The diameter of the isomorphism class of a Banach space

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1 Annals of Mathematcs, 162 (2005), The dameter of the somorphsm class of a Banach space By W. B. Johnson and E. Odell* Dedcated to the memory of V. I. Gurar Abstract We prove that f X s a separable nfnte dmensonal Banach space then ts somorphsm class has nfnte dameter wth respect to the Banach-Mazur dstance. One step n the proof s to show that f X s elastc then X contans an somorph of c 0. We call X elastc f for some K< for every Banach space Y whch embeds nto X, the space Y s K-somorphc to a subspace of X. We also prove that f X s a separable Banach space such that for some K< every somorph of X s K-elastc then X s fnte dmensonal. 1. Introducton Gven a Banach space X, let D(X) be the dameter n the Banach-Mazur dstance of the class of all Banach spaces whch are somorphc to X; that s, D(X) = sup{d(x 1,X 2 ):X 1,X 2 are somorphc to X} where d(x 1,X 2 ) s the nfmum over all somorphsms T from X 1 onto X 2 of T T 1. It s well known that f X s fnte (say, N) dmensonal, then cn D(X) N for some postve constant c whch s ndependent of N. The upper bound s an mmedate consequence of the classcal result (see e.g. [T-J, p. 54]) that d(y,l N 2 ) N for every N dmensonal space Y. The lower bound s due to Gluskn [G], [T-J, p. 283]. It s natural to conjecture that D(X) must be nfnte when X s nfnte dmensonal, but ths problem remans open. As far as we know, ths problem was frst rased n prnt n the 1976 book of J. J. Schäffer [S, p. 99]. The problem was recently brought to the attenton of the authors by V. I. Gurar, who *Johnson was supported n part by NSF DMS , Texas Advanced Research Program , and the U.S.-Israel Bnatonal Scence Foundaton. Odell was supported n part by NSF DMS and was a partcpant n the NSF supported Workshop n Lnear Analyss and Probablty at Texas A&M Unversty.

2 424 W. B. JOHNSON AND E. ODELL checked that every nfnte dmensonal super-reflexve space as well as each of the common classcal Banach spaces has an somorphsm class whose dameter s nfnte. To see these cases, note that f X s nfnte dmensonal and E s any fnte dmensonal space, then t s clear that X s somorphc to E 2 X 0 for some space X 0. Therefore, f D(X) s fnte, then X s fntely complementably unversal; that s, there s a constant C so that every fnte dmensonal space s C-somorphc to a C-complemented subspace of X. Ths mples that X cannot have nontrval type or nontrval cotype or local uncondtonal structure or numerous other structures. In partcular, X cannot be any of the classcal spaces or be super-reflexve. In hs unpublshed 1968 thess [Mc], McGugan conjectured that D(X) must be larger than one when dm X > 1. Schäffer [S, p. 99] derved that D(X) 6/5 when dm X>1as a consequence of other geometrcal results contaned n [S], but one can prove drectly that D(X) 2. Indeed, t s clearly enough to get an approprate lower bound on the Banach-Mazur dstance between X 1 := Y 1 R and X 2 := Y 2 R when Y s a nonzero Banach space. Now X 1 has a one dmensonal subspace for whch every two dmensonal superspace s sometrc to l 2 1. On the other hand, every one dmensonal subspace of X 2 s contaned n a two dmensonal superspace whch s sometrc to l 2 2. It follows that d(x 1,X 2 ) d(l 2 1,l2 2 )= 2. The Man Theorem n ths paper s a soluton to Schäffer s problem for separable Banach spaces: Man Theorem. If X s a separable nfnte dmensonal Banach space, then D(X) =. Part of the work for provng the Man Theorem nvolves showng that f X s separable and D(X) <, then X contans an somorph of c 0. Ths proof s nherently not local n nature, and, strangely enough, local consderatons, such as those mentoned earler whch yeld partal results, play no role n our proof. We do not see how to prove that a nonseparable space X for whch D(X) < must contan an somorph of c 0. Our proof requres Bourgan s ndex theory whch n turn requres separablty. Our method of proof nvolves the concept of an elastc Banach space. Say that X s K-elastc provded that f a Banach space Y embeds nto X then Y must K-embed nto X (that s, there s an somorphsm T from Y nto X wth y Ty K y for all y Y ). Ths s the same (by Lemma 2) as sayng that every space somorphc to X must K-embed nto X. X s sad to be elastc f t s K-elastc for some K<. Obvously, f D(X) < then X as well as every somorph of X s D(X)- elastc. Thus the Man Theorem s an mmedate consequence of

