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Minorant properties Weisenhofer, Stephen 1993

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MINORANT PROPERTIES By Stephen Weissenhofer B. Sc. (Honours) The Flinders University of South Australia  A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY  in THE FACULTY OF GRADUATE STUDIES MATHEMATICS  We accept this thesis as conforming to the required standard  THE UNIVERSITY OF BRITISH COLUMBIA March 1993 © Stephen Weissenhofer, 1993  In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the head of my department or by his or her representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my written permission.  Mathematics The University of British Columbia 2075 Wesbrook Place Vancouver, Canada V6T 1Z1  Date:  tictpel, I , 1 ctc13  Abstract  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 < p < 2(n — 1) and p is not even. We also consider versions of the minorant properties for LP-spaces of function on some commutative topological groups. We show that the positive minorant property fails on the cyclic group of order n when 0 < p < 2[(n — 1)/2] and p is not even. This relation between 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 suitably related. We completely determine the combinations of p and n for which the minorant properties hold on some low-order groups. For compact abelian groups, we show that the positive minorant property is equivalent to a certain function operating on the class of positive definite functions. We deduce that if LP(G) has the positive minorant property, then so does LP+2 (G) . 11  Table of Contents  Abstract^  ii  Acknowledgement^  v  1 Introduction.^  1  1.1 Notation.  ^1  1.2 Majorants on Groups.  ^2  1.3 Majorants and Matrices  ^7  2 General Facts About Minorant Properties.^ 2.1 Reductions.  11 ^11  2.2 The positive minorant property^  17  3 Counterexamples.^ 3.1 Groups and matrices  23 ^23  3.2 Cyclic groups ^  27  3.3 Beyond cyclic groups. ^  36  4 Special Cases.^  38  4.1 Minorant properties on low-order groups and matrices. ^ 38 4.2 Functions that operate on finite groups. ^  55  5 Failure of the Positive Minorant Property for Matrices. ^56 5.1 Statement of the main theorem ^  111  56  5.2 Eigenvalues and eigenvectors of A. ^  58  5.3 When the last entry is zero. ^  60  5.4 When the last entry is not zero^  62 65  Bibliography^  iv  Acknowledgement  First and foremost I wish to thank my research supervisor Dr. John Fournier. I am greatly indebted to him for his help, support and guidance. I especially appreciated our regularly scheduled meetings, they helped me to be more disciplined and motivated during those long periods when nothing would seem to work. I am also grateful for the time and effort he spent in correcting rItXing errors in the initial typed version of my thesis. 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 in typing my thesis quickly and accurately. I appreciate his willingness to learn TEX and mathematical symbols way beyond what would have been reasonable to expect. His assistance was also instrumental in getting my thesis prepared on schedule.  I thank my many friends for their moral support and encouragement, especially Everett 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 and Dr. 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 times I've spend at U.B.C. and Canada. Finally I wish to thank my parents for their support in all my endeavours. Financial support from the Canadian Commonwealth Scholarship and Fellowship Plan, and from Teaching and Research Assistantships in the Department of Mathematics at U.B.C. are gratefully acknowledged. v  Chapter 1  Introduction.  In this chapter we will introduce notation and definitions to be used throughout this thesis. We also provide some of the history of the majorant problem. As well, we explain why the consideration of finite abelian groups and matrices is of interest in connection with the majorant problem.  1.1 Notation. Our standard reference for notation and terminology related to Fourier analysis is Rudin's book [25]. We let G denote an abelian group which in most cases will be compact. A character of G is a continuous complex valued function y defined on G for which j y(x)I = 1 -  -  for all x E G and y(x + y) = -y(x)-y(y) for all x, y E G. Under pointwise multiplication -  the set of all characters of G forms a group called the dual of G and will be denoted by  Ô. The Fourier transform of f E L' (G) is the function j defined on G by (-y)  =  L  f (x) y(x) dx -  for all y E Ô. As an example, consider G = T, the circle group. The characters of T are -  7(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'' de where (1/271 - )de is the Haar measure.  1  Chapter 1. Introduction.^  2  1.2 Majorants on Groups. The majorant problem was introduced by Hardy and Littlewood [12] who asked what comparisons can be made between the LP norms of two integrable functions when one is a majorant of the other, that is, when the Fourier coefficient of one dominates the Fourier coefficients of the other. Their work was done on the circle group T, but we will make our definitions in the more general setting of compact abelian groups equipped with the normalized Haar measure. Definition 1.1 Let G be a compact abelian group. If f,g E L1(G) and .4(7) > (7)1 for every 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 majorant of f, then 11/11p < 11911p• However they showed that this inequality was not true in general by exhibiting trigonometric polynomials f and g with the properties that g majorizes f but 11f 113> 119113. Could it be possible though, that whenever g majorizes f,  Ilf Ilp 5- AplIglip for some number Ap dependent only on p? Boas [2] proposed saying that the Banach space 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 the upper majorant property (with constant Ap) if there exists a constant Ap such that  Ilf II,,^AplIgIlp whenever g is a majorant off.  Chapter 1. Introduction.^  3  Also of interest to Hardy and Littlewood was what Boas called the "lower majorant property." Definition 1.3 Let G be a compact abelian group. The space L(G) is said to have the lower majorant property (with constant Aq) if there exists a constant Aq such that every f E L(G) has a majorant g E L(G) for which  ligliq 5- Aql While 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 did show that if LP(T) has the upper majorant property, then 11(11') has the lower majorant property when p-1 + q-1 = 1 (and hence they proved the nontrivial result that D(T) has the lower majorant property for q = p' = (2k)' = 2k I (2k — 1), k E N). Boas [2] showed that 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, the space LP(G) has the upper majorant property if and only if Lq(G), where p-1 + q-1 = 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. For instance, 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 and Sunouchi [17] proved the "only if' part for the locally compact abelian groups. Further generalizations to Banach spaces satisfying certain conditions were made by DechampsGondim, Lust-Piquard, 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 that  Chapter 1. Introduction.^  4  if 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) to prove that LP(T) did not have the upper majorant property with unit constant when p is greater 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 of argument, 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, or  p = oo. It is natural to ask what happens if G is a finite abelian group. Of course such groups will always have the upper majorant property regardless of p because if the order of G is n, and if 11(7)1 < L,(-y), then  II! II  =  -n E if(x)iP xEG  E f(7)7(x) -yEa-  = g (0)P  E Ig(x)IP  xEG  So 17(G) has the upper majorant property with constant n1/P. On the other hand the validity of the upper majorant property with unit constant is not trivial. For any fixed value of p that is not even, the methods used in [9] show that this property fails on finite abelian groups with large enough order.  Chapter 1. Introduction.