J. Math. Anal. Appl. 340 (2008) 1312–1325 www.elsevier.com/locate/jmaa
Nonlinear structure of some classical quasi-Banach spaces and F-spaces ✩
F. Albiac a,b,∗
a Department of Mathematics, University of Missouri, Columbia, MO 65211, USA b Departamento de Matemáticas, Universidad Pública de Navarra, Pamplona 31006, Spain Received 9 January 2007 Available online 29 September 2007 Submitted by J. Diestel
Abstract
We investigate the Lipschitz structure of p and Lp for 0 Keywords: Banach space; F-space; Uniform homeomorphism; Lipschitz 1. Introduction Very little is known about the nonlinear structure of (infinite dimensional) quasi-Banach spaces and F-spaces. In general, in the few results found in the literature, authors have treated uniform structure rather than Lipschitz structure. For example, Enflo [10] proved that L p[0, 1] is not uniformly homeomorphic to Lq [0, 1] if 0 ✩ The research reported in this paper was carried out during the author’s stay in the Department of Mathematics at the University of Missouri– Columbia as a Visiting Research Scholar funded by a Richard Miller Scholarship, in the Fall of 2006. * Address for correspondence: Departamento de Matemáticas, Universidad Pública de Navarra, Campus de Arrosadía, Pamplona 31006, Spain. E-mail address: [email protected]. 0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.09.052 F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 1313 As regards Lipschitz classification, Albiac and Kalton have shown in [3] that the Lipschitz structure of a separable quasi-Banach space which is not a Banach space need not determine its linear structure. This remains an important open question for separable Banach spaces. A Lipschitz function from a separable Banach space X to a Banach space Y with the Radon–Nikodym property is (Gâteaux) differentiable at least at one point. This important theorem, proved independently by Aronszajn, Chris- tensen, and Mankiewicz [5,8,14], in combination with the simple fact that if a Lipschitz embedding between Banach spaces is differentiable at some point, then its derivative at this point is a linear into isomorphism, proves the impos- sibility of Lipschitz embedding q in p if 1 q = p<∞. It also shows that Lq is not Lipschitz isomorphic to Lp when 1 q = p<∞. To see that the differentiation tool is no longer available for F-space valued functions, one can consider, for instance, the function ϕ(t) = χ[0,t] (i.e., the characteristic function of the interval [0,t]) from the metric p space [0, 1] into Lp for 0 functional analysis, and many basic but fundamental questions remain open. We hope that the remarks contained in this paper bring the reader’s attention to this interesting and genuine subject. 2. Preliminaries An F-space is a complete metric linear space. The uniform structure of an F-space X can be given by an F-norm, i.e., a function ·F : X →[0, ∞) with the following properties: (i) xF > 0ifx = 0; (ii) αxF |α|xF, whenever |α| 1 and x ∈ X; (iii) limα→0 αxF = 0 for all x ∈ X; (iv) x + yF xF +yF, for all x,y ∈ X. 1314 F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 Note that axioms (ii) and (iv) imply that αxnF → 0 for any scalar α provided xnF → 0. If ·F is an F-norm on X, then dF(x, y) =x − yF is a translation invariant metric on X. A countable base