On Uniform Homeomorphisms of the

arXiv:math/9207208v1 [math.FA] 21 Jul 1992On Uniform Homeomorphisms of theUnit Spheres of Certain Banach LatticesF. ChaatitAbstractWe prove that if X is an infinite dimensional Banach lattice witha weak unit then there exists a probability space (Ω, Σ, µ) so thatthe unit sphere S(L1(Ω, Σ, µ) is uniformly homeomorphic to the unitsphere S(X) if and only if X does not contain ln∞’s uniformly.1IntroductionRecently E.Odell and Th.Schlumprecht [O.S] proved that if X is an infinitedimensional Banach space with an unconditional basis then the unit sphere ofX and the unit sphere of l1 are uniformly homeomorphic if and only if X doesnot contain ln∞uniformly in n. We extend this result to the setting of Banachlattices.

In Theorem 2.1 we obtain that if X is a Banach lattice with a weakunit then there exists a probability space (Ω, Σ, µ) so that the unit sphereS(L1(Ω, Σ, µ) is uniformly homeomorphic to the unit sphere S(X) if and onlyif X does not contain ln∞uniformly in n. A consequence of this -Corollary2.11- is that if X is a separable infinite dimensional Banach lattice then S(X)and S(l1) are uniformly homeomorphic if and only if X does not contain ln∞uniformly in n. Quantitative versions of this corollary are given in Theorem2.2 and Theorem 2.3. A continuous function f : [0, ∞) →[0, ∞) withf(0) =0 is a modulus of continuity for a function between two metric spaces F :(A, d1) →(B, d2) if d2(F(a1), F(a2)) ≤f(d1(a1, a2)) whenever a1, a2 ∈A.Theorem 2.2 says that if X and Y are separable infinite dimensional Banachlattices with Mq(X) < ∞and Mq′(Y ) < ∞for some q, q′ < ∞then thereexists a uniform homeomorphism F : S(X) →S(Y ) such that F and F −1have modulus of continuity f where f depends solely on q, q′, Mq(X) andMq′(Y ).

Here Mq(X) is the q-concavity constant of X and will be definedbelow.1

Central in defining these homeomorphisms is the entropy map, consideredin [G] and [O.S]. We refer the reader to [B] and its references for a survey ofsome results concerning uniform homeomorphisms between Banach spaces.In particular it is interesting to note Enflo’s result that l1 and L1 are notuniformly homeomorphic [B] while their unit spheres are.

Also we refer to[L.T] for facts related to the theory of Banach lattices.Aknowledgements: I would like to express my gratitude to ProfessorsE.Odell and Th.Schlumprecht for proposing this work to me and for providingme with valuable suggestions and references. Thanks are also due to ProfessorV.Mascioni for simplifying the proof of lemma 2.6.After this work was done, Nigel Kalton discovered a proof of our mainresult using complex interpolation theory [K].NotationLet us start by recalling some definitions and well known facts.

A nonnegative element e of a Banach lattice X is a weak unit if e∧x = 0 for x ∈Ximplies that x = 0. Every separable Banach lattice has a weak unit [L.T, p9].

A Banach lattice is order continuous if and only if every increasing, orderbounded sequence is convergent. By a general representation theorem (see[L.T, p 25]) any order continuous Banach lattice with a weak unit can berepresented as a Banach lattice of functions.

More precisely:1. there exist a probability space (Ω, Σ, µ) and an ideal fX of L1(Ω, Σ, µ),along with a lattice norm ∥· ∥eXon fXso that X is order isometric to (fX, ∥· ∥eX)2. fX is dense in L1(Ω, Σ, µ) and L∞(Ω, Σ, µ) is dense in fX.3. ∥f∥1 ≤∥f∥eX ≤2∥f∥∞for all f ∈L∞(Ω, Σ, µ).Moreover fX∗= {g : Ω−→R : ∥g∥eX∗< ∞} is isometric to X∗, where∥g∥eX∗= sup{Zfgdµ; ∥f∥eX ≤1}2