^  5  The results for the upper majorant problem were extended to noncompact locally compact abelian groups by Rains [22] and by Lee and Sunouchi [16] in the special case of tP.  Theorem 1.5 Let G be a noncompact locally compact abelian group. Then LP(G) has the 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 was prompted 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 showed that if tP(Z) has the upper majorant property with unit constant, then Fp(z) = lz IP-lsgn z operates on the positive definite sequences, that is, the composition Fp 0 0 is positive definite whenever 0 is a positive definite sequence. Rudin [24] had characterized the real-valued functions on the interval ] — 1, 1[ that operate on real-valued positive definite sequences as being those of the form 00  F(x) =  E cne, cn > o.  n=0  Since the restriction of Fp(z) to ]1, 1[ only has this form when p is even, Lee and Sunouchi had a proof that the upper majorant property with unit constant failed in EP(Z) whenever p was not even. It again followed that the upper majorant property also failed in these cases. The connection between the upper majorant property and functions that operate on positive definite sequences is enlightening. The earlier results in [9] on the failure of the upper 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 remark in which he referred to the work of Rudin [24], Herz [13], Rider [23], and Graham [10] on  Chapter 1. Introduction.^  6  functions that operate, but that comment was not made in regards to the upper majorant property 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 and only if the function Fp(z) = I zIP-lsgn z operates on the positive definite functions on G. By the "positive upper majorant property with unit constant" we mean the version of the upper majorant property where, if f, g E LP(G) with g a majorant of f and if f has nonnegative Fourier coefficients, then  lifIlp liglip.  We are overdue for some simpler terminology. The results of Fournier and Rains mentioned above tell us that on infinite locally compact abelian groups, LP(G) has the upper majorant property if and only if p is an even integer. However, when p is an even integer, we really have the upper majorant property with unit constant. This along with the triviality of the upper majorant property for finite groups suggests that what we should 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 the lower and upper majorant properties when generalizations to the certain Banach spaces were studied. Oberlin [19] said a space B has the "majorant property" if for every x E B there exists some X E B satisfying I-±(n)1 < X(n) for every n. Bachelis had shown that this is equivalent to the lower majorant property in the LP(G) setting. DechampsGondim, Lust-Piquard and Queffelec [7] also used the term "majorant property" with this meaning, and they defined what they called the "minorant property" on certain Banach spaces. In the LP(G) setting the latter property is equivalent to the upper majorant property. Later, Dechamps-Gondim, Lust-Piquard and Queffelec [8] again used the term minorant property for matrices but changed the meaning slightly by defining it to be the upper majorant property with unit constant. Hence, in like manner we make the following definitions.  Chapter 1. Introduction.^  7  Definition 1.6 Let G be a compact abelian group. We say that LP(G) has the minorant property 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 for every f, g E LP(G) with 0 < f(y) < "(y) for all y E 6, then  111111 ^lIglIp. With this terminology, we can restate the link between the upper majorant property and functions that operate by saying that LP(G) has the positive minorant property if and only if Fp(z) = I zIP-Isgn 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) has the 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 some low order groups, by showing that Fp(z) operates on the positive definite trigonometric polynomials. In doing so, we are able to add to the examples of Rider [23] and Graham  [10] (or see Graham and McGehee [11] p. 278-279) of functions that operate on positive definite functions on some groups but do not come under the classifications that have been studied thus far. In Chapter 3 we look into the matter of when we can say the minorant 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 with complex entries equipped with the p norm ii 'Hp. Any notation used that is associated  8  Chapter 1. Introduction.^  with matrices and which is not explicitly defined will follow Horn and Johnson [15]. For 1 < p < oo, define the p norm of A E M(C) by  11Allp = (tr [(A*A)i]) . Alternatively if ai (A) > 0-2(A) > ^> o(A) > 0 denote the singular values of A, then  1 A l l p = (E n u ( A ) P) i=1 For 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, . . . , n and 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, it differs from the convention in operator theory where lAl would denote the matrix fie—V— `A. We also differ from Dechamp-Gondim, Lust-Piquard and Queffelec [8] who no doubt wish to 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 (or  that B majorizes A). Definition 1.9 We say that (Mn(C),11 • lip) has the minorant property  if Mb^liBlip  whenever B is a majorant of A. As is usual, it is easy to prove that the minorant property holds when p is an even positive integer. That is, if p = 2k, where k E N and if lAl < B, then IA*Al < B*B and thus 1.24*Alk < (B*B)k so that  MAU,^tr RA*A) tr RA*A)k] tr [(B*B)k] •  Chapter 1. Introduction.^  9  Feller [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 is thus analogous to the one for LP(G) spaces when G is an infinite locally compact abelian group. As with groups, it is interesting to study what happens in the finite cases, that is, 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 not have the minorant property. Letting N(p) denote the smallest n for which the minorant property fails, Simon had shown that  N(p) < 2 [] + 5. Dechamps-Gondim, Lust-Piquard 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 the matrices we consider, the more likely it is that the minorant property will fail, or that for a fixed n, the larger a noneven p is, the more likely (Mn(C), ii • 11p) is to have the minorant property. A specific example of this tendency was demonstrated by DechampsGondim, Lust-Piquard 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. We will consider the positive minorant property on matrices. Definition 1.10 The space (Mn(C),Vilp) is said to have the positive minorant property if 11Allp < liBlin whenever 0 < A < B. Dechamps-Gondim, Lust-Piquard and Queffelec showed that the positive minorant  Chapter 1. Introduction.^  10  property 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 when G 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 — 2 and p 0 2k. For n odd and n > 5, the space (Mn(C), 11 ' 11p) does not have the positive minorant property if 1 < p < n — 3 and p 0 2k. In Chapter 5 we will improve this using an entirely new method to show that (Mn(C), 11 • 11) does not have the positive minorant property if 0 < p < 2(n — 1) and p 0 2k. Our method actually shows that the positive minorant property fails in the subspace of symmetric matrices in (M (C), 11 ' 11p)-  Chapter 2  General Facts About Minorant Properties.  For this chapter, G will denote a compact abelian group. We present some apparently weaker 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 the function Fp(z) =iziP lsgn z operating on the positive definite trigonometric polynomials. -  Also, we prove some theorems of the type: LP(G) has the (positive) minorant property implies 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 LP-norm, 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  majorized  by g (for the positive minorant version of the proof i( y) > 0 for all -y E as well). Take -  an approximate identity of trigonometric polynomials MI with /5:„ > 0 (see [14] page 88, 11  Chapter 2. General Facts About Minorant Properties. ^  12  Theorem 28.53 and in particular note (i) and (v) for existence of such an approximate identity). Then there is a sequence {Pn} which is a subset of {Pa} for which Pn * f —* f and Pn * g -4 g in 17-norm (see [14] page 273, Remark (b) of 32.33). So for each E > 0 there exists N such that n > N implies  Ilf —Pn*fllp < lig Pn * gill, < —  E  E.  