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Let $(E, \|\cdot\|)$ be a real normed vector space such that for any $a,b\in E$, $$ \|x +y\|^2 + \|x-y\|^2 \geq 4 \|x\|\cdot \|y\| $$ I want to show that the norm is induced by an inner product. Any suggestion or references would be helpful.

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    $\begingroup$ This is not the right site for this question; you might ask instead at math.stackexchange. But you should ask yourself: how is the usual inner product defined in terms of the usual Euclidean norm, and vice-versa? This should give you an idea of what to try. $\endgroup$ – Todd Trimble Sep 7 '13 at 15:53
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    $\begingroup$ See Theorem 2 of Schoenberg, A remark on M. M. Day's characterization of inner-product spaces and a conjecture of L. M. Blumenthal, Proc. Amer. Math. Soc. 3 (1952), 961-964. @Todd Trimble: I think the question is a bit more subtle than what you suggest in that the parallelogram law requires an equality rather than an inequality. $\endgroup$ – Martin Sep 7 '13 at 17:43
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    $\begingroup$ @Martin: You're right. My response was a bit Pavlovian, shall we say. Apologies then to the OP (and I will move to reopen, as this deserves to be made public and not locked away in a journal). $\endgroup$ – Todd Trimble Sep 7 '13 at 18:30
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    $\begingroup$ I also voted to reopen it; thanks Martin for pointing to another nice paper by Schoenberg! $\endgroup$ – Suvrit Sep 7 '13 at 19:24
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If $E$ is to be a Hilbert space, a proof must establish more or less directly that the inequality implies the parallelogram law $\lVert x + y\rVert^2 + \lVert x-y\rVert^2 = 2\lVert x\rVert^2 + 2\lVert y\rVert^2$ for all $x,y \in E$. Since both, your hypothesis and the parallelogram law, are conditions on all $2$-dimensional subspaces of $E$, one can assume that $\dim E = 2$.

I. J. Schoenberg proves the following stronger result as Theorem 2 in A remark on M. M. Day's characterization of inner-product spaces and a conjecture of L. M. Blumenthal, Proc. Amer. Math. Soc. 3 (1952), 961-964:

If the inequality $\lVert x + y \rVert^2 + \lVert x-y\rVert^2 \geq 4$ holds for all unit vectors $x,y \in E$ then $E$ is an inner-product space.

After reducing to the case $\dim E = 2$, the proof proceeds by showing that the inequality forces the curve $\Gamma$ described by $\lVert x \rVert = 1$ to be an ellipse. To achieve this, Schoenberg shows that $\Gamma$ must coincide with the boundary of the John ellipse of the unit ball by a nice geometric argument. The proof is then completed by appealing to Day's theorem that the parallelogram law $\lVert x + y\rVert^2 + \lVert x-y\rVert^2 = 4$ for unit vectors characterizes inner-product spaces, see Theorem 2.1 in Some characterizations of inner-product spaces, Trans. Amer. Math. Soc. 62 (1947), 320-337.


Added: Your hypothesis appears as condition $(7)$ in Schoenberg's article. It is pointed out in footnote 5 that the parallelogram law and your condition are equivalent.

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