BROWN, Dean Raymond, 1939- B-CONVEXITY IN BANACH SPACES.

The Ohio State University, Ph.D., 1970 Mathematics

V University Microfilms, A XEROX Company, Ann Arbor, Michigan\

T H T Q n-TSSPRTATTON HAS BEEN MICROFILMED EXACTLY AS RECEIVED B-CONVEXITY IN BANACH SPACES

DISSERTATION

Presented in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the Graduate School of The Ohio State University

By

Dean Raymond Brown, B.S., M.S., M.S.

******

The Ohio State University 1970

Approved By

Adviser Department of Mathematics ACKNOWLEDGMENTS

I would like to thank Dr. David W. Dean, my adviser, for his guidance and encouragement. X would also like to express appreciation for the help of Dr. William J. Davis and for the financial assistance of the National Science Foundation.

11 VITA

April 18, 1939 . . . Born - Detroit, Michigan

1960 ...... B.S., Rose Polytechnic Institute, Terre Haute, Indiana

1960-1964..... Electrical Engineer, General Electric Company, Schenectady, New York

1964 ...... M.S. (Electrical Engineering), Rensselaer Polytechnic Institute, Troy, New York

1964-1966...... Teaching Assistant, Department of Mathematics, The Ohio State University, Columbus, Ohio

1966 ...... M.S. (Mathematics), The Ohio State University, Columbus, Ohio

1966-1970...... Teaching Assistant and Research Assistant, Department of Mathematics, The Ohio State University, Columbus, Ohio

FIELDS OF STUDY

Major Field: Mathematics

Studies in Algebra. Professor W. Kappe

Studies in Analysis. Professors R. Chacon, F. Carroll, P. Reichelderfer

Studies in Topology. Professor N. Levine

Studies in Functional Analysis. Professors D. Dean, W. Davis

iii TABLE OF CONTENTS

Page

ACKNOWLEDGMENTS ...... 11

VITA ...... ill

INTRODUCTION ...... 1

Chapter

I. PRELIMINARIES...... 5

II. THE PROPERTY OPPOSITE TO B- C O N V E X I T Y ...... 17

III. B-CONVEXITY AND BASES ...... 35

IV. P-CONVEXITY AND OTHER PROPERTIES RELATED TO B-CONVEXITY ...... 50

V. SOME UNSOLVED PROBLEMS ...... 65

BIBLIOGRAPHY ...... 74

Iv INTRODUCTION

The notion of a B-convex Banach space was introduced in [2] by A. Beck as a characterization of those Banach spaces X having the property that a certain strong law of large numbers holds for X-valued random variables. B-convexity is a generalization of uniform con­ vexity , as will be apparent from the definition, and is, by virtue of the characterization of Beck, invariant under equivalent renorming.

Definition; A Banach space X is said to be B-convex if

there is some positive integer k, and some 6>o so that for

any x^,...,Xjt ; ||x^|| < 1, i“l,...,k there is a set of

signs so that ||2j[Li | < k(l - s).

The definition of a uniformly convex space [5] can be phrased:

Definition: A Banach space X is said to be uniformly

convex if for each number e>o there is a number 6>o so

that if llxjl < 1, 11x^11 < 1 then either

- | Ix j ^ l | < 1 - 6 or | Ixj-Xgl | < e .

A lengthy discussion of the strong law of large numbers is contained in [3] in which the following question is raised; Can every B-convex space be renormed to be uniformly convex? This question is still open.

Further study of B-convexity was done by R. C. James

[14,15,17] and D. P. Giesy [12] from a functional analytic point of view. Giesy showed that B-convex spaces have many of the properties of reflexive spaces. James conjectured that all B-convex spaces are reflexive, and proved that In case k ■ 2 the conjecture is true. Both

James and Glesy showed that the conjecture also holds for B-convex spaces having an unconditional basis. Later, C. A. Kottman [20,21] showed that another subclass of B-convex spaces are reflexive; these are said to be F-convex. Examples are known of spaces which are reflexive but not B-convex.

Chapter I below lists some of the known facts about B- convexity and extends one of them (Proposition 1.20). The chapter includes several examples of Banach spaces, most of which are from the literature, illustrating some of the relationships between

B-convexity and other properties.

Chapter II takes up the property opposite to B-convexity.

A characterization of this property by Giesy is weakened, and a family of four new characterizations is presented. A technique used by

James in [14] is extended In a complicated way to give a condition sufficient for non-B-convexity. It is shown that a strengthened version of this condition resembles one of the family of four con­ ditions referred to above and thereby is sufficient for non-B-convexity in a much simpler way. These sufficient conditions hold for c q , and any space containing either of them. It is not known whether these sufficient conditions are also necessary.

Chapter III takes up questions involving B-convexity and the theory of Schauder bases. A. Felczynski [26] has proved that a

Banach space is reflexive if every subspace having a Schauder basis is reflexive. The main theorem for this chapter is an analogous

result for B-convex spaces: A space is B-convex if every subspace

having a Schauder basis is B-convex. Also in this chapter a different

proof is given of the known result that a B-convex space with uncon­

ditional basis is reflexive, using one of the family of four

characterizations of non-B-convexity given in Chapter II, and using

the unconditional basis in an elementary way.

Chapter IV investigates three properties related to B-

convexity. Kottman [20,2l] defined a space to be P-convex if

P(n,X) < \ for some n, where P(n,X) is the supremum of numbers r so

that there is a set of n pairwise-disjoint closed balls of radius r

in the closed unit ball of X. P-convexity is a generalization of

uniform convexity. Kottman showed that all P-convex spaces are also

B-convex and reflexive. It is not known whether there is a space which is B-convex but not P-convex. (Some examples of Kottman are

') given in Chapter V to illustrate the nature of this problem.) It is

known that the direct sum of two B-convex spaces is B-convex. The

analogous problem for P-convex spaces seems more difficult, partly

because it is not known whether P-convexity is preserved under

equivalent renorming. A partial solution to the direct sum problem,

two special cases, is given using a theorem of combinatorics. One of

these special cases is critical to the proof of a result similar to

the main theorem of Chapter III: A space is P-convex if each subspace

having a Schauder decomposition into finite dimensional subspaces

is P-convex.

Another generalization of uniform convexity is the Banach-

Saks property. A space is said to have the Banach-Saks property if every bounded sequence of elements of the space has a subsequence whose (C,l) means converge strongly. An example is given of a space, shown by Nishiura and Waterman [24] to have the Banach-Saks property, which is not B-convex. The converse question to this remains open and is of interest since it has been shown by Nishiura and Waterman

[24] that all Banach-Saks spaces are reflexive.

The last part of Chapter IV shows that a property proved in

[30] for spaces termed "near convex" actually holds for all B-convex spaces and all reflexive spaces.

Chapter V lists some unsolved problems and remarks. CHAPTER I

PRELIMINARIES

The symbol X will always represent a real Banach space. The closed unit ball of X, {x : ||x|| < l} , will be denoted U(X). We r in denote the closed span of that is, the smallest closed sub­ space containing these elements, by We will write {x^} and

[x^] instead of and ^xi^i«l * For * — p < ^p wil1 denote the Banach space of n-tuples with norm given by ||(x^,...,xn )( | «*

space of n-tuples with norm given by 11 (x^,...,xn )| | = max [|x,J : 1= 1,...,n}. and are the usual sequence spaces denoted by these symbols.

1.1 Definition; Let k be a positive integer and e a positive number. X is said to be k.e-convex if for any {x^,...,Xjt} C U(X) there is some choice of signs §^,...,1^ so that 11 x^| | < k (1-e).

X is said to be B-convex if it is k,e-convex for some choice of k and e .

The terminology in the definition is that of Giesy [12],

James [14]] has defined a space to be uniformly non— if it is n,e-convex for some e. A space which is uniformly non— Mas said to be uniformly non-square.*"

Geometrically, we can say that if X is 2,e-convex then the largest "square" which can be inscribed in a plane cross section

^In James' terminology the subscript and superscript on <6 are reversed from our convention. 5 of U(X) has side length, in terms of the norm of X, less than 2 - e.

Listed below are a number of facts about E-convexity, most

of which will be referred to later. The first group of facts can be

proved by elementary applications of the definition. Except for fact

1.6i they were proved by Giesy [12].

1.2; If a space is k,e-convex and 6 < e then it is

k,6-convex.

1.3: If a space is k,e-convex then every subspace is

k,e-convex.

1.4: Uniformly convex spaces are 2,e-convex for some e.

1.5: If the dimension of X is less than k, X is

k,^-convex.

Fact 1.2 shows that the number e can be made smaller. It is

also true that the number k can be made larger if e is appropriately

changed. A simple result in this direction is

1.6; If X is k,e-convex then it is k+1, s - -convex.

■ Proof: Given { x ^ . . . ^ ^ } c U(X) there is a set of signs

so that x x + ... + Sk x j | < k (1 - e)

from which

+ ... + 5k ^ | < k(l - e) + 1 =

A key fact is that by making k large enough, we can provide

an e nearly one:

1.7: If X Is k,e-convex and n is any positive integer,

for any {x.,...,x _} c u(X) there is some choice of signs 1 k

3 0 that - „ k n . ii . ,.n „ .i If we let e =* 1 - (l-e)n we have that X is kn .e -convex and n x ' * n lim = 1. The proof of 1.7 is obtained by inductively applying the

definition of k,e-convex and is contained in Giesy's proof of our

Lemma 1.9 ([12], Lemma 1.4).

We now list several deeper results about B-convexity.

Except for Proposition 1.20, which is new, and except where noted

otherwise, these were proved by Giesy.

1.8 Definition; For a space X we define the sequence

a ^ X ) , k > 2, by

ak (X) = sup{min[£|| £|| | : « ±l}: c U(X)} . i=l

From the definitions, for any X, a^CX) < 1 * If X is B-convex

there is some k so that a^CX) < 1 . If X is not B-convex, a^CX) = 1

for all k. Fact 1.7 shows that if X is k,e-convex, a^n(X) < (l-e)n .

Further, Giesy proves the following.

1.9 Lemma; If X is B-convex, lim a^(X) = 0 .

As noted, if X is not B-convex the limit is 1.

The following lemma is a useful characterization of non-

B-convexity giving insight to James' term "uniformly non-t^^".

Giesy proved this by a geometrical method.

1.10 Lemma: X is not B-convex if and only if for every k > 2, l>e>o there are x^,...,Xjt Ci U(X) so that k k k

(1 - e) £ |or1| < \ \Y Qf± x j | < £ lorj . i=l i=l i=l

If we define an e-isometry to be an isomorphism T so that

| |t| | < 1 + e and | |t"^| | < ■,— - ■— , then the above lemma shows that " i " 6 8 a space is not B-convex if and only if for arbitrarily large n and small e there is an e isometry from some subspace to 4^.

Several theorems of Giesy on the preservation of B-convexity under various operations are listed below.

