SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES

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1 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES A. DALET AND G. LANCIEN Absrac. We sudy he srucure of he space of coarse Lipschiz maps beween Banach spaces. In paricular we inroduce he noion of norm aaining coarse Lipschiz maps. We exend o he case of norm aaining coarse Lipschiz equivalences, a resul of G. Godefroy on Lipschiz equivalences. This leads us o include he non separable versions of classical resuls on he sabiliy of he exisence of asympoically uniformly smooh norms under Lipschiz or coarse Lipschiz equivalences. 1. Inroducion In a recen paper [6] G. Godefroy sudied various noions of norm aaining Lipschiz funcions. If (M, d) and (N, δ) are wo meric spaces and f : M N is Lipschiz, i is naural o say ha f aains is norm a he pair (x, y) in M M wih x y if δ(f(x), f(y)) = Lip (f), d(x, y) where Lip (f) denoes he Lipschiz consan of f. In [6], G. Godefroy inroduced he following weaker form of norm aaining vecor valued Lipschiz funcions. Le (M, d) be a meric space, (Y, Y ) a Banach space and f : M Y a Lipschiz map. We say ha f aains is norm in he direcion y S Y, where S Y denoes he uni sphere of Y, if here exiss a sequence (s n, n ) n=0 in M M wih s n n and such ha f(s n ) f( n ) lim = y Lip (f). n d(s n, n ) One of he main resuls of [6] is ha if a Lipschiz isomorphism f beween wo Banach spaces X and Y aains is norm in he direcion y S Y, hen here exiss a consan c > 0 such ha ρ Y (y, c) 2ρ X (), where ρ denoes he modulus of asympoic uniform smoohness (see definiions in secion 6). Then, noicing ha his is impossible if one of he spaces is asympoically uniformly fla and he oher has a norm wih he Kades-Klee propery, he provides examples of pairs of Banach spaces (X, Y ) for which he se of norm aaining Lipschiz maps, in his weaker sense, is no dense Mahemaics Subjec Classificaion. Primary 46B80; Secondary 46B03, 46B20. Key words and phrases. coarse Lipschiz maps and equivalences, asympoically uniformly smooh norms. 1

2 2 A. DALET AND G. LANCIEN The saring poin of his work was o noice ha his argumen could be adaped o he seing of coarse Lipschiz maps beween Banach spaces X and Y. This space of funcions is a vecor space on which a naural semi-norm is given by he Lipschiz consan a infiniy of a coarse Lipschiz map. I is hen naural o work wih he corresponding quoien space ha we shall denoe CL(X, Y ). In secion 2 we inroduce hese basic definiions as well as he analogue of Godefroy s definiion for norm aaining coarse Lipschiz maps. In secion 3 we define he noion of coarse Lipschiz equivalen Banach spaces, or quasi-isomeric Banach spaces in he erminology inroduced by M. Gromov in [10]. In Proposiion 3.3, we gaher some characerizaions of he coarse Lipschiz equivalence beween Banach spaces ha were essenially known. In paricular we describe he link wih he noion of ne equivalence of Banach spaces. We also insis on he exisence of coninuous represenaives of coarse Lipschiz equivalences. This will be crucial in our furher use of he Gorelik principle. In secion 4 we address he quesion of he compleeness of our normed quoien space CL(X, Y ). In Proposiion 4.5 we give a sufficien condiion for CL(X, Y ) o be complee. We also describe siuaions when he coarse Lipschiz equivalences can be viewed as an open subse of our quoien space. In secion 5 we gaher he necessary background on he so-called Gorelik principle. Firs, we recall is classical version for uniform homeomorphisms and Lipschiz isomorphisms. Then we prove in Theorem 5.3 a version of he Gorelik principle which is a varian of Theorem 3.8 in [8] saed in erms of coarse Lipschiz equivalences insead of ne equivalences of Banach spaces. Secion 6 is devoed o he sudy of he preservaion of he asympoic uniform smoohness under Lipschiz isomorphisms and coarse Lipschiz equivalences. Firs we recall he definiions of he relevan moduli and heir relaionships. The sabiliy of he exisence of an equivalen asympoically uniformly smooh norm was proved in [7] in he separable case. We ake in Theorem 6.3 he opporuniy o deail is proof in he non separable case ha we have no found in he lieraure. In Theorem 6.5 we deail a precise quaniaive version of he sabiliy of asympoically uniformly smooh renormings under coarse Lipschiz equivalences, again in he general case. This resul was menioned in [8] wih only a very brief ouline of he proof. Moreover he deails of his proof will be used in our las secion. Finally, in secion 7 (Theorem 7.1), we exend Godefroy s resul o our seing of norm aaining coarse Lipschiz equivalences and we give examples of siuaions when i can be properly saed ha he se of norm aaining coarse Lipschiz maps beween wo Banach spaces X and Y is no dense in he quoien space CL(X, Y ). 2. Norm aaining coarse Lipschiz maps Definiion 2.1. Le (M, d) and (N, δ) be wo meric spaces and a map f : M N. If (M, d) is unbounded, we define { δ((f(x), f(y)) } s > 0, Lip s (f) = sup, d(x, y) s and Lip (f) = inf d(x, y) Lip s(f). s>0

