The Resolvent Average: An ExpansiveAnalysis of Firmly NonexpansiveMappings and Maximally MonotoneOperatorsbySarah Michelle MoffatB.Sc. Mathematics Honours, The University of British Columbia, 2007M.Sc., The University of British Columbia, 2009A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinTHE COLLEGE OF GRADUATE STUDIES(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Okanagan)December 2014c© Sarah Michelle Moffat, 2014AbstractMonotone operators and firmly nonexpansive mappings are essential tomodern optimization and fixed point theory. Minty first discovered thelink between these two classes of operators; every resolvent of a monotoneoperator is firmly nonexpansive and every firmly nonexpansive mapping isa resolvent of a monotone operator.This thesis provides an in-depth study of the relationship between firmlynonexpansive mappings and maximally monotone operators. First, corre-sponding properties between maximally monotone operators and their re-solvents are collected. Then a new method of averaging monotone operatorsis presented, called the resolvent average, which is based on the convex com-bination of the resolvents of monotone operators. Several new results aregiven concerning the asymptotic regularity of compositions and convex com-binations of firmly nonexpansive mappings. Finally, the resolvent averageis studied with respect to which properties the average inherits from theaveraged operators.iiPrefaceThis thesis is based on the papers [15, 18–21]. The papers [18–21], whichform the basis of Chapters 4–6 and Chapter 8 are based on joint work withmy supervisors, Dr. Heinz H. Bauschke and Dr. Xianfu Wang. The paper[15] which is the basis of Chapter 7 is based on joint work with Dr. HeinzH. Bauschke, Dr. Victoria Martin-Marquez, and Dr. Xianfu Wang.For all co-authored papers, each author contributed equally.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . ivList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiiGlossary of Notation . . . . . . . . . . . . . . . . . . . . . . . . . ixAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . xiDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiiChapter 1: Introduction . . . . . . . . . . . . . . . . . . . . . . . 1Chapter 2: Preliminary Details . . . . . . . . . . . . . . . . . . 32.1 Normed vector spaces . . . . . . . . . . . . . . . . . . . . . . 32.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.1 Single-valued operators . . . . . . . . . . . . . . . . . 42.2.2 Set-valued operators . . . . . . . . . . . . . . . . . . . 72.3 Convex analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3.1 Convex sets . . . . . . . . . . . . . . . . . . . . . . . . 82.3.2 Convex functions . . . . . . . . . . . . . . . . . . . . . 122.4 Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.4.1 Arithmetic and harmonic averages . . . . . . . . . . . 172.4.2 Geometric mean . . . . . . . . . . . . . . . . . . . . . 172.4.3 Proximal average . . . . . . . . . . . . . . . . . . . . . 18Chapter 3: Nonexpansive Mappings and Monotone Operators 23ivTABLE OF CONTENTS3.1 Nonexpansive mappings . . . . . . . . . . . . . . . . . . . . . 233.2 Fixed points and asymptotic regularity . . . . . . . . . . . . . 263.3 Monotone operators . . . . . . . . . . . . . . . . . . . . . . . 293.4 Rectangular monotone operators . . . . . . . . . . . . . . . . 34Chapter 4: Correspondence of Properties . . . . . . . . . . . . 384.1 Maximally monotone operators and firmly nonexpansive map-pings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.2 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.3 Reflected resolvents . . . . . . . . . . . . . . . . . . . . . . . . 56Chapter 5: The Resolvent Average of Monotone Operators . 645.1 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . 645.2 The resolvent average of positive semidefinite matrices . . . . 705.2.1 Inequalities among means . . . . . . . . . . . . . . . . 715.2.2 A comparison to weighted geometric means . . . . . . 77Chapter 6: Near Equality, Near Convexity, Sums of Maxi-mally Monotone Operators, and Averages of FirmlyNonexpansive Mappings . . . . . . . . . . . . . . . 796.1 Near equality and near convexity . . . . . . . . . . . . . . . . 796.2 Maximally monotone operators . . . . . . . . . . . . . . . . . 886.3 Firmly nonexpansive mappings . . . . . . . . . . . . . . . . . 90Chapter 7: Compositions and Convex Combinations of Asymp-totically Regular Firmly Nonexpansive Mappings 957.1 Properties of the operator M . . . . . . . . . . . . . . . . . . 967.2 Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . 1017.3 Asymptotic regularity . . . . . . . . . . . . . . . . . . . . . . 1047.4 Convex combination . . . . . . . . . . . . . . . . . . . . . . . 105Chapter 8: Inheritance of Properties of the Resolvent Aver-age of Monotone Operators . . . . . . . . . . . . . 1098.1 Dominant properties . . . . . . . . . . . . . . . . . . . . . . . 1098.1.1 Single-valuedness of Rµ(A, λ) . . . . . . . . . . . . . . 1098.1.2 Domain and range . . . . . . . . . . . . . . . . . . . . 1128.1.3 Strict monotonicity . . . . . . . . . . . . . . . . . . . . 1138.1.4 Banach contraction . . . . . . . . . . . . . . . . . . . . 1148.1.5 Rectangularity and paramonotonicity . . . . . . . . . 1158.2 Dominant or recessive properties . . . . . . . . . . . . . . . . 116vTABLE OF CONTENTS8.2.1 Strong monotonicity . . . . . . . . . . . . . . . . . . . 1168.2.2 γ-cocoercive . . . . . . . . . . . . . . . . . . . . . . . . 1178.3 Recessive properties . . . . . . . . . . . . . . . . . . . . . . . 1188.3.1 Maximality and linearity . . . . . . . . . . . . . . . . 1188.3.2 Rectangularity and paramonotonicity . . . . . . . . . 1188.3.3 k-cyclical monotonicity . . . . . . . . . . . . . . . . . 1248.3.4 Displacement mappings . . . . . . . . . . . . . . . . . 1298.3.5 Nonexpansive monotone operators . . . . . . . . . . . 1318.4 Indeterminate properties . . . . . . . . . . . . . . . . . . . . . 138Chapter 9: Conclusion . . . . . . . . . . . . . . . . . . . . . . . . 1409.1 Key results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1409.2 Future work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152Appendix A: Uniformly Convex Banach Spaces . . . . . . . . . . . 153Appendix B: Maple Code . . . . . . . . . . . . . . . . . . . . . . . 156B.1 Code to verify Example 8.35 . . . . . . . . . . . . . . . . . . . 156B.2 Code to verify Example 8.42 . . . . . . . . . . . . . . . . . . . 157B.3 Code to verify Remark 8.47 . . . . . . . . . . . . . . . . . . . 159viList of TablesTable 4.1 Summary of self dual properties between monotone op-erators and their resolvents. . . . . . . . . . . . . . . . . 57Table 8.1 Summary of completely classified properties of the re-solvent average. . . . . . . . . . . . . . . . . . . . . . . 139Table 8.2 Summary of incompletely classified and indeterminantproperties of the resolvent average. . . . . . . . . . . . . 139viiList of FiguresFigure 4.1 Duality of a monotone operator, A, and its associatedresolvent, T = JA. . . . . . . . . . . . . . . . . . . . . . 49Figure 4.2 Duality of a monotone operator A, its associated re-solvent, T , and its reflected resolvent, N = 2T − Id. . . 57Figure 8.1 Resulting angle γ of the rotation of R1(A,λ) of Ex-ample 8.42 with Aα = Aθ+pi2 . . . . . . . . . . . . . . . . 134viiiGlossary of NotationB(H1,H2) The set of continuous linear mappings fromH1 to H2. 5H A real Hilbert space. 3N Strictly positive integers, 1, 2, 3, · · · . 3Q Rational numbers. 3R Real numbers. 3R+ Nonnegative real numbers. 3R++ Strictly positive real numbers. 3Rn n-dimensional Euclidean space. 3SN Space of N ×N real symmetric matrices. 4SN+ Set of N×N real symmetric positive semidef-inite matrices. 4SN++ Set of N ×N real symmetric positive definitematrices. 4xn → x Strong convergence. 4xn ⇀ x Weak convergence. 4A(A,λ) Arithmetic average. 17G(x,λ) Geometric average. 18H(A,λ) Harmonic average. 17Pµ(f ,λ) Proximal average. 18Rµ(A,λ) Resolvent average. 64Γ0(H) The class of proper lower semi-continuousconvex functions from H → ]−∞,+∞]. 12∇f(x) The Gaˆteaux gradient of f at x, unless other-wise specified to be the Fre´chet gradient. 7∂f Subdifferential of f . 12epi f Epigraph of f . 12f∗ Fenchel conjugate of f . 12α ? f Epi-multiplication. 12ixGlossary of Notationfg Infimal convolution of f and g. 12proxf x The proximal mapping of f at x. 15〈·, ·〉 Inner product. 3‖x‖ The norm of x. 3A : H1 ⇒ H2 Set valued operator A from H1 to H2. 7A∗ The adjoint of A. 5A+ The symmetric part of an N ×N matrix A. 4λA Yosida λ-regularization of A. 31A† Moore-Penrose inverse. 11domA The domain of a set valued operator. 7FixT Fixed points of operator T . 26graA Graph of the operator A. 7Id Identity mapping. 5JA The resolvent of the monotone operator A. 31kerA The nullspace of the operator A. 5q The quadratic form q = qId = 〈x, x〉. 6qA The quadratic form qAx = 〈x,Ax〉. 6ranT The range of a single valued operator. 4ranA The range of a set valued operator. 7aff C The affine hull of the set C. 8C The closure of C. 8coneC The conical hull of the set C. 8convC The convex hull of the set C. 8C⊥ The orthogonal complement of C. 5ιC The indicator function of the set C. 10NCx The normal cone to the set C at x. 11PCx The projection of x onto the set C. 10riC The relative interior of the set C. 8spanC The span of the set C. 8A ≈ B The set A is nearly equal to the set B. 79B A A−B ∈ SN+ . 4B ≺ A A−B ∈ SN++. 4xAcknowledgementsFirst and foremost, I would like to thank my supervisors, Dr. HeinzBauschke and Dr. Shawn Wang, for all of their help and guidance through-out my graduate studies. They have been excellent mentors and withoutthem this thesis would not have been possible. I would also like to thankDr. Yves Lucet and Dr. Stephen Simons for serving on my committee andproviding me with valuable insights, and the Natural Sciences and Engineer-ing Research Council of Canada (NSERC) for funding this research.Finally, I would like to recognize the support of all of the staff of Unit5, who have made it a pleasure to work at the university; in particular, mylab mates Walaa Moursi and Liangjin Yao who supported me along thisjourney.xiDedicationFor Jim, Zoe, and Hailey.xiiChapter 1IntroductionThe study of firmly nonexpansive mappings and their connection tomonotone operators is motivated by the large number of problems to whichthese types of operators have been applied. Signal processing, image restora-tion, and phase retrieval problems are examples that can be solved usingprojection methods, where projections are a type of firmly nonexpansiveoperator. The general problem is to find a point x in the intersection of nconvex subsets of a Hilbert space. That is, for convex subsets C1, . . . , Cnand C =n⋂i=1Ci 6= ∅,Find x ∈ C. (1.1)Equation (1.1) is referred to as the convex feasibility problem. Numerousalgorithms have been created to solve these types of problems and thosealgorithms make use of the operators studied in this thesis. The simplestexample of such an algorithm is the method of alternating projections, whereC1 and C2 are convex sets with C1 ∩ C2 6= ∅ and the update formula isxk+1 = PC1 ◦ PC2xk,where PC denotes the projection operator, discussed in more detail in Chap-ter 2. The method of alternating projections, and variations thereof, was thedriving force behind the study of compositions and convex combinations offirmly nonexpansive mappings. The majority of the background theory usedin this thesis can be found in Rockafeller’s Convex Analysis, [58]; Rockafellerand Wet’s Variational Analysis, [59]; and Bauschke and Combette’s ConvexAnalysis and Monotone Operator Theory in Hilbert Spaces, [11].The rest of this thesis is organized as follows:Chapter 2 gives notations and background information on operators,convex analysis, and methods of averaging operators. Chapter 3 coversknown results on nonexpansive mappings and monotone operators.My contribution begins in Chapter 4, with an in-depth look at whichproperties of resolvents correspond to properties of their associated mono-tone operator. Dual and self-dual properties are also identified. The materialin this chapter is based on [19].1Chapter 1. IntroductionChapter 5 then introduces the resolvent average, a new method of aver-aging monotone operators based on the convex combination of the resolventsof the operators. Basic properties of the resolvent average of monotone op-erators are gathered and properties specific to positive semidefinite matricesare also derived. Results in this chapter can be found in [18] and [21]Chapter 6, based on the paper [20], uses the notions of near equality andnear convexity to study convex combinations of monotone operators andfirmly nonexpansive mappings.In Chapter 7, it is shown that compositions and convex combinationsof asymptotically regular mappings maintain asymptotic regularity. Thischapter is based on [15].Chapter 8, based on [21], looks at how properties of monotone operatorsand their resolvents extend to the resolvent average. Properties are classifiedaccording to whether they are(i) dominant, i.e. only one averaged operator needs the property to ensurethe average maintains the property, or(ii) recessive, i.e. all average operators need to have the property to ensurethe average has the property.Finally, the key results of this thesis are summarized in Chapter 9.2Chapter 2Preliminary Details2.1 Normed vector spacesWe work in a variety of spaces throughout this thesis, most commonlyHilbert and Euclidean spaces, which are both subclasses of the class of Ba-nach spaces.Definition 2.1. A Banach space, X , is a complete normed vector space.Definition 2.2. A Hilbert space, H, is a complete inner product space.Let H denote a real Hilbert space, with inner product 〈·, ·〉 and inducednorm ‖ · ‖. The n-dimensional Euclidean space, Rn, is a classic example ofa Hilbert space. The real numbers, nonnegative real numbers, and strictlypositive real numbers are indicated by R, R+, and R++ respectively. Wealso denote the strictly positive integers, 1, 2, 3, . . . by N and the rationalnumbers by Q. Let I be an index set with I = {1, 2, . . . ,m} for someinteger m and letHm = {x = (xi)i∈I | (∀i ∈ I) xi ∈ H},denote the Hilbert product space with inner product 〈x,y〉 =∑i∈I xiyi.Clearly, every Euclidean space is a Hilbert space and every Hilbert spaceis a Banach space.Example 2.3. [37, Example 1.19(2)] The space of square summable se-quences,`2(N) = {(xn)n∈N |∞∑n=1|xn|2 <∞},with inner product 〈x, y〉 =∞∑n=1xnyn is a Hilbert space.Example 2.4. [32, pg. 2] Let n ≥ 2. Then Rn with the infinity norm,‖x‖∞ = max{|x1|, . . . , |xn|},32.2. Operatorsis a Banach space that is not a Hilbert space since ‖x‖∞ is not induced byan inner product.A sequence (xn)n∈N in H converges strongly to a point x iflimn→∞‖xn − x‖ = 0.This is written xn → x. The sequence converges weakly to x, or xn ⇀ x, iffor every u ∈ H,limn→∞〈xn, u〉 = 〈x, u〉 .In the space SN of N ×N real symmetric matrices, SN+ denotes the setof N ×N positive semidefinite matrices, and SN++ the set of positive definitematrices. For A,B ∈ SN , we write B A if A − B ∈ SN+ and B ≺ A ifA−B ∈ SN++.Example 2.5. [11, Example 2.4] SN with inner product 〈A,B〉 = tr(AB)is a Hilbert space, where tr is the trace function defined by trA =N∑i=1aii.A 2× 2 matrix A is called a rotation matrix if A is of the formA =(cos θ − sin θsin θ cos θ),for some angle θ. A matrix A is orthogonal if A−1 = AT , where AT denotesthe transpose of A.Fact 2.6. [48, 3.7.16] For an invertible matrix A,(A−1)T=(AT)−1.The symmetric part of an N ×N matrix A isA+ =12(A+AT ).2.2 Operators2.2.1 Single-valued operatorsLet H1 and H2 be real Hilbert spaces with D ⊆ H1. Let T : D → H2denote an operator (or mapping) T that maps every point x ∈ D to a pointTx ∈ H2. The range of T isranT = {y ∈ H2 | ∃x ∈ H1 with Tx = y}.42.2. OperatorsWe setB(H1,H2) = {T : H1 → H2 | T is linear and continuous}.Fact 2.7. [37, 8.25] For T ∈ B(H1,H2), the adjoint of T is the uniqueoperator T ∗ that satisfies(∀x ∈ H1)(∀y ∈ H2) 〈Tx, y〉 = 〈x, T∗y〉 .Example 2.8. Let R be the cyclic right-shift operator,R : Hm → Hm : (x1, x2, . . . , xm) 7→ (xm, x1, . . . , xm−1).Let x = (x1, . . . , xm) and y = (y1, . . . , ym) ∈ Hm, then R∗ satisfies〈Rx, y〉 = 〈x,R∗y〉 ⇔ 〈(xm, x1, . . . , xm−1), (y1, y2, . . . ym)〉 = 〈x,R∗y〉⇔ 〈x,R∗y〉 = y1xm + y2x1 + . . .+ ymxm−1⇔ 〈x,R∗y〉 = x1y2 + x2y3 + . . .+ xm−1ym + xmy1.Thus R∗ is the cyclic left-shift operatorR∗ : Hm → Hm : (x1, x2, . . . , xm) 7→ (x2, x3, . . . , xm, x1).The kernel of T is kerT = {x ∈ H | Tx = 0}.Fact 2.9. [37, Lemma 8.33(2)] Let T ∈ B(H1,H2). Then(kerT )⊥ = ranT ∗,where (kerT )⊥ denotes the orthogonal complement of kerT , i.e.,(kerT )⊥ = {u ∈ H1 | (∀x ∈ kerT ) 〈x, u〉 = 0}.If H2 ⊆ H1, then Tnx denotes the n-fold composition of T . The identitymapping is the operator Id : H → H : x 7→ x.Definition 2.10. Let T : H → H. T is Lipschitz continuous with constantβ if(∀x ∈ H)(∀y ∈ H) ‖Tx− Ty‖ ≤ β‖x− y‖.If β ∈ ]0, 1[, then T is called a Banach contraction.52.2. OperatorsDefinition 2.11. T is sequentially weakly continuous if for every sequence(xn)n∈N in H such that xn ⇀ x, then Txn ⇀ Tx.Definition 2.12. A mapping T : H → H is an isometry if(∀x ∈ H)(∀y ∈ H) ‖Tx− Ty‖ = ‖x− y‖. (2.1)When T : H → H is linear, the quadratic form qT : H → R is defined byqT (x) = 12 〈Tx, x〉 ∀x ∈ H,and qId = q is used interchangeably.Fact 2.13. [11, Corollary 15.34] Let H1 and H2 be Hilbert spaces and T ∈B(H1,H2). Then ranT is closed if and only if ranT ∗ is closed.Fact 2.14 (Closed Range Theorem). [37, Theorem 8.18] Let H1 and H2 beHilbert spaces and T ∈ B(H1,H2) \ {0}. Then the following are equivalent:(i) T has closed range;(ii) There exists ρ > 0 such that ‖Tx‖ ≥ ρ‖x‖ for all x ∈ (kerT )⊥;(iii) ρ := inf{‖Tx‖ | x ∈ (kerT )⊥, ‖x‖ = 1} > 0.Definition 2.15 (Gaˆteaux differentiability). Let T : C → X , with C ⊆ Hand X a real Banach space. Let x ∈ C be such that (∀y ∈ H)(∃α ∈ R++)[x, x+ αy] ⊆ C. Then T is Gaˆteaux differentiable at x if there exists anoperator DT (x) ∈ B(H,X ), called the Gaˆteaux derivative of T at x, suchthat(∀y ∈ H) DT (x)y = limα→0+T (x+ αy)− T (x)α.Remark 2.16. Unless otherwise noted, when differentiability is mentionedthen Gaˆteaux differentiability is assumed.Definition 2.17 (Fre´chet differentiability). Let x ∈ H and let T : U → X ,where U is an open subset of H and X is a real Banach space. Then Tis Fre´chet differentiable at x if there exists an operator DT (x) ∈ B(H,X ),called the Fre´chet derivative of T at x such thatlim06=‖y‖→0‖T (x+ y)− Tx−DT (x)y‖‖y‖= 0.62.2. OperatorsFact 2.18 (Fre´chet-Riesz Representation Theorem). [37, Theorem 6.10] Letf ∈ B(H,R). Then there exists a unique vector u ∈ H such that(∀x ∈ H) f(x) = 〈x, u〉 .Moreover, ‖f‖ = ‖u‖.Let C ⊆ H, f : C → R and suppose that f is Fre´chet differentiable atx ∈ C. Then by Fact 2.18, there exists a unique vector ∇f(x) ∈ H suchthat(∀y ∈ H) Df(x)y = 〈y,∇f(x)〉 .We call ∇f(x) the Fre´chet gradient of f at x. If f is Fre´chet differentiableon C the gradient operator is∇f : C → H : x 7→ ∇f(x).The Gaˆteaux gradient is defined similarly.2.2.2 Set-valued operatorsAn operator A : H1 ⇒ H2 is set-valued if (∀x ∈ H1) Ax ⊆ H2. Forset-valued operators,domA = {x | Ax 6= ∅},andranA =⋃x∈domAAx.A set-valued operator A is characterized by its graphgraA = {(x, u) ∈ H1 ×H2 | u ∈ Ax}.The set-valued inverse A−1 of A is defined by(y, x) ∈ graA−1 ⇔ (x, y) ∈ graA.The operator A is at most single-valued if Ax is a singleton or Ax = ∅. Thesum of operators,A+B : x 7→ Ax+Bx,and therefore gra(A+B) = {(x, u+v) ∈ H1×H2 | (x, u) ∈ graA and (x, v) ∈graB} and dom(A+B) = domA ∩ domB.72.3. Convex analysisDefinition 2.19. Let A : H ⇒ H. Then A is a linear relation if graA isa linear subspace of H×H. Similarly, A is an affine relation if graA is anaffine subspace of H×H, i.e. ifgraA 6= ∅ and (∀λ ∈ R) graA = λ graA+ (1− λ) graA.See [36] for more on linear relations.Fact 2.20. [36, I.2.3 and I.4] Let A,B be linear relations on H. Then A−1and A+B are linear relations.Definition 2.21. An operator A : H⇒ H is disjointly injective if(∀x ∈ H)(∀y ∈ H) x 6= y ⇒ Ax ∩Ay = ∅. (2.2)2.3 Convex analysis2.3.1 Convex setsA subset C of H is convex if for all x, y ∈ C and λ ∈ ]0, 1[,λx+ (1− λ)y ∈ C.The closure of C is denoted by C. A set C is sequentially weakly closed ifevery weakly convergent sequence (xn)n∈N in C has its weak limit x also inC.The intersection of all the convex sets containing C is called the convexhull of C, and is denoted by convC. The intersection of all affine subspacescontaining C is likewise called the affine hull of C and is denoted by aff C.A subset C of H is a cone if C = R++C. That is, x ∈ C and λ > 0implies λx ∈ C. A convex cone is a set that is both convex and a cone. Theconical hull of C, coneC, is the intesection of all the cones in H containingC. The smallest linear subspace of H containing C is spanC. The interiorof C is the largest open set contained in C,intC = {x | ∃ > 0, B(x, ) ⊆ C}.The relative interior of C isriC = {x ∈ aff C | ∃ > 0, B(x, ) ∩ aff C ⊆ C},where B(x, ) is a ball centered at x with radius .82.3. Convex analysisLemma 2.22. Let A and B be subsets of Rn such that A ⊆ B and aff A =aff B. Then riA ⊆ riB.Proof. This follows directly from the definition.Fact 2.23. [58, pg. 44] Let A be a subset of Rn. Then A ⊆ aff A.Lemma 2.24. Let A and B be subsets of Rn such that A = B. Thenaff A = aff B.Proof. Let x ∈ aff A. Then x = λ1a1 + · · · + λmam for ai ∈ A, λi ∈ R,i = 1, . . . ,m, andm∑i=1λi = 1. Clearly, each ai ∈ A = B and by Fact 2.23,B ⊆ aff B, so x is an affine combination of elements in aff B. Hence x ∈aff B. Altogether, aff A ⊆ aff B. Similarly, you can show aff B ⊆ aff A, andthus aff A = aff B.Fact 2.25 (Rockafellar). Let C and D be convex subsets of Rn, and letλ ∈ R. Then the following hold.(i) riC and C are convex.(ii) C 6= ∅ ⇒ riC 6= ∅.(iii) riC = C.(iv) riC = riC.(v) aff riC = aff C = aff C.(vi) riC = riD ⇔ C = D ⇔ riC ⊆ D ⊆ C.(vii) riλC = λ riC.(viii) ri(C +D) = riC + riD.Proof. (i)&(ii): See [58, Theorem 6.2]. (iii)&(iv): See [58, Theorem 6.3].(v): See [58, Theorem 6.2]. (vi): See [58, Corollary 6.3.1]. (vii): See [58,Corollary 6.6.1]. (viii): See [58, Corollary 6.6.2].Fact 2.26. [58, Theorem 6.5] Let Ci be a convex set in Rn for i = 1, . . . ,msuch thatm⋂i=1riCi 6= ∅. Thenm⋂i=1Ci =m⋂i=1Ci,92.3. Convex analysisandrim⋂i=1Ci =m⋂i=1riCi.Fact 2.27. [59, Proposition 2.40] Let C 6= ∅ be a convex subset of Rn. ThenriC is nonempty and convex withriC = C.Definition 2.28. Let C be a convex subset of H, the indicator function ofC at x isιC(x) ={0 if x ∈ C,+∞ if x /∈ C.(2.3)Fact 2.29 (projection). [11, Definition 3.7]Let C be a nonempty, closed,convex subset of H and let x ∈ H. Then there exists a unique vector p ∈ Csuch that‖x− p‖ = infy∈C‖x− y‖,and p is called the projection of x onto the set C, denoted by PCx.Fact 2.30 (projection characterization). [11, Theorem 3.14] Let C be aclosed convex subset of H. For every x and p in H, p = PCx if and only ifp ∈ C and (∀y ∈ C) 〈y − p, x− p〉 ≤ 0. (2.4)Example 2.31. Let x ∈ R2 and C = {(x1, x2) ∈ R2 | x1 = x2}. ThenPC(x) = 12(x1 + x2, x1 + x2).Proof. Let x = (x1, x2) and y = (y1, y1) ∈ C. Clearly,p =12(x1 + x2, x1 + x2) ∈ C,and〈y − p, x− p〉=〈(y1, y1)−12(x1 + x2, x1 + x2), (x1, x2)−12(x1 + x2, x1 + x2)〉=12y1(x1 − x2)−14(x1 + x2)(x1 − x2)−12y1(x1 − x2)−14(x1 + x2)(x2 − x1)= 0.Thus by Fact 2.30, p = 12(x1 + x2, x1 + x2) = PC(x).102.3. Convex analysisExample 2.32. Let ∆ ⊆ Hm with ∆ ={x = (x)i∈I∣∣ x ∈ H}and x ∈ Hm,then P∆x satisfies‖x− P∆x‖2 = infy∈∆‖x− y‖2= infy∈∆((x1 − y)2 + (x2 − y)2 + · · ·+ (xm − y)2) . (2.5)Since ‖ · ‖2 is convex, differentiating (2.5) with respect to y, setting it equalto zero and solving for y yields−2(x1 − y)− 2(x2 − y)− · · · − 2(xm − y) = 0⇔ y =1mm∑i=1xiThus, P∆x =(1mm∑i=1xi, 1mm∑i=1xi, . . . , 1mm∑i=1xi).Definition 2.33. Let C be a nonempty convex subset of H and x ∈ H.Then the normal cone operator to C at x isNCx ={{u ∈ H | sup 〈C − x, u〉 ≤ 0} if x ∈ C;∅ otherwise.Definition 2.34. Let H1 and H2 be real Hilbert spaces, let T ∈ B(H1,H2),let x ∈ H1 and y ∈ H2. Then x is a least-squares solution to the equationTz = y if‖Tx− y‖ = minz∈H1‖Tz − y‖.Fact 2.35. [11, Proposition 3.25] Let H1 and H2 be real Hilbert spaces, letT ∈ B(H1,H2) be such that ranT is closed, and let y ∈ H2. Then theequation Tz = y has at least one least-squares solution. Moreover, for everyx ∈ H1, the following are equivalent:(i) x is a least-squares solution,(ii) Tx = PranT y,(iii) T ∗Tx = T ∗y (normal equation).Definition 2.36 (Moore-Penrose inverse). Let H1 and H2 be real Hilbertspaces, let T ∈ B(H1,H2) be such that ranT is closed and for every y ∈ H2,set Cy = {x ∈ H1 | T ∗Tx = T ∗y}. The Moore-Penrose inverse of T isT † : H2 → H1 : y 7→ PCy0.112.3. Convex analysisSee [41] for more on the Moore-Penrose inverse.Fact 2.37. [11, Proposition 3.28(v)] Let H1 and H2 be real Hilbert spaces,let T ∈ B(H1,H2) be such that ranT is closed. ThenranT † = ranT ∗.2.3.2 Convex functionsA function f : H → ]−∞,+∞] = R ∪ {+∞} is said to be convex if its(essential) domain, dom f = {x ∈ H | f(x) < +∞}, is a convex set and∀ x, y ∈ H, 0 < λ < 1,f(λx+ (1− λ)y) ≤ λf(x) + (1− λ)f(y), (2.6)with f being strictly convex if (2.6) becomes a strict inequality wheneverx 6= y. A function f is proper if(∀x ∈ H) f(x) > −∞ and (∃x0 ∈ H) such that f(x0) < +∞.A function f is lower semi-continuous if for every sequence (xn)n∈N in H,xn → x⇒ f(x) ≤ lim infn→∞f(xn).The epigraph of f isepi f = {(x, r) ∈ H × R | f(x) ≤ r}.For α > 0, epi-multiplication isα ? f = αf(·/α).The lower semi-continuous hull of f is the function whose epigraph is theclosure in H× R of the epigraph of f .The class of proper lower semi-continuous convex functions from H →]−∞,+∞] will be denoted by Γ0(H). For f ∈ Γ0(H), ∂f denotes its convexsubdifferential,∂f(x) = {x∗ ∈ H : f(y) ≥ f(x) + 〈x∗, y − x〉 ∀y ∈ H}.If f is continuous and differentiable at x, then ∂f(x) = {∇f(x)}, see [63,Theorem 2.4.4(i)]. The function f∗ denotes its Fenchel conjugate given by(∀x∗ ∈ H) f∗(x∗) = supx{〈x∗, x〉 − f(x)}.If f, g ∈ Γ0(H), fg stands for the infimal convolution of f with g givenby(∀x ∈ H) (fg)(x) = inf{f(x1) + g(x2) : x1 + x2 = x}.122.3. Convex analysisFact 2.38. [11, Example 16.12] Let C be a convex subset of H. Then∂ιC = NC .Example 2.39. Set C = {0} and let x ∈ H. Then by Fact 2.38, we have∂ι{0}(x) = N{0}x={{u ∈ H | sup 〈0, u〉 ≤ 0} if x ∈ {0};∅ otherwise.