# Prove that the norm of $E$ is generate by the inner product $\langle x,y \rangle =\frac{1}{4}\left(||x+y|^2-||x-y||^2\right)$ [duplicate]

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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 JanjigianJul 1 '14 at 19:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – user99680 Jun 30 '14 at 21:53
• 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. – Ivo Terek Jun 30 '14 at 21:53
• Is true!!! Thanks – Valerin Jun 30 '14 at 22:04
• See the changes a minus in the inner product definition. – Valerin Jun 30 '14 at 22:05
• This is more or less the same question as here math.stackexchange.com/questions/21792/… – PhoemueX Jul 1 '14 at 18:10