Low distortion embeddings between C(K) spaces joint work with Luis Sánchez González
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1 Low distortion embeddings between C(K) spaces joint work with Luis Sánchez González Tony Procházka Université de Franche-Comté August 2014 First Brazilian Workshop in Geometry of Banach Spaces Maresias
2 Introduction If K and L are homeomorphic, then C(K) and C(L) are isometrically isomorphic.
3 Introduction If K and L are homeomorphic, then C(K) and C(L) are isometrically isomorphic. If C(K) and C(L) are isometrically isomorphic, then K and L are homeomorphic. (Banach-Stone, 1937).
4 Introduction If K and L are homeomorphic, then C(K) and C(L) are isometrically isomorphic. If C(K) and C(L) are isometrically isomorphic, then K and L are homeomorphic. (Banach-Stone, 1937). If there is a linear isomorphism f : C(K) C(L) such that f f 1 < 2, then K and L are homeomorphic (Amir-Cambern, 1965).
5 Introduction If K and L are homeomorphic, then C(K) and C(L) are isometrically isomorphic. If C(K) and C(L) are isometrically isomorphic, then K and L are homeomorphic. (Banach-Stone, 1937). If there is a linear isomorphism f : C(K) C(L) such that f f 1 < 2, then K and L are homeomorphic (Amir-Cambern, 1965). If there is a Lipschitz homeomorphism f : C(K) C(L) such that f Lip f 1 Lip < 1 + ε, then K and L are homeomorphic (Jarosz, 1989).
6 Introduction If K and L are homeomorphic, then C(K) and C(L) are isometrically isomorphic. If C(K) and C(L) are isometrically isomorphic, then K and L are homeomorphic. (Banach-Stone, 1937). If there is a linear isomorphism f : C(K) C(L) such that f f 1 < 2, then K and L are homeomorphic (Amir-Cambern, 1965). If there is a Lipschitz homeomorphism f : C(K) C(L) such that f Lip f 1 Lip < 1 + ε, then K and L are homeomorphic (Jarosz, 1989). efinition (Lipschitz embedding/homeomorphism) Let (M, d) be a metric space and (X, ) a Banach space. We denote f : M X if d(x, y) f(x) f(y) d(x, y). Equivalent to f Lip f 1 Lip up to a scalar multiple of f.
7 Introduction If K and L are homeomorphic, then C(K) and C(L) are isometrically isomorphic. If C(K) and C(L) are isometrically isomorphic, then K and L are homeomorphic. (Banach-Stone, 1937). If there is a linear isomorphism f : C(K) C(L) such that f f 1 < 2, then K and L are homeomorphic (Amir-Cambern, 1965). If there is a Lipschitz homeomorphism f : C(K) C(L) such that f Lip f 1 Lip < 17 16, then K and L are homeomorphic (utrieux, Kalton, 2005). efinition (Lipschitz embedding/homeomorphism) Let (M, d) be a metric space and (X, ) a Banach space. We denote f : M X if d(x, y) f(x) f(y) d(x, y). Equivalent to f Lip f 1 Lip up to a scalar multiple of f.
8 Introduction If K and L are homeomorphic, then C(K) and C(L) are isometrically isomorphic. If C(K) and C(L) are isometrically isomorphic, then K and L are homeomorphic. (Banach-Stone, 1937). If there is a linear isomorphism f : C(K) C(L) such that f f 1 < 2, then K and L are homeomorphic (Amir-Cambern, 1965). If there is a Lipschitz homeomorphism f : C(K) C(L) such that f Lip f 1 Lip < 6 5, then K and L are homeomorphic (Górak, 2011). efinition (Lipschitz embedding/homeomorphism) Let (M, d) be a metric space and (X, ) a Banach space. We denote f : M X if d(x, y) f(x) f(y) d(x, y). Equivalent to f Lip f 1 Lip up to a scalar multiple of f.
9 A countable compact K is always homeomorphic to [0, ω α n] where K (α+1) = and 1 n = K (α) (Mazurkiewicz-Sierpiński).
10 A countable compact K is always homeomorphic to [0, ω α n] where K (α+1) = and 1 n = K (α) (Mazurkiewicz-Sierpiński). Górak for countable compacts: If α β or n m, then there is no Lipschitz homeomorphism f : C([0, ω α n]) C([0, ω β m]) such that f Lip f 1 Lip < 6 5. Theorem Let α < ω 1 and β < ω α. Then there is no Lipschitz embedding f : C([0, ω α ]) C([0, β]) such that f Lip f 1 Lip < 2.
11 A countable compact K is always homeomorphic to [0, ω α n] where K (α+1) = and 1 n = K (α) (Mazurkiewicz-Sierpiński). Górak for countable compacts: If α β or n m, then there is no Lipschitz homeomorphism f : C([0, ω α n]) C([0, ω β m]) such that f Lip f 1 Lip < 6 5. Theorem Let α < ω 1 and β < ω α. Then there is no Lipschitz embedding f : C([0, ω α ]) C([0, β]) such that f Lip f 1 Lip < 2. In particular if α β < ω 1 then, for any n, m N, there is no Lipschitz homeomorphism f : C([0, ω α m]) C([0, ω β n]) such that f Lip f 1 Lip < 2.
12 Another angle Aharoni (1974), 2 M separable metric sp.: M c 0.
13 Another angle Aharoni (1974), 2 M separable metric sp.: M c 0. Kalton-Lancien (2008) = 2.
14 Another angle Aharoni (1974), 2 M separable metric sp.: M c 0. Kalton-Lancien (2008) = 2. Baudier (2013): what if we replace c 0 by C(K)?
