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Let $V$ be a normed space with a norm $\|\cdot\|.$

The polarisation identity Wiki page states that

In a normed space $(V,\|\cdot\|),$ if the parallelogram law holds, then there is an inner product $\langle\cdot,\cdot\rangle$ on $V$ such that $\|x\|^2 = \langle x,x\rangle $ for all $x\in V$.

This post provided a sketch of a proof to the statement above.

I would like to ask whether the induced equation $\|x\|^2 = \langle x,x\rangle$ for inner product is unique, that is,

Question: If a norm satisfies the parallelogram law, then does there exist an inner product $\langle\cdot,\cdot\rangle'$ on $V$ such that $\|x\|^2 \neq \langle x,x\rangle'$ for some $x\in V?$

In other words, if a norm induces an inner product, is the inner product unique in the sense that $\|x\|^2 = \langle x,x\rangle?$

Idonknow
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    Do you mean to ask, if two inner products induce the same norm, are the inner products necessarily the same? If so, the answer is yes, and this follows from the polarization identity itself. – Travis Willse Sep 25 '18 at 13:46
  • Are you asking if there are inner products that don't induce the same norm? – Arnaud D. Sep 25 '18 at 13:47
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    The polarization equation defines the inner product via the norm. So the inner product (if it exists) is unique. – mfl Sep 25 '18 at 13:49
  • @Travis Yes, that is my question. More precisely, if a norm induces two inner products, must the norm related to the two inner product via $|x|^2 = \langle x,x\rangle?$ – Idonknow Sep 25 '18 at 14:19
  • Yes. How else would they be 'induced' by the norm? – Berci Sep 25 '18 at 14:23
  • There seems to be confusion surrounding the terminology. Inner products induce norms, not vice versa. That said, the parallelogram law allows you to identify the inner product that induced the norm (if it exists). As mentioned by @mfl, that inner product (again, if it exists!) is unique. – David Kozak Nov 12 '18 at 18:06

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