MINORANT PROPERTIESByStephen WeissenhoferB. Sc. (Honours) The Flinders University of South AustraliaA THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinTHE FACULTY OF GRADUATE STUDIESMATHEMATICSWe accept this thesis as conformingto the required standardTHE UNIVERSITY OF BRITISH COLUMBIAMarch 1993© Stephen Weissenhofer, 1993In presenting this thesis in partial fulfilment of the requirements for an advanced degree atthe University of British Columbia, I agree that the Library shall make it freely availablefor reference and study. I further agree that permission for extensive copying of thisthesis for scholarly purposes may be granted by the head of my department or by hisor her representatives. It is understood that copying or publication of this thesis forfinancial gain shall not be allowed without my written permission.MathematicsThe University of British Columbia2075 Wesbrook PlaceVancouver, CanadaV6T 1Z1Date:tictpel, I , 1 ctc13AbstractWe study the minorant property and the positive minorant property for norms on spacesof matrices and norms on spaces of functions. A matrix is said to be a majorant ofanother if all the entries in the first matrix are greater than or equal to the absolutevalues of the corresponding entries in the second matrix. The Cp norm of a matrix is thetP norm of its singular values. The space of n x n matrices, with this norm, is said to havethe minorant property provided that the norm of each nonnegative matrix is greater thanor equal to the norm of every matrix that it majorizes. Similarly, if the norm of eachnonnegative matrix is greater than or equal to the norm of every nonnegative matrix thatit majorizes, then the space of matrices is said to have the positive minorant property.It is easy to verify that these properties hold if p is even. We show that the positiveminorant property fails on n x n matrices with the Cp norm when 0 < p < 2(n — 1) andp is not even.We also consider versions of the minorant properties for LPspaces of function on somecommutative topological groups. We show that the positive minorant property fails onthe cyclic group of order n when 0 < p < 2[(n — 1)/2] and p is not even. This relationbetween p and n is different from the one for matrices, and yet both seem to be optimal.We also show that the minorant property fails on cyclic groups when p and n are suitablyrelated. We completely determine the combinations of p and n for which the minorantproperties hold on some loworder groups.For compact abelian groups, we show that the positive minorant property is equivalentto a certain function operating on the class of positive definite functions. We deduce thatif LP(G) has the positive minorant property, then so does LP+2 (G) .11Table of ContentsAbstract^ iiAcknowledgement^ v1 Introduction. 11.1 Notation. ^11.2 Majorants on Groups. ^21.3 Majorants and Matrices ^72 General Facts About Minorant Properties.^ 112.1 Reductions. ^112.2 The positive minorant property^ 173 Counterexamples.^ 233.1 Groups and matrices ^233.2 Cyclic groups^ 273.3 Beyond cyclic groups. ^ 364 Special Cases.^ 384.1 Minorant properties on loworder groups and matrices. ^ 384.2 Functions that operate on finite groups. ^ 555 Failure of the Positive Minorant Property for Matrices.^565.1 Statement of the main theorem^ 561115.2 Eigenvalues and eigenvectors of A. ^ 585.3 When the last entry is zero. 605.4 When the last entry is not zero^ 62Bibliography^ 65ivAcknowledgementFirst and foremost I wish to thank my research supervisor Dr. John Fournier. I amgreatly indebted to him for his help, support and guidance. I especially appreciatedour regularly scheduled meetings, they helped me to be more disciplined and motivatedduring those long periods when nothing would seem to work. I am also grateful for thetime and effort he spent in correcting rItXing errors in the initial typed version of mythesis. Without this help I would have never made the deadlines I wanted to meet.I would like to thank my good friend Everett Ofori who did an excellent job intyping my thesis quickly and accurately. I appreciate his willingness to learn TEX andmathematical symbols way beyond what would have been reasonable to expect. Hisassistance was also instrumental in getting my thesis prepared on schedule.I thank my many friends for their moral support and encouragement, especiallyEverett Ofori, Douglas Williams, Yoshie Akai, April Nagy, Art and June Le Patourel,and Gordon Coulson. Thanks also goes to Dr. E. E. Granirer, Dr. Z. A. Melzak andDr. J. E. Coury, whose courses were a pleasure to take, and in various ways better prepared me when it came to writing this thesis. I am grateful to Dr. B. Abrahamson,Dr. G. Gaudry, and Dr. W. Cornish who encouraged me to go overseas to do my Ph.D.It has been a decision I'll never regret, and I'll always have fond memories of the timesI've spend at U.B.C. and Canada. Finally I wish to thank my parents for their supportin all my endeavours.Financial support from the Canadian Commonwealth Scholarship and FellowshipPlan, and from Teaching and Research Assistantships in the Department of Mathematicsat U.B.C. are gratefully acknowledged.vChapter 1Introduction.In this chapter we will introduce notation and definitions to be used throughout thisthesis. We also provide some of the history of the majorant problem. As well, we explainwhy the consideration of finite abelian groups and matrices is of interest in connectionwith the majorant problem.1.1 Notation.Our standard reference for notation and terminology related to Fourier analysis is Rudin'sbook [25]. We let G denote an abelian group which in most cases will be compact. Acharacter of G is a continuous complex valued function y defined on G for which j y(x)I = 1for all x E G and y(x + y) = y(x)y(y) for all x, y E G. Under pointwise multiplicationthe set of all characters of G forms a group called the dual of G and will be denoted byÔ. The Fourier transform of f E L' (G) is the function j defined on G by(y) = L f (x) y(x) dxfor all y E Ô. As an example, consider G = T, the circle group. The characters of T are7(0) = e"0, where n E Z. Hence It Z and for f e Ll(T), the Fourier transform of f is(n) =^f (9)e'' dewhere (1/271 )de is the Haar measure.1Chapter 1. Introduction.^ 21.2 Majorants on Groups.The majorant problem was introduced by Hardy and Littlewood [12] who asked whatcomparisons can be made between the LP norms of two integrable functions when one isa majorant of the other, that is, when the Fourier coefficient of one dominates the Fouriercoefficients of the other. Their work was done on the circle group T, but we will makeour definitions in the more general setting of compact abelian groups equipped with thenormalized Haar measure.Definition 1.1 Let G be a compact abelian group. If f,g E L1(G) and .4(7) > (7)1 forevery 7 in a", then we say that g is a majorant off (or that g majorizes f). Furthermore,if"a(7) = 'f(y) I for every y in 6, we say g is the exact majorant off.Hardy and Littlewood observed that when p is an even integer and g is a majorantof f, then 11/11p < 11911p• However they showed that this inequality was not true in generalby exhibiting trigonometric polynomials f and g with the properties that g majorizes fbut 11f 113> 119113. Could it be possible though, that whenever g majorizes f,Ilf Ilp 5 AplIglipfor some number Ap dependent only on p? Boas [2] proposed saying that the Banachspace LP(T) would have the "upper majorant property" in this situation.Definition 1.2 Let G be a compact abelian group. The space LP(G) is said to have theupper majorant property (with constant Ap) if there exists a constant Ap such thatIlf II,,^AplIgIlpwhenever g is a majorant off.Chapter 1. Introduction.^ 3Also of interest to Hardy and Littlewood was what Boas called the "lower majorantproperty."Definition 1.3 Let G be a compact abelian group. The space L(G) is said to have thelower majorant property (with constant Aq) if there exists a constant Aq such that everyf E L(G) has a majorant g E L(G) for whichligliq 5 AqlWhile Hardy and Littlewood were unable to give confirmation as to whether the LP (T)spaces have the upper majorant property or not (with the exception of p even), they didshow that if LP(T) has the upper majorant property, then 11(11') has the lower majorantproperty when p1 + q1 = 1 (and hence they proved the nontrivial result that D(T) hasthe lower majorant property for q = p' = (2k)' = 2k I (2k — 1), k E N). Boas [2] showedthat the converse was true. This is part of what we will call the duality theorem.Theorem 1.4 (Duality) Let G be a locally compact abelian group. For 1 < p < oo, thespace LP(G) has the upper majorant property if and only if Lq(G), where p1 + q1 = 1,has the lower majorant property with the same constant.Here the definition of majorant has to be extended appropriately to locally compact abelian groups. Different cases of the theorem were proved by various people. Forinstance, Bachelis [1] proved that the duality holds with the same constant on compact abelian groups. Rains [21] proved the "if' part of the theorem, while Lee andSunouchi [17] proved the "only if' part for the locally compact abelian groups. Furthergeneralizations to Banach spaces satisfying certain conditions were made by DechampsGondim, LustPiquard, and Queffelec [7].The duality theorem allows us to now focus on just the upper majorant property.Bachelis [1] using an argument shown to him by Y. Katznelson was able to show thatE f(7)7(x)yEaChapter 1. Introduction.^ 4if LP(T) had the upper majorant property, then it had the property with unit constant.Boas [2] had earlier generalized Hardy and Littlewood's counterexample on L3(T) toprove that LP(T) did not have the upper majorant property with unit constant when p isgreater than 2 and not an even integer. Hence Bachelis had proven that for the same p's,the space LP(T) does not have the upper majorant property. Using the same lines ofargument, Fournier [9] generalized this result to all infinite compact abelian groups G,showing that LP(G) has the upper majorant property only if p is an even integer, orp = oo.It is natural to ask what happens if G is a finite abelian group. Of course such groupswill always have the upper majorant property regardless of p because if the order of Gis n, and if 11(7)1 < L,(y), thenII! II =  E if(x)iPn xEG= g (0)PE Ig(x)IPxEGSo 17(G) has the upper majorant property with constant n1/P. On the other hand thevalidity of the upper majorant property with unit constant is not trivial. For any fixedvalue of p that is not even, the methods used in [9] show that this property fails on finiteabelian groups with large enough order.Chapter 1. Introduction.