Also  (Pn* g)"ey) = Pneyrgey) ?- 1371(7)1/(7)1 as .0(7) -.  If(7)1  = l(Pn * fY(7)1 which implies that IIPn * glip > iiPn * fl1p since we have the minorant property for trigonometric polynomials. Hence  lIglIp — IlfIlp = 11 911 p — 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. 0 The proposition above is stated explicitly in Rains [21] for the minorant property and was 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.  13  Chapter 2. General Facts About Minorant Properties.^  Definition 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, then  iiffip 5_ Nip. Theorem 2.3 Let G be a compact abelian group. For p > 1 the space LP(G) has the minorant 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 minorant property, Proposition 2.1 says it suffices to prove that (T ri g (G) , II• 11) has the minorant property. Let f,g E Trig(G) with for all -y E  O.  Let E = {-y EÔ : :g(7) > 0}. The set E is finite, as g is a trigonometric polynomial. For each -y E E, let a7 be the unique number in [0, 7/2] for which  cos 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 that kl(7) = :9(7)ei(°^I±a7) .1c2(7) = .4(7)e'  ^for all -y E  Then  ki (7) + k2(7) = :0 (7)(e1(e-f +ao + ei(e,-")) = .4(-y)[cos(07 + ay) + cos(97 — ^i(sin(07 + a7) + sin(07  —  a7))]  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) =^2 and  kJ. + k2 . f - ^ 2  Thus i  II f Ilp =^Illci + k21Ip 1^1 5_-PIG+ -11k21Ip by the triangle inequality 2^2  1^1 —blip ± —11.911p as g is an exact majorant of k1 and k2 2^2^  = kip and hence LP(G) has the minorant property. 0 Notation. Let E be a finite subset of  d and h E Ll(G). We define hE to be the  trigonometric polynomial for which  hE(7) =  1 h(7) if 7 E E 0 otherwise.  Theorem 2.4 The space LP(G) for 1 < p < co has the positive minorant property if and only if for every h E LP(G) with -1(7) > 0 for all 7 E  we have  IlhEllp < IlhIlp.  a and every finite subset E of a  Chapter 2. General Facts About Minorant Properties.^  15  Proof. If L-P(G) has the positive minorant property and h E LP(G) and E any finite subset of  6, then hE is such that 0 _< 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 have  IlhEllp^11h11, Then by Proposition 2.1, to prove LP(G) has the positive minorant property it suffices to prove that whenever f, h E Trig(G) and  0 < (-y) < it(7), then II/11p 5Let E = {-y e^:  In.  h('y) > 0}. Label the elements of E so that if E = {71,-y2,^,-ym}  and we define ai = 1(71)  then ai < a2 <^< am. We may assume that am = 1, since if am < 1, we can replace  h with anzh and still have that 0 < j('y) < it(-y) for all -y E O. Define subsets Ei of E so that Et^7i+11  •••7  7n1}.  We show that it is possible to choose real numbers Ai > 0 with Ell,. Ai = 1 and  I = E71_1 AihEs. In fact, Ai = ai — =^ai-i)hE„ then  "(7/c) =  E(ai — i=1  E(ai — ai-1)11(7k) 1=1  ^where we define ao to be 0. Indeed, if  Chapter 2. General Facts About Minorant Properties. ^  16  = Ral — ao) + (a2 — ai) +^+ (ak — ak--1)111(7k) = s'Yk, ) = 1(7k), so that = f and g = f. Note also that Er_i Ai = Urn = 1. Finally,  IlfIlp =  E  1=1  E i= 1  AihEi  Ai ilhEi lip  EAdhllp  for p > 1 by assumption  i=1  = Ilhilp. 0 The theorem above is a generalization of Hardy and Littlewood's Theorem 4. The next theorem uses a generalization of a Hardy and Littlewood argument, where they showed 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) for  the positive minorant property). Then fn , gn E LP(G) and (remove I • I throughout to prove the positive minorant property),  11n(7)1 = li* • • • * i(7)I Ita • • •  fa kr^— A2 — . • • — An_oi(An_i) • • • .f(Ai) dAn-i • • • dAil  < fa... fa 'kr — A1 — A2 —^— An-1)1If (An-1)I^If (A1)IdAn-i^dAl <^. . . fa :0(-y—^— A2 —^— An-1).4(An-1) • • • .4(Al)dAn-1 ^dAl =^*^* 4(7) (convolution of n terms)  =  17  Chapter 2. General Facts About Minorant Properties. ^  As LP(G) has the (positive) minorant property, it follows that  and so  or  as required. 0 2.2 The positive minorant property.  The derivative of F(t) = II f + t-yllp where f E Trig(G) and -y EÔ is computed in Lemma 2 of Hardy and Littlewood [12] for G = T and also in the proof of Lemma 2 in Bachelis' paper [1] for compact abelian groups. We will compute a little more, the derivative of F(t) II f thilp where f,h E T ri g (G) as this will be required for the next theorem. First though, we borrow some notation from Bachelis. Definition 2.6 For! E LP(G), with 1 < p < oo and q so that p-1 q-1 = 1, let Aq(f)(x) = fix)IP-1 sgn f (x) f for all x E G, where  sgn z =  zilzi if z^0 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 real-valued function F(t) = JJJ + thlip for t E R. The function F is continuous and differentiable when 1 <p < co and F'(t)^  fa Re [Aq(f thr(-y)h(-y)] d-y.  ^ ^  Chapter 2. General Facts About Minorant Properties. ^  Proof. The function F is continuous since for real numbers IF(t) - F(s)I  18  s and t,  = 1111 + thllp — Ill + shllp I  11(i + th) - ( f + sh)lip  = il(t - s)hiip =  It — sl IlhIlp-  S o i t — si < 6/11 hIl p implies IF(t) - F(s)I < 0. As  F(t) we have  I'M  I  = (LI f (x) + th(x)IP dx)  1 P ,  1 d thr—P— I If (x) + th(x)IP dx , = —1If + P^P dt G  and provided (d dt)I f (x) + th(x)IP is continuous we may differentiate under the integral sign (G is compact). Since  If (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))) l  x (h(x)(f(x) + th(x)) + (f (x) + th(x))h(x)) =^If (x) + th(x)IP-2 (h(x)(f(x) . P2 I f (x) + th(x)IP 1 -  + th(x)) + (1(x) + th(x))h(x))  -  x (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 well-defined if p> 1. Hence  F' (t) =  1,  d  HI f + Mr ^I — I f (x) + th(x)IP dx p^P^G dt  Chapter 2. General Facts About Minorant Properties.^  19  . ;-- 1G (Aq(f + th)(x)h(x) + Aq(f + th)(x)h(x)) dx Aq(f + th)(x) jak-y)-y (x) d-y dx +..2- JG Aq(f + th)(x) fait(-y)-y (x) d-y dx as h(x) = fa ii(-y)7(x) dx and this is in fact just a finite sum as h E T rig (G) . Thus we -  may interchange the order of integration to get  1 F' (t) = ,f, LAq(f +th)(x)7(x) dxh(y) d-y + fa. IGAg(f 1-th)(x)7(x)dxit(7)d-y 1 ^ =^ja A q( f + th)" (-y)hey) + Aq( f + th) (7)y) d-y LRe(Aq(i -f-th)^(7)it(7))&y.  0  Remark (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 is bounded away from zero. As f and h are continuous and G is compact, it then suffices that 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 positive minorant property on LP(G) is equivalent to saying that the functional Aq(•) operates on the positive definite trigonometric polynomials. Theorem 2.8 Let G be a compact abelian group. Then LP(G) with 1 < p < oo has  the positive minorant property if and only if Aq(f)" > 0 whenever f is a trigonometric polynomial with f  > 0.  Proof. Suppose LP(G) has the positive minorant property and let f E Trig(G) with  I > 0. Putting g = f + tx for some x E d, we have that g majorizes f so  II/11p 5_ Ilf + tx11p.  when t > 0 and  Chapter 2. General Facts About Minorant Properties. ^  Let F(t) =  If + txlip- By Lemma 2.7, the function  20  F is differentiable and  F'(t) = LRe(Aq(f -Ftx)-(7)5,(7),d-y = Re [Aq(f + tX)-(X)] (for proof when p = 1, use Remark (a) above and read "F'(t)" as "F'(t+)" in what follows). 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. So  AqUY (X) = Re(Aq(1)-(X)) = F'(0) .  hill  Ilf + txllp — Ilfilp  t-4o^  > 0  since  t  Iv + txllp — IlfIlp > o for t > O.  Conversely, suppose Aq(•) operates on positive definite trigonometric polynomials. To show LP(G) has the positive minorant property, it suffices to show that it has the property for trigonometric polynomials by Proposition 2.1. Let f, g E Trig(G) with 0 < f('y) < .4(7) for all 7 E 0 and define  F(t) = Ilf + t(g — Plip for t E [0,1]. By Lemma 2.7, the function F is differentiable and  Pm = fa  Re [(Aa + t(g — .f))-(7)] (9 — f)-(7)d7  > 0 since (g — f(-y) > 0 and (f -Ft(g — f))-(-y) ..> 0, so that Aq(9 + t(g — f))-(-y) > 0 by the assumption that Aq(•) operates on positive definite trigonometric polynomials.  