Israel Journal of Mathematics, Volume 4, Issue 4 (1966)

FARTHEST POINTS IN REFLEXIVE LOCALLY UNIFORMLY ROTUND BANACH SPACES BY

EDGAR ASPLUND ABSTRACT

If S is a bounded and closed subset of a Banach space B, which is both reflexive and locally uniformly rotund, then, except on a set of first Baire category, the points in B have farthest points in S. In the note [3] by Edelstein it is shown that, if B is a uniformly rotund Banach space and S a bounded and strongly closed subset of B, then the set

a(S) = { c ~ B : 3 s ~ S such that l l c - s t l > I I c - x l l V x ~ S } i.e., the set of all points in B that have farthest points in S, is dense in B. Here we will show, by a different method, that the restriction on B can be slightly relaxed and at the same time the condition on a(S) strengthened. We will require B to be reflexive and locally uniformly rotund (LUR), a condition first introduced by Lovaglia [4]: (LUR) Given e > 0 and x in B, with Hx I[ = 1, there exists a 6(e,x) such that

' -II (x + z)/2 II >for all ~ in B such that I1= II =< 1 and II x-=

'~

II-->="

Then a(S) will be shown to contain the intersection of a denumerable family of open and dense subsets of B. By the Baire category theorem, the intersection is itself dense in B. Such a set will be called a fat subset of B. We begin with a lemma on convex functions in arbitrary Banach spaces. LEMg_A 1.

I f f is a convex function which satisfies a Lipschitz condition

f ( x ) - f ( y ) 0 there exist z in B and real numbers a, k satisfying Received July 19, 1966 213

214

EDGAR ASPLUND

[December

f<x) _= II y - ~ II - II y - ~ II with equality in the above case. Thus

216

EDGAR ASPLUND

- ~ Irp II Ify - z l[ s 82/k + k ( - 2 ~ + ~2)11 y - z II~ Again put 2 = 8/k y - z , this is possible since by (2) and the condition

5 ~ IIp 11/4 we h a v e . ~ k IIy - z [I We get an inequality which combined with (2) gives (3)

[ I]P][-2k][y-z[I

[ < 28.

Since e > 0 can be chosen arbitrarily small, this proves Lemma 3. To see that the Corollary follows from Lemma 3, we use the estimate of Lemma 2 instead of the subgradient relation in (1) and proceed as before, the result is (2')

1 _-- 1 whereas C = 1 in the Lipschitz condition for r proves the opposite inequality. We can now prove the statement announced in the beginning. THEOREM. I f the reflexive Banach space B is (LUR) and S is a bounded and closed subset of B , then a(S) is f a t in B. Proof. We will in fact prove that the set a(S) contains the earlier constructed set E for the function r on B. Suppose, then, that y is in E and take p in ar(y). The functional p assumes its minimum (since B is reflexive) on B(y,r(y)) at a point x. We will show that x is in S and hence a farthest point to y in S. With no loss of generality we assume that y = 0 and r ( y ) = 1, i.e., 1]x ]l = 1. Consider now the function r on the segment from 0 to - x . By the corollary to Lemma 3 the increase of r on this segment is exactly one. Hence the ball B ( - x , 2) is the smallest ball with center - x that contains S, so we may find points z in S very near to the boundary of this ball, i.e.

1 - I[(x + z)/2 II can be made arbitrary small. Say that it is made smaller than the quantity 6(e, x) in the condition (LUR); it then follows that l[ x - z [! < 5 because IIz 11 0 . n=l

To conclude the proof, we show that, for any given m, Em+~ is not essentially contained in any finite union of translates of Am, let alone of Era. 3. More specifically, we claim that for any given positive integers m, k and any tl, ..., t k ~ R we have k

(5)

# ( [ 1 - 2-k, 1] r~Em+,) > p ( [ 1 - - 2-~,1] N iU=l(t,+Am)),

which implies the desired conclusion. Indeed, since p(m + 1 ; n) > p(1; 1) = 2 for all n, [1 - 2 -k, 1] does not meet any holes of Am+ 1 of any order strictly smaller than k; and it contains, for each n > k, 2 n - k holes of order n and total m e a s u r e 2n-k2 -n-v(m+l; n) = 2 - k - v ( m + l ; n) < 2 -"-v<m+l;k) (on account of (4)). Therefore (6)

p([1-2-k,

1] x E m + l ) = # ( [ 1 - 2 - k ,

1]nAm+l)