of neighborhoods at the origin for the topology defined on X by the F-norm is given by sets of the form Vn ={x ∈ X: xF < 1/n}. A sequence ∞ ∈ − → (xn)n=1 converges to x X if and only if xn x F 0. A quasi-normed space X is a locally bounded topological vector space. This is equivalent to saying that the topol- ogy on X is induced by a quasi-norm, i.e., a map ·:X →[0, ∞) satisfying (i) x=0 if and only if x = 0; (ii) αx=|α|x if α ∈ R,x∈ X; (iii) there is a constant κ 1 so that for any x and y in X we have x + y κ x+y . (2.1) If it is possible to take κ = 1 we obtain a norm. A quasi-norm ·is called p-norm (0 Proposition 2.1 (Aoki–Rolewicz). Let (X, ·) be a quasi-normed space and 0 Definition. Let (X, ·X) and (Y, ·Y ) be (real) quasi-normed spaces. We say that a map f : X → Y is Lipschitz if there exists a positive constant C so that − − ∈ f(x) f(y) Y C x y X,x,yX. The least such C is called the Lipschitz norm of f and it is denoted by f Lip. Alternatively, f Lip can be defined by f(x)− f(y)Y f Lip = sup : x,y ∈ X, x = y . x − yX f Lip =∞iff f is not Lipschitz. If f : X → Y is one-to-one and Lipschitz, and f −1 : f(X)→ X is Lipschitz, then f is called a Lipschitz embedding and we say that X Lipschitz embeds in Y .Iff is a bijection, and both f and f −1 are Lipschitz, then we say that f is bi-Lipschitz, and X and Y are called Lipschitz isomorphic. A (not necessarily linear) map U from a quasi-Banach space (X, ·X) into a quasi-Banach space (Y, ·Y ) is an isometry if − = − ∈ U(x) U(y) Y x y X,x,yX. F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 1315 A classical theorem of Mazur and Ulam from 1932 [15] establishes that a surjective isometry between two real Banach spaces that takes 0 to 0 is linear, i.e., the linear structure of a Banach space is completely determined by its isometric structure. The generalization of this result to quasi-Banach spaces can be deduced from the following theorem of Rolewicz from 1968: Theorem 2.2. (See [21, Theorem 2].) Let X and Y be two real locally bounded F-spaces with arbitrary F-norms ·F. If U : X → Y is an onto isometry with U(0) = 0 such that − = − tU(x) tU(y) F tx ty F, (2.2) for all positive t and all x,y in X, then U is a linear operator. Corollary 2.3. Suppose that X is a p-Banach, that Y is a q-Banach space and that 0 Proof. Since p q, the quasi-norm on Y is also p-subadditive. If U(x)− U(y)Y =x − yX whenever x,y ∈ X, ·p ·p then U is also an isometry between the F-spaces (X, X) and (Y, Y ) which satisfies (2.2). Using Theorem 2.2 we conclude that U must be linear. 2 Corollary 2.4. Suppose X is a quasi-Banach space. If X is isometric to p (respectively Lp) for some 0 In order to obtain the impossibility of having isometries between some classical F-spaces we need another theorem of Rolewicz on isometries between F-spaces, where the F-norms in question are now concave. Following [22], an F- norm · on X is said to be concave if the function tx is concave for all x in X and all positive t, i.e., F F + t1 t2 1 x t1xF +t2xF ,t1,t2 0. 