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Consider a Banach space $B$. I'm trying to prove or disprove the following:

If every separable subspace of $B$ is isomorphic to a Hilbert space then, the space $(B, ||\cdot||_B)$ is isomorphic to a Hilbert space in the category of Banach spaces and bounded linear maps?

My initial idea is to think about the union of separable subspaces that contain the entire Banach space.

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    Fréchet-von Neumann-Jordan theorem – Anne Bauval Aug 04 '24 at 22:06
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    @Anne: that answers the question with "isomorphic" replaced by "isometrically isomorphic," but I assume the OP means the usual notion of isomorphic (in the category of Banach spaces and bounded linear maps). – Qiaochu Yuan Aug 04 '24 at 23:14
  • @QiaochuYuan Yeah this is right. I meant just the isomorphism in this case. It is not clear to me how to proceed with the Parallelogram law. My initial idea is to think about the union of separable subspaces that contains the entire Banach space. –  Aug 04 '24 at 23:24
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    This is a very reasonable question and I neither understand nor agree with users who have closed is. – Jochen Aug 06 '24 at 06:03
  • I don't understand why was it closed. If there is something I need to edit I can add that. –  Aug 06 '24 at 19:18
  • @supernova I don't agree with closing this question either, and it seems to have been re-opened. One small advice though: you seemed to have never accepted any of the answers to your questions, despite some questions receiving multiple good answers. I'd suggest actually accepting those answers you find satisfactory. It increases your own reputation on this site, accrues good will for you in this community, and makes people more willing to answer your questions. – David Gao Aug 07 '24 at 08:52
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    @DavidGao Thanks for the remark! Will keep this in mind! –  Aug 07 '24 at 19:09

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Here http://matwbn.icm.edu.pl/ksiazki/or/or2/or214.pdf Theorem IV.9.12 states the following criterion for a Banach space $E$ to be isomorphic to a Hilbert space, which can be thought as a generalization of the parallelogram law:

There exists a constant $A > 0$ such that $$A^{-1}\sum_{k = 1}^n \|x_k\|^2 \le \frac{1}{2^n}\sum_{\varepsilon_1, \dots, \varepsilon_n \in \{-1, 1\}}\|\sum_{k = 1}^n \varepsilon_k x_k\|^2 \le A\sum_{k = 1}^n \|x_k\|^2$$ for all finite collections $x_1, \dots, x_n \in E$.

I claim that this property holds for $E$ if it holds for all its separable closed subspaces. Let $A_L$ denote the infimum of the constants $A$ that satisfy the inequalities above for a separable subspace $L \subset E$. For a future reference we note that $L \subset N$ implies $A_L \le A_N$. Let $A = \sup_{L} A_L$, where $L$ ranges over all separable closed subspaces of $E$. If $A < +\infty$, then, clearly, it satisfies the inequalities above for the whole space $E$. Assume on the contrary that $A = +\infty$, so for all $n \in \mathbb N$ there exists separable $L_n$ such that $A_{L_n} \ge n$. If $$L = \overline{\mathrm{span}\left(\bigcup_n L_n\right)},$$ then $L$ is separable and $L_n \subset L$ for all $n$. Thus, $A_{L_n} \le A_L$, so $A_L \ge n$ for all $n \in \mathbb N$. We arrived at a contradiction.

Matsmir
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    I think the middle term should be multiplied with $\frac{1}{2^n}$, otherwise, it doesn't hold even for Hilbert norms. –  Aug 07 '24 at 19:10
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    Yes, it should be. Thanks for the correction! @supernova – Matsmir Aug 08 '24 at 11:19
  • Is there an example of a Banach space that satisfies the property above? It is stated in Kwapien's paper, a reference in the paper you mentioned, but nowhere does it concretely point out an example. –  Aug 09 '24 at 19:46
  • @supernova I can't remember a "natural" example of a Banach space that is isomorphic to a Hilbert space. In practice all examples of such spaces that come to mind (except finite dimensional) are Hilbert, i.e. are endowed with a dot product in definition. I think, Kwapien's criterion is quite nice to prove the opposite statement, i.e. that a Banach space is not isomorphic to a Hilbert space, since all you need is to construct a sequence that does not satisfy some inequalities. – Matsmir Aug 09 '24 at 22:23
  • I think $L^1(0,1)$ can be shown to be isomorphic to $L^2(0,1)$ using the Hamel basis argument (not entirely sure). However, it is not clear how you could use this criterion to show something like that. There is some discussion here: https://math.stackexchange.com/questions/3093759/if-a-is-isomorphic-to-a-hilbert-space-h-can-we-show-that-a-is-a-hilbert/3093783#3093783 –  Aug 10 '24 at 00:00
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    These spaces are not isomorphic (e.g. because $L^1$ is not reflexive). They are isomorphic only in algebraic sense, but not as Banach spaces. @supernova – Matsmir Aug 10 '24 at 11:45