3 THE ISOMORPHISM CLASS OF A BANACH SPACE 425 Theorem 1. If X s a separable Banach space and there s a K so that every somorph of X s K-elastc, then X s fnte dmensonal. A key step n our argument nvolves showng that an elastc space X admts a normalzed weakly null sequence havng a spreadng model not equvalent to ether the unt vector bass of c 0 or l 1. To acheve ths we frst prove (Theorem 7) that f X s elastc then c 0 embeds nto X. It s reasonable to conjecture that an elastc nfnte dmensonal separable Banach space must contan an somorph of C[0, 1]. Theorem 1 would be an mmedate consequence of ths conjecture and the arbtrary dstortablty of C[0, 1] proved n [LP]. Our dervaton of Theorem 1 from Theorem 7 uses deas from [LP] as well as [MR]. Wth the letters X, Y, Z,... we wll denote separable nfnte dmensonal real Banach spaces unless otherwse ndcated. Y X wll mean that Y s a closed (nfnte dmensonal) subspace of X. The closed lnear span of the set A s denoted [A]. We use standard Banach space theory termnology, as can be found n [LT]. The materal we use on spreadng models can be found n [BL]. For smplcty we assume real scalars, but all proofs can easly be adapted for complex Banach spaces. 2. The man result The followng lemma [Pe] shows that the two defntons of elastc mentoned n Secton 1 are equvalent. Lemma 2. Let Y (X, ) and let be an equvalent norm on (Y, ). Then can be extended to an equvalent norm on X. Proof. There exst postve reals C and d wth d y y C y for y Y. Let F CB X be a set of Hahn-Banach extensons of all elements of S (Y, ) to all of X. Forx X defne x = sup { f(x) : f F } d x. Let n N and K <. We shall call a basc sequence (x ) block n-uncondtonal wth constant K f every block bass (y ) n =1 of (x ) s K- uncondtonal; that s, n n ±a y K a y =1 =1 for all scalars (a ) n =1 and all choces of ±. The next lemma s essentally contaned n [LP]. In fact, by usng the slghtly more nvolved argument n [LP], the concluson wth constant 2 can

4 426 W. B. JOHNSON AND E. ODELL be changed to wth constant 1 + ε, whch mples that the constant n the concluson of Lemma 4 can be changed from 2 + ε to 1 + ε. Lemma 3. Let X be a Banach space wth a bass (x ). For every n there s an equvalent norm n on X so that n (X, n ), (x ) s block n-uncondtonal wth constant 2. Proof. Let (P n ) be the sequence of bass projectons assocated wth (x n ). We may assume, by passng to an equvalent norm on X, that (x n ) s bmonotone and hence P j P = 1 for all <j. Let S n be the class of operators S on X of the form S = m k=1 ( 1)k (P nk P nk 1 ) where 0 n 0 < <n m and m n. Defne x n := sup{ Sx : S S n }. Thus x x n n x for x X. It suffces to show that for S S n, S n 2. Let x span (x n ) and Sx n = TSx for some T S n. Then snce TS S 2n S n + S n, TSx 2 x n. Lemma 4. For every separable Banach space X and n N there exsts an equvalent norm on X so that for every ε>0, every normalzed weakly null sequence n X admts a block n-uncondtonal subsequence wth constant 2+ε. Proof. Snce C[0, 1] has a bass, the lemma follows from Lemma 3 and the the classcal fact that every separable Banach space 1-embeds nto C[0, 1]. Lemma 4 s false for some nonseparable spaces. Partngton [P] and Talagrand [T] proved that every somorph of l contans, for every ε>0, a1+ε-sometrc copy of l and hence of every separable Banach space. Our next lemma s an extenson of the Maurey-Rosenthal constructon [MR], or rather the footnote to t gven by one of the authors (Example 3 n [MR]). We frst recall the constructon of spreadng models. If (y n )sa normalzed basc sequence then, gven ε n 0, one can use Ramsey s theorem and a dagonal argument to fnd a subsequence (x n )of(y n ) wth the followng property. For all m n N and (a ) m =1 [ 1, 1], f m 1 < < m and m j 1 < <j m, then m m a k x k a k x jk <ε m. k=1 k=1 It follows that for all m and (a ) m =1 R, m m lm... lm a k x k a k x k 1 m k=1 k=1