oo

> 2-~ _ ~ 2-"-v("+l;k) = 2-k(1 _ 21-v(m+1;~)). n=k

On the other hand, [1 - 2 -k, 1], being of length 2-*, contains at least one interval of the form 01 + 2 - k - l r , tl + 2 -~- ~(r + 1)) (r an integer), and therefore contains one complete hole of t, + A., of order k + 1 and length 2 -C*+*)-v°":*+*) = 2-q("'*;k); this hole again contains a complete hole of t2 + A , . of order q(m, 1;k) + 1 and length 2 -q(m'2;*) (we here use (2)); continuing in this fashion, we find that [ 1 - 2 -k, 1] contains an open interval of length 2 -~(''k;*) that is

1966]

SETS WITH HOLES O N THE REAL LINE

227

a hole of tk + A= contained in a hole of tk_ 1 + Am'." contained in a hole of tx + A=, and thus disjoint from [.JR 1 (q + At,). Therefore, using (3), k

/~([1 - 2 - k ] n ~=,= (fi +

Am)) =< 2 -k -- 2 -q(''k:k) = 2-k(t

- 22-p(m+l;k)).

This, together with (6), implies (5), thus establishing our claim. 3. The application. R+ denotes the positive real half-line. Measure, measurable, null, all refer to Lebesgue measure. L~ is the usual Banach lattice of (equivalence classes modulo null sets of) essentially bounded measurable real valued functions on R+, with the essential-supremum norm I]" I]" A constant in L =, is denoted by its constant value, the characteristic function of the measurable set E c R + by X~. For f e L °°, s e R, we define the translate T~fe L °° by Tj(t)

Yf ( t [ 0

s)

t > max{0, s}

0 2 there is a finite-dimensional Banach space which is universal for the class of all j-dimensional spaces. This question was solved negatively for j = 2 (and hence for every j > 2) by C. Bessaga [1]. He showed that the set of all subspaces of a finite-dimensional space cannot include all two-dimensional spaces since it has a too low dimension. (Since we do not use this fact here we do not enter into the precise definition of the notions involved in it.) Klee [4] has extended considerably the argument of Bessaga and obtained for example numerical estimates for the dimension of Banach spaces universal for all j-dimensional spaces whose unit cell is a polyhedron with r vertices. Our first aim in this paper is to present a different method for proving the non-existence of certain kinds of universal finite-dimensional Banach spaces. Our method seems to be conceptually more elementary than that of Bessaga and Klee. We exhibit for every n a set of 2" polyhedral 3-dimensional Banach spaces such that no Banach space of dimension < cn/log n has all those spaces as subspaces. Our aim is mainly to present the method and we have not tried to get sharp estimates. From the results of Bessaga and Klee it follows by a routine compactness argument that for every j and n there are k(j, n)j-dimensional spaces such that no n-dimensional space is universal for them. It seems to us that from the point of view of estimates for k(j, n) our method will probably give better results than those obtainable by [4]. We have not checked Received October 7, 1966. * The research reported in this document has been sponsored by the Air Force Office of Scientific Research under Grant AF EOAR 66-18, through the European Office of Aerospace Research (OAR) United States Air Force. 235

236

JORAM LINDENSTRAUSS

[December

this and moreover we are sure that also the estimates of k(j,n) which can be obtained by our method are rather crude. The method used here is closely related to the one used in [7, pp. 98-99], for proving a result on the extension of operators. As far as Mazur's problem is concerned, this method has one obvious disadvantage in comparison with Bessaga's. It cannot be used in the case j = 2, it works only for j > 3. It is hoped that the method can be modified so as to solve some open problems concerning infinite-dimensional universal spaces (see section 5(a)). At the end of his paper [4] Klee mentions some infinite-dimensional questions. One of the questions (originally posed by S. Mazur) is whether there exists a separable reflexive Banach space which is universal for all separable reflexive spaces. This problem was solved negatively by W. Szlenk (private communication 1965). Szlenk showed in fact that there is no Banach space X with a separable conjugate such that every separable reflexive space is isomorphic to a subspace of X. We present here an extremely simple solution of Mazur's problem. Our method is not powerful enough to derive Szlenk's result but it applies in many situations in which Szlenk's method does not apply. Section 4 is devoted to the solution of another problem mentioned at the end of [4]. Klee called a Banach space X polyhedral if the unit cell of every finitedimensional subspace of X is a polyhedron. He showed that Co is polyhedral and asked whether there is an infinite-dimensional reflexive polyhedral space. By using a method due to Klee himself [6] we prove in section 4 that every infinite-dimensional space has a two-dimensional quotient space whose unit cell is not a polygon. Hence no infinite-dimensional conjugate space is polyhedral. We conclude the paper with a section devoted to open problems. All Banach spaces are taken over the reals. Let X be a Banach space. We denote by Sx(xo, r) the cell {x; IIX - X o II < r}. The unit cell Sx(O, 1)is denoted also by

Sx. Section 2.

