On an isomorphic BanachMazur rotation problem and maximal norms in Banach spaces


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1 On an isomorphic BanachMazur rotation problem and maximal norms in Banach spaces B. Randrianantoanina (joint work with Stephen Dilworth) Department of Mathematics Miami University Ohio, USA Conference on Geometric Functional Analysis and its Applications Besançon, France October 2731, 2014
2 BanachMazur Rotation Problem 1932 Problem Is every separable Banach space with a transitive group of isometries isometrically isomorphic to a Hilbert space? Definition A norm on X is called transitive if the orbit under the group of isometries of any element x in the unit sphere of X is equal to the unit sphere of X. i.e. x, y S X, T Isom(X) : Tx = y. Throughout this talk isometry will mean linear surjective isometry.
3 Theorem (Mazur) If a finite dimensional Banach space X is transitive then it is isometric to a Hilbert space.
4 Theorem (Mazur) If a finite dimensional Banach space X is transitive then it is isometric to a Hilbert space. Theorem (PełczyńskiRolewicz) There exist nonseparable transitive Banach spaces which are not isomorphic to a Hilbert space.
5 Theorem (Mazur) If a finite dimensional Banach space X is transitive then it is isometric to a Hilbert space. Theorem (PełczyńskiRolewicz) There exist nonseparable transitive Banach spaces which are not isomorphic to a Hilbert space. Theorem (Lusky 1979) Every separable Banach space (X, ) is complemented in a separable almost transitive space (Y, ). Definition A norm on X is called almost transitive if the orbit under the group of isometries of any element x in the unit sphere of X is norm dense in the unit sphere of X. i.e. x, y S X, ε > 0 T Isom(X) : Tx y < ε.
6 Problem of Deville, Godefroy and Zizler 1993 Problem Does every superreflexive space admit an equivalent almost transitive norm?
7 Problem of Deville, Godefroy and Zizler 1993 Problem Does every superreflexive space admit an equivalent almost transitive norm? The study of isometry groups of renormings of X is equivalent to the study of bounded subgroups of the group GL(X) of all isomorphisms from X onto X. Indeed, if G is a bounded subgroup of GL(X) (i.e. sup g G g < ) we can define an equivalent norm G on X by x G = sup gx. g G Then G is a subgroup of Isom(X, G ), and Isom(X, G ) is a bounded subgroup of GL(X).
8 Problem of Deville, Godefroy and Zizler 1993 Problem Does every superreflexive space admit an equivalent almost transitive norm? The study of isometry groups of renormings of X is equivalent to the study of bounded subgroups of the group GL(X) of all isomorphisms from X onto X. Indeed, if G is a bounded subgroup of GL(X) (i.e. sup g G g < ) we can define an equivalent norm G on X by x G = sup gx. g G Then G is a subgroup of Isom(X, G ), and Isom(X, G ) is a bounded subgroup of GL(X). Theorem (Ferenczi and Rosendal 2013) No.
9 Problem of Deville, Godefroy and Zizler 1993 Problem Does every superreflexive space admit an equivalent almost transitive norm? The study of isometry groups of renormings of X is equivalent to the study of bounded subgroups of the group GL(X) of all isomorphisms from X onto X. Indeed, if G is a bounded subgroup of GL(X) (i.e. sup g G g < ) we can define an equivalent norm G on X by x G = sup gx. g G Then G is a subgroup of Isom(X, G ), and Isom(X, G ) is a bounded subgroup of GL(X). Theorem (Ferenczi and Rosendal 2013) No. There exists a complex superreflexive HI space which does not admit an equivalent almost transitive renorming.
10 Theorem The spaces l p, 1 < p <, p 2, and all infinitedimensional subspaces of their quotient spaces do not admit equivalent almost transitive renormings.