={H if x = 0;∅ otherwise.Definition 2.40. Let f : H → ]−∞,+∞] be proper and let β ∈ R++.Then f 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).Fact 2.41. [11, Proposition 10.6] Let f : H → ]−∞,+∞] be proper and letβ ∈ R++. Then f is strongly convex with constant β if and only if f − βqis convex.Fact 2.42. [11, Lemma 2.13] Let (xi)i∈I and (ui)i∈I be finite families in H,and let (αi)i∈I be a family in R such that∑i∈I αi = 1. Then the followingholds〈∑i∈Iαixi,∑j∈Iαjuj〉+∑i∈I∑j∈Iαiαj 〈xi − xj , ui − uj〉 /2 =∑i∈Iαi 〈xi, ui〉 .In particular, ‖ · ‖2 is strongly convex and‖∑i∈Iαixi‖2 =∑i∈Iαi‖xi‖2 −∑i∈I∑j∈Iαiαj‖xi − xj‖2/2. (2.7)Fact 2.43. [10, Theorem 2.1] Let f ∈ Γ0(H) and let β ∈ R++. Then thefollowing are equivalent(i) f is Fre´chet differentiable on H and ∇f is β-Lipschitz continuous.132.3. Convex analysis(ii) f∗ is 1β -strongly convex.Definition 2.44. [8, 58] A proper convex function f on RN is essentiallystrictly convex if f is strictly convex on every convex subset of dom ∂f .Definition 2.45. [8, 58] A proper convex function f on RN is essentiallysmooth if it satisfies the following conditions for C := int(dom f):(i) C is not empty;(ii) f is differentiable throughout C;(iii) limn→∞|∇f(xn)| = +∞ whenever (xn)n∈N is a sequence in C convergingto a point x ∈ bdryC := C \ intC.Definition 2.46. Let f ∈ Γ0(RN ). Then f is Legendre if f is essentiallysmooth and essentially strictly convex.Fact 2.47. [58, Theorem 26.1] Let f be a closed proper convex function.Then ∂f is a single-valued mapping if and only if f is essentially smooth.Fact 2.48. [58, Theorem 26.3] A closed proper convex function f is essen-tially strictly convex if and only if f∗ is essentially smooth.Definition 2.49. Let f : H → [−∞,+∞]. Then f is coercive iflim‖x‖→+∞f(x) = +∞,and f is supercoercive iflim‖x‖→+∞f(x)‖x‖= +∞.Fact 2.50. [11, Proposition 12.15] Let f ∈ Γ0(H). Then the infimal convo-lution,fq : H → ]−∞,+∞] : x 7→ infy∈H(f(y) + q(x− y)) ,is convex, real-valued, continuous, and the infimum is uniquely attained.Remark 2.51. In Fact 2.50 the existence of a minimizer follows from thesupercoercivity of q while the uniqueness follows from the strict convexityof q. This motivates the next definition.142.3. Convex analysisDefinition 2.52 (proximal mapping). [11, Definition 12.23] Let f ∈ Γ0(H)and let x ∈ H. Then proxf x is the unique point in H which satisfiesminy∈H(f(y) +12‖x− y‖2)= f(proxf x) +12‖x− proxf x‖2.The operator proxf x : H → H is the proximal mapping or proximity opera-tor of f .Fact 2.53. [11, Proposition 16.34] Let f ∈ Γ0(H) and let x, p ∈ H. Thenp = proxf x⇔ x− p ∈ ∂f(p).In other words,proxf x = (Id +∂f)−1.Fact 2.54. [59, Exercise 11.27] or [11, Remark 14.4] Let f ∈ Γ0(H). Thenproxf = ∇(f∗ q),where ∇(f∗ q) is the Fre´chet gradient of f∗ q.Example 2.55. Set f = ‖ · ‖, thenproxf x ={(1− 1‖x‖)x if ‖x‖ > 1;0 if ‖x‖ ≤ 1.Proof. Set g(y) = ‖y‖ + 12‖x − y‖2. If x = 0, then clearly y = 0 is theminimizer of g(y). We consider two cases: Case 1: ‖x‖ ≤ 1. If y = 0,g(0) = 12‖x‖2. If ‖y‖ = ‖x‖, we haveg(y) = ‖y‖+12‖x‖2 − 〈x, y〉+12‖y‖2≥ ‖y‖+12‖x‖2 − ‖x‖‖y‖+12‖y‖2= ‖x‖+12‖x‖2 − ‖x‖‖x‖+12‖x‖2 = ‖x‖≥12‖x‖2,so any y such that ‖y‖ = ‖x‖ is not the minimizer. Clearly if ‖y‖ > ‖x‖then g(y) > g(0). Finally, if ‖y‖ = λ‖x‖ for some λ ∈ ]0, 1[ we haveg(y) ≥ ‖y‖+12‖x‖2 − ‖x‖‖y‖+12‖y‖2= λ‖x‖+12‖x‖2 − λ‖x‖2 +λ22‖x‖2= λ(‖x‖ − ‖x‖2) +12(1 + λ2)‖x‖2.152.3. Convex analysisThe first term is greater than or equal to zero, while the second term isgreater than 12‖x‖2. Altogether y = 0 is the minimizer if ‖x‖ < 1.Case 2: ‖x‖ > 1. When ‖x‖ > 1, g(x) < g(0). Thus y = 0 is not theminimizer and since g(y) is convex, differentiating with respect to y andsettling equal to zero will yield the minimizer. Doing this, we havey‖y‖− (x− y) = 0⇔ x = y(1‖y‖+ 1). (2.8)Taking norms of each side of (2.8) gives,‖x‖ = ‖y‖(1‖y‖+ 1)⇔ ‖x‖ = 1 + ‖y‖ ⇔ ‖y‖ = ‖x‖ − 1.So if ‖x‖ > 1, (2.8) becomes y = x(1− 1‖x‖)and if ‖x‖ ≤ 1, y = 0 is theminimizer.Fact 2.56. [58, Theorem 16.4] Let f1, · · · , fn be proper convex functions onRN . Then(f1 · · ·fn)∗ = f∗1 + · · ·+ f∗n. (2.9)If the sets ri(dom fi), i = 1, · · · , n have a point in common, then(f1 + · · ·+ fn)∗(x∗) = infx∗1+···+x∗n=x∗(f∗1 (x∗1) + · · ·+ f∗n(x∗n)) , (2.10)where for each x∗ the infimum is attained.Fact 2.57. [58, page 108] Let A ∈ SN++. Then(qA)∗ = qA−1 .Fact 2.58. [58, Theorem 12.3] Let A ∈ SN++ be an injective linear operator,a and b ∈ RN and r ∈ R. Setf(x) = qA(x− a) + 〈x, b〉+ r,Thenf∗(x∗) = qA−1(x∗ − b) + 〈x∗, a〉 − 〈a, b〉 − r.162.4. AveragesFact 2.59. [58, Theorem 23.5] Let f ∈ Γ0(Rn). Then∂f∗ = (∂f)−1.Fact 2.60. [58, Theorem 25.7] Let C be a nonempty open convex subset ofRN , and let f be a convex function which is finite and differentiable on C.Let f1, f2, . . . , be a sequence of convex functions finite and differentiable onC such that limi→∞fi(x) = f(x) for every x ∈ C. Thenlimi→∞∇fi(x) = ∇f(x), ∀x ∈ C.In fact, the sequence of gradients ∇fi converges to ∇f uniformly on everycompact subset of C.2.4 AveragesThere are many methods of averaging; this section gathers the definitionsof some methods that will be of interest.2.4.1 Arithmetic and harmonic averagesThe most commonly used averages are the arithmetic, harmonic, and ge-ometric averages. Let Ai, i = 1, . . . , n be N ×N positive semidefinite matri-ces, λi be strictly positive real coefficients withn∑i=1λi = 1, A = (A1, . . . , An),and λ = (λ1, . . . , λn).Definition 2.61. (Arithmetic average) The λ-weighted arithmetic averageof A isA(A,λ) = λ1A1 + · · ·+ λnAn. (2.11)Definition 2.62. (Harmonic average) The λ-weighted harmonic average ofA isH(A,λ) = (λ1A−11 + · · ·+ λnA−1n )−1. (2.12)2.4.2 Geometric meanFor matrices A,B ∈ SN++, the geometric mean is defined byA]B = A12(A−12BA−12) 12A12 .There have been several suggestions for how to define the geometric meanof A1, . . . , An ∈ SN+ for n ≥ 3, [1, 44, 51, 54].172.4. AveragesDefinition 2.63. (Geometric mean) Let x = (x1, . . . , xn) ∈ Rn with xi > 0for all i = 1, . . . , n. The λ-weighted geometric average of x isG(x,λ) = xλ11 xλ22 · · ·xλnn .The weighted geometric mean always has the following properties:Fact 2.64. Let x = (x1, . . . , xn) and y = (y1, . . . , yn) ∈ Rn such that(∀ i) xi > 0, yi > 0, and x−1 = (x−11 , . . . , x−1n ). Let λi ∈ R++ such thatn∑i=1λi = 1. Then we have(i) (harmonic-geometric-arithmetic mean inequality):(λ1x−11 + · · ·+ λnx−1n)−1≤ G(x,λ) ≤ λ1x1 + · · ·+ λnxn.Moreover, G(x,λ) = λ1x1 + · · ·+ λnxn if and only x1 = · · · = xn.(ii) (self-duality): [G(x,λ)]−1 = G(x−1,λ).(iii) If x = (x1, . . . , x1), then G(x,λ) = x1.(iv) If z = (x1, x−11 , x2, x−12 , . . . , xn, x−1n ) and µ = (12n , . . . ,12n), thenG(z, µ) = 1.(v) The function x 7→ G(x,λ) is concave on R++ × · · · × R++.(vi) If x y, then G(x,λ) ≥ G(y,λ).Proof. (i): See [58, page 29]. (ii)-(iv) and (vi) are simple. (v): See [59,Example 2.53].2.4.3 Proximal averageOne key tool used later is the proximal average of convex functions, whichfinds its roots in [16, 50, 52], and which has been further systematicallystudied in [12–14, 22].Definition 2.65 (proximal average). Let (∀i) fi ∈ Γ0(H) and λi be strictlypositive real numbers withn∑i=1λi = 1. The λ-weighted proximal average off = (f1, . . . , fn) with parameter µ > 0 is defined byPµ(f ,λ) =(λ1(f1+ 1µq)∗+λ2(f2+ 1µq)∗+· · ·+λn(fn+ 1µq)∗)∗− 1µq. (2.13)182.4. AveragesThe function Pµ(f ,λ) is a proper lower semi-continuous convex functionon H, and it inherits many desirable properties from each underlying func-tion fi; see [12, 13]. The next fact is a fundamental property of the proximalaverage.Fact 2.66. [12, Theorem 5.1](Pµ(f ,λ))∗= Pµ−1(f∗,λ).Lemma 2.67. [11, Lemma 2.13(ii)] Let x1, . . . xn ∈ RN and λi ∈ R++ suchthatn∑i=1λi = 1. Then the following identity holds:n∑i=1λiq(xi)− q(n∑i=1λixi) =14n∑i=1n∑j=1λiλj‖xi − xj‖2.Proof. From Fact 2.42, we have‖n∑i=1λixi‖2 −n∑i=1λi‖xi‖2 =12n∑i=1n∑j=1λiλj‖xi − xj‖2. (2.14)Multiplying (2.14) by 12 on each side gives the desired identity.The following reformulation of the proximal average will be useful.Proposition 2.68. Let f1, . . . , fn ∈ Γ0(RN ) and λ1, . . . , λn > 0 with∑ni=1 λi = 1. Then for every x ∈ RN ,Pµ(f ,λ)(x)= minx1+···+xn=x{λ1(f1 +1µq)(x1λ1) + · · ·+ λn(fn +1µq)(xnλn)}−1µq(x)(2.15)= minx1+···+xn=xλ1f1(x1λ1) + · · ·+ λnfn(xnλn) +14µn∑i=1n∑j=1λiλj‖xiλi−xjλj‖2(2.16)= minλ1y1+···+λnyn=x{λ1f1(y1) + · · ·+ λnfn(yn) +1µ[λ1q(y1) + · · ·+ λnq(yn)−q(λ1y1 + · · ·+ λnyn)]}(2.17)192.4. Averages= minλ1y1+···+λnyn=xλ1f1(y1) + · · ·+ λnfn(yn) +14µn∑i=1n∑j=1λiλj‖yi − yj‖2(2.18)= minx1+···+xn=x{λ1f1(x1λ1) + · · ·+ λnfn(xnλn) +1µ[λ1q(x−x1λ1) + · · ·+λnq(x−xnλn)]}. (2.19)Proof. Indeed, as (fi +1µq)∗= f∗i (µq),it is finite-valued everywhere, we writef = λ1 ? (f1 +1µq) · · ·λn ? (fn +1µq)−1µq,by Fact 2.56. That is, for every x,f(x) = infx1+···+xn=x{λ1(f1 +1µq)(x1λ1) + · · ·+ λn(fn +1µq)(xnλn)}−1µq(x),and the infimum is attained, again by Fact 2.56. Hence, replacing inf withmin we get (2.15).Now rewrite (2.15) asminx1+···+xn=x{λ1f1(x1λ1) + · · ·+ λnfn(xnλn) +1µ[λ1q(x1λ1) + · · ·+ λnq(xnλn)−q(x1 + · · ·+ xn)]}, (2.20)= minλ1y1+···+λnyn=x{λ1f1(y1) + · · ·+ λnfn(yn) +1µ[λ1q(y1) + · · ·+ λnq(yn)−q(λ1y1 + · · ·+ λnyn)]}.Thus, (2.16)–(2.18) follow by using Lemma 2.67. Next, recall thatx = x1 + · · ·+ xn,202.4. Averagesand observe that by expanding and simplifying we getλ1q(x1 + · · ·+ xn −x1λ1) + · · ·+ λnq(x1 + · · ·+ xn −xnλn)=λ12∥∥∥∥x−x1λ1∥∥∥∥2+ · · ·+λn2∥∥∥∥x−xnλn∥∥∥∥2=n∑i=1λi2〈x−xiλi, x−xiλi〉=n∑i=1λi2(‖x‖2 − 2〈x,xiλi〉+ ‖xiλi‖2)=12‖x‖2 − ‖x‖2 +n∑i=1λiq(xiλi)= λ1q(x1λ1) + · · ·+ λnq(xnλn)− q(x1 + · · ·+ xn),thus we have (2.19) by (2.20).Fact 2.69 (inequalities). [12, Theorem 5.4](λ1f∗1 + · · ·+ λnf∗n)∗ ≤ Pµ(f ,λ) ≤ λ1f1 + · · ·+ λnfn.Fact 2.70. [12, Example 4.5] Let α1, . . . , αn be strictly positive real numbersand suppose that (∀i) fi = αiq. ThenPµ−1(f ,λ) =(n∑i=1λi(αiq + µq)∗)∗− µq =(n∑i=1λiαi + µq)∗− µq=(n∑i=1λiαi + µ)−1q− µq.And thus,Pµ(f ,λ) =(n∑i=1λiαi + µ−1)−1− µ−1 q.Fact 2.71. [12, Corollary 7.7] Suppose that at least one function fi is es-sentially smooth and that λi > 0. Then Pµ(f ,λ) is essentially smooth.Fact 2.72. [12, Theorem 8.5] Let x ∈ RN . Then the function]0,+∞[→ ]−∞,+∞] : µ 7→ Pµ(f ,λ)(x) is decreasing. (2.21)212.4. AveragesConsequently, limµ→0+ Pµ(f ,λ)(x) and limµ→+∞ Pµ(f ,λ)(x) exist. In fact,limµ→0+Pµ(f ,λ)(x) = supµ>0Pµ(f ,λ)(x) =(λ1f1 + · · ·+ λnfn)(x) (2.22)andlimµ→+∞Pµ(f ,λ)(x) = infµ>0Pµ(f ,λ)(x) =(λ1 ? f1 · · ·λn ? fn)(x). (2.23)We have now covered the building blocks needed for the main focus ofthis thesis, nonexpansive mappings and monotone operators. In the nextchapter, we introduce several different notions of “nonexpansiveness” andmonotonicity and cover many of the known results about these kinds ofoperators.22Chapter 3Nonexpansive Mappings andMonotone OperatorsThis chapter contains a collection of known results involving nonexpan-sive mappings and monotone operators. We begin with the concept of anonexpansive mapping.3.1 Nonexpansive mappingsDefinition 3.1. Let D be a nonempty subset of H. A mapping T : D → His(i) nonexpansive, or Lipschitz continuous with constant 1, if(∀x ∈ D)(∀y ∈ D) ‖Tx− Ty‖ ≤ ‖x− y‖; (3.1)(ii) strictly nonexpansive if(∀x ∈ D)(∀y ∈ D) x 6= y ⇒ ‖Tx− Ty‖ < ‖x− y‖; (3.2)(iii) firmly nonexpansive if(∀x ∈ D)(∀y ∈ D) ‖Tx−Ty‖2 +‖(Id−T )x− (Id−T )y‖2 ≤ ‖x−y‖2;(3.3)(iv) a Banach contraction, or Lipschitz continuous with constant β, if thereexists β ∈ [0, 1[ such that(∀x ∈ D)(∀y ∈ D) ‖Tx− Ty‖ ≤ β‖x− y‖; (3.4)(v) strongly nonexpansive if T is nonexpansive and whenever (xn)n∈N and(yn)n∈N are sequences in D such that (xn − yn)n∈N is bounded and‖xn−yn‖−‖Txn−Tyn‖ → 0, it follows that (xn−yn)−(Txn−Tyn)→0.233.1. Nonexpansive mappingsRemark 3.2. Clearly, both firmly nonexpansive and strongly nonexpansiveimply nonexpansive. And in Hilbert spaces, Bruck and Reich showed thatfirmly nonexpansive implies strongly nonexpansive, see Fact 3.23. The op-posite implication does not hold, see Example 3.6. Thus we havefirmly nonexpansive⇒ strongly nonexpansive⇒ nonexpansive.Fact 3.3. [11, Proposition 4.2] Let D be a nonempty subset of H andT : D → H. Then the following are equivalent:(i) T is firmly nonexpansive.(ii) Id−T is firmly nonexpansive.(iii) 2T − Id is nonexpansive.(iv) (∀x ∈ D)(∀y ∈ D) ‖Tx− Ty‖2 ≤ 〈x− y, Tx− Ty〉.(v) (∀x ∈ D)(∀y ∈ D) 0 ≤ 〈Tx− Ty, (Id−T )x− (Id−T )y〉.Example 3.4. The identity mapping is both strongly nonexpansive andfirmly nonexpansive. However, when T = − Id, T is nonexpansive butit fails to be strongly nonexpansive, and consequently fails to be firmlynonexpansive.To see that, let x, y ∈ H. To see that T = − Id is not strongly nonex-pansive, set xn = Txn−1 = Tnx0 and yn = Tyn−1 = Tny0. Then (xn − yn)is bounded and for all n ∈ N,‖xn − yn‖ − ‖Txn − Tyn‖ = ‖xn − yn‖ − ‖ − xn + yn‖ = 0.But,(xn − yn)− (Txn − Tyn) = (xn − yn)− (−xn + yn) = 2(xn − yn)= 2(Tnx0 − Tny0),which only goes to zero if x0 = y0, so T is not strongly nonexpansive, andconsequently not firmly nonexpansive.Example 3.5. [11, Proposition 4.8] Let C be a nonempty closed convexsubset of H. Then the projection operator PC is firmly nonexpansive.Example 3.6. Let x, y ∈ R2, let C = R×{0} and D = {x ∈ R2 | x1 = x2}.Clearly, PC(x) = (x1, 0) and by Example 2.31 PD(x) = 12(x1 + x2, x1 + x2).243.1. Nonexpansive mappingsBy Example 3.5, PC and PD are firmly nonexpansive. Then consider T (x) =PCPD(x) = 12(x1 + x2, 0) with the points x = (1, 2) and y = (1, 3),‖Tx− Ty‖2 =∥∥∥∥12(3, 0)−12(4, 0)∥∥∥∥2=14and〈x− y, Tx− Ty〉 =〈(1, 2)− (1, 3),12(3, 0)−12(4, 0)〉=〈(0,−1), (−12, 0)〉= 0.Thus ‖Tx − Ty‖2 > 〈x− y, Tx− Ty〉, so by Fact 3.3(iv) T is not firmlynonexpansive.T is strongly nonexpansive though. Let (xn)n∈N and (yn)n∈N be se-quences in R2 such that (xn−yn) is bounded and ‖xn−yn‖−‖Txn−Tyn‖ →0. Set xn = (xn1 , xn2 ), yn = (yn1 , yn2 ), dn = xn1 − yn1 and en = xn2 − yn2 . Now,‖xn − yn‖ − ‖Txn − Tyn‖ → 0⇔ ‖xn − yn‖2 − ‖Txn − Tyn‖2 → 0⇔ (xn1 − yn1 )2 + (xn2 − yn2 )2 −(12(xn1 − yn1 + xn2 − yn2 ))2→ 0⇔ d2n + e2n −14(dn + en)2 → 0⇔34d2n +34e2n −12dnen → 0⇔14(2d2n + 2e2n + (dn − en)2)→ 0.Thus we havee2n → 0 and (dn − en)2 → 0. (3.5)And we see that(xn − yn)− (Txn − Tyn) = (dn, en)− (12dn +12en, 0)= (12dn −12en, en).Taking the norm,‖(12dn −12en, en)‖2 =14(dn − en)2 + e2n,which goes to zero by (3.5). Thus T is strongly nonexpansive.253.2. Fixed points and asymptotic regularityRemark 3.7. Example 3.6 shows both that strongly nonexpansive does notimply firmly nonexpansive and that the composition of two firmly nonex-pansive operators may fail to be firmly nonexpansive.Definition 3.8. Let D ⊆ H with D 6= ∅ and T : D → H be nonexpan-sive. Let α ∈ ]0, 1[. Then T is averaged with constant α if there exists anonexpansive operator N : D → H such that T = (1− α) Id +αN .Fact 3.9. T is firmly nonexpansive if and only if T is 1/2-averaged.Proof. This follows directly from Fact 3.3(iii).Definition 3.10. Let D ⊆ H with D 6= ∅ and T : D → H and let β ∈ R++.Then T is β-cocoercive if βT is firmly nonexpansive. That is,(∀x ∈ D)(∀y ∈ D) 〈x− y, Tx− Ty〉 ≥ β‖Tx− Ty‖2.Remark 3.11. T being β-cocoercive is the same as T−1 being β stronglymonotone, see Definition 3.31(iv). Thus β-cocoercive is also referred to asbeing β-inverse strongly monotone.Fact 3.12 (Baillon-Haddad Theorem). [4, Corollaire 10] or [11, Corollary18.16] Let f : H → R be a Fre´chet differentiable convex function and letβ ∈ R++. Then ∇f is β-Lipschitz continuous if and only if ∇f is (1/β)-cocoercive. In particular, ∇f is nonexpansive if and only if ∇f is firmlynonexpansive.Remark 3.13. For more on the Baillon-Haddad theorem, see [4] and [10].Definition 3.14. T is cyclically firmly nonexpansive if for every set of points{x1, . . . , xn} ⊆ H, where n ∈ {2, 3, . . .} and xn+1 = x1, we haven∑i=1〈xi − Txi, Txi − Txi+1〉 ≥ 0. (3.6)3.2 Fixed points and asymptotic regularitySeveral problems in science and engineering can be formulated as fixedpoint problems, where the set of desired solutions is the set of fixed pointsof T ,FixT :={x ∈ H∣∣ x = Tx}. (3.7)263.2. Fixed points and asymptotic regularityIf T is firmly nonexpansive and FixT 6= ∅, then the sequence of iterates(Tnx)n∈N (3.8)converges weakly to a fixed point [29]. The iterates xn+1 = Txn, for alln ∈ N, are referred to as Banach-Picard iterates. However, if the mappingis simply nonexpansive then this result does not hold. For example, T = − Idis nonexpansive with FixT = {0}, but (Tnx)n∈N converges only if you beginat the fixed point x = 0.Fact 3.15. [11, Corollary 4.15] Let C be a nonempty closed convex subsetof H and let T : C → H be nonexpansive. Then FixT is closed and convex.Fact 3.16. [62, Lemma 1.8, Corollary 2] Let C be a closed convex subset ofH and let T : H → H be a firmly nonexpansive mapping such thatranT ⊆ FixT = C.Then T = PC .Proof. Let x ∈ H and y ∈ C. Then,Tx ∈ ranT ⊆ C and y = Ty ∈ C = FixT.Since T is firmly nonexpansive, by Fact 3.3(v)0 ≤ 〈Tx− Ty, (x− Tx)− (y − Ty)〉⇔ 0 ≤ 〈Tx− y, x− Tx〉⇔ 〈y − Tx, x− Tx〉 ≤ 0.Thus by Fact 2.30, Tx = PCx.Definition 3.17. A mapping T : H → H is asymptotically regular if(∀x ∈ H) Tnx− Tn+1x→ 0.T is weakly asymptotically regular if the convergence is weak.Fact 3.18. [3, Theorem 1.2] Let T : H → H be a nonexpansive mapping.Then (Tnx)n∈N converges weakly to a fixed point of T if and only if FixT 6=∅ and T is weakly asymptotically regular.Fact 3.19. [3, Corollary 2.2] Let C be a closed convex subset of H. LetU : C → H be an averaged nonexpansive mapping. Then FixU = ∅ if andonly if limn→∞‖Unx‖ =∞ for all x in C.273.2. Fixed points and asymptotic regularityFact 3.20. [3, Corollary 2.3] Let C be a closed convex subset of H and letU : C → X be an averaged nonexpansive mapping. Then for each x ∈ Climn→∞(Unx− Un+1x)→ v,where v is the element of least norm in ran(Id−U).Remark 3.21. Facts 3.18 - 3.20 were originally formulated in a Banach spacewith additional structure. Details for how those results apply in Hilbertspaces are provided in Appendix A.Remark 3.22. Suppose T : H → H is asymptotically regular. Then, for everyx ∈ H,Tnx− Tn+1x→ 0⇔ (Id−T )Tnx→ 0and hence 0 ∈ ran(Id−T ). The opposite implication fails in general (con-sider T = − Id), but it is true for strongly nonexpansive mappings, seeFact 3.24.The next result illustrates that strongly nonexpansive mappings gener-alize the notion of firmly nonexpansive mappings. In addition, the class ofstrongly nonexpansive mappings is closed under compositions.Fact 3.23 (Bruck and Reich). [30, Proposition 2.1 and Proposition 1.1] Ina Hilbert space H, the following hold.(i) Every firmly nonexpansive mapping is strongly nonexpansive.(ii) The composition of finitely many strongly nonexpansive mappings isalso strongly nonexpansive.In contrast, the composition of two (necessarily firmly nonexpansive)projectors may fail to be firmly nonexpansive, see Example 3.6The sequences of iterates and of differences of iterates have striking con-vergence properties as we shall see now.Fact 3.24 (Bruck and Reich). [30, Corollary 1.5, Corollary 1.4, and Corol-lary 1.3] Let S : H → H be strongly nonexpansive and let x ∈ H. Then thefollowing hold.(i) The sequence (Snx− Sn+1x)n∈N converges strongly to the unique ele-ment of least norm in ran(Id−S).283.3. Monotone operators(ii) If FixS = ∅, then ‖Snx‖ → +∞.(iii) If FixS 6= ∅, then (Snx)n∈N converges weakly to a fixed point of S.Fact 3.25. [57, Corollary 2] Let D be a subset of H and let T : D → D befirmly nonexpansive. Set d = infy∈D‖y − Ty‖, then for each x ∈ D,limn→∞‖Tn+1x− Tnx‖ = d.3.3 Monotone operatorsWe now look at known results for monotone operators.Definition 3.26. A set-valued operator A : H⇒ H is monotone if(∀(x, u) ∈ graA)(∀(y, v) ∈ graA) 〈x− y, u− v〉 ≥ 0. (3.9)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)Lemma 3.27. Let A : H → H be linear. Then A is monotone if and onlyif(∀z ∈ H) 〈z,Az〉 ≥ 0.Proof. Since A is linear it is single-valued, thus (3.9) becomes(∀x ∈ H)(∀y ∈ H) 〈x− y,Ax−Ay〉 ≥ 0.Set z = x− y and by linearity we get〈z,Az〉 ≥ 0.Lemma 3.28. Let A : H⇒ H and λ ∈ R++. Then(x, u) ∈ graA⇔ (x, λu) ∈ graλA.Proof. Take (x, u) ∈ graA. Then u ∈ Ax ⇒ λu ∈ λAx, i.e. (x, λu) ∈graλA. On the other hand, let (x, λu) ∈ graλA, then λu ∈ λAx⇒ u ∈ Ax.Altogether, (x, u) ∈ graA⇔ (x, λu) ∈ graλA.293.3. Monotone operatorsProposition 3.29. Let A : H ⇒ H be maximally monotone and λ ∈ R++.Then λA is maximally monotone.Proof. Let (x, u) ∈ graλA, then by Lemma 3.28 (x, λ−1u) ∈ graA. Since Ais maximally monotone, (3.10) gives(x, λ−1u) ∈ graA⇔ (∀(y, λ−1v) ∈ graA)〈x− y, λ−1u− λ−1v〉≥ 0.Then for all (y, v) ∈ graλA,〈x− y, u− v〉 = λ〈x− y, λ−1u− λ−1v〉≥ 0.Conversely, let (x, u) ∈ H×H such that (∀(y, v) ∈ graλA) 〈x− y, u− v〉 ≥0. Then〈x− y, u− v〉 = λ〈x− y, λ−1u− λ−1v〉≥ 0⇒〈x− y, λ−1u− λ−1v〉≥ 0.That is, for every (x, λ−1u) ∈ H × H and for every (y, λ−1v) ∈ graA,〈x− y, λ−1u− λ−1v〉≥ 0. Thus by (3.10), (x, λ−1u) ∈ graA and therefore(x, u) ∈ graλA.Fact 3.30 (monotonicity versus convexity). [59, Theorem 12.17] Let H befinite dimensional and let f ∈ Γ0(H). Then ∂f is maximal monotone, andf is essentially strictly convex if and only if ∂f is strictly monotone.Definition 3.31. An operator A : H⇒ H is(i) paramonotone if it is monotone and(∀(x, u) ∈ graA)(∀(y, v) ∈ graA)〈x− y, u− v〉 = 0⇒ (x, v) ∈ graA.(ii) strictly monotone if(∀(x, u) ∈ graA)(∀(y, v) ∈ graA) x 6= y ⇒ 〈x− y, u− v〉 > 0.(iii) uniformly monotone with modulus φ : R+ → [0,+∞] if φ is increasing,vanishes only at zero, and(∀(x, u) ∈ graA)(∀(y, v) ∈ graA) 〈x− y, u− v〉 ≥ φ (‖x− y‖) .(iv) strongly monotone with constant β ∈ R++ if A − β Id is monotone.That is,(∀(x, u) ∈ graA)(∀(y, v) ∈ graA) 〈x− y, u− v〉 ≥ β‖x− y‖2.303.3. Monotone operatorsExample 3.32. Let z ∈ R+ and β ∈ R++. Set φ(z) = βz2, then it isclear that every operator that is strongly monotone with constant β is alsouniformly monotone with modulus φ.Example 3.33. [11, Example 22.3(iv)] Let f : H → ]−∞,+∞] be properand strongly convex with constant β ∈ R++. Then ∂f is strongly monotonewith constant β.Definition 3.34. Let A : H → H and α ∈ R. A is hemicontinuous if forevery (x, y, z) ∈ H3,limα→0+〈z,A(x+ αy)〉 = 〈z,Ax〉 .Fact 3.35. [11, Example 22.9(iii)] Let A : H → H be strongly monotoneand hemicontinuous, and let r ∈ H. Then the equation Ax = r has exactlyone solution.Now let A be a monotone operator from H⇒ H and denote the associ-ated resolvent byJA = (Id +A)−1. (3.11)For λ > 0, the Yosida λ-regularization of A is,λA = λ−1(Id−JλA). (3.12)The resolvent satisfies the useful resolvent identity,JA = Id−JA−1 , (3.13)which allows for the Minty parametrizationgraA ={(JAx, x− JAx)∣∣ x ∈ dom JA}(3.14)of the graph of A, which provides the bijection x 7→ (JAx, x − JAx) fromdom JA onto graA, with inverse (x, u) 7→ x + u. The Yosida regularizationis related to the resolvent through the following identity,λA = (λ Id +A−1)−1 = λ−1[Id−(Id +λA)−1]. (3.15)When A = ∂f for some f ∈ Γ0(H) then Fact 2.53 yields thatJ∂f = proxf . (3.16)Minty observed that JA is in fact a firmly nonexpansive operator fromH to H and that, conversely, every firmly nonexpansive operator arises thisway:313.3. Monotone operatorsFact 3.36. (See [38] and [50].) Let T : H → H and let A : H ⇒ H. Thenthe 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, and is single-valued and firmly nonexpansive, and A = J−1A − Id.Definition 3.37. [11, Definition 21.9] Let A : H ⇒ H and x ∈ H. ThenA is locally bounded at x if there exists δ ∈ R++ such that A(B(x; δ)) isbounded, where B(x; δ) is the closed ball centered at x with radius δ.Fact 3.38. [11, Corollary 21.19] Let A : H ⇒ H be maximally monotone.Then A is surjective if and only if A−1 is locally bounded everywhere on H.Fact 3.39. [11, Corollary 21.21] Let A : H ⇒ H be maximally monotonewith bounded domain. Then A is surjective.Fact 3.40. [11, Proposition 20.22] Let A : H⇒ H be maximally monotone,let u and z be in H and γ ∈ R++. Then A−1 and x 7→ u + γA(x + z) aremaximally monotone.Fact 3.41. [11, Example 20.41] Let C be a nonempty closed convex subsetof H. Then NC is maximally monotone.Fact 3.42. [11, Example 23.4] Let C be a nonempty closed convex subset ofH. ThenJNC = (Id +NC)−1 = proxιC = PC .Fact 3.43 (Minty’s Theorem). [11, Theorem 21.1] Let A : H ⇒ H bemonotone. Then A is maximally monotone if and only if ran(Id +A) = H.Remark 3.44. Minty’s Theorem provides a characterization for maximalmonotonicity which allows for determining maximality without having toshow graph inclusions.Fact 3.45. [11, Proposition 23.11] Let A : H ⇒ H be monotone and letβ ∈ R++. Then A is strongly monotone with constant β if and only if JAis (β+ 1)-cocoercive, in which case JA is Lipschitz continuous with constant1/(β + 1) ∈ ]0, 1[.Fact 3.46. [11, Example 20.26] Let T : H → H be nonexpansive and letα ∈ [−1, 1]. Then Id +αT is maximally monotone.323.3. Monotone operatorsWhile the sum of two monotone operators is still monotone, the sum oftwo maximally monotone operators can fail to be maximally monotone.Example 3.47. Let H = R2 and set C to be the closed unit ball centeredat (−1, 0) and D be the closed unit ball centered at (1, 0). By Fact 3.41,both NC and ND are maximally monotone and we havedomNC ∩ domND = {(0, 0)} 6= ∅.But given Fact 3.39 and the fact thatran(NC +ND) = R× {0},NC +ND is not maximally monotone.The next fact gives some constraint qualifications under which the sumis maximally monotone.Fact 3.48 (Rockafellar). [59, Theorem 12.44] and [11, Corollary 24.4] LetA and B be maximally monotone on H. Suppose one of the following holds:(i) domA ∩ int domB 6= ∅.