- Lipschitz continuous).(iii) Solving (9.22) is equivalent to solve(Id− PC+∆)(12Id−T)y = y. (9.75)Proof. (i): It follows from Fact 3.2.(ii): From (i) and Fact 3.10(ii) we have Id− PC+∆ is firmly nonexpansive.Therefore, it is nonexpansive (1- Lipschitz continuous).(iii): From (9.22), we obtain −12y − Ty ∈ N−1C+∆(y) ⇔ y ∈ NC+∆( −12y − Ty) ⇔ y + ( − 12y − Ty) − ( − 12y − Ty) ∈ NC+∆( − 12y − Ty) ⇔12y−Ty ∈(Id +NC+∆)(− 12y−Ty)⇔ −12y−Ty = PC+∆ (12y−Ty)⇔−12y−Ty + y− y = PC+∆(12y−Ty), and hence,(Id− PC+∆)(12Id−T)y = y.1739.10. Finding the difference vectors by forward-backwardWe could solve (9.75) by using Banach contraction principle as long asthe norm of 12Id−T less than 1. The calculation of 12Id−T norm is givenin the next example. The code that can be used to verify this example canbe found in Appendix B.3.m eigenvalues of (12Id−T)∗(12Id−T) ‖12Id−T‖2 14 (twice)12 = 0.53 13 (twice),141√3 ≈ 0.584 12 (twice),14 (twice)1√2 ≈ 0.715 12 +12√5 (twice),12 − 12√5 (twice),14√12 +12√5 ≈ 0.856 1 (twice), 13 (twice),14 (twice) 1Table 9.1: Computing the operator norm of 12Id−T.9.10 Finding the difference vectors byforward-backwardProposition 9.21. The following hold:(i) Solving (9.22) is equivalent to solving0 ∈ N−1C+∆(y)+(12Id + T)(y). (9.76)(ii) Let (9.76) be the Attouch-The´ra primal problem for the pair(N−1C+∆,12Id+T). Then the Attouch-The´ra dual problem is0 ∈ NC+∆(x)+(12Id + T)−1(x). (9.77)Proof. (i): Using (9.22) gives 0 ∈ ∂σC+∆(y)+(12Id + T)(y) ⇔ 0 ∈∂ι∗C+∆(y)+(12Id+T)(y)by Fact 2.35⇔ 0 ∈ (∂ιC+∆)−1(y)+(12Id+T)(y)by Fact 2.27 ⇔ 0 ∈ N−1C+∆(y)+(12Id + T)(y)by Fact 2.36.1749.10. Finding the difference vectors by forward-backward(ii): Using the Attouch-The´ra duality with A = N−1C+∆ and B =12Id + T,and the linearity of T by Theorem 7.6(i), we obtain−Id ◦(12Id + T)−1 ◦ (− Id) = (12Id + T)−1.For(N−1C+∆)−1 = NC+∆ and the linear relation (12Id + T)−1, we can for-mulate the Attouch-The´ra dual problem0 ∈ NC+∆(x)+(12Id + T)−1(x).Proposition 9.22. Let y be the solution of (9.76), x be the solution of(9.77), and e be as in (9.23). Then the following hold:(i) The solution set of (9.77) is a singleton and unique.(ii) x = e.(iii) We have{e} = ∆⊥ ∩ Fix (PC+∆ R) (9.78)(iv) x ∈ C + ∆ ∩∆⊥.Proof. (i): Let y be the unique solution of (9.76). Then the set of solutionsof the (9.77) viaN−1C+∆(y) ∩ −(12Id + T)(y),is nonempty by Fact 3.51(iii). From Proposition 9.3 and because T is single-valued by Theorem 7.6 this implies the unique solution to (9.77).(ii): From (i) and (9.23), we obtainx = N−1C+∆(y) ∩ −(12Id + T)(y) = −12y−Ty = e.(iii): It follows from (ii), (9.77), and Lemma 9.6(ii) that0 ∈NC+∆(e)+(12Id + T)−1(e)= NC+∆(e)+(Id−R + 2 P∆)(e)(From Lemma 7.13(v))= NC+∆(e)+(Id−R)(e).Hence, Re ∈ (e +NC+∆)(e). Therefore, we obtaine = PC+∆(Re) ∈ C + ∆ ∩∆⊥.(iv): It follows from (ii), (iii), and Lemma 9.6(ii).1759.10. Finding the difference vectors by forward-backwardTheorem 9.23. Recall the Attouch-The´ra dual problem0 ∈ NC+∆(x)+(12Id + T)−1(x). (9.79)Let γ ∈ ]0, 1[, let x0 ∈H, and generate a sequence (xn)n∈N viaxn+1 = PC+∆(xn − γ(12Id + T)−1xn). (9.80)Then the following hold:(i) We havexn+1 = PC+∆((1− γ)xn + γRxn − 2γ P∆ xn). (9.81)(ii) Rxn − xn − 2 P∆ xn → y.(iii) xn → e.(iv) R∗xn − xn → v.Proof. Set A := NC+∆ and B =(12Id + T)−1. Then B is β := (1/2)-cocoercive by Lemma 7.13(viii) and γ ∈ ]0, 1[ = ]0, 2β[. Set δ := 2 −γ/(2β) = 2 − γ > 1 and λ := λn = 1. Then λn(δ − λn)= δ − 1 > 0 and∑n∈N λn(δ − λn)= +∞. Next applying Fact 3.59 givesxn+1 = xn + λn(JγA(xn − γBxn)− xn) (9.82)= xn + JγNC+∆(xn − γ(12Id + T)−1xn)− xn (9.83)= JNC+∆(xn − γ(12Id + T)−1xn)(9.84)= PC+∆(xn − γ(12Id + T)−1xn), (9.85)by Fact 3.34(ii).(i): From (9.85) and Lemma 7.13(v), we havexn+1 = PC+∆(xn − γ(12Id + T)−1xn)= PC+∆(xn − γ(Id−R + 2 P∆)xn)= PC+∆((1− γ)xn + γRxn − 2γ P∆ xn).1769.10. Finding the difference vectors by forward-backward(ii): It follows from Proposition 9.22(i)&(ii) that zer(A + B)= {e} andfrom Fact 3.59(i) that xn will converge weakly to a point in zer(A + B).Hence, xn ⇀ e. Next, Fact 3.59(ii) givesBxn =(12Id + T)−1xn →(12Id + T)−1e= e−Re + 2 P∆ e= e−Re + 0 (From Lemma 9.6(ii))= −y,by Lemma 9.6(iv). Note that our dual problem differs from the one inFact 3.59 by a negative sign. Hence we have−Bxn = Rxn − xn + 2 P∆ xn → Re− e = y.(iii): It follows from (ii) that xn ⇀ e and from Theorem 7.6(i) that T iscontinuous and single-valued. Thenxn =(12Id + T)(12Id + T)−1xn →(12Id + T)(12Id + T)−1e = e.(iv): It follows from (iii) that xn → e. Now apply R∗ − Id, we obtain(R∗ − Id)xn → (R∗ − Id)e= R∗e− e = v,by (9.26).177Chapter 10Projection mappings,Generalized Cycles, and GapVectors10.1 OverviewThe results in this chapter are new. Recall from (8.3)C := C1 × C2 × · · · × Cm,where C1,C2, . . . ,Cm are nonempty closed convex subsets of H. Recall from(8.2)∆ ={x =(x)i∈I∣∣ x ∈ H},and from (8.8)∆⊥ ={x =(xi)i∈I∣∣∣ ∑i∈Ixi = 0}.We summarize our main results in this chapter as follows:• We introduce the generalized cycle, which is a variant of the classicalone (9.2) (see Section 10.2 for details).• We prove that the generalized cycle has a unique solution and wederive a formula for it (see Theorem 10.2(ii)).• The relationship between the generalized cycles and the generalizedgap vectors is given in Theorem 10.2(i).• Under the assumption x ∈∆⊥ , we prove that finding the generalizedcycle is equivalent to solving the fixed point equation x = PC+∆(Rx)(see Lemma 10.1).• The relationship among the classical cycles, generalized cycles and thegeneralized gap vectors is given in Theorem 10.3.17810.2. Extending the primal approach• We illustrate our results by Example 10.4 which illustrates the con-cepts of generalized cycles and generalized gap vectors when the clas-sical one do not exist.10.2 Extending the primal approachWe can start from Attouch-The´ra primal (9.2), which can be writtenequivalently as:0 ∈ ∂ιC(u)+(Id−R)(u). (10.1)Since(Id −R) is full domain and maximally monotone by Theorem 7.1(i)and ιC is closed convex and lower semicontinuous so ∂ιC is maximally mono-tone by Example 3.25. This implies that ∂ιC +(Id − R) is maximallymonotone by Fact 3.44. Therefore,ran(Id +(∂ιC +(Id−R))) =H.Therefore, we cannot find enlargement of (10.1). We need to rewrite it inan equivalent form. Since −(Id −R)u ∈ ∆⊥ by Theorem 7.1(v), then forall d ∈∆−(Id−R)u ∈ ∂ιC(u)∈ ∂ιC(u) ∩∆⊥∈ ∂ιC(u) ∩ ∂ι∆(d), where d ∈∆ (From Proposition 8.4(ii))= ∂(ιCι∆)(u + d)= ∂ιC+∆(u + d)(From Proposition 8.5(ix))⊆ ∂ cl(ιCι∆)(u + d) = ∂ιC+∆(u + d).From Theorem 7.1(ii), we have ker(Id−R) = ∆. Hence, (Id−R)(d) = 0.Therefore, we obtain0 ∈ ∂ιC+∆(u + d)+(Id−R)(u + d), (10.2)with x := u + d ∈ dom ∂ιC+∆ ⊆ dom ιC+∆ = C + ∆. Therefore,x ∈ C + ∆.Moreover, with d = −P∆(u) ∈∆ and x = u + d = P∆⊥ (u) ∈∆⊥. Thusx ∈∆⊥. (10.3)17910.3. The Relationship between the generalized cycles and the generalized gap vectorsCombining all these together and using the fact that ∂ιC+∆ = NC+∆, wewrite (10.2) as0 ∈ NC+∆(x)+(Id−R)(x), where x ∈∆⊥ ∩C + ∆. (10.4)The solution x given by (10.4) is called a generalized cycle of ιC+∆.Lemma 10.1. Assume that x ∈ ∆⊥. Then solving (10.4) is equivalent tosolving x = PC+∆(Rx).Proof. Let x ∈∆⊥ and x solves (10.4). Then 0 ∈ NC+∆(x)+(Id−R)(x)⇔Rx ∈ (Id +NC+∆)(x)⇔ x = (Id +NC+∆)−1(Rx). Hence,x = PC+∆(Rx),by Fact 3.34(ii).10.3 The Relationship between the generalizedcycles and the generalized gap vectorsIn this section we will show that the generalized cycles and the gener-alized gap vectors can be put into the frame work of the Attouch-The´raduality.Theorem 10.2. Consider the following0 ∈ NC+∆(x)+(Id−R)(x), where x ∈∆⊥. (P˜)and0 ∈ ∂σC+∆(y)+ 12y + Ty. (D)Then(i)(P˜)is the Attouch-The´ra dual of(D), and(D)has a unique solution.(ii)(P˜)has a unique solution x:x = −12y−Ty.18010.4. The Relationship among the classical cycles, generalized cycles and the generalized gap vectorsProof. (i): Starting from(D), the dual pairs of(D)are(∂σC+∆,12y + Ty)and (∂σC+∆)−1 = (∂ι∗C+∆)−1 = ∂ιC+∆ = NC+∆.Moreover, by Lemma 7.13(iv) we obtain(12y + Ty)−1 = Id−R + 2 P∆,combine these together, we obtain0 ∈ NC+∆(x)+ x−Rx + 2 P∆(x)(10.5)From lemma 8.6 ranNC+∆ =(C + ∆)⊥ = C⊥ ∩∆⊥ ⊆ ∆⊥ and ran (Id−R)= ∆⊥. Moreover, 2 P∆(x) ∈∆∩∆⊥, so P∆ (x) = 0. Therefore, (10.5)can be written as0 ∈ NC+∆(x)+(Id−R)(x), where x ∈∆⊥,which implies that(P˜)is the Attouch-The´ra dual of(D). From Theo-rem 9.5(i) we obtain(D)always has a unique solution.(ii): Since y be the unique solution of(D), then the set of solutions of(P˜)via∂σC+∆(y) ∩ −(12Id + T)(y) 6= ∅is nonempty by Proposition 9.3 and Fact 3.51(i). Because T is a single-valued by Theorem 7.6, this implies that the unique solution to(P˜)isx = ∂σC+∆(y) ∩ −(12Id + T)(y) = −12y−Ty.10.4 The Relationship among the classical cycles,generalized cycles and the generalized gapvectorsIn this section we will study the relationship between the solution of thegeneral cycles(P˜), the classical cycle defined by (9.2) and the generalizedgap vectors.Theorem 10.3. With x and y given in Theorem 10.2, e be as in (9.23),and Z be as in (9.3). The following hold:18110.4. The Relationship among the classical cycles, generalized cycles and the generalized gap vectors(i) x, y ∈∆⊥.(ii) y = Rx− x.(iii) x = −12y−Ty.(iv) x = e.(v) We haveZ =(x + ∆) ∩ ∂σC(Rx− x) = (Id−R)−1(− y) ∩ ∂σC(y)=(e + ∆) ∩ ∂σC(Re− e) = (Id−R)−1(− y) ∩ ∂σC(y).(vi) Z ⊆ (F1 × F2 × · · · × Fm) ∩ (Id−R)−1(− y).Proof. (i): It follows from Lemma 9.6(i) and (10.4).(ii): Set q = Rx− x. This implies that −x = (Id−R)−1(q). Then0 ∈ NC+∆(q)+(Id−R)−1(q)= NC+∆(q)+ q2 + Tq +N∆⊥(q)( From Theorem 7.7)= NC+∆(q)+ q2 + Tq + ∆= NC+∆(q)+ q2 + Tq.This implies that q ∈ ∆⊥ because q ∈ NC+∆(x) ⇔ x ∈ N−1C+∆(q) =∂σC+∆(q). Thus,x ∈ ∂σC+∆(q),which implies that q ∈ dom ∂σC+∆ ⊆ dom σC+∆ ⊆∆⊥.Hence, it solves (9.22) and by Theorem 9.5(i), we have q = y. Therefore,y = Rx− x,by Proposition 9.3(ii).(iii): It follows from Theorem 10.2(ii).(iv): It follows from (iii) and (9.23).18210.5. Example(v): Let Z 6= ∅. From Fact 3.51(iii) and Theorem 7.7, we obtainz ∈ Z ⇔ z ∈ N−1C(y) ∩ (Id−R)−1(− y)⇔ z ∈ ∂ι∗C(y) ∩ −(Id−R)−1(y) (From )⇔ z ∈ ∂σC(y) ∩ −(Id−R)−1(y) (From )⇔ z ∈ ∂σC(y)and z ∈ −(Id−R)−1(y)⇔ z ∈ ∂σC(y)and z ∈ (− 12y−Ty−N∆⊥(y)),⇔ z ∈ ∂σC(y)and z ∈ (e + ∆), (From (9.23))⇔ z ∈ ∂σC(y) ∩ (e + ∆).Hence,Z = ∂σC(y) ∩ (e + ∆).By (ii), (iii), and (9.23), we obtainZ = ∂σC(Rx− x) ∩ (x + ∆).(vi): We have from Lemma 9.4(v) that z ∈ Z⇒ z ∈ (F1×F2×F3×· · ·×Fm).Moreover, we obtain from (v) thatz ∈ Z⇔ z ∈ ∂σC(y) ∩ (x + ∆)⇒ z ∈ ∂σC(y) and z ∈ x + ∆.Combining these together, we obtain thatz ∈ (F1 × F2 × F3 × · · · × Fm) ∩ (x + ∆).Hence,Z ⊆ (F1 × F2 × F3 × · · · × Fm) ∩ (x + ∆)=(F1 × F2 × F3 × · · · × Fm) ∩ (− 12y−Ty−N∆⊥(y)) (From (iii))=(F1 × F2 × F3 × · · · × Fm) ∩ (Id−R)−1(− y),from Theorem 7.7.10.5 ExampleExample 10.4. Let α ≥ 0. ConsiderC1 = epi(exp)={(x, r) | exp (− x)+ α ≤ r and x ∈ R}. (10.6)18310.5. ExampleandC2 = R× {0}.Then the following hold(i) We haveNC1(x, exp(−x)) =⋃λ≥0λ(− exp(−x),−1).(ii) ιC has neither a cycle nor a gap vector.(iii) ιC+∆ has both a generalized cycle and a generalized gap vector, namelyx = (0, α/2, 0,−α/2) ∈ R4,andy = (0,−α, 0, α) ∈ R4.Proof. Note that C = C1 × C2 and ∆ ={(u, v, u, v) | u, v ∈ R} are inR2 × R2 = R4.(i): From Definition 2.18, we haveNC1(x, exp(−x)) = {{u ∈ H | sup 〈C1 − (x, exp(−x)), u〉 ≤ 0}, if x ∈ C1;∅, otherwise.From (10.6), we obtain(x, exp(− x)) ∈ C1 if and only ifexp(− x)+ α ≤ exp (− x),which clearly happen if and only if α = 0. Therefore, NC1(x, exp(−x)) = ∅whenever α > 0. Assume that α = 0, and fix x ∈ R. Then u ∈ NC1(x, exp(−x))if and only if0 ≥ 〈(q, r)− (x, exp (− x)), u〉 = 〈(q − x, r − exp (− x)), u〉 ,for all(q, r) ∈ C1.”