11 Theorem The spaces l p, 1 < p <, p 2, and all infinitedimensional subspaces of their quotient spaces do not admit equivalent almost transitive renormings. Key Lemma Suppose that a Banach space X with a Schauder basis contains an almost transitive subspace Y. Then, given δ > 0, y 0 Y, and N N, there exists y Y, with PN X (y) < δ, such that for all scalars a, b, we have ( (1 δ) a 2 y b 2) 1 2 ( ay 0 + by (1+δ) a 2 y b 2) 1 2.
12 Proof of Key Lemma By compactness n := n(n, δ) such that if (y i ) n i=1 B Y then 1 i < j n such that P X N (y j y i ) < δ.
13 Proof of Key Lemma By compactness n := n(n, δ) such that if (y i ) n i=1 B Y then 1 i < j n such that P X N (y j y i ) < δ. By Dvoretzky s theorem and almost transitivity of Y, (y i ) n i=1 S Y such that for all scalars a, b 1,..., b n, we have (1 δ) ( a 2 y ) 1/2 n b i 2 ay 0 + i=1 (1 + δ) n b i y i i=1 ( a 2 y ) 1/2 n b i 2 i=1
14 Proof of Key Lemma By compactness n := n(n, δ) such that if (y i ) n i=1 B Y then 1 i < j n such that P X N (y j y i ) < δ. By Dvoretzky s theorem and almost transitivity of Y, (y i ) n i=1 S Y such that for all scalars a, b 1,..., b n, we have (1 δ) ( a 2 y ) 1/2 n b i 2 ay 0 + i=1 (1 + δ) n b i y i i=1 ( a 2 y Choose 1 i < j n such that P X N (y j y i ) < δ and set y = (1/ 2)(y j y i ). ) 1/2 n b i 2 i=1
15 Theorem Suppose that a Banach space X with a Schauder basis contains an almost transitive subspace Y. Let r FR(Y ) := {1 r : l r is finitely representable in Y }. Then, given ε > 0 and any sequence (a i ) of nonzero scalars, there exists a normalized block basis (x i ) in X such that, for all m 1 and all scalars b, we have ( m ) 1/r (1 ε) a k r + b r m a k x k + bx m+1 k=1 k=1 ( m ) 1/r (1 + ε) a k r + b r. k=1
16 Theorem No subspace of l p, 1 p <, p 2, or of c 0, admits an almost transitive renorming.
17 Theorem No subspace of l p, 1 p <, p 2, or of c 0, admits an almost transitive renorming. Proof: By previous theorem for all n N and any ε > 0 there exists a normalized block basis (x k ) n k=1 in X such that, (1 ε)n 1/2 n x k (1 + ε)n 1/2. k=1
18 Theorem No subspace of l p, 1 p <, p 2, or of c 0, admits an almost transitive renorming. Proof: By previous theorem for all n N and any ε > 0 there exists a normalized block basis (x k ) n k=1 in X such that, (1 ε)n 1/2 n x k (1 + ε)n 1/2. k=1 But in X = l p, 1 p <, every block basis is isometrically equivalent to the standard basis of l p,
19 The same argument also works in asymptoticl p spaces.
20 The same argument also works in asymptoticl p spaces. Recall: X is an asymptoticl p space if there exists C > 0 such that for all n x 1 < < x n, we have ( n ) 1/p ( 1 x k p n n ) 1/p x C k C x k p. k=1 k=1 k=1
21 The same argument also works in asymptoticl p spaces. Recall: X is an asymptoticl p space if there exists C > 0 such that for all n x 1 < < x n, we have ( n ) 1/p ( 1 x k p n n ) 1/p x C k C x k p. k=1 k=1 k=1 In the previous theorem we can start the block basis after n and we get that for all n N and any ε > 0 there exists a normalized block basis n < (x k ) n k=1 in X such that, (1 ε)n 1/2 n x k (1 + ε)n 1/2. k=1 and the resulting contradiction proves the following corollary.