(ii) If H = RN , ri domA ∩ ri domB 6= ∅.Then A+B is maximally monotone.Definition 3.49. Let A : H ⇒ H and let n ∈ N be such that n ≥ 2.Then A is n-cyclically monotone if, for every (x1, . . . , xn+1) ∈ Hn+1 and(u1, . . . , un) ∈ Hn,(x1, u1) ∈ graA, . . . , (xn, un) ∈ graA, xn+1 = x1 ⇒n∑i=1〈xi+1 − xi, ui〉 ≤ 0.If A is n-cyclically monotone for every integer n ≥ 2, then A is cyclicallymonotone. If A is cyclically monotone and there exists no cyclically mono-tone operator B : H ⇒ H such that graB properly contains graA, then Ais maximally cyclically monotone.Fact 3.50 (Rockafellar). [11, Theorem 22.14] Let A : H ⇒ H. Then A ismaximally cyclically monotone if and only if there exists f ∈ Γ0(H) suchthat A = ∂f .333.4. Rectangular monotone operatorsFact 3.51. [5, Theorem 6.6] Suppose H is a real Hilbert space and letT : H → H. Then T is the resolvent of the maximally cyclically mono-tone operator A : H ⇒ H if and only if T has full domain, T is firmlynonexpansive, and T is cyclically firmly nonexpansive. That is, for everyset of points {x1, . . . , xn} where n ∈ N, n ≥ 2 and xn+1 = x1, one hasn∑i=1〈xi − Txi, Txi − Txi+1〉 ≥ 0. (3.17)3.4 Rectangular monotone operatorsThe notion of rectangularity for monotone operators requires the use ofthe Fitzpatrick function.Definition 3.52 (Fitzpatrick function). (See [40], [31] or [47].) Let A : H⇒H. Then the Fitzpatrick function associated with A isFA : H×H → [−∞,+∞] :(x, x∗) 7→ sup(a,a∗)∈graA(〈x, a∗〉+ 〈a, x∗〉 − 〈a, a∗〉)(3.18)= 〈x, x∗〉 − inf(a,a∗)∈graA(〈x− a, x∗ − a∗〉) . (3.19)Example 3.53 (energy). [17, Example 3.10] The Fitzpatrick function ofthe identity operator isFId : H×H → R : (x, x∗) 7→ 14‖x+ x∗‖2.Definition 3.54 (Bre´zis-Haraux). (See [28].) Let A : H → H be mono-tone. Then A is rectangular (which is also known as star-monotone or 3∗monotone), ifdomA× ranA ⊆ domFA. (3.20)Remark 3.55. If A : H ⇒ H is maximally monotone and rectangular, thenone obtains the “rectangle” domFA = domA × ranA, which promptedSimons [61] to call such an operator rectangular. Such operators are alsoreferred to as star-monotone in [53] or (BH)-operators in [33].Proposition 3.56. A monotone operator A : H⇒ H is rectangular if(∀(x, y∗) ∈ domA× ranA) sup(z,z∗)∈graA〈x− z, z∗ − y∗〉 < +∞.343.4. Rectangular monotone operatorsProof. This follows from (3.20) and (3.19).Fact 3.57 (Rank-Nullity Theorem). [48, (4.4.15)] Let A : RN → RN be anN ×N matrix. Thendim ranA+ dim kerA = N.Fact 3.58. Let A : RN → RN be a linear maximally monotone operator.Then the following hold:(i) A is paramonotone if and only if A is rectangular;(ii) A is paramonotone if and only if rankA = rankA+ if and only ifranA = ranA+.Proof. (i) See [9, Remark 4.11] or [24, Corollary 4.11]. (ii) Since A is mono-tone, we have ranA+ ⊆ ranA. Thus, the result follows from [24, Corollary4.11] and Fact 3.57.Fact 3.59. [11, Proposition 24.15] Let A : H⇒ H be monotone. Then A isrectangular ⇔ A−1 is rectangular.Fact 3.60. [11, Proposition 24.18] Let A and B be monotone operatorsfrom H ⇒ H such that (domA ∩ domB) × H ⊆ domFB. Then A + B isrectangular.Example 3.61. (See [28, Example 3] or [2, Example 6.5.2(iii)].) LetA : H⇒H be maximally monotone. Then A + Id and (A + Id)−1 are maximallymonotone and rectangular.Proof. Combining Fact 3.60 and Example 3.53, we see that A + Id is rect-angular. Furthermore, A + Id is maximally monotone by Fact 3.48. UsingFact 3.59, we see that (Id +A)−1 is maximally monotone and rectangu-lar.Proposition 3.62. [17, Proposition 4.2] Let A and B be monotone on H,and let (x, x∗) ∈ H ×H. Then FA+B(x, x∗) ≤(FA(x, ·)FB(x, ·))(x∗).Lemma 3.63. [20, Lemma 3.11] Let A and B be rectangular on H. ThenA+B is rectangular.353.4. Rectangular monotone operatorsProof. Clearly, dom(A+B) = (domA)∩(domB), and ran(A+B) ⊆ ranA+ranB. Take x ∈ dom(A + B) and y∗ ∈ ran(A + B). Then there exista∗ ∈ ranA and b∗ ∈ ranB such that a∗ + b∗ = y∗. Furthermore, (x, a∗) ∈(domA) × (ranA) ⊆ domFA and (x, b∗) ∈ (domB) × (ranB) ⊆ domFB.Using Proposition 3.62 and the assumption that A and B are rectangular,we obtainFA+B(x, y∗) ≤ FA(x, a∗) + FB(x, b∗) < +∞. (3.21)Therefore, dom(A+B)×ran(A+B) ⊆ domFA+B and A+B is rectangular.Fact 3.64. [24, Theorem 6.1] Let T : H → H be nonexpansive and definethe corresponding displacement mapping byA = Id−T.Then the following hold:(i) A is maximally monotone.(ii) A is 12 -cocoercive, i.e.12A is firmly nonexpansive.(iii) A is rectangular.(iv) A−1 is strongly monotone with constant 12 .(v) A−1 is strictly monotone.(vi) A is paramonotone.Example 3.65. [24, Example 6.2] Let N be a strictly positive integer andletR : HN → HN : (x1, . . . , xN ) 7→ (xN , x1, . . . , xN−1),be the cyclic right-shift operator in HN . Since ‖Rx‖ = ‖x‖ for all x ∈ H, Ris nonexpansive and therefore by Fact 3.64, Id−R is maximally monotone,rectangular, and paramonotone.Fact 3.66 (Bre´zis-Haraux). [2, Theorem 6.5.1(b) and Theorem 6.5.2] LetA and B be monotone on a Hilbert space H such that A + B is maximallymonotone. Suppose that one of the following holds.(i) A and B are rectangular.(ii) domA ⊆ domB and B is rectangular.363.4. Rectangular monotone operatorsThen ran(A+B) = ranA+ ranB, int(ran(A + B)) = int(ranA + ranB),and if H is finite dimensional ri conv(ranA+ ranB) ⊆ ran(A+B).In this chapter we have seen many properties of firmly nonexpansivemappings and monotone operators. We have also seen how the two conceptsare linked through the resolvent of a maximally monotone operator. Thiswill be fundamental to the results in chapters 4–8.37Chapter 4Correspondence ofPropertiesThis chapter contains new results concerning the closely-knit nature offirmly nonexpansive mappings and maximally monotone operators and isbased on [19].4.1 Maximally monotone operators and firmlynonexpansive mappingsThe first result in this section provides a comprehensive list of corre-sponding properties of firmly nonexpansive mappings and maximally mono-tone operators, building on Minty’s Fact 3.36.Theorem 4.1. Let T : H → H be firmly nonexpansive, let A : H ⇒ Hbe maximally monotone, and suppose that T = JA or equivalently thatA = T−1 − Id. Then the following hold:(i) ranT = domA.(ii) T is surjective if and only if domA = H.(iii) Id−T is surjective if and only if A is surjective.(iv) T is injective if and only if A is at most single-valued.(v) T is an isometry if and only if there exists z ∈ H such that A : x 7→ z,in which case T : x 7→ x− z.(vi) T satisfies(∀x ∈ H)(∀y ∈ H)Tx 6= Ty ⇒ ‖Tx− Ty‖2 < 〈x− y, Tx− Ty〉 (4.1)384.1. Maximally monotone operators and firmly nonexpansive mappingsif and only if A is strictly monotone, i.e.,(∀(x, u) ∈ graA)(∀(y, v) ∈ graA)x 6= y ⇒ 〈x− y, u− v〉 > 0. (4.2)(vii) T is strictly monotone if and only if A is at most single-valued.(viii) T is strictly firmly nonexpansive, i.e.,(∀x ∈ H)(∀y ∈ H) x 6= y ⇒ ‖Tx− Ty‖2 < 〈x− y, Tx− Ty〉 (4.3)if and only if A is at most single-valued and strictly monotone.(ix) T is strictly nonexpansive, i.e.,(∀x ∈ H)(∀y ∈ H) x 6= y ⇒ ‖Tx− Ty‖ < ‖x− y‖ (4.4)if and only if A is disjointly injective, i.e.,(∀x ∈ H)(∀y ∈ H) x 6= y ⇒ Ax ∩Ay = ∅. (4.5)(x) T is injective and strictly nonexpansive, i.e.,(∀x ∈ H)(∀y ∈ H) x 6= y ⇒ 0 < ‖Tx− Ty‖ < ‖x− y‖ (4.6)if and only if A is at most single-valued and disjointly injective.(xi) Suppose that ε ∈ ]0,+∞[. Then (1 + ε)T is firmly nonexpansive if andonly if A is strongly monotone with constant ε, i.e., A− ε Id is mono-tone, in which case T is a Banach contraction with constant (1 + ε)−1.(xii) Suppose that γ ∈ ]0,+∞[. Then (1+γ)(Id−T ) is firmly nonexpansiveif and only if A is γ-cocoercive , i.e.,(∀(x, u) ∈ graA)(∀(y, v) ∈ graA) 〈x− y, u− v〉 ≥ γ‖u− v‖2. (4.7)(xiii) Suppose that β ∈ ]0, 1[. Then T is a Banach contraction with constantβ if and only if A satisfies(∀(x, u) ∈ graA)(∀(y, v) ∈ graA)1− β2β2‖x− y‖2 ≤ 2 〈x− y, u− v〉+ ‖u− v‖2. (4.8)394.1. Maximally monotone operators and firmly nonexpansive mappings(xiv) Suppose that φ : [0,+∞[→ [0,+∞] is increasing and vanishes only at0. Then T satisfies(∀x ∈ H)(∀y ∈ H) 〈Tx− Ty, (x− Tx)− (y − Ty)〉 ≥ φ(‖Tx−Ty‖)(4.9)if and only if A is uniformly monotone with modulus φ, i.e.,(∀(x, u) ∈ graA)(∀(y, v) ∈ graA) 〈x− y, u− v〉 ≥ φ(‖x− y‖).(4.10)(xv) T satisfies(∀x ∈ H)(∀y ∈ H)‖Tx− Ty‖2 = 〈x− y, Tx− Ty〉 ⇒{Tx = T(Tx+ y − Ty)Ty = T(Ty + x− Tx)(4.11)if and only if A is paramonotone, i.e.(∀(x, u) ∈ graA)(∀(y, v) ∈ graA)〈x− y, u− v〉 = 0 ⇒ {(x, v), (y, u)} ⊆ graA. (4.12)(xvi) (Bartz et al., [5]) T is cyclically firmly nonexpansive, i.e.,n∑i=1〈xi − Txi, Txi − Txi+1〉 ≥ 0, (4.13)for every set of points {x1, . . . , xn} ⊆ H, where n ∈ {2, 3, . . .} andxn+1 = x1, if and only if A is a subdifferential operator, i.e., thereexists f ∈ Γ0(H) such that A = ∂f .(xvii) T satisfies(∀x ∈ H)(y ∈ H) infz∈H〈Tx− Tz, (y − Ty)− (z − Tz)〉 > −∞(4.14)if and only if A is rectangular, i.e.,(∀x ∈ domA)(∀v ∈ ranA) inf(z,w)∈graA〈x− z, v − w〉 > −∞. (4.15)(xviii) T is linear if and only if A is a linear relation, i.e., graA is a linearsubspace of H×H.404.1. Maximally monotone operators and firmly nonexpansive mappings(xix) T is affine if and only if A is an affine relation, i.e., graA is an affinesubspace of H×H.(xx) (Zarantonello) ranT = FixT := C if and only if A is a normal coneoperator, i.e., A = ∂ιC ; equivalently, T is a projection (nearest point)mapping PC .(xxi) T is sequentially weakly continuous if and only if graA is sequentiallyweakly closed.Proof. Let x, y, u, v be in H.(i): Clear.(ii): This follows from (i).(iii): Clear from the Minty parametrization (3.14).(iv): Assume first that T is injective and that {u, v} ⊆ Ax. Then{x+ u, x+ v} ⊆ (Id +A)x,and hencex = T (x+ u) = T (x+ v).Since T is injective, it follows that x + u = x + v and hence that u = v.Thus, A is at most single-valued.Conversely, let us assume that A is at most single-valued and that Tu =Tv = x. Then{u, v} ⊆ (Id +A)x = x+Ax,and hence{u− x, v − x} ⊆ AxSince A is at most single-valued, we have u−x = v−x and so u = v. Thus,T is injective.(v): Assume first that T is an isometry. Then by (2.1) and (3.3),‖Tx− Ty‖2 = ‖x− y‖2 ≥ ‖Tx− Ty‖2 + ‖(Id−T )x− (Id−T )y‖2.Thus,0 ≥ ‖(Id−T )x− (Id−T )y‖2.It follows that there exists z ∈ H such that T : w 7→ w−z. Thus, T−1 : w 7→w + z. On the other hand, T−1 = Id +A : w 7→ w + Aw. Hence A : w 7→ z,414.1. Maximally monotone operators and firmly nonexpansive mappingsas claimed. Conversely, let us assume that there exits z ∈ H such thatA : w 7→ z. Then Id +A : w 7→ w + z and henceT = JA = (Id +A)−1 : w 7→ w − z.Thus, T is an isometry.(vi): Assume first that T satisfies (4.1), that {(x, u), (y, v)} ⊆ graA, andthat x 6= y. Set p = x+ u and q = y + v. Then(x, u) = (Tp, p− Tp),and(y, v) = (Tq, q − Tq).Since x 6= y, it follows that Tp 6= Tq and therefore that‖Tp− Tq‖2 < 〈p− q, Tp− Tq〉 ,because T satisfies (4.1). Hence0 < 〈(p− Tp)− (q − Tq), Tp− Tq〉 = 〈u− v, x− y〉 .Thus, A is strictly monotone. Conversely, let us assume that A is strictlymonotone and that x = Tu 6= Tv = y. Then {(x, u−x), (y, v− y)} ⊆ graA.Since x 6= y and A is strictly monotone, we have〈x− y, (u− x)− (v − y)〉 > 0⇔ ‖x− y‖2 < 〈x− y, u− v〉⇔ ‖Tu− Tv‖2 < 〈Tv − Tu, u− v〉 .Thus, T satisfies (4.1).(vii): In view of (vi) it suffices to show that T is injective if and onlyif T is strictly monotone. Assume first that T is injective and that x 6= y.Then Tx 6= Ty and hence0 < ‖Tx− Ty‖2 ≤ 〈x− y, Tx− Ty〉 .Thus, T is strictly monotone. Conversely, assume that T is strictly monotoneand that x 6= y. Then 〈x− y, Tx− Ty〉 > 0 and hence Tx 6= Ty. Thus, Tis injective.(viii): Observe that T is strictly firmly nonexpansive if and only if T isinjective and T satisfies (4.1). Thus, the result follows from combining (iv)and (vi).424.1. Maximally monotone operators and firmly nonexpansive mappings(ix): Assume first that T is strictly nonexpansive, that x 6= y, and thatu ∈ Ax ∩Ay. Thenx+ u ∈ (Id +A)x and y + u ∈ (Id +A)y;equivalently,T (x+ u) = x 6= y = T (y + u).Since T is strictly nonexpansive, we have‖x− y‖ = ‖T (x+ u)− T (y + u)‖ < ‖(x+ u)− (y + u)‖ = ‖x− y‖,which gives a contradiction. Thus, A is disjointly injective. Conversely,assume that A is disjointly injective, that u 6= v, and that ‖Tu − Tv‖ =‖u− v‖. Since T is firmly nonexpansive, we deduce thatu− Tu = v − Tv.Assume that x = u− Tu = v − Tv. Then,Tu = u− x and Tv = v − x;equivalently,u ∈ (Id +A)(u− x) and v ∈ (Id +A)(v − x).Thus, x ∈ A(Tu) ∩ A(Tv), which contradicts the assumption on disjointinjectivity of A.(x): Combine (iv) and (ix).(xi): Assume first that (1 + ε)T is firmly nonexpansive and that{(x, u), (y, v)} ⊆ graA.Then x = T (x+ u) and y = T (y + v). Hence by Fact 3.3(iv),〈(x+ u)− (y + v), x− y〉 ≥ (1 + ε)‖x− y‖2⇔ 〈x− y, u− v〉 ≥ ε‖x− y‖2.Thus, A− ε Id is monotone. Conversely, assume that A− ε Id is monotoneand that{(x, u), (y, v)} ⊆ graT .Then {(u, x− u), (v, y − v)} ⊆ graA and hence〈u− v, (x− u)− (y − v)〉 ≥ ε‖u− v‖2⇔ 〈x− y, u− v〉 ≥ (1 + ε)‖u− v‖2.434.1. Maximally monotone operators and firmly nonexpansive mappingsThus, (1+ε)T is firmly nonexpansive. Alternatively, this result follows fromFact 3.45.(xii): Applying (xi) to Id−T and A−1, we see that (1+γ)(Id−T ) is firmlynonexpansive if and only if A−1 − γ Id is monotone, which is equivalent toA being γ-cocoercive.(xiii): Assume first that T is a Banach contraction with constant β andthat {(x, u), (y, v)} ⊆ graA. Set p = x+ u and y = y + v. Then(x, u) = (Tp, p− Tp),(y, v) = (Tq, q − Tq), and‖Tp− Tq‖ ≤ β‖p− q‖,i.e.,‖x− y‖2 ≤ β2‖(x+ u)− (y + v)‖2 = β2‖(x− y) + (u− v)‖2 (4.16)= β2(‖x− y‖2 + 2 〈x− y, u− v〉+ ‖u− v‖2).Thus, (4.8) holds. The converse is proved similarly.(xiv): The equivalence is immediate from the Minty parametrization(3.14).(xv): Assume first that T satisfies (4.11) and that {(x, u), (y, v)} ⊆ graAwith 〈x− y, u− v〉 = 0. Set p = x+ u and q = y + v. Then(x, u) = (Tp, p− Tp) and (y, v) = (Tq, q − Tq),and we have〈Tp− Tq, (p− Tp)− (q − Tq)〉 = 0⇔ ‖Tp− Tq‖2 = 〈p− q, Tp− Tq〉 .By (4.11),Tp = T (Tp+ q − Tq)⇔ x = T (x+ v)⇔ x+ v ∈ x+Ax⇔ v ∈ Ax.And similarly,Tq = T (Tq + p− Tp)⇔ y = T (y + u)⇔ y + u ∈ y +Ay ⇔ u ∈ Ay.444.1. Maximally monotone operators and firmly nonexpansive mappingsThus, A is paramonotone. Conversely, assume that A is paramonotone, that‖Tu− Tv‖2 = 〈u− v, Tu− Tv〉, that x = Tu, and that y = Tv. Then,{(x, u− x), (y, v − y)} ⊆ graA,and〈x− y, (u− x)− (v − y)〉 = 0.Since A is paramonotone, we deduce thatv − y ∈ Ax⇔ x− y + v ∈ (Id +A)x⇔ x = T (x− y + v)⇔ Tu = T (Tu+ v − Tv).And similarly,u− x ∈ Ay ⇔ y − x+ u ∈ (Id +A)y⇔ y = T (y − x+ u)⇔ Tv = T (Tv + u− Tu).Thus, T satisfies (4.11).(xvi): This follows from Fact 3.51.(xvii): The equivalence is immediate from the Minty parametrization(3.14).(xviii): Indeed,T = JA is linear⇔ (A+ Id)−1 is a linear relation,⇔ A+ Id is a linear relation,⇔ A is a linear relation.(xix): This follows from (xviii).(xx): Assume C := FixT = ranT . Fact 3.15 yields C is a closed convexset and by Fact 3.16, T = PC . On the other hand, if A = NC , then byFact 3.42, T = JNC = PC which gives FixT = C and ranT = C. The factthat NC = ∂ιC follows from Fact 2.38.(xxi): Assume that T is sequentially weakly continuous. Let (xn, un)n∈Nbe a sequence in graA that converges weakly to (x, u) ∈ H × H. Then(xn + un)n∈N converges weakly to x + u. On the other hand, Id−T issequentially weakly continuous because T is. Altogether,(xn, un)n∈N = (T (xn + un), (Id−T )(xn + un))n∈N⇀ (T (x+ u), (Id−T )(x+ u)).454.1. Maximally monotone operators and firmly nonexpansive mappingsBut (xn, un) ⇀ (x, u) and thus (x, u) = (T (x+ u), (Id−T )(x+ u)) ∈ graA.Therefore graA is sequentially weakly closed. Conversely, let us assume thatgraA is sequentially weakly closed. Let (xn)n∈N be a sequence in H thatis weakly convergent to x. Our goal is to show that Txn⇀Tx. Since T isnonexpansive, the sequence (Txn)n∈N is bounded. After passing to a sub-sequence and relabeling if necessary, we can and do assume that (Txn)n∈Nconverges weakly to some point y ∈ H. Now (Txn, xn−Txn)n∈N lies in graA,and this sequence converges weakly to (y, x− y). Since graA is sequentiallyweakly closed, it follows that (y, x− y) ∈ graA. Therefore,x− y ∈ Ay ⇔ x ∈ (Id +A)y ⇔ y = Tx,which implies the result.Example 4.2. Concerning items (xi) and (xiii) in Theorem 4.1, it was previ-ously known that if A is strongly monotone, then T is a Banach contraction,see Fact 3.45. The converse, however, is false. Consider the case H = R2and setA =(0 −11 0). (4.17)Then (∀z ∈ H) 〈z,Az〉 = 0 so A cannot be strongly monotone. On the otherhand,T = JA = (Id +A)−1 =12(1 1−1 1)(4.18)is linear and ‖Tz‖2 = 12‖z‖2, which implies that T is a Banach contractionwith constant 1/√2.Corollary 4.3. Let A : H → H be continuous, linear, and maximally mono-tone. Then the following hold.(i) If JA is a Banach contraction, then A is (disjointly) injective.(ii) If ranA is closed and A is (disjointly) injective, then JA is a Banachcontraction.Proof. The result is trivial if H = {0} so we assume that H 6= {0}. Let xand y be in H.(i): Assume that JA is a Banach contraction, with constant β ∈ [0, 1[. Ifβ = 0, then JA ≡ 0 ⇔ A = N{0}, which contradicts the single-valuedness ofA. Thus, 0 < β < 1. By Theorem 4.1(xiii),(∀x ∈ H)(∀y ∈ H)1− β2β2‖x− y‖2 ≤ 2 〈x− y,Ax−Ay〉+ ‖Ax−Ay‖2.(4.19)464.1. Maximally monotone operators and firmly nonexpansive mappingsIf x 6= y, then the left side of (4.19) is strictly positive, which implies thatAx 6= Ay. Thus, A is (disjointly) injective.(ii): Let us assume that ranA is closed and that A is (disjointly) injective.Then kerA = {0} and hence, by Fact 2.14, there exists ρ ∈ ]0,+∞[ suchthat (∀z ∈ H) ‖Az‖ ≥ ρ‖z‖. Thus,(∀z ∈ H) ‖Az‖2 − ρ2‖z‖2 ≥ 0. (4.20)Set β = 1/√1 + ρ2 and z = x − y. Then ρ2 = (1 − β2)/β2 and hence by(4.20) and Lemma 3.27,(∀x ∈ H)(∀y ∈ H)1− β2β2‖x− y‖2 ≤ ‖Ax−Ay‖2 ≤ 2 〈x− y,Ax−Ay〉+ ‖Ax−Ay‖2.(4.21)Again by Theorem 4.1(xiii), JA is a Banach contraction with constant β ∈]0, 1[.Example 4.4. Suppose that H = `2(N), the space of square-summablesequences, i.e., x = (xn) ∈ H if and only if∑∞n=1 |xn|2 < +∞, and setA : H → H : (xn) 7→(1nxn). (4.22)Then A is continuous, linear, maximally monotone, and ranA is a dense,proper subspace of H that is not closed. The resolvent T = JA isT : H → H : (xn) 7→(nn+1xn). (4.23)Now denote the nth unit vector in H by en (i.e. en has a one at position nand zeros otherwise). Then ‖Ten − T0‖ = nn+1‖en − 0‖. Sincenn+1 → 1, itfollows that T is not a Banach contraction.Remark 4.5. When A is a subdifferential operator, then it is impossible toget the behavior witnessed in Example 4.2, as we see next in Proposition 4.6.Proposition 4.6. Let f ∈ Γ0(H) and let ε ∈ ]0,+∞[. Then (1 + ε) proxfis firmly nonexpansive if and only if proxf is a Banach contraction withconstant (1 + ε)−1.474.2. DualityProof. Set β = (1 + ε)−1. It is clear that if (1 + ε) proxf is firmly non-expansive, then (1 + ε) proxf is nonexpansive and hence, for x and y inH‖(1 + ε) proxf x− (1 + ε) proxf y‖ ≤ ‖x− y‖⇔ ‖proxf x− proxf y‖ ≤ (1 + ε)−1‖x− y‖,thus proxf is a Banach contraction with constant β. Conversely, assume thatproxf is a Banach contraction with constant β. Since proxf is the Fre´chetgradient mapping of the continuous convex function f∗12‖ · ‖2 : H → R(see Fact 2.54), the Baillon-Haddad theorem (Fact 3.12) guarantees thatβ−1 proxf is firmly nonexpansive.Remark 4.7. If n = 2, then (4.13) reduces to〈x1 − Tx1, Tx1 − Tx2〉+ 〈x2 − Tx2, Tx2 − Tx1〉 ≥ 0⇔ 〈(Id−T )x1 − (Id−T )x2, Tx1 − Tx2〉 ≥ 0i.e., to firm nonexpansiveness of T (see Fact 3.3(v)).4.2 DualityThere is a natural duality for firmly nonexpansive mappings and maxi-mally monotone operators; namely,T 7→ Id−T and A 7→ A−1,respectively. Note that the dual of the dual is the original property, e.g.Id−(Id−T ) = T . Every property considered in Theorem 4.1 has a dualproperty. We have considered all dual properties and we shall explicitlysingle those out that we found to have simple and pleasing descriptions.Among these properties, those that are “self-dual”, that is the property isidentical to its dual property, stand out even more. First, we more explicitlydefine the notion of dual properties.Definition 4.8 (dual and self-dual properties). Let (p) and (p∗) be prop-erties for firmly nonexpansive mappings defined on H. If, for every firmlynonexpansive mapping T : H → H,T satisfies (p) if and only if Id−T satisfies (p∗), (4.24)484.2. Dualitythen (p∗) is dual to (p), and hence (p) is dual to (p∗). If (p) = (p∗), we saythat (p) is self-dual. Analogously, let (q) and (q∗) be properties of maximallymonotone operators defined on H. IfA satisfies (q) if and only if A−1 satisfies (q∗) (4.25)for every maximally monotone operator A : H⇒ H, then (q∗) is dual to (q),and hence (q) is dual to (q∗). If (q) = (q∗), we say that (q) is self-dual.Figure 4.1: Duality of a monotone operator, A, and its associated resolvent,T = JA.Theorem 4.9. Let T : H → H be firmly nonexpansive, let A : H ⇒ H bemaximally monotone, and suppose that T = JA or equivalently that A =T−1 − Id. Then the following are equivalent:(i) T is surjective.(ii) A has full domain.(iii) A−1 is surjective.Thus for maximally monotone operators, surjectivity and full domain areproperties that are dual to each other. These properties are not self-dual;for example, A = 0 has full domain while A−1 = ∂ι{0} does not.494.2. DualityProof. (i)⇔(ii): Theorem 4.1(ii). (ii)⇔(iii): Obvious.Theorem 4.10. Let T : H → H be firmly nonexpansive, let A : H ⇒ H bemaximally monotone, and suppose that T = JA or equivalently that A =T−1 − Id. Then the following are equivalent:(i) T is strictly nonexpansive.(ii) A is disjointly injective.(iii) Id−T is injective.(iv) A−1 is at most single-valued.Thus for firmly nonexpansive mappings, strict nonexpansiveness and injec-tivity are dual to each other; and correspondingly for maximally monotoneoperators disjoint injectivity and at most single-valuedness are dual to eachother. These properties are not self-dual, T ≡ 0 is strictly nonexpansive,but Id−T = Id is not. Correspondingly, A = ∂ι{0} is disjointly injective butA−1 = 0 is not.Proof. We know that (i)⇔(ii) by Theorem 4.1(ix). We also know that(iii)⇔(iv) by Theorem 4.1(iv) (applied to A−1 and Id−T ). It thus suf-fices to show that (ii)⇔(iv). Assume first that A is disjointly injective andthat {x, y} ⊆ A−1u. Then u ∈ Ax ∩ Ay. Since A is disjointly injective, wehave x = y. Thus, A−1 is at most single-valued. Conversely, assume thatA−1 is at most single-valued and that u ∈ Ax ∩ Ay. Then {x, y} ⊆ A−1uand so x = y. It follows that A is disjointly injective.Theorem 4.11. Let T : H → H be firmly nonexpansive, let A : H ⇒ H bemaximally monotone, and suppose that T = JA or equivalently that A =T−1 − Id. Then the following are equivalent:(i) T satisfies (4.1) i.e.,Tx 6= Ty ⇒ ‖Tx− Ty‖2 < 〈x− y, Tx− Ty〉 .(ii) A is strictly monotone.(iii) Id−T satisfies(∀x ∈ H)(∀y ∈ H)(Id−(Id−T ))x 6=(Id−(Id−T ))y⇒ ‖(Id−T )x− (Id−T )y‖2 < 〈x− y, (Id−T )x− (Id−T )y〉 .(4.26)504.2. Duality(iv) A−1 satisfies(∀(x, u) ∈ graA−1)(∀(y, v) ∈ graA−1) u 6= v ⇒ 〈x− y, u− v〉 > 0.(4.27)Thus for firmly nonexpansive mappings, properties (4.1) and (4.26) are dualto each other; and correspondingly for maximally monotone operators strictmonotonicity and (4.27) are dual to each other. These properties are notself-dual; T = 0 trivially satisfies (4.1), but Id−0 = Id does not.Proof. (i)⇔(ii): Theorem 4.1(vi). (i)⇔(iii): Indeed, (4.26) and (4.1) areequivalent as is easily seen by expansion and rearranging. (ii)⇔(iv): Clear.Theorem 4.12 (self-duality of strict firm nonexpansiveness). Let T : H →H be firmly nonexpansive, let A : H ⇒ H be maximally monotone, andsuppose that T = JA or equivalently that A = T−1 − Id. Then the followingare equivalent:(i) T is strictly firmly nonexpansive.(ii) A is at most single-valued and strictly monotone.(iii) Id−T is strictly firmly nonexpansive.(iv) A−1 is at most single-valued and strictly monotone.Consequently, strict firm nonexpansive is a self-dual property for firmly non-expansive mappings; correspondingly, being both strictly monotone and atmost single-valued is self-dual for maximally monotone operators.Proof. Note that T is strictly firmly nonexpansive if and only if(∀x ∈ H)(∀y ∈ H) x 6= y ⇒ 0 < 〈Tx− Ty, (Id−T )x− (Id−T )y〉 ,(4.28)which is obviously self-dual. In view of Theorem 4.1(viii), the correspondingproperty for A is being both at most single-valued and strictly monotone.Theorem 4.12 illustrates the technique of obtaining self-dual propertiesby fusing any property and its dual. Here is another example of this type.Theorem 4.13 (self-duality of strict nonexpansiveness and injectivity). LetT : H → H be firmly nonexpansive, let A : H ⇒ H be maximally monotone,and suppose that T = JA or equivalently that A = T−1 − Id. Then thefollowing are equivalent:514.2. Duality(i) T is strictly nonexpansive and injective.