1.11 Theorem: X is k, e-convex if and only if X** is k, e-convex.

1.12 Theorem; X is B-convex if and only if X* is B-convex.

In this theorem the k and e for X and X* may be different.

1.13 Theorem: If X is B-convex, and Y is isomorphic to X, then Y is also B-convex.

1.14 Theorem: If Z is a closed subspace of X, then X is

B-convex if and only if Z and X/Z are B-convex.

1.15 Theorem: If X is the linear span of two of its closed subspaces Y and Z, then X is B-convex if and only if Y and Z are B-convex.

1.16 Theorem: X is B-convex if and only if each separable closed subspace of X is B-convex.

As was mentioned in the introduction, James has conjectured that all B-convex spaces are reflexive. Theorem 1.11 above shows that if X is B-convex, X** is not only B-convex, but is so with the same k and c. Theorems 1.12 through 1.16 remain valid if the words "B- convexity" are replaced with the word "reflexive”. As has been mentioned before, the following have been proved.

1.17 Theorem: (James) If X is 2, e-convex it is reflexive.

1.18 Theorem: (James and Giesy) If X is B-convex and has an uncord itional basis, it is reflexive.

The subcase P-convex will be discussed in Chapter IV.

Kottman [21] has proved that all P-convex spaces are reflexive. 9 Lee S be an index set, X a Banach space of real valued functions of S, {x a family of Banach spaces. As in [fi], page 31, S S6 5 or [25], we define the product space to be the space of all functions x on S so that

i) x eXc for all seX s * ii) The real valued function Q defined on S by g(s) = I|x || 8 ' is in X.

If X satisfies the condition

iii) If £eX, and 7] is a real valued function on S so that

h(»>l < I C(s)| for all seS

then

7)ex and ||7l|| < lldl , then PjjXg is a Banach space.

Giesy has shown

1.19 Theorem: Let S, X, Cx_} be as above. If X is also uniformly convex and X is k ,e -convex where sup k < », inf e > 0, 8 S S 8 S then P„X is B-convex. X s Fact 1.7 listed before can be written in a form for sequences in a k,e-convex space.

1.20 Proposition: If X is k,e-convex and {xn}cU(X) then there is a sequence of signs so that

lim ir*® the proof given is similar to Giesy1s proof of our Fact 1.7 ([12],

Lemma 1.4). 10 Proof: We construct signs in groups

Ca n »•••» CTn+i^»*** ^ induction as follows, k +1 k let

yl = k (?1X1 + + y2 “ k ^ + 1^+1 + *•* + 52kX2k^ » and so on where the signs are chosen so that

I |yJ 11 ^ 1 - e for all j. Let a% = ***>°k = ^k

For the second induction step let

2 1 ,-2 1 , , .2 1 . 2 _ 1 ,.2 1 . . _2 1 . yl k (?ly l + •** + y2 " k ^k+lyk+l + ••• + ^2ky2k

and so on. Since | jv"- y 4 11 5 1 ^or each j, the signs can be chosen © J t o 0 0 so that | y j 1 | £ 1 - e, so that 1 |y^ [ | £ (1 - e) . Further they 2 can be chosen so that = + 1.

We can write

y l “ J T (51^1X 1 + ••• + S l V ' k + ^2^fcflxW-l + *'* + ^ ^ 2 ^ 2 *

2 Since ^ = +1, the signs associated with are

Let j th® signs associated with ,,x k k

Continue the construction with = +1 for all i, and

I b[ I i = 1 ^ x + ... + ct ±x )|| £ <1 - e)1 1 k- k k 1 which gives the desired result.

We close Chapter I with a listing of examples, most of which 11 are from the literature. The first group are of spaces which are B- convex. The first two were mentioned above and are included here for completeness. a) All finite dimensional spaces are B-convex.

b) All uniformly convex spaces are 2, e-convex for some e.

c) A space which is 2, e-convex but neither finite dimensional nor uniformly convex was described by Giesy. We give here a geometrical description of that example. For a more complete proof and other pro­ perties of a generalization of this see Giesy [12], page 142.

We construct a sequence of spaces Xr which are all 2, e-convex for the same e» hence X = P X is 2, e-convex. Referring to the & n definition as given on page 1 , we show X is not uniformly convex by showing there is 1] > o so that for any 6 > o there are x,yeU(X) so that both 1/2 | |x + y| | ^ 1 - 6 and | |x - y 11 a T\, Let Xn be a sequence of

two dimensional normed linear spaces so that, if C is the Euclidean unit circle and H is a regular hexagon inscribed in the circle,

C => U(XX) 3 UCXg) 3.. .3 H .

CD and 0 U(X ) « H. n=li n

Letting |j* (| be the norm in X^j ||*||^ be the Euclidean norm, and

j]•||h be the hexagonal norm, lim||x||n = (|x|for all x. n-w

Take e so that ^/Z/3 < 1 - e < 1. We show X^ is 2, e-convex by showing that if a square is inscribed in U(Xn), its side length is less than or equal to 2.t l !3 in II’ISn* a square is inscribed in

U(Xn ), it is inside C so its side length must be less than or equal

to f l in || *||Since U(X ) is outside H, and so outside a circle 12 of radius *j3/2, we have I I'll £ Hrj | • | („, so that the length of the n c side of the square must be less than or equal to 2.^/3 In Il*!ln*

To find T|, x, y, choose any x / y on one face of the hexagon.

Since U(Xn )^I we have | |x| |n, | |y| | <: 1 for all n. Since C 3 U(Xn ),

II*- y I ln * I lx “ y| lc; let T1 = llx - y| |c. Since lim | |x + y| |n = n-«

| |x + y | = 2 , we have for n sufficiently large, | |x + y | |n s 2 - 2fi.

Thus,regarding x and y in X^, which is a subspace of X, we have the required inequalities.

Remark: No space is known which is 2, e-convex but not isomorphic to any uniformly convex space. James [15] has conjectured that any space isomorphic to a 2 , e-convex space is also isomorphic to a uniformly convex space.

d) A space which is B-convex but neither 2, e-convex for 2 any e nor finite dimensional is X = P, X where X is 1 . Each X J j&2 n n 1 n is 3, l/3-convex since it is 2 dimensional, so X is 3, 1/3-convex.

But X has a subspace which is not 2, e-convex for any e» namely X^ for any n, so is not 2, e-convex.

e) The spaces Xp ^ defined by Lindenstrauss and Felczynski in [22] are B-convex if 1 < p < ®. For li p S o, X is said to be an

Xp ^ space if every finite dimensional subspace B is contained in another finite dimensional subspace E so that d(E,£p) £ A, where n is the dimension of E, and d(X,Y) is defined to be the infimum of

I IT11*|1 1 for all linear bounded invertable mappings T from X into

Y.

That these spaces are B-convex for l

Is a measure p, and a complemented subspace of L^Cy.) which is Isomorphic to X. It also can be seen by the following elementary proof. We first note that, since c £ , there is a 6 (depending on p) so that is

2,6-convex for all n. Let X be X ,. Choose m f e; o

{Tx.}k c U(tn), so by Fact 1.7, there is a choice of signs • •»§!_ i 1=1 P * so that k

|| Z 5.Tx 11 £ k(l-6)m , i=l 1 1 and

1 E,X3j"j i " i £Xk ( l - 8> " £ k(l-e) • - i=l

The second group of examples are not B-convex.

f) is not B-convex. Let x^,...,^ be the first k usual norm 1 basis vectors. Then for any choice of signs k k

II Z f^xjl = k 1 1

so that is not k,e-convex for any e. th g) c q is not B-convex. Let e^ be the i usual basis vector and let 14

X 1 el " e2 + e3 ” e4 + e5 “ •” " e2k

x2 ” el + e2 “ e3 " e4 + e5 + *** e2k

X3 = el + e2 + e3 + e4 - e5 ' — " % k

*k “ el + e2 + * *’ + e2k-l " e2k-l+1 " *'• ’ e2k

Clearly | |xjJ | - 1; i *= 1, ..., k. For any choice of signs Ic there is some j, l ^ j £ 2 , so that the sign on e^ in the x^ sum is

^ for each i. Therefore

k k

I I Zjy ^.x. I I ^ I I — y e4 I I ~ k so that cQ is not k, e-convex i=l 1 1 i=l J for any e.

h) A space which is reflexive but not B-convex is P . 2 1 Day [7] has shown that the product P X^ is reflexive if each X^ is P 'reflexive, so that this space is reflexive. Since it contains isometric

copies of JL™ for each n, it is not B-convex.

James [14] and Giesy [12] have both given other examples of

spaces which are reflexive but not 2,e-convex, resp., B-convex.

i) A space which is locally uniformly convex but not B-convex 15 was given by Giesy. (A space is locally uniformly convex if for each x » I lx 11 ~ 1» and each e > o there is a number 6 > o depending on e and x so that if [|y|| = 1 then either |jx-y|| i e or l/2 ||x+y|| £ 1 - 6.) n i The space is P. (£ ) where p.| 1 and n t “• Lovaglia [23] has shown 2 ^i 1 1 that this product of locally uniformly convex spaces is locally uni- ni formly convex. Since the spaces $, become arbitrarily good approxima- ^i ^ tions of arbitrarily large dimensional the space is not B-convex by Lemma 1.10.

j) Giesy and James have proved that the example of James in

[16] of a space which is isometric to its second dual but not reflexive is also not B-convex. The proof of Giesy, which is rather complicated, is described in [12].

k) All infinite dimensional .C^ ^ spaces (see example f above for definition) are not B-convex since Lindenstrauss and Pelczynski have shown ([22] Prop. 7.3) that infinite dimensional ^ spaces have a sub space isomorphic to j£^.

Remark: Because of the use of in the definition of

. and in Giesy's characterization of non-B-convexity (lemma 1.10), *> A one might ask whether all non-B-convex spaces are Z-, ,. The answer to this is no, as is shown by example.

4) © J&2 *s not B-convex and not y The space is not

B-convex since it contains SLy Suppose it is y Since the property

^ for some \ is invariant under isomorphism we may assume is a subspace of © l y Ta^e so that dim(B) = m. There is a sub­ space E 3 B and an isomorphism T:£^ -♦ E, for some n, having

||T [j* ||t * || £ and without loss of generality we may assume -1 16 ||x|I = 1, ||t 11 £ X. let the images in E of the jfc” basis vectors

be x ^ x^. These form a basis for E and satisfy the inequality

n n n

X ^ K l * II ^ aixill ^ K l » A i=l 1 i=l 1 1 i=l 1

for all fajK-l* Since B is an m dimensional subspace of it is

isometric to X? and we may let e,, ...*e be the usual X? basis vectors. £ l m £ If we write

n

= Lt a,x4 , i = I,.**, m, we have, by the above inequality, j = i J n n 1 = I Iei I I 83 II ^ ajxjH ^ la j l * 3=1 j=l J

Therefore* using the inequality again*

m m n m n

= | | Z, e | | = 11 X X ajx j 11 * X ^ ^ laj I * X i=l 1 i=l j=l J 3 A i=l 3=1 J A

Since this must be true for all m, we have a contradiction. CHAPTER II

THE PROPERTY OPPOSITE TO B-CONVEXITY

In this chapter we give five conditions equivalent to non-B-convexity and two conditions sufficient for non-B-convexity,

As stated in Lemma 1.10, Giesy has shown that a space is non-B-convex if and only if for every positive integer n £ 2 and every 1 > e > o there is a subspace e-lsometric to We weaken this condition to get

2.1 Proposition: A space X is not B-convex if and only if there is some 1 > s > o so that for every k s 2 there are x^,...,x^ crU(X) satisfying

k . k k

(1-e) Z lofj J £ 11 X ofjX. 11 £ X ]or. j » i=l i=l i=l for all scalars

Proof: If X is not B-convex by Lemma 1.10 this condition holds.

