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If we have $X$ a real or complex IPS with inner product $\langle \cdot, \cdot \rangle_1$ and associated norm $\left\| \cdot \right\|_1$. If we have a norm $\left\| \cdot\right\|_2$ on $X$ which is equivalent to the norm $\left\| \cdot \right\|_1$, must $\left\| \cdot\right\|_2$ also come from an inner product??
I have tried messing around with the parallelogram law for the $2^{\text{nd}}$ norm but I couldn't seem to get anywhere. Thanks in advance!

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On a finite dimensional space any tow norms are equivalent. However, the p-norm satisfies parallelogram law only for $p=2$.

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  • $\begingroup$ Of course thank you! $\endgroup$ – SEWillB May 30 '17 at 21:33
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The answer is No. In $R^n$ Take usual (dot) inner product which generates Eucledean norm, i.e., $l_2$. Now take any other $l_p$-norm, $p \neq 2$. we know that only $l_p$-norm comes from inner product is $l_2$.

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