and if g ∈fX∗and f ∈fX theng(f) =Zfgdµ.If X is a Banach lattice which is not order continuous then X containsc0 ([L.T, pages 6–7]).A Banach lattice X is q-concave if there exists a constant Mq < ∞suchthat(nXi=1∥xi∥q)1q ≤Mq∥(nXi=1|xi|q)1q ∥(⋆)(resp.p-convex if there exists Mp < ∞so that∥(nXi=1|xi|p)1p∥≤Mp(nXi=1∥xi∥p)1p (⋆⋆))for all n ∈N and xi ∈X , 1 ≤i ≤n.Mq(X) is the smallest constant satisfying (⋆) and Mp(X) is the smallestconstant that satisfies (⋆⋆).Given a Banach lattice of functions X, the p-convexification X(p) of X isgiven byX(p) = {f : Ω−→R : |f|p ∈X}with|||f||| = ∥|f|p∥1p.The space X(p) is a Banach lattice with Mp(X(p)) = 1 ([L.T, p 53]).We will also need the following result. If X is r-convex and s-concave,for 1 ≤r, s ≤∞then X(p) is pr-convex and ps-concave withMpr(X(p)) ≤(Mr(X))1pandMps(X(p)) ≤(Ms(X))1p .

(See [L.T, p 54]).We will use standard Banach space notations, BaX = {x ∈X : ∥x∥≤1}will denote the unit ball of X and S(X) = {x ∈X : ∥x∥= 1} the unit sphereof X. If h is a real function on Ω, then supph = {ω ∈Ω: h(ω) ̸= 0} is thesupport of h. If B ⊂Ω, then Bh(ω) = h(ω)χB(ω) where χB is the indicatorfunction of B.3

2The main result:We now state the main result of this workTheorem 2 .1 Let X be an infinite dimensional Banach lattice with a weakunit. Then there exists a probability space (Ω, Σ, µ) so that S(L1(Ω, Σ, µ))is uniformly homeomorphic to S(X) if and only if X does not contain ln∞uniformly in n.Our proof of Theorem 2.1 will yield two quantitative results:Theorem 2 .2 If X and Y are separable infinite dimensional Banach lat-tices with Mq(X) < ∞and Mq′(Y ) < ∞for some q, q′ < ∞then there existsa uniform homeomorphism F : S(X)−→S(Y ) such that F and F −1 havemodulus of continuity α where α depends solely on q, q′, Mq(X) and Mq′(Y ).Theorem 2 .3 If X and Y are both uniformly convex and uniformly smoothseparable infinite dimensional Banach lattices then there exists a uniformhomeomorphism F : S(X)−→S(Y ) such that F has modulus of continuityf where f depends solely on the modulus of uniform convexity of Y and themodulus of uniform smoothness of X, and F −1 has a modulus of continuity gdepending solely on the modulus of uniform smoothness of Y and the modulusof uniform convexity of X.The proofs will involve a sequence of steps similar to those in [O.S].

Webegin with a simple extension of Proposition 2.8 of [O.S]. Recall that X(p) isthe p-convexification of X.Proposition 2 .4 Let X be a Banach lattice of functions on a set Ωand let1 < p < ∞.

Then the mapGp : S(X(p))−→S(X)given by Gp(f) = |f|psignf is a uniform homeomorphism. Furthermore themoduli of continuity of Gp and (Gp)−1 are functions solely of p.4

Proof Clearly Gp maps S(X(p)) one-to-one onto S(X). Let f and g be inS(X(p)) with 1 > δ = ∥f −g∥X(p) = ∥|f −g|p∥1pX.As in [O.S] we shall show that there exist two functions H and F such thatH(δ) ≤∥Gp(f) −Gp(g)∥≤F(δ)where F(δ) = 2(1−(1−δ1p)p)+δp−1+δp and H(δ) =12p−1δp.

The propositionthen follows.LetΩ+ = {ω ∈Ω: signf(ω) = signg(ω)}andΩ−= {ω ∈Ω: signf(ω) ̸= signg(ω)}.We then have:∥Gp(f) −Gp(g)∥=∥||f|psignf −|g|psigng|∥=∥||f|p −|g|p|χΩ+ + (|f|p + |g|p)χΩ−∥.But ap −bp ≥(a −b)p and ap + bp ≥21−p(a + b)p for a ≥b ≥0.Thus,∥Gp(f) −Gp(g)∥≥||f| −|g||p χΩ+ +12p−1(|f| + |g|)pχΩ−≥12p−1 ||f| −|g||p χΩ+ +12p−1(|f| + |g|)pχΩ−=21−p∥|f −g|p∥=21−p∥f −g∥pX(p)So we obtain H(δ) =12p−1δp as a lower estimate. For the upper estimate wehave:∥Gp(f) −Gp(g)∥=∥||f|p −|g|p| χΩ+ + (|f|p + |g|p)χΩ−∥≤∥||f|p −|g|p|χΩ+∥+ ∥(|f|p + |g|p)χΩ−∥First we note that since(|f|p + |g|p)χΩ−≤(|f| + |g|)pχΩ−≤|f −g|p χΩ,5

