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A useful property of a norm induced by an inner product is the parallelogram equality: $2\|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2$. This is trivial to show. It is also the case that if the parallelogram equality holds for every pair of vectors in the space, then the norm is necessarily induced by an inner product. This fact is both harder to show and less useful than its converse.

I noticed that $$\begin{split}\text{RHS}&=\langle x+y, x+y\rangle + \langle x-y, x-y\rangle\\&=\langle x, x\rangle+\langle y, y\rangle +2\langle x, y\rangle+\langle x, x\rangle+\langle y, y\rangle -2\langle x, y\rangle\\&=2\langle x, x\rangle + 2\langle y, y\rangle\\&=\text{LHS}\end{split}$$

completing the proof of the parallelogram equality if the norm is induced by inner product.

I was wondering what the proof of "if paralellogram equality is satisfied then the norm is induced by an inner product" is. I'm looking at the second answer in here and don't understand the first step. Where did $\langle x,y\rangle = \frac 1 4 \left(\|x+y\|^2-\|x-y\|^2\right)$ come from? It looks like they are just proving that $\langle x, y\rangle$ is an inner product if it's defined in terms of the norm that satisfies the parallelogram equality. How does this show that the norm must be induced by an inner product?

Vons
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    Note that "the norm is induced by an inner product" is not the same thing as "the norm is the $L_2$ norm". The statement "if paralellogram inequality is satisfied then the norm is the $L_2$ norm" is false. – Ben Grossmann Jul 20 '21 at 22:35
  • @BenGrossmann I suspected that might not be true, but doesn't $|x|_2^2=\langle x, x\rangle$? I am assuming real numbers. What's the difference? – Vons Jul 20 '21 at 23:02
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    The dot-product is an example of an inner product, but it is not the only inner product. For instance, if $x\in \Bbb R^n$, then one example of an inner product is the function defined by $$ \langle x,y \rangle = x^Ty, $$ but this is not the only possible definition of $\langle x,y \rangle$. For example, with $n = 2$, the function $$ \langle x,y \rangle = 2x_1y_1 + 2x_2y_2 - x_1y_2 - x_2y_1 $$ is also an inner product. – Ben Grossmann Jul 20 '21 at 23:07
  • @BenGrossmann Ohh okay thank you – Vons Jul 20 '21 at 23:22
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    To add on @BenGrossmann comment, $\langle x,y\rangle$ is an inner product on $\Bbb R^n$, if and only if there exists positive definite $A\in\Bbb{R}^{n \times n}$ such that $ \langle x,y\rangle= x^\top Ay$. – Surb Jul 21 '21 at 00:05
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    When you have a norm $|\cdot|$ that satisfies the parallelogram law, then $\langle \cdot, \cdot \rangle$ defined by $\langle u,v \rangle := \tfrac14(|u+v|^2-|u-v|^2)$ is an inner product (the proof of this is the linked question) that induces $|\cdot|$ because $\langle x,x \rangle = \tfrac14(|x+x|^2-|x-x|^2) = \tfrac14 |2x|^2 = \tfrac14(2|x|)^2 = |x|^2$, so $|x| = \sqrt{\langle x,x \rangle}$. – azif00 Jul 21 '21 at 00:06
  • @azif00 So the linked post doesn't answer the question - why must the norm be induced by an inner product, just that it can be induced by an inner product. – Vons Jul 21 '21 at 00:09
  • Wait, actually, I think it does answer the question. Never mind. – Vons Jul 21 '21 at 00:16

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