Let $X$ be a Banach space and suppose that $X$ is isometric to its continuous dual space $X^*$. Must $X$ be hilbertable in the sense that there exists an inner product which induces the norm on $X$? The converse of this statement is the Riesz representation theorem for hilbert spaces; I am wondering if the theorem can be stengthened to "if and only if".

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    $\begingroup$ If you take any reflexive space $X$, then $X\oplus_{2} X^*$ is isometrically isomorphic to its dual, but in general won't be isomorphic to a Hilbert space. $\endgroup$ – Philip Brooker Sep 18 '11 at 22:29
  • $\begingroup$ @Philip: Ah, you beat me to it! $\endgroup$ – t.b. Sep 18 '11 at 22:32
  • $\begingroup$ @Theo: well, you took the time to give more detail, so you deserve the points! Tangentially related: I believe it is an open problem whether $X$ and $X^{\*\*}$ are necessarily isomorphic whenever $X$ and $X^{\*\*}$ are isomorphic to complemented subspaces of one another. Meanwhile, it is known that $X$ and $X^{\*\*}$ needn't be isomorphic if $X^{\*\*}$ is isomorphic to a complemented subspace of $X$. $\endgroup$ – Philip Brooker Sep 18 '11 at 22:39

Here's a construction:

Take any reflexive Banach space $X$. Take the direct sum $E = X \oplus X^{\ast}$ and equip it with the norm $\|(x,x^{\ast})\|_E = \left(\|x\|_{X}^2 + \|x^{\ast}\|_{X^{\ast}}^2\right)^{1/2}$. This space is usually denoted by $E=X \oplus_2 X^{\ast}$ for short.

Then $E$ is isometric to its dual space $E^{\ast} = X^{\ast} \oplus_2 X^{\ast\ast}$: An isometric isomorphism is given by $(x,x^{\ast}) \mapsto (x^{\ast},i_{x})$ where $i:X \to X^{\ast\ast}$ is the canonical inclusion (by reflexivity of $X$ the map $i$ is an isometric isomorphism).

The space $E$ won't be isomorphic to a Hilbert space unless $X$ is itself isomorphic to a Hilbert space.

In fact, a (real) Banach space is “Hilbertable” if and only if every closed subspace has a complement by a (very deep) Theorem of Lindenstrauss and Tzafriri.

So: If $X$ is not isomorphic to a Hilbert space, it has a subspace that isn't complemented, and thus so has $E$.

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    $\begingroup$ What does $\oplus_2$ mean? $\endgroup$ – Qiaochu Yuan Sep 18 '11 at 22:44
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    $\begingroup$ @Qiaochu: Direct sum equipped with the $2$-norm (the one I gave) as opposed to other ways of choosing a norm on the direct sum. $\endgroup$ – t.b. Sep 18 '11 at 22:48

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