Show