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Let $X$ and $Y$ be Banach spaces. The notion of the embedded spaces was introduced by D.S. Djordjevic. Say that $X$ embed in $Y$, and write $X \preceq Y$, if there exists a left invertible operator $J:X \rightarrow Y$. My question is the following；

If $X$ embed in $Y$, and $Y$ embed in $X$, then $X$ isomorphic onto $Y$？

PS: The answer is positive when $X$ and $Y$ are Hilbert spaces. But for general Banach space,I can not find a solution. Is there exists a couterexample?

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  • $\begingroup$ The answers below show that this is false but it is perhaps worth mentioning that, on the positive side, there are versions under additional conditions which are correct and these are frequently used in the isomorphic theory of Banach spaces. They go under the collective name of "Pelczynski decomposition method". $\endgroup$
    – jbc
    Commented Apr 4, 2013 at 6:45

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This is called the Schroeder-Bernstein problem, and for Banach spaces there are constructions of nonisomorphic Banach spaces which embed into each other.

W. T. Gowers, "A Solution to the Schroeder-Bernstein Problem for Banach Spaces" Bull. London Math. Soc. (1996) 28 (3): 297-304.

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    $\begingroup$ There is a noteworthy special case of this problem that is still open: if $X$ and $X^{\ast \ast}$ are isomorphic to complemented subspaces of one another, are they in fact isomorphic? $\endgroup$
    – Philip Brooker
    Commented Jun 16, 2012 at 13:31
  • $\begingroup$ Thanks Douglas for you reminding me the Schroeder-Bernstein problem. $\endgroup$
    – Qingping Zeng
    Commented Jun 17, 2012 at 3:11
  • $\begingroup$ @Philip: I did not know this was open and I'm surprised it is not obviously true. I suppose one might construct a counterexample using a non-reflexive (so-called Jamesification, coined by Spiros) version of Gowers' space. Also, it's at least possible that Spiros has already constructed a counterexample, perhaps without knowing, in one of his papers. Where did you see this question? $\endgroup$
    – Kevin Beanland
    Commented Jun 18, 2012 at 0:17
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    $\begingroup$ @Kevin: I first came across the question in a paper of Plichko and Wojtowicz, Note on a Banach space having equal linear dimension with its second dual, Extracta Mathematica 18(3) (2003), p.311--314 (in particular, see the final remark of the paper). I wrote to one of the authors and also to Galego a couple of years ago enquiring as to the status of the problem, and at that time it was still open as far as they knew. More generally, I think it is open whether there exists an integer $n\geq 3$ and a Banach space $X$ such that $X$ is isomorphic to its $n$th dual but not to its $j$th dual $\endgroup$
    – Philip Brooker
    Commented Jun 18, 2012 at 12:47
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    $\begingroup$ for any $1\leq j \leq n-1$; if such $n$ and $X$ do exist, then $X$ and $X^{\ast\ast}$ are nonisomorphic Banach spaces that are isomorphic to complemented subspaces of one another. $\endgroup$
    – Philip Brooker
    Commented Jun 18, 2012 at 12:48
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The Schroeder-Bernstein problem asked about complemented subspaces. It is a much stronger property for a subspace to be complemented. The only spaces in which every subspace is complemented are isomorphic to Hilbert space (Lindenstrauss and Pelczynski). Also, Hilbert space (and its isomorphs) is the only space isomorphic to all of its subspaces. Therefore, if you consider any space which is not isomorphic to Hilbert space but embedds into its subspaces (e.g. $\ell_p $ for $p \not= 2$; in general, these are called minimal spaces), any non-isomorphic subspace gives a counterexample.

Edit: I guess I was confused by the term left invertible. I'll leave my answer up in case someone doesn't know how to do it in this weaker case.

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    $\begingroup$ @Kevin: The OP's terminology is confusing and, I think, differs from standard Banach space terminology; what he calls an embedding is what I think you or I would call a complemented embedding. In particular, I presume that, in the OP's terminology, a left inverse for $J$ would be a map $K: Y\longrightarrow X$ such that $KJ$ is the identity on $X$, hence $JK$ is an idempotent operator on $Y$ with range isomorphic to $X$. $\endgroup$
    – Philip Brooker
    Commented Jun 16, 2012 at 23:12
  • $\begingroup$ Thanks for clarifying. I should have assumed others would have corrected! $\endgroup$
    – Kevin Beanland
    Commented Jun 16, 2012 at 23:16
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    $\begingroup$ On another note. I'm a bit surprised at how popular this question (and answer) is given that the solution is simply to quote a major theorem in Banach spaces that (as it happens) was part of the work that won Gowers the Fields. I suppose Banach Theory needs better PR. $\endgroup$
    – Kevin Beanland
    Commented Jun 16, 2012 at 23:31
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    $\begingroup$ @Kevin, as someone who still gets upvotes for an answer to one question which consisted solely of pointing to work of Knutson and Tao ... so it goes $\endgroup$
    – Yemon Choi
    Commented Jun 17, 2012 at 0:15
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    $\begingroup$ @Kevin: The theorem you cite about complemented subspace is not by Lindenstrauss and Pelczynski, but by Lindenstrauss and Tzafriri. Precisely in Lindenstrauss, J. and Tzafriri, L. On the complemented subspaces problem, Israel J. Math. 9 (2) (1971) 263-269. $\endgroup$
    – Valerio Capraro
    Commented Jun 17, 2012 at 5:07

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