⊆”: Suppose u = λ(− exp (− x),−1). Then for all (q, r) ∈ C1 we have〈(q − x, r − exp (− x)), u〉 = 〈(q − x, r − exp (− x)), λ(− exp (− x),−1)〉= λ(− (q − x) exp (− x)− (r − exp (− x)))= λ((x− q) exp (− x)+ ( exp (− x)− r))≤ λ((x− q) exp (− x)+ exp (− x)− exp (− q))≤ λ exp (− x)((x− q)+ 1− exp (x− q)).18410.5. ExampleClearly,(x− q)+ 1− exp (x− q) determines weather or not this is less thanor equal to zero. Let z = x− q, and consider the function f(z) = z+ 1− ez.We havef′(z)= 1− ez f ′′(z) = −ez < 0.Hence, f(z)has only one critical point ( at z = 0 ), and f(z)is concave downeverywhere, so the critical point is a maximum. Since f(0)= 0 + 1− 1 = 0,we have f(z) ≤ 0 for all z. In other words,〈(q − x, r − exp (− x)), u〉 ≤ λ exp (− x)((x− q)+ 1− exp (x− q)) ≤ 0,Therefore, u ∈ NC1(x, exp(− x)). Thus,⋃λ≥0λ(− exp(−x),−1) ⊆ NC1(x, exp(− x)). (10.7)”⊇”: Suppose u = (a, b) ∈ NC1 . Then0 ≥ 〈(q − x, r − exp (− x)), (a, b)〉 = (q − x)a− exp (− x)b+ rb.The only restriction on r is that it must be greater than or equal to exp(−q).If b were positive. Then for all q we could find r such thatrb > |(q − x)a− exp (− x)b|,which means that the inner product is positive, contradictingu ∈ NC1(x, exp(− x)).Thus, b < 0 . By a similar argument, since we can choose q > x, we musthave a < 0.Finally, it remains to show that a = b exp( − x). Since, the inner productinequality holds for all(q, r) ∈ C1, we can pick (q, r) such that x > q andr = exp(− q). Then0 ≥ 〈(q − x), exp (− q)− exp (− x), (a, b)〉0 ≥ (q − x)a+ ( exp (− q)− exp (− x))b− (q − x)a ≥ ( exp (− q)− exp (− x))ba ≥ −b(exp (− q)− exp (− x)q − x)18510.5. ExampleNote that the sign does not change because −(q − x) > 0 and the fractionon the right hand side is the formula for average slope of − exp ( − x). Wecan take the limit as q → x from the bellow:a ≥ −b · limq→x−exp(−q)− exp(−x)q − x= b · ddx(− exp(−x))=(− b)(− exp (− x))= b exp(− x).Hence,a ≥ b exp (− x). (10.8)If we repeat the same steps with q > x, we will get the opposite inequality( Since −(q − x) < 0 means the sign does change direction). Hence,a ≤ b exp (− x). (10.9)Combining (10.8) and (10.9), we havea = b exp(− x).Now set λ = −b. Thenu =(a, b)=(− λ exp (− x),−λ) = λ(− exp (− x),−1).Hence,NC1(x, exp(− x)) ⊆ ⋃λ≥0λ(− exp(−x),−1). (10.10)Combining (10.7) and (10.10), we haveNC1(x, exp(− x)) = ⋃λ≥0λ(− exp(−x),−1).(ii): Let z =(z1, z2), where z1, z2 ∈ R2. Then from (10.1), we obtain(0, 0) ∈ ∂ιC(z)+ (Id−R)(z)⇒ (0, 0) ∈ ∂ιC1(z1)× ∂ιC2(z2)+ (Id−R)(z1, z2).Hence, we need to show that(0, 0) ∈ ∂ιC1(z1)× ∂C2(z2) + (Id−R)(z1, z2) (10.11)18610.5. Examplehas no solution. Suppose (10.11) has a solution. Then there exist z1 ∈ C1,z2 ∈ C2, andz2 − z1 ∈ NC1(z1), z1 − z2 ∈ NC2(z2) = {0} × R.Hence, z2 − z1 = (0, y) ∈ NC1(z1) for some y ∈ R. BecauseNC1(x, exp(−x)) =⋃λ≥0λ(− exp(−x),−1),and NC1(z1) = {0} if z1 ∈ int C1, this implies y = 0 so that z1 = z2. Thiscontradicts C1 ∩C2 = ∅. Because (10.11) has no solution, so ιC has no gapvector.(iii): ιC+∆ has a generalized cycle. We show that(0, 0) ∈ NC+∆(x) + (Id−R)(x) and x ∈∆⊥ (10.12)has a solution (0, α/2, 0,−α/2). Indeed, since(0, exp(−x) + α/2, 0,−α/2) = (x, exp(−x) + α, x, 0) + (−x,−α/2,−x,−α/2)∈ C + ∆,by sending x → ∞ we have (0, α/2, 0,−α/2) ∈ C + ∆ ∩∆⊥. With z1 =(0, α/2) = −z2, we show that x = (z1, z2) solves (10.12), i.e.,(0, 0) ∈ NC+∆(0, α/2, 0,−α/2) + (0, α, 0,−α).This holds because (0, α, 0,−α) ∈ ∆⊥, and for r ≥ exp(−x) + α, x, x˜ ∈ R,(u, v, u, v) ∈∆,〈(0, α, 0,−α), ((x, r, x˜, 0) + (u, v, u, v))− (0, α/2, 0,−α/2)〉= 〈(0, α, 0,−α), (x, r, x˜, 0)− (0, α/2, 0,−α/2)〉= α(r − α/2)− α(α/2) = α(r − α) ≥ 0.The gap vector is given by y = Rx− x = (0,−α, 0, α).187Chapter 11Attouch-The´ra Duality andFixed Point Sets ofCompositions of ProximalMappings11.1 OverviewThis chapter is based on our paper [5]. Even though the main resultsare formulated in the given Hilbert space H, the key space to work in is theproduct space H. Suppose (fi)mi=1 be a finite family in Γ0(H) and recallfrom (8.22) its separable sum:f := f1 ⊕ f2 ⊕ · · · ⊕ fm :H→]−∞,+∞] : x 7→ m∑i=1fi(xi),where x =(x1, x2, . . . , xm) ∈ H. We summarize our main results in thischapter as follows:• We prove that the cycles that are defined by proximal mappings canbe formulated in the Hilbert product space as solutions to a fixed pointequation (see Proposition 11.4).• We show that the set of the classical cycles that is defined by us-ing proximal mappings is at most singleton and possibly empty (seeProposition 11.6).• We study the set of the classical cycles that is defined by using theproximal mappings and the set of classical gap vectors (see Proposi-tion 11.7).• The relationship among the set of cycles and fixed point sets for compo-sitions of proximal mapping is given in Lemma 11.8 and Lemma 11.9.18811.2. Important setup!• The enlargement of the dual problem and the general gap vectors aregiven in Section 11.6.• We prove the uniqueness of the solution to the enlarged dual problemand we show that if the enlarged dual solution exists, then it is exactlythe solution to the classical dual problem (see Theorem 11.12).• We derive a formula for the gap vector of the enlarged dual problem(see Theorem 11.12(i)).11.2 Important setup!The result from now in this chapter rely on the following assumptions:1.(fi)mi=1 are in Γ0(H), and f is given by (8.22).2. dom(f∗ + ι∗∆)= dom(f∗ + ι∆⊥) 6= ∅ , equivalently,dom f∗ ∩∆⊥ 6= ∅. (11.1)Proposition 11.1. If dom(f∗+ ι∗∆) 6= ∅, then fι∆ is proper, convex, andpossess a continuous minorant.Proof. By Fact 2.24, fι∆ is possess a continuous minorant ⇔(fι∆)∗ 6≡+∞⇔ f∗ + ι∗∆ 6≡ +∞ by Proposition 8.8(ii).Lemma 11.2. We have(i) dom(f∗ + ι∗∆)= dom(f∗ + ι∆⊥) 6= ∅.(ii) There exists y∗ =(y∗1, y∗2, . . . , y∗m) ∈∆⊥ and f∗1 (y∗1), f∗2 (y∗2), . . . , f∗m(y∗m)∈ R such that (∀xi ∈ H) fi(xi) ≥ 〈y∗i , xi〉 − f∗i (y∗i ),for every i = 1, 2, . . . ,m.Then (i) is equivalent to (ii).Proof. Let y∗ ∈ dom (f∗ + ι∗∆). This implies that y∗ ∈ ∆⊥ by Propo-sition 8.8(i). By (2.16) and Proposition 11.1, we have +∞ > f∗(y) =∑mi=1 supxi∈H( 〈y∗i , xi〉 − fi(xi)) = ∑mi=1 f∗i (y∗i ). Hence,f∗i(y∗i) ≥ 〈y∗i , xi〉 − fi(xi) ⇔ fi(xi) ≥ 〈y∗i , xi〉 − f∗i (y∗i ).for every i = 1, 2, . . . ,m and for every xi ∈ H.18911.2. Important setup!Lemma 11.3. Suppose that one of the following hold:(i) There exists y∗ ∈∆⊥ and σ1, σ2, . . . , σm ∈ R such that(∀xi ∈ H) (∀1 ≤ i ≤ m) fi(xi) ≥ 〈y∗i , xi〉 − σi. (11.2)(ii) One of the functions fi is supercoercive, i.e, having a bounded domain.(iii) inf fi > −∞ for every i = 1, 2, . . . ,m.Then (11.1) holds.Proof. (i): Suppose that (i) holds. Then we obtain(∀x1 ∈ H) f1(x1 − d) ≥ 〈y∗1, x1 − d〉 − σ1 = 〈y∗1, x1〉 − 〈y∗1, d〉 − σ1,(∀x2 ∈ H) f2(x2 − d) ≥ 〈y∗2, x2 − d〉 − σ2 = 〈y∗2, x2〉 − 〈y∗2, d〉 − σ2,...(∀xm ∈ H) fm(xm − d) ≥ 〈y∗m, xm − d〉 − σm = 〈y∗m, xm〉 − 〈y∗m, d〉 − σm.Summing up these m inequalities givesm∑i=1fi(xi − d) ≥ m∑i=1〈y∗i , xi〉 −m∑i=1σi= 〈y∗,x〉 −m∑i=1σi.Taking the infimum over d ∈ H gives(fι∆)(x) ≥ 〈y∗,x〉 − m∑i=1σi.(ii): Suppose (ii) holds. So fm is supercoercive, then by Fact 2.42 for each1 ≤ i ≤ m− 1 it exists y∗i ∈ H and σi ∈ R such that〈xi, y∗i 〉 − σi ≤ fi(xi) (∀xi ∈ H).Since y∗ =(y∗1, y∗2, . . . , y∗m) ∈ ∆⊥, then y∗1 + y∗2 + · · · + y∗m = 0 ⇒ y∗m =−(y∗1 + y∗2 + · · ·+ y∗m−1). Also, fm is supercorecive, thenfm −〈−(y∗1 + y∗2 + · · ·+ y∗m−1), ·〉19011.3. The fixed point sets of the compositions of the proximal mappings and the set of cyclesis coercive. Then it attained its minimum at −σm ∈ R. Therefore,fm − 〈y∗m, xm〉 ≥ −σm(∀xm ∈ H).Hence, (i) applies.(iii): Let y∗ =(0, 0, . . . , 0), then from (i) we have(1 ≤ i ≤ m) (∀xi ∈ H) fi(xi) > 0− σi.Summing up these m-inequalities givesm∑i=1fi(xi)> −m∑i=1σi.Taking the infimum for all xi ∈ H, then we have(fι∆)(x) ≥ − m∑i=1σi.11.3 The fixed point sets of the compositions ofthe proximal mappings and the set of cyclesDefine the fixed point sets of the cyclic compositions of proximal map-pings byFm := Fix(Proxfm Proxfm−1 · · ·Proxf1),Fm−1 := Fix(Proxfm−1 Proxfm−2 · · ·Proxfm),...F1 := Fix(Proxf1 Proxfm · · ·Proxf2).(11.3)(11.4)(11.5)(11.6)Let z1 ∈ F1, and set z2 := Proxf2 z1, . . . , zm−1 := Proxfm−1 zm−2, andzm := Proxfm zm−1. Then z =(z1, z2, · · · , zm)is a cycle, by Definition 5.1.11.4 Classical cycles, gap vectors, and the set ofall the cyclesProposition 11.4. Let z ∈H is a cycle. Thenz = Proxf(Rz). (11.7)19111.4. Classical cycles, gap vectors, and the set of all the cyclesMoreover, solving (11.7) is equivalent to solve the following monotone inclu-sion problem0 ∈ ∂f(z)+ (Id−R)(z). (11.8)Proof. Since z =(z1, z2, . . . , zm−1, zm) ∈H is a cycle, then z1 = Proxf1 zm,z2 = Proxf2 z1, . . . , zm−1 = Proxfm−1 zm−2, and zm = Proxfm zm−1. Thisimplies thatz =(z1, z2, . . . , zm−1, zm)=(Proxf1 zm,Proxf2 z1, . . . ,Proxfm−1 zm−2,Proxfm zm−1)= Proxf(R(z1, z2, . . . , zm−1, zm))(From Fact 8.7(iii) and (8.5))= Proxf(Rz).Moreover, z = Proxf(Rz)⇔ z = (Id + ∂f)−1(Rz) by Fact 3.35(ii). Hence,Rz ∈ z + ∂f(z) and 0 ∈ ∂f(z)+ (Id−R)(z).Define the set of all the cycles byZ := Fix Proxf R. (11.9)Lemma 11.5. Let γ ∈ ]0, 1[. ThenZ = Fix Proxf R = Fix Proxγf((1− γ)Id + γR). (11.10)Proof. From Proposition 11.4, we havex = Proxf Rx⇔ 0 ∈ ∂f(x)+ x−Rx⇔ 0 ∈ γ∂f(x)+ γ(x−Rx)⇔ −γx + γRx ∈ ∂(γf)(x)⇔ (1− γ)x + γRx ∈ (Id + ∂(γf))(x)⇔ x = Proxγf((1− γ)x + γRx).By fact 3.48 we view (11.8) as primal (Attouch-The´ra) problem for thepair(∂f , Id−R). Since R is linear, then the dual pair:(∂f , Id−R)∗=((∂f)−1, (Id−R)−>)=((∂f)−1, (Id−R)−1).Hence, by Fact 3.49 the Attouch-The´ra dual is0 ∈ (∂f)−1(y)+ (Id−R)−1(y). (11.11)19211.5. The fixed point sets of cyclic proximal mappingsProposition 11.6. The following hold:12Id +(N∆⊥ + T +(∂f)−1) is strongly monotone.Moreover, the solution set of (11.11) is at most a singleton (possibly empty).Proof. The proof is similar to Proposition 9.2 applies to((∂f)−1, (Id−R)−1)instead of(N−1C , (Id−R)−1).Proposition 11.7 (cycle and gap vector). Consider the sets of classicalcycles and classical gap vectors defined respectively byZ :={z ∈H | 0 ∈ ∂f(z)+ (Id−R)(z)}, (11.12)K :={y ∈H | 0 ∈ (∂f)−1(y)+ (Id−R)−1(y)}. (11.13)We have(i) Z = ∪y∈K(Id−R)−1(− y) ∩ (∂f)−1(y).(ii) K = ∪z∈Z{Rz− z | z ∈ Z}. If K 6= ∅, thenK is a singleton, y ∈∆⊥, and y = Rz− z for every z ∈ Z.Proof. The proof is similar to Proposition 9.3 applies to(∂f , Id −R) and((∂f)−1, (Id−R)−1) instead of (NC, Id−R) and (N−1C , (Id−R)−1) respec-tively .11.5 The fixed point sets of cyclic proximalmappingsSome basic properties and relationship among the Fi’s, and Z are givenbelow.Lemma 11.8. Suppose that U := ∩mi=1 argmin fi 6= ∅. Then the followingare hold:(i) For every 1 ≤ i ≤ m, we have Fi = U .(ii) Z = {(z, z, · · · , z) | z ∈ U} = Um ∩∆.19311.5. The fixed point sets of cyclic proximal mappingsProof. (i): Proxfi is firmly nonexpansive for every 1 ≤ i ≤ m andFix(Proxfi)= argmin fi.Then by Fact 3.15, we have(∀(1 ≤ i ≤ m)), Fix (Proxfi Proxfi−1 · · ·Proxf1 Proxfm · · ·Proxfi+1 ) = U.(ii): Let z =(z1, z2, · · · , zm) ∈ Z. Thenz1 = Proxf1 Proxfm Proxfm−1 · · ·Proxf2 z1 ⇔ z1 ∈ F1 = Uz2 = Proxf2 Proxf1 Proxfm · · ·Proxf3 z2 ⇔ z2 ∈ F2 = U...zm = Proxfm Proxfm−1 Proxfm−2 · · ·Proxf1 zm ⇔ zm ∈ Fm = U.Hence,z =(z1, z2, · · · , zm) ∈ F1 × F2 × · · · × Fm = U × U × U × · · · × U = Um.Moreover,z =(z1, z2, · · · , zm)=(z, z, · · · , z) ∈∆.Therefore,z ∈ Um ∩∆.In the following Lemma we will study the relationship among Fi’s andthe set of the cycles. LetQi :H→ H : z 7→ zi.Lemma 11.9. For every 1 ≤ i, j ≤ m, the following hold:(i) Fi are closed and convex. Moreover,Fm =(Proxfm Proxfm−1 . . .Proxf3 Proxf2)(F1)= · · · = Proxfm(Fm−1).Fm−1 =(Proxfm−1 . . .Proxf3 Proxf2 Proxf1)(Fm)= · · · = Proxfm−1(Fm−2)....F2 =(Proxf2 Proxf1 Proxfm . . .Proxf4)(F3)= · · · = Proxf2(F1).F1 =(Proxf1 Proxfm . . .Proxf4 Proxf3)(F2)= · · · = P1(Fm).(ii) ∩mi=1 argmin fi ⊆ ∩mi=1Fi and if Fi = ∅, then ∩mi=1 argmin fi = ∅.19411.5. The fixed point sets of cyclic proximal mappings(iii) For 1 ≤ i ≤ m − 1, Proxfi+1(Fi)=(Fi+1), and Proxf1(Fm)= F1.Therefore,Proxf R(F1 × F2 × · · · × Fm)= F1 × F2 × · · · × Fm. (11.14)(iv) Fi 6= ∅ if and only if Fj 6= ∅ if and only if Z = ∅.(v) Z is closed and convex, and Z ⊆ F1 × F2 × · · · × Fm.(vi) The mapping Qi|Z : Z→ Fi is bijective and Qi(Z)= Fi.Proof. The proof is similar to Lemma 9.4, applied to (11.9) andFi := Fix(Proxfi · · ·Proxf1 Proxfm · · ·Proxfi+1)instead of (9.3) and Fi := Fix(Pi · · ·P1 Pm · · ·Pi+1).Lemma 11.10. Denote by y =(y1, y2, · · · , ym)the unique solution of(11.11). Then the following hold:(i) The mapping Proxf1 : Fm → F1 is bijective on Fm and it is givenby Proxf1(z)= z − y1. Moreover, for 1 ≤ i ≤ m − 1 the mappingProxfi+1 : Fi → Fi+1 is bijective and is given by Proxfi+1(z)= z−yi+1.