We bring first three simple and well known lemmas.

LEMMA 2.1. Let X be a Banach space and let {Sx (x~,ri)} be a family (finite or in[inite) of mutually intersecting cells in X . Then there is a Banach space Y = X with d i m Y [ X = 1 and a point y e Ysuch that [ l Y - x, it 0 and let {xi},= k 1 be a set of points in Sx such that IIx, - x j tl >=2fi for i ~ j . Then k ~ (1 .-[- £~-1)m.

1966] ON KLEE'S PAPER "POLYHEDRAL SECTIONS OF CONVEX BODIES"

237

Proof. The interiors of the cells Sx(xi,6) are mutually disjoint and all are contained in Sx(O, 1 + 6). By considering the volumes of the cells the result follows. LEMMA 2.3. Let B(n), n > 3, be the two-dimensional Banach space whose unit celt is the regular 2n-gon. Then there are vectors {bi}~= 1 in B(n) such that llbill=2n2/n 2 for every i and I l b , + b j l l < = l [ b , H + l l b j [ I - 2 for i # j . Proof. Take n consecutive sides of the boundary of the unit cell of B(n). Let bi be the vector of norm 2n2/~ 2 in the direction of the middle of the ith side. Then for i ~ j ,

tl b,___b,I [ __ 3. Then there exist 2" three-dimensional Banach spaces {Co} such that for every O, Sco is a polyhedron with 4n + 2 vertices and such that there exists no m-dimensional space which is universal for all the C o if 2" > 3m(2n 2 + 1) TM . Proof. Let B = B(n) and {b~}~'=1 be as in Lemma 2.3. For every choice of n signs 0 = (Oa,Oz,...,O,) , i.e. 0 i = _+ 1 for every i, there is by Lemma 2.1 a Banach space C o = B and a vector uo•Co such that Itu0t] < 1 and 11uo-O,b, b, 1 for e v e r y / . By the proof due to Griinbaum of Lemma 2.1 we can take as C o a space whose unit cell has at most 4n + 2 extreme points. We may even assume that Sco has exactly 4n + 2 extreme points (otherwise approximate the unit cell by a polyhedron with 4n + 2 vertices and use such an approximation as a new unit cell. It will be evident that by doing this we do not effect the argument below).

II- 1. Hence, by Lemma 2.2, there are at most 3" different 0 for which there is an isometry To:Co ~ X such that the restriction of To to B satisfies (2.1) for a given j (observe that [1Touo II --< 1 for every 0 and use the lemma with 6 = 1/2). Since the number of the indices j is k < (2n 2 + 1) 2m we get that 2" < 3"(2n 2 + 1) 2'n. Q.E.D. Section 3. We present now a simple method for proving the non-existence of certain infinite-dimensional universal spaces. If X and Y are Banach spaces and l < p < o o we denote by ( X @ Y ) p the space of all pairs (x,y) with

II~x, y)I[ -- (11x I[p + IIY 1[')I/p if p < oo and = max(] I x II, I! y and y e Y). The one dimensional space is denoted by R.

