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Let $(X,||.||)$ be a real Banach space , then is it true that there exists a norm $||.||_1$ on $X$ coming from an inner product such that $||.|| , ||.||_1$ generates same topology on $X$ or atleast $(X,||.||)$ and $(X,||.||_1)$ are homeomorphic ?

3 Answers3

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This is not the case. If it were, then $\mathrm{id}: (X,\|\cdot\|)\to (X,\|\cdot\|_1)$ would be a linear isomorphism and $(X,\|\cdot\|_1)$ a reflexive space, hence, as reflexivity is preserved by linear isomorphism, $(X,\|\cdot\|)$ would be a reflexive space, but this is not the case for all Banach spaces.

Sebastian Bechtel
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No, not every Banach space is isomorphic to a Hilbert space. Specific example is $c_0$.

As the OP is interested in non-linear homeomorphisms, then the answer is positive. See my answer here.

Community
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Tomasz Kania
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The following answers your last question. You can obtain this way a Hilbert space $(X,||.||_1)$ such that the spaces $(X,||.||)$ and $(X,||.||_1)$ are indeed homeomorphic (not by the identity map, of course). This follows

H. Toruńczyk, Characterizing Hilbert space topology. Fund. Math. 111 (1981), no. 3, 247–262.

In the separable setting this was proven earlier by M.I.Kadec:

"A proof of the topological equivalence of all separable infinite-dimensional Banach spaces", Funct. Anal. and Appl., Vol. 1 (1967).

Moishe Kohan
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  • Essentially, all Banach spaces of the same density are homeomorphic as spaces, and at least one of them (some $\ell_2(D)$) is a Hilbert space. Use the homeomorphism to esentially copy the latters inner product to the Banach space. Of course there is no relation whatsoeever to the original norm of the Banach space... – Henno Brandsma Jan 06 '17 at 05:58
  • @HennoBrandsma: Right. Toruńczyk, proves even more, namely homeomorphism of Frechet spaces. – Moishe Kohan Jan 06 '17 at 06:34