Chapter 2. General Facts About Minorant Properties. ^  21  Hence,  lIglIp - Ilf Ilp ^F(1) — F(0) F'()(1 — 0) for some 4" E]0, 1[, by the mean value theorem F' (e) >0 and so  If Ilp < lIglIp as required. 0  Remark. When 0 < p < 1, the first half of Theorem 2.8 still works for trigonometric polynomials f with no zeros (see Remark (b) following the proof of Lemma 2.7). Thus the positive minorant property on LP(G) implies that Aq(f)- > 0 for all nowhere-zero trigonometric polynomials f with j > 0. Corollary 2.9 Let G be an infinite compact abelian group. Then LP(G) does not have the positive minorant property for noneven p > 1. Proof. The proof divides up into two cases, when G is a non-exceptional group and when it is exceptional. Any locally compact abelian group is exceptional if it is the direct product 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 positive definite functions on a non-exceptional group must be infinitely differentiable. When p is not even, the function Fp(z) = Izi sgn z is not smooth at zero, and therefore does not operate on the class of positive definite functions. By the theorem above, LP(G) does not have the positive minorant property. If G is an exceptional group, then it contains a subgroup isomorphic to Z3") because G is infinite. Hence the generalization of Theorem 4 of Herz [13] applies. That is, any function defined on the interval [ —1, 1] which operates on the positive definite functions is smooth near zero. So again, Fp cannot operate on the class of positive definite functions  Chapter 2. General Facts About Minorant Properties. ^  22  defined on G when p is noneven and so the space LP(G) doesn't have the positive minorant property. 0 Another 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 > 0 whenever f E Trig(g) and .1. > 0. Now )tq(f)(x)  = if (x)IP-1 sgn 1(x) II f 1113;1  so that Fp(z) = I zIP-isgn z operates on P(G) = If E Trig(G) : i > 01. We claim that F+2(z) = IzIP-Fisgn 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 positive definite (the latter because  Fp operates on P(G)) and the product of positive definite  functions is positive definite, that is, if 0, 0 E P(G) then  (Y e) = = 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 polynomials  f with f > 0. So by Theorem 2.8 this means that LP+2(G) has the positive minorant property. 0  Chapter 3  Counterexamples.  Let G be a finite abelian group. The focus of this chapter will be to determine some  11(G) (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 compact abelian group G, the space 11(G) (G) has the minorant property if and only if p is an even integer 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 holds for all p > 2. We get the same type of behaviour for matrices. If  cp(e) denotes the Banach space of  all 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 other hand, Dechamps-Gondim, Lust-Piquard, 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 it should be noted that whenever D '(G) fails to have the minorant or positive minorant property for a finite abelian group G of order n, then (M (C), ' 11) also fails to have the property. This is because we can embed 11(G) into (M„(C), li ' lip). Simon [28] used this fact for cyclic groups. Before proving that there is such an embedding in general, first consider a couple of examples. 23  Chapter 3. Counterexamples.^  24  Example 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 + b-y + cry2 + dry3 E LP(Z4), define T: L7'(Z4)^(M4(C),II • IIp) such that ladcb\ T(f) =  badc cbad \d c b al  Example 2. Let  4 = {1,x, y, xy : x2 = y2 = 1} and 222 = 11,7,X,7X 7(x) = —1,7(0 = 1,x(x) = 1,X(Y) = —11. For each f=a+ fry cx  ct yx, define T : LP(4) ---+ (M4(C), II 11p) such that -  la b c d\ T(f) =  badc dab \dcba  In both cases / < .4 implies that T(f) < T(g). In the proof of the next proposition we show 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 mapping  T : LP(G)^(M„(C),II lip) which is injective, preserves order among norms, and if 1/(7)I^.07), then  IT(f)I^T(g).  25  Chapter 3. Counterexamples.^  Proof. Let -y E  6 and define the operator T7 : 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 compute the matrix of T-y, denoted by m(T7). As Tai = -y-yi E  = {71, , NI we can  6, the ith column of the  matrix of Ty has all entries equal to zero except for one which is 1. Also, as Ty-y1 = T77; if and only if i = j, the matrix of T7 is a permutation matrix. Moreover, if y, x E  6 we  have Tr-yk = Tx-yk if and only if -y = x, so for i and j, m(T7)1i = 1 means m(Tx)ii = 0 when 7 x. Thus the mapping T : Trig(G) M(C) defined by  T(i)(E^f(7)-y) 7E-a  =^f(7)Tri(T-r) 7EG  is injective. Also, if for f, g E Trig(G) the function g is a majorant of f, then clearly T(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 mapping T. 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) and T( f) are when G is a cyclic group). In fact, if G = , g,}, then the eigenvalues of m(T7) are the diagonal entries in Ary = diag (7(g1), • • • 7(gn))  ^  •^ •  Chapter 3. Counterexamples.^  26  and with corresponding eigenvectors being the columns of  X = (xii) = 1 Note 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 that  m (T7)X = XA7. Now ^(771(Ty)A  j  = ^m(T)ikxki  k=1 = Xt where "Y'Yi =  1 \Ft^71(gi) 1  ^VT17(gi)-Yi(gi) = -y(gi)xii •  (X A7)ii.  Also X-1 = X* since ^(XX*)2^=^1 7,^'Yz (gle )73 (gk) — k=1  =  We compute  X-1T(f)X = X-1 (E ky)m(71y)) X  =E ,E  -a  7E5  hy)X-1m(T7)X  "Ye  Chapter 3. Counterexamples. ^  = E  27  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),.  •• , E :/(7)7(gn))  ^yEa  ^yE&  = diag (.f(gi), • , f(g)). Taking the conjugate transpose of the last equation gives  X*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 and  IIT(f) II  and thus T preserves order among norms. 0 3.2 Cyclic groups. It is known [25, p. 255, Theorem B8} that every finite abelian group is a direct sum  of cyclic groups. Rains [22] proves the following proposition, which also holds for the positive minorant property. Proposition 3.2 Suppose that G1 and G2 are locally compact abelian groups and suppose either LP(G1) or LP(G2) fails to have the minorant property. Then LP(Gi x G2) does not have the minorant property.  Chapter 3. Counterexamples.^  28  Hence a good place to start in finding LP(G) spaces which do not have the minorant property is with the cyclic groups. The following theorem is a modification of Theorem 1(a) of Dechamps-Gondim, Lust-Piquard and Queffelec [8]. Theorem 3.3 For p > 0 and noneven, the space LP(Z) does not have the minorant  property for n = [p12] ± 2. Equivalently, for fixed n, the space LP(Z) does not have the minorant property for 2(n  —  2) < p < 2(n  —  1).  Proof. Suppose n = [0] + 2 and let k and m be nonnegative integers such that n = 2km, where in is odd.^ Put  c = e2 - k7ri  ^  (3.1)  (3.2)  and choosing a value of t E]0, 1[ and any 7 E Zn which is element of order n, let  f = V1 — t2 + tcy ,  (3.3)  g = 1/1 — t2 + try . Then clearly g is a majorant of f and we will show that We will need to compute the  1,13  IlfIlp > lIglIp.  norm of a function of the form h = V1 — t2 + tu  where lui 17- 1. Now  Th = (-V1 — t2 + ti7)(V1 — t + tu) = 1 — t2 + tV1 — t2(u + rt) + t2  ^  (3.4)  = 1 + p(u + ri) where  p = tV1 — t2 < 1/2^  (3.5)  ^  29  Chapter 3. Counterexamples.^  if t < 1/2 and thus plu + iI < 2p < 1. Thus ▪  — 1 I Crth)P/2dX 71 G  (3.6)  =— 1 I (1 p (u ii))P/2 dx.^ n G  Recall that for a real number a and an integer s, (as) = 1-4 (a i=o  and  (1 + z) =^(a)zs s=o ■s/ always converges whenever Izi < 1. Hence, due to (3.5) we can write oo  (1  ± P(tt +17))a = E  s=o  (lps(u  + T).  (3.7)  where a =1..  (3.8)  Note that as p is noneven, (a) (a) ())  ( a —1  > 0,  but  (a)  n)  <0  (3.9)  by our choice of n = [p/2] + 2, and sgn I) n+r  =  for r > 0.  (3.10)  Consider  f [(1+ p(-y + -TO — (1+ p(cy + c7))P12] dx using (3.6) ^11g11';, - 11f1';, =^ • G (a)Ps((Y + 7)8 (cT + c7)8)) dx using (3.7) =^ G s =0 S n I^  t  (a) Ps f^(s.)  n s=o s^G  e7i(7y-)s-j — (c-)i cc—Ty1  dx  ^ ^  •  30  Chapter 3. Counterexamples.^  a)p^4 s^(s) ^(s  (1./. 72i-8 — s=0 _^j=0^n G oo^s^s  - (1  (c7)2j--sdx)  1^2.; — c2i-81^  s P^ (n 7'0 ( as) =0 (i)^  JG^8(x) dx) .  (3.11)  We claim this is a sum of negative terms. To see this, first note that 1 72-i-8(x)dx = 1 — nG • if 2j — s 0 (mod n), and that otherwise the integral is zero since 7 was chosen to be an element of order n. When 2j — s^0 (mod n), there is an integer t such that 2j — s = ne.^  (3.12)  Then 1 — C2i-8 = 1 — •  1 — e2-knilr1 by (3.2)  •  1 — em^by (3.1)  •  1 — (-1)t as m is odd 2 if t is odd 0 if t is even.  (3.13)  So all the terms in (3.11) are nonnegative with the exception of (:). By (3.9), for  s = 0,1, ... , n — 1 we have that (:) > 0, but then —s<2j—s<s as 0 <j<s so that —(n — 1) < 2j — s < n — 1,  Chapter 3. Counterexamples.^  31  and the only time 2j — s^0 (mod n) is when 2j — s = 0. But then (3.13) says 1 — c2j-s = 0. Thus none of the terms (a9) survive when s = 0,1, ... , n — 1. Also, (3.10) says ( a ) = for r > 0. sgn (-1Y+1 n+r When s = n + r and r is odd, then (a) > 0. For the positive term (a) s^ s to survive in the sum (3.11), we need (3.12) to hold, and then 2j  —  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), but then ne is odd so that s = 2j — ne is odd, but s = n + r is even, so that again (:) does not 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 — c2i-8 = 2, so that  Or  as 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 when n — 2 = [12] 2  —1 < n — 2 < p — 2 < 2n —4 < p,  Chapter 3. Counterexamples. ^  32  that is, when 2n-4 < p < 2n-2. 0 Corollary 3.4 For noneven p > 0, the space LP(ZN) does not have the minorant property for N = n + 2k, k E {0} U N, where n = [p12] + 2. Alternatively, for fixed N, the space LP(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 the  same, noting that ( a )^( a ) sgn^= sgn N + r^n+2k+r) . (_ 1 )2k-Er+1 = (- 1)r+1.  Alternatively, if we fix N and k E {0} U N, then LP(ZN) does not have the minorant property for indices p such that P N= {j+2+2k, 2  that is, P N — 2 — 2k = [—  2]  Or  P  — 1 < N — 2 — 2k < P  Or  p — 2 < 2N — 4 — 4k < p,  and finally 2(N — 2) — 4k < p < 2(N -- 1) — 4k. 0  Chapter 3. Counterexamples.^  33  Given a cyclic group of order N, the above corollary gives us intervals of length 2 in which LP(ZN) does not have the minorant property, but these intervals for p are separated by other intervals of the same length in which we don't know anything with regards to the minorant property. However, Theorem 2.5 can sometimes help to fill in some of the gaps. For example, for LP(Z4), Corollary 3.4 tells us that these spaces do not have the minorant property for p E] 0, 2[ U] 4, 6[. Theorem 2.5 tells us that if ./2(ZN) had the minorant 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 minorant property. Theorem 4(b) of Dechamps-Gondim, Lust-Piquard, and Queffelec [8] says that (M (C), II IIp) does not have the positive minorant property for noneven p with 1 < p < n — 2 for even n > 4, and that the property fails with 1 < p < n — 3 for odd n > 5. We shall use the same argument as Dechamps-Gondim, Lust-Piquard and Queffelec, but we show that LP(ZN) (and hence (MN(C), II• Hp)) does not have the positive minorant 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 the positive 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 and 0 < p < 2RN — 1)/4  Proof. Let f^ r2 + r-y where 0 < r < 1 and -y E ZN is the character for which —  ,y(k) = e2irki/N for k = 0, 1,^, N — 1. Put g = V1 — r2 + rry + t-ym where t > 0 and m = [p/2] + 1.  34  Chapter 3. Counterexamples. ^  Clearly 0 < I <"j and we will show that  IIfIIp > 11g1Ip for some choices of r and t. By  Lemma 2.7 and as in Theorem 2.8, for F(t) = Ilf  +t-rmllp,  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 > 0 by the Remark (b) following Lemma 2.7. By showing that F'(0) = Aq(f)eym) <0 we will prove that  lIglIp  —  II/11p = Ill + t7mIlp — II/11p < o  for small positive t. Now  hi, ^+^— II/11p =  Aq (f)- (m)  t-40  and noting that f has the same form as h appearing in (3.4), and that here p = r-V1  —  r2  then 1J fyp12)-1 tym ^ dx G Mir  Aq(freym)  1  JG (1 + p(-y + ry))"3 f-ym dx where /3= (p/ 2) — 1  1 nlIfIlr  jG [ :t°0  CS) Ps (7 ± 77)3]  when p < frym dx^  1  1 ^f [ct (0) ps^(s.)72j d ( 1 r2 + r-y)-ym dx fir G 3_0 S^_0 nlif^ 1 nIlf As 0 = (p/2)  —  Ilr :E3=o (i3s)Psi=o (si) IG(Vi^  272j 3+m ± ni2j-8+m+1) dx. (3.14)  1 and p noneven,  (0)^0(0 — 1) ... (0 — )^s!  s + 1) > 0  for s = 0,1, ... [p/2] but () < 0 for s = m = [p/2] +1. Note that (V1 — r2,72i- 8+m 7.)(23-s+m+1)dx = 0  Chapter 3. Counterexamples. ^  for s= 0, ... ,[p/2] as  L  35  -yn(x)dx = 0  except when n 0 (mod N), but N > 2[p/2]+ 3 whereas 1 < 2j — s + m < 2[p/2] + 1 because for s < [p/2], and 0 < j < s, we have 2j — s+m > — s+m  — [p/2]+ [p/2] + 1 =1 and 2j—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^  r72j-8i-m+1)  - r272j-s+-  dx  j=0^jG  r272.i r72J+1) dx (3.15)  +r if N = 2[p/2]+ 3 if N > 2[p/2] + 3 as -y2i^1 in the sum (3.15) only when j = 0, and 72-i+1 may be 1 when j = s if N = 2[p/2] + 3, since then 2j + 1 = 2s + 1 = 2[p/2] + 3. Hence (3.14) becomes lim  Ilf + t-rmllp — IlfIlp  t_,0^  —  ^1  IffIlrl  (le n^— r2m+1 + o (rm+1)  Chapter 3. Counterexamples.^  36  as p = 7-V1 — r2, so that for sufficiently small r > 0 we have lim  Ilf + t7mllp - VII,' < 0  t_,0^t  since (S) m < 0. Hence for sufficiently small t > 0, we have  if + tlirnlip = Ilgiip < Ilf Ilp as required. That is, LP(ZN) does not have the positive minorant property whenever  N > 2[p/2] + 3. Alternatively, for fixed N, the space LP(ZN) does not have the property if p satisfies  N — 1 > 2 [1-23-] + 2 N-1 > [1]+1EN so 2  [N 2— 1] > [i] + 1 > I--) as x — 1 < [x] for all x, and hence^p < 2  [N; 1] .  0  3.3 Beyond cyclic groups. All of the counterexamples considered so far have been for cyclic groups. Let G be a finite abelian group of order n. If p is a prime number such that p divides n, then G will 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. Let  m = Hp.  (p) .  Pi n  Then Zni is isomorphic to the biggest cyclic subgroup of G, and in fact G'-'-'-' , Zm x G1 for some group GI. Thus Proposition 3.2 applies so that failure of the minorant properties  Chapter 3. Counterexamples.^  37  on LP(Z,L) will mean failure on LP(G). However, for some finite abelian groups G there are LP(G) spaces which will not have either the minorant or positive minorant property but for which LP(Z,z) has the corresponding property. In this sense finding the biggest cyclic subgroup of G does not provide the best counterexamples. Another way to find counterexamples to the minorant property when G is not cyclic is found in Fournier's paper [9]. There we have counterexamples of the minorant property for 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 the power a. However, for some groups G containing a relatively high number of copies of a cyclic group, the method of Fournier may produce better counterexamples than those provided by finding the biggest cyclic subgroup of G.  Chapter 4  Special 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 this seems to be a difficult problem. To illustrate, note that to date the question about when the minorant property holds for matrices equipped with the p-norms has only been answered for 3 x 3 matrices [8]. As the size of matrices or the order of groups increases, it becomes increasingly more difficult to determine exactly when the properties hold. On the other hand, as we consider the counterexamples from the last chapter along with some of the positive results we can find for low-order cases, a definite pattern seems to emerge in the relation between p and the order of the group or size of the matrix being considered. Before we make a conjecture about this pattern, we present some results on a few low-order groups and matrices.  4.1 Minorant properties on low-order 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 minorant property 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.  38  ^  Chapter 4. Special Cases.