we get(|f|p + |g|p)χΩ− ≤∥f −g∥pX(p) = δp.Next we estimate ∥||f|p −|g|p| χΩ+∥. For this purpose we split Ω+ into Ω1+and Ω2+ whereΩ1+ = {ω ∈Ω+ : |f(ω) ≤q|g(ω)| or |g(ω)| ≤q|f(ω)|}andΩ+2 = Ω+ ∼Ω1+and q = 1 −δ1p.Note that if C = (1 −q)−p then||f|p −|g|p|χΩ1+ ≤C|f −g|p.Indeed,C|f −g|p −|g|p + |f|p ≥C|g −qg|p −|g|p = 0in case |f| ≤q|g| (the proof is similar if |g| ≤q|f|).Thus∥χΩ1+||f|p −|g|p|∥≤C∥χΩ+|f −g|p∥≤C∥|f −g|p∥=C∥f −g∥pX(p)=Cδp=(1 −q)−pδpAnd since (1 −q)−p = δ−1, we obtain∥χΩ1+||f|p −|g|p|∥≤δp−1.Finally we have on Ω2+ :∥||f|p −|g|p|χΩ2+∥≤(1 −qp)∥|f|p + |g|p|∥≤2(1 −(1 −δ1p)p).6

SoF(δ) = 2(1 −(1 −δ1p)p) + δp−1 + δpand as p > 1, F(δ)−→0 when δ−→0•Throughout the rest of the paper, X will denote a Banach lattice with therepresentation as a lattice of functions on (Ω, Σ, µ) satisfying the conditionsmentionned in the introduction. The next step in proving Theorem 2.1 willbe to produce a uniform homeomorphismFX : S(L1(Ω, Σ, µ))−→S(X)in the case where our lattice X is uniformly convex and uniformly smooth.In order to do this we need first to define the entropy function E(h, f).Let h ∈(L∞(µ))+ and define E(h, ·) : X−→¯R byE(h, f) =Zh log |f|dµfor f ∈X, (we use the convention that 0 log 0 ≡0) and more generally,E(h, f) = E(|h|, |f|)if h ∈L∞(µ).The entropy map was considered in [G] and in the sequel we use argumentsof both [O.S] and [G].Proposition 2 .5 Suppose X is uniformly convex.

Let h ∈(L∞(µ))+ andsetλ ≡supf∈BaXZh log |f|dµ.Then −log 2 ≤λ ≤∥h∥∞and if h ̸= 0 there exists a unique f ∈S(X)+ sothat λ = E(h, f). Moreover suppf = supph.Proof First we note that λ ≤∥h∥∞.

To see this it suffices to observe thatλ=supf∈BaX+Zh log |f|dµ7

≤supf∈BaX+Zh|f|dµ≤supf∈BaX+ ∥h∥∞∥f∥L1≤supf∈BaX+ ∥h∥∞∥f∥X≤∥h∥∞.Also λ ≥−log 2 since χΩ/2 ∈Ba(X)+. Next let (fn) ⊆(BaX)+ be such thatE(h, fn) ≥λ−2−n.

Since X is uniformly convex, by passing to a subsequence,we can suppose that fn converges weakly to f ∈(BaX)+. Let (un) be asequence of “far-out” convex combinations of fn, such that (un) convergesto f in norm, thus un = Ppn+1i=pn+1 cifi where p1 < p2 < ... < pn < ....ci ≥0,Ppn+1i=pn+1 ci = 1 and ∥un −f∥X−→0 as n−→∞.We next note that if (gi)ni=1 ⊆BaX, and (di)ni=1 ⊆(R)+ with Pni=1 di = 1thenE h,nXi=1digi!≥nXi=1diE(h, gi).Moreover if B = supph and Bgi ̸= Bgj for some i, j thenE h,nXi=1digi!>nXi=1diE(h, gi)This follows from the strict concavity of the logarithm function.Thereforelimn→∞E(h, un) = λ.Claim: E(h, f) = λNote that∥un −f∥L1(µ) ≤∥un −f∥X →0and so in order to prove the Claim, it suffices to prove the following lemma:Lemma 2 .6 Let λ ∈R, h ∈L+1 (µ), (un) ⊆L+1 (µ) and suppose un−→f inL1(µ).

