Parallelogram law is a necessary condition for a Banach space to be Hilbert, but it is not sufficient. Can anyone give an example of that kind of normed space, which satisfies $$ \|x+y\|^2+\|x-y\|^2=2\big(\|x\|^2+\|y\|^2\big) $$ but is not an Hilbert space?
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1Take any incomplete inner product space (so necessarily infinite-dimensional), for example $C([0,1])$ with the usual integral inner product $\langle f,g\rangle=\int_0^1f(x)\overline{g(x)},dx$ (and the corresponding induced norm $|f|=\sqrt{\langle f,f\rangle}$). – peek-a-boo Aug 28 '23 at 10:34
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1None of the proofs require completeness of the vactor space (or at least they shouldn't) so the norm satisfying the parallelogram law is equivalent to having an inner product space. – Bruno B Aug 28 '23 at 10:38
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$$ \text{paralellogram law}\quad\Longleftrightarrow\quad \text{inner product space} \\ \text{Hilbert space}\quad\Longleftrightarrow\quad \text{inner product space and complete} $$
so the answer would be: an inner product space which is not compete.
GEdgar
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If you look at this thread, you will see that for a complete normed space it is actually sufficient using Polarization identities. Therefore, the only way to find your sort of example is by finding a non complete normed space, which should also be an innner product space.
Keen-ameteur
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