(ii) The fixed point sets F1 = Fm − y1 and Fi+1 = Fi − yi+1.Proof. (i): Let z and z˜ be in Fm satisfying that Proxf1 z = Proxf1 z˜. Ourgoal is to show that Proxf1 is injective on Fm. Then we havez = Proxfm Proxfm−1 · · ·Proxfi Proxfi−1 · · ·Proxf1 zandz˜ = Proxfm Proxfm−1 · · ·Proxfi Proxfi−1 · · ·Proxf1 z˜.Since Proxf1 z = Proxf1 z˜, then we obtain z = z˜. Thus, Proxf1 is an injectivemapping on Fm. Moreover, Lemma 11.9(iii) shows that Proxf1 is a surjectivemapping on Fm. Therefore, Proxf1 is a bijective mapping on Fm. For everyz ∈ Fm, we havez = Proxfm Proxfm−1 · · ·Proxfi Proxfi−1 · · ·Proxf1 z. (11.15)Set z1 = Proxf1 z, z2 = Proxf2 z1, · · · , zm−1 = Proxfm−1 zm−2, zm =Proxfm zm−1. Therefore, using (11.15), we have z = Proxfm zm−1 and z =(z1, z2, · · · , zm−1, z). Therefore, Proposition 11.7(ii) gives(y1, y2, · · · , ym)=(z, z1, z2, · · · , zm−1)− (z1, z2, · · · , zm−1, zm).19511.5. The fixed point sets of cyclic proximal mappingsHence, y1 = z − z1 ⇒ z1 = z − y1 ⇒ Prox1 z = z − y1. The proof of Proxfiis the same as Proxf1 .(ii): It follows from (ii) that for every z ∈ Fm, we obtain Proxf1 z = z − y1.Then by Lemma 11.9(iii) we have F1 = Fm − y1. The proof for Fi+1 =Fi − yi+1 is the same as F1.Proposition 11.11. Suppose one of the fixed point set Fis is nonemptyand bounded. Then(i) The dual problem (11.11) has a unique solution y.Moreover, the following are equivalent(ii) F1 = F2 = · · · = Fm 6= ∅.(iii) ∩mi=1 argmin fi 6= ∅.Under either (ii) or (iii) hold, we haveF1 = F2 = · · · = Fm =m⋂i=1argmin fi =m⋂i=1Fix(Proxfi) 6= ∅. (11.16)Proof. (i): From Lemma 11.9(iv), we have Fi 6= ∅⇔ Z 6= ∅. Now, applyingProposition 11.7 gives that the dual problem (11.11) has a unique solutiony.(ii)⇒(iii): Because that Fi = Fi+1 = Fi − yi+1 by Lemma 11.10(ii), andthat Fi is bounded, closed and convex. Then Fact 2.17 implies yi+1 = 0 fori = 1, . . . ,m, so y = 0. From Proposition 11.7(ii), we have x is the solutionto (11.8). This implies thatRx− x = y⇒ Rx− x = 0.Then (xm, x1, . . . , xm−1)− (x1, x2, . . . , xm) = (0, 0, . . . , 0)Hence, x1 = · · · = xm. By (11.8) and Fact 8.7(i), we obtain(0, . . . , 0) ∈ ∂f1(x1)× · · · × ∂fm(x1),so x1 ∈ argmin f1, x1 ∈ argmin f2, . . . , x1 ∈ argmin fm. Hence,x1 ∈m⋂i=1argmin fi ⇒m⋂i=1argmin fi 6= ∅.Therefore, (iii) holds.(iii)⇒(ii): It follows from Lemma 11.8(i). Clearly, Equation (11.16)follows from 11.8(i) and argmin fi = Fix Proxfi .In the next section we will get the generalized gap vectors.19611.6. The generalized gap vectors11.6 The generalized gap vectorsFrom Theorem 7.7, we have(Id−R)−1 = 12Id + N∆⊥ + T.Moreover, because ∂ι∗∆ = ∂ι∆⊥ = N∆⊥ , we have(∂f)−1 + (Id−R)−1 = ∂f∗ + 12Id + T + ∂ι∗∆ = ∂f∗ + ∂ι∗∆ + 12Id + T⊆ ∂(f∗ + ι∗∆)+ 12Id + T= 12[Id +(2T + 2∂(f∗ + ι∗∆))].Hence, instead of working with (11.11), which has no solution if there areno cycles, we will work with the enlarged dual is0 ∈ ∂(f∗ + ι∗∆)(y)+ 12y + Ty. (11.17)Theorem 11.12. The following hold:(i) The enlarged dual problem (11.17) always has a unique solution ysatisfies thaty = J2T+2∂(f∗+ι∗∆)(0).(ii) y ∈∆⊥.(iii) If (11.11) has a solution, then it is exactly the solution of (11.17).Proof. (i): T is full domain and maximally monotone by Theorem 7.6(iv)and ∂(f∗+ι∗∆)is maximally monotone by Example 3.25. Then 2T+2∂(f∗+ι∗∆)is maximally monotone by Fact 3.44. Therefore,ran(Id + 2T + 2∂(f∗ + ι∗∆))=H.Moreover, from (11.17) we have0 ∈ y + 2Ty + 2∂(f∗ + ι∗∆)(y)⇔ 0 ∈ (Id + (2Ty + 2∂(f∗ + ι∗∆)))(y).Hence,y =(Id +(2Ty + 2∂(f∗ + ι∗∆)))−1(0)⇔ y = J2T+2∂(f∗+ι∗∆)(0).19711.6. The generalized gap vectors(ii): We havey ∈ dom ∂(f∗ + ι∗∆) ⊆ dom (f∗ + ι∗∆)= dom f∗ ∩ dom ι∗∆= dom f∗ ∩ dom ι∆⊥⊆ dom ι∆⊥ = ∆⊥.(iii): This is clear since(∂f)−1 + (Id−R)−1 = ∂f∗ + ∂ι∗∆ + 12Id + T⊆ 12Id + T + ∂(f∗ + ι∗∆).198Chapter 12Proximal Mappings,Generalized Cycles, and GapVectors12.1 OverviewThis chapter is based on our paper [5]. The space that we will work inis the Hilbert product space H. Suppose (fi)mi=1 be a finite family in Γ0(H)and recall from (8.22) its separable sum:f := f1 ⊕ f2 ⊕ · · · ⊕ fm :H→ ]−∞,+∞] : x 7→m∑i=1fi(xi),where x =(x1, x2, . . . , xm) ∈ H. We used the assumptions as in Sec-tion 11.2. We summarize our main results in this chapter as follows:• We present the generalized cycle which is a variant of the classical one(11.8) (see Section 12.2 for details).• We prove that the generalized cycle is unique and we derive a formulafor it (see Theorem 12.2(ii)).• The relationship between the generalized cycles and the generalizedgap vectors is given in Theorem 12.2(i).• Under the assumption x ∈ ∆⊥ we prove that finding the gener-alized cycle is equivalent to solving the fixed point equation x =Proxcl(fι∆)(Rx)(see Lemma 12.1).• The relationship among the classical cycles, generalized cycles and thegeneralized gap vectors is given in Theorem 12.5.• We find the difference vectors by using the forward-backward algo-rithm (see Theorem 12.9).19912.2. The generalized cycles• We illustrate our results by Example 12.11 which characterizes whenthe set of cycles is a singleton or infinite for a finite number of lines ina Hilbert space.12.2 The generalized cyclesRecall from (11.8) that0 ∈ ∂f(u)+ (Id−R)u.Note that(Id − R) is full domain and maximally monotone by Theo-rem 7.1(i) and ∂f is maximally monotone by Fact 3.25. Then ∂f +(Id−R)is already maximally monotone by Fact 3.44. Hence, there is no need forenlargements. We will rewrite (11.8) in an equivalent form. In view of−(Id−R)u ∈ ∂f(u), where − (Id−R)u ∈∆⊥ (From Theorem 7.1(v)).By Proposition 8.4(ii), Lemma 8.12(iii) and for all d ∈∆, we have−(Id−R)u ∈ ∂f(u) ∩∆⊥= ∂f(u) ∩ ∂ι∆(d) ⊆ ∂(fι∆)(u + d)⊆ ∂[cl(fι∆)](u + d), where d ∈∆.Hence, we have0 ∈ ∂[cl(fι∆)](u + d)+ (Id−R)(u + d),since(Id−R)(d) = 0. Moreover, withd =(− m∑i=1ui/m, . . . ,−m∑i=1ui/m) ∈∆,we havex = u + d ∈∆⊥. (12.1)Hence,0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x), and x ∈∆⊥. (12.2)The solution x given by (12.2) is called the generalized cycle of cl(fι∆).Lemma 12.1. Assume that x ∈ ∆⊥. Then solving (12.2) is equivalent tosolve x = Proxcl(fι∆)(Rx).20012.3. The primal-dual approachProof. Let x ∈ ∆⊥ and x solves (12.2). Then 0 ∈ ∂[cl(fι∆)](x) + (Id −R)(x)⇔ Rx ∈ (Id + ∂[cl(fι∆)])(x)⇔ x = (Id + ∂[cl(fι∆)])−1(Rx).Hence,x = Proxcl(fι∆)(Rx),by Fact 2.33.12.3 The primal-dual approachIn this section, we study the relationship between the generalized cyclesand the generalized gap vectors (11.17) as well as their solutions. The gen-eralized cycles and gap vectors of cl(fι∆)can be put into the frame workof Attouch-The´ra duality.Theorem 12.2. Consider the following(P˜ ) 0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x), and x ∈∆⊥,(D˜) 0 ∈ ∂(f∗ + ι∗∆)(y)+ 12y + Ty.Then(i)(P˜)is the Attouch-The´ra dual of(D˜), and(D˜)has a unique solution.(ii)(P˜)has a unique solution x;x = −12y−Ty,where y is the unique solution to (11.17).Proof. (i): Starting from(D˜), the dual pairs of(∂(f∗ + ι∗∆),(12Id + T))is(∂(f∗ + ι∗∆),(12Id + T))∗=((∂(f∗ + ι∗∆))−1,(12Id + T)−1).We have (∂(f∗ + ι∗∆))−1 = ∂(f∗ + ι∗∆)∗ = ∂[cl(fι∆)].Moreover, by Lemma 7.13(iv) we obtain(12y + Ty)−1 = Id−R + 2 P∆,20112.4. The relationship between classical cycles and generalized cyclescombine these together, we obtain0 ∈ ∂[cl(fι∆)](x)+ x−Rx + 2 P∆ (x) (12.3)From lemma 8.9 ran ∂[cl(fι∆)] ⊆ ∆⊥ and ran (Id − R) = ∆⊥ by The-orem 7.1(v). Moreover, 2 P∆(x) ∈ ∆ ∩∆⊥, so P∆ (x) = 0. Therefore,(12.3) can be written as0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x), where x ∈∆⊥,which implies that(P˜)is the Attouch-The´ra dual of(D˜). From Theo-rem 11.12(i) we obtain(D˜)always has a unique solution.(ii): Since y be the unique solution of(D˜), then the set of solutions of(P˜)via∂(f∗ + ι∗∆)(y) ∩ −(12Id + T)(y) 6= ∅is nonempty by Proposition 11.7 and Fact 3.51(i). Because T is a single-valued by Theorem 7.6, this implies that the unique solution to(P˜)isx = ∂(f∗ + ι∗∆)(y) ∩ −(12Id + T)(y) = −12y−Ty.12.4 The relationship between classical cyclesand generalized cyclesTheorem 12.3. Let x be the generalized cycle of cl(fι∆)such that0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x), and x ∈∆⊥.If cl(fι∆)(x)=(fι∆)(x)and fι∆ is exact at x, then∃u ∈H such that x = u + v, v = −P∆(u) ∈∆ and x = P∆⊥ (u).Moreover, u is a classical cycle for f . i.e.,0 ∈ ∂f(u)+ (Id−R)u.Proof. Since fι∆ is exact at x, then there exists u ∈ H and v ∈ ∆ suchthat x = u + v and(fι∆)(x)= f(u)+ ι∆(v). If cl(fι∆)(x)=(fι∆)(x), then∂[cl(fι∆)](x)= ∂(fι∆)(x), (12.4)20212.4. The relationship between classical cycles and generalized cyclesby Lemma 8.10. From Lemma 8.12 and (12.4), we have∂[cl(fι∆)](x)= ∂(fι∆)(x)= ∂f(u) ∩ ∂ι∆(v) (From Lemma 8.12)= ∂f(u) ∩∆⊥ (From Proposition 8.4(ii)).Because ker(Id − R) = ∆ by Theorem 7.1(ii). Then (Id − R)(u + v) =(Id−R)(u) since v ∈∆. Hence,0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x)simplifies to0 ∈ ∂f(u)+ (Id−R)u.Because x = u + v, x ∈∆⊥ and v ∈∆. Then0 = P∆ x= P∆(u)+ P∆(v)= P∆(u)+ v.Hence, v = −P∆(u).Theorem 12.4. Let u be a classical cycle for f , i.e.,0 ∈ ∂f(u)+ (Id−R)u.Set v = −P∆(u) ∈∆ and x = u + v = P∆⊥ (u). Then(i) fι∆ is lower semicontinuous and exact at x.(ii) x ∈∆⊥ and x solves0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x).(iii) P∆⊥(Z)= {x}.Proof. From Theorem 7.1(v), we have ran(Id−R) = ∆⊥. Hence, −(Id−R)(u) ∈∆⊥.(i): Starting from the classical cycle, we obtain that for all d ∈ ∆−(Id−R)(u) ∈ ∂f(u)= ∂f(u) ∩∆⊥= ∂f(u) ∩ ∂ι∆(d) (From Proposition 8.4(ii))= ∂f(u) ∩ ∂ι∆(d) 6= ∅.20312.4. The relationship between classical cycles and generalized cyclesSince ∂f(u) ∩ ∂ι∆(d) 6= ∅, then fι∆ is exact at x by Lemma 8.12(ii).Hence, (fι∆)(x)= f(u)+ ι∆(v).and∂(fι∆)(x)= ∂f(u) ∩∆⊥ 6= ∅, (12.5)by Lemma 8.12(ii) and Proposition 8.4(ii). Since ∂(fι∆)(x) 6= ∅, then(fι∆)(x)is lower semicontinuous and exact at x by Lemma 8.11(i).(ii): From (ii), we have(fι∆)is lower semicontinuous and exact at x.Hence,(fι∆)(x)= cl[(fι∆)](x). Therefore,∂[cl(fι∆)](x)= ∂(fι∆)(x)= ∂f(u) ∩∆⊥, (12.6)by Lemma 8.11(ii) and (12.5). Also,x = u + v = u− P∆(u)= P∆⊥(u).Hence,x ∈∆⊥. (12.7)Moreover,(Id−R)(x) = (Id−R)(u + v)=(Id−R)(u)+ (Id−R)v=(Id−R)(u)+ 0 (From Theorem 7.1(ii))=(Id−R)(u) ∈∆⊥,by Theorem 7.1(v). Using (12.7) and (12.5), we could rewrite (11.8) as0 ∈ ∂f(u)+ (Id−R)(u)⇒ 0 ∈ ∂f(u) ∩∆⊥ + (Id−R)(x)∈ ∂(fι∆)(x)+ (Id−R)(x)= ∂[cl(fι∆)](x)+(Id−R)(x).Hence, we obtain0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x), where x ∈∆.(iii): It is clear from (ii).20412.5. The relationship among the classical cycles, generalized cycles and the generalized gap vectors12.5 The relationship among the classical cycles,generalized cycles and the generalized gapvectorsIn this section we will study the relationship between the solution of thegeneralized cycles (12.2), the classical cycles (11.8) and the generalized gapvectors (11.17).Theorem 12.5. With x and y given in Theorem 12.2 and Z be as in (11.9).The following hold:(i) x, y ∈∆⊥.(ii) y = Rx− x.(iii) x = −12y−Ty.(iv) Z =(x + ∆) ∩ (∂f)−1(Rx− x) = (Id−R)−1(− y) ∩ (∂f)−1(y).(v) Z ⊆ (F1 × F2 × · · · × Fm) ∩ (Id−R)−1(− y).Proof. (i): It follows from Theorem 11.12(ii) that y ∈ ∆⊥ and from (12.1)that x ∈∆⊥.(ii): Set v = Rx− x. This implies that −x = (Id−R)−1(v). Then0 ∈ ∂(f∗ + ι∗∆)(v)+ (Id−R)−1(v)= ∂(f∗ + ι∗∆)(v)+ v2 + Tv +N∆⊥(v)( From Theorem 7.7)=[∂(f∗ + ι∗∆)(v)+ ∆]+ v2 + Tv= ∂(f∗ + ι∗∆)(v)+ v2 + Tv.This implies that v ∈∆⊥ becausev ∈ dom ∂(f∗ + ι∗∆) ⊆ dom (f∗ + ι∗∆)= dom f∗ ∩ dom ι∗∆= dom f∗ ∩ dom ι∆⊥⊆ dom ι∆⊥ = ∆⊥.Then it solves (11.17) and by Theorem 12.2(i), we have v = y. Therefore,by Theorem 11.12(iii) and Proposition 11.7(ii), we havey = Rx− x.20512.5. The relationship among the classical cycles, generalized cycles and the generalized gap vectors(iii): It follows from Theorem 12.2(ii).(iv): Let Z 6= ∅. From Fact 3.51(iii) and Theorem 7.35, we obtainz ∈ Z ⇔ z ∈ (∂f)−1(y) ∩ (Id−R)−1(− y)⇔ z ∈ (∂f)−1(y) ∩ −(Id−R)−1(y)⇔ z ∈ (∂f)−1(y) and z ∈ (− 12y−Ty−N∆⊥(y)),⇔ z ∈ (∂f)−1(y) and z ∈ (x + ∆), (From (iii))⇔ z ∈ (∂f)−1(y) ∩ (x + ∆).Hence,Z =(∂f)−1(y) ∩ (x + ∆).By (ii) and (iii), we obtainZ =(∂f)−1(Rx− x) ∩ (x + ∆).(v): We have from Lemma 11.9(v) that z ∈ Z⇒ z ∈ (F1×F2×F3×· · ·×Fm).Moreover, we obtain from (iv) thatz ∈ Z⇔ z ∈ (∂f)−1(y) ∩ (x + ∆)⇒ z ∈ (∂f)−1(y) and z ∈ x + ∆.Combining these together, we obtain thatz ∈ (F1 × F2 × F3 × · · · × Fm) ∩ (x + ∆).Hence, from Theorem 7.35 we haveZ ⊆ (F1 × F2 × F3 × · · · × Fm) ∩ (x + ∆)=(F1 × F2 × F3 × · · · × Fm) ∩ (− 12y−Ty−N∆⊥(y))=(F1 × F2 × F3 × · · · × Fm) ∩ (Id−R)−1(− y),from Theorem 7.7.Lemma 12.6. Let x be the unique solution of0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x), and x ∈∆⊥.Define the parallel sum of ∂f and ∂ι∆ by(∀w ∈H) (∂f∂ι∆)(w) = ⋃w=u+v∂f(u)+ ∂ι∆(v). (12.8)Then the following are equivalent:20612.5. The relationship among the classical cycles, generalized cycles and the generalized gap vectors(i) Z 6= ∅.