Norm induced by inner product?

1. Apr 4, 2012

owlpride

On a finite-dimensional vector space over R or C, is every norm induced by an inner product?

I know that this can fail for infinite-dimensional vector spaces. It just struck me that we never made a distinction between normed vector spaces and inner product spaces in my linear algebra course on finite-dimensional vector spaces.

Why I actually care about it: I wonder why the unit sphere in a finite-dimensional normed vector space is weakly closed. Obviously the statement should be that a sequence in a finite-dimensional space converges weakly if and only if it converges strongly, but I'm not sure how to go about this without using an inner product.

2. Apr 4, 2012

morphism

No. The parallelogram law can and does fail for a variety of norms (e.g. all the p-norms except when p=2).

On a finite-dimensional vector space, all norm topologies are equivalent. So you can always assume your norm topology comes from a norm that is induced by an inner product. (In fact it's also true that in this case the weak topology and any norm topology are the same.)

3. Apr 6, 2012

wisvuze

It should be noted that if a norm satisfies the parallelogram law, then you can always uncover an inner product from the norm via the "polarization identity"

1/4 ( | v + w |^2 - | v - w |^2 ) = < v , w >

4. Apr 6, 2012

morphism

Just for the sake of completeness: that form of the polarization identity is only valid if we're working over R; over C, the identity becomes $$\frac14 (\|v+w\|^2 -\|v-w\|^2 + i\|v+iw\|^2 - i\|v-iw\|^2) = \langle v, w\rangle.$$