To prove the converse suppose the condition holds but X is B-convex.

Then there is k and 6 > e so that X is k,6-convex by Lemma 1.7 or 1.9.

If x^,...,x^ are as asserted above, there is some choice of signs

» • • • > that

k k

k(l-8) a || Z 11 a (1-e) I \g. \ = k(l-e) , i=l 1 1 i=l giving 6 ^ e which is a contradiction.

17 18 This proposition shows that if for some e there is an

e-isometric image of every j i n X, then for any e, n there is an

e-isometric image of in X. James [14] proved a similar lemma

for If X contains an isomorphic image of JL^ (hence an e-isometric

image of for some e), then for any e, X contains an_e-isometric

image of The proof of James lemma is quite different.

By using the result in Chapter X we get a family of

conditions equivalent to non-B-convexity which involve both points and functionals.

2.2 Proposition: The following four conditions are equivalent to non-B-convexity:

(1) For some o < e < 1* o < C, pa positive integer,

and any k £ 2, U(X) contains x^,...,x^ so that for

any choice of signs there is a functional

f» | |f || s C(lnk)p , such that fC^x^) > e, i = l,...,k.

(1 ,) For some o < e <1, and any k ^ 2, U(X) contains

x l****,xk 80 t^at ^or any choice of signs

there is a functional f, ||f|| ^ 1, such that

f(§iXi> > e, i ~ 1 ,..., k.

(2) For some o < e < 1» o

for any k & 2, U(X*) contains f^,...,f^ so that for

any choice of signs there is an xeX,

||x|| £ C(ln k)P , such that f ^ ^ x ) > e, i =* l,...,k.

(27) For some o < e < 1* and any k ^ 2, U(X*) contains

f^,...,fk so that for any choice of signs ^,..., 1^,

there is an xgX, ||x|| £ 1, such that f^(l^x) > e»

i=l,...,k. 19 Remark; Conditions (1) and (2) are weakenings o£ Conditions

(1') and (2'). Conditions (2) and (2*) are duals of Conditions (1) and (l')» Condition (2') will be used later In this chapter.

Condition (I7) will be used as a lemma In Theorem 3.10.

Proof; We will show: X not B-convex -» (1 /) -♦ (1) -* X not

B-convex,and (1-* (2 f) -» (2) -* X not B-convex.

First let 6 be any number so that o < 8 < 1 and choose e < 1 - 6. Since X is not B-convex, by Lemma 1.10 we have (for any k k k) x^,... ^ C U(X) so that (1-6) 'S 1^1 £ || II j | for all a^,...,^.

Therefore, x^,...,^ are linearly independent and there are f^,...,^, functionals defined on so that f^(x^) = 6.^ the Kronecker delta. For a given set of signs let f be an extension to X 1 * k without increase of norm of the functional (1-6) SJ 5-i^.r* Than i=l 1 1 | |f 11 « sup £ |f (x) | ;xeU([x^] j_=j_) }• If x eU(tx^] jB^) we have k " k i x = S a.x. for some a.,...,a, so that (1-6) E |a.j £ 1. For such x 1=1 1 1=1 k k k

If (x) I = (1-6) | ( X *)( £ a^x t) I ^ (1“6) ^ |atl ^ 30 | |f | | £ 1 . j=l 3 3 1=1 1=1

k Also f d ^ ) = (1-6) (S |.f,)(g.x.) = 1 - 6 > e so that (1 ') holds. j=1 n ■ To see thatt (1 ) implies (1) let C = — r, Pp “= 1-1. Then f

To see that (1) implies that X is not B-convex, suppose X is n,6-convex for some n,6. With the C,p of (1),lim (l-6)m C (In nm )^ = o, ra-»a>

Therefore with e of (1), m can be chosen so that (1—6)m C (In < e- Let x^,...,x m be as given by (1). By Lemma 1.7 there is some set of n signs ^ , . . . , 1 m so that n m n

I 5.x..-I. 11 < n (1-6) .*■ i=l 1

Therefore, for the functional f asserted by (1), m m n n

nm e < f (X llfll I! ^ ?ix i 11 C ^ 111 tim)P *nm (l-6 )m < n

which is a contradiction.

We now have (I7) is equivalent to non-B-convexity. Using

this we show (17) implies (27): If X satisfies (17) then it is not

B-convex; by Theorem 1.12 X* is not B-convex so (l7) holds for X*.

Therefore, for some o < e < 1 and any k s 2, U(X*) contains

so that for any choice of signs there is cpeX**, | |cp| | ^ 1

such that cpC^^) > e» i = l,...,k.

The set {>eX**: ^(g±f - cp | < f > 1 =

(2) is an X* neighborhood of tp in X**. .Since the usual imbedding of

U(X) in U(X**) is dense in the X* topology of X**, there is some

xeU(X) so that its image is in this neighborhood, i.e.,

1 ^ 0 0 ~ | ^ jT» ^ =

therefore f^(^x) i-l»..#»k .

(2) We use the terminology of Dunford and Schwartz, [10]. 21 To see that (2') Implies (2) let C p “ 1*

To see that (2) implies non-B-convexity note that if X has

property (2) then X* has property (1) by letting the required

functional in X** be the natural imbedding of the point X asserted by

(2). Thus X* is not B-convex and by Theorem 1.12 X is not B-convex.

In the proof that 2,e-convex spaces are reflexive, James defines for a Banach space X a sequence of numbers Kn> He shows that

if X is not reflexive then K £ 2n, and in that case X cannot be n 2,e**convex. The next result is an extension of the methods of the

second part of James' result to show essentially that if Kr is a bounded sequence, then X cannot be B-convex. For reasons associated with Condition (3) below, we change slightly the definition of and call the redefined sequence K ^ .

After the result discussed above, we will present another condition, to be called Condition (3), which is a strengthening of

the condition k 7 bounded. Condition (3) will trivially imply

Condition (2 r), and so will imply non-B-convexity in a much simpler way.

2.3 Definition: For CU(X*), pl*****p2n an *ncreas“

ing set of positive integers, let

i I 3 S(p1»...,P2n; f^}) “ [xtC-D1- fk (x) a ^ for P2i_1 ^ P2i»i=l»...»n},

sPi ‘ inf nM!:*eSCPl • 22 (If S(plt...,p2 8 0 then let s „ rf i a “) » x i ”l*“"**^2n^ 1-^

K(n,{f.}) 8 11m ... s , , ^ ifcT" 1* * *" * 2n* ^ and

Kn / = inf {K(n, {f^}) : fft} <= U(X*)f .

James definition of is the same except that he lets

S(p1,...,P2n;{fi}) = {x:|- 5 (-l)i_1fk (x) £ 1 for p^^s&sip^i8!,...,n} ,

and requires ||f^|| 8 1. i“l,2 ,...

Clearly K * <: K . n n

—2.4 .. Lemma: '■ For any X, n, K n 1 £ Kn*r± •

Proof: From the definitions, for any p^,...»P2n+2 »^j.3*

Jpl»* * * »p2n* i ' P1*’* **P2n+2, i^

Taking 2n limits, for any Pj»P2 we have

lim ... 11m s rc t ^ lim ... lim s r <; ■> pl-» p2n-« pl*****p2n* *■ V p3-*» p2n+2~*co 1 ..... P2n+2,[fi-f *

so that KCn,^}) S lW .lfjH il/ SKj+1 23 2,5 Theorem: Let be defined for a space X as in Definition

2.3. If there is a number M so that s M for all n, then X is not

B-convex.

Proof: Given k, 6, we will show that X is not k,6-convex by showing there are x^,...,x^ c U(X) so that for any choice of signs

W e have k

X e.x. 11 > k(l-6) . i=l 1 1

Choose m so that ^2m > 1 - £ . (2.5-1) 3

This is possible since the sequence is bounded and monotone.

6(K' k_)2 Choose p, so that o < p , < — rrp . (2.5-2) zm

Choose cU(X*) so that

K3*2k *m + » > K O ^ V C ^ } ) • (2.5-3)

Choose e so that .2 5(K'. k y

0 < e < T • (2.5-4) 3(K2m + K • k } 3*2 m

Using these four inequalities we obtain 24 * 2m - « K(3*2km,{fi3) + e

^ 2m e (by inequality 2.5-3) K . + \j, + e 3-2 m

^2m *L 7— K K . + n + e K 3*2km 3*2 m 3-2kmJ

- eK * 2 m k ; 2 ^ - ^a ■ ^ m 2m + eK' . 3*2 m « ' „k )z + M*' k 3*2 m 3*2 m 3*2 m

,k + M ^ a + ^ m 2m 3*2 m k (K' fc )2 3*2 m 3*2 m

K^m * ; a + ^ < m V> * k (K' fc )2

^ K ' 2 k 6 3*2 ro 3*2"m 3*2 m 1 - L 3 IV (K' 3*2 km >

(by inequalities 2.5-1, 4 and 3)

1 - 6

We will choose k increasing sets of integers

.,p* . }, i = l,...,k, having the following properties: 5*2 m 25 (1) For each i ** 1,... ,k there is u^eSCp^,...,? ^ 5(^3) 6*2 m

of minimal norm, that is, ||u£|| ^ K(3*2km, {f^})■+ e .

(2) For each choice of signs ^,...,5^ there is an

increasing set of integers

{^,...,0^ } c { p*:j ** l,...,6*2km; i = l,...,k],

depending on the choice of signs, so that k 'T (2a) £ Zj e S(ct^,... {f^}) ,

and so that

(2b) any element of S(o^m » {f^}) will have norm

greater than K(2m,{f^]) - e

The procedure for choosing these integers will be described later. Having chosen the integers, let

x, u ., i — 1,•..,k. K(3*2m,{fi))+ e

By (1) above jjx^lj £ 1. By (2) above, for any choice of signs we have

1 „ Y ,, K(2m,{f }) - e k 11L I * — e— ;— >i-s , 1=1 1 K<3-2 c which will prove the theorem. 26 We now describe the choice of the integers i k {Pj: j = 1,...,6*2 m;i = 1.... k}. They will be chosen in m blocks

of 6*2 *k integers each. The first of these blocks will be i k {Pj:J “ 1,...,6*2 ,i = l,...,k}. The choice for this first block will be described. The other m -1 blocks of integers are chosen in

the same way, choosing the integers of each block larger than those

already chosen.

