Proof inner-product on square-integrable products is always well defined?

Question

The inner product of any two square-integrable functions $f$, $g$ is defined to be $\int f^*(x) g(x) dx$ (forgive my ugly maths layout). But how do we know this is never ill-defined? Why does the existence of $\int |f(x)|^2 dx$ and $\int |g(x)|^2 dx$ ensure this? Obviously when this does exist, we have a sensible and finite inner product, but couldn't a nasty choice of $f$ and $g$ result in some function $f^* g$ which doesn't even have an infinite integral?

Answer 1

Suppose $f,\,g$ are square integrable. For $n\in\Bbb Z$,$$\begin{align}0&\le\int|f+i^ng|^2dx\\&\le\int(|f|+|g|)^2dx\\&=\int(2(|f|^2+|g|^2)-|f-g|^2)dx\\&\le2\int(|f|^2+|g|^2)dx\end{align}$$(the second $\le$ uses the triangle inequality), so $f+i^ng$ is square-integrable. The polarization identity states $\langle f,\,g\rangle=\frac14\sum_{n=0}^3i^{-n}|f+i^ng|^2$.