Series in Banach Spaces: Conditional and Unconditional Convergence

Operator Theory Advances and Applications Vol. 94 Editor I. Gohberg

Editorial Office: School of Mathematical Sciences Tel Aviv University Ramat Aviv, Israel Editorial Board: J. Arazy (Haifa) A. Atzmon (Tel Aviv) J.A. Ball (Blackburg) A. Ben-Artzi (Tel Aviv) H. Bercovici (Bloomington) A. Bottcher (Chemnitz) L. de Branges (West Lafayette) K. Clancey (Athens, USA) L.A. Coburn (Buffalo) K. R. Davidson (Waterloo, Ontario) R.G. Douglas (Stony Brook) H. Dym (Rehovot) A. Dynin (Columbus) P.A. Fillmore (Halifax) C. Foias (Bloomington) P.A. Fuhrmann (Beer Sheva) S. Goldberg (College Park) B. Gramsch (Mainz) G. Heinig (Chemnitz) J.A. Helton (La Jolla) M.A. Kaashoek (Amsterdam)

T. Kailath (Stanford) H.G. Kaper (Argonne) S. T. Kuroda (Tokyo) P. Lancaster (Calgary) L.E. Lerer (Haifa) E. Meister (Darmstadt) B. Mityagin (Columbus) V. V. Peller (Manhattan, Kansas) J.D. Pincus (Stony Brook) M. Rosenblum (Charlottesville) J. Rovnyak (Charlottesville) D. E. Sarason (Berkeley) H. Upmeier (Marburg) S.M. Verduyn-Lunel (Amsterdam) D. Voiculescu (Berkeley) H. Widom (Santa Cruz) D. Xia (Nashville) D. Yafaev (Rennes)

Honorary and Advisory Editorial Board: P.R. Halmos (Santa Clara) T. Kato (Berkeley) P.O. Lax (New York) M.S. Livsic (Beer Sheva) R. Phillips (Stanford) B. Sz.-Nagy (Szeged)

Series in Banach Spaces Conditional and Unconditional Convergence

Mikhail I. Kadets Vladimir M. Kadets

Translated from the Russian by Andrei lacob

Birkhauser Verlag Basel· Boston· Berlin

M.l. Kadets and Y.M. Kadets Pr. Pravdy 5. apt. 26 Kharkov 310022 Ukraine

1991 Mathematics Subject Classification 46815. 46B20. 460 10. 42C20. 52A20. IOE05. 60B 12

A C1P catalogue record for this book is available from Ihe Library of Congress. Washington D.C .• USA

Deutsche Bibliothek Cataloging·in·Publication Data

Kadec, Midsaill.: Series in Banach spaces: conditional and unconditional convergence' Mikhail I. Kadets ; Vladimir M. Kadets. Transl. from the Russ. by Andrei lacob. - Basel; Boston; Berlin: Birkhliuser. 1997 (Operator theory ; Vol. 94) ISBN 3·7643-5401·1 (BaseL) ISBN 0-8176-5401-1 (Boston) NE: Kadec. Vladimir M.:; GT

This work is subject to copyright. All rights are reserved. whelher the whole or pan of the material is concerned. specifically Ihe rights of translation. reprinting. re·use of ilIusU3tions. recitation. broadcasting. reproduction on microfilms or in other ways. and storage in data banks. For any kind of use Ihe permission of Ihe copyright holder must be obtained. 1997 Birkhluser Verlag. P.O. Box 133. CH-4010 Basel. Switzerland Printed on acid·free paper produced from chlorine-free pulp. TCF 00 Cover design: Heinl Hiltbrunner. Basel Printed in Germany ISBN 3-7643-5401-1 Q

CONTENTS

Introduction .................................................

vii

Notations ....................................................

1

Chapter 1. Background Material §l. Numerical Series. Riemann's Theorem .................... §2. Main Definitions. Elementary Properties of Vector Series .......................................... §3. Preliminary Material on Rearrangements of Series of Elements of a Banach Space .............................. Chapter 2. Series in a Finite-Dimensional Space §1. Steinitz's Theorem on the Sum Range of a Series ......... §2. The Dvoretzky-Hanani Theorem on Perfectly Divergent Series.......................................... §3. Pecherskii's Theorem..................................... Chapter 3. Conditional Convergence in an InfiniteDimensional Space §1. Basic Counterexamples ................................... §2. A Series Whose Sum Range Consists of Two Points ....... §3. Chobanyan's Theorem.................................... §4. The Khinchin Inequalities and the Theorem of M. I. Kadets on Conditionally Convergent Series in Lp ................. Chapter 4. Unconditionally Convergent Series §1. The Dvoretzky-Rogers Theorem......... ................. §2. Orlicz's Theorem on Unconditionally Convergent Series in Lp Spaces ............................................. §3. Absolutely Summing Operators. Grothendieck's Theorem Chapter 5. Orlicz's Theorem and the Structure of Finite-Dimensional Subspaces §l. Finite RepresentabiIity ................................... §2. The space Co, C-Convexity, and Orlicz's Theorem. . . . . . . . . §3. Survey on Results on Type and Cotvpe .. . . . . . . . . . . . . . . . . .

5 7

9 13

21 23

29 32 36 39 45

49 52

59 62 fi7

vi

CONTENTS

Chapter 6. Some Results from the General Theory of Banach Spaces §l. Frechet Differentiability of Convex Functions ............. §2. Dvoretzky's Theorem .................................... §3. Basic Sequences .......................................... §4. Some Applications to Conditionally Convergent Series

71 73

79 82

Chapter 7. Steinitz's Theorem and B-Convexity §l. Conditionally Convergent Series in Spaces with Infratype ........................................... 87 §2. A Technique for Transferring Examples with Nonlinear Sum Range to Arbitrary Infinite-Dimensional Banach Spaces... 93 §3. Series in Spaces That Are Not B-Convex ................. 97 Chapter 8. Rearrangements of Series in Topological Vector Spaces §l. Weak and Strong Sum Range............................ 101 §2. Rearrangements of Series of Functions ..... , ... ...... . .. .. 106 §3. Banaszczyk's Theorem on Series in Metrizable Nuclear Spaces ........................................... 110 Appendix. The Limit Set oC the Riemann Integral Sums of a Vector-Valued Function §l. FUnctions Valued in a B-Convex Space................... §2. The Example of Nakamura and Amemiya ................. §3. Separability of the Space and the Structure of I(f) ........ §4. Connection with the Weak Topology .....................

120 122

127 131

Comments to the Exercises ................................... 141 References. .. .. . . .. . . ...... .. ....... . . .. ..... . ... . ...... ..... 149 Index ....................................................... , 155

INTRODUCTION

Series of scalars, vectors, or functions are among the fundamental objects of mathematical analysis. When the arrangement of the terms is fixed, investigating a series amounts to investigating the sequence of its partial sums. In this case the theory of series is a part of the theory of sequences, which deals with their convergence, asymptotic behavior, etc. The specific character of the theory of series manifests itself when one considers rearrangements (permutations) of the terms of a series, which brings combinatorial considerations into the problems studied. The phenomenon that a numerical series can change its sum when the order of its terms is changed is one of the most impressive facts encountered in a university analysis course. The present book is devoted precisely to this aspect of the theory of series whose terms are elements of Banach (as well as other topological linear) spaces. The exposition focuses on two complementary problems. The first is to characterize those series in a given space that remain convergent (and have the same sum) for any rearrangement of their terms; such series are usually called unconditionally convergent. The second problem is, when a series converges only for certain rearrangements of its terms (in other words, converges conditionally), to describe its sum range, i.e., the set of sums of all its convergent rearrangements. Recall that for series of real nwnbers both problems are settled by a wellknown theorem of lliemann (1867): a series converges uncOnditionally if and only if it converges absolutely; if the series converges conditionally, then its sum range is the whole real line. Difficulties arise when one considers more general series, for instance, series in a finite--dimensional space, or even series of complex numbers. To be more precise, the answer to the first problem is readily obtained, as before: in any finite-dimensional vector space the commutativity law (invariance under rearrange-ments) holds only for absolutely convergent series. By contrast, the solution to the second problem-the description of the set of elements to whlch a series can converge for a suitable rearrangement of its terms-requires considerably more efforts. For series of complex numbers such a description was given by P. Levy in 1905150J, while for case of series in finite-dimensional spaces it was provided by E. Steinitz

[871· The study of rearrangements of series in infinite--dimensional spaces was initiltt.pn hv W ()r-liM Ih~ 1'><11 "n~ ~lthc ..r"' .. ntl" -t .." .. I"",..,l h" ""<> .. " ...... +ho~~.:,..:~n~ :n

viii

INTRODUCTION

different countries, and continues to be an active field of research. However, until now this theme found almost no exposition in textbooks and monographs. Several years ago we undertook the task of partially filling this gap by writing a textbook Rearrangements oj Series in Banach Spaces 1431, intended for students at Tartu University. However, the beautiful results of D. V. Pecherskii, S. A. Chobanyan, W. Banaszczyk, and others, which appeared after that textbook was published, changed in essential manner the state of the art of the theory of series, have led to the development of new approaches, and in some instances even to a new point of view on the subject. For this reason we have decided to address again, but on a more modern level, the same topics, the result being the present book. One of our goals was to bring the topics under consideration and the I.heory of Banach spaces to the attention of young mathematicians. This explains the large number of exercises, among which are elementary as well as very difficult ones, and the fact that numerous open problems are pointed out. With t.he same goal in mind we included an Appendix titled The limit set of the Riemann integral sums 0/ a vector-valued function, which, although does not belong directly to the main subject of the book, is closely related to it through the methods used in proofs and contains readily formulated open problems. We hope that our efforts will prove fruitful for beginner mathematicians as well for specialists in functional analysis and related areas of mathematics. We deal here mostly with general series, but beyond the topics covered in our exposition there are many other situations where the idea of considering rearrangements of series has interesting applications: basic expansions (the theory of unconditional bases /12], [53],183], (101)), orthogonal systems of functions, Fourier analysis, ctc. As a sample we wish to mention here the following brilliant theorem of Sz. Gy. R.ev5z 11001: for every 21T-periodic continuous function f it is possible to rearrange its Fourier series in such a way that some subsequence of the sequence of partial sums of the rearranged series will converge to f uniformly. For the understanding of the material presented in the book familiarity with standard courses of mathematical analysis and linear algebra and also with the elementary notions and results of the theory of Banach spaces is assllmed (the content of the books [45J, \551 or 1781 is more than sufficient). The junior of the authors has used the book to teach a special course at Khar'kov University. Discussions with the participants in the course were extremely useful in preparing the final version of the manuscript. Our work was supported in part by the AMS !SU Aid Grant.s and, ill the final stage, by ISSEP Grants SPU 061025 and APU 06140. We wish to express our gratitude to W. Banaszczyk, 1. S. Belov, W. A. Beyer, E. A. Gorin, V. S. Grinberg, and especially S. A. Chobanyan for useful discussions. to 1. Halperin and T. Ando for kindly sending us supplementary bibliography on rearrangement of series, and to D. L. Dun, the spouse of the senior and the mother of the junior of the authors, who took upon herself the worries of everyday lifewithout the freedom she thus provided we would have never decided to undertake the work whose results are offered now to the reader.

NOTATIONS

In the book "Lemma 3.1.2" means "Lemma 2 in Section 1 of Chapter 3." Theorems, definitions, and exercises are numbered in a similar fashion, each type with its independent numbers, i.e., in the text one can encounter both, say, Definition 2.1.1 and Theorem 2.1.1. The proof of any assertion starts with the word "Proof" and ends with the symbol O. If 0 appears immediately after the formulation of an assertion, this means that the assertion is given without proof or that the proof follows from the preceding arguments. Spaces and operators are as a rule denoted by upper case Italic letters, whereas elements of a space and linear functionals are denoted by lower case Italic letters. If not otherwise stipulated, "number" means "real number", and linear spaces are considered over the field R of real numbers. The space R n of rows of the form x = (XI, X2,' .• , x n ), Xi E R, is called here the n-dimensional coordinate space. The vectors el = (1,0,0, ... ,0), e2 = (0,1,0, ... ,0), ... , en == (0,0, ... ,0,1) form the standa.rd basis in an. In infinitedimensional spaces of sequences the unit (norm-one) vectors of the standard basis are also denoted by en. Whenever it is clear what norm we speak about, we use the simple notation 11·11. When several norms in t.he same space are considered. we use notations like II·ll p , II· IIx, etc. In the following list "-" stands for "denotes."

X· - the dual (conjugate space) of the space X.

X·· - the second dual of the space X. (x,f) - the value of the functional f E X· on the element x E X, or the inner product of two elements of Euclidean space.

Lin A -

the linea.r span of the set A.

Lin A - the closed linear span of the set A.

B(X) - the unit ball of the normed space X, i.e., the set of elements whose norm is smaller than or equal to 1.

NOTATIONS

2

S(X) - the unit sphere of the normed space X. dim E - the dimension of the linear space E. l~n), where 1 ::; p norm

< 00

-

the n-dimensional coordinate space, equipped with the

where Xi are the coordinates of the vector X; l~n) will be called the Euclidean space.

ll,::) - the n-dimensional coordinate space, equipped with the norm

IIxli oo = I~i~n max Ix;!. C\O,I] - the space of continuous functions on the segment [0,1]' with the norm 11/11 =maxtEIO.11I/(t}l· (0, E, I-') - a measure space, i.e., a set n with a o--algebra 1: of measurable susbsets and a measure 1-'.

=

Lp Lp(n, 1:, I-'}, where 1 ::; p < 00 R for which

11/11 =

-

the space of measurable functions

(10 IIIP dp.)

I/p

I:

n

-+

-+

R,

< 00.

Loo = Loo (n, E, p.) - the space of bounded measurable functions equipped with the norm 11/11 = esssupl/(t)l· tEn

I:

n

(Recall that when one deals with· spaces of measurable functions one identifies functions that differ on a set of measure zero.) Here

ess supf(t) = inf{sup f(t): U tEf!

E

E, p.(n \ U) = OJ.

tEU

C{) the space of all numerical sequences x :::: (Xl, xz, ... ) that converge to 0, equipped with the norm IIxll max IXil.

=

NOTATIONS

3

lp, where 1 ::; p < 00 - t.he space of all numerical sequences for which 00, equipped with the norm

/lrll

loo -

=(

2:::

1

Ix,lP <

t; Ix,P' )I/P oc

the space of all bounded numerical sequences, equipped with the norm

IIxl! = sup Ix;!. 0:::0=1 Xn)p - the lp-sum of the spaces Xn, i.e., the space of all sequences x = .,xn , ..• ), Xi E Xi, for which L::=llIxnIl P < 00, equipped with the

(XI,X2,'"

norm

IIxli = (L::=I IIxnIlP)l/p.

SR(L~I

Xk) -

the sum range of the series L~I

Xk

(Definit.ion 2.1.1).

d(X, Y) - the Banach-Mazur distance between the spaces X and Y (Definition 5.1.1).

X.!... Y - the space X is finitely representable in the space Y (Definition 5.1.2). X

J. Y -

the space Y is finitely saturated by t.he space X (Definition 6.3.4).

r;; f(x) - the Frechet derivative of the function f at the point. x (Definition 6.1.2).

CHAPTER 1 BACKGROUND MATEruAL In this short chapter we collected facts that are sufficiently well-known, the consideration of which may be regarded as an introduction to the main text.

§l. Numerical Series. Riemann's Theorem Since the basic properties of numerical series are well known from the calculus course, we will give here only the most needed definitions and facts. DEFINITION 1.1.1. A series E~I XI" vergent if the series E~I IXlcl converges.

XIc

E

R, is said to be absolutely con-

DEFINITION 1.1.2. A series E~I XIc, XIc E R, is said to be conditionally convergent if it converges, but the series E~I IXlcl diverges. DEFINITION 1.1.3. A permutationofa set A is a bijective mapping of A onto itself. THEOREM 1.1.1. Suppose that all XIc are nonnegative numbers and the series XIc converges to a number s. Then for any permutations 1r of the set of natural numbers the series L~1 X,,(Ic) converges and its sum is equal to s.

L~l

PROOF. Let us denote by Sn the partial sum of the rearranged series: Sn = X,,(Ic)· Since the terms are nonnegative, it is readily seen that SI ::; 82 :::; S3 ~ •.. Sn :::; •.• and SUPIc Sic :::; 8. Hence, the sequence Sn converges to some number 5 :::; s, i.e., E~1 X,,(Ic) = 5 :::; s. Switching the roles of the series E~l XIc and E~l X,,(k) and arguing in the same way as above, we obtain the opposite inequality E~l XIc S :::; 5 L~l X,,(Ic). Therefore, s s. 0 L~=1

=

=

=

THEOREM 1.1.2. Suppose that the series E~1 XIc is a.bsolutely convergent. Tben E~l XIc converges; moreover, for any permutation 1r oltbe natural numbers the rearranged series E~=l X"CIc) converges and L~=l Xic = L~1 X"Ck)· PROOF. For any real number x set x+ = x if x > 0 and x+ = 0 if x :s 0, and set x- = x+ - x. C()nsider E~l xt and L~l x; . These are series of nonnegative numbers; they are convergent, being dominated by the series E~llxlcl. Denote L~I x; = s+ and E~1 S-. Then, by Theorem 1.1.1, for any permutation . ,",00 + d ,",00 ,",00 + + an d 1r th e serles L-k=1 X"'(Ic) an L-Ic=1 X,,(Ic) converge; moreover, L-k=1 X,,(Ic) = s

x; =

6

CHAPTER 1.

BACKGROUND MATERIAL

E:;-=1 X~(k) s-. Consequently, the series E~l I,,(k) == E~I «(k) - x~(k») converges. and E;;"'=I X,.(k) = s+ - S-. Clearly, the initial series E~l Ik also converges, and its sum is equal to s+ - 05-. 0 Tm:oREM 1.1.3 (RIEMANN'S THEOREM). Let Er=l x" be a cOllditionally convergent series of real numbers. Then: 1) [or any s E Rone can find a permutation n stich tha.t E~l xr.(k-) == 05; 2) one can find a permutation 0 such tllal. I:~l Xa(k) == 00. PROOF. Let liS partition the se't N of nat.ural numbers into two sets, A == {a),a2,"'} and B == {bl,bz,···}, such that X"k ~ 0, Xb k < 0 for all kEN. But if L::I X Ok < 00 and L~I Xbk > -00, then the series I:;:C=I IXkl converges. But if L~1 x"> = 00 and L~I Xbk > -00, or E~I X"k < oc and L~I Xb. = -00, then the series L~I Xk diverges. Since, by the hypothesis of the theorem, the series L~=I Ik converges, but not absolutely, we see that L~l X"k = 00 and L~I Xbk = -00. Now let us prove assertions 1) and 2). 1) Let. s ~ O. We construct. the needed permutation n as follows: 1T( 1) == ai, 71"(2) = Q2,"" n(j) = aj and so on, up to the number i il for which the partial sum Sj == L{=l x n (") of the rearranged series becomes larger t.han s: Sk ~ S for 0 < k ~ il - I, and si) > s. Then we start to add negat.ive t.erms: n(jl + 1) == bl> n(jl + 2) =~, ... lip to t.he moment when the partial slim sJ) +i2 b~omes for the first times smaller than s. Then we again add positive terms, then again negative ones, and so on. It is readily seen that the partial sums S1 constructed in t.his manner tend to s. In the case s < 0 the construction is almost identical. 2) This time the permut.ation 0 is constructed as follows: o(l) = a .. 0(2) = Q2, ., .• up to j] for which sj, becomes larger than 1 - Xb), and we put o(il) == bl . Then we again add positive terms for a.c; long as sh is not larger t.han 2 - Xb" and we put a(h + 1) = ~, and we continue in this way ad infinitum. Then Sj > k for all j > i", which implies that Sj ...... 00. A similar pro('edure yields a permutation for which the permuted series diverges to -00. 0

=

EXERCISE 1.1.1. Clarify whether Riemann's theorem is valid for series of complex numbers.

The result.s of this section, and among them the main one - Riemann's theorem -, give information on how a series of numbers behaves wit.h resp~t to different permutations of its terms. At the same times there exists another, relatively poorly studied circle of problems, dealing with how one and the same permutation acts 011 different series. The exercises given below concern precisely this kind of problems. DEFINITION 1.1.4. A permutation n: N ..... N is said to be convergencemodifying if there exists a convergent numerical series E~I X~ for which the series E~I x .. (~) diverges.

§2. MAIN DEFINITIONS EL£MENTARY PROPERTIES OF VECTOR SERIES

7

EXERCISE 1.1.2. Charac.terize the convergence-modifying permutations in terms of the "degree to which they mix" the index set. DEFINITION 1.1.5. A permutation 1T: N - N is said to he sum-modifying if there exists a convergent numerical sE'ries E~l :tk for which the series E::1 X,,(k) converges, but E~, X" i= L~=I ·'I:,,(kj·

•

EXERCISE

1.1.3. Characterize t.he sum-modifying permutations.

EXERCISE

1.1.4. Show that every sum-modifying permutation is conver-

gence-modifying. EXERCISE 1.1.5. Let 1T be a permutation such that. 7T(3k) = 2k, 1T(3k + j) =: 4k - 2j + 1 for kEN and j = 1,2. Further, let u be a permut.a.t.ioll such that u(5k} = 2k, u(5k + j) = 8k + 2j - 1 for kEN and j = 1.2,3,4. Show that if the series 2:~=1 XI: and 2::-=1 .r,,(k) converge and 2:~=1 XI: = 0, then the series 2:~1 Xo(k) also converges, and 2:;:1 x o (") = 2 2:~=1 X,,(k)'

§2. Main Definitions. Elementary Properties of Vector Series In this section we included those definitions and properties of series of elements in Banach spaces which do not differ in essential manner from the corresponding definitions and properties of ordinary numerical series. DEFINITION 1.2.1. A series of elements of a Banach space X is an expression having t.he form of the sum of an infinite number of t.erms that belong to X: XI

+ X2 + ... + .'l:n + . .. .

(1)

To simplify notation of series one uses the symbol 2:~1 XI:. One has t.o remember that the expression (1) is not a sum in the usual sense of the word, because in a Banach space addition is defined only for a finite number of terms. DEFINITION Sn

1_2.2. The n-th partial sum of OLe series

= E~=I It of its first n (n < (0) terms.

E::l X/r

is the sum

DEFINITION 1.2.3. A series is said to be convergent if the sequence of its partial sums converges in the norm of the space X. The limit of that sequence is called the sum of the series: s = lim n _ oo Sn. Whenever we write s = E:::1 Xk we mean that the seriE'.s E:"l XI: converges and its sum is equal to s. DEFINITION

element Inti

Tn

=S-

1.2.4. The remainder of the convergent series E:"I Xk is the Sn. In other words, the remainder Tn is the sum of the series

+ Xn+2 + ....

When n grows the remainder tends to zero. Sometimes the t.erm "remainder" is used not for the sum of the series E:::n+ I Xb but for the series x nt I + Xn +2 + ... itself.

CHAPTER

8

1.

BACKGROUND MATERIAL

DEFINITION 1.2.5. A segment of a series is a sum of a finite munber of consecutive terms: L:;=m+l Xk = Sn - Sm' Sometimes by a "segment" one means the set {xdk';'m+l itself. THEOREM 1.2.1 (THE CAUCHY CONVERGENCE CRITERION). The series E:l Xk converges if and only jf the sequence (more precisely - the net) of its segments converges to zero:

PROOF. Suppose the series converges: Sn -

t II

Xtll = II sn - 8mll $

118n

-

8

as n

-+ 00.

811 + IIsm -

Then

sll -+ 0,

"=m+l i.e., Cauchy'S condition is satisfied. Conversely, suppose Cauchy'S condition holds: limm,n_oo IIsn - smll = O. This means that the partial sums form a Cauchy (fundamental) sequence. Since the space X is complete, any fundamental sequence in X converges; hence, the sequence of partial sums Sn converges. 0 DEFINITION 1.2.6. The series E~l x" is said to be absolutely convergent if E:ll1x,,11 < 00. THEOREM 1.2.2. IE the series E~n+l

x" converges absolutely, then it con-

verges. PROOF. Since E:I

IIx,,1I < 00, the Cauchy criterion yields n

lim m,n-oo

L

IIxkll = O.

"=m+l

By the triangle inequality,

lI"fl "f+l

IlXkll

Xkll $ , whence limm,n_oo II E;=m+l x,,11 = o. By Cauchy's criterion, this means that the series E:l x" converges.

0

It is not difficult to show (this is left as an exercise (or the reader) that if the series E:l x" and E:l y" converge, then the series E : 1 (ax" + by,,) also converges; and if a" E R+, k = 1,2, ... , E:l < 00, and IIx,,1I $ for all k, then the series L:~1 x" converges. Further, if T: X - Y is a continuous of elements of X the series linear operator, then for any convergent series L:~I L::l TXk also converges. Many other classical convergence tests for numerical series can also be readily generalized to the case of series in a Banach space.

a"

a"

x"

EXERCISE 1.2.1. Show that if Theorem 1.2.1 or Theorem 1.2.2 holds in the normed space X, then X is complete (i.e., a Banach space).

§3

PRELIMINARY MATERIAL ON REARRANGEMENTS

9

§3. Preliminary Material on Rearrangements of Series of Elements of a Banach Space From this moment on we shall address problems in which the specific features of the underlying space, its geometric and linear-topological properties, can play an important role. DEFINITION 1.3.1. A series ~~l XIc is said to be unconditionally convergent if it converges for any rearrangement of its terms.

A numerical series is unconditionally convergent if and only jf it is absolutely cOllvergent. From Theorem 1.2.2 it readily follows that in one direction this conne<:tion remains valid in the general case as well: absolute convergence in a Banach space implies unconditional convergence. The converse is false, as the next example shows. EXAMPLE 1.3.1. Let X = l2 and XI; = (0,0, ... ,0, k- 1 , O, ... ), where the nonzero coordinate is the kth. Then the series E~l x" converges to the element 5 = (1,2- 1 ,3- 1 , ... , n -I, ... ) for any rearrangement of its term'>, but does not converge absolutely, since ~~l IIxkll = L~l k- I 00.

=

A little down the road we shall see (the Dvoret.sky-Rogers theorem in Chapter 4) that. in every infinite-dimensional Banach space one can construct an unconditionally convergent series which is not absolutely convergent. DEFINITION 1.3.2. A series L~l XI; is said to be conditionally convergent if it converges, but not unconditionally, i.e., among its rearrangements there are divergent ones. Notice that for numerical series we gave a different definition of the notion of conditional convergence (Definition 1.1.2); however, since for numerical series absolute and unconditional convergence are equivalent, those two definitions are equivalent. THEOREM 1.3.1. If tile series L~I XI; in the Banach spac(' X is unconditionally convergent. then all its rearrangements have the same sum. PROOF. We argue by reductio ad absurdum. Let E~I Xk = 5 and suppo:;e that for some permutation 11" the sum 5' of the rearranged series is not equal to s. Pick a functional 1 E X' such that 1(5) f= 1(5/). Then the numerical serie.., ~~l l(xIc) will not converge absolutely, since the permutation 11" changes its sum. Hence, by Rjemann's theorem, there exists a permutation (T for which the series ~~l I(Xa(Ic») diverges. Then the series ~~I Xo(k) will also diverge, which contradicts the unconditional convergence of the series ~~J Xk. 0 The study of unconditionally convergent series based directly on Definition 1.3.1 encounters some difficulties. For this reason we shall presently give another, more convenient definition, which is equivalent to t.he basic definition of unconditional convergence.

CHAPTER 1.

BACKGROUND MATERlAL

10

ON DEFINITI

1.3.3. A series 2:~1 Xk of elements of a Banach space is said ergent if the series L~I akXk converges ror any choice of the

perfectly conti . a::: :I: 1coefficIents • oc . . 32 For series Lk=1 Xk In a Banach space X the followmg THEOREM 1.. , .. equivalent: condItIOns are. converges unconditionally; 1) tbe ~:o(tbe form :tnt +Xn2 +:tn3 + ... (subseries of the series E:I XIo), 2) all sen converD"e' 0 and numbers ml < Tl < m2 < T2 < ... such that nl < n2 criterion, the,r1e eJCl Denote the terms of the series E: , Xk that do not appear in "r~ X > E. t..-i:=m~ n. - ents {x,.,} ~~m~ by y., Y2, Y3, .... Now construct a rearrangement of any of ~he se~ XII as follows: write first the segment {Xn.} ~; ... \' then the term Yl, the serIes EII::1 {"" }:2 then the term Y2, and so on. By Cauchy's criterion, ent ... n; ,=m,' then the segm ries will diverge, which contradicts the unconditional convergence the rearranged ~ of the initial ~es·pose 1) does not hold, and let X .. (Io) be a divergent re2) ::} 1). r : series. By Cauchy's criterion, there exist disjoint segments 6. j , arrangement 0 0 anged series for which the infimum of the norms is larger than i E N, of the ~he terms in each 6. j again, rearranging them in the order of £ > O. Permu~. dices (Le., in the original order). Denote the index of the first . A"~ of theIr In A C {}rl . mcr~ . A., by m,- Iresp. T; I: L.1j X/c /c-m· p assmg, 1·r necessary, to t rmlDU -; [resp. lastI e one can easily choose 6.. so that TI < m2 < T2 < m3 < T3 < .... a subseq~e~ce, ccessively all terms from 6 1, then from 6.2, from 6.3, and so on, Then, w~t1ng sU of the series E~l XIo that diverges, contrary to condition bseries ODe obtalns a su 2}. Let {adf be an arbitrary sequen.ce of ±l. P~i:ion the natu~al 2) ::} 3). _ AU B, where A = {nl' "2, •.. } IS the set of mdlces n for which numbers as N-=- {m m2,'''} is the set of indices n for which an = -1. Then It . 00 d ,,00 d h . an _- 1 and 'sB both series L:.I:=I :en. an [..,j=1 xm; converge, an so t e senes by bypoth~ {"'co Xn _ L:~I xm; will also converge. "co l.Ji=l a'X" - l.-II=I Let n1 <~ n2 <1"3 < ... bearb·ltrary .md'ICes. Put { nl, n2, n3,· .. } = 3) ~ 2). _ B. eonsider two sequences of coefficients: a; = 1 for all i E N A and N \ A -: E A {3; = -1 for i E B. By hypothesis, the series E:l ajXj and rt · X n , + X n2 + X n , + ... = and{3; -1fo e 'and consequent1y so d oes t eh series 2:"" 1 PiXi converg • 0 ,,~ !(a-Xi +PiXi)' t..-i=1 2 ' h ve established the equivalence of unconditional and perfect conThus. we a 'II on we shall make no distinction between these two notions. vergence. From no . . . EM 1.3.5. Let X be an n-dImenSIonal normed space. Then In X any THEOR nvergent series is absolutely convergent .

to be

"

E:,

. _ .J;.:~"./lv CO

§3. PRELIMINARY MATERIAL ON REARRANGEMENTS

11

PROOF. Since in a finite-dimensional space all norms are equivalent to one another, one can assume that X = l~n). Denote by /;, i = 1, ... , n, the coordinates Cunctionals, which assign to each vector its i-th component. Let E~l Xk be an unconditionally convergent series in X. Then the numerical series E~l !i(XIc) are also unconditionally convergent, and hence E~ll"(xlc)1 < 00 for i = 1, ... , n. Therefore, :F

~ IIXkll = ~ (~I/;(Xk)l) <

00

o

x" converges absolutely.

and the series E~l

The next assertion shows that unconditionally convergent series E:l Xi not only have the property that all series E:l x, with 0i = ±1 converge, but this convergence is in a certain sense uniform with respect to the coefficients ai. LEMMA 1.3.1. Suppose the series E:l Xi in a Banach space X converges unconditionally. Then for any g > 0 there exists a number N = N( E N such that

sup {II% aix,ll: mEN, PROOF.

e

0,

= ±1} < t.

Suppose that the assertion of the lemma is false, i.e., there exist

> 0 and natural numbers ml < rl < m2 < r2 < . .. such that, for any j EN,

max{

t

(XiX,

: ai

= ±1}

~ e.

(1)

k=m;

Define a sequence

/3i

=±1, i EN, in such a way that

L rj

PiXi

= max

{

L rj

(XiX,

: 0i

=

}

±l ,

k=m;

/t=m;

Le., for i such that m; ~ i ~ rj take for the {3i those ai for which the maximum in (1) is attained, and for the remaining values of i take, say, {3i = 1. By hypothesis, the series E:l {3iXi must converge (the series E:l Xi is unconditionally convergent); but inequality (1) shows that the constructed series diverges. The contradiction we have reached proves the lemma. 0 THEOREM

1.3.4

(GEL'FAND'S THEOREM).

Let X be a Banach space and

L~=l Xi be an unconditionaJ1y convergent series in X. Then the collection of the sums s(a) Q"iXi, where 0 {Oi}~l runs through all sequences of ±l,

= E:l

=

forms a compact set in X. PROOF.

Equip the set of sequences

osrv of r.nnrnin:.t.p.wic;.p

t""nnv~rapn,...p.

Q

= {(X;}~1

in question with the topol-

The "Dc"l+,"", +,... ... "'1.,... _: ........ I ... _ .........

! ........... - - - - - ' "

12

CHAPTE R 1.

BACKGROUND MATER IAL

1.3.1, space homeomorphic to the Tikhonov product K ::= {-I, l} N. By Lemma y. topolog this in ous the mapping 0 ...... 5(0) = E~1 CtiX, from K to X is continu a of image the that g assertin To complete the proof it remains to use the theorem 0 t. compac is compact space under a continuous mapping ct EXERCISE 1.3.1. Determine for which of the spaces lp one can constru the g followin series, ent. exa.mples of unconditionally, but not absolutely converg pattern of Example 1.3.1. converEXERCISE 1.3.2. Construct an unconditionally, but not absolutely gent serics in the space LIIO, IJ. Consider other classical Banach spaces. inEXERCISE 1.3.3. Show that if, in the definition of perfect convergence, ents coeffici of ns collectio all for Ctixi stead of the convergence of the series L:~1 8jXi for all collections OJ == ±1, one requires the convergence of the series that contains at least two set d of coefficicnts Ok E T, where T C R is a bounde ence. converg points, then one obtains an equivalent notion of

2::'1

spaees EXERCISE 1.3.4. Investigate for which classes of linear topological . one can successfully generalize the definitions given in the last two sections

CHAPTER 2 SERIES IN A FINITE-DIMENSIONAL SPACE

We shall now turn to the exposition of the main material, starting with the "simplest" case where the space is finite-dimensional. Already here there arise rather nontrivial problems, which continue to be the subject of fruitful investigations. Analogies with the material studied in the present chapter will play an important role in the sequel, in the examination of features that are specific for the infinite-dimensional casco

§l. Steinitz's Theorem on the the Sum Range or a Series DEFINITION 2.1.1. Let E~l XI; be a series in a Banach space X. We say that a point X belongs to the sum range of the series if there exists a permutation 1r for which the rearranged series E:1 Xtr(l;) converges to x. The set of all such points, called the sum range of the series E~l XI;, will be denoted by SR(E~=1 Xk).

=

By Riemann's theorem, if X = R, then SR(E~l XI;) X for any conditionally convergent series in X. If the dimension of the space X is larger than one, then there exist conditionally convergent series whose sum range does not coincide with the whole space. For example, if all terms of a series are collinear to some vector e, then for any rearrangement the sum of the series will also be collinear to e. In this example the sum range of the series is the line passing through 0 and the point e. Let X be a finite-dimensional space. For any given subspace Y of X the reader will find no difficulty in constructing a series 2::::1 XI; in X for which SR(2:~l XI;) = Y. FUrther, observing that the relation SR(xo

+

00

<>Co

1;=1

k=1

2::>1;) = Xo + SR(I>I;)

holds for any Xo EX, one concludes that in a finite-dimensional space the sum range of a series ma.y be any translated (affine) subspace. As it turns out, this exhausts all possible situations, i.e., if 2:~1 XI; is a conditionally convergent series in a finite-dimensional space, then its sum range is an affine subspace of dimension larger than or equal to one. In the two-dimensional case this fact was first established by P. Levy in 1905 [50]; a proof for arbitrary finite-dimensional spaces was

14

CHAPTER 2.

SERIES IN A FINITE-DIMENSIONAL SPACE

given by E. Steinitz [87J. In this section we shall prove the theorem formulated above, following mainJy Steinitz's scheme. For the proof we need several auxiliary assertions. The first of these can be formulated as follows: the number of faces that meet at any given vertex of a polyhedron is not smaller than the dimension of the polyhedron. LEMMA 2.1.1. Let K be a polyhedron in R n given by a system of linear equalities and inequalities:

fi(X) = O-i,

i

= 1.2, ... ,p,

{ 9j(x):5 bj , j == 1,2 ..... q.

where Ii and 9j are linear functionals. Let Xo be a vertex (extreme point) of K and A == {j: 9j (xo) =bj }. Then the number of elements in A is not smaller than n-p. PROOF.

Suppose the contrary holds. Then the system of linear equations

'i(X) ==0, { 9j(X) == 0,

i.=1.2, ... ,P, J E A,

has less equations than unknowns (the coordinates of the vector x). and hence has a nonntrivial solution Xl' For sufficiently small E > 0 the points Xo ± EXI will belong to K, which contradicts the fact that Xo is a vertex of the polyhedron. 0 LEMMA 2.1.2 (ROUNDING-OFF-COEFFICIENTS LEMMA). Let {xd~=1 be a finite subset of an m-dimensional normed space, P;}f=l be a set of scalar coefficients, 0 :5 Ai :5 1, and X = E:':l AjXi. Then there exists a set of coefficients {ei }i'::I' each OJ equal to 0 or 1 (a set of rounded off coefficients) such that

Ilx- t PROOF.

8i

OiXi

ll:5; .mFlixill.

If n :5 m, then it suffices to take 8,

(I)

= 0 whenever Ai

:5 1/2 and

= 1 whenever Ai > 1/2. Now consider the case n > m. Let us introduce the

auxiliary space of coefficients Rn and the polyhedron K in Rn given by the system of inequalities 0 :5 ti :5 1, i = 1,2, ...• n, and equalities x = E:':I tiXi. where (tl.h, .... t n ) are the coordinates of a vector in K. Since the polyhedron K is nonempty and bounded. there exists a vertex T = (i l ,t2 •... ,In) E K. Notice that the vector equality x = E:'..l tiXi is a system of m scalar equations. Hence, by Lemma 2.1.1. among the coordinates of the point T there are n - m that are equal to 0 or 1. Now define the numbers fJi as follows: if ti = 0 or l, 1, then fJ, = ti; if 0< ti ::; 1/2. then 9i = 0; finally. if 1/2 < t; < 1. then fJ i = 1. We have

=

Ilt

,=1

fJiXi

-xii = Ilt fJiXi - ttiXill:5 (t l -til) .mr-xllxiI/. 8i

,=1

,=1

t=1

§l STEINITZ'S THEOREM

15

Since n - m of the numbers 19, -1il are equal to zero, and the remaining ones are no larger t.han 1/2, we conclude that

o EXERCISE 2.1.1. Show t.hat if X l~nI), then the constant m/2 in the righthand side of inequality (I) cl\n be replaced by j1ii/2.

=

LEMMA 2.1.3 (REARRANGEMENT LEMMA). Suppose that in the m-dimensional normed space X there is given a finite set {.ri}i=l of vectors, whose sum is denoted by x. Then one call rearrange t.lle elements of this set ill such a way that

for any natural k

Sn

the [ollowing inequality holds:

(2)

where 7r is the corresponding permutation of the index set {l, 2 .... , n}. PROOF. Since inequality (2) is homogeneous, i.e., is prcserved when all the vectors Xi are multiplied by the same constant, we may assume that the norms of the elements are bounded by 1: maXi IIxdl 1. It also clear t.hat we only need to prove inequa.lity (2) for n > In, since for n ::; In it holds for any permutation. Let us construct. by induction a chain of sets {I, 2•... , n} An ~ An - 1 ~ •• , Am and numbers >.~ (k = m, m + 1, ... ,n; i E "h), with the following propert.ies:

=

=

card Ak

= k;

0::;

>'1: ~

1,

L >.~ = k - m;

~.

iE.4k

iEA.

~

k-m >'l.xi = -'-- x.

(3)

n

=

The set An is already given. Indeed, for >.~ = (n - m) In properties (3) with k n hold. The induction is carried out by decreasing k. Suppose the set Ak+l and the set of coefficients {>.~+l }iEAkH are a.lready construct.ed. Consider the set K of collections of numbers {lAi, i E AIc + I} that satisfy the conditions

o$

Ili

$ 1;

L iEAHI

Iii

=k -

m;

L iEA.+,

k-m

l'iJ.'j

= --x.