^ 5The results for the upper majorant problem were extended to noncompact locallycompact abelian groups by Rains [22] and by Lee and Sunouchi [16] in the special caseof tP.Theorem 1.5 Let G be a noncompact locally compact abelian group. Then LP(G) hasthe upper majorant property if and only if p is an even integer or p = oo.Of interest here is a method used to prove part of this theorem, employed independently in the thesis of Rains [21] and in the paper of Lee and Sunouchi [16] which wasprompted by the work of Civin [5]. They both exploited a characterization of the functions that operate on positive definite sequences. For instance, Lee and Sunouchi showedthat if tP(Z) has the upper majorant property with unit constant, then Fp(z) = lz IPlsgn zoperates on the positive definite sequences, that is, the composition Fp 0 0 is positivedefinite whenever 0 is a positive definite sequence. Rudin [24] had characterized therealvalued functions on the interval ] — 1, 1[ that operate on realvalued positive definitesequences as being those of the form00F(x) = E cne, cn > o.n=0Since the restriction of Fp(z) to ]1, 1[ only has this form when p is even, Lee and Sunouchihad a proof that the upper majorant property with unit constant failed in EP(Z) wheneverp was not even. It again followed that the upper majorant property also failed in thesecases.The connection between the upper majorant property and functions that operate onpositive definite sequences is enlightening. The earlier results in [9] on the failure of theupper majorant property with unit constant could also have been proved by this method.It is worth noting that in Shapiro's paper [27] on majorant problems there was a remarkin which he referred to the work of Rudin [24], Herz [13], Rider [23], and Graham [10] onChapter 1. Introduction.^ 6functions that operate, but that comment was not made in regards to the upper majorantproperty itself.Actually, a stronger connection between the two topics will be shown in Chapter 2,namely that LP(G) has the positive upper majorant property with unit constant if andonly if the function Fp(z) = I zIPlsgn z operates on the positive definite functions on G.By the "positive upper majorant property with unit constant" we mean the version ofthe upper majorant property where, if f, g E LP(G) with g a majorant of f and if f hasnonnegative Fourier coefficients, then lifIlp liglip.We are overdue for some simpler terminology. The results of Fournier and Rainsmentioned above tell us that on infinite locally compact abelian groups, LP(G) has theupper majorant property if and only if p is an even integer. However, when p is an eveninteger, we really have the upper majorant property with unit constant. This along withthe triviality of the upper majorant property for finite groups suggests that what weshould be studying is the upper majorant property with unit constant.This name is so cumbersome that we will change to the names that were given to thelower and upper majorant properties when generalizations to the certain Banach spaceswere studied. Oberlin [19] said a space B has the "majorant property" if for every x E Bthere exists some X E B satisfying I±(n)1 < X(n) for every n. Bachelis had shownthat this is equivalent to the lower majorant property in the LP(G) setting. DechampsGondim, LustPiquard and Queffelec [7] also used the term "majorant property" with thismeaning, and they defined what they called the "minorant property" on certain Banachspaces. In the LP(G) setting the latter property is equivalent to the upper majorantproperty. Later, DechampsGondim, LustPiquard and Queffelec [8] again used the termminorant property for matrices but changed the meaning slightly by defining it to bethe upper majorant property with unit constant. Hence, in like manner we make thefollowing definitions.Chapter 1. Introduction.^ 7Definition 1.6 Let G be a compact abelian group. We say that LP(G) has the minorantproperty if for every 1,9 E LP(G) with g a majorant off,Ilfilp^Ilgilp.Definition 1.7 We say that the space LP(G) has the positive minorant property if forevery f, g E LP(G) with 0 < f(y) < "(y) for all y E 6, then111111^lIglIp.With this terminology, we can restate the link between the upper majorant propertyand functions that operate by saying that LP(G) has the positive minorant property ifand only if Fp(z) = I zIPIsgn z operates on positive definite trigonometric polynomials.As a consequence of this, in Chapter 2 we will easily prove a new result, that if LP(G) hasthe positive minorant property, then L"2(G) has the positive minorant property. Also,in Chapter 4 we use this to prove that the positive minorant property holds on somelow order groups, by showing that Fp(z) operates on the positive definite trigonometricpolynomials. In doing so, we are able to add to the examples of Rider [23] and Graham[10] (or see Graham and McGehee [11] p. 278279) of functions that operate on positivedefinite functions on some groups but do not come under the classifications that havebeen studied thus far. In Chapter 3 we look into the matter of when we can say theminorant properties fail to hold on LP(G) by finding specific counterexamples.1.3 Majorants and Matrices.As mentioned earlier, the majorant problem can be considered on certain Banach spaces([19], [7]). One such Banach space is (Mn(C), li • lip), the set of n x n matrices withcomplex entries equipped with the p norm ii 'Hp. Any notation used that is associatedChapter 1. Introduction.^ 8with matrices and which is not explicitly defined will follow Horn and Johnson [15]. For1 < p < oo, define the p norm of A E M(C) by11Allp = (tr [(A*A)i]) .Alternatively if ai (A) > 02(A) >^> o(A) > 0 denote the singular values of A, then11 A l l p = (En u (A )P) i=1For A E Mn(C), define IA) to be the entrywise absolute value of A, that is, the matrix(Jaij J). For A, B E Mn (11Z), write A < B or B > A if and only if ai < bi j for i = 1, . . . , nand j = 1, , n. In particular, A > 0 means that all the entries of A are nonnegative.We warn the reader that while this notation is consistent with Horn and Johnson, itdiffers from the convention in operator theory where lAl would denote the matrix fie—V—`A.We also differ from DechampGondim, LustPiquard and Queffelec [8] who no doubt wishto emphasize the parallelism with LP(G) and so write A < h.' to mean a j <Definition 1.8 If A, B E M(C) with IAI < B, then we say B is a majorant of A (orthat B majorizes A).Definition 1.9 We say that (Mn(C),11 • lip) has the minorant property if Mb^liBlipwhenever B is a majorant of A.As is usual, it is easy to prove that the minorant property holds when p is an evenpositive integer. That is, if p = 2k, where k E N and if lAl < B, then IA*Al < B*Band thus 1.24*Alk < (B*B)k so thatMAU,^tr RA*A)tr RA*A)k]tr [(B*B)k]•Chapter 1. Introduction.^ 9Feller [20] showed that for certain infinite matrices (the compact operators on £2)the minorant property fails when p is not a positive even integer. This conclusion isthus analogous to the one for LP(G) spaces when G is an infinite locally compact abeliangroup. As with groups, it is interesting to study what happens in the finite cases, thatis, for n x n matrices. Simon [28] using matrices based on the Boas [2] counterexample,proved that if p is not an even integer and n = 2[p/2] + 5, then (Mn(C), ii dip) does nothave the minorant property. Letting N(p) denote the smallest n for which the minorantproperty fails, Simon had shown thatN(p) < 2 [] + 5.DechampsGondim, LustPiquard and Queffelec [8] improved on this by showing (Theorem 1)N(p) < 2 [] +2,or that (Mn(C), II • lip) fails to have the minorant property for noneven 1 < p < 2(n — 1).The general tendency appears to be that for a fixed positive noneven p, the larger thematrices we consider, the more likely it is that the minorant property will fail, or thatfor a fixed n, the larger a noneven p is, the more likely (Mn(C), ii • 11p) is to have theminorant property. A specific example of this tendency was demonstrated by DechampsGondim, LustPiquard and Queffelec [8] for 3 x 3 matrices. Their Theorem 1 showed(M3(C), II II) does not have the minorant property for 1 < p < 4, except when p = 2,while their Theorem 2 said (M3(C), If • lip) has the minorant property for all p > 4. Wewill consider the positive minorant property on matrices.Definition 1.10 The space (Mn(C),Vilp) is said to have the positive minorant propertyif 11Allp < liBlin whenever 0 < A < B.DechampsGondim, LustPiquard and Queffelec showed that the positive minorantChapter 1. Introduction.^ 1 0property on (Mn(C), 11 • 11) is equivalent to the condition that (B*B)(P/2)1B* _> 0 whenever B > 0. This is the matrix counterpart of what we will prove for LP(G) spaces whenG is a compact abelian group. Theorem 4(b) of [8] says that: For n even and n > 4,the space (Mn(C), 11 • 11p) does not have the positive minorant property if 1 < p < n — 2and p 0 2k. For n odd and n > 5, the space (Mn(C), 11 ' 11p) does not have the positiveminorant property if 1 < p < n — 3 and p 0 2k. In Chapter 5 we will improve this usingan entirely new method to show that (Mn(C), 11 • 11) does not have the positive minorantproperty if 0 < p < 2(n — 1) and p 0 2k. Our method actually shows that the positiveminorant property fails in the subspace of symmetric matrices in (M (C), 11 ' 11p)Chapter 2General Facts About Minorant Properties.For this chapter, G will denote a compact abelian group. We present some apparentlyweaker but as it turns out, equivalent conditions to the (positive) minorant property.Moreover, we show that the positive minorant property on LP(G) is equivalent to thefunction Fp(z) =iziPlsgn z operating on the positive definite trigonometric polynomials.Also, we prove some theorems of the type: LP(G) has the (positive) minorant propertyimplies Lr(G) has the (positive) minorant property for suitably related p and r.2.1 Reductions.We first show that it suffices to have the (positive) minorant property hold for trigonometric polynomials, with the LPnorm, in order for it to hold on the whole LP(G) space.We let Trig(G) denote the space of all trigonometric polynomials on G.Proposition 2.1 For 1 < p < oo, the following are equivalent.