I am trying to work out the details of the statement:
A norm $\|{\cdot}\|$ is induced by inner product if and only if it satisfies the parallelogram law.
For the converse direction, the inner product is defined as \begin{align*} \langle x, y \rangle =\frac{1}{4} \left[ \|x+y\|^2 - \|x-y\|^2 + i\|x+iy\| - i \|x-iy\|^2 \right]. \end{align*}
Basically I am following the proof posted in this post Norms Induced by Inner Products and the Parallelogram Law by t.b. but with different strategy. That is I am attacking the complex case directly. It does not make a very big difference until the homogeneity in the first argument, i.e., $\varphi(\alpha) = \langle \alpha x, y \rangle - \alpha \langle x, y \rangle = 0$ for all $\alpha \in \mathbb C$. I have shown this equality holds for all $\alpha = r+ i s$ where $r, s \in \mathbb{Q}$ and $i$ is the imaginary symbol. I make the following claims:
(1) $\varphi \colon \mathbb C \to \mathbb C$ is a continuous map for fixed $x, y \in V$. The argument is for fixed $x, y \in V$, $\varphi$ can be viewed as following composition: \begin{align*} \alpha \mapsto (\alpha x, x, y) \mapsto \langle \alpha x, y \rangle + \alpha \langle x, y \rangle, \end{align*} The first map is continuous as $V$ is a topological vector space and the second is also continuous because inner product is defined as sum of norm functions.
(2) $\mathbb Q + i \mathbb Q$ is dense in $\mathbb C$ and we know the restriction of $\varphi|_{\mathbb Q + i\mathbb Q}$ is identically $0$. So the continuous extension $\varphi$ must be identically $0$.
I haven't taken too much complex analysis before. So (1): could anyone point out whether there are flaws in my argument; (2) Is $\mathbb Q + i \mathbb Q$ standard notation in complex analysis; (3) My thinking for $\mathbb Q + i \mathbb Q$ being dense in $\mathbb C$ is if we identify $\mathbb C$ by $\mathbb R \times \mathbb R$, then $\mathbb Q \times \mathbb Q$ is dense in $\mathbb R \times \mathbb R$. Is this OK or there exists more standard argument?
Many thanks in advance.