(4)

n

Then K is llonempty (one can take Jl.. = k~-::~l >'~+l) and is a convex polyhedron in the space Rk+l of vectors Jl. = {lAi: i E Ak+d. It. is readily seen that the condit.ions of Lemma 2.1.1 are satisfied with p = Tn + 1 (t.he last vect.or eq\lali~y in (4) is equivalent to Tn scalar equalities) and q = 2(k + 1). The polyhedron [( is a bounded set, since allili belong to the segment \0,1]. It. follows that K ha'i vertices. Let Ii = iii. ; t= ~ • • \ h .. roM ,,( thnrn ..,,_.;nnn

16

CHAPTER 2.

SERIES IN A FINITE-DIMENSIONAL SPACE

By Lemma 2.1.1, the set A of those i for which lJ.i is equal to 0 or 1 has at least (k + 1) - (m + 1) = k - m elements. We claim that at least one of the numbers Tii is equal to zero. Indeed, if Tii = 1 for all i E A, then by the first two conditions in (4), card A = k - m, and the remining numbers J.Li (i E AH 1 \ A) are equal to O. If 'iIi 1 does not hold for all i E A, then again among the numbers 'iIi there are some equal to O. Let j be an index such that 'iIi = O. Set Ali: = Ak+l \ {j} and >'k = Tii (i E Ak). It is readily verified that conditions (3) are satisfied. Hence, our construction is complete. The sought-for permutation 11' is defined as follows: for i = m+ 1, m+ 2, ... , n set lI'(i) equal to the index j that was eliminated from the set Ai in the inductive construction of the set Ai - l , and for the remaining values of i define lI'(i) in arbitrary manner. Let us check that the permutation 11' constructed above satisfies inequality (2). For k ~ m this is obvious. For k > m, by conditions (3), we have

=

=

11.l: (1 - >'~)Xill ~ .2: (1- >.~) = k - (k - m) =m. leA.

o

leA.

The proof of Lemma 2.1.3 given above belongs to V. S. Grinberg and S. V. Sevast'yanov 1271. REMARK 2.1.1. If we remove the second term from the left-hand side of inequality (2), we obtain the following inequality, which is more convenient for our

purposes:

Ir\

IIt.XW(il l ~ m· mF IIx;l1 + (m + 1) IIt.Xill·

(5)

REMARK 2.1.2. The last lemma is often formulated as follows: Let X be a finite-dimensional space, and let {Xk}k=l be a set of vectors such that EZ=l Xk = O. Then there exists a permutation 11' of the first n natural numbers such that

(6) where K depends only on the space X. In this formulation Lemma 2.1.3 is known as Steinitz's lemma. The Steinitz constant of the space X is defined as the infimum K (X) of the constants K that can be taken in inequality (6). From Lemma 2.1.3 it follows that K(X) ~ dimX.

§l. STEINITZ'S THEOREM

17

It is probable that by using the specific structure of a space X one could better estimate the quantity K(X). For instance, it is not known what is the Steinitz constant of a Euclidean space; one would guess that K(l~n» :5 ..;n. It even not known what is the exact value of the Steinitz constant of a three-dimensional Euclidean space (see [11, [21 for the best known values). DEFINITION 2.1.2. Let X be a Banach space and let E:'i Xk be a conditionally convergent series in X. A linear functional f E X· is called a convergence functional for this series if E:'II/(x,JI < 00. The set of all convergence functionals of a series will be denoted by r. Also, we will denote by r.l c X the annihilator of the set of convergence functionals: r.l

= {x EX: f(x) = 0 for all fEr}.

Clearly, rand r.l are linear subspaces of X· and X, respectively. In the infinite-dimensional case r is not necessarily closed. Next let us introduce the following two sets: P({Xk}~l)

= {XiI + Xi, + ... + Xip: il < i2 < ... < ip;

pEN},

Q({xd~') = {tAiXi: 0 $ Ai $ 1, N = 1,2, ... }. 1=1

The elements of P({Xk}k'=l) will be referred to as partial sums. P({Xk}k::l) c Q({Xk}k'=l); also, the set Q({Xk}~l) is convex. We let Q denote the closure of Q({Xd~l)' LEMMA 2.1.4. Let X be an arbitary Banach space, and let E:'l Xk be a conditionally convergent series in X. Then for any x E Q the set x+ rl. is contained in Q.

PROOF. Pick an arbitrary functional! E X· \ r. Clearly, for such an f the numerical series E~=l !(Xk) converges and its sum equals J(E~=l Xk), but the convergence is not absolute: E~l If(Xk)1 = 00. It follows that among the partial sums of the series E~l !(XI.,) there are arbitrarily large ones "in both directions,"

i.e., sup{J(y): y E P({Xk}~l)} = inf{J(y): y E P({Xk}k=l)}

00,

= -00.

Since Q( {Xk }k=l) :::> P({Xk }~1)' we see that sup{J(y): y

E

Q({Xk}k=l)} =

00.

(7)

Now suppose the conclusion of the lemma is false, i.e., there exist x E Q and z E rl. such that x + z ¢ Q. Then, by the Hahn-Banach theorem, one can separate x + z from the set Q by some linear functional ! EX·: sup{f(y): y E Q} < f(x

+ z).

18

CHAPTER 2.

SERIES IN A

FINIT&DI~IENSIONAL

SPACE

If fEr, t.hen the last condition cannot be satisfied, since then f(z) = 0, and consequently f(x) = I(x + z). If now 1 f. r, then again the last condition cannot be satisfk'
=

Lemma 2.1.4 is taken from V. P. Fonf's paper [23].

2.1.3. Let 5' be an arbitrary finite subset of terms of the series Xl:, and let y be an arbitrary element of t~ space X. Then t~e conclusion of the preceding lemma remains valid if we replace Q( {Xl:}f:I) by y+Q( {Xl: }~l \5'). REt.·IARK

L:~I

We proved the last lemma under more general assumptions t.han we actually need 1I0W, with the fut.ure aim of establishing an infinite-dimensional generalization of Steillitz's theorem. In the present. section Lemma 2.1.4 will actually be lIsed only for series in an m.-dilllcnsionlll space. THEOREM 2.1.1 (STEINITZ'S THEOREM). Let L~l X" be a convergent series in all rn-dimensional spa.ce E, and let L~l Xk s. Then tile sum raIlge of the series is tile affine subspace 5 + r.l, where r.l is the a.nnihilator of the set r of convergence functiona/s: SR(L~1 Xk) = 5 + r.l.

=

X·

PROOF. Let f E be a convergcnce fUllctional for L~l Xk· By definition, L~I !(Xk) converges absolutely. lienee, for any permutation 71', if the series [:"1 X,,(k) converges, then

f

(~X"(k») = ~f(XTf(k») = 1(5).

Therefore, SR(L~I Xk) - s C Kel' f. Since the last inclusion holds for all 1 E r, we conclude that SR(L~Il'k) - 5 C rl., i.e., SR(L~l l'k) C s + r.l. The proof of the opposite inclusion SR(L~) Xk) :::> 5 + r.l is considerably more difficult and will be broken into two steps. First we will show that for any element 5' E s + r.l one call find a permut.ation 71'0 of the given series and a sequence of indices 7l-1 < HZ < ... such that

i.e., t.hat some subsequence of the sequence of partial sums of the rearranged series convergp.s to 5'. Only after that we will const-ruct a "correct.ed·' permu tation 71' for whidl the series L~I :r.,,(k) will indeed converge to s'. The set Q contains 5 and. by Lemma 2.14, it also eontains 5' (5' E 5 + 1'1 C Q). Take a sequence of numbers ell '\,. O. Approximate 5' by an elelIIent ql E Q({Xk}t;.): I/s' - qlll = /Is' - L -'i.r.1I < e). Next, applying the Rounding-olf-Coefficients Lemma (Lemma 2.1.2), approximate ql by an element

§1. STEINITZ'S THEOREM

19

IIql - I::lhrd\ ::; m· maxk~1 IIIkll, where m = dim E and the coefficients 9; are equal to 0 or 1. Consider the set of those elements Xk for which 9k = 1 and adjoin to it the element XI, it XI is not. already there. Denote t.he resulting set by 51 and the sum of its elements by 51. Then

PI E P({xdr':I) :

Now consider the set. SI + Q({Xd~1 \ 5t}. It contains s, and, by Remark 2.1.3, s' E 51 + Q({Xk}~l \ Sd· Further, approximate 5' - 51 by an element 92 E Q({xdf:1 \St}: lis' -sl-Q211 = lis' -51 - E>'jx,lI:5 £2· Now approximate q2 by an element P2 E P( {l=df:1 \ Sd: \lq2 - 1'211 = 1192 - E9i xdl ::; m.. maxk~2I1xkl\, where the 9; 's are equal to 0 or 1. Adjoin to S) all elements X, from the last sum for which OJ = 1 and also the element X2, if X2 is not already in 51 or among the newly adjoined elements. Denote the result.ing set. hy S2 and the sum of its elements by 52' Then

Continuing this construction indefinitely we obt.ain a sequence of finite set.s

n

such t.hat the sums s" of t.heir elements satisfy the inequality

lis' - snll

::;!n

+ m· max 1I·1:kll + IIxnll· kln

If we now writ.e t.he elements of the sets SI. 52 \ SI, S3 \ 52, S4 \ S3, . " one after another, then, by the estimates derived above, we obtain the desired rearrangement. of the series E~I Tk· Now let lIS proceed to t.he second st.ep of our proof. We have produced a series (for convenience we will denote it again by L:"1 Xk) whose general term COllVE.'rges to zero (t.hanks t.o the cOllvergence of t.he initial, ll11rearranged series). and for which a subsequence of its partial sums converges to s':

From the last relation it follows t.hat

"J~ II k=~"1 Xk = O. I

20

CHAPTER 2.

SERJES IN A FINITE-DIMENSIONAL SPACE

Apply to each of the sets {Xk}~;'+~}+1 the Rearrangement Lemma in the form given in Remark 2.1.1, and denote the resulting permutation of the set N by 7r. Then nj

nj

LX"(k)

= LXk,

k=1

j

= 1,2, ... ,

k=1

njH

I ' " x1I"(k) ~ k=n;+1

:5 m· n·
-,

2:

Xk

k=n;+l

Let r be an arbitrary natural number, r > nl, and let j be such that nj < r :5 nj+l' Then

It l

k=1

x

1I"(k) -

s'll :5lltX1I"(k) -s'll + k=1

t

X1I"(k)

k=n;+1

Consequently,

r~~ II~

X .. (k) -

s'll = o.

This means that the series L~i X,,(k) converges and its sum is equal to s'.

0

REMARK 2.1.4. From the last theorem it follows that the sum range of a conditionally convergent series in a finite-dimensional space cannot reduce to a single point. We leave the proof of this assertion to the reader.

In this section we have encountered a group of problems that could be regarded as belonging more to combinatorial geometry than to functional analysis, which, generally speaking, should come as no surprize. Indeed, here the main subject of our investigations is the permutation of vectors, i.e., a combinatorial notion. If one follows such an terminology, then the main object studied in this book is combinatorial geometry in an infinite-dimensional space. As is sometimes the case, results of some or another abstract branch of mathematics turn out to be useful in applied problems. Thus, to us it came as a complete surprize the fact that the Rounding-off-Coefficients Lemma and the Rearrangement Lemma are of essential use in a number of optimal design problems (see 18J, [82J, and also 143, pp. 21-22]). From this point of view the estimation of the quantity K(X) (see Remark 2.1.2) is not only an interesting and difficult problem, but also one of practical importance.

§2. THE DVORETZKY-HANANI THEOREM

21

EXERCISE 2.1.2. Show that if a series 2::=1 Xk in a Danach space X converges, if 2:~1 Xk = 0, and if for some permutation 11" one has that·l::=1 X"'(k) = x, then one can find a permutation (J and a sequence of indices nl < 112 < n3 < ... such that limj_oo 2:;~1 X,,(k) = 2:r. EXERCISE 2.1.3. Empolying only the preceding exercise and t.he Rearrangement Lemma, show that in a finite-dimensional space the sum'range of a series is an affine subspace. EXERCISE 2.1.4. Characterize the sequences {xdr:1 in R7I which can be ordered in such a way that their elements will form a convergent series. Solve first the one-dimensional version of the problem. §2. The Dvoretzky-Hanani Theorem on Perfectly Divergent Series In this section we shall consider the series that from the point. of view of their propert.ies are the farthest from the unconditionally convergent series. DEFINITION 2.2.1. A series 2:~=1 is said to be perfectly divergent if for any arrangement of signs the series XI ± X2 ± X3 ± ... diverges.

It is clear that. if the general term of a series does not. tend to zero, then the series is perfectly divergent. The simple example of the series 2:::'~1 n- I / 2c,,, where en is an orthonormal sequence in a Hilbert space, shows that failure of the general term to tend to zero is not the only reason for the perfect divergence of a series. Nevertheless, in a finite-dimensional space the perfectly divergent series are precisely the series whose general term does not tend t.o zero. Our next goal is to prove this result, obt.ained by A. Dvoretzky and C. Hanani in 1947118). LEMMA 2.2.1. Let {xd:~1 be eJement.s of a space X, dimX there exist coefficients Ui = ± 1, i = 1, ...• n, such that

= m.

Then

PROOF. (The proof given here belongs to V. S. Crinberg.) If the number n of elements does not exceed the dimension m of the space, then by t.he triangle inequality the above inequality holds for any choice of signs 0i. Now let n > m. Consider the polyhedron Kl in R",+l given by the system of equalities 2::~i 1 tiX; 0 and inequalities -1 ::; ti ::; 1, i 1, ... , m + 1, where (td::'i I denote the coordinates of a vector in K I. This bounded polyhedron is not empt.y (indeed, E Kd· Hence, by Lemma 2.1.1, there exists a collection (t!)::.1 1 E KI such that at least one of the coordinates, which we denote by ti" is equal to 1 or -1. Now consider the polyhedrom /(2 in Rm+2 given by the equalities til = 1:,_ 2:;~i2 tiX; = 0, and the inequalities 1 ::; ti ::; 1, i E {I, ... , 111 + 2} \ {i d. By Lemma 2.1.1, there exists a collection (tf)~12 E /(2 for which at least one of

=

°

=

CHAPTER 2.

22

SERIES IN A FINITE-DIMENSIONAL SPACE

the coordinates, which we denote by t~2' is equal to 1 or -1 (i2 i= il)' Moreover, t2 t,l I by construction. 'I Repeating this process n - m times, we obtain indices il, i2, ... , i n- m and k k collections of coefficients (tn?::t such that ik :5 m + k, t~xi 0, lt~1 :5 1, It~.1 = 1, and = for k > i. The sought-for collection of signs 0i can now t~. ' and if i is different from all the numbers ik, put be defined as follows: Oi. 0, = 1. Then for any i :::; m we have

=

tt tt,

E:;i

=

=

Now suppose i > m; set i = m + 1. By construction, the sum E:'::i'(Oi - tDXi contains no more than m nonzero terms. Consequently, if we take into account ~m+1 tiXi 'O ' that L...i=t = ,web 0 tam

Hence, for any

i :::; n,

lit I : ; °iXi

2m .

mF IIxi II, 0

as claimed.

REMARK. As in Steinitz's Lemma, here the best value of the constant that one can put instead 2m in the right-hand side of the inequality obtained above is not known. The case X l~m) is of special interest.

=

THEOREM 2.2.1 (THE DVORETZKY-HANANI THEOREM). 1£ a series in a linite-dimensional normed space is perfectly divergent, then its general term does not tend to zero. PROOF. We shall proceed by reductio ad absurdum. Let X be an m dimensional space, and let E~=l Xn be a series in X such that lim n_ oo Xn O. We need to exhibit an arangement of signs for which the series E~=l ±xn will converge. Denote dn = sUPi>n IIxdl. Clearly, dn ! O. Consider a sequence of indices 0= nl < n2 < n3 < '" such that E~l dn. < 00. Using Lemma 2.21, for each of the segments {Xi}~~~+l one can choose signs ti such that

=

Then, if

nk

< it < 12. we have IIE1~jl tix;11 :5 4m Ej:k dj . That is to say, for

our choice of signs the series E:l ~~.

tiX;

satisfies the Cauchy criterion, and hence

0

§3. PECHERSKII'S THEOREM

23

In any infinite-dimensional Banach space there are perfectly divergent series whose general term tends to zero. The reader will be able to prove this result. on his own by solving the next exercise and using Dvoretzky's theorem on almost Euclidean subspaces (Theorem 6.2.1). EXERCISE 2.2.1. For a Banach space X the following two assertions are I equivalent: 1) for any perfectly divergent series in X the general term does not tend to zero; 2) there exists a constant C > 0 such that, for any finite set of elements {xi}i=1 C X there exists a set of coefficients ti ±1 (depending on the set {xi}i=I)' for wroch

=

The following interesting problem remains open: Does the Dvoretzky-Hanani theorem extend to series in nuclear Frtkhet spaces? EXERCISE

2.2.2. Does the assertion of Exercise 2.2.1 hold true for noncom-

plete nonned spaces?

§3. Pecherskii's Theorem In 1988 D. V. Pecherskii \68J published the following result which, in particular, establishes a connection between the Dvoretzky-Hanani and the Steinitz theorems. THEOREM 2.3.1. Let X be an arbitrary Banach space, 2::=1 xn be a conditionally convergent series in X, and Xo = 2::=1 x n . fUrther, suppose that no rearrangement makes the series perfectly divergent. Then RS(2::=1 xn) is the closed affine subspace Xo + rl., where rl. is the anihilator in X of the set reX· of convergence functionals (the terminology is taken from Definitions 2.1.1 and 2.1.2).

This assert.ion provides the most general of the known sufficient conditions for linearity of the sum range of a series in an infinite-dimensional space. In the finite-dimensional case Theorem 2.3.1 is identical to Steinitz's theorem, because, by the Dvoretzky-Hanani theorem, the perfect divergence of a series is not affected by rearrangements of its terms. Pecherskii's theorem was later, although independently, also proved by S. A. Chobanyan. Since Chobanyan's proof is simpler than Pecherskii's proof, we shall follow here his scheme. In the proof of Steinitz's theorem the finite-dimensionality of the space was used twice: in the Rearrangement Lemma 2.3.1 and in the Rounding-off-Coefficients Lemma 2.1.2. Analyzing that proof, one can derive the following assertion, which alrearlv h()lti~ in .. nv R"no,.l. ~n~~~

24

CHAPTER 2.

SERIES IN A FINITE-DIMENSIONAL SPACE

LEMMA 2.3.1. Suppose the series L::"::1 Xn in a Banach space X has the following two properties: (A) [or any I:: > 0 there exist N == N(c) and 6 > 0 such that if {Y, }:'::l is a finite set of terms of the series, {Y;}:'::l C {X;}~N' IIL~l y;\1 :::; 6, then one can find a permutation 1f of the first n natural numbers for lI'hich

(B) [or any £ > 0 there exists a number M = M(c) such that i[ {Yj }i:1 is a finite set of terms o[ the series, {Yi }i:l C {Xi }~JIf' and j[O ::; Ai ::; 1, i = 1, ... ,n, then one can find a set of coefficients {(Ji} i: I' (Ji E {O. I}, for which

Tllen for any

X

E

RS( E::'=l xn) it holds that RS( E::"=l Xn) = x + r1.

0

The aim of the following lemmas is to show that under the assumptions of Theorem 2.3.1 the series 2::'=1 In satisfies conditions (A) and (B) of Lemma 2.3.l. LEMMA 2.3.2. §uppose that there is no rearrangement that makes the series perfectly ~ergcllt. Then for any E > 0 there exists a number N N(c) sud) tllat, for any finite collection, written in arbitrary order, of clements

2::'1 Xn

=

{YI, Y2,··· Yn} of the set {XN, XN+), XN+2,· .. } one can find a collection of signs ±l for wiJich

OJ ::;:

PROOF. Suppose the assertion of the lemma is false. Then one can find an c > 0 sl1ch that, for any N E N, the set {Xi}i:.1 contains a finite subset {y,}i::1 with the property that

for any choice of :o;igns Q; == ±l. Since such a subset {Yi}:'=1 exists in a tail of arbitrarily high order of the series, one can find numbers 1 = NI < N2 < ... and collections of terms of the series {yf}:.!1 C {Xd~N~-\ such that

lit

min J5.N. max ;=1 <>,=,%1

(1)

CtiYf11 > E.

Now write the terms of the series in the following order: { 1IiI } 'II i=="

i=I' {Yi2}n, i=I'

\ { I}TI' { X, } N,-l N. Yi

N3-1 \ { 2}n, { XI } N, Yi

i=I'

§3 PECHERSKII'S THEOREM

25

where the elements of the sets {yn?,!\ are written in exactly the order for which {yn:~\ are written inequality (1) holds, while the elements ofthe sets in arbitrary order. Denote the resulting rearrangement of the series E:l Xi by 7f. By inequality (1), the series L~ \ O,X"(i) diverges for any choice of signs 0" which contradicts the hypothesis. 0

{X.}Z:+I-\

LEMMA 2.3.3. [Jet ~ be an arbitrary positive number, arid let {x;}f:., 1 be a set of elements of the normed space X, with the property that for any finite subset {Y.}~. C {Xi};;'. there exist signs 0, = ±1 such that II E:: 1 oiy.1I $~. Then for any set of coefficients {Ai} ~, 0 $ Ai $ 1 there exist "rounded off" coefficients (Ji E {O, ]} for which

PROOF'. From considerations of continuity it is clear that it suffices to prove the lemma in the case when the coefficients A, are finite dyadic fractions:

1 1 ] >"=A'0+A'I'-+A'2'-+ • I, I, 2 I, 4 .'. +A I," ' 2"'

(2)

where the Ai,4' are equal to 0 or 1. The number of terms in the representation (2) will be called the length of the fraction. It is clear that if Ai.O = 1, then because of the inequalities 0 $ Ai $ 1 all the remaining Ai,k'S in (2) are equal to zero, and also that all Ai may be regarded as dyadic fractions of the same length (adding, if needed, more zero terms). Let us show, by induction on n, that if Ai are fractions of length n, then one can choose Oi E to, I} such that

(3)

This will complete the proof of the lemma. n = 1. In this case the Ai themselves are 0 or 1, and we obtain zero in the left-hand side of inequality (3) jf we take Oi :; Ai. The n ..... n + 1 step. Let It; be fractions of length n + 1, i.e., Iti = + .2- 1 + ... + 2-". Denote by A the set of all indices i $ N for which = 1. By hypothesis, there exist signs for which \I L'EA oixill $ ~. Introduce auxiliary numbers Ai as follows: if i ¢ A, put Ai = Ili; if i E A, put >'. ::: Ili + 0i ·2-". Then

Jji,l Jji,n

lti,O

Iti,n .

li t

1=1

{Oi}iEA

tltiXil1 $ 2nII?=OiXill $~' 2~' .=1 .eA 1

AiXi -

CHAPTER 2.

SERlES IN A FINITE-DIMENSIONAL SPACE

Ai

the other hand, the are dyadic fractions of length n, and by the inductive On otbesis there exist numbers 8; E {O, I} such that inequality (3) holds. This hyP yields

t

t

8iXi li $ lit Ai x; - 8iXi li + Ilt IJiXi -tA;Xill tJJiXi Il j:::1

.-1

.-1

.-1

.-1

.-1

o Combining lemmas 2.3.2 and 2.3.3, we conclude that under the assumptions rTheorem 2.3.1 condition (B) of Lemma 2.3.1 -an analogue of the Rounding·off~ef!icients Lemma-holds. It remains to establish property (A)-the analogue of the Rearrangement Lemma. LEMMA 2.3.4 (CHOBANYAN'S LEMMA). Let {Xi}:'=l beeJements o[thespace X, with E;'.,1 Xi = O. Then there exists a pennutation (1 such that, for any choice

{signs OJ

=±l,

~lltaix"i'll ~~lltx"i'll.

o PROOF.

(4)

Take (J to be the permutation for which

. ttained. Let Oi be an arbitrary fixed collection of signs. We claim that (4) holds. ~~~eed, denote by A [resp. BJ the set of indices i for which = 1 [resp. = -1 J, anrlputAk::An{l, ... ,k}, Bk = Bn{l, ... ,k}, B~ = {l, ... ,n} \Bk. Then

eli

eli

ret: lit eliX.,(i)II = T~ II.I: X"(i) - .2: X<1(i) II· .=! .EA.

Consequently,

'EB.

~ lit. aix.,i' I +~ II~ "'i,11

~ max {T~ Ili~. I ' T~ Ili~' X<1(i) I } 2

X.,(i)

~ , rna> {T~ Ili~' X',i, II· T~ i~: X',i, }

(5)

§3. PECHERSKJI'S THEOR EM

27

(to obtain the last equality we used the condition L~;I Xi == 0). Since we have the inclusions Al C A2 C .. , C An C B~ C ~_I C ... c B~, of which n are equalities, there exists a permutation /.I such that (6)

By the definition of the permutation

(1,

(7) Combining (5), (6), and (7) we c{)llc\ude t.hat

which in turn yields the needed inequality (4). LEMMA 2.3.5. Let {Xl}~1

permlltat.ion

/.I

eX,

1\

L::':.I xi/I

there exists a choice of signs

Then there exists a permut ation

q such

0

Qj

~ t, and assume that for allY

= ±l

for .....hich

t.hat maxk~ .. 11 L::=1

;/:"(i)1I

:5 3c.

L:;:"] Xi· Then L:'::11Xi == 0, and we can apply the

PROOF. Denot.e Xn+1 == preceding lemma to the set {Xi} ~Il . This yield') a permut ation {1, ... , n + I} such that

max

k:5,,+1

{I ..... n + I} .....

X7
If we now discard the element which

as claimed.

7r :

X,,+t. ..... e

obtain a permutat.ion

q

of {I, ... , n},

fOI'

o

Combining lemmas 2.3.2 and 2.3.5, we conclude that under the assump tions of Theorem 2.3.1 an analogue of the Rearrangement Lemma -asserti on (A) of u>tnma 2.3.1-h olds true. Since property (B) has been established earlier, this completes the proof of Pecherskij':; t.heorem. EXERCI SE 2.3.1. Give an example of a convergent series in a Hilbert space which after a rearrangement becomes perfectly divergent.

CHAPTER 3 CONDITIONAL CONVERGENCE IN AN INFINITE-DIMENSIONAL SPACE Throughout this chapter we shall consider mainly the classical Lp spaces (with the exception of §3). The transition to a more general situation will be made later, after an exposition of the necessary supplementary material.

§1. Basic Counterexamples Now that in the preceding chapter we gave a complete description of the sum range of a conditionally convergent series in a finite-dimensional space, it is only natural to attempt to extend those results to the infinite-dimensional setting. However, as is most frequently the case, here the extension to the infinite-dimensional object of the properties of its finite-dimensional analogue cannot be achieved in full. For this reason in what follows, in order to obtain analogues of Steinitz's theorem, we will have (as, for instance, in Pecherskii's theorem) to impose additional restrictions on series. The main goal of this section is to construct examples demonstrating that without additional restrictions it is not possible to extend Steinitz's theorem to series in an infinite-dimensional Banach space. We have already remarked that if L::=l XIc is a conditionally convergent series in a finite-dimensional space, then SR(L:~l Xi:) is an infinite set; more precisely, it contains a whole line. The next examples shows that for a conditionally convergent series in an infinite-dimensional space the sum range may consist of a single point. EXAMPLE 3.1.1. The construction will be carried out in the Hilbert space l2' Let {elc}f:,l be the standard orthonormal basis in h. Consider the series 1 1 1 1 1 1 el + -e2 - -e2 + -e3 - -e3 + -e3 - -e3

2

244

4

4

111111111 +ge4 - ge4 + Se4 - ge4 + Se4 - ge4 + ge4 - Se. + 16 es - ....

This series converges and its sum is equal to el' However, it does not converge unconditionally, since the series in which all the signs are replaced by + diverges (recall tha.t, by Theorem 1.3.2, perfect and unconditional convergence COincide). On the other hand, the sum range of our series reduces to the pointel: indeed, the projection of the series on any coordinate axis contains only a finite number of nonzero terms, and a finite sum does not change under rearrangements.

30

CHAPTER 3.

CONDITIONAL CONVERGENCE IN AN co-DIM SPACE

EXERCISE 3.1.1. Show that in any infinite-dimensional Banach space there exists a conditionally convergent series whose sum range consists of a single point. EXERCISE 3.1.2. Show that the expansion of the function In(1 + x) ill a Maclaurin series is a conditionally convergent series in the space C[O, 1] and that the sum range of this series is a single point of qo, 1]. EXERCISE 3.1.2. Solve Exercise 2.3.1 by a construction analogous to that of Example 3.1.1. The main part of the assertion of Steinitz's t.heorem is the linearity of the set SR(E;;:\ Xk), i.e., the fact that SR(E;;:! Xk) contains, together with any two distinct points, the line that connects them. Below we shall discuss an example which shows that in an infinite-dimensional Hilbert space there are series with a nonlinear sum range. It is a known historical fact that the mathematicians of the Lvov school had the habit of meeting in the "Scottish Cafe" and discuss there mathematical problems. At that. time the theory of Banach spaces has been just created, and during these table talks the Lvov mathematicians, under the leadership of St.efan Banach, laid the fOllndations of modern functional analysis. When mathematicians discuss some topic, they often need to write down formulas, draw figures, et.c. During the discussions at the "Scottish Cafe" such notes were written in indelible ink right on a marble table. After that the waiters would wipe off the notes. Some results of Banach's school were perhaps lost only because they were not copied from table to paper. This difficulty was solved by Lucja, Banach's wife. She bought a thick hardbound book note and gave it to the cafe's cashier. (According to another version, Banach himself bought the booknote.) From that moment on people started to write the problems proposed (and, if they were successfully solved, sketches of their solutions) in the booknote. Any mathematician visiting the cafe could leave a note in the book. For the solution of some problems their authors would set a prize-from a small cup of coffee to a goose. In this way a manuscript w~ created, named the "Scottish Book" [811. After the 2nd World War this book was published in the USA by one of Banach's students, S. Ulam. Some problems in the "Scottish Book" were solved after its publication, while others remain open even today; part of the unsolved problems are passed on as "folklore" and mnny people are not aware of the fact that the original source is the "Scottish Book." The topics dealt with in the "Scottish Book"determined in many respects the directions of further development of the theory of Banach spaces. In problem 106 of the "Scottish Book" Banach proposed t.o prove that for any series in a Banach space its sum range is a linear set. A prize was offered for the solution of this problem - a bottle of wine. The solution also appears in the Book; it presumably belongs to J. Marcinkiewicz (authorship established according to handwriting). As already mentioned, the answer provided was negative, namely, a counterexample ill the space L2[0, 1]. Unfortunately, Soviet mathematicians were not aware of this result, and it was rediscovered by E. M. Nikishin, who constructed a more complicated example.

§l. BASIC COUNTEREXAMPLES

31

It has to be said, however, that Nikishin's example simultaneously solved a considerably more difficult problem posed by Banach and Orlicz, concerning the sum range of a series of functions in the sense of almost.-everywhere convergence. We shall return to this problem in the last chapter. Now let us describe Marcinkiewicz's construction. Denote by Xja,bl the function that takes the value 1 on ~the segment [a, bJ and the value 0 on its complement (the indicator or characteristic function of the segment [a, bj). 3.1.2. Consider the series with the terms

EXAMPLE

where 0::; i the order

< 00, 0::; k < 2i; these are elements of L2[0, 1]. If we write its sum in XO,O

+ Yo,o + X),O + YI,O + +XI.\ + Yl,l + X2,O + ....

then the series will converge to zero. If we now change the order to XO,O

+ X1.0 + Xl,) + Yo,o + X2.0 + X2.) + Yl.O + X2,2 + X2,3 + Y),I + .. , ,

then the series will converge to the function identically equal to 1 on [0. 1]. However, no rearrangement will make the series converge to the function identically equal to 1/2 on 10,1], because any of the partial sums of the series is an integer-valued function. The next example, which belongs to M. I. Ostrovskii [65], shows that the set SR(E~=1 XIc) is not necessarily closed. EXAMPLE 3.1.3. The example lives in the Hilbert space L2([0, 1] x [0, 11) of square-integrable functions on the square 0 ::; X ::; 1, 0 ::; y ::; 1. Let E~=l IIc(t) be the series constructed in the preceding example. Our new series is constructed as follows: 00

L9k(X,y)

= h(x) + .fih(y) + h(x) + ...fih(y) + ...

k=l

It is not hard to show that a function u(x, y) == const belongs to SR(E~l 9k) if and only if u(x, y) ::: n + v2m, where n and m are integers. Hence, the set SR(E~) 9k) is not closed. EXERCISE 3.1.4. Show that the sum range of the series constructed in Example 3.1.2 contains all integer-valued functions. EXERCISE 3.1.5. Show that if X is a Banach space and Y is a separable subspace of X, then there exists a series E~=I Xk for which SR(E:=I Xk) = Y.

32

CHAPTER 3.

CONDITIONAL CONVERGENCE IN AN oo-DIM SPACE

EXERCISE 3.1.6. Show that by means of an appropriate rearrangement the series of Example 3.1.2 can be made to converge weakly to any prescribed element of £2' EXERCISE 3.1.7. Determine for which function spaces on the segment [0,1] does the series of functions given in Example 3.1.2 have a convergent rearrangement. EXERCISE 3.1.8. In each of the spaces lp and Co exhibit examples series with nonlinear sum range. REMARK 3.1.1. There is no known example of a series in a Hilbert space whose sum range is a linear, but nonclosed subspace. REMARK 3.1.2. As we will show in the last chapter, there are infinitedimensional linear topological spaces in which Steinitz's theorem is valid. In the case of normed spaces the validity of Steinitz's theorem implies the finite-dimensionality of the space (Chapter 7, § 2). EXERCISE 3.1.9. Show that Steinitz's theorem holds in the space of all numerical sequences with the topology of coordinate-wise convergence. §2. A Series Whose Sum Range Consists of Two Points In this section we shall give an example of a series in a space Lp (which consisting of the same functions for all values of p, 1 ~ P < 00), whose sum range consists of two points - the functions identically equal to 0 and to 1. This example disproves, in particular, H. Hadwiger's conjecture that the sum range of any conditionally convergent series is the coset of some additive subgroup of the space under consideration. The construction given here belongs to M. I. Kadets; its justification was first obtained independently, and about the same time, by P. A. Kornilov {48J and K. Wozniakowski (see (34]). It is interesting that similar constructions were proposed at least by two other mathematicians. A. Dvoretzky told us that many years ago he had such an example, but he, too (like M. 1. Kadets), was not able to find a justification. P. Enfto constructed an example with a complete proof at about the same time {48J and [34J were written, but he did not publish it because l. Halperin informed him about the preprint containing the example presented below. Denote by Q (0,1)'" the infinite dimensional cube, i.e., the product of a countable number of unit segments, equipped with the standard product probability measure. The example will be constructed in Lp(Q), 1 ~ 'P < 00. Our series consists of funct.ions of two kinds. The functions of the first kind are defined as follows:

=

if

m-1 n

< tn < m n

otherwise.

(1)

§2. A SERIES WHOSE SUM RANGE CONSISTS OF TWO POINTS

for n EN, m = 1,2, ... , n. For each n E N set Fn functions of the second kind are defined as products: n 9m.j

= - fn"'. f jn+ 1

for n EN,

1~ m ~

= {J,r;.:

11,

33

1 ~ m ~ n}. The

l~j~n+l.

(2)

=

For each n E N set C n = {9:!t,j: 1 ::; m ::; n, 1 ::; j ::; n + I}. Clearly, IIf::'1I n- I / P , 119~)1 = (n(n + 1»-1/". The functions thus defined are connected by the following obvious relations: 1'1

n+l

gn . I~ = - L,9~,j' 1]1'1+ 1 = _ '" ~ m,J' n

L, m=1

n

I~

(3)

",=1

j=1

1'1+1

= - L, L 9~,J = 1.

(4)

m=1 j=1

The sought-for series is obtained by writing after each element of the set Fn t.he corresponding row of the set Cn: 1 I 2 2 2 2 2 + 91,2 + 121 + 91.1 + 91,2 + 91.3 + f22 + 92,1 + 92,2 + .... 111 + g1.1

In what follows we shall write this series as

E:"=I hn.

THEOREM 3.2.1. The series E::l hn converges to 0 in tIle metric of any of the spaces L,,(Q), 1 ::; p < 00. There exists a permutation 7r such that the rearranged series E~=I h,,(n) converges to 1. The sum of any convergent rearrangement of the original series, ko = E~=l ha(n). is equal to either 0 or 1. PROOF. The convergence of our series to 0 is a consequence of relations (1)(4). The rearranged series E~=I h,,(n) is const.ructed as follows: first write the element II, then after each element from Fn (n > I) write the corresponding column of the set C n _ l :

/ I1 + 121

1 1 2 2 2 2 + 91.1 + /22 + 91,2 + f31 + 91.1 + 92,1 + f32 + 91,2 + 92.2 + f33 +

This series converges to 1, thanks to the same relations (1)-(4). Now let us turn to the last, main assertion of the theorem. It clearly suffices to consider the case of the weakest L,,-topology, i.e., the case p = 1. Notice that if he = L:~=1 ha(n) is the sum of a convergent rearrangement, t.hen, as seen from (3), for any kEN all partial sums, starting with one of them, will not depend on the coordinate tk' Since all terms take only integer values, he is an integer constant. In view of this observations it remains to show that

34

CHAPTER 3.

CONDITIONAL CONVERGENCE IN AN oo-DIM SPACE

If ho = I, then (*) holds. Eliminating this case, we have arbitrary 0 > O. Take KEN such that, for any N ~ K,

llho - 111

~ 1.

Fix an

liho - ~ h~(n)1I $ 0,

(5)

II~h~(n)11 ~ o.

(6)

and for any m > I > K,

Denote the partial sum E:=I h~{nl by h. In addition to the sets Fn and Gn introduced above, consider also the sets Vn = U:=I (FA: u G k ). Choose an index MEN such that horn) E VM U FM+1 for all n ~ K. Define the functions h~,

h"

hn,

h" as follows:

= {h~(nl

if h~{nl E VM U FM+I, otherwise,

0

n

-

hn=

{ho(n)

o

n> K,

if h~(n) E G M +1 otherwise,

= E';=K+I h~. From (4) it follows that h + h" = 1. Hence, IIh·1I = IIh - 111 ~ llho - 111 -lIh1J - hll ~ 1- o. Denote no = K and for j = 0, I, 2,3 set

and h·

niH

56 ~ . ~ =mm. { n: 4"1 - "4 ~

o} .

h"i ~ 4"1 - 4"

(7)

(8)

,=nj+l

The fact that the indices niH are well defined follows from (6) and (7). For j 0,1; 2, 3 define the following functions:

=

h;~l =

nl+1

L

h~, hi+1 =

n=nj+l

nj+1

L

fti+l

ii,., h.j + 1 =

L

horn},

n=nJ+l

n=nj+1

s"

rj

and for j = 1,2,3,4 set = hi - hj" -hj. Let h = E:O=n.+1 h~. Notice that rj (1 ~ j $ 4) is the sum of all functions h~(n) such that (a) nj_1 < n $ nj, and (b) horn) ¢ VM U FM + 1 U G M + I. The remaning part of the proof of inequality (*) goes as follows: take the partial sum H = L::!:1 h~(n) of the series under consideration, write it in the form 444

H= h+

Lhi· + Lhj + Lrj,

;=1

;=1

;=1

and estimate IIH - ~II, and then IIho - ~II· We shall need the following auxiliary assertion, the proof of which we omit.

§2. A SERIES WHOSE SUM RANGE CONSISTS OF TWO POINTS

35

LEMMA 3.2.1. Let (X,X,~) and (Y,Y,IJ) be measure spaces with probability measures. Let I(x,y) and g(x,y) be (unctions in L1(X xV), each o(which depends on only one variable: I(x, y) = i(x), g(x, y) = g(y). Then

III + gil ~ 11/11 + IIgl\ll- 2(JL x II)(SUppf)J,

,

where JL x IJ is the product measure on X x Y and Supp I denotes the support oE the (unction I, i.e., the set oE the points (x, y) E (X, Y) such that I(x, y) f. O. 0

Returning now to the proof of the theorem, let us estimate Irkjll from below. Applying Lemma 3.2.1 to the functions hj* and Tj (they depend on different coordinates), we obtain

(9) (indeed, since the function hi" is integer valued, the condition IIhtli :5 ~ implies meas(supphr) :s l). On the other hand, from (6) is follows that IIk;1I :5 6. By (8) and (9),

IIhjll = IIh;· - T; - kjll ~ IIh;" + Till-lIkjll ~ II hi·II-lIkj ll ~ ~ -

9:.