3 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES 3 Then f is said o be coarse Lipschiz if Lip (f) <. The se of coarse Lipschiz maps from M o N is denoed CL(M, N). The following equivalen formulaions are easy o verify. Proposiion 2.2. Le X and Y be wo Banach spaces and le f : X Y be a mapping. Then he following asserions are equivalen. (i) The map f is coarse Lipschiz. (ii) There exis A and θ in [0, + ) such ha x, x X x x θ f(x) f(x ) A x x. (iii) There exis A and B in [0, + ) such ha x, x X f(x) f(x ) A x x + B. Noe ha in he above saemen, Lip (f) coincide wih he infimum of all A 0 such ha (ii) is saisfied for some θ 0 and also wih he infimum of all A 0 such ha (iii) is saisfied for some B 0. Suppose now ha (M, d) is an unbounded meric space and (Y, Y ) is a Banach space. Then i is easy o see ha CL(M, Y ) is a vecor space on which Lip is a semi-norm ha we shall also denoe CL(M,Y ) or simply CL if no confusion is possible. Then we denoe N (M, Y ) = {f CL(M, Y ), Lip (f) = 0} and CL(M, Y ) he quoien space CL(M, Y )/N (M, Y ). The semi-norm Lip induces a norm on CL(M, Y ) ha will also be denoed Lip, CL(M,Y ) or CL. We shall ry o avoid as much as possible he confusion beween elemens of CL(M, Y ) and elemens of CL(M, Y ). We now inroduce he noion of norm aaining coarse Lipschiz maps. Definiion 2.3. Le (M, d) be an unbounded meric space and (Y, Y ) a Banach space. Assume ha f : M Y is coarse Lipschiz. We say ha f aains is norm in he direcion y S Y if here exiss a sequence of pairs of disinc poins (s n, n ) in M such ha lim d(s n, n ) = + and n f( n ) f(s n ) lim = y Lip (f). n d(s n, n ) Remark. Noe ha he above definiion is only ineresing when Lip (f) 0, ha is when f 0 in he quoien space CL(M, Y ). Noe also ha if f CL(M, Y ) aains is norm in he direcion y S Y and g : M Y is such ha Lip (f g) = 0, hen g also aains is norm in he direcion y. Therefore, his noion is well defined for an elemen f of he quoien space CL(M, Y ). 3. Coarse Lipschiz equivalence of meric spaces Definiion 3.1. Le (M, d) and (N, δ) be wo unbounded meric spaces and f : M N be a coarse Lipschiz map. We say ha f is a coarse Lipschiz equivalence

4 4 A. DALET AND G. LANCIEN from M o N, if here exiss a coarse Lipschiz map g : N M and a consan C 0 such ha x M d ( (g f)(x), x ) C and y N δ ( (f g)(y), y ) C. We denoe CLE(M, N) he se of coarse Lipschiz equivalences from M o N. If CLE(M, N) is non empy, we say ha M and N are coarse Lipschiz equivalen and denoe M CL N. This noion of coarse Lipschiz equivalen meric spaces is exacly he same as he noion of quasi-isomeric meric spaces inroduced by Gromov in [10] (see also he book [5] by E. Ghys and P. de la Harpe). Remark. I is easy o check, for insance using he characerizaion (iii) in Proposiion 2.2, ha CL is an equivalence relaion beween Banach spaces. We now urn o he noion of ne in a meric space. Definiion 3.2. Le 0 < a b. An (a, b)-ne in he meric space (M, d) is a subse M of M such ha for every z z in M, d(z, z ) a and for every x in M, d(x, M) < b. Then a subse M of M is a ne in M if i is an (a, b)-ne for some 0 < a b. Le us now give wo echnical equivalen formulaions of he noion of coarse equivalence beween Banach spaces, ha we shall use laer. The main resul, which is he fac ha (ii) implies (iii) is essenially conained in he proof of Theorem 3.8 in [8]. Proposiion 3.3. Le X and Y be wo Banach spaces and le f : X Y be a coarse Lipschiz map. The following asserions are equivalen. (i) The map f belongs o CLE(X, Y ) (ii) There exis A 0 > 0 and K 1 such ha for all A A 0 and all maximal A-separaed subse M of X, N = f(m) is a ne in Y and x, x M 1 K x x f(x) f(x ) K x x. (iii) There exis wo coninuous coarse Lipschiz maps ϕ : X Y and ψ : Y X and a consan C 0 such ha ϕ(x) f(x) C for all x in X and x X (ψ ϕ)(x) x C and y Y (ϕ ψ)(y) y C. Proof. (i) (ii). Assume ha here exis g : Y X and consans C, D, M > 0 such ha x X (g f)(x) x C, y Y (f g)(y) y C. and x, x X f(x) f(x ) D + M x x, y, y Y g(y) g(y ) D + M y y.

5 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES 5 Le A 0 = (2C + D)(M + 1), A A 0 and M be a maximal A-separaed subse of X. Noe ha M is a (A, A)-ne of X. Le now x x M, y = f(x) and y = f(x ). Then f(x) f(x ) D + M x x A + M x x (M + 1) x x. On he oher hand g(y) x C and g(y ) x C, which implies ha g(y) g(y ) x x 2C and herefore x x 2C + D + M y y A M M y y x x M M y y. I follows ha x x (M + 1) y y. So f is a Lipschiz isomorphism from M ono N = f(m) and K = M + 1 saisfies he required inequaliies. In paricular N is a-separaed, wih a = A(M + 1) 1. Finally le z Y. There exiss x M such ha x g(z) A. Then f(x) z f(x) f(g(z)) + C D + MA + C = b. We have shown ha N is an (a, b)-ne in Y, which finishes he proof of his implicaion. (ii) (iii) For A A 0, we pick (x i ) i I a maximal A-separaed subse of X. Noe ha (x i ) i I is an (A, A)-ne in X. For i I, le y i = f(x i ). Then, by assumpion, (y i ) i I is an (a, b)-ne in Y, for some 0 < a b, and we have 1 i, j I K x i x j y i y j K x i x j. Then we can find a coninuous pariion of uniy (f i ) i I subordinaed o he open cover (B X (x i, A)) i I of X and a coninuous pariion of uniy (g i ) i I subordinaed o he open cover (B Y (y i, b)) i I of Y and we se x X ϕ(x) = i I f i (x) y i and y Y ψ(y) = i I g i (y) x i. Noe firs ha ϕ and ψ are coninuous. Le x X and pick i I such ha x x i A. Now, if f j (x) 0, hen x x j A and x i x j 2A. I follows ha ϕ(x) y i = f j (x) (y j y i ) 2AK. j,f j (x) 0 Le now x X and j I so ha x x j A. Then we have ϕ(x) ϕ(x ) 4AK + y i y j 4AK + K x i x j 6AK + K x x. This shows ha ϕ is coarse Lipschiz and Lip (ϕ) K and a similar proof yields ha he same is rue for ψ. For x X, pick again i I such ha x x i A. If g j (ϕ(x)) 0, hen ϕ(x) y j b and y i y j ϕ(x) y i + ϕ(x) y j 2AK + b. Therefore ψ(ϕ(x)) x i = g j (ϕ(x)) (x j x i ) K(2AK + b). j,g j (ϕ(x)) 0

