In the Eculidean Space, one can automatically define a norm if a specific scalar product is given. On the contrary, it's not always true. A p-norm is a scalar product if and only if p=2. My question is what condition do we need in order to move back?


I'm not sure if this is what you're asking:

A norm on a vector space is induced from a scalar product if and only if the parallelogram law $\| x - y\|^{2} + \|x + y\|^{2} = 2\|x\|^{2} + 2\|y\|^{2}$ is satisfied.

One direction is obtained by expanding the scalar products. If the parallelogram law holds, then one can verify that the expression given by polarization is a scalar product inducing the norm: \[ \langle x, y\rangle = \frac{1}{4}( \|x + y\|^{2} - \|x - y\|^{2}) \] in the real case and \[ \langle x, y\rangle = \frac{1}{4} \sum_{k = 1}^{4} i^{k} \|x + i^{k} y\|^{2} \] in the complex case.

Added much later: For a good outline of the somewhat painful proof of the non-trivial direction, see Arturo's answer here.


Maybe more generally, while every norm gives us a metric, a metric is generated by a norm, I think, if the norm is translation-invariant.

  • $\begingroup$ This is not true (I guess you mean to say if the metric is translation invariant). For instance $d(f,g) = \sum_{n=0}^{\infty} 2^{-n} \frac{\|f^{(n)} - g^{(n)}\|_{\infty}}{1 + \|f^{(n)} - g^{(n)}\|_{\infty}}$ is a metric on the space of bounded smooth functions on $\mathbb{R}$ that cannot come from a norm. $\endgroup$ – t.b. Feb 4 '11 at 9:13
  • $\begingroup$ More generally, a vector space equipped with a countable family of semi-norms separating points can be equipped with a translation invariant metric. The topology can be induced by a norm if and only if there is a bounded convex neighborhood of $0$. This is proved using Minkowski functionals, see e.g. Rudin FA, Thm 1.39. $\endgroup$ – t.b. Feb 4 '11 at 9:14
  • $\begingroup$ Sorry, in my first comment I should have said $C^{\infty}([0,1])$, the space of smooth functions on the unit interval. $\endgroup$ – t.b. Feb 4 '11 at 9:33
  • $\begingroup$ Yes, of course, I meant the metric is translation-invariant. I have been having a senior day today; getting a head-start (way too soon) on becoming senile. Sorry. $\endgroup$ – user6600 Feb 4 '11 at 9:45

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