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Polarisation identity states that $\langle x, y\rangle = \frac{1}{4}\|x+y\|^2 - \frac{1}{4} \| x - y \|^2$. And this is proven by expanding the terms on the right using $\|x\|^2 = \langle x,x\rangle $ which assumes that the norm on that inner product space is defined that way.

Then we have the parallelogram law that states that $\|x+y\|^2 = \|x\|^2 + \|y\|^2$ which is also proven using $\|x\|^2 = \langle x,x\rangle $.

And then we have this statement that says let $\| . \|$ be a norm on a real vector space $V$ satisfying the parallelogram law and define $\langle x, y\rangle = \frac{1}{4}\|x+y\|^2 - \frac{1}{4}\| x - y \|^2$. Then $\langle\ ,\ \rangle$ defines an inner product on $V$ such that $\|x\|^2 = \langle x,x\rangle $.

I find it the last statement to be a circular argument because parallelogram is proven using $\|x\|^2 = \langle x,x\rangle $ and now it says if the norm satisfy parallelogram law, then $\|x\|^2 = \langle x,x\rangle $. What am I misunderstanding here?

Is it true that if parallelogram law holds, then $\|x\|^2 = \langle x,x\rangle $ if and only if polarisation identity holds?

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Is it true that if parallelogram law holds, then ∥x∥^2=⟨x,x⟩ if and only if polarisation identity holds?

This is not what the question is asking. The question is the following: If the parallelogram law holds for some norm on our vector space and we define a binary operation on our vector space to be ⟨x,y⟩=1/4∥x+y∥^2−1/4∥x−y∥^2 then will our operation define an inner product on V such that ∥x∥^2=⟨x,x⟩? The answer yes but to prove it we need to show that if you take ⟨x,y⟩=1/4∥x+y∥^2−1/4∥x−y∥^2 with x=y and assuming the parallelogram law, you will get back ∥x∥^2, thus ∥x∥^2=⟨x,x⟩.

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  • $\begingroup$ But it is also true that if ||x||^2 = <x,x> , then ⟨x,y⟩=1/4∥x+y∥^2−1/4∥x−y∥^2 right? Because we just have to expand the RHS using ||x||^2 = <x,x> , so gives me an if and only if relationship. Isn't it? $\endgroup$ – user10024395 Oct 31 '14 at 7:51

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