We now describe the choice of integers for the first block. Tr There are 2 choices of signs We will successively choose

a set of integers for each choice of signs. The number of integers

chosen for each choice of signs depends on the signs; 4k are chosen

for each plus sign and 8k for each minus sign. Since ^ is plus for

exactly half of the choices, and so on for §2»****?k» we choose k altogether (in the first block) 6*k*2 integers.

We first show in detail the choice of integers for the signs

5^ = ... = = +1, In each step below*the chosen integer is to be

larger than all previously chosen integers. Choose p* so that

— ... — fs . V K(3*2k *m,ff }) + e. '2 6.2k . A V '2.....* / 6‘2 -

(2.5-5)

(To simplify notation we omit "-w" from the limit inf and from

the index of s.)

Choose p^, i = 2,...,k - 1 in increasing order so that i l k 2.5-5 holds with p^ replacing p^. Choose p^ so that 2.5-5 holds 27 fc 1 with replacing p^, and also so that

lim lim f \ *2 " • r j

Choose p2 so that

lim lim f \ «S K(3-2km,):f } + 2J^— . (2.5-7) r„ *'* r . Is 1 i ) 6*2 *m 3 6-2k - m W ’P2*r 3 ..... r ..,k / o 2 »m

and so that

(2.5-8) i3 f - i f4m\kp f k1,p 2 i ,r3,...,r^m/

Choose p2 » i = 2,...,fc so that inequality 2.5-7 is satisfied with

p^,P2, replacing Choose p^, i = l,...,k - 1, so that

i*2 ... ^ \ s K ( 3 - 2 k.m,{£ )) + — 3^— 4 6*2k.m \plp2p3r4 % ,k / 5'2 ■m 6*2 *ra (2.5-9)

v Choose p^ so that inequality 2.5-9 holds and also so that

i4 f - i f 4 m f ^ pk lfp2 i ,p3>r4k ,...,r4m >J *«-.c*ti> -Is • <2-5-10> 28 Choose so that

lim f \ TT —*•* Tjr(a „k (s 11 1 1 .11 .11 )

<2.5-11) and so that

4e (2.5-12) f 5 ••• k i k i K <2m «ffi » - 4m 5 4m^p1,p2,p3,p4,r5,...,r4]i/V

Choose p4 , i = 2,...,k so that inequality 2.5-11 is satisfied if i 1 1 1 Pj>••• J^®pl^ces p^,. • • »P4 *

This completes the step for * +1.

Inequalities 2.5-5, 2.5-7, 2.5-9, and 2.5-11 are chosen so that (1) will hold. To see how (2) will hold let

k 1 _ k 1 ° l “ pl* ct2 " p2* a3 " p3* °4 *" 4

Inequalities 2.5-6, 2.5-8, 2.5-10, and 2.5-12 will provide (2b).

We will see that (2a) follows from the order of choosing used. The order of choosing can be illustrated by the diagram below. Illustration of choice of integers p^, 1 « l,...,k;j = 1, for all plus signs.

In this diagram, a to the right of b means a > b.

To see why (2a) will hold (after completing all m blocks) note that if

A A eS(p^»..•»P^»r^»• • ^ for any r^,***,r ^ , 6*2 »ra 6*2 m

then ^ % for S ^ £ p2 and ~fj ^ui^ & £ for P3 £ ^ ^ p4 *

k 1 3 Hence if = p^ j £ Pg = we ^ave ^j a 4 f°r i 30 For choices of signs containing one or more minus signs the construction is modified so that (2a) will hold with the new choice of signs. The construction is tedious and the same ideas are used as above, so only the modifications required by the minus signs will be described.

For illustration, suppose the second choice of signs

H * ’***^k has ^i “ and ^i “ +1 for 1 * io o

Choose integers in the order indicated in the diagram on the following page, satisfying inequalities similar to those above. k 1 k For this choice of signs let ^ a p 5, a2 = ^6* CT3 ” ^7*

= Pg. By using inequalities similar to those above, (1) and (2b) can be provided. To see why (2a) will hold, suppose that,

for i ^ sS(r^,••.,r^,p^,••.,Pg,Tg,...,r ^ ;{f^3) for an 6*2 m arbitrary choice of r^,...,r^; r^,...,r ^ . If p^ £ j £ pg we have 6*2 ra

f^ (u^) s 3/4 and if p* *s j «s p* we have -f(u^) s 3/4. On the other

hand, if i = i^ and u^ 5S (r^,•#•,r^,p^ ,...,p ,r^g,...,r ^ 'o 12 6-2 m and p_ £ j £ Pg we have -f,(u^ ) = f (E^ u^ ) a 3/4 and for J A J o rt o i i Pg° ^ j £ P^q we have f^(u^ ) = -f^(£^ ^ 3/4. Therefore we have,

1 k if ffj s j s 0*2, s 3/4 for a11 1 and fj a 3/4.

1 k If tig £ j £ we have s 3/4 for all I so -f^ (^ 2 s 3/4.

1 k Thus £ 2 ?^u^cS(o^,•••,c^i{f^})• 31

Illustration of Choice of Integers p* for = -1, ZL = +1 for 1 4 i(

nr°2 - n4*°4 «- 6 1 1 5 P1 p7 p8 2 2 2 5 p6 p7 p8

£o *o 5 p6 8 p10 P11 P12 1 +1 1 +1 1 +1 1 +1 o o o o p5 p6 p8

■8 iD ic ’ 32 The additional integers P ^ » P^2 are Provided so that if k Pq s pg and j S pQ than z 3/4 for all i. Thus integers for

other choices of sign may be chosen by the same method. This completes

the proof of Theorem 2.5.

We now present Condition (3). f -I » (3) For some o < e < 1, U(X*) contains a sequence

so that for any k 2 1 and any choice of signs there is xeX,

| |x|| s i , such that > e, i = l,...,k.

As discussed before Theorem 2.5, we will see that Condition

(3) is a strengthening of bounded. It can also be seen that Condition

(3) is an "infinite version" of Condition (2r), and so Condition (3)

implies non-B-convexity in a much simpler way than the condition bounded.

2.6 Theorem: If X satisfies Condition (3), then it also

satisfies Condition (2') and so is not B-convex. In addition is bounded.

Proof: The first statement is obvious by inspection and by

Proposition 2.2.

To show the second it is sufficient to show that for ff^} 3 and any p,,...,p.>„» s ff 1 5 F • This wil1 i^piy that I**"- 2n i q Kn ** 4e * G:Lven pl * " * ,p2n* ctl0osc = 1,***,p2n so that

P2^_^ 5 i £ p2j, j=l,...,n then g^ = (-l)^*”1. Let x be as asserted 3 by Condition (3). Let y = ^ x. Then if P2j«x ^ 5 p2j** * *,n*

then ( - D ^ f ^ y ) > | , so that y cS (p ^ ... ,p2^, {ft}) and ||y|| . 33 We do not know whether non-reflexivity implies bounded or

Condition (3). It would also be of interest to know whether K / bounded n or Condition (3) is equivalent to non-B-convexity. This would be true

if Condition (2/) implied Condition (3).

We do have, however, a large class of examples which satisfy

Condition (3).

2.7 Proposition: The space cQ satisfies Condition (3).

Proof: Take any o < e < 1. Let (e^,f^) be the usual biorthogonal system in c q , where f^(e^) =* Clearly j|f^|[ = 1.

Given a set of signs 5^,...,^, let

X ~ c 0 •

Then 1|x11 £ 1 and

fi e for all i=l k .

2.8 Proposition: The space satisfies Condition (3),

Proof: Take any o < e < 1. Let (e^,g^) be the usual

biorthogonal system in where-g^e^) = 6^. Let

“ 6^ “ S2 8^ " 8^ 8g “ • • •

^2 ” 8^ S2 ” 8j — S4 8^ • • •

^2 6^ S2 8j S4 ” 85 ” ... 3 4 Clearly | (^ 11 = 1, i-1,2,... .

For any choice of signs 5^,...,^ there is some j so that the sign on g^ in the sum for f^ is Let x = e^. Then f^(x) = so f^CSjX) = 1 > e for i=l,...,k.

2.9 Proposition: If Y satisfies Condition (3) and X ID Y then

X satisfies Condition (3).

Proof: Extend the asserted functionals in U(Y*) to all of X by Hahn Banach theorem.

These propositions show any Banach space containing cq or

satisfies Condition (3). All of these spaces are non-B-convex and non-reflexive (since they contain a non-B-convex space and a non­ reflexive subspace). CHAPTER III

B-CONVEXITY AND BASES

The main theorem of this chapter shows that X is B-convex

if every subspace having a Schauder basis is B-convex. Following this we include a different proof of the known fact that every B-convex

space with unconditional basis is reflexive.

3.1 Definition; A sequence {x^} C X is said to be a (Schauder) basis for X if for each xeX there is a unique sequence of numbers {a^}

so that

n

lim 11 X a.x. - x 11 = 0 . n-*s> i=l

A sequence {x^} is said to be a basic sequence if it is a

basis for its span [x^]• We will use the following well known

3.2 Lemma; A sequence {x^} is a basis for [x^] if (and

only if) there is some number K so that for any integers n and q and

any sequence of numbers {a^} we have

n n+q

V V L a x || £ K|| L. a±x || . i=l i=l

3.3 Remarks: We will prove the announced theorem, Theorem

3.9, by constructing a basic sequence in an arbitrary non-B-convex

space in such a way that the span of this sequence is not B-convex. 36 The technique for construction of this basic sequence is an adaptation of the method of Day [8] and Gelbaum [11].

In the proof of Theorem 3.9 we shall need the following technical lemma.

3.4 Lemma; If {6^} is a sequence of numbers, 1 > 6^ o, and p(m) is an increasing sequence of integers, then there are sequences and so that

(1) e± -* o

<2>

(3) if 1 £ n £ p(l) we have

U + V sl + 6n

(4) if p(m) < n ^ p(m+-l) we have

1 + \ + <2 + \ + \ n > 3 + «n •

Proof: let v = inff5_ : n = p(m) + 1,... ,p (nri-1)], m=o,l,..., - 'm n where we let p(o) = o.