(10)

s"

Next, let us estimate IIh "1I from above. Suppose IIh II > 116. Starting with this assumption, choose an index ns such that k = 'E:~n4+1 h~ obeys the bounds

s

s"

IIhs"U :5 116. Take h5 = 'E:;'n4+1 lin. Similarly to estimate (10), we obtain IIh511 ~ IIks"1I - 6. But 1 ~ 'E~=ll1linll ~ 'E~=lllhill > 1 - 96 + 96 = 1. The

106 <

contradiction we arrived at establishes the estimate Consider now E~=l IITill· We have

1~

IIhs"1I :5 116.

4

4

4

4

j=1

i=1

i=1

i=1

l: IIh,lI = L IIhi - hi" - Till ~ l: IIh;" +Till-l: II hi II·

Combining (9), (10), and (8), we obtain 1 ~ 1 - 56 +

1

4

-l: IITill- 40, 2 i=1

whence L~=ll1r,1I :S 186. Now let us estimate

IIH - 411·

We have

36

CHAPTER 3

CONDITIONAL CONVERGENCE IN AN oo-DIM SPACE 4

4

)=1

)=1

= h + h' - ~ + Lhj + L

T) -

hS'

Finally,

1 110 - ~II SIiH - ~II + IlIto - HII S ~ + 296 + 6 = ~ + 306, which in view of the arbitrariness of 6 completes the proof of t.he theorem.

0

The proof given above belongs to K. Wozniakowski. P. A. Kornilov's proof uses a somewhat modified construction: essentially, a transition from Q to the segment 10,1\ is effect.ed by means of a measure-preserving mapping. EXERCISE 3.2.1. Show that in a Hilbert space for any n EN there exists a series whose sum range consists of exactly n points.

It is not known whether any set in a Hilbert space can serve as the sum range of a series. This problem is still open even for the case of finite sets. DEfiNITION 3.2.1. A point is said to be a limit point for the series E:'=l Xk if it is the limit of some subsequence of the sequence of partial sums of some rearrangement of the series. The set of all such points, called the limit-point range of the series, will be denoted by LPR(I:::1 Xt).

Clearly. LPR(I:~l It) is a closed set, and SR(E::I It) C LPR(E::1 XI,,). EXERCISE 3_2.2. Show that not every closed subset of a Hilbert space is the limit-point range of a series.

§3. Chobanyan's Theorem In the two preceding sections we have demonstrated in a convincing manner that Steinitz's theorem does not carryover to the infinite-dimensional case. The relevant cxanlples were const.rueted in L" function spaces, but, as will be shown in Chapter 7, exanlples with similar properties can be constmcted in any infinitedimensional Banach space. We shall now adrress the qup.stion of what conditions imposed 011 the terms of a series in an infinite-dimensional space are sufficient to guarantee that the sum range is linear and has t.he properties stated in Steinitz's Lheorem. We have already encountered an assertion of this kind, namely, Pecherskii's theorem (Theorem 2.3.1) which, as the reader undboubtcdly recalls, holds true in Lhe finite-dimensional as well as the infinite-dimensional case. The condition discussed in the present section represents the most elegant connection between

§3. CHOBANYAN'S THEOREM

37

the problems we are interested in and the theory of random series in Banach spaces. The proof will rely on Theorem 2.3.1, although Chobanyan's result !10] was obtained before that theorem. From this moment on, TI, T2,' .. will denote a sequence of independent random variables that take the values ±l with equal probabilities: P!Tj I] Ph = -11 1/2. Let Xlo X2, . .. be elements of a Banach space X. COJlsider the random

= =

=

variables Sk

= \I L:7=1 Tix,lI·

LEMMA 3.3.1.

For any t > 0, P [sup Sk I:~n

> t]

~ 2PISn > tJ.

fl be the probability space on which the random variables Ti are defined. Define sets Ai C n as follows: AI = lSI > tJ, A2 = lSI ~ t] n {S2 > tJ, ... , An = n~:/ lSi ~ tJ n ISn > tl· It is readily seen that PROOF. Let

(2)

Denote Sn,k \I E~=I T,Xi - I:;'..k+1 TiXdl· By the triangle inequality, we have Sk ~ ~(Sn + Sn,k)' Consequently,

=

Since the probabilities in the right-hand side of this inequality coincide, P(Ak) ~ 2P(A k n ISn > tl). Using inequality (2) and the fact that the sets AI: are disjoint, we conclude that

o Inequality (1) established above is a particular case of Levy's inequality !92, Chapter 5, § 21. An obvious consequence is the follOWing inequality for mean values: LEMMA

3.3.2. E (SUPI:
PROOF. It suffices to use the formula E(v)

a positive random variable and inequality (1).

= 10

00

P!II

> tl dt

for the mean of

0

38

CHAPTER 3.

CONDITIONAL CONVERGENCE IN AN oo-DlM SPACE

LEMMA 3.3.3 (CHOBANYAN'S INEQUALITY). Let {x'}?:1 be elements of a norrned space, with L:::':1 = O. Then there exists a permutation q for wlJich

