TITLE: Norms Induced by Inner Products and the Parallelogram Law
QUESTION [273 upvotes]: Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$.
It's not hard to show that if $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality
$$ 2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2 $$
holds for all pairs $u, v \in V$.
I'm having difficulty with the converse.  Assuming the parallelogram identity, I'm able to convince myself that the inner product should be
$$ \langle u, v \rangle = \frac{\lVert u\rVert^2 + \lVert v\rVert^2 - \lVert u - v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u\rVert^2 - \lVert v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u - v\rVert^2}{4} $$
I cannot seem to get that $\langle \lambda u,v \rangle = \lambda \langle u,v \rangle$ for $\lambda \in \mathbb{R}$.  How would one go about proving this?

REPLY [10 votes]: One should also convince oneself that:
$$\langle\cdot,\cdot\rangle\to\|\cdot\|\to\langle\cdot,\cdot\rangle$$
$$\|\cdot\|\to\langle\cdot,\cdot\rangle\to\|\cdot\|$$
(Otherwise really bad things could happen...)
Luckily, this can be checked rather easily:
$$\|x\|'=\sqrt{\frac{1}{4}\left(\|x+x\|^2-\|x-x\|^2\right)}=\|x\|$$
$$\langle x,y\rangle'=\frac{1}{4}\left(\sqrt{\langle x+y,x+y\rangle}^2-\sqrt{\langle x-y,x-y\rangle}^2\right)=\langle x,y\rangle$$