II) ir p = ~ (x ~ x

LEMMA 3.1. Let X be a Banach space and let 1 < p < oo. Assume that (X O)R)p is isometric to a subspace of X . Then X has a subspace isometric to Ip if p < ov and to co if p = o o . Proof. We consider the case p < oo only, the proof for p = oo is similar. By our assumption X contains a vector x 1 with xl ] = 1 and a subspace Y1 which is isometric to X such that x 1 + y p = 1 + y p for every y E I71. Continuing inductively we get for every n a vector x, in Y,_ a with x, = 1 and a subspace Y, of Y,-1 which is isometric to X and such that IIx, + y ~ = 1 + I y IIp for every y ~ Y,. It follows easily that for every real {2~)7=x we have " [~,I p" Hence the subspace of X spanned by {x ,},=1 oo has 1[ Z ,n= I 2ix, ][P = ~,=1 the desired properties. TrrEOR~M 3.1. The following sets of Banach spaces do not have a member which is universal for all the spaces in this set. (i) All reflexive spaces of a given density character .,¢g. (ii) All spaces whose conjugate is separable. (iii) All conjugate separable spaces. (iv) All spaces of a given density character ~ which do not contain an infinite-dimensional reflexive subspace. Proof. All these facts follow easily from Lemma 3.1. (i) follows since Co and 11 are not reflexive. (ii) is a consequence of the fact that l* is not separable. To derive (iii) we have to use a result of Bessaga and Petczyfiski [2] that no separable conjugate space contains a subspace isomorphic to Co. Finally, (iv) follows by using the lemma for 1 < p < oo. Section 4. In this section we prove that there is no conjugate infinite-dimensional polyhedral space. First some notations and conventions. Let P be a sym-

1966] ON KLEE'S PAPER "POLYHEDRAL SECTIONS OF CONVEX BODIES'"

239

metric convex body in the plane R 2 . The boundary of P will be denoted by 0P and for 0 < e < 1 the interior of the set (1 + e)P ,,, ( 1 - e)P will be denoted by nbd 8P. We shall consider in the sequel many norms in R 2 . However, for the sake of definiteness we single out one norm (an inner product norm, say) and this will be used whenever the symbol II[I is used for a vector in R z or an operator into R 2 . All other norms in the plane will appear only implicitly through their unit cell and the symbol I1" I1 will not be used for them. If Y c X and T is an operator on X its restriction to Y will be denoted by Tit. -

LEMMA 4.1. Let X be a finite-dimensional Banach space whose unit cell is a polyhedron. Let e > 0 and let T be a linear operator f r o m X onto R 2. Assume that there is a subspace Y of X of co-dimension 1 such that T S x = T S r . Then there is an operator T : X . . . , R 2 such that a~Sx~ e - nbdaTS x and such that T S x has more extreme points than T S x .

[Izl,- lYlt 0, let T~ be an operator from X into R 2 such that T~ is one to one on extS x and II Zll < ~ it is easily checked that for small enough 6 the points {___T~Ai)~= 1 are extreme points of T~Sx. Take one such 6 = rio < e/2. If TaoSx has more than 2n extreme points we can take T = T~o. If {_ T~oA~),"=~ are all the extreme points of T~oSx then by the argument of Klee in [6] there is a ~:X--~ R 2 such that T i t = Too It, TSx has more than 2n extreme points and 8TS x ~ el2 - nbdOTooS x. This operator 7~ has all the desired properties. LEMMA 4.2. Let P be a symmetric polygon in the plane. Then there is e > 0 such that every symmetric convex body C in the plane f o r which aC ~ e - nbdOP has at least as m a n y extreme points as P. Proof.

Obvious.

THEOREM 4.1. Let X be an infinite-dimensional Banach space. Than X has a two-dimensional quotient space whose unit cell is not a polygon. Proof. We assume that X is an infinite-dimensional Banach space all whose two-dimensional quotient spaces are polygonal and argue to a contradiction. By the results of [5] every finite-dimensional quotient space of X is polyhedral. Let T~ be a bounded linear operator from X onto R 2 and let T~Sx have 2n~ extreme points. Let T I : X ~ R "'+~ and U1 :R " ~ + ~ R z be linear operators such

240

JORAM LINDENSTRAUSS

[December

that T; is bounded and onto and such that T1 = U1T;. Let Za be the (n a + 1)dimensional Banach space whose unit cell is T'aSx. The operator U a maps Szl onto T1Sx and by the choice of na there is subspace Y1 of Z1 of codimension 1 such that U1Sy~, contains all the extreme points of T1Sx and hence all of T1Sx. Let now el > 0 be such that Lemma 4.2 holds with 2el if TISx is taken as P. Choose next points ,fxl"lnl ~ ~~ = 1 in Sx and a 61 > 0 such that Y~x~ e Y1 for every i and such that whenever []c~-Tlx][] 2nl extreme points, [l(.71T;x~-Tlx,'l[ < 6 j 2 for every / a n d Ot71T[Sx e 1 - nbdOT1S x. Put T2 = ~1T1'. Next we choose T~ : X ~ R "~+"~+ 1 and U2: R "~+"' + 1 ._, R2such that T2 = UzT~. We take Y2, a subspace of deficiency 1 in Z2 ( = R "~+"' + 1 with unit cell T2Sx) such that T~xt, e Y2 for every i and U2Sy, = U2Sz,. Let ez > 0 be such that Lemma 4.2 holds for 2e2, with P = T2Sx, and such that e 2 - n b d T 2 S x < e l - nbdOTaSx'. We choose 6 2 > 0 and t~'x2/"~ , ,~=1 ~ Sx as in the first step and then (by Lemma 4.1) we choose a 02 such that ]1UzTzxk--T2xk[] 2n k.