2 F 2 Theorem 2.5. (See [21, Theorem 1], [22, Theorem 9.3.12].) Let X and Y be two real locally bounded F-spaces with arbitrary F-norms ·F. Let the F-norms in X and Y be concave. Then each isometry U : X → Y such that U(0) = 0 is linear. As an immediate consequence we can now state the following corollary1: Corollary 2.6. Let 0 (i) (Lp,dp) and (Lq ,dq ) are not isometric. (ii) ( p,dp) and ( q ,dq ) are not isometric. Proof. It is straightforward to check that the F-norm in Lp (respectively p) is concave and so an isometry between Lp and Lq (respectively p and q ) must be linear, an absurdity. 2 If X is a quasi-normed space, we will consider, as for Banach spaces, the Lipschitz dual of X,Lip0(X), the vector space of all real-valued Lipschitz functions f defined on X such that f(0) = 0, endowed with the Lipschitz norm |f(x)− f(y)| f Lip = sup : x,y ∈ X, x = y . x − yX · It can be readily checked that (Lip0(X), Lip) is a Banach space, and hence we omit the details. The proof of Proposition 2.8 will rely on the following well-known fact. We recall that a metric space (M, d) is = = 1 said to be metrically convex if for any x and y in M there is z in M with d(x,z) d(y,z) 2 d(x,y). Such z is a metric midpoint between x and y in M. 1 The author was made aware by N.J. Kalton that this result appears in [17], where it was obtained in a completely different manner. 1316 F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 Lemma 2.7. Let (X, d) be a metrically convex metric space and let (Y, ρ) be any metric space. Suppose f : X → Y satisfies ρ f(x),f(y) Cdα(x, y), x, y ∈ X, (2.3) for some constants C>0 and α>1. Then f is constant. Proof. Let C be the smallest constant so that (2.3) holds. Given x = y in X we pick z, a metric midpoint between x and y. Then, by the hypothesis, ρ f(x),f(z) Cdα(x, z), and ρ f(z),f(y) Cdα(y, z), which yields 2C ρ f(x),f(y) ρ f(x),f(z) + ρ f(z),f(y) dα(x, y). 2α Notice that 2C/2α ={ } Proposition 2.8. If 0 Proof. Suppose that f ∈ Lip (Lp), i.e., f(0) = 0 and 0 f(x)− f(y) f Lipx − yp,x,y∈ Lp. p Let dp(x, y) denote the distance in Lp induced by the p-norm, i.e., dp(x, y) =x − yp. Then, − p p ∈ f(x) f(y) f Lipdp(x, y), x, y Lp, and so 1/p f(x)− f(y) f Lipdp (x, y), x, y ∈ Lp. It is a known result that (Lp,dp) is metrically convex: given x and y in Lp, the continuity of the function t [0, 1]→R,t→ x(s) − y(s) p ds 0 t0 1 1 ∈ | − |p = | − |p = [ ] + [ ] yields t0 (0, 1) such that 0 x y 2 0 x y ; thus z yχ 0,t0 xχ t0,1 is a metric midpoint between x and y in (Lp,dp). Using Lemma 2.7 we deduce that f must be constant and, since f(0) = 0, it must be f = 0. 2 3. Lipschitz structure of the quasi-Banach spaces p and Lp for 0 Proposition 3.1. Let (X, ·X) and (Y, ·Y ) be quasi-normed spaces. If X Lipschitz embeds in Y and Y is q- normable for some 0 Proof. Suppose that f : X → Y is a Lipschitz embedding with f(0) = 0. Let K be the maximum of the Lipschitz { }n = +···+ = constants of f and its inverse. Given any xi i=1 in X,for1 k n let zk x1 xk and put z0 0. Then q n = − q xi zn z0 X i=1 X q − q K f(zn) f(z0) Y F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 1317 q n = q − K f(zk) f(zk−1) k=1 Y n q q − − K f(zk) f(zk 1) Y k=1 n 2q − q K zk zk−1 X k=1 n = 2q q K xk X, k=1 and so by Proposition 2.1 X is q-normable. 