Choose e^,i-l,2,... positive decreasing to zero so that

* 1 L. (3.4-1) 1 _ Vn

and Then choose ^,1=1,2,..., so that

s oin [ ^ ] . (3.4-3)

Property (1) is clearly true. To see that property (2) holds, use inequality 3.4-3 to get

^m-1 8 \ £ < 6p(m) •

To see that property (3) holds, note that

1 - > ---- — by inequality 3.4-1, so that 1 +r

1-e- 4

V Y By inequality 3.4-3, we have 1+1|^1+^<1+^ ,

so that, using the fact that y < 1 ,

2 2 p i - d ^ X (l+jfe) =l+^ + Y2-

For property (4), observe by inequality 3.4-2 that 38 1 1 1 - £ 2 * 8 0 YI— ^ Therefore, using inequality 3,4-3 we have in

\ ( 1 + i d -7 ) s3\ fir em+l

Also by 3.4-3

*%ri~l _ 'Ym lmV i £ ^ s 3 ' and by 3.4-2, 1 - ^ a , 6 so that

1-V i 3

Therefore

1 + \ + <2 + \ + W

= 1 + n_ + .. + --- 2 tn 1- e ... 1-e .. l-i EnH*l em + l etn+l

S l + 5 s + ri + 2 + Js=3 + Y-^3 + J 3 3 3 Tm n

The following lemma is important to the Day-Gelbaum technique.

3.5 Lemma: If X is finite dimensional, for any e > ° there are c U(X*) such that for any x

x < (1+e) max {f^(x) : i=l,...,n} . 39

Proof: Let S = {x : jjx|| - 1}. For any xgS there is fx eX* so that | |fj | = fx (x) = 1. The sets x eS, form a family of open (relative to S) sets covering S, which is compact, so that a finite subfamily of them cover S. Let the corresponding functionals be • Then for any xeX there is some i so that .

1 ,-1 ( 1 1 \ TFIT * e 1 ^ l+e * 1-e / *

and so

£i ( TFTT x) * h S ’ or

(1+e) ft (x) a | |x 11 .

As usual, we have

3.6 Definition; The subspace Y c X is said to be of finite codimension in X if there is a finite dimensional subspace Z so that

X « Y © Z.

We will need the following well known lemma. A proof for this lemma can be found in, for example, [11] Lemma 5a. n 3.7 Lemma: If c x* and Y = (°) then Y is a space of finite codimension.

A property of B-convexity needed for Theorem 3.9 is contained 40 in the following lemma, which is a simple corollary of a result of

Giesy. In Chapter IV, Corollary 4.10, this property is shown to hold also for P-convexity by a quite different method.

3*8 Lemma: jf x is not B-convex, then all subspaces of finite codimension are not B-convex.

Proof; Let X = Y © Z where Z is finite dimensional. If Y is B-convex, by Theorem 1.15, X is B-convex which is a contradiction.

We now present the main theorem: If all subspaces of X having a basis are B-convex, then X is B-convex. As noted earlier, this will be proved by showing any non-B-convex space has a basic sequence whose span is not B-convex. Certain additional facts about this basic sequence are revealed in the statement of the theorem.

3.9 Theorem: If X is not B-convex, for any sequence of numbers 1 > 6^ -* o, and sequence of integers there is a sequence {e^t -♦ o, a sequence of integers p(m), and a sequence of vectors [x^ cz U(X) so that:

0 0 (1) The space L = [x.]. . is not B-convex, in particular, 1 i=1 P(m) < 1 0 For each m=i,2,..., the space [ x ^ = + i is k an ^-isometric image of , and

p(m) p(m) p(m)

(1-e ) X ja | s 11 Y fijX. 11 ja. | , i=p (m-1 )+l i=p (m-1 )+l i=p (m-1 )+l

and

(2) {x^} is a basis for L, in particular, 41 (2#) For any fa^, n, q

n n+q

II ^ aixill ^ (3 + 6 ) 11 I* i=l 1=1 and if n = p(m), m=l,2,...

ii n+q

I I ^ aixi 11 * <1 + 8n) 11 £ a x j | . i-1 1=1

Proof: (1') implies (1) by Lemma 1.10 and (2') implies (2) by Lemma 3.2. Therefore it is sufficient to construct {x^} so that

(!') and (2 ') hold.

For m=l,2,... let p(m) = S k. and let p(o) = o. Let i=l

^i ,0l as in LenBna 3.4. We will construct induction on m. We denote [x.]?^ by L , For each m we will choose i 1=1 p (m)

from a previously constructed subspace so that (1 7) is satisfied. We then will construct a subspace Am so that A^ H ” ° and the projection Pm : - Lp(m) satisfies | |pj | £ 1 +

The x^ chosen in subsequent induction steps will be taken from A^.

Following the induction we will show that (2*) is satisfied.

Let m = 1. Since X is not B-convex, we can choose k x=p(l) c U(X) satisfying (l')* To construct A^ choose l-^i^i-1^ C ^^ptl)^ ^ Lemma 3.5 and extend them to X without increase of norm so that if xeL

| |x| | £ (1+TIjl) max { ^ 6 0 s i=l,... ,q(l)l q d ) .1 ^2 Let A^ fjo). By Lemma 3.7, is of finite codimension.

^*p(l) ^ ~ °* s*nce 3Ce^*p(l)* x ^ °* c^en by the above inequality

there is some i so that f^(x) > o. Therefore there is a projection

Pj : Lp(l) ® ^1 "* Lp(l)* To see t*iat I !^i 11 £ 1 + T|^ we have for any

X ®LP(1)’ *eAl’ and some | |p (x^X) 11 = ]|x|| £ (1+H1)fi (x)

= (i+'n1)f1 c»hx) € | |x^\| | .

Now suppose we have satisfying (1/),

{ f i } ^ " 1^ c U(X*), and for n=l,2,...,m-l, An -**0 ^ ^ ( o ) , and

H Pnll s 1 + \ uhere Pn ! Lp(n) - Lp(n)' Since V l ls o£ finite codimension, by Lemma 3.8 it is not B-convex so that there are

{x.|i}!ip(in_1)+1 c satisfying (l7). By Lemma 3.5 choose

ffi*i-q(oi-l)+l C U(X* ) 80 that if XeLp(m)' llXH * ( H V m qfa) {f. (x) : i = q(m - 1) + l,...,q(m)}. Let A = Pi f4 (o). Then m i-1 exactly as in the m = 1 case, L ^ ^ HA^ = o and | |Pm | ] S 1 + ,

To show (27) holds we first observe, for any a^,i’*l,2,...,

P(m) p (m)+q

A. || Z a.x | | £ C l ^ )||Z 11 for any q=l,2,..., i=l i=l 1 1

p(m)+q p(m)

since Pm ( aixi V i H Pmll 5 1 + V

Further, we observe

B. If p(ra-l) < n £ p(m) then

II X Vi" * v Z l " I “i*1" ’ i-p(m-l)+l i=p(m-l)+l since

aixilI ^ I Kl ^ 1 K l i=p(m-l)+l i=p(m-l)+l i= p(m-1) +1 (m)

II 1 aix i II * m p(m-l)+l

using the inequalities of (1 ').

We now prove (2') in four cases using A and B.

Case 1: 1 < n < n+q < p(l).

Using B, with m*l, and Lemma 3.4-(3), we obtain

n n+q n+q

Case 2: 1 < n < p(l) < n+q.

Using B, A, where m=l, and Lemma 3.4-(3), we obtain

n P<1) n+q

n+q

< a + v l I 1 atx i 11 • i=l

Case 3: There is m so that p(m) < n < n+q < p(m+l).

Using A, and B with m replaced with m+1; we obtain n p(m) n II I V i 11^11 Z V i ll+ll I aixi II i°l i=l i=p(m)+l

n+q n+q < 0+\>l I Z aixi.l I + I T T - 11 I V i 'm+1 i=l i=p(m)+l

n+q n+q P(m) < (i+i1m)l I Z aixil I + TT II Z v j I + 11 I ai m+1 i=l 1=1 i=i

and using A and Lemma 3.4-(4) we continue

n+q

s [ 1+\ + i r r - o +m v ] II Z aixi H* 1 1=1

n+q

< <3+^ n > II X aix i II * 1=1

Case 4: There is m so that p(m) < n < p(m+l) < n+q.

Using A, and B with m+1 replacing m, we obtain

n P(m ) n Z atxi II ^ M Z aiXi z V i i=l i=l i=p(m)+l

n+q p(m+l> <

n+q p(m+l) p(m) < d+\)ll Z aixi 11 + TTT II Z aixilMI Z m+1 i=l i=l i=l 45 and using A directly and with m+1 replacing m we continue

n+q

- [ 1+\ + T^cTTTnrt-l <1+W 1+V ] II . 1 - V i 11 * i=i

and finally by Lemma 3.4-(4) ,

n+q < (3+6n ) 11 £ 11 , 1=1

proving the first part of (2*). The second part of (2*) holds by using

directly A and Lemma 3.4-(2). This completes the proof of Theorem 3.9.

We now present the second theorem about B-convexity and basis.

3.10 Theorem; If X has an unconditional basis and is B-convex,

then it is reflexive.

This theorem has been proved by both James [ 14] and Giesy [12].

Both of these proofs use the fact, from James [16], that a non-reflexive

space with an unconditional basis contains an isomorphic copy of either

c or -t.. Giesy uses the facts that c and £. are not B-convex, that o 1 o i ' B-convexity is preserved by isomorphism, and that every subspace of a

B-convex space is B-convex. James works directly from the B-convex

definition using the lemma mentioned in Chapter II, namely, that the

existence of an isomorphic copy of implies, for any e, the existence

of an e-isometric copy of and a similar lemma for c q .

The advantage of the proof presented here is that the uncon­ ditional property of the basis is used directly without invoking the

theorem quoted above. 46 3.11 Definition; A basis {x^} of X is said to be an uncon­

ditional basis if the convergence in Definition 3.1 is unconditional,

that is, the limit is zero for any rearrangement of terms a^x^.

0 0 fx.} is said to be boundedly complete if E aix^ converges for 1 i=l each {a.3 such that sup || 2 a*x, 11 < “ • 1 n i=l 1 1 {x.} is said to be shrinking if for each f c X*, lim ||f || “ 0, 1 fl-«C0 “ where llfll is the norm of f on Cx.1 " . n 1 i=n

The following lemma contains the facts of basis theory we will use.

3.12 Lemma; Let [x^} be a basis for X and { f^} the associated

biorthogonal sequence of functionals. Then

(1) If X is not reflexive, either {x^} is not shrinking

or not boundedly complete,

(2) If {xjJ is shrinking, then is a boundedly

complete basis of X*,

(3) If {x,j3 is shrinking and X is equivalently renormed,

then {xj3 is shrinking in the new norm,

(4) If {x^ is not shrinking, there is 6>0, f e X*,

{p(a)3, an increasing sequence of integers, and

{a.} so that if y = ^ . a.x. then i i=p (n)+l i *■

yn ( | = 1 and £(yn ) > 6 f°r n»l, 2 , ... ..

Proof; Statement (1) is Theorem 1 of James [16], Statement

(2) is Theorem 3 of the sane paper.

