The Jordan-von Neumann Theorem states that "If a normed space satisfies the parallelogram law
$2||x||^2 + 2||y||^2 = ||x + y||^2 + ||x - y||^2$
then we can define a inner product for it by the polarisation identity
$\langle x, y \rangle = 1/4(||x + y||^2 - ||x - y||^2 +i||x + iy||^2 - i||x - iy||^2)$
and this is the unique inner product which induces the original norm $||x|| = \langle x, x \rangle^{1/2}$ ".
So, knowing that $L^2$ is Banach with a norm $||f|| = (\int |f|^2)^{1/2} $ (where $||f||$ is the norm of the equivalence class represented by $f$) can we show that it satisfies the parallelogram law and then construct the inner product ? I think so, but I haven't found a proof of this anywhere so I would appreciate confirmation or correction of my own proof which follows.
(1) The sum of two (Lebesgue) integrable functions is integrable and the integral of the sum = sum of the integrals.
(2) Re-express the norm: $|f|^2 = ff^*$ (where $f^*$ is the complex conjugate of $f$). So if $|f|^2$ is integrable then so is $ff^*$ and then $||f|| = (\int ff^*)^{1/2} $
(3) Parallelogram Law: Consider $||f + g||^2 + ||f - g||^2 $ Since $f, g$ are elements of Banach $L^2$ then so are $f + g, f - g$ and the norms and corresponding integrals exist.
$||f + g||^2 + ||f - g||^2 = \int(f + g)(f + g)^* + \int(f - g)(f - g)^*$
And by (1), $ = \int(f + g)(f + g)^* + (f - g)(f - g)^*$
Now I can expand and simplify the integrand without worrying if individual terms are integrable knowing that the total remains integrable, and when I do it reduces to $ \int (2ff^* + 2gg^*)$
And since $ff^*, gg^*$ are integrable (being the norms of $f, g$) then again by (1) $||f + g||^2 + ||f - g||^2 = \int (2ff^* + 2gg^*) = 2\int ff^* + 2\int gg^* = 2||f||^2 + 2||g||^2$
So the parallelogram law is satisfied.
(4) Polarisation: by Jordan-von Neumann Theorem, I can now define an inner product
$\langle f, g \rangle = 1/4(||f + g||^2 - ||f - g||^2 +i||f + ig||^2 - i||f - ig||^2)$ where $\langle f, g \rangle$ denotes the inner product of the equivalence classes $[f], [g]$.
Applying linerity of integrals and expansion and simplification of terms which is too tedious to type in here, this expression reduces (not surprisingly) to $\int fg^*$
(5) Hilbert: starting from $L^2$ being Banach it now has an inner product $\langle f, g \rangle = \int fg^*$ and is therefore a Hilbert space.