2 Proposition 3.2. (i) If 0 Proof. (i) If p Lipschitz embedded into q , then p would be q-normable, which is impossible (see [13, Lemma 2.7]). The corresponding statement for the spaces Lp goes along the same lines since Lp is not q-normable for any p< q 1 [13, Theorem 2.2]. (ii) Suppose that there is f : L → L satisfying p q − − ∈ f(x) f(y) q C x y p,x,yLp, for some C>0. This implies − q q − q = q − p q/p ∈ f(x) f(y) q C x y p C x y p ,x,yLp. Since q/p > 1 and the metric space (Lp,dp) is metrically convex, Lemma 2.7 yields that f is constant. : → = ∗ ∈ ∗ ∗ ◦ ∈ (iii) Let f Lp q be a Lipschitz map, and assume that f(0) 0. Then for any x q we have that x f ∗ ◦ = ∈ ∗ Lip0(Lp). Proposition 2.8 implies that x f(x) 0 for all x Lp.But q separates the points of q , which forces f(x)to be 0 for all x ∈ Lp. Property (iv) is an immediate consequence of Proposition 3.1. 2 Remark. Using negative definite kernels, Aharoni, Maurey, and Mityagin [1] have obtained that the quasi-Banach space Lp embeds isometrically into the quasi-Banach space Lr if 0 4. Lipschitz structure of ( p,dp) for 0 Proposition 4.1. Let 0 (i) Any Lipschitz function from ( q ,dq ) to ( p,dp) is constant. In particular, the metric space ( q ,dq ) does not Lipschitz embed in ( p,dp). (ii) ( p,dp) Lipschitz embeds in ( q ,dq ). 1318 F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 Proof. (i) Suppose there is ϕ : q → p satisfying − p − q ∈ ϕ(x) ϕ(y) p K x y q ,x,y q , ∗ ∈ ∗ ∗= for some K>0. Then, for any x p with x 1wehave ∗ ◦ − ∗ ◦ ∗ − 1/p − q/p ∈ x ϕ(x) x ϕ(y) x ϕ(x) ϕ(y) p K x y q ,x,y q . Fix x ∈ q with xq = 1 and consider the metric space Rx ={rx: r ∈ R} with the distance ρ between points sx and tx in Rx defined as ρ(sx,tx)=sx − txq =|s − t|,s,t∈ R. If we restrict the map x∗ ◦ ϕ to Rx we obtain ∗ ∗ x ◦ ϕ(tx) − x ◦ ϕ(sx) K1/p|t − s|q/p,t,s∈ R. Notice that (Rx,ρ) is metrically convex and that q/p > 1, so by Lemma 2.7, x∗ ◦ ϕ is constant on Rx. R x Now take any two different nonzero elements x and y in q . Both 0 and x belong to x . Analogously, both 0 R y and y belong to y . Thus, ∗ ∗ ∗ ∗ ∗ ∗ x ◦ ϕ(x) − x ◦ ϕ(y) x ◦ ϕ(x) − x ◦ ϕ(0) + x ◦ ϕ(0) − x ◦ ϕ(y) = 0, ∗ ◦ = ∗ ◦ ∗ = i.e., x ϕ(x) x ϕ(y).But p separates the points of p, thus ϕ(x) ϕ(y), i.e., ϕ must be constant. (ii) Let us see that for every 0 Let k ∈ Z be so that either s or t or both are in the support of fN,k. Then for the integers n N we write n(1−p) n(1−p) p 1−p fn,k(s) − fn,k(t) = 2 |s − t|=2 |s − t| |s − t| − − − − = 2n(1 p)|s − t|pα1 p2 (1 p)N − − − = 2(1 p)(n N)α1 p|s − t|p. If n>N we have −np fn,k(s) − fn,k(t) 2 − − = 2(N n)p2 Np − − = 2(N n)pα p|s − t|p. F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 1319 Hence, for each k so that either s or t or both belong to the support of fn,k, ∞ fn,k(s) − fn,k(t) = fn,k(s) − fn,k(t) + fn,k(s) − fn,k(t) n=−∞ nN n>N − − − − − α1 p 2(1 p)(n N) + α p 2(N n)p |s − t|p nN n>N 21−p 2p + |s − t|p. 21−p − 1 2p − 1 Notice that for each fixed n ∈ Z, the points s and/or t belong to the support of a finite and uniform number N0 8of functions f ,f ,...,f . Thus we obtain a Lipschitz estimate for f , n,k1 n,k2 n,kN0 ∞ ∞ = − = − d1 f(s),f(t) f(s) f(t) 1 fn,k(s) fn,k(t) n=−∞ k=−∞ 21−p 2p N + |s − t|p. 0 21−p − 1 2p − 1 To obtain the reverse Lipschitz estimate, let us pick the largest k ∈ Z so that both s and t belong to the support of fN,k. By symmetry, without loss of generality we can assume that k = 1, and to simplify our discussion we will suppose that s The proof of the Lipschitz embedding of ( p,dp) in ( q ,dq ) for p Corollary 4.2. The spaces ( p,dp) and ( q ,dq ) are not Lipschitz isomorphic when 0 Remark. Notice that if 0 5. Uniform nonequivalence of L0 and Lp (0 Let L0 be the space of Lebesgue-measurable functions on [0, 1], with the usual convention of identifying functions equal almost everywhere, equipped with the F-norm |f(t)| f = dt. 0 1 +|f(t)| The topology induced by the L0-metric is just the topology of convergence in measure. Clearly, L0 is a nonlocally bounded F-space. Lemma 5.1. Given δ>0 there is an integer N = N(δ) such that for any f and g in L0 there exist functions f = f0,f1,...,fN−1,fN = g in L0 such that fj − fj−10 <δ,forj = 1,...,N. Proof. Without loss of generality we take f = 0. Given δ>0wepickN =[1/δ]+1 and functions f0 = 0 and = fj gχ[ j ] if 1 j N. Then 0, N j N |g(t)| 1 f − f − = dt <δ. 2 j j 1 0 1 +|g(t)| N j−1 N Proposition 5.2. Let X be a locally bounded topological vector space. If ϕ : L0 → X is a uniformly continuous map, then the range of ϕ is a bounded set in X. Proof. The metric topology on X is induced by a quasi-norm ·, which can be assumed p-subadditive for some 0 0 so that for any f1,f2 ∈ L0 with p f1 − f20 <δwe have ϕ(f1) − ϕ(f2) < 1. Given any f ∈ L0, by Lemma 5.1 we take 0 = f1,f2,...,fN = f ∈ L0 (where N depends only on δ) with fj − fj−10 <δfor j = 1,...,N. Then p N N p = − − p 2 ϕ(f) ϕ(fj−1) ϕ(fj ) ϕ(fj ) ϕ(fj−1) Corollary 5.3. L0 is not uniformly homeomorphic to any locally bounded topological vector space. In particular L0 is not uniformly homeomorphic to Lp for 0 6. The spaces Lp for 0 Ultraproduct techniques appear in a natural way, and have played a decisive role, in the reduction of some questions on uniform classification of Banach spaces to the Lipschitz category, as can be appreciated by reading [6, Chapter 10, Section 1]. To define the ultrapower of a p-Banach space (0 Schreiber [24] proved that the Lp spaces are stable under ultrapowers when 0 Proposition 6.1. (See [24, Proposition 3.6].) Let 0 The next result is a generalization to a certain class of F-spaces of a theorem by Heinrich and Mankiewicz [12]. r Theorem 6.2. Let (X, ·) be an r-Banach space (0 Proof. Let f : X → Y be a uniform homeomorphism, and assume that f(0) = 0. Since both (X, dX) and (Y, dY ) are metrically convex, by the Corson–Klee lemma both f and f −1 satisfy a Lipschitz condition for large distances, i.e., there is a constant K>0 so that 1 x − yr f(x)− f(y) p Kx − yr (6.1) K for all x,y ∈ X with x − yr 1. For each n ∈ N let us consider the function 1 f (x) = f n1/rx ,x∈ X. n n1/p The modulus of continuity of fn is bounded by that of f . Indeed, by the uniform continuity of f ,given >0 there is r p δ = δ( ) so that whenever a − b <δ, then f(a)− f(b) < . Since (X, dX) is metrically convex, given any x,y 1/r 1/r r r in X we can find points n x = x0,x1,...,xn = n y in X with xk − xk−1 =x − y <δfor each k = 1,...,n. Then 1 f (x) − f (y) p = f n1/rx − f n1/ry p n n n 1322 F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 p 1 n = f(x − ) − f(x ) n k 1 k k=1 n 1 p f(x − ) − f(x ) < . n k 1 k k=1 ∞ This actually shows that (fn)n=1 is equicontinuous. Let x and y in X and suppose that x − yr 1/n for some n ∈ N, i.e., n1/rx − n1/ryr 1. Then, Eq. (6.1) yields f n1/rx − f n1/ry p K n1/rx − n1/ry r = Knx − yr , or equivalently, 1/r 1/r p f(n x) f(n y) r − Kx − y . n1/p n1/p We