(a) The space LP(G) has the minorant property (positive minorant property).(b) For all f, g E Trig(G) with g a majorant off (with 0 5_ J(7) :4(7) for all y E a),IlfIlp^lIglIp.Proof. Obviously (a) implies (b). Conversely, suppose f,g E LP(G) with f majorizedby g (for the positive minorant version of the proof i(y) > 0 for all y E as well). Takean approximate identity of trigonometric polynomials MI with /5:„ > 0 (see [14] page 88,1 1Chapter 2. General Facts About Minorant Properties.^ 12Theorem 28.53 and in particular note (i) and (v) for existence of such an approximateidentity). Then there is a sequence {Pn} which is a subset of {Pa} for which Pn * f —* fand Pn * g 4 g in 17norm (see [14] page 273, Remark (b) of 32.33). So for each E > 0there exists N such that n > N impliesIlf —Pn*fllp < Elig — Pn * gill, < E.Also(Pn* g)"ey) = Pneyrgey)? 1371(7)1/(7)1 as .0(7) . If(7)1= l(Pn * fY(7)1which implies that IIPn * glip > iiPn * fl1p since we have the minorant property fortrigonometric polynomials. HencelIglIp — IlfIlp = 11 911p — IIPn *gIIp — ( 1 1fIlp — I IPn* PIP) ± IIPn * glIP — IIPn* PIP> —E. — E. ± 11Pn* AP — 11Pn * fllp> —2s.As E is arbitrary, we get ilglIp — Ilf Ilp > 0 as required. For the positive minorant property,remove I • 1 in this argument. 0The proposition above is stated explicitly in Rains [21] for the minorant property andwas used implicitly in Hardy and Littlewood [12]. The following theorem is a generalization of Hardy and Littlewood's Theorem 2. Recall that g is the exact majorant of f if:0(7) = 1 j(y)I for all 7.Chapter 2. General Facts About Minorant Properties.^ 13Definition 2.2 We say LP(G) has the exact minorant property if for all f,g E LP(G)with g an exact majorant of f, theniiffip 5_ Nip.Theorem 2.3 Let G be a compact abelian group. For p > 1 the space LP(G) has theminorant property if and only if LP(G) has the exact minorant property.Proof. Trivially, the minorant property implies the exact minorant property.Suppose LP(G) has the exact minorant property. To prove LP(G) has the minorantproperty, Proposition 2.1 says it suffices to prove that (T ri g (G) , II• 11) has the minorantproperty. Let f,g E Trig(G) withfor all y E O.Let E = {y EÔ : :g(7) > 0}. The set E is finite, as g is a trigonometric polynomial. Foreach y E E, let a7 be the unique number in [0, 7/2] for whichcos c 7 — li(7)1 7 — .07)^(2.1)and if y E, put al. = 0. Also, when j(y) 0 0, let 07 be such that (7) =otherwise let 07 = 0 .Define k1 , k2 E Trig(G) such thatkl(7) = :9(7)ei(°^I±a7).1c2(7) = .4(7)e' ^for all y EThenki (7) + k2(7) = :0 (7)(e1(ef +ao + ei(e,"))= .4(y)[cos(07 + ay) + cos(97 —^i(sin(07 + a7) + sin(07 — a7))]1 h(7) if 7 E EhE(7) =0 otherwise.Chapter 2. General Facts About Minorant Properties.^ 14= .4(7) [2 cos (1.), cos ay + 2i sin 07 cos ad= 2.4(7) cos ayei°'= 2/(7) by (2.1).So ki (7) 4 .k2 (7) f (7) =^2andkJ. + k2f  ^ .2ThusiII f Ilp =^Illci + k21Ip1^15_PIG+ 11k21Ip by the triangle inequality2^21 1—blip ± —11.911p2^2^as g is an exact majorant of k1 and k2= kipand hence LP(G) has the minorant property. 0Notation. Let E be a finite subset of d and h E Ll(G). We define hE to be thetrigonometric polynomial for whichTheorem 2.4 The space LP(G) for 1 < p < co has the positive minorant property if andonly if for every h E LP(G) with 1(7) > 0 for all 7 E a and every finite subset E of awe haveIlhEllp < IlhIlp.Chapter 2. General Facts About Minorant Properties.^ 15Proof. If LP(G) has the positive minorant property and h E LP(G) and E any finitesubset of 6, then hE is such that0 _< ri;(y)^h('y)for all y E Ô and so IlhEllp < Ilhilp.Conversely, suppose that for every h E LP(G) and E a finite subset of Ô we haveIlhEllp^11h11,Then by Proposition 2.1, to prove LP(G) has the positive minorant property it sufficesto prove that whenever f, h E Trig(G) and0 < (y) < it(7), then II/11p 5 In.Let E = {y e^: h('y) > 0}. Label the elements of E so that if E = {71,y2,^,ym}and we define1(71) ai =then ai < a2 <^< am. We may assume that am = 1, since if am < 1, we can replaceh with anzh and still have that 0 < j('y) < it(y) for all y E O.Define subsets Ei of E so thatEt^7i+11 • • • 7 7n1}.We show that it is possible to choose real numbers Ai > 0 with Ell,. Ai = 1 andI = E71_1 AihEs. In fact, Ai = ai — ^where we define ao to be 0. Indeed, if=^aii)hE„ then"(7/c) = E(ai —i=1E(ai — ai1)11(7k)1=1Chapter 2. General Facts About Minorant Properties.^ 16= Ral — ao) + (a2 — ai) +^+ (ak — ak1)111(7k)s'Yk,= )= 1(7k),so that = f and g = f. Note also that Er_i Ai = Urn = 1. Finally,IlfIlp = E AihEi1=1E Ai ilhEi lipi= 1 for p > 1EAdhllpi=1= Ilhilp. 0by assumptionThe theorem above is a generalization of Hardy and Littlewood's Theorem 4. Thenext theorem uses a generalization of a Hardy and Littlewood argument, where theyshowed that since L2(T) has the minorant property, so does L2k(T) where k E N.Theorem 2.5 Let p > 1. If LP(G) has the (positive) minorant property then LnP(G)also has the (positive) minorant property for all n E N.Proof. Suppose f,g E LnP (G) with 1/(7)1 < j(7) for all E Ô (or 0 f(7) ...4(7) forthe positive minorant property). Then fn , gn E LP(G) and (remove I • I throughout toprove the positive minorant property),11n(7)1 = li* • • • * i(7)IIta • • • fa kr^— A2 — . • • — An_oi(An_i) • • • .f(Ai) dAni • • • dAil< fa... fa 'kr — A1 — A2 —^— An1)1If (An1)I^If (A1)IdAni^dAl<^. . . fa :0(y—^— A2 —^— An1).4(An1) • • • .4(Al)dAn1^dAl=^*^* 4(7) (convolution of n terms)=Chapter 2. General Facts About Minorant Properties.^ 17As LP(G) has the (positive) minorant property, it follows thatand sooras required. 02.2 The positive minorant property.The derivative of F(t) = II f + tyllp where f E Trig(G) and y EÔ is computed inLemma 2 of Hardy and Littlewood [12] for G = T and also in the proof of Lemma 2in Bachelis' paper [1] for compact abelian groups. We will compute a little more, thederivative of F(t) II f thilp where f,h E T ri g (G) as this will be required for the nexttheorem. First though, we borrow some notation from Bachelis.Definition 2.6 For! E LP(G), with 1 < p < oo and q so that p1 q1 = 1, letAq(f)(x) = fix)IP1 sgn f (x) ffor all x E G, wherezilzi if z^0sgn z =0 if z = 0.Note that Aq(i) E Lq(G), that liAq(f )iiq = 1, and that fG f (x)Aq(i)(x) dx = II/11p.Lemma 2.7 For f,h E Trig(G), define the realvalued function F(t) = JJJ + thlip fort E R. The function F is continuous and differentiable when 1 <p < co andF'(t)^fa Re [Aq(f thr(y)h(y)] dy.Chapter 2. General Facts About Minorant Properties.^ 18Proof. The function F is continuous since for real numbers s and t,IF(t)  F(s)I = 1111 + thllp — Ill + shllp I11(i + th)  (f + sh)lip= il(t  s)hiip= It — sl IlhIlpSo i t — si < 6/11 hIlp implies IF(t)  F(s)I < 0.As1F(t) = (LI f (x) + th(x)IP dx) P ,we have1I'M = —1If + thr—P—d I If (x) + th(x)IP dx ,P^P dt Gand provided (d I dt)I f (x) + th(x)IP is continuous we may differentiate under the integralsign (G is compact).SinceIf (x) + th(x)IP = ((1(x) + th(x))(f (x) + th(x))) ,^lif (x) + th(x)IP =^((1(x) + th(x))(i (x) + th(x))) lx (h(x)(f(x) + th(x)) + (f (x) + th(x))h(x))=^If (x) + th(x)IP2 (h(x)(f(x) + th(x)) + (1(x) + th(x))h(x))P. 2 I f (x) + th(x)IP 1x (h(x)sgn (f (x) + th(x)) + sgn (1(x) + th(x))h(x))= l' Ilf + thr (Aq(f + th)(x)h(x) + Aq(/ + th)(x)h(x)) ,which is continuous on G x R since f, h E Trig(G) and which is welldefined if p> 1.Hence1 ,F' (t) = HI f + Mr ^ —I d I f (x) + th(x)IP dxp^P^G dtChapter 2. General Facts About Minorant Properties.^ 19. ; 1G (Aq(f + th)(x)h(x) + Aq(f + th)(x)h(x)) dxAq(f + th)(x) jaky)y (x) dy dx +..2 JG Aq(f + th)(x) fait(y)y (x) dy dxas h(x) = fa ii(y)7(x) dx and this is in fact just a finite sum as h E T rig (G) . Thus wemay interchange the order of integration to get,f, LAq(f +th)(x)7(x) dxh(y) dy + fa. IGAg(f 1th)(x)7(x)dxit(7)dy11 ^=^ja A q( f + th)" (y)hey) + Aq( f + th) (7)y) dyLRe(Aq(i fth)^(7)it(7))&y. 0Remark (a) The conclusion of the lemma also holds for the one sided derivative F' (e)when p = 1. The proof now requires the use of the dominated convergence theorem.Remark (b) The conclusion of the lemma will hold for 0 < p < 1 provided f + th isbounded away from zero. As f and h are continuous and G is compact, it then sufficesthat f (x) + th(x) be nonzero for all x E G for the conclusion of Lemma 2.7 to hold.We now prove the main theorem in this chapter. It shows that asserting the positiveminorant property on LP(G) is equivalent to saying that the functional Aq(•) operates onthe positive definite trigonometric polynomials.Theorem 2.8 Let G be a compact abelian group. Then LP(G) with 1 < p < oo hasthe positive minorant property if and only if Aq(f)" > 0 whenever f is a trigonometricpolynomial with f > 0.Proof. Suppose LP(G) has the positive minorant property and let f E Trig(G) withI > 0. Putting g = f + tx for some x E d, we have that g majorizes f when t > 0 andsoII/11p 5_ Ilf + tx11p.F' (t) =Chapter 2. General Facts About Minorant Properties.^ 20Let F(t) = If + txlip By Lemma 2.7, the function F is differentiable andF'(t) = LRe(Aq(f Ftx)(7)5,(7),dy= Re [Aq(f + tX)(X)](for proof when p = 1, use Remark (a) above and read "F'(t)" as "F'(t+)" in whatfollows).Now because j > 0, in particular j is real, and therefore f (—x) = f (x) for all x E G.Thus Aq(f)(—x) = Aq(f)(x) for all x E G, and therefore Aq(f)^ is real. SoAqUY (X) = Re(Aq(1)(X))= F'(0).hill Ilf + txllp — Ilfilp t4o^t> 0 since Iv + txllp — IlfIlp > o for t > O.Conversely, suppose Aq(•) operates on positive definite trigonometric polynomials. Toshow LP(G) has the positive minorant property, it suffices to show that it has the propertyfor trigonometric polynomials by Proposition 2.1.Let f, g E Trig(G) with 0 < f('y) < .4(7) for all 7 E 0 and defineF(t) = Ilf + t(g — Plip for t E [0,1].By Lemma 2.7, the function F is differentiable andPm = fa Re [(Aa + t(g — .f))(7)] (9 — f)(7)d7> 0since (g — f(y) > 0 and (f Ft(g — f))(y) ..> 0, so that Aq(9 + t(g — f))(y) > 0 by theassumption that Aq(•) operates on positive definite trigonometric polynomials.Chapter 2. General Facts About Minorant Properties.^ 21Hence,lIglIp  Ilf Ilp^F(1) — F(0)F'()(1 — 0) for some 4" E]0, 1[, by the mean value theoremF' (e)>0and so If Ilp < lIglIp as required. 