5 THE ISOMORPHISM CLASS OF A BANACH SPACE 427 exsts. The sequence ( x ) s then a bass for the completon of (span ( x ), ) and ( x ) s called a spreadng model of (x ). If (x ) s weakly null, then ( x ) s 2-uncondtonal. One shows ths by checkng that ( x ) s suppresson 1-uncondtonal, whch means that for all scalars (a ) m =1 and F {1,...,m}, m a x a x. F Also, (x ) s 1-spreadng, whch means that for all scalars (a ) m =1 and all n(1) < <n(m), m m a x = a x n(). =1 It s not dffcult to see that, when (x ) s weakly null, ( x ) s not equvalent to the unt vector bass of c 0 (respectvely, l 1 ) f and only f lm m m =1 x = (respectvely, lm m m =1 x /m = 0). All of these facts can be found n [BL]. Lemma 5. Let (x n ) be a normalzed weakly null basc sequence wth spreadng model ( x n ). Assume that ( x n ) s not equvalent to ether the unt vector bass of l 1 or the unt vector bass of c 0. Then for all C < there exst n N, a subsequence (y ) of (x ), and an equvalent norm on [(y )] so that (y ) s -normalzed and no subsequence of (y ) s block n-uncondtonal wth constant C for the norm. Proof. Recall that f (e ) n 1 s normalzed and 1-spreadng and bmonotone then n 1 e n 1 e 2n where (e )n 1 s borthogonal to (e ) n 1 [LT, p. 118]. n 1 Thus e = e n 1 n 1 e s normed by f = e n n =1 e, precsely f(e) =1= e, and f 2. These facts allow us to deduce that there s a subsequence (y ) of (x ) so that f F N s admssble (that s, F mn F ) then f F F y ( ) y F satsfes f F 5 and f F (y F ) = 1, where F y F y F y. Indeed, ( x ) s 1-spreadng and suppresson 1-uncondtonal (snce (x ) s weakly null). Gven 1/2 > ε > 0 we can fnd (y ) (x ) so that f F N s admssble then (y ) F s 1+ε-equvalent to ( x ) F =1. Furthermore we can choose (y ) so that f F s admssble then for y = a y, F a y (2 + ε) y (for a proof see [O] or [BL]). Hence f F (2 + ε) f F [y] F < 5 for suffcently small ε by our above remarks. We are ready to produce a Maurey-Rosenthal type renormng. Choose n so that n>7c and let ε>0 satsfy n 2 ε<1. We choose a subsequence =1 =1 F

6 428 W. B. JOHNSON AND E. ODELL M =(m j ) j=1 of N so that m 1 = 1 and for j and for all admssble sets F and G wth F = m and G = m j, a) k F y k k G y k <ε,fm <m j and b) k F y k k G y k m j m <ε,fm >m j. Indeed, we have chosen (y ) so that 1 F F x y 2 x 2 =1 F and smlarly for G. Snce ( x ) s not equvalent to the unt vector bass of c 0 (and s uncondtonal) lm m m 1 x = so that a) wll be satsfed f (m k ) ncreases suffcently rapdly. Furthermore, snce ( x ) s not equvalent to the m unt vector bass of l 1, lm x 1 m m = 0 and so b) can also be acheved. For N set A = {y F : F s admssble and F = m } and A = {f F : F s admssble and F = m }. Let φ be an njecton nto M from the collecton of all (F 1,...,F ) where <nand F 1 <F 2 < <F are fnte subsets of N. Here F<Gmeans max F<mn G. Let { n F = : F 1 < <F n, F 1 = m 1 =1, each F s admssble =1 f F =1 } and F +1 = φ(f 1,...,F ) for 1 <n. For y [(y )] let { } y F = sup f(y) : f F and set y = y F ε y. Ths s an equvalent norm snce for f F, f 5n. Also, y = 1 for all. Note that f f F A and y G A j wth j then f F (y G ) = k F y k ( y k G y ) k k m k F k G y k F y k k m m j k G y. k m If m <m j then f F (y G ) <εby a). If m >m j then f F (y G ) <εby b). It follows that f y = n =1 y F and f = n =1 f F F then y f(y) =n and f z = n =1 ( 1) y F, then for all g = n j=1 f G j F, g(z) 6+n 2 ε<7. Indeed, we may assume that g f and f F 1 G 1 then G j F for all 1 < n and 1 j n and so by a), b), n n ) n n g(z) = f Gj( ( 1) y F f Gj (y F ) <n 2 ε. j=1 =1 j=1 =1

