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First, I would like to thank the organizers as well as the people who participate in this forum Second, Let $H$ be an infinite dimensional Hilbert space and $I$ a closed interval, $C\left(I,H\right)$ is the space of Continuous function My question is as follows: Is the space of Continuous functions $C\left(I,H\right)$ is homeomorphic to $H$?

daw
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zakaria hamidi
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  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Aug 18 '18 at 18:07
  • Certainly no if $H$ is finite-dimensional - in that case $H$ and $C(I,H)$ are vector spaces, but $C(I,H)$ has infinite dimension. Presumably you meant for $H$ to be an infinite-dimensional Hilbert space. – David C. Ullrich Aug 18 '18 at 18:14
  • thank you for your comment. $H$ is supposed an infinite dimensional Hilbert space – zakaria hamidi Aug 18 '18 at 18:29
  • I guess it depends on what kind of interval you are taking. If you take an open interval and $H=l^2(\mathbb{N})$, then $H$ is separable, but $C(I, H)$ isn't and therefore they are not isomorphic. But I guess you want to take a closed interval. – Severin Schraven Aug 18 '18 at 20:10
  • thank you Severin Schraven for your comment. yes $I$ is supposed closed – zakaria hamidi Aug 18 '18 at 20:19
  • The question is general, we can treat both cases, the case where $H$ is reflexive and the case where $H$ is not reflexive the question is to find a sufficient conditions such that $H$ and $C\left(I,H\right)$ are Homoeomorphic – zakaria hamidi Aug 19 '18 at 08:30
  • I believe a useful result is that two Banach spaces are homeomorphic if and only if they have the same density. – Daron Aug 20 '18 at 11:35

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Theorem 6.1 of this paper implies homoemorphism classes of Banach spaces are determined entirely by their density character. For each cardinality $\alpha$ there is a unique Hilbert space $\ell^2(\alpha)$ for which $\alpha$ is the density character. So each Banach space $V$ is topologically equivalent to $\ell^2(\alpha)$ for $\alpha$ the density of $V$.

This answer shows the special case that $C(I,\ell^2(\mathbb N))$ is separable hence homeomorphic to $\ell^2(\mathbb N)$. The proof for the more general $\ell^2(\alpha)$ should be almost identical.

Daron
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