22 Corollary No subspace of any asymptoticl p space, 1 p <, p 2, admits an almost transitive renorming.
23 Corollary No subspace of any asymptoticl p space, 1 p <, p 2, admits an almost transitive renorming. For 1 < p <, let C p denote the class of Banach spaces which are isomorphic to a subspace of an l p sum of finitedimensional normed spaces.
24 Corollary No subspace of any asymptoticl p space, 1 p <, p 2, admits an almost transitive renorming. For 1 < p <, let C p denote the class of Banach spaces which are isomorphic to a subspace of an l p sum of finitedimensional normed spaces. Corollary Let 1 < p <, p 2. If X C p then X does not admit an equivalent almost transitive norm.
25 Corollary No subspace of any asymptoticl p space, 1 p <, p 2, admits an almost transitive renorming. For 1 < p <, let C p denote the class of Banach spaces which are isomorphic to a subspace of an l p sum of finitedimensional normed spaces. Corollary Let 1 < p <, p 2. If X C p then X does not admit an equivalent almost transitive norm. It is known that C p contains every infinitedimensional subspace of a quotient space of l p [Johnson, Zippin 1974].
26 Corollary No subspace of any asymptoticl p space, 1 p <, p 2, admits an almost transitive renorming. For 1 < p <, let C p denote the class of Banach spaces which are isomorphic to a subspace of an l p sum of finitedimensional normed spaces. Corollary Let 1 < p <, p 2. If X C p then X does not admit an equivalent almost transitive norm. It is known that C p contains every infinitedimensional subspace of a quotient space of l p [Johnson, Zippin 1974]. Thus no infinitedimensional subspace of a quotient space of l p admits an equivalent almost transitive norm.
27 Recall the notion of asymptotic structure introduced by Maurey, Milman, and TomczakJaegermann (1992): A basis (b i ) n i=1 of unit vectors for an ndimensional normed space belongs to {X, (e i )} n if
28 Recall the notion of asymptotic structure introduced by Maurey, Milman, and TomczakJaegermann (1992): A basis (b i ) n i=1 of unit vectors for an ndimensional normed space belongs to {X, (e i )} n if ε > 0 m 1 x 1 > m 1 s.t. m 2 x 2 > m 2 s.t.... m n x n > m n such that there exist c and C, with 0 < c C and C/c < 1 + ε, such that for all scalars (a i ) n i=1, we have n n n c a i b i a i x i C a i b i. i=1 i=1 i=1
29 Recall the notion of asymptotic structure introduced by Maurey, Milman, and TomczakJaegermann (1992): A basis (b i ) n i=1 of unit vectors for an ndimensional normed space belongs to {X, (e i )} n if ε > 0 m 1 x 1 > m 1 s.t. m 2 x 2 > m 2 s.t.... m n x n > m n such that there exist c and C, with 0 < c C and C/c < 1 + ε, such that for all scalars (a i ) n i=1, we have n n n c a i b i a i x i C a i b i. i=1 i=1 i=1 In this language our result states that if X has a Schauder basis (e i ) and contains an almost transitive subspace Y then for all r FR(Y ), the unit vector basis of l 2 r belongs to {X, (e i )} 2.
30 In case when X has an unconditional basis (instead of just a Schauder basis) we can strengthen the main theorem to the following version of Krivine s Theorem:
31 In case when X has an unconditional basis (instead of just a Schauder basis) we can strengthen the main theorem to the following version of Krivine s Theorem: Theorem Suppose that X has an unconditional basis (e i ). If X contains an almost transitive subspace Y, then there exist p [p X, p Y ] and q [q Y, q X ] such that, for r = p and r = q and for all n 1 and ε > 0, there exist disjointly supported vectors (x i ) n i=1 X such that (x i ) n i=1 is (1 + ε)equivalent to the unit vector basis of l n r. In particular, if Y = X, then p = p X and q = q X.
