Problem: Let $(X, || \cdot ||)$ be a normed space. Prove that the following are equivalent:
i) There exists an inner product $< \cdot , \cdot >$ on $X$ such that $||x|| = \sqrt{< x, x >}$ for every $x \in X$.
ii) The $|| \cdot ||$ satisfies the parallelogram law, i.e., \begin{align} ||x + y||^2 + ||x - y||^2 = 2||x||^2 + 2||y||^2 for every $x,y \in X$. \end{align} (For the implication $(ii) \Rightarrow (i)$, set $< x,y > = \frac{1}{4}(||x + y||^2 - ||x - y||^2)$, for every $x,y \in X$.)
I've been given a hint by my professor for proving (ii) implies (i), which involves defining an inner product based on the norm using the formula: <x,y>=41(∣∣x+y∣∣2−∣∣x−y∣∣2) for every x,y∈X.
However, I'm struggling to complete the proof in both directions. Here's what I've done so far:
(i) implies (ii): I'm not sure how the properties of an inner product would lead to the parallelogram law. Can someone guide me through this implication?
(ii) implies (i): I understand the concept behind the given formula for the inner product, but how do I rigorously show that this actually defines a valid inner product? I need to prove that it satisfies the following properties for all x,y,z∈X and scalar α:
Positive definiteness: <x,x>≥0 and <x,x>=0 if and only if x=0.
Symmetry: <x,y>=<y,x>.
Linearity in the first argument: <αx,y>=α<x,y>.
Any help or guidance on completing this proof would be greatly appreciated!