(ii) cl(fι∆)(x)=(fι∆)(x)and fι∆ is exact at x.(iii)(∂f∂ι∆)(x) 6= ∅.Proof. (i)⇒(ii): Z 6= ∅, then 0 ∈ ∂f(u) + (Id − R)(u) has a solution by(11.9) and Proposition 11.4. Then it follows from Theorem 12.4(i) fι∆ islower semicontinuous at x, i.e.,cl(fι∆)(x)=(fι∆)(x).Moreover, fι∆ is exact at x = u + v, i.e.,(fι∆)(x) = f(u)+ ι∆(v).(ii)⇒(i): Since cl(fι∆)(x) = (fι∆)(x) and fι∆ is exact at x, then itfollows from Theorem 12.3 that the generalized cycles (12.2) can be simpli-fied to the classical cycles (11.8). Hence, ∂f(u)+(Id−R)(u) has a solution.Then Z 6= ∅ by (11.9) and Proposition 11.4.(ii)⇒(iii): Note that x solves (12.2) implies that∂[cl(fι∆)](x) 6= ∅. (12.9)Since cl(fι∆)(x)=(fι∆)(x), then∂[cl(fι∆)](x)= ∂(fι∆)(x). (12.10)Since fι∆ is exact at x, then there exists u ∈ H and v ∈ H such thatx = u + v and∂(fι∆)(x)= ∂f(u) ∩ ∂ι∆(v), (12.11)by Lemma 8.12(i). Combining (12.8), (12.9), (12.10), and (12.11) gives(∂f∂ι∆)(x)= ∂f(u) ∩ ∂ι∆(v), (From (12.8))= ∂(fι∆)(x)(From (12.11))= ∂[cl(fι∆)](x)(From (12.10))6= ∅ (From (12.9)).(iii)⇒(ii): [∂f∂ι∆](x) 6= ∅ implies that there exist u ∈ H and v ∈ Hsuch that x = u + v and∂f(u) ∩ ∂ι∆(v) 6= ∅,by (12.8). Then fι∆ is exact at x and ∂(fι∆)(x)= ∂f(u)∩∂ι∆(v) 6= ∅by Lemma 8.12(ii). Hence, fι∆ is lower semicontinuous at x by Lemma 8.11(i).20712.6. More on generalized cycles12.6 More on generalized cyclesLemma 12.7. Consider (12.2):0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x), (12.12)without the restriction x ∈∆⊥. Then the Attouch-The´ra dual of (12.12) is0 ∈ ∂(f∗ + ι∗∆)(y)+ 12y + Ty.Proof. The Attouch-The´ra dual of (12.12) for the pair(∂[cl(fι∆)], Id−R)is0 ∈ ∂(f∗ + ι∗∆)(y) + (Id−R)−1(y). (12.13)Moreover, (Id −R)−1 = 12Id + T + N∆⊥ by Theorem 7.7 and ι∗∆ = ι∆⊥ ,(12.13) becomes0 ∈ ∂(f∗ + ι∆⊥)(y) +y2 + Ty +N∆⊥(y) (12.14)= ∂(f∗ + ι∆⊥)(y) +N∆⊥(y) +y2 + Ty. (12.15)Because dom(f∗+ ι∆⊥) = dom f∗∩∆⊥ ⊆∆⊥, N∆⊥(y) = ∆ if y ∈∆⊥ and∅ if y 6∈∆⊥, and 0 ∈∆, we have∂(f∗ + ι∆⊥)(y) +N∆⊥(y) = ∂(f∗ + ι∆⊥)(y). (12.16)Hence combining (12.15) and (12.16), we obtain0 ∈ ∂(f∗ + ι∆⊥)(y) +12y + Ty.Theorem 12.8. Let x be given in Theorem 12.2. Then(i) E = x + ∆. Consequently, the problem (12.12) always has infinitelymany solutions.(ii) P∆⊥(E) = {x}.Proof. (i): ”⊆”: Let d ∈∆. From Theorem 7.1(ii), we have ker (Id−R) =∆. Hence, (Id−R)(x + d) = (Id−R)(x),20812.7. Finding the difference vectors by forward-backwardand by using Lemma 8.10(iv), we obtain∂[cl(fι∆)](x + d)= ∂[cl(fι∆)](x).This implies that x + d ∈ E. Hencex + ∆ ⊆ E. (12.17)”⊇”: Let x˜ ∈ E. By Theorem 12.2 and Lemma 12.7, the uniqueness of thesolution to (11.17) by Theorem 11.12(i) implies that y = x−Rx = x˜−Rx˜,from which x˜− x ∈ ker (Id−R) = ∆. Hence, x˜ ∈ x + ∆. Therefore,E ⊆ x + ∆. (12.18)Combing (12.17) and (12.18) givesE = x + ∆.(ii): It is clear from (i).12.7 Finding the difference vectors byforward-backwardTheorem 12.9. Recall from (12.2) that the Attouch-The´ra pimal problemis0 ∈ ∂[cl(fι∆)](x)+ (Id−R)(x),and from (11.17) that the Attouch-The´ra dual problem is0 ∈ ∂(f∗ + ι∗∆)(y)+ 12y + Ty.Let γ ∈ ]0, 1[, let x0 ∈H, and generate a sequence (xn)n∈N viaxn+1 = Proxγcl(fι∆)(xn − γ(Id−R)xn). (12.19)Then the following hold:(i) We havexn+1 = Proxγcl(fι∆)((1− γ)xn + γRxn). (12.20)(ii) xn ⇀ x.20912.7. Finding the difference vectors by forward-backward(iii) Rxn − xn → y.(iv) P∆⊥ xn → P∆⊥ x.Proof. Set A := ∂[cl(fι∆)]and B =(Id − R). Then B is β := (1/2)-cocoercive by Lemma 7.13(viii) and γ ∈ ]0, 1[ = ]0, 2β[. Set δ := 2 −γ/(2β) = 2 − γ > 1 and λ := λn = 1. Then λn(δ − λn)= δ − 1 > 0 and∑n∈N λn(δ − λn)= +∞. Next applying Fact 3.59 givesxn+1 = xn + λn(JγA(xn − γBxn)− xn) (12.21)= xn + Jγ∂cl(fι∆)(xn − γ(Id−R)xn)− xn (12.22)= Jγ∂cl(fι∆)(xn − γ(Id−R)xn) (12.23)= Proxγcl(fι∆)(xn − γ(Id−R)xn), (12.24)by Fact 3.35(ii).(i): From (12.24), we havexn+1 = Proxγcl(fι∆)(xn − γ(Id−R)xn)= Proxγcl(fι∆)((1− γ)xn + γRxn).(ii): It follows from Theorem 12.2(ii) that zer(A + B)= {x} and fromFact 3.59(i) that xn will converges weakly to a point in zer(A + B). Hence,xn ⇀ x.(iii): It follows from (ii) that xn ⇀ x. Next, Fact 3.59(ii) givesBxn =(Id−R)xn →(Id−R)x= x−Rx= −y,by Theorem 12.5(ii). Note that our dual problem differs from the one inFact 3.59 by a negative sign. Hence we have−Bxn = Rxn − xn → Rx− x = y.(iv): Note that(12Id + T)−1= Id −R + 2 P∆ by Lemma 7.13(v), that it21012.7. Finding the difference vectors by forward-backwardis linear. Moreover, ker T = ∆ by Theorem 7.6(v). ThenId =(12Id + T)(12Id + T)−1=(12Id + T)(Id−R + 2 P∆)(From Lemma 7.13(v) )=(12Id + T)(Id−R)+ 2(12Id + T)P∆=(12Id + T)(Id−R)+ P∆ +2T P∆=(12Id + T)(Id−R)+ P∆ +0 (ker T = ∆, by Theorem 7.6(v))=(12Id + T)(Id−R)+ P∆ .Hence,P∆⊥ = Id− P∆ (12.25)=(12Id + T)(Id−R). (12.26)By using (iii) and (12.26), we obtainP∆⊥(xn)=(12Id + T)(xn −Rxn)→ (12Id + T)(− y) = x,by Theorem 12.5(iii).Theorem 12.10. Let γ ∈ ]0, 1[. Recall from Lemma 11.5 thatxn+1 = Proxγf((1− γ)xn + γRxn). (12.27)Then the following hold:Case 1: If Z 6= ∅, then(i) xn ⇀ x for some x ∈ Z.(ii) Rxn − xn → y.Case 2: If Z = ∅, then ‖xn‖ → +∞.Proof. Set A := ∂f and B =(Id−R). Then B is β := (1/2)-cocoercive byLemma 7.13(viii) and γ ∈ ]0, 1[ = ]0, 2β[. Set δ := 2 − γ/(2β) = 2 − γ > 1and λ := λn = 1. Then λn(δ−λn)= δ−1 > 0 and ∑n∈N λn(δ−λn) = +∞.Next applying Fact 3.59 givesxn+1 = Proxγf((1− γ)xn + γRxn).21112.8. ExampleIf the Z 6= ∅, then(i): It follow from Proposition 9.22(i)&(ii) that zer(A+B)= {x} and fromFact 3.59(i) that xn will converges weakly to a point in zer(A + B). Hence,xn ⇀ x.(ii): It follows from (i) that xn ⇀ x. Next, Fact 3.59(ii) givesBxn =(Id−R)xn →(Id−R)x= x−Rx= −y,by Theorem 12.5(ii). Note that our dual problem differs from the one inFact 3.59 by a negative sign. Hence we have−Bxn = Rxn − xn → Rx− x = y.If Z = ∅, then by Fact 3.12, we have‖xn‖ → +∞.12.8 ExampleExample 12.11. Given m sets in H: Ci = {ai+ tibi | ti ∈ R} where ai ∈ Hand bi ∈ H \ {0} for i = 1, . . . ,m. Then the following hold:(i) ιC always has a classical cycle, i.e., Z 6= ∅.(ii) ιC has a unique classical cycle if and only if the set of vectors {bi | i =1, 2, . . . ,m} is not parallel.(iii) ιC has infinitely many classical cycles if and only if the set of vectors{bi | i = 1, 2, . . . ,m} is parallel.Proof. (i): It is well known that if we take the cartesian product of subspacesand add it to the diagonal subspace, we will get a closed subspace. Thenwe shift it, we will get affine closed subspace. Hence, C + ∆ is closed inH. Then ιCι∆ = ιC+∆ is lower semicontinuous and exact, so Z 6= ∅ by21212.8. ExampleLemma 12.6(ii)&(i).Hence, ιC has a classical cycle.(ii): Write u = (u1, . . . , um). u is a cycle of ιC. Then by (11.8), we have0 ∈ ∂ιC(u) +(Id−R)(u)⇒ (R − Id)(u) ∈ ∂ιC(u),which is equal to(um, u1, . . . , um−1)− (u1, u2, . . . , um) ∈ (∂ιC1 × · · · × ∂ιCm)(u1, . . . , um).Hence, we obtainum − u1 ∈ ∂ιC1(u1), u1 − u2 ∈ ∂ιC2(u2), . . . , um−1 − um ∈ ∂ιCm(um).As ∂ιCi(ui) = {x ∈ H | 〈bi, x〉 = 0} for ui ∈ Ci, we have〈b1, u1 − um〉 = 0, 〈b2, u2 − u1〉 = 0, . . . , 〈bm, um − um−1〉 = 0.As ui = ai + tibi, the above equations become〈b1, a1 − am〉+ 〈b1, b1〉 t1 − 〈b1, bm〉 tm = 0 (12.28)...〈bi, ai − ai−1〉 − 〈bi, bi−1〉 ti−1 + 〈bi, bi〉 ti = 0 (12.29)...〈bm, am − am−1〉 − 〈bm, bm−1〉 tm−1 + 〈bm, bm〉 tm = 0. (12.30)In the form of At = b with t = (t1, . . . , tm)ᵀ andb = (〈b1, a1 − am〉 , 〈b2, a2 − a1〉 , . . . , 〈bm, am − am−1〉)ᵀ, the linear systemof equations (12.28)–(12.30) has A =〈b1, b1〉 0 0 · · · 0 −〈b1, bm〉− 〈b2, b1〉 〈b2, b2〉 0 · · · 0 00 −〈b3, b2〉 . . . · · · 0 00 0 . . . · · · 0 00 0 −〈bi, bi−1〉 〈bi, bi〉 0 0...... . . . . . . . . . 00 0 · · · 0 −〈bm, bm−1〉 〈bm, bm〉,and is consistent by (i). Using co-factor expansions, we obtain the determi-nantdetA =m∏i=1〈bi, bi〉 − 〈b1, bm〉m∏i=2〈bi, bi−1〉=m∏i=1〈bi, bi〉(1− 〈b1, bm〉‖b1‖‖bm‖m∏i=2〈bi, bi−1〉‖bi‖‖bi−1‖).21312.8. ExampleThen detA = 0 if and only if {bi/‖bi‖|i = 1, . . . ,m} is a collection of parallelvectors. Because At = b has a unique solution if and only if detA 6= 0, theresult follows.(iii): This follows from (ii).214Chapter 13On the Approximation Curve13.1 OverviewMoudafi and The´ra [41] proposed two splitting methods for the problemof finding a zero of two maximally monotone operator with a not necessarilymaximally monotone sum. The first one is based on a regularization of oneof the operators. Instead of considering the sum problemfind x ∈ (A+B)−1(0), (13.1)they define the regularized versionfind x ∈ ( αA+B)−1(0), (13.2)whereαA = 1α(x− JαA).Proposition 13.1. Define:0 ∈ αA(x) +B(x). (13.3)Thenx solves (13.3) ⇔ x ∈ Fix JαBJαA. (13.4)Proof. x solves (13.3) ⇔ 0 ∈ αAx+Bx⇔ 0 ∈ 1α(x−JαAx)+Bx⇔ 0 ∈ x−JαAx+αBx⇔ JαAx ∈(Id +αB)x⇔ x = JαBJαAx⇔ x ∈ Fix JαBJαA.We summarize our main results in this chapter as follows:• We derive formulas for JγNC , JγN−1C ,γNC, and γ(NC)−1, where γ ∈R++ (see Theorem 13.2).• We prove that the composition JγM−1JγN−1C is a Banach contractionwith constant (2/(2 + γ)) < 1 (see Lemma 13.3).21513.2. Useful properties and the composition JγM−1JγN−1C• When m = 2, we derive a formula for the fixed point xγ of JγM−1JγN−1C(see Section 13.3). We also prove that (xγ) belongs to ∆⊥ (see Lemma13.4 for more details).• When m = 2, we prove that every weak cluster point of (xγ) coincideswith PC1−C2(0) (see Theorem 13.5(ii)).• When m = 2 and both sets C1 and C2 are bounded, we prove that(xγ) converges weakly to the gap vector v = PC1−C2(0), where γ → 0+(see Theorem 13.5(iii)).• When m = 3, we derive a formula for the fixed point xγ of JγM−1JγN−1C(see Section 13.5).• We illustrate our results by giving two examples. The first one iswhen m = 2 and both the two sets are intervals (see Lemma 13.6).The second one is when m = 3 one set is interval and the other twosets are singleton (see Lemma 13.9).To solve (13.3), Moudafi and The´ra suggest to iteratexn+1 = JγnBJγnA, (13.5)where γn → 0+. Under certain assumptions, they prove that(xn)convergesto a zero the extended sum of A and B. All this suggests we should computeJγA and γA for γ > 0 and A taken from M, NC, M−1 and N−1C . Chapter 7covers the formulas and properties of M, M−1, JγM, JγM−1 , γM, and γM−1for γ > 0. In the next section we will collect some formulas and propertiesof JγNC , γNC, γ(N−1C), and JγN−1C . Recall from (8.3) thatC := C1 × C2 × · · · × Cm,and from (7.2) thatM = Id−R,13.2 Useful properties and the compositionJγM−1JγN−1CTheorem 13.2. Let γ > 0. Then the following hold:(i) JγNC = PC.21613.2. Useful properties and the composition JγM−1JγN−1C(ii) γNC = γ−1(Id− JγNC)=(Jγ−1N−1C) ◦ (γ−1Id) = γ−1(Id− PC ).(iii) γ(N−1C)= γ−1(Id− JγN−1C)= Jγ−1NC ◦(γ−1Id)= γ−1 PγC.(iv) JγN−1C = Id− PγC.(v) JγN−1C is firmly nonexpansive.Proof. (i): We have JγNC = JNC = ProxιC = PC by Fact 3.34(ii).(ii): This follows directly from (3.14) and (i) thatγNC = γ−1(Id− JγNC)= γ−1(Id− PC).Moreover, we have(x,u) ∈ gra γNC ⇔ u ∈ γNCx ⇔ x − γu ∈ JγNCx ⇔u ∈ NC(x − γu) ⇔ x ∈ γu + N−1C u ⇔ 1γx ∈ (Id + 1γN−1C )u ⇔ u ∈Jγ−1N−1C(γ−1x). Therefore, we haveγNC =(Jγ−1N−1C) ◦ (γ−1Id).(iii): By doing the same process as (ii), we obtainγ(N−1C ) = Jγ−1NC ◦ (γ−1Id) = JNC ◦ (γ−1Id).Moreover, from (3.14) and (ii)γ(N−1C ) = γ−1(Id− JγN−1C ) = γ−1 PγC .(iv): By using (iii), we haveId = JγN−1C + γ(γN−1C)= JγN−1C + γJγ−1NC ◦(γ−1Id)= JγN−1C + γJNC(γ−1Id).Hence, (i) givesJγN−1C= Id− γJNC(γ−1Id)= Id− γ PC(γ−1Id)= Id− PγC . (13.6)Hence, (3.14) and (13.6) giveγ(N−1C ) = γ−1(Id− (Id− PγC )) = γ−1 PγC .