ThenZh log undµ−→λ impliesZh log fdµ ≥λ.8

Proof By passing to a subsequence we may assume that un →f a.e. Thus(log un)−→(log f)−a.e.

and soZh(log f)−dµ ≤lim infn→∞Zh(log un)−dµ.by Fatou’s lemma. Therefore(⋆) lim supn→∞Z−h(log un)−dµ ≤Z−h(log f)−dµ.On the other hand, one has also the inequality:(⋆⋆) lim supn→∞Zh(log un)+dµ ≤Zh(log f)+dµ.Indeed, fix ε > 0.

Since 0 ≤(log un)+ ≤un, and (un) is uniformly integrable,there exists δ > 0 so that µ(A) < δ impliesfor all n,ZA(log un)+dµ < ε andZA(log f)+dµ < ε. ((log f)+ is integrable since 0 ≤(log f)+ ≤f.) Now h(log un)+−→h(log f)+a.e: So by Egoroff’s theorem, there exists a set C with µ(C) < δ such thath(log un)+−→h(log f)+uniformly except perhaps on C. More exactly, for ε > 0, there exist n(ε) ∈Nand a set C with µ(C) < δ such that for any n ≥n(ε) we havesupω∈Cc |h(log un)+ −h(log f)+| < ε.ThusZh(log un)+dµ≤Z|h(log un)+ −h(log f)+|dµ +Zh(log f)+dµ.=ZC |h(log un)+ −h(log f)+|dµ+ZCc |h(log un)+ −h(log f)+|dµ +Zh(log f)+dµ.<2ε + ε +Zh(log f)+dµ9

Solim supn→∞Zh(log un)+dµ ≤Zh(log f)+dµ.Now adding (⋆) and (⋆⋆) yieldsλ ≤Zh log fdµ,which proves Lemma 2.6 •Note that since λ ≥E(h, f), we get E(h, f) = λ, proving the Claim.Now we prove that f is unique. Indeed, let f ̸= g with E(h, f) = E(h, g)and ∥f∥= ∥g∥= 1.

Thus by uniform convexity f+g2 < 1 and so f+g2cannotmaximize the entropy, while clearly supph ⊆suppg a.e and soλ = 12 (E(h, f) + E(h, g)) ≤E(h, f + g2) < λ,a contradiction.Let now B = supph. In order to obtain suppf = B a.e consider first g =Bf in what preceeds to get f = Bf.

Then observe that trivially suppBf ⊂Ba.e, while if the previous inequality was strict, then there exists a set A ⊂Bwith µ(A) > 0 such that f|A = 0. Thus−∞= E(h, f) ≥E(h, χΩ/2) = −log 2;a contradiction.

Hence suppf = suppBf = B. •Thus under the assumption that X is uniformly convex we can defineFX : S(L1(µ))+ \L∞(µ)−→S(X)+by FX(h) = f where f ∈S(X)+ is such thatE(h, f) =maxg∈(BaX)+Zh log |g|dµ = EX(h)We then defineFX : S(L1(µ))\L∞(µ)−→S(X)10

by FX(h) = (signh)FX(|h|).We shall show that FX is uniformly continuous, and thus extends to acontinuous function on S(L1(µ)). To do so we will need a proposition similarto Proposition 2.3.C of [O.S].The proof is nearly the same, adapted tofunction spaces.Proposition 2 .7 Let h1, h2 be in S(L1(µ))+ T L∞(µ) with ∥h1 −h2∥1 ≤1.Let x1 = FX(h1), and x2 = FX(h2).

Thenx1 + x22 ≥1 −∥h1 −h2∥121 .Proof Letx1+x22 = 1 −2ε. We need to show that2ε ≤∥h1 −h2∥12.We may assume ε > 0.

Define fx1 = x1 + εx2 and fx2 = x2 + εx1. Thensuppfx1 = suppfx2 = supph1 ∪supph2 ≡B,andfx1 + fx22 ≤x1 + x22 + ε = 1 −εWith this we can prove thatε ≤| log(1 −ε)| ≤12{E(h1, fx1) −E(h1, fx2)} (⋆)Indeed, since fx1 ≥x1, we clearly haveE(h1, fx1)≥E(h1, x1)≥E h1,fx1 + fx22(1 −ε)!sinceex1+ ex22(1−ε) ∈BaX and x1 maximizes the entropy.

