Re: CBS for Banach spaces

From: Stephen Montgomery-Smith (stephen_at_math.missouri.edu)
Date: 12/28/04 

Date: Tue, 28 Dec 2004 02:01:36 GMT

Stephen Montgomery-Smith wrote:
> Gonçalo Rodrigues wrote:
>
>> Hi all,
>>
>> Is Cantor-Bernstein-Schroeder true for Banach spaces? IOW let A and B
>> be Banach spaces. A <= B iff there is an *isometric embedding* A->B.
>> Is it true that if A<=B and B<=A then A and B are isometrically
>> isomorphic? And if we relax to topological embeddings/isomorphisms?
>> Counter-examples?
>>
>> TIA, best regards,
>> G. Rodrigues
>
>
> The question for linear isomorphisms is interesting and difficult. The
> CSB is true if A is linearly isomorphic to A+A (the direct sum of A with
> itself), and similarly for B. I think this is due to Pelczinski, and
> that the short proof can be found, for example, in the book by Joe
> Diestel on Banach spaces.
>
> However if there is no such condition "A linearly isomorphic to A+A",
> the answer is negative. This is related to the counterexample of Gowers
> and Maurey, and is definitely difficult and deep mathematics.
>
> Stephen

OK, I got this wrong. Both results cited above are concerned with the
following version of CBS for Banach spaces. If A and B are Banach
spaces such that A embeds complementedly into B and vice versa, then is
A isomorphic to B? We say that A embeds complementedly into B is B is
(linearly and topologically) isomorphic to A+C (here + means direct sum).

THe counterexample when we don't have the "complementedly" might be L_1
(i.e. L_1(0,1)) and L_1+l_2, since l_2 embeds into L_1 via sums of
independent gaussian random variables. Presumably L_1+l_2 is not
isomorphic to L_1 - perhaps this follows because perhaps L_1 has no
complmented subspaces isomorphic to l_2, but I am not expert enough to
say for sure.

The Pelczinski argument is rather nice. So A=B+C, B=A+D and A=A+A and
B=B+B.

A=B+C=B+B+C=A+D+B+C=A+D+A=A+A+D=A+D=B.