0Remark. When 0 < p < 1, the first half of Theorem 2.8 still works for trigonometricpolynomials f with no zeros (see Remark (b) following the proof of Lemma 2.7). Thusthe positive minorant property on LP(G) implies that Aq(f) > 0 for all nowherezerotrigonometric polynomials f with j > 0.Corollary 2.9 Let G be an infinite compact abelian group. Then LP(G) does not havethe positive minorant property for noneven p > 1.Proof. The proof divides up into two cases, when G is a nonexceptional group andwhen it is exceptional. Any locally compact abelian group is exceptional if it is the directproduct of a finite group with exponent greater than two and a group of exponent two.In Theorem 1.2 of [23], Rider shows that functions that operate on the class of positivedefinite functions on a nonexceptional group must be infinitely differentiable. When pis not even, the function Fp(z) = Izi sgn z is not smooth at zero, and therefore doesnot operate on the class of positive definite functions. By the theorem above, LP(G) doesnot have the positive minorant property.If G is an exceptional group, then it contains a subgroup isomorphic to Z3") becauseG is infinite. Hence the generalization of Theorem 4 of Herz [13] applies. That is, anyfunction defined on the interval [ —1, 1] which operates on the positive definite functionsis smooth near zero. So again, Fp cannot operate on the class of positive definite functionsChapter 2. General Facts About Minorant Properties.^ 22defined on G when p is noneven and so the space LP(G) doesn't have the positive minorantproperty. 0Another consequence of Theorem 2.8 is the following.Theorem 2.10 If LP(G) has the positive minorant property for some p with 1 < p < oo,then so does LP+2(G).Proof. As LP(G) has the positive minorant property, the theorem above says Aq(fr > 0whenever f E Trig(g) and .1. > 0. NowII f 1113;1so that Fp(z) = I zIPisgn z operates on P(G) = If E Trig(G) : i > 01. We claim thatF+2(z) = IzIPFisgn z = IzI2Fp(z) = z..h'p(z)operates on P(G) also. This is because if q5 E P(G), then q5, q and Fp(0) are positivedefinite (the latter because Fp operates on P(G)) and the product of positive definitefunctions is positive definite, that is, if 0, 0 E P(G) then(Ye) == jaisx7—x),Rx)dx> . as 0,0 E P(G).Thus F+2(z) operates on P(G) implies A(p+2),(fr > 0 for all trigonometric polynomialsf with f > 0. So by Theorem 2.8 this means that LP+2(G) has the positive minorantproperty. 0)tq(f)(x) = if (x)IP1 sgn 1(x) Chapter 3Counterexamples.Let G be a finite abelian group. The focus of this chapter will be to determine some11(G)) spaces for which the minorant or positive minorant property fails to hold.3.1 Groups and matrices.The work of Fournier [9] and Rains [22] has shown that for an infinite locally compactabelian group G, the space 11(G)) has the minorant property if and only if p is an eveninteger or is infinite. However, the story for finite groups is quite different. For example,in Chapter 4 we will show that for Z2, the group of order 2, the minorant property holdsfor all p > 2.We get the same type of behaviour for matrices. If cp(e) denotes the Banach space ofall compact operators C on 12 for which the singular values of C belong to tP, Peller [20]showed that Cp(e2) fails to have the minorant property for noneven p > 1. On the otherhand, DechampsGondim, LustPiquard, and Queffelec [8] have shown that the space(M3(C), 11'11p) has the minorant property for all p > 4.We will give more consideration to the space (Mn(C),11 • 11p) in Chapter 5, but itshould be noted that whenever D '(G) fails to have the minorant or positive minorantproperty for a finite abelian group G of order n, then (M (C), ' 11) also fails to havethe property. This is because we can embed 11(G) into (M„(C), li ' lip). Simon [28] usedthis fact for cyclic groups. Before proving that there is such an embedding in general,first consider a couple of examples.23Chapter 3. Counterexamples.^ 24Example 1. Let Z4 = {0, 1, 2,3} and let= {1,7,72,73 7(0) =1,7(1) = i,7(2) = 1,7(3) = i}.For each f = a + by + cry2 + dry3 E LP(Z4), define T: L7'(Z4)^(M4(C),II • IIp) such thatT(f) =ladcb\badccbad\d c b alExample 2. Let4 = {1,x, y, xy : x2 = y2 = 1} and222 = 11,7,X,7X 7(x) = —1,7(0 = 1,x(x) = 1,X(Y) = —11.For each f=a+ fry cx ctyx, define T : LP(4) + (M4(C), II 11p) such thatla b c d\T(f) =badcd a b\dcbaIn both cases / < .4 implies that T(f) < T(g). In the proof of the next proposition weshow that 11T(i)11C, = nilf II; so that^blip implies IIT(f)IIp^IIT(9)11pProposition 3.1 Let G be a finite abelian group of order n. There exists a mappingT : LP(G)^(M„(C),II lip) which is injective, preserves order among norms, and if1/(7)I^.07), thenIT(f)I^T(g).Chapter 3. Counterexamples.^ 25Proof. Let y E 6 and define the operatorT7 : Trig(G) Trig(G)to be such that, Ty( f) = 'yf, for all f E LP(G) = Trig(G); that is, Ty ( f)(x) = 7(x) f (x)for all x E G. The operator T7 is linear and so choosing the basis O = {71, , NI we cancompute the matrix of Ty, denoted by m(T7). As Tai = yyi E 6, the ith column of thematrix of Ty has all entries equal to zero except for one which is 1. Also, as Tyy1 = T77;if and only if i = j, the matrix of T7 is a permutation matrix. Moreover, if y, x E 6 wehave Tryk = Txyk if and only if y = x, so for i and j, m(T7)1i = 1 means m(Tx)ii = 0when 7 x. Thus the mappingT : Trig(G) M(C)defined byT(i)(E^f(7)y)7Ea=^f(7)Tri(Tr)7EGis injective. Also, if for f, g E Trig(G) the function g is a majorant of f, then clearlyT(g) is a majorant of T(f) in the matrix sense.It remains for us to show that the order among norms is preserved under the mappingT. Before we can do this, we need to know what the eigenvalues and eigenvectors of m(T7)are (this is done in Theorem 3.2.2 of [6] for circulant matrices, which is what m(T7) andT( f) are when G is a cyclic group). In fact, if G = , g,}, then the eigenvalues ofm(T7) are the diagonal entries inAry = diag (7(g1), • • • 7(gn))Chapter 3. Counterexamples.^ 26and with corresponding eigenvectors being the columns ofX = (xii) = 1Note that X is independent of y so that all the matrices m(T7), for y E 0, are simultaneously diagonalized by X. To prove these claims we need to show thatm (T7)X = XA7.Now^(771(Ty)A j =^m(T)ikxkik=1= Xt where "Y'Yi = "Ye1^\Ft•^71(gi)1^VT^17(gi)Yi(gi)= y(gi)xii• (XA7)ii.Also X1 = X* since^(XX*)2^=^17,^'Yz (gle )73 (gk)— k=1=We computeX1T(f)X = X1 (E ky)m(71y)) X7E5= E hy)X1m(T7)X,E aChapter 3. Counterexamples.^ 27= E i(7)A7^yEa= E /(7)diag (7(gi), • • •,7(gn))yEa= E diag (/(7)7(gi), • • • f(y )7(g))•yEa.diag (E j(7)y(gi),.^yEa•• , E :/(7)7(gn))^yE&= diag (.f(gi), • , f(g)).Taking the conjugate transpose of the last equation givesX*T(f)* X = diag^f (gn)).Thus X*T(f)*T(f)X = diag (gi)12,^,If(gn)12) so that the singular values of T(f)are If (m)1,^ If (gn)I andIIT(f) IIand thus T preserves order among norms. 03.2 Cyclic groups.It is known [25, p. 255, Theorem B8} that every finite abelian group is a direct sumof cyclic groups. Rains [22] proves the following proposition, which also holds for thepositive minorant property.Proposition 3.2 Suppose that G1 and G2 are locally compact abelian groups and supposeeither LP(G1) or LP(G2) fails to have the minorant property. Then LP(Gi x G2) does nothave the minorant property.Chapter 3. Counterexamples.^ 28Hence a good place to start in finding LP(G) spaces which do not have the minorantproperty is with the cyclic groups. The following theorem is a modification of Theorem 1(a) of DechampsGondim, LustPiquard and Queffelec [8].Theorem 3.3 For p > 0 and noneven, the space LP(Z) does not have the minorantproperty for n = [p12] ± 2. Equivalently, for fixed n, the space LP(Z) does not have theminorant property for 2(n — 2) < p < 2(n — 1).Proof. Suppose n = [0] + 2 and let k and m be nonnegative integers such thatn = 2km, where in is odd.^ (3.1)Putc = e2 k7ri^(3.2)and choosing a value of t E]0, 1[ and any 7 E Zn which is element of order n, letf = V1 — t2 + tcy ,g = 1/1 — t2 + try .(3.3)Then clearly g is a majorant of f and we will show that IlfIlp > lIglIp.We will need to compute the 1,13 norm of a function of the form h = V1 — t2 + tuwhere lui 17 1. NowTh = (V1 — t2 + ti7)(V1 — t + tu)= 1 — t2 + tV1 — t2(u + rt) + t2^(3.4)= 1 + p(u + ri)wherep = tV1 — t2 < 1/2^ (3.5)Chapter 3. Counterexamples.^ 29if t < 1/2 and thus plu + iI < 2p < 1. Thus▪ —1 I Crth)P/2dX71 G= —1 I (1 p (u ii))P/2 dx.^ (3.6)n GRecall that for a real number a and an integer s,(as) = 14 (ai=oand(1 + z) =^(a)zss=o ■s/always converges whenever Izi < 1. Hence, due to (3.5) we can writeoo(1 ± P(tt +17))a = Es=owhere(lps(u + T). (3.7)a =1.. (3.8)Note that as p is noneven,(a) (a)())( a— 1 > 0,but(a) < 0n)(3.9)by our choice of n = [p/2] + 2, andsgn =I)n + r for r > 0. (3.10)Consider^11g11';,  11f1';, =^f [(1+ p(y + TO — (1+ p(cy + c7))P12] dx using (3.6)• G^=^I^(a)Ps((Y + 7)8 (cT + c7)8)) dx using (3.7)n G s =0 St (a) Ps f^(s.) e7i(7y)sj — (c)i cc—Ty1 dxn s=o s^GChapter 3. Counterexamples.^ 30^(s• a)p^4s (s) (1./. 72i8 — (c7)2jsdx)s=0 _^j=0^n Goo^s^s^P^ (1 — c2i81^21^.;07' (sas) =0 (i)^(n JG 8(x) dx) .We claim this is a sum of negative terms. To see this, first note that1— G 72i8(x)dx = 1n (3.11)• if 2j — s 0 (mod n), and that otherwise the integral is zero since 7 was chosen to bean element of order n. When 2j — s^0 (mod n), there is an integer t such that2j — s = ne.^ (3.12)Then1 — C2i8 = 1 —• 1 — e2knilr1 by (3.2)• 1 — em^by (3.1)• 1 — (1)t as m is odd2 if t is odd(3.13)0 if t is even.So all the terms in (3.11) are nonnegative with the exception of (:). By (3.9), fors = 0,1, ... , n — 1 we have that (:) > 0, but then—s<2j—s<s as 0 <j<sso that—(n — 1) < 2j — s < n — 1,Chapter 3. Counterexamples.^ 31and the only time 2j — s^0 (mod n) is when 2j — s = 0. But then (3.13) says1 — c2js = 0. Thus none of the terms (a9) survive when s = 0,1, ... , n — 1. Also, (3.10)sayssgn( a )= (1Y+1 for r > 0.n + rWhen s = n + r and r is odd, then (a) > 0. For the positive term (a) to survive in thes^ ssum (3.11), we need (3.12) to hold, and then2j — s = ne.If n is even, this says s must be even, but s = n + r is odd so (as) does not appear in(3.11). If n is odd, we need to be odd for the sum to survive because of (3.13), butthen ne is odd so that s = 2j — ne is odd, but s = n + r is even, so that again (:) doesnot appear in (3.11).Thus no positive term ("s) survives. On the other hand, when s = n, and j = s = n,condition (3.12) is satisfied with t = 1, and (3.13) gives 1 — c2i8 = 2, so thatOras required.So, we have shown that LP(Z) does not have the minorant property for n = [51+ 2.Alternatively, for fixed n, the space LP(Z) does not have the minorant property whenn — 2 = [12]2—1 < n — 2 <p — 2 < 2n —4 < p,Chapter 3. Counterexamples.