AndE h1,fx1 + fx22(1 −ε)!=E h1,fx1 + fx22!+ | log(1 −ε)|≥12E(h1, fx1) + 12E(h1, fx2) + | log(1 −ε)|11

Similarly we haveε ≤| log(1 −ε)| ≤12{E(h2, fx2) −E(h2, fx1)} (⋆⋆).Then by averaging (⋆) and (⋆⋆) we getε ≤14[E(h1, fx1) −E(h1, fx2) + E(h2, fx2) −E(h2, fx1)].Soε≤14ZB(h1 −h2)(log fx1 −log fx2)dµ≤14ZB |h1 −h2|logfx1fx2 dµButlogfx1fx2 ≤log 1ε on Bforfx1fx2= x1 + εx2x2 + εx1=x1 + εx2ε(x1 + ε−1x2) ≤1ε.and similarlyfx2fx1≤1ε.Since log 1ε ≤1ε, we finally getε ≤14∥h1 −h2∥11ε.Hence2ε ≤∥h1 −h2∥121 •Proposition 2 .8 Let X be uniformly convex. ThenFX : S(L1(µ))\L∞(µ)−→S(X)is uniformly continuous and hence extends to a uniformly continuous mapFX : S(L1(µ))−→S(X).

Moreover the modulus of continuity of FX dependsonly on the modulus of uniform convexity of X.12

Proof Recall that X is uniformly convex if and only ifδX(ε) = inf1 −x + y2 : ∥x∥= ∥y∥= 1, ∥x −y∥≥ε> 0.We first observe that FX : S(L1(µ))+−→S(X) is uniformly continuous.Indeed, by Proposition 2.7, if h1 and h2 are in S(L1(µ))+ T L∞(µ) and∥h1 −h2∥1 ≤1 thenFX(h1) + FX(h2)2 ≥1 −∥h1 −h2∥121or1 −FX(h1) + FX(h2)2 ≤∥h1 −h2∥121 .So if ∥FX(h1) −FX(h2)∥≥ε then ∥h1 −h2∥≥(δX(ε))2. Thus there existsη(ε) = (δX(ε))2 so that ∥h1 −h2∥< η(ε) implies ∥FX(h1) −FX(h2)∥≤ε.Letting η(0) = 0,the function η is continuous and strictly increasing on [0, 2].

So η has aninverse g depending only on the modulus of uniform convexity of X, and∥FX(h1) −FX(h2)∥≤g(∥h1 −h2∥).For the general case let h1, h2 in S(L1(µ))T L∞(µ) and setxi = FX(hi) = signhi · FX(|hi|)for i = 1, 2. Then∥x1 −x2∥≤∥FX(|h1|) −FX(|h2|)∥+ ∥χD(FX(|h1|) + FX(|h2|))∥whereD = {ω ∈Ω: signh1(ω) ̸= signh2(ω)}.By what we observed in the beginning of the proof,∥FX(|h1|) −FX(|h2|)∥< g(ε) whenever∥|h1| −|h2|∥≤∥h1 −h2∥< ε.13

Our next step is to estimate ∥χDFX(|hi|)∥, for i = 1, 2. To do so, we notethat∥χDFX(|h1|)∥= ∥DFX(|h1|)∥≤FX(|h1|) −FX Dc|h1|∥Dc|h1|∥!

.We are then lead to estimateh1 −Dch1∥Dch1∥≤D(h1 −Dch1∥Dch1∥) +Dc(h1 −Dch1∥Dch1∥)=∥Dh1∥+Dch1 −Dch1∥Dch1∥ .We first get that∥Dh1∥= ∥D|h1|∥≤∥D(|h1| + |h2|)∥≤∥h1 −h2∥< ε;and, since ∥h1∥= ∥Dh1 + Dch1∥= 1 and ∥Dh1∥< ε, an easy computationyieldsDch1 −Dch1∥Dch1∥ ≤∥Dh1∥< ε.Soh1 −Dch1∥Dch1∥ < 2ε and thus∥DFX(|h1|)∥≤FX(|h1|) −FX Dc|h1|∥Dc|h1|∥!≤g(2ε).Similarly ∥DFX(|h2|)∥≤g(2ε). Hence ∥FX(h1) −FX(|h2|)∥≤g(ε) + 2g(2ε).Therefore FX extends uniquely to a uniformly continuous map, that westill denote FX, from S(L1(µ)) to S(X), and the modulus of continuity ofFX depends only on the modulus of uniform convexity of X.

