# Question on Norm and Inner Product

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?