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I know that $\ell_2$ is the only separable Hilbert space of infinite dimension up to isometric isomorphism, so in particular, any separable Hilbert space of infinite dimension is isomorphic to $\ell_2$.

So my question is, can someone give an example of a Banach space isomorphic to $\ell_2$ but not isometrically isomorphic to it?

I know, for what is said above, that this space cannot be a Hilbert space but I can't think of any example.

Arctic Char
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Eparoh
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    Take a direct sum with a finite dimensional non-Hilbert space. – David Mitra May 21 '21 at 10:38
  • Could you elaborate the answer a bit more, please? I'm not sure why the resulting space is not Hilbert – Eparoh May 21 '21 at 10:48
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    It contains a subspace that is not a Hilbert space. – David Mitra May 21 '21 at 10:53
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    For finite dimensional case, use $\mathbb R^2$ with any norm whose unit ball is not an ellipse. The parallelogram law fails, so it is not isometrically Hilbert; but of course it is isomorphically Hilbert. – GEdgar May 21 '21 at 11:19

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For example, let $X=\ell_2(\{0,1,2,\dots\})$ and define a norm on $X$ by $$||x||=|x_0|+\left(\sum_{n=1}^\infty|x_n|^2\right)^{1/2}.$$

The identity map is an isomorphism onto $\ell_2$. A simple way to make certain that $X$ is not isometric to $\ell_2$ is to verify that the parallelogram law fails: $$||e_0+e_1||^2+||e_0-e_1||^2=8,$$while $$2||e_0||^2+2||e_1||^2=4.$$

quid
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David C. Ullrich
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