(ii) A is at most single-valued and disjointly injective.(iii) Id−T is strictly nonexpansive and injective.(iv) A−1 is at most single-valued and disjointly injective.Consequently, being both strictly nonexpansive and injective is a self-dualproperty for firmly nonexpansive mappings; correspondingly, being both dis-jointly injective and at most single-valued is self-dual for maximally mono-tone operators.Proof. Clear from Theorem 4.1(x).Remark 4.14. In Theorem 4.12 and Theorem 4.13, arguing directly (or byusing the characterization with monotone operators via Theorem 4.1), it iseasy to verify the implicationT is strictly firmly nonexpansive⇒ T is injective and strictly nonexpansive. (4.29)The converse of implication (4.29) is false in general, see Example 4.15.In contrast, we see in Corollary 4.17 that when H is finite-dimensional andT = JA is a proximal mapping (i.e., A is a subdifferential operator), thenthe converse implication of (4.29) is true.Example 4.15. Consider H = R2, and let A denote the counter-clockwiserotation by pi/2, which we utilized already in (4.17). Clearly, A is a linearsingle-valued maximally monotone operator that is (disjointly) injective, butA is not strictly monotone. Accordingly, T = JA is linear, injective andstrictly nonexpansive, but not strictly firmly nonexpansive.Lemma 4.16. Suppose that H is finite-dimensional and let f ∈ Γ0(H).Then the following are equivalent:(i) ∂f is disjointly injective.(ii) (∂f)−1 = ∂f∗ is at most single-valued.(iii) f∗ is essentially smooth.(iv) f is essentially strictly convex.(v) ∂f is strictly monotone.524.2. Duality(vi) proxf is strictly nonexpansive.(vii) (∀x ∈ H)(∀y ∈ H) proxf x 6= proxf y⇒ ‖proxf x− proxf y‖2 <〈x− y,proxf x− proxf y〉.Proof. “(i)⇔(ii)”: Theorem 4.10. “(ii)⇔(iii)”: Fact 2.47. “(iii)⇔(iv)”:Fact 2.48. “(iv)⇔(v)”: Fact 3.30. “(i)⇔(vi)”: Theorem 4.1(ix) and Fact 2.53.“(v)⇔(vii)”: Theorem 4.1(vi).Lemma 4.16 admits a dual counterpart that contains various characteri-zations of essential smoothness. The following consequence of these charac-terizations is also related to Remark 4.14. Recall that for a finite-dimensionalH, a function f ∈ Γ0(H) is Legendre if it is both essentially smooth andessentially strictly convex.Corollary 4.17 (Legendre self-duality). Suppose that H is finite-dimensionaland let f ∈ Γ0(H). Then the following are equivalent:(i) ∂f is disjointly injective and at most single-valued.(ii) ∂f is strictly monotone and at most single-valued.(iii) f is Legendre.(iv) proxf is strictly firmly nonexpansive.(v) proxf is strictly nonexpansive and injective.(vi) ∂f∗ is disjointly injective and at most single-valued.(vii) ∂f∗ is strictly monotone and at most single-valued.(viii) f∗ is Legendre.(ix) proxf∗ is strictly firmly nonexpansive.(x) proxf∗ is strictly nonexpansive and injective.Proof. Combine Theorem 4.13 and Lemma 4.16 using Fact 2.53.Theorem 4.18 (self-duality of paramonotonicity). Let A : H⇒ H be max-imally monotone, let T : H → H be firmly nonexpansive, and suppose thatT = JA or equivalently that A = T−1 − Id. Then A is paramonotone ifand only if A−1 is paramonotone; consequently, T satisfies (4.11) if and534.2. Dualityonly if Id−T satisfies (4.11) (with T replaced by Id−T ). Consequently, be-ing paramonotone is a self-dual property for maximally monotone operators;correspondingly, satisfying (4.11) is a self-dual property for firmly nonex-pansive mappings.Proof. Self-duality is immediate from the definition of paramonotonicity,and the corresponding result for firmly nonexpansive mappings follows fromTheorem 4.1(xv).Theorem 4.19 (self-duality of cyclical firm nonexpansiveness and cyclicalmonotonicity).Let T : H → H be firmly nonexpansive, let A : H⇒ H be maximally mono-tone, let f ∈ Γ0, and suppose that T = JA or equivalently that A = T−1−Id.Then the following are equivalent:(i) T is cyclically firmly nonexpansive.(ii) A is cyclically monotone.(iii) A = ∂f .(iv) Id−T is cyclically firmly nonexpansive.(v) A−1 is cyclically monotone.(vi) A−1 = ∂f∗.Consequently, cyclic firm nonexpansiveness is a self-dual property for firmlynonexpansive mappings; correspondingly, cyclic monotonicity is a self-dualproperty for maximally monotone operators.Proof. The fact that cyclically maximal monotone operators are subdiffer-ential operators is due to Rockafellar and well known, see Fact 3.50, as isthe identity (∂f)−1 = ∂f∗, see Fact 2.59. The result thus follows fromTheorem 4.1(xvi).Theorem 4.20 (self-duality of rectangularity). Let T : H → H be firmlynonexpansive, let A : H ⇒ H be maximally monotone, and suppose thatT = JA or equivalently that A = T−1−Id. Then the following are equivalent:(i) T satisfies (4.14).(ii) A is rectangular.(iii) Id−T satisfies (4.14).544.2. Duality(iv) A−1 is rectangular.Consequently, rectangularity is a self-dual property for maximally monotoneoperators; correspondingly, (4.14) is a self-dual property for firmly nonex-pansive mappings.Proof. It is obvious from the definition that either property is self-dual; theequivalences thus follow from Theorem 4.1(xvii).Theorem 4.21 (self-duality of linearity). Let T : H → H be firmly nonex-pansive, let A : H ⇒ H be maximally monotone, and suppose that T = JAor equivalently that A = T−1 − Id. Then the following are equivalent:(i) T is linear.(ii) A is a linear relation.(iii) Id−T is linear.(iv) A−1 is a linear relation.Consequently, linearity is a self-dual property for firmly nonexpansive map-pings; correspondingly, being a linear relation is a self-dual property formaximally monotone operators.Proof. It is clear that T is linear if and only if Id−T is; thus, the resultfollows from Theorem 4.1(xviii).Theorem 4.22 (self-duality of affineness). Let T : H → H be firmly nonex-pansive, let A : H ⇒ H be maximally monotone, and suppose that T = JAor equivalently that A = T−1 − Id. Then the following are equivalent:(i) T is affine.(ii) A is an affine relation.(iii) Id−T is affine.(iv) A−1 is an affine relation.Consequently, affineness is a self-dual property for firmly nonexpansive map-pings; correspondingly, being an affine relation is is a self-dual property formaximally monotone operators.Proof. It is clear that T is affine if and only if Id−T is; therefore, the resultfollows from Theorem 4.1(xix).554.3. Reflected resolventsRemark 4.23 (projection). Concerning Theorem 4.1(xx), note that being aprojection is not a self-dual: indeed, suppose that H 6= {0} and let T be theprojection onto the closed unit ball. Then Id−T is not a projection sinceFix(Id−T ) = {0} $ H = ran(Id−T ).Theorem 4.24 (self-duality of sequential weak continuity). Let T : H → Hbe firmly nonexpansive, let A : H⇒ H be maximally monotone, and supposethat T = JA or equivalently that A = T−1 − Id. Then the following areequivalent:(i) T is sequentially weakly continuous.(ii) graA is sequentially weakly closed.(iii) Id−T is sequentially weakly continuous.(iv) graA−1 is sequentially weakly closed.Consequently, sequential weak continuity is a self-dual property for firmlynonexpansive mappings; correspondingly, having a sequentially weakly closedgraph is a self-dual property for maximally monotone operators.Proof. Since Id is weakly continuous, it is clear that T is sequentially weaklycontinuous if and only if Id−T is; thus, the result follows from Theo-rem 4.1(xxi).The self dual properties of this section are summarized in Table 4.1.4.3 Reflected resolventsIn the previous two sections, the correspondence between firmly nonex-pansive mappings and maximally monotone operators was extensively uti-lized. However, Fact 3.3 provides another correspondence with nonexpansivemappings:T is firmly nonexpansive if and only if N = 2T − Id is nonexpansive.(4.30)Note that N is also referred to as a reflected resolvent. The correspondingdual of N within the set of nonexpansive mappings is simply−N. (4.31)564.3. Reflected resolventsTable 4.1: Summary of self dual properties between monotone operators andtheir resolvents.Monotone Operator ResolventA and A−1 T and (Id−T )At most single-valued andstrictly monotoneStrictly firmly nonexpansiveAt most single-valued and Strictly nonexpansivedisjointly injective and injectiveParamonotone Satisfies (4.11)Cyclically monotoneCyclically firmlynonexpansiveLinear relation LinearAffine relation AffineSequentially weakly closedSequentially weaklycontinuousFigure 4.2: Duality of a monotone operator A, its associated resolvent, T ,and its reflected resolvent, N = 2T − Id.Thus, all results have counterparts formulated for nonexpansive map-pings. These counterparts are most easily derived from the firmly nonex-pansive formulation, by simply replacing T by 12 Id +12N .Theorem 4.25 (strict firm nonexpansiveness). Let T : H → H be firmly574.3. Reflected resolventsnonexpansive, let N : H → H be nonexpansive, and suppose that N =2T − Id. Then T is strictly firmly nonexpansive if and only if N is strictlynonexpansive.Proof. Let x and y be in H. T is strictly firmly nonexpansive if x 6= yimplies‖Tx− Ty‖2 + ‖(Id−T )x− (Id−T )y‖2 < ‖x− y‖2⇔14‖(Id +N)x− (Id +N)y‖2 +14‖(Id−N)x− (Id−N)y‖2 < ‖x− y‖2.Now expand and simplify to yield the result.Remark 4.26.(i) We know from Theorem 4.12 that strict firm nonexpansiveness is aself-dual property with respect to monotone operators and firmly non-expansive mappings. This can also be seen within the realm of nonex-pansive mappings since N is strictly nonexpansive if and only if −Nis.(ii) Furthermore, combining Theorem 4.12 with Theorem 4.25 yields thefollowing: a maximally monotone operator A is at most single-valuedand strictly monotone if and only if its reflected resolvent 2JA − Idis strictly nonexpansive. This characterization was observed by Rock-afellar and Wets; see [59, Proposition 12.11].(iii) In passing, we note that when H is finite-dimensional, the iterates ofa strictly nonexpansive mapping converge to the unique fixed point(assuming it exists). For this and more, see, e.g., [39].Theorem 4.27 (strong monotonicity). Let A : H⇒ H be maximally mono-tone, let N : H → H be nonexpansive, suppose that N = 2JA − Id and thatε ∈ ]0,+∞[. Then A is strongly monotone with constant ε if and only ifε Id +(1 + ε)N is nonexpansive.Proof. We know from Theorem 4.1(xi) that A is strongly monotone withconstant ε if and only if (1 + ε)T is firmly nonexpansive. This is equivalentto2(1 + ε)T − Id = (1 + ε)(2T − Id) + ε Id,is nonexpansive.584.3. Reflected resolventsTheorem 4.28 (reflected resolvent as Banach contraction). Let A : H⇒ Hbe maximally monotone, let T : H → H be firmly nonexpansive, and letN : H → H be nonexpansive. Suppose that T = JA, that N = 2T − Id, andthat β ∈ [0, 1]. Then the following are equivalent:(i) (∀(x, u) ∈ graA)(∀(y, v) ∈ graA)(1− β2)(‖x− y‖2 + ‖u− v‖2) ≤ 2(1 + β2) 〈x− y, u− v〉(ii) (∀x ∈ H)(∀y ∈ H)(1− β2)‖x− y‖2 ≤ 4 〈Tx− Ty, (Id−T )x− (Id−T )y〉(iii) (∀x ∈ H)(∀y ∈ H)‖Nx−Ny‖ ≤ β‖x− y‖.Proof. In view of the Minty parametrization, (3.14), item (i) is equivalentto(∀x ∈ H)(∀y ∈ H) (1− β2)(‖Tx− Ty‖2 + ‖(x− Tx)− (y − Ty)‖2)≤ 2(1 + β2)〈Tx− Ty, (x− Tx)− (y − Ty)〉 . (4.32)Simple algebraic manipulations show that (4.32) is equivalent to (ii), whichin turn is equivalent to (iii).It is clear that the properties (i)–(iii) in Theorem 4.28 are self-dual (forfixed β). The following result is a simple consequence.Corollary 4.29 (self-duality of reflected resolvents that are Banach con-tractions). Let A : H⇒ H be maximally monotone, let T : H → H be firmlynonexpansive, let N : H → H be nonexpansive, and suppose that T = JAand N = 2T − Id. Then the following are equivalent:(i) inf{〈x− y, u− v〉‖x− y‖2 + ‖u− v‖2∣∣∣∣ {(x, u), (y, v)} ⊆ graA, (x, u) 6= (y, v)}> 0.(ii) inf{〈Tx− Ty, (Id−T )x− (Id−T )y〉‖x− y‖2∣∣∣∣ {x, y} ⊆ H, x 6= y}> 0.(iii) N is a Banach contraction.Furthermore, these properties are self-dual for their respective classes of op-erators.594.3. Reflected resolventsRemark 4.30. Precisely when A : x 7→ x − z for some fixed vector z ∈ H,we compute T : x 7→ (x + z)/2 and therefore we reach the extreme case ofCorollary 4.29 where N : x 7→ z is a Banach contraction with constant 0.Corollary 4.31. Let A : H ⇒ H be maximally monotone, let T : H → Hbe firmly nonexpansive, let N : H → H be nonexpansive, and suppose thatT = JA and N = 2T − Id. Then the following are equivalent:(i) Both A and A−1 are strongly monotone.(ii) There exists γ ∈ ]1,+∞[ such that both γT and γ(Id−T ) are firmlynonexpansive.(iii) N is a Banach contraction.Proof. Let us assume that A and A−1 are both strongly monotone; equiv-alently, there exists ε ∈ ]0,+∞[ such that A − ε Id and A−1 − ε Id aremonotone. Let {(x, u), (y, v)} ⊆ graA. Then {(u, x), (v, y)} ⊆ graA−1 and〈x− y, u− v〉 ≥ ε‖x− y‖2 and 〈u− v, x− y〉 ≥ ε‖u− v‖2. (4.33)Adding these inequalities yields 2 〈x− y, u− v〉 ≥ ε(‖x − y‖2 + ‖u − v‖2).Thus, item (i) of Corollary 4.29 holds. Conversely, if item (i) of Corol-lary 4.29 holds, then both A and A−1 are strongly monotone. Therefore,by Corollary 4.29, (i) and (iii) are equivalent. Finally, in view of Theo-rem 4.1(xi), we see that (i) and (ii) are also equivalent.Additional characterizations are available for subdifferential operators:Proposition 4.32. Let f ∈ Γ0(H). Then the following are equivalent:(i) f and f∗ are strongly convex.(ii) f and f∗ are everywhere differentiable, and both ∇f and ∇f∗ areLipschitz continuous.(iii) proxf and Id−proxf are Banach contractions.(iv) 2 proxf − Id is a Banach contraction.Proof. It is well known that for functions, strong convexity is equivalent tostrong monotonicity of the subdifferential operators, see Example 3.33. Inview of Proposition 4.6 and Corollary 4.31, we obtain the equivalence ofitems (i), (iii), and (iv). Finally, the equivalence of (i) and (ii) follows fromFact 2.43.604.3. Reflected resolventsWe now turn to linear relations.Proposition 4.33. Let A : H → H be a maximally monotone linear rela-tion. Then the following are equivalent:(i) Both A and A−1 are strongly monotone.(ii) A is a continuous surjective linear operator on H andinfz∈Hr{0}〈z,Az〉‖z‖2 + ‖Az‖2> 0.(iii) 2JA − Id is a Banach contraction.If H is finite-dimensional, then (i)–(iii) are also equivalent to(iv) A : H → H satisfies (∀z ∈ Hr {0}) 〈z,Az〉 > 0.Proof. “(i)⇔(iii)”: Clear from Corollary 4.31.“(i)⇒(ii)”: By Fact 3.35 A and A−1 are single-valued surjective operatorswith full domain. Since A and A−1 are linear, Fact 3.38 implies that A andA−1 are continuous. Thus, (ii) holds.“(i)⇐(ii)”: (ii) implies that item (i) of Corollary 4.29 holds. Thus, (i)follows from Corollary 4.29 and Corollary 4.31.“(ii)⇒(iv)”: Clear.“(ii)⇐(iv)”: Since A is injective and H is finite-dimensional, A is bijec-tive and continuous. To see that the infimum in item (ii) is strictly positive,note that we may take the infimum over the unit sphere, which is a compactsubset of H.Example 4.34. In Proposition 4.33(iv), if H is infinite dimensional thenthe equivalence does not hold. Consider again the case in Example 4.4 whereH = `2(N) and A : H → H : (xn) 7→ ( 1nxn). Then for x ∈ `2(N) \ {0},〈x,Ax〉 =∞∑n=11nx2n > 0,so (iv) holds. But take the unit vectors en and en+1 and we see that〈en − en+1, Aen −Aen+1〉 =1n+1n+ 1=2n+ 1n2 + n→ 0.614.3. Reflected resolventsSo @β ∈ R++ such that 〈x− y,Ax−Ay〉 ≥ β‖x− y‖2 ∀x, y ∈ `2(N), thus Ais not strongly monotone and thus (i) does not hold. Similarly,inf〈en, Aen〉‖en‖2 + ‖Aen‖2=1n1 + 1n2→ 0,so (ii) does not hold. Finally, we have Ten = JAen = ( nn+1en) and thus‖(2T − Id)en‖ =∥∥∥∥2nn+ 1en − en∥∥∥∥ =∥∥∥∥n− 1n+ 1en∥∥∥∥→ 1,and so 2T − Id is not a Banach contraction and (iii) does not hold.We shall conclude this chapter with some comments regarding applica-tions of the above results to splitting methods. See also [11] for furtherinformation and various variants. Here is a technical lemma, which is wellknown and whose simple proof is omitted.Lemma 4.35. Let T1, . . . , Tn be finitely many nonexpansive mappings fromH to H, and let λ1, . . . , λn be in ]0, 1] such that λ1 + · · ·+λn = 1. Then thefollowing hold:(i) The composition T1T2 · · ·Tn is nonexpansive.(ii) The convex combination λ1T1 + · · ·+ λnTn is nonexpansive.(iii) If some Ti is strictly nonexpansive, then T1T2 · · ·Tn is strictly nonex-pansive.(iv) If some Ti is strictly nonexpansive, then λ1T1 + · · · + λnTn is strictlynonexpansive.(v) If some Ti is a Banach contraction, then T1T2 · · ·Tn is a Banach con-traction.(vi) If some Ti is a Banach contraction, then λ1T1+ · · ·+λnTn is a Banachcontraction.Corollary 4.36 (backward-backward iteration). Let A1 and A2 be two max-imally monotone operators from H to H, and assume that one of these is dis-jointly injective. Then the (backward-backward) composition T1T2 is strictlynonexpansive.Proof. Combine Theorem 4.1(ix) and Lemma 4.35.624.3. Reflected resolventsCorollary 4.37 (Douglas-Rachford iteration). Let A1 and A2 be two max-imally monotone operators from H to H, and assume that one of these isboth at most single-valued and strictly monotone (as is, e.g., the subdiffer-ential operator of a convex Legendre function when H is finite-dimensional;see Corollary 4.17). Denote the resolvents of A1 and A2 by T1 and T2,respectively. Then the operator governing the Douglas-Rachford iteration,i.e.,T := 12(2T1 − Id)(2T2 − Id) +12 Id, (4.34)is not just firmly nonexpansive but also strictly nonexpansive; consequently,FixT is either empty or a singleton.Proof. In view of Theorem 4.12 and Theorem 4.25, we see that 2T1− Id and2T2−Id are both nonexpansive, and one of these two is strictly nonexpansive.By Lemma 4.35(iii), (2T1 − Id)(2T2 − Id) is strictly nonexpansive. Hence,by Lemma 4.35(iv), T is strictly nonexpansive.Remark 4.38. Consider Corollary 4.37, and assume that Ai, where i ∈ {1, 2},satisfies condition (i) in Corollary 4.29. Then 2Ti−Id is a Banach contractionby Corollary 4.29. Furthermore, Lemma 4.35 now shows that the Douglas-Rachford operator T defined in (4.34) is a Banach contraction. Thus, FixTis a singleton and the unique fixed point may be found as the strong limitof any sequence of Banach-Picard iterates for T .This chapter gave a comprehensive list of how properties of firmly non-expansive mappings translate to the corresponding maximally monotoneoperators. The duality of these properties was also examined, and thoseproperties that are self-dual were identified. Finally, some applications tooperators occurring in splitting methods, including reflected resolvents, weregiven.63Chapter 5The Resolvent Average ofMonotone OperatorsThis chapter is based on the papers [18] and [21]. We begin this chapterwith a new method of averaging monotone operators.Definition 5.1 (Resolvent average). Let Ai, i = 1, . . . , n be monotone op-erators, λi > 0 with∑ni=1 λi = 1, and µ > 0. For A = (A1, . . . , An) andλ = (λ1, . . . , λn) the resolvent average of A is,Rµ(A,λ) :=[λ1(A1+µ−1 Id)−1+· · ·+λn(An+µ−1 Id)−1]−1−µ−1 Id . (5.1)The name “resolvent average” is motivated from the fact that when µ = 1(R1(A,λ) + Id)−1= λ1(A1 + Id)−1+ · · ·+ λn(An + Id)−1, (5.2)which says that the resolvent of R1(A,λ) is the arithmetic average of resol-vents of the Ai, with weight λ = (λ1, . . . , λn). The resolvent average pro-vides a novel averaging technique, and having the parameter µ in Rµ(A,λ)will allow us to take limits which compare the resolvent average with thearithmetic and harmonic averages.5.1 Basic propertiesIn this section, we give some basic properties of Rµ(A,λ).Proposition 5.2. We haveJµRµ(A,λ) = λ1JµA1 + · · ·+ λnJµAn , (5.3)µ(Rµ(A,λ))= λ1µA1 + · · ·+ λnµAn. (5.4)645.1. Basic propertiesProof. It follows from (5.1) thatµRµ(A,λ) + Id =[λ1(µA1 + Id)−1 + · · ·+ λn(µAn + Id)−1]−1.Then (5.3) follows by taking inverses on both sides and using the definitionof the resolvent, (3.11).By (5.3), we obtain that(Id−JµRµ(A,λ)) = λ1(Id−JµA1) + · · ·+ λn(Id−JµAn).Dividing both sides by µ,µ−1(Id−JµRµ(A,λ)) = λ1µ−1(Id−JµA1) + · · ·+ λnµ−1(Id−JµAn).Then apply the definition of the Yosida regularization, (3.12).Theorem 5.3. For all i ∈ I, let Ai be a monotone operator from H ⇒ H.Then Rµ(A,λ) is monotone. Moreover,dom JµRµ(A,λ) = dom JµA1 ∩ · · · ∩ dom JµAn , i.e., (5.5)ran(µRµ(A,λ) + Id) = ran(µA1 + Id) ∩ · · · ∩ ran(µAn + Id).Consequently, Rµ(A,λ) is maximal monotone if and only if (∀i) Ai is max-imal monotone.Proof. Since Ai is monotone, (µAi + Id)−1 is firmly nonexpansive, so thereexists a nonexpansive mapping Ni such that JµAi =Ni+Id2 . Thenλ1JµA1 + · · ·+ λnJµAn =(λ1N1 + · · ·+ λnNn) + Id2,is firmly nonexpansive, since λ1N1 + · · · + λnNn is nonexpansive. Thismeans that there exists a monotone operator B such that (µB + Id)−1 =λ1JµA1 + · · ·+ λnJµAn . ThenµB = (λ1JµA1 + · · ·+ λnJµAn)−1 − Id = µRµ(A,λ),therefore Rµ(A,λ) = B is monotone. Since JµRµ(A,λ) = λ1JµA1 + · · · +λnJµAn , this givesdom JµRµ(A,λ) = dom JµA1 ∩ · · · ∩ dom JµAn ,which is (5.5). If each Ai is maximal monotone, then µAi is maximal mono-tone, and thus by Fact 3.43 dom JµAi = H. By (5.5), dom JµRµ(A,λ) = Hand since µRµ(A,λ) is maximal monotone, so is Rµ(A,λ). On the otherhand, if Rµ(A,λ) is maximal monotone, then dom JµRµ(A,λ) = H. It fol-lows from (5.5) that (∀i ∈ I) domJµAi = H, thus µAi must be maximalmonotone and therefore Ai is maximal monotone.655.1. Basic propertiesProposition 5.4. For all i ∈ I, let Ai be a maximally monotone operatorfrom H ⇒ H. Let A = (A1, A−11 , . . . , Am, A−1m ), λ = (12m ,12m , . . . ,12m), andµ = 1. Then Rµ(A,λ) = Id.Proof. This follows directly from the definition of Rµ(A,λ), (5.2), and theresolvent identity, (3.13).Proposition 5.5. Let A = (A1, . . . , A1). Then Rµ(A,λ) = A1.Proof. We haveRµ(A,λ) =((λ1 + · · ·+ λn)(A1 + µ−1 Id)−1)−1− µ−1 Id=((A1 + µ−1 Id)−1)−1− µ−1 Id = A1 + µ−1 Id−µ−1 Id = A1,which proves the result.For clarification, in the following result we write Rµ(A1, λ1, · · · , An, λn)for Rµ(A,λ).Proposition 5.6 (recursion). We haveRµ(A1, λ1, . . . , An, λn) = Rµ(Rµ(A1,λ11−λn, . . . , An−1,λn−11−λn), 1− λn, An, λn).In particular, for λ1 = · · · = λn = 1n one hasRµ(A1, 1n , · · · , An,1n)= Rµ(Rµ(A1, 1n−1 , . . . , An−1,1n−1), 1− 1n , An,1n).Proof. This follows from the definition of Rµ(A,λ). Indeed,Rµ(A,λ) =[λ1(A1 + µ−1 Id)−1 + · · ·+ λn(An + µ−1 Id)−1]−1− µ−1 Id=[(1− λn)(λ11− λn(A1 + µ−1 Id)−1 + · · ·+λn−11− λn(An−1 + µ−1 Id)−1)+ λn(An + µ−1 Id)−1]−1− µ−1 Id=[(1− λn)(Rµ(A1, λ1/(1− λn), · · · , An−1, λn−1/(1− λn)) + µ−1 Id)−1+ λn(An + µ−1 Id)−1]−1− µ−1 Id= Rµ(Rµ(A1,λ11−λn, . . . , An−1,λn−11−λn), 1− λn, An, λn).665.1. Basic propertiesProposition 5.7 (Minty parametrization of Rµ(A,λ)). For all i ∈ I, letAi be a maximally monotone operator from H⇒ H. Then for every x ∈ H,we have(JµRµ(A,λ)(x), x− JµRµ(A,λ)(x)) =λ1(JµA1(x), x− JµA1(x)) + · · ·+ λn(JµAn(x), x− JµAn(x)). (5.6)Consequently,graµRµ(A,λ) ⊂ λ1 graµA1 + · · ·+ λn graµAn.In particular,graR1(A,λ) ⊂ λ1 graA1 + · · ·+ λn graAn.Proof. As Minty’s parametrization of µRµ(A,λ) isgraµRµ(A,λ) ={(JµRµ(A,λ)(x), x− JµRµ(A,λ)(x))∣∣ x ∈ H},then applying (5.3) andId−JµRµ(A,λ) = λ1(Id−JµA1) + · · ·+ λn(Id−JµAn),we havegraRµ(A,λ)={(n∑i=1λiJµAix,n∑i=1λi(Id−JµAi)x)∣∣ x ∈ H}={(λ1JµA1x, λ1(Id−JµA1)x) + · · ·+ (λnJµAnx, λn(Id−JµAn)x)∣∣ x ∈ H}={λ1 (JµA1x, (Id−JµA1)x) + · · ·+ λn (JµAnx, (Id−JµAn)x)∣∣ x ∈ H}⊂ λ1 graµA1 + · · ·+ λn graµAn.Theorem 5.8 (self-duality). For all i ∈ I, let Ai be a monotone operatoron H and µ > 0. Assume that∑ni=1 λi = 1 with λi > 0. Then(Rµ(A,λ))−1 = Rµ−1(A−1,λ), i.e., (5.7)[(λ1(A1 + µ−1 Id)−1 + · · ·+ λn(An + µ−1 Id)−1)−1− µ−1 Id]−1=(λ1(A−11 + µ Id)−1 + · · ·+ λn(A−1n + µ Id)−1)−1− µ Id .675.1. Basic propertiesProof. By (3.15) we have,(Ai + µ−1 Id)−1 = µ(Id−(Id +µ−1A−1i )−1).This and the fact that∑ni=1 λi = 1 givesRµ(A,λ)=[λ1µ(Id−(Id +µ−1A−11 )−1)+ · · ·+ λnµ(Id−(Id +µ−1A−1n )−1)]−1− µ−1 Id=[µ(n∑i=1λi Id−n∑i=1λiJµ−1A−1i)]−1− µ−1 Id=(Id +(−n∑i=1λiJµ−1A−1i))−1◦ (µ−1 Id)− µ−1 Id .By (3.13) we have,(Id +(−n∑i=1λiJµ−1A−1i))−1= Id−Id +(−n∑i=1λiJµ−1A−1i)−1−1.Then,Rµ(A,λ) =Id−Id +(−n∑i=1λiJµ−1A−1i)−1−1 ◦ (µ−1 Id)− µ−1 Id= µ−1 Id−Id +(−n∑i=1λiJµ−1A−1i)−1−1◦ (µ−1 Id)− µ−1 Id= −Id +(−n∑i=1λiJµ−1A−1i)−1−1◦ (µ−1 Id).685.1. Basic properties= −µId +(−n∑i=1λiJµ−1A−1i)−1−1= −µ Id +µ(−n∑i=1λiJµ−1A−1i)−1−1= −µ Id +((−n∑i=1λiJµ−1A−1i)◦ (µ−1 Id))−1−1= −[µ Id +((−λ1(Id +µ−1A−11 )−1 ◦ (µ−1 Id) − · · ·− λn(Id +µ−1A−1n )−1 ◦ (µ−1 Id)))−1]−1= −[µ Id +(−λ1(µ(Id +µ−1A−11 ))−1− · · ·−λn(µ(Id +µ−1A−1n ))−1)−1]−1= −[µ Id +(−λ1(µ Id +A−11)−1− · · · − λn(µ Id +A−1n)−1)−1]−1.To continue, we writeRµ(A,λ)= −[µ Id +(−(λ1(µ Id +A−11)−1+ · · ·+ λn(µ Id +A−1n)−1))−1]−1= −[µ Id +(λ1(µ Id +A−11)−1+ · · ·+ λn(µ Id +A−1n)−1)−1◦ (− Id)]−1=[(µ Id +(λ1(µ Id +A−11)−1+ · · ·+ λn(µ Id +A−1n)−1)−1◦ (− Id))◦ (− Id)]−1=[−µ Id +(λ1(µ Id +A−11)−1+ · · ·+ λn(µ Id +A−1n)−1)−1]−1=(Rµ−1(A−1,λ))−1,which gives Rµ(A,λ) =(Rµ−1(A−1,λ))−1. Taking inverses on both sideswe obtain (5.7).Corollary 5.9. Let Ai : H⇒ H be monotone operators for all i = 1, . . . , n,λi be strictly positive real numbers such that∑ni=1 λi = 1, and µ > 0. SetA = (A1, . . . , An) and A−1 = (A−11 , . . . , A−1n ). ThenJ(µRµ(A,λ))−1 = Jµ−1Rµ−1 (A−1,λ) = λ1Jµ−1A−11+ · · ·+ λnJµ−1A−1n . (5.8)695.2. The resolvent average of positive semidefinite matricesIn particular,JR1(A,λ)−1 = λ1JA−11+ · · ·+ λnJA−1n . (5.9)Proof. Combine (5.3) with Theorem 5.8.5.2 The resolvent average of positive semidefinitematricesThis section covers results specific to positive definite and positive semidef-inite N ×N matrices. Recall the following fact for these types of matrices:Fact 5.10. [42, Corollary 7.7.4.