Xi

PROOF. Take for

C1

the permutation given by Lemma 2.3.4. Then

~~~ lit. x~(il I ~ E (~~~ lit. TiX~(il II) ~ 2E (lit. TiXn(jl l) ~ 2E (lit TiXil ) .

o

THEOREM 3.3.1. Let {x;}~1 be elements of a Banach space with the pro~ = O. Then the assertion of Steinitz's tJJeorem holds for the series E:l

erly that limm>n ..... oo E (IIE::n TiXi II)

Xi·

PROOF. Let G = {n, n + 1, ... ,m} and A c G. By the triangle inequality,

E(lit rixill) ~ ~ (E ( L .=1

TiXi

.eA

+E (

?=riXi - .

.eA

L

'EG\A

TiXi ) )

+ .L

.eG\A

=E

riXj )

(iltrixill)· .=1

Then, by Lemma 3.3.3 [resp. Lemma 2.3.3) condition (A) [resp. (B)) of Lemma 2.3.1 is satisfied. Tha.t is, analogues of the Permutation Lemma and the Roundingoff-Coefficients Lemma hold true. By Lemma 2.3.1, this implies the assertion of Steinitz's theorem. 0 COROLLARY 3.3.1 (CHOBANYAN'S THEOREM). Let X be a Banach space. Then for the assertion of Steinitz's theorem to hold for a series E:=1 Xn in X it is 5uflicient that the series E:=l TnXn converge almost everywhere, or, in other words, that the series E:'I ±xn converge {or almost aU c1Joices of signs. PROOF. It is well known (see (92), Chapter 5, § 5) that the condition

is equivalent to the almost everywhere convergence of the series

E:=I TnXn·

§4. KHINCHIN INEQUALITIES. THEOREM OF M. I. KADETS

39

EXERCISE 3.3.1. Prove directly that the condition of almost everywhere convergence of the series E::'=l TnXn is preserved for any rearrangement of its terms. Show that Corollary 3.3.1 follows from this conclusion and Pecherskii's theorem. REMARK 3.3.1. S. A. Chobanyan also proved (see [95J) that under the conditions of Lemma 3.3.3 the two mean values

EI

:=

~ L sup

n.

If

k:5n

lit I ;=1

X,,(i)

and

are equivalent, i.e., C 1 EI ~ E2 S C2EI, where C\ and C2 are constants independent of {Xi} and n. §4. The Khinehin Inequalities and the Theorem of M. I. Kadets on Conditionally Convergent Series in L" Conditions for the linearity of the sum range of a conditionally convergent series in an infinite-dimensional space were obtained for the first time in 1954 by M. I. Kadets [31J for the case of the L" spaces. Various generalizations and sharpened forms of t.hese results were proved by E. M. Nikishin [621, S. L. Troyanskii [901, V. P. Fonf [23J, D. V. Pecherskii [67J, V. M. Kadets [35J, M. I. Ostrovskii [65J, and others. The culmination of these generalizations are the t.heorems of Peche\'skii and of Chobanyan, already familiar t.o the reader. A large part of the preceding results can be be deduced from these, strongest theorems. Here we shall use Chobanyan's theorem to establish a sufficient condition, due to M. I. Kadets. for the Iinearit.y of the sum range of a series in Lp. This result, which historically was t.he first in this direction, asserts that, in a space L p , if 1 < P < 2, then for thc set SR(E~ I Xk) to be linear it suffices that E~I IIXkllP < 00 (condition of p-absolut.e convergcnce of the series), while if p ~ 2, p f:. 00 it suffices that L:~\ IIXkll2 < 00. For the proof we need some special inequalities, known as the Khinchin inequalit.ies, the exposition of which is our next objective. Let XI, .•. ,Xn be arbitrary fixed real numbers. Consider the quantity Alp = (EI E~I rixiIP)I/", where r, are independent random variables t.hat take the values ±l with equal probabilities; Mp is the p-mcan value of the 2n combinations of the form 10\xI + ... + onxnl, where the coefficients 0, take the values ±l. If we ptlt p == 2 and then open the parentheses and reduce the like terms in the expression

40

CHAPTER

CONDITIONAL CONVERGENCE IN AN oo-DIM SPACE

3.

we see that M2 = (2:~=1IxiI2)1/2. It directly follows from the properties of pmeans that Mp does not decrease with the growth of p: Ml :5 Mp :5 Mq for 1 :5 p :5 q < 00. By the Khinchin inequalities one usually means the inequalities

a,

(t Ix;I') .,' ,; (t IX;I') .,' , M, ,;

(t IXd')'" ,;

M, ,; A,

(t IX;I')'" ,

where the coefficients 0 < a p < 1 :5 AI' <

00

1 :5 p :5 2,

(1)

2:5 p < 00,

(2)

depend on P, but not on n or

PROOF OF THE KHINCHIN INEQUALITIES. THE CASE 1 :5 p :5 2. The right inequality is a direct consequence of the relation Mp :5 M2 and the expression of M2 . Since Ml :5 Mp , it suffices to establish the left inequality for MI' Thus, we have to show that

(3)

for some a > 0 that does not depend on Xi or on n. Let us divide both sides of inequality (3) by (E~l IXill)I/2 and denote ti = xiI (E~1 IXiII)1/2. Then (3) can be recast as (4) Using the inequality

ItI 2: 1 -

cos t, t E R, and the identity

we conclude that

2: 2-

n

L Qi=±l

(1- (t cos

Qiti))

i=1

=1-

To prove inequality (4) it remains to show that

n n

min { 1 -

.

cos ti:

t; t~ = n

}

1

ITi=1 costi'

§4. KHINCHIN INEQUALITIES. THEOREM OF M. I. KADETS

41

is bounded from below by a positive number that does not depend on n. Omitting the details, let us observe that the minimum in question is attained for tl == •.. == tn == n- I/ 2 and is equal to 1 - (cosn-I/ 2 )", which in turn is estimated from below by 1 - e- I/ 2 • Thus, Khinchin's inequality is established for 1 $ P $ 2, with a constant ap ;::: 1 - e- I/ 2 • THE CASE 2 < P < 00. The left inequality follows from the fact that Mp does not decrease when p is increased. To prove the right inequality let us again divide both sides by (L~I Ix;j2) 1/2 and denote ti == xiI (L~I Ix;j2) 1/2. Then the inequality becomes

"

I>~ = 1.

(5)

;=1

Here we use the inequality

IW $

Cp(cosht - 1),

t E R,

which holds for any p ~ 2, with the constant Cp depending on p. Performing the corresponding substitution in the left-hand side of inequality (5) we obtain, after simplifications,

As in the preceding case,

max

{n

cosh t; - 1:

t t~ =

1}

is attained for tl = ... == t" == n- 1/ 2 and is equal to (coshn-l/2)" - 1, which in turn is bounded above by el/2 - 1. This completes the proof of the Khinchin 0 inequality for the case 2 $ p < 00. The sharp values of the constants a" and Ap are presently known (see [701, 1881, [94]). In particular, al == 1/../2. It is possible to introduce the quarttities Mp == (E IIE~=l TiXdIP)l/P for Xi belonging to an arbitrary normed space (an not only for Xi E R). The surprizing inequality of Kahane (see [99]) shows that in this case the p-mean values Mp with different p's are also equivalent. DEFINITION 3.4.1. A normed space X is said to have type p with constant C if, for any finite set {xi}i=1 of elements of X the following inequality holds:

By the triangle inequality, any space has type p == 1. Hence, of interest are only the spaces of type p > i. We shall discuss such spaces in more detail in Chapter 5; here we need only their definition.

42

CHAPTER 3.

CONDITIONAL CONVERGENCE IN AN oo-D1M SPACE

THEOREM 3.4.1 (S. A. CHOBANYAN). Let the space X have type p and Jet the series l:::1 x" be such that l:::l IIxkI!J' < 00. Then the assertion of Steinitz's

theorem holds for SR(r::.\ x,,).

o

PROOF. This is an immediate consequence of Theorem 3.3.1.

To obtain the theorem of M. 1. Kadets it remains to deduce from the Khinchin inequalities that Lp spaces have a type. THEOREM 3.4.2.

PROOF. Let

The spaces Lp(O, IL} with 2 < p < 00 have type 2.

h,· .., In E Lp(O,'l}. Then

E(lit. ,;/,10 S (E (lit. ';/,11'))'" ~ (E lit.,;/,(tf d")", ~

~ (l

E

r

It."",tf d")", A, (l (t.IMt)I')", d" S

(in t.he last step we used the Khinchin inequality). If one regards the functions lfi(tW as elements of the space Lp / 2(O, IL}, then the triangle inequality gives

Consequently,

o THEOREM 3.4.3.

The spaces Lp(O, "') with 1 $ p $ 2 have type p.

PROOF. As in the preceding assertion, the Khinchin inequality yields

Using the inequality

which expresses the monotone dependence of the lp-norm on p, we obtain

E

(lit. ,,"ID (In t.1J;1' $

dp )

'I, ~ (t.IIJ;II') 'I,

0

§4. KHINCHIN INEQUALITIES. THEOREM OF M. 1 KADETS

43

COROLLARY 3.4.1 (M. 1. KADETS' THEOREM). Let 1 :5 p < 00 and r = min{2,p}. Then in order for the sum range of a series L:~I Xl: in Lp to be Jjnear it suffices that L:~I IIxl:IIT < 00. 0 REMARK 3.4.1. In the preceding assertion SR(L::'I Xl:) has all the properties stated in Steinitz's theorem, i.e., not only the set SR(L~I Xl:) is linear, but it also has the form Xo + r.lj this fact was first observed by V. P. Fonf in 1972 123). More precisely, FOllf's theorem gives a sufficient condition for the equality SR(L:~I Xk) = Xo + r.l to hold for series in uniformly smooth spaces: the condition is that L:~I p(lIxklD < 00, where p(t) is the modulus of smoothness of the space (for the definition of this notion see 113]). EXERCISE 3.4.1. Carry out the calculations needed to solve the extremum problems involved in the proof of the Khinchin inequalities. EXERCISE 3.4.2. Show that theorems 3.4.1,3.4.2, and 3.4.3 are sharp, i.e., the exponents appearing in these theorems cannot be improved. REMARK 3.4.2. An analysis of the foregoing arguments yields the following, weaker sufficient condition: for the equality SR(L:r..l Xl:) = Xo + rl. to hold in an

Lp space it is sufficient that g(t) = (L~l IXl:(t)l2) 1/2 E Lp. This result, which is stronger than M. 1. Kadets' theorem, belongs to Nikishin 162\, and has a latticetheoretic, rather than a Banach space· theoretic character. EXERCISE 3.4.3. Verify that none of the sufficient conditions for the linearity of the sum range given above is necessary.

CHAPTER 4 UNCONDITIONALLY CONVERGENT SERIES

Starting with this chapter we move deeper and deeper into the realm of the concepts, repesentations, and methods of the contemporary theory of Banach spaces.

§1. The Dvoretzky-Rogers Theorem In this section we shall establish the following general fact: in any infinitedimensional Banach space there are series that converge unconditionally, but not absolutely. Thus, the equivalence between unconditional and absolute convergence characterizes t.he finite-dimensional spaces within the class of all Banach spaces. LEMMA 4.1.1 (THE DVORETZKy-ROGERS LEMMA). Let X be an n-dimensional normed space. Then there exist elements {X;}i=l C S(X) such that

(1)

for any scalars

{t;}~l and any m, 1 ~ m ~ n.

PROOF. In the unit ball U(X) of the space X pick an ellipsoid E of maximal volume. (Recall that in an n-dimensional linear space an ellipsoid is defined as the set of all points for which the value of some positive-definite quadratiC form, which defines the ellipsoid, is smaller than or equal to 1.) Equip X with the inner product generated by the ellipsoid E (or, equivalently, by the quadratic form that defines E). Denote by 1\ . liE the corresponding norm, with respect to which E is the unit ball of X. Obviously, IIxll :5 IIxllE for all :r: EX. Since the notion of volume in a finite-dimensional space is defined up to a constant factor, we may assume that the volume of E is equal to one. Then the volume of any ellipsoid will be the product of the lengths of its principal semi-axes, calculated in the metric

II· \IE. Let us construct, by induction on m, an orthonormal (with respect to the inner product introduced above) basis {Ui}i:! and a set of vectors {:cdr=!, with the following properties:

46

CHAPTER 4.

UNCONDITIONALLY CONVERGENT SERIES

(A) IIx",1I = IIXmllE = 1, 1 :5 m :5 n; (B) Xm =E;:I a...,jUj, am,m > 0; (C) a;',1 + a;',2 + ... + a;',m-l = 1 - a;',m :5 m;l. Take for UI = XI an arbitrary point at which the ellipsoid E touches the unit ball U(X), i.e., a point a such that lIali lIaliE = 1. The existence of such points on the unit sphere S(X) is guaranteed by the finite-dimensionality of X and the maximality of E. Suppose we have already found vectors {u.;}i:.1 and {xi}f=l satisfying the requirements (Am) IIx;!1 IIxdlE 1, 1:5 i :5 mj (Bml Xi = L~=I Cli,jUj, O-i,i > 0; (cm) a~'I ,+'a~,2 + ... + a~",.-I = 1 - a~·t,t :5 i-I. n Complete the set {Ui}i:.1 in an arbitrary way to an orthonormal basis by adding vectors Vm +1,"" Vn. Now let us carry out the next inductive step. Given an arbitrary E > 0, consider the perturbed ellipsoid Et formed by the points X = (gt. ... , gn) whose coordinates in the orthonormal basis constructed above satisfy the inequality

=

=

=

(1 +e)n-m(g~ +... + g!.)

+ (1 + e + e2)-m(g!'+l + ... + g~)

:5 1.

(2)

The volume Yt: of Et is larger than the volume of the basic ellipsoid E:

V.t --

(1

+ e + e2 )m(n-m) (1

+ e)(n-m).m > 1.

Consequently, one can find a point x(e) E Et such that x(e) E S(X) (even more, one can find points in Et that lie outside U(X)). Since IIx(e)IIE :::: IIx(e)1I = I, for this point we have gl + ... + g! : : 1. Substracting the last inequality from (2), we see that the coordinates of x(e) satisfy the inequality [(I tEt-'" -l)(g~ + ... + g~)

+ [(1 + e + e2 )-m -I](g~+1 + ... + g~) :5 O.

(3)

Let us divide both sides of (3) by e and pass to the limit for a sequence of numbers E\.O for which the corresponding sequence of vectors x(e) converges to some vector y E S(X). This yields the following inequality for the coordinates of y: (n - m)(g~ + ...

+ g!.) - m(g!'+1 + ... + g!) :5 O.

(4)

Now set X m +1 = y. Since x m +! E S(X) n E, IIxm+tll=lIxm+lI1E = 1. Notice that from (4) it follows that for m < n the vector Xm+l is linearly independent with the vectors Ut. •.• , Urn· Now choose a vector u m +! E S(X) in the linear span Of Ul, ... ,U""X m +l so that Urn+! is orthogonal to Ul, ... ,U rn and the coefficient 4m+l.m+l in the decomposition m+l Xm+l

=

L

j=\

am+l,juj

§l. THE DVORETZKY-ROGERS THEOREM

41

is positive. Then the vectors X m +l and U m +1 satisfy conditions (Am+d and (Brn+d. Let us verify that condition (em+d is also satisfied. By construction, the numbers 9i figuring in inequality (4) are connected with the numbers 4m+1,j by the relations 4m+1,; = 9; for j ~ m, and a~+l,m+l 1:~1 Consequently, we can rewrite (4) in the form

=

9l·

(n - m)(a~+l,1 Of,

+ '" + a~+I,m) -

s

ma~+l,m+1 ~ 0,

equivalently,

In conjunction with the relation IIxm+lll~

= Ei=i1 a~.j = 1,

this yields the

required condition 2 (lm+l,l

22 + .,. + am+l,m = 1 - am+ 1,m+l

m

~ -

n

Thus, by completing the induction process we produce an orthonormal set {Udi:l and a set {Xi}i:l of norm-one elements, which satisfy the conditions (A), (B), and (e). We claim that Lhe set {xilf=l satisfies inequality (1). Indeed,

5

~ t.tl + t.lt.I'lIu. -%.11.5 ~t.tl (1 +

The norm

IIUi -

(5)

xills is estimated by \lui -

xill~

= (1

-

2 (l . . ) tit

= a~,l + a~,2 + ... + a~,i_t + (1 + ( 1 - au) 2 < 2(1 I

-

2 (l . . ) 'It

ai,i)2

< 2(i n- 1) . -

(6)

Here we made use of all the conditions (A), (B), (e), including t.he requirement that Cli,i > O. Substituting (6) in (5), we obt.ain the needed inequality (1). 0 COROLLARY 4.1.1. Let n ~ m(m - 1), and let X be an arbitrary n-dimensional normed space. Then on the unit sphere S(X) one can lind elements {xtJ?;t with the property that

for any choice

of coefficients {ti} ~ 1 .

o

48

CHAPTER 4.

UNCONDITIONALLY CONVERGENT SERIES

TEO REM 4.1.1 (THE DVORETZKY-ROCERS THEOREM). Let X be an infinite-dimensional Banach space. Then for any sequence {ai} ~1 of positive numbers satisfying the condition E~1 < 00 there exists an unconditionally convergent series E:l Xi in X for which II xiii = ai, i = 1,2, ....

at

Divide the sequence {a;}~1 into segments {ail;';;:;>]> 0 < ... , such that

PROOF. mI

< m2

j

= rno

<

= 1,2, ....

i=m,+1 Choose the first ml vectors Xi in arbitrary way, subject only to the requirement that IIxili = ai, 1 $ i $ mi. Next, take a sequence of subspaces Xj of X of dimensions dimXj (mj+l - mj)(mj+1 - mj - 1). By Corollary 4.1.1, in each Xj one can find a set of norm-one elements {X;}:;;:;:+1 for which

=

for any choice of coefficients ti' Set Xi inequality yields

= aixi' For the choice Itil = at the preceding

max

(7)

Q,=±1

The sequence {X;}~l constructed in the indicated manner satisfies the condition IIx,1I = Clj, i = 1,2, .... It remains to show that the series E:l Xi is unconditionally convergent. To this end let us take an arbitrary segment E:=l X, of the series and, using inequality (7), let us estimate from above the norm II E~"'+l Qixill. Let j be a number such that m; $ m :5 mj+l. Then 00

~

L 2-

k +1

= 2-;+2

k=)

(here we used the fact that if A

c

BeN, then

which in turn is an immediate consequence of the triangle inequality). Since j grows when m does, according to Cauchy'S criterion our series diverges for any IIIranEement of signs Q, = ±l, i.e., unconditionally. 0

§2. ORLlez's THEOREM

49

Since in the formulation of the last theorem the sequence {Uj} ~ I does not necessarily satisfy the condition Uj < the Dvoretzky-Rogers theorem establishes, in particular, the existence, in any infinite-dimensional Banach space, of unconditionally convergent series that are not absolutely convergent. A proof of this fact based on the theory of absolutely summing operators (see Definition 4.3.1) can be found ill 114J.

L::l

00,

EXERCISE 4.1.1. Investigate the possibility of choosing elements {x;}~l in any n-dimensional Banach space X, such that IIxill = 1, i = 1, ... ,m, and

for all coIlections {t;} ~I' Find the relation between m and n.

§2. Orlicz's Theorem on Unconditionally Convergent Series in Lp Spaces As we already know, in finite-dimensional spaces unconditional convergence implies absolute convergence (Theorem 1.3.3). In infinite-dimensional Banach spaces this is no longer the case: by the Dvoretzky-Rogers theorem, in each such space there is a series that converges unconditionally. but not absolutely. One can then ask whether unconditional convergence of a series of elements in a Banach space imposes any constraints whatsoever on t.he behavior of the numerical series formed by the norms of its terms. A trivial constraint, which holds not only for unconditionally convergent, but also for ordinarily convergent series, is that the general term must tend to zero. Modifying the construction given in Example 1.3.1 one can readily see that in the space Co the norms of the terms of an unconditionally convergent series are not requied to satisfy any nont.rivial conditions. However, there are spaces whose nature imposes rather strong constraints on the behavior of unconditionally convergent series. The first result in this direction was obtained by W. Orlicz in 1930.

r::l

THEOREM 4.2.1. Let the series Xi o[ elements o[ a space Lp converge unconditionally. Then [or 1 $ p $ 2 the series r:~l IIxill2 converges, and [or 2$ p< the series IIXi II" converges.

00

r::l

We will reduce the proof of Orlicz's theorem to the proof of an assertion on the sum of a finite number of elements of a Banach space. DEFINITION 4.2.1. The space X is said to have the Orlicz property with exponent r if for any unconditionally convergent series Xi of elements of X the series E:l IIxilir converges.

r::l

DEFINITION 4.2.2. The space X is said to have M -cotype r with constant "Y

> 0 if the inequality (1)

50

CHAPTER 4.

UNCONDITIONALLY CONVERGENT SERIES

holds for any fini te set {x;} of elements of X. THEOREM 4.2.2. The following assertions are equivalent: (A) the space X has the Orlicz property with exponent r; (B) the space X has M -cotype r (i.e., there exists a -y > 0 such that X has M-cotype r with constant -y). PROOF. First let us prove the implication (A) ~ (B). Suppose (B) does not hold. Then there exist a sequence {Xi}:~:~+1 of finite subsets of the space X, 1 = no < n1 < n2 < ... , for which nk+1

2:

IIxdl r ~

1,

i=nk+ 1

The series E:l Xi formed by these elements converges unconditionally, but E:l IIXilir == 00. Thus, negation of (B) implies negation of (A), which is equivalent to (A) :} (B). Now let us show that (B) ~ (A). Suppose there exists an unconditionally convergent series E:l Xi for which L:l IIxdl r = 00. From Gel'fand's theorem (Theorem 1.3.4) it follows that the sums :E:':m+l aiXi are uniformly bounded, i.e.,

max{

t

lIaixill: ai = ±1,

m,n E

N} ~ C<

00.

i=m+l

On the other hand, from the fact that L:1 IIxillr == a sequence of segments of the series such that

00

it follows that there exists

nl

L

IIxdl r

--+ 00,

1 ~ ml

< nl < m2 < n2 < ....

i=m~+l

This clearly contradicts inequality (1), i.e., (B) does not hold. This establishes the 0 implication (B) ~ (A).

=

REMARK 4.2.1. The finite-dimensional spaces have M -cotype r 1, whereas by the Dvoretzky-Rogers theorem, in the infinite-dimensional case one necessarily has r ~ 2. To complete the proof of Theorem 4.2.1 it remains to show that the spaces Lp have the corresponding M -cotype. This fact will be derived from the Khinchin

inequalities. PROOF OF ORLICZ'S THEOREM. Consider first the case 1 ::; P ::; 2. Let us pick a finite collection {Xi}i':l of elements in Lp(O,Jl) and, in accordance with Theorem 4.2.2, let us estimate from below the quantity

§2. ORLICZ'S THEOREM

51

To this end let us first replace the maximum by the mean value over t.he and then apply the Khinchin inequalit.y:

0:;

= ±1

" Ixl(tW )1'/2 dp.(t). In E It;"11' TiXi(t) dp.(t) ~ (ap)P In ( ~ The integrand in the last integral can be regarded as the norm of the value at the point t. of the vector- function

in the coordinate space l~i!:

By the triangle inequality, the integral of the norm is not. smaller than the norm of the integral:

f

10

(t

IXI(t)1 2)P/2 dp.(l)

;=1

=

(t, (In lx,

=f

1n

(tjIPdp(tj)

lIy(t)l12/p dIL(tj

"P) pi' =

~ II 10f y(t)dIL(t)11

2/p

(t,IIXillt.) PI'

This finally yields the inequality

which, by Theorem 4.2.2. proves Orlicz's theorem for the case I ~ P ~ 2. Now let. us consider the case 2 ~ p < 00. As in the first. part of the proof, replacing the maximum by the mean value and applying the Khillchin inequality we obtain

52

CHAPTER 4.

UNCONDITIONALLY CONVERGENT SERIES

Now using the inequality

(l:>n

1/2

(P

~ ~tf

)l/P

,

C; ~

0,

p~ 2

(whkh expresses the monotonicity of the lp-norm) we obtain the needed estimate

This completes the proof of OrIicz's theorem.

o

EXERCISE 4.2.1. We have proved that the spaces Lp have M-cotype r max{p,2}. Show that if s < r, then Lp does not have M-cotype s.

=

EXERCISE 4.2.2. Show that the Dvoretzky-Rogers theorem is sharp, in the < 00 cannot be weakened. sense that in its statement the condition L~I

a1

EXERCISE 4.2.3. Clarify whether some analogue of OrIicz's theorem holds in the spaces C[O, 1) and Loo. EXERCISE 4.2.4. Use the Khinchin inequality to show that in any of the spaces Lp[O,l), 1 ::; p < 00, the closure of the linear span of the sequence of Rademacher functions is a subspace isomorphic to 12 (the Rademacher functions rn{t) = sign sin (2nll't) constitute one of the simplest models of a sequence of independent Bernoulli random variables). EXERCISE 4.2.5. Prove that in the space Loo[O, I) the preceding assertion is false. More precisely, in this space the closure of the linear span of the sequence of Rademacher functions {rn} is isomorphic to the space LI .

§3. Absolutely Summing Operators. Grothendiec:k's Theorem One of the possible ways of generalizing the results of the preceding section is to consider the convergence of a series in one topology, but compute the norms of its elements in another, weaker topology. This approach is quite productive and lies at the foundation of the deep and well developed theory of absolutely summing operators. We shall present here some ideas of this theorYi for a more detailed exposition the reader should consult the monograph 17o) and the references given therein. For applications to Banach spaces and function theory see (14J, (52), (1011. THEOREM 4.3.1. The canonical embedding operator Loo[O,I] -+ L p IO,I] (1 ::; p < (0) maps unconditionally convergent series into p-absolutely conver-

gent ones, (i.e., series with the property that

E:l IIXillP < 00).

PROOF. To simplify the arguments we remark that the assertion of the thenrem is equivalent to the following one: for any finite collection {Xi} of elements

§3. ABSOLUTELY SUMMING OPERATORS

53

of L",,[O, 1] one has the inequality

m_ax

a,-±I

IlL Ckixill

Loo

;::: 'Y (2: IIxilit ,. ) 1/1',

(1)

where'Y > 0 does not depend on the choice of the elements Xi. Thus, pick a finite set of elements Xi(t) of L,,(O,1]. Using the fact that the norm of an element in lp Iresp. L,,) decreases !resp. increseas] when p is increased, we obtain the following chain of inequalities, starting from the left-hand side of

(1 ): max

aj=±l

=

IlL CkiXi11 L"" = Q,=±l max sup ILCkiXi(t)1 tEIO,l)

sup

max

tEIO,I) Q;=±I

IL CkiXi(t)1 = tEIO,ll sup ~)xi(t)1

1/1' (1 L IXi(t)lp dt (L IXi(t)lp);::: 1

;::: sup tEIO,I)

)

1/1'

0

=

(L IIxill~) 1/" . p

o

This proves (1), and hence the assertion of the theorem.

EXERCISE 4.3.1. Show that in the preceding theorem the requirement of p-absolute convergence cannot be replaced by r-absolute convergence with r < p. DEFINITION 4.3.1. A linear operator T: X -+ Y is said to be absolutely Xi, Xi EX, into summing if it maps any unconditionally convergent series an absolutely convergent series: IITxi II <

1::1

1::1

00.

As in many of the problems considered earlier in this book, the last definition can be translated into the language of finite co1\ections of elements: DEFINITION 4.3.2. A linear operator T: X summing if there exists a constant K such that

-+

Y is said to be absolutely

(2) for any finite collection {Xi} of elements of X. The smallest value of the constant K is called the absolutely summing norm of the operator T, and is denoted by 7r(T}. If the operator T is not absolutely summing, we put 7r(T) = 00. In other words,

EXERCISE

of operators.

4.3.2. Show that the two preceding definitions give the same class

54

CHAPTER 4.

UNCONDITIONALLY CONVERGENT SERIES

4.3.3. Show that any absolutely summing operator is bounded S 1T(T).

EXERCISE

and that

IITII

4.3.4. Show that the set f1(X, Y) of absolutely summing operators Y is a Banach space with respect to the norm 1T(T) whenever the normed

EXERCISE

T: X

-+

spaces X and Yare complete. EXERCISE 4.3.5. Calculate the absolutely summing norm of the identity (embedding) operator Loo -+ L,.

4.3.6. Show that the identity (embedding) operator II absolutely summing. EXERCISE

-+

12 is

The fact observed in the last exercise is a simple particular consequence of the following more general assertion. which lies at the foundation of the theory of absolutely summing operators. THEOREM

ator T: h

--t

4.3.2

(GROTHENDIECK'S THEOREM).

Any bounded linear oper-

h is absolutely summing; moreover, (4)

1T(T) $ KIITII,

where K does not depend on T. PROOF.

By Definition 4.3.2. we need to show that

L IITXt II S KilT\!· Q~=a;,IIL OiXill· In the left-hand side of this inequality let us replace \!TXili by the inner product (Txi'Y')' where Yi E i2 , lIydl $ 1:

II)Txi,Yi)1 $ KIITII· Q~a;,112:0iXill· Inequality (4) is equivalent to the last inequality being valid for all Xi E II and al\ E 12 , lIydl < 1. Let us decompose the elements Xi. Yi and the operator T with respect to the standard bases {ej} and {ej} in II and 12 • respectively:

Yi

Xi = LXi,je;.

Te'J

= ~tL",J'e" , ~

I:

j

Yi

= LYi,k e".

(5)

"

With no loss of generality we may assume that aU the sums in (5) have only finitely many nonzero terms (thanks to the convergence of series, the passage to an infinitely number of elements can be accomplished by continuity). We shall assume also that IITII S 1, i.e., Lk $ 1 for all j E N. Then, in the notations of (5), inequality (4) takes on the form

tL

LX.,] t",jYi,k S K max . . Q.=±l

.,],k

where [I: Yl,k S 1 for all i and

I:. \I: ]

•

O,Xi,; \

= K Q"I3,=±1 max

Lie tL S 1 for all j.

I: ',j

Oi X i,j{3;,

(6)

§3. ABSOLUTELY SUMMING OPERATORS

55

Thus, all reduces to proving inequality (6). First., let us rename the quantities involved to arrive a.t the traditional form, namely, let us ma.ke the substitutions

Then (6) becomes

Lk y;,,, ::;

=

where Lir X;,k $ I, 1. If we now put Xi Lit xi,lre" and Yi Lit Yj,ke", where {elt} now denotes the standard orthonormal basis in 12 , we arrive at the following inequality. known as the Grothendieck inequality:

=

(7) where (~,i) is an arbitrary real n x n matrix (which can be assumed to be square by adding zeros for the missing entries) and where Xi and Yj are elements of tlle unit sphere of the space l~n). PROOF OF GROTHENDlECK'g INEQUALITY. First of all let us remark that, thanks to the homogeneity of (7), it suffices to examine the case of matrices (a',i) for which

max

L

s.,t;=:H . ,

ai,jSitj

= 1,

',}

or, equivalently,

-1 < 'st, < 1 - "a· ~ '.) ' ) for all

Si.

t)

(8)

= ± 1. Consider the integral

Is

sign(x. u) , sign(y, u) dm(u) ,

(9)

where S is the unit sphere of l~n), m is the normalized Lebesgue measure on S, x, yare fixed elements of S, u E S is the integration variable, and (.,.) denotes the inner product. Let us pass to spherical coordinates: if u 2:~=1 = 1, then

u?

= 2:~1 Uiei, Ui = (u, e,),

UI

= sinc,ol'

sinc,02''' sinc,on-2' sin c,o,,-l ,

U2

= cos c,ol'

sin c,02 .. , sinrpn-2' sin c,on-l,

U3

=

COSc,02''' sinc,on_z' sin c,on-I ,

Un-l

=

56

CHAPTER 4.

UNCONDITIONALLY CONVERGENT SERlES

with 0 :S !PI :S 211", 0 :S !Pi :S of coordinates equals

11",

i

= 2, ... ,n. The Jacobian of this transformation

. . 2 . n-2 J() 'P = sm 'P2 sm 'P3····· sm 'Pn-\· With no loss of generality we may assume (thanks to the invariance of the Lebesgue measure under rotations of the sphere) that x == e\, Y == cosO e\ + sinO e2, where () is the angle made by the vectors x and y. After the change of variables the integral (9) becomes

where

f2"

r r

c= io io ""]0

J('P)d'P\ ... d'Pn-l.

After the usual simplifications (in particular, cancellation of all factors that do not contain 'PI), we find that

1 f2" sign[sin!PI'sin(!pl +8)ld!pI 271" io i.e.,

= 1- 28 = ~arcsin(x,y), 71"

71"

isf sign (x, u) . sign(y, u) dm(u) = 7r~ arcsin(x, y).

(10)

Now, armed with the auxiliary formula (10), let us return to Grothendieck's inequality. Since the maximum of the left-hand side of (7), taken for fixed aj.; over

x"y; E B(l~n», can be attained only for Ilxil/ = lIy;1I = 1, we shall assume that Xi, y; E S(= S(l~n»)). Given an arbitrary number mEN, let us go to inequality (8), which is supposed to hold for all

Si, tj

= ±1, substitute in it

m

Si

= IT sign(x" k=1

m

u(k»,

t;

= II sign(y;, u(k», k=l

and then integrate with respect to the variables (10), we see that, for any mEN, -1 $

u(k)

E S, k

~Ili.; [~arcsin(Xj,y;)]m :51.

t.,

= 1, ... , m.

Using

(11)

Now let us multiply each inequality (11) with m odd by (7r/2)m( _I)m-I /m!, take into account the sign of the factor, and add the resulting two-sided inequalities. We obtain ~

~sin(arcsin(xi' y;». ai'; i,j

•

11"

:5 smh 2'

§3. ABSOLUTELY SUMMING OPER.ATORS

57

or, finally,

Thus, we have established Grothendieck's inequality with the constant K = sinh ~. Since Grothendieck's inequality is equivalent to inequality (4)/this completes the proof of the theorem. 0 The exact value of the constant K (Grothendieck 's constant K c) is not known. It is only known that (see 170\) 7r

7r

-
trace(ABC) :5 KcIlAII·IIBII·IICII· DEFINITION 4.3.3. A locally convex topological linear space E is said to be nuclear if for any continuous semi norm p on E there exists a continuous seminorm q > p on E for which the identity operator I: (E,q) ..... (E,p) is absolutely

summing. EXERCISE 4.3.8. Show that the space of functions analytic in an open domain, equipped with the topology of uniform convergence on compact sets, is nuclear. EXERCISE 4.3.9. Show that in a nuclear space any unconditionally convergent series L:f=l XIc converges absolutely, i.e., L:~1 p(x,,) < 00 for any continuous seminorm p.

CHAPTER 5 ORLICZ'S THEOREM AND THE STRUCTURE OF FINITE-DIMENSIONAL SUBSPACES In the preceding chapters we have often encountered assertions of the following type: from a certain property of the sum of a finite number of elements follows a certain theorem on series in an infinite-dimensional space. Since any finite set of elements lies in some finite-dimensional subspace, in all such assertions only special features of the structure of the finite-dimensional subspaces of the ambient infinite-dimensional space considered play a role. In the first section of this chapter we will present a modern convenient language for describing what kind of finite-dimensional subspaces a given infinite-dimensional space has. In the sequel we shall actively employ this language and, in particular, give a complete description of the spaces in which an analogue of Orlicz's theorem on unconditionally convergent series holds.

§l. Finite Representability DEFINITION 5.1.1. Let X and Y be Banach spaces. The Banach·Mazur dis· tance between X and Y is the number d(X, Y) = infT {IITII,I\T-ll\}, where the infimum is taken over all isomorphisms T: X -+ Y. If X and Yare not isomorphic, we put d(X, Y) = +<x).

Let us list the simplest properties of the Banach-Mazur distance. 1) d(X, Y) ~ 1 for any X, Y. If X and Y are isometrically isomorphic, then d(X, Y) = 1. 2) The Banach-Mazur distance is symmetric: d(X, Y) = d(Y, X). 3) The multiplicative triangle inequality holds:

d(X, Y) . d(Y, Z)

~

d(X, Z).

The last property follows from the definition of d(·,·) and the inequality IIBAI! ~ IIBII·IIAII, which holds for all bounded operators A: X -+ Y and B: Y -+ z. DEFINITION 5.1.2. A Banach space X is said to be

finitely representable in

a space Y (and one writes X .!... Y) if for any e > 0 and any finite-dimensional subspace Z c X there exists a finite-dimensional subspace ZI C Y such that d(Z,Zd < 1 +e.

CHAPTER 5

60

ORLICZ'S THEOREM AND SUBSPACES

Notice that. if X admits an isometric embedding in Y, then X 2. Y. Hence, for any p 2: 1 the space lp is finitely representable in LpIO,I]. However, finite reprenscntability is not equivalent to the existence of an isometric embedding. PROPOSITION 5.1.1. LplO,lJ

2.l p

for any p

~ 1.

PROOF. Let Z C Lp[O, I] be a subspace, dim Z

= n.

Pick a ba.<;is {Zk}k=l in

Z. Since all the norms on a finite-dimensional space are equivalent to one another,

there exists a constant C such that

II~ tkzkll ~ C ~ Itkl for all collections of numbers {tk}k=l' A measurable function will be said to be simple if it takes only a finite number of values. Take an arbitrary e > 0 and choose a 6 > 0 such that (1 + 6)/(1 - 6) < 1 +t:. Approximate each element Zk in the Lp-metric by a simple function Yk such that IIZk - Ykll $ Co. Set Y ::: Lin {YI,"" Yn}. Let us show that d(Z, Y) $ 1 + e. To this end let us introduce the operator T: Z ..... X by the rule

T

(ttkZk) = ttkYk' k=l

k=1

Then

liT (t. t", )

I lit. t,y,11 ~ lit. t, ',II + t.1t,IIIY, -"II =

$II~ tkzkll + C6~ Itkl which mcans that

$ (1

+ 6)

II~ tkzkll,

IITII $ 1 + O. On the other hand,

Hence, T is an isomorphism,

d(Z, Y)

IITII $ 1 + 6, and liT-III $ $IITII·IIT- 1 1I

$

1/(1 - 6). Consequently,

~ ~: < 1 + e.

It remains to show that Y embeds isometrically in lp. Since the functions Yk

=

are simple, we can find disjoint measurable sets A;, j 1•... , N, such that and ~(A;) > 0 for j = 1, .... N, and on each Aj each of the ru~ctions Yk is constant. Let XA, denote the indicator function of the set Ai' Then Lin{XAI" .. ,XA,..} is a subspace of LpIO, 1] that is isometric to I~N). Since Y C Lin{x.otp··· .XA,..}, we conclude that Y is isometrically ismorphic to a subspace of lp, which completes the proof of the proposition. 0

UN=! At = 10,1]

§l. FINITE REPRESENTABILITY

61

PROPOSITION 5.1.2. The space G[O,11 is finitely representable in the

space CQ. PROOF. On proceeds as in the proof of Proposition 4.1.1, but use piecewiselinear functions instead of simple ones. 0 From the last proposition and the theorem on the univer§ality of the space

G[O.11 (see \551) we obtain COROLLARY 5.1.1.

space

AII'y

Banach space is finitely representable in the

CQ.

EXERCISE 5.L1. Show that if X..!.. Y and Y ~ Z, then X

..!.. Z.

EXERCISE 5.1.2. Give an example of spaces X and Y such t.hat X it is not true that Y

..!.. Y,

but

..!.. X.

EXERCISE 5.1.3. Show that if X a Hilbert space.

!. h, then X

is isometrically ismorphic to

EXERCISE 5.1.4. Give an example of spaces X and Y that are not isometrically isomorphic, and such that d{X, Y) = 1. EXERCISE 5.1.5. Show that if X metrically in Lp[O, 1].

..!.. Lp[O, 11,

then the space X embeds iso-

EXERCISE 5.1.6. Show that if X is a finite-dimensional space and the unit sphere S(X) is a polyhedron, then X embeds isometrically in CQ. EXERCISE 5.1.7. Show that any two-dimensional normed space is finitely representable in II . EXERCISE 5.1.8. Without relying on Exercise 5.1.5, show that Orlicz's theorem on unconditionally convergent series in Lp remains valid for any space that is finitely representable in Lp EXERCISE 5.1.9. Calculate d{l~n) .l~n»). EXERCISE 5.1.10. Show that the set of all n-dimensional normed spaces, equipped with the metric 10gd{X, V), is a compact space (here isometrica.lly isomorphic spaces are identified). EXERCISE 5.1.11. Show that for any nonseparable space X there exists a r r separable space Y such that X -+ Y and Y -. X. EXERCISE 5.1.12. Prove the principle of local reflexivity, i.e .. that X·' ~ X for any space X.

CHAPTER 5.

62

ORLICZ'S THEOREM AND SUBSPACES

§2. The Space Co, C-Convexity, and Orlicz's Theorem Let us consider the Banach space Co, that is, the space of all sequence x = (Xl, X2,· •. ), Xi E a, such that Xi -+ 0, equipped with the norm IIxll = max; IXj I. Let {ek}f:l be the standard basis of Co. Clearly, if ak are positive numbers such that ak -+ 0 when k -+ 00 and if Ok = ±l, then

when n -+ 00. Hence, the series E~l a"e" converges unconditionally for any sequence of numbers {a,,}r..l such that a" ...... O. We see that in Co the unconditional convergence of a series E~l y" imposes no constraints on the norms of its terms, except for the trivial condition limk .... oo lIy,,1I = O. It follows that no analogue of Orlicz's theorem holds in CO. We shall demonstrate now that this is the case not only for the space Co, but also for any space in which Co is finitely representable. r THEOREM 5.2.1. Let X be a Banach space such that Co -+ X. Let {a,,}r..l be a sequence of positive numbers that converges to zero. Then in X there eJcists an unconditionally convergent series E~l Xk such that IIXkll = a" for all k. PROOF. ai

Choose an increasing sequence of natural numbers

< 2-" for all i > n". For such a choice the series

n" such that (1)

converges. Since l~) is a finite-dimensional subspace of CO and Co is finitely representable in X, it follows that one can find subs paces in X that are almost isometric to the spaces In other words, one can exhibit a sequence of vectors {ei}~l in X such that the inequalities

will hold for any i E N and any collection of numbers ti' This implies, in particular, that 1 :5 lIedl :5 2. Now we define the sought-for series E~l Xk by the rule Xk = ake,,/lIekli. k = 1,2, .... Obviously, IIXkll = ak, and so it remains to verify that our series converges unconditionally. Let Ok = ± 1 be an arbitrary collection of signs and let nand m be natural numbers, with n > nt. Then

§2. Co. C-CONVEXITY. ORLICZ'S THEOREM

63

Thanks to the convergence of the series (1), we have

which by the Cauchy criterion means that the series L~l QjX; converges.

0

Thus, we have exhibited a class of spaces in which for sure no analogue of Orlicz's theorem is valid, namely, the class of spaces in which C() is finitely representable. It is natural to look now at t.he opposite class. DEFINITION 5.2.1. A space X is said to be C-convex if representable in X.

C()

is not finitely

The next simple lemma, which is a reformulation of the definition of Cconvexity, is given without proof. LEMMA 5.2.1. For the space X to be C-convex it is necessary and sufficient that there exist a natural number n > 0 and a number E > 0 such that

o

for any n-dimensional su bspace A eX.

LEMMA 5.2.2. Let X be a Banach space. The following two conditions are equivalent: (a) There exist n E N and E > 0 such that

for any n-dimensional subspace A eX. (b) There exist n E Nand €I > 0 such that, for any collection of elements {xili=l' IIx,1I 2: 1. one can find signs Ilk = ±1 for which

PROOF. First let us show that (b) => (a). Let nand El be as in condition (b) and A be an n-dimensional subspace of X. Furt.her, let T: l~) -+ A be an arbitrary isomorphism between l~) and A. Denote bye. the vectors of the standard basis of l~). Then 1

IITedl

~ ~.

i

= 1.

TI

64

CHAPTER 5.

Choose signs

Vi

ORLICZ'S THEOREM AND SUBSPACES

= ± 1 such that

Thanks to the relation

II E~I v,edloo = 1, this yields

IIT1! ;::= IIT(E]-l v,ei)lIx = IltviTeil1 ;::= 1 +_~I . II Ei=1 v,e,lIoo i=1 x liT II Therefore,

IITII·IIT- I II ~ 1+ £ll which in view of the arbitrariness of the operator

T means that d (l~)

,A) ~ 1 + £) for any n-dimensional subspace AeX.

(a) => (b). We shall argue by reductio ad absurdum. Suppose that for any n E N and any £ > 0 there exists a collection of elements {Xi }i=I' IIx, II ;::= 1, i = 1, ... , n, such that the inequality

holds for any choice of signs /I, = ±1. Consider the operator T: l~) - Lin {xtl i.: I' defined by Tei = Xi, i = 1, ... , n. We claim that IITII ~ 1 + e. Indeed, let X be an element of the unit ball of l~). Write x as a convex combination of elements of the form E~l vie" Vi ±1:

=

where

>.VI ..... V~

;::= 0

and

L

.,,=:!:I

>'''10' ... v"

= 1.

The existence of such a representation is guaranteed by the fact that the unit ball of l~) is the polyhedron (more precisely, the cube) with the vertices E~I We have

Viei.

IITxll

=

I .,,=:1 L (>.." .....

v"

tv,Te i ) ,=1

11$ .,.L=:!: 1 (>.VI ......," Iltvieill) $1 +e. ,=1

Since X E B(l~») is arbitrary, we conclude that UTI! $; 1 + e. Now let us show that liT-III $ (1 - e)-1. Let x = E?:1 QiXi and set aio = maxi lail. Consider the auxiliary vector y = E~I biX" where b, = ai if i :F io and b,o = -Q;o. Then

§2. co, C·CONVEXITY, ORLICZ'S THEOREM

Consequently,

IIxll

~ 2112::':1 aieill

lIyll =

65

-llyll, and since

II~biXill = liT (~biei) II

$ (1 +E)

1\~bie;l\

= (1 + c) mF lail = (1 + c) II~ a;eill, we conclude that IIxll ~ (1 (1 - c)-l. Therefore,

c)1I L~=1 aieill = (1 -

d (l~) ,Lin{xdf::l)

$

c)IIT-l(x)lI, whence

IITII· IIT- l "

$

IIT-lli $

!~:, o

contrary to (a).

DEFINITION 5.2.2. The measure of C -convexity of the space X is the function C(n, X) : N -+ ~ defined by the formula

It is readily seen that C(n, X) ~ 1 for all n. From the two preceding lemmas it follows that for a space X to be C-convex it is necessary and sufficient that the strict inequality C(n, X) > 1 holds for some n = n(X) E N. THEOREM 5.2.2. The sequence C(n, X} is nondecreasing; the numbers C(n, X} form a "semimultiplicative sequence,n i.e.,

C(n· m,X) ~ C(n,X}· C(m,X}. PROOF. The first assertion is an immediate consequence of the inequality max{lIx + yilt IIx -till} ~ IIzll· Let us prove the semimultiplicativity property. Let {Xi}~ be elements of X, IIxili ~ 1. Write the set {xi}i:1 as a rectangular table: Yr./' where r runs from 1 to n and l runs from 1 to m. For each l choose numbers IIr,l = ±1, r = 1, ... , n, such that

II~ Vr,1Yr,111 ~ C(n, X), and denote the sum 2:~=1 Vr,IYr.1 by max

9i =±1

ZI.