conclude that p r r fn(x) − fn(y) Kx − y , whenever x − y 1/n. (6.2) We can now show that if (xn)n∈N is a bounded sequence in X, then (fn(xn))n∈N is a bounded sequence in Y . ∞ Suppose that supn xn 1. Then, on one hand, thanks to the equicontinuity of (fn)n=1, there exists δ>0 so that p r p r fn(xn) 1 whenever xn <δ. On the other hand, by (6.2), fn(xn) K if xn 1/n. The only elements ∞ ={ ∈ N r from the sequence (xn)n=1 that have not yet been taken care of are the ones from the set F n : δ xn < 1/n}, which is finite; put c = max ∈ f (x )p. This way we have n F n n p fn(xn) max{1,K,c},n∈ N. Thus given a free ultrafilter U on N, the natural map between the ultrapowers of X and Y through U, : → ∈N → F (X)U (Y )U ,(xn)n fn(xn) n∈N is well defined. We will show that F is a Lipschitz map (with constant at most K)from(X)U to (Y )U regarded as metric spaces, with the distances induced by their respective quasi-norms. = ∞ = ∞ To that end, let x (xn)n=1 and y (yn)n=1 be two different points in (X)U . Let us pick δ>0 so that p r r f(a)− f(b) K limU xn − yn whenever a − b <δ. As we have seen before, this also implies that p r r fn(a) − fn(b) K limU xn − yn for all n.Theset{n ∈ N: δ xn − yn 1/n} does not belong to the ultrafilter U.Ifn is such that x − y r <δwe argued that n n f (x ) − f (y ) p K lim x − y r , n n n n U n n and if x − y r 1/n, by (6.2) we have n n p r fn(xn) − fn(yn) Kxn − yn . Whence, F(x)− F(y) p = lim f (x ) − f (y ) p K lim x − y r = Kx − yr . U n n n n U n n To complete the proof, we would follow the previous steps and define G : (Y )U → (X)U via the functions 1 − g (y) = f 1 n1/py ,y∈ Y. n n1/r G is a Lipschitz map (with constant no bigger than K) from (Y )U to (X)U regarded as metric spaces, with the distances induced by their respective quasi-norms, and it is the inverse of F . 2 We can now use this result to reduce the problem of uniformly classifying the spaces Lp and Lq when 0 Theorem 6.3. The spaces Lp and Lq (0 Proof. Assume Lp and Lq are uniformly homeomorphic. Then, by Theorem 6.2, given a nonprincipal ultrafilter U on N, (Lp)U is Lipschitz isomorphic to (Lq )U . Proposition 6.1 yields nonatomic measure spaces (Ω,Σ,μ) and (Ω ,Σ ,ν)such that (Lp)U and (Lq )U are linearly isomorphic and order isometric to Lp(μ) and Lq (ν), respectively. Hence there is a bijection and bi-Lipschitz map S : Lp(μ) → Lq (ν). Now we first pick a closed nonatomic separable sublattice X1 of Lp(μ). It follows that X1 must be isometrically isomorphic to Lp[0, 1] and that S(X1) is a separable subset of Lq (ν).LetY1 be the closed nonatomic (separable) sublattice of Lq (ν) generated by the functions in S(X1). Y1 is isometrically isomorphic to Lq [0, 1]. Next we take the −1 smallest closed nonatomic separable sublattice X2 in Lp(μ) that contains the subset S (Y1) and repeat this procedure ⊂ ⊂··· recurrently. This way we obtain two increasing sequences of closed nonatomic separable sublattices X1 X2 in ⊂ ⊂··· ∞ = ∞ Lp(μ) and Y1 Y2 in Lq (ν) so that S( n=1 Xn) n=1 Yn. Hence we have a Lipschitz isomorphism between ∞ ∞ [ ] [ ] 2 n=1 Xn and n=1 Yn, which are separable sublattices isometric to Lp 0, 1 and Lq 0, 1 , respectively. Aharoni, Maurey, and Mityagin [1] have shown that a linear metric space embeds uniformly into a Hilbert space if and only if it is linearly