7 THE ISOMORPHISM CLASS OF A BANACH SPACE 429 Otherwse there exsts 1 j 0 <nso that F j = G j for j j 0, F j0+1 = G j0+1 and F G j for j 0 +1<,j n. Usng f Gj0 +1(z) 5+nε we obtan j 0 g(z) f Gj (z) + f n G j0 +1(z) + f Gj (z) j=1 j>j 0+1 < 1+5+nε +(n (j 0 + 1))nε < 6+n 2 ε. Hence z 7 follows and the lemma s proved snce n/7 >Cand such vectors y and z can be produced n any subsequence of (y ). Our next lemma follows from Proposton 3.2 n [AOST]. Lemma 6. Let X be a Banach space. Assume that for all n, (x n ) =1 s a normalzed weakly null sequence n X havng spreadng model ( x n ) whch s not equvalent to the unt vector bass of l 1. Then there exsts a normalzed weakly null sequence (y ) X wth spreadng model (ỹ ) such that (ỹ ) s not equvalent to the unt vector bass of l 1. Moreover, there exsts λ>0 so that for all n, λ2 n a x n a ỹ for all (a ) R. Theorem 7. Let X be elastc, separable and nfnte dmensonal. Then c 0 embeds nto X. We postpone the proof to complete frst the Proof of Theorem 1. Assume that X s nfnte dmensonal and every somorph of X s K-elastc. Then by Theorem 7, c 0 embeds nto X. Choose k n so that 2 n k n. Usng the renormngs of c 0 by { } (a ) n = sup a : F = kn F and that X s K-elastc we can fnd for all n a normalzed weakly null sequence (x n ) =1 X wth spreadng model ( xn ) =1 satsfyng a x n K 1 (a ) n and moreover each ( x n ) s equvalent to the unt vector bass of c 0. Thus by Lemma 6 there exsts a normalzed weakly null sequence (y )nx havng spreadng model (ỹ ) whch s not equvalent to the unt vector bass of l 1 and whch satsfes for all n, k n ỹ λk 1 2 n k n. 1 Thus (ỹ ) s not equvalent to the unt vector bass of c 0 as well.

8 430 W. B. JOHNSON AND E. ODELL By Lemmas 2 and 5, for all C< we can fnd n N and a renormng Y of X so that Y contans a normalzed weakly null sequence admttng no subsequence whch s block n-uncondtonal wth constant C. By the assumpton on X, the space Y must K-embed nto every somorph of X. But f C s large enough ths contradcts Lemma 4. It remans to prove Theorem 7. We shall employ an ndex argument nvolvng l -trees defned on Banach spaces. If Y s a Banach space our trees T on Y wll be countable. For some C the nodes of T wll be elements (y ) n 1 Y wth (y ) n 1 bmonotone basc and satsfyng 1 y and n 1 ±y C for all choces of sgn. Thus (y ) n 1 s C-equvalent to the unt vector bass of ln. T s partally ordered by (x ) n 1 (y ) m 1 f n m and x = y for n. The order o(t ) s gven as follows. If T s not well founded (.e., T has an nfnte branch), then o(t )=ω 1. Otherwse we set for such a tree S, S = {(x ) n 1 S :(x ) n 1 s not a maxmal node}. Set T 0 = T, T 1 = T and n general T α+1 =(T α ) and T α = β<α T β f α s a lmt ordnal. Then o(t ) = nf{α : T α = φ}. By Bourgan s ndex theory [B1], [B2] (see also [AGR]), f X s separable and contans for all β<ω 1 such a tree of ndex at least β, then c 0 embeds nto X. We now complete the Proof of Theorem 7. Wthout loss of generalty we may assume that X Z where Z has a bmonotone bass (z ). Let X be K-elastc. We wll often use sem-normalzed sequences n X whch are a tny perturbaton of a block bass of (z ) and to smplfy the estmates we wll assume below that they are n fact a block bass of (z ). For example, f (y ) s a normalzed basc sequence n X then we call (d ) a dfference sequence of (y )fd = y k(2) y k(2+1) for some k 1 <k 2 <.We can always choose such a (d ) to be a sem-normalzed perturbaton of a block bass of (z ) by frst passng to a subsequence (y )of(y ) so that lm zj (y ) exsts for all j, where (z ) s borthogonal to (z ), and takng (d )tobea sutable dfference sequence of (y ). We wll assume then that (d ) s n fact a block bass of (z ). We nductvely construct for each lmt ordnal β<ω 1, a Banach space Y β that embeds nto X. Y β wll have a normalzed bmonotone bass (y β ) that can be enumerated as (y β ) =1 = {yβ,ρ,n : ρ C β, n N, N} where C β s some countable set. The order s such that (y β,ρ,n ) =1 s a subsequence of (yβ ) for fxed ρ and n. Before statng the remanng propertes of (y β ) we need some termnology. We say that (w )sacompatble dfference sequence of (y β ) of order 1f(w )