^ 32that is, when2n4 < p < 2n2. 0Corollary 3.4 For noneven p > 0, the space LP(ZN) does not have the minorant propertyfor N = n + 2k, k E {0} U N, where n = [p12] + 2. Alternatively, for fixed N, the spaceLP(ZN) does not have the minorant property for p E] 2(N  2)  4k, 2(N — 1) — 4k [.Proof. Proceed exactly as above except now using N instead of n. Everything is thesame, noting that( a )^( a )sgn^= sgnN + r n+2k+r). (_ 1 )2kEr+1= ( 1)r+1.Alternatively, if we fix N and k E {0} U N, then LP(ZN) does not have the minorantproperty for indices p such thatPN= {2j+2+2k,that is,PN — 2 — 2k = [—2]OrP — 1 < N — 2 — 2k < POrp — 2 < 2N — 4 — 4k < p,and finally2(N — 2) — 4k < p < 2(N  1) — 4k. 0Chapter 3. Counterexamples.^ 33Given a cyclic group of order N, the above corollary gives us intervals of length 2 inwhich LP(ZN) does not have the minorant property, but these intervals for p are separatedby other intervals of the same length in which we don't know anything with regards tothe minorant property. However, Theorem 2.5 can sometimes help to fill in some of thegaps. For example, for LP(Z4), Corollary 3.4 tells us that these spaces do not have theminorant property for p E] 0, 2[ U] 4, 6[. Theorem 2.5 tells us that if ./2(ZN) had theminorant property for p E] 2, 3[, then it would also have it for p E]4, 6[, a contradiction.Hence LP(ZN) does not have the minorant property for p e] 0, 2[U ] 2, 3[ U] 4, 6[.We now turn our attention to cases when LP(ZN) does not have the positive minorantproperty. Theorem 4(b) of DechampsGondim, LustPiquard, and Queffelec [8] saysthat (M (C), II IIp) does not have the positive minorant property for noneven p with1 < p < n — 2 for even n > 4, and that the property fails with 1 < p < n — 3 forodd n > 5. We shall use the same argument as DechampsGondim, LustPiquard andQueffelec, but we show that LP(ZN) (and hence (MN(C), II• Hp)) does not have the positiveminorant property for noneven p with 0 < p < 2[(N — 1)/2].Theorem 3.5 When 0 < p < oo and p is noneven, the space LP(ZN) does not have thepositive minorant property for all integers N > 2[p/2] + 3. Alternatively, for fixed N,the space LP(ZN) does not have the positive minorant property when p is noneven and0 < p < 2RN — 1)/4Proof. Let f^— r2 + ry where 0 < r < 1 and y E ZN is the character for which,y(k) = e2irki/N for k = 0, 1,^, N — 1. Putg = V1 — r2 + rry + tymwhere t > 0 and m = [p/2] + 1.Chapter 3. Counterexamples.^ 34Clearly 0 < I <"j and we will show that IIfIIp > 11g1Ip for some choices of r and t. ByLemma 2.7 and as in Theorem 2.8, for F(t) = Ilf +trmllp,F'(0) = Aq(f)(7"").Since for r small enough, f(k) 0 for all k E ZN, the formula above is true for all p > 0by the Remark (b) following Lemma 2.7.By showing that F'(0) = Aq(f)eym) <0 we will prove thatlIglIp — II/11p = Ill + t7mIlp — II/11p < ofor small positive t. Nowhi, ^+^— II/11p = Aq (f) (m)t40and noting that f has the same form as h appearing in (3.4), and that here p = rV1 — r2then1J fyp12)1 tymAq(freym)G Mir^ dx1 JG (1 + p(y + ry))"3 fym dx where /3= (p/ 2) — 11 when p <nlIfIlr jG [ :t°0 CS) Ps (7 ± 77)3] frym dx^11 ^f [ct (0) ps^(s.)72j ( 1 r2 + ry)ym dxnlifdfir G 3_0 S^_01 nIlf Ilr :E3=o (i3s)Psi=o (si) IG(Vi^272j 3+m ± ni2j8+m+1) dx. (3.14)As 0 = (p/2) — 1 and p noneven,(0)^0(0 — 1) ... (0 — s + 1) > 0) s!for s = 0,1, ... [p/2] but () < 0 for s = m = [p/2] +1. Note that(V1 — r2,72i8+m 7.)(23s+m+1)dx = 0Chapter 3. Counterexamples.^ 35for s= 0, ... ,[p/2] asL yn(x)dx = 0except when n 0 (mod N), but N > 2[p/2]+ 3 whereas1 < 2j — s + m < 2[p/2] + 1because for s < [p/2], and 0 < j < s, we have2j — s+m > —s+m— [p/2]+ [p/2] + 1=1and2j—s+m < 2s—s+m< s + m< [p/2] + [p/2] + 1= 2[p/2] + 1.On the other hand, when s = m = [p/2] +1,1 t`^f^ r272js+j=0^jGr72j8im+1) dxr272.i r72J+1) dx (3.15)+r if N = 2[p/2]+ 3if N > 2[p/2] + 3as y2i^1 in the sum (3.15) only when j = 0, and 72i+1 may be 1 when j = s ifN = 2[p/2] + 3, since then 2j + 1 = 2s + 1 = 2[p/2] + 3. Hence (3.14) becomeslim Ilf + trmllp — IlfIlp — ^1 (le n^— r2m+1 + o (rm+1)t_,0^ IffIlrlChapter 3. Counterexamples.^ 36as p = 7V1 — r2, so that for sufficiently small r > 0 we havelim Ilf + t7mllp  VII,' < 0t_,0^tsince (S) < 0. Hence for sufficiently small t > 0, we havemif + tlirnlip = Ilgiip < Ilf Ilpas required. That is, LP(ZN) does not have the positive minorant property wheneverN > 2[p/2] + 3. Alternatively, for fixed N, the space LP(ZN) does not have theproperty if p satisfiesN — 1 > 2 [123] + 2N1 > [1]+1EN so2[N 2— 1] > [i] + 1 > I) as x — 1 < [x] for all x,and hence^p < 2 [N; 1]. 03.3 Beyond cyclic groups.All of the counterexamples considered so far have been for cyclic groups. Let G be afinite abelian group of order n. If p is a prime number such that p divides n, then Gwill have some cyclic subgroups of the form Zpb where b is a positive integer. Let a(p)be such that Zpa(p) is the biggest such subgroup of G. Letm = Hp. (p) .Pi nThen Zni is isomorphic to the biggest cyclic subgroup of G, and in fact G'''', Zm x G1 forsome group GI. Thus Proposition 3.2 applies so that failure of the minorant propertiesChapter 3. Counterexamples.^ 37on LP(Z,L) will mean failure on LP(G). However, for some finite abelian groups G thereare LP(G) spaces which will not have either the minorant or positive minorant propertybut for which LP(Z,z) has the corresponding property. In this sense finding the biggestcyclic subgroup of G does not provide the best counterexamples.Another way to find counterexamples to the minorant property when G is not cyclic isfound in Fournier's paper [9]. There we have counterexamples of the minorant propertyfor powers of a cyclic group, that is Z. These are not the best possible counterexamples(except maybe when r = 2) as they do not rely on the order of Zr but rather on thepower a. However, for some groups G containing a relatively high number of copies ofa cyclic group, the method of Fournier may produce better counterexamples than thoseprovided by finding the biggest cyclic subgroup of G.Chapter 4Special Cases.In the previous chapter we discussed counterexamples to the minorant and positive minorant properties for LP(G) when G is a finite abelian group, principally for cyclic groups.Of more interest though are the cases where the properties do hold; unfortunately thisseems to be a difficult problem. To illustrate, note that to date the question about whenthe minorant property holds for matrices equipped with the pnorms has only been answered for 3 x 3 matrices [8]. As the size of matrices or the order of groups increases, itbecomes increasingly more difficult to determine exactly when the properties hold. Onthe other hand, as we consider the counterexamples from the last chapter along withsome of the positive results we can find for loworder cases, a definite pattern seems toemerge in the relation between p and the order of the group or size of the matrix beingconsidered. Before we make a conjecture about this pattern, we present some results ona few loworder groups and matrices.4.1 Minorant properties on loworder groups and matrices.Proposition 4.1 For 0 < p < oo, the spaces LP(Z2) and (M2(C),ii ' 11p) have the minorant property if and only if p > 2.Proof. When 0 < p < 2, Theorem 3.3 tells us that LP(Z2) does not have the minorantproperty as [p12] + 2 = 2. By Proposition 3.1, the space L7'(Z2) can be embedded in(M2(C), ll ' 11p), so that (M2(C), II • 11p) does not have the minorant property either.38Chapter 4. Special Cases.^ 39Conversely, if p > 2, it follows from this embedding that to prove the minorantproperty on LP(Z2) it will suffice to prove it on (M2(C), • Hp). To do this we follow thesame lines of argument given by DechampsGondim, LustPiquard and Queffelec [8] intheir Theorem 2 for 3 x 3 matrices.Let A,B E M2(C) with IAI < B. Using the same type of construction used inTheorem 2.3, we may assume IAI = B. Let Al > A2 and respectively pi > //2 denote theeigenvalues of A*A and B*B. We wish to prove that 104 < II BII, in other words thattr ((A*A)P1'2) < tr ((B*B)1)/2), orÁT/2 ± AP/2^P/2 , p/22^Pi I 1/2 . (4.1)Let g be the continuous function defined by^g(r) A7; + A; —^— it;for all r> 0.Suppose that for some p> 2 the minorant property fails to hold, that is (4.1) is false.Then for q = p/2^g(q) > 0.^ (4.2)Let v be an integer for which v > q. As (M2(C), 11 112) and (M2(C), II • 112,) have theminorant property, condition (4.1) holds andg(1) < 0, g(v) < 0.By the intermediate value theorem there exists q' in the interval ]q, v] for which^g(V) = 0.^ (4.3)Since IA1 = B, if follows that tr (A*A) = tr (B* B) so that IIA113 = IIBII3 and+ A2 = P1 + P2^ (4.4)Chapter 4. Special Cases.^ 40so thatg(1) = 0.^ (4.5)We now prove that g has too many zeros, that is, the function g contradicts Lemma 2of DechampsGondim, LustPiquard and Queffelec [8] which generalizes Descartes' ruleof signs. The lemma says that if 0 < al < < a„ and^P(x) ao +^+ + anxanwith the ai real and not all zero, then the number of positive zeros of P is less than orequal to the number of sign changes in the coefficients.Note that A1 < 121 since A1 = 11A11.0 = sup{ I1Av112 : v E 1R2, II112 < 11 andIlAvI12^IIBlvi 112^1113110011v1I2^pi when 11v112 < 1. In fact A1 < pi, since otherwisevg(r) a 0 which would contradict (4.2). Thus (4.4) says A2 > p2 and we getP2 < A2 < Al <^(4.6)If p2 > 0, we may assume that p2 > 1 by multiplying A and B by an appropriateconstant. Then0< ln /22 < ln A2 < ln Ai < ln piand letting al = in p2, a2 = ln A2, a3 = ln A1, a4 = ln pi and x = er, we get^g(r) P(x) =^+ x2 + xa3 — x.4Thus P(x) has zeros at el ,e (by (4.3) and (4.5) respectively) and also at 1. Since thereare only two sign changes in the coefficients, the lemma implies that there are at most 2positive roots, a contradiction.If p2 = 0, then we can ensure that A2> 1 and we now get.C(r) = P (x) = x"2 + x"3 — xa4.Chapter 4. Special Cases.^ 41Now P(x) still has zeros at eq' and e but has only one sign change in the coefficients,again contradicting the lemma.These contradictions were a result of assuming that (M2(C), ii ' 11p) does not have theminorant property for p> 2 and so the proof is complete. 