•Proposition 2 .9 Let X be uniformly convex and uniformly smooth. ThenFX : S(L1(µ))−→S(X) is a uniform homeomorphism.

Moreover (FX)−1 :S(X)−→S(L1(µ)) has modulus of continuity depending only on the modulusof uniform smoothness of X. Furthermore (FX)−1(x) = |x∗| · x where x∗∈S(X∗) is the unique supporting functional of x.14

Proof Our goal now is to show that the map FX previously defined is in-vertible and that (FX)−1 has the described form and is uniformly continuous.Claim 1: Let h ∈S(L1(µ)) T L∞(µ). Then g = FX(h)−1 · h ∈S(X∗) where· denotes the pointwise product.Note that suppFX(h) = supph and we define FX(h)−1 · h to be 0 offthesupport of h. Assume Claim 1 for the moment.For x ∈S(X), define G(x) = |x∗| · x, where x∗is the unique supportingfunctional of x.

Let h ∈S(L1(µ)) T L∞(µ). Since signFX(h) = signh,ZhFX(h)|FX(h)|dµ =Z|h|dµ = 1.Thus from Claim 1 it follows thathFX(h) = |FX(h)|∗= |FX(h)∗|.Hence G(FX(h)) = |FX(h)|∗· FX(h) = h for any h ∈S(L1(µ)) T L∞(µ).Furthermore G is uniformly continuous.

Indeed, the support functionalx 7→x∗is uniformly continuous since X is uniformly smooth, and sinceG(xi) = |x∗i | · xi i = 1, 2 we have∥G(x1) −G(x2)∥=∥|x∗1| · x1 −|x∗2| · x2∥≤∥|x∗1| · (x1 −x2)∥+ ∥(|x∗1| −|x∗2|) · x2∥≤∥x1 −x2∥+ ∥x∗1 −x∗2∥.Thus G is uniformly continuous. Moreover since the modulus of continuityof x 7→x∗depends only on the modulus of uniform smoothness of X, thesame is valid for G. Thus G(FX(h)) = h for all h ∈S(L1(µ)).Claim 2: G is one-to-one.15

It then follows that G = (FX)−1. We now prove Claim 1Proof of Claim1: We will follow the path of [G].

Let h ∈S(L1(µ))T L∞(µ)and suppose x = FX(h). We can assume that h ∈S(L1(µ))+ T L∞(µ).

Thensuppx = supph ≡B and x ∈S(X)+. Let k ∈X+ be arbitrary, then∞> E(h, x) ≥Zh log x + k∥x + k∥dµ.So writing x + k = x(1 + kx) for x ∈B yieldsE(h, x) ≥E(h, x) +ZB h log(1 + kx−1)dµ −log ∥x + k∥.This gives:ZB h log(1 + kx−1)dµ≤log ∥x + k∥≤log(∥x∥+ ∥k∥)=log(1 + ∥k∥).SoZB h log(1 + kx−1)dµ ≤∥k∥(⋆).Thus on B, kx−1 is finite µ-almost everywhere.

Letσn = {ω ∈B : k(ω)x−1(ω) ≤n}and χn = χσn then χn ր χB, pointwise µ-a.e; and since t ≤log(1 + t) + 12t2holds for all t ≥0 we have for 0 < s < ∞sZB hx−1kχndµ≤ZB h log(1 + skx−1χn)dµ + 12s2ZB k2(x−1)2χnhdµ≤ZB h log(1 + skx−1)dµ + 12s2n2≤s∥k∥+ 12s2n2 by (⋆).16

Thus dividing by s and letting s go to 0, we obtain for all n ∈NZhx−1kχndµ ≤∥k∥;and therefore by the monotone convergence theorem,ZB hx−1kdµ ≤∥k∥.Now let g = hx−1. The previous equality yields ∥g∥X∗≤1.

On the otherhand1 =Zhdµ=Zg · xdµ≤∥x∥X∥g∥X∗So ∥g∥X∗= 1 which proves Claim 1. •Proof of Claim 2: Let h = |x∗1| · x1 = |x∗2| · x2 be a member of S(L1(µ))with x∗i (xi) = 1, xi ∈S(X) and x∗i ∈S(X∗) for i = 1, 2.

We first note thatsupph = suppxi for i = 1, 2. Indeed supph ⊂suppxi is clear, and in case theinclusion is strict let us consider B|xi| where B = supph.