(v): Clear.21713.2. Useful properties and the composition JγM−1JγN−1CLemma 13.3. Recall from (7.2) thatM = Id−R,and from Theorem 13.2(iv) thatJγN−1C= γ(Id− PC)(γ−1Id).Then the compositionT := JγM−1JγN−1C = JγM−1(γ(Id− PC)(γ−1Id))is a Banach contraction with constant α = 2/(2 + γ) < 1 and thus has aunique fixed point x¯.Proof. Set x˜ := γ((γ−1x)− PC (γ−1x)) and y˜ := γ((γ−1y)− PC (γ−1y)).Take u = T(x)and v = T(y). Then u = T(x) ⇔ u ∈ JγM−1JγN−1C (x) ⇔u ∈ JγM−1(γ((γ−1x) − PC (γ−1x))) ⇔ γ((γ−1x) − PC (γ−1x)) ∈ u +γM−1u⇔ γ((γ−1x)− PC (γ−1x))− u ∈ γM−1u. Hence,x˜− u ∈ γM−1u. (13.7)Similarly, v = T(y)givesy˜− v ∈ γM−1v. (13.8)From Lemma 7.8 (iv) we obtain〈(x˜− u)− (y˜− v),u− v〉 ≥ γ2 ‖u− v‖2 ⇔〈x˜− y˜,u− v〉 − ‖u− v‖2 ≥ γ2 ‖u− v‖2 . This is equivalent to〈x˜− y˜,u− v〉 ≥ γ + 22 ‖u− v‖2 .By using Cauchy-Schwarz inequality we have‖x˜− y˜‖ ‖u− v‖ ≥ 〈x˜− y˜,u− v〉 ≥ γ + 22 ‖u− v‖2 .Hence, we obtain‖x˜− y˜‖ ≥ γ + 22 ‖u− v‖ . (13.9)This leads to‖x˜− y˜‖2 ≥ (γ + 2)24 ‖u− v‖2 .21813.2. Useful properties and the composition JγM−1JγN−1CThenγ2∥∥∥(Id− PC)(γ−1x)− (Id− PC)(γ−1y)∥∥∥2 ≥ (γ + 2)24 ‖u− v‖2 , (13.10)and by using Theorem 13.2(v), we obtainγ2∥∥(Id− PC)(γ−1x)− (Id− PC)(γ−1y)∥∥2 ≤ γ2 ∥∥γ−1x− γ−1y∥∥2− γ2 ∥∥PC(γ−1x)− PC(γ−1y)∥∥2 ,hence,γ2∥∥∥(Id− PC)(γ−1x)− (Id− PC)(γ−1y)∥∥∥2 ≤ γ2 ∥∥∥γ−1x− γ−1y∥∥∥2 . (13.11)substituting (13.11) back into (13.10), we have(γ + 2)24 ‖u− v‖2 ≤ γ2 ∥∥γ−1x− γ−1y∥∥2 .Hence‖T(x)−T(y)‖ ≤ 22 + γ ‖x− y‖ . (13.12)Therefore, it has a unique fixed point.For fixed γ, we iterateT = JγM−1JγN−1C . (13.13)Comparing (13.13) with (13.5), A = N−1C and B = M−1. Here A is theregularized one and B is untouched one. Let xγ be the unique fixed pointof γ, we would like to answer the following questions− What is xγ?− What happens if γ → 0+?− Is there any monotonicity for ‖xγ‖ ?− Is xγ bounded as γ varies?− When m = 2, does xγ converge to PC2−C1(0)as γ → 0+ ?Lemma 13.4. Recall thatT = JγM−1JγN−1C = JγM−1(γ(Id− PC)(γ−1Id)).Then the following hold:21913.3. Finding the approximation curve for m = 2(i) ran T ⊆∆⊥.(ii) xγ ∈∆⊥.Proof. (i): We haveran JγM−1 = dom γM−1 = dom M−1 = ran M = ∆⊥,by Theorem 7.1(v).(ii): It is clear from (i).13.3 Finding the approximation curve for m = 2It turns out that (thinking of block matrices)JγM−1 =22 + γ P∆⊥ =12 + γ(Id −Id−Id Id). (13.14)Hence,JγM−1 : H2 → H2 :(x1, x2) 7→ (1/(2 + γ))(x1 − x2, x2 − x1). (13.15)andJγN−1C: H2 → H2:(x1, x2) 7→ (x1 − γ PC1(x1/γ), x2 − γ PC2(x2/γ)). (13.16)Then from (13.15) and (13.16) the composition operator isJγM−1JγN−1C:(x1x2)7→ 12 + γ((x1 − γ PC1(x1/γ))− (x2 − γ PC2(x2/γ))(x2 − γ PC2(x2/γ))− (x1 − γ PC1(x1/γ)))(13.17)Clearly,x1 = −x2.Therefore, we are going to focus on the first equation. Write xγ instead of x1to indicate the dependency on γ. Then because there is only one parameter,the set of fixed points is a curve xγ . Thenxγ =12 + γ(xγ − γ PC1(xγ/γ)−(− xγ − γ PC2(−xγ/γ)))= 12 + γ(2xγ − γ PC1(xγ/γ) + PC2(−xγ/γ)).22013.3. Finding the approximation curve for m = 2This gives(2 + γ)xγ − 2xγ = γ PC2(−xγ/γ)− γ PC1(xγ/γ).Hence,xγ = PC2(−xγ/γ)− PC1(xγ/γ). (13.18)and by using the dilation formula we obtainxγ = −γ−1(PγC1(xγ) + P−γC2(xγ)). (13.19)Theorem 13.5. The following hold:(i) Let c ∈ H. Thenmaxx∈H〈c− x, x〉 = 14 ‖c‖2 . (13.20)(ii) Every weak cluster point of(xγ)coincides with PC2−C1(0).(iii) If C1 and C2 are both bounded, then (xγ) converges weakly tov = PC2−C1(0),when γ → 0+.Proof. (i): Set f(x) := 〈c− x, x〉 = 〈c, x〉− ‖x‖2 and hence ∇f(x) = c− 2x.Setting this to 0 and solve for x gives the unique maximum value x = c2 .Therefore,maxx∈H〈c− x, x〉 =〈c− c2 ,c2〉= 12 ‖c‖2 − 14 ‖c‖2 = 14 ‖c‖2 .(ii): Set p2 = PC2(−xγ/γ) and p1 = PC1(xγ/γ). Clearly, p1 and p2 dependon γ and from (13.18), we havexγ = p2 − p1 ∈ C2 − C1. (13.21)By the projection theorem, we have〈c1 − p1, (xγ/γ)− p1〉 ≤ 0(∀c1 ∈ C1). (13.22)and〈c2 − p2, (−xγ/γ)− p2〉 ≤ 0(∀c2 ∈ C2). (13.23)22113.4. Examples on m = 2 caseAdding (13.22) and (13.23), gives that for all c1 ∈ C1 and c2 ∈ C2;〈(c2 − c1)− (p2 − p1),−(xγ/γ)〉 ≤ 〈c2 − p2, p2〉+ 〈c1 − p1, p1〉 . (13.24)Moreover, by (13.21) and Theorem 13.5 (i), we obtain〈(c2 − c1)− (p2 − p1), 0− (xγ/γ)〉 ≤ 〈c2 − p2, p2〉+ 〈c1 − p1, p1〉⇔ 1γ〈(c2 − c1)− xγ , 0− xγ〉 ≤ 〈c2 − p2, p2〉+ 〈c1 − p1, p1〉 (by (13.21))⇒ 1γ〈(c2 − c1)− xγ , 0− xγ〉 ≤ 14 ‖c1‖2 + 14 ‖c2‖2 (by Theorem 13.5 (i)),multiply both side by γ and simplify, we obtain‖xγ‖2 − 〈c2 − c1, xγ〉 =〈(c2 − c1)− xγ , 0− xγ〉 ≤ γ4 ( ‖c1‖2 + ‖c2‖2 ).It is known that ‖·‖2 is convex and lower semi-continuous ⇔ ‖·‖2 is convexand weakly lower semi-continuous. Assume that x¯ is a weak cluster pointof the curve(xγ), as γ → 0+. Then it follows that‖x¯‖2 − 〈c2 − c1, x¯〉 ≤ lim infγ→0+( ‖xγ‖2 − 〈c2 − c1, xγ〉 )≤ lim infγ→0+γ4( ‖c1‖2 + ‖c2‖2 )≤ 0,therefore, 〈(c2 − c1)− x¯, 0− x¯〉 ≤ 0.Moreover, x¯ ∈ C2 − C1 because xγ ∈ C2−C1. Hence, by projection theoremwe havex¯ = PC2−C1(0).(iii): It follows from (ii).13.4 Examples on m = 2 case13.4.1 Two intervalsLemma 13.6. Let H = R, C1 = [a, b] and C2 = [c, d], where a, b, c, d ∈ Rand a < b < c < d. Then the following hold;22213.4. Examples on m = 2 case(i) If d is negative, thenxγ =d− b, if and only if γ ∈ ( bd − 1,∞);−γ(1+γ)b, if and only if γ ∈ [ bc − 1, bd − 1];c− b, if and only if γ ∈ (0, bc − 1).(ii) If a is positive, thenxγ =c− a, if and only if γ ∈ ( ca − 1,∞);γ(1+γ)c, if and only if γ ∈ [ cb − 1, ca − 1];c− b, if and only if γ ∈ (0, cb − 1).(iii) If c is negative and d is not negative, thenxγ =−γ(1+γ)b, if and only if γ ∈[bc − 1,∞);c− b, if and only if γ ∈ (0, bc − 1).(iv) If b is negative and c is not negative, thenxγ = c− b if and only if γ ∈(0,∞).(v) If a is negative and b is not negative, thenxγ =−γ(1+γ)c, if and only if γ ∈[cb − 1,∞);c− b, if and only if γ ∈ (0, cb − 1).Moreover, xγ is continuous curve and when γ → 0+,xγ → v := PC2−C1(0)=c− b.Proof. Clearly,PγC1(xγ) =γa, if xγ < γaxγ , if γa ≤ xγ ≤ γbγb, if xγ > γb(13.25)(i): Since d is negative, then a < b < c < d < 0, then we only have threepossibilities:possibility 1: If PγC1(xγ)= xγ , then P−γC2(xγ) = −γd.22313.4. Examples on m = 2 casepossibility 2: If PγC1(xγ)= γa, then P−γC2(xγ) = −γd.possibility 3: If PγC1(xγ)= γb, then we obtain 3 cases of P−γC2(xγ);P−γC2(xγ) =−γd, if xγ < −γdxγ , if − γd ≤ xγ ≤ −γc−γc, if xγ > −γc(13.26)”⇒”: Using (13.19), gives the following:(a) If xγ satisfies possibility 1, thenxγ =−1γ(xγ − γd)⇒ xγ = γ(1 + γ)d.But xγ = γ1+γd > γd > γb for all γ. This is contradicting the assumptionγa ≤ xγ ≤ γb.(b) If xγ satisfies possibility 2, thenxγ =−1γ(γa− γd)⇒ xγ = d− a.Obviously, xγ = d−a > 0 > γa for all γ. Therefore, this case is contradictingthe assumption xγ < γa.(c) If xγ satisfies possibility 3, thenxγ = d− b,xγ =−γ(1 + γ)b,orxγ = c− b.Case 1: Let xγ = d− b. Clearly, xγ = d− b > 0 > γb for all γ. Moreover,xγ < −γd⇒ d− b < −γd⇒ 1− bd < −γ ⇒ γ > bd − 1. Hence,if xγ = d− b, then γ ∈( bd− 1,∞). (13.27)22413.4. Examples on m = 2 caseCase 2: If xγ = −γ(1+γ)b. It is obvious that−γ(1+γ)b > γb, for all γ. Moreover,− γd ≤ xγ ≤ −γc⇔ −γd ≤ −γ(1 + γ)b ≤ −γc⇔ d ≥ 1(1 + γ)b ≥ c⇔ bc− 1 ≤ γ ≤ bd− 1.Therefore,if xγ =−γ(1 + γ)b, then γ ∈ [bc− 1, bd− 1]. (13.28)Case 3: If xγ = c− b. Clearly, c− b > 0 > γb, for all γ andxγ > −γc⇔ c− b > −γc⇔ bc− 1 > γ.Since γ > 0⇒ bc > 1, then we obtainif xγ = c− b, then γ ∈(0, bc− 1). (13.29)”⇐”: Actually, the three cases we considered covered all the possible valueof γ, so we have three intervals:Interval 1:(bd − 1,∞).Interval 2: [ bc − 1, bd − 1].Interval 3:(0, bc − 1).Suppose γ ∈ Interval 3. If xγ is not in case 3, i.e., it would be in case 1 orcase 2, then γ would be in interval 1 or 2, which is contradiction becauseintervals do not intersect. Hence,if γ ∈ (0, bc− 1), then xγ = c− b. (13.30)Combining (13.29) and (13.30), we havexγ = c− b if and only if γ ∈(0, bc− 1).22513.4. Examples on m = 2 caseUsing the same argument can show thatxγ =−γ(1 + γ)b if and only if γ ∈ [bc− 1, bd− 1],andxγ = d− b if and only if γ ∈( bd− 1,∞).Next, we will show the continuity of the curve:limγ→((b/c)−1)+xγ =−((b/c)− 1)b1 +((b/c)− 1)=b(1− (b/c))(b/c)= b(1− (b/c))(c/b).Hence,limγ→((b/c)−1)+xγ = c− b. (13.31)Also,limγ→((b/c)−1)−xγ =−((b/c)− 1)b1 +((b/c)− 1)=b(1− (b/c))(b/c)= b(1− (b/c))(c/b).Therefore,limγ→((b/c)−1)−xγ = c− b. (13.32)So we obtainlimγ→((b/c)−1)+xγ = x(b/c)−1 = limγ→((b/c)−1)−xγ ,which proves that the curve is continuous at the point (b/c) − 1. Next, weneed to show the continuity of the curve at (b/d)− 1.limγ→((b/d)−1)+xγ =−((b/d)− 1)b1 +((b/d)− 1)=b(1− (b/d))(b/d)= b(1− (b/d))(d/b).22613.4. Examples on m = 2 caseHence,limγ→((b/d)−1)+xγ = d− b. (13.33)Also,limγ→((b/d)−1)−xγ =−((b/d)− 1)b1 +((b/d)− 1)=b(1− (b/d))(b/d)= b(1− (b/d))(d/b).Therefore,limγ→((b/d)−1)−xγ = d− b. (13.34)So we obtainlimγ→((b/d)−1)+xγ = x(b/d)−1 = limγ→((b/d)−1)−xγ ,which proves that the curve is continuous at the point (b/d)− 1.(ii): The proof is similar to (i).(iii): Since c is negative, then a < b < c < 0 and d ≥ 0, then we only havetwo cases:Case 1: γC1 ∩ −γC2 = ∅. This could happen when1. a < b < c < −d < 0 ≤ d.2. a < b < −d < c < 0 ≤ d.3. a < b < −d = c < 0 ≤ d.We will prove the first subcase and the proof of other subcases will be thesame. Let a < b < c < −d < 0 ≤ d. Because of (13.25), we have thefollowing possibilities:Possibility 1: If PγC1(xγ)= xγ , then P−γC2(xγ) = −γd.Possibility 2: If PγC1(xγ)= γa, then P−γC2(xγ) = −γd.Possibility 3: If PγC1(xγ)= γb, then we obtain 3 cases of P−γC2(xγ);P−γC2(xγ) =−γd, if xγ < −γdxγ , if − γd ≤ xγ ≤ −γc−γc, if xγ > −γc(13.35)”⇒”: Using (13.19), gives the following:22713.4. Examples on m = 2 case(a) If xγ satisfies possibility 1, thenxγ =−1γ(xγ − γd)⇒ xγ = γ(1 + γ)d.Hence, xγ = γ1+γd > 0 > γb. However, this contradicts the assumptionγa ≤ xγ ≤ γb.(b) If xγ satisfies possibility 2, thenxγ =−1γ(γa− γd)⇒ xγ = d− a.Obviously, xγ = d − a > 0 > γa. Therefore, this case is contradicting theassumption xγ < γa.(c) If xγ satisfies possibility 3, thenxγ = d− b,xγ =−γ(1 + γ)b,orxγ = c− b.Let xγ = d− b. Clearly,xγ = d− b > 0. (13.36)Therefore xγ > γb. But (13.36) contradicts the assumption xγ < −γd < 0.Next, let xγ = −γ(1+γ)b. Obviously,xγ =−γ(1 + γ)b > 0. (13.37)Therefore, xγ > γb. Moreover,xγ ≤ −γc⇔ −γ(1 + γ)b ≤ −γc⇔ 1(1 + γ)b ≥ c⇔ bc− 1 ≤ γ.22813.4. Examples on m = 2 caseTherefore,if xγ =−γ(1 + γ)b, then γ ∈[bc− 1,+∞). (13.38)Finally, let xγ = c− b. Clearly, c− b > 0 > γb andxγ > −γc⇔ c− b > −γc⇔ bc− 1 > γ.Since γ > 0 and bc > 1, then we obtainif xγ = c− b, then γ ∈(0, bc− 1). (13.39)”⇐”: Actually, the cases we considered covered all the possible value of γ,so we have two intervals:Interval 1:[bc − 1,+∞).Interval 2:(0, bc − 1).Suppose γ ∈ Interval 2. If xγ 6= c− b , i.e., xγ = −γ(1+γ)b. Then γ would be ininterval 1, which is contradiction because intervals do not intersect. Hence,if γ ∈ (0, bc− 1), then xγ = c− b. (13.40)Combining (13.39) and (13.40), we havexγ = c− b if and only if γ ∈(0, bc− 1).Using the same argument can show thatxγ =−γ(1 + γ)b if and only if γ ∈[bc− 1,+∞).Case 2: γC1 ∩ −γC2 6= ∅. This could happen when1. a < −d < b < c < 0 ≤ d.2. a < b = −d < c < 0 ≤ d.3. a = −d < b < −d = c < 0 ≤ d.22913.4. Examples on m = 2 case4. −d < a < b < c < 0 ≤ d.We will prove 1 and 4. The other subcases will be the same. Leta < −d < b < c < 0 ≤ d.Because of (13.25), we have the following possibilities:Possibility 1: If PγC1(xγ)= xγ , thenP−γC2(xγ) ={−γd, if xγ < −γdxγ , if − γd ≤ xγ ≤ −γc(13.41)Possibility 2: If PγC1(xγ)= γa, then P−γC2(xγ) = −γd.Possibility 3: If PγC1(xγ)= γb, then we obtain 2 cases of P−γC2(xγ);P−γC2(xγ) ={xγ , if − γd ≤ xγ ≤ −γc−γc, if xγ > −γc(13.42)”⇒”: Using (13.19), gives the following:(a) If xγ satisfies possibility 1, thenxγ =γ(1 + γ)d > 0,orxγ = 0.Let xγ = γ(1+γ)d > 0. Hence, xγ > 0 > γb. However, this contradicts theassumption γa ≤ xγ ≤ γb. Moreover, if xγ = 0, then xγ = 0 > γb. Thiscontradicts the assumption γa ≤ xγ ≤ γb.(b) If xγ satisfies possibility 2, thenxγ =−1γ(γa− γd)⇒ xγ = d− a.Obviously, xγ = d − a > 0 > γa. Therefore, this case is contradicting theassumption xγ < γa.(c) If xγ satisfies possibility 3, thenxγ =−γ(1 + γ)b,orxγ = c− b.23013.4. Examples on m = 2 caseLet xγ = −γ(1+γ)b. Obviously,xγ =−γ(1 + γ)b > 0. (13.43)Therefore, xγ > γb. Moreover,xγ ≤ −γc⇔ −γ(1 + γ)b ≤ −γc⇔ 1(1 + γ)b ≥ c⇔ bc− 1 ≤ γ.Therefore,if xγ =−γ(1 + γ)b, then γ ∈[bc− 1,+∞). (13.44)Finally, let xγ = c− b. Clearly, c− b > 0 > γb andxγ > −γc⇔ c− b > −γc⇔ bc− 1 > γ.Since γ > 0 and bc > 1, then we obtainif xγ = c− b, then γ ∈(0, bc− 1). (13.45)”⇐”: Actually, the cases we considered covered all the possible value of γ,so we have two intervals:Interval 1:[bc − 1,+∞).Interval 2:(0, bc − 1).Suppose γ ∈ Interval 2. If xγ 6= c− b , i.e., xγ = −γ(1+γ)b. Then γ would be ininterval 1, which is contradiction because intervals do not intersect. Hence,if γ ∈ (0, bc− 1), then xγ = c− b. (13.46)Combining (13.39) and (13.40), we havexγ = c− b if and only if γ ∈(0, bc− 1).23113.4. Examples on m = 2 caseUsing the same argument we can show thatxγ =−γ(1 + γ)b if and only if γ ∈[bc− 1,+∞).Now, let−d < a < b < c < 0 ≤ d.Because of (13.25), we have the following possibilities:Possibility 1: If PγC1(xγ)= xγ , then P−γC2(xγ) = xγ .Possibility 2: If PγC1(xγ)= γa, then P−γC2(xγ) = −γd.Possibility 3: If PγC1(xγ)= γb, then we obtain 2 cases of P−γC2(xγ);P−γC2(xγ) ={xγ , if − γd ≤ xγ ≤ −γc−γc, if xγ > −γc(13.47)”⇒”: Using (13.19), gives the following:(a) If xγ satisfies possibility 1, thenxγ = 0.Hence, xγ > γb, which contradicts the assumption γa ≤ xγ ≤ γb.