Equivalent conditions for a topological vector space to admit an inner-product

Question

Let $V$ be a topological vector space.

Inspired by this list

• Every inner-product induces a (unique?) norm
• Every norm induces a (unique?) metric
• Every metric induces a (unique?) uniform structure
• Every uniform structure induces a (unique?) topology,

I'm trying to complete the list

• $V$ admits an inner-product iff ???
• $V$ admits a norm iff it is Hausdorff and has a convex bounded neighborhood of zero (Kolmogorov's theorem)
• $V$ admits a metric iff it is Hausdorff and has a countable base of neighborhoods of zero
• $V$ admits a (unique) uniform structure, for free

and the list

• $V$ is a Euclidean space iff ???
• $V$ is a Hilbert space iff it has a complete inner-product and the scalars are $\Bbb R$ or $\Bbb C$
• $V$ is a Banach space iff it has a complete norm and the scalars are $\Bbb R$ or $\Bbb C$
• $V$ is a ??? iff it has a complete metric

Namely, what are equivalent conditions for $V$ to admit an inner-product? How do Euclidean spaces $\textbf{R}^n$ fit into this scheme? Does $V$ have a special name when it has a complete metric, or are these spaces not important enough? In the first list, is the induced structure always unique?

Answer 1

I'll assume TVS includes that $X$ is $T_1$ (so Tychonoff) Some random relevant facts:

A normed space $(X,||.||)$ has its norm induced by an inner product iff the paralellogram law holds:

$$\forall x,y \in X: 2||x||^2 + 2||y||^2 = ||x-y||^2 +||x+y||^2$$

e.g. see this question and its answers. The inner product is uniquely determined by the norm.

A TVS is normable iff it is locally convex and it has a base at $0$ of bounded sets (in the sense that $A$ is bounded iff for every neighbourhood $V$ of $0$ there is some scalar $c$ with $A \subseteq cV$). See any good textbook. The norm can always be scaled etc. so it's not quite unique.

$V$ is a Euclidean space iff it has finite dimension (if the scalar field is $\mathbb{R}$ or $\mathbb{C}$, of course).

A TVS that has a complete and compatible metric is called a Fréchet space. They can be topologically characterised by $V$ with a countable local base at $0$ which is also uniformly complete in the induced uniformity. See these notes for more info.

The uniformity on $V$ is never unique (but it exists for all $T_1$ TVS's). Spaces with unique uniformities are "almost-compact" and TVS's never are.