We then note that∥B|x|∥< 1 by uniform convexity. Also|x∗|(B|x|)=Z|x∗|B|x|dµ=ZB |x∗||x|dµ=Z|h|dµ=1, a contradiction.Also suppx∗i = B since X∗is uniformly convex.

Now as in [G] we observethat there exists a measurable function θ of modulus one so that x∗2 = θx∗1.Indeed define θ = x∗2x∗1 on B and θ = 1 on Bc. ThenZ|h||θ|dµ=Z|x1||x∗2|dµ17

≤∥|x∗2|∥X∗∥|x1|∥X=1Similarly,R |h||θ−1|dµ ≤1. SoZ|h|{|θ| + |θ−1|}dµ ≤2.And since t + t−1 ≥2 for t > 0 we getZ|h|{|θ| + |θ−1|}dµ ≥2Z|h|dµ = 2.Thus |θ| + |θ−1| = 2, but this cannot happen unless |θ| = 1.

Thus |x∗1| = |x∗2|.Now suppxi = supph and h = |x∗1| · x1 = |x∗2| · x2 yields that x1 = x2. •We are now ready to give a proof of the main result of this work.Proof of Theorem 2.1: Suppose thatX contains ln∞uniformly in n. ThenS(X) is not homeomorphic to S(L1((Ω, Σ, µ))) for any measure space (Ω, Σ, µ).Indeed this follows, as in [O.S], from Enflo’s result [E] that the setsS(ln∞), n ∈N cannot be uniformly embedded into S(L1).For the converse assume that X does not contain ln∞uniformly in n. ThenX must be order continuous since X does not contain c0 [L.T].

Then theproof goes as in [O.S]. By a theorem of Maurey and Pisier [M.P] X must havea finite cotype q′.

Thus X is q-concave, in fact for all q > q′ ([L.T, p 88]).Renorm X by an equivalent norm for which Mq(X) = 1 and such that Xhas the same lattice structure (see [L.T, p 54]). Then the 2-convexificationX(2) of X in this norm satisfiesM2q(X(2)) = 1 = M2(X(2))([L.T, p 54] ).

This implies that X(2) is uniformly convex and uniformlysmooth ([L.T, p 80]), and soFX(2) : S(L1(µ))−→S(X(2))is a uniform homeomorphism by Proposition 2.9. ThereforeG2 ◦FX(2) : S(L1(µ))−→S(X)18

is a uniform homeomorphism by Proposition 2.4. •Remark 2.10 [O.S] If S(X) is uniformly homeomorphic to S(Y ) then BaXand BaY are uniformly homeomorphic.Corollary 2.11: If X is a separable infinite dimensional Banach latticethen S(X) and S(l1) are uniformly homeomorphic if and only if X does notcontain ln∞uniformly.Proof: By Theorem 2.1, S(X) is uniformly homeomorphic to S(L1(µ)) forsome probability space (Ω, Σ, µ) where L1(µ) is separable.

By standard rep-resentation theorems either L1(µ) ∼= l1 or L1(µ) ∼= (L1[0, 1] ⊕l1(I))1 where Iis countable. So S(X) is uniformly homeomorphic to S((L1[0, 1] ⊕l1(I))1).Then one can defineH : S((L1[0, 1] ⊕l1(I))1)−→S((l1 ⊕l1(I))1)as follows: Let F be a uniform homeomorphism between S(L1) and S(l1).

(Such homeomorphism exists by [O.S]). If (g, x) ∈S(L1[0, 1] ⊕l1(I))1 thendefine H(g, x) =∥g∥Fg∥g∥, xfor g ̸= 0 and H(0, x) = (0, x).

It is easilychecked that H is a uniform homeomorphism and now, since I is countable,l1 ⊕l1(I) ≡l1 which proves the Corollary. •Remark 2.12: In [R], Y.Raynaud already obtained that if the unit ball ofa Banach space E, embeds uniformly into a stable Banach space F, then Edoes not contain c0.

He also proved that if F is supposed superstable thenE does not contain ln∞uniformly. Since L1 is superstable, we could get onedirection of Theorem 2.1 in the separable case using the result of [R].Remark 2.13: If X is q-concave with constant 1, then X(2) satisfiesM2q(X(2)) = M2(X(2)) = 1,19

([L.T, p 54]) and as we noted before, X(2) is uniformly convex and uniformlysmooth ([L.T, p 80]). We then proved thatFX(2) : S(L1(µ))−→S(X(2))is a uniform homeomorphism with modulus of continuity of FX(2) dependingonly on the modulus of uniform convexity δX(2)(ε) of X(2) (which in turn isof power type 2, i.e for some constant0 < K < ∞,δX(2)(ε) ≥Kε2.