9 THE ISOMORPHISM CLASS OF A BANACH SPACE 431 s a dfference sequence of (y β ) that can be enumerated as follows, (w )={w β,ρ,n : ρ C β,n, N} and such that for fxed ρ and n, (w β,ρ,n ) s a dfference sequence of (y β,ρ,n+1 ). If (v ) s a compatble dfference sequence of (w ) of order 1, n the above sense, (v ) wll be called a compatble dfference sequence of (y β ) of order 2, and so on. (y β ) wll be sad to have order 0. Let (v ) be a compatble dfference sequence of (y β ) of some fnte order. We set T ( (v ) ) = { (u ) s 1 : the u s are dstnct elements of {v } 1, possbly n dfferent order, and s 1 } ±u = 1 for all choces of sgn. T ( (v ) ) s then an l -tree as descrbed above wth C = 1. The nductve condton on Y β, or should we say on (y β,ρ,n ), s that for all compatble dfference sequences (v )of(y β,ρ,n ) of fnte order, o(t ((v ))) β. Before proceedng we have an elementary but key Sublemma. Let C< and let (w ) be a block bass of a bmonotone bass (z ) wth 1 w C for all and let A = {F N : F s fnte and F ±w C for all choces of sgn}. Then there exsts an equvalent norm on [(w )] so that (w ) s a bmonotone normalzed bass such that for all F A, ±w =1. Proof. Defne a w = (a ) C 1 a w. F We begn by constructng Y ω. Let (x ) X be a normalzed block bass of (z ). For n N, let n be an equvalent norm on [(x )] gven by the sublemma for C =2 n.thus F ±x n =1f F 2 n. Snce X s K-elastc, for all n, ([(x )], n ) K-embeds nto X. We thus obtan for n N, a sequence (x n ) X wth 1 x n K for all and such that 1 F ±xn K for all F 2n and all choces of sgn. Furthermore

10 432 W. B. JOHNSON AND E. ODELL (x n ) s K-basc. By standard perturbaton and dagonal arguments we may for each n pass to a dfference sequence (d n ) of (x n+1 ) so that enumeratng, (d )={d n : n, N} s a block bass of (z ) wth 1 d K and wth each (d n ) beng a subsequence of (d ). We have that for F 2 n and all sgns, 1 F ±d n K. We renorm [(d )] by the sublemma for C = K and let the ensung space be Y ω. We change the name of (d )to(y ω ) n ths new norm and let (yω,1,n ) = (d n ).(y ω) has the property that f (w ) s a compatble dfference sequence of (y ω) of fnte order, then o(t ((w ))) ω. Indeed, f (w )={w ω,1,n : n N, N} then for F 2 n, F ±wω,1,n =1. Assume that Y β has been constructed for the lmt ordnal β wth bass (y β )={yβ,ρ,n : p C β, n, N} wth the requste propertes above. Let U : Y ω X and V : Y β X be K-embeddngs. Snce n total we are dealng wth a countable set of sequences, namely (y ω,1,n ) for n N and (y β,ρ,n ) for p C β, n N, by dagonalzaton and perturbaton we can fnd a compatble dfference sequence (w ω) of (y ω) of order 1 and a compatble dfference sequence (w β ) of (y β ) of order 1 so that under a sutable reorderng, (d )=(Uw ω) (Vw β ) s a block bass of (z ). Moreover each (Uw ω,1,n ) and (Vw β,ρ,n ) s a subsequence of (d ). Adjon a new pont p 0 to C β and set C β+ω = C β {ρ 0 }. Let d β+ω,ρ,n = Vw β,ρ,n for ρ C β and d β+ω,ρ0,n = Uw ω,1,n.for F 2 n and G C β N N for whch = 1, we have by the trangle nequalty that ( ) (ρ,n,) G F ±w β,ρ,n ±d β+ω,ρ0,n + ±d β+ω,ρ,n (ρ,n,) G 2K. It follows that f we let (y β+ω ) be the bass (d β+ω ), renormed by the sublemma for C =2K, that Y β+ω =[(y β+ω )] has the requred propertes. Indeed ( ) yelds that o(t ((y β+ω ))) β +2 n for n N and so o(t ((y β+ω ))) β + ω. Moreover by our nducton hypothess and choce of (d β+ω,p0,n ) the analogue of ( ) holds for a compatble dfference sequence of order 1 or of any fnte order. Thus o(t (z β+ω )) β + ω for any compatble dfference sequence of (y β+ω ) of fnte order. If β s a lmt ordnal not of the form α + ω we let U α : Y α X be a K-embeddng for each lmt ordnal α<β. We agan dagonalze to form compatble dfference sequences (w α) of (y α ) of order 1 for each such α so that (U α w α) α, s a block bass of (z ) n some order. We let C β be a dsjont