32 In case when X has an unconditional basis (instead of just a Schauder basis) we can strengthen the main theorem to the following version of Krivine s Theorem: Theorem Suppose that X has an unconditional basis (e i ). If X contains an almost transitive subspace Y, then there exist p [p X, p Y ] and q [q Y, q X ] such that, for r = p and r = q and for all n 1 and ε > 0, there exist disjointly supported vectors (x i ) n i=1 X such that (x i ) n i=1 is (1 + ε)equivalent to the unit vector basis of l n r. In particular, if Y = X, then p = p X and q = q X. Proof relies on a result of Sari (2004) about disjoint envelope functions.
33 In case when X has an unconditional basis (instead of just a Schauder basis) we can strengthen the main theorem to the following version of Krivine s Theorem: Theorem Suppose that X has an unconditional basis (e i ). If X contains an almost transitive subspace Y, then there exist p [p X, p Y ] and q [q Y, q X ] such that, for r = p and r = q and for all n 1 and ε > 0, there exist disjointly supported vectors (x i ) n i=1 X such that (x i ) n i=1 is (1 + ε)equivalent to the unit vector basis of l n r. In particular, if Y = X, then p = p X and q = q X. Proof relies on a result of Sari (2004) about disjoint envelope functions. Question Fix m N, ε > 0. Let X have an uncoditional basis such that ({x i } m i=1 ) is (1 + ε)equivalent to the standard unit basis of l m r, with r 2. Can X be transitive?
34 Recall that a basis satisfies (p, q)estimates, where 1 < q p <, if there exists C > 0 such that ( n ) 1/p ( 1 x k p n n ) 1/q x C k C x k q, k=1 k=1 k=1 whenever x 1 < x 2 < < x n.
35 Recall that a basis satisfies (p, q)estimates, where 1 < q p <, if there exists C > 0 such that ( n ) 1/p ( 1 x k p n n ) 1/q x C k C x k q, k=1 k=1 k=1 whenever x 1 < x 2 < < x n. Corollary Suppose that a Banach space X with a Schauder basis (e i ) contains a subspace Y which admits an equivalent almost transitive norm. If (e i ) satisfies (p, q)estimates, then q 2 p.
36 Recall that a basis satisfies (p, q)estimates, where 1 < q p <, if there exists C > 0 such that ( n ) 1/p ( 1 x k p n n ) 1/q x C k C x k q, k=1 k=1 k=1 whenever x 1 < x 2 < < x n. Corollary Suppose that a Banach space X with a Schauder basis (e i ) contains a subspace Y which admits an equivalent almost transitive norm. If (e i ) satisfies (p, q)estimates, then q 2 p. Proof. Let be the equivalent almost transitive norm on Y. Then extends to an equivalent norm on X. Thus the Main Theorem gives the desired conclusion.
37 Corollary Suppose that X is an Orlicz sequence space l M (where M is an Orlicz function). Then X contains a subspace Y which admits an almost transitive norm if and only if X contains a subspace isomorphic to l 2.
38 Corollary Suppose that X is an Orlicz sequence space l M (where M is an Orlicz function). Then X contains a subspace Y which admits an almost transitive norm if and only if X contains a subspace isomorphic to l 2. Proof. Previous Corollary implies that the MatuszewskaOrlicz indices of M satisfy α M 2 β M, which in turn implies by a theorem of Lindenstrauss and Tzafriri that l M contains a subspace isomorphic to l 2.
39 Corollary Suppose that X is an Orlicz sequence space l M (where M is an Orlicz function). Then X contains a subspace Y which admits an almost transitive norm if and only if X contains a subspace isomorphic to l 2. Proof. Previous Corollary implies that the MatuszewskaOrlicz indices of M satisfy α M 2 β M, which in turn implies by a theorem of Lindenstrauss and Tzafriri that l M contains a subspace isomorphic to l 2. Conversely, if X contains a subspace Y isomorphic to l 2 then Y admits an equivalent transitive norm.