^  39  Conversely, if p > 2, it follows from this embedding that to prove the minorant property on LP(Z2) it will suffice to prove it on (M2(C), • Hp). To do this we follow the same lines of argument given by Dechamps-Gondim, Lust-Piquard and Queffelec [8] in their Theorem 2 for 3 x 3 matrices. Let A,B  E  M2(C) with IAI < B. Using the same type of construction used in  Theorem 2.3, we may assume IAI = B. Let Al > A2 and respectively pi > //2 denote the eigenvalues of A*A and B*B. We wish to prove that 104 < II BII, in other words that tr ((A*A)P1'2) < tr ((B*B)1)/2), or ^P/2 , p/2 ÁT/2 ± AP/2 2^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 the minorant property, condition (4.1) holds and g(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)  40  Chapter 4. Special Cases.^  so that g(1) = 0.^  (4.5)  We now prove that g has too many zeros, that is, the function g contradicts Lemma 2 of Dechamps-Gondim, Lust-Piquard and Queffelec [8] which generalizes Descartes' rule of signs. The lemma says that if 0 < al < < a„ and  ^P(x)  ao +^+ + anxan  with the ai real and not all zero, then the number of positive zeros of P is less than or equal to the number of sign changes in the coefficients. Note that A1 < 121 since A1 = 11A11.0 = sup{ I1Av112 :  v E 1R2,  II112 < 11 and v  IlAvI12^IIBlvi 112^1113110011v1I2^pi when 11v112 < 1. In fact A1 < pi, since otherwise g(r) a- 0 which would contradict (4.2). Thus (4.4) says A2 > p2 and we get  P2 < A2 < Al <  ^  (4.6)  If p2 > 0, we may assume that p2 > 1 by multiplying A and B by an appropriate constant. Then 0< ln /22 < ln A2 < ln Ai < ln pi and 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.4  Thus P(x) has zeros at el ,e (by (4.3) and (4.5) respectively) and also at 1. Since there are only two sign changes in the coefficients, the lemma implies that there are at most 2 positive 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 — xa 4. (  Chapter 4. Special Cases.^  41  Now 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 the minorant property for p> 2 and so the proof is complete. 0 Since we've just seen that the argument of Dechamps-Gondim, Lust-Piquard and Queffelec works for 2 x 2 matrices as well as 3 x 3 matrices, the obvious question arises of the suitability of the method for general n x n matrices. Unfortunately, as soon as we consider 4 x 4 matrices or bigger, the "polynomial" P(x) in the argument can have four or more sign changes in the coefficients. This means that unless a way can be found to produce 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 be considerably easier thanks to Theorem 2.8.  Proposition 4.2 For 0 < p < oo, the space LP(Z2) has the positive minorant property if and only if p> 1. Proof. Recall that by Theorem 2.8, when p > 1 the space LP(Z2) has the positive minorant property if and only if Aq(h)-(-y) > 0 for all h E LP(Z2) with 4-y) > 0 for all  -y E 2-2. Let h = a + br where r E 2-2 with r(0) = land r(1) = —1 and a, b> 0. We may assume a > b as h may be replaced with rh if otherwise. Thus, Aq(h)(0) =  = and Aq(h)(1) =  I h (0) IP- 1 sgn h (0)  iihiir,-1 (a + b)P-1  iihiri (a b)P-1 —  IlhIlf, --1^  (= 0 if a = b).  • Chapter 4. Special Cases.^  42  Consequently,  1  Aq(h)-(1)  211h1r1  [(a + b)'-1 ± (a _ b)P-11  >  0  •  1 ^ [(a + b)"' — (a — b)P-1] 211hIlrl 0 ifp—l>0  and Aq(h)^(r)  while 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 2.8 applies since a br is never zero (see Remark which immediately follows the proof of this theorem), and so LP(Z2) does not have the positive minorant property. 0 Dechamps-Gondim, Lust-Piquard, and Queffelec [8] remarked that it would be interesting to know if the positive minorant property is actually weaker than the minorant property. The last two propositions demonstrate that the properties are indeed different for 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. ) does not have the positive minorant property for 0 < p <1. The main theorem in Chapter 5 will further show that the property also fails to hold for 1 < p < 2 while Proposition 4.1 confirms that (M2 (C), II • lip) has the minorant, and hence the positive minorant property for p> 2. Hence we have proved  Proposition 4.3 For 0 < p < co, the space (M2(C),11 • ilp) has the positive minorant property if and only if p> 2.  Chapter 4. Special Cases.^  43  Proposition 4.4 For 0 < p < oo, the spaces LP(Z3) and (M3(C),11 ' 11) have the mino-  rant 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 the minorant property. As 2[p/2] + 3 = 3 for 0 < p < 2, Theorem 3.5 says LP(Z3) does not even 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 minorant property for 0 < p < 4 and p 0 2. Conversely, the space (M3(C), II • lip) P has the minorant property (and hence so does LP(Z3)) for p > 4 by Theorem 2 in the paper by Dechamps-Gondim, Lust-Piquard, and Queffelec [8]. 0 As with (M2(C), li • 11p), the minorant and positive minorant properties are equivalent for the space (M3(C) ,II • lip) • Proposition 4.5 For 0 < p < oo, the space (11/13(C),11 • 11) has the positive minorant  property 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 Proposition 4.4 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 the matrix 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 if  B > 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 > 0  44  Chapter 4. Special Cases.^  and thus  Ea B* = (B*13)(P12)-1B* > 0. 0 The following lemmas will assist us in proving the positive minorant on some of the  LP(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 define the function H R n^Ill by H (Xi, . Xn)  101177: where h = Exai. i=1  Then for p > 1 the function H is convex in xi and if p > 2, the partial derivative 01110xi is 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 > 1 and Ak < 1 for k = 1, 2. Since H is convex in xi,  82H> 0  axF —  when the partial derivative exists (that is when p > 2) and so 01110x2 is an increasing -  function. 0  Lemma 4.7 Let G be a finite abelian group and p > 2. If for all h E LP(G) with it(-y) > 0 for every 7 E Ô and it(1) = 0 we have Aq(h)- (1) > 0, then LP(G) has the positive minorant property.  Chapter 4. Special Cases.^  45  Proof. 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 > 0 since if Aq(h)^(-y) <0 for some -y E 6, then If?' E LP(G) with (h-7) > 0 and Aq(h7yr(1) =  L;I:Aq(117)7)(x) zEG 1^ih(x)7(x)17-Isgn (h(x)3)  — E n xEc^Hh11 1^Ih(x)1P-isgn (h(x)) =— E 7(x) zEG^Ilh11;^ Aq(h)^(y)  <0. Let G =^7n1 with a- 1. Let h = ExEGxi-yi where xi > 0. By the lemma above, since p > 2, the function n  = 111111;  is such that OH/493cl is increasing in xi and hence OH,^OH Ox1  kX1, X2,  • • •  xn)  in  >^kV, X2, • • • aCn)• aTi  (4.7)  But 1 x-, 11(Xi, X21 • • • xn) =^2_, xEG  xi-yi(x)  n  and so OH  (T1,...,xn)  =^E  p-1  Exi7i(x)^sgn  sEG i=1  = Pii  1111r Aq(hr (1).  (E xi-yi(x)) 1=1  Thus to show Aq(h)-(1) > 0 for all it > 0 it suffices by (4.7) to show this for it > 0 and h(1) = O.^0  ^  46  Chapter 4. Special Cases.^  Proposition 4.8 For 0 < p < oo, the space LP(Z3) has the positive minorant property if and only if p > 2. Proof. Theorem 3.5 tells us that for 0 < p < 2 the space LP(7Z3) does not have the  positive minorant property, since 2 [12] + 3 = 3. 2 To show LP(Z3) does have the positive minorant property for p > 2, the lemma above shows it will suffice to prove that Aq(h)"(1) > 0 whenever h > 0 with ii(1) = 0. Let  = {12: (0),7(1) = P =  -1+^i O  2^,'Y(2) = P2}  Then for h E LP(Z3) with 14-0 , it(-y2) > 0 and ii(1) = Owe have, Aq(h)-(1) =^[Aq(h)(0) + q(h)(1) + Aq(h)(2)] ^1^-^^ [1h(7) + her2)1P-1 311hlir _wimpjj (y 2 )p2P_l sgn (it(.y)p ii(72)p2) +111(7)p2+11(72)pr-lsglick-op2+11(72)p)]  ^31I h'^ Po) + 11(72))P-1 —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 simple application of the triangle inequality shows that lit(y)p it(-y2)p21 5_ h(y) + il(-y2), we get Aq(h) (1) = 11(7) +1472) PO) 311/1111r1 >O as p > 2. 0  + 1/(72))P-2 —11/(7)P + 11(72)P2IP-21  Proposition 4.9 For 0 < p < oo, the space LP(4) has the positive minorant property if and only if p > 2.  47  Chapter 4. Special Cases.