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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1$\begingroup$ If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $\ell^2$ $\endgroup$– user124910Apr 7, 2019 at 21:35
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1$\begingroup$ @user124910 We can extend this to non-separable as well. See my answer. $\endgroup$– Henno BrandsmaApr 7, 2019 at 21:37
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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.
answered Apr 7, 2019 at 21:36
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$\begingroup$ Do you know of a reference with the proof of this? $\endgroup$ Apr 8, 2019 at 0:22
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