(4.5)

t~if"k~"~si= t c S x and whenever llci - Tkx~[ I < 6 k f o r i = l , . . . , n 0 (convex hull of {_+ ci}~,k I) = ek-nbdOTkSk.

(4.6)

t[Tkx[- T~_lx[[] k. Again 2.1 yields a desired Y. So assume that p ~ F . Any line in F through q has at least one point of X on each side of q. Let xl and xz be two such points of X such that q e c o n { x l , x 2 } . Then pos({xl,x2} u T) = pos({q} U T) = L and, since x 1 and x z are in

248

WILLIAM E. BONNICE AND JOHN R. REAY

F c L, pos ({xl,x2} UT) = L. With j + 1 denoting card T, since peL.,,F, j+l=dimL>dimF+l=k+l. Applying 2.1 with j + l playing the role of d and f = card({xl,x2} u T) (thus j + 1 < f < j + 3) there is a Y c X such that p ~ int con Y and with card

Y=< 2 [ n - (j + 1)] +f O

LEMMA 1. (Kadec [3], Klee [4]). There is a homeomorphism h: X

onto

~1

such that d , ( x ) = d,(hx) for n = 1,2,.... This homeomorphism has obviously the property: h(B x) = B' and h(TX(e))= Tt,(e). Since the set Tt,(e) is contained in the e-neighbourhood of the compact set B zc3 {4 e l: ~k = 0 for k > n}, we get LEMMA 2. The set Tt,(e) admits a finite 2e-net. LEMMA 3. I f (pn) is a sequence of positive integers ,(F,) is a decreasing sequence of closed sets with F k c TVk(1/k), k = 1,2,... and V is either l or X, then the set F = N°~= 1 f k is non empty and compact. Proof. 1° Let V= I. Let x, e Fn. By Lemma 2, (x,) is totally bounded, hence it has a cluster point Xo = 0,~°-- 1F,, i.e. F # ~ . From Lemma 2 it also follows that F is totally bounded, and since F is closed, it must be compact. 2 ° In the case V = X the assertion follows from Lemma 1 and from 1°. LEMMA 4. I f K is a closed convex subset of X with supx ~K I1x I[ = M < 1, then for every ~ > 0 there exists a closed face F 1 in K and a positive integer n such that F1 ~ TX(e). Proof. Take an y in K such that I]Yll > = M - e]4. Let n be such that d,(y) >_-M - e/2. Let f be a linear functional such that f ( x ) < d,(x) for all x ~ X and f ( y ) = d,(y) ( f is a supporting functional of the "cylinder" {x:d,(x) 6 1} at the point y/d,(y). By the Bishop Phelps theorem [1], there exists a g ~ X* with I I g - f l l < e / 4 M such that the face F~={xeK:g(x)= sup,~x g(u)} is nonempty. For x E F t we have

264

C. BESSAGA AND A. PELCZYI(ISKI

>__s(x) ->__

- Ils- g II !1 II ->- g(Y) -- lif- g I111x

>f(s)-llf-gll(ll~ll+jlylJ) =

>M =

~ 2

[I

~>]lxll 2

=

-~.

Hence F1 c TX(e). Q.E.D. Proof of the proposition. Without loss of generality we may assume that K c B x. By L e m m a 4, there is a sequence (Fn) of closed faces of K and a sequence

of integers (pn) such that Fn+ 1 is a face o f F , and F~ c TX(1/n). Hence, by L e m m a 3 F = ( ' ~ o~= 1 F , is a nonempty compact face of K. By the Krein Milman theorem, ext F # ~ and hence ext K # ~ . This concludes the proof. PROBLEM. Let X be a (separable) Banach space with the property that every bounded closed convex subset of X is the closed convex hull of its extreme points. Must X be isomorphic with a closed linear subspace of a (separable) conjugate space?