^  Proof. Let 4 = {1, x, y, xy : x2 . y2 = 1} and  13 =  {  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)  - h(7x) h(xy)^= h(1) - h(7) - h(x) + 11(7X). h(Y)^= 11(1) + h(7) — h(x)  (4.8)  Lemma 4.7 above shows that it will suffice to show Aq(h)-(1) > 0 whenever Ii > 0 with h(1) = 0. We have 1 Aq(h) (1) =^-1 [0(1)11-1sgn h(i) + Ih(x)IP Isgn h(x) 411h1r, -  +1h(y)risgnh(y) + ih(xy)I0-1 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) > 0 Case 2. When one of h(x), h(y), h(xy) < 0 Case 3. When two of h(x), h(y), h(xy) < 0 Case 4. When h(x) , h(y), h(xy) < 0 In Case 1, with h(1) > 0, h(x) > 0, h(y) > 0, and h(xy) > 0, the conclusion that Aq(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 SO that if h(z) <0, then --11/(z)IP-2h(z) < h(1)P-1 for p > 1  Chapter 4. Special Cases.^  48  or h(1)P-1 -I- Ih(z)IP-2h(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 by symmetry. Recalling that we may assume h(1) = 0 in (4.8) we have ^h(y)  ^  147) — h(x) — herx) = 147) —11(x) +14-rx) — 211(7x) =  ^and  h(xy)  —h(x)  ^  — 2h(7x)  —147) — h(x ) + 147X)  = —10) — 11(X) — 11(7X) -I- 211(7X) = —h(1) + 2h,(-yx). Also, since h(1) — h(x) = 2-h('y) + 2h(7 X),, we have 0 < 2h(-y) _5_ h(1) — h(x). Hence Aq(h)-(1) =  ohr [h(1)P-1 ± h(x)P-1 — (h(x) + 211(7X))P-1 — (h(1) — 2147 X))P-11 -  So Aq(h)-(1) > 0 provided we can show that the function F (a, b, c) = aP-1 + bP-1 — (a — c)7-1 — (b + 01 > 0  when 0 < c < a — b (that is, let a = h(1),b = h(x), c = 211('Y X)) • Now OF  Oc  and  02F  (a b c) = 0 + 0 + (p — 1)(a — c)'2 — (p — 1) (b + c)P-2  ' '  (a ' b' c) = — (p — 1)(p — 2)(a — c)P-3 — (p — 1)(p — 2) (b + c)P-3 . Oc2  Chapter 4. Special Cases.^  49  Since a2Fl0c2 < 0, to show that F > 0 it suffices to check that F(a,b, 0) > 0 and F(a,b, a — b)> 0. Computing we discover that F(a,b, 0) = aP-1 +bP-1 — aP-1 —bP-1 . 0 F(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. So 1 Aq(h)-(1) = 411h111 [h(1)P-1 — Ih(x)IP-1 — ihMIP-1 — Ih(xY)IP-11 11,> ^1 max {1h(x)IP-2, Ih(Y)IP-2, Ih(xY)IP-2}[h(1) + h(x) + h(Y) + h(xY)1 411h^ 1r1 since 1h(z)1P-2 < h(1)P-2 when p > 2. Using equations (4.8) with 1' .41) = 0, we see that h(1) + h(x) + h(y) + h(xy) = 0, and thus get Aq(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). Then h(1) = 3, h(x) = h(y) = h(xy) = —1, so h(z) is nonzero for all z E  4 and  A(h)-(1)  1 ^1-1 = 411h1 r1 1-3P-1 — — 1 — 1] =  1 [  3P-1 — 3] 411h1r1 < 0 when 0 < p < 2.  Hence by Theorem 2.8 and the Remark following it for the case when p < 1, the positive minorant property fails for 0 < p < 2. 0  50  Chapter 4. Special Cases. ^  Proposition 4.10 For 0 < p < oo, the space LP(Z4) has the positive minorant property  if and only if p > 2. Proof.  Let Z4 = {0, 1, 2, 3} and let 24 = {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. Let 3  h(X)  =E  11(73)73 (x).  .7=1  Then h(0) = ii,(7) +102) + h( y3) -  h(1)  = —h(-y2) + i 0(7) — h(-y3))  h(2)  = —11(7) + 11(72) — 1473)  (4.9)  h(3) = 4(72) — i (11(7) —103)) -  Using the fact that 0(1)1 = Ih(3)1 in the following, we get 1^3  Aq(h)-(1) — ohiri .01h(i)119-1sgn 14 j)  = =  1  41  11111 7r1 ploy-1 + ( h(1) + h(3))1h(1)IP-2 + h(2)1h(2)1P- 2 1 1  411hli^ rl  [h(0)73-1 - 2/02) Ih(1) IP-2 + /(2) Ih(2) IP-2] .  Case 1. Suppose h(2) > 0. Using (4.9) note that h(0) + h(2) = 2./(72) h(0) — h(2) = 211(7) + 2h(73) so that^Re h(1) =^ and^Im h(1) = h(-y) — h,(73) =  (h(0) -I- h(2)  2^)  (h(0) — h(2))  2^)  (4.10)  Chapter 4. Special Cases.^  so that^Ih(1)1 =  h(0) h(2)  51  )2+  (h(0) —2 h(2) 2ii(73))  2.  Also notice that 0 < 2i/(73) < 2k-y)+ 2-h(73) = h(0) — h(2). Hence to show 1 Ap(h)-(1) = ^ "' 41Ihr > 0  ^— 2h(-y2)1h(1)'2 + h(2)1h(2)1P-2]  "  it suffices to show (let a = h(0),b = h(2c = 21(73))  F(a,b,c) =  ap_, _ (a ± ((a b )  2  ± (a  p-2 2) 2  —2 b  + bp-1  > 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 will suffice to show that F(a,b, 0) > 0. Now  F(a,b, 0) = F(a,b, a — 2  )2) ^a+b^ a— b _ + ((^ a^ 2 2  = aP-1 + bP-1 — (a + b)(  2  Ezz a2 b2) 2  2  ^.  Note 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 0-1 and putting t = bl a we wish to prove that  1 + t2)Y 1 + t > (1 + t) ( 2 or that (taking logs) ln(1 +t"') > ln(1+ t) + (P2 2) ln  2  52  Chapter 4. Special Cases.^  or that  G(t,p) = 21n(1 + 91 —21n(1 + t)  —  (p  —  1+ t2) 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. Now  aG _ ^2  (1 + t2) 1 ln(t)tP-1 In Op — 1 + tP-^2 ) = 21n(t) (t1-1 + 1i\ i —In  1 + t2 2  Also a2G^  ap2  \ -2  = —21n(t)2 (tl-P + I) (-1)tl-P  = 2t1-P ln(t)2 (ti-P + 1)-2 >0. Hence to show OG/ap > 0, it will suffice to show that (OG/Op)(t, 2) > 0, that is, to show  H(t) =  1 + t2 OG (t, 2) = 21n(t)(t-1 + 1)-1 — In ( ) > 0. 2 op^  —  Since H(1) = 21n(1)(1 + 1)-1— ln(1) = 0, to show H(t) > 0, it suffices show that H'(t) < 0. We have 2 2(t 1 + 1) 1^ t H'(t) = ^ +2111(t)(t-1 +1)-2(-1)(-1)t-2 + 1 + t2 t^ = 2(t +1) + 21n(t)^2t (1 + t)2^+ 1 + t2. -  -  Chapter 4. Special Cases.^  53  To show H'(t) < 0 it suffices to show that  K(t) =  (1 + t)2  H'(t) < 0.  2  Now  K(t)^1 + t + ln(t) = ln(t) +  ln(t) +  t(1 + t)2 1 +t2  (1 + 0(1 + t2) — t — 2t2 — t3 1 + t2 1 — t2 1 + t2  Since K(1) = 0, to show K(t) < 0 it suffices to show K'(t) > 0. Now 1^—2t(1 +  t2) — (1 — t2)(2t)  K' (t) = - + t (1 + t2)2 1 —2t — 2t3 — 2t + 2t3 (1 + t2)2 1 + 2t2 + t4 — 4t2 t(i + t2)2 ( 1 -t2)2 t(1 + t2)2 0 as required. Consequently, Aq(h)-(1) > 0 when p > 2.  Case 2. Suppose h(2) <0. Then from (4.10) we get  1  Aq(h)-(1)^omir {h(0)P-1 + (-2(-y2) + h(2)) max fik1w-2,1h(2)1p-21]  > ^ ih(0)_2i,(72)+ 417 10111  h(2)] max flh(1)IP-2, 0(2)JP 2}  when p > 2, and since Ih(x)1 < h(0) for all x E Z4. Using equations (4.9) we see that h(0) — 2i1(-y2) + h(2) = 0 so we finally have Aq(h)^(1) = 0.  Chapter 4. Special Cases.^  54  Hence in both cases, Aq(h)-(1) > 0 when p > 2 and so LP(7L4) has the positive minorant property when p > 2. By Theorem 3.5, as 4 > 2[p/2] + 3 = 3 when 0 < p < 2, the space LP(Z4) does not have the positive minorant property for 0 < p < 2. 0 It is clearly becoming more difficult to prove the positive minorant property as the order 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(7-1) and if 0 5_ f(-y) _< .4(-y) for all 7 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) with it > 0 and h even, that is h('y) = it(-y-1) for all -y E 25. This is because this proves that for two even functions f, g with 0 < j < .0, then Ye -=  If IIp < lIglIp. Now if g is not even, then  g+g 2  is even, and ge still majorizes f, so that  Ilf Ilp^Ilgellp  = 1 g +2 g  P  1^1  lIglIp +^11:ollp  =^lIglIp. To prove Aq(h)- > 0, may still apply Lemma 4.6 to assume it(1) = 0 but may not use Lemma 4.7 and so must check Aq(h)^(7) > 0 for each -y E 25. If 7 E 25 and 7 0 1, then the condition that Aq(h)^(7) > 0 is the same as Aq(h)'(74) > 0, and similar to the proof that Aq(h)-(-y2) > 0 which is the same condition as Aq(h)^(73) > 0. The hardest part is to prove that Aq(h)-(1) > 0, in spite of it being a two variable problem and the others a three variable problem, but it is not as bad as the proof in the LP(Z4) proposition. 0  55  Chapter 4. Special Cases.^  The 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 be found. 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] or p 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 the form  E 00  F(z) =  anz,„zinr  ^  (4.11)  mtn=0  where anz,n > 0^  (4.12)  and  E 00  ani,n < 00.^  (4.13)  m,n=0  However these results did not apply to positive definite functions defined on finite groups (nor some exceptional infinite groups). For matrices the classification was for functions that 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 of the form (4.11) but fail to satisfy one of the conditions (4.12) or (4.13). In some of the propositions above, to find when LP(G) has the positive minorant property we showed that Aq(.) operates on the positive definite functions. This means that the function Fp(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 5  Failure of the Positive Minorant Property for Matrices.  