(b) If xγ satisfies possibility 2, thenxγ =−1γ(γa− γd)⇒ xγ = d− a.Obviously, xγ = d − a > 0 > γa. Therefore, this case is contradicting theassumption xγ < γa.(c) If xγ satisfies possibility 3, thenxγ =−γ(1 + γ)b,orxγ = c− b.Let xγ = −γ(1+γ)b. Obviously,xγ =−γ(1 + γ)b > 0. (13.48)23213.4. Examples on m = 2 caseTherefore, xγ > γb. Moreover,xγ ≤ −γc⇔ −γ(1 + γ)b ≤ −γc⇔ 1(1 + γ)b ≥ c⇔ bc− 1 ≤ γ.Therefore,if xγ =−γ(1 + γ)b, then γ ∈[bc− 1,+∞). (13.49)Finally, let xγ = c− b. Clearly, c− b > 0 > γb andxγ > −γc⇔ c− b > −γc⇔ bc− 1 > γ.Since γ > 0 and bc > 1, then we obtainif xγ = c− b, then γ ∈(0, bc− 1). (13.50)”⇐”: Actually, the cases we considered covered all the possible value of γ,so we have two intervals:Interval 1:[bc − 1,+∞).Interval 2:(0, bc − 1).Suppose γ ∈ Interval 2. If xγ 6= c− b , i.e., xγ = −γ(1+γ)b. Then γ would be ininterval 1, which is contradiction because intervals do not intersect. Hence,if γ ∈ (0, bc− 1), then xγ = c− b. (13.51)Combining (13.39) and (13.40), we havexγ = c− b if and only if γ ∈(0, bc− 1).Using the same argument can show thatxγ =−γ(1 + γ)b if and only if γ ∈[bc− 1,+∞).23313.4. Examples on m = 2 caseNext, we will show the continuity of the curve:limγ→((b/c)−1)+xγ =−((b/c)− 1)b1 +((b/c)− 1)=b(1− (b/c))(b/c)= b(1− (b/c))(c/b).Hence,limγ→((b/c)−1)+xγ = c− b. (13.52)Also,limγ→((b/c)−1)−xγ =−((b/c)− 1)b1 +((b/c)− 1)=b(1− (b/c))(b/c)= b(1− (b/c))(c/b).Therefore,limγ→((b/c)−1)−xγ = c− b. (13.53)So we obtainlimγ→((b/c)−1)+xγ = x(b/c)−1 = limγ→((b/c)−1)−xγ .(iv): Since b is negative, then a < b < 0 and c ≥ 0, then we only have twocases:Case 1: γC1 ∩ −γC2 = ∅. This could happen when1. a < b < −d < −c ≤ 0 < d.2. −d < −c < a < b < −c ≤ 0 < d.We will prove the first subcase and the proof of other subcases will be thesame. Let a < b < −d < −c ≤ 0 < d. Because of (13.25), we have thefollowing possibilities:Possibility 1: If PγC1(xγ)= xγ , then P−γC2(xγ) = −γd.23413.4. Examples on m = 2 casePossibility 2: If PγC1(xγ)= γa, then P−γC2(xγ) = −γd.Possibility 3: If PγC1(xγ)= γb, then we obtain 3 cases of P−γC2(xγ);P−γC2(xγ) =−γd, if xγ < −γdxγ , if − γd ≤ xγ ≤ −γc−γc, if xγ > −γc(13.54)”⇒”: Using (13.19), we obtain:(a) If xγ satisfies possibility 1, thenxγ =−1γ(xγ − γd)⇒ xγ = γ(1 + γ)d.Hence, xγ = γ1+γd > 0 > γb. However, this contradicts the assumptionγa ≤ xγ ≤ γb.(b) If xγ satisfies possibility 2, thenxγ =−1γ(γa− γd)⇒ xγ = d− a.Obviously, xγ = d − a > 0 > γa. Therefore, this case is contradicting theassumption xγ < γa.(c) If xγ satisfies possibility 3, thenxγ = d− b,xγ =−γ(1 + γ)b,orxγ = c− b.Let xγ = d− b. Clearly,xγ = d− b > 0. (13.55)Therefore xγ > γb. But (13.55) contradicts the assumption xγ < −γd < 0.Next, let xγ = −γ(1+γ)b. Obviously,xγ =−γ(1 + γ)b > 0. (13.56)23513.4. Examples on m = 2 caseTherefore, xγ > γb. However, xγ > −γc, which contradicts the assumptionxγ ≤ −γc.Finally, let xγ = c− b. Clearly, c− b > 0 > γb andxγ > −γc⇔ c− b > −γc⇔ bc− 1 < γ.Since γ > 0 and bc − 1 < 0, then we obtainif xγ = c− b, then γ ∈ (0,+∞) . (13.57)”⇐”: Actually, the case we considered covered all the possible value of γ, sowe have only one interval: (0,+∞).with xγ = c− b.if γ ∈ (0,+∞), then xγ = c− b. (13.58)Combining (13.57) and (13.58), we havexγ = c− b if and only if γ ∈(0,+∞).Case 2: γC1 ∩ −γC2 6= ∅. This could happen when1. a < −d < b < −c ≤ 0 < d.2. a < b = −d < −c ≤ 0 < d.3. −d < −c = a < b < 0 < d.4. −d < a < −c < b < 0 < d.5. a < −d < −c < b < 0 < c < d.6. a < −d < −c = b < 0 < c < d.7. −d = a < −c < b < 0 < c < d.8. −d < a < b < −c < 0 < c < d.23613.4. Examples on m = 2 caseWe will prove 1, 2 , and 5. The other subcases will be the same. Leta < −d < b < −c ≤ 0 < d.Because of (13.25), we have the following possibilities:Possibility 1: If PγC1(xγ)= xγ , thenP−γC2(xγ) ={xγ , if − γd ≤ xγ ≤ −γc−γd, if xγ < −γd(13.59)Possibility 2: If PγC1(xγ)= γa, then P−γC2(xγ) = −γd.Possibility 3: If PγC1(xγ)= γb, then we obtain 2 cases of P−γC2(xγ);P−γC2(xγ) ={xγ , if − γd ≤ xγ ≤ −γc−γc, if xγ > −γc(13.60)”⇒”: Using (13.19), gives the following:(a) If xγ satisfies possibility 1, thenxγ = 0,orxγ =γ(1 + γ)d > 0.Let vγ = 0. Then xγ > γb, which contradicts the assumption γa ≤ xγ ≤ γb.Moreover, xγ > −γc, which contradicts the assumption −γd ≤ xγ ≤ −γc.Next, let xγ = γ(1+γ)d > 0. Then xγ > γb, which contradicts the assumptionγa ≤ xγ ≤ γb and the assumption xγ < −γd.(b) If xγ satisfies possibility 2, thenxγ =−1γ(γa− γd)⇒ xγ = d− a.Obviously, xγ = d − a > 0 > γa. Therefore, this case is contradicting theassumption xγ < γa.(c) If xγ satisfies possibility 3, thenxγ =−γ(1 + γ)b,orxγ = c− b.23713.4. Examples on m = 2 caseLet xγ = −γ(1+γ)b. Obviously,xγ =−γ(1 + γ)b > 0. (13.61)Therefore, xγ > γb. However, xγ > −γc. This is contradicting the assump-tion xγ ≤ −γc.Finally, let xγ = c− b. Clearly, c− b > 0 > γb andxγ > −γc⇔ c− b > −γc⇔ bc− 1 > γ.Since γ > 0 and bc − 1 < 0, then we obtainif xγ = c− b, then γ ∈ (0,+∞) . (13.62)”⇐”: Actually, the case we considered covered all the possible value of γ, sowe have one interval: (0,+∞),with xγ = c− b. Hence,if γ ∈ (0,+∞), then xγ = c− b. (13.63)Combining (13.62) and (13.63), we havexγ = c− b if and only if γ ∈(0,+∞).Leta < b = −d < −c < 0 < dBecause of (13.25), we have the following possibilities:Possibility 1: If PγC1(xγ)= xγ , then P−γC2(xγ) = −γd.Possibility 2: If PγC1(xγ)= γa, then P−γC2(xγ) = −γd.Possibility 3: If PγC1(xγ)= γb, then we obtain 2 cases of P−γC2(xγ);P−γC2(xγ) ={xγ , if − γd ≤ xγ ≤ −γc−γc, if xγ > −γc(13.64)”⇒”: Using (13.19), gives the following:23813.4. Examples on m = 2 case(a) If xγ satisfies possibility 1, thenxγ =γ(1 + γ)d > 0.Then xγ > γb, which contradicts the assumption γa ≤ xγ ≤ γb and theassumption xγ < −γd.(b) If xγ satisfies possibility 2, thenxγ =−1γ(γa− γd)⇒ xγ = d− a.Obviously, xγ = d − a > 0 > γa. Therefore, this case is contradicting theassumption xγ < γa.(c) If xγ satisfies possibility 3, thenxγ =−γ(1 + γ)b,orxγ = c− b.Let xγ = −γ(1+γ)b. Obviously,xγ =−γ(1 + γ)b > 0. (13.65)Therefore, xγ > γb. However, xγ > −γc. This is contradicting the assump-tion xγ ≤ −γc.Finally, let xγ = c− b. Clearly, c− b > 0 > γb andxγ > −γc⇔ c− b > −γc⇔ bc− 1 > γ.Since γ > 0 and bc − 1 < 0, then we obtainif xγ = c− b, then γ ∈ (0,+∞) . (13.66)”⇐”: Actually, the case we considered covered all the possible value of γ, sowe have one interval: (0,+∞),23913.4. Examples on m = 2 casewith xγ = c− b. Hence,if γ ∈ (0,+∞), then xγ = c− b. (13.67)Combining (13.66) and (13.67), we havexγ = c− b if and only if γ ∈(0,+∞).Leta < −d < −c < b < 0 < c < dBecause of (13.25), we have the following possibilities:Possibility 1: If PγC1(xγ)= xγ , thenP−γC2(xγ) =−γd, if xγ < −γdxγ , if − γd ≤ xγ ≤ −γc−γc, if xγ > −γc(13.68)Possibility 2: If PγC1(xγ)= γa, then P−γC2(xγ) = −γd.Possibility 3: If PγC1(xγ)= γb, then P−γC2(xγ) = −γc.”⇒”: Using (13.19), gives the following:(a) If xγ satisfies possibility 1, thenxγ =γ(1 + γ)d > 0,xγ = 0,orxγ =γ(1 + γ)c > 0Clearly, xγ ≥ 0 > γb, which contradicts the assumption γa ≤ xγ ≤ γb.(b) If xγ satisfies possibility 2, thenxγ =−1γ(γa− γd)⇒ xγ = d− a.Obviously, xγ = d − a > 0 > γa. Therefore, this case is contradicting theassumption xγ < γa.24013.4. Examples on m = 2 case(c) If xγ satisfies possibility 3, thenxγ = c− b.Let xγ = −γ(1+γ)b. Obviously,xγ =−γ(1 + γ)b > 0. (13.69)Clearly, c− b > 0 > γb andxγ > −γc⇔ c− b > −γc⇔ bc− 1 > γ.Since γ > 0 and bc − 1 < 0, then we obtainif xγ = c− b, then γ ∈ (0,+∞) . (13.70)”⇐”: Actually, the case we considered covered all the possible value of γ, sowe have one interval: (0,+∞),with xγ = c− b. Hence,if γ ∈ (0,+∞), then xγ = c− b. (13.71)Combining (13.70) and (13.71), we havexγ = c− b if and only if γ ∈(0,+∞).13.4.2 Linear subspace and single setExample 13.7. Suppose C1 = {c1} and C2 = c2 + L, where L is closedlinear subspace, and c2 ∈ L⊥. Thenxγ =−γ(1 + γ) PL(c1)+ c2 − PL⊥(c1)(13.72)is bounded and (xγ) converges strongly to the gap vector as γ → 0+.24113.4. Examples on m = 2 caseProof. By using the translation formula, we obtainPC2(−xγ/γ) = Pc2+L (−xγ/γ) = c2 +PL ((−xγ/γ)−c2) = c2− 1γ PL (xγ),(13.73)andPC1(xγ/γ)= c1. (13.74)By using (13.73) and (13.74) in (13.18), we havexγ = c2 − c1 − 1γPL(xγ). (13.75)Hence,PL(xγ)= −PL(c1)− 1γPL(xγ)and so PL(xγ)= −γ(1 + γ) PL (c1).(13.76)Also,PL⊥(xγ)= c2 − PL⊥(c1). (13.77)Hence, from (13.76) and (13.77)xγ =−γ(1 + γ) PL (c1)+ c2 − PL⊥ (c1).Moreover,‖xγ‖2 =∥∥∥∥ −γ(1 + γ) PL (c1)+ c2 − PL⊥ (c1)∥∥∥∥2=∥∥∥∥ −γ(1 + γ) PL (c1)∥∥∥∥2 + ∥∥∥∥c2 − PL⊥ (c1)∥∥∥∥2− 2〈−γ(1 + γ) PL (c1), c2 − PL⊥ (c1)〉=∥∥∥∥ −γ(1 + γ) PL (c1)∥∥∥∥2 + ∥∥∥∥c2 − PL⊥ (c1)∥∥∥∥2+ 2〈γ(1 + γ) PL (c1), c2 − PL⊥ (c1)〉=∥∥∥∥ −γ(1 + γ) PL (c1)∥∥∥∥2 + ∥∥∥∥c2 − PL⊥ (c1)∥∥∥∥2=( 1(1 + 1γ))2 ∥∥PL (c1)∥∥2 + ∥∥c2 − PL⊥ (c1)∥∥224213.5. Finding the approximation curve for m = 3= 1(1 + 1γ)2 ∥∥PL (c1)∥∥2 + ∥∥c2 − PL⊥ (c1)∥∥2 .which shows that the curve xγ is bounded and ‖xγ‖ decreases as γ →0+.13.5 Finding the approximation curve for m = 3Thinking of block matrices, we haveJγM−1 =1γ2 + 3γ + 3 2 + γ −1 −(1 + γ)−(1 + γ) 2 + γ −1−1 −(1 + γ) 2 + γ .Let β := γ2 + 3γ + 3. ThenJγM−1 =1β((2 + γ)x1 − x2 − (1 + γ)x3,−(1 + γ)x1 + (2 + γ)x2 − x3,− x1 − (1 + γ)x2 + (2 + γ)x3), (13.78)andJγN−1C: H3 → H3:(x1, x2, x3) 7→ (x1 − γ PC1(x1/γ), x2 − γ PC2(x2/γ), x3 − γ PC3(x3/γ)).(13.79)Then the composition is(γ2 + 3γ + 3)x1 =(2 + γ)(x1 − PγC1(x1))− (x2 − PγC2(x2))− (1 + γ)(x3 − PγC3(x3)), (13.80)(γ2 + 3γ + 3)x2 =(2 + γ)(x2 − PγC2(x2))− (x3 − PγC3(x3))− (1 + γ)(x1 − PγC1(x1)), (13.81)and (γ2 + 3γ + 3)x3 =(2 + γ)(x3 − PγC3(x3))− (x1 − PγC1(x1))− (1 + γ)(x2 − PγC2(x2)). (13.82)Thanks to the symmetry in the three equations, if we add them together weobtain (γ2 + 3γ + 3)(x1 + x2 + x3)= 0,24313.5. Finding the approximation curve for m = 3which reduces tox1 + x2 + x3 = 0. (13.83)We also can combine the equations in different ways. We have a system ofthree nonlinear equations in 3 unknown variables. Therefore, we can do rowreduction to simplify.If we do (1 + γ)× (13.80) − (13.82), we get(1 + γ)βx1 − βx3 = β(x1 − PγC1(x1))+ 0(x2 − PγC2(x2))− β(x3 − PγC3(x3))γβx1 = −β PγC1(x1) + β PγC3(x3)γx1 = −PγC1(x1) + PγC3(x3)Doing symmetric operations, we get another set of three equations:γx1 = −PγC1(x1) + PγC3(x3). (13.84)γx2 = −PγC2(x2) + PγC1(x1). (13.85)γx3 = −PγC3(x3) + PγC2(x2). (13.86)Corollary 13.8. A triple x1, x2, x3 satisfies (13.84)-(13.86) if and only if itsatisfies the original system of equations (i.e., (13.80)-(13.82) ).Proof. Rearrange (13.80) to have all of the non-projected terms on the leftand all of the projected terms on the right:(γ2 + 3γ + 3)x1 + x2 + (1 + γ)x3 =−(2 + γ)PγC1(x1) + PγC2(x2)+(1 + γ)PγC3(x3)Since we derived (13.84)-(13.86) from (13.80)-(13.82), it is clear that anysolution to the original system will satisfy (13.84)-(13.86).Suppose x1, x2, x3 satisfy (13.84)-(13.86). Then ( multiplying (13.84)-(13.86)by coefficients matching the ones from the rearranged equation (13.80)), wehave(γ2 + 2γ + 1)x1 = −(γ + 2 + γ−1)PγC1(x1) +(γ + 2 + γ−1)PγC3(x3)x2 = −γ−1 PγC2(x2) + γ−1 PγC1(x1)(1 + γ)x3 = −(γ−1 + 1)PγC3(x3) +(γ−1 + 1)PγC2(x2)Adding these together, we obtain(γ2 + 3γ + 3)x1 + x2 + (1 + γ)x3 =−(2 + γ)PγC1(x1) + PγC2(x2)+(1 + γ)PγC3(x3)which is exactly the rearranged version of (13.80) we were aiming form.Therefore, x1, x2, x3 satisfies (13.80). A symmetric argument shows thatx1, x2, x3 also satisfy (13.81) and (13.82), concluding the proof.24413.6. Examples on m = 3 case13.6 Examples on m = 3 case13.6.1 One intervalLemma 13.9. Let H = R, C1 = [a, b] and C2 = {c}, and C3 = {e} wherea, b, c, e ∈ R and a < b < c < e. The following hold;(i) if a is positive, then(x1, x2, x3)=(e− b, b− c, c− e), if γ ∈ (0, eb − 1);( γ1+γ e,11+γ e− c, c− e), if γ ∈ [ eb − 1, ea − 1];(e− a, a− c, c− e), if γ ∈ ( ea − 1,∞).