isomorphic to a subspace of L0. In particular, Lp embeds uniformly into L2 if and only if 0 Theorem 6.4 (Schoemberg’s Theorem). Let N(x,y) be a real-valued kernel on a set X satisfying N(x,x) = 0 for every x ∈ X. Then N is negative definite if and only if there is a Hilbert space H and a map T from X into H such that N(x,y) = T(x)− T(y) 2 for all x,y ∈ X. Proposition 6.5. Suppose 0 Proof. We will see that (i) Lp embeds isometrically in L1, and that (ii) L1 embeds isometrically in Lq . (i) The function N(s, t) such that N(s, t) =|s − t|2 is a normalized negative definite kernel on R × R, hence for each 0 7. Final remarks and open problems It is clear that our work leaves many questions unanswered. Some of them are very well-known problems to the analysts working in the nonlinear classification and have already been formulated and published previously. In any case, we have decided to restate them here for reference, since they motivated our results and seem the natural road to take for further research in this direction. Proposition 3.2(iv) immediately yields that a nonnormable quasi-Banach space cannot be Lipschitz isomorphic to a normed space. In connection with this, one of the most important questions is: 1324 F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 Problem 7.1. (See [6, p. 407].) Can a nonnormable quasi-Banach space be uniformly homeomorphic to a Banach space? Enflo [10] proved that the answer to this problem is negative when the Banach space has roundness greater than one. Prassidis and Weston [18] have replaced roundness by the more general notion of metric type and showed that if a Banach space X with metric type greater than one is uniformly homeomorphic to a locally bounded linear space Y , the uniform Banach group structure induced on X by the uniform homeomorphism is uniformly dissipative and Y is normable. A different problem is suggested by Corollary 2.4: Problem 7.2. Assume X is a quasi-Banach space Lipschitz isomorphic to p (respectively Lp). Is X linearly isomor- phic to p (respectively Lp)? To complete the list (which we started in Proposition 3.2) of the q quasi-Banach spaces that embed into another p quasi-Banach space we need to answer the following question: Problem 7.3. Does the quasi-Banach space p for 0 In Section 4 we saw that the metric spaces ( p,dp) are not Lipschitz isomorphic for different p<1. The following are important well-known related problems: Problem 7.4. (See [6, p. 407].) Are the spaces p for 0 Problem 7.5. Does the metric structure of p for 0 Problem 7.6. (See [6, p. 407].) Are the spaces (Lp,dp) for 0 We came up with this last problem in Section 6 when we transferred to the Lipschitz category the question of whether the Lp spaces for p<1 are mutually uniformly homeomorphic. Problem 7.7. Does the Lipschitz structure of (Lp,dp) for 0 Acknowledgment The author is grateful to Professor Nigel J. Kalton for being his Miller Scholar sponsor during his stay in the Department of Mathematics at the University of Missouri–Columbia in the Fall Semester of 2006 and for helpful discussions that contributed to shape this paper. References [1] I. Aharoni, B. Maurey, B.S. Mityagin, Uniform embeddings of metric spaces and of Banach spaces into Hilbert spaces, Israel J. Math. 52 (3) (1985) 