This chapter is devoted to improving the known results on when (Mii(R), II* iiP) fails to have 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 for noneven 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 fact this 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 for 0 < 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 for compact abelian groups LP(G) has the positive minorant property if and only if the function Fp(z) = IzIP-Isgn z operates on positive definite polynomials. The following statement is proved in Theorem 3(1) in the paper [8] by Dechamps-Godim, Lust-Piquard, and Queffelec. Theorem 5.2 For 1 < p < oo, the space (Mp(R), II • II) has the positive minorant property 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 B is not invertible. To prove Theorem 5.1 for p> 1 it suffices for us to find a nonnegative 56  Chapter 5. Failure of the Positive Minorant Property for Matrices. ^57  invertible matrix B for which (B*B)(P/2)-113* has a negative entry. Specifically, we show this for the matrix A = Ei 1 + J(0) + 4,(0)T, where Eli is the matrix with all zero entries except for a single 1 in the upper left-hand corner, and Jn (0) is the Jordan matrix with all zero entries except for l's just above the main diagonal. That is,  o  101 010 A=  o  010 101 010  \  /  This matrix has the property that the last entry in (A*A)(P/2)-1A* is negative for 2(n — 2) < p < 2(n — 1). To deal with smaller values of p we build other nonnegative matrices which we denote by Bk and which have the property that (BZ.Bk)(P/2)-1BZ has a negative last entry for 2(k — 2) < p < 2(k — 1) for k = 2, 3, ... , n — 1. Let A, denote the leading r x r principal submatrix of A, and let Bk be the matrix with blocks (Ink  0 )  0^fik where Jr is the r x r identity matrix. Then Bk E M(1I) and  (BI:Bk)(7312)-1BZ = (  1.7"  0  0^(AZAk)(P/2)-1AZ^.  ^  ^Xi  Chapter 5. Failure of the Positive Minorant Property for Matrices. ^58  Our 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 Bk by adding E in the last entry has a strictly smaller Cp norm than Bk does, although Ek majorizes Bk. We cannot use Theorem 5.2 when 0 < p < 1. Instead we use some facts from Chapters 3 and 4. For n > 3 the embedding of LP(Z) into (Mn(C), II 11) and our Theorem 3.5 apply when 0 < p < 2. When n = 2 Proposition 4.2 applies provided 0 < p < 1. When n = 2 and p = 1 simple calculations show that the matrices E2 with E = 1 and B2 defined above 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 an orthonormal basis of eigenvectors of A. We will show that none of the eigenvalues is 0 and that each eigenspace is one-dimensional. List the eigenvalues as Ai > A2 >^>^An. Letting xi denote an eigenvector associated with the eigenvalue Aj, then Axj = Ajxj for j = 1, , n or by equating each TOW j  ^Xj_i  +  X2  j  Xj+i  j j  = =  AiXi AiXi  j  ^  j^i = 27^, n — 1  (5.1) ^  Xn—i j = Aixn j.^  (5.2)  (5.3)  If Aj = 0, equation (5.3) says xn_1 = 0, but then equation (5.2) with i = n —2 says xn_3 j xn_i j = 0 so xn_3 j = 0 which in turn says xn_5 j = 0 etc., until we either get  Chapter 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 so we 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 that Xn-2 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 eigenspace is one-dimensional and the eigenvalues are distinct. Equation (5.1) turns out to be redundant when Ai is an eigenvalue. Defining polynomials ro(A) = 1 ri (A) = A  rk(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 and  j = 1, . . , n. That is xi = (rn_i (Ai), , ro(AMT. Let pi = xj/lixill2 be normalized eigenvectors of A and put P = (pi, , pn) = (pi ;), and A = diag (A1, , )n). Then  AP = PA. Since A is symmetric and the eigenspaces are one-dimensional, P is an orthonormal matrix.  Chapter 5. Failure of the Positive Minorant Property for Matrices. ^60  5.3 When the last entry is zero. We now consider (A*A)(P/2)-1A* and will show that the last entry of this matrix is negative for 2(n — 2) < p < 2(n — 1), that is  R A* A) F 1,41„ n < o. To do this we first show that this entry is 0 for certain even integer values of p. Now A*A = A2 =  pA2pT ,  (A*A)f-i = p(A2)f—lpT = plAlp-2pT  where IAl ^(hag(JAi..., IA.1). Thus [(A*A)-1Aln n  = [PIAIP-2API]n  =  EPn i[lAIP-2Ap7]  j=1  =E j=1  Pn j l Aj I2AjpTn  E P2n I Ai IP-lsgn Ai.  J=1 We define f(p) = RA*A)1-1Aln  IP lsgn Ai J=1 and as noted earlier Ai 0, so f is well-defined for all real numbers p. n =^jPti  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 and  f (2) =^ A. j=i  Chapter 5. Failure of the Positive Minorant Property for Matrices. ^61  But P is an orthonormal matrix, so that if p(i) denotes the i-th row of P, 0 = p(n) pi)  = EPT j=1  ji)n-1 j j  ^ = EN, ,=1^Ilx,112 1  =^Pn ^ ri(Ai) j=1^lixi112  =^  as Xn—i j = rn—(n-1)(Aj) Xn j  jAi  as pn = ilxiI12 j=1^  =  _  1 and ri (Ai) = Ilx;112  1(2).  Suppose that f (2k) = 0 for k = 1, . . . , — 1. Then f (2) =^ilAirt-lsgn Aj ,,2 12t-1 Z_,Pn j'j • j=1  Again, as P is orthonormal, 0 = p (n —t-I- 1) p ( n —t)  EPn-t+1 jPn—t j  E  Xn—t-f-1 j Xn—t j  = j=1 11302 x—, 71^1^1  ▪  ^ rt_i(Ai)rt(Ai) as Ilx./112^Ilx./112  xn_i j =  r1 (A)  • E p2n jri_i (Ai ) rt (Ai )  Recall that r22 is an even monic polynomial of degree 2i and polynomial of degree 2i + 1 so that  7-.6_1 •  is an odd monic  rt is an odd monic polynomial of degree 2f — 1.  If (A) (A) =A 2e-1 + bt_1A21-3 ^bjA2i-1  ^  ^„.,2  Chapter 5. Failure of the Positive Minorant Property for Matrices. ^62  then 0 = p(n-e+i) on-t)  • Egz Jr1-1(Aj)ri(Aj) j=1  n  n  E /3! jAr-3 + . . • + bi E Pn2 JAJ  —2 2f-1 -^  = i Pn jAi + bf-1  J=1^J=1^J=1  12e-1 •  = E .7=^  +  0  Z_, Pn f‘j  +^+ 0  j=1  2k-1 as (2k)^ for k = 1, . . . , — 1. But then  ^f  (2) =^A' =0, J=1  completing the induction, and proving that f (2k) = 0 for k = 1, . . . , n — 1. 0 5.4 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 the following lemma. Lemma 5.4 Given two sequences of real constants ^and Icil with t j's positive and distinct and not all cj = 0, define the function h(a) = E73= ! 1 c.t" 3 • Then h has at most  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 zeros  when c and t nonzero. Suppose when k> 1 that all functions of the form Ei1=1 c3t9 3 have at most k — 2 distinct zeros and consider h(a) = cjt. . Then k-1  ti  g(a) = h(a) — = c; — t" 3=1^tk  and ^k  a  + ak  a dg k-1 (L) (ti \ In ^ cu =^ da^ ^t ^tk ) i=1  •  Chapter 5. Failure of the Positive Minorant Property for Matrices.^63  By the inductive hypothesis, dgicla has at most k — 2 distinct zeros, and so by the contrapositive of Rolle's theorem g has at most (k — 2) + 1 = k —1 distinct zeros. The functions h and g have the same zeros, so h has at most k —1 distinct zeros, completing the 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 derivative of the corresponding function g would also have a zero at this point. By Rolle's theorem  g' 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 first part of the lemma, since g' is a sum of n — 1 terms. 0 We 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 the individual coefficients cj cannot be 0 because pn = 1iiix2ii2 and none of the eigenvalues is O. While the eigenvalues are distinct, it may appear possible that their absolute values might not be distinct. If IA.; I = I  Aji  for distinct j and j', then we can combine the  corresponding terms in the sum for f (p) and get something of the same form with fewer terms. We need to show that this expression does not reduce to the zero function. Since the trace of the matrix A is 1, the list of eigenvalues cannot just consist of pairs with the same 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 —1 zeros in Lemma 5.3 they must all be simple. It follows that the sign of f changes at each of 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 > 0 when p = 2k. More precisely, since 1(2k) = 0 when k = 1, ,n — 1, we must have  1(2k) > 0 for all integers k > n . It follows that f(p) > 0 for all p> 2(n — 1), since it  Chapter 5. Failure of the Positive Minorant Property for Matrices. ^64  has no zeros in this interval and is positive at some points in the interval. So the sign change of f (p) at p = 2(n — 1) must go from negative to positive. Hence f (p) < 0 for all p 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), 119-128. 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[27] Shapiro, H.S., Majorant problems for Fourier coefficients, Quart. J. Math. Oxford (2) 26 (1975), 9-18. [28] Simon, B., Pointwise domination of matrices and comparison of Ip norms, Pacific J. Math. 97 (1981), 471-475.  

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