6 6 A. DALET AND G. LANCIEN Finally, we ge ha ψ(ϕ(x)) x ψ(ϕ(x)) x i + x x i K(2AK + b) + A = C 1. Similarly, we ge ha here exiss C 2 0 such ha for all y Y, ϕ(ψ(y)) y C 2. Finally, recall ha f is coarse Lipschiz. So, here exis D, E 0 such ha for all x, x X, f(x) f(x ) D x x + E. Since x X ϕ(x) f(x) = f j (x) (f(x j ) f(x)), j,f j (x) 0 and x j x A, whenever f j (x) 0, we obain ha x X, ϕ(x) f(x) DA + E = C 3. We conclude he proof of his implicaion by aking C = max{c 1, C 2, C 3 }. (iii) (i) is clear. Remark. The main informaion of Proposiion 3.3 is ha for any f in CLE(X, Y ), here exiss ϕ which is a coninuous represenaive of he equivalence class of f in CL(X, Y ) and also a coarse Lipschiz equivalence wih a coninuous coarse Lipschiz inverse ψ. This will be crucial when we shall apply he Gorelik principle whose proof is based on Brouwer s fixed poin heorem. Le us noice ha, using for insance he characerizaion (ii) of Proposiion 3.3, he following is immediae. Corollary 3.4. Le X, Y be wo Banach spaces and f CLE(X, Y ). Then, for any λ 0, λf CLE(X, Y ). 4. On he compleeness of CL(X,Y) Definiion 4.1. Le X and Y be wo Banach spaces and M be a ne in X. We say ha (M, X, Y ) has he Lipschiz exension propery if any Lipschiz funcion from M o Y admis a Lipschiz exension from X o Y. We say ha he pair (X, Y ) has he ne exension propery (in shor NEP) if here exiss a ne M in X such ha (M, X, Y ) has he Lipschiz exension propery Lemma 4.2. Assume ha X and Y are Banach spaces and M is a ne in X such ha (M, X, Y ) has he Lipschiz exension propery, hen here exiss λ 1 such ha any Lipschiz funcion f : M Y admis an exension g : X Y wih Lip (g) λlip (f). Proof. We may and do assume ha 0 M and f(0) = 0. Then he conclusion follows from a sraighforward applicaion of he open mapping heorem o he resricion operaor o M defined from Lip 0 (X, Y ) ono Lip 0 (M, Y ), where Lip 0 (X, Y ) is he Banach space of all Lipschiz funcions from X o Y ha vanish a 0 equipped wih he norm f L = Lip (f).

7 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES 7 Definiion 4.3. Le X and Y be wo Banach spaces and le µ 1. We say ha (X, Y ) has he µ-lipschiz represenaion propery if (in shor µ-lrp) if for any f CL(X, Y ) and any c > Lip (f), here exiss g Lip 0 (X, Y ) so ha Lip (g) < µc and f g is bounded. Proposiion 4.4. Assume ha X and Y are Banach spaces such ha (X, Y ) has he ne exension propery. Then here exiss µ 1 such ha (X, Y ) has he µ-lrp. Proof. Le f CL(X, Y ) such ha Lip (f) < c. Pick M be a ne in X and λ 1 such ha he conclusion of Lemma 4.2 is saisfied. I follows from an easy change of variable argumen ha for any A 1, he ne AM also saisfies he conclusion of Lemma 4.2 wih he same consan λ. Then for A large enough, he resricion of f o AM is c-lipschiz. So i admis an exension g : X Y such ha g is λc-lipschiz. Since f and g are boh coarse Lipschiz and coincide on a ne, i is no difficul o see ha f g is bounded. By adding a consan o g, we may also assume ha g(0) = 0, which concludes he proof. Remark. We do no know if he converse of his las proposiion is rue. However, i is no difficul o check ha he exisence of µ 1 such ha (X, Y ) has he µ-lrp is equivalen o he exisence of λ 1 such ha (X, Y ) has he λ-anep. Here, λ-anep sands for λ-almos ne exension propery, which is formally weaker han NEP and has he following ad hoc meaning: here exiss a ne M in X such ha for any Lipschiz funcion f : M Y, here exiss g : X Y Lipschiz so ha Lip (g) λlip (f) and f g is bounded on M. Proposiion 4.5. Assume ha X and Y are Banach spaces such ha (X, Y ) has he µ-lrp for some µ 1. Then (CL(X, Y ), CL ) is a Banach space Proof. Le (f n ) n=1 be a sequence in CL(X, Y ) such ha n=1 f n CL <. Then for any n in N, here exiss g n Lip 0 (X, Y ) such ha g n belongs o he equivalence class of f n and Lip (g n ) µ f n CL + 2 n. Then using he compleeness of (Lip 0 (X, Y ), L ) we ge ha here exiss g Lip 0 (X, Y ) such ha lim g N g n L = 0. N n=1 I follows ha lim N g N n=1 f n CL = 0, which concludes our proof. Remark. We do no know if (CL(X, Y ), CL ) is a Banach space wihou any assumpion on he Banach spaces X and Y. We conjecure ha i is no he case, bu a counerexample sill has o be consruced. Proposiion 4.6. Assume ha X and Y are Banach spaces such ha (X, Y ) and (Y, X) have he µ-lrp for some µ 1. Then for any f CLE(X, Y ), here exiss ε > 0 such ha f u CLE(X, Y ), whenever u : X Y is such ha Lip (u) < ε. Proof. Since f CLE(X, Y ), here exiss C 1 and g CLE(Y, X) so ha Lip (f) < C, Lip (g) < C and x X (g f)(x) x C and y Y (f g)(y) y C.