251–265. [2] F. Albiac, N.J. Kalton, Topics in Banach Space Theory, Grad. Texts in Math., vol. 233, Springer, New York, 2006. [3] F. Albiac, N.J. Kalton, Lipschitz structure of quasi-Banach spaces, Israel J. Math., in press. [4] T. Aoki, Locally bounded linear topological spaces, Proc. Imp. Acad. Tokyo 18 (1942) 588–594. [5] N.J. Aronszajn, Differentiability of Lipschitzian mappings between Banach spaces, Studia Math. 57 (2) (1976) 147–190. [6] Y. Benyamini, J. Lindenstrauss, Geometric Nonlinear Functional Analysis, vol. 1, Amer. Math. Soc. Colloq. Publ., vol. 48, Amer. Math. Soc., Providence, RI, 2000. [7] H. Corson, V. Klee, Topological classification of convex sets, in: Proc. Sympos. Pure Math., vol. VII, Amer. Math. Soc., Providence, RI, 1963, pp. 37–51. [8] J.P.R. Christensen, Measure theoretic zero sets in infinite dimensional spaces and applications to differentiability of Lipschitz mappings, in: Actes du Deuxième Colloque d’Analyse Fonctionnelle de Bordeaux, I, Univ. Bordeaux, 1973, Publ. Dép. Math. (Lyon) 10 (2) (1973) 29–39. [9] J. Diestel, H. Jarchow, A. Tonge, Absolutely Summing Operators, Cambridge Stud. Adv. Math., vol. 43, Cambridge Univ. Press, Cambridge, 1995. [10] P. Enflo, Uniform structures and square roots in topological groups. I, II, Israel J. Math. 8 (1970) 230–252; Israel J. Math. 8 (1970) 253–272. [11] S. Heinrich, Ultraproducts in Banach space theory, J. Reine Angew. Math. 313 (1980) 72–104. F. Albiac / J. Math. Anal. Appl. 340 (2008) 1312–1325 1325 [12] S. Heinrich, P. Mankiewicz, Applications of ultrapowers to the uniform and Lipschitz classification of Banach spaces, Studia Math. 73 (3) (1982) 225–251. [13] N.J. Kalton, N.T. Peck, J.W. Rogers, An F-Space Sampler, London Math. Lecture Notes, vol. 89, Cambridge Univ. Press, Cambridge, 1985. [14] P. Mankiewicz, On the differentiability of Lipschitz mappings in Fréchet spaces, Studia Math. 45 (1973) 15–29. [15] S. Mazur, S. Ulam, Sur les transformations isométriques d’espaces vectoriels normés, C. R. Acad. Sci. Paris 194 (1932) 946–948. [16] M. Mendel, A. Naor, Metric cotype, Ann. of Math., in press. [17] A. Naor, Geometric problems in nonlinear functional analysis, Master’s thesis, Hebrew University, 1998. [18] E. Prassidis, A. Weston, Uniform Banach groups and structures, C. R. Math. Acad. Sci. Soc. R. Can. 26 (1) (2004) 25–32 (with English and French summaries). [19] M. Ribe, On uniformly homeomorphic normed spaces, Ark. Mat. 14 (1976) 237–244. [20] S. Rolewicz, On a certain class of linear metric spaces, Bull. Acad. Polon. Sci. Cl. III 5 (1957) 471–473, XL (with Russian summary). [21] S. Rolewicz, A generalization of the Mazur–Ulam theorem, Studia Math. 31 (1968) 501–505. [22] S. Rolewicz, Metric Linear Spaces, second ed., Math. Appl. (East European Series), vol. 20, D. Reidel Publishing Co., Dordrecht, 1985. [23] I.J. Schoenberg, Metric spaces and positive definite functions, Trans. Amer. Math. Soc. 44 (3) (1938) 522–536. [24] M. Schreiber, Quelques remarques sur les caractérisations des espaces Lp,0 p<1, Ann. Inst. H. Poincaré Sect. B (N.S.) 8 (1972) 83–92 (with English summary). [25] A. Weston, Some non-uniformly homeomorphic spaces, Israel J. Math. 83 (3) (1993) 375–380. [26] A. Weston, Linearization of certain uniform homeomorphisms, J. Aust. Math. Soc. 82 (2007) 1–9.