0Since we've just seen that the argument of DechampsGondim, LustPiquard andQueffelec works for 2 x 2 matrices as well as 3 x 3 matrices, the obvious question arisesof the suitability of the method for general n x n matrices. Unfortunately, as soon as weconsider 4 x 4 matrices or bigger, the "polynomial" P(x) in the argument can have fouror more sign changes in the coefficients. This means that unless a way can be found toproduce more zeros, we can no longer get a contradiction from the generalized Descartes'rule of signs.Next we determine when LP(Z2) has the positive minorant property. This will beconsiderably easier thanks to Theorem 2.8.Proposition 4.2 For 0 < p < oo, the space LP(Z2) has the positive minorant propertyif and only if p> 1.Proof. Recall that by Theorem 2.8, when p > 1 the space LP(Z2) has the positiveminorant property if and only if Aq(h)(y) > 0 for all h E LP(Z2) with 4y) > 0 for ally E 22. Let h = a + br where r E 22 with r(0) = land r(1) = —1 and a, b> 0. We mayassume a > b as h may be replaced with rh if otherwise. Thus,I h (0) IP 1 sgn h (0)Aq(h)(0) =iihiri(a — b)P1and Aq(h)(1) = IlhIlf, 1^(= 0 if a = b).=iihiir,1(a + b)P1Chapter 4. Special Cases.^ 42Consequently,1 Aq(h)(1)211h1r1 [(a + b)'1 ± (a _ b)P11> 01and Aq(h)^(r) ^ 211hIlrl [(a + b)"' — (a — b)P1]• 0 ifp—l>0while Aq(h)(r) • 0 if p—l<0.Thus for p > 1, the space LP(Z2) has the positive minorant property. For p < 1,we can choose a,b > 0 with a 0 b so that Aq(h)(r) < 0. The proof of Theorem 2.8applies since a br is never zero (see Remark which immediately follows the proof of thistheorem), and so LP(Z2) does not have the positive minorant property. 0DechampsGondim, LustPiquard, and Queffelec [8] remarked that it would be interesting to know if the positive minorant property is actually weaker than the minorantproperty. The last two propositions demonstrate that the properties are indeed differentfor the LP(G) spaces when G is a finite abelian group. When 1 < p < 2 the space LP(Z2)has the positive minorant property but does not have the minorant property. However,it is still an open problem whether the properties are equivalent or not on (Mn(C),For example, Proposition 4.2 and the embedding argument show that (M2(C), j. ) doesnot have the positive minorant property for 0 < p <1. The main theorem in Chapter 5will further show that the property also fails to hold for 1 < p < 2 while Proposition 4.1confirms that (M2 (C), II • lip) has the minorant, and hence the positive minorant propertyfor p> 2. Hence we have provedProposition 4.3 For 0 < p < co, the space (M2(C),11 • ilp) has the positive minorantproperty if and only if p> 2.Chapter 4. Special Cases.^ 43Proposition 4.4 For 0 < p < oo, the spaces LP(Z3) and (M3(C),11 ' 11) have the minorant property if and only if p> 4 or p = 2.Proof. As {p/2] + 2 = 3 for 2 < p < 4, Theorem 3.3 says L1'(Z3) does not have theminorant property. As 2[p/2] + 3 = 3 for 0 < p < 2, Theorem 3.5 says LP(Z3) does noteven have the positive minorant property, so it cannot have the minorant property. Thus,via the embedding argument, both LP(Z3) and (M3(C), ll ' lip) do not have the minorantproperty for 0 < p < 4 and p 0 2.Conversely, the space (M3(C), II • lip)P has the minorant property (and hence so doesLP(Z3)) for p > 4 by Theorem 2 in the paper by DechampsGondim, LustPiquard, andQueffelec [8]. 0As with (M2(C), li • 11p), the minorant and positive minorant properties are equivalentfor the space (M3(C) ,II • lip) •Proposition 4.5 For 0 < p < oo, the space (11/13(C),11 • 11) has the positive minorantproperty if and only if p > 4 or p = 2.Proof. In Chapter 5 we will show that for 0 < p < 4 and p 0 2, the space (11/13(c), II . ilp)does not have the positive minorant property. For p > 4, we can simply use Proposition4.4 above. Alternatively, the following argument is of interest.We use Theorem 5 in [8], due to B. Virot, which shows E" > 0 if a > 1 and thematrix E is positive definite with nonnegative entries. Also use Theorem 3(1) of [8],which says that (M,i(C), II • lip) has the positive minorant property if and only ifB > 0 implies (B*B)(P/2)1B* > 0.Then E= B*B > 0 and it is positive definite. For p > 4 and a = p/2 — 1 > 1,Ea > 0Chapter 4. Special Cases.^ 44and thusEa B* = (B*13)(P12)1B* > 0. 0The following lemmas will assist us in proving the positive minorant on some of theLP(G) spaces to be considered next.Lemma 4.6 Let G be a finite abelian group of order n. Let Ô = 171, • • • 7 77,1 and definethe function H R n^Ill byH (Xi, . Xn) 101177: where h = Exai.i=1Then for p > 1 the function H is convex in xi and if p > 2, the partial derivative 01110xiis an increasing function in the variable xi.Proof. To prove convexity, note that if A1, A2 > 0 and A1 + A2 = 1 then(Xi . . .^A2Xli , • • • X n)^(X1 , • • • ,^• • • , Xn)^A2(Xi, • • .^. . . Xn).Now apply H and then the triangle inequality, and finally note that Ali: < Ak as p > 1and Ak < 1 for k = 1, 2. Since H is convex in xi,82H> 0axF —when the partial derivative exists (that is when p > 2) and so 01110x2 is an increasingfunction. 0Lemma 4.7 Let G be a finite abelian group and p > 2. If for all h E LP(G) withit(y) > 0 for every 7 E Ô and it(1) = 0 we have Aq(h) (1) > 0, then LP(G) has thepositive minorant property.But1 x,11(Xi, X21 • • • xn) =^2_,n xEGxiyi(x)Chapter 4. Special Cases.^ 45Proof. Theorem 2.8 shows that it will suffice to show Aq(h)" > 0 whenever it > 0.However, it actually suffices to show that Aq(h)(1) > 0 for all h in LP(G) with it > 0since if Aq(h)^(y) <0 for some y E 6, then If?' E LP(G) with (h7) > 0 andL;I:Aq(117)7)(x)zEG1^ih(x)7(x)17Isgn (h(x)3) — En xEc^Hh111^Ih(x)1Pisgn (h(x)) = — EzEG^Ilh11;^7(x)Aq(h)^(y)<0.Let G =^7n1 with a 1. Let h = ExEGxiyi where xi > 0. By the lemmaabove, since p > 2, the functionn = 111111;is such that OH/493cl is increasing in xi and henceOH,^OH inOx1kX1, X2, • • • xn) >^kV, X2, • • • aCn)•aTi(4.7)Aq(h7yr(1) =and soOH(T1,...,xn)p1=^E Exi7i(x)^sgn (E xiyi(x))sEG i=1 1=1= Pii 1111r Aq(hr (1).Thus to show Aq(h)(1) > 0 for all it > 0 it suffices by (4.7) to show this for it > 0 andh(1) = O.^0Chapter 4. Special Cases.^ 46Proposition 4.8 For 0 < p < oo, the space LP(Z3) has the positive minorant propertyif and only if p > 2.Proof. Theorem 3.5 tells us that for 0 < p < 2 the space LP(7Z3) does not have thepositive minorant property, since2 [12] + 3 = 3.2To show LP(Z3) does have the positive minorant property for p > 2, the lemma aboveshows it will suffice to prove that Aq(h)"(1) > 0 whenever h > 0 with ii(1) = 0.Let O = {12: (0),7(1) = P = 1+^i2^,'Y(2) = P2}Then for h E LP(Z3) with 140 , it(y2) > 0 and ii(1) = Owe have,Aq(h)(1) =^[Aq(h)(0) + q(h)(1) + Aq(h)(2)]^1^^311hlir^ [1h(7) + her2)1P1_wimpjj (y2 )p2P_l sgn (it(.y)p ii(72)p2)+111(7)p2+11(72)prlsglickop2+11(72)p)]^31I h'^ Po) + 11(72))P1 —11(7)P + 11(72)P21P•2(11(7) +11(72))1^since 111(7)P + 11('Y2)P21^111(7)P2 + 11(72)P1 and p2^p = —1. Also, since a simpleapplication of the triangle inequality shows that lit(y)p it(y2)p21 5_ h(y) + il(y2), we getAq(h) (1) = 11(7) +1472) PO) + 1/(72))P2 —11/(7)P + 11(72)P2IP21311/1111r1>O as p > 2. 0Proposition 4.9 For 0 < p < oo, the space LP(4) has the positive minorant propertyif and only if p > 2.Chapter 4. Special Cases.^ 47Proof. Let4 = {1, x, y, xy : x2 . y2 = 1} and13 = {1,7, x, n : 7(x) = 1,'y(y) = 1, (x) = 1, x(Y) = —11.For h E LP(4),h(1)^= it(1) + h('Y) +11(X) + h(7X)h(x)^= h(1) — h(7) + it(X) — 11(7X) (4.8)h(Y)^= 11(1) + h(7) — h(x)  h(7x)h(xy)^= h(1)  h(7)  h(x) + 11(7X).Lemma 4.7 above shows that it will suffice to show Aq(h)(1) > 0 whenever Ii > 0 withh(1) = 0. We have[0(1)111sgn h(i) + Ih(x)IP Isgn h(x)+1h(y)risgnh(y) + ih(xy)I01 sgn h(xy)] .Note that h(1) > 0, and so to prove Aq(h)^(1) > 0 we divide the work into 4 cases.Case 1. When h(x), h(y), h(xy) > 0Case 2. When one of h(x), h(y), h(xy) < 0Case 3. When two of h(x), h(y), h(xy) < 0Case 4. When h(x) , h(y), h(xy) < 0In Case 1, with h(1) > 0, h(x) > 0, h(y) > 0, and h(xy) > 0, the conclusion thatAq(h)'(1) > 0 is trivial.In Case 2 one of h(x), h(y) , or h(xy) is negative. But Ih(z)l < h(1) for all z E 4 SOthat if h(z) <0, then11/(z)IP2h(z) < h(1)P1 for p > 11 Aq(h) (1) =^1411h1r,Chapter 4. Special Cases.^ 48or h(1)P1 I Ih(z)IP2h(z) _?.. 0.Thus it is clear that Aq(h)(1) > 0.In Case 3 two of h(x), h(y) and h(xy) are negative. We assume that h(1) > 0,that h(x) > 0, and that h(y) < 0, h(x y) < 0, the other possibilities being the same bysymmetry. Recalling that we may assume h(1) = 0 in (4.8) we have^h(y)^147) — h(x) — herx)= 147) —11(x) +14rx) — 211(7x)= —h(x) — 2h(7x)^and h(xy)^—147) — h(x) + 147X)= —10) — 11(X) — 11(7X) I 211(7X)= —h(1) + 2h,(yx).Also, since h(1) — h(x) = 2h('y) + 2h(7 X),, we have0 < 2h(y) _5_ h(1) — h(x).HenceAq(h)(1) = ohr [h(1)P1 ± h(x)P1 — (h(x) + 211(7X))P1 — (h(1) — 2147 X))P11 So Aq(h)(1) > 0 provided we can show that the functionF (a, b, c) = aP1 + bP1 — (a — c)71 — (b + 01 > 0when 0 < c < a — b (that is, let a = h(1),b = h(x), c = 211('Y X)) • NowOFOc (a' b' c) = 0 + 0 + (p — 1)(a — c)'2 — (p — 1) (b + c)P2and02F (a ' b' c) = — (p — 1)(p — 2)(a — c)P3 — (p — 1)(p — 2) (b + c)P3 .Oc2 Chapter 4. Special Cases.^ 49Since a2Fl0c2 < 0, to show that F > 0 it suffices to check that F(a,b, 0) > 0 andF(a,b, a — b)> 0. Computing we discover thatF(a,b, 0) = aP1 +bP1 — aP1 —bP1 . 0F(a,b, a — b) = aP 1 +bp_i. _bp_i. _ a1 =0.Consequently, Aq(h)(h) > 0 when p > 2 and this completes Case 3.In Case 4, we have that h(x),h(y),h(xy) < 0. So1Aq(h)(1) = 411h111 [h(1)P1 — Ih(x)IP1 — ihMIP1 — Ih(xY)IP1111,> ^1411h^max {1h(x)IP2, Ih(Y)IP2, Ih(xY)IP2}[h(1) + h(x) + h(Y) + h(xY)11r1since 1h(z)1P2 < h(1)P2 when p > 2. Using equations (4.8) with '1.41) = 0, we see thath(1) + h(x) + h(y) + h(xy) = 0, and thus getAq(h)'(1) > 0 when p > 2.In all cases we find that Aq(h)(1) > 0 for ii > 0 and 1'41) = 0, when p > 2, and so LP(4)has the positive minorant property for p> 2.When 0 < p < 2, let h(z) = 7(z) + x(z) + 7x(z). Thenh(1) = 3, h(x) = h(y) = h(xy) = —1,so h(z) is nonzero for all z E 4 and1 1^1A(h)(1)= — 1 — 1]411h1 r1 13P1 —= 1 3]411h1r1 [3P1 —< 0 when 0 < p < 2.Hence by Theorem 2.8 and the Remark following it for the case when p < 1, the positiveminorant property fails for 0 < p < 2. 0Chapter 4. Special Cases.