8 8 A. DALET AND G. LANCIEN Le us now fix u CL(X, Y ) such ha Lip (u) < (µ 2 C) 1. I follows from our assumpions ha here exis K > 0, ϕ Lip 0 (X, Y ), ψ Lip 0 (Y, X) and v Lip 0 (X, Y ) so ha Lip (ϕ) < µc, Lip (ψ) < µc, Lip (v) < (µc) 1 and such ha f ϕ, g ψ and u v are bounded. Noe firs ha i is no difficul o deduce ha ψ ϕ I X is bounded on X and ϕ ψ Id Y is bounded on Y. So le K > 0 be such ha f ϕ, u v, g ψ, ψ ϕ I X and ϕ ψ Id Y are bounded by K on heir respecive domains. We now exhibi a coarse Lipschiz inverse G of f u as follows. For y Y and x X, we define L y (x) = ψ(y + v(x)). Since Lip (L y ) < 1, he map L y admis a unique fixed poin in X ha we denoe G(y), which is hus defined by he equaion (1) G(y) = ψ ( y + v G(y)) ). Classical elemenary manipulaions of he above equaion yield ha G is Lipschiz and more precisely ha Lip (G) Lip (ψ) ( 1 Lip (v)lip (ψ) ) 1. I remains o show ha (f u) G I Y and G (f u) I X are bounded. Since G is Lipschiz, i is enough o show ha (ϕ v) G I Y and G (ϕ v) I X are bounded. Le us firs fix y Y. Then using (1) we ge (ϕ v) G(y) y = ϕ ψ ( y + v G(y) ) v ψ ( y + v G(y) ) y Consider x X. Then K + y + v G(y) v G(y) y = K. G (ϕ v)(x) x ψ ( (ϕ v)(x) + v G ( (ϕ v)(x) ) ψ ϕ(x) + K I follows ha Lip (ψ) v G ( (ϕ v)(x) ) v(x) + K Lip (ψ)lip (v) G (ϕ v)(x) x + K. G (ϕ v)(x) x K ( (1 Lip (v)lip (ψ) ) 1. We have proved ha f u CLE(X, Y ). Remarks. Noe ha for X and Y Banach spaces and f : X Y coarse Lipschiz, f CLE(X, Y ) if and only if all he elemens of is equivalence class in CL(X, Y ) belong o CLE(X, Y ) (his is a consequence of Proposiion 3.3). So, in he paricular siuaion described in Proposiion 4.6, we can denoe CLE(X, Y ) he se of equivalen classes of elemens of CLE(X, Y ) and sae ha i is open in he quoien space CL(X, Y ). In his work we have chosen o follow Gromov s definiion for he inverible elemens of CL(X, Y ). One of he advanages of his definiion is o coincide wih he noion of ne equivalence for Banach spaces. However, in pursuing he sudy of our normed quoien space, i could be more naural o say ha f CL(X, Y ) is inverible if here exiss g CL(Y, X) such ha Lip ( (f g) IdY ) = Lip ( (g f) IdX ) = 0.

9 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES 9 5. Background on he Gorelik principle. The ool ha we shall now describe is he Gorelik principle. I was iniially devised by Gorelik in [9] o prove ha l p is no uniformly homeomorphic o L p, for 1 < p <. Then i was developed by Johnson, Lindensrauss and Schechman [12] o prove ha for 1 < p <, l p has a unique uniform srucure. We now recall he crucial ingredien in he proof of he Gorelik Principle (see sep (i) in he proof of Theorem in [2]). This saemen relies on Brouwer s fixed poin heorem and on he exisence of Barle-Graves coninuous selecors. We refer he reader o [1] or [2] for is proof. Proposiion 5.1. Le X 0 be a finie-codimensional subspace of a Banach space X and le 0 < c < d. Then, here exiss a compac subse A of db X such ha for every coninuous map φ : A X saisfying φ(a) a c for all a A, we have ha φ(a) X 0. Le us now sae he Gorelik principle as i can be found in [1], [2] or [7]. Theorem 5.2. Le X and Y be wo Banach spaces and le f be a homeomorphism from X ono Y whose inverse is uniformly coninuous. Le b, d > 0 so ha ω(f 1, b) < d, where ω(f 1,.) is he modulus of uniform coninuiy of f 1. Assume ha X 0 is a closed finie codimensional subspace of X. Then here exiss a compac subse K of Y so ha bb Y K + f(2db X0 ). In paricular, if f is a Lipschiz isomorphism such ha Lip(f) 1 and Lip(f 1 ) M, he condiion ω(f 1, b) < d is saisfied when Mb < d. We will now sae a version of he Gorelik principle ha will be used o sudy coarse equivalen Banach spaces. For he sake of compleeness we shall reproduce he proof ha can be found in [8] (Theorem 3.8) wih more aenion given on keeping opimal esimaes and wih slighly weaker assumpions. Theorem 5.3. Le X and Y be wo Banach spaces. Assume ha f : X Y and g : Y X are coninuous, and ha here exis consans C, D, M > 0 such ha and y, y Y g(y) g(y ) D + M y y x X (g f)(x) x C and y Y (f g)(y) y C. Le λ < 1. Then for any α α 0 = 2(C + D)(1 λ) 1 and any finie codimensional subspace X 0 of X, here is a compac subse K of Y so ha λα M B Y K + CB Y + f(2αb X0 ). Proof. Le µ = 1+λ 2, α 0 = C+D µ λ and α α 0. Le also X 0 be a finie codimensional subspace of X. I follows from Proposiion 5.1 ha here exiss a compac subse A of αb X such ha for every coninuous map φ : A X saisfying φ(a) a µα for all a A, we have ha φ(a) X 0. = 2(C+D) 1 λ

