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This is a follow-up to my question here. A topological vector space is normable, i.e. its topology is induced by some norm on the vector space, if and only if it is Hausdorff and the $0$ vector has a bounded convex neighborhood. My question is, under what circumstances is a topological vector space “inner product-able”, i.e. when is its topology induced by some inner product on the vector space?

Note that this is not the same as asking under what circumstances a norm is induced by some inner product. The answer to that question is when the norm obeys the parallelogram law. But we could have a situation where the topology is induced by multiple norms, one which is not induced by any inner product and another which is induced by some inner product. In any case, another way to phrase my question is, under what circumstances is a given norm equivalent to some norm obeying the parallelogram law?

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  • $\begingroup$ What does 'bounded' mean in your first paragraph? $\endgroup$
    – JonathanZ
    Dec 18, 2018 at 16:03
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    $\begingroup$ @JonathanZ A subset $B$ of a topological vector space $X$ is said to be bounded if every neighborhood $U$ of the $0$ vector can be stretched to include $B$ as a subset. Stretching the set $U$ means taking the set $\lambda U = \{\lambda x: x\in U\}$ where $\lambda$ is some real number greater than 1. $\endgroup$
    – Keshav Srinivasan
    Dec 18, 2018 at 18:19
  • $\begingroup$ Apparently, any Banach space in which every closed subspace is complemented is isomorphic to a Hilbert space ieeexplore.ieee.org/document/325179 $\endgroup$
    – SmileyCraft
    Dec 20, 2018 at 19:59

1 Answer 1

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In his dissertation (Topological characterization of an inner product space), Howard Lambert proves the following:

Theorem 12 Let $X$ be a topological linear space such that there exists a Hamel basis $H$ with the property that the set \begin{equation} C = \left\{x : x=\sum_{i=1}^n \alpha_i x_i, x_i\in H ,\sum_{i=1}^n |\alpha_i|^2<1,n \text{ arbitrary} \right\} \end{equation}
is open and bounded, then $X$ is innerproductable.

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  • $\begingroup$ Doesn't this basically follow directly directly from the definitions? And does this theorem only apply to finite dimensional vector spaces? $\endgroup$
    – SmileyCraft
    Dec 20, 2018 at 20:08
  • $\begingroup$ It is not only for finite dimensional spaces. $\endgroup$
    – Dunham
    Dec 20, 2018 at 21:05

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