(ii) If e is negative, then(x1, x2, x3)=(e− b, b− c, c− e) if γ ∈ (0,+∞).(iii) If c is negative and e is not negative, then(x1, x2, x3)=(e− b, b− c, c− e) if γ ∈ (0,+∞).(iv) If b is negative and c is not negative, then(x1, x2, x3)=(e− b, b− c, c− e) if γ ∈ (0,+∞).(v) If a is negative and b is positive, thenxγ :=(x1, x2, x3)={(e− b, b− c, c− e), iff γ ∈ (0, eb − 1);( γ1+γ e,11+γ e− c, c− e), iff γ ∈ [ eb − 1,+∞) .Proof. (i):”⇒”: Clearly, PγC2 = γc and PγC3 = γe, which simplifies equa-tion (13.84)-(13.86) considerably,γx1 = −PγC1(x1) + γe. (13.87)γx2 = −γc+ PγC1(x1) (13.88)γx3 = −γe+ γc (13.89)First , we see this system only has a solution if x3 = c − e, which we canplug into (13.85).x1 + x2 + c− e = 0 so x2 = e− c− x1 (13.90)Therefore, we only have one free variable: x1. We can break into cases basedon the value of x1.24513.6. Examples on m = 3 case1. Suppose x1 > γb. Then PγC1(x1) = γb. Plugging this into (13.87), wehaveγx1 = −γb+ γeBecause xγ = e− b > γb, we haveγ <eb− 1.It then follows from (13.90) thatx2 =(e− c)− (e− b) = b− c.Hence,xγ :=(x1, x2, x3)=(e− b, b− c, c− e). (13.91)2. Suppose γa ≤ x1 ≤ γb. Then PγC1(x1) = x1. Plugging this into(13.87), we haveγx1 = −x1 + γe so x1 = γ(1 + γ)eBecause γa ≤ γ1+γ e ≤ γb, we haveea− 1 ≥ γ ≥ eb− 1.It then follows from (13.90) thatx2 =(e− c)− γ1 + γ e = 11 + γ e− c.Hence,xγ :=(x1, x2, x3)=( γ1 + γ e,11 + γ e− c, c− e).3. Suppose x1 < γa. Then PγC1(x1) = γa. Plugging this into (13.87),we haveγx1 = −γa+ γeBecause e− a < γa, we haveγ >ea− 1.It then follows from (13.90) thatx2 =(e− c)− (e− a) = a− c.Hence,xγ :=(x1, x2, x3)=(e− a, a− c, c− e).24613.6. Examples on m = 3 case”⇐”: Actually, the three cases we considered covered all the possible valueof γ, so we have three intervals:Interval 1:(0, eb − 1).Interval 2: [ eb − 1, ea − 1].Interval 3:(ea − 1,+∞).Suppose γ ∈ Interval 1. If xγ 6=(e − b, b − c, c − e), i.e., it would bexγ =( γ1+γ e,11+γ e − c, c − e)or xγ =(e − a, a − c, c − e), then γ would bein interval 2 or 3, which is contradiction because intervals do not intersect.Hence,if γ ∈ (0, eb− 1), then xγ = (e− b, b− c, c− e). (13.92)Combining (13.92) and (13.91), we havexγ =(e− b, b− c, c− e) iff γ ∈ (0, eb− 1).Using the same argument can show thatxγ :=(x1, x2, x3)=( γ1 + γ e,11 + γ e− c, c− e)iff γ ∈[eb− 1, ea− 1],andxγ :=(x1, x2, x3)=(e− a, a− c, c− e), iff γ ∈ ( ea− 1,+∞).Plugging these together, we have the following values for the solutions tothe system of equations:xγ :=(x1, x2, x3)=(e− b, b− c, c− e), iff γ ∈ (0, eb − 1);( γ1+γ e,11+γ e− c, c− e), iff γ ∈ [ eb − 1, ea − 1];(e− a, a− c, c− e), iff γ ∈ ( ea − 1,∞).Note that It is easy to check that at the boundary cases, the transition iscontinuous: If γ ∈ [ eb − 1, ea − 1], then xγ =( γ1+γ e,11+γ e− c, c− e). Whenγ = eb − 1 or γ = ea − 1, then we havelimγ→((e/b)−1)+xγ =( eb − 11 +(eb − 1)e, 11 +(eb − 1)e− c, c− e)=((eb− 1)(be)e,(be)e− c, c− e)=(e− b, b− c, c− e),24713.6. Examples on m = 3 caselimγ→((e/b)−1)−xγ =( eb − 11 +(eb − 1)e, 11 +(eb − 1)e− c, c− e)=((eb− 1)(be)e,(be)e− c, c− e)=(e− b, b− c, c− e).So we obtainlimγ→((e/b)−1)+xγ = x(e/b)−1 = limγ→((e/b)−1)1xγ .Next,limγ→((e/a)−1)+xγ =( ea − 11 +(ea − 1)e, 11 +(ea − 1)e− c, c− e)=(( ea− 1)(ae)e,(ae)e− c, c− e)=(e− a, a− c, c− e),andlimγ→((e/a)−1)−xγ =( ea − 11 +(ea − 1)e, 11 +(ea − 1)e− c, c− e)=(( ea− 1)(ae)e,(ae)e− c, c− e)=(e− a, a− c, c− e),Hence,limγ→((e/a)−1)+xγ = x(e/a)−1 = limγ→((e/a)−1)−.Therefore, the curve xγ is continuous.(ii): Here e is negative, then a, b, c are all negative. We have the followingcases:(a) Suppose x1 > γb, then PγC1(x1)= γb. Plugging this in (13.87), wehavex1 = e− b. (13.93)Because e− b > γb, we haveγ >eb− 1.It then follows from (13.90) thatx2 = (e− c)− (e− b) = b− c.24813.6. Examples on m = 3 caseHence,xγ =(e− b, b− c, c− e).(b) Suppose that γa ≤ x1 ≤ γb, then PγC1(x1)= x1. Plugging this into(13.87), we obtainγx1 = −x1 + γe so x1 = γ1 + γ e. (13.94)Because γa ≤ γ1+γ e ≤ γb, we haveea− 1 ≤ γ ≤ eb− 1.But, this is contradiction because b is negative and e is not negative.Hence, eb − 1 < 0. However, γ > 0 by assumption. So this contradictsthe assumption γ ≤ eb − 1.(c) Suppose x1 < γa, then PγC1(x1)= γa. Plugging this in (13.87), wehaveγx1 = −γa+ γe so x1 = e− a.Because e− a < γa, we haveγ <ea− 1.But this is impossible because ea − 1 < 0 and γ > 0.Hence, combining this together, we only have one case for xγ which isxγ =(e− b, b− c, c− e) with γ ∈ (0,+∞).(iii): We have e is not negative, but a, b, c are all negative. We have thefollowing cases:(a) Suppose x1 > γb, then PγC1(x1)= γb. Plugging this in (13.87), wehavex1 = e− b. (13.95)Because e− b > γb, we haveγ >eb− 1.It then follows from (13.90) thatx2 = (e− c)− (e− b) = b− c.Hence,xγ =(e− b, b− c, c− e).24913.6. Examples on m = 3 case(b) Suppose that γa ≤ x1 ≤ γb, then PγC1(x1)= x1. Plugging this into(13.87), we obtainγx1 = −x1 + γe so x1 = γ1 + γ e > 0. (13.96)Because x1 > 0 > γb, which contradicts the assumption x1 ≤ γb.(c) Suppose x1 < γa, then PγC1(x1)= γa. Plugging this in (13.87), wehaveγx1 = −γa+ γe so x1 = e− a > 0.Because e− a < γa, we haveγ <ea− 1.But this is impossible because ea − 1 < 0 and γ > 0. Moreover,x1 > 0 > γa, which contradicts the assumption x1 < γa.Hence, combining this together, we only have one case for xγ which isxγ =(e− b, b− c, c− e) with γ ∈ (0,+∞).(iv): Let c be not negative, but a, b are all negative. We have the followingcases:(a) Suppose x1 > γb, then PγC1(x1)= γb. Plugging this in (13.87), wehavex1 = e− b. (13.97)Because e− b > γb, we haveγ >eb− 1.It then follows from (13.90) thatx2 = (e− c)− (e− b) = b− c.Hence,xγ =(e− b, b− c, c− e).(b) Suppose that γa ≤ x1 ≤ γb, then PγC1(x1)= x1. Plugging this into(13.87), we obtainγx1 = −x1 + γe so x1 = γ1 + γ e > 0. (13.98)25013.6. Examples on m = 3 caseBecause γa ≤ γ1+γ e ≤ γb, we haveea− 1 ≥ γ ≥ eb− 1.But, this is contradiction because x1 > 0 > γb, which contradicts theassumption x1 ≤ γb. Moreover, a is negative and e is not negative.Hence, ea − 1 < 0. However, γ > 0 by assumption. So this contradictsγ ≤ ea − 1.(c) Suppose x1 < γa, then PγC1(x1)= γa. Plugging this in (13.87), wehaveγx1 = −γa+ γe so x1 = e− a > 0.Because e− a < γa, we haveγ <ea− 1.But this is impossible because ea − 1 < 0 and γ > 0. Moreover,x1 > 0 > γa, which contradicts the assumption x1 < γa.Hence, combining this together, we only have one case for xγ which isxγ =(e− b, b− c, c− e) with γ ∈ (0,+∞).(v): Let b be not negative, but a is a negative. We have the following cases:(a) Suppose x1 > γb, then PγC1(x1)= γb. Plugging this in (13.87), wehavex1 = e− b. (13.99)Because e− b > γb, we haveγ <eb− 1.It then follows from (13.90) thatx2 = (e− c)− (e− b) = b− c.Hence,xγ =(e− b, b− c, c− e).25113.6. Examples on m = 3 case(b) Suppose that γa ≤ x1 ≤ γb, then PγC1(x1)= x1. Plugging this into(13.87), we obtainγx1 = −x1 + γe so x1 = γ1 + γ e. (13.100)Because x1 = γ1+γ e ≤ γb, we haveγ1 + γ e ≤ γb⇔ 11 + γ e ≤ b⇔ 11 + γ ≤be⇔ eb≤ (1 + γ).Hence,eb− 1 ≤ γ.It then follows from (13.90) thatx2 =(e− c)− γ1 + γ e = 11 + γ e− c.Hence,xγ :=(x1, x2, x3)=( γ1 + γ e,11 + γ e− c, c− e).(c) Suppose x1 < γa, then PγC1(x1)= γa. Plugging this in (13.87), wehaveγx1 = −γa+ γe so x1 = e− a.Because e− a < γa, we haveγ <ea− 1.But this is impossible because ea − 1 < 0 and γ > 0.”⇐”: Actually, the two cases we considered covered all the possible value ofγ, so we have three intervals:Interval 1:(0, eb − 1).Interval 2:[eb − 1,+∞).25213.6. Examples on m = 3 caseSuppose γ ∈ Interval 1. If xγ 6=(e − b, b − c, c − e), i.e., it would bexγ =( γ1+γ e,11+γ e− c, c− e), then γ would be in interval 2, which is contra-diction because intervals do not intersect. Hence,if γ ∈ (0, eb− 1), then xγ = (e− b, b− c, c− e). (13.101)Combining (13.101) and (13.91), we havexγ =(e− b, b− c, c− e) iff γ ∈ (0, eb− 1).Using the same argument, we can show thatxγ :=(x1, x2, x3)=( γ1 + γ e,11 + γ e− c, c− e)iff γ ∈[eb− 1,+∞).Plugging these together, we have the following values for the solutions tothe system of equations:xγ :=(x1, x2, x3)={(e− b, b− c, c− e), iff γ ∈ (0, eb − 1);( γ1+γ e,11+γ e− c, c− e), iff γ ∈ [ eb − 1,+∞) .Next, check the continuity of the curve. Let γ = eb − 1. Then we havelimγ→((e/b)−1)+xγ =( eb − 11 +(eb − 1)e, 11 +(eb − 1)e− c, c− e)=((eb− 1)(be)e,(be)e− c, c− e)=(e− b, b− c, c− e),limγ→((e/b)−1)−xγ =( eb − 11 +(eb − 1)e, 11 +(eb − 1)e− c, c− e)=((eb− 1)(be)e,(be)e− c, c− e)=(e− b, b− c, c− e).So we obtainlimγ→((e/b)−1)+xγ = x(e/b)−1 = limγ→((e/b)−1)1xγ .253Chapter 14ConclusionIn this thesis we advanced the analysis of the Arago´n Artacho-Campoyalgorithm (AACA). We studied the fixed point set of compositions of projec-tion mappings and the geometry conjecture was resolved completely. Thenwe extended the results from projectors to proximal mapping. The work inthis thesis is based on the author’s joint work that appears in the publica-tions [1], [2], [3], [4], and [5]. We start by summarizing the main contribu-tions of this thesis.14.1 Main resultsWe developed the analysis of the Arago´n Artacho-Campoy algorithm(AACA) by demonstrating that the underlying curve converges to the near-est zero of the sum of the two operators. We completed the analysis of the(AACA) using the proximal and resolvent average point of view. We studiedthe fixed point set of compositions of linear reflectors and provided positiveresults. We used the framework of monotone operator theory to resolve thegeometry conjecture completely. We obtained alternative descriptions of theset of cycles that is defined using projection mappings. We also sketchednumerical approaches for the computation of the difference vector by usingSeeger’s algorithm. Using the Attouch-Ther´a duality, we studied the cy-cles that are defined using proximal mappings, gap vectors and fixed pointsets of compositions of proximal mappings. A primal and dual frameworkprovides an exact relationship between the cycles and gap vectors. We alsointroduced the generalized cycle and gap vectors to tackle the case when theclassical ones do not exist. We proved that the composition of the resolventof the inverse of the displacement mapping and the resolvent of the inverseof the normal cone on a nonempty closed convex set is Banach contractionwith constant less than 1 and we derived a formula for the fixed point of thiscomposition. We also proved that every cluster points of the curve inducedby the fixed point coincides with the projection of the closure of the nearestpoint set.25414.2. Further results14.2 Further resultsWe sketched the connection between the cycles and the fixed point sets ofcomposition of projection mappings. We studied in details the displacementmappings of isometries of finite order. We derived a formula for its inverseand its Yosida approximation. In terms of convex analysis, we gave sufficientconditions under which the infimal convolution is closed and convex. Westudied the relationship among the fixed point set of cycles that are definedusing projection mappings respectively, proximal mappings and the fixedpoint set for composition of projectors respectively proximal mappings. Totackle the case when the classical cycles do not exists, we introduced thegeneralized cycles which is a variant of classical ones that are defined usingprojectors (that are defined using proximal mappings).14.3 Future workIn the following we discuss some possible directions of future research:Q1: For Section 9.4, we define ∂σC+∆ to be the extension of N−1C +N∆⊥ .Moreover, in Section 11.6, we define ∂(f∗ + ι∗∆) to be the extensionof ∂f∗ + ∂ι∗∆. Set S := N−1C + N∆⊥ and S˜ := ∂f∗ + ∂ι∗∆. Are Sand S˜ premaximal monotone operators? In other words, are ∂σC+∆and ∂(f∗ + ι∗∆) unique maximally monotone extension of S and S˜respectively?Q2: What are the maximally monotone enlargements of A−1 +N∆⊥ ?Q3: Is it true in general that that the set of classical cycles Z satisfiesZ = e+(∆∩dom A−e), where A is a product of maximally monotoneoperators ?Q4: In Chapter 13, if m = 2 and one of the set is bounded, does it thenfollow that xγ converges to the gap vector?255Bibliography[1] S. Alwadani, H. H. Bauschke, W. M. Moursi, and X. Wang,On the asymptotic behaviour of the Arago´n Artacho–Campoy algorithm,Operations Research Letters, 46 (2018), pp. 585–587. → pages 33, 254[2] S. Alwadani, H. H. Bauschke, J. P. Revalski, and X. Wang, Re-solvents and Yosida approximations of displacement mappings of isome-tries, Set-Valued and Variational Analysis, (2020). → pages 93, 94, 254[3] S. Alwadani, H. H. Bauschke, J. P. Revalski, and X. 