^ 50Proposition 4.10 For 0 < p < oo, the space LP(Z4) has the positive minorant propertyif and only if p > 2.Proof. Let Z4 = {0, 1, 2, 3} and let24 = {1,7,72,73 : 7(0) = 1,7(1) = i, 7(2) = 1,7(3) = il .As usual, we show Aq(h)"(1) > 0 whenever it > 0 with it(1) = 0. Let3h(X) = E 11(73)73 (x)..7=1Thenh(0) =h(1) =h(2) =h(3) =ii,(7) +102) + h(y3)—h(y2) + i 0(7) — h(y3))—11(7) + 11(72) — 1473)4(72) — i (11(7) —103)) Using the fact that 0(1)1 = Ih(3)1 in the following, we get1^3Aq(h)(1) — ohiri .01h(i)1191sgn 14 j)141 111117r11 =411hli^[h(0)731  2/02) Ih(1) IP2 + /(2) Ih(2) IP2] .rlCase 1. Suppose h(2) > 0.Using (4.9) note thath(0) + h(2) = 2./(72)h(0) — h(2) = 211(7) + 2h(73)so that^Re h(1) =^(h(0) I h(2) 2^)and^Im h(1) = h(y) — h,(73) = (h(0) — h(2))2^)=(4.9)(4.10)ploy1 + (h(1) + h(3))1h(1)IP2 + h(2)1h(2)1P 2 12Chapter 4. Special Cases.^ 51h(0) h(2) ) 2 + (h(0) —2 h(2) 2ii(73)) 2.so that^Ih(1)1 =Also notice that0 < 2i/(73) < 2ky)+ 2h(73) = h(0) — h(2).Hence to showAp(h)(1) = ^1" "'^— 2h(y2)1h(1)'2 + h(2)1h(2)1P2]41Ihr> 0it suffices to show (let a = h(0),b = h(2c = 21(73))p2 2) 2ap_, _ (a ± ((a b ) 2 ± (a —2 b + bp1F(a,b,c) => 0 when 0 < c < a — b.Notice that for p > 2, F(a,b,c) > F(a,b, 0) = F(a,b, a — b) for 0 < c < a — b, so it willsuffice to show that F(a,b, 0) > 0. NowF(a,b, 0) = F(a,b, a —2 2^a+b^a—2b)2)a^_ + ((^2 Ezza2 b2) 2= aP1 + bP1 — (a + b)( ^.2Note that a > b > 0 and if a = b, then F(a,a,0) = 0. So we may assume a> b > 0,and then by dividing through by 01 and putting t = bl a we wish to prove that1 + t > (1 + t) ( 1 + t2)Y2 or that (taking logs)ln(1 +t"') > ln(1+ t) + (P2 2) lnChapter 4. Special Cases.^ 52or that1+ t2)G(t,p) = 21n(1 + 91 —21n(1 + t) — (p — 2) ln ( 2> 0 for p> 2, 0 < t < 1.In particular,G(t, 2) = 2 ln(1 + t) — 2 ln(1 + t) — (0) In (1 + t2)2 )=0.So it suffices to show that OG/ap > 0. NowaG _ ^2 1 ln(t)tP1 In (1 + t2)Op — 1 + tP^2 )1 + t2= 21n(t) (t11 + 1i\ i —In2Alsoa2G^ \ 2= —21n(t)2 (tlP + I) (1)tlPap2= 2t1P ln(t)2 (tiP + 1)2>0.Hence to show OG/ap > 0, it will suffice to show that (OG/Op)(t, 2) > 0, that is, to showH(t) = —OG (t, 2) = 21n(t)(t1 + 1)1 — In ( 1 + t2 ) > 0.op^ 2SinceH(1) = 21n(1)(1 + 1)1— ln(1) = 0,to show H(t) > 0, it suffices show that H'(t) < 0. We haveH'(t) = ^2(t1 + 1)  1^ 2t^+2111(t)(t1 +1)2(1)(1)t2 + 1 + t2t2(t +1) + 21n(t)^2t (1 + t)2^+ 1 + t2.=Chapter 4. Special Cases.^ 53To show H'(t) < 0 it suffices to show thatK(t) = (1 + t)2 H'(t) < 0.2NowK(t)^1 + t + ln(t) t(1 + t)2 1 +t2(1 + 0(1 + t2) — t — 2t2 — t3= ln(t) +1 + t2ln(t) + 1 — t21 + t2Since K(1) = 0, to show K(t) < 0 it suffices to show K'(t) > 0. Now1^—2t(1 + t2) — (1 — t2)(2t) K' (t) = t+(1 + t2)21 —2t — 2t3 — 2t + 2t3 (1 + t2)21 + 2t2 + t4 — 4t2t(i + t2)2( 1 t2)2t(1 + t2)20 as required.Consequently, Aq(h)(1) > 0 when p > 2.Case 2. Suppose h(2) <0. Then from (4.10) we get1Aq(h)(1)^omir {h(0)P1 + (2(y2) + h(2)) max fik1w2,1h(2)1p21]> ^10111 ih(0)_2i,(72)+ h(2)] max flh(1)IP2, 0(2)JP 2}417when p > 2, and since Ih(x)1 < h(0) for all x E Z4. Using equations (4.9) we see thath(0) — 2i1(y2) + h(2) = 0 so we finally haveAq(h)^(1) = 0.Chapter 4. Special Cases.^ 54Hence in both cases, Aq(h)(1) > 0 when p > 2 and so LP(7L4) has the positive minorantproperty when p > 2. By Theorem 3.5, as 4 > 2[p/2] + 3 = 3 when 0 < p < 2, the spaceLP(Z4) does not have the positive minorant property for 0 < p < 2. 0It is clearly becoming more difficult to prove the positive minorant property as theorder of the group increases. I have been unable to completely determine when LP(Z5)has the positive minorant property. However, the following is a partial result.Proposition 4.11 If f, g E LP(74) with l('y) = 1(71) and if 0 5_ f(y) _< .4(y) for all7 E 25, then IIf HP 5 liglip for all p > 4.Outline of proof. First it will suffice to show that Aq(h)^ > 0 for all h E LP(Z5) withit > 0 and h even, that is h('y) = it(y1) for all y E 25. This is because this proves thatfor two even functions f, g with 0 < j < .0, then If II p < lIglIp. Now if g is not even, theng + gYe = 2is even, and ge still majorizes f, so thatIlf Ilp^Ilgellp = 1 g +2 g P1^1lIglIp +^11:ollp=^lIglIp.To prove Aq(h) > 0, may still apply Lemma 4.6 to assume it(1) = 0 but may not useLemma 4.7 and so must check Aq(h)^(7) > 0 for each y E 25. If 7 E 25 and 7 0 1, thenthe condition that Aq(h)^(7) > 0 is the same as Aq(h)'(74) > 0, and similar to the proofthat Aq(h)(y2) > 0 which is the same condition as Aq(h)^(73) > 0. The hardest part isto prove that Aq(h)(1) > 0, in spite of it being a two variable problem and the others athree variable problem, but it is not as bad as the proof in the LP(Z4) proposition. 0Chapter 4. Special Cases.^ 55The examples considered in this chapter for the positive minorant property on LP(G)show that the corresponding counterexamples in Chapter 3 were the best that could befound. While they are little to go on, it seems reasonable to conjecture that for n > 3,the space LP(Z) has the positive minorant property if and only if p > 2[(n — 1)/2] orp is even.4.2 Functions that operate on finite groups.In the study of the classification of functions which operate on positive definite sequences[24], functions [10, 13, 23], or matrices [26, 3, 4] it was generally found they are of theform00F(z) = E anz,„zinr^ (4.11)mtn=0whereanz,n > 0^ (4.12)and00E ani,n < 00.^ (4.13)m,n=0However these results did not apply to positive definite functions defined on finite groups(nor some exceptional infinite groups). For matrices the classification was for functionsthat operate simultaneously on positive definite matrices of all sizes.In the group setting, Rider [23], Moran [18], and Graham [10] give examples of functions defined on exceptional groups that operate in some sense. These functions are ofthe form (4.11) but fail to satisfy one of the conditions (4.12) or (4.13). In some of thepropositions above, to find when LP(G) has the positive minorant property we showedthat Aq(.) operates on the positive definite functions. This means that the functionFp(z) = Izrisgn z operates on the positive definite functions when p is large enough.This function is not of the form (4.11) when p is noneven, in fact it is not even smooth.Chapter 5Failure of the Positive Minorant Property for Matrices.This chapter is devoted to improving the known results on when (Mii(R), II* ii ) fails toPhave the positive minorant property.5.1 Statement of the main theorem.In Chapter 3 we showed that LP(7L) does not have the positive minorant property fornoneven p such that 0 < p < 2[(n — 1)/2]. Proposition 3.1 allows us to conclude that(Mn (R), ilJ. 11p) does not have the positive minorant property for these values of p. In factthis failure occurs for a larger set of p's.Theorem 5.1 The space (M„(R),11.11p) does not have the positive m,inorant property for0 < p < 2(n — 1) and p not an even integer.When p > 1, we use the matrix counterpart of Theorem 2.8, which says that forcompact abelian groups LP(G) has the positive minorant property if and only if thefunction Fp(z) = IzIPIsgn z operates on positive definite polynomials. The followingstatement is proved in Theorem 3(1) in the paper [8] by DechampsGodim, LustPiquard,and Queffelec.Theorem 5.2 For 1 < p < oo, the space (Mp(R), II • II) has the positive minorantproperty if and only if (B*B)(P12)1B* > 0 whenever B > 0.The authors of [8] explain what they mean by (B*B)(P/2)1/3* when 1 < p < 2 and Bis not invertible. To prove Theorem 5.1 for p> 1 it suffices for us to find a nonnegative56Chapter 5. Failure of the Positive Minorant Property for Matrices.^57invertible matrix B for which (B*B)(P/2)113* has a negative entry. Specifically, we showthis for the matrixA = Ei 1 + J(0) + 4,(0)T,where Eli is the matrix with all zero entries except for a single 1 in the upper lefthandcorner, and Jn (0) is the Jordan matrix with all zero entries except for l's just above themain diagonal. That is,101010oA=010101010 /This matrix has the property that the last entry in (A*A)(P/2)1A* is negative for2(n — 2) < p < 2(n — 1). To deal with smaller values of p we build other nonnegativematrices which we denote by Bk and which have the property that (BZ.Bk)(P/2)1BZ hasa negative last entry for 2(k — 2) < p < 2(k — 1) for k = 2, 3, ... , n — 1. Let A, denotethe leading r x r principal submatrix of A, and let Bk be the matrix with blocks(Ink 0 )0^fikwhere Jr is the r x r identity matrix. Then Bk E M(1I) and0(BI:Bk)(7312)1BZ = ( 1.7"0^(AZAk)(P/2)1AZ^.\oChapter 5. Failure of the Positive Minorant Property for Matrices.^58Our analysis of the full matrix An = A will show that for k > 2,[(AZ Ak)(PI2)1 Ask]k k < 0 for 2(k — 2) < p < 2(k — 1),so that [(BZBk)(P/2)1Bnn n <0 when 2(k — 2) < p < 2(k — 1), for k = 2, 3, ... , n.It follows that for small enough positive values of e the matrix Ek obtained from Bkby adding E in the last entry has a strictly smaller Cp norm than Bk does, although Ekmajorizes Bk.We cannot use Theorem 5.2 when 0 < p < 1. Instead we use some facts from Chapters3 and 4. For n > 3 the embedding of LP(Z) into (Mn(C), II 11) and our Theorem 3.5apply when 0 < p < 2. When n = 2 Proposition 4.2 applies provided 0 < p < 1. Whenn = 2 and p = 1 simple calculations show that the matrices E2 with E = 1 and B2 definedabove provide the needed counterexample.5.2 Eigenvalues and eigenvectors of A.To begin with, A is a symmetric matrix so its eigenvalues are real, and there is anorthonormal basis of eigenvectors of A. We will show that none of the eigenvalues is 0and that each eigenspace is onedimensional.^List the eigenvalues as Ai > A2 >^>^An. Letting xi denote an eigenvectorassociated with the eigenvalue Aj, then Axj = Ajxj for j = 1, , n or by equating eachTOW^Xi j + X2 j = AiXi j^ (5.1)^Xj_i j Xj+i j = AiXi j^i = 27^, n — 1^(5.2)Xn—i j = Aixn j. (5.3)If Aj = 0, equation (5.3) says xn_1 = 0, but then equation (5.2) with i = n —2 saysxn_3 j xn_i j = 0 so xn_3 j = 0 which in turn says xn_5 j = 0 etc., until we either getChapter 5. Failure of the Positive Minorant Property for Matrices.