40 We combine our results with known results about the structure of subspaces of L p, 2 < p <, and we obtain that if X is a subspace of L p, 2 < p <, such that every subspace of X admits an equivalent almost transitive norm, then X is isomorphic to a Hilbert space (this also is true in the noncommutative case).
41 We combine our results with known results about the structure of subspaces of L p, 2 < p <, and we obtain that if X is a subspace of L p, 2 < p <, such that every subspace of X admits an equivalent almost transitive norm, then X is isomorphic to a Hilbert space (this also is true in the noncommutative case). This suggests the following question. Problem Suppose that every subspace of a Banach space X admits an equivalent almost transitive (or transitive) renorming. Is X isomorphic to a Hilbert space?
42 We combine our results with known results about the structure of subspaces of L p, 2 < p <, and we obtain that if X is a subspace of L p, 2 < p <, such that every subspace of X admits an equivalent almost transitive norm, then X is isomorphic to a Hilbert space (this also is true in the noncommutative case). This suggests the following question. Problem Suppose that every subspace of a Banach space X admits an equivalent almost transitive (or transitive) renorming. Is X isomorphic to a Hilbert space? YES, if such a space X is a stable and admits a C 2 bump.
43 Maximal and nonmaximal norms Definition (Pełczyński and Rolewicz 1962) A Banach space (X, ) is called maximal if whenever is an equivalent norm on X so that Isom(X, ) Isom(X, ), then Isom(X, ) = Isom(X, ).
44 Maximal and nonmaximal norms Definition (Pełczyński and Rolewicz 1962) A Banach space (X, ) is called maximal if whenever is an equivalent norm on X so that Isom(X, ) Isom(X, ), then Isom(X, ) = Isom(X, ). A maximal isometry group of a renorming of X is the same as a maximal bounded subgroup of GL(X) (where GL(X) denotes the group of linear surjective isomorphisms on X). Any bounded subgroup G of GL(X) is a subgroup of the isometry group of (X,. ) where x = sup gx. g G
45 Rolewicz 1984 proved that all nonhilbertian 1symmetric spaces are maximal.
46 Rolewicz 1984 proved that all nonhilbertian 1symmetric spaces are maximal. In 1982 Wood posed a problem whether every Banach space admits an equivalent maximal norm, that is whether GL(X) always has maximal bounded subgroups.
47 Rolewicz 1984 proved that all nonhilbertian 1symmetric spaces are maximal. In 1982 Wood posed a problem whether every Banach space admits an equivalent maximal norm, that is whether GL(X) always has maximal bounded subgroups. In 2013 Ferenczi and Rosendal answered this problem negatively by constructing a complex superreflexive space and a real reflexive space, both without an equivalent maximal renorming.
48 In 2006 Wood asked, what he called a more natural question, whether for every Banach space there exists an equivalent maximal renorming whose isometry group contains the original isometry group, that is, whether every bounded subgroup of GL(X) is contained in a maximal bounded subgroup of GL(X).
49 In 2006 Wood asked, what he called a more natural question, whether for every Banach space there exists an equivalent maximal renorming whose isometry group contains the original isometry group, that is, whether every bounded subgroup of GL(X) is contained in a maximal bounded subgroup of GL(X). We show that even in Banach spaces which admit an equivalent maximal norm there can also exist equivalent norms with isometry groups which are not contained in any maximal isometry group.