(a)] and [46, Section 16.E] or [26, page 55].Let A,B ∈ SN++, we haveA B ⇔ A−1 B−1 (5.10)andA B ⇔ A−1 ≺ B−1; (5.11)Proposition 5.11. Assume that (∀ i) Ai, Bi ∈ SN+ and Ai Bi. ThenRµ(A,λ) Rµ(B,λ). (5.12)Furthermore, if additionally some Aj Bj, then Rµ(A,λ) Rµ(B,λ).Proof. Note that ∀ µ > 0,Ai + µ−1 Id Bi + µ−1 Id 0,so that0 ≺ (Ai + µ−1 Id)−1 (Bi + µ−1 Id)−1,by (5.10). As SN+ and SN++ are convex cones, we obtain that0 ≺n∑i=1λi(Ai + µ−1 Id)−1 n∑i=1λi(Bi + µ−1 Id)−1. (5.13)Using (5.10) on (5.13), followed by subtracting µ−1 Id, gives[n∑i=1λi(Ai + µ−1 Id)−1]−1− µ−1 Id [n∑i=1λi(Bi + µ−1 Id)−1]−1− µ−1 Id,which establishes (5.12). The “furthermore” part follows analogously using(5.11).705.2. The resolvent average of positive semidefinite matricesTheorem 5.12. Assume that (∀ i) Ai ∈ SN+ . Then Rµ(A,λ) ∈ SN+ . Fur-thermore, if additionally some Aj ∈ SN++, then Rµ(A,λ) ∈ SN++.Proof. This follows from Proposition 5.11 (with each Bi = 0) and Proposi-tion 5.5.5.2.1 Inequalities among meansIn this section, we derive an inequality comparing the resolvent averageto the arithmetic and harmonic averages when (∀ i) Ai ∈ SN++. We startby computing the proximal average of general linear-quadratic functionsthereby extending Fact 2.70.Lemma 5.13. Let Ai ∈ SN+ , bi ∈ RN , ri ∈ R. If each fi = qAi + 〈bi, ·〉+ ri,i.e., linear-quadratic, then ∀x∗ ∈ RN ,Pµ(f ,λ)(x∗)= qRµ(A,λ)(x∗) +〈x∗,(n∑i=1λi(Ai + µ−1 Id)−1)−1n∑i=1λi(Ai + µ−1 Id)−1bi〉+ q(∑ni=1 λi(Ai+µ−1 Id)−1)−1(n∑i=1λi(Ai + µ−1 Id)−1bi)−n∑i=1λi(q(Ai+µ−1 Id)−1(bi)− ri). (5.14)In particular, if (∀i) fi is quadratic, i.e., bi = 0, ri = 0, then Pµ(f ,λ) isquadratic withPµ(f ,λ) = qRµ(A,λ);If (∀i) fi is affine, i.e., Ai = 0, then Pµ(f ,λ) is affine.Proof. We have fi +µ−1q = q(Ai+µ−1 Id) + 〈bi, ·〉+ ri and applying Fact 2.58and then expanding we get(fi + µ−1q)∗(x∗) = q(Ai+µ−1 Id)−1(x∗ − bi)− ri= q(Ai+µ−1 Id)−1(x∗)−〈x∗, (Ai + µ−1 Id)−1bi〉+ q(Ai+µ−1 Id)−1(bi)− ri.715.2. The resolvent average of positive semidefinite matricesThen(λ1(f1 + µ−1q)∗ + · · ·+ λn(fn + µ−1q)∗)(x∗) =n∑i=1λi(q(Ai+µ−1 Id)−1(x∗)−〈x∗, (Ai + µ−1 Id)−1bi〉+ q(Ai+µ−1 Id)−1(bi)− ri)= q∑ni=1 λi(Ai+µ−1 Id)−1(x∗)−〈x∗,n∑i=1λi(Ai + µ−1 Id)−1bi〉+n∑i=1λi(q(Ai+µ−1 Id)−1(bi)− ri).It follows again from Fact 2.58 thatPµ(f ,λ)(x∗) = q[∑ni=1 λi(Ai+µ−1 Id)−1]−1(x∗ +n∑i=1λi(Ai + µ−1 Id)−1bi)−n∑i=1λi(q(Ai+µ−1 Id)−1(bi)− ri)− qµ−1 Id(x∗).Sinceq[∑ni=1 λi(Ai+µ−1 Id)−1]−1(x∗ +n∑i=1λi(Ai + µ−1 Id)−1bi)= q[∑ni=1 λi(Ai+µ−1 Id)−1]−1(x∗)+〈x∗, [n∑i=1λi(Ai + µ−1 Id)−1]−1n∑i=1λi(Ai + µ−1 Id)−1bi〉+ q[∑ni=1 λi(Ai+µ−1 Id)−1]−1(n∑i=1λi(Ai + µ−1 Id)−1bi),we obtain thatPµ(f ,λ)(x∗) = q[∑ni=1 λi(Ai+µ−1 Id)−1]−1−µ−1 Id(x∗)+〈x∗, [n∑i=1λi(Ai + µ−1 Id)−1]−1n∑i=1λi(Ai + µ−1 Id)−1bi〉+ q[∑ni=1 λi(Ai+µ−1 Id)−1]−1(n∑i=1λi(Ai + µ−1 Id)−1bi)−n∑i=1λi(q(Ai+µ−1 Id)−1(bi)− ri),725.2. The resolvent average of positive semidefinite matriceswhich is (5.14). The remaining claims are immediate from (5.14) and thatRµ(A,λ) = 0 when (∀ i) Ai = 0 by Proposition 5.5.We are ready for the main result of this section:Theorem 5.14 (harmonic-resolvent-arithmetic average inequality and lim-its).Let A1, . . . , An ∈ SN++. We have(i)H(A,λ) Rµ(A,λ) A(A,λ); (5.15)In particular, Rµ(A,λ) ∈ SN++.(ii) Rµ(A,λ)→ A(A,λ) when µ→ 0+.(iii) Rµ(A,λ)→ H(A,λ) when µ→ +∞.Proof. (i): According to Fact 2.69,(λ1f∗1 + · · ·+ λnf∗n)∗ ≤ Pµ(f ,λ) ≤ λ1f1 + · · ·+ λnfn. (5.16)Let fi = qAi . Using (qAi)∗ = qA−1i (by Fact 2.57) and Lemma 5.13 we have(λ1f∗1 + · · ·+ λnf∗n)∗ = (λ1qA−11 + · · ·+ λnqA−1n )∗ = (qλ1A−11 +···+λnA−1n )∗= q(λ1A−11 +···+λnA−1n )−1 = qH(A,λ). (5.17)λ1f1 + · · ·+ λnfn = qλ1A1+···+λnAn = qA(A,λ), (5.18)Pµ(f ,λ) = qRµ(A,λ). (5.19)Then (5.16) becomesqH(A,λ) ≤ qRµ(A,λ) ≤ qA(A,λ).As qX ≤ qY ⇔ X Y , (5.15) is established. Since Ai ∈ SN++, A−1i ∈SN++, λ1A−11 + · · · + λnA−1n ∈ SN++, we have H(A,λ) = (λ1A−11 + · · · +λnA−1n )−1 ∈ SN++, thus Rµ(A,λ) ∈ SN++ by (5.15). (Alternatively, applyTheorem 5.12.)(ii) and (iii): Observe that (∀i)(λi ? fi)∗ = λif∗i = λiqA−1i has fulldomain. By Fact 2.56,(λ1f∗1 + · · ·+ λnf∗n)∗ = (λ1 ? f1) · · · (λn ? fn) .735.2. The resolvent average of positive semidefinite matricesBy Fact 2.72, ∀x ∈ RN one haslimµ→0+Pµ(f ,λ)(x) = (λ1f1 + · · ·+ λnfn)(x),limµ→+∞Pµ(f ,λ)(x) = (λ1f∗1 + · · ·+ λnf∗n)∗(x).Since (∀i) fi, f∗i are differentiable on RN , so is Pµ(f ,λ) by Fact 2.71. Ac-cording to Fact 2.60, for all xlimµ→0+∇Pµ(f ,λ)(x) = λ1∇f1(x) + · · ·+ λn∇fn(x), (5.20)limµ→+∞∇Pµ(f ,λ)(x) = ∇(λ1f∗1 + · · ·+ λnf∗n)∗(x). (5.21)Moreover, the convergences in (5.20)-(5.21) are uniform on every closedbounded subset of RN . Now it follows from (5.17)-(5.19) that∇Pµ(f ,λ) = Rµ(A,λ),∇(λ1f1 + · · ·+ λnfn) = A(A,λ),∇(λ1f∗1 + · · ·+ λnf∗n)∗ = H(A,λ).(5.20)–(5.21) becomeslimµ→0+Rµ(A,λ)x = A(A,λ)x, (5.22)limµ→+∞Rµ(A,λ)x = H(A,λ)x, (5.23)where the convergences are uniform on every closed bounded subset of RN .Hence (ii) and (iii) follow from (5.22) and (5.23).Note that in Theorem 5.14(ii) and (ii), there is no ambiguity since allnorms in finite dimensional spaces are equivalent.Definition 5.15. A function g : D→ SN , where D is a convex subset of SN ,is matrix convex if ∀A1, A2 ∈ D, ∀λ ∈ [0, 1],g(λA1 + (1− λ)A2) λg(A1) + (1− λ)g(A2).Matrix concave functions are defined similarly.It is easy to see that a symmetric matrix valued function g is matrixconcave if and only if ∀x ∈ RN the function A 7→ qg(A)(x) is concave.Similarly, g is matrix convex if and only if A 7→ qg(A)(x) is convex. Someimmediate consequences of Theorem 5.14 on matrix-valued functions are:745.2. The resolvent average of positive semidefinite matricesCorollary 5.16. Assume that (∀i) Ai ∈ SN++ and∑ni=1 λi = 1 with λi > 0.Then(λ1A1 + · · ·+ λnAn)−1 λ1A−11 + · · ·+ λnA−1n .Consequently, the matrix function X 7→ X−1 is matrix convex on SN++.Proof. Apply (5.15) with A = (A−11 , · · · , A−1n ).Corollary 5.17. For every µ > 0, the resolvent average matrix functionA 7→ Rµ(A,λ) given by(A1, · · · , An) 7→ [λ1(A1 + µ−1 Id)−1 + · · ·+ λn(An + µ−1 Id)−1]−1 − µ−1 Idis matrix concave on SN++ × · · · × SN++. (5.24)For each λ = (λ1, · · · , λn) with∑ni=1 λi = 1 and λi > 0 ∀i, the harmonicaverage matrix function(A1, · · · , An) 7→ (λ1A−11 + · · ·+ λnA−1n )−1 is matrix concave (5.25)on SN++ × · · · × SN++. Consequently, the harmonic average function(x1, · · · , xn) 7→1x−11 + · · ·+ x−1nis concave (5.26)on R++ × · · · × R++.Proof. Set fi = qAi . Then ∀x ∈ RN , we have from (2.17)Pµ(f ,λ)(x) = minλ1x1+···+λnxn=x((λ1qA1(x1) + · · ·+ λnqAn(xn))+(µ−1λ1q(x1) + · · ·+ µ−1λnq(xn)))− µ−1q(x).Since for each fixed (x1, . . . , xn),(A1, · · · , An) 7→(λ1qA1(x1) + · · ·+ λnqAn(xn)),is affine, being the minimum of affine functions we have that ∀x the function(A1, · · · , An) 7→ Pµ(f ,λ)(x),is concave. As Pµ(f ,λ)(x) = qRµ(A,λ)(x) by Lemma 5.13, this shows that∀x ∈ RN the functionA = (A1, · · · , An) 7→ qRµ(A,λ)(x) is concave,755.2. The resolvent average of positive semidefinite matricesso A 7→ Rµ(A,λ) is matrix concave.Now by Theorem 5.14(iii), Rµ(A,λ) → H(A,λ) when µ → +∞. Thisand (5.24) implies thatA 7→ H(A,λ),is also matrix concave. Equation (5.26) follows from (5.25) by setting N = 1and λ1 = · · · = λn = 1/n.Remark 5.18. Corollary 5.16 is well-known, cf. [59, Proposition 2.56]. Equa-tion (5.26) is also well-known, cf. [27, Exercise 3.17].The next theorem provides a simplified proof of Theorem 5.8 when theoperators are positive definite matrices.Theorem 5.19 (self-duality). Let (∀ i) Ai ∈ SN++ and µ > 0. Assume that∑ni=1 λi = 1 with λi > 0. Then[Rµ(A,λ)]−1= Rµ−1(A−1,λ), (5.27)i.e.,[(λ1(A1 + µ−1 Id)−1 + · · ·+ λn(An + µ−1 Id)−1)−1− µ−1 Id]−1=(λ1(A−11 + µ Id)−1 + · · ·+ λn(A−1n + µ Id)−1)−1− µ Id .In particular, for µ = 1, [R1(A,λ)]−1 = R1(A−1,λ).Proof. Let fi = qAi . By Fact 2.66,(Pµ(f ,λ))∗= Pµ−1(f∗,λ), taking sub-gradients on both sides, followed by using Fact 2.59, we obtain that∂(Pµ(f ,λ))∗=(∂Pµ(f ,λ))−1= ∂(Pµ−1(f∗,λ)).By Lemma 5.13, Pµ(f ,λ) = qRµ(A,λ),Pµ−1(f∗,λ)) = qRµ−1 (A−1,λ), we have∂Pµ(f ,λ) = Rµ(A,λ),∂Pµ−1(f∗,λ) = Rµ−1(A−1,λ).Hence[Rµ(A,λ)]−1= Rµ−1(A−1,λ),as claimed.Remark 5.20. Although the harmonic and arithmetic average lack self-duality,they are dual to each other:[H(A,λ)]−1 = λ1A−11 + · · ·+ λnA−1n = A(A−1,λ),[A(A,λ)]−1 =[λ1(A−11 )−1 + · · ·+ λn(A−1n )−1]−1 = H(A−1,λ).765.2. The resolvent average of positive semidefinite matrices5.2.2 A comparison to weighted geometric meansTo compare the resolvent average with the well-known geometric mean,we restrict our attention to non-negative real numbers (1 × 1 matrices).When A = x = (x1, . . . , xn) with xi ∈ R+ and µ = 1, we writeR(x,λ) = Rµ(A,λ) =(λ1(x1 + 1)−1 + · · ·+ λn(xn + 1)−1)−1 − 1,and x−1 = (1/x1, . . . , 1/xn) when (∀i) xi ∈ R++.Proposition 5.21. Let (∀ i) xi > 0, yi > 0. We have(i) (harmonic-resolvent-arithmetic mean inequality):(λ1x−11 + · · ·+ λnx−1n)−1≤ R(x,λ) ≤ λ1x1 + · · ·+ λnxn. (5.28)Moreover, R(x,λ) = λ1x1 + · · ·+ λnxn if and only if x1 = · · · = xn.(ii) (self-duality): [R(x,λ)]−1 = R(x−1,λ).(iii) If x = (x1, . . . , x1), then R(x,λ) = x1.(iv) If x = (x1, x−11 , x2, x−12 , . . . , xn, x−1n ) and λ = (12n , . . . ,12n), thenR(x,λ) = 1.(v) The function x 7→ R(x,λ) is concave on R++ × · · · × R++.(vi) If x y, then R(x,λ) ≥ R(y,λ).Proof. (i): For (5.28), apply Theorem 5.14(i) with µ = 1. NowR(x,λ) = λ1x1 + · · ·+ λnxnis equivalent to(λ1(x1 + 1)−1 + · · ·+ λn(xn + 1)−1)−1 = λ1x1 + · · ·+ λnxn + 1, (5.29)As∑ni=1 λi = 1, (5.29) is the same asλ11(x1 + 1)+ · · ·+ λn1(xn + 1)=1λ1(x1 + 1) + · · ·+ λn(xn + 1).Since the function x 7→ 1/x is strictly convex on R++, we must havex1 = · · · = xn.(ii): Theorem 5.19. (iii): Proposition 5.5. (iv): Proposition 5.4. (v):Corollary 5.17. (vi): Proposition 5.11.775.2. The resolvent average of positive semidefinite matricesProposition 5.21 and Fact 2.64 demonstrate that R(x,λ) and G(x,λ)have strikingly similar properties. Are they the same?Example 5.22.(i) Let λ = (12 ,12). When x = (0, 1), G(x,λ) = 0 but R(x,λ) =13 , soG(x,λ) 6= R(x,λ).(ii) Is it right that G(x,λ) ≤ R(x,λ) for all x ∈ R2++? The answer is alsono. Assume that G(x,λ) ≤ R(x,λ), ∀ x ∈ R++ × R++. Taking theinverse of both sides, followed by applying the self-duality of G(x,λ)and R(x,λ), givesG(x,λ)−1 ≥ R(x,λ)−1 = R(x−1,λ) ≥ G(x−1,λ) = G(x,λ)−1,and this gives that G(x,λ)−1 = R(x,λ)−1 so that G(x,λ) = R(x,λ).This is a contradiction to (i), thus R(x,λ) is distinct from G(x,λ).78Chapter 6Near Equality, NearConvexity, Sums ofMaximally MonotoneOperators, and Averages ofFirmly NonexpansiveMappingsIn this chapter, based on [20], we introduce near equality for sets andshow that this notion is useful in the study of nearly convex sets. Theseresults are the key to study ranges of sums of maximal monotone operatorsin the next section. Recall that I denotes an index setI = {1, 2, . . . ,m},for a strictly positive integer m.6.1 Near equality and near convexityDefinition 6.1 (near equality). Let A and B be subsets of Rn. We say thatA and B are nearly equal , ifA = B and riA = riB. (6.1)and denote this by A ≈ B.Remark 6.2. The following holds:A ≈ B ⇒ intA = intB. (6.2)Observe that if intA 6= ∅ then there exists x ∈ A such that B(x, ) ⊆ A forsome > 0. This is an n-dimensional convex set in Rn and thus aff A = Rn796.1. Near equality and near convexityand so intA = riA = riB. Thus, B(x, ) ⊆ B and so intA = intB.Similarly, intB 6= ∅⇒ intA 6= ∅, thus intA = ∅⇔ intB = ∅. Altogether,A ≈ B ⇒ intA = intB.Proposition 6.3 (equivalence relation). The following hold for any subsetsA, B, C of Rn.(i) A ≈ A.(ii) A ≈ B ⇒ B ≈ A.(iii) A ≈ B and B ≈ C ⇒ A ≈ C.Proposition 6.4 (squeeze theorem). Let A,B,C be subsets of Rn such thatA ≈ C and A ⊆ B ⊆ C. Then A ≈ B ≈ C.Proof. By assumption, A = C and riA = riC. Thus A = B = C and byLemma 2.24, aff(A) = aff(A) = aff(C) = aff(C). Hence aff A = aff B =aff C and so, by Lemma 2.22, riA ⊆ riB ⊆ riC. Since riA = riC, wededuce that riA = riB = riC. Therefore, A ≈ B ≈ C.The equivalence relation “≈” is best suited for studying nearly convexsets (defined next), as we do have that, e.g., Q ≈ RrQ!Definition 6.5 (near convexity). [59, Theorem 12.41] Let A be a subset ofRn. Then A is nearly convex if there exists a convex subset C of Rn suchthat C ⊆ A ⊆ C.Lemma 6.6. Let A be a nearly convex subset of Rn, say C ⊆ A ⊆ C, whereC is a convex subset of Rn. ThenA ≈ A ≈ riA ≈ convA ≈ ri convA ≈ C. (6.3)In particular, the following hold.(i) A and riA are convex.(ii) If A 6= ∅, then riA 6= ∅.Proof. We haveC ⊆ A ⊆ convA ⊆ C and C ⊆ A ⊆ A ⊆ C. (6.4)Since C ≈ C by Fact 2.25(iv), it follows from Proposition 6.4 thatA ≈ A ≈ convA ≈ C. (6.5)806.1. Near equality and near convexityThis impliesri(riA) = ri(riC) = riC = riA (6.6)andriA = riC = C = A (6.7)by Fact 2.25(iii). Therefore, riA ≈ A. Applying this to convA, which isnearly convex, it also follows that ri convA ≈ convA. Finally, (i) holdsbecause A ≈ C while (ii) follows from riA = riC and Fact 2.25(ii).Remark 6.7. The assumption of near convexity in Lemma 6.6 is necessary:consider R with A = Q. Then riQ = ∅ but Q is obviously not. Thus (6.3)fails, although (i) still holds.Lemma 6.8 (characterization of near convexity). Let A ⊆ Rn. Then thefollowing are equivalent.(i) A is nearly convex.(ii) A ≈ convA.(iii) A is nearly equal to a convex set.(iv) A is nearly equal to a nearly convex set.(v) ri convA ⊆ A.Proof. “(i)⇒(ii)”: Apply Lemma 6.6. “(ii)⇒(v)”: Indeed, ri convA =riA ⊆ A. “(v)⇒(i)”: Set C = ri convA. By Fact 2.25(iii), C ⊆ A ⊆convA = ri convA = C. “(ii)⇒(iii)”: Clear. “(iii)⇒(i)”: Suppose thatA ≈ C, where C is convex. Then, using Fact 2.25(iii), riC = riA ⊆ A ⊆A = C = riC. Hence A is nearly convex. “(iii)⇒(iv)”: Clear. “(iv)⇒(iii)”:(The following simple proof was suggested by a referee of [20]) SupposeA ≈ B, where B is nearly convex. Then, applying the already verified im-plications “(i)⇒(ii)” and “(ii)⇒(iii)” to the set B, we see that B ≈ C forsome convex set C. Using Proposition 6.3(iii), we conclude that A ≈ C.Remark 6.9. The condition appearing in Lemma 6.8(v) was also used byMinty [49] and named “almost-convex”.Remark 6.10. Bre´zis and Haraux [28] define, for two subsets A and B of Rn,A ' B ⇔ A = B and intA = intB. (6.8)(i) In view of (6.2), it is clear that A ≈ B ⇒ A ' B.816.1. Near equality and near convexity(ii) On the other hand, A ' B 6⇒ A ≈ B: indeed, consider R2 withA = Q × {0}, and B = R × {0}. Then intA = intB = ∅, butriA 6= riB.(iii) The implications (iii)⇒(i) and (ii)⇒(i) in Lemma 6.8 fail for': indeed,consider R2 with A = (R r {0}) × {0} and C = convA = R × {0}.Then C is convex and A ' C. However, A is not nearly convex becauseriA 6= riA.Proposition 6.11. Let A and B be nearly convex subsets of Rn. Then thefollowing are equivalent.(i) A ≈ B.(ii) A = B.(iii) riA = riB.(iv) convA = convB.(v) ri convA = ri convB.Proof. “(i)⇒(ii)”: This is clear from the definition of ≈. “(ii)⇒(iii)”: riA =riA and riB = riB by Lemma 6.6. “(iii)⇒(iv)”: riA = convA and riB =convB by Lemma 6.6. “(iv)⇒(v)”: ri convA = ri convA and ri convB =ri convB. “(v)⇒(i)”: Lemma 6.6 gives that ri convA = riA and ri convB =riB so that riA = riB, ri convA = convA = A and ri convB = convB = Bso that A = B. Hence (i) holds.The next results generalize Rockafellar’s Fact 2.26 to nearly convex sets.Lemma 6.12. Let A1 and A2 be nearly convex sets in Rn such that riA1 ∩riA2 6= ∅. Then A1 ∩A2 is nearly convex andri(A1 ∩A2) = riA1 ∩ riA2.Proof. From Lemma 6.8(v) and the definition of the convex hull, we haveri(convA1) ⊆ A1 ⊆ convA1ri(convA2) ⊆ A2 ⊆ convA2.This implies thatri(convA1) ∩ ri(convA2) ⊆ A1 ∩A2 ⊆ convA1 ∩ convA2. (6.9)826.1. Near equality and near convexityFrom Lemma 6.6 we have riA1 = ri convA1 and riA2 = ri convA2 and thusri convA1 ∩ ri convA2 6= ∅. Then we can apply Fact 2.26 and Lemma 6.6(i)to get,ri(convA1 ∩ convA2) = ri(convA1) ∩ ri(convA2),so by (6.9),ri(convA1 ∩ convA2) ⊆ A1 ∩A2 ⊆ convA1 ∩ convA2.Thus by Fact 2.27, A1 ∩A2 is nearly convex and by Lemma 6.6,A1 ∩A2 ≈ ri(convA1 ∩ convA2).Using (6.6), this meansri(A1 ∩A2) = ri (ri(convA1 ∩ convA2)) = ri(convA1 ∩ convA2). (6.10)Now we also have, A1 ≈ convA1 and A2 ≈ convA2 by Lemma 6.8(ii). Thatand Fact 2.26 yield,riA1 ∩ riA2 = ri convA1 ∩ ri convA2 = ri(convA1 ∩ convA2). (6.11)Combining (6.10) and (6.11) we getri(A1 ∩A2) = ri(convA1 ∩ convA2) = riA1 ∩ riA2,which proves the result.Theorem 6.13. Let Ai be a nearly convex set in Rn for i = 1, . . . ,m suchthatm⋂i=1riAi 6= ∅. Thenm⋂i=1Ai is nearly convex andm⋂i=1riAi = rim⋂i=1Ai.Proof. Clearly, when m = 1 this holds. When m = 2, by Lemma 6.12 wehave A1 ∩A2 is nearly convex andri(A1 ∩A2) = riA1 ∩ riA2.Thus we proceed via induction and assume that whenm⋂i=1riAi 6= ∅,m⋂i=1riAi = rim⋂i=1Ai andm⋂i=1Ai is nearly convex.836.1. Near equality and near convexityThen considerm+1⋂i=1riAi such thatm+1⋂i=1riAi 6= ∅. We havem+1⋂i=1riAi =m⋂i=1riAi ∩ riAm+1 6= ∅. (6.12)Sincem+1⋂i=1riAi 6= ∅ we must havem⋂i=1riAi 6= ∅. Thus, by the inductivehypothesis (6.12) becomes,m+1⋂i=1riAi = rim⋂i=1Ai ∩ riAm+1 6= ∅, (6.13)and sincem⋂i=1Ai is nearly convex we can apply Lemma 6.12 to the setsm⋂i=1Aiand Am+1 to getm⋂i=1Ai∩Am+1 =m+1⋂i=1Ai is nearly convex and (6.13) becomesm+1⋂i=1riAi = ri(m⋂i=1Ai ∩Am+1)= rim+1⋂i=1Ai.Lemma 6.14. Let A1 and A2 be nearly convex sets such that riA1∩ riA2 6=∅. ThenA1 ∩A2 = A1 ∩A2.Proof. Clearly we always haveA1 ∩A2 ⊆ A1 ∩A2.On the other hand, by Fact 2.25(iii) and Fact 2.26,A1 ∩A2 = riA1 ∩ riA2 = riA1 ∩ riA2 ⊆ A1 ∩A2.Thus A1 ∩A2 = A1 ∩A2.Theorem 6.15. Let Ai be a nearly convex set in Rn for i = 1, . . . ,m suchthatm⋂i=1riAi 6= ∅. Thenm⋂i=1Ai =m⋂i=1Ai. (6.14)846.1. Near equality and near convexityProof. This clearly holds when m = 1. When m = 2, (6.14) holds byLemma 6.14. Continuing via induction, we assume that whenm⋂i=1riAi 6= ∅,we havem⋂i=1Ai =m⋂i=1Ai.Then we considerm+1⋂i=1Ai such thatm+1⋂i=1riAi 6= ∅. We havem+1⋂i=1Ai =m⋂i=1Ai ∩Am+1. (6.15)Sincem+1⋂i=1riAi 6= ∅, then clearlym⋂i=1riAi 6= ∅. Thus, by the inductivehypothesis, (6.15) becomesm+1⋂i=1Ai =m⋂i=1Ai ∩Am+1. (6.16)Now, by Theorem 6.13,m⋂i=1Ai is nearly convex and rim⋂i=1Ai =m⋂i=1riAi so,ri(m⋂i=1Ai)∩ riAm+1 =m⋂i=1riAi ∩ riAm+1 =m+1⋂i=1riAi 6= ∅.Thus, apply Lemma 6.14 to the setsm⋂i=1Ai and Am+1, and (6.16) becomes,m+1⋂i=1Ai =m⋂i=1Ai ∩Am+1 =m+1⋂i=1Ai,which proves the desired result.In order to study addition of nearly convex sets, we require the followingresult.Lemma 6.16. Let (Ai)i∈I be a family of nearly convex subsets of Rn, andlet (λi)i∈I be a family of real numbers. Then∑i∈I λiAi is nearly convex,and ri(∑i∈I λiAi) =∑i∈I λi riAi.856.1. Near equality and near convexityProof. For every i ∈ I, there exists a convex subset Ci of Rn such thatCi ⊆ Ai ⊆ Ci. We have∑i∈IλiCi ⊆∑i∈IλiAi ⊆∑i∈IλiCi ⊆∑i∈IλiCi, (6.17)which yields the near convexity of∑i∈I λiAi and∑i∈I λiAi ≈∑i∈I λiCi byLemma 6.6. Moreover, by Fact 2.25(vii)&(viii) and Lemma 6.6,ri(∑i∈IλiAi)= ri(∑i∈IλiCi)=∑i∈Iri(λiCi)=∑i∈Iλi riCi =∑i∈Iλi riAi.(6.18)This completes the proof.Theorem 6.17. Let (Ai)i∈I be a family of nearly convex subsets of Rn, andlet (Bi)i∈I be a family of subsets of Rn such that Ai ≈ Bi, for every i ∈ I.Then∑i∈I Ai is nearly convex and∑i∈I Ai ≈∑i∈I Bi.Proof. Lemma 6.8 implies that Bi is nearly convex, for every i ∈ I. ByLemma 6.16, we have that∑i∈I Ai is nearly convex andri∑i∈IAi =∑i∈IriAi =∑i∈IriBi = ri∑i∈IBi. (6.19)Furthermore,∑i∈IAi =∑i∈IAi =∑i∈IBi =∑i∈IBi (6.20)and the result follows.Remark 6.18. Theorem 6.17 fails without the near convexity assumption:indeed, consider R and m = 2, with A1 = A2 = Q and B1 = B2 = R r Q.Then Ai ≈ Bi, for every i ∈ I, yet A1 +A2 = Q 6≈ R = B1 +B2.Theorem 6.19. Let (Ai)i∈I be a family of nearly convex subsets of Rn,and let (λi)i∈I be a family of real numbers. For every i ∈ I, take Bi ∈{Ai, Ai, convAi, riAi, ri convAi}. Then∑i∈IλiAi ≈∑i∈IλiBi. (6.21)Proof. By Lemma 6.6, Ai ≈ Bi for every i ∈ I. Now apply Theorem 6.17.866.1. Near equality and near convexityCorollary 6.20. Let (Ai)i∈I be a family of nearly convex subsets of Rn,and let (λi)i∈I be a family of real numbers. Suppose that j ∈ I is such thatλj 6= 0. Then(intλjAj)+∑i∈Ir{j}λiAi ⊆ int∑i∈IλiAi; (6.22)consequently, if 0 ∈ (intAj) ∩⋂i∈Ir{j}Ai, then 0 ∈ int∑i∈I λiAi.Proof. By Theorem 6.19, ri(λjAj +∑i∈Ir{j} λiAi) = ri∑i∈I λiAi. Since(intλjAj)+∑i∈Ir{j}λiAi ⊆ ri(λjAj +∑i∈Ir{j}λiAi), (6.23)and(intλjAj)+∑i∈Ir{j} λiAi is an open set, (6.22) follows. In turn, the“consequently” follows from (6.22).We develop a complementary cancelation result whose proof relies onR˚adstro¨m’s cancelation:Fact 6.21. (See [55].) Let A be a nonempty subset of Rn, let E be anonempty bounded subset of Rn, and let B be a nonempty closed convexsubset of Rn such that A+ E ⊆ B + E. Then A ⊆ B.Theorem 6.22. Let A and B be nonempty nearly convex subsets of Rn,and let E be a nonempty compact subset of Rn such that A + E ≈ B + E.Then A ≈ B.Proof. We have A + E ⊆ A+ E = B + E = B + E. Fact 6.21 impliesA ⊆ B; hence, A ⊆ B. Analogously, B ⊆ A and thus A = B. Now applyProposition 6.11.Finally, we give a result concerning the interior of nearly convex sets.Proposition 6.23. Let A be a nearly convex subset of Rn. Then intA =int convA = intA.Proof. By Lemma 6.6, A ≈ B, where B ∈{A, convA}. Now recall (6.2).876.2. Maximally monotone operators6.2 Maximally monotone operatorsFact 6.24 (Minty). [59, Theorem 12.41] Let A : Rn ⇒ Rn be maximallymonotone. Then domA and ranA are nearly convex.Theorem 6.25. Let A and B be monotone on Rn such that A + B ismaximally monotone. Suppose that one of the following holds.(i) A and B are rectangular.(ii) domA ⊆ domB and B is rectangular.Then ran(A+B) is nearly convex, and ran(A+B) ≈ ranA+ ranB.Proof. The near convexity of ran(A + B) follows from Fact 6.24. UsingFact 3.66 and Fact 2.25(iii),ri conv(ranA+ ranB) ⊆ ran(A+B)⊆ ranA+ ranB⊆ conv(ranA+ ranB)= ri conv(ranA+ ranB).Proposition 6.4 and Lemma 6.6 imply ran(A + B) ≈ ranA + ranB ≈ri conv(ranA+ ranB).Remark 6.26. Considering A + 0, where A is the rotator by pi/2 on R2which is not rectangular, we see that A+B need not be rectangular underassumption (ii) in Theorem 6.25.If we let Si = ranAi and λi = 1 for every i ∈ I in Theorem 6.27, thenwe obtain a result that is related to Pennanen’s [53, Corollary 6].Theorem 6.27. Let (Ai)i∈I be a family of maximally monotone rectangularoperators on Rn with⋂i∈I ri domAi 6= ∅, let (Si)i∈I be a family of subsetsof Rn such that(∀i ∈ I) Si ∈{ranAi, ranAi, ri(ranAi)}, (6.24)and let (λi)i∈I be a family of strictly positive real numbers. Then∑i∈I λiAiis maximally monotone, rectangular, and ran∑i∈I λiAi ≈∑i∈I λiSi isnearly convex.886.2. Maximally monotone operatorsProof. We have I = {1, . . . ,m}. To see that∑i∈I λiAi is maximally mono-tone we proceed using induction on m. When m = 1, since λi ∈ R++ andAi is maximally monotone, then λiAi is maximally monotone by Proposi-tion 3.29. When m = 2, by assumption we haveri domA1 ∩ ri domA2 6= ∅⇒ ri domλ1A1 ∩ ri domλ2A2 6= ∅.so by Fact 3.48(ii) λ1A1 + λ2A2 is maximally monotone. Now assumethis holds for λ1A1 + . . .+ λmAm withm⋂i=1ri domAi 6= ∅. Then considerλ1A1 + · · ·+ λm+1Am+1 = (λ1A1 + · · ·+ λmAm) + λm+1Am+1.By the inductive hypothesis, λ1A1 + . . . + λmAm is maximally monotone.We haveri dom(λ1A1 + · · ·+ λmAm) ∩ ri domλm+1Am+1= ri(m⋂i=1domλiAi)∩ ri domλm+1Am+1. (6.25)By Fact 6.24, domλiAi is nearly convex for all i ∈ I, so apply Lemma 6.13and (6.25) becomesri dom(λ1A1 + · · ·+ λmAm) ∩ ri domλm+1Am+1=(m⋂i=1ri domλiAi)∩ ri domλm+1Am+1 =m+1⋂i=1ri domλiAi 6= ∅.Thus by Fact 3.48(ii), λ1A1+· · ·+λm+1Am+1 is maximally monotone. UsingLemma 3.63 and induction we have λ1A1 + · · ·+ λm+1Am+1 is rectangular.With Theorem 6.17, Fact 3.48 and Lemma 3.63 in mind, Theorem 6.25(i)and induction yields ran∑i∈I λiAi ≈∑i∈I λi ranAi and the near con-vexity. Finally, as ranAi is nearly convex for every i ∈ I by Fact 6.24,ran∑i∈I λiAi ≈∑i∈I λiSi follows from Theorem 6.19.The main result of this section is the following.Theorem 6.28. Let (Ai)i∈I be a family of maximally monotone rectangularoperators on Rn such that⋂i∈I ri domAi 6= ∅, let (λi)i∈I be a family ofstrictly positive real numbers, and let j ∈ I. SetA =∑i∈IλiAi. (6.26)Then the following hold.896.3. Firmly nonexpansive mappings(i) If∑i∈I λi ranAi = Rn, then ranA = Rn.(ii) If Aj is surjective, then A is surjective.(iii) If 0 ∈⋂i∈I ranAi, then 0 ∈ ranA.(iv) If 0 ∈ (int ranAj) ∩⋂i∈Ir{j} ranAi, then 0 ∈ int ranA.Proof. Theorem 6.27 implies that ran∑i∈I λiAi ≈∑i∈I λi ranAi is nearlyconvex. Henceri ranA = ri ran∑i∈IλiAi = ri(∑i∈Iλi ranAi)=∑i∈Iλi ri ranAi (6.27)andranA = ran∑i∈IλiAi =∑i∈Iλi ranAi . (6.28)(i): Indeed, using (6.27),Rn = riRn = ri∑i∈Iλi ranAi = ri ranA ⊆ ranA ⊆ Rn.(ii): Clear from (i). (iii): It follows from (6.28) that0 ∈∑i∈IλiranAi ⊆∑i∈IλiranAi = ranA.(iv): By Fact 6.24, ranAi is nearly convex for every i ∈ I. Thus, 0 ∈int∑i∈I λi ranAi by Corollary 6.20. On the other hand, (6.27) implies thatint∑i∈Iλi ranAi ⊆ ri∑i∈Iλi ranAi = ri ranA.Altogether, 0 ∈ ri ranA = int ranA because int ranA 6= ∅.6.3 Firmly nonexpansive mappingsIn this section, we apply the result of Section 6.2 to firmly nonexpansivemappings.Corollary 6.29. Let T : Rn → Rn be firmly nonexpansive. Then T is max-imally monotone and rectangular, and ranT is nearly convex.Proof. Combine Example 3.61, Fact 3.36(i), and Fact 6.24.906.3. Firmly nonexpansive mappingsIt is also known that the class of firmly nonexpansive mappings is closedunder taking convex combinations. For completeness, we include a shortproof of this result.Lemma 6.30. Let (Ti)i∈I be a family of firmly nonexpansive mappings onRn, and let (λi)i∈I be a family of strictly positive real numbers such that∑i∈I λi = 1. Then∑i∈I λiTi is also firmly nonexpansive.Proof. Set T =∑i∈I λiTi. By Fact 3.3, 2Ti − Id is nonexpansive for everyi ∈ I, so 2T−Id =∑i∈I λi(2Ti−Id) is also nonexpansive. Applying Fact 3.3once more, we deduce that T is firmly nonexpansive.We are now ready for the first main result of this section.Theorem 6.31 (averages of firmly nonexpansive mappings). Let (Ti)i∈I bea family of firmly nonexpansive mappings on Rn, let (λi)i∈I be a family ofstrictly positive real numbers such that∑i∈I λi = 1, and let j ∈ I. SetT =∑i∈I λiTi. Then the following hold.