IIE8iXili ~ max IlfOIZll1 ~ ;=1


1=1

Then C(m,X)· min IIllil

which in view of the arbitrariness of the elements muItipIicativity.

I

Xk

~ C(m,X)· C(n, X),

completes the proof of semi0

CHAPTER 5.

66

COROLLARY

ORLICZ'S THEOREM AND SUBSPACES

5.2.1. If X is a C-convex space. then Iimn _

oo

C(n, X)

= 00.

COROLLARY 5.2.2. Let X and Y be isomorphic BaJJach spaces. Then from the C-conl'exity of one of them follows the C-convexity of the other (stability of the class of C-convex spaces under isomorphisms).

Recall that a Banach space is said to have M -cotype p with constant C if the inequality

>0

holds for ally finite collection of elements {x;} i= 1 . EXERCISE 5.2.1. Show that if a space X has .fI{-cotype. then X is C-convex; tha.t if 1'2 > PI and X has M-cotype Pit then it also has cotype 1'2; that the Mcotype of any space cannot be smaller than 1; and that the M-cotype of a space X is inherited by all spaces that are finitely representable in X.

In the Chapter 4 we have shown (Theorem 4.2.2) that a space has M-cotype p if alld only if Orlicz's theorem holds with exponent P in that space. Hence. in

order to prove Orlicz's theorem for C-convex spaces is suffices to show that these spaces have an M -cotype. THEOREM

pEN and C

5.2.3. Let X be

Ii

C-convex space. Then there exist numbers

> 0 such that X has II{ -cotype p with constant C.

PROOF. Choose a natural number nl > 1 such that C(nt, X) = 1 + 6 > 1. Set no = I. nit = (nd k , k = 1,2.3, .... By Theorem 5.3.1. C(nk.X) ~ (1 +6)". Pick a natural number p such that (1 + 6),,-1 ~ nl. Let us show that X has M-cotype P with the constant C = l/nl. Take an arbitary collection {X;}:'::I of elements of X. Partition the index set {I, ... ,n} into diSjoint sets A". k = 0.1,2 •... as follows:

Ak ~f

{j: (L::':llIxiIlP)I/p ~ I/x;1I > (L:7-11I Xi Il P)IIP}. nk

ttk+l

Denote by tnk the number of dements in Ak. Then

whence

t

j=1

!lxil/" =

f L I/Xjll" :s f. m" E:'::l l/:iIl k=OjEA~ k=O

P •

(nk)

It follows tha.t E~o Ink/{n,,)" ~ 1, or, equivalently. E:':o mk/nk" ~ 1. This shows that there exists a number k = ko for which mko ~ nko(p-I)' (Indeed. if ntk < nk(p-I) for all k, then mo = 0, mklnk" < l/(nl)", and E:':o m"ln,,-p < 1.)

§3. RESULTS ON TYPE AND COTYPE

Recall that max ",,=±1

II~ ~ o:.Xill ~ i=1

max "'I=±I

67

II~ ~ O:iXi11 iEA

for any set A C {I, ... , n}. We are now ready to esta.blish the inequality appearing in the definition of the M -cotype:

=

((1 + 6)P-) I

nl

k{)

. ...!... L IIx.lIp (

nl

n

) 1/1'

i=1

~~

(

I

L IIx;!l n

p

) I/p

i=1

o

Thus, X has M -cotype p with constant l/nl, as claimed. Combining this result with Theorem 5.2.1 we obtain

COROLLARY 5.2.3. If in a Banaei) space unconditional convergence imposes some constraint on the character of the convergence to zero of the general term of the series, then in that space unconditional convergence imposes a constraint of p-absolute c:onvergence type.

EXERCISE 5.2.2. Show that if for some n one has the inequality C(n, X) n l / p , then X has M-cotype p + e: for any ~ > O. EXERCISE 5.2.3. Provide an example of a space which has Af-cotype 2

for any e

~

+€

> 0, bllt does not have M-cotype 2.

The results 011 C-convexity given in this section belong to B. Maurey and G. Pisier 157\. The connection between these notions and the theory of series was also studied by S. A. Rakov 174\, 175\. §3. Survey of Results on Type and Cotype We have shown t.hat in a Banach space the validity of an a.nalogue of Orlicz's theorem on unconditional convergence of series is equivalent to C-convexity as well as to the existence of an M-cotype. Chobanyan's result (Theorem 3.4.2) and the results of Cha.pter 7 exhibit rela.tionships hetween conditional convergence and concepts similar to t.hose of C-convexity and M-cotype, namely, B-convexity and infratype. The theory of type and cotype is one of the modern directions in Banach space theory, direction that is closely connected with interesting domains of functional

68

CHAPTER 5.

ORLICZ'S THEOREM AND SUBSPACES

a Banach space, or the theory of Banach-space valued stochastic processes. For this reason, the present section may be regarded as a link between the theory of series and many other actual problems of Banach space theory. To keep within the framework of our monograph, we wrote this section as a brief survey. Although sketches of proofs will be provided for some of the results, the bulk of the results is given without proofs. Many of the results listed below can be found in 199J or 110 11. DEFINITION 5.3.1. A space X is said to have cotype l' with constant C > 0 if the inequality

holds for any finite collection of elements {Xi}i..l eX. It is readily seen that if the space X has cotype 1', then it has M-cotype p. It is not know whether for a infinite-dimensional spaces the existence of an Mcotype l' is equivalent to the existence of a cotype p. The only known result in this direction is the following assertion. THEOREM 5.3.1 [57]. Suppose the space X has M-cotype p. Then X has cotype PI for any PI > p. 0 Recall that a space X is said to have type p with constant C if the inequality Ell E:=1 rix,lI $ C (E~=l II xiIl P)l/p holds for any collection oC elements {xdi..1 C

X. DEFINITION 5.3.2. A space X is said to have infratype p > I with constant C if the inequality

~l~

lit I ,=1

O:iX,

$ C

(t,=1

UXiIlP)

l/p

holds for any finite collection oC elements {X.}i..l eX. The type and the infratype are related in the same way as the cotype and the M -cotype: THEOREM 5.3.2. If a space has type 1', then it also has infratype p. If a space has iniratype 1', then it has type l' - e for any E > 0 (see [57]). 0 Recently M. Talagrand proved the following result [89J: for p < 2, a space has infratype p if and only if it has type p. Let us list some elementary properties of the type and cotype. The type, infratype, cotype, and M -cotype of a space X are inherited by all it subspaces as well as by all spaces that are finitely representable in X. The type and infratype of a space are inherited by quotient spaces, whereas the cotype and M -cotype are not.

§3. RESULTS ON TYPE AND COTYPE

69

Finite-dimensional spaces have M-cotype I, cotype 2, type 2, and any infratype. The spaces Lp with 1 < P < 00 have type = rotype = lIlin{2,p}. The cotype and M cotype of the spaces Lp with 1 ~ p < 00 coincide and are equal to max {2, p}. Dvoretzky's theorem (see Chapter 6) has the following simple consequence: THEOREM 5.3.3. Let X be all infinite-dimensional Banach space. If X has infratype p, then p ~ 2, and if X has M -cotype q, then q ~ 2.$ 0

Hilbert spaces can be characterized in term of type and cotype. THEOREM 5.3.4 1491. An infinite-dimensional Banach space has simultaneously type 2 and cotype 2 if and only if it is isomorpllic to a Hilbert space.

D THEOREM 5.3.5 1571. Let X be an infinite-dimensional Banach space and let px denote the supremum of the types of X. Tllell for any p E Ip x, 21 the space lp is finitely representable in X. D

DEFINITION 5.3.3. The space X is said to be 8-convex if the space i l is not finitely representable in X.

If the space X is B-convex, then so are its dual X·, all spaces that are finitely representable in X, and all subspaces as well as all quotient spaces of X. The next result can be established in much the same way as the corresponding assertion concerning C-convexity. THEOREM 5.3.6. A space X is B-convex if and only if there exist n E N and e: > 0 such that the inequality min",=±1 II I:~I v;xill $ (1 - ~)n holds for any collection of elements {xdf=l C B(X). D

COROLLARY.

1£ a space has infrat.ype p > 1, then it is B-convex.

D

As measures of B-convexity of a space X we will take the quantities

From Theorem 5.3.6 it follows that for the space X to be B-convex it is necessary and sufficient that there exist n EN such that b(n, X) < n. THEOREM

5.3.7. [711. Let X be a B-convex space. Then X has an infratype

0

p> 1. To prove this assertion one needs first to establish the inequality b(n· m,X) ~ b(n,X)· b(m,X),

and then follow the scheme of proof of Theorem 5.2.3. Between type and cotype there is a certain duality (see 153, page 791):

0

CHAPTER 5.

70

ORLICZ'S THEOREM AND SUBSPACES

5.3.8. Let X have type p > 1. Then X· has cotype q. where 0 q This duality is not complete: the space Ll has cotype 2, but (L)· = Loo is THEOREM

1+1:;:1 . p

not B-convex. The concept of B-convexity was introdu ced by A. Beck in 1962 171 and was bsequently investigated by many authors (see 125), 1261}. For a long time it was su t latown whether a non reflexive space can be B-convex. The first such example constructed by R. C. James in 1974 129).

::s

5.3.1. Prove theorems 5.3.3 and 5.3.6--5.3.8. EXERCISE 5.3.2. Show that any B-convex space is C-convex.

EXERCISE

EXERCISE

5.3.3. Show that the C-convex space I) has CO among its quotien t

spaces. EXERCISE 5.3.4. Let X and Y be a C-convex and respectively a B-convex space. Show that the quotient space X/V is C-convex. EXERCISE 5.3.5. Without using the results of the present section, show that every C-convex space has a cotype. EXERCISE 5.3.6. Let X be a B-convex space. Show that 12(X) is also Bconvex (here '2(X) denotes the space of sequences x = (Xl,X2, ... ), x" E X for

which

Ilxll =(E:'1Ilx",1I 2) 1/2 < 00, equipped with the norm I\xl\).

EXERCISE

5.3.7. Let X be a C-convex space. Show that 12(X) is also C-

convex. DEFINITION

5.3.4. The function

I\x+ylI 6x (t) = .mf { 1 - 2 - : x, y E B(X), !Ix - yll ~ t } , defined for t E 10,21, is called the modulus 0/ convexity 0/ the space X. The space X is said to be IJni/onnlll convex if 6x(t) > 0 for all t > O. EXERCISE

with 1 < p < 00.

5.3.8. Prove that the spaces Lp are uniformly convex for any p

EXERCISE 5.3.9. Let X be a uniformly convex space. Show that for any unconditionally convergent series L~=l x" in X the numerical series L~1 6x(lIx"I

converges.

EXERCISE 5.3.10. Show that any uniformly convex space is reflexiv e.

D

CHAPTER 6 SOME RESULTS FROM THE GENERAL THEORY OF BANACH SPACES

To continue our exposition, in particular, to prove the theorem of V. M. Kadets on the existence, in any infinite-dimensional Banach space, of series with nonlinear sum range, we need two theorems on the structure of infinite-dimensional spaces: Dvoretzky's theorem on almost-Euclidean sections and Mazur's theorem on basic sequences. Since these deep results are not incorporated in the standard functional analysis courses, their proof will be provided here in detail for the reader's convenience. §1. Fredlet Difl'erentiabillty of Convex Functions The definitions and results collected in this section are well known, but, unfortunately, they are seldom included in the university analysis courses. For this reason the authors found themselves in a difficult position: not including these results would make the understanding of the already complicated next section even more difficult, while including them would mean overlapping known textbooks. We decided for the following method of exposition: give all statements without proofs, but in the order in which the reader, with a certain amount of effort. will be able to reconstruct the proofs on his own. A detailed exposition can be found, for example, in 177). Throughout this section / will denote a real-valued fWlction, defined on the Banach space X. Whenever we write X = R n we mean tha.t X is the n-dimensional coordinate space, equipped with some norm. In this case / will alternatively be regarded as a function of n-variables: f(x) = /(Xl,' .. , Xn).

f(AX

DEFINITION 6.1.1. The function f is said to be convex if the inequality + Ilx) $ Af(x) + Il/(Y) holds for all x, y E X and all A, Il E R+ such that

).+Il=l.

It is clear that if f is convex and Ale are positive numbers satisfying E~=l Ale = 1, then

72

CHAPTER 6.

RESULTS mOM GENERAL THEORY

LEMMA 6.1.1. Let X = R", and let 1 be a convex [unction that is differentiable at z
!~o lIylI

= O.

DEFINITION 6.1.2. The function f is said to he Frechet differentiable at the point x E X if there exists a linear functional 'iJ J(x) E X· such that

J(z) - f(x)

= {'iJ J(x), z -

x)

+ o(lIz -

xU),

or, in other words, such that lim J(z) - f(x) - ('iJ J(x), z - x}

liz - xII

z-r

= O.

In this case the element 'iJJ(x) E X· is called the Frechet differential (derivative) of the function J at the point x. It is readily verified that this definition is correct, in the sense that a function cannot have two different Frechet derivatives as a given point.

=

THEOREM 6.1.1. kt X Rn. TheIl i[ the convex function J is differentiable with respect to each varible at the point x EX, then J is Freehet differentiable at x and

'iJ f(x)

of (x), ... , oX Of) = ( OXI (x) . n

(Hint: reduce this assertion to Lemma 6.1.1.) THEOREM 6.1.2. Let X = an and let f be a convex function. Then f is Freebet differentiable almost everyv.·here (with respect to the Lebesgue measure). THEOREM 6.1.3. Let. {fn}~=1 be a sequence of convex Functions that converges uniformly to f on some open set D. Suppose that all In and f are Freehet differentiable at the point Xo E D. Then

lim

n-oo

II" fn{xo)

- "f(xo)1I

= o.

REMARK 6.1.1. Suppose that some norm p on the space X is Frechet differentiable at a point Xo. Then

('iJp(Xo), y)

~

p(y)

for all y EX. Indeed, 'iJ ( ) ) - r p(xo ( p Xo , Y - .>.~o

+ AY) A

p(xo) <

r

- .>.~

p(xo)

+ Ap(y) A

p(xo) _ ( ) - PY .

§2. DVORETZKY'S THEOREM

73

REMARK 6.1.2. Many of the properties of ordinary differentiability remain valid for Frechet differentiability. In particular, the rule for differentiating composite functions holds: if a(t) is a mapping of the segment 10,11 into X, a(t} is differentiable at a point to, f(x) is a real-valued function on X, and f(x) is Frtkhet differentiable at the point a(to), then the function f(a(t)) is differentiable at to and

[J(a(to))I' EXERCISE

= (\1 f(a(to}), a'(to)).

6.1.1. Show that without the convexity assumption on f Theorem

6.1.1 is false. EXERCISE 6.1.2. Let f be a convex function on X and Xo E X. Show that there exists a functional 9 E X· such that (g, y - xo) $ !(y) - !(xo) for all y E X. (Such a 9 is called a supporting functional of f at the point xo.) Prove that if the convex function! is Frechet differentiable at the point Xo, then \1 '(xo) is the unique supporting functional of ! at xo.

=

=

EXERCISE 6.1.3. Let 1 < p < 00, X lp, f(x) IIxli. Show that! is Frechet differentiable at every point x I- O. Calculate \1 !(x). Show that in the spaces II and CO the norm is not Frechet differentiable. DEFINITION 6.1.3. The modulus of smoothness of the Banach space X is the function px(t), defined by the formula

px(t)

= sup {IIX + tyl\ ; IIx - tYIl

_ 1: x, y E S(X)} .

The space X is said to be unifonnly smooth if px(t)/t

-+

0 when t

-+

O.

EXERCISE 6.1.4. Show that in a uniformly smooth space the norm is Frechet differentiable at every point x t- O. EXERCISE

6.1.5. Show that the spaces Lp(n, p.) with 1 < p <

00

are uni-

formly smooth. EXERCISE 6.1.6. Let X be a uniformly smooth space and let L:::'I Xk be a conditionally convergent series in X. Show that if E~I Px (lIxkll) < 00, then the assertion of Steinitz's theorem holds true for E:'I XA;. EXERCISE

6.1.7. Let the series L:~l

Xk

in X be perfectly divergent. Then

L::" l px(lIxkll) = 00. §2. Dvoretzky's Theorem The aim of this section is to prove that a Hilbert space is finitely representable in any infinite-dimensional Banach space. To understand the proof the reader must be familiar with the theory of integration of functions of several variables at the

CHAPTER 6.

74

RESULTS FROM GENERAL THEORY

level found in the modern analysis courses. In the sequel we shall use the result obtained, but not the method for proving it. The main idea of the proof is to introduce a certain auxiliary measure of closeness of an infinite-dimensional space to a Euclidean space (Definition 6.2.2). This measure of closeness proves to be more convenient in the kind of problems considered here than the Banach-Mazur distance. DEFINITION 6.2.1. A norm p given on the coordinate space X is sait to belong to the class Pn if dim X n (Le., X = Rn) and the inequality !lIxlloo :5 p(x) :5 IIxll2 holds for all x EX. (Recall that II . lip denotes the norm of the space l~n) .)

=

LEMMA 6.2.1. Let X be a finite-dimensional normed space, dim X = n ~ 4m2. Then there exists an m-dimensional subspace Y c X with a basis {e~,} k!:1' such that, for any x = L~l tke,,:.

~1I(tk)1"lIoo :5l1xllx :51\(tk)1"i12 (in other words, Y belongs to the class Pm). PROOF. In the course of the proof of the Dvoretzky-Rogers lemma (Lemma 4.1.1) we established the existence oCa basis {ek}i:=l and ofelements {Xk}i:=I' such that if one regards X as a coordinate space with basis {ek}i:=1> then 1I·lIx :5 11·112 and the following conditions are satisfied:

(A) Ilxillx = Ilx;1l2 = 1, i = 1, ... , n; (B) Xi = L~=I Qi,jej, i = 1, ... , n; ~i-I 2 - 1 2 < i-I . - 1 (C) ""k=l ai,k - ai,i _ n' l , ...• n. Recall that a supporting functional of the ball B at a point x is a functional x· such that (x·, x) :5 IlYlls for all y E X (here II . liB denotes the norm in which B is the unit ball). The existence of a supporting functional is guaranteed by the Hahn-Banach theorem. Generally speaking, such a functional is not unique. be a supporting functional of the unit ball of X at the point Xi. Since Let II· IIx :5 II ·112 and II xilix = II xill2, xi is also a supporting functional of the unit ball of l~n). Hence, xi is the functional defined by the l~n) -inner product with the element Xi: x;(x) = (Xi, x) for all x E X. Consequently, since Ix;(x)1 :5 Ilxllx, one has that I(Xi,X}1 :5 IIxlix for all x EX. Now take for Y the linear span of the vectors {ek}k!:,. Let x == E;'..I tkek and IIxkllx = 1. By construction, IIxllx :5 lI(tk)rIl2' It remains to show that ~ II(tk)r 1100 :5 IIxllx, i.e., that Itk 1:5 2 for all k :5 m. We proceed by induction. Since Xl = al,lel and lal,ll = 1, we have

x;

ltd

= I(x,el}! = l(x,xI) :5l1xllx = 1.

Further, suppose one knows that Itkl :5 2 for any k :5 i. Then

75

§2. DVORETZKY'S THEOREM

~ ~ (1 +2il/2 (~)1/2)2 ~ n-

n

t

1

4m2

.(2m+2i)2

- t

~

4{2m-l)2 $ 4m 2 - m + 1

4. 0

Let X be the Euclidean space l~n) or a subspace of l~n). Denote x.l

= {y EX: (x, y) = O},

= {{x,y): x E S{X),

E(X)

Y E S(xJ.)}.

Further, denote by Ax and ax rotation-invariant Borel probability measures defined on S(X) and E{X), respectively. LEMMA

6.2.2. Let X

= l~n) and let f

[

is(x) PROOF.

be a linear functional on X. Then

If(x)1 2 dAx(x)

= ~ IIfll~· n

Since the measure Ax is invariant under rotations,

for any k :5 n. Hence,

1 = -lIfll~ n as claimed.

1

s(X)

IIxll~dAx(x)

1 = -Ilfll~, n

o

CHAPTER 6.

76

RESULTS FROM GENERAL THEORY

LEMMA 6.2.3. Let /(x.y) be a bounded measurable function on E(X). and let dim X == n. Then for any k n one can find a subspace E eX, dim E = k,

:s

such that (

f(X.y)daE(X,y) $ (

f(x,y) dax (x.y).

lr:.(x)

jI;(X}

PROOF. Let r denote the set of all k-dimensional subspaces of X, equipped with the natural topology. Introduce a rotation-invariant Borel probability measure '1 on r. The expression

(d'1(E) (

lr

g(x,y)cJqe(x,y)

lr:.(E)

defines a rotation-invariant integral on E(X); moreover,

(d-y(E) {

lr

ldaE(X,y)=l

lE(E}

Thanks to the uniqueness of the rotation-invariant Borel probability measure, the following identity holds:

(d-y(E) (

k

h(~

f(X,y)daE(X,y)

=(

h(~

f(x,y)dax(x,y).

Assume that the assertion of the lemma is false for the given k. Then (

f(x.y)dax(x,y)

It(x)

lr

> ( d-y(E) (

ir

= [ d"(E)

f(x,y) dax (x, y)

l'E(x)

[

!(X,y)daE(X,y)

lE(E)

=[

f(x,y)dax(x,y).

lr:.(x)

0

The contradiction we have reached proves the lemma.

DEFINITION 6.2.2. Let X c l~N), dim X = n, and let another norm p be given on X. The integral measure of closeness of the norm p to the Euclidean norm is defined as

Notice that 1In(tp)

= 1In(p), and that if p is the Euclidean norm of the space

I~N) and (x,y) E E(X), then

'Vp(x)

= x,

y e x.l,

(P(x) , y)

= 0;

hence, for the Euclidean norm vn{P) = O. The next lemma shows that Vn (P) can be indeed regarded as a measure of I. .."" ,.1n.c;P, the norm p is to an Euclidean norm.

77

§2. DVORETZKY'S THEOREM

6.2.4. For any kEN and any E > 0 there exists a 6(k,E) > 0 such 6(k, E), that if some norm p on the space X = l~k) obeys the condition Vk(P) then d(X,p( '))' (X, 1\'112)) 1 + E. LEMMA

:s

:s

Suppose that the assertion of the lemma is false .•Take a sequen ce of norms qn on X = l~k) and an e > 0 such that PROOF .

(1)

max qn{X)

~ES(X)

= 1,

min qn{x):s 1 - e: for a.ll n E N.

~ES(X)

\45\ one Since all the qn have the same Lipschitz constan t 1, by Anela's theorem Denote S(X). on ly uniform es can choose a subsequence {qn} C {qn} which converg its limit by q. Then max q(x) = 1,

:r:ES(X)

min q{x):S 1 - e. :r:eS(X)

qn and q are Denote by A the set of all points x E SeX) at which all the norms Ax{A) = 1. neous, homoge and convex are q Frechet differentiable. Since qn and By (1), (2)

als 'Vq.. (x} almost everywhere on E(X). By Remark 6.1.1, the sequence of function converges e sequenc this A E is bounded in norm. Next, by Theore m 6.1.3, for x (q(x), y) product inner the to V'q(x). Therefore, from rela.tion (2) it follows that Let is equal to zero for almost all pairs (x, y) E E(X).

B

= {x E A: (q(x),y) = 0

for all y E xol}

[0,1] --+ SeX) Since the function y ..... (q{x), y) is continuous, >'x(B) = 1. Let g: (get), get») = Since be a smooth curve such that get} E B for almost all t E [0, I}. the set B, of n definitio 1 for all t, it follows that get) .1 g'{t) and, by the 6.1.2, Remark (V'q(g(t»,g'(t)} = 0 for almost aU t. Further , by q(g(l» - q(g(O»

=

11

[q(g(t»1' dt

=

11

(Vq(g(t », g' (tm dt

= o.

Hence, the function q is constan t on S(X) and 1=

max q(x) =

~ES(X)

we have reached a contradiction.

min q(x)

%ES(X)

= l-e;

o

CHAPTER 6

78

RESULTS FROM GENERAL THEORY

LEMMA 6.2.5. Let P n be the class of norms introduced ill Definition 6.2.1. Then limn_co suppEP .. vn(p) = O. PROOF.

= l~n),

Let X

Vn(p)

=

p

r (r

JS(X)

E

JS(:rJ.)

~ ~l n-

r

P n . By Lemma 6.2.2,

I('Vp(x), Y)12 dA:rJ. (y»)

lS()O

[P(x)t 2dA.dx)

(lI'Vp(x)1I2)21P(xW 2dAx(x).

Dy Remark 6.1.1, ('Vp(x),y) ~ p(y) for all y; also, p(y) ~ Consequently, lI'Vp(x)112 ~ 1. It follows that

Vn(p) ~ - 1

n- 1

1

Ip(x) I- 2 dAx(x) ~ - 1

n- 1

SIX)

1 [IIxli 4

lIylll

for any p E Pn .

=

oo ]-2 dAx(x) def an·

SIX)

The last qllantity does not depend on p, and so suppEP .. vn(p) ~ a". Let liS estimate the behavior of an when n -+ 00. To do this il. is convenient to ensure that the integration will be carried over the same measure space for all values of n. To this end, choose in some probability space (n, E, I') a sequence of independent, normally distributed random variables {Vk}f:t with the same distribution and mean value 7,ero. Fix n for the moment and consider the random variables tk = V,.. I ( E:'..l IViI2) 1/2. The mapping w ...... (tt (w), ... I tn(w» transforms the measure It into the measure Ax on S(X). Consequently,

n - 1 an -4.n

=

1

1 ( max Itkl )-2 -1 [lIxll'lOj -2 dAx(x) = -E n l::;k::;n

SIX) 7l.

Thus, we managed t.o reduce the prohlem to the estimat.ion of mean values on a single probability space. The quantit.y R.,. = IVll2 . ( maxi Sk::;,. Il'd) -2 is monotonically nonincreasing in 11. Moreover,

for allY A > O. We see I.hat the sequence Rn converges 1.0 0 with probabili~y 1. lIcllce, by the dominated convcrgence t.heorem, Iim .. _ co an = Iim .. _ oo E(R~I) = O.

o

§3. BASIC SEQUENCES

79

THEOREM 6.2.1 (DVORETZKY'S THEOREM). Let k be an arbitrary natural number and let e > o. Then there exists a number n(k, e) such that, for any narmed space X with dim X > n(k, e-) there is a k-dimensional subspace Y C X such that d ( y,l~ < 1 + €:.

It»)

PROOF. Using Lemma 6.2.5, put n(k,€:) = 4m 2 , in which m is such that sUPpEP", vm(p) < 6(k,€:), where 6(k,€) is as in Lemma 6.2.4. By Lemma 6.2.1, there exist an m-dimensional subspace X. c X and elements {ei};';. in Xl> such that

By construction,

\I. \Ix,

E

Pm and vm(II·llx,) < 6(k, e). Then Lemma 6.2.3 yields a

subspace Y C XI sllch that 1Ik(\I·!lY) ~ 6(k,e}. By Lemma 6.2.4, d (y,l~k»)

::; I+€:. o

COROLLARY. The space 12 is finitely representable ill allY infinite-dimensional Banach space. Let us point out that, by Exercise 5.1.3, only Hilbert space can posc;css this universality property. Our exposition above followed the work ofT. Figiel {21J. A detailed exposition of the beautiful modern theory connected wit.h Dvoret.zky's theorem can be found in the very useful monograph of V. D. Milman and G. Schechtman 199]. EXERCISE 6.2.1. Show that the space Co cont.ains no infinit.e-dimensional subspaces isomorphic to 12. EXERCISE 6.2.2. Est.imate from above the numbers n(k.e). §3. Basic Sequences DEFINITION 6.3.1. A sequence {ek}f: I in an infinite-dimensional Banach space is called a basis if every element x E X has a unique representation as a series ill the elements elt: 00

.t

= 2:: ak(x)ek,

alt{x) E R.

k=.

This definition is t.he natural generalization of the notion of a "basis in a finite-dimellsional space." Obviously, bases defilled ill this manner can exist only in a separable space. All classical separable Banach spaces posess bases. The first example of a separable space that has no ba.<;is WllS rnnc:tr .. ,.tcA h .. P J;'"A~:_ " ' " ' ' ,",,,

CHAPTER 6.

80

RESULTS FROM GENERAL THEORY

DEFINITION 6.3.2. A sequence of elements {ekH';1 of a Banach space is called a basic sequence if {et}f=1 is a basis in Lin {et}41'

The next theorem is given here without a proof, for which the reader is referred to the books 155, Chapter 3, § 61 and 153, page 21· THEOREM

6.3.1 (S.

BANACH).

In order for a sequence {ek}r.:J! ek f. 0, to

be a basic sequence it is necessary and sufficient that there exist a constant K

such that

m+n

L ajej

>0

m

~K

;=1

La;ej

(1)

;=1

o

for all m,n E N and aj E R.

DEFINITION 6.3.3. The basic constant of the basic sequence {ek}f=1 is the supremum of all constants K for which inequality (1) holds.

Notice that if K is the basic constant of the sequence {ek}r.:l' then for any all'" ,an and any n EN we have

(2) Indeed, for j

:5 n relation (1) gives

THEOREM 6.3.2 (THE KREIN-MIL'MAN-RuTMAN THEOREM). Let {ed~1 C SeX) be a basic sequence with basic constant K and let L~1 lie, - y,1I = Kl < K/2. Then {Yi}~l is also a basic sequence. PROOF. Define an operator T: Lin {ed -+ Lin {Yi} by the rule Tei = Yi, i = 1,2, .... To prove that {Yd~l is a basic sequence it suffices to show that T is an isomorphism ofnormed spaces. First let us check that T is bounded. Let 2:7=1 be an arbitrary element in Lin Then

a;e,

{ei}

:51!~ Gie;!! +

t

lail·lIy; - edl

:5

lit aie;11

+ Kl

:511~ aie,ll (1 + 2~ ) :5 211~ a,e,ll·

l~~n la;1

§3. BASIC SEQUENCES

81

Thus, we have shown that IITI! ~ 2. Let us establish the boundedness of the operator T- 1 , and hence the invertihility of T. We have

o

Thus, T is an isomorphism.

The result just esta.blished asserts that the property of being a basic sequence is preserved under small perturbations. Below we shall prove Mazur's theorem on the existence of basic sequences in any infinite-dimensional Banach space. To facilitate the exposition, let us introduce the following definition: a subspace Z is said to be E-011.hogonal to a subspace Y if lIy + zll ~ (1 - E)lIyll for all y E Y and all z E Z. LEMMA 6.3.1. Let X be a Banach space, Y a fillite-dimensional subspace of X, and Z an infinite-dimensional subspace of X. Then for any e > 0 there exists a subspace FeZ of finite codimension in Z, such that F is e-orthogonal to Y.

PROOF. Pick a finite e-net {adf=\ on the unit sphere of the subspace Y and denote by Ii a supporting functional at the point a; (i.e., !lfdl = 1 and f.(aj) == 1). Then the subspace E = n~=l Ker fi is e-orthogonal t.o Y. Indeed, let y E Y and e E E. Choose an element ai such that lI(y/llyID - aill < e. Then

lIy + ell ~ h(y + e) = li(Y) = lIyll

= lIyll

(1 + Ii (II~II -

aj )

(I (II~II - ai) + f(ad) )

~ lIyll(l - e).

Since E has finite codimension in X, the subspace F properties.

= En Z has the

required 0

THEOREM 6.3.3. Let X be an infinite-dimensional Banach space. Then for anye > 0 there exists a basic sequence {en}~=l C X whose basic constant is not

smaller than 1 - e. PROOF. Pick an arbitrary element el E S(X) and put Y1 = Lin{eJ}' Lemma 6.3.1 yields a finite-codimensional subspace Fl C X that is E-orthogonal to Y\. Then pick an arbitrary element e2 E S(Fd, put Y2 = Lin{e),e2}, and take a subspace F2 C Fl of finite codimension in FJ that is e-orthogonal to Y2 . Continuing this process, we obtain a sequence of elements {en} ~= \ and sequences of su bspaces Yn = Lin{e,.)i:=l and Fn :) {ed~=n+l' such that Fn is E-orthogollal to Y". Let us

82

CHAPTER 6.

RESULTS FROM GENERAL THEORY

verify that {en}~l is a basic sequence. Let {ai}?:t be arbitrary numbers. Then y = E~l aiei E Yn , Z = E:':n':l aiei E Fn , and

DEFINITION 6.3.4. A Banach space X is said to be finitely satumted by a Banach space Y (and one writes Y finite-codimensional subspace of X.

:b

X) if Y is finitely representable in any

DEFINITION 6.3.5. A sequence {Xi}~l of subs paces ofthe space X is called a basic sequence of subspaces if there exists a constant K > 0 such that the inequality

holds for all n, mEN and all Xi E Xi, i = 1,2, ... , n + m. The supremum of all such constants is called the basic constant of the sequence {Xi}~l' The next theorem will play an essential role in further applications. THEOREM 6.3.4. Let X and Y be Banach spaces, Y

J. X

and let {Yi}~1

be an arbitrary sequence of finite-dimensional subspaces of Y. Then for any e > 0 there exists a basic sequence ofsubspaces (Xi}~l in X with basic constant 1- e, such that d(Xi. Vi) $ 1 + E for all i. PROOF. We argue in much the same way as in the proof of Theorem 6.3.3. with the difference that instead of an element en E Fn - 1 we must choose a subspace Xn c Fn-l such that d(Xn, Y n) < 1 + e, and then define Yn as LinU7.:1 Xi. 0 REMARK 6.3.1. If {X;}~l is a basic sequence of subspaces with basic constant K and if Xi E Xi, i E N, then by analogy with inequality (2) one can readily show that

EXERCISE 6.3.1. Show that the spaces Co. ClO, I). and l". L,,[O. I). with 1 $ p < 00, have bases.

§4. Some Applications to Conditionally Convergent Series In this section we collected a number of interesting facts that are somewhat beyond the general scope of our exposition. The proofs of these results rely on the technique of singling out basic sequences.

§4.

APPLICATIONS

83

THEOREM 6.4.1. In any infinite-dimensional Banach space there are series whose sum range reduces to a single point, but which are not unconditionally convergent. PROOF. Let {en}~1 be a basic sequence in the space X, 1, 2 .... The series el - el

1

+ -e2 2 -

1 -e2 2

1

+ -e2 2 -

1 1 1 -e2 2 + -e3 4 - -e3 4

lIe,,1I = 1,

+ ...

,

n =

(1)

where each group of 2n terms of the form ±2-n + Ien mutually annihilate, converges to zero. If the terms of the series (1) are rearranged so that it remains convergent, then its sum will be again zero. since the projection on any vector in the basis is zero. At the same time, if in (1) we replace all signs ± by +, we obtain a divergent series, because the series segments formed by the vectors 2- n+1 en give the vector 2en • which does not converge to zero, contrary to the Cauchy criterion. 0 Notice that series (1) obeys the assertion of Steinitz's theorem: SR(E~I :c",)

= 0 + rl.; indeed, here the set of absolute convergence functionals includes the coordinate functionals as well as the functionals that annihilate all vectors en. Below we shall consider series for which any bounded functional is an absolutely convergence functional. DEFINITION 6.4.1. A series E~I Xk of elements of a Banach space X is said to be weakly absolutely convergent if E~I If(Xk)! < 00 for all f E X'. Let us remark that an 'unconditionally convergent series is taken by any continuous linear functional into an unconditionally convergent numerical series; and since for nwnerical series unconditional and absolute convergence coincide, every unconditionally convergent series is weakly absolutely convergent. EXAMPLE 6.4.1. In the space eo consider the se~es E~=J ek formed by the unit vectors of the standard basis. This series diverges, since its general term does not converge to zero. Take an arbitrary linear functional fEll = (eo)"; then f = (11,h,···), j" E R, Er=tlfk! = 11/11 < 00. We have Er=1 I(ek) = E~l!!k! < 00, i.e., our series is weakly absolutely convergent. Thus, we constructed a series in eo that is weakly absolutely convergent, but diverges in the norm topology. Morevoer, this series diverges in the weak topology as well: if it were convergent, the coordinates of its sum would all be equal to 1, but Co contains no such element. COROLLARY 6.4.1. If the Banach space X contains an isomorphic copy of Co, then in X there exists a weakly absolutely convergent series which diverges (in

the strong as well as in the weak topology).

As we shall see below, in spaces that contain no isomorphic copy of co, weak absolute convergence of a series implies its unconditional convergence. To prove this theorem of Bessaga and PelczyJiski we need the following assertions, which are of indeoenclp.nt. int"r...,t

CHAPTER 6.

84 LEMMA

there exists

Ii

RESULTS FROM GENERAL THEORY

6.4.1. Let the series E:'I Xk be weakly absoJutel.Y convergent. Then constant A > 0 such that L::;"=I If(Xk)1 ~ Aliflil {or all f E X·.

PROOF'. Let {e.k}k:l be the standard basis of the space I,. Define operators Sn: X· ~ II by the formula Sn(f) = E~=I !(Xk)Ck. These operators are continuous and, for any fixed lEX', the numbers IISn(J)1I = L:~=, If(xk)1 are bounded by L~J If(Xkll· By the Banach-Steinhaus theorem, this means t.hat SUPn IIS"II = A < 00. Therefore, L~J If(Xk)\ SUPn IISn(/)1\ ~ AII/II· 0

=

Let X be a Banach space, Y a finite-dimensional subspace a sequence of elements of X such that Xi ~ 0 weakly and infi \lXi\! = > O. Then for any e > 0 there exists a number 71 such t.hat the subspace Lin {Xn} is e-orthogonal to Y. LEMMA 6.4.2.

of X, and

{xd~1

c

PROOf'. Pick a finite e/2-net {Yi}f..l on S(Y) and denote by Ii a supporting functional at the point Yi. Take 0 < C!/4 and choose a number n such that maxt
=

Hence, we may assume that

lal < 2/lIxn \I. Pick an element Yk such that lIy - yk\1 <

e/2. Then

e

e

e

26

= 1 - 2 + alk(xn) 2: 1 - 2 - lal6 ~ 1 - 2 - IIxnll

~ 1-

e26 2 - C 2: 1 - e. 0

THEOREM 6.4.2. Let {xd~, be a sequence of elements of a Banach space X such that Xi --+ 0 weakly and inCi IIxdl > O. Then for any! > 0 there exists a subsequence {Yi} of {x;} which is a basic sequence with basic constant larger than or equal to 1 - e.

0:':1

PROOF. Let ei > 0 be numbers such that (1 - e;) > 1 - e. Set YI = XI. Next, using the preceding lemma, pick successively Y2, Y3,··· E {XI. X2, .. . } such that Lin {Yn+d will be etl-ort.hogonal to Lin {y" ... , Yn} for each n. Then, Cor any coefficients ai Ilnd any n, m. EN,

§4 APPLICATIONS

85

THEOREM 6.4.3 (THE BESSAGA-PELCZYNSKI THEOREM). The (oJ/owing assertions are equivalent: (a) The Banach space X cOlltains no subspaces isomorphic to Co. (b) Every weakly absolutely convergent series in X is weakly convergent. (c) Every weakly absolutely convergent series in X is unconditionally convergent. (d) Every weakly absolutely convergent series in X is norm convergent. PROOF. (d) =? (c). Let the series E:I Xj in X be weakly absolutely convergent. Then all series E: 1OjIi, OJ = ±l, are weakly absolutely convergent as well. By (d), this means that all series E:I OjIi are norm-convergent, which means precisely that the series E~I Xi is unconditionally convergent. The implication (c) => (b) is obvious, while the implication (b) =? (a) follows from Corollary 6.4.1. It remains to prove that (a) => (d). Suppose that in X there exists a divergent series which is weakly absolutely convergent. Using the fact that the series is divergent one can group its terms so that the resulting series with terms Yle = E7~~:+1 Xi, 0 = n, < n2 < ... will satisfy the condition infle IIYIeIl = 6 > o. The series E:I Yle is again weakly absolut.ely convergent. Consequently, the sequence {YIe} converges weakly to zero. Extract from {YIe} a basic subsequence {ZIe}. The series 1 Zle is also weakly absolutely convergent and, by Lemma 6.4.1. there exists a constant A> 0 such that E~IIf(zle)1 ~ AII/II for all I E X·. Define an operator T: Co -> X by the rule Tele = Zit, where {ed is the standard basis of co. Let tl,' .. , tn be arbitrary numbers. Then

E::

l (t I = lit I = T i

1e=1

~ max It"l· Ie

tleCIe)

sup /ES(X')

tic Zic

1e=1

t

Ie=)

l/(z,,)1

sup.

t

tlc/(Zk)

fES(X ) 1e=1

~ A m:x Itlcl = A lit helel · 1e=1

Hence, the operator T is well defined and continuous. Let us show that T is an isomorphic embedding of CO into X. By Remark 6.3.1, there exists a number K > 0 such that Conscq uently,

o EXERCISE 6.4.1. Show that a series E~I if and only if

XI;

is weakly absolutely convergent

CHAPTER 6.

86

RESULTS FROM GENER AL THEOR Y

EXERCISE 6.4.2. Suppose lim1c .... oo I/X1cl/ = 0 and the series E~=l Xk is weakly absolutely convergent. Does this imply that the series is convergent? DEFINITION 6.4.2. A series E:'I Xk is said to be weakly unconditionally convergent if the series L~l akXk is weakly convergent for any choice of signs Ok

=±l.

EXERCISE 6.4.3. Proceeding by analogy with the proof of Theorem 6.4.3, prove Orlicz's theorem on the coincidence of the notions of weak uncond itional convergence and ordinary unconditional convergence.

CHAPTER 7 STEINITZ'S THEOREM AND B-CONVEXITY The main results presented in this chapter were obtained in the period 19841986.

§1. Conditionally Convergent Series in Spaces with Infratype In Chapter 2 we presented Pecherskii's theorem, and all further result.s on the linearity of the sum range were derived in this book from that theorem. Among such results is Chobanyan's theorem (Theorem 3.4.1): In a space of type p the condition E:I IIXkllP < 00 is sufficient in order for the assertion of Steinit.z's theorem to hold for the sum range SR(E:I Xk). In this section we shall strengthen Theorem 3.4.1, replacing the assumption that the space has type p by the assumption that it has infratype p (see Definition 5.3.2). This result of V. M. Kadets arid M. I. Ostrovskii, obtained about the same time as Theorem 3.4.1, is one of the few that up to now was not shown to be deducible from Pecherskii's theorem. In accordance with Lemma 2.3.1, we only need to establish some analogue of the Permutation Lemma (in order to obtain condition (A)) and some analogue of the Rounding-olf-Coefficients Lemma (to obtain condit.ion (B)). DEFINITION 7.1.1. A space X is said to have property Al with exponent p E N and any collection {Xi} i: I of elements of X there exists a permutation (1 of the set {I, ... , n} sllch that

and constant C if for any n

Tt:

lit

x,'') -

~ t, x·11 ,; c (t,nx.Hr'

Condition AI is an analogue of the Steinitz lemma. Clearly, in a space with property Al from the assumption that E~I \lxkllP < 00 it follows that the series E~=I Xk satisfies condition (A) of Lemma 2.3.1. The aim of the following arguments is to show that property AI is equivalent to the exist.ence of an infratype. DEFINITION 7.1.2. A space X is said to have property A2 with exponent p and constant C if for any mEN and any collectioll {Xi H:I of elements of X there exists a permutation

(1

such that

m

II

LXa(k) -

~ LXk 2m

II

(m

:5 C L I\xkll P

\ 1/1'

I

88

CHAPTER 7.

STEINITZ'S THEOREM AND B-CONVEXITY

Obviously, property AI implies property A2 (it suffices to replace n by 2m and j by m in Definition 7.1.1). Furthermore, since

property A'J. with exponent p implies that the space has infratype p. LEMMA 7.1.1. lithe space X has infratype p with constant C, then it has property Al with the same parameters. PROOF. Let {Xi n~ be an arbitrary collection of elements of X. Consider the auxiliary elements Yle = X21e - X21e-l and choose ale = ±1 so that

Then

';:C

(~IIX'II·r +C (~II"II·r =2C (~II"II')'"

On the other hand, m

LaIeY/; 1e=1

m

= 2:>~IeX21e 1e=1

m

La/;x2k-1 1e=1

2m

= L.8i X" i=1

where .8i = ±1 and there are as many values of i for which .8i = 1 as there are for which .8, = -1. Hence, if the permutation u is taken so that .8a(le) = 1 for k 1, ... , n, then

=

§l. SERIES IN SPACES WITH INFRATYPE

89

LF:MMA 7.1.2. Suppose that X has property A2 with exponent p and constant C. Then for any n E N and any collection of elements {XtH:\ C X there exists a permutation q such that, for each j E {I, 2, ... , 2n },

= (1/2)1 /P •

where g

=

PROOF. We proc(.-OO by induction on n. For n 1 the assertion of the lemma is true. Suppose it is true for n = N - 1, and let us show that it remains true for N. Pick an arbitrary col1ection of elements of X. Using the inductive hypothesis and property A2 one can construct a permutation q with the following properties:

n=

{xkH:t

(a)

IlL!:;' ~ L!:l xkll $ C (2:!:1 IIxkllP) 112:!:2N-'+1 (k)lr $ ~ 2:!:1 II x kll P. XI7(k) -

(b) X 17 (c) for any j E {I, ... , 2N -

.

LX k=1

t1 (k)

-

2N -

1 },

2N-'

1

J

I

LX

17 (k)

(2 $ C

k=l

(d) for any j, 2N -

1

lip;

N

-'

L

) lip

II xl7(k)II P

k=l

(1-

j

4

•

€ -

2N -

2 -

2) ;

< j < 2N , 2N

L

k=2

X17 (k)

N - 1 +l

Let us explain the order in which the indicated permutation q is constructed. Property A2 yields a partition of the index set {I, ... , 2N} into subsets h and h each consisting of 2N - 1 elements, such that

From the sets h and h pick the one (say, 12) for which 2:i E/ l x dl P $ ~ L~=l IIxi liP. To obtain the desired permutation q, write first the elements {X;}iE11' permuted

90

CHAPTER 7.

STEINITZ'S THEOREM AND B-CONVEXITY

in accordance with the inductive hypothesis, and then the elements {Xi}.e/" also permuted in accordance with the inductive hypothesis. For this construction properties (a)-(d) are automatically satisfied. Let. us verify that inequality (1) holds for the permutation (J constructed above. Consider first the case i < 2N - 1 • Using the triangle inequality and then properties (a) and (c), we obtain the following estimates: j

2"

.

LX

C7 (k) - iN

LXk

j

~

l:>C7(k) 1.=1

k=1

k=1

.

+

2;-1

2"-'

.

2"-1

.

L Xcr(k) k=1

2;-1 L

Xcr(k)

k=1

2"

iN LXk k=1

~ ,Lc (~II",II') "' +c (~",,",) '/'. C~, -,L -,) = c (~IIXkW»)I/P. L

k=1

Now consider the case i > 2N (a) and (d), we obtain:

I

•

(_4 - _ i - 2). l-c

2N -

1

Using the triangle inequality and then properties

2"

L

+

XC7(k)

k=2"-I+1

2N

+C L (

)

/IX C7 (k)/IP

k=2 N -'+1

IIp

(1 ~ , - 2L2 - 2) .

By property (b), the second term in the last sum is bounded by

C~ Iixk/lP L "

(

k=1

)

IIp

.c

(_4__ 2_ 2), J_' -

1-,

N- 2

§l. SERIES IN SPACES WITH INFRATYPE

91

and the whole slim is no larger than

(

N

~lIxkllP L2

G

)

IIp

k=l

Recalling that e is bounded by

4~

'(-+2---~--~). 1- e 2N - I 2N-2 j

j