REFERENCES 1. E. Bishop and R. R. Phelps, The support functionals of a convex set, Proc. Symposia in Pure Math. VII (Convexity), Amer. Math. Soc. 00 (1963), 27-37. 2. I. M. Gelfand, Abstrakte Funktionen und lineare Operatoren, Mat. Sb. 4 (46) (1938), 235-286. 3. M. I. Kadec, On connection between weak and strong convergence (Ukrainian), Dopovidi Akad. Nauk Ukrain. R.S.R. 9 (1959), 456-468. 4. V. L. Klee, Mapping into normedlinear spaces, Fund. Math. 49 (1960), 25-34. 5. J. Lindenstrauss, On extreme points in 11, Israel J. Math. 4 (1966), 59-61. 6. A. Petczyfiski, On the impossibility of embedding of the space L in certain Banaeh spaces, Colloq. Math. 8 (1961), 199-203. UNIVERSITYOF WARSAW, WARSAW,POLAND

O N PERFECTLY H O M O G E N E O U S BASES IN B A N A C H SPACES BY

M. ZIPPIN* ABSTRACT

It is proved that a Banach space is isomorphic to co or to lp if and only if it co has a normalized basis {X i}i=l which is equivalent to every normalized co 1" block-basis with respect to {X i}i= 1. Introduction. F. Bohnenblust gave in [2] an axiomatic characterization of c o and Ip. The following proposition follows easily f r o m his proof: oo PROPOSITION 1.1. Let {X i}/=1 be a normalized basis in a Banach space X . I f f o r every normalized block-basis { Y , } i ~ (with respect to {xi},~l) and any a oo real sequence { i}i=1

i=1

a:,lI =

"=

m f o r all natural n then the basis {X i}i=l is equivalent to the unit-vectors basis in c o or in lp f o r some p > 1. Moreover, this equivalence of the bases induces an isometric isomorphism of X onto c o or Ip. T h e following natural question arises: I f we assume only that all normalized block-bases with respect to { x ~ } ~ are equivalent, is {x~}~=~ equivalent to the unit-vectors basis in Co or Ip? In this p a p e r we show that the answer is positive and the equivalence of all normalized block-bases characterizes the unit-vectors bases in c o and l~. After the preliminary l e m m a s of Section 2, we use the m e t h o d of the p r o o f o f L e m m a 4.3 o f [2] to prove our main result, T h e o r e m 3.1. A r e m a r k concerning a result of A. Petczyfiski and I. Singer [6] concludes Section 3.

DEFINmONS AND NOTATIONS. A basis {xi}~= ~ in a Banach space is called normalized if x, = 1 for every i. The sequence {Yi}i:l is called a block-basis + 1) ~ajxj, where {p( )}i=~ with respect to the basis {x~}i~o = ~ if for every i Yi = ~v ,j ( i=pci)+

II II

Received December 21, 1966. * This is part of the author's Ph.D. thesis prepared at the Hebrew University of Jerusalem under the supervision of Prof. A. Dvoretzky and Dr. J. Lindenstrauss. The author wishes to thank Dr. Lindenstrauss for his helpful guidance and for the interest he showed in the paper, and the referee for his valuable remarks. 265

266

M. ZIPPIN

[December

is an increasing sequence of nonnegative integers. In this paper we discuss only one basis {X i}~oo= 1 in the Banach space X ; all block-bases mentioned are assumed to be block-bases with respect to the basis {xi}~= 1. A basis {xi}~°:1 in X is equia oo valent to a basis {z i}~=1 ~ in a Banach space Z if for every real sequence { i)i=t ~i ~=1 aixi converges if and only if ~ t aiz~ converges. The closed subspace oo t in X is denoted by [yj~O . A sequence which is spanned by a sequence {Y..}~= {y,}~= ~ in X is called a basic sequence if it forms a basis in [yi]~o=;. (Every blockbasis is a basic sequence in X.) Following C. Bessaga and A. Petczyfiski [11 we call a basis {xi}~ 1 perfectly homogeneous if it is normalized and every normalized block-basis {z~}~°:t with respect to {xi}~°=l is equivalent to the basis {xi}~°:~. 2. Preliminary lemmas. Let {xi}i~=t be a normalized basis in a Banach space X and let {f~}~; denote its biorthogonal sequence in X*. In the sequel we shall consider the following property: (a)

If S and T are disjoint finite sets of positive integers and E a i x i + t " ~, aix i teS

>=

leT

~, aix t + s ieS

Itl ___lsithen

~, aix i ieT

for every real {a~}, i ~ S U T. LEMMA 2.1. Proof.