(i) T is firmly nonexpansive and ranT ≈∑i∈I λi ranTi is nearly convex.(ii) If Tj is surjective, then T is surjective.(iii) If 0 ∈⋂i∈I ranTi, then 0 ∈ ranT .(iv) If 0 ∈ (int ranTj) ∩⋂i∈Ir{j} ranTi, then 0 ∈ int ranT .Proof. By Corollary 6.29, each Ti is maximally monotone, rectangular andranTi is nearly convex. (i): Lemma 6.6, Lemma 6.30, and Theorem 6.27.(ii): Theorem 6.28(ii). (iii): Theorem 6.28(iii). (iv): Theorem 6.28(iv).The following averaged-projection operator plays a role in methods forsolving (potentially inconsistent) convex feasibility problems because itsfixed point set consists of least-squares solutions; see, e.g., [7, Section 6],[23] and [35] for further information.Example 6.32. Let (Ci)i∈I be a family of nonempty closed convex subsetsof Rn with associated projection operators Pi, and let (λi)i∈I be a family ofstrictly positive real numbers such that∑i∈I λi = 1. Thenran∑i∈IλiPi ≈∑i∈IλiCi. (6.29)Proof. This follows from Theorem 6.31(i) since (∀i ∈ I) ranPi = Ci.916.3. Firmly nonexpansive mappingsRemark 6.33. Let C1 and C2 be nonempty closed convex subsets of Rnwith associated projection operators P1 and P2 respectively, and—insteadof averaging as in Example 6.32—consider the composition T = P2 ◦ P1,which is still nonexpansive. It is obvious that ranT ⊆ ranP2 = C2, butranT need not be even nearly convex: indeed, in R2, let C2 be the unit ballcentered at 0 of radius 1, and let C1 = R×{2}. Then ranT is the intersectionof the open upper halfplane and the boundary of C2, which is very far frombeing nearly convex. Thus the near convexity part of Corollary 6.29 has nocounterpart for nonexpansive mappings.Remark 6.34. Let T : Rn → Rn be firmly nonexpansive. Recall the set offixed points is denoted byFixT ={x ∈ Rn∣∣ x = Tx}, (6.30)and that T is asymptotically regular if there exists a sequence (xn)n∈N in Rnsuch that xn − Txn → 0.If the sequence (xn)n∈N converges to a point, say x¯, then continuity ofT implies that x¯ ∈ FixT .The next result is a consequence of fundamental work by Baillon, Bruckand Reich [3] .Theorem 6.35. Let T : Rn → Rn be firmly nonexpansive. Then T is asymp-totically regular if and only if for every x0 ∈ Rn, the sequence defined by(∀n ∈ N) xn+1 = Txn (6.31)satisfies xn − xn+1 → 0. Moreover, if FixT 6= ∅, then (xn)n∈N convergesweakly to a fixed point; otherwise, ‖xn‖ → +∞.Proof. T is firmly nonexpansive ⇔ T is 12 -averaged, so by Fact 3.20, Tnx−Tn+1x → v where v is the element of minimum norm in ran(Id−T ). SinceT is asymptotically regular, v = 0 and thus xn−xn+1 → 0. By Fact 3.18, ifFixT 6= ∅, then (xn)n∈N ⇀ x ∈ FixT . And by Fact 3.19 if FixT = ∅, then‖xn‖ → +∞.Here is the second main result of this chapter.Theorem 6.36 (asymptotic regularity of the average). Let (Ti)i∈I be afamily of firmly nonexpansive mappings on Rn, and let (λi)i∈I be a familyof strictly positive real numbers such that∑i∈I λi = 1. Suppose that Ti isasymptotically regular, for every i ∈ I. Then∑i∈I λiTi is also asymptoti-cally regular.926.3. Firmly nonexpansive mappingsProof. Set T =∑i∈I λiTi. ThenId−T =∑i∈Iλi(Id−Ti).Since each Id−Ti is firmly nonexpansive and 0 ∈ ran(Id−Ti) by the asymp-totic regularity of Ti, the conclusion follows from Theorem 6.31(iii).Remark 6.37. Consider Theorem 6.36. Even when FixTi 6= ∅, for everyi ∈ I, it is impossible to improve the conclusion to Fix∑i∈I λiTi 6= ∅.Indeed, in R2, set C1 = R × {0} and C2 = epi exp. Set T = 12PC1 +12PC2 .Then FixT1 = C1 and FixT2 = C2, yet FixT = ∅.The proof of the following useful result is straightforward and henceomitted.Lemma 6.38. Let A : Rn ⇒ Rn be maximally monotone. Then JA isasymptotically regular if and only if 0 ∈ ranA.We conclude this chapter with an application to the resolvent average ofmonotone operators.Corollary 6.39 (resolvent average). Let (Ai)i∈I be a family of maximallymonotone operators on Rn, let (λi)i∈I be a family of strictly positive realnumbers such that∑i∈I λi = 1, let j ∈ I, and setR1(A,λ) =(∑i∈Iλi(Id +Ai)−1)−1− Id . (6.32)Then the following hold.(i) R1(A,λ) is maximally monotone.(ii) domR1(A,λ) ≈∑i∈I λi domAi.(iii) ranR1(A,λ) ≈∑i∈I λi ranAi.(iv) If 0 ∈⋂i∈I ranAi, then 0 ∈ ranR1(A,λ).(v) If 0 ∈ int ranAj ∩⋂i∈I\{j} ranAi, then 0 ∈ int ranR1(A,λ).(vi) If domAj = Rn, then domR1(A,λ) = Rn.(vii) If ranAj = Rn, then ranR1(A,λ) = Rn.936.3. Firmly nonexpansive mappingsProof. Observe thatJR1(A,λ) =∑i∈IλiJAi (6.33)andJR1(A,λ)−1 =∑i∈IλiJA−1i(6.34)by using (3.13). Furthermore, using (3.14), we note thatran JR1(A,λ) = domR1(A,λ) and ranJR1(A,λ)−1 = ranR1(A,λ).(6.35)(i): This follows from (6.33) and Fact 3.36. (ii): Apply Theorem 6.31(i)to (JAi)i∈I , and use (6.33) and (6.35). (iii): Apply Theorem 6.31(i) to(Id−JAi)i∈I , and use (3.13) and (6.35). (iv): Combine Theorem 6.36 andLemma 6.38, and use (6.33). (v): Apply Theorem 6.31(iv) to (6.34), anduse (6.35). (vi) and (vii): These follow from (ii) and (iii), respectively.Remark 6.40 (proximal average). In Corollary 6.39, one may also start froma family (fi)i∈I of functions on Rn that are convex, lower semi-continuous,and proper, and with corresponding subdifferential operators (Ai)i∈I =(∂fi)i∈I . This relates to the proximal average, P, of the family (fi)i∈I ,where ∂P is the resolvent average of the family (∂fi)i∈I . See [12] for fur-ther information and references. Corollary 6.39(vii) essentially states thatP is supercoercive provided that some fj is. Analogously, Corollary 6.39(v)shows that that coercivity of P follows from the coercivity of some functionfj .94Chapter 7Compositions and ConvexCombinations ofAsymptotically RegularFirmly NonexpansiveMappingsIn this chapter, based on [15], we extend some of the results in Chapter 6into a Hilbert space setting. Even though the main results are formulatedin the given Hilbert space H, it will turn out that the key space to work inis the product space,Hm ={x = (xi)i∈I∣∣ (∀i ∈ I) xi ∈ H}, (7.1)where m ∈ {2, 3, 4, . . .} and I = {1, 2, . . . ,m}. This product space containsan embedding of the original space H via the diagonal subspace∆ ={x = (x)i∈I∣∣ x ∈ H}. (7.2)We also assume that we are given m firmly nonexpansive operatorsT1, . . . , Tm; equivalently, m resolvents of maximally monotone operatorsA1, . . . , Am. We now define various pertinent operators acting on Hm. Westart with the Cartesian product operatorsT : Hm → Hm : (xi)i∈I 7→ (Tixi)i∈I (7.3)andA : Hm ⇒ Hm : (xi)i∈I 7→ (Aixi)i∈I . (7.4)Denoting the identity on Hm by Id, we observe thatJA = (Id +A)−1 = T1 × · · · × Tm = T. (7.5)957.1. Properties of the operator MOf central importance will be the cyclic right-shift operatorR : Hm → Hm : (x1, x2, . . . , xm) 7→ (xm, x1, . . . , xm−1) (7.6)and for convenience we setM = Id−R. (7.7)We also fix strictly positive convex coefficients (or weights) (λi)i∈I , i.e.,(∀i ∈ I) λi ∈ ]0, 1[ and∑i∈Iλi = 1. (7.8)Let us make Hm into the Hilbert product spaceH = Hm, with 〈x,y〉 =∑i∈I〈xi, yi〉 . (7.9)Fact 7.1. [11, Proposition 25.4(i)] Set ∆ ={x = (x)i∈I∣∣ x ∈ H}.The or-thogonal complement of ∆ with respect to this standard inner product is∆⊥ ={x = (xi)i∈I∣∣∑i∈Ixi = 0}. (7.10)7.1 Properties of the operator MIn this section, we collect several useful properties of the operator M,including its Moore-Penrose inverse. To that end, the following result willbe useful.Proposition 7.2. Let Y be a real Hilbert space and let B be a continuouslinear operator from H to Y with adjoint B∗ and such that ranB is closed.Then the Moore-Penrose inverse of B satisfiesB† = PranB∗ ◦B−1 ◦ PranB. (7.11)Proof. Take y ∈ Y . Define the corresponding set of least squares solutions(see Fact 2.35) by C = B−1(PranBy). By Fact 2.13, since ranB is closed, sois ranB∗; hence, by Fact 2.9 and setting U = (kerB)⊥ we haveU = (kerB)⊥ = ranB∗ = ranB∗.Thus,C = B†y + kerB = B†y + U⊥.967.1. Properties of the operator MTherefore, since by Fact 2.37 ranB† = ranB∗,PU (C) = PUB†y = B†y,as claimed.Theorem 7.3. DefineL : ∆⊥ →H : y 7→m−1∑i=1m− imRi−1y. (7.12)Then the following hold.(i) M is continuous, linear, and maximally monotone with dom M = H.(ii) M is rectangular.(iii) ker M = ker M∗ = ∆.(iv) ran M = ran M∗ = ∆⊥ is closed.(v) ran L = ∆⊥.(vi) M ◦ L = Id |∆⊥.(vii) M−1 : H⇒H : y 7→{Ly + ∆, if y ∈∆⊥;∅, otherwise.(viii) M† = P∆⊥ ◦ L ◦ P∆⊥ = L ◦ P∆⊥.(ix) M† =m∑k=1m− (2k − 1)2mRk−1.Proof. (i): Clearly, dom M = H and (∀x ∈ H) ‖Rx‖ = ‖x‖. Thus, R isnonexpansive and therefore by Fact 3.46, M = Id−R is maximally mono-tone.(ii): This is Fact 3.65.(iii): The definitions of M and R and the fact that R∗ is the cyclic leftshift operator (see Example 2.8) readily imply thatker M = {x ∈H |Mx = 0}= {x ∈H | Id−R = 0},977.1. Properties of the operator Mwhich yields x1 = xm and xi = xi+1 for all i ∈ I. That is, ker M = ∆.Similarly, ker M∗ = ∆ and thus,ker M = ker M∗ = ∆.(iv), (vi), and (vii): Let y = (y1, . . . , ym) ∈ H. Assume first thaty ∈ ran M. Then there exists x = (x1, . . . , xm) such that y1 = x1 − xm,y2 = x2 − x1, . . . , and ym = xm − xm−1. It follows that∑i∈I yi = 0, i.e.,y ∈ ∆⊥ by Fact 7.1. Thus,ran M ⊆∆⊥. (7.13)Conversely, assume now that y ∈∆⊥. Now setx := Ly =m−1∑i=1m− imRi−1y. (7.14)Thenx =m− 1mR0y +m− 2mRy +m− 3mR2y + · · ·+1mRm−2y=m− 1m(y1, . . . , ym) +m− 2m(ym, y1, . . . , ym−1) + · · ·+1m(y3, . . . , y1, y2).It will be notationally convenient to wrap indices around, i.e., ym+1 = y1,y0 = ym and likewise. We then get(∀i ∈ I) xi =m− 1myi +m− 2myi−1 + · · ·+1myi+2. (7.15)Therefore,∑i∈Ixi =m− 1m∑i∈Iyi +m− 2m∑i∈Iyi + · · ·+1m∑i∈Iyi=m(m− 1)−m−1∑iim∑i∈Iyi=m(m− 1)− m(m−1)2m∑i∈Iyi=m− 12∑i∈Iyi = 0.987.1. Properties of the operator MThus x ∈∆⊥ andran L ⊆ ∆⊥. (7.16)Furthermore, (∀i ∈ I)xi − xi−1 =(m− 1myi +m− 2myi−1 + · · ·+1myi+2)−(m− 1myi−1 + · · ·+1my(i−1)+2)=m− 1myi +((m− 2)− (m− 1)m)yi−1 + · · ·+((m− (m− 2))− 1m)yi+2 −1myi+1=m− 1myi −1myi−1 −1myi−2 − · · · −1myi+1= yi −1m∑j∈Iyj = yi.Hence Mx = x−Rx = y and thus y ∈ ran M. Moreover, in view of (iii),M−1y = x + ker M = x + ∆. (7.17)We thus have shown∆⊥ ⊆ ran M. (7.18)Combining (7.13) and (7.18), we obtain ran M = ∆⊥. We thus have verified(vi), and (vii). Since ran M is closed, so is ran M∗, by Fact 2.13. Thus (iv)holds.(viii)&(v): We have seen in Proposition 7.2 thatM† = PranM∗ ◦M−1 ◦ PranM. (7.19)Now let z ∈H. Then, by (iv),y := PranMz = P∆⊥z ∈∆⊥.By (vii), M−1y = Ly + ∆. So,M†z = PranM∗M−1PranMz = PranM∗M−1y= P∆⊥(Ly + ∆) = P∆⊥Ly = Ly= (L ◦ P∆⊥)z,997.1. Properties of the operator Mbecause ran L ⊆ ∆⊥ by (7.16). Hence (viii) holds. Furthermore, by (iv)and Fact 2.37, ran L = ran L ◦ P∆⊥ = ran M† = ran M∗ = ∆⊥ and so (v)holds.(ix): Note that P∆⊥ = Id−P∆ and from Example 2.32,P∆ = m−1∑j∈IRj .Hence,P∆⊥ = Id−1m∑j∈IRj . (7.20)Thus, by (viii) and (7.12),M† = L ◦ P∆⊥ =1mm−1∑i=1(m− i)Ri−1 ◦(Id−1m∑j∈IRj)=1mm−1∑i=1(m− i)Ri−1 −1m2m−1∑i=1(m− i)∑j∈IRi+j−1=1mm−1∑i=1(m− i)Ri−1 −1m2(m− 1)m∑j=1Rj++(m− 2)m∑j=1Rj+1 + · · ·+ (1)m∑j=1Rm−2+j .Using the fact that Rm = R0, Rm+1 = R1, etc., and noting thatm−1∑i=1(m− i)Ri−1 =m∑i=1(m− i)Ri−1,we getM† =1m((m− 1)R0 + (m− 2)R1 + · · ·+ Rm−2)−1m2((m− 1)(R1 + · · ·+ Rm) + (m− 2)(R2 + · · ·+ Rm+1) + · · ·+ (Rm−1 + · · ·+ R2(m−1)))=m∑k=1(m− km−1m2m−1∑i=1(m− i))Rk−11007.2. Composition=m∑k=1(m− km−m(m− 1)m2+1m2m−1∑i=1i)Rk−1=m∑k=1(2m(m− k)2m2−2m(m− 1)2m2+m(m− 1)2m2)Rk−1=m∑k=1m− (2k − 1)2mRk−1Thus,M† = (Id−R)† =m∑k=1m− (2k − 1)2mRk−1. (7.21)Remark 7.4. Suppose that L˜ : ∆⊥ →H satisfies M ◦ L˜ = Id |∆⊥ . ThenM−1 : H⇒H : y 7→{L˜y + ∆, if y ∈∆⊥;∅, otherwise.(7.22)One may show that M† = P∆⊥ ◦ L˜◦P∆⊥ and that P∆⊥ ◦ L˜ = L (see (7.12)).Concrete choices for L˜ and L are∆⊥ →H : (y1, y2, . . . , ym) 7→ (y1, y1+y2, . . . , y1+y2+y3+· · ·+ym); (7.23)however, the range of the latter operator is not equal to ∆⊥ whenever H 6={0}.Corollary 7.5. The operator A+ M is maximally monotone andran(A+ M) = ∆⊥ + ranA.Proof. Since each Ai is maximally monotone and recalling Theorem 7.3(i),we see thatA and M are maximally monotone. On the other hand, dom M =H. Thus, by Fact 3.48, A + M is maximally monotone. Furthermore, byTheorem 7.3(ii) and (iv), M is rectangular and ran M = ∆⊥. The resulttherefore follows from Fact 3.66(ii).7.2 CompositionWe now use Corollary 7.5 to study the composition.1017.2. CompositionTheorem 7.6. Suppose that (∀i ∈ I) 0 ∈ ran(Id−Ti). Then the followinghold.(i) 0 ∈ ran(A + M).(ii) (∀ε > 0) (∃(b,x) ∈H×H) ‖b‖ ≤ ε and x = T(b + Rx).(iii) (∀ε > 0) (∃(c,x) ∈H×H) ‖c‖ ≤ ε and x = c + T(Rx).(iv) (∀ε > 0) (∃x ∈H) (∀i ∈ I)‖Ti−1 · · ·T1xm − TiTi−1 · · ·T1xm − xi−1 + xi‖ ≤ (2i− 1)ε,where x0 = xm.(v) (∀ε > 0) (∃x ∈ H) ‖x− TmTm−1 · · ·T1x‖ ≤ m2ε.Proof. (i): The assumptions and the Minty parametrization (3.14) implythat (∀i ∈ I),0 ∈ ran(Id−Ti)⇔ ∃xi ∈ H such that (JAixi, 0) ∈ graAi⇔ 0 ∈ ranAi.Hence, 0 ∈ ran A. Obviously, 0 ∈ ∆⊥. It follows that 0 ∈ ∆⊥ + ran A.Thus, by Corollary 7.5, 0 ∈ ran(A + M).(ii): Fix ε > 0. In view of (i), there exists x ∈ H and b ∈ H suchthat ‖b‖ ≤ ε and b ∈ Ax + Mx. Hence b + Rx ∈ (Id +A)x and thusx = JA(b + Rx) = T(b + Rx).(iii): Let ε > 0. By (ii), there exists (b,x) ∈H×H such that ‖b‖ ≤ εand x = T(b+Rx). Set c = x−T(Rx) = T(b+Rx)−T(Rx) Then, sinceT is nonexpansive, ‖c‖ = ‖T(b + Rx)−T(Rx)‖ ≤ ‖b‖ ≤ ε.(iv): Take ε > 0. Then, by (iii), there exists x ∈ H and c ∈ H suchthat ‖c‖ ≤ ε and x = c + T(Rx). Let i ∈ I. Then xi = ci + Tixi−1. Since‖ci‖ ≤ ‖c‖ ≤ ε and Ti is nonexpansive, we have‖TiTi−1 · · ·T1x0 − xi‖ ≤ ‖TiTi−1 · · ·T1x0 − Tixi−1‖+ ‖Tixi−1 − xi‖≤ ‖TiTi−1 · · ·T1x0 − Tixi−1‖+ ε≤ ‖Ti−1 · · ·T1x0 − xi−1‖+ ε≤ ‖Ti−1 · · ·T1x0 − Ti−1xi−2‖+ ‖Ti−1xi−2 − xi−1‖+ ε≤ ‖Ti−2 · · ·T1x0 − xi−2‖+ 2ε.Continuing similarly, we thus obtain‖TiTi−1 · · ·T1x0 − xi‖ ≤ iε. (7.24)1027.2. CompositionHence,‖Ti−1 · · ·T1x0 − xi−1‖ ≤ (i− 1)ε. (7.25)Adding (7.24) and (7.25), and recalling the triangle inequality and the factthat xm = x0,‖Ti−1 · · ·T1xm − TiTi−1 · · ·T1xm − xi−1 + xi‖≤ ‖Ti−1 · · ·T1x0 − xi−1‖+ ‖TiTi−1 · · ·T1x0 − xi‖≤ (i− 1)ε+ iε = (2i− 1)ε,as stated.(v): Let ε > 0. In view of (iv), there exists x ∈H such that(∀i ∈ I) ‖Ti−1 · · ·T1xm − TiTi−1 · · ·T1xm − xi−1 + xi‖ ≤ (2i− 1)ε (7.26)where x0 = xm. Now set,(∀i ∈ I) ei = Ti−1 · · ·T1xm − TiTi−1 · · ·T1xm − xi−1 + xi.Then (∀i ∈ I) ‖ei‖ ≤ (2i− 1)ε. Set x = xm. Thenm∑i=1ei =m∑i=1Ti−1 · · ·T1xm − TiTi−1 · · ·T1xm − xi−1 + xi (7.27)= x− TmTm−1 · · ·T1x. (7.28)This, (7.26), and the triangle inequality imply that‖x− TmTm−1 · · ·T1x‖ ≤m∑i=1‖ei‖ ≤m∑i=1(2i− 1)ε = m2ε. (7.29)This completes the proof.Remark 7.7. When m = 2, then Theorem 7.6(v) also follows from [56,p. 124].Corollary 7.8. Suppose that (∀i ∈ I) 0 ∈ ran(Id−Ti). Then0 ∈ ran(Id−TmTm−1 · · ·T1).Proof. This follows from Theorem 7.6(v).1037.3. Asymptotic regularityRemark 7.9. The converse implication in Corollary 7.8 fails in general: in-deed, consider the case when H 6= {0}, m = 2, and v ∈ H \ {0}. Nowset T1 : H → H : x 7→ x + v and set T2 : H → H : x 7→ x − v. Then0 /∈ ran(Id−T1) = {−v} and 0 /∈ ran(Id−T2) = {v}; however, T2T1 = Idand ran(Id−T2T1) = {0}.Remark 7.10. Corollary 7.8 is optimal in the sense that even if (∀i ∈ I) wehave 0 ∈ ran(Id−Ti), we cannot deduce that 0 ∈ ran(Id−TmTm−1 · · ·T1):indeed, suppose that H = R2 and m = 2. Set C1 := epi exp and C2 :=R × {0}. Suppose further that T1 = PC1 and T2 = PC2 . Then (∀i ∈ I)0 ∈ ran(Id−Ti); however, 0 ∈ ran(Id−T2T1) \ ran(Id−T2T1).7.3 Asymptotic regularityIn this section we show that the composition of asymptotically regularmappings is still asymptotically regular. The following results are corollariesto Bruck and Reich’s Fact 3.24.Corollary 7.11. Let S : H → H be strongly nonexpansive. Then S isasymptotically regular if and only if 0 ∈ ran(Id−S).Proof. “⇒”: Recall that S is asymptotically regular if(∀x ∈ H) Snx− Sn+1x→ 0⇔ Snx− S(Snx)→ 0⇒ 0 ∈ ran(Id−S)“⇐”: Fact 3.24(i).Remark 7.12. Under the assumption that T is firmly nonexpansive, theprevious result also follows from Fact 3.25.Corollary 7.13. Set S = TmTm−1 · · ·T1. Then S is asymptotically regularif and only if 0 ∈ ran(Id−S).Proof. Since each Ti is firmly nonexpansive, it is also strongly nonexpansiveby Fact 3.23(i). By Fact 3.23(ii), S is strongly nonexpansive. Now applyCorollary 7.11. Alternatively, 0 ∈ ran(Id−S) by Corollary 7.8 and againCorollary 7.11 applies.We are now ready for our first main result.Theorem 7.14. Suppose that each Ti is asymptotically regular. Then thecomposition TmTm−1 · · ·T1 is asymptotically regular as well.1047.4. Convex combinationProof. Theorem 7.6(v) implies that 0 ∈ ran(Id−TmTm−1 · · ·T1). The con-clusion thus follows from Corollary 7.13.Remark 7.15. (i) When m = 2, then the conclusion of Theorem 7.14 alsofollows from [56, p. 124].(ii) As an application of Theorem 7.14, we obtain the main result of [6],Example 7.16.Example 7.16. Let C1, . . . , Cm be nonempty closed convex subsets of H.Then the composition of the corresponding projectors, PCmPCm−1 · · ·PC1 isasymptotically regular.Proof. For every i ∈ I, the projector PCi is firmly nonexpansive, hencestrongly nonexpansive, and FixPCi = Ci 6= ∅. Suppose that (∀i ∈ I)Ti = PCi , which is thus asymptotically regular by Corollary 7.11. Nowapply Theorem 7.14.7.4 Convex combinationIn this section, we use our fixed weights (λi)i∈I to turn Hm into a Hilbertproduct space different from H considered in the previous sections. Specif-ically, we setY = Hm with 〈x,y〉 =∑i∈Iλi 〈xi, yi〉 (7.30)so that ‖x‖2 =∑i∈I λi‖xi‖2. We also setQ : Hm → Hm : x 7→ (x¯)i∈I , where x¯ =∑i∈Iλixi. (7.31)Fact 7.17. [11, Proposition 28.13] In the Hilbert product space Y, we haveP∆ = Q.Corollary 7.18. In the Hilbert product space Y, the operator Q is firmlynonexpansive and strongly nonexpansive. Furthermore, Fix Q = ∆ 6= ∅,0 ∈ ran(Id−Q), and Q is asymptotically regular.Proof. By Fact 7.17, the operator Q is equal to the projector P∆ and hencefirmly nonexpansive. Now apply Fact 3.23(i) to deduce that Q is stronglynonexpansive. It is clear that Fix Q = ∆ and that 0 ∈ ran(Id−Q). Finally,recall Corollary 7.11 to see that Q is asymptotically regular.1057.4. Convex combinationProposition 7.19. In the Hilbert product space Y, the operator T is firmlynonexpansive.Proof. Since each Ti is firmly nonexpansive, (∀x = (xi)i∈I ∈ Y) (∀y =(yi)i∈I ∈ Y) we have‖Tixi − Tiyi‖2 ≤ 〈xi − yi, Tixi − Tiyi〉 ,which implies,‖Tx−Ty‖2 =∑i∈Iλi‖Tixi − Tiyi‖2≤∑i∈Iλi 〈xi − yi, Tixi − Tiyi〉= 〈x− y,Tx−Ty〉 .Thus T is firmly nonexpansive.Theorem 7.20. Suppose that (∀i ∈ I) 0 ∈ ran(Id−Ti). Then the followinghold in the Hilbert product space Y.(i) 0 ∈ ran(Id−T).(ii) T is asymptotically regular.(iii) Q ◦T is asymptotically regular.Proof. (i): This follows because (∀x = (xi)i∈I)‖x−Tx‖2 =∑i∈Iλi‖xi − Tixi‖2.(ii): Combine Fact 3.23(i) with Corollary 7.11.(iii): On the one hand, Q is firmly nonexpansive and asymptoticallyregular by Corollary 7.18. On the other hand, T is firmly nonexpansiveand asymptotically regular by Proposition 7.19 and Theorem 7.20(ii). Al-together, the result follows from Theorem 7.14.We are now ready for the second main result of this chapter, whichconcerns convex combinations of asymptotically regular mappings.Theorem 7.21. Suppose that each Ti is asymptotically regular. Then∑i∈IλiTi,is asymptotically regular as well.1067.4. Convex combinationProof. Set S =∑i∈I λiTi. Fix x0 ∈ H and set (∀n ∈ N) xn+1 = Sxn.Set x0 = (x0)i∈I ∈ Hm and (∀n ∈ N) xn+1 = (Q ◦ T)xn. Then (∀n ∈ N)xn = (xn)i∈I . Now Q ◦ T is asymptotically regular by Theorem 7.20(iii);hence, xn−xn+1 = (xn−xn+1)i∈I → 0. Thus, xn−xn+1 → 0 and thereforeS is asymptotically regular.Remark 7.22. Theorem 7.21 extends Theorem 6.36 from Euclidean to Hilbertspace.Remark 7.23. Similarly to Remark 7.10, one cannot deduce that if each Tihas fixed points, then∑i∈I λiTi has fixed points as well: indeed, considerthe setting described in Remark 7.10 for an example.We conclude this chapter by showing that it was necessary to work inY and not in H; indeed, viewed in H, the operator Q is generally not evennonexpansive. The following fact is needed:Fact 7.24. [11, Proposition 25.4(iii)] In the Hilbert product space H, set∆ ={x = (x)i∈I∣∣ x ∈ H}and j : H →∆ : x 7→ (x, . . . , x). ThenP∆x = j(1m∑i∈Ixi).Theorem 7.25. Suppose that H 6= {0}. Then the following are equivalentin the Hilbert product space H.(i) (∀i ∈ I) λi = 1/m.(ii) Q coincides with the projector P∆.(iii) Q is firmly nonexpansive.(iv) Q is nonexpansive.Proof. “(i)⇒(ii)”: Fact 7.24. “(ii)⇒(iii)”: Clear. “(iii)⇒(iv)”: Clear.“(iv)⇒(i)”: Take e ∈ H such that ‖e‖ = 1. Set x := (λie)i∈I and y :=∑i∈I λ2i e. Then Qx = (y)i∈I . We compute ‖Qx‖2 = m‖y‖2 = m(∑i∈I λ2i)2and ‖x‖2 =∑i∈I λ2i . Since Q is nonexpansive, we must have that ‖Qx‖2 ≤‖x‖2, which is equivalent tom(∑i∈Iλ2i)2≤∑i∈Iλ2i (7.32)1077.4. Convex combinationand tom∑i∈Iλ2i ≤ 1. (7.33)On the other hand, applying the Cauchy-Schwarz inequality to the vectors(λi)i∈I and (1)i∈I in Rm yields1 = 12 =(∑i∈Iλi · 1)2≤∥∥(λi)i∈I∥∥2∥∥(1)i∈I∥∥2 = m∑i∈Iλ2i . (7.34)In view of (7.33) and the Cauchy-Schwarz inequality, (7.34) is actually anequality which implies that (λi)i∈I is a multiple of (1)i∈I . We deduce that(∀i ∈ I) λi = 1/m.In this chapter, we have show that the composition TmTm−1 · · ·T1 andthe convex combination∑i∈I λiTi of asymptotically regular firmly nonex-pansive mappings in a Hilbert space are asymptotically regular (Theorem 7.6and Theorem 7.21). Theorem 7.21 also extended a previous result, Theo-rem 6.36, into the more general Hilbert space setting. In the next chapterwe continue with the notion of averages “inheriting” properties from theaveraged operators with a look at the resolvent average.108Chapter 8Inheritance of Properties ofthe Resolvent Average ofMonotone OperatorsThe resolvent average was previously defined in Chapter 5. In this chap-ter, based on [21], we determine which properties the average inherits fromthe averaged operators and provide new results in monotone operator the-ory. Specifically, we cover the properties provided in Theorem 4.1, as wellas k-cyclic monotonicity, orthogonality, and difference maps.Definition 8.1 (inheritance of properties). For all i ∈ I, let Ai : H⇒ H bemaximally monotone operators, and A = (A1, . . . , An). A property (P ) is:(i) Dominant if for some j ∈ I, Aj has property (P ) implies thatRµ(A,λ)has property (P );(ii) Recessive if for all i ∈ I, Ai has property (P ) implies that Rµ(A,λ)has property (P ), but property (P ) is not dominant;(iii) Indeterminant if the property is neither dominant nor recessive.All of the theorems in this chapter build upon Theorem 5.3, whichshowed Rµ(A,λ) maintains monotonicity when all of the averaged oper-ators are monotone.8.1 Dominant properties8.1.1 Single-valuedness of Rµ(A, λ)Lemma 8.2. For all i ∈ I, let Ti be a firmly nonexpansive mapping and letT =∑i∈I λiTi. Then for every x and y in H, we have1098.1. Dominant properties‖Tx− Ty‖2=∑iλi‖Tix− Tiy‖2 − 12∑i,jλiλj‖Tix− Tiy − Tjx+ Tjy‖2 (8.1)≤ ‖x− y‖2 − ‖x− Tx− y + Ty‖2 −∑i,jλiλj‖Tix− Tiy − Tjx+ Tjy‖2(8.2)≤ ‖x− y‖2 − ‖x− Tx− y + Ty‖2. (8.3)Consequently, T is firmly nonexpansive.Proof. Let x and y be in H. By (2.7) and since each Ti is firmly nonexpan-sive, we have‖Tx− Ty‖2 = ‖∑i λi(Tix− Tiy)‖2=∑i λi‖Tix− Tiy‖2 − 12∑i,j λiλj‖Tix− Tiy − Tjx+ Tjy‖2≤∑i λi(‖x− y‖2 − ‖(x− Tix)− (y − Tiy)‖2)− 12∑i,j λiλj‖Tix− Tiy − Tjx+ Tjy‖2= ‖x− y‖2 −(∑i λi‖(x− Tix)− (y − Tiy)‖2+ 12∑i,j λiλj‖Tix− Tiy − Tjx+ Tjy‖2)= ‖x− y‖2 −(‖∑i λi(x− Tix− y + Tiy)‖2+∑i,j λiλj‖Tix− Tiy − Tjx+ Tjy‖2)= ‖x− y‖2 − ‖x− Tx− y + Ty‖2−∑i,jλiλj‖Tix− Tiy − Tjx+ Tjy‖2≤ ‖x− y‖2 − ‖x− Tx− y + Ty‖2,and the result follows.Corollary 8.3. For all i ∈ I, let Ti be firmly nonexpansive on H, λi bestrictly positive real numbers such that∑i∈I λi = 1, and set T =∑ni=1 λiTi.Let x and y be in H such that Tx = Ty. Then (∀i ∈ I) Tix = Tiy. Conse-quently, if some Ti is injective, so is T .1108.1. Dominant propertiesProof. By Lemma 8.2, we have0 = ‖Tx− Ty‖2 =∑iλi‖Tix− Tiy‖2 − 12∑i,jλiλj‖Tix− Tiy − Tjx+ Tjy‖2≤ ‖x− y‖2 − ‖x− y‖2 −∑i,jλiλj‖Tix− Tiy − Tjx+ Tjy‖2= −∑i,jλiλj‖Tix− Tiy − Tjx+ Tjy‖2 ≤ 0.