In chapter 9 of Rudin's "Real and Complex Analysis", the formula is stated without proof after he shows that $f\mapsto \hat f$ is an isometry of $L^2(\mathbb R)$ onto $L^2(\mathbb R),$ with Lebesgue measure.
Is Plancherel's formula an immediate consequence of Rudin's proof? If so, how? I know that isometries preserve inner products, but I think it requires at least a mention, if not an argument.
The usual proof (at least, the only one I have seen) of this identity without using directly the above-mentioned fact is a somewhat messy calculation, which I doubt fits into the situation presented by Rudin.
My thought is: $[f,g]=\int \hat f \overline {\hat g}\ $ defines an inner product, such that $[f,f]=\|f\|^2\ $ (using the isometry). From here, an application of the polarization identity gives the result: $\int f\overline g=\int \hat f \overline {\hat g}.$