Let $H$ be a Banach space with associated norm $\|-\|.$ Suppose that for any $x,y\in H,$ we have:
$$\|x+y\|^2+\|x-y\|^2=2\left(\|x\|^2+\|y\|^2\right),$$
which we call the parallelogram law.
Then it is a well-known standard fact that $H$ becomes a Hilbert space. This is true both for real and complex coefficients.
I managed to prove this fact for an Hilbert space over $\mathbb{R},$
defining the inner product
$$(x,y) \mapsto \langle x,y\rangle=\frac{\|x+y\|^2-\|x-y\|^2}{4}= \frac{\|x+y\|^2-\|x\|^2-\|y\|^2}{2}.$$
Question
How to prove this for the complex case?
The inner product should be in this case
$$(x,y)\mapsto\ \alpha(x,y)= \frac{1}{4} \left(\|x + y\|^2 – \|x-y\|^2 + i\|x + iy\|^2 -i\|x-iy\|^2 \right)= \langle x,y \rangle + i \langle x,iy \rangle$$
but I am not able to replicate the proof of the real case.
It would be awesome if there was a slick proof that used the real case to deduce the complex case, but I would still be happy with any kind of direct proof too. Since this is a standard result, if you can provide a reference where a detailed proof is given, that would be excellent too.
Best Answer
Let $\|\cdot\|: H\to \mathbb{R}$ be a norm in a complex Banach space $H$ which fulfills for all $x,y\in H$ the parallelogram law \begin{align*} \|x+y\|^2+\|x-y\|^2=2\left(\|x\|^2+\|y\|^2\right)\tag{1} \end{align*} We show the map $\alpha:H\times H\to\mathbb{C}$ defined by \begin{align*} \alpha(x,y)= \frac{1}{4} \left\{\|x + y\|^2 - \|x-y\|^2 + i\|x + iy\|^2 -i\|x-iy\|^2 \right\}\tag{2} \end{align*} fulfills for all $x,y,z \in H$ and $c\in\mathbb{C}$ \begin{align*} c\alpha(x,y)=\alpha(cx,y)\tag{3} \end{align*}
Comment:
In (4) we use the definition of $\alpha$ from (2).
In (5) we do some preparatory work in order to apply the parallelogram law.
In (6) we apply the parallelogram law (1).
Case $c\geq 0$:
Case $c\in\mathbb{R}$:
Case $c\in\mathbb{C}$:
Note: This post follows closely the proof provided in section 1.2 of Linear Operators in Hilbert Spaces by J. Weidmann.