# TVS: Topology vs. Scalar Product

It is clear that every scalar product induces a norm and that a metric and that finally a topology.

Turning this argument around we know:
Not every topology induces a metric only the metrizable ones, not every metric induces a norm only the homogeneous translational invariant ones and finally not every norm induces a scalar product only the ones that obeys the parallelogram law.

In between we know that:
A metrizable topology admits many metrics, a metric space admits at most one norm as well as a normed space admits at most one scalar product. So from many to less!

So, my thoughts were:
1.) Does every metrizable topology admit at least one norm?
2.a) Is that norm necessarily unique? 2.b) Are there several norms?
3.) Is there necessarily one that admits a scalar product?
4.a) Is that scalar product necessarily unique? 4.b) Are there several scalar products?

...I would always like to start from some given topology:

On the one hand, there are metrisable topological vector spaces that are not normable, for example the space of continuous functions on $\mathbb{R}^n$ endowed with the topology of locally uniform convergence is a Fréchet space that is not normable (it has no bounded neighbourhood of $0$). There are many more examples of metrisable but not normable topological vector spaces, e.g. the $L^p$ spaces for $0 < p < 1$ are (completely) metrisable spaces, but not locally convex, hence not normable.

If a topological vector space is normable, the norm inducing the topology is not unique (unless the space is the trivial vector space). Aside from the trivial non-uniqueness due to the possible choice of constant multiples of a norm, for any normable space of dimension $\geqslant 2$, there are norms inducing the topology that are not constant multiples of each other - choose a nonzero continuous linear form $\lambda$, and let $\lVert x\rVert' := \lVert x\rVert + \lvert\lambda(x)\rvert$, for example.

Not every normable topology can be induced by an inner product, since Hilbert spaces are reflexive. So a necessary condition for a topology to be induced by an inner product is that the completion of the space with respect to the norm is reflexive. (I think, but can't quote a theorem at the moment, that reflexivity of the completion is not sufficient.)

If a topology can be induced by an inner product, that inner product is not unique (except again in the trivial case), a constant positive multiple of an inner product induces the same topology, and if the dimension is $\geqslant 2$, that is again not the only non-uniqueness, $\langle x,y\rangle':=\langle x,y\rangle + \lambda(x)\cdot \overline{\lambda(y)}$ is an inner product that is not a constant positive multiple of $\langle\,\cdot\,,\,\cdot\,\rangle$ and induces the same topology for every nonzero continuous linear form $\lambda$.

• AWESOME ANSWER!!! Favorisable. – C-Star-W-Star Feb 3 '14 at 14:45
• Thanks for always considering tvs – C-Star-W-Star Feb 3 '14 at 14:47

1) is simple. Every metric clearly does not admit at least one norm as a metric does not need to be defined on a vector space, while a norm does.

2) If you have a metric $d$ on a vector space that originates in a norm $|| . ||$, then the equation $d(x,y)=||x-y||$ defines the norm of any vector $x$ by $||x||=||x-0||=d(x,0)$

• Yes, but thats not the end of the story since there are hell lot of equivalent metrics – C-Star-W-Star Feb 3 '14 at 14:50
• Well it's the end story in the sense that if I give you a metric which also defines a norm, there is only one possible candidate for this norm. – 5xum Feb 3 '14 at 14:53
• But the (corrected) question is not given a metric but a metrizable topology ...but you're right if a metric is normable then there's one and only one corresponding norm...the same for given a norm there can be only one and only one scalar product inducing the norm - however that is standard questions in every fa-book ;) – C-Star-W-Star Feb 3 '14 at 16:52