([L.T, p 80])) and the modulus of continuityof (FX(2))−1 depending only on the modulus of uniform smoothness ρX(2)(τ)of X(2) (which in turn is of power 2q i.e.for some constant 0 < K <∞, ρX(2)(τ) ≤Kτ 2q) [L.T, p 80]. )Proof of Theorem 2.2: We first observe that X and Y must have weakunits, since they are separable [L.T, p 9]; and are order continuous since theyboth don’t contain c0.

In fact, since q < ∞and q′ < ∞, X and Y don’t con-tain ln∞. So, by Corollary 2.11, S(X) and S(Y ) are uniformly homeomorphicto S(L1).

Let ¯X be X endowed with an equivalent norm and the same order,for which Mq( ¯X) = 1, and let ¯Y be Y with an equivalent norm and the sameorder, for which Mq′( ¯Y ) = 1. With the previous notations used throughoutthis work, we have the following diagram:S(X) u−1−→S( ¯X)(G ¯X,2)−1−→S( ¯X(2))(F ¯X(2))−1−→S(L1)F ¯Y (2)−→S( ¯Y (2))G ¯Y ,2−→S( ¯Y )v−→S(Y )where v is a uniform homeomorphism from S( ¯Y ) to S(Y ) with a modulus ofcontinuity a depending solely on Mq′(Y ), and u−1 is a uniform homeomor-phism from S(X) to S( ¯X) with a modulus of continuity f depending onlyon Mq(X).LetF = v ◦G ¯Y ,2 ◦F ¯Y (2) ◦(F ¯X(2))−1 ◦(G ¯X,2)−1 ◦u−1,then F is clearly a homeomorphism andF −1 = u ◦G ¯X,2 ◦F ¯X(2) ◦(F ¯Y (2))−1 ◦(G ¯Y ,2)−1 ◦v−1.Let b, c, d and e be respectively the modulus of continuity of respectivelyG ¯Y ,2, F ¯Y (2), (F ¯X(2))−1, (G ¯X,2)−1.

b and e are functions solely of 2 by Proposi-tion 2.4 while c and d are functions of q′ and q by Proposition 2.9, Proposition20

2.8, and the remark 2.13 above. Then the modulus of uniform continuity αof F is of the form α = a ◦b ◦c ◦d ◦e ◦f and is a function solely ofq, q′, Mq(X), Mq′(Y ).

Note that the modulus of continuity of F −1 is also givenby a ◦b ◦c ◦d ◦e ◦f. •Proof of Theorem 2.3: The proof is exactly the same as in Theorem2.2 with the only difference that F = FY ◦(FX)−1.

Indeed we have nowthe diagram: S(X)(FX)−1−→S(L1)FY−→S(Y ). We then let F = FY ◦(FX)−1and use Proposition 2.9 to get that the modulus of continuity of F dependssolely on the modulus of uniform convexity of Y and the modulus of uniformsmoothness of X.

•References[B]Y. Benyamini,The uniform classification of Banach spaces,’Longhorn Notes 1984-85’, The University of Texas at Austin.[E]P. Enflo, On a problem of Smirnov, Ark.

Mat. 8 (1969), 107-109.[G]T.A.

Gillespie Factorization in Banach function spaces, Indaga-tiones Math. bf 43 (1981), 287-300.[K]N.J.

Kalton Uniform homeomorphisms and complex interpolation,in preparation.[L.T]J. Lindenstrauss and L. Tzafriri “Classical Banach spaces” Vol 2,Springer Verlag NY.

(1979)[M.P]B.MaureyandG.PisierS´eriesdevariablesal´eatoiresind´ependantes et propri´et´es g´eom´etriques des espaces de Banach.Studia Math. 58(1976), 45-90.[O.S]E.

Odell and Th. Schlumprecht The distortion problem Preliminaryversion.21

[R]Y. Raynaud Espaces de Banach superstables, distances stables ethom´eomorphismes uniformes Israel J. Math.

44 (1983) 33-52.F. ChaatitDepartment of MathematicsThe University of Texas at AustinAustin, TX 78712-1082 U.S.A.chaatit@math.utexas.edu22

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