11 THE ISOMORPHISM CLASS OF A BANACH SPACE 433 unon of the C α s and n the manner above obtan Y β. Agan by our nducton hypothess we wll have that o(t ((y β ))) α for all lmt ordnals α<βand the same holds for all compatble dfference sequences of (y β ) of fnte order. 3. Concludng remarks and problems C[0, 1] s, of course, 1-elastc. By vrtue of Lemma 4 (or Theorem 1), for all K, C[0, 1] can be renormed to be elastc but not K-elastc. Are there other examples of separable elastc spaces? Problem 8. Let X be elastc (and separable, say). nto X? Does C[0, 1] embed Usng ndex arguments, we have the followng partal result. Proposton 9. Let X be a separable Banach space, and suppose that Y = X n s a symmetrc decomposton of a space Y nto spaces unformly somorphc to X. If Y s elastc, then C[0, 1] embeds nto X. In partcular, f 1 p< and l p (X) s elastc, then C[0, 1] embeds nto X. Proof. Let us frst observe that f C[0, 1] embeds nto Y, then C[0, 1] embeds nto X. Snce ths s surely well known, we just sketch a proof (whch, ncdentally, uses only that the decomposton Y = X n of Y s uncondtonal): Let P n be the projecton from Y onto X n and let W be a subspace of Y whch s somorphc to C[0, 1]. By a theorem of Rosenthal s [R], t s enough to show that for some n, the adjont P n W of the restrcton of P n to W has nonseparable range. Ths wll be true f there s an m so that S m W has nonseparable range, where S m := m =1 P. Let Z be a subspace of W whch s somorphc to l 1. If no such m exsts, then for every m, the restrcton of S m to Z s strctly sngular (that s, not an somorphsm on any nfnte dmensonal subspace of Z), and t then follows that Z contans a sequence (x n ) of unt vectors whch s an arbtrarly small perturbaton of a sequence (y n ) whch s dsjontly supported. The sequence (y n ), a fortor (x n ), s then easly seen to be equvalent to the unt vector bass of l 1 and ts closed span s complemented n Y snce the decomposton s uncondtonal. It follows that l 1 s somorphc to a complemented subspace of C[0, 1], whch of course s false. To complete the proof of Proposton 9, we assume that Y s K-elastc and prove that C[0, 1] embeds nto Y. The proof s smlar to, but smpler than, the proof of Theorem 7. Frst we recall the defnton of certan canoncal trees T α of order α for α<ω 1 (see e.g. [JO]). These form the frames upon whch we wll hang our bases. The tree T 1 s a sngle node. If T α has been defned, we choose a new node z T α and set T α+1 := T α {z}, ordered by z<tfor