= (1/2)I/p is positive, we see that the coefficient in parentheses 4e

j

l-e +2-

2N-I

j

4

= l-e -

2N - J -2.

o

This completes the proof of the lemma.

Jn what follows instead of (1) we shall use the inequality

where Gp

= I~£

-

2.

THEOREM 7.1.1 (V. M. KADETS [35]). A space X has infratype p witb constant G if and onl,\' if it has property Al witll exponent p and some constant

K. PROOF. We have already proved the implications Al :::} A2 :::} infratype :::}

A2 . Here we shall lise Lemma 7.1.2 to prove the implication

A2:::} AI'

Consider an arbitrary finite collection {xdr=1 of elements of X. Let n be such that 2" > m. Introduce a.uxiliary elements {Yk}r~1 as follows: _

{ Xk -

Obviously,

~ I::':I Xi

0

Yk -

if 1 $ k $ m < k $ 2".

if m

I:!:I y" = O. Lemma 7.1.2 yields a permutation

II

J { ; Yu(k)

(1

such that

I $ G (2" t; IIYle liP) I/p p

for all j $ 2". Since Yk = 0 for k > m, we conclude that there exists a permutation 71' of the set {l, ... , Tn} such that

92

CHAPTER 7.

STEINITZ'S THEOREM AND B·CONVEXITY

for all j $ m. For this permutation, we have

II~X*) -~ t,Xil s c, (~("X," + ! IIt.x.ll),}'I,

t; IIxkll m

;5 Cp (

) P

I/p

m

+ Cpm~-I { ; IIxkl/

for all j ;5 m. Using the Holder inequality, we finally obtain

o Thus, we have shown that the validity of one of the version of Permutation Lemma (condition AI) is equivalent to the space having an infratype. To prove a full analogue of Steinitz's theorem we also need a Rounding-off-Coefficients Lemma. LEMMA 7.1.3 (M. I. OSTROVSKII 165]). Let the Banach space X have infratype p with constant C. FUrther, let {Xi}~l be an arbitrary finite collection of elements of X, {Ai }f=l be a collection of numerical coefficients, 0 ;5 Ai ;5 1, and x = E~=l >'iXi. Then there exists coefficients {8i}~1' each equal to 0 or 1, such that

PROOF. This is a direct consequence of Lemma 2.3.3.

o

We can now state the main result of this section. THEOREM 7.1.2. Let the space X have infratype p, let E~l Xk be a series in X, and let s = E~l Xk· If the terms of the series satisfy the condition ~:lllxkIlP < 00, then SR(E~l Xk) is the affine subspace s+r.L. where r..l c X is the annihilator of the set reX· of convergence functionals. 0 EXERCISE 7.1.1. Show that if the space X does not have infratype p, then the assertion of Lemma 7.1.3 is false. EXERCISE 7.1.2. Suppose that in the space X the series E;;:l IIxkllP is divergent whenever the series E;;:l Xk is perfectly divergent. Show that X has infratype p. CONJECTURE. If a space X has infratype P, then for any perfectly divergent series E~l Xk the series E~l IIxkliP diverges.

§2 TRANSFERRING EXAMPLES

93

§2. A Technique for Transferring Examples with Nonlinear Sum Range to Arbitrary Infinite-Dimensional Banach Spaces In this section we shall prove a theorem of V. M. Kadets 011 the existence of series with nonlinear sum range in any infinite-dimensional Banach space. The auxiliary results needed for this theorem will be stated in general form, to facilitate their use in the next section. The first of these results is based"on Lemma 6.3.4. THEOREM 7.2.1. Let X and Y be Banach spaces and suppose that X :b Y. Then there exists a constant D, 0 :5 D :5 12, such that, for each basic secJuence {ek}f:1 in X and each sequence {Adf:l of posith'e 1Iumbers tllat l.e1lds monotonically to infinity one can find a basic sequence {gdf:1 in Y sucll that

(1)

for all n and all collections of numbers {tdk=I' 0=

PROOF. Let C be t.he basic constant oCthe sequence {edk"=I' Choose int.egers < 112 < ... such that An~ 2: 21:+7 Ie for k = 1.2 •.... Denote

no < nl

Ej

= Lin{en,+l, ... ,en,+l}'

j

= 0,1,2 .....

Using Theorem 6.3.4, pick a basic sequence {V; }~o of subspaces of Y with basic constant l-c, such that d(Ej, l'j) :5 l+c, j 0, 1,2, .... Thanks to this inequality, one can find in ea.ch subspace Yj a basis {Yi}~~~+1 with the property that the double inequality

=

n,+1

n,+l

L

i=n,+1

t.e, <

2::

11;+1

t,Yi

i="1 +t

:5(1+0)

2::

t,e.

(2)

k=nj+1

holds for any choice of numbers ti. Define the sought-for sequence {gi}~t as follows: gi == 2j +3Yi for i E {nj + 1, ... , nJ+t}, j 0, 1,2, .... Put D 8(1 + c). Then for n :5 nl inequality (1) follows from (2). Now let us prove (1) for n > nt. Let j be such t.hat nj < n $ nJ+l' and let {td:=t be arbitrary numbers. For the sake of convenience, put tk = 0 for k > n. Using the definition of the elements gj and inequality (2), we obtain

=

=

94

CHAPTER 7.

STEINITZ'S THEOREM AND B-CONVEXITY

By Remark 6.3.1. the last quantity is estimated from above as follows:

Thus, one part of inequality (I) is proved. Let us prove the second part. By construction, {Y; }~o is a basic sequence of subspaces with basic constant 1- E > 1/2. Using Remark 6.3.1, the definition of 9j, and inequality (2), we obtain

lit. I ~ ~ mrx {11,f., l,g,ll} I,g,

mr+tf" ',g,II} .

~ ~ mrx {2i+'II,I "g,1I} ~ 2 00

00

i=O

k=O

L: lail·lbd :5 mF 11li1· L: Ibkl, the preceding inequality yields

0

which completes the proof of the theorem. LEMMA

7.2.1. Let X be a Banach space with a basis {e"Jr.:l and let

{xli;lf:l C X be an arbitrary countable set. Tben one can find a basis {ek}f=t in X such tbat {Xi}r:t C Lin{ei}r:l' or, in otber words, the decomposition of each element Xi in the basis {ek}~J is a finite sum. PROOF. We shall use the Krein-Mil'man-Rutman theorem (Theorem 6.3.2). Let C be the basic constant of the basis {ek}~l' With no loss of generality we may assume that lIe;1I == I, i = 1,2, .... Let X; :F 0 for all i. Write the decomposition of XI in the given basis: Xl = E:l il be a number for which

Define ei for i

:5 nl as follows: e, = e; for i =I: i l , and 1

00

e,. = e'l + -a;, ;__L:.... , a,e,.

§2. TRANSFERRING EXAMPLES

95

Then lIei - eill :5 C2- i - 3 , i = 1, ... , nh and, by the Krein-Mil'man-Rutman theorem, the system el, ... ,enl'ent+l,enl+2,'" is a basis in X. Moreover, Xl = ~nl

-

L..-i=1 a,ei·

We will now proceed by induction. Specifically, let us show that one can choose a sequence of natural numbers nl < n2 < ... and a system of vectors {ei}~1 C X such that, for any i E N, (3) Thus, suppose that we already found numbers nl, . .. , ni and vectors el,' .. , en, such that relations (3) hold. Then, by the Krein-Mil'man-Rutman theorem, the system is a basis in X. Let

US

decompose the vector n, Xi+!

with respect to this basis:

00

= L)kek + k=1

Xi+1

L

bkek.

k=n,+l

If only finitely many of the numbers bk there are different from zero, we can take for 1ti+1 a natural number such that bk = 0 for k > 11;H, and take the vectors ek with 1ti < k ~ 1ti+ 1 equal to the corresponding vectors Ck. Hence, we need to examine only the case when infinitely many bk'S are different from zero. Let r > 11; be an index such that br ¥ O. Take for nH 1 a natural number such that niH > r

and 00

L

bkek ~ Cbr 2- r - 3 .

k=n'+I+l

Now define f" for nj

< k ~ niH

as follows: ek

= ek for k ¥ r, and

=

Then X1+1 E~!:i' b"e" and lie" - e,,11 :5 C2-"-3 for k :5 1ti+1' Continuing this induction proces, we obtain a system {e,,}k'=I' By the Krein-Mil'man-Rutman theorem, this system is a basis in X and Xi E Lin{e,,} for all i. 0 THEOREM

7.2.2. Let X and Y be Banach spaces, X

::!.

Y. Suppose that

X has a basis {e,,}r:l and let E~=l x" be a series in X such that SR(E:'l x,,) is not a linear set. Then for any monotone sequence of positive numbers {a,,}f:l with ak -+ 00, k -+ 00, tbere exists a series E~I y" in Y such that SR(L:~l y,,) is not a linear set and liy"n :5 aklix"n for alJ kEN. PROOF. ,",10 .......L:_~ __

~

To simplify the arguments we will assume that al > 12. This can

L __

96

CHAPTER 7.

STEINITZ'S THEOREM AND B·(;ONVEXITY

dividing all the vect.ors Xj by K. Since SR(I:~1 xd is not a line
AU + (1 - A)V

f/. SR(L: XIe). 1e=1

With no lol)S of generality we may assume that all the vectors X" tL, and v belong to Lin{elc}~1 (by Lemma 7.2.1, this can be ensured by a small perturbation of the initial basis). Let 11' and (1 be permutat.ions such that 00

00

LXtr(lc) 1e=1

=u,

LX(7(k)

=v.

1.:=1

Denote

=

Clearly, limj-oo bj O. Pick numbers each j the elements

nl

< n2 < ... , sufficient.ly

j

u-

LX"(k),

1e=1

large so that for

j

v-

LX(7(Ie),

1e=1

belong to I.he linear span of t.he corresponding set {el, ... , en,}. Let {Ad f: 1 be a monotone sequence of positive numbers such that Ale --+ 00,

Now let us use Theorem 7.2.1 and select a set of vectors {gk}k~1 C Y such that, for any collection of numbers tic,

The linear operator T: Lin{gIcH~1 ..... Lin{ek}f:1 acting as Tg" = ele, kEN, is continuous and injective. Now define the terms Yk of the sought-for series by Yk =T-1x/c. Also, denote ii = T-I u , ti = T-i tl . Since Xk E Lin{ek}~~I' and hence Yk E Lin{gd~:'I' we have IIYkll $ (12 + An.)lIxkll $ akllxk!l. In a similar manner we can show that

§3. SERIES IN SPACES THAT ARE NOT B-CONVEX

97

and since the right.- hand side of this illequality t.ends t.o zero, the vectors u and SRO=~l Yk). We claim that Xu + (1 - ~)v ~ SR(L~l Yk). Indeed, suppose that. the opposit.e holds, i.e., for some permutation IJ the series L~=l Y,,(k) converges to AU + (1 - >.)v. Thanks to the continuity of T,

u belong to

00

.x>

k=1

k=1

L .r,,(,,) = LTy,,(,,) = T(>.u + (1 - >.)v) = >.u + (1' - >.)v which contradicts the as.c;umption that >.u + (1 - >'lv ~ SR(L~I x,,). Therefore, 0 the constructed series L~l Yk satisfies all t.he requirements of t.he theorem. COROLLARY 7.2.1. In any infinite-dimensional Banach space X thel'e are series with a nonlinear sum range.

l2 is finitely represent.able in any infinitedimensional space. Consequently, L210, 1] :b X. Since in L210, 1] there are series with nonline.ar sum range, it follows that such series exists ill X as weJl. 0 PROOF. By Dvoretzky's theorem,

COROLLARY 7.2.2. In any illfinite-dimensional Banl\Ch space there are series whose sum rallge consists of two points. PROOF. Notice that if in Theorem i.2.2 SR(L~l x,,) consists of two points, then the same is t.rue about SR(L~l yd. Indeed. it is readily seen that the operator T constructed in the proof of the theorem extends to an injective operator T: Lin{g,,}f..l ~ X. Since SR.(L~l x,,) :J T(SR(L~l y,,», we see that SR(L~l Yk) cannot contain points different from u and V. To prove the corollary it suffices to use Dvoretzky's theorem and the existence in L2 of a series with a t.wo-point sum range. 0 EXERCISE 7.2.1. In Theorem 7.2.1 we omitted to show t.hat {gk}~l is a basic sequence (we did not lise that property further). Complete tht' omitted part of the proof and estimate the basic constant of {gJ;}f=I' EXERCISE 7.2.2. Under the assumptions of t.he Krein-Mil'man-Rutman theorem, show that. if {ei}~l is a basis ill X, then {Yi}f:1 is also a basis (we did use this fact, wit.hout. explicit mention, in the proof of Lemma 7.2.1). EXERCISE 7.2.3. Show that for p > 2 there is no infinite-dimensional Banach spl\Ce in which the condition L::"I IIxkllP < 00 is sufficient for the linearity of the set SR(L:~Il'k)' EXERCISE 7.2.4. Find a more precise estimate of the constant.

D appearing

in Theorem 1.2.1. §3. Series iu Spaces That Are Not B-Convex In this section we will show that results analogous to those obtained in §l do not hold in spaces that are not B-convex.

98

to be

CHAPTER 7.

STEINITZ'S THEOREM AND B-CONVEXITY

DEFINITION 7.3.1. A series E~l XIc of elements of a Banach space is said dominated by a numerical series E::l ak if IIxkll $ ale for all kEN.

DEFINITION 7.3.2. We say that Steinitz's theorem holds in the space X relative to the sequence {ale}t;1 o/positive numbers, and we write X E S({ak}r:l)' if any series in X that is dominated by the series E::l ak has a linear sum range.

2::0=1

ale < 00, then Steinitz's theorem relative to the Needless to say, if sequence {ak} holds in any Banach space. Our goal is to prove that if ale diverges and X is not a B-wnvex space, then X ¢ S( {ak}r:l)' We will start by proving this assertion for the case X = LIlO, 1], which was dealt with by P. A. Kornilov 146J in a somewhat djfferent formulation.

r:l

2::1

THEOREM 7.3.1. Suppose that {ak}r:lis a monotonically non increasing ~ quence of positive numbers and 2::1 ale = 00. Then in Ll [0,1] there exist a series E:l Xk dominated by 2::1 ak, such that SR(2::1 Xk) is not a linear set. PROOF. Pick positive numbers {bk}r:l such that bk $ min{a2k,a21e-l}, limk-<>ooble = 0, and E:l ble = 00. Denote the segment 10,1] by 10· Choose a sequence of segments lie, k ~ I, and numbers nle, = nl < n2 < .... , such that Ilie I < ble , the segments {lk}:~+':j+1 are pairwise disjoint, and for j one has that

°

= U:~~J+l lie' This selection is carried out by means of the following inductive process: first partition the segment 10 into a finite number of disjoint segments lie, then partition each lie into a finite number of disjoint segments, and so on. The condition E:l ble = 00 guarantees that one can satisfy the inequalities Ilkl $ bk , k = 1,2, .... Now define the sought-for series E : 1 Xk as follows:

Ij-l

= { ° ift ¢ I j , I

X2j(t)

if t E Ij

and Since IIx2;1I = IIx2;-1I1 = IIjl < b;, we see that limk-ooo IIxleli = 0. The inequality bj < min{a2j,a2j_l} shows that the series E:l Xle is dominated by the series E~1 ak· It is clear that Xle = and that, by analogy with Marcinkiewicz's example (Example 3.1.2), one can rearrange the terms of our series so that it will converge to the function identically equal to 1. At the same time, the terms of the constructed series are integer-valued functions, and consequently the series cannot be rearranged so that it will converge to the function identically equal to 1/2. 0

2:::1

°

LEMMA 7.3.1. Suppose that the space X is not B-convex. Then X has no

finite-codimensional B-convex subspaces. PROOF. It suffices to prove the lemma for subspaces of codimension 1. We shall argue by reductio ad absurdum. So, let Y be a codimension-l subspace of X which is a B-convex space. Then Y has some type p, 1 < p $ 2, with constant C.

§3. SERIES IN SPACES THAT ARE NOT B·CONVEX

99

We claim than in this case X also has type p. Indeed, let lEX· be a functional such that. Ker f Y and 11/11 = 1, and let e E S(X) be such that f(e) > 1/2. FUrther, let {X;}~=l be an arbitrary finite collection of elements of X. Write each Xi as Xi = ale + Yi. where ai E R, Yi E Y. Then

=

la,l

21(f(a,e)1 = 21/(aie + y,)1

~

and lIydl ~ IIx;!1 + I/a l el/ ~ inequality, we obtain

3UXil ·

~

211x;ll

F

Using the definition of the type and Khinchin's

E{II~TiX'11} ~ E{II~riYill} + E{I~riail} $C

$ 3C Therefore,

(t,lIlhll.y" + K (t, 1..1')'"

(t, IIx; II')'" +

K

(t,la.I')'"

$ (3C + 2K)

(t, I x. I ')'"

X has type p > 1, which contradicts our hypothesis.

o

Finally, let us prove the main result of this section. THEOREM 7.3.2 (V. M. KADETS). Suppose that the space Y is not Bconvex and Jet {ak}r=l be a sequence ofpositive numbers such that E~I ak = 00. Then Y ~ S({ak}f:,d. PROOF. We need to exhibit a series

1::"1 Yk in Y, dominated by the series

2::"1 ak, for which the set SR(1::1 Yk) is not linear. Since rearranging the terms of a series does not change its sum range, and since by making the numbers ai smaller we would only make our problem more difficult, it suffices to examine the case when the sequence {ad~1 converges monotonically to zero. Choose two sequences of positive numbers {bk} and {Ck} such that ak = bkCk, bk T 00, Ck ! 0, and E:O=l Ck 00. By Theorem 7.3.1, in the space Ltl0,11 one can exhibit a series Er..l Xk, dominated by the series E~1 Ck, such that the set

=

SR(E~l

Xk)

is not linear. By Proposition 5.1.1, LdO, 111.1\. Hence. by Lemma

7.3.1, LdO, 11 J. Y. Now using Theorem 7.2.2 we see that in Y there exists a series Er..1 Yk with nonlinear sum range, dominated by the series E~=l bkllxkll· Since bkck = ak, the series E~l Yk is dominated by the series E~1 ak. 0 Combining this result with the results of §1, we conclude the the analogues of Steinitz's theorem considered here hold in the B-convex spaces and only in them. EXERCISE 7.3.1. Prove Lemma 7.3.1 wit.hout using the notions of type and infratype.

100

CHAPTER 7.

STEINITZ'S THEOREM AND B·CONVEXITY

EXERCISE 7.3.2. Let P E [1,2]' ak > 0 for all kEN, and L~1 a~ Show that under these assumptions L"IO, IJ ¢. S( {ak }k:l). EXERCISE 7.3.3. Suppose the space

L:~l

a: =00. Show that X ¢. S( {at} f= 1)·

= 00.

X does not have infratype p and

We do not know the answers to the following problems: PROBLEM 7.3.1. Suppose the space X is not B-convex and {ak}k:l is a ak = Does there exist a series Xk sequence sllch that ak 1 0 and in X with nonlinear sum range and such that IIxkll ale for all k?

r::l

PROBLEM

infratype p?

00.

=

r::l

7.3.2. Can Theorem 7.1.2 hold in spaces X that do not have

CHAPTER 8 REARRANGEMENTS OF SERIES IN TOPOLOGICAL VECTOR SPACES There are many types of convergence that cannot be defined in terms of a norm. In this chapter we will examine the problem of the validity of Steinitz's theorem when the convergence of series is understood in the sense of the weak topology of a normed space (§1), or when one deals with convergence in measure for series of functions (§2), and finally when one deals with convergence of series in a nuclear Frecbet space (§3). Steinitz-type problems in topological vector spaces are relatively poorly investigated, and we collected here results that are the most definitive in nature. In the first two sections we construct a number of counterexamples, and in the third section we give a result of W. Banaszczyk which extends Steinitz's theorem from finite-dimensional spaces to nuclear Frechet spaces.

§l. Weak and Strong Sum Range Let 2::1 XIc be s series of elements of a Banach space X. By analogy with the notion of the sum range, we introduce the notion of the weak sum range as follows: WSR(E~=1 XI,:) is the set of elements to which a rearrangement of the series 2::1 Xk converges in the weak topology. REMARK 8.1.1. Let 2::1 XIc be a series of elements in a Banach space

X, let reX· be the set of convergence functionals of this series, and let a E WSR(Er=l

XIc).

Then the following obvious inclusions hold: 00

00

SR(Lxk) c WSR(Lxk) C a + r.L. k=l

k=1

Hence, if Steinitz's theorem holds for the sum range, i.e., SR(E:"1 XIc) then it also holds for the weak sum range.

= a + r.L,

The weak topology is much closer in its properties to a topology of finitedimensional space than the norm topology. This is particularly evident in reflexive spaces, where from any bounded sequence one can extract a weakly convergent subsequence. Nevertheless, the weak sum range of a series can fail to be a linear set even in such a nice space as Hilbert space. A counterexample was provided by V. M. Kadets 141]; its construction is technically quite difficult. Before we turn

CHAPTER 8.

)02

REARRANGEMENT OF SERIES IN TVS

to the exposition of t.hat construction, let us point out t.hat in Marcinkiewicz's example the weak sum range is linear (the reader is invited to check t.his fact by himself). To facilitate the exposition, let us introduce a number of auxiliary definitions. An ordered set A with an order relation F will be denoted by (A, F). The relation "(11 succeeds a2 in the order F' will be written as al > F a2. We say that the element al is a direct successor of the element a2 if al i- a2, al > F a2, and the relations QI >F a, a > F a2 imply that either a = al or a = a2. The set of direct successors of the element a in the order F will be denoted by F(a). If Be A, then we put F(B) = UbEB F(b). We define the set F"(a) as follows:

FI(a)

= F(a). F2(a) = F(F(a», ... ,F"(a) = F(Fn-l(a».

DEFINITION 8.1.1. An order relation F on the set N of natural numbers is said to be tree-like if: a) 1 is the smallest element in (N, F); b) the set F(n) is finite for any n E N; c) F(n) n F(m) = 0 for n i- m; d) U~J Fk(l) = N \ {l}. DEFINITION

8.1.2. A sequence {Xn }~=1 of elements of a Hilbert space is said

to be branche.d with r('spect to the tree-like order relation F (or simply F-branched) if the following conditions are satified: 1) Xn = LiEF(n) Xi for any n E N; 2) {XI}'eF~(1) is a linearly independent set of vectors for any n EN. Let x be all arbitrary element of the linear span of an F-branched sequence By condition 1), for any sufficiently large 71 EN one can represent x. in lhe form x = L:iEF"(I) ti X ,. By condition 2), for fixed n such a representation is unique. Let n > m, k E F"'(l), and {Xn}~=1'

x

=

L

(1)

tiXi·

iEF"(I)

Put Tk(x)

=

tiX,.

i>rk. iEF"(1)

Using the uniqueness of representation (1) for each fixed n and condition 1) of Definition 8.1.2, which allows us to relat.e the represent.ations (1) for different values of n, we deduce that Tdx) does not depend 011 n and in this way a linear operator Tk (possibly unbounded) is defined on the linear span of the sequence lor _ \ ~_,. Let us list a number of element.ary properties of the operators Tic.

§1. WEAK AND STRONG SUM RANGE

103

LEMMA 8.1.1. a) If:1: is a linear combination with integer coefficients of elements of the F-branclled sequence {In }~=I' then for any n all coefficients ti in representation (1) and in the definition ofTk(x) are also integers. b) The identity EiEpn(1) 1j(x) = x holds for any n E N and any x E Lin{xn}~=l· 0 DEFINITION 8.1.3. A system {X;'}~=I of elements of a Hilbert space is said to be conjugate to the F-branched sequence {Xn}~=1 if a) for any n EN and any i,j E F"(I) one has that (Xi, x;> Oij, where Oi1 is the Kronecker symbolj b) Lin{x;}iEP"(l) = Lin{xd'EF"(I)' n = 1,2, ....

=

LEMMA 8.1.2. For any F-brandled sequence {xn}~=1 tJlere exists a unique conjugate system {X~}~=I· The operator Pn acting as Pn(x) = LiEF-(1) (x, xi> Xi is the orthogonal projection of the ambient space onto Lin{xi liEF-(I). PROOF. The existence of a conjugate system follows from the linear independence of the set {XdiEFn(l). while its uniqueness follows from condition b) of Definition 8.1.3. Further, the fact that P,.. is the project.ion onto Lin{xi}iEFnO) is a consequence of condition a) of Definition 8.1.3, and the orthogonality of Pn is a consequence of condition b) of the same definition. 0

In what follows the notation Pn will be reserved for the projections defined in the preceding lemma.

r:l

DEFINITION 8.1.4. An F·branched sequence {xd is said to be F . widely spread if the following conditions are satisfied: I) for any n EN, the angle between any two elements of the set {Xi LE P" (1) is obtusej 2) ifi >F j, then (xj,Xi) > OJ 3) if j E P(1) and i >F j, then Pn(Xi) is collinear with Xi; 4) if jl,h E F"(1), i l >F jl, i2 >p h, and iJ =1= h. then the vectors XiI - Pn(Xil) and Xi2 - Pn(Xi 2 ) form a right angle.

x

LEMMA 8.1.3. Suppose that {xdr:1 is aJl F.widel.y spread sequence lwd Jet. Then IITk(x)1I ~ IIxkll· minltil.

= EiEpm(l) xiti' n < m and k E Fn(l).

PROOF. From condition 1) of Definition 8.1.4 it follows that

It;lx;

Using conditions 3) and 2) of Definition 8.1.4 we obtain

CHAPTER 8.

104

REARRANGEMENT OF SERlES IN TVS

(In the last [resp. last but ~~eJ step we used condition a) [resp. condition I)J of Definition 8.1.3 Iresp. Defimtlon 8.1.2J.) 0 LEMMA 8.1.4. Let X denote the linear span of an F-widely spread sequence

{Xi}f:l' Let {dk}k:l be a sequence of elements of X tbat converges weakly to zero. Then, for any n E N,

PROOF. Consider the equality dk = LiEF"(l) 1i(dk), take its square, and use condition 4) of Definition 8.1.4 to transform the result as follows:

iEF"(I)

==

L

111i(dk)1I 2 +

iEF"(!)

i,jEF"(1). if.;

L

(Pn1i(dk), PnTj(dk)).

(2)

i,jEF"(l),i#;

By the definition of the projection Pn and condition 3) of Definition 8.1.4, PnTi (dk) == (x. , Ti(dk)) Xi. i E ~(1); further, by the definition of the operator Ti and conditio~ 4) of Definition 8.1.4, PnTi(d k) = (xi, dk)Xi. i E Fn(1). Since the sequence

{dk}f:l converges weakly to zero,

which in conjunction with relation (2) completes the proof of the lemma.

0

LBMMA 8.1.5. Let F be a tree-like order relation and {ak}~l be a sequence of positive numbers. For the existence of an F-widely spread sequence {Xk}k'== such tbat \Ixtll == at, kEN, it is necessary and sufficient that a~ :5 LkEF(i) a!

(or alii

e N.

PROOF. Necessity is a consequence of condition 1) of Definition 8.1.4. Sufficiency follows from the fact that we are working in an infinite-dimensional space and when we construct the elements {Xi}iEP"+I(l) we can add to Lin{xi}iEF"(l) enough dimensions to ensure that all requirements of definitions 8.1.2 and 8.1.4 are fulfilled. 0

§1. WEAK AND STRONG SUM RANGE

105

THEOREM 8.1.1. Let H be an infinite-dimensional Hilbert space. Then (or any sequence {ak}k:J o( pO-')itive numbers satisfying the conditions limk_co Uk = o and l:~1 a~ = 00 there exists a series l:~1 Zk of elements of H such that IIzkll = Uk for all kEN and WSR(l:~1 Zk) is not a linear set. More precisely, [or some rearrangement o[ terms the series converges to 0, for other it converges to some element Z E H, z :I 0, and there is no rearrrangement [9r which the series converges weakly to tz witll t not an integer.

PROOF'.

lows:

n. =

1;

Construct by induction a sequence of natural numbers {n.i} as folis the smallest number for which

n.k+!

".+1

L

a~ > 2a~.

i=".+1

Define the tree-like order relation F by F( k) F enjoys the following properties:

2a,~ _ < "~

= {nk + 1, nk + 2, ... , nk+ d. Then ak! 2

(3)

2

(4)

= O.

(5)

kEF(i)

2

"~ a, > 2"a I' iEP(l)

lim

"~

n-+oo

a~

iEF(n)

Using Lemma 8.1.5, pick an F-widely spread sequence {X;}~I C Ii such that IIXill :;:: ai, i E N. Now as the terms of the sought-for series 2::1 Zj take ±Xj where the sign is U_" for i E F 2n-I(l), n E N, and u+" for the remaining i E N. If we write the elements Zj in the order

L L Xi-"E 00

n=O iEF2"(1)

(

XI< )

kEF(i)

(pO(l) = {I}), where one first calculates the inner sums, and then the outer sum, then thanks to conditions 1) in definitions 8.1.2 and 8.1.4 and to relation (5), the series will converge in norm to O. If now the series is rewritten in the order

then it will converge to

XI.

106

CHAPTER 8.

REARRANGEMENT OF SERIES IN TVS

Let us show that if the series L::l Z,.(i) converges weakly to tXI, then t is necessarily an integer. To this end let us denote by e the distance from t to the nearest integer. Further, let Sn denote the partial sums of the series L::I Z"(i), and put Dn = Sn - tXI' By hypothesis, the sequence Dn converges weakly to O. Since for m large enough we cart write Sn as a linear combination L:iEPIft(1) tiXi with integer coefficients ti, and since Xl = EiEF"'(I) Xi, by assertion a) of Lemma 8.1.1 and Lemma 8.1.3 we have IITIc(Dn)1I ~ Ellxlcll = ealc· Denote SUPn IIDnll by D. By Lemma 8.1.4,

D2 ~ l~~~f

L iEF"(I)

IITi(Dk)II ~ 2

E

L

a~ ~ Ea~ • 2n

iEF"(l)

for all n EN. Since n is arbitrary, this means that

E

= 0 and t is an integer.

0

REMARK 8.1.2. An examination of the preceding proof shows that no subsequence Sn~ of the sequence Sn of partial sums of the series L:I Z"(i) can converge weakly to a point of the form tXI with t not an integer. EXERCISE 8.1.1. Show that series with nonlinear weak sum range exists in any infinite-dimensional Banach space.

The authors cannot provide an example of a series in a Hilbert space with two-point weak sum range. §2. Rearrangements of Series of Functions

In addition to the already familiar to the reader Problem 106 of the Scottish Book, solved by J. Marcinkiewicz, there is the following similar problem that goes back to S. Banach and W. Orlicz: does Steinitz's theorem extend to almosteverywhere convergence of series .of functions? This turned to be a considerably more difficult problemj a counterexample was constructed by E. M. Nikishin 161) in 1971. In this ~ion we will give a more convenient method for constructing similar examples, proposed by V. M. Kadets !39J, which allows one to solve some other problems as well. Thus, we will construct a series of bounded functions on the segment 10, 1] with the property that for one rearrangement of its terms the series converges uniformly to zero, for another rearrangement it converges uniformly to a function u(t), but there is no rearrangement for which the series will converge, even in the sense of convergence in measure, to a third function, different from zero and u(t). Recall that for a subset A c [0,1] we denote by XA(t) the indicator function of A. The construction will make use of two classical orthonormal systems in L2[0, II-the Rademacher system rn(t) = sign(sin(2n1l't)), and the Haar system. The Rademacher system, which is the simplest example of a sequence of independent random variables that take the values ±I with equal probabilities, was already mentioned before. Let us recall the definition of the Haar system:

§2. REARRANGEMENTS OF SERlES OF FUNCTIONS

107

....................................... ,

where n E Nand 1 :5 k :5 2n - 1. The Haar system is known to be a.basis in L2[0, I]. Let us denote by LR the class of measurable functions on [0, I] that are sums of series of the form En~1 enrn(t), Clc E R, which converge in the metric of 12[0, I]. Since the system {Tn}~=1 is orthonormal, the convergence of such a series is equivalent to the condition 2::=1 ~ < 00. LEMMA 8.2.1. Let I be an arbitrary measurable subset of the segment [0, I] of measure m{l) > 2/3. Then there exists a constant K = K(m(J)) E R+, such that the inequality

t [f(x)]2dx $ K sup[f(x)]2 %EJ

(1)

10 bolds for all f E LR.

PROOF. Consider the system of functions r,,;(x) = T,(X)Tj(X), i > j, and adjoin to it the function 1 identically equal to 1. The resulting system is an orthonormal sequence on [0, I]. By the Parseval inequality,

Hence, (2) Let f(x) = 2::=1 enrn(x), where en = 0 for all n larger than or equal to some number N. Then using the relation Tn . Tn == 1 we have

11!(x)]2 dx

= m(l) f>~ + 2 Le;c;

I

n=1

1

r,,;(x)dx.

I

i>j

Estimating the last term by the Cauchy-Bunyakovskii inequality and applying inequality (2), we obtain

!

l!(xWdx 2: m(l) ~ c!

- 2 ( ~(e;Cj)2 )

1/2

(m(l) - [m(l)]2) 1/2 2:

108

CHAPTER 8.

~ m(I)

REARRANGEMENT OF SERlES IN TVS :xl

2: C~

00

-

v'2 (m(l)

- [m(JW) 1/2

n=\

2: C~ n=\

(3) By an obvious passage to the limit one can show that inequality (3) holds for arbitrary {unctions f E LR, which are not necessarily finite linear combinations

of Rademacher functions. Denote

1- (2 I::~g» 1/2 by K. Then

1 2 1 ( 10 2 suplf(x)J ~ m(I) JrlJ(xWdx ~ K L C! = K lJ(x)J dx. 00

r

xE 1

n=1

o

0

COROLLARY 8.2.1. The claBs LR is closed in the sense of convergence in measure; moreover, any sequence of elements of LR t.hat converges in measure also converges ill the metric of L2[0, 1J.

PnooF. Let fn(x) a sequence of functions from LR which converges in measure to a function f(x). By the definition of convergence in measure, there exist measurable sets In C [0, IJ, m(In) -+ 1, for which lim sup Ifn(x) - f(x)1

"-00 rEIn Denoting I" n 1m

= O.

= [n.m we obtain

By Lemma 8.2.1, this means that {In} is a Cauchy sequence ill t.he metric of L2[O, IJ, and consequently it converges in L210, IJ to some function g(x). It is clear that f(x) = g(x) almost everywhere and that f belongs to the closure of t.he class LR in '''2 [0, 11· Hence, f E LR. 0 LEMMA

8.2.2. Let f(x)

= E::O=l cnTn(x). Then 00

sup If(x)1 rEtO.11

= L len Ii ,,=1

moreover. if one of the sides of this relation is equal to infinity, then so is the

secolld.

Pnoof'. It suffices to calculate directly the left-ha.nd side of the equality. 0

§2. REARRANGEMENTS OF SERIES OF FUNCTIONS

109

THEOREM 8.2.1. In the space il there exists a series L::=I Zn such that [or one rearrangement the series converges to zero, for another it converges to some v E ll' and there is no rearrangement for which the series converges, even in the metric of h, to a third element of l2. PROOF. We shall make use of the construction of a series with two-point sum range given in §2 of Chapter 3; however, to avoid any coMusion, the series constructed t.herein will be denoted here by 2::'1 cJ.... Define a system of functions cr.; on the cube Q = 10, IJ'" as follows: eLI == 1, Ci~j(tl,t2'''') = ~(tn)hi(tn+l)' where i,j, n E N, i 1= 1, and hi are the Haar funct.ions. These functions form an orthonormal system in L 2 (Q), and each term of the series 2:::0=1 d,. can be represented as a finite linear combination of functions cr.]. Denote the closed linear span of the system {cr.;} by H. Define an isomorphism T: H --+ l2 by choosing some bijective correspondence between the unit basis vectors of l2 and the functions cr.i' Put Zn = T(d n ). v = T(cl,l)' Clearly, t.he series 2:~=1 Z.. has the property that for one rearrangement it converges in l2 to 0, for another it converges to v, and there is no rearrangement for which it converges ill l2 to some element different from 0 and v. The clements Zn are linear combinations of unit basis vectors, and so Zn Ell. We claim that for the rearrangements of the series 2::: 1dn constructed in §2 of Chapter 3, for which that series converges to 0 or to 1, the series L:~= I Zn converges in II t.o 0 or to tl, respectively. Indeed, for those rearrangements of the series 2::=1 dn, the difference between the sum and the partial sum has the form

where one of the segments [al,bll, [a2,~1 has the form [m/k,(m + I)/kj, and the length 11k of this segment t.ends to 0 when the index of the partial sum tends to infinity. Expanding R(t} with respect to the functions ciJ, we find that the sum of the absolute magnitudes of the coefficients of the expansion does not exceed A In kl.fk. We thus conclude that the images under the map T of the rearrangements of the series E:='=I dn discussed in §2 of Chapter 3 converge not only in the metric of l2, but also in the metric of i l . 0 THEOREM 8.2.2. There exists a series E:='=l fn(x) of bounded measurable functions on [0, 1\ such that [or one rearrangement of its terms the series converges uniformly to zero, for another it converges uniformly to a function u(x), but even with respect to convergence in measure there is no rearrangement [or which it converges to a function that differs from zero and u(x). PROOf. Let P: l2 ..... LR denote the linear isomorphism defined by the rule anrn· Now take for the series n whose existence is asserted in the theorem the image under P of the series Zn constructed in Theorem 8.2.1: In P(zn}. By Lemma 8.2.2, the functions fn(x) are bounded on

P«(a}, a2, al,·· .)

= 2::=1

=

L::='=l/ L::='=I

llO

CHAPTER 8.

REARRANGEMENT OF SERIES IN TVS

10, I), and if for some permutation 11' the series E:=l z,,(n) converges in ll' then the series E:=l I,,(n) (x) converges uniformly on 10,1). Hence, by Theorem 8.2.1, there exist two rearrangements of the series E:'l In for which it converges uniformly to zero and to u = P(v), respectively. To complete the proof it remains to use 0 Lemma 8.2.1 and the orthonormality of the sequence {Tn(X}}. REMARK 8.2.1. If instead of the Rademacher functions we use the lacunary trigonometric system {sin(2n1l'x}}, we can take the functions fn(x) in Theorem 8.2.2 to be trigonometric polynomials.

EXERCISE 8.2.1. In his PhD dissertation P. A. Komilov has shown that if one expands the function in Marcinkiewicz's example with respect to the orthonormal Maar systems, then that example works in the space What results on the sum range of series of functions can be derived from this?

'1'

EXERCISE 8.2.2. From Corollary 8.2.1 one can obtain another prooC of the Khinchin inequalities for 1 ::5 P ::5 2 (and even for 0 < p < 1 !). One can also prove Lemma 8.2.1, Corollary 8.2.1, and Khinchin inequalities Cor the functions sin(2"1I'x) instead of the Rademacher functions. Do it!

§3. Banaszczyk's Theorem

00

Series in Metrizable Nuclear Spaces

For this section we are assuming that the reader is familiar with the basic definitions of the theory of locally convex spaces (if not, she is referred to the books 145 , Chapter 3, §5J, 176J or [781, which will suffice for our purposes). On the other hand, the notion of nuclear space, despite its extreme usefulness (for instance, the spaces of analytic functions with the topology of convergence on compact sets, the spaces of infinitely differentiable functions, and others, are nuclear), is still insufficiently popular, and is usually not incorporated in the university courses. A good (even if somewhat old) source for learning the theory of nuclear spaces is A. Pietsch's monograph 169). To facilita.te the understanding of the ensuing material we shall give all the necessary definitions. Let E be a metrizable locally convex linear space, i.e., the topology of E is given by a countable family of semi norms P1 ::5 1'2 ::5 P3 ::5 .... Denote by Bn = Bp" the unit balls with respect to these seminorms. Convergence in E means convergence in all the semi norms Pn. A continuous linear functional fEE' is 8 linear Cunctional that is continuous with respect at least one of the seminorms pn, or, equivalently, boundt.'
§3. BANASZCZYK'S THEOREM

III

dependently the necessary arguments. we will give here only the statement of the theorem. omitting the proof. THEOREM 8.3.1. Suppose that in the metri?able locally convex space E the following conditions are satisfied: (A) Rounding-off-Coefficients Lemma. For any continuous seminorm p on E there exists a continuous seminorm q ~ p with the property that, for any elements Xl •...• X n , n E N, of the ball Bq and any set of coefficients Ill.··· .I'n, 0 S 11.; $ 1, one can lind coefficients 9; E {O, 1} such that

p

n) S 1. " - L9;x; (LII.;x; i=1

i=1

(B) Permutation Lemma. For any continuous seminorrn p on E t.here exists a continuous seminorm q ~ p with the property that, for allY linite set of elements XI .... ,Xn of the ball Bq satisfying 2:~ IIi = 0, one can clloose a permutation 7i of the lirst n natural numbers such that

Then for any conditionally convergent series 2:::0=1 Xk in the space E one has that SR(2::=1 x,,) = a + r 1 , where a = 2::'=1 x". 0 EXERCISE 8.3.1. Show that in the above theorem condition (A) is a consequence of condition (B), and hence the statement of the theorem can be simplified. DEFINITION 8.3.1. Let A and B be two convex subsets of a linear space E. The Kolmogorov widths of the set A with respect 1.0 the set B are the numbers

dlc(A, B)

= infinf{e > 0: A C :r + L + eB}, L.7

where the second infimum is taken over all subspaces LeE with dim L < k and over all X E E. Notice that d,.(A, B) does not change when the sets A and B undergo parallel translation, does not decrease when A is enlarged, and does not increase when B is enlarged; moreover, d,. does not increase when k is increased. In what follow Un will dcnot.e the standard Euclidean ball ill Rn: n

Un

= {x = (Xl," .• Xn): I)rd 2 S l} . • =1

112

CHAPTER 8.

REARRANGEMENT OF SERIES IN TVS

EXERCISE 8.3.2. Show that if A is an ellipsoid in Rn with principal semi-axes ..\1 ~ ..\2 ~ ... ~ An, then dk(A, Un) = Ak. In what follows orthogonality in Rn will be understood as orthogonality with respect to the standard inner product, associated with the ball Un. DEFINITION 8.3.2. A semi norm p on a linear space E is said to be Euclidean ifit is given by some nonnegative bilinear form Q(x,y): p(x) (Q(X,X))1/2.

=

In a finite-dimensional space the unit ball of such a semi norm is either an ellipsoid, or an elliptic cylinder, i.e., the direct sum of an ellipsoid and a subspace. DEFINITION 8.3.3. A space E is said to be countably Euclidean if its topology can be given by a countable family of Euclidean norms. DEFINITION 8.3.4. A countably Euclidean space E is said to be nuclear if for any continuous Euclidean seminorm 'P on E there exists a continuous Euclidean seminorm q such that 00

L dHB

q,

B,,) ::; 1.

k;l

Since we are not dealing here with the nonmetrizable case, in what follows countably-Euclidean nuclear spaces will be simply referred to as nuclear spaces. REMARK 8.3.1. In Definition 4.3.3 we introduced the concept of a nuclear space in a somehwat different way. In the metrizable case the two definitions are equivalent; in particular, any nuclear space in the sense of Definition 4.3.3 is countably Euclidean. For more on this see 169}.

The main goal of this section is to show that Steinitz's theorem holds in nuclear spaces. We shall follow the original proof of this result, obtained by W. Banaszczyk in 1989 12}. LEMMA 8.3.1. Let C be an n-dimensional ellipsoid in Rn with principal semi-axes AI, . .. , An, and Jet P be an n-dimensional right-angled parallelepiped circumscribed to C. Then n

diamP

=2 ( ~A~

) 1/2

k=l

an .....

PROOF. We may assume that the center of C is at the origin. Let T: Rft be a linear operator for which T(Un ) C. Pick an orthonormal basis {ek}k;1 in Rft whose vectors are parallel to the edges of the parallelepiped P. Then

=

§3. BANASZCZYK'S THEOREM

where

SA:

> 0 is haH o[ the length of the corresponding edge. By hypothesis, 51:

= sup (Tu,el:) uEU"

and

113

= sup (u, T"el:) = llT"ekll uEU"

~d;am P= Ct, si)'" ~ (t,II7"••11')'"

It remains to observe that the right. hand side in the last equality is precisely the Hilbert-Schmidt norm of the operator T, and hence is equal to (L~=I A~) 1/2. 0 LEMMA 8.3.2. Let CeRn be an n-dimensional ellipsoid centered at the origin whose principal semi·axes AI, ... ,>'n satisfy the condition

>'1- 2 + ... + An-2 -< 1,

(I)

and let M be an (n - I).dimensional affine hyperplane that intersects Un. Then C n M is 80 (n -I)-dimensional ellipsoid whose principal semi-axes ILl,"" Jl.n-I satiSfy the condition -2 < 1 (2) ILl-2 + ... + ILn-1 - . PROOF. By condition (1), Un C C, and consequently C n M principal semi·axes of C in increasing order: >'1 ~ ... :::; An. Then

:f:. 0.

Label the

(3) Similarly, denoting by u the center of the ellipsoid C n M, setting Mo and labeling the semi-axes ILI!'" ,ILn-1 in increasing order, we obtain

=M

- u,

(4) Let T: Rn -+ Rn be a linear operator such that T(C) = Un. Denote N = T(M), No = T(Mo), D = T(Un ), and "k = >.;1; then 1I1, .•. ,lIn are the principal semiaxes of the ellipsoid D. By condition (1), II~

+ ... + II;

~ 1.

(5)

Let P be a right-angled parallelepiped circumscribed to D such that one of the (n-l}-dimensional faces of P is parallel to N. From (5) and Lemma 8.3.1 it follows that P C Un. By construction, D C P and NnD:f:. 0. Let n denote the orthogonal projection of Rn onto the hyperplane N. Then Il(P) is an (n - I)-dimensional right-angled parallelepiped circumscribed to the ellipsoid Il(D); moreover, n(p) c

CHAPTER 8.

114

U

n N.

REARRANGEMENT OF SERIES IN TVS

If we equip N with a Hew Euclidean metric in which Un :s 2, and. by Lemma 8.3.1,

b~l: then in this metric diam n(p)

n N is the unit

71-1

(6)

LdHn(D),U,.nN) $1, k=1

because d,,(TI(D), Un n N) is the length of the k-th principal semi-axis of the \1ipsoid £I(D) in the new metric. e Since n(D n No) can be obtained from D n No by parallel translation and since n(D n No) c n(D), inequality (6) gives n-I

(7)

[.d%(DnNo,UnnN) $ 1.

"=1 Next, since l' is a.n isomorphism, dk(TA, TB)

= d,,(A, B)

for any two sets A, B.

Therefore, which in view of rela.tions (4) and (7) yields the needed inequality (2).

0