(8.4)Thus∑i,j λiλj‖Tix−Tiy−Tjx+Tjy‖2 = 0 so we must have (∀i ∈ I)(∀j ∈ I)Tix− Tiy = Tjx− Tjy and therefore0 = Tx− Ty =n∑i=1λiTix−n∑i=1λiTiy = Tix− Tiy.Thus Tix = Tiy and the result follows.Corollary 8.4. For all i ∈ I, let Ti be firmly nonexpansive on H, λi bestrictly positive real numbers such that∑i∈I λi = 1, and set T =∑ni=1 λiTi.Let z, u, v in H be given such that u = T (u+ z) and (∀i ∈ I) v = Ti(v + z).Then v = T (v + z) and (∀i ∈ I) Ti(u+ z) = u.Proof. It is clear that v = T (v + z). Now set x = u + z and y = v + z inLemma 8.2 to deduce‖u− v‖2 ≤ ‖u− v‖2 −∑i,jλiλj‖Ti(u+ z)− Tj(u+ z)‖2. (8.5)Hence (∀i ∈ I)(∀j ∈ I) Ti(u+ z) = Tj(u+ z). Since T (u+ z) = u, we musthaveT (u+ z) =n∑i=1λiTi(u+ z) = Ti(u+ z) = u.Thus (∀i ∈ I) Ti(u+ z) = u.We are now ready for the main result of this section.Theorem 8.5 (single-valuedness is dominant). For all i ∈ I, let Ai : H⇒ Hbe maximally monotone and assume that some Aj is at most single-valued.Then Rµ(A,λ) is also at most single-valued.Proof. By Theorem 4.1(iv), a maximally monotone operator is at mostsingle-valued if and only if its resolvent is injective. Hence JµAj is injec-tive. By Corollary 8.3 and (5.3), JµRµ(A,λ) is injective. Hence µRµ(A,λ)and therefore Rµ(A,λ) is at most single-valued.1118.1. Dominant properties8.1.2 Domain and rangeProposition 8.6 (resolvents are rectangular). Let T : H → H be firmlynonexpansive. Then T is rectangular.Proof. Since T is firmly nonexpansive, it is the resolvent of a monotone op-erator, A, such that T = (Id +A)−1. From the definition, it follows thatT is rectangular if and only if T−1 = Id +A is rectangular. Now by Ex-ample 3.53, FId(x, x∗) = 14‖x + x∗‖2. We know that FA(x, x∗) = 〈x, x∗〉if (x, x∗) ∈ graA. To show that Id +A is rectangular, we must show thatdom(A+ Id)× ran(A+ Id) ⊆ domFA+Id, which by Fact 3.43 means thatdomA×H ⊆ domFA+Id.To this end, let (x, u) ∈ domA×H and take x∗ ∈ Ax. Then FA+Id(x, u) ≤(FA(x, ·)FId(x, ·))(u) ≤ FA(x, x∗) + FId(x, u − x∗) < +∞. Hence (x, u) ∈domFA+Id and thus A+ Id is rectangular.Proposition 8.7. Let A : H ⇒ H be maximally monotone, and let γ > 0.Then A is rectangular if and only if γA is rectangular.Proof. Use the definitions and FγA(x, x∗) = γFA(x, x∗/γ).Proposition 8.8 (surjective). For all i ∈ I, let Ti be a firmly nonexpansive.If some Tj is surjective, then so is T =∑ni=1 λiTi.Proof. First, consider the case where T = λ1T1 + λ2T2. Without loss ofgenerality, assume that T1 is surjective. By Proposition 8.6, since T is firmlynonexpansive, it is a resolvent and hence rectangular. Each Ti is rectangularas well, as is each λiTi by Proposition 8.7. Using Fact 3.66(i),int ran(λ1T1 + λ2T2) = int(ranλ1T1 + ranλ2T2),we see that T is surjective. The case n > 2 now follows inductively.Theorem 8.9 (full domain is dominant). For all i ∈ I, let Ai be max-imally monotone and suppose that for some j ∈ I, domAj = H. ThendomRµ(A,λ) = H.Proof. Since domAj = H, then domµAj = H. By Theorem 4.1(ii),domµAj = H if and only if JµAj is surjective. Then by (5.3) and Propo-sition 8.8, JµRµ(A,λ) =∑i∈IλiJµAi is surjective and thus domRµ(A,λ) =H.1128.1. Dominant propertiesTheorem 8.10 (surjectivity is dominant). For all i ∈ I, let Ai be maximallymonotone and suppose that for some j ∈ I, Aj is surjective. Then R1(A,λ)is surjective.Proof. By Theorem 4.1(iii) and (3.13),Aj is surjective ⇔ Id−JAj is surjective.⇔ JA−1jis surjective.Thus, by Corollary 5.9, and Proposition 8.8, JR1(A,λ)−1 is surjective. Ap-plying Theorem 4.1(iii) and (3.13) again, we haveJR1(A,λ)−1 is surjective ⇔ Id−JR1(A,λ) is surjective.⇔ R1(A,λ) is surjective.8.1.3 Strict monotonicityLemma 8.11. For all i ∈ I, let Ti be firmly nonexpansive and λi be strictlypositive real numbers such that∑i∈Iλi = 1. If there exists j ∈ I such that Tj isstrictly firmly nonexpansive then T =∑i∈IλiTi is strictly firmly nonexpansive.Proof. We know from Lemma 8.2 that T is firmly nonexpansive. To showT is strictly firmly nonexpansive we need to show that for u, v ∈ domT ifTu 6= Tv then ‖Tu − Tv‖2 < 〈Tu− Tv, u− v〉. Suppose to the contrarythatTu 6= Tv and ‖Tu− Tv‖2 = 〈u− v, Tu− Tv〉 . (8.6)We know from (8.1) and the firm nonexpansiveness of the Ti that‖Tu− Tv‖2 = ‖∑i∈Iλi(Tiu− Tiv)‖2 ≤∑i∈Iλi‖Tiu− Tiv‖2≤∑i∈Iλi 〈u− v, Tiu− Tiv〉 = 〈u− v, Tu− Tv〉 . (8.7)Since ‖Tu− Tv‖2 = 〈u− v, Tu− Tv〉, (8.7) yields‖Tu− Tv‖2 = ‖∑i∈Iλi(Tiu− Tiv)‖2 =∑i∈Iλi‖Tiu− Tiv‖2=∑i∈Iλi 〈Tiu− Tiv, u− v〉 (8.8)1138.1. Dominant propertiesSince ‖ · ‖2 is strictly convex, we haveTiu− Tiv = Tju− Tjv (∀i ∈ I). (8.9)Since each Ti is firmly nonexpansive, ‖Tiu− Tiv‖2 ≤ 〈Tiu− Tiv, u− v〉 andthe third equality of (8.8) gives‖Tiu− Tiv‖2 = 〈Tiu− Tiv, u− v〉 (∀i ∈ I). (8.10)By definition of T , (8.9) and the fact that Tu 6= Tv we also haveTu− Tv =∑i∈Iλi(Tiu− Tiv) = Tju− Tjv 6= 0.Then ‖Tju − Tjv‖2 < 〈Tju− Tjv, u− v〉, since Tj is strictly firmly nonex-pansive. But this contradicts (8.10), and thus (8.6) is false. Therefore,‖Tu− Tv‖2 < 〈Tu− Tv, u− v〉 whenever Tu 6= Tv,and hence T is strictly firmly nonexpansive.Theorem 8.12 (strict monotonicity is dominant). For all i ∈ I, let Ai bemonotone and additionally assume that some Aj is strictly monotone. ThenR1(A,λ) is strictly monotone.Proof. By Theorem 4.1(vi), since Aj is strictly monotone then JAj is strictlyfirmly nonexpansive, thus by Lemma 8.11 we have JR1(A,λ) =∑i∈IλiJAiis strictly firmly nonexpansive. Apply Theorem 4.1(vi) again to see thatR1(A,λ) is strictly monotone.8.1.4 Banach contractionProposition 8.13 (Banach contraction). For all i ∈ I, let Ti be a firmlynonexpansive mapping and λi ∈ R++ such that∑ni=1 λi = 1. If some Tjis a Banach contraction with constant β, then T =∑i∈I λiTi is a Banachcontraction with constant (1− λj(1− β)).Proof. Suppose that Tj is β-Lipschitz, with 0 ≤ β < 1. Let x and y be in1148.1. Dominant propertiesH. Then‖Tx− Ty‖ ≤∑i∈Iλi‖Tix− Tiy‖≤∑i 6=jλi‖x− y‖+ λjβ‖x− y‖=∑i∈Iλi‖x− y‖ − λj(1− β)‖x− y‖= (1− λj(1− β))‖x− y‖.Theorem 8.14. For all i ∈ I, let Ai be maximally monotone operators fromH ⇒ H and assume that for some j ∈ I and JAj is a Banach contractionwith constant β. Then JR1(A,λ) is a Banach contraction with constant γ =(1− λj(1− β)) and R1(A,λ) satisfies(∀(x, u) ∈ graR1(A,λ)) (∀(y, v) ∈ graR1(A,λ))1− γ2γ2‖x− y‖2 ≤ 2 〈x− y, u− v〉+ ‖u− v‖2,Proof. Since, JAj is a Banach contraction with constant β, applying Propo-sition 8.13 and (5.3), JR1(A,λ) =∑i∈IλiJAi is a Banach contraction withconstant γ = (1 − λj(1 − β)). Therefore, Theorem 4.1(xiii) yields thatR1(A,λ) satisfies(∀(x, u) ∈ graR1(A,λ)) (∀(y, v) ∈ graR1(A,λ))1− γ2γ2‖x− y‖2 ≤ 2 〈x− y, u− v〉+ ‖u− v‖2,where γ = (1− λj(1− β)).8.1.5 Rectangularity and paramonotonicityWhile rectangularity and paramonotonicity are not typically dominantproperties, in the special case where the operators are linear on RN , theyare dominant.Theorem 8.15 (linear rectangularity and paramonotonicity are dominanton RN ). Assume that (∀i ∈ I) Ai : RN → RN is linear and at least oneAi is paramonotone (equivalently rectangular). Then R1(A,λ) is linear andparamonotone (equivalently rectangular).1158.2. Dominant or recessive propertiesProof. This is Theorem 8.24(iii) combined with Fact 3.58(i).8.2 Dominant or recessive propertiesIn this section we gather the properties that are at least recessive, butfor which there is not yet a proof or counterexample for dominance.8.2.1 Strong monotonicityTheorem 8.16. For all i ∈ I, let Ti be (1 + i) firmly nonexpansive withi ≥ 0. Then T =∑i∈IλiTi is (1+) firmly nonexpansive, where = mini∈I i.Proof. Ti is (1 + ) firmly nonexpansive with ≥ 0 gives‖Tix− Tiy‖2 ≤ (1 + )−1 〈Tix− Tiy, x− y〉 . (8.11)By the convexity of ‖ · ‖2, and (8.11), we have‖Tx− Ty‖2 ≤∑i∈Iλi‖Tix− Tiy‖2≤∑i∈Iλi(1 + i)−1 〈Tix− Tiy, x− y〉≤n∑i=1λi(1 + )−1 〈Tix− Tiy, x− y〉= (1 + )−1 〈Tx− Ty, x− y〉 .Thus T is (1 + ) firmly nonexpansive, where = mini∈I i.Theorem 8.17 (strong monotonicity). For all i ∈ I, let Ai be stronglymonotone with constant i. Then R1(A,λ) is strongly monotone with con-stant = mini∈I i.Proof. By Theorem 4.1(xi), JAi is (1 + i) firmly nonexpansive for all i ∈I. Then by Theorem 8.16,∑i∈IλiJAi is (1 + ) firmly nonexpansive where = mini∈I i. Thus by (5.3) and Theorem 4.1(xi), R1(A,λ) is stronglymonotone.1168.2. Dominant or recessive properties8.2.2 γ-cocoerciveTheorem 8.18 (γ-cocoercive). For all i ∈ I, let Ai be maximally monotoneand γi-cocoercive for γi > 0 and λi ∈ R++ such that∑ni=1 λi = 1. ThenR1(A,λ) is γ- cocoercive, where γ = mini∈Iγi.Proof. By Theorem 4.1(xii), (3.13) and Theorem 4.1(xi),Ai is γi-cocoercive⇔ (1 + γi)(Id−JAi) is firmly nonexpansive;⇔ (1 + γi)JA−1iis firmly nonexpansive.⇔ (∀x ∈ H)(∀y ∈ H)‖JA−1ix− JA−1iy‖2≤ (1 + γi)−1〈x− y, JA−1ix− JA−1iy〉.Then by (8.1), (∀x ∈ H)(∀y ∈ H)‖∑i∈IλiJA−1ix−∑i∈IλiJA−1iy‖2 (8.12)≤∑i∈Iλi‖JA−1ix− JA−1iy‖2≤∑i∈Iλi(1 + γi)−1〈x− y, JA−1ix− JA−1iy〉≤∑i∈Iλi(1 + γ)−1〈x− y, JA−1ix− JA−1iy〉= (1 + γ)−1〈x− y,∑i∈IλiJA−1ix−∑i∈IλiJA−1iy〉, (8.13)where γ = mini∈Iγi. Applying Theorem 4.1(xi), Theorem 5.19, and Theo-rem 4.1(xii) to (8.13) we have, (∀x ∈ H)(∀y ∈ H)‖∑i∈IλiJA−1ix−∑i∈IλiJA−1iy‖2≤ (1 + γ)−1〈x− y,∑i∈IλiJA−1ix−∑i∈IλiJA−1iy〉⇔ (1 + γ)JR1(A,λ)−1 is firmly nonexpansive;⇔ (1 + γ)(Id−JR1(A,λ)) is firmly nonexpansive;⇔ R1(A,λ) is γ-cocoercive.1178.3. Recessive properties8.3 Recessive properties8.3.1 Maximality and linearityTheorem 8.19 (maximal monotonicity is recessive). Rµ(A,λ) is maxi-mally monotone if and only if for all i ∈ I, Ai is maximally monotone.Proof. This is a consequence of Theorem 5.3.Theorem 8.20 (linear relations are recessive). For all i ∈ I, let Ai be amaximal monotone linear relation. Then Rµ(A,λ) is a maximal monotonelinear relation.Proof. Since each Ai is a linear relation, Fact 2.20 shows that Ai + Id is alinear relation and therefore (Ai+Id)−1 is a linear relation. The maximalityof Ai and Fact 3.43 imply that (Ai + µ−1 Id)−1 is single-valued and full-domain. Thus, (Ai + µ−1 Id)−1 is a linear mapping. Using(Rµ(A,λ) + µ−1 Id)−1 = λ1(A1 + µ−1 Id)−1 + · · ·+ λn(An + µ−1 Id)−1,we see that (Rµ(A,λ) + Id)−1 is linear, therefore Rµ(A,λ) + Id is a linearrelation. Then Rµ(A,λ) = (Rµ(A,λ) + Id)− Id is a linear relation.The next example shows that linearity is not a dominant property.Example 8.21. Set f = ‖ · ‖ and A1 = ∂f and A2 = 0. Then by (3.16)and Example 2.55JA1x = proxf x ={(1− 1‖x‖)x, if ‖x‖ > 1;0, if ‖x‖ ≤ 1,and we have JA2 = Id. Then JA1 is not linear and JA2 is linear. However,JR1(A,λ)x = λJA1x+ (1− λ)JA2x ={(1− λ 1‖x‖)x, if ‖x‖ > 1;(1− λ)x, if ‖x‖ ≤ 1,which is not linear. Thus R1(A,λ) is not a linear relation.8.3.2 Rectangularity and paramonotonicityTheorem 8.22 (rectangularity is recessive). Assume that for each i ∈ I,Ai : H⇒ H is a rectangular maximally monotone operator. Then R1(A,λ)is rectangular.1188.3. Recessive propertiesProof. By Theorem 4.1(xvii), we need to show (∀x ∈ H)(∀y ∈ H),infz∈H〈JR1(A,λ)x− JR1(A,λ)z, (y − JR1(A,λ)y)− (z − JR1(A,λ)z)〉> −∞.(8.14)Using JR1(A,λ) =∑ni=1 λiJAi , (8.14) becomes (∀x ∈ H)(∀y ∈ H),infz∈H〈n∑i=1λi(JAix− JAiz),n∑j=1λj [(y − z)− (JAjy − JAjz)]〉> −∞.(8.15)Since Ai is rectangular, by Theorem 4.1(xvii) we have for each i ∈ I,(∀x ∈ H)(∀y ∈ H) infz∈H〈JAix− JAiz, (y − JAiy)− (z − JAiz)〉 > −∞.(8.16)Setting xi = JAix− JAiz, ui = (y − z)− (JAiy − JAiz), andcij = −〈JAix− JAjx, JAiy − JAjy〉+14‖(JAix− JAjx) + (JAiy − JAjy)‖2,we have 〈xi − xj , ui − uj〉 =〈(JAix− JAjx)− (JAiz − JAjz),−(JAiy − JAjy) + (JAiz − JAjz)〉= −〈JAix− JAjx, JAiy − JAjy〉+〈(JAix− JAjx) + (JAiy − JAjy), JAiz − JAjz〉− ‖JAiz − JAjz‖2= −〈JAix− JAjx, JAiy − JAjy〉(8.17)−∥∥∥∥(JAix− JAjx) + (JAiy − JAjy)2− (JAiz − JAjz)∥∥∥∥2+‖(JAix− JAjx) + (JAiy − JAjy)‖24= cij −∥∥∥∥(JAix− JAjx) + (JAiy − JAjy)2− (JAiz − JAjz)∥∥∥∥2. (8.18)Then for given x, y ∈ H, using Fact 2.42, (8.18) and (8.16),infz∈H〈n∑i=1λi(JAix− JAiz),n∑j=1λj [(y − z)− (JAjy − JAjz)]〉= infz∈H[ n∑i=1λi 〈JAix− JAiz, (y − z)− (JAiy − JAiz)〉−n∑i=1n∑j=1λiλj12(cij −∥∥∥∥(JAix− JAjx) + (JAiy − JAjy)2− (JAiz − JAjz)∥∥∥∥2)]1198.3. Recessive properties≥ infz∈H(n∑i=1λi 〈JAix− JAiz, (y − z)− (JAiy − JAiz)〉)−n∑i=1n∑j=1λiλjcij2≥n∑i=1λi infz∈H〈JAix− JAiz, (y − z)− (JAiy − JAiz)〉 −n∑i=1n∑j=1λiλjcij2=n∑i=1λi infz∈H〈JAix− JAiz, (y − JAiy)− (z − JAiz)〉 −n∑i=1n∑j=1λiλjcij2> −∞.Hence (8.15) holds and R1(A,λ) is rectangular.Remark 8.23. Theorem 8.22 is not true if only one Ai is rectangular. SeeExample 8.27 for a counterexample.Theorem 8.24. Let Ai be maximally monotone operators from H⇒ H forall i ∈ I. The following hold:(i) Assume that (∀i ∈ I) Ai is paramonotone. Then R1(A,λ) is para-monotone.(ii) Assume that at least one Ai is paramonotone and at most single-valued.Then R1(A,λ) is paramonotone and at most single-valued.(iii) Assume that (∀i ∈ I) Ai is linear and at least one Ai is paramonotone.Then R1(A,λ) is linear and paramonotone.Proof. By Theorem 4.1(xv), we need to show that (∀x ∈ H)(∀y ∈ H),‖JR1(A,λ)x− JR1(A,λ)y‖2 =〈x− y, JR1(A,λ)x− JR1(A,λ)y〉(8.19)⇒{JR1(A,λ)x = JR1(A,λ)(JR1(A,λ)x+ y − JR1(A,λ)y)JR1(A,λ)y = JR1(A,λ)(JR1(A,λ)y + x− JR1(A,λ)x).(8.20)Using JR1(A,λ) =∑ni=1 λiJAi , (8.19) becomes‖n∑i=1λi(JAix− JAiy)‖2 =n∑i=1λi 〈x− y, JAix− JAiy〉 . (8.21)By the strong convexity of ‖ · ‖2, (2.7) with xi = JAix− JAiy, gives‖n∑i=1λi(JAix− JAiy)‖2 =n∑i=1λi‖JAix− JAiy‖2−n∑i=1n∑j=1λiλj‖(JAix− JAiy)− (JAjx− JAjy)‖22. (8.22)1208.3. Recessive propertiesThen it follows from (8.21) and (8.22) that (8.19) is equivalent ton∑i=1λi‖JAix− JAiy‖2 −n∑i=1n∑j=1λiλj‖(JAix− JAiy)− (JAjx− JAjy)‖22=n∑i=1λi 〈x− y, JAix− JAiy〉 , (8.23)That is,n∑i=1λi(‖JAix− JAiy‖2 − 〈x− y, JAix− JAiy〉)=n∑i=1n∑j=1λiλj‖(JAix− JAiy)− (JAjx− JAjy)‖22≥ 0. (8.24)Since JAi is firmly nonexpansive, Fact 3.3(iv) gives(∀i ∈ I) ‖JAix− JAiy‖2 − 〈x− y, JAix− JAiy〉 ≤ 0 (8.25)Then because λi > 0, (8.24) and (8.25) indicates that(∀i ∈ I) ‖JAix− JAiy‖2 = 〈x− y, JAix− JAiy〉 , (8.26)and(∀ i ∈ I)(∀j ∈ I) JAix− JAiy = JAjx− JAjy = d (8.27)where d ∈ H. In particular, multiplying (8.27) by λi, followed by summation,givesJR1(A,λ)x− JR1(A,λ)y = d. (8.28)With these, we are ready to show:(i) If each Ai is paramonotone, then (8.26) and (4.11) gives(∀i ∈ I) JAix = JAi(JAix+ y − JAiy),(∀ i ∈ I) JAiy = JAi(JAiy + x− JAix).In view of (8.27) this is,(∀i ∈ I) JAix = JAi(d+ y), (8.29)(∀i ∈ I) JAiy = JAi(−d+ x). (8.30)1218.3. Recessive propertiesThen multiplying (8.29) and (8.30) by λi, taking summations and using(8.28) leads ton∑i=1λiJAix =n∑i=1λiJAi(d+ y) = JR1(A,λ)(JR1(A,λ)x+ y − JR1(A,λ)y),n∑i=1λiJAiy =n∑i=1λiJAi(−d+ x) = JR1(A,λ)(JR1(A,λ)y + x− JR1(A,λ)x),which is (8.20). Thus R1(A,λ) is paramonotone.(ii) If at least one Ai is paramonotone, say Ai, then for this Ai, (8.26)and (4.11) givesJAix = JAi(JAix+ y − JAiy),JAiy = JAi(JAiy + x− JAix).By Theorem 4.1(iv), since Ai is at most single-valued then JAi is injective,thereforex = JAix+ y − JAiy,y = JAiy + x− JAix. (8.31)Both of which imply,x− y = JAix− JAiy. (8.32)Note that (8.27) and (8.32) together signify(∀j ∈ I) x− y = JAjx− JAjywhich gives us(∀ j ∈ I) x = JAjx+ y − JAjy,(∀ j ∈ I) y = JAjy + x− JAjx.Multiplying both sides by λj , followed by summation, givesx =n∑j=1λjJAjx+ y −n∑j=1λjJAjy = JR1(A,λ)x+ y − JR1(A,λ)y,y =n∑j=1λjJAjy + x−n∑j=1λjJAjx = JR1(A,λ)y + x− JR1(A,λ)x.Hence (8.20) follows immediately. Therefore, R1(A,λ) is paramonotone.By Theorem 8.5 R1(A,λ) is at most single-valued.(iii) follows from (ii).1228.3. Recessive propertiesRemark 8.25. Theorem 8.15 gives an improved version of Theorem 8.22when each Ai is linear and monotone.Remark 8.26. Example 8.27 demonstrates that Theorems 8.22 and 8.24 arealmost optimal, and they cannot be significantly improved.Example 8.27. Define A1 : R2 7→ R2 to be the normal cone operator ofthe set R× {0}. That is,A1 := NR×{0}.Let A2 : R2 → R2 be the skew operator such thatA2 =(0 1−1 0)and JA2 = (Id +A2)−1 =(12 −121212).Then by Fact 3.42, JA1 is the projector on R× {0},JA1 = PR×{0} =(1 00 0)Then for λ1 = λ2 = 12 we have,R1(A,λ) = (12JA1 +12JA2)−1 − Id=(12(1 00 0)+12(12 −121212))−1−(1 00 1)=(34 −141414)−1−(1 00 1)=(1 1−1 3)−(1 00 1)=(0 1−1 2),R1(A,λ)+ =R1(A,λ) +R1(A,λ)ᵀ2=(0 00 2).Clearly, rankR1(A,λ) = 2 while rankR1(A,λ)+ = 1. As rankR1(A,λ) 6=rankR1(A,λ)+, Fact 3.58(ii) implies that R1(A,λ) is not paramonotoneand equivalently R1(A,λ) is not rectangular.Remark 8.28. Note that in Example 8.27 we have demonstrated the follow-ing:1238.3. Recessive properties(i) A1 is rectangular and A2 is not rectangular, and we have R1(A,λ) isnot rectangular. Therefore the requirement for all Ai to be rectangularin Theorem 8.22 is optimal.(ii) A1 is paramonotone andA2 is not paramonotone, and we haveR1(A,λ)is not paramonotone. This implies that Theorem 8.24(i) is optimal.(iii) A1 is paramonotone but A1 is not single valued, and R1(A,λ) is notparamonotone. Thus Theorem 8.24(ii) is optimal.Theorem 8.29. For every i ∈ I suppose there exists z ∈ H such thatAi : x 7→ zi. Then R1(A,λ) : x 7→∑i∈Iλizi.Proof. By Theorem 4.1(v) there exists zi ∈ H such that Ai : x 7→ zi if andonly if JAi is an isometry, in which case JAi : x 7→ x− zi. Then,JR1(A,λ)x =∑i∈IλiJAix=∑i∈Iλi(x− zi) = x−∑i∈Iλizi.Thus JR1(A,λ) is an isometry, so R1(A,λ) : x 7→∑i∈Iλizi.8.3.3 k-cyclical monotonicityRecall that for an operator A : H⇒ H, A is k-cyclically monotone if forall (x1, u1), . . . , (xk, uk) ∈ graA and xk+1 = x1 one hask∑i=1〈ui, xi+1 − xi〉 ≤ 0. (8.33)The operator A is cyclically monotone if ∀k ∈ {2, 3, . . .}, A is k-cyclicallymonotone.Example 8.30. [5, Example 4.6] Let H = R2 and let n ∈ {2, 3, . . .}. Denotethe matrix corresponding to the counter-clockwise rotation by pi/n by Rn.That is,Rn =(cos pin − sinpinsin pin cospin).Then Rn is maximally monotone and n-cyclically monotone, but Rn is not(n+ 1)-cyclically monotone.1248.3. Recessive propertiesLemma 8.31. Let A and B be k-cyclically monotone operators from H ⇒H. Then(i) αA is k-cyclically monotone for α > 0.(ii) A+B is k-cyclically monotone;(iii) A−1 is k-cyclically monotone.Proof. (i): Let (xi, ui) ∈ graαA for i = 1, . . . , k + 1 with xk+1 = x1. Then(xi, α−1ui) ∈ graA for i = 1, . . . , k + 1, and we havek∑i=1〈xi+1 − xi, ui〉 = αk∑i=1〈xi+1 − xi, α−1ui〉≤ 0,by the k-cyclical monotonicity of A. Thus αA is k-cyclically monotone.(ii): Let (xi, ui + vi) ∈ gra(A + B) for i = 1, . . . , k + 1 with xk+1 = x1,(xi, ui) ∈ graA, (xi, vi) ∈ graB. Since A and B are k-cyclic, by definitionwe have,k∑i=1〈xi+1 − xi, ui〉 ≤ 0, andk∑i=1〈xi+1 − xi, vi〉 ≤ 0.Adding these two inequalities yields,∑ki=1 〈xi+1 − xi, ui + vi〉 ≤ 0. ThusA+B is k-cyclic.(iii): Let (ui, xi) ∈ graA−1 for i = 1, . . . , k. Then (∀i) (xi, ui) ∈ graA.By the k-cyclical monotonicity of A, one can do the k-cyclical summationfor points arranged in(xk+1, uk+1) = (x1, u1), (xk, uk), (xk−1, uk−1), · · · , (x2, u2),to obtain that2∑i=k+1〈xi−1 − xi, ui〉 ≤ 0 ⇔k∑i=1〈xi − xi+1, ui+1〉 ≤ 0. (8.34)Nowk∑i=1〈xi − xi+1, ui+1〉 =k∑i=1〈xi, ui+1〉 −k∑i=1〈xi+1, ui+1〉=k∑i=1〈xi, ui+1〉 −k∑i=1〈xi, ui〉 =k∑i=1〈xi, ui+1 − ui〉 ,so (8.34) transpires to∑ki=1 〈xi, ui+1 − ui〉 ≤ 0. Hence A−1 is k-cyclicallymonotone.1258.3. Recessive propertiesFact 8.32. [5, Theorem 6.6] Let T : H → H. Then T is the resolvent ofthe maximal monotone and k-cyclic operator A : H ⇒ H if and only if Thas full domain, T is firmly nonexpansive, and the mapping Tx 7→ x − Txis k-cyclic, i.e., for every set of points {x1, . . . , xk}, where xk+1 = x1, onehask∑i=1〈xi − Txi, Txi − Txi+1〉 ≥ 0.Proposition 8.33. Suppose that A1 and A2 are two maximal monotoneand k-cyclical mappings from H ⇒ H. Further, let α ∈]0, 1[. Then thereexists a k-cyclical monotone operator B such that(Id +B)−1 = α(Id +A1)−1 + (1− α)(Id +A2)−1. (8.35)Hence the set of resolvents{JA : A is k-cyclically monotone},is a convex set.Proof. Set T1 = JA1 and T2 = JA2 . Then T1 and T2 are firmly nonexpansivewith full domain as they are the resolvents of maximally monotone operators.Let α ∈]0, 1[, then T := αT1 + (1 − α)T2 is firmly nonexpansive with fulldomain, and is thus the resolvent of a maximally monotone operator, B. Toshow that B is k-cyclically monotone, by Fact 8.32 we need to show thatk∑i=1〈xi − (αT1xi + (1− α)T2xi), (αT1xi + (1− α)T2xi)− (αT1xi+1 + (1− α)T2xi+1)〉 ≥ 0. (8.36)1268.3. Recessive propertiesFor each i we have,〈xi − (αT1xi + (1− α)T2xi), (αT1xi + (1− α)T2xi)− (αT1xi+1 + (1− α)T2xi+1)〉= α2 〈xi − T1xi, T1xi − T1xi+1〉+ (1− α)2 〈xi − T2xi, T2xi − T2xi+1〉+ α(1− α) 〈xi − T1xi, T2xi − T2xi+1〉+ α(1− α) 〈xi − T2xi, T1xi − T1xi+1〉=(α2 + α(1− α))〈xi − T1xi, T1xi − T1xi+1〉+((1− α)2 + α(1− α))〈xi − T2xi, T2xi − T2xi+1〉+ α(1− α) 〈T1xi − T2xi, (T1xi − T1xi+1)− (T2xi − T2xi+1)〉= α 〈xi − T1xi, T1xi − T1xi+1〉+ (1− α) 〈xi − T2xi, T2xi − T2xi+1〉+ α(1− α) 〈T1xi − T2xi, (T1xi − T1xi+1)− (T2xi − T2xi+1)〉 . (8.37)By Fact 8.32,k∑i=1〈xi − T1xi, T1xi − T1xi+1〉 ≥ 0 andk∑i=1〈xi − T2xi, T2xi − T2xi+1〉 ≥ 0.(8.38)Since Id is cyclically monotone, then any points x1, . . . , xk satisfyk∑i=1〈xi, xi − xi+1〉 ≥ 0,where xk+1 = x1. Thus,k∑i=1〈T1xi − T2xi, (T1xi − T1xi+1)− (T2xi − T2xi+1)〉 ≥ 0. (8.39)Altogether, (8.37), (8.38), and (8.39) yield,k∑i=1〈xi − (αT1xi + (1− α)T2xi), (αT1xi + (1− α)T2xi)− (αT1xi+1 + (1− α)T2xi+1)〉 ≥ 0,which is (8.36). The convexity of C := {JA : A is k-cyclically monotone}then follows from induction. Clearly, if n = 1 then JA1 ∈ C. Assume that1278.3. Recessive propertiesλ1JA1 + · · ·+ λn−1JAn−1 ∈ C, thenλ1JA1 + · · ·+ λnJAn= λ1JA1 + · · ·+ λn−1JAn−1 + λnJAn= (1− λn)(λ1λ1 + · · ·+ λn−1JA1 + · · ·+λn−1λ1 + · · ·+ λn−1JAn−1)+ λnJAn= (1− λn)JA˜ + λnJAnWe know that A˜ is k-cylic by the induction assumption, thus apply (8.35)to get λ1JA1 + · · ·+ λnJAn ∈ C.Theorem 8.34 (k-cyclic monotonicity is recessive). For all i ∈ I, let Ai bemaximal monotone and k-cyclic, then Rµ(A,λ) is k-cyclic. In particular,Rµ(A,λ) is cyclic if each Ai is cyclic.Proof. By Theorem 8.33, µRµ(A,λ) is k-cyclic since,JµRµ(A,λ) = λ1JµA1 + · · ·+ λnJµAn ,Apply Lemma 8.31(i) with α = µ−1 to get Rµ(A,λ) is k-cyclic.To see that k-cyclic monotonicity is not dominant, we look at the fol-lowing example.Example 8.35. Let H = R2 and set λ = (1/2, 1/2). DefineA1 =(cos pi2 − sinpi2sin pi2 cospi2)and A2 =(cos pi3 − sinpi3sin pi3 cospi3).By Example 8.30, A1 is 2-cyclically monotone, but not 3-cyclically mono-tone and A2 is 3-cyclically monotone. Then R1(A,λ) is not 3-cyclicallymonotone, as can be verified using (8.33) and the pointsx1 =(00), x2 =(10), x3 =(01), and x4 = x1.The code that can be used to verify this example can be found in Ap-pendix B.1.1288.3. Recessive properties8.3.4 Displacement mappingsTheorem 8.36. Let T be a mapping from H to H. Then T ◦ (2 Id) is adisplacement mapping, i.e.T ◦ (2 Id) = Id−N,for some nonexpansive mapping N : H → H if and only if T is firmlynonexpansive.Proof. Assume first that T ◦ (2 Id) = Id−N . ThenTx = (Id−N)(12x) =x2−N(12x) =x− 2N(12x)2.As N is nonexpansive, we have (∀x ∈ H)(∀y ∈ H) ‖Nx − Ny‖ ≤ ‖x − y‖.So,∥∥∥∥N(12x)−N(12y)∥∥∥∥ ≤∥∥∥∥12x−12y∥∥∥∥⇔∥∥∥∥2N(12x)− 2N(12y)∥∥∥∥ ≤ ‖x− y‖ .So 2N ◦(12 Id)is nonexpansive, hence by Fact 3.3(iii) T is firmly nonexpan-sive.Conversely, assume T is firmly nonexpansive. ConsiderN = (Id−T ◦ (2 Id)),we will show N is nonexpansive.‖Nx−Ny‖2 = ‖(x− T (2x))− (y − T (2y))‖2= ‖(x− y)− (T (2x)− T (2y))‖2= ‖x− y‖2 − 2 〈x− y, T (2x)− T (2y)〉+ ‖T (2x)− T (2y)‖2= ‖x− y‖2 −(〈2x− 2y, T (2x)− T (2y)〉 − ‖T (2x)− T (2y)‖2)≤ ‖x− y‖2,since T is firmly nonexpansive. Thus N is nonexpansive andT ◦ (2 Id) = Id−N.1298.3. Recessive propertiesTheorem 8.37. Let T : H → H be a displacement mapping, i.e. T = Id−Nfor some nonexpansive mapping N : H → H. Then T = 2JA for somemonotone operator A : H⇒ H.Proof. By Fact 3.3(iii) and Fact 3.36, N = 2JA − Id for some monotoneoperator, A. Then using the resolvent identity we have,T = Id−(2JA − Id) = 2(Id−JA) = 2JA−1 .Theorem 8.38. A : H ⇒ H is 12 -strongly monotone if and only if A−1 =Id−N for some nonexpansive mapping, i.e. A−1 is a displacement mapping.Proof. Assume A is 12 -strongly monotone. Then A =12 Id +B for somemonotone operator B, and we haveA−1 = (12Id +B)−1 =(12(Id +2B))−1= (Id +2B)−1 ◦ (2 Id) = J2B ◦ (2 Id).Thus by Theorem 8.36, A−1 is a displacement mapping.On the other hand, assume A−1 is a displacement mapping. Then byTheorem 8.37, A−1 = Id−N = 2JB for some monotone operator B andA−1 = 2(Id +B)−1 ⇔ A = (Id +B) ◦ (12Id) =12Id +B(12Id)⇔ B(12Id)= A−12Id .Since B is monotone, A is 12 -strongly monotone.Theorem 8.39. Assume that for all i ∈ I, Ai is a displacement mapping,i.e. Ai = Id−Ni for some nonexpansive Ni. ThenR1(A,λ) = (λ1JA1 + · · ·+ λnJAn)−1 − Id,is a displacement mapping, i.e. R1(A,λ) = Id−N for some nonexpansivemapping N .Proof. Using Corollary 5.9,J(R1(A,λ))−1 = λ1JA−11+ · · ·+ λnJA−1n .1308.3. Recessive propertiesBy Theorem 8.38, A−1i is12 -strongly monotone for all i ∈ I, and there-fore by Theorem 4.1(xi) JA−1iis (1 + 12)-firmly nonexpansive. By Theo-rem 8.16, J(R1(A,λ))−1 =n∑i=1λiJA−1iis (1 + 12)-firmly nonexpansive. ThenTheorem 4.1(xi) gives R1(A,λ)−1 is 12 -strongly monotone and thus Theo-rem 8.38 yields that R1(A,λ) is a displacement mapping.8.3.5 Nonexpansive monotone operatorsTheorem 8.40 (nonexpansiveness is recessive). For all i ∈ I, let Ai be anonexpansive monotone mapping, i.e. Ai = 2Ti − Id and Ti = JBi for somemonotone operator Bi. Then R1(A,λ) is nonexpansive andR1(A,λ) = 2T − Id,where T = JB, B =n∑i=1λiBi, and Bi is nonexpansive for all i ∈ I.Proof. We haveJAi = (Id +2JBi − Id)−1 = (2JBi)−1 = (Id +Bi) ◦ (12Id). (8.40)Thus JAi ◦ (2 Id) = Id +Bi, and using Theorem 8.36Bi = −Ni, (8.41)for some nonexpansive mapping Ni. So we haveJR1(A,λ) =n∑i=1λiJAi =n∑i=1λi(Id +Bi) ◦ (12Id) =n∑i=1λi(Id−Ni) ◦ (12Id)(8.42)Or, JR1(A,λ) ◦ (2 Id) =n∑i=1λi(Id−Ni). On the other hand, by Theorem 8.36JR1(A,λ) ◦ (2 Id) = Id−N. (8.43)for some nonexpansive N . Then we have N =n∑i=1λiNi.We also have, from (8.43)(Id +R1(A,λ))−1 = (Id−N) ◦ (12Id)⇔ Id +R1(A,λ) =((Id−N) ◦ (12Id))−1= 2(Id−N)−1 (8.44)⇔ R1(A,λ) = 2(Id−N)−1 − Id . (8.45)1318.3. Recessive propertiesBut N =n∑i=1λiNi = −n∑i=1λiBi, thus we haveR1(A,λ) = 2(Id +n∑i=1λiBi)−1 − Id = 2JB − Id,where B =n∑i=1λiBi.Lemma 8.41. Let A be a real 2×2 matrix of the form A =[a −bb a]. ThenA is a rotation matrix if and only if AT = A−1, i.e. A is an orthogonalmatrix.Proof. First, assume AT = A−1 and A =[a −bb a]. Then ATA = Id ⇒a2 + b2 = 1, i.e. a and b lie on the unit circle. Therefore, converting to polarcoordinates using a = cos θ and b = sin θ gives that A is a rotation matrix.On the other hand, assume A is a rotation matrix, thenA =[cosα − sinαsinα cosα],andATA =[cosα sinα− sinα cosα] [cosα − sinαsinα cosα]=[1 00 1].Thus, AT = A−1.Example 8.42 (2 × 2 rotation matrices). Let Aα and Aθ be the 2 × 2rotation matrices,Aα =[cosα − sinαsinα cosα]Aθ =[cos θ − sin θsin θ cos θ],with α, θ ∈ [−pi/2, pi/2]. Then using mathematical software, it is easy toverify that R1(A,λ) is a matrix of the form[a −bb a], with R1(A,λ)T =R1(A,λ)−1 (see Appendix B.2), thus by Lemma 8.41, R1(A,λ) is a rotationmatrix.1328.3. Recessive propertiesRemark 8.43. Example 8.42 shows that the resolvent average of two rotatorsis also a