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It is not hard to give examples of normed spaces which are not inner product spaces. Now let $(V, \|\cdot\|)$ be a normed space.

Is it always possible to construct an inner product on $V$ which gives the same topology on $V$ as before? (As pointed out in comments, in finite dimensional case it it always the case).

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  • $\begingroup$ All norms of $\mathbb{R}^n$ are equivalent. And only a few (relatively speaking) are induced by inner products. $\endgroup$ Dec 10, 2014 at 18:45
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    $\begingroup$ The second question is true for $\mathbb{R}^n$, i.e. all finite-dimensional cases. But is fails for infinite-dimensional ones. $\endgroup$ Dec 10, 2014 at 18:47
  • $\begingroup$ @HennoBrandsma Thanks for the comments. I have edited the question. Do you have a counterexample for the infinite dimensional setting? $\endgroup$
    – Philip M
    Dec 10, 2014 at 18:50

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An inner product space (i.e. the topological vector space induced by an inner product) is reflexive (even: $X$ and $X^\ast$ are isometric), at least if $X$ is complete. And e.g. $\ell_1$ does not have this property.

So at least you need to demand that $X$ and its dual are isometric (in the complete case). And even then there probably are counterexamples.

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