LEMMA 8.3.3. Let P be an n-dimellsional parallelepiped in Rn such that the lenghts of its edges l1rc 110 larger than I, let C be an n-dimensional ellipsoid in tile same spat.'(' whose principal semi-axes AI.' ..• An obey the inequality Al2 + ... + A~2 ::; 1. 8l1d suppose the cellter of C lies ill P. Then one of the edges of P

;s cont.ained ;n C. PROOF. With no loss of generality we may assume that C is centered at the origin. Then 0 EPal\(1 0 E C. To prove t.he lemma we proceed by induction on the dimension 11 of the space. For -n = 1 there is nothing to prove. Suppose the assert.ion of the lemma is already proved in dimension -n - 1 and let us show that it holds true in dimension n. Denote sup{ r > 0: rC C P} by s. Then the ellipsoid sC is tangent to olle of t.he (n - i)-dimensional faces F of the parallelepiped P at some point UI. Let M be the hyperplane containing F. Siuce 0 E P and the lengths of the edges of P are not larger than 1, it follows that Un n AI :f; 0. By Lemma 8.3.2. the pricipal semi-axes IL\ •... ,JLn-l of t.he ellipsoid CnAl satisfy the condition IL12 + ... + IL;;~ 1 ::; i. Since IV is the center of the ellipsoid C n AI and Ul E F. the inductive hypothesis says that olle of the edges of the parallelepiped F is contained in en AI. But this edge is also an edge of the original parallelepipied P and lies in C. 0 Let liS denote by I. t.he collectioll of all subsets of the set {I, ...• s}. LEMMA 8.3.4 (ROUNDING-OFF-COEFFICIENTS LEMMA IN NUCLEAR SPACES). Let p ~ q be two Euclidean seminorms on a vector space E satis(ying ~

Ld~(Bp.B,,) $ 1. ~=I

(8)

§3. BANASZCZYK'S THEOREM

Let bEE,

t'I,""V.

E Bp,

S

~ 2, and Jet A

=

liS

{b+L:=lt.vi:O$L, $l}.

Suppose 0 E A. Then there exists J E 1$, 1 $ card J $ s-l, such that b+ LIe J Vi E

Bq • PROOF. Since we are dealing with finitely many vectors, confining our considerations to their linear span we may assume that the space E is finite-dimensional. Further, with no loss of generality we may assume that Bp and Bq are ellipsoids (and not elliptic cylinders) and that the vectors {Vi}:=1 are linearly independent (the general case is readily derived from this case by a limit process). Finally, choosing an appropriate isomorphism we may assume that E = RS and Bp = Us. Then condition (8) says that the principal semi-axes ),1, ... , >'71 of the ellipsoid Bq satisfy the inequality ),\2 + ... + >.;2 ~ 1. The set A is a parallelepiped that contains the point 0, i.e., the center of the ellipsoid Bq . By Lemma 8.3.3, one of t.he edges of .4. is contained in Bq . Since s ~ 2, one of the vertices of that edge is different from both band b + Vi· Since every vertex of the parallelepiped A has the form b + EiEJ tJ;, this completes the proof of the lemma. 0

E:=1

REMARK 8.3.2. Here the Rounding-off-Coefficients Lemma is just all intermediate step in t.he proof of the Permutation Lemma in a nuclear space. Otherwise, in view of Exercise 8.3.1 we would not need this lemma.

LEw>'IA 8.3.5. Let Bee c D be three n-dimensional ellipsoids with common center (tile origin) and such that n

Ld~(B,C) ~ 1

(9)

k=1

and (10)

Then, givun arbitrary elements ttl, ..• , 'U~ E Band Ct E C sat.is~l"iIJg a + L;=1 C, olle can clloose a permuat.ioll a of the set {1, .... s} sudJ that i

a+

L

tL,,(il

E D,

j

= L ... , s.

Ui

E

(11)

;=1

PROOF. We proceed by simultaneous induction on n and s. For the cases (n = 1, sEN arhitrClIY) and (s = 1, n E N arbitrary) the assertion of the lemma is obvious. Suppose it is true for all pairs (ii, s) sitch that n < nand s $ s, or n = nand s < .s, and let us verify that it is true for the pail' (n,s). Thus, we need to establish the existence of a permutation a such that (ll) holds. We shall cOl1!';ic\pr IWi) <:pn"r"t., """""..

116

CHAPTER 8.

REARRANGEMENT OF SERIES IN TVS

I) Suppose 0 E A == {a + l::=l titLi: 0::; t; ::5 I}. Then. by Lemma 8.3.4, there exists a set J E [6. 1 ::; card J ::; s - 1. for which

a+

LUi EG.

(12)

iEJ

For the sake of convenience we shall assume, with no loss of generality, that J == {l, ... , z}, with z ~ 8-1. Applying the inductive hypothesis to the set J we obtain a permutation u of J such that

i

a +

L

UC7(i) E

D,

j

= 1, ... , Z.

i=1 Hence, it remains to permute in the needed way the elements u.z+l.' .. , U 3 • If we denotea+E::l Ui by b, then bEG (as shown by relation (12)), andb+E:=.z+1 Ui E C (by hypothesis). Applying again the inductive hypothesis, but now to the indices {z + 1, ... , 8}, we extend the permutation u to this set of indices in such a manner that ; ;

b+

L

U C7 (i) = a + L UC7 (i)

;= .. +1

ED,

j=z+I, ...• s.

i=1

This completes the construction of the required permutation u. 2) Suppose now that 0 ¢ A. Then there exists a linear functional such that I(u) > 0 for all U E A.

I

on Rn

(13)

Denote h = sup{J(u): U E G}. The set A is centrally symmetric about the point x = a + E:=1 Ui. Since x is the midpoint of the segment connecting a and a + 2::=1 Uj, it follows that x E C. Next, since the center of symmetry of A lies in G, condition (13) implies that

!

sup{J(u):

U

E A} ~ 2h.

(14)

With no loss of generality we may assume that D = Un. Let H = Ker I (then = n - 1), and let n denote the orthogonal projection onto H. Inequality (9) obviously yields

dimH

n-l

L d~(n(B), I1(G» ::5 1.

(15) "=1 Circumscribe an n-dimensional right-angled parallelepiped P to C such that one of the (n - 1)-dimensional faces of P is parallel to H. By Lemma 8.3.1 and relation (10), 2P C Un· From ~his and also from the construction of P is follows that 2n(p) C W ~f I1(Un n rl(2h».

(16)

§3. BANASZCZY"'S THEOREM

117

n(p) is an (n - I)-dimensional parallelepiped circumscribed to t.he ellipsoid n( C). By Lemma 8.3.1. inclusion (16) means that n-I

L dhn(c). W) ~ ~.

(17)

1:=1

Let liS apply the inductive hypothesis to the ellipsoids n(8) c n(C} c tV and the elements band {tJ.}i::I' defined as b = n(a), Vi = n(Ui) (that we are allowed to do this is ensured by inequalities (15) and (17)). We obt.ain a permutat.ion q such that j

n(a) +

L n(U,,(i») E W.

j

= 1. ... , s.

;=1

By relations (13) and (14), this means that j

a

+ '~::>O"(i)

E n-I(W) n

r

1

(10. 2h]),

i=1

which by the definition of the set \·V implies that (11) holds.

o

Using the fact that the Permutation Lemma deals with a finite number of vectors, we can derive from the preceding finite-dimensional variant the needed Permutat.ion Lemma in nuclear spaces, which in t.urn implies Banas::czyk's theorem

(see 12J, 131): THEOREM 8.3.2. In metrizable nuclear spaces the assertion ofSteinitz's theorem on the slim range of conditionally converge/lt series llOlds true. EXERCISE 8.3.3. Show that in the nonmetrizable nuclear space Rlo,11 St.einitz's theorem fails. EXERCISE 8.3.3. Give all example of a nonmetrizable nuclear space in which Steinitz's theorem does hold.

APPENDIX THE LIMIT SET OF THE RIEMANN INTEGRAL SUMS OF A VECTOR-VALUED FUNCTION Consider a bounded function f: 10, 11 - t X, where X is a Banach space. Denote M(J) = sup{lIf(t)II,t E 10. I]}. Let r = {Ai}f=1 be a partition of the segment 10,IJ into segments Ai lai,biJ. 0 al < b1 = a2 < ~ = a3 < ba = .,. < bn 1. Take a set of points T {tdi'::l' ti E Ai. It is natural to define the Riemann integral sum of the function f as

=

=

=

=

n

S(f. r, T)

= L f(t,)(b i -

n

a,) =

L f(t,)IA,I. i=1

i=1

IA,I denotes the length of the segment Ai.* Call d(r) = max, lAd the diameter of the partition r. As in the usual real-valued case. we will say that f is Riemann integrable if the integral sums S(J. r. T) tend to some limit when d(r) -+ O. The limit is called the integral of the function f.

where

If for real-valued functions Riemann integrability is a very restrictive condition and the Lebesgue integral, thanks to its considerably more general nature, turns out to be an incomparably more convenient object, for Banach space-valued functions the situation is more complicated. EXAMPLE I **. Let h[O,I] be the space of real-valued functions that are different from zero in at most countably many points of the segment [0.1] and obey the condition EtE[O,I)lg(t)1 2 < 00. Equip this space with the norm IIgll = (EtE(O,l) Ig(t)j2) 1/2. Defined in this way, 12[0, I] is a nonseparable Hilbert space with a noncountable orthonormal basis ex, x E 10,1]. where ex{t) = 1 if t = x, and ex(t) 0 if t =F x. Consider the function f: [0, I] -+ 1210, I] defined by I(x) = e",. Then any integral sum of f is estimated as

=

118(1. r. T)II =

lit,,. 1 .111 (t, 1 .1') 'I' 6

·To keep in agreement with the notation to be multiplied by scalars to the right . •• T}'U:a. """" ....D .. ; " - ....

; . . . . " ' ...

A ____ ~:.o :w

o.

.I

=

6

f f dx for the integral, we will allow here vectors

120

APPENDIX.

LIMIT SET OF RIEMANN INTEGRAL SUMS

We see that when d(f) -+ 0 the integral sum also tends to zero, and hence the constructed function / is Riemann integrable. At t.he same time / cannot be represented as the limit of an almost everywhere convergent. sequence of simple functions (i.e .. functions, each of which takes only a finite number of values). Thus, we produced an example of a function that is Riemann integrable, but not measurable. As a matter of fact, our Riemann-integrable function is discontinous at any point. For functions that are not Riemann integrable the role of an integral may be played by the limit of some sequence S(f, r n, Tn) of integral sums, where the partitions r n are increasingly finer: d(r n) -+ 0 as n -+ 00. For a nonintegrable function there may be many such limits. It therefore becomes necessary to consider not a single limit, bllt rather the set of all snch limits. DEFINITION 1. For a bounded function f: 10, 1\ ..... X we call the set of all points of X that can be represented as the limit of a sequence of integral sums S(f, r n, Tn) with d(r n) -+ 00 the set of limits oj Riemann integral sums oj the function f and denote it by I(f).

Obviously, I(f) is a closed set contained in the ball of radius M(f). In this appendix we will discuss various results connected with the description of the structure of the set 1(/). This topic has many aspects that are similar t.o results and methods of the theory of conditionally convergent series, but is considerably less developed and does not have such a definitive character as the latter.

§1. Functions Valued in a B-Convex Space For real-valued functions f the set I(f) is the segment whose endpoints are the lower and upper Riemann integrals. This simple fact is readily generalized to the case of functions taking values ill a finite-dimensional space: compactness considerations show that 1(f) is a nonempty connected set. The following theorem, proved in 1947 by P. Hartman 13\,* should be regarded as the first nontrivial result from which the theory expounded below originated: for any bounded function taking values in a finite-dimensional space the set I(f) is convex. This theorem plays the same role in the theory under consideration as Steinitz's theorem in the theory of series. However, in contrast to Steinitz's theorem, Hart.man's theorem admits a direct generali7..ation to some infinite-dimensional spaces. Thus, in 1954 I. Halperin and N. Miller [2] extended Hartman's theorem to functions with values in a Hilbert space. Developing these ideas, in 1966 1'\"1. Nakamura and l. Amemiya [10] generalized the rcsult to B-convex spaces. Since the proofs in all thE'.se cases ••

. •. - • __ M"I;v "'n

will .. Iwavs

refer to the Jist of "Supplementary References

n

§J. FUNCTIONS VALUED IN A 8-CONVEX SPACE

I'll

can be <:arried out by completely similar means, here we shall prove directly the last, lIlost general version, following in our exposit.ion the paper [81, where the Nakamura-Amemiya theorem has been rediscovered (with a delay of 20 years!) by the authors of the present. book. THEOREM 1. Let X be a B-convex normed space and f: [0,11 bounded (unction. Then tile set IU) is convex.

X be a

PROOF'. Recall (Chapter 5, §3) that X, being a B-convex space, has some infratype p > I, i.e., there exists a constant C > 0 such t.hat the inequality

(1) holds for any finit.e collection {x;}f=l of elements of X. Let Xl, X2 be two points in l(f). Since the set IU) is closed, to prove the theorem it suffices to show that ~(Xl + X2) E l(f). To this end fix N E Nand choose sequences of partitions and of collections of points (r~, T~) and (r~, T;) such that lim d(r~) 0, r~, T~) ~ Xi> i = 1,2. n-oc

=

su,

With no loss of generality one can assume that. both partitions r~ and r~ include as division points all points of the form k / N, k = 0, 1, ... ,N (indeed, by adjoining these points to f:. one adds N segments whose lenghts do not exceed d(f~), and hence, for any way of selecting the necessary supplementary points~, the integral sum is modified by a term bounded by 2Nd(f~)M(f), which tends to zero when 71 ..... (0). Pick e > 0 and denote by Slone of the integral sums SU, r~, T~) for which IIS l - XIII < ~, and similarly take S2 such that [lS2 - X2[1 < e. Next, divide the segment 10.11 into N segments {~d~=l of equal length and denote by i = 1,2, the part of the integral sum Si corresponding t.o the segments of the partition that lie in ~k' Then Si = L~=l i = 1,2. If now for each of the segments t.k we choose in arbitrary manner either the slim st or the sum s~, we obtain 2N different. integral sums of the function f. These sums ca.n be formally written as

SL

st,

N

N= )~ " ( -2-SIr 1 + Ok I 1 - Ok Sk2) + -2.

S ({Oklk=l

where Ok = ±l are arbitrary. We claim that among these sums there is one that lies close enough to + X2). Indeed,

HXl

122

APPENDIX.

LIMIT SET OF RIEMANN INTEGRAL SUMS

Using inequality (1) and the relation

IISLII ::; M(f)IN,

1 min 11-2 (Xl +X2) - S ({QkH'=I)II::;

Qk=±1

E:

we obtain

+ CN~-l M(f).

Since E: can be made arbitrarily small and N arbitrarily large, the last relation means precisely that the point ~ (Xl + X2) can be arbitrarily well approximated by arbitrarily fine integral sums of the function f. 0 We wish to emphasize an important feature of the above theorem: we have shown that the set I(f) is convex, but nothing was said about whether this set is nonempty. This question will be addressed in Section 3 of the Appendix.

§2. The Example of Nakamura and Amemiya In functional analysis one often encounters theorems that generalize some or another result of classical analysis to functions with values in a Banach space. For some theorems of analysis such a generalization turns out to be possible for functions taking values in any space, regardless of the structure of that space. In other cases, however, whether the generalization is possible depends on the geometry of the space. It is precisely this situation that is of interest in Banach space theory, since any such theorem of analysis becomes a tool for investigating the structural properties of Banach spaces. From this point of view, the example constructed in the aforementioned paper 110] of a space for which Theorem 1 fails is of exceptional importance. This example, which will be given here is a somewhat modified form, yields the "limiting case" of nonconvex set: for the function f in question, I(f) consists of exactly two points. Recall that IdO,IJ is the space of real-valued functions defined on the segment [O,IJ and such that E'E(O,lllg(-Y)1 < 00 (from this condition it follows, in particular, that the function 9 takes nonzero values for at most countably many points). The norm in It/O,IJ is given by the formula 11911 = L,E(o.lllg(-Y)I. The unit basis vectors of the space idO, IJ are defined to be the elements et given by et(-y)

={

I

if'Y

=t

° if'Y =I t.

Then lIedl = 1 for all t E 10, IJ. Any element 9 E IdO, IJ can be written in the form 9 = E~I aket., and 112:~=1 aket.1I = 2:~llakl· EXAMPLE 2. For any n E N consider the finite subsets En and Gn of the segment \0, 1J defined as follows: En consists of the 2" points of the form

_k_ ±

?n-l

v'2 ~n'

n

k=O,1,2, ... ,2

1 -,

§2. THE EXAMPLE OF NAKAMURA AND AMEMIYA

123

while G n consists of the 2n points of the form

_k_± J2 n 2 -1

IOn'

k=0,1,2 .... ,2"-1

(we have 2n and not 2n + 1 points, because two of the points of the indicated form fall outside 10,11). Ii Define the function f: ldO, IJ -+ ldO, IJ by the rule if t ~ (Un

et

=

f(t)

{

eo

+ .}. l:aEE.. ect -

el

+ 21.

et

LaEC" ea - e,

E.. ) u (Un G,d

if tEEn if t E G n .

Clearly, f is bounded and Af(f) :5 3. If we partition the segment 10,11 into 2n equal segments (call this partition r .. ), then each of them cont.ains exact.iy one point from En. It follows that

Similarly. 5'(f,rll,G n ) THEOREM 2.

sists exactl'y of the

= el.

Hence, the points eo and ('I belong to I(f).

For the function f COIlRtructed ill Example points eo and el·

2

the set lU)

COIl-

t.\\'O

PROOF. Since any unit vector ea with Q f:. 0,1 appears in the expression of only finitely many values of f, the expansion of an arbitrary x E I(J) may contain only the vectors eo and el. In other words, any element x E I(f) has thl? form ~eo + P.el. We need to show that either ~ = 1 and JJ. 0, or >. 0 and it = 1. Suppose that some sequence S(f, ~., T~) of integral sums of the function f, with lim,,~'X) d(r~) 0, converges to >.eo + liel· Passing to a subsequence (r 'I, T,,) of the sequence (r~, T~), one can ensure that the integral slims S,. S(f. r n, T,,) will obey the inequalities

=

=

=

=

(2) Let An denote the set of those segments Ak in r" for which t·k ¢ (Un E")u(U,, G,,). From the definition of the norm in the space I dO, 1J it readily follows that

2-

2n

2: IISn - >.eo - t,e] II 2:

II L e,. Ale I = L A~

Hence, LAtEA .. ;\11 thp

tprm~

IAlcl -+ 0 when n

,..1"\,. ..o" ..... ,U'"A:~ .......... L_

EA..

IAA:I·

A. EA ..

~ 00. Therefore, if we discard from the sum --_...

"

_.

•

....

S"

124

APPENDIX.

LIMIT SET OF RIEMANN INTEGRAL SUMS

that we require to be satisfied for the sequence of integral sums are preserved. Further, if from the integral sum Sn we discard all the terms corresponding to the segments that contain points from Ele or Gle with k ~ n, then the sum does not change by more than M(f)2 n+ 1d(rn) < 3· 2- n+1. Hence, the described procedure ill not affect the limit relations. W A segment in the partition r n will be said to be "squeezed" if it lies between points of the form j

a = 2n +I<-1

-

1 4n+lc

1

j

and b = 2n+Ie -

1

+ 4n+1e

0:

for some j and k. The totallen~h all squeezed segments in r n do~ not exceed 2-n+2. For this reason from thIs pomt on we shall assume that the mtegral sum does not contain terms corresponding to squeezed segments. After all segments listed above are discarded, inequality (2) may deteriorate and becOme

(3)

where en is some sequence that converges to O. Denote by en,i [resp. Dn,j] the set of indices k for which the points tk in the collection Tn belong to Ej [resp. Gjl· Also, set 00

Cn

U

=

00

en,j,

Dn =

U DnJ . i=n+l

j=n+l

Since the "superfluous segments" were discarded from the integral sum, Sn = !(tle)IAIeI + EkED.. !(tle)IAlel· Writing inequality (3) in more detail and ~ng that the expressions of the form !(tk) - eo with kEen and !(tk) - el :ith k E Dn contain the unit vectors eo and el with the coefficient zero and that for k E Dn and k E Cn these expressions contain different unit vectors, we obtain

E

+

L: [!(tle) - eoJlAkl + L = I~ -

+

I!(tk) -

edlAkll1

kED ..

leEC..

L IAkl1 + IJl- L IAkl1

kEC..

kED ..

II.~ [f(tk) - eollAkll1 + Ilk~. U(tk) - eollAklll·

§2. THE EXAMPLE OF NAKAMURA AND AMEMIYA

125

Consequently,

and (5)

Since the segments Ak with kEen and k E Dn form almost the entire partition r n, inequality (5) implies that

>. + Jl = nlim ~ I~kl + n-oo lim ~ I~kl = l. ...... OO ~ ~ ieEe"

kEDn

It remains to show that A and /.I cannot be simultaneously different from zero. To this end we need to examine more closely the structure of the sets en and Dn. Let us fix n for the moment and define the numbers R(j) as the difference between 2i and the number of elements in the set en,;. Let us write in more detail the first term of inequality (4):

Combining this with inequality (4), we obtain

Let U denote the set of indices j for which R{j) < En2i. By the last inequality,

LIMIT SET OF RIEMANN INTEGRAL SUMS

APPENDIX.

126

which means that the set U is "hig" in I.he sense that

L L

16;1 ~c".

(6)

ir/.U ;EC".)

Since the integral sum involves only a finite number of segments, U is a finite set. Letjo max{j: j E U}. Let us show that the set U \ {jo} is "small" in a sense analogous to inequality (6). To this end let us recall how the set En. was constructed: it consists of 2m points of the segm~nt 10, ~ j, and to the left. a~d to the right of eac~ point of the form k/2 m - 1 there IS a pomt from Em at the distance 8- mJ2. That IS to say, each point of the (or",l k/2 m - 1 is "sq~leezed" bet.ween two points of Em· Let j E U\ {jo}. Then Gn,j consists of 2) - R(}) ~ 2) (1 - ~n) elements and hence it "squeezes" between the points tic with k E Cn,j at least 2}-I(l - 2€n) points of the form k/2n-'. Since all the "squeezed" segments were discarded from our integral sum, R(jo) ~ 2j(1-2~n)' Further, since jo E U, combining the inequality R(jo) < €n 230 from the definition of the set U and the last inequality, we conclude that

=

.

2]0

.

~n

1 _ 2~n ~ 2},

j E U \ {jo}.

(7)

Now let us estimate the sum LjEU\{jo} LkeCn.) 16k l· To this end we shall break it into two sums, denoted EI and E2: tl~e first \resp. second] involves the segments of length smaller !resp. largerj than 2- 30 +2 • Then inequality (7) yields EI <

L: 2:

2- io +2

~ 2· 2io 1 ~;!n

. 2-}0+2

< lO~n'

jElJ\{Jo) keC ... )

Further, each segment 6k fro~ E2 contains at least 16kl· 2)0- 2 points of the set £)0' Consequently, R(jo) 2: 2]0-2E2, and since jo E U, E2 < 4e n · Hence,

2: 2:

lAic I = EI + E2 < 14e".

(8)

jEU\!io} kEC ... )

Combining the inequalities (5), (6), and (8), we conclude that LkECn.,o IAkl is almost equal to A. In much the same way one can show t.hat there exists an index i, for which the sum LkEDn.;1 IA4'1 is almost equal to J.L. Suppose jo 2: jl' Then, using estimates similar to those used in the proof of inequality (8), we conclude that the quantity LkEDn.)1 16kl is small. If jo < j\, n" c 16kl is smal\. This means that either), or J.L is estimated from above the L..kE •. )0 by a quantity that tends to zero when n - 00. Therefore, >.eo + J.Le 1 is equal to either eo or e,0 REMARK 1. By modifying the foregoing arguments one can readily produce a function I such that J(f) consists of any given finite number of points. In this connection one can formulate the follOWing QUESTION. Can any bounded closed subset of IdO: 1] serve as the set I(!) (", ~me function?

§3. SEPARABILITY AND STRUCTURE OF I(f)

127

§3. Separability of the Space and the Structure of J(f)

The example constructed in the preceding section, of a function I for which 1(f) consists of two points, lives in a nonseparable Hilbert space. It is natural to ask whether the construction can be carried over to some simpler, separable space. Let us try to achieve this, modifying the first row of the definition of J(t) in Example 2: for t ¢ (Un En) U (Un Gn ) take f(t) equal to some fixed basis vector, say el/2' The rest of the construction remains essentially the same, the resulting function takes values in a separable space, and 1{f) remains nonconvex: as before, a point of the form >.eo + IJel can belong to I(f) only if >. = I, 11 = 0, or >. = 0, 11 = 1. However, I{f) is no longer a two-point set: it contains "parasitic elements" of the form >.eo + (1 - >')el/2 and Aet + (1 - >.)el/2, 0 $ A $ 1. Below it will be shown that this complication is not a drawback of the construction used, but rather a general feature for functions with values in a separable space. More precisely, we will prove that in the separable case the set IU) is star-shaped (Le., there exists an x E IU) such that AX + (1- >')y E lU) for all y E 1(/) and all >. E 10, In. This will also show that in this case l(f) is not empty. A partition is said to be a refinement of the partition r if all the partition points of r are partition points for

t

r.

DEFINITION 2. An integral sum SU, r, T) is said to be ~-mixed if, for any refinement t of the partition r, any choice of points ti, and any >. E 10,1] one can find a third partition, r", and a collection of points T", such that d(r,,) $ d(r) and

t

118U, f", T.d -1>.8U, r, T) + (I - >')8U, t, T)]II < ~. In what follows we shall denote by IL' Iresp. IL.) the outer Iresp. inner} Lebesgue measure on the segment 10,1]. LEMMA 1. Let X be a separable Banach space, f: {0,1] -+ X a bounded function. Then for any ~ > 0 the fUllction I has ~-mixed integral sums.

PROOF. All values of I lie in the ball B of radius M = M(f) centered at zero. Partition B into disjoint sets {Bi}~l of diameters smaller than or equal to ~/4 (this means that if x, y E Bi for some i, then IIx - yll "$ ~/4). The existence of such a partition of B is guaranteed by the separability of t.he space X. Consider the preimages J- I (B;). This is a system of disjoint, generally speaking non measurable sets, which cover the segment 10, II. Choose a number n such that the outer measure of the set A = U~=I I-I (B k ) is larger than I - ~/(8MT/.). Cover AI :: l (Bd by a countable set of disjoint segments whose t.otal length equals I,·{A,) + e/(8Mn). Choose among these a finite number of segments ~I" .. ,~ml whose total lengt.h l I~il is larger than 1l"(AtJ· Denote the part of the set (B 2 ) that. is not contained ill U~l~" by A 2 · Clearly, I(A 2 ) C 8 2 . Proceeding with A2 in the , . ,. "'_. c-:.n ... ", ............. -. ................ .J:J ..• :a.L

r

L::\

r

A

128

~·(A2) <

APPENDIX.

m2

2:

i=m+l+l

LIMIT SET OF RIEMANN INTEGRAL SUMS

IAil < ~·(A2) + 8~n'

Make this selection 50 that all segments ~i with i E [1, m2) are disjoint. Repeat inductively this process n times, where at the jth step the set Aj is defined as /-l(Bj)\U;,;jl ~Ic' This yields a set of segments {~Ic:}~l' oftotallength smaller that 1 - f./(8M), such that for each k there exists a number j(k) E 1, ... , n for which m"

I>· (~Ic: k=l

\rl(Bj{k»)) <

8~'

Next, adjoin to {~k};:';l a finite number of segments {~k}:=m .. +l so that the system {~Ic:}:=l will be a partition of [0, I). Then E~=m"+ll~lcl :5 f./(8M). Choose j(k) for k > Tnn in arbitrary manner. Then N

Lit. (~Ic: \rl(Bj(Ic:»)) < 4~'

(9)

Ic:=l

Denote the partition formed by the segments ~Ic by r. Now pick a collection T of points tic: E ~k as follows: if ~k n rl(Bj{k» -::f:. 0, take tk E !-l(Bj (k», while if the above intersection is empty take for tic: any point in ~k' We claim that S(f, r, T) is an E-mixed integral sum. Indeed, let t be a partition of [0,1) into segments Ci , with t a refinement of r, and let t be an arbitrary collection of points ti E Ci • Then SU, t, T) = E !(ti)ICil. Let us construct the needed partition r.\. To this end we replace each segment Ci by a segment C; C Ci such that IG;I = (1 - >')ICd and ti E Each set ~k \ Ui C: consists of a finite number of segments Rj • Now let r,\ be the partition consisting of the segments and R;. Obviously,

C:.

2: C:Cbk

IG;I = (1- >')IAkl,

c:

L

IRjl = >'I~kl·

RjCbk

.

The needed collection - of points T.\ is selected as follows. In each segment C! take the same point ti that we took in Ci • Let Uk denote the set of indices j for which Rj C Ale: and Rj n,-1 (Bj(k» -::f:. 0. For j E Uk pick a point e Rj that also lies in r1(Bj(k»' whereas for j ~ Uk pick an arbitrary E R j • By (9),

t;

t;

(10)

A~D

§3. SEPARABILITY

STRUCTURE Of' l(f)

129

Now let us write the constructed integral sum in detail: .tV

S(j.rA,TA} = Lf(t;)IC:1 + N

= (l- )")S(I,t, T) + L

L l: f(t;lIRjl + L

L

L

f(tj)IRjl +

f(tjllR;1

=

f(tjllRjl·

iEUt'... Uk

k=1 jeUk

By inequality (10), we have

/18(1, r A, TA )

~

~

t. [;~.

t (L k=\

-

>..5(1, f, T) - {l - )")S(I, t, T)II

f(tj)lRjl- )"f(t k )l.1 kl )

(I(t;) - f(t.))

$

+M

jEl'k

IR,I + f(t.)

t. [;~y(t;) -

(,E

f(t.))

Since f(t,,) and I(t;) lie in Bj(k) whenever j E is not larger than €/4, one has the bound

e:

IR;I- >'IA.I) ] +-4

IR;I] +

Uk,

4~f

~

and since the diameter of

Bj(k)

whence

IIS(f.rA, T A )

-

)"S(f, f, T) - (1 - ),,)5(/,1\ f) II

< E.

o

as claimed. REMARK 2. The above construction shows t.hat t.he part.ition can be taken as fine as one wishes.

r

in Lemma 1

LEMMA 2. Let X be a separable space, f: 10,11 -> X a bounded function. Theil tllere exists an element x E I(!) such that. for any e > 0, x b~ "r/Jit.rarify well approximated b.y £-mixing integral sums of f that correspond to arbitrarily fine partir.iolls of tile segment [0,11.

PROOF. Set M = M(f) and pick a sequence En '\. 0 such that L::=1 En < 00. We will construct by induction a sequence of integral slims Sn SU, f", Tn) with

=

130

APPENDiX.

LIMIT SET OF RlEMANN INTEGRAL SUMS

the following properties: (1) 8 n is an en-mixed integral sum with d(rn) < en, and (2) 118n+1 - 8n ll $ en for all n. Once this construction is successfully carried out, condition (2) guarantees that the sequence 8n converges to some x EX, and since limn-'oo d(f n) := 0, it follows that x E I(f). Now to the induction process. Construct the sum SI 8U, f 1, T I ) by repeating the arguments of Lemma 1 with E EI, taking care that the inequality d(fl) < El will be satisfied (which is possible according to Remark 2). Then the sum 51 will be EI-mixing. Let N, Bi, AI:, and j(k) be as in the proof of Lemma 1. Then (9) holds with E EI· To obtain the partition r 2 , we proceed by analogy with Lemma 1 for each individual segment AI:. Thus, partition each set Bi into a countable number of subsets of diameters no larger than E2/4. Reindex these sets and denote them by Choose the new partition of (0,1], consisting of the segments {AD~ I' so that it will enjoy the following property: for any index i one can find an index j'(i) such that

=

=

=

B:.

r2

N'

I>. (A~ \ r 1(B}'(i») < 4E~,

(11)

i=1

and for all indices i for which a~ c AI: the set B},(;) is "almost always" a subset of B'(I')' Here "almost always" is understood in the following sense: if HI: denotes the ~t of all i for which a~ c A k , but Bhk) i. Bj(k)' then N

I: L IA~I < 4~' k=lieH.

(12)

That this can be achieved follows from inequality (9). Now choose the collection T2 of points ~ so that whenever A~ n r l (Bj'(i» -I- 0 the point ti belongs to

rl(Bj'(i»' Condition (12) and the fact that d(Bj(k»

< El/4

yield

118(J,f},Tt} -s(J,r2,T2)II < EI,

=

and inequality (11) shows that 52 S(J,r2,T2) is an E2-miXed integral sum. Next, carrying out. the construction of Lemma 1 on each segment of the partition r2, we produce an E3-mixed integral sum S3 such that 1153 - S211 $ E2· Continuing in this manner, we obtain the desired sequence {8n }. 0 Notice that the last lemma shows, in particular, that for functions with values in II. separable space the set I(f) is nonempty. This result was first obtained by H. W. Ellis {II. Repeating the construction of Example 1 for the space '1{0, 11, it can be readily verified that for arbitrary nonseparable spaces Ellis' theorem fails. Nevertheless, as the results of the following section imply, for certain nonseparable spaces (for instance, nonseparable L" spaces) Ellis' theorem does remain valid. In this connection there arises the following question, which is still open: How to characterize the spaces for which the assertion of Ellis' theorem holds true?

§4. CONNECTION WITH THE WEAK TOPOLOGY

131

THEOREM 3 (V. M. KADETS [7]). Let f be a bounded function on the segment [0,1] with values in a separabJe space X. Then IU) is a star-shaped set. PROOF. Let x E IU) be the element provided. by Lemma 2, i.e., x can be approximated bye-mixing integral sums of f. We claim that IU) is star-shaped with respect to x. Indeed, let y E IU) and >. E (0,1). We need to show that >.x + (1 - >.)y E 1(f)- Take an arbitrary e > 0 and pick an (e/4,-mixed integral sum S% S(f, r, T) such that d(r) < e and IIx - S% II S; e/ 4. Let n be the number of segments in the partition r. Pick an integral sum SII = S(f, r l , TI ) such that d(rl) < e/(SnM(f» and lIy- SIIII S; e/4. Add to r l all the endpoints of the segments of partition r and denote the resulting partition by On the segments of that coincide with segments of r I preserve the choice of the points ti E T I , while on the remaining segments choose the points in arbitrary manner. Denote the resulting collection of points by t. The total length of the segments from r 1 that contain no endpoints of segments from r is no larger than nd{rd < e/{SM(f)). Hence, the integral sum 811 = S(f, T) satisfies /lSII- 811 11 S; 2M(f) 'e/(SM(f) = e/4 and lIy - 811 11 S; £/2. Moreover, is a refinement of r. Therefore. since the sum Sz is (e/4)-mixed, there exist a partition r A and a collection of points TA such that d(r A) S; d{r) S; e and IIS(f, r A, TA) - >'Sz - (1- >')811 11 S; e/4. Thus, we have produced an integral sum SA = S(f, r A, TA) with the property that

=

r.

r

r,

liSA -

>.x - {1 - >')yll S;

liSA -

r

>'Sz - (1 - >')511 11 + >.lIx -

e

e

szll + {1 - >')lIy -

slIlI

e

< - + - + - =e. - 4 4 2 Thanks to the arbitrariness of e, this means that >.x+ (1- >.)y belongs to l(f), as claimed. 0

§4. Connection with the Weak Topology DEFINITION 3. We say that Ban~ space X belongs to the convex class (and write X E CONY) if for any bounded function f: [0, 1J -+ X the set IU) is convex.

In §l we have shown that the B-convex spaces belong to CONY, whereas, as it became clear in §2 and the beginning of §3, the space ll' and hence all the spaces that contain an isomorphic copy of ll. do not belong to CONY. One is led to the problem of giving a full characterization of the class CONY. This problem turned out to be quite challenging, and a complete solution is not yet available. Moreover. we do not know whether such a classical space as Co belongs to CONY. However, substantial progress on the path to a full description of the class CONY can be achieved by bringing into consideration weakly convergent sequences of integral sums. DEFINITION 4. The set of weak limits of Riemann integral sums of a bounded function f: (0, 1\ -+ X is the set WI(f) of all points of X that are limits of weakly convergent sequences of integral sums correspondine: to finer ancl finpr nutiti",.,.,

APPENDIX.

132

LIMIT SET OF RIEMANN INTEGRAL SUMS

DEFINITION 5. We say that a space X belongs to the weakly convex class (and write X E WCONV) if the set WJ(f) is convex for any bounded function f: [0,1] ..... X. Obviously, 1(1) C WI(f) for any function f· In this section we will show that actually for l11 any spaces X the sets I(~) and WI(f) coincide for anyfu~clion f' 10 1] -+ X with M(J) < 00. We call this property of a space the comctdence . 'rty. and use the notation X E COIN. Using the complete description of the prope WCONV that will be given below and the coincidence property, we shall c aSS that X E CONY for a wide class of spaces X. All the results presented here prove belong the V. M. Kadet.s .15, 61·. . We next turn to the mvt'.stlgatlon of the structure of the set WI(f). We begin b listing two elementary propert.ies: if A is a convex set and I: 10,11 ~ A, then ~l(J) c A; if X is reflexive and the fUllction I: 10,11 -> X is bounded, then

Wl(f~ ~!~ction 1 will be said to be weakly Riemann-integrable if any sequence

f' legral slims of 1 corresponding to increasingly finer partitions is weakly cono Inent It is obviouS that for slich a function 1 the set WI (f) reduces to a single verg . point. THEOREM 4. In order tllat the Banadl space X be reflexive it is necessary and sullicient. that /lny boun~t'(/ (unc~ioll I: [0, 11 -+ X (or which WIU) consists of Ii single point be weakly Riemann-mtegrable. PROOF'. NECf,sSITY. Suppose that X is reflexive and WI(/) consists of a

.
~~r

)__

SUFFICIENCY. Suppose X is not reflexive. Then according to a result in

[91 one can choose a basic sequence {ek} k"= 1 of norm-one elements of X that is se~arl\ted from ~ro by a common !inear fUllctional x' EX': inf{x'(ek)} = a> O.

=

Define the function I: 10.11 --> X as follows: 1(1/2) = el, f(1/4) = f(3/4) !(1/8) f{3/8) = f(5/8) = f(7/8) = e3 , ... , f«2k - 1)/211 ) = en for ~2~ 1,2,.",2n-I, n E N, and f(t) = 0 for the remaining values of t. Suppose that some sequence Sn = SU, r n, Tn) of integral sums of f for which d(rn) ~ 0 converges weakly to some element x = L::'l (lkek· Since for each element ek the coefficient. wit.h whirll it may appear in Sn is not larger than 2k d{r n), we see tha.t 0 for all k. Therefore, WI(f) = {O}. On the other hand, if we take the /t~tiliOIl I"n obtained ~Y dividing /O,lJ into 2n- 1 equal parts and the collection ~ consisting of the pomts tk = (2k - 1)/2n, then S(J, r~, T~) = e". Since the ~uence {en} does not covel'ge. weakly to zero, the constructed fUllction f is not _'Mltlv Riemann-integrable, which completes the proof of the theorem. 0

=

=

§4. CONNECTION WITH THE WEAr; TOPOLOGY

133

THEOREM 5. The following two conditions are equivalent: 1) X does not contain an isomorphic copy ofl l ; 2) X E WCONV. PROOF. As we have observed earlier, modifying slightly the construction of Example 2 one ca.n built a bounded function g: 10,11 -. h for which I(g) is not convex. In II every weakly convergent sequence converges strongly, and so WI(g) = l(g). Therefore WI (g) is not convex and II rt WCONV. Since the prdperty WCONV is obviously inherited by subspaces, from the assumption that X E WCONV it follows that X does not contain isomorphic copies of II' Now let us prove the implication 1) ~ 2). Let. f: [0,11-- X be a bounded function and x, y E WI(f). Pick an arbitary >. E [0, IJ. We need to show t.hat >.x + (I - >.)y E WI(J). Suppose S(f, r n, Tn) -. :r. weakly, d(rn) -- 0, S(f, r~, T~) -> x weakly, d(r~) -> O. Further, let. {X;}bl be arbitrary elements of X'. Define the operator U: X __ l~n) by the rule Ux = (xj(x), ... ,x:n(:r.)). Consider the auxiliary funet.ion 10,1\ -> I~l given by 7(t) = U f(t). Since the operator U is continuous, Ux and Uy belong to 1(f). Next, since the space l~n) is finite-dimensional, the set 1(/) is convex. Consequently, there exist a partition r~ and a collection of points T~ such that S(f, r~, T~) -+ U(..\x+(l-"\)y), d(r::) -- 0, and T~ c U:I (TnUTn) (that t.he last condition can be satisfied follows from an examination of the proof of Theorem 1). Recalling the definitions of U and of the norm in I~), we obtain

7:

lim

(max Ix; (>'x + (1 -

n-oo I::;.:;:m

..\y) - S(J, r~,

T;nl) + d(r~) = O.

(13)

Consider the auxiliary Banach space X consist.ing of the pairs (x, d), where I EX, d E R, with the norm lI(x, d) II = IIxli + Idl· Thanks to the arbitrariness of the functionals xi, relation (13) means that (>.x + (1 - ~)y, 0) is a limit point in the weak topology for the set V C X of the points of the form (SU, r, T),d(r)) with T c U:"I (Tn U T~). The last condition ensures that the set V is separable. By a theorem of H. P. Rosethal 113], in a space that does not contain a. copy of II, the weak closure and the weak sequential closure of a bounded separable set coincide. Hence, there exists a pair r:', for which the sequence (S(f,r'~',r,:'),d(r~'») converges weakly to (>'x + (1 - ..\)y, 0) in X. But this means that SU, r:', T:a") -+ AX + (1- >.)y weakly and d(r~') ...... 0, i.e .. >.:1' + (I - >.}y E WI{f), as needed. 0

T:t

Thus, we have obtained a complete description of the class WCONV. Now our task is to study the coincidence property. Clearly, spaces which, like II, have the Schur property (i.e., in which weak and strong convergence of sequences coincide), belong to the class COIN. Our immediate objective is to show that the spaces that are the "fart.hest." from ll' namely, the B-convex spaces. also belong to COIN. The following assertion, which is of independent interest as well, will play here an auxiliary role.

LEMMA 3. Suppose the Banach space X l,as infratype p > 1 with constant C > O. Let {Ad:':1 be an arbitrary fillite collectioll of bounded subsets of X, d i =

APPENDIX.

134

LIMIT SET OF RIEMANN INTEGRAL SUMS

diarn Ai, i = I, ... , n, and Jet {b i } r=l be points such that bi E cony Ai, i = 1, ... , n. Then one c8.11 choose points ai E Ai, i 1, ... n, such that IIL~l (ai - bi )1I :5 "~ d':)I/P where C = 2CLoc:. 2(~-I)t. 1 (

=

CI

lJ.=I'

t-I

'

Using the assumption that bi E conv Ai, let us choose numbers 0= mo < ml < m2 < ... < m.., elements {Xt};;':l' and numbers {At};:':.. with the foJlowing properties: Xle E Ai for ffl;-1 < k :5 ffl;i At > 0 for all k; PROOF.

m,+l

L:

mi+1

At

L

= 1;

AleXic

= bit

i

=0,1, ... ,n -

1.

le=m,+1

le=m,+l

Our proof will rely on the same idea (borrowed from the paper [111) as the proof of Lemma 2.3.3. Specifically, we shall assume that all Ale are finite dyadic fractions (this assumption does not result in a loss of generality, since it can be realized by slightly perturbing the numbers Ale). Let N be the largest length of the dyadic decomposition of the numbers At. Then each Ale admits can be written as At = EN Ell: ,2-', where the ek,1 are equal to 0 or 1. I:=OLet k(l) < k(2) < ... < k(r) be the numbers for which Ek(I),N = 1. Since t"'mHI All: = I, the number ri of indices k(l) for which mi < k(l) < mi+1 L..k::m;+l is even for any i. For the same reason, rj :5 2N for all i < n. Introduce the auxiliary elements g, = Xle(21-1) - Xle(21), I = I, ... , r/2. Clearly, IIgll1 :5 d;+1 for m; < k(21) < ffi;+1' Since X has infratype p, we can choose signs 0, = ±I such that

r'

~o,g, "C (~IIg,II'

(14)

Now introduce the new numbers {At •.};:':1 as follows: for k ~ {k(l), ... , kern put N All:, 1 =All:, while for k E {k(l), ... , kern put AA:(21-1),1 = AII:(21-1) + 012- and N AII:(2I),1 = AII:(21) - 012- . Then AIe,1 are dyadic fractions of length N - 1 which satisfy the relations AII:,1 ~ 0 and L;;~~+l AII:,1 = 1. Let us use inequality (14) to estimate the quantity IIL;:':1 AleXA: - L:::l Ale,IXt II: r/2

L2- Nol(Xle(21_1) - Xle(21») /=1

r/2

=2- N

L

/=1

O '91

§4. CONNECTION WITH THE WEAK TOPOLOGY

Now apply to Ak,l the same procedure that we applied to the numbers yields dyadic fractions AI;,2 ~ 0 of length N 2 , such that

135

AI;.

This

m,+1

L

).k,2

= I,

i =O,I, ... ,n-lj

k=m,+l

Iterating this procedure N times, we obtain dyadic fractions ).k,N of length zero, i.e., either zeros or ones. Because of the relation E;;~:i +1 Ak,N = 1, we conclude that for k E {mi + 1, .... mi+l} exactly one of t.he numbers AIc,N is equal to one, while the remaining ).k,N are equal to zero. For each i denote by s(i) E {mi + I, ...• m;+ I} t.he index for which AS{i),N 1, and put at = X'(i}' Then

=

o

which completes the proof of the lemma. THEOREM 6. If X has illfratype p > 1. then X E COIN.