10 10 A. DALET AND G. LANCIEN Consider now y λα clearly coninuous and M B Y and define φ : A X by φ(a) = g(y + f(a)). Then φ is a A, φ(a) a C+ g(y+f(a)) g(f(a)) C+D+M y C+D+λα µα. Hence, here exiss a A so ha φ(a) X 0. Since a α and φ(a) a µα, we have ha φ(a) 2αB X0. Finally, we use he fac ha (f g)(y + f(a)) (y + f(a)) C o conclude ha y K + CB Y + f(2αb X0 ), where K = f(a) is a compac subse of Y. Remark. Lipschiz. Noe ha in he above resul we have no assumed ha f is coarse 6. Asympoic uniform smoohness and coarse Lipschiz equivalence We now recall he definiions of he modulus of asympoic uniform smoohness of a norm and he modulus of weak asympoic uniform convexiy of a dual norm. They are due o V. Milman [13] and we follow he noaion from [11]. So le (X, ) be a Banach space. For > 0, and x S X we define ρ X (x, ) = inf Y sup ( x + y 1), y S Y where Y runs hrough all closed subspaces of X of finie codimension. Then ρ X () = sup x S X ρ X (x, ). The norm is said o be asympoically uniformly smooh (in shor AUS) if ρ lim X () = 0. 0 We say ha he norm is asympoically uniformly fla if Now, for > 0, and x S X 0 (0, + ) [0, 0 ] ρ X () = 0. we define θ X (x, ) = sup E inf y S E ( x + y 1), where E runs hrough all finie dimensional subspaces of X. Then θ X () = inf θ X (x, ). x S X The norm of X is said o be weak asympoically uniformly convex (in shor w - AUC) if > 0 θ X () > 0. The dualiy beween hese wo moduli is now well undersood. complee and precise saemen is aken from [4] Proposiion 2.1. Proposiion 6.1. Le X be a Banach space and 0 < σ, τ < 1. (a) If ρ X (σ) < στ 6, hen θ X(τ) > στ 6. (b) If θ X (τ) > στ, hen ρ X (σ) < στ The following

11 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES 11 As an immediae consequence we have ha X is AUS if and only if X is w -AUC. Le us also deail a few oher classical consequences. Firs we recall ha for a funcion f which is coninuous monoone non decreasing on [0, 1] and such ha f(0) = 0, is dual Young funcion is denoed f and defined by s [0, 1] f (s) = sup{s f(), [0, 1]}. As a corollary of he previous proposiion we obain. Corollary 6.2. Le X be a Banach space. Then s [0, 1] (θ X ) ( s) (s) ρ X and 2 (θ X ) ( s) ρx (s). 6 Proof. Consider firs = 2 s ρ X( s 2 ) [0, 1]. Then ρ X( s 2 ) = s 2. So i follows from Proposiion 6.1 (b) ha θ X () s 2. Therefore (θ X) (s) s θ X () s 2 = ρ X( s 2 ). Assume now ha (θ X ) ( s 6 ) > ρ X(s). Then here exiss [0, 1] such ha s 6 θ X() > ρ X (s). Thus θ X () < s 6 ρ X(s) s 6. I now follows from Proposiion 6.1 (a) ha ρ X (s) s 6. Bu his implies ha θ X() < 0, which is impossible. The following heorem saes ha he exisence of an asympoically uniformly smooh norm is sable under Lipschiz isomorphisms and appeared firs in [7], in a separable seing. Is proof can also be found in he recen exbook [1] (see paragraph 14.6). The general case can be deduced by rouine argumens of separable sauraion and separable deerminaion of he moduli. However, we shall deail here he direc proof in he general case. The only modificaion is ha we deal wih he definiion of he asympoic moduli insead of using weak -null or weakly null sequences. Theorem 6.3. Le X and Y be wo Banach spaces and assume ha f : X Y is a bijecion such ha Lip(f) 1 and Lip(f 1 ) M. Then here exiss an equivalen norm on Y such ha Y M Y and ( ) [0, 1], θ () θ X. 4M Proof. Le { f(x) f(x ) } C = conv x x, x x X. Clearly, C is closed convex symmeric and C B Y. Le now y Y such ha y = 1 M. For [0, + ), denoe x = f 1 (y). We have ha x 1 x 0 1 and x M x 0 1. So, here exiss [1, M] such ha x x 0 = 1. I follows ha y C. Since C is convex and symmeric, we deduce ha 1 M B Y C. So, if we denoe he Minkowski funcional of C, we have ha is an equivalen norm on Y such ha Y M Y. Is dual norm is given by { y y Y y, f(x) f(x ) = sup x x x x }.