12 434 W. B. JOHNSON AND E. ODELL all t T α and wth T α preservng ts order. If β<ω 1 s a lmt ordnal, we let T β be the dsjont unon of {T α : α<β}. Then f s, t are n T β, we say that s t f and only f s, t are both n T α for some α<βand s t n T α. We shall prove by transfnte nducton that f (x ) =1 s any normalzed bmonotone basc sequence and α<ω 1, there s a Banach space Y α = Y α (x ) wth a normalzed bmonotone bass (yt α ) t Tα so that f (γ ) n =1 s any branch n T α then (yγ α ) n =1 s 1-equvalent to (x ) n =1 and for all scalars (a r) t Tα, n =1 a γ yγ α t T α a t yt α. Furthermore, each Y α wll K-embed nto Y. Just as n the proof of Theorem 7, t then follows from ndex theory that the Banach space spanned by (x ) =1 embeds nto Y. Fx any normalzed bmonotone basc sequence (x ) =1. Suppose that β s a lmt ordnal and Y α = Y α (x ) =1 has been defned for all α<β. In vew of the hypotheses, the space Y has a symmetrc decomposton nto spaces unformly somorphc Y, whch we can ndex as Y = α<β X α. For each α, there s an somorphsm L α from Y α nto X α so that L α = 1 and L 1 α s bounded ndependently of α. We can put an equvalent norm on Y = α<β X α to make each L α an sometry and make the decomposton 1-uncondtonal (but not necessarly 1-symmetrc). Defne Y β to be the closed lnear span of {L α Y α : α<β} n Y wth ts new norm. The space Y β has the desred bass ndexed by T β and Y α must K-embed nto Y wth ts orgnal norm because Y s K-elastc. If β = α +1,weletY β (x ) =1 = R Y α(x ) =2 wth the norm gven by (a, y) := y sup{ ax 1 + n =2 y α γ (y)y α γ : (γ ) n =2 s a branch n T α }. Agan t s clear that Y β must K-embed nto Y and that Y β has the desred bass. Gven metrc spaces X and Y, the Lpschtz dstance d L (X, Y ) between them s defned to be the nfmum of Lp (f) Lp (f 1 ), where the nfmum s taken over all blpschtz mappngs f from X onto Y (the nfmum of the empty set s ). Call a Banach space X Lpschtz K-elastc provded that every somorph of X has Lpschtz dstance at most K to some subset of X, and say that X s Lpschtz elastc f X s Lpschtz K-elastc for some K<. It s nterestng to note that the analogue of Theorem 1 for Lpschtz elastcty s false. Indeed, Aharon [A], [BeL, Theorem 7.11] proved that every separable metrc space has, for each ɛ>0, Lpschtz dstance less than 3 + ɛ to some subset of c 0, whle James [J], [BeL, Proposton 13.6] proved that for each ɛ>0, the space c 0 s 1 + ɛ-equvalent to a subspace of every somorph of c 0. Consequently, any separable Banach space whch contans an somorphc copy of c 0 s Lpschtz 3 + ɛ-elastc.

13 THE ISOMORPHISM CLASS OF A BANACH SPACE 435 Problem 10. If X s a separable, nfnte dmensonal Banach space and there s a K so that every somorph of X s Lpschtz K-elastc, must X contan a subspace somorphc to c 0? We also do not know whether the Man Theorem has a Lpschtz analogue: Problem 11. If X s a separable Banach space and there exsts K< so that d L (X, Y ) <Kfor all spaces Y whch are somorphc (or blpschtz equvalent) to X, then must X be fnte dmensonal? The space c 0 does not have the property mentoned n Problem 11 (use, for example, the fact [GKL] that for every K< there exsts K < so that f a Banach space X s K blpschtz equvalent to c 0 then d(x, c 0 ) K ). Snce Schäffer s problem remans open n the nonseparable settng, we menton t explctly as: Problem 12. Does there exst a nonseparable Banach space X for whch D(X) <? Also open s: Problem 13. Does there exst a nonseparable Banach space X and a K< so that all somorphs of X are K-elastc? Under the generalzed contnuum hypothess, t follows from [E] that for every cardnal ℵ there s a Banach space X of densty character ℵ so that every Banach space of densty character ℵ s sometrcally somorphc to a subspace of X. These provde the only known examples of elastc Banach spaces. Problem 14. If X s elastc and nfnte dmensonal, does X contan an somorph of every Banach space whose densty character s the same as the densty character of X? As was mentoned earler, the space l s, for every ɛ>0, 1+ɛ-somorphc to a subspace of every somorph of tself (see [P] and [T]). However, l s not elastc. To see ths, take a borthogonal system (x α,x α) α<2 ℵ 0 n l wth x α = 1 and x α bounded. Gven a natural number m, renorm l by ( ) 1/2 1 x m := max m x, sup x α(x) 2 σ =m. It s not hard to see that (l, m ) s not better than m-normed by any countable set of norm one functonals, whch mples that d((l, m ),Y) m for any subspace Y of l. A smlar argument shows that l (ℵ) s not elastc for any nfnte cardnal number ℵ. α σ

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