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Let $E$ a normed linear space such that: $$\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2 $$ Prove that the norm of $E$ is generate by the inner product $$\langle x,y \rangle =\frac{1}{4}\left(\|x+y\|^2-\|x-y\|^2\right)$$ To prove this we have to prove three things: Bilinearity, symmetry, and positive definite.

My approach: Two of this parts are easy to prove. But the Bilinearity I think is difficult. Any hint is welcome. Thanks.


marked as duplicate by PhoemueX, Hayden, drhab, user21467, Chris Janjigian Jul 1 '14 at 19:04

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    $\begingroup$ See : en.wikipedia.org/wiki/Parallelogram_law $\endgroup$ – user99680 Jun 30 '14 at 21:53
  • $\begingroup$ Are you sure that right at the beginning is $\|x + y \|^2 - \|x - y \|^2$? I believe that it should be $\|x+y \|^2 + \|x - y\|^2$. The parallelogram law states that "the sum of the diagonals's squares equals the sum of the squares of the sides". A norm is generated by an inner product if and only if it satisfies the parallelogram law. $\endgroup$ – Ivo Terek Jun 30 '14 at 21:53
  • $\begingroup$ Is true!!! Thanks $\endgroup$ – Valerin Jun 30 '14 at 22:04
  • $\begingroup$ See the changes a minus in the inner product definition. $\endgroup$ – Valerin Jun 30 '14 at 22:05
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    $\begingroup$ This is more or less the same question as here math.stackexchange.com/questions/21792/… $\endgroup$ – PhoemueX Jul 1 '14 at 18:10