Prove that a norm that satisfies the parallelogram inequality defines an inner product In the opening chapter of a functional analysis book I had this question:
Prove that for a norm $||\cdot||$ that if for all vectors $u$ and $v$ it is true that $2||u||^2 + 2||v||^2 = ||u+v||^2 + ||u-v||^2$ then there is an inner product that results in this norm. (That is $\left<u,u\right> = ||u||^2$ ). 
So, I figured that if there is an inner product it ought to be definable by $\left<u,v\right> = \frac{||u+v||^2 - ||u-v||^2}{4}$ and I would proceed to show that this satisfies the conditions of inner products. 
However, when I go to prove that $\left<\lambda u, v\right> = \lambda \left<u,v\right>$ I'm completely stuck. I can't see any way of demonstrating this.
Does anyone know how I can complete this proof or of an alternate method of proving this?  
 A: Here are some hints.
The inner product is definable by means of the polarization identity,
$$
\langle x,y \rangle = \frac{1}{4}\left(||x+y||^2 - ||x-y||^2\right) + \frac{\rm i}{4}\left(||x+{\rm i}y||^2 - ||x-{\rm i}y||^2\right) .
$$
Showing strict positivity & symmetry is easy. Showing linearity under addition is half a page, but very doable (use the parallelogram law).
Showing linearity under scalar multiplication calls for a far deeper insight - at least the way I did it; maybe there are easier ones. Essentially, you have to build your way up from showing such linearity over $\mathbb{Q}_+$ to showing it for $\mathbb{R}_+$ (closure - use continuity), then to $\mathbb{R}$ (enough to see what happens under multiplication by $-1$) & then to $\mathbb{C}$ (split into real & imaginary parts; see what happens under multiplication by $\rm i$). So why is linearity over $\mathbb{Q}_+$ doable? Well, because it boils down to using additive linearity which you showed in half a page above :).
NB: I'd be very interested to know whether someone knows of a simpler proof.
A: Hint: Consider the case that $\langle u, u \rangle = ||u||^{2}$ and expand out $||u+v||^{2} + ||u-v||^{2}$ to see what you get
