Let $B$ be a Banach space. It is not necessarily true that there exists a Hilbert space $H$ linearly isometric to $B$.

However, is it true that there exists a Hilbert space $H$ homeomorphic to $B$?

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    $\begingroup$ If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $\ell^2$ $\endgroup$
    – user124910
    Apr 7, 2019 at 21:35
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    $\begingroup$ @user124910 We can extend this to non-separable as well. See my answer. $\endgroup$ Apr 7, 2019 at 21:37

1 Answer 1


Yes, but this is quite a deep result. Two infinite-dimensional Banach spaces $X$ and $Y$ are homeomorphic iff $d(X)=d(Y)$, where the density $d(X)$ is the minimal size of a dense subset of $X$.

So any separable infinite-dimensional Banach space is homeomorphic to the Hilbert space $\ell^2$ (and even to $\mathbb{R}^\omega$, because the result extends to locally convex completely metrisable TVS's as well). And for higher densities we have Hilbert spaces $\ell_2(\kappa)$ as models. Finite dimensional we only have the $\mathbb{R}^n$ up to homeomorphism, which are already Hilbert spaces.

  • $\begingroup$ Do you know of a reference with the proof of this? $\endgroup$
    – user156213
    Apr 8, 2019 at 0:22
  • 2
    $\begingroup$ @user156213 This post has a reference. $\endgroup$ Apr 8, 2019 at 11:42

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