50 Theorem Spaces l p, 1 < p <, p 2, have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming.
51 Theorem Spaces l p, 1 < p <, p 2, have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming. Consider renormings of the form ( ) Z = l kn 2 n=1 where k n is a sequence of natural numbers, so that n N, n 1 k j < k n. j=1 l p
52 Theorem Spaces l p, 1 < p <, p 2, have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming. Consider renormings of the form ( ) Z = l kn 2 n=1 where k n is a sequence of natural numbers, so that n N, n 1 k j < k n. j=1 It is well known that Z is isomorphic to l p. l p
53 Theorem Spaces l p, 1 < p <, p 2, have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming. Consider renormings of the form ( ) Z = l kn 2 n=1 where k n is a sequence of natural numbers, so that n N, n 1 k j < k n. j=1 It is well known that Z is isomorphic to l p. Isometries of this kind of spaces were described by H. Rosenthal. l p
54 Theorem (Rosenthal 1986) X a pure space with a 1unconditional basis E = {e γ } γ Γ (i.e. γ 1 γ 2, span{e γ1, e γ2 } is not cannonically isometric to l 2 2 ) (H γ ) γ Γ Hilbert spaces all of dimension at least 2, Z = ( γ Γ H γ) E is called a functional hilbertian sum.
55 Theorem (Rosenthal 1986) X a pure space with a 1unconditional basis E = {e γ } γ Γ (i.e. γ 1 γ 2, span{e γ1, e γ2 } is not cannonically isometric to l 2 2 ) (H γ ) γ Γ Hilbert spaces all of dimension at least 2, Z = ( γ Γ H γ) E is called a functional hilbertian sum. Let P(Z ) be the set of all bijections σ : Γ Γ so that (a) {e σ(γ) } γ Γ is isometrically equivalent to {e γ } γ Γ, and (b) H σ(γ) is isometric to H γ for all γ Γ.
56 Theorem (Rosenthal 1986) X a pure space with a 1unconditional basis E = {e γ } γ Γ (i.e. γ 1 γ 2, span{e γ1, e γ2 } is not cannonically isometric to l 2 2 ) (H γ ) γ Γ Hilbert spaces all of dimension at least 2, Z = ( γ Γ H γ) E is called a functional hilbertian sum. Let P(Z ) be the set of all bijections σ : Γ Γ so that (a) {e σ(γ) } γ Γ is isometrically equivalent to {e γ } γ Γ, and (b) H σ(γ) is isometric to H γ for all γ Γ. Then T : Z Z is a surjective isometry if and only if σ P(Z ) and surjective isometries T γ : H γ H σ(γ) (γ Γ), so that for all z = (z γ ) γ Γ in Z, and for all γ Γ, (Tz) σ(γ) = T γ (z γ ).
57 Let p 2 and Z = ( n=1 l kn 2 where k n N are such that k 1 2 and n N, n 1 j=1 k j < k n. ) l p,
58 Let p 2 and Z = ( n=1 l kn 2 where k n N are such that k 1 2 and n N, n 1 j=1 k j < k n. Then Rosenthal s theorem applies and the set P(Z ) consists of the identity map only. ) l p,
59 Let p 2 and Z = ( n=1 l kn 2 where k n N are such that k 1 2 and n N, n 1 j=1 k j < k n. Then Rosenthal s theorem applies and the set P(Z ) consists of the identity map only. Thus m N ( ) Z m = (l d 2 ) p l kn 2 n=m where d = m j=1 k j, is isomorphic to Z, P(Z m ) also consists of the identity only, and ) l p, l p, Isom(Z m ) Isom(Z m+1 ).
60 Using the same method we obtain Theorem The following spaces have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming. Spaces l p, 1 < p <, p 2,
61 Using the same method we obtain Theorem The following spaces have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming. Spaces l p, 1 < p <, p 2, the 2convexified Tsierelson space T (2),
62 Using the same method we obtain Theorem The following spaces have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming. Spaces l p, 1 < p <, p 2, the 2convexified Tsierelson space T (2), and the space U with a universal unconditional basis,
63 Using the same method we obtain Theorem The following spaces have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming. Spaces l p, 1 < p <, p 2, the 2convexified Tsierelson space T (2), and the space U with a universal unconditional basis,
64 Using the same method we obtain Theorem The following spaces have a continuum of renormings none of whose isometry groups is contained in any maximal isometry group of an equivalent renorming. Spaces l p, 1 < p <, p 2, the 2convexified Tsierelson space T (2), and the space U with a universal unconditional basis, Question: Does X = T (2), or X a general weak Hilbert space, other than l 2, have a maximal bounded subgroup of GL(X)?