PROOF. Let C l be the constant provided for X by Lemma 3. Further, let. f: 10, 1J -+ X be an arbitrary bounded function, AI (f) = M, and x E WI(f). Pick a sequence (rn,Tn ) with d(rn) ...... 0, such that Sn = S(J,rn,Tn ) ..... x weakly. Take an arbitrary ~ > 0 and choose mEN such that 4Mm~-\Cl < ~. Divide 10,11 into m equal segments and denote by S",n the part of the int.egral sum Sn which corresponds to the segments to. that lie completely inside the interval I(k - I)/m, klml. As we already did several times before, without affecting the weak convergence Sn ...... x we may assume that all the points kim, k = 1, ... , m are partition points for r n' Then Sn = E:=\ Sk,n' Now we lise a theorem of Mazur which asserts that for any sequence {y,,} that converges weakly to x there exists a sequence of convex combinations of the elements Yk that converges to x in norm. Pick coefficients Ai ~ 0 such that E~l Ai = 1 and II E~l )..S; - xII ~ f./2. Set b," = )"'!" ~ >.._8 •. _ ~nr4 AI = I~. \N r'1"'" ...1.....,. __ A 00 • •

APPENDIX.

LIMIT SET OF RIEMANN INTEGRAL SUMS

136

of the set Ale does not exceed 2M/n. Using Lemma 3, choose numbers n(k) such tha.t tSIe.n(le) -

Il 11:=1

tblell

$

1e=1

CI2Mm~-1 < ~.

Since 5 == E;'=1 SIe,n(le) is an integral sum (composed of "pieces" of the sums Sn) d since the element E~=1 ble = E~l ).S; is close to x, the last inequality means a~ t IIx _ 511 < t, i.e., x can be arbitrarily well approximated in norm by integral !~ of the function f. Therefore, x E l(f). 0 Thus, we have reached the conclusion that classes of spaces that in some sense opposite (the B-convex spaces and the spaces with the Schur property) belong :eCOIN. Before we give new examples of spaces with the coincidence property, let US shoW that there are spaces that do not have this property. EXAMPLE 3. Consider again the sets En C [0,11 constructed in Example 2. Let t t4 e, be the bijective mapping of the set E =U:=I En onto the set of the unit vectors of the standard basis of the space II, and let {9n}~=1 be an othonormal basis in l2' Define the function I: 10, 1] ..... l\ e 12 as follows:

I(t)

1

= 9n+l + 2n

L eo - et oEEn

if tEEn' and I(t) = 91 if t ¢ E. Obviously, M(f) ~ 3. If one takes r n to be h partition of 10, 1] into 2n equal segments, then each of these segments will ~~tain a point from En. Consequently, S(f, r n, En) = 2~ EtEEn I(t) = 9n+l· Since gn -+ 0 weakly, it follows that 0 ¢ WI(f). On the other hand, arguing much in the same way as in the proof of Theorem 2 we conclude that 0 ¢ IU)· Therefore,

h $h ~COIN. CONJECTURE 1. If X contains an isomorphic copy of 11 but does not have the Schur property, tben X ¢ COIN.

Combining theorems 5 and 6 we recover the already familiar Theorem 1: the class CONY contains all B-convex spaces. To ehxibit new representatives of CONY we shall introduce a new property of Banach spaces, called weak B-convexity, then study it and prove that the reflexive weakly B-convex spaces belong to COIN. DEFINITION 6. A Banach space X is said to be weakly B-convex if there exist numbers n E N and e > 0 such that, for any finite collection {A i }i':l of weakly convergent to zero sequences of elements of the unit ball of X one can find elements {Xi}f=p Xi e Ai, i = 1, ... , n, for which II E~=, xiii ~ n(l - t}.

In what follows, if A C X is a bounded set we will denote sup{ IIx II: x E A} ,. \..•• 11 All

§4. CONNECTION WITH THE WEAK TOPOLOGY

137

DEFINITION 7. A Banach space X is said to have weak type p with constant C if for any finite collection {Ai}~l of weakly convergent to zero sequences of elements of X one can find elements {Xi}~l' Xi E Ai, i = 1, ... ,n, for which

Obviously, a space that has weak type p > 1 is weakly B-convex. Following the scheme used by G. Pisier in [12J to prove that every B-convex spaces has an infratype p > lone can obtain the following result: THEOREM 7. Every

wea.kly B-convex space has a weak type p > 1.

To justify the term ''weak B-convexity" let us show that this notion is indeed weaker than that of B-convexity. THEOREM

8. If X bas iniratype p > 1, then X bas weak type p.

PROOF. Let Cl be the constant appearing in Lemma 3, and let {Ai}i=l be an arbitrary finite collection of weakly convergent to zero sequences of elements of X. By Mazur's theorem on weakly convergent sequences, for any E > 0 there exist elements bi E conY Ai, i = 1, ... , n, such that IIbili < E for all i. Lemma 3 yields elements Xi E Ai such that

Then

Since e is arbitrary, this means that X has weak type p with constant 3Cl .

0

The class of weakly B-convex spaces is considerably wider than that of B-convex spaces: for instance, it contains all the spaces with the Schur property as well as the space CO. Let us give an example showing that among the reflexive spaces there are many weakly B-convex spaces that are not B-convex. EXAMPLE 4. Let Xl"'" X n , .•. be a sequence of finite-dimensional normed spaces. Consider the space Y = CE X n )2' i.e., the space of sequences of the form a = (at. ... ,an , ... ), where ai E Xi, r::ll1a.1I 2 < 00, equipped with the norm

I\all =

(E:lllaiIl2)1/2. Y is a reflexive space. At the same time, a space of this

type is not necessarily B-convex (for example, take Xn = lin». Let us show that Y is necessarily weakly B-convex with n = 2 and any e < 1 - .;2/2.

APPENDIX.

138

LIMIT SET OF RIEMANN INTEGRAL SUMS

Choose a 6> 0 such that V2+ -/21J < 2(I-e}. Let A and B be two sequences of elements of Y that converge weakly to zero, IIAII = IIBII = 1, and let a = (ai)f:l be an arbitrary element of A. Fix a number N such that 2::N+lllaiIl2 < 6. Since the sequence B is weakly convergent, one can choose an element b E B, b = (bi)f:J' such that E~ll1bill < 6 (weak convergence implies coordinate-wise convergence). Then

lIa + bll = (

s

(t, ;t n4;II' +

$

t; lIa + 00

i

Ilbdl') 'I' +

)

1/2

bi 112

Ct t, 114;11' +

lib; II') 'I'

J2+ v'26 < 2(1- e).

This means that the space Y is weakly B-convex. A more careful examination shows that Y has weak type p = 2. THEOREM

9. Let X have weak type p > 1 with constant C. It X is reflexive,

then X E COIN. PROOF. Let f: \0,11 -+ X be a bounded function, MU) = M, x E Wl(f), and let S.. = S(f, r n, Tn), d(rn) -+ 0, be a sequence of integral sums of f that converges weakly to x. As in the proof of Theorem 6, let us divide 10, 1I into m equal segments a.nd denote by Sk,n the part of the integral sum that corresponds to the segments /).i C I(k - l)/m, kim]. Using the reflexivity of X we can choose a sequence of indices nj such that the weak limit w - limj_oo Sk.n, will exist for all k 1, ... , m. Denote w - Iimj_:>o Sk,n, by Xk· Clearly, E:;'=1 Xk = x. Further,

=

since we can choose numbers fiA: such that

11t.(Sk.n

k -

Xk)1I $ C2

~. m

1 p / •

As in the proof of Theorem 6, the quantity 2:;::"1 Sk,Hk can be regarded as some integral sum Sm of the function f. Then the last inequalit.y reads IISm - xII $ 2CMm~-1 and since p > I, we conclude that Iim m_oo IISm - xII = O. This means that x E 1(/), which completes the proof of the theorem. 0 Combining theorems 5, 7, and 9 we see deduce that every reflexive weakly B-convex space has the CONY property. Notice also that the property CONY is inherited by subspaces as well as by quotient spaces. CONJECTURE ()lI~STION.

2. The properties CONY and WCONV are equivalent.

Does the space Co have the property CONY or WCONV?

SUPPLEMENTARY LIST OF REFERENCES

139

Exercises 1. Show that if X is infinite-dimensional, then from the fact that I(f) consists of a single point it does not necessarily follow that f is Riemann-integrable. 2. Show that if f take values in some compact set K eX, then I(f) is a convex set. 3. Show that any convex, closed, bounded subset of a Banacb space with the power of the continuum can serve as the set IU) for some function f. 4. Give an example of a space that is not weakly 8-convex. 5. Let X E WCONV be a reflexive space. Show that any quotient space of X also belongs to WCONV. 6. Using the COIN property, provide an example of a nonseparable space X such that for any function f: \0, 1J ..... X with M U) < 00 the set lU) is nonempty. Supplementary List of References 11J [2J !3J [4J [5J

161

[7J

18J

[9J 1101

!IlJ

H. W. Ellis. On the limits 0/ Riemann sums, J. London Math. Soc. 34 (1959), no. 1,93-100 [MR 20 #7084J. I. Halperin and N. Miller, An inequality of Steinitz and the limits 0/ Riemann sums, Trans. Roy. Soc. Canada Sect. III. (3) 48 (1954), 27-291MR 16, 596aJ. P. Hartman, On the limits of Riemann approximating sums, Quart. J. Math. Oxford Ser. 18 (1947), 124-127 [MR 9, 137fJ. R. L. Jeffery, Limit points oj Riemann sums, Trans. Roy. Soc. Canada Sect. III. (3) 44 (1950), 43-49 [MR 12, 487dJ. V. M. Kadets, Riemann integral sums and the geometry oj a Banach space, Dokl. Akad. Nauk Ukrain. SSR Ser. A (1987), no. 12, 12-14 (Russian) IMR 89d:46018J. V. M. Kadets, The domain oj weak limits of Riemann integral sums oj an abstract function, Izv. Vyssh. Uchebn. Zaved. Mat. (1988), no. 9,39-46 (Russian) IMR 9Ok:46099J. V. M. Kadets, The starness of the domain of limits of Riemann integral sums of a vector-valued function, in: Operator Theory, Subharmonic Functions. "Naukova Dumka", Kiev, 1991, pp. 60-67 (Russian) IMR 93m:260461· V. M. Kadets and M. I. Kadets, Conditions for the convexity of the set of limits of Riemann sums of a vector-valued function, Mat. Zametki 35 (1984), no. 2, 161-167 (Russian) IMR 85k:46010J. D. P. Mil'man and V. D. MiI'man, Some properties of non-reflexive Banach spaces, Mat. Sb. 65(101) (1964), no. 4, 486-497 (RUSSian) IMR 30 #1383]. M. Nakamura and I. Amemiya, On the limits of Riemann sums of functions in Banach spaces, J. Fac. Sci. Hokkaido Univ. Ser. I 19 (1966), no. 3,4, 135-145 IMR 34 #6023J. M. I. Ostrovskii, Domains of sums of conditionally convergent series in Banach spaces, Teor. F'tmktsii Funktsional. Anal. i Prilozhen. 46 (1986), no. 2, 77_QC, IDu ...... !_ ... \ I.,n

nn~

.,.."n,l

APPENDIX. 140

LIMIT SET OF RIEMANN INTEGRAL SUMS

.. r Sur les espaces de Banach qui ne contiennent pas uniformement de

G pisle,

llZ\ (il c. R. Acad. Sci. Paris, Ser. ~-B 27.7 (1.973), ~991-A994IMR 48 #11998). I.. ' R,oSeJlthal, Some recent dsscovenes In the ssomorphic theory oj Banach 113] ~s, Bull. Amer. Math. Soc. 84 (1978), no. 5,803-831 IMR 8Od:46023J.

COMMENTS ON THE EXERCISES

Chapter 1 Ex. 1.1.1. False; for more details, see §I of Chapter 2. Ex. 1.1.2. Let A C N. Denote by p(A) the smallest number s of segments of the form [nl,mll = {nl,nl + l,nl + 2, ... ,ml - l,md such that A can be written as U:=dnj. mil. For a permutation 11': N -+ N denote p(7r) = sup{p(1I'-1!n, mJ): n, mEN, n ~ m}. THEOREM (A. S. KRONROD, 1946). The permutation 11' is not convergencemodifying if and only if p( 11') < 00.

Let us sketch the proof proposed in 1985 by V. M. Ogranovich, one of the participants in the spccial course read that year by M. I. Kadets at Khar'kov University. Let E be the space of numerical sequences e = {III, 112, •.. } for which the series E:lllt converges. Equip E with the norm

llell = sUP{l~lIjl

:n,m

E

N,

n~ m}.

Then E is a Banach space (which, incidentally, is isomorphic to eo). Let D1< be the set of all elements e E E for which the series L:III1«i) converges. Define an operator TlI : D1< -+ E by the rule T,..((1I1, 112, ••• = (111<(1),111<(2),"')' Then T,.. is closed; its domain D,.. contains all the finite sequences and hence is dense in E. Thus, ror DlI to coincide with the whole space E (Le., for 11' to be not convergencemodifying) it is necessary and sufficient that T" be a bounded operator. It remains to show that IIT,..II = p(1I'). Suppose that n < m and the set 11'-1 In, mJ is represented in the form

»

p(lI- l ln.m))

U

=

[nj,m;j, nl:5 ml < n2:5 m2 < .... j=l Consider the element c E E for which 11m; =1, IIm,+l =-1, i=I, .... p(7I'-I[n,mJ), and the remaining coordinates are equal to O. Then lIeli 1 and 71'-1 [n,mJ

=

m

IIT"ell ~ ~1I,,(j) 1

I

p(".-llll,mJ)

=

~

11m ,

1 =p(1T- [n,mj).

COMMENTS ON THE EXERCISES

142

Since the segment [n, mJ is arbitrary, we conclude that liT" II ~ p('n} The opposite inequality liT"" $ p(tr) is an immediate consequence of the definition of the norm

inE. Ex. l.l.3. A permutation is sum-modifying iflim n..... oo p(tr- 1 [I,nJ) = 00 (in the notations of the preceding comment). We cannot offer a "nice" proof. Ex. 1.1.4. Combine the answers to the two preceding exercises. One can

also give a direct proof (M. I. Kadets): Let E and T" be as in the comment to Exercise 1.1.2. Define a functional fEE" by f{e} = E:ll1,. If the permutation is not convergence-modifying, then f 0 T" is a bounded functional. Since f 0 T" = f for all elements with finitely many nonzero coordinates, and since the subset of these elements is dense in E, f(T,,(e» = f(e} for all e E E, i.e., L::111,,(;) = L::ll1, for any convergent series

2::1 11;.

2::'1

EZ=1

Ex. 1.1.5. Let Xk = O. Denote Un = L:Z=1 X2k and Vn = X2k-l· Then limn..... oo(un+vn } = O. For the permutation tr we have E:~I X,..(k) = Un+V2n. Hence, limn_oo(Un + V2n} = x. For the permutation (1 we have 5n

LXO'(k) k=1

= Un + V4n =

= 2:::1

(un + V2n) + (U2n + V4n) - (U2n

+ V2n).

=

Since limn ..... oo(Un +V2n) x, limn_ 00 (u2n +V4n} = x, and lim n.... 00 (u2n +V2n) 0, it follows that limn-oo XO'(k) = 2x. Since the general term of the series tends to 0, we conclude that the series E~=1 XO'(k) converges and its sum is equal to 2x (V. M. Kadets). Ex. 1.3.2. Show that the closure ofthe linear span in Ll [0,1] ofthe sequence rn(t) = sign sin(2"trt) is isomorphic to 12 (use Khinchin's inequality from §4 of Chapter 3). Using this fact, transfer Example 1.3.1 to L1 [0, 1). In the space Lp[O, 1) with 1 :5 p < 00 one can argue in the same manner; Loo[O, 1] and C[O, 1] also contain isomorphic copies of 12. The most difficult case is that of the space 11. Chapter 2 Ex. 2.1.1. Modify the method of choosing the coefficients 8; for which 8; =F Ex. 2.1.2. By assumption, there are permutations tr, (1, and sequence of numbers n(l) < n(2) < n(3) < ... , m(l) < m(2) < m(3) < ... , such that

143

COMMENTS ON THE EXERCISES

Using this fact, choose sets of indices Ml C Nl C M2 C N2 C ... for which

Consider the sets Rl = Ml U(M2 \Nil, R2 = M3U(M4 \N3), R3 = MsU(Ms \Ns), ... , Rn = M2n - 1 U(M2n \N2n-l), .... Clearly, Rl C R2 C R3 C ... , U:'l Rt N, and

=

X-

E

Xi

1

L

+

iEM2n-l

Xi

iEN2n_l

1

1

<--+--+- 22n - 1 22n-1 2 2n '

(1)

Define a new permutation v as follows: write first the elements of R 1 , then those of R2 \ R 1 , R3 \ R2, and so on. Denote by Ik the number of elements in Rk. Then E!:'1 X,,{i) = EiERt Xi, k = 1,2,3, ... and (1) implies that

Ex. 2.1.3. See, e.g., 130j. Yet another proof of Steinitz's theorem in this simplified formulation is based on P. Levy's idea ISO} of considering the set of "limit directions of a series." A correct exposition can be found in Steinitz's classical work or in D. O. ShklyarskH's paper [85}. Ex. 2.1.4. The characterization is that the following two conditions are simultaneously satisfied: a) limr._oo Xk 0; b) for any functional J both series E~.If(xr.)J+ and E::llf(xr.)t .either converge, or diverge. In the infinitedimensional case the situation is quite different (see 142)).

=

Ex. 2.2.2. In Exercise 2.2.1 each of assertions 1) and 2) is equivalent to the space being finite-dimensional, independently of the completeness assumption. In contrast to Exercise 2.2.1, the present problem is quite delicate. We only know a proof based on Dvoretzky's Theorem 6.2.1. Ex. 2.3.1. Let {er.}f..l be an orthonormal sequence in a Hilbert space. Then the needed example is provided by the series 1

el - el

+ ,tt2-

1

1

,r/2 + v'3e3 -

1 v'3e3

+ ....

If here one selects a large group of terms with the sign +, then one with the sign -, then another large group with the sign +, then one with the sign -, and so on, one obtains a perfectly divergent series.

COMMENTS ON THE EXERCISES

144

Chapter 3 Ex. 3.1.1. See Theorem 6.4.1. There also exists a proof that does not resort to the theorem on basic sequences. Ex. 3.1.2. The series converges absolutely at all points x < 1, but converges conditionally at x = 1. Ex. 3.1.4. This result was obtained by N. N. Bogdan in her MSc Thesis (Zaporozhye University, Ukraine, 1992, not published). Ex. 3.1.5. On the unit sphere of the space Y pick a countable dense set of elements {ydf:I' Let L~I E:i be an arbitrary conditionally convergent numerical series. The set of terms of the sought-for series is the union of the following sets:

{E:kydf:l'

{~£kY2}f:I""

1 { n E:kYn}f:I' .... 2

The reader will encounter no difficulty in proving that for this series it holds that

SR([~I

Xk)

= Y.

Ex. 3.1.6. A weakly convergent sequence in L2 may diverge at every point: the Rademacher functions rn take the values ±1, but the weak limit is identically equal to zero. To solve the exercise one should use precisely such sequences of partial sums that "converge in the mean." Ex. 3.1.8. Such examples were constructed for the first time in P. A. Kornilov's PhD dissertation. A proof that works for arbitrary spaces is given in Chapter 7. Ex. 3.1.9. See

I90J·

Ex. 3.2.1. See P. A. Kornilov's paper [48J or V. M. Kadets' paper 140J. The results were obtained at about the same time and independently, but Kornilov's work was published earlier. Ex. 3.2.2. LPR has a "group structure" (H. Hadwiger [97]): if x, y, z E LPR, then x + y - z E LPR. In other words, LPR is a shifted group with respect to addition. The proof is similar to the comment on Exercise 2.1.2.

If one considers only series whose general term tends to zero, LPR has another interesting property (Y. M. Kadets): if LPR contains a subspace HI of finite codimension, then LPR itself is a subspace. PROOF. In view of the first part of this comment, it suffices to show that for any a eLPR\H 1 the entire line Aa, A E R is contained in LPR. Consider the canonical projection T: X - X/H\. If 1 Xi is the original series, then L:~ T(x,) is already a series in a finite-dimensional space. From Steinitz's lemma it ';"~i1Y follows that SR(E:I T(xd) = LPR(E:1 T(xd)· Consequently, the 'f(x;) has a rearrangement that converges to 0, as well as one that

2::

'-- ,,00

COMMENTS ON THE EXERCISES

145

converges to T(a), and, by Steinitz's theorem, there exist a permutation 7r of the terms for which the series L~I T(X"(i») converges to T(>.a). Hence, since HI C LPR(L:'I x,), the element >.a can be arbitrarily well approximated by partial sums of the form LiEA Xi; moreover, the index set A can be selected to contain any prescribed finite set of indices. This means that {>.a: >. E R} C LPR We cannot provide a precise characterization of the sets that can serve as LPR for a series whose general term tends to zero. Chapter 4

Ex. 4.1.1. Proceed by analogy with the Dvoretzky-Rogers theorem, but instead of the inscribed ellipsoid of maximal volume consider the cir(:umscribed ellipsoid of minimal volume. Ex. 4.2.3. Consider series of functions with disjoint supports. Ex. 4.2.4. Yet another proof of this fact can be found in §2 of Chapter 8. Ex. 4.2.5. Consider the operator T that maps the unit vectors of the standard basis of II into the functions Tn: Ten Tn· Show that T is an isometry.

=

Exs. 4.3.8 AND 4.3.9. For a detailed exposition of the theory of nuclear spaces consult the monograph 1691. Chapter :;

Ex. 5.1.2. Take X

= h and Y = co·

Ex. 5.1.3. Show that the parallelogram equality holds in X. Ex. 5.1.4. Let S be the set of all two-dimensional normed spaces, equipped with the metric Ind(X, V). Let {Ai}~t and {Bi}~t be two dense sequences in S such that the unit sphere of Ai Iresp. B i ] is a polyhedron Iresp. a smooth curve] for all i ~ 1. Then X = (L~t Ai }2 and Y = (L:t Bi )2 provide the needed spaces. Ex. 5.1.5. See 1521. Ex. 5.1.6. Represent the unit ball of the space X as the intersection of a finite number of half-spaces. Correspondingly, the norm in X can be written as IIxli = maxl
= nH-~1

(see Ill], 128]).

COMMENTS ON THE EXERCISES

146

Ex. 5.l.10. These metric compact spaces Mn are known as the Minkowski

compacta. An interesting question is what is the diameter dn of Mn ? E. D. Gluskin has shown that log n - dn ..... 0 as n ..... 00. Ex. 5.1.11. Choose a countable set .A of finite-dimensional subspaces of

X

'th the property that, for any finite-dimensional subspace V of X there are of A arbitrarily close to V in the Banach-Mazur distance. Now take eeme I WI nts

y :::: Lin UAE.A A. Ex. 5.1.12. See 186J.

Ex. 5.2.3. Take

X= (E:=II~:"k)2' where the numbers kn are large enough

(say, len :::: 2n). Ex. 5.3.2. If the space ~ is not C-convex, then CO f I"ft I1-+"'"

We conclude that

!.. x. By Corollary 5'.1.1,

II . . . X, and hence Xis not B-convex.

Ex. 5.3.3. One can actually prove the following more general fact: any sepble Banach space is a quotient space of 11. Here is a sketch of a proof: ara Let X be separable. Choose a dense sequence {Xdr:l in S(X). Define an tor T: 11 -+ X by the rule Tei = Xi, where {ei}r:l is the standard basis of 0, per~en UTU == 1 and T is a surjection. Consequently, II JKer T X.

=

I'

Ex. 5.3.4. See 173J. If you find a relatively elementary proof, then you should

publish it.

Ex. 5.3.5. This is analogous to the corresponding result on the M-cotype. Ex. 5.3.6. Show that the spaces X and ' 2 (X) have the same type. Ex. 5.3.7. X and h(X) have the same cotype. Ex. 5.3.8. See 113, Chapter 3J. Ex. 5.3.9. This is a theorem of M. I. Kadets 132J; see [13, Chapter 3J. Ex. 5.3.10. See 113J. Chapter 6

Ex. 6.1.2. Use the fact that for a convex function f the set {(x, t): x E

X, t E R} is convex. Apply the Hahn-Banach theorem. EXS. 6.1.3-6.1.7. See 113J. Ex. 6.2.1. If X is an infinite-dimensional subspace of eo, then X contains a uenee of elements from S(eo) that is arbitrarily close to being disjoint. Therefore ~ntains asu~pa.c~ iso~orph~c to eo, and hence X ~s not reflexive. An analogous rlion (every mfirute-dlmenslonal subspace contams another subspace that is :orphiC to the whole space) holds true for the space I" with p < 00, p '# 2.

COMMENTS ON THE EXERCISES

147

Ex. 6.2.2. See 199]. Ex. 6.3.1. In Co and lp the standard basis is a basis; in CIO, I] the Schauder system is a basis; and in L"IO, IJ the Haar system is a basis. For more details see 155, Chapter 3, §6}; the most fundamental exposition is found in 183]. Ex. 6.4.1. Sufficiency is an obvious corollary of Riemann'!i theorem. Necessity follows from the Banach-Steinhaus theorem.

Ex. 6.4.2. No. If {en} is the standard basis in Co, then a counterexample is provided by the series

Ex. 6.4.3. One needs to show that every weakly unconditionally convergent series in a Banach space converges unconditionally in norm. The proof can be carried out by reductio ad absurdum. Suppose Xi is weakly unconditionally convergent but diverges in the strong topology. By analogy with the proof of the implication (a) :} (d) in Theorem 6.4.3, one obtains series Yn and E~l Zn which, in addition to the properties from the proof of that implication, are also weakly convergent. Since the sequence {zd is equivalent to the standard basis of Co, we reach a contradiction.

E:'t

E:'t

Chapter 7

Ex. 7.1.1. Take >'i = ~. Then x - E~1

= E~I ±~Xi'

B,Xi

Ex. 7.1.2. The following two assertions are equivalent: (1) For any perfectly divergent series L~I Xi in X the series [:1 diverges. (2) There exists a C

x,

> 0 such that, for any n

min max v.=±IISA:Sn

lit I ViZ;

i=1

$ C

E N and any collection

IIxillP

{Xi}

'=

1

C

(tllx.IIP)IIP ;=1

Ex. 7.2.2. Define an operator T: X .... X by the rule Te, = Yi. Then T is bounded; moreover, liT - III < 1 (here I is the identity operator). Hence, T is an isomorphism and since the image of a basis under an isomorphism is a basis, we conclude that {Yi}~1 is a basis. Ex. 7.2.3. First prove that this is true for Hilbert spaces, and then use Theorem 7.2.2. Ex. 7.3.3. Use the fact that in this case lp the list of results on type and cotype)

:&

X, and hence L"

:&

X (see

COMMENTS ON THE EXERCISES

148

Cbapter 8

Ex. 8.1.1. Use Theorem 7.2.2 and the fact that a strongly continuous operator is weakly continuous. Then the example of a series in 12 for which {O, tL} c SR, but f/:. WSR can be carried over to any infinite-dimensional Banach space.

!

Ex. 8.2.1. One can construct an example similar to Nikishin's example. Ex. 8.3.1. See §1 of Chapter 7 (property (AI) implies the existence of an infratype) and Lemma 2.3.3. Ex. 8.3.3. Show that Rlo. l ] contains a subspace isomorphic to l2, equipped with the weak topology.

Ex. 8.3.4. A space with the strongest locally-convex topology.

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(7) [8)

(9)

\10]

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[IS) 116\ (17)

[18J

W. Banaszayk. The Steinitz comtant of the plane. J. Reine Angew. Math. 373 (1987). 218-220 (MR 88e:52016). W. Ban&5zczyk. The Steinitz theorem on reon-cngement of sene" for nuclear spaces. J. Reine Angew. Math. 403 (1990). 187-200 (MR 9Ok:46010). W. Banaszczyk. Additive Subgroups of Topological Vector Spo~s. Lecture Notes in Math .• Vo\. 1466. Springer-Verlag. Berlin. 1991 (MR 93b:46005). 1. Barany. A vector-"um theorem and its appli.cGtiom to improving flow shop guarantees. Math. Oper. Res. 6 (1981). no. 3. 445-452 (MR 82i:90057). l. Barany and V. S. Grinberg. On some combinatorial questiom in finite-dimensional ~. Linear Algebra. Appl. 41 (1981). 1-9 [MR 84(:05001). B. Beauzamy, Introduction to Banch Spo~ and their Geometry, 2nd edition. NorthHolland. Amsterdam-New York. 1985 (MR 88f:46021). A. Beck. A COnt/aity condition in Banach spa~ and the strong law of large numbers. Proc. Amer. Math. Soc. 13 (1962). 329-334 (MR 24 #A3681). I. S. Belov and Ya. N. Stolin. An algorithm for the single-route scheduling problem, in: Ma.thema.tica1 Economics and Functional Analysis. "Nauka.". Moscow. 1974. pp. 248-257 (Russian); (MR 56 #17780). S. A. Chobanyan. The structure of the set of sums of a conditionally convergent series in a Banach space, Dokl. Akad. Nauk. SSSR 278 (1984). no. 3. 556-559 (Russian); (MR 86h:46020) English trans!. in Soviet. Math. Ookl. 30 (1984). S. A. Chobanyan. The structure of the set of "ums of a conditionally convergent series in a Banach space, Mat. Sb. 128(110) (1985). no. 1.50-65 (Russian); (MR 87c:46014) English trans\. in Math. USSR Sb. W. J. Davis. V. D. Milman. and N. Tomcuk-Jaegermann. The distan~ between certain n-dimensional BlJnach ~. Israel J. Math. 39 (1981). no. 1-2. 1-15 (MR 82h:46024). M. M. Day. Normed Linear S~. 3rd edition. Springer-Verlag. New York-Heidelberg. 1973 (MR 49 #9588). J. Diestel. Geometry oj Banach S~. Selected Topic8. Lecture Notes in Math .• Vol. 485. Springer-Verlag. Berlin-New York. 1980 (MR 57 #1079). J. Dieste1. Sequen~ and Series in Banach Spaces. Springer-Verlag. New York-Berlin. 1984 (MR 85i:46020). V. Dobric. A retTl4rk on Dvoretzky-Rogers theorem, GI&5. Mat. Ser. III 19 (1984). no. 1. 139-142 (MR 86e:46012). V. Drobot. ReafTQngements of series of /unctiON. Tr&n8. Amer. Math. Soc. 142 (1969). 239-248 (MR 39 #7328). A. Dvoretzky. Some ruulU on convu bodiu and Banach spaces. in: Proc.lnternat. Sympos. Linear Spaces (Jerusalem. 1960). Jerusalem Academic Press. Jerusalem; Pergamon. Oxford. 1961. pp. 123-160 (MR 25 #2518J. A. Dvoretzky a.nd C. Hanani. Sur lu changements des signes du termu d'une serie Ii termu complaes. C. R. Acad. Sci. Paris 255 (1947).516-518 [MR 9. 139gl.

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150 A ovoretzky and C. A. Rogers. Absolute and uncondItional convergence in normed linear [191 spa~ . ••s, ProC. Nat. Acad. Sci. USA 36 (1950). 192-197 [MR 11. 525al· [2 1 p. Enll o , A counterexample to the approximation problem, Acta. Math. 130 (1973), 309-311 0 (MR 53 #6288). 21) T. Figiel, A short proof of Dvoretzky's theorem on almost spherical sections 0/ convex [ bodies, Compositio Math. 33 (1976). no. 3, 297-301 [MR 58 #70301· [22) T. Figiel, J. Lindenstrauss, and V. D. Milman, The dimension of almost spherical sections 0/ convex bodies, Acta. Math. 139 (1977), no. 1-2,53-94 IMR 56 #3618). V. P. FonC, Conditionally convergent series in a uniformly smooth Banach space, Mat. (231 Zametki 11 (1972), 110. 2. 209-214 (Russian); IMR 22 #10884) English trans\. in Math.

Notes n (1972). 1 ) I. Gel'Cand, Abstraktc Funktionen und lineare Operatoren, Mat. Sb. 4(46) (1938). no. 2.

24

235-286. 1 1 O. P. Giesy, B·convexity and reflexivity, Israel J. Math. 15 (1973), 430-436 [MR 48 #46981· 25 O. P. Giesy and R. C. James, Uniformly non·1 1 and B·convex Banach spaces, Studia Math. [26) 48 (1973), 61-691MR 48 #119941· V. S. Grinberg and S. V. Sevast'yanov, Value of the Steinitz constant, Funkstiona!' Anal. i [21} Prilozhen. 14 (1980), no. 2, 56-57 (Russian); [MR 81h:52008} English trans!' in Functional Anal. ApP\. 14 (1980), no. 2. V. \. Gurarii, M.1. Kadets, and V.1. Mat5a.ev, Distances between finite· dimensional analogs [28} 0/ the Lp'spaces, Mat. Sb. 70(112) (1966), no. 4, 481-489 (Russian); IMR 33 #4649} English trans\. in Amer. Math. Soc. Trans!. (2) 76 (1968). [29J R. C. James, A nonn:flexive Banach space that is uniformly nonoctahedral, Israel J. Math. 18 (1914), 145-155 [MR 50 #8012}. [301 M. I. Kadets, On a property of broken lines in n·dimensional space, Uspekhi Mat. Nauk 8 (l!I53), no. \, 139-143 (Russian); IMR 14, 866d1· 1 ) M. I. Kadets, On conditionally convergent series in the space L p , Uspekhi Mat. Nauk 9 31 (1!J!i4), no. I, 107-110 (Russian); [MR 15, .s02aJ. [321 M. I. Kadets, Unconditionally convergent series in a uniformly convex space, Uspekhi Mat. Nauk 11 (1956), no. 5, 185-190 (Russian); IMR 18, 733eJ. ) M. I. Kadets, Geomdry of normed spaces, in: Itogi Nauki i Tekhniki. Mat. Analiz 13. [33 VINITI, Moscow, 1975, pp. 99-127 (Russian); 1l\IR 58 #300641; English trans\. in J. Soviet Math. [34J M. 1. Kadets and K. Wozniakowski, On series whose permutations have only two sums, Bull. Polish Acad. Sci. Math. 37 (1989), no. 1·6, 15-21 IMR 92d:46028}. 1 1 V. M. Kadets, B·convexity and the Steinitz lemma, h;v. Scvero-Kav~. Nauchn. Tsentra 35 Vyssh. Shkoly Estestv. Nauki (1984), no. 4, 27-29 (Russian); IMR 87a:460321· 1 V. M. Kadets, A problem 0/ S. Banach (problem 106 from the "Scottish Book n ), Funkstio136 nal. Anal. i Prilozhen. 20 (1986), no. 4, 74-15 (Russian); [MR 88£:460281 English trails\. in Functional Ana\. App\. 20 (1986), no. 4. V. M. Kadets, The Steinitz theorem and B-convexity, Izv. Vyssh. Uchebn. Zaved. 12 (1986), 1371 32-36 (Russian); [MR 88e:460131; English transl. in Soviet Math. (Izv. VUZ) (1986). 1 V. M. Kadets, A domain 0/ sums of w~akly convergent series, Funkstional. Anal. i Prilo138 then. 23 (1989), no. 2. 60-62 (Russian); (MR 9Oj:400041 English trans\. in Functiona.l Anal. ,A,ppl. 23 (1989), no. 2. I v. M. Kadets, ReafTUngement of function series and different forms of convergence, Ook\. 139 ,A,\
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M. Klldets, Weak and strong domcins 0/ sums 0/ a serie" In a Banach space, Mat. Zametki 48 (1990), no. 2, 36-44 (Russian); [MR 91j:460161 English trans\. in Math. Notes 48 (1990), no. 1-2. V. M. Kadets, On a problem 0/ the uistence 0/ a converyent rearrangement, Izv. Vysh. Uchebn. Zaved. (1992), no. 3, 7-9 (Russian); [MR 94e:400131. V. M. Kadets and M. I. Kadets. Reoml1lgements 0/ Series in Banach Spacu, T&rtu Gos. Univ., Tartu. 1988 [MR 90b:460381; English trans\. Translations of Matb. Monographs, 86, Amer. Math. Society, Providence, R. I., 1991 [MR 92d:460291. J. P. Kahane, Some Random Seriu 0/ FUnctions, Heath, Lexington, MA, 1973 [MR 40 #80951· A. N. Kolrnogorov and S. V. Fornin, Elements of the Theory of FUnctions and FUnctional Analysis. 2nd ed., "Nauka", Moscow, 1988 IMR :S8 #2559J; English trans\. of 1st ed., vols. 1,11, Graylock Press, Albany, New York, 1957, 196\. P. A. Kornilov, Rearrangements 0/ conditionally convergent functional series, Mat. Sb. 113(155) (1980), no. 4,598-616 (Russian); [MR 82b:400041 English trans!' in Ma.th. USSR Sb. 41 (1982). P. A. Kornilov, On the linearity 0/ the set 0/ sums 0/ a functional series, Uspekhi Mat. Nauk 37 (1982). no. 2,205-206 (Russian); [MR 83g:400151. P. A. Kornilov, The set 0/ sums 0/ a conditionally cont.erging function senes, Mat. Sb. 137(179) (1988), no. I, 114-127 (Russian); [MR 89k:40002J English t.rans\. in Math. USSR Sh .. S. Kwapieli. IsomorphiC CM'"(lcte,uations 0/ Inner product spaces by orthogonal series with vector t.alued coefficients, Studia Math. 44 (1972).583-595 [MR 49 #5789J. P. Levy, Sur les serie" semi-convergentes, Nouv. Ann. of MaUl. 5 (1905), 506-511. J. Lindenstrauss, On the modulus 0/ smoothness and divergent series in Banach ~aces, Michigan Math. J. 10 (1963), 241-252 [MR 29 #6316}. J. Linden~trauss and A. Pelczynski, Absolutely summing operators in Lp-spaces and their applications, Studia Math. 29 (1968), 275-326 [MR 37 #67431. J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces I, Springer-Verlag. Berlin, 1977 [MR 58 #17766J. J. Lindellstrauss and L. Tzafriri, Classical Banach Spaces II. Springer-Verlag, Berlin, 1977 !MR 81c:460011. L. A. Lyusternik and V. l. Sobolev, Elements 0/ F\mchonal AnalYSIS. 2nd revised edition, "Nauka", Moscow, 1965 (MR 35 #6981; English trans!. Himlustan Pub!. Co., Delhi; Halstead Press, New York, 1974 [MR ~o #28521. R. P. Maleev, Conditionally convergent seriu in some Banach lattices, C. R. Acad. Bulgare Sci. 32 (1979), no. 8, 1015-1018 (Russian); IMR 8Ig:46029I. B. Maurey and G. Pisier, Series de variables aliatoires indeependentes et. propriites geometriqu~s de.! espaccs de Banach. Studia Math. 58 (1976), no. I, 45-90 [MR 58 #1388]. V. D. MiI'man, A new proof 0/ A. Dvoretzl.:y's theorem on cross-sections 0/ C01\l/U bodies, Funkstional. Anal. i Prilozhen. 5 (1971). no. 4, 28-37 (Russian); IMR 45 #24511 English trans!. in Functional Ann!. App\. 4 (1971), no. 4. V. D. l\-1iImlLll and M. Sharir, A new proof 0/ the Maurey-Pi3ie,· theorem, Israel J. Math. 33 (1979), no.!. 73-87[MR 83b:460221. E. M. Nikishin, The set of sums ola function series. Mat. Zametki 7 (1970), no. 1.403-410 (Ru.qsian); IMR 42 #65571 English trans!. in Math. Notes 7 (1970). E. M. Nikishin, Rearrangemeflts 01 function series, Mat. Sb. 85(127) (1971), no. 2, 272-286 (Russian); !MR 44 #19661 English trans!. in Math. lISSR Sb. 14 (1971). E. M. Nikishin, Rearrangements of se,"ies in Lp. Mat. Zametki 14 (1973), no. I. 31-38 (Russian); {MR 48 # 9172) English trans!. in Math. Notes 14 (1973). w. Or1ic~, Be\trage zu Theorie der Orthogonalentwicklungen, 11. Studia Math. (1929), ... '"

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.. unbedingte Konvergenz in fUnktionraumen, Studia Math. (1930). no. 1, W. OrliCZ, Viler l641 83-85. ki" Domain.J 0/ $Urn.! 0/ conditionally covergents seriu in Banach spaces. 651 M. I. Ostrov~. ~nktsional. Anal. i Prilozhen. 46 (1986), 77-85 (Russian); IMR 88i:4602 41· 1 Tear. F\lnkts ll ., Reommgements 0/ the term.! in /unction series, Ook!. Aka.d. Nauk. SSSR D. V. Pecherskll 1'285-1287 (Russian); [MR 41 #9120J English trans!. in Soviet. Math. 1661 209 (1973), no. • Dokl. 14 (1973J: A theorem on projections 0/ permuted series with term.! belonging to Lp , [67J D. v. Peche;~~ SSSR Ser. Mat. 41 (1917), no. I, 203-214 (Russian); [MR 55 #10901J by· ~Jcad. I. in Math. USSR Izv. 11 (l~17~. English t~.. &arrangements of senes In 8aooch spaces and ammgements 0/ signs, 68] D. v. Pech. d1;) (1988), po. 1, 24-35 (Russian); [MR 89f:460231 English trans!. in Math. 1 Mat. Sb.l~ USSR Sb.. N clear LocJJ/ly Conva Spaces, Springer-Verlag, New York-Heidelberg, 1972 A. pjetSCb, u 1691 IMR.50 *~Ior Ideo/.!, North-Holland, Amsterdam-New York, 1980 (MR 81j:47001J. 1 1 A. Pletseb, Iu upoCU de Banach qui ne contiennent pas uniformement de l~, C. R. 70 1111 G. Pisier: 271 (1913), A991-A994 [MR 48 #119981· AeJIl.~· M0rfin4gGles with vGluu in uniformly convez spocu, Israel J. Math. 20 (1975), [72\ G· PisIel', 350 (MR 52 #14940). no.~, ~olienlS 01 Banach spaces of cotype q, Proc. Amer. Math. Soc. 85 (1982), no. 1, 113] G. PlSler, 84d:46019]. 32-36[MR C/lUSU of Banach spacu that are connected with unconditio7i4l con vergence, 1141 S. A. RaJtoV'"jfunktsional . Anal. i Prilozhen. 25 (1976), 106-112 (Russian); (MR 55 #3763]. 'Thor. f\JPktsIlJ4nIIch ,ptlCtS in which Or/ia's theorem is not valid, Mat. Zametki 14 (1973), [75J S. A. ~;06 (RUSSian); [MR 48 #9373] Eng~h transl. in Math. Notes .l~ (1973). nO. 1, ~D and W. Robertson, Topological Vector Spaces, 2nd edition, Cambridge 116\ A..P. l.oI1doo- New York, 1973 (MR 50 #2854). VOIV. ~~rrll-- (JonvtS Analysis, Princeton Univ. Pr_, Princeton, N. J., 1970 (MR 43 It T.IW"""'.-o, [11] ,4451.. ~ AnGlysis, Second edition, McGraw-Hili, Inc., New York, 1991 IMR 178) w. Rudio, 92k: 4600 s!erer Topologiocll Vector Spo.cu, 3rd printing, corrected: Springer-Verlag, New 119] H. H. . 1911 (MR~' #7122J. York-)lerIiD, GtlJllltlTJ and Probability in 8aooch Spaces, Lecture Notes in Math., Vol. ISO\ J,. Sch~.verlB8' New York-Berlin, 1981 [MR 83j:46003J. 852, Sp fli,A Book, Birkbiuser, Boston, MA, 1981 [MR 84m:OOOI5). 181 \ ,77at Sco av, ApproziJll4te ,olution of lome problems in scheduling theory, in: MethI82J s. V:;:::: Analysis in Ole Design oC Control Systems, Novosibirsk, 1978, pp. 66-75 ods . ). [MR 8211l:900931· ~'B u in Bonoch Spocu I, Springer-Verlag, New York-Berlin, 1970 (MR 45 1831 I. SiJIgef, M ,7451\. lI43u in StmGCh Spocu 11, Edit. Acad. R S. Romania, Bucharest; Springer1841 I. SiJICel' lin-New York, 1981 (MR 45 #74511· verlag.~, Conditionall¥ CORwergent ,mea of veetON, Uspekhi Mat. Nauk (1944), 185\ D. O. 1-59 (RUS'iaD); [MR T, 121· po.l0~, A proo/o/the principle of local re/fezivity, Proc. Amer. Math. Soc. 18 (1980), 186\ C.,,· 54-156IMR 81e:46012J. pO. 1, .1. BtIlingIlonuergenle Reihen und konveze S"steme, J. Reine Angew. Math. 143 ~~ E. SWD~175; 144 (1914), 1-49; 148 (1916), 68-111 . ~9:3~, On ~.~I.:,n:ton!s in !he Khinchin inequality, Studia Math. 58 (1976),

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M. Talagrand, On the best con.stants m Khmehm's mequahty, Sturlia Math. S8 (19;6), 110.2, 195-208 (MR 55 #36721· S. Trojanski, Con.dltionally convergent .~enes m certain F ·spaces, Teor. f'ullktsii F'unktsiona!. Ana!' i Prilozhen. S (1967), 102-107 (Russian); [MR 37 #3245J. S. Ulam, A Collection of Mathemahcal Problems, Interscience Pub!', New York-London, 1960 [MR 22 #10884J. N. N. Vakhaniya, V. I. Tarieladze, and S. A. Chobanyan, ProbabiJity Dtstribuhons in Banach Spaces, "Nauka", Moscow, 1985 (Russian); English trans!., 2nd edition, SpringerVerlag, New York-Berlin, 1989 [MR 86j:600141. Y. Yamasaki, A simple proof of K1/Iapieri's theorem, Pub!. Res. Inst. Math. Sci. 20 (1984), no. 6, 1247-1251 [MR 86j:46025]. R. M. G. Young, On the best possible constants in the Khintchme inequality, J. London Math. Soc. (2) 14 (1976), no. 3,496-504 [MR 55 #11,0081. Added in proor

[95] S. A. Chobanyan and G. J. Georgobialli, A problem on reamlngement of summancLs m normed spaces and Rademacher sums, in: Probability Theory on Vector Spaces, IV (Lancut, 1987), Lecture Notes in Math., Vol 1391, Springer-Verlag, Berlin-New York, 1989, pp. 33-46; [MR 9Om:460181· 196] E. D. Gluskin, The dillmeter of the Minkowsl.;i compactum i.~ roughly equal to n, Funktiona!. Anal. i Prilozhen. tr. (1981), no. 1,72-73 (Russian); [MR 83d:46026J Enlish Trans!. in Functional Anal. App!. 15 (1981), no. 1. [971 H. Hadwiger, U&er das Umordnungsproblem im Hilbertschen Raum, Math. Zeitschrift 46 (1!)40), 7Q-79[MR I, 2421. [98] A. S. Kronrod, On permutation of terms of numerical series, Mat. Sb. 18(60) (1946), 237280 (Russian); [MR 7, 432]. [991 V. D. Milman and G. Schechtman, Asymptotic Theory of Finite-Dimensional Normed. Spaces, Lecture Notes in Math., Vol 1466, Springer-Verlag, Berlin-New York, 1986 [MR 87tn:460381· [100] Sz. Gy. R.evesz, ReofTOngement of Fourier series, J. Apflrox. Theory 60 (1990), no. I, 101-121 [MR 90m:420421· [101] P. Wojtasuzyk. Banach Spaces Jor Analysts, Cambridge University Press, Cambridge, 1986 IMR 93d:46001J.

INDEX

absolutely summing norm, 53 absolutely summing operator, 53 B-convex space, 69 Banach-Mazur distance, 59 Banaszczyk's theorem, 117 basic constant, SO, 82 basic sequence, 80 basic sequence of subspaces, 82 basis, 79 Bessaga-Pelczynski theorem, 85 C-convex space, 63 Cauchy's criterion, 8 Chobanyan's inequality, 38 Chobanyan's lemma, 26 Chobanyan's theorem, 38 class P n , 74 coincidence property COIN, 132 convergence functional, 17 convex class CONY, 131 convex function, 71 cotype p, 68 diameter of partition, 119 direct successor, 102 Dvoretzky's theorem, 79 Dvoretzky-Hanani theorem, 22 Dvoretzky-Rogers lemma, 45 Dvoretzky-Rogers theorem, 48 e-orthogonal subspace, 81 finite representability, 59 finitely saturated space, 82 Frechet differentiability, 72 Frechet differential, 72 function, convex, 71 function, Frechet differentiable, 72 function, Riemann-integrable, 110

,~.,

r, the set of convergence functionals, 17

ri , the annihilator of the set of convergence functionals, 17 Gel'fand's theorem, 11 Grothendieck's constant, 57 Grothendieck's inequality, 55 Grothendieck's theorem, 54 infratype p, 68 integral, 119 - measure of closeness, 76 - sum, 119 - sum, e-mixed, 127 Kadets's theorem, 43 Khinchin inequalities, 40 Kolmogorov width, III Krein-Mil'man-Rutman theorem, 80 limit point, 36 limit-point range, 36 M-cotype, 49, 66 measure of C-convexity, 65 modulus of convexity, 70 modulus of smoothness, 73 nuclear space, 57 OrHcz's property, 49 Orlicz's theorem, 49 partial sum, 17 permutation, 5 -lemma, III -, convergence-modifying, 6

INDEX 156

, 'Ie of local reflexivity, 61 prlllCIP 87 ropertY AI, p A roperty 2, 87 p gem ent Lemma, 15 Rearra.n of parti tion, 127 ent refinea.nrn II integrability, 119 tu em 's theorem, 6 tuelnsd'nnng_off-Coefficlents ' Rou n 1 11 114 lel nnls, 14, 1 , , orrn Euclidean, 112 ~'elllln cc 'F_branched, 102 _t>f1Uen , F-widely spread, 103 """ sequence, seriesabSOlutely convergen, t 5, 8 -, ditiollally cOllvergent, 5, 9 _ con _' convergent, 7 , _ 'd01ninsted by a numencal I series, 98 _ nth partial sum, 7 , knlutely convergent, 52 - , JralP" rfectly convergent, 10 -_: pe perfectly divergent, 21 _, remainder, 7 _, segment, 8 _ SUJll,7

_' sum range, 13 _ , uneonditionally convergent, 9 , weakly absolutely convergent, 83 :' weakly unconditionally , convergent, 86 Ill) of limits of ruemann integral set urns of a function, 120 sWIll) of weak limits of Riemann se\ntegral sums of s function, 131 simple function, 60 space _, 8.eo nvex , 69 _, C-eonveY., 63 _, countably Euclidean, 112 _, finitely saturated, 82

--, nuclear, 112 -, uniformly convex, 70 -, weakly B-convex, 136 squeezed segment, 124 Steinitz - constant, 16 -'s lemma, 16 -'s theorem, 18 -'s theorem relative to a sequence, 98 sum range, 13 supporting functional, 73, 74 system conjugate to an F-branched sequence, 103 theorem -, Banaszczyk, 117 -, Bessaga-Pelczynski, 85 -, Chobanyan, 38 --, Dvoretzky, 79 -, Dvoretzky-Hanani, 22 -, Dvoretzky-Rogers, 48 -, Gel'fand's, 11 -, Grothendieck, 54 -, Kadets (M.l.), 43 -, Krein-Mil'man-Rutman, 80 -, Orlicz, 49 -, Riemann, 6 -, Steinitz, 18 -, Steinitz's - relative to a sequence, 98 tree-like order, 102 type p, 41, 68 uniformly smooth space, 73 weak absolute convergence, 83 weak sum range WSR, 101 weak type, 137 weakly convex class WCONV, 132 weakly unconditional convergence, 86

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