# Conditions for a topological vector space to be an inner product space

Are there any conditions on a (real, Hausdorff) topological vector space $\mathcal{V}$ that guarantees the existance of an inner product on that space which induces the same topology?

I'm looking for conditions that depend only on the topology (possibly using the dual $\mathcal{V}^*$, of course).

Thank you very much in advance :)

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Looking only at the topology is not enough because all separable Banach spaces are homeomorphic to $\ell^2$. The pair $\mathcal{V}=L^4$ and $\mathcal{V}^*=L^{4/3}$ looks just like the pair $\mathcal{V}=L^2$, $\mathcal{V}^*=L^2$ until you bring in the linear structure. –  user53153 Feb 10 '13 at 16:48
It would first, have to be normable. Then that norm would have to satisfy the parallelogram law. –  David Mitra Feb 10 '13 at 16:51
@DavidMitra: There seems to be a step missing from being normable to admitting a Hilbert space norm in your description. There are many complete non-Hilbert norms on a Hilbert space. The solution to the complemented subspace problem would be a way to bridge the gap. –  Martin Feb 10 '13 at 17:23
@5PM: But that isn't really a problem, because the linear structure will appear when we demand the topology on $\mathcal{V}$ to be induced by an inner product (in the usual way). The homeomorphism given in your example won't induce an inner product in $L^4$. –  Luiz Cordeiro Feb 10 '13 at 18:08
@DavidMitra: That is a good solution. It's easy to show (using Minkowsky functionals) that a t.v.s. is normable iff it is locally convex and locally bounded. Using that fact and taking completions when necessary, I think you got a nice solution! –  Luiz Cordeiro Feb 10 '13 at 18:08