65 Maximal renormings Theorem For 1 < p <, p 2, l p admits a continuum of renormings whose isometry groups are maximal and are not pairwise conjugate in the isomorphism group of l p.
66 Maximal renormings Theorem For 1 < p <, p 2, l p admits a continuum of renormings whose isometry groups are maximal and are not pairwise conjugate in the isomorphism group of l p. Take Z m = where k n = m for all n N. ( n=1 l kn 2 ) l p,
67 Maximal renormings Theorem For 1 < p <, p 2, l p admits a continuum of renormings whose isometry groups are maximal and are not pairwise conjugate in the isomorphism group of l p. Take Z m = ( n=1 l kn 2 where k n = m for all n N. Then P(Z m ) consists of all permutations, and the group Isom(Z m ) is maximal ) l p,
68 To get continuum mutually noncojugate renormings let J N, with min J 2, and let N = j J A j, where A j are disjoint infinite subsets of N. For k A j, let H k = l j 2, and ( ) Z J = Z J (E) = H k. k=1 E
69 To get continuum mutually noncojugate renormings let J N, with min J 2, and let N = j J A j, where A j are disjoint infinite subsets of N. For k A j, let H k = l j 2, and ( ) Z J = Z J (E) = H k This works for any space k=1 Z = ( l k 2 ) E, k=1 where E = {e k } k=1 is a pure 1symmetric basis for some X which is not isomorphic to l 2. E.
70 To get continuum mutually noncojugate renormings let J N, with min J 2, and let N = j J A j, where A j are disjoint infinite subsets of N. For k A j, let H k = l j 2, and ( ) Z J = Z J (E) = H k This works for any space k=1 Z = ( l k 2 ) E, k=1 where E = {e k } k=1 is a pure 1symmetric basis for some X which is not isomorphic to l 2. It is wellknown that if (k j ) j=1 N is any unbounded sequence then ( j=1 lk j 2 ) E is 4isomorphic to Z. Thus Z J is 4isomorphic to Z for all J. E.
71 Proposition Let X be a space with a nonhilbertian symmetric pure basis E such that X(X) (i.e., the Esum of infinitely many copies of X) is isomorphic to X, and X contains uniformly complemented and uniformly isomorphic copies of l n 2. Then X is isomorphic to Z = ( k=1 lk 2 ) E and hence X has a continuum of renormings whose isometry groups are maximal and are not pairwise conjugate in the isomorphism group of X.
72 Proposition Let X be a space with a nonhilbertian symmetric pure basis E such that X(X) (i.e., the Esum of infinitely many copies of X) is isomorphic to X, and X contains uniformly complemented and uniformly isomorphic copies of l n 2. Then X is isomorphic to Z = ( k=1 lk 2 ) E and hence X has a continuum of renormings whose isometry groups are maximal and are not pairwise conjugate in the isomorphism group of X. Question Does there exist a separable Banach space X with a unique, up to conjugacy, maximal bounded subgroup of GL(X)?
73 Proposition Let X be a space with a nonhilbertian symmetric pure basis E such that X(X) (i.e., the Esum of infinitely many copies of X) is isomorphic to X, and X contains uniformly complemented and uniformly isomorphic copies of l n 2. Then X is isomorphic to Z = ( k=1 lk 2 ) E and hence X has a continuum of renormings whose isometry groups are maximal and are not pairwise conjugate in the isomorphism group of X. Question Does there exist a separable Banach space X with a unique, up to conjugacy, maximal bounded subgroup of GL(X)? If yes, does X have to be isomorphic to a Hilbert space?
74 Thank you.
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