rotator. The same is not true of the arithmetic average. Consider,A = λAα + (1− λ)Aθ =[λ cosα+ (1− λ) cos θ −λ sinα− (1− λ) sin θλ sinα+ (1− λ) sin θ λ cosα+ (1− λ) cos θ].Then we haveATA =[2(λ− λ2) cos(α− θ) + 2λ2 − 2λ+ 1 00 2(λ− λ2) cos(α− θ) + 2λ2 − 2λ+ 1].By Lemma 8.41, A is a rotator if and only if ATA = Id. This implies thatcos(α− θ) = 1. That is, α = θ + 2kpi for k = 0, 1, . . .. So the arithmetic av-erage of two rotation matrices only produces another rotation matrix undervery specific circumstances.Remark 8.44. Although we can see that R1(A,λ) is a rotation matrix, evenin very simple cases it is difficult to see the relationship between the originalrotation matrices and the resulting rotation matrix. See Figure 8.1 to seehow the angle of rotation varies with certain values of θ and α.Theorem 8.45 (orthogonality is recessive). Let Ai be monotone orthogonalmatrices for all i ∈ I. Then R1(A,λ) is an orthogonal matrix, i.e. A−1 =AT .Proof. By Theorem 5.19, the orthogonality of each Ai and Fact 2.6, we have(R1(A,λ))−1 = R1(A−1,λ)= R1(AT , λ)=(λ1(Id +AT1 )−1 + · · ·+ λm(Id +ATm)−1)−1 − Id=(λ1(IdT +AT1 )−1 + · · ·+ λm(IdT +ATm)1)−1 − Id=(λ1((Id +A1)T )−1 + · · ·+ λm((Id +Am)T )−1)−1− Id=(λ1((Id +A1)−1)T + · · ·+ λm((Id +Am)−1)T)−1− Id=((λ1(Id +A1)−1 + · · ·+ λm(Id +Am)−1)T)−1− Id=((λ1(Id +A1)−1 + · · ·+ λm(Id +Am)−1)−1)T− Id= R1(A,λ)T .Thus R1(A,λ) is orthogonal.1338.3. Recessive propertiesFigure 8.1: Resulting angle γ of the rotation of R1(A,λ) of Example 8.42with Aα = Aθ+pi2 .Remark 8.46. As you would expect, Theorem 8.45 only holds when µ = 1.For example, take A1 = Id, A2 be the 2 × 2 rotation by pi/2, λ1 = λ2 = 12and µ = 2, then Rµ(A,λ) is not orthogonal.Remark 8.47 (Pythagorean triples). As an interesting aside, note that whenwe set α = pi/2 and θ = 0 in Example 8.42 then any rational value for λproduces a pythagorean triple, i.e. three numbers x, y, and z such thatx2 + y2 = z2. We begin with the matrix,R1(A,λ) =1−λ21+λ2−2λ1+λ22λ1+λ21−λ21+λ2 .1348.3. Recessive propertiesSubstituting λ = ab , where a and b are integer values, we haveR1(A,λ) =b2−a2a2+b2−2aba2+b22aba2+b2b2−a2a2+b2 .Theorem 8.45 shows that R1(A,λ) maintains orthogonality and by Exam-ple 8.42 R1(A,λ) is a rotation matrix. So the entries correspond tocosβ =b2 − a2a2 + b2and sinβ =2aba2 + b2,for some β ∈ [0, pi/2]. Thus the angles formed in R1(A,λ) are the angles inright triangles with all integer sides such that(b2 − a2)2+ (2ab)2 =(a2 + b2)2. (8.46)In fact, all possible triples can be generated this way. By [60, Theorem 11.1],all pythagorean triples can be generated by relatively prime integers m andn such thatx = m2 − n2, y = 2mn, z = m2 + n2,then x2 + y2 = z2, which is exactly (8.46). The code used to generate thisexample can be found in Appendix B.3.Lemma 8.48. For all i ∈ I, let Ai be monotone, and let some Aj be stronglymonotone with constant β ∈ R++. Then A =n∑i=1λiAi is strongly monotonewith constant λjβ.Proof. Since Aj is strongly monotone with constant β then Aj − β Id ismonotone. Thenn∑i=1λiAi−λjβ Id is monotone, as it is the sum of monotoneoperators. Thus, A is strongly monotone with constant λjβ.Lemma 8.49. Let β ∈ R++. An operator A is strongly monotone withconstant β ⇔ A ◦(12 Id)is strongly monotone with constant β2 .Proof. Let u ∈ A(12x)and v ∈ A(12y), then(x, u) ∈ graA ◦(12Id)and (y, v) ∈ graA ◦(12Id).As well,(12x, u) ∈ graA and (12y, v) ∈ graA.1358.3. Recessive propertiesSince A is strongly monotone with constant β, we have〈12x−12y, u− v〉≥ β‖12x−12y‖2⇔12〈x− y, u− v〉 ≥β4‖x− y‖2⇔ 〈x− y, u− v〉 ≥β2‖x− y‖2.Thus A ◦(12 Id)is strongly monotone with constant β2 .Lemma 8.50. Let β > 0 and α ≥ 1. Then A is strongly monotone withconstant β ⇔ αA is strongly monotone with constant αβ.Proof. Let (x, u) and (y, v) ∈ graA. Then (x, αu) and (y, αv) ∈ graαA.Since A is strongly monotone with constant β,〈x− y, αu− αv〉 = α 〈x− y, u− v〉 ≥ αβ‖x− y‖2.Thus αA is strongly monotone with constant αβ.Lemma 8.51. Assume A is both nonexpansive and strongly monotone withconstant β. Then A−1 is strongly monotone with constant β.Proof. Let (x, u) ∈ graA and (y, v) ∈ graA. Then (u, x) ∈ graA−1 and(v, y) ∈ graA−1. By the strong monotonicity and then nonexpasiveness ofA, we have〈x− y, u− v〉 ≥ β‖x− y‖2 ≥ β‖u− v‖2,i.e. A−1 is strongly monotone with constant β.Theorem 8.52. Let Ai = 2Ti − Id be monotone for all i ∈ I and Ti = JBifor a monotone operator Bi. Additionally, assume some Aj = 2Tj − Id is aBanach contraction. Then R1(A,λ) is a Banach contraction.Proof. By Corollary 4.31, Aj is a Banach contraction ⇔ Bj and B−1j arestrongly monotone. Using Theorem 8.40,R1(A,λ) = 2JB − Id,1368.3. Recessive propertieswhere B =n∑i=1λiBi. Setting N = 2JR1(A,λ) − Id, we haveN = 2(Id +2JB − Id)−1 − Id= 2(Id +B) ◦(12Id)− Id= Id +2B ◦(12Id)− Id= 2n∑i=1λiBi ◦(12Id).Combining Lemma 8.48, Lemma 8.49, and Lemma 8.50 we haveN is stronglymonotone. By Theorem 8.40, Bi is nonexpansive for all i = 1, . . . , n there-foren∑i=1λiBi is nonexpansive and we get‖n∑i=1λiBi ◦(12x)−n∑i=1λiBi ◦(12y)‖ ≤ ‖x2−y2‖⇔ ‖2n∑i=1λiBi ◦(12x)− 2n∑i=1λiBi ◦(12y)‖ ≤ ‖x− y‖.Thus, N is both nonexpansive and strongly monotone, so by Lemma 8.51,N−1 is strongly monotone. Therefore by Corollary 4.31, 2JN−Id is a Banachcontraction. Now,2JN − Id = 2(Id +2JR1(A,λ) − Id)−1 − Id= 2(2JR1(A,λ))−1 − Id= 2(Id +R1(A,λ)) ◦(12Id)− Id= 2R1(A,λ) ◦(12Id).So 2R1(A,λ) ◦(12 Id)is a Banach contraction, i.e. there exists β ∈ [0, 1[such that for all x, y ∈ H,‖2R1(A,λ) ◦(12x)− 2R1(A,λ) ◦(12y)‖ ≤ β‖x− y‖⇔ ‖R1(A,λ) ◦(12x)−R1(A,λ) ◦(12y)‖ ≤ β‖x2−y2‖.Thus R1(A,λ) is a Banach contraction.1378.4. Indeterminate properties8.4 Indeterminate propertiesWe conclude this chapter with a couple of examples of properties thatdo not satisfy the definition of dominant or recessive.Example 8.53 (projections are indeterminant). Let A1 and A2 be the pro-jections in R2 onto R× {0} and {0} × R, respectively. That is,A1 =(1 00 0)and A2 =(0 00 1)Then A1 and A2 are both projections, butR1(A,λ) = (λJA1 + (1− λ)JA2)−1 − Id=( λ2−λ 00 1−λλ+1),is not a projection, since (R1(A,λ))2 6= R1(A,λ) unless λ = 0 or λ = 1.Example 8.54 (normal cones are indeterminant). Let A1 = NC1 and A2 =NC2 be the normal cones operators of C1 = R × {0} and C2 = {0} × R.Then,JA1 =(1 00 0)and JA2 =(0 00 1)A1 and A2 are both normal cones, but R1(A,λ) is not a normal cone byTheorem 4.1(xx), sinceJR1(A,λ) =(λ 00 (1− λ)),and thus ranJR1(A,λ) 6= FixJR1(A,λ).Remark 8.55. In Examples 8.53 and 8.54 we have shown that the resolventaverage is not necessarily a projection (or normal cone) even if all of theaveraged operators are projections (normal cones).The classifications of each of the properties considered in this chapterare summarized in Table 8.1 and Table 8.2.1388.4. Indeterminate propertiesTable 8.1: Summary of completely classified properties of the resolvent av-erage.Dominant Properties Recessive PropertiesSingle-valuedness Maximal monotonicityFull domain LinearitySurjectivity k-cyclic monotonicityStrict monotonicity Displacement mappingsBanach contraction OrthogonalityLinear rectangularity Nonlinear rectangularityLinear paramonotonicty Nonlinear paramonotonicityNonexpansivenessTable 8.2: Summary of incompletely classified and indeterminant propertiesof the resolvent average.Dominant or RecessivePropertiesIndeterminant PropertiesStrong monotonicity Projection operatorsCocoercivity Normal cone operators139Chapter 9Conclusion9.1 Key resultsThis thesis has provided a comprehensive study of the relationship be-tween maximally monotone operators and firmly nonexpansive mappings aswell as defining a new method for averaging monotone operators. The keyresults presented are outlined below.Theorem 4.1 lists the corresponding properties between maximally mono-tone operators and their associated resolvents. This theorem covers twenty-one properties of interest in monotone operator theory.Definition 5.1 describes the resolvent average of monotone operators, anew average that maintains several desirable properties that the arithmeticaverage does not.Theorem 5.3 demonstrates that the resolvent average is maximally mono-tone if and only if all of the averaged operators are maximally monotone.This is a stronger result than for the arithmetic average, which requires theadditional constraint qualifications found in Fact 3.48.The resolvent average satisfies the beautiful duality presented in Theo-rem 5.8,(Rµ(A,λ))−1 = Rµ−1(A−1,λ).Theorem 5.14 develops an inequality between the arithmetic, resolvent,and harmonic averages for positive semidefinite matrices,H(A,λ) Rµ(A,λ) A(A,λ),and shows the limitsRµ(A,λ)→ A(A,λ) when µ→ 0+,andRµ(A,λ)→ H(A,λ) when µ→ +∞.Theorem 6.28 provides results on the range of convex combinations ofrectangular maximally monotone operators on Rn.1409.1. Key resultsTheorem 6.31 and Theorem 6.36 give results on convex combinations offirmly nonexpansive mappings in Rn.Theorem 7.14 shows that the composition of asymptotically regular map-pings is again asymptotically regular in a Hilbert space.Theorem 7.21 derives the asymptotic regularity of a convex combinationof asymptotically regular mappings, extending Theorem 6.36 to a Hilbertspace setting.Chapter 8 classified properties of monotone operators and/or their re-solvents as dominant, recessive, or indeterminant. Dominant properties in-clude:(i) single valuedness,(ii) full domain,(iii) surjectivity,(iv) strict monotonicity,(v) Banach contraction,(vi) linear paramonotonicity (equivalently rectangularity).Dominant or recessive properties are:(i) γ-cocoercive, and(ii) strong monotonicity.Recessive properties are:(i) maximal monotonicity,(ii) linear relations,(iii) rectangularity (except as noted above),(iv) paramonotonicity (except as noted above),(v) k-cyclic monotonicity,(vi) displacement mappings, and(vii) orthogonality.Altogether, these results have expanded on the known theory regardingmaximally monotone operators and firmly nonexpansive mappings.1419.2. Future work9.2 Future workAreas to consider for future research include specializing the inheritanceproperties of Chapter 8 to positive semidefinite matrices, similar to Sec-tion 5.2. 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Bertsekas, On the Douglas-Rachford split-ting method and the proximal point algorithm for maximal monotoneoperators, Math. Prog. Ser. A, 55 (1992), pp. 293–318. → pages 32[39] L. Elsner, I. Koltracht, and M. Neumann, Convergence of se-quential and asynchronous nonlinear paracontractions, Numer. Math.,62 (1992), pp. 305–319. → pages 58[40] S. Fitzpatrick, Representing monotone operators by convex functions,Proc. Centre Math. Anal. Austral. Nat. Univ., 20 (1988), pp. 59–65. →pages 34[41] C. W. Groetsch, Generalized Inverses of Linear Operators: Repre-sentation and Approximation, Marcel Dekker, 1977. → pages 12[42] R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge Uni-versity Press, 1985. → pages 70[43] S. Kim, J. Lawson, and Y. Lim, The matrix geometric mean of pa-rameterized, weighted arithmetic and harmonic means, Linear AlgebraAppl., 435 (2011), pp. 2114–2131. → pages 142[44] F. Kubo and T. Ando, Means of positive linear operators, Math.Ann., 246 (1980), pp. 205–224. → pages 17[45] S. Kum, Resolvent average on second-order cone, Taiwanese J. Math.,15 (2011), pp. 2733–2750. → pages 142146Bibliography[46] A. W. Marshall and I. Olkin, Inequalities: Theory of Majorizationand Its Applications, Academic Press, 1979. → pages 70[47] J.-E. Mart´ınez-Legaz and M. The´ra, A convex representation ofmaximal monotone operators, J. Nonlinear Convex Anal., 2 (2001),pp. 243–247. → pages 34[48] C. D. Meyer, Matrix Analysis and Applied Linear Algebra, SIAM,2000. → pages 4, 35[49] G. J. Minty, On the maximal domain of a ‘monotone’ function, Mich.Math. J., 8 (1961), pp. 135–137. → pages 81[50] , Monotone (nonlinear) operators in Hilbert spaces, Duke Math.J., 29 (1962), pp. 341–346. → pages 18, 32[51] M. Moakher, A differential geometric approach to the geometric meanof symmetric positive-definite matrices, SIAM J. Matrix Anal. Appl.,26 (2005), pp. 735–747. → pages 17[52] J.-J. Moreau, Proximite´ et dualite´ dans un espace hilbertien, Bull.Soc. Math. France, 93 (1965), pp. 273–299. → pages 18[53] T. Pennanen, On the range of monotone composite mappings, J. Non-linear Convex Anal., 2 (2001), pp. 193–202. → pages 34, 88[54] D. Petz, Means of positive matrices: geometry and a conjecture, Ann.Math. Inform., 32 (2005), pp. 129–139. → pages 17[55] H. Ra˚dstro¨m, An embedding theorem for spaces of convex sets, Proc.Am. Math. Soc., 3 (1952), pp. 165–169. → pages 87[56] S. Reich, On the asymptotic behavior of nonlinear semigroups and therange of accretive operators, J. Math. Anal. Appl., 79 (1981), pp. 113–126. → pages 103, 105[57] S. Reich and I. Shafrir, The asymptotic behavior of firmly nonex-pansive mappings, Proc. Am. Math. Soc., 101 (1987), pp. 246–250. →pages 29[58] R. T. Rockafellar, Convex Analysis, Princeton University Press,1970. → pages 1, 9, 14, 16, 17, 18[59] R. T. Rockafellar and R. J-B Wets, Variational Analysis,Springer-Verlag, 1998. → pages 1, 10, 15, 18, 30, 33, 58, 76, 80, 88147Bibliography[60] K. H. Rosen, Elementary Number Theory and its Applications,Addison-Wesley, 1986. → pages 135[61] S. Simons, LC-functions and maximal monotonicity, J. Nonlinear Con-vex Anal., 7 (2006), pp. 123–137. → pages 34[62] E. H. Zarantonello, ed., Contributions to Nonlinear FunctionalAnalysis, Academic Press, 1971, ch. Projections on convex sets inHilbert space and spectral theory I. Projections on convex sets, pp. 237–341. → pages 27[63] C. Za˘linescu, Convex Analysis in General Vector Spaces, World Sci-entific Publishing Co. Inc., 2002. → pages 12148Index3∗ monotone, see rectangularadjoint, 5affine hull, 8affine relation, 8, 41, 55affine subspace, 8arithmetic average, 17asymptotic regularity, 104asymptotically regular, 27, 28, 92weakly, 27averaged operator, 26, 27averagesarithmetic, 17geometric, 18, 77harmonic, 17harmonic-resolvent-arithmeticinequality, 73, 77proximal, 18resolvent, 64backward-backward iteration, 62Baillon-Haddad theorem, 26, 48Banach contraction, 5, 23, 39, 46,47, 59, 60, 114Banach space, 3, 153Banach-Picard iterates, 27, 63cocoercive, 26, 32, 39, 117coercive, 14, 94compositionof asymptotically regular map-pings, 105of Banach contractions, 62of firmly nonexpansive map-pings, 102of nonexpansive mappings, 62of strongly nonexpansive map-pings, 28concavematrix, 74cone, 8convex, 8conical hull, 8convergencestrong, 4weak, 4convexcone, 8essentially strictly, 14function, 12matrix, 74set, 8strictly, 12strongly, 13, 14, 60convex combinationof asymptotically regular map-pings, 106of asypmtotically regular map-pings, 92of Banach contractions, 62of firmly nonexpansive map-pings, 91of nonexpansive mappings, 62convex hull, 8cyclically firmly nonexpansive, 26,149Index34cyclically monotone, 33, 124–126,128disjointly injective, 8, 39, 46, 50,52, 53displacement mapping, 36, 54, 129,130dominant or recessive property, 116dominant property, 109Douglas-Rachford iteration, 63dual properties, 48self, 49epi-multiplication, 12epigraph, 12essentially smooth, 14, 52essentially strictly convex, 14, 52Fenchel conjugate, 12, 16, 52, 54,60of a quadratic, 16of a subdifferential, 17firmly nonexpansive, 23and averaged operators, 26characterizations, 24cyclically, 26, 34, 40, 54strictly, 39, 51Fitzpatrick function, 34, 35fixed points, 26, 92Fre´chet gradient, 48Fre´chet derivative, 6Fre´chet gradient, 7, 15Gaˆteaux derivative, 6Gaˆteaux gradient, 7geometric mean, 18gradient, 7, 15graph, 7harmonic average, 17hemicontinuous, 31Hilbert space, 3identity, 5indeterminant property, 109indicator function, 10infimal convolution, 12, 16inheritance, 109isometry, 6, 38, 124kernel, 5least-squares, 11, 96Legendre, 14, 53Legendre function, 14, 53linear relation, 8, 40, 55, 61Lipschitz continuous, 5, 23, 32, 60locally bounded, 32lower semi-continuous, 12lower semi-continuous hull, 12Minty parametrization, 31Minty’s Theorem, 32monotone, 29n-cyclically, 33cyclically, 33, 54maximally, 29paramonotone, 30strictly, 30, 39, 50–52, 114strongly, 30, 32, 39, 58, 60, 61,116uniformly, 30, 40monotonicityof normal cone, 32of subdifferential, 30Moore-Penrose inverse, 11, 96nearly convex, 80characterization, 81nearly equal, 79nonexpansive, 23150Indexfirmly, 23strictly, 23, 39, 50, 52, 53strongly, 23, 28normal cone, 11, 13normal cone operator, 41, 138orthogonal complement, 5, 96paramonotone, 30, 35, 40, 53, 118,120projection, 10, 24, 27, 32, 41, 56,92, 138proper function, 12proximal average, 18, 94proximal mapping, 15, 47, 53, 60,118of indicator function, 32of subdifferential operator, 31recessive property, 109rectangular, 34, 35, 40, 54, 112,118reflected resolvent, 56, 58–61relative interior, 8resolvent, 31of normal cone operator, 32resolvent average, 64, 93resolvent identity, 31right-shift operator, 36, 96rotation matrix, 4, 52, 124, 132self-dual, 48sequentially weakly closed, 8, 41,56sequentially weakly continuous, 6,41, 56set-valued, 7inverse, 7single-valued, 7, 50–53, 111subdifferential operator, 12, 40, 52–54, 60maximal cyclical monotonicity,33supercoercive, 14uniformly convex, 153uniformly Gaˆteaux differentiable,154Yosida regularization, 31of resolvent average, 64151Appendices152Appendix AUniformly Convex BanachSpacesA Banach space is a complete normed linear space, whereas a Hilbertspace is a complete inner product space. Because each Hilbert space has anorm induced by its inner product, every Hilbert space is a Banach space.The dual space of an inner product space X is the set X∗ of all boundedlinear functionals on X. The dual space of a Hilbert space is isomorphic tothe original space [37, Theorem 6.10]. That is, H∗ = H.Definition A.1. [32, Equation 11.1] A normed linear space X is uniformlyconvex if, for each > 0, ∃δ = δ() > 0 such that‖x‖ < 1, ‖y‖ < 1, ‖x− y‖ > ⇒∥∥∥∥x+ y2∥∥∥∥ ≤ 1− δ.Lemma A.2. [parallelogram identity] Let x, y ∈ H. Then‖x− y‖2 + ‖x+ y‖2 = 2‖x‖2 + 2‖y‖2.Proof.‖x− y‖2 + ‖x+ y‖2 = 〈x− y, x− y〉+ 〈x+ y, x+ y〉= ‖x‖2 − 2 〈x, y〉+ ‖y‖2 + ‖x‖2 + 2 〈x, y〉+ ‖y‖2= 2‖x‖2 + 2‖y‖2.Lemma A.3. Every Hilbert space is uniformly convex.Proof. Let x, y ∈ H such that ‖x‖ < 1, ‖y‖ < 1 and ‖x − y‖ ≥ . ByLemma A.2 we have‖x+ y‖2 = 2‖x‖2 + 2‖y‖2 − ‖x− y‖2≤ 4− 2.153Appendix A. Uniformly Convex Banach SpacesSet δ = 1− 12√4− 2, then‖x+ y‖2 ≤ 4− 2 ⇔∥∥∥∥x+ y2∥∥∥∥ ≤ 1− δ.Since the parallelogram identity holds for every Hilbert space, every Hilbertspace is uniformly convex.Definition A.4. [34, pg. 126]A normed linear space X has a uniformlyGaˆteaux differentiable norm if for each y ∈ X and each > 0, there existsδ(, y) > 0 such that for every x ∈ X, ‖x‖ = 1, there is a continuous linearfunctional fx on X and∣∣∣∣‖x+ ty‖ − ‖x‖t− fx(y)∣∣∣∣ < for all 0 < t < δ(, y).Fact A.5. [34, pg. 127] Every Hilbert space has a uniformly Gaˆteaux dif-ferentiable norm.Fact A.6. [3, Theorem 1.2] Let T be a nonexpansive mapping and let Xbe a uniformly convex Banach space with a weakly sequentially continuousduality map, then (Tnx)n∈N converges weakly to a fixed point of T if andonly if FixT 6= ∅ and T is weakly asymptotically regular.Fact A.7. [3, Corollary 2.2] Let X be a Banach space and C be a closedconvex subset of X. Let U : C → X be an averaged nonexpansive mapping.If X is uniformly convex, then FixU = ∅ if and only if limn→∞|Unx| =∞ forall x in C.Definition A.8. [25] Let C be a nonempty closed convex subset of a Banachspace X and let D be a nonempty subset of C. A retraction from C to Dis a mapping T : C → D such that Tx = x for all x ∈ D.Definition A.9. [25] A retraction T : C → D is sunny if it satisfies thepropertyT (Tx+λ(x−Tx)) = Tx for x ∈ C and λ > 0 whenever Tx+λ(x−Tx) ∈ C.A retraction T : C → D is sunny nonexpansive if it it both sunny andnonexpansive.Fact A.10. [3, Corollary 2.3] Let X be a Banach space and C be a closedconvex subset of X. Let U : C → X be an averaged nonexpansive mapping.Suppose that the norm of X is uniformly Gaˆteaux differentiable while the154Appendix A. Uniformly Convex Banach Spacesnorm of X∗ is Fre´chet differentiable. If C is a sunny nonexpansive retractof X, then for each x ∈ Climn→∞(Unx− Un+1x)→ v,where v is the element of least norm in ran(Id−U).Remark A.11. Sunny nonexpansive retracts are unique, if they exist. If C isa nonempty closed convex subset of a Hilbert space H then the projectionoperator PC is the sunny nonexpansive retraction [25].155Appendix BMaple CodeThe following sections provide the code that was used to verify examples.All code was run using Maplesoft’s Maple 15 software.B.1 Code to verify Example 8.35> restart: with(LinearAlgebra):> A1 :=[cos (α) − sin (α)sin (α) cos (α)][cos (α) − sin (α)sin (α) cos (α)]> A2 :=[cos (θ) − sin (θ)sin (θ) cos (θ)][cos (θ) − sin (θ)sin (θ) cos (θ)]> Id :=[1 00 1][1 00 1]> A1 := subs(alpha = (1/2)*Pi, A1);[0 −11 0]> A2 := subs(theta = (1/3)*Pi, A2);[1/2 −1/2√31/2√3 1/2]156B.2. Code to verify Example 8.42> R := simplify(MatrixInverse((1/2)*MatrixInverse(Id+A1)+(1/2)*MatrixInverse(Id+A2))-Id);−−4+√38+√3−6+2√38+√36+2√38+√3−−4+√38+√3> x1 :=[00]; x2 :=[10]; x3 :=[01]; x4 := x1;[00][10][01][00]> u1 := Multiply(R, x1);u2 := Multiply(R, x2);u3 := Multiply(R, x3);[00]−−4+√38+√36+2√38+√3−6+2√38+√3−−4+√38+√3> simplify(sum((xi+1[1] − xi[1]) ∗ ui[1] + (xi+1[2] − xi[2]) ∗ ui[2], i =1..3)); evalf(%);2−1 + 2√38 +√30.5063889748B.2 Code to verify Example 8.42> restart: with(LinearAlgebra):157B.2. Code to verify Example 8.42> A1 :=[cos (α) − sin (α)sin (α) cos (α)][cos (α) − sin (α)sin (α) cos (α)]> A2 :=[cos (θ) − sin (θ)sin (θ) cos (θ)][cos (θ) − sin (θ)sin (θ) cos (θ)]> Id :=[1 00 1][1 00 1]> JA1 := MatrixInverse(Id+A1)1+cos(α)1+2 cos(α)+(cos(α))2+(sin(α))2sin(α)1+2 cos(α)+(cos(α))2+(sin(α))2− sin(α)1+2 cos(α)+(cos(α))2+(sin(α))21+cos(α)1+2 cos(α)+(cos(α))2+(sin(α))2> JA2 := MatrixInverse(Id+A2)1+cos(θ)1+2 cos(θ)+(cos(θ))2+(sin(θ))2sin(θ)1+2 cos(θ)+(cos(θ))2+(sin(θ))2− sin(θ)1+2 cos(θ)+(cos(θ))2+(sin(θ))21+cos(θ)1+2 cos(θ)+(cos(θ))2+(sin(θ))2> R := simplify(MatrixInverse(lambda∗JA1+(1− lambda)∗JA2)−Id)158B.3. Code to verify Remark 8.47[[− (λ+ cos (θ) + cos (α) cos (θ)− λ cos (θ) + λ cos (α)− λ cos (α) cos (θ)+λ2 cos (θ) cos (α)− λ sin (α) sin (θ) + λ2 sin (α) sin (θ)− λ2)/(−λ cos (α) cos (θ) + λ2 cos (θ) cos (α)− λ cos (θ) + λ cos (α)− 1 + λ+λ2 sin (α) sin (θ)− λ2 − cos (α)− λ sin (α) sin (θ)),− ((−λ sin (α)− λ sin (α) cos (θ)− sin (θ)− sin (θ) cos (α) + sin (θ)λ+ sin (θ)λ cos (α)) /(−λ cos (α) cos (θ) + λ2 cos (θ) cos (α)− λ cos (θ)+λ cos (α)− 1 + λ+ λ2 sin (α) sin (θ)− λ2 − cos (α)− λ sin (α) sin (θ)))],[(−λ sin (α)− λ sin (α) cos (θ)− sin (θ)− sin (θ) cos (α) + sin (θ)λ+ sin (θ)λ cos (α)) /(−λ cos (α) cos (θ) + λ2 cos (θ) cos (α)− λ cos (θ)+λ cos (α)− 1 + λ+ λ2 sin (α) sin (θ)− λ2 − cos (α)− λ sin (α) sin (θ)),− (λ+ cos(θ) + cos(α) cos(θ)− λ cos(θ) + λ cos(α)− λ cos(α) cos(θ)+λ2 cos(θ) cos(α)− λ sin(α) sin(θ) + λ2 sin(α) sin(θ)− λ2/ (−λ cos (α) cos (θ)+λ2 cos (θ) cos (α)− λ cos (θ) + λ cos (α)− 1 + λ+ λ2 sin (α) sin (θ)−λ2 − cos(α)− λ sin(α) sin(θ))]]> R[1, 1]-R[2, 2];0> R[1, 2]+R[2, 1];0> simplify(Multiply(Transpose(R), R))[1 00 1]B.3 Code to verify Remark 8.47Define A1, A2, Id and R as in Section B.2.> R2 := factor(simplify(subs([theta = 0, alpha = (1/2) ∗ Pi], R)));−−1+λ21+λ2 −2λ1+λ22 λ1+λ2 −−1+λ21+λ2> factor(simplify(subs(lambda = a/b,R2)));159B.3. Code to verify Remark 8.47−−b2+a2b2+a2 −2abb2+a22 abb2+a2 −−b2+a2b2+a2160
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The resolvent average : an expansive analysis of firmly nonexpansive mappings and maximally monotone… Moffat, Sarah Michelle 2014
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Title | The resolvent average : an expansive analysis of firmly nonexpansive mappings and maximally monotone operators |
Creator |
Moffat, Sarah Michelle |
Publisher | University of British Columbia |
Date Issued | 2014 |
Description | Monotone operators and firmly nonexpansive mappings are essential to modern optimization and fixed point theory. Minty first discovered the link between these two classes of operators; every resolvent of a monotone operator is firmly nonexpansive and every firmly nonexpansive mapping is a resolvent of a monotone operator. This thesis provides an in-depth study of the relationship between firmly nonexpansive mappings and maximally monotone operators. First, corresponding properties between maximally monotone operators and their resolvents are collected. Then a new method of averaging monotone operators is presented, called the resolvent average, which is based on the convex combination of the resolvents of monotone operators. Several new results are given concerning the asymptotic regularity of compositions and convex combinations of firmly nonexpansive mappings. Finally, the resolvent average is studied with respect to which properties the average inherits from the averaged operators. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2014-12-18 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NonCommercial-NoDerivs 2.5 Canada |
DOI | 10.14288/1.0074402 |
URI | http://hdl.handle.net/2429/51593 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Irving K. Barber School of Arts and Sciences (Okanagan) Computer Science, Mathematics, Physics and Statistics, Department of (Okanagan) |
Degree Grantor | University of British Columbia |
GraduationDate | 2015-02 |
Campus |
UBCO |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nc-nd/2.5/ca/ |
AggregatedSourceRepository | DSpace |
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