12 12 A. DALET AND G. LANCIEN Le (0, 1] and assume as we may ha θ X ( 4M ) > 0. So le y Y such ha y = 1 and η > 0. We can pick x x X such ha y, f(x) f(x ) (1 η) x x. We may assume ha x = x and f(x ) = f(x), so ha we have (2) y, f(x) (1 η) x. Pick 0 < δ < θ X ( 4M ). I follows from saemen (b) in Proposiion 6.1 ha ρ X ( 4Mδ ) < δ. So, here exiss a finie codimensional subspace X 0 of X such ha (3) z 4Mδ x B X0, x + z (1 + δ) x. Pick b < 4δ x. I now follows from he Gorelik principle (Theorem 5.2) ha here exiss a compac subse K of Y such ha ( 8Mδ x (4) bb Y K + f B X0 ). Fix now ε > 0, consider a finie ε-ne F of K and denoe E he finie dimensional subspace of Y spanned by F {f(x)}. For any z E such ha z =, we have z and, if ε > 0 was iniially chosen small enough, by (4) we deduce ha (5) z 8Mδ x B X0 z, f(z) (b η). I now follows from he fac ha y = 1 and (3) ha y, f(x) + f(z) = y, f(z) f(x ) x z (1 + δ) x. Then (2) implies ha y, f(z) (δ + η) x. Combining his las inequaliy wih he fac ha z, f(x) = 0 and (2), (3) and (4), we obain ha y + z, f(x) f(z) (1 η) x (δ + η) x + (b η). Using again he definiion of and (3) we ge ( )( 1. y + z (1 η) x (δ + η) x + (b η) (1 + δ) x ) Leing b end o 4δ x and η end o 0, we deduce ha θ (y, ) 1 + 3δ 1 + δ 1 δ. In he above esimae, which does no depend on y in he uni sphere of, we le δ end o θ X ( 4M ) o conclude our proof. Corollary 6.4. Le X and Y be wo Banach spaces and assume ha f : X Y is a bijecion such ha Lip (f) 1 and Lip (f 1 ) M. Then here exiss an equivalen norm on Y such ha Y M Y and ( ) [0, 1] ρ ρx (). 48M

13 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES 13 Proof. Le f, g be coninuous monoone non decreasing on [0, 1] wih f(0) = g(0) = 0. If here exiss a consan C 1 such ha for all [0, 1], f() g(/c), hen i is clear ha for all [0, 1], f (/C) g (). Since he norm given by Theorem 6.3 saisfies ( [0, 1] θ () θ X 4M ), he conclusion of he proof follows now direcly from Corollary 6.2. We now urn o he sudy of he preservaion of he modulus of weak asympoic uniform convexiy, up o renorming, under coarse Lipschiz equivalence. The following precise quaniaive saemen is a sligh modificaion of Theorem 3.12 in [8], in which he proof is only very briefly oulined. I will also be crucial for us o use he deails of he consrucion of his equivalen norm in our las secion. Theorem 6.5. Le X and Y be wo Banach spaces and M > 1. Assume ha f : X Y and g : Y X are coninuous wih Lip (f) 1, Lip (g) < M and ha here exiss a consan C 0 such ha x X (g f)(x) x C and y Y (f g)(y) y C. Then for any ε in (0, 1), here exiss an equivalen norm on Y such ha ε ( ) Y M Y and [0, 1] θ () θ X ε. 48 M 2 Proof. We will adap he proof of Theorem 5.3 in [7]. For k N, we define { f(x) f(x ) C k = conv x x, x x 2 k}. Then (C k ) k=1 is a decreasing sequence of closed convex and symmeric subses of Y. Since Lip (f) 1, we have ha C k (1 + ε k )B Y, where (ε k ) k=1 is a sequence of posiive numbers ending o 0. In paricular here exiss k 0 N such ha k k 0 C k (1 + ε 16M )B Y (1 + ε)b Y 2B Y. Fix now k N, y S Y and denoe y 0 = f(0). I follows easily from our assumpions ha lim g(y) =. Recall also ha f(g(y)) = y + u, wih u C. So, for large enough f(g(y)) y 0 g(y) = y g(y) + u y 0 g(y) C k. I follows from he assumpion ha Lip (g) < M ha here exis α 1 M and a sequence ( n ) n ending o + such ha n g( ends o α. Since C ny) k is closed, we obain ha αy C k. Finally, we use he fac ha C k is convex and symmeric o deduce ha 1 M y C k and hus ha 1 M B Y C k. So, if we denoe k he Minkowski funcional of C k, we have ha for all k k 0,

14 14 A. DALET AND G. LANCIEN k is an equivalen norm on Y such ha (1 + ε 16M ) 1 Y k M Y. I will be useful o describe he dual norm of k, also denoed k, as follows { y y Y y, f(x) f(x } ) k = sup x x, x, x X, x x 2 k. Noe ha our assumpions also imply he exisence of D 0 such ha (6) y, y Y g(y) g(y ) D + M y y. This will enable us o apply he Gorelik principle as i is saed in Theorem 5.3. The key lemma is he following. ( ) Lemma 6.6. Le (0, 1] and assume ha θ X 48M > 0. Le y Y such ha 2 y M, ε > 0 and k 1 N such ha k 1 k 0, 24M 2 ( ) θ X 2 k 1 48M 2 > 4(C + D) and 2 k 1 (CM + 1) ε 8. Then here exiss a finie dimensional subspace E of Y so ha for all k k 1 and all z E such ha 2 z M, we have (7) y + z k 2 y k+1 y k + θ X ( 48M 2 ) ε 2. ε Proof. Le η = 16M and pick 0 < δ < θ X( ) such ha 48M 2 (8) 24M 2 δ2 k 1 > 4(C + D). Le k k 1 and choose x x X such ha x x 2 k+1 and y, f(x) f(x ) (1 η) y k+1 x x. We may assume ha x = x and f(x ) = f(x), so ha we have (9) y, f(x) (1 η) y k+1 x. Since 0 < δ < θ X ( ). I follows from saemen (b) in Proposiion 6.1 ha 48M 2 ρ X ( 48M 2 δ ) < δ. So, here exiss a finie codimensional subspace X 0 of X such ha (10) z 48M 2 δ x B X0 x + z (1 + δ) x and x + z x 2 k. From (6), (8) and Theorem 5.3, applied wih λ = 1 2 and α = 1 24M 2 δ x, we infer he exisence of a compac subse K of Y such ha 12M δ x ( 48M 2 δ x B Y K + CB Y + f B X0 ). As in he previous proof, fix η > 0, pick a finie η -ne F of K and le E be he linear span of F {f(x)}. Le now z E such ha 2 z M. Then, (11) z 48M 2 δ x B X0 z, f(z) 6Mδ x (CM + 1), if η was iniially chosen small enough. We hen deduce from (10) ha (12) y, f(x) + f(z) = y, f(z) f(x ) y k x z (1 + δ) y k x.