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Za˘linescu, Convex Analysis in General Vector Spaces, World Sci-entific, 2002. → pages 1260Appendix AMatlab code for Example 5.91 % setup !23 wa=[1 ,−1 ,5 ] ;4 wc=[1 ,1 ,−5] ;5 wb= [ 1 , 0 , 0 ] ;67 x = [ − 1 0 : 0 . 0 1 : 1 0 ] ;8 a= x∗wa(2) /wa(1 ) +5;9 b = x∗wb(2) /wb(1) ;10 c= x∗wc (2) /wc (1 ) +5;1112 % Plot the l i n e s !13 p lo t (x , a , ’ b lack ’ , ’ LineWidth ’ , 5 ) ;14 hold on15 p lo t (x , b , ’ b lue ’ , ’ LineWidth ’ , 5 ) ;16 hold on17 p lo t (x , c , ’ green ’ , ’ LineWidth ’ , 5 ) ;18 hold on1920 % Alte rnat ing p r o j e c t i o n a lgor i thm21 x0 =[−3.54 , 2 . 9 ] ;22 x k = ze ro s (2 , 50) ;23 x k ( : , 1 ) = x0 ;2425 maincycl = [ ] ;26 f o r i = 2 : 3 :2127 c y c l e= ze ro s (2 , 3) ;28 x k ( : , i ) = pro j (wa , x k ( : , i −1) ’ ) ;29 x k ( : , i +1) = projn (wb, pro j (wa , x k ( : , i −1) ’ ) ) ;30 x k ( : , i +2) = pro j (wc , projn (wb, pro j (wa , x k ( : , i −1)’ ) ) ) ;261Appendix A. Matlab code for Example 5.931 i f sum( abs ( x k ( : , i ) − x k ( : , i +3) ) ) < 1e−5 . . .32 && sum( abs ( x k ( : , i +1) − x k ( : , i +4) ) ) < 1e−5 . . .33 && sum( abs ( x k ( : , i +2) − x k ( : , i +5) ) ) < 1e−534 break35 end36 end37 c y c l e = [ x k ( : , i ) x k ( : , i +1) x k ( : , i +2) ] − c y c l e ;38 maincycl = [ maincycl ; c y c l e ] ;394041 % Evaluat ing us ing an o b j e c t i v e func t i on42 XX = [ c y c l e ( 1 , 1 ) , c y c l e ( 2 , 1 ) ; c y c l e ( 1 , 2 ) , c y c l e ( 2 , 2 ) ] ;43 YY= [ c y c l e ( 1 , 2 ) , c y c l e ( 2 , 2 ) ; c y c l e ( 1 , 3 ) , c y c l e ( 2 , 3 ) ] ;44 ZZ= [ c y c l e ( 1 , 3 ) , c y c l e ( 2 , 3 ) ; c y c l e (1 , 1 ) , c y c l e ( 2 , 1 ) ] ;45 va rph i cyc l e = pd i s t (XX, ’ euc l i d ean ’ )+ pd i s t (YY, ’euc l i d ean ’ ) . . .46 + pd i s t (ZZ , ’ euc l i d ean ’ ) ;47 v a r t h e t a c y c l e= (1/2) ∗ pd i s t (XX, ’ euc l i d ean ’ ) ˆ2 . . .48 + (1/2) ∗ pd i s t (YY, ’ euc l i d ean ’ )ˆ2+ (1/2) ∗ pd i s t (ZZ , ’euc l i d ean ’ ) ˆ2 ;4950 % Compare a c y c l e with a po int51 z 0= [−1.32 , 3 . 7 ] ;52 z 1= pro j (wa , z 0 ) ;53 z 2= projn (wb, z 1 ) ;54 z 3= pro j (wc , z 2 ) ;5556 X = [ z 1 (1 ) , z 1 (2 ) ; z 2 (1 ) , z 2 (2 ) ] ;57 Y= [ z 2 (1 ) , z 2 (2 ) ; z 3 (1 ) , z 3 (2 ) ] ;58 Z= [ z 3 (1 ) , z 3 (2 ) ; z 1 (1 ) , z 1 (2 ) ] ;59 varph ipo int = pd i s t (X, ’ euc l i d ean ’ )+ pd i s t (Y, ’euc l i d ean ’ ) . . .60 + pd i s t (Z , ’ euc l i d ean ’ ) ;6162 var the tapo in t= (1/2) ∗ pd i s t (X, ’ euc l i d ean ’ ) ˆ2 . . .63 + (1/2) ∗ pd i s t (Y, ’ euc l i d ean ’ ) ˆ2 + (1/2) ∗ pd i s t (Z , ’euc l i d ean ’ ) ˆ2 ;262Appendix A. Matlab code for Example 5.91 f unc t i on [ ProjPointn ] = projn ( l i n e , q )2 ProjPointn = ( ( q (1 ) ∗ l i n e (1 ) + q (2) ∗ l i n e (2 )+ l i n e (3 ) )/( l i n e (1 ) ∗ l i n e (1 )+l i n e (2 ) ∗ l i n e (2 ) ) ) ∗ [ l i n e (1 ) , l i n e(2 ) ] ;3 end1 f unc t i on [ ProjPoint ] = pro j ( l i n e , q )2 ProjPoint = q− ( ( q (1 ) ∗ l i n e (1 ) + q (2) ∗ l i n e (2 )+ l i n e (3 ) )/( l i n e (1 ) ∗ l i n e (1 )+l i n e (2 ) ∗ l i n e (2 ) ) ) ∗ [ l i n e (1 ) , l i n e(2 ) ] ;3 end263Appendix A. Matlab code for Example 5.91 %% Answers2 >> c y c l e34 c y c l e =56 0 0 2 .50007 5 .0000 0 2 .500089 >> XX1011 XX =1213 0 514 0 01516 >> YY1718 YY =1920 0 021 2 .5000 2 .50002223 >> ZZ2425 ZZ =2627 2 .5000 2 .500028 0 5 .00002930 >> va rph i cyc l e3132 va rph i cyc l e =3334 12.07113536 >> v a r t h e t a c y c l e3738 v a r t h e t a c y c l e =39264Appendix A. Matlab code for Example 5.940 254142 >> z 04344 z 0 =4546 −1.3200 3 .70004748 >> z 1= pro j (wa , z 0 )4950 z 1 =5152 −1.3100 3 .69005354 >> z 2= projn (wb, z 1 )5556 z 2 =5758 −1.3100 05960 >> z 3= pro j (wc , z 2 )6162 z 3 =6364 1 .8450 3 .15506566 >> X = [ z 1 (1 ) , z 1 (2 ) ; z 2 (1 ) , z 2 (2 ) ]6768 X =6970 −1.3100 3 .690071 −1.3100 07273 >> Y= [ z 2 (1 ) , z 2 (2 ) ; z 3 (1 ) , z 3 (2 ) ]7475 Y =7677 −1.3100 078 1 .8450 3 .155079265Appendix A. Matlab code for Example 5.980 >> Z= [ z 3 (1 ) , z 3 (2 ) ; z 1 (1 ) , z 1 (2 ) ]8182 Z =8384 1 .8450 3 .155085 −1.3100 3 .69008687 >> varph ipo int8889 varph ipo int =9091 11.35199293 >> var the tapo in t9495 var the tapo in t =9697 21.8822266Appendix BMathematica codeThe following sections provide the Mathematica code that was used toverify examples. All code was run using Mathematica software.B.1 Code to verify Example 7.12A1:=m−1∑k=1γk(γ + 1)−k+m−1MatrixPower[R, k]f1(m ,R , d ):=(γ + 1)m−1IdentityMatrix[d] + A1(γ + 1)m − γmA2:=m−1∑k=1(1γ)k (1γ+ 1)−k+m−1MatrixPower[R, k]f2(m ,R , d ):=IdentityMatrix[d]−(1γ + 1)m−1IdentityMatrix[d] + A2(1γ + 1)m − ( 1γ)mA3:=m−1∑k=1γk(γ + 1)−k+m−1MatrixPower[R, k]f3(m ,R ,d ):=IdentityMatrix[d]− (γ+1)m−1IdentityMatrix[d]+A3(γ+1)m−γmγA4:=m−1∑k=1(γ + 1)−k+m−1MatrixPower[R, k]f4(m ,R , d ):=(γ + 1)m−1IdentityMatrix[d] + A4(γ + 1)m − 1267B.1. Code to verify Example 7.12Simplify[f1(2,RotationMatrix[2pi2], 2)]( 12γ+1 00 12γ+1)Simplify[f2(2,RotationMatrix[2pi2], 2)]( 2γ+2 00 2γ+2)Simplify[f3(2,RotationMatrix[2pi2], 2)]( 22γ+1 00 22γ+1)Simplify[f4(2,RotationMatrix[2pi2], 2)]( 1γ+2 00 1γ+2)Simplify[f1(3,RotationMatrix[2pi3], 2)] 3γ+26γ2+6γ+2 − √3γ6γ2+6γ+2√3γ6γ2+6γ+23γ+26γ2+6γ+2Simplify[f2(3,RotationMatrix[2pi3], 2)] 3(γ+2)2(γ2+3γ+3) √3γ2γ2+6γ+6−√3γ2γ2+6γ+63(γ+2)2(γ2+3γ+3)Simplify[f3(3,RotationMatrix[2pi3], 2)] 6γ+36γ2+6γ+2 √36γ2+6γ+2−√36γ2+6γ+26γ+36γ2+6γ+2268B.2. Code to verify Example 7.14Simplify[f4(3,RotationMatrix[2pi3], 2)] 2γ+32γ2+6γ+6 − √32γ2+6γ+6√32γ2+6γ+62γ+32γ2+6γ+6B.2 Code to verify Example 7.14A1:=m−1∑k=1γk(γ + 1)−k+m−1MatrixPower[R, k]f1(m ,R , d ):=(γ + 1)m−1IdentityMatrix[d] + A1(γ + 1)m − γmA2:=m−1∑k=1(1γ)k (1γ+ 1)−k+m−1MatrixPower[R, k]f2(m ,R , d ):=IdentityMatrix[d]−(1γ + 1)m−1IdentityMatrix[d] + A2(1γ + 1)m − ( 1γ)mA3:=m−1∑k=1γk(γ + 1)−k+m−1MatrixPower[R, k]f3(m ,R ,d ):=IdentityMatrix[d]− (γ+1)m−1IdentityMatrix[d]+A3(γ+1)m−γmγA4:=m−1∑k=1(γ + 1)−k+m−1MatrixPower[R, k]f4(m ,R , d ):=(γ + 1)m−1IdentityMatrix[d] + A4(γ + 1)m − 1Simplify[f1(2,(0 11 0), 2)]269B.2. Code to verify Example 7.14(γ+12γ+1γ2γ+1γ2γ+1γ+12γ+1)Simplify[f2(2,(0 11 0), 2)]( 1γ+2 − 1γ+2− 1γ+2 1γ+2)Simplify[f3(2,(0 11 0), 2)]( 12γ+11−2γ−11−2γ−112γ+1)Simplify[f4(2,(0 11 0), 2)](γ+1γ2+2γ1γ2+2γ1γ2+2γγ+1γ2+2γ)Simplifyf13, 0 0 11 0 00 1 0 , 3(γ+1)23γ2+3γ+1γ23γ2+3γ+1γ(γ+1)3γ2+3γ+1γ(γ+1)3γ2+3γ+1(γ+1)23γ2+3γ+1γ23γ2+3γ+1γ23γ2+3γ+1γ(γ+1)3γ2+3γ+1(γ+1)23γ2+3γ+1Simplifyf23, 0 0 11 0 00 1 0 , 3γ+2γ2+3γ+3 − 1γ2+3γ+3 − γ+1γ2+3γ+3− γ+1γ2+3γ+3γ+2γ2+3γ+3 − 1γ2+3γ+3− 1γ2+3γ+3 − γ+1γ2+3γ+3 γ+2γ2+3γ+3Simplifyf33, 0 1 00 0 11 0 0 , 3270B.3. Code to verify Section 9.92γ+13γ2+3γ+1 − γ3γ2+3γ+1 − γ+13γ2+3γ+1− γ+13γ2+3γ+1 2γ+13γ2+3γ+1 − γ3γ2+3γ+1− γ3γ2+3γ+1 − γ+13γ2+3γ+1 2γ+13γ2+3γ+1Simplifyf43, 0 0 11 0 00 1 0 , 3(γ+1)2(γ+1)3−11(γ+1)3−1γ+1(γ+1)3−1γ+1(γ+1)3−1(γ+1)2(γ+1)3−11(γ+1)3−11(γ+1)3−1γ+1(γ+1)3−1(γ+1)2(γ+1)3−1B.3 Code to verify Section 9.9F (m ,R ):=IdentityMatrix[m]2 −∑m−1k=1 (m− 2k)MatrixPower[R, k]2mEigenvaluesF (2,( 0 11 0))†.F(2,(0 11 0)){14 ,14}∥∥∥∥∥F(2,(0 11 0))∥∥∥∥∥212EigenvaluesF3, 0 0 11 0 00 1 0†.F3, 0 0 11 0 00 1 0{13 ,13 ,14}271B.3. Code to verify Section 9.9∥∥∥∥∥∥∥F3, 0 0 11 0 00 1 0∥∥∥∥∥∥∥21√3EigenvaluesF4,0 0 0 11 0 0 00 1 0 00 0 1 0†.F4,0 0 0 11 0 0 00 1 0 00 0 1 0{12 ,12 ,14 ,14}∥∥∥∥∥∥∥∥∥F4,0 0 0 11 0 0 00 1 0 00 0 1 0∥∥∥∥∥∥∥∥∥21√2EigenvaluesF5,0 0 0 0 11 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 0†.F5,0 0 0 0 11 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 0{ 110(√5 + 5),110(√5 + 5),110(5−√5),110(5−√5),14}∥∥∥∥∥∥∥∥∥∥∥F5,0 0 0 0 11 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 0∥∥∥∥∥∥∥∥∥∥∥2272B.3. Code to verify Section 9.9√110(√5 + 5)EigenvaluesF6,0 0 0 0 0 11 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 0†.F6,0 0 0 0 0 11 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 0{1, 1, 13 ,13 ,14 ,14}∥∥∥∥∥∥∥∥∥∥∥∥∥F6,0 0 0 0 0 11 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 0∥∥∥∥∥∥∥∥∥∥∥∥∥21.273Indexα-averaged operator, 19β-cocoercive operator, 18λ-Weighted proximal average, 15Arago´n Artacho-Campoy algorithm,28convex feasibility problem, 66epigraph of a function, 10a polyhedral convex function, 141a polyhedral convex set, 14adjoint operator, 6affine hull, 8Arago´n Artacho-Campoy algorithm,33Attouch-The´ra dual problem, 145,175, 208, 209Attouch-The´ra dual problem for theorder pair(A,B), 26Attouch-The´ra pimal problem, 179,209Attouch-The´ra primal problem, 144,200averaged mapping, 19averaged operator, 48Banach contraction, 6, 101, 112, 173,215best approximation problem, 2, 28biconjugate of function, 11canonical hull, 136Cauchy-Schwarz inequality, 112, 218circular left shift operator, 6, 152, 153closure of set, 8cocoercive operator, 112coercive function, 10cone of set, 8conical hul of a set, 8continuous minorant, 189convex cone of a set, 8convex feasibility problem, 1, 27convex function, 10, 123, 150, 179,189convex hull, 7convex set, 7, 194convex subdifferential of function, 11cycle, 67, 72, 74, 75, 144cyclic projection method, 67, 74, 142cyclic right-shift operator, 6, 115, 121,142diagonal subspace, 121difference vectors, 67, 74, 173, 174,209displacement mapping, 93, 216domain of set-valued operator, 7Douglas-Rachford algorithm, 27dual solutions to the dual problem,26, 193enlarged dual problem, 149, 197Euclidean plane, 86exact infimal convolution, 133, 136,141, 203, 207Fenchel conjugate of function, 11firmly nonexpansive operator, 17, 217fixed point set, 17, 154, 156, 191, 194,195274INDEXfixed point sets of cyclic compositions,66Forward-backward method, 174, 209gap vector between two sets, 70general gap vectors, 180, 196generalized cycles, 180generalized solutions sets, 70geometry conjecture, 70, 76, 120, 154graph of set-valued operator, 7Hilbert product space, 5, 9, 121, 142,188, 199Hilbert space, 5indicator function, 10infimal convolution, 10, 125, 129, 131inner w-perturbation, 24interior of set, 8isometry operator, 6, 18, 92, 93, 95kernel, 6linear relation, 7Lipschitz continuous, 6, 17, 18, 104,112lower semicontinuous, 10, 122–124,132, 136, 141, 150, 179maximally monotone operator, 20,21, 25, 38, 43, 44, 47, 94, 95,99, 107, 150, 179, 200modified reflected resolvent, 22monotone operator, 20, 22, 43, 46,111Moore-Penrose inverse, 9, 113negatively averaged operator, 46, 48nonexpansive operator, 17–19, 48, 77,93, 99normal cone operator, 9, 216operator inverse, 110pairwise orthogonal spaces, 83, 84paramonotone operator, 20periodic operator, 86primal problem for the order pair(A,B), 26primal solutions to the primal prob-lem, 26, 193projection operator, 8, 17proper function, 10, 150, 189proximity operator, 12, 191quasinonexpansive operator, 17range of set-valued operator, 7rational rotation, 114reflected resolvent, 22, 34reflector operator, 92relative interior of a set, 8resolvent, 22, 27, 34, 93, 98, 216resolvent average, 16, 25, 38rotation operator, 92self-adjoint operator, 84separable sum, 128set of cycles, 144, 145, 152, 192, 194,207set of gap vectors, 145set of zeros, 7, 24, 39, 43, 145set-valued inverse, 7span of a set, 8strictly monotone operator, 20strong convergent, 5, 175, 209, 211strong relative interior of a set, 8strongly convex function, 10strongly monotone, 111strongly monotone operator, 20, 39,43, 112strongly nonexpansive operator, 19subdifferential, 125, 129, 132support function, 12, 122weak convergent, 5, 19, 175, 209, 211275INDEXYosida approximation, 104Young inequality, 132276
Thesis/Dissertation
eng
Mathematics
Vancouver : University of British Columbia Library
Attribution-NonCommercial-NoDerivatives 4.0 International
Graduate
Bauschke, Heinz H.
On the behaviour of algorithms featuring compositions of projectors and proximal mappings with no solutions
Text