15 Another angle Aharoni (1974), 2 M separable metric sp.: M c 0. Kalton-Lancien (2008) = 2. Baudier (2013): what if we replace c 0 by C(K)? A.P., L. Sánchez (2013): M countable metric sp. s.t. M X, < 2 X not Asplund.
16 Another angle Aharoni (1974), 2 M separable metric sp.: M c 0. Kalton-Lancien (2008) = 2. Baudier (2013): what if we replace c 0 by C(K)? A.P., L. Sánchez (2013): M countable metric sp. s.t. M X, < 2 X not Asplund. Where does M live?
17 Another angle Aharoni (1974), 2 M separable metric sp.: M c 0. Kalton-Lancien (2008) = 2. Baudier (2013): what if we replace c 0 by C(K)? A.P., L. Sánchez (2013): M countable metric sp. s.t. M X, < 2 X not Asplund. Where does M live? Proposition For α < ω 1 M α C([0, ω α ]) countable uniformly discrete s.t. M C(K), < 2 K (α).
18 Another angle Aharoni (1974), 2 M separable metric sp.: M c 0. Kalton-Lancien (2008) = 2. Baudier (2013): what if we replace c 0 by C(K)? A.P., L. Sánchez (2013): M countable metric sp. s.t. M X, < 2 X not Asplund. Where does M live? Proposition For α < ω 1 M α C([0, ω α ]) countable uniformly discrete s.t. M C(K), < 2 K (α). Corollary If M ω α X, < 2, then Sz(X) ω α+1.
19 More about M
20 More about M Theorem There exists a countable metric space M such that if M X, < 2, then l 1 X.
21 More about M Theorem There exists a countable metric space M such that if M X, < 2, then l 1 X. Also, if M l 1, then 2.
22 More about M Theorem There exists a countable metric space M such that if M X, < 2, then l 1 X. Also, if M l 1, then 2. But...
23 More about M Theorem There exists a countable metric space M such that if M X, < 2, then l 1 X. Also, if M l 1, then 2. But... there is an equivalent norm on l 1 such that M 1 (l 1, ).
24 The unwieldy metric space M
25 The unwieldy metric graph M
26 The unwieldy metric graph M {1, 3, 2, 4} {1, 3, 2} {1, 2, 4} {1, 3, 4} {3, 2, 4} {1, 4} {1, 2} {1, 3} {3, 2} {2, 4} {3, 4} M = { } N F where F = {A N : 2 A < }
27 The unwieldy metric graph M {1, 3, 2, 4} {1, 3, 2} {1, 2, 4} {1, 3, 4} {3, 2, 4} {1, 4} {1, 2} {1, 3} {3, 2} {2, 4} {3, 4} M = { } N F where F = {A N : 2 A < } (a, b) is an edge iff a = and b N
28 The unwieldy metric graph M {1, 3, 2, 4} {1, 3, 2} {1, 2, 4} {1, 3, 4} {3, 2, 4} {1, 4} {1, 2} {1, 3} {3, 2} {2, 4} {3, 4} M = { } N F where F = {A N : 2 A < } (a, b) is an edge iff a = and b N or a N, b F and a b
29 Proof of M X, < 2 l 1 X
30 Proof of M X, < 2 l 1 X Assume f : M X, < 2.
31 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N
32 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy.
33 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy. Indeed, let (k n ) N be given.
34 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy. Indeed, let (k n ) N be given. Put A N = {k 2n : n N} and B N = {k 2n 1 : n N} N N.
35 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy. Indeed, let (k n ) N be given. Put A N = {k 2n : n N} and B N = {k 2n 1 : n N} N N. Put X a,b = {x B X : x, f(a) f(b) 4 2}.
36 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy. Indeed, let (k n ) N be given. Put A N = {k 2n : n N} and B N = {k 2n 1 : n N} N N. Put X a,b = {x B X : x, f(a) f(b) 4 2}. Then n N K n := a A n,b B n X a,b.
37 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy. Indeed, let (k n ) N be given. Put A N = {k 2n : n N} and B N = {k 2n 1 : n N} N N. Put X a,b = {x B X : x, f(a) f(b) 4 2}. Then n N K n := a A n,b B n X a,b. (K n ) is a decreasing sequence of non-empty w -compacts
38 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy. Indeed, let (k n ) N be given. Put A N = {k 2n : n N} and B N = {k 2n 1 : n N} N N. Put X a,b = {x B X : x, f(a) f(b) 4 2}. Then n N K n := a A n,b B n X a,b. (K n ) is a decreasing sequence of non-empty w -compacts x n=1 K n
39 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy. Indeed, let (k n ) N be given. Put A N = {k 2n : n N} and B N = {k 2n 1 : n N} N N. Put X a,b = {x B X : x, f(a) f(b) 4 2}. Then n N K n := a A n,b B n X a,b. (K n ) is a decreasing sequence of non-empty w -compacts x n=1 K n x, x k2n x k2n+1 4 2
40 Proof of M X, < 2 l 1 X Assume f : M X, < 2. Put x k := f(k) k N Claim: No subsequence of (x k ) is weakly Cauchy. Indeed, let (k n ) N be given. Put A N = {k 2n : n N} and B N = {k 2n 1 : n N} N N. Put X a,b = {x B X : x, f(a) f(b) 4 2}. Then n N K n := a A n,b B n X a,b. (K n ) is a decreasing sequence of non-empty w -compacts x n=1 K n x, x k2n x k2n (x kn ) n is not weakly Cauchy+Rosenthal s theorem l 1 X
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