15 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES 15 Thus, combining he above informaions we ge ( ) y + z, f(x) f(z) 2(1 η) y k+1 (1 + δ) y k + 6Mδ x (CM + 1). Using again (10) we hen have y + z k 1 ( ) 2(1 η) y k+1 (1 + δ) y k + 6Mδ 2 k 1 (CM + 1). 1 + δ So, i follows from our iniial choice of k 1 and he fac ha δ 1 and M 1 ha y + z k 2(1 η)(1 δ) y k+1 y k + (2M + 1)δ ε 8. Noe ha y M implies ha y k+1 (1 + So we obain ha ε 16M )M and recall ha η = ε y + z k 2(1 η) y k+1 y k 2Mδ + (2M + 1)δ ε 8. Then wih our choice of η implies ha y + z k 2 y k+1 y k + δ 3ε 8. So, if δ was iniially chosen close enough o θ X ( ), we obain 48M 2 16M. y + z k 2 y k+1 y k + θ X ( 48M 2 ) ε 2. End of proof of Theorem 6.5. Noe ha a simple convexiy argumen shows ha for any space Z, he funcion ( 1 θ Z () is increasing on (0, 1]. Assume firs ha θ 1 ) X 48M ε 2 2. Then for any (0, 1] we have ha ( 1 ) ( ) θ Y () 0 > θ X ε θx ε, 48M 2 48M 2 and he original norm on ( Y works. Assume now ha θ 1 ) X 48M > ε 2 2. Since θ X is coninuous, here exiss 0 (0, 1) ( so ha θ 0 ) X 48M = ε 2 2. As above, we easily have ha for any equivalen norm N ( ) on Y and any (0, 0 ], θ N () 0 θ X 48M ε. So we only have o rea he 2 problem for [ 0, 1]. Le us pick k 1 N saisfying he assumpions of Lemma 6.6 for 0. I hen follows from he monooniciy of 1 θ X () ha he conclusion of Lemma 6.6 applies for any [ 0, 1] and any k k 1. Pick now N N such ha 4M N < ε 2 and define which is a dual norm on Y wih y = 1 N M 1 y y (1 + k 1 +N k=k 1 +1 y k ε 16M ) y (1 + ε) y 2 y.

16 16 A. DALET AND G. LANCIEN Le y Y, wih y = 1. I follows from Lemma 6.6 ha for any [ 0, 1], here exiss a finie dimensional subspace E of Y so ha for all k [k 1, k 1 + N] and all z E such ha z =, we have y + z k 2 y k+1 y ( ) ε k + θ X 48M 2 2, which implies, summing over k, ha y + z y + 2 ( y k1 +N+1 y ) ( ) ε k θx N 48M 2 2. Since y = 1, we have ha y M and y k+1 2M. So y + z y ( ) ε + θ X 48M 2 2 4M N ( ) y + θ X ε. 48M ( ) 2 This shows ha for all [ 0, 1], θ () θ X 48M ε and concludes our proof. 2 Corollary 6.7. Le X and Y be wo Banach spaces and M > 1. Assume ha f : X Y and g : Y X are coninuous wih Lip (f) 1, Lip (g) < M and ha here exiss a consan C 0 such ha x X (g f)(x) x C and y Y (f g)(y) y C. Then for any ε in (0, 1), here exiss an equivalen norm on Y such ha ε ( ) Y M Y and [0, 1] ρ ρx 576 M 2 () + ε. Proof. Le ϕ, ψ be coninuous monoone non decreasing on [0, 1] wih ϕ(0) = ψ(0) = 0. If here exiss D 1 and ε > 0 such ha for all [0, 1], ϕ() ψ(/d) ε, hen i is clear ha for all [0, 1], ϕ (/D) ψ () + ε. Then we can apply Corollary 6.2 o ge ha if is he norm given by Theorem 6.5, hen for all [0, 1]: ( ) ρ (θ 576 M 2 ) ( ) (θx 288 M 2 ) ( ) + ε ρx () + ε Applicaion o norm aaining coarse Lipschiz maps In his secion, we will exend o he seing of coarse Lipschiz maps and equivalences, he resuls obained by G. Godefroy in [6] on norm aaining Lipschiz maps. Our firs resul is he analogue of Theorem 3.2 of [6]. Theorem 7.1. Le X and Y be wo Banach spaces and M > 1. Assume ha f : X Y and g : Y X are coninuous wih Lip (f) = 1, Lip (g) < M and ha here exiss a consan C 0 such ha x X (g f)(x) x C and y Y (f g)(y) y C. Assume also ha f aains is norm f CL = 1 in he direcion y S Y. Then (0, 1] ρ Y ( y, 576M 3 ) ρx ().

