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?
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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. |
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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. |
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