To prove Statement (3) let the new norm |||*jj| be related to

the old norm by k| j x| | < 111 x| | | < K 11 x| | for all x e X then 47

£x:||x|| £ 1] =) £x: | ||x| 11 s k} so

llflln " sup[|f(x)|:xe[xi]i“n ,||x|| £ 1}

& sup£|f (x) | 11 M 11 * kl

= k sup{|f(x)|:xe[xi]i®n , 11 |x| 11 s; 1} = k| | |f 11 |n ,

from which (3) follows directly.

To prove (4), since [x^ is not shrinking there is f so that lim||f|| ^ o and ||f|| =* 1. Therefore there is p(l), p(2), and zl c^Xi^p(l)+l 80 fc^at llzill ^ 1 + ^ an<^ ^(z^) > '4*' By the same method there are z^, with these properties and y^ = — rr z^ will n • Zn^ ' n satisfy (4).

The following lemma contains the facts about unconditional bases we will need.

3.13 Lemma: If X has unconditional basis £x^} with biorthogonal functionals we have

(1) X can be equivalently renormed so that every

rearrangement of {x^} is a monotone basis for

X. (A basis is said to be monotone if the K of

Lemma 3.2 is 1.)

(2) If {x±} is shrinking, {f^} is an unconditional

basis for X*.

Proof; Statement (1) is Theorem IV-4-1 of Day [6] and statement (2) is Theorem 3c of James [16]. The proof of both of these theorems are based on elementary facts of unconditional basis theory. 48

The construction for the announced theorem is primarily in

the following

3.14 Lemma: If X has an unconditional, non-shrinking basis

{x^}, it is not B-convex.

Proof: Renorm X so that every rearrangement of {x^T is a monotone basis by Lemma 3.13-(1). By Lemma 3.12- (3), {x^l is non­

shrinking in the new norm so there is f, y , p(n), as in Lemma 3.12-(4).

We will construct, for any k a 2 and any choice of signs

functional satisfying Proposition 2.2-(lO, namely, for this choice of

signs there is a functional g so that ||g[|£l and g(5nyn ) > j, n = 1

...,k. By that proposition, this will show X is not B-convex in the new norm, and consequently not in the original norm.

Given the signs let g be defined by

i gnf(x±) if 1 - p(n)+l p(n+l), n=l k

■r f(x.) for other i P t a i n

Then

P (n+1)

i=p(n)+l

To show | |g | | ^ 1 let

P = {i:p(n) < i £ p(n+l) for some n so that = +1}

N = [i:p(n) < i £ p(n+l) for some n so that = -1}

I = [i:i /Sp u n ] CO 49

Take any x so that | |x| | 1. Write x ** aix j_ ant* ^■e*: i=l

Y y y y = 4i.a.x. - Li a.x. + L> a.x. ieP ieN * 1 iel 1 i

Since the rearrangement of {x^}, ^xi^iePUl^ *S a inonotone basis, we have

|| £ aixill <: ||x|| <; 1 so that ieN

I b11 = Ilx-2 X aixi 11 * 3 ieN

and |g(x) | = |-f(y) | £ |||f 11 | |y 11 <; 1 .

Proof of Theorem 3.10: If X is not reflexive, {x^} is either non-shrinking or shrinking and non-boundedly-complete. If the former, the proof is complete by Lemma 3.14. If the latter, the biorthogonal functionals form an unconditional non-shrinking basis for X* (unconditional by Lemma 3.13-(2), since {x^l is shrinking, and non-shrinking by Lemma

3*12-(2) since X* is not reflexive). Thus by Lemma 3.14, X* Is not B- convex, so neither is X. CHAPTER IV

P-CONVEXITY AND OTHER PROPERTIES RELATED TO B-CONVEXITY

In this chapter we consider three properties of Banach spaces, and the relations between these properties and B-convexity.

The first of these properties, P-convexity, was introduced

in [20, 21] by C. Kottman.

4.1 Definition: For a Banach space X, let P(n,X) be the supremum of all numbers r so that there is a set of n pairwise-disjoint

closed balls of radius r inside U(X). We say X is P-convex if

P(n,X) < ^ for some n.

Note that P(n,X) £ | if n > 1 and P(2,X) = for all X.

Another way to express the property P-convexity is found in

4.2 Lemma: P(n,X) < ^ holds if and only if there is some

e,o < e < 2, so that if are distinct points in U(X), then there are i^j so that ||x^ - x ^ 11 < 2 - e .

This is essentially Kottman's Corollary II.4 and Remark II.5.

A set 80 that | (x^ - | | a 2 - e for all i,j=l,...,n,

i^j, is called a 2 - e separated set of order n. P(n,X) < ^ means that for some e there is no 2 - e separated set of order n in U(X).

Using this lemma it is easy to see (Kottman's Theorem IV.24) that if a space is P-convex it is also B-convex; in particular, if

P(n,X) < ^ then X is n, e-convex for some e > o .

To see this observe that if x^,...,xn are distinct points in

50 51

U(X)» by Lemma 4.2 there is 6, i» j so that ||x^ - x^ | ( < 2 - 6

Hence

n n II ' xjH s II I *JI + IK ' *jll

Note that the signs on xm ,tnj£L,j, are in fact completely arbitrary, so that for many different choices of signs the required inequality holds, including a choice with only one minus sign and a choice with only one change of signs.

It is not known whether there is a space which is P-convex but not B-convex. This problem will be discussed further in Chapter V.

Kottman proved that all P-convex spaces are reflexive. The proof uses the device of James discussed in Chapter II to show that if

^ 2n then the space cannot be P-convex.

To compare P-convexity with B-convexity, we will discuss analogs of facts about B-convexity which are listed in Chapter I.

Analog to 1.3: If P(n,X) < ^ and Y c X then P(n,Y) < ^ .

This is obvious using Lemma 4.2.

Analog to 1.4; If X is uniformly convex, then P(3,X) < ^ .

This is Kottman's Theorem IV.31.

Analog to 1.5: If X is finite dimensional it is P-convex.

This follows from the compactness of U(X). No relationship has been shown between n so that P(n,X) < ^ and the dimension of the space. The smallest n so that P(n,j&™) < ^ is greater than 2m since the 2m points of the form Eg1•* *» w ^ere take on all possible 52

combinations of +1 and -1, form a 2 separated set of order n. It is

not known whether there is an m dimensional space X having P(2m+l,X) =

Analog to 1.6: P(n+l,X) £ P(n,X) for any X.

Analog to 1.7 and 1.9; The P-convex analog to these properties would be: If X is P-convex then lim P(n,X) =» o. While this is true for n-*<» finite dimensional spaces (our Corollary 4.8), it is false in general. 7*2 f-u For example, in let x. = ___ !___ e. where e. is the i usual basis 1 J ~ 2 + l 1 1 vector. For any n,e the balls

Bi = {xej&2 x —x , | | £ "■ — s] t i”l ,... ,n] ,/2+l

are disjoint (since ||x,-x. || = ---— and twice the radius is J J 2 + 1 2 J~2+1 ” an(* *n (since if xeB^ then

x| | ^ | |x-x. || + | |x. 11 = — 1— - e + = 1 - e) J 2+1 J 2+1

so that for any n

P ( n , 0 £ J 2+1

Analog to 1.10: No e-isometric imbedding property is known

to characterize P-convex spaces.

Analog to l.lli P(n,X) = P(n,X**).

This holds if X is P-convex since in that case X is reflexive.

If X is not P-convex, F(n,X) = 1, and using the cannonical imbedding

of X in X** and the Analog to 1,3 P(n,X**) = 1 . 53

Analog to 1.12; The property dual to P-convexity has been described by Kottman. However, it is not known whether this dual property is equivalent to P-convexity.

Analog to 1.13: It is not known whether P-convexity is preserved under isomorphism.

Analog to 1.14: It is not known whether P-convexity of

Z and X/Z implies P-convexity of X.

Analog to 1.15: We will present two theorems (4.6 and 4.9) showing under certain conditions if X ■ Y © Z and Y and Z are P-convex then X is P-convex. The question is not solved in general, however.

Analog to 1.16: We will show (Proposition 4.11) that a space is P-convex if each separable subspace is P-convex.

We will also present an analog to our Theorem 3.9 for P- convexity.

The problem of proving that the direct sum of two P-convex spaces is P-convex seems more difficult than that for B-convex spaces.

We will prove this for the following special cases: If X = Y @ Z, and either

(1) The norming of X is ||(y,z)|| = max{||y||, ||z[|], or (2) Y is finite dimensional and Y and Z are subspaces of X.

These results are based on a theorem of combinatorics proved in 1930 by Ramsey [27]. Ramsey's theorem can be stated as follows.

4.3 Theorem (Ramsey): Let p, q, and r be integers so that p,q a r > 1. Then there is a number n(p,q,r) having the following property. Let S be a set having n(p,q,r) or more elements. Let the 54

family of all r-subsets of S (where an r-subset is a subset having r

elements) be divided into two disjoint families, a and 3 . Then either

(1) There is A c S, a subset with p elements, so that

any r subset of A is in a, or

(2) There is BcS, a subset of q element, so that

any r' subset of B is in 3 *

We use this theorem to prove the following lemma.

4.4 Lemma: Let A and B be sets, P^ a property which a pair

of points (a^,a^) in A may have, and P^ a property on pairs of points

(b^,bj) in B. Suppose there is an integer so that if a^,...,an ,nsN^,

are distinct points of A, then there is i,j so that (a^,a^> has P^, and

there is Ng with the corresponding property for B. Then there is an

integer so that if n a n a b * ai,’*’*an distinct Points of A,b^,...,bn

distinct points of B, then there is i,j so that both (a^a^) has P^

and (b. ,b.) has P„. (By "pair1* we mean unordered pair.) i j B

Proof: Let Nq = max (N^»^) and let = n(No ,NQ ,2) from

Ramsey's theorem. For n S N^g let S = fl,...,n}. Given let

Of = £(i,j) : (a^a^) does not have-PA }

B - £(i»j) i (bt ,bj) does not have PB ; (i,j) k or]

Now suppose there is no i,j as asserted in the lemma. Then a U S is

the set of all pairs of elements of S. Also a H B = so Ramsey's n theorem applies. If Conclusion 1 holds, there is fi 1 . c S so that n ■ n=l n each (i ,i ) e at . Thus fa. 1 . is a set of N. or more points, no nm L x n=l A r 55 pair of which has P^, which is a contradiction.

Lemma 4.4 will be incorporated into the following lemma for ordered pairs (a,b)eAxB

4.5 Lemma: Let A.B.P. ,P„,N. be as in Lemma 4.4 with A B A B AB the additional property that a pair having the same first and second elements of a, (a^,a^), always has P^, and the corresponding property for B. Then if £ *s a set distinct pairs from AxB,

(i.e., any two pairs differ in the first or second entries, or both) and n ^ ^AB^A^B* t*ien there is i,j so that both (a^,a^) has P^ and

(b^b.) has Pfi. 1 2 rA Proof: Let a ,a ,...,a be the distinct values of

and write the sets

x { ( a ^ b ^ : a± = a1}, {(ai,bi> : a± = a2],... ,{(ai,bi) : a± = a A } .

t”h If one of these sets, say the K , has N_ or more pairs, then for these £ w pairs {b^ : a^ = a } are distinct and so there is i,j so that (b^jb^) K K has Pp. By hypothesis (a^,a^) =(a ,a ) has P^ so the conclusion of

the lemma holds. Otherwise each of the sets have less than Ng pairs,

so that the total number of pairs in all of the sets is n < Nfir^.