17 SOME PROPERTIES OF COARSE LIPSCHITZ MAPS BETWEEN BANACH SPACES 17 Proof. Le us fix ε in (0, 1) and denoe he norm consruced in Theorem 6.5. There exiss sequences (x n ) n=1, (x n) n=1 in X such ha lim x n x f(x n ) f(x n n = and lim n) n x n x = y. n f(x Noe ha for any k in N, n) f(x n ) x n x n C k for n large enough. So y C k and y k 1. Therefore, y 1. On he oher hand y (1 + ε) 1 1 ε. Denoe u = y y. I follows from Corollary 6.7 ha ρ ( ) u, 576M 2 ρx () + ε. Then here exiss a finie codimensional subspace E of Y such ha for all v E wih v, we have 576M 2 u + v 1 + ρ X () + 2ε. I follows ha for all v E wih v, 576M 3 y + v (1 + ε)( y u + u + v ) (1 + ε)(1 + ρ X () + 3ε). Since ε > 0 is arbirary in he above inequaliy, his concludes our proof. Le us now recall ha a Banach space X has he Kades-Klee propery if he norm and weak opologies coincide on he uni sphere of X. The following corollaries are he coarse Lipschiz analogues of Corollary 3.5 in [6]. Corollary 7.2. Le X and Y be wo infinie dimensional Banach spaces such ha X is asympoically uniformly fla and Y has he Kades-Klee propery and assume ha f : X Y is a coarse Lipschiz equivalence. Then f does no aain is norm in any direcion in S Y. Proof. Assume on he conrary ha f : X Y is a coarse Lipschiz equivalence and ha f aains is norm in he direcion y S Y. Assume also, as we may by Corollary 3.4, ha f CL = 1. Since X is asympoically uniformly fla, i follows from Theorem 7.1 ha here exis 0 > 0 so ha for all [0, 0 ], ρ Y (y, ) = 0. Consider now a weakly null ne (y α ) α A in Y such ha y α = 0 for all α in A. Then ( y + y α ) α A ends o 1, which conradics he assumpion ha Y has he Kades-Klee propery. Corollary 7.3. There exiss a pair of Banach spaces (X, Y ) such ha he norm aaining coarse Lipschiz maps are no dense in CL(X, Y ). Proof. Consider X = (c 0, ) and Y = (c 0, Y ), where Y is an equivalen norm on c 0 wih he Kades-Klee propery. We recall ha such an equivalen norm exis on any separable Banach space (see for insance he book [3] and references herein). Moreover X is clearly asympoically uniformly fla. Since X is an absolue rerac (see Example 1.5 of Chaper 1 in [2]), we have in paricular ha (X, Y ) and (Y, X) have he ne exension propery. Therefore, by Proposiions 4.4 and 4.6, CLE(X, Y ) can be viewed as an open subse of CL(X, Y ). Since i conains he ideniy map on c 0, i is a non empy open subse of CL(X, Y ). Combining his wih Corollary 7.2 finishes our proof. Aknowledgemens. The auhors wish o hank G. Godefroy, M. Marin, A. Procházka and A. Valee for fruiful discussions on he subjec of his paper.

18 18 A. DALET AND G. LANCIEN References [1] F. Albiac and N. J. Kalon, Topics in Banach space heory - Second Ediion, Graduae Texs in Mahemaics, vol. 233, Springer, New York, [2] Y. Benyamini and J. Lindensrauss, Geomeric nonlinear funcional analysis. Vol. 1, American Mahemaical Sociey Colloquium Publicaions, vol. 48, American Mahemaical Sociey, Providence, RI, [3] R. Deville, G. Godefroy, and V. Zizler, Smoohness and renormings in Banach spaces, Piman Monographs and Surveys in Pure and Applied Mahemaics, vol. 64, Longman Scienific & Technical, Harlow, [4] S. J. Dilworh, D. Kuzarova, G. Lancien, and L. Randrianarivony, Equivalen norms wih he propery β of Rolewicz, Revisa de la Real Academia de Ciencias Exacas, Físicas y Naurales. Serie A. Maemáicas 111 (2017), no. 1, [5] E. Ghys and P. de la Harpe, Sur les groupes hyperboliques d après Mikhael Gromov, Progress in Mahemaics, vol. 50, Birkhäuser, [6] G. Godefroy, On norm aaining Lipschiz maps beween Banach spaces, Pure and Applied Funcional Analysis 1 (2016), [7] G. Godefroy, N. J. Kalon, and G. Lancien, Szlenk indices and uniform homeomorphisms, Trans. Amer. Mah. Soc. 353 (2001), [8] G. Godefroy, G. Lancien, and V. Zizler, The non linear geomery of Banach spaces afer Nigel Kalon, Rocky Mounain J. of Mah 44 (2014), no. 5, [9] E. Gorelik, The uniform nonequivalence of L p and l p, Israel J. Mah. 87 (1994), 1 8. [10] M. Gromov, Hyperbolic groups, Essays in group heory, Mah. Sci. Res. Ins. Publ. (1987), [11] W. B. Johnson, J. Lindensrauss, D. Preiss, and G. Schechman, Almos Fréche differeniabiliy of Lipschiz mappings beween infinie-dimensional Banach spaces, Proc. London Mah. Soc. 84 (2002), no. 3, [12] W. B. Johnson, J. Lindensrauss, and G. Schechman, Banach spaces deermined by heir uniform srucures, Geom. Func. Anal. 6 (1996), [13] V. D. Milman, Geomeric heory of Banach spaces. II. Geomery of he uni ball, Uspehi Ma. Nauk 26 (1971), (Russian). English ranslaion: Russian Mah. Surveys 26 (1971), Univ. Bourgogne Franche-Comé, Laboraoire de Mahémaiques de Besançon UMR 6623, 16 roue de Gray, Besançon Cedex, FRANCE. address: aude.dale@univ-fcome.fr Univ. Bourgogne Franche-Comé, Laboraoire de Mahémaiques de Besançon UMR 6623, 16 roue de Gray, Besançon Cedex, FRANCE. address: gilles.lancien@univ-fcome.fr