I f a basis {x i}i=1 ~ s a t i s a e s (a) then

II/,ll >--Y,(~,)=

liT, II =

l for

every i.

1. On the other hand /co

,I,,il-- o up

Ilj=l'~xJll-~l

,~=,

\

o xj _- o sup ,

II~J=, ,,x~ll~l

l a, l 1

_"

~

]

1966]

PERFECTLY HOMOGENEOUS BASES IN BANACH SPACES

269

k

(3.1)

2k =

~ x~ t == 1

It follows from (a) that for k > 1 (3.2)

2k+i > 2,.

By (b), for every increasing sequence {p( i)}i"= 1 of positive integers

(3.3) It follows that n

ilk

Ilk- [

[=

(We substitute for z i in the right side inequality of (b) the normalized block [I ~i=1 .k-1 xy+(,_l)ilk- , []-1 . ( ~ yIlk-, = l xj+(,-i),k-~) and use (3.3).) Using 3.4 we can prove by induction that for every natural n and k (3.5)

M2k.

xi k >

xi .

i=l

i=

On the other hand, by the left-side inequality of (b) and by (3.3) n

ii k

~ l x'

< M2"

i-

ilk - 1

ZlXJ " i=

~1 X/ -1

.

j=

Again it follows by induction that for every natural n and k (3.6)

~ 1 x'

< M2k"

ii,,Yl x, "

(3.1), (3.5) and (3.6) yield (3.7)

M -2k" 2nk < 2~ < M 2k" 2ilk for every n and k.

For any natural N, n and k let h = h ( N , n, k) be the non-negative integer for which Nh M-6"(~ ki)l/P'm-l/'2M -6(~ )'/" I

--

=

--

= 1 ri

"

Hence, using (a), it is easily proved that for any real ai, a2, "", a.

(3.14)

II ;~ a'x'll ~

M-6"

(n)

~ la'l' 1,,.

i=1

t

1

Similar arguments yield the following

1

t=I

1. For 2 ~ p > 1 one can construct in lp a subspace isomorphic to the space (El 0) E2 ~)...)p, where Ek denotes the k-dimensional euclidean space. This can be done without using [4] (see e.g. [5].) The space Y = (El @ E2 0)"')p has an unconditional basis non-equivalent to the unit-vectors basis in Ip, l < p # 2 . In fact, if ts~'~"("+t) ""iS i = {r(n- 1 )n+ 1 plays the r61e of the unit vectors basis in E, c Y, n = 1, 2,..., the sequence {X i}ioo= 1 forms an unconditional normalized basis in Y. If it were equivalent to the unit

272

M. ZIPPIN

oo vectors basis {e ~}i=1 in lp we would get that, for some fixed M > 1, every natural n and any real a l , a 2 , . . . , a n n ae II Z,=I ,,

_-< I1 Z7=1 a,x,+~,n 1,o II~ M rl Z7~1 aiei [1' which is known to be false, since ~ ,n= l a i e ,ll = ~ r = l l o , lP) ~'' while II is equal to ( ~ 7 = 1 [a,l~) ~ It follows that there exist normalized unconditional basic sequences in Ip which are not equivalent to the unit vectors basis. Similar basic sequences can be easily constructed in c o and in It. It follows that E is isomorphic to 12. M-

1

~7=la,x,+~,~-.olJ

REFERENCES 1. C. Bessage and A. Petczyfiski, On bases and unconditional convergence of series in Banach spaces, Studia Math. 17 (1958); 151. 2. F. Bohnenblust, Axiomatic characterization of Lv-spaces, Duke Math. J. 6 (1940), 627. 3. M. M. Day, Normed Linear Spaces, Springer-Verlag, 1962. 4. A. Dvoretzky, A theorem on convex bodies and applications to Banach spaces, Nat. Acad. Sci. U.S.A. 45 (1959), 223. 5. A. Petczyfiski, Projections in certain Banach spaces, Studia Math. 19 (1960), 209. 6. A. Pelczyfiski and I. Singer, On non-equivalent bases and conditional bases in Banach spaces. Studia Math. 25 (1964--5), 5. THE HEBREW UNIVERSITY OF JERUSALEM