Hilbert spaces possess some special properties, such as being uniformly convex and isometrically isomorphic to its dual space. Why does the assumption of an inner product lead to such strong statements, going so far as to identify all (separable) Hilbert spaces with $l_2(\mathbb{N})$? As a student, and at first glance, one would not perceive that they would differ so dramatically from the Banach spaces considered in the previous chapters with the weaker assumption of a norm.
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On the contrary, for one thing, the ability to define orthogonality gives Hilbert spaces a very special structure. – Matematleta Nov 17 '20 at 18:56
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One possible answer is that out of the $L_p$ spaces, the p-norm $\|\cdot\|_p$ satisfies the Parallelogram Law:
$$2\|u\|^2+2\|v\|^2=\|u+v\|^2+\|u-v\|^2$$
only for $p=2$. It is a well-known result that by satisfying the above, one can define an inner product via:
$$(u,v) := \frac{1}{2}\left(\|u\|^2+\|v\|^2-\|u-v\|^2\right).$$
Conversely, you can show that the induced norm $\|u,u\|:=\sqrt{(u,u)}$ must satisfy the Parallelogram Law.
Alex R.
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