^59= 0 Or X2 j = 0. Then equation (5.1) says xi + x2 = 0 so both xi i and x2 j = 0,and so using equations (5.2) we can conclude xi = 0. However, xi is an eigenvector sowe have a contradiction, and none of the eigenvalues of A can be zero.Turning our attention now to the eigenvectors of A, we suppose xn j = 0. Then equation (5.3) says xn_1 = 0 and then equations (5.2) working backwards give in turn thatXn2 j = 0, xn_3 j = 0, • . • , X1 j = 0. Again our supposition has produced a contradiction,so xn 0 0.Putting xn = 1 then by (5.3) xn_i j = Ai and by (5.2)xi_1 j = Axi — xj4.1 j for i = 2, .. . , n — 1.Hence we can solve for xi using this recursive method. It shows that each eigenspaceis onedimensional and the eigenvalues are distinct. Equation (5.1) turns out to beredundant when Ai is an eigenvalue.Defining polynomialsro(A) = 1ri (A) = Ark(A) = Ark_i(A) — rk_2(A)for k > 2 we have that r2k (A) is an even monic polynomial of degree 2k, while r2k+1 (A)is an odd monic polynomial of degree 2k + 1, and ri(Ai) = xi.; for i = 1, . . . , n andj = 1, . . , n. That is xi = (rn_i (Ai), , ro(AMT. Let pi = xj/lixill2 be normalizedeigenvectors of A and put P = (pi, , pn) = (pi ;), and A = diag (A1, , )n). ThenAP = PA.Since A is symmetric and the eigenspaces are onedimensional, P is an orthonormalmatrix.Chapter 5. Failure of the Positive Minorant Property for Matrices.^605.3 When the last entry is zero.We now consider (A*A)(P/2)1A* and will show that the last entry of this matrix isnegative for 2(n — 2) < p < 2(n — 1), that isRA*A) F1,41„ n < o.To do this we first show that this entry is 0 for certain even integer values of p. NowA*A = A2 = pA2pT ,(A*A)fi = p(A2)f—lpT= plAlp2pTwhere AlI ^(hag(JAi..., IA.1).ThusWe define[(A*A)1Aln n = [PIAIP2API]n= EPn i[lAIP2Ap7]j=1= E Pn j l Aj I2AjpTnj=1E P2n I Ai IPlsgn Ai.J=1f(p) = RA*A)11Aln n =^jPti IP lsgn AiJ=1and as noted earlier Ai 0, so f is welldefined for all real numbers p.Lemma 5.3 The quantity f (p) = 0 if p = 2k, for k = 1, 2, . . . , n — 1.Proof. We use induction on k. When k =1,p= 2 andf (2) =^ A.j=iChapter 5. Failure of the Positive Minorant Property for Matrices.^61But P is an orthonormal matrix, so that if p(i) denotes the ith row of P,0 = p(n) pi)= EPT ji)n1 jj=1= EN,^j,=1^Ilx,1121=^Pn ^ ri(Ai)j=1^lixi112as Xn—i j = rn—(n1)(Aj)Xn j =^jAi as pn =j=1 ilxiI12= 1(2)._ 1Ilx;112and ri (Ai) =Suppose that f (2k) = 0 for k = 1, . . . , — 1. Thenf(2) =^ilAirtlsgn Aj,,2 12t1Z_,Pn j'j •j=1Again, as P is orthonormal,0 = p (n —tI 1) p (n —t)EPnt+1 jPn—t jE Xn—tf1 j Xn—t j= j=1 11302x—,71^1^1^ rt_i(Ai)rt(Ai) as xn_i j = r1 (A)▪ Ilx./112^Ilx./112• E p2n jri_i (Ai ) rt (Ai )Recall that r22 is an even monic polynomial of degree 2i andpolynomial of degree 2i + 1 so that 7.6_1 • rt is an odd monic polynomial of degree 2f — 1.If(A) (A) =A 2e1 + bt_1A213^bjA2i1is an odd monicChapter 5. Failure of the Positive Minorant Property for Matrices.^62then0 = p(ne+i) ont)• Egz Jr11(Aj)ri(Aj)j=1n n—2 2f1 ^ 2= i Pn jAi + bf1 E /3! jAr3 + . . • + bi E Pn JAJJ=1^J=1 J=1^„.,2 12e1 + 0 +^+ 0• Z_, Pn f‘jj=1= E .7=^2k1as (2k)^for k = 1, . . . , — 1. But then^f (2) =^A' =0,J=1completing the induction, and proving that f (2k) = 0 for k = 1, . . . , n — 1. 05.4 When the last entry is not zero.We will show that the zeros specified in Lemma 5.3 are the only zeros of f. We use thefollowing lemma.Lemma 5.4 Given two sequences of real constants^and Icil with t j's positive anddistinct and not all cj = 0, define the function h(a) = E7 1 3! c.t" • Then h has at most3= n — 1 distinct zeros. Moreover, if h has n —1 distinct zeros they must all be simple.Proof. We proceed by induction on n. If n = 1, then h(a) = ct", which has no zeroswhen c and t nonzero. Suppose when k> 1 that all functions of the form Ei1=1 c3 3t9 haveat most k — 2 distinct zeros and consider h(a) = cjt. . Thenk1g(a) = h(a)— = c; ti— a + akt" 3=1^tkanddg k1da^In (L) (ti \=^cu^a^i=1 t^k ^tk) •Chapter 5. Failure of the Positive Minorant Property for Matrices.^63By the inductive hypothesis, dgicla has at most k — 2 distinct zeros, and so by thecontrapositive of Rolle's theorem g has at most (k — 2) + 1 = k —1 distinct zeros. Thefunctions h and g have the same zeros, so h has at most k —1 distinct zeros, completingthe induction for the first part of the lemma.If h actually has n —1 distinct zeros and one of them is not simple, then the derivativeof the corresponding function g would also have a zero at this point. By Rolle's theoremg' would also have at least one zero strictly between each pair of consecutive zeros of h.This would provide at least 1 + (n — 2) = n — 1 distinct zeros of g', contradicting the firstpart of the lemma, since g' is a sum of n — 1 terms. 0We need to confirm that f (p) = iP A.i isgn Aj is of the form E7=1 c3t7 specified in Lemma 5.4. Let a = p — 1, and cj = p isgn Aj and ti = lAj I for all j. First theindividual coefficients cj cannot be 0 because pn = 1iiix2ii2 and none of the eigenvaluesis O.While the eigenvalues are distinct, it may appear possible that their absolute valuesmight not be distinct. If IA.; I = I Aji for distinct j and j', then we can combine thecorresponding terms in the sum for f (p) and get something of the same form with fewerterms. We need to show that this expression does not reduce to the zero function. Sincethe trace of the matrix A is 1, the list of eigenvalues cannot just consist of pairs with thesame absolute value and opposite signs. So f(p) cannot be identically equal to 0.Applying Lemma 5.4 to f shows that it has at most n —1 zeros. Since we found n —1zeros in Lemma 5.3 they must all be simple. It follows that the sign of f changes at eachof its zeros.The space (Ain (IR), lip) has the positive minorant property when p = 2k so that(B*B)(P/2)1B* > 0 for matrices B > 0 and in particular, f (p) = RA*A)(P/2)1Aln n > 0when p = 2k. More precisely, since 1(2k) = 0 when k = 1, ,n — 1, we must have1(2k) > 0 for all integers k > n . It follows that f(p) > 0 for all p> 2(n — 1), since itChapter 5. Failure of the Positive Minorant Property for Matrices.^64has no zeros in this interval and is positive at some points in the interval. So the signchange of f (p) at p = 2(n — 1) must go from negative to positive. Hence f (p) < 0 for allp E] 2(n — 2), 2(n — 1)[. This completes the proof of Theorem 5.1.Bibliography[1] Bachelis, G.F., On the upper and lower majorant properties in LP(G), Quart. J.Math. Oxford (2), 24 (1973), 119128.[2] Boas, R.P., Majorant problems for Fourier series, J. d'Analyse Math. 10 (19623),253271.[3] Christensen, J.P.R. and Ressel, P., Positive Definite Kernels on the Complex HilbertSphere, Math. Z. 180 (1982), 193201.[4] Christensen, J.P.R. and Ressel, P., Functions operating on positive definite matricesand a theorem of Schoenberg, Trans. Amer. Math. Soc. 243 (1978), 8995.[5] Civin, P., Fourier coefficients of dominant functions, Duke Math. J. 13 (1946), 17.[6] Davis, P.J., Circulant Matrices, John Wiley & Sons, Inc., 1979.[7] DechampsGondim, M., LustPiquard F., and Queffelec, H., La propriete du minorant dans les espaces de Banach, Publ. Math. Orsay 6 (198081), V111.[8] DechampsGondim, M., LustPiquard, F., and Queffelec, H., On the minorant properties in CP(H), Pacific J. Math. 119 (1975), 89101.[9] Fournier, J.J.F., Majorants and LP norms, Israel J. Math. 18 (1974), 157166.[10] Graham, C.C., Functional calculus and positivedefinite functions, Trans. Amer.Math. Soc. 231 (1977), 215231.[11] Graham, C.C. and McGehee, 0.C., Essays in Commutative Harmonic Analysis,SpringerVerlag New York Inc., 1979.[12] Hardy, G.H. and Littlewood, J.E., Notes on the theory of series (XIX): A problemconcerning majorant of Fourier series, Quart. J. Math., Oxford 6 (1935), 304315.[13] Herz, C.S., Fonctions operant sur les fonctions definespositives, Ann. Inst. Fourier(Grenoble) 13 (1963), 161— 180.[14] Hewitt, E. and Ross, K.A., Abstract Harmonic Analysis II, SpringerVerlag NewYork Inc., 1970.[15] Horn, R.A. and Johnson, C.R., Matrix Analysis, Cambridge University Press, 1985.65Bibliography^ 66[16] Lee, E.T.Y. and Sunouchi, G., On the majorant properties in tP, Bull. Inst. Math.Academia Sinica, 4 (1976), 327336.[17] Lee, E.T.Y. and Sunouchi, G., On the majorant properties in LP(G), TOhoku Math.J. (2) 31 (1979), 4148.[18] Moran, W., The individual symbolic calculus for measures, Proc. London Math. Soc.(3) 31 (1975), 385417.[19] Oberlin, D.M., The majorant problem for sequence spaces, Quart. J. Math. Oxford(2), 27 (1976), 227240.[20] Feller, V., Smooth Hankel operators and their applications (the ideals Sp, Besovclasses, and random processes) Dokl. Akad. Nauk. Math., 252 (1980), 43— 47 (inRussian); translated as Soviet Math. Dokl. 21 (1980), 683688.[21] Rains, M., Majorant problems in harmonic analysis, Ph.D. Thesis, University ofBritish Columbia, 1976.[22] Rains, M., On the upper majorant property for locally compact abelian groups,Canadian J. Math., 30 (1978), 915925.[23] Rider, D., Functions which operate on positive definite functions, Proc. Camb. Phil.Soc. 69 (1971), 8797.[24] Rudin, W., Positive definite sequences and absolutely monotonic functions, DukeMath. J. 26 (1959), 617622.[25] Rudin, W., Fourier Analysis on Groups, Interscience Publishers, New York, 1962.[26] Schoenberg, I.J., Positive definite functions on spheres, Duke Math. J. 9 (1942),96108.[27] Shapiro, H.S., Majorant problems for Fourier coefficients, Quart. J. Math. Oxford(2) 26 (1975), 918.[28] Simon, B., Pointwise domination of matrices and comparison of Ip norms, PacificJ. Math. 97 (1981), 471475.
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Title  Minorant properties 
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Weisenhofer, Stephen 
Date Issued  1993 
Description  We study the minorant property and the positive minorant property for norms on spaces of matrices and norms on spaces of functions. A matrix is said to be a majorant of another if all the entries in the first matrix are greater than or equal to the absolute values of the corresponding entries in the second matrix. The Cp norm of a matrix is the tP norm of its singular values. The space of n x n matrices, with this norm, is said to have the minorant property provided that the norm of each nonnegative matrix is greater than or equal to the norm of every matrix that it majorizes. Similarly, if the norm of each nonnegative matrix is greater than or equal to the norm of every nonnegative matrix that it majorizes, then the space of matrices is said to have the positive minorant property. It is easy to verify that these properties hold if p is even. We show that the positive minorant property fails on n x n matrices with the Cp norm when 0

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Language  eng 
Date Available  20080916 
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DOI  10.14288/1.0079653 
URI  http://hdl.handle.net/2429/2045 
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Science, Faculty of Mathematics, Department of 
Degree Grantor  University of British Columbia 
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