Since na b NANB £ ^ we have rA > NABNA* By choosin6 one Pair from each rA of the sets, we get a family of pairs {(a ,b^. )}n„, having distinct n 1 2 rB first elements, i.e., if n ^ m then a r a • Now let b ,b ,... ,b rA be the distinct values of {b^ )n_^ and write the sets n ~ *

1 T, rB {(aIl,bi ) : b - b 1},...,[(aIl,bi ) : bf = b } . n n n n 56 th If any one of these sets has NA or more elements, say the K , then

{a11 : = b^} are distinct and there is j,k so that (a^,a^) has PA

and (b. ,b. ) = (bK ,b^) has P_. Since (a^,b. ) and (a^,b ) in the j k B \ original set of pairs { che conclusion of the lemma holds.

Otherwise each of the sets have less than pairs, so that the total

number of pairs in all the sets is r^ < Since we showed

NABNA < rA we have r^ > NAfl. Take one pair from each of the sets to

get

{(a ^»b^)}j®lt a subset of f »

ii • ■ * n * so that if j ^ k then a * a and b** ^ b . Thus fa },=i end 1 rB are distinct points of A and B. Applying Lemma 4.4 to these

pairs concludes the proof.

4.6 Theorem: Let Y (£> Z be the direct sum of two P-convex

Banach spaces normed by ||(y,z)|| = max (||y||>||z||)» Then Y © Z is

P-convex.

Proof: By Lemma 4.2, there is so that if are distinct points in U(Y) and n ^ ny we have some i,j so that | ly^Yj ! I

< 2-Cy. Similarly there is nz ,ez with this property for points in

U(Z). Let A = U(Y). Say (y^y^) has PA if | Jy^-y^ 11 < 2-e, where

e = min(b ,ez). Let NA = Similarly let B ® U(Z) and define P^ and Ng. Let £ * z±^^±=1 ^^st^nct Pflirs in U (Y + Z), n ^ NABNA^B*

Then c A, fzj^i=i By hemma 4.5 there is i,j so that

(^•yj) has PA ; i.e., | |yi-yj 11 < 2- e 57

and (zi »2j) has Pgj i.e., | |z±-z^| | < 2-e

and thus

11

To prove the second direct sum theorem we need

4.7 Lemma: If X is finite dimensional and e>o there is N N so that if are distinct points in U(X), there is i,j such that

I K " xj I I < e.

Proof: For each y e U(X), let S = {x : | |x-y| | < * y ^ The family {S^ : y e U(X) } covers U(X), which is compact, so we may find

^i^i”l 80 fc^at £Sy : cover U(X). Then if Cx£}^_^ c U(X)

there is some i,j ^ N, i , so that x.,x. e S , and ||x^-x.|| < e . c j y ^ 1 o

We note that this trivially provides a partial analog to

B-convex facts 1.7 and 1.9:

4.8 Corollary; If X is finite dimensional, lim P(n,X) = o n-*» 4.9 Theorem: Let Y and Z he subspaces of X so that X = Y © Z.

If Y is finite dimensional and Z is P-convex then X is P-convex.

Proof: Since Z is P-convex there is nz»6 so that if are distinct points in U(Z) and n £ n^ then there is i,j such that

[Jzj-Zj | | <2-6. Since Y is finite dimensional, by Lemma 4.7 there is 11^ so that if are distinct points in U(Y) and n & n^ then there is i,j such that | ly^-y^ | | < ^ • Let u 0 0 = A, say (y^y^) has

PA if I lyi"y j I I 2* and let NA = "y* Let ” B * say (z ± * z p has PB if | |z^"Zj || <2 - 6, and let Let distinct 58

pairs in TJ(X), n a Then c ^ an<* ^zi^i-l c B*

By Lemma 4.5 there is i,j so that

(y^Yj) has PA ; ie | ly^y^ 11 < | ,

and (zitZj) has PB ; ie [\ z ^~ z ^ | j <2-6

Thus

I l C y ^ ) - (yyt-z^H * Ibi-yjH + I I2i+Zj 11 < 2 - | *

4.10 Corollary: If X is not P-convex, and Y is a subspace of

X of finite codimension, then Y is not P-convex.

Proof: Write X = Y © Z where Z is finite dimensional and use

Theorem 4.9.

In analog to B-convex Fact 1.16 we prove

4.11 Proposition: X is P-convex if each of its separable

subspaces is P-convex.

Proof: If X Is not P-convex, choose a sequence en -* o of

positive numbers less than 2. For each n, by Lemma 4.2, there is x” ,...,x^ c=U(X) which is a 2-en separated set. Let Yr = [x” ,...,x^]

and let Y = [Y^Y^...]. Y is clearly separable. It is also not

P-convex, since for any n,e we can select m so that em < e, m £ n,

making {x™,... ,x™3 a 2-e separated set of order n in U(Y).

Improving upon this proposition, we have in analog to Theorem

3.9 for B-convex spaces, the fact that a space is P-convex if each sub­

space having a Schauder decomposition into finite dimensional subspaces

is P-convex. We will use the following

4.12 Definition: A sequence [M^] of closed subspaces of a

Banach space E is a Schauder decomposition of [M^] if every element u of [M.] has a unique, norm-convergent expansion u = ^ u ., where i=l U j ^ for i=l,2,...

Grinblyum [13] has characterized Schauder decompositions as follows.

4.13 Lemma: A sequence [M^] of closed subspaces of E is a

Schauder decomposition of [M^] if and only if there is a constant K such that for all integers m,n and all sequences [u.] with u .eM. we have n n+m

We are now ready for the analog to Theorem 3.9, namely, a space is P-convex if every subspace having a Schauder decomposition into finite dimensional subspaces is P-convex.

4.14 Theorem: If X is not P-convex, for any sequences of numbers {e^J 1 > 6^ -♦ o, l > e i -»o, and any sequence of integers

{k^}, k^ -* oo, there is a sequence of integers p(m) and a sequence of vectors [x^3 C U ( X ) so that

(1) The space L = [x^] is not P-convex, in particular

(1') For each m=l,2,... there is a set of distinct k points f y ™ } ^ <= u ^ xi3iipXm -i)+i> 80 that

||y“ - yjll > 2 - ^ for i ? j, i,j=i,...,km .

and (2) L has a Schauder decomposition into finite

dimensional subspaces, 60

in particular

(2 0 Letting Lm = [xjjip(m.1)+1» for any n »(l» C^}

where u^eL*, i=l,2,..., we have

n n+q I! S u ill £ (1+6 ! I ^ U 11 . 1=1 1 1=1 1

Proof: (1 0 implies (1) by Lemma 4.2. (2 ') implies (2) by Lemma 4.13. We will prove (1/) and (2'). As in the proof of

Theorem 3.9, we will construct fx^] in blocks {x£li-m (m i)+i by p (m) induction on m. Denote by bm- For each m we will choose

and K }iip(m-1)+1 from a previously constructed subspace so

that they satisfy (I7). We will then construct a subspace of finite

codiraension in X,Am » so that A^flL^ = o and so that the projection

P : L + A_ -* I* has norm less than or equal to 1 + 6 . The m m m m ’ m t {x^l, {yj}n “l chosen in subseauentsubsequent induction steps are taken in jA^

Let m = 1. Since X isLs not P-convex, by Lemma 4.2 tlthere is 1 kl a set of distinct points fy. 1 C U(X) so that (l7) is satisfied forEoi kk. m=l. Let be a linearly independent set spanning L^ *» [y^] j—l *

To construct A, choose C U(L*) by Lemma 3.5 and extend them

to X so that if xgL^ j|x|J £ (1+6^) max {^(x) : i=l,... ,q ( 1 ) } . •l(l) _ Define A-. - 0 fj (o). Exactly as in Theorem 3.9, we have 1 1 = 1 X

Pj : L r ^ Ax -» Lx and | |Pt | | s 1 + 6r

The induction step is similar to that of 3.9. Since A . is m- i of finite codimension, it is not P-convex by Corollary 4.10, and so

contains fy™}^ satisfying (l7). Taking (m_i)+i to be a k linearly independent set spanning the induction step is 61 completed exactly as In the B-convex proof.

To see that (2 ') Is satisfied, observe that, for u^eL** i=l,2,... and any n,q,

- n+q n

Pn(I ui)"Zui and H Pnllsl + 6n • i=l 1=1

The second property to be discussed in connection with B- convexity is known as the Banach-Saks property. A space is said to have the Banach-Saks property if each bounded sequence of elements has a subsequence whose (C,l) means converge strongly. Banach and Saks [1J proved that this property holds in L (o,l) and j?. (p>l). The proof of P P Kakutani [18], with the fact that all uniformly convex spaces are reflexive, shows that all uniformly convex spaces have the Banach-

Saks property. Nishiura and Waterman have proved recently in [24] that all spaces having the Banach-Saks property are reflexive. It is not known whether there are any reflexive spaces which do not have the

Banach-Saks property.

It would be interesting to know whether B-convexity implies the Banach-Saks property since this latter property implies reflexivity.

This is not known. However, the converse question is known to be false.

4.15 Proposition: There is a space which is not B-convex but has the Banach-Saks property.

Proof: Nishiura and Waterman [24] have shown, for p > 1, that P has the Banach-Saks property. But this space is not B- P convex since, as shown for c q , example g of Chapter I, for any k,e we can find x^,...,x^ violating the B-convexity inequality.

t 62

Nishiura and Waterman used the following property, which they called (*):

(*) There is 9e(o,l) so that if {xn } c U(X), x r converges weakly

to zero, there is i, j so that Ijx^ + x^ | [ < 20.

Kakatutani, [18], using an equivalent formulation of (*), showed every uniformly convex space satisfies (*) and that every reflexive space satisfying (*) is Banach-Saks. It is interesting to compare (*) with the following formulations of P-convexity and 2,e-convexity.

P-convexity: There is N and 0e(o,l) so that if

C U(X) there is i,j so that

I 11 < 8 •

2,e-convexity: There is 0e(o,l) so that if x,y c U(X)

either ]|x-y|| < 20 or ||x+yj| < 20 .

The known relationships between these properties are summarized in the following diagram: if space is reflexive

2,e-convex P-convex B-convex Banach-Saks

All of these conditions are generalizations of uniform convexity.

All except B-convexity and (*) are known to imply reflexivity. The last property to be discussed was called "near convex" by D. R. Smart in [30], As was noted in the review of this paper,

[31], near-convexity is the same as 2,e-convexity. We say a series CO ^ x is absolutely bounded if there is a number N so that for any n=l

finite set of distinct integers [n.} we have | x || £ N. l n t

Smart showed that if a space is 2,e-convex then each absolutely bounded

series converges. 63

It is natural to ask whether this also holds for any B-convex space or for any reflexive space. We will shot? that the answer to both questions is yes. 00 4.16 Definition: E x is weakly unconditionally convergent n 00 I if for any permutation of integers k the sequence E x. converges n - ie n«l n weakly.

00 4.17 Lemma: In any Banach space, T] is absolutely bounded if and only if it is weakly unconditionally convergent.

Proof: Bessaga and Pelczynski ([4] Lemma 2) have shown GO that E x is weakly unconditionally convergent if and only if there n n=l is a constant C so that for every bounded real sequence {t^} n

sup L,* 5 C sup |t.j i i n i

CO Thus if S x is weakly unconditionally convergent it is absolutely I n CO bounded. To see the converse suppose E x is absolutely bounded with n ^ l n constant N but not weakly unconditionally convergent. Then there is a permutation k and a functional f so that E fCxt. ) does not converge. n=l n Hence there are sequences and a number e>o so that

**m

I > o m = 1,2,... £(xk > n=P, n

Without loss of generality %

It > o TO — 1)2)* i f(xk > n*Pi inm 11 64 (Otherwise, if positive for infinitely many m we may reindex to omit those n for which the sum is negative, or if positive for only finitely many m, change the sign on f and reindex as above.)

For any k, we have

k k «m r-i k e < f ') f N 1 L *k J * 11*11 L> 1 *k n n hf*i1 n=p tn=l n=p *m *m since £ x^ is absolutely bounded by N, which is a contradiction if k > M l i . e

Theorem 5 of [4] states that the following are equivalent:

(1) There is a sequence in X which is weakly unconditionally

convergent but not unconditionally convergent,

(2) X contains a subspace isomorphic to c q .

This gives directly, with the lemma,

4.18 Theorem: The following are equivalent:

(1) Every absolutely bounded series in X converges

unconditionally,

(2) X has no subspace isomorphic to c q .

Since reflexive spaces and B-convex spaces have no subspace

isomorphic to c q , Smart's result generalizes as desired.

The idea of this proof is due to Prof. W. J. Davis. CHAPTER V

SOME UNSOLVED PROBLEMS

In.this chapter some unsolved problems related to the topics of this dissertation are discussed. Some of the problems have also been treated in earlier chapters. The first group of problems relate to B-convexity.

Problem 1: Is every B-convex space reflexive?

This was conjectured to be true by James in [14] and [15].

To support this conjecture there are the three subclasses of B-convex spaces known to be reflexive (2,e-convex spaces, spaces with uncon­ ditional basis, and P-convex spaces) and the list of properties

(Theorems 1-12 through 1-16, and Theorem 3-9) which remain true if the words "B-convex" are replaced with "reflexive".

In [3] Beck asked

Problem la: Is every B-convex space isomorphic to a uni­ formly convex space? and in [12] Giesy asked

Problem lb: Is every B-convex space isomorphic to a

2,e-convex space?

If either of these problems could be answered yes, Problem

1 would be solved.

In [15] James conjectured that Problems la and lb are equivalent: 66

Problem 2: If a Banach space is isomorphic to a 2,e-convex

space, is it also isomorphic to a uniformly convex space?

Ihe converse to Problem 2 is trivially true. James points t* out that if Problem 2 can be answered yes, then the existence of an

isomorphism to a uniformly convex space would be equivalent to the

existence of an isomorphism to a uniformly smooth space. This follows

since uniform smoothness and uniform convexity are dual properties [9] while 2,e-convexity is self dual [17]. Further discussion of Problem

2 can be found in [17].

Other questions which would support the truth of Problem 1

are

Problem 3: If a space is B-convex and separable, is either

the first or second dual separable?

Problem 4: Is every B-convex space weakly complete?

The most important question regarding P-convexity is

Problem 5: Is every B-convex space isomorphic to a P-convex

space?

If the answer to Problem 5 is yes, Problem 1 is solved.

Kottman, in some private communications, has given a set of examples which illustrate the difficulty of this problem by showing that there

is no simple relationship between k for which a space X is k,e-convex

and n so that P(n,X) < ^ . He provides, for each n, a finite dimen­

sional space B for which there is e so that B is 2,e convex, but n n n n having P(n,Bn) = ^ . However, any space containing all the B^ is not

B-convex. Kottman1s examples and their properties are described in detail below. 67

5.1 Definition: Let be the space of n-tuples with norm

defined by

For example, if n = 2 we have

9

so that U ^ ) an hexagon*

5.2 Proposition; For each n there is en so that B^ is

2,en~convex.

Proof; We first prove it is sufficient to show there is no x,yeU(Bn ) so that ||x-y|[ = ||xfy|| - 2 : If is not 2,e-convex for any e, for each m we can find xm »ym c ^(Bn > so that for both signs,

||xmdym |j >2(1-^). Since UCB^) is compact, there is 0^}; x,yeU(Bn), so that x^ -♦ x and y^ -* y. For any fi > o we can choose j so that

2 g and — <■£•, giving, 9 m. j j j

which is a contradiction. — —

Now suppose there is x,yeU(Bm ) so that |Jx±y|| = 2. Then there is k, 1 £ p1 < ... < pt £ n, so that Thus equality holds and sgn (x ) = sgn (y ) for all i such that pi p i both and y n are nonzero. Also we have p i p i k . k , * IMI-- i-i i*l 2 1 i=l 2 x

Now If there is i 7 so that x^ = o, then x^ = o for all p i 7 p j j > i 7, for otherwise we have

|x I |x I |x I 1 P1 ' P2 . ' V x I I = ---— * + ----:— +...+ 2° 2 1 2k

lxp I lxp , I lxp , I K I 1 , , pi -1 , pi 7+l , , pk * '-2 T P ‘ „k-l 2° 2" "*• 2s- r lxp ! Ixp , I lxp ,1 lxp 1 . P 1 . , pi -1 , Pi +1 , , Pk „ i[ . i C ^ +...+ 2i'_2 + 2i'-l "* 2k-2 I ■ 9

which is a contradiction. Therefore, there is k 7, k 77 so that 1 / A * * y Ktl y 11* 1! = i). —- & ± ■ r . Ilyll Iv I -L= L - £ • , i=l 2 i=l 2 L

t / and x ^ o,i=l,,..,k7 , y^ ^o,i=l,...,k , p i pi

and we can assume k 77 £ k 7.

In the same way, using I |x-y I J, we can get m,l^q^<...<=qin

= -sgn(Xj). But since the norm of y is achieved with both the sets

/;] and fq^,...,^}, there is some j so that |y^ |=maxfy^}^^r lc and j must be in [p.,...,p T fl (q.,... ,q "I. (If not, k" or m < n *> 4 // 1 k m and the p^ or q^ can be rechosen to include j and thereby produce a larger sum.)

5.3 Proposition: P(n,Bn) = ^

Proof; We will exhibit n distinct points c ^(Bn) so that |Jx^-x^|| = 2 for i 4 j.

Let x. = (l,o,....,o)

t I 1 n xn ("2 *••••

k lXPi To see that x. eU(X), the largest sum — r—r is obtained with k = i 1 i*=l 2 and p^,...,?^ = l,...,i, in which case the sum is so that ||xt !! = 1 , i-l,...,n. To see that |jx^-x^|| = 2, suppose i < j . Then the number in the i place of x, - x, is ± ^ , in the

In view of Propositions 5.2 and 5.3, one might attempt to use

Theorem 1.19 of Giesy to show that P . B is 2,e-convex for some e but ju n P not P-convex. However, the following proposition shows that this cannot be done.

5.4 Proposition; If X is a space containing isometrically

B ,n=l,2,..., then X is not B-conve:c. n* 1 ’ Proof: We first observe two elementary facts.

A. Let a be any number, m s n. If x is an n-tuple

having a in m of the n entries

This holds since we can let p^jp^,...,pm be the indices of the entries which are a. Then

m m-1

B. Let b be any number.

This fact is clear since the largest possible sum — i=l 2 71

We now construct, for any k,e, c U00 80 that for any choice of signs we have

> k(l-c) i i i=l

1 k Choose m so that — < e. Let n *= m'2 , Let A be the m-tuple 2

^ aTU* ^orm c^ e n_tuPle

xx — (A,-A,A,-A,... ,A) ,

using 2 m-tuples A. Similarly let

Xj ~ (A,A,“A ,“A ,A , * • •,A)

x^ = (A,A,...,A^-A,-A,...,-A) ,

in each case using 2 m-tuples A. The pattern of signs is the same as in example g of Chapter I, c q. From Fact B we have ||x^ll £ 1. k For any choice of signs, the sum ?, £.x. will be an n-tuple containing i-1 the m-tuple kA, so by Fact A above k r-*

£ « i « i - ^ ) > k • 72

A b was pointed out in Chapter IV, many facts known true for

B-convex spaces are not known for P-convex spaces. The truth of these would support the truth'of Problem 5.

Problem 6: Is P-convexity self dual?

Problem 7: Is P-convexity preserved under isomorphism?

Problem 8; If Z is a closed subspace of X so that both Z and X/Z are P-convex, is X P-convex?

Problem 9: If Y and Z are subspaces spanning X, does

P-convexity of Y and Z imply P-convexity of X?

Problem 10i If every subspace of X having a basis is P-convex, is X P-convex?

Note that an attempt to extend the proof of Theorem 4.14 to obtain this result, would need a number M and a family of projections from each Ln onto spans of some of the basis elements of Ln such that the norms of all the projections are less than M.

It is known that for a given finite dimensional space such projections can be found with norm less than or equal to the dimension of the space, for example [32], but we know of no bound independent of the dimension of the space.

Problem 11: Is there a number N(d) so that

if dim N(d^ P(n,X) < j 1

As was pointed out in the discussion of the P-convex analog to 1.5,

N(d) would have to be at least 2^.

Related to this, we have 73

Problem 12: Is there a number D(n) so that

if P(n,X) = | then dim (X) > D(n)? J , 'V The major problem regarding the Banaclni-Saks property and

B-convexity is

Problem 13: Does every B-convex space have the Banach-Saks property?

If the answer to Problem 13 is yes, Problem 1 is solved.

The converse to Problem 13 is false, as was pointed out in Proposition

4.15.

Related to this we have

Problem 14: Is there a reflexive space which does not have the Banach-Saks property?

If the answer to Problem 14 is no, Problem 13 is equivalent to Problem 1.

Problem 14 was first raised by Sakai in [28]. In [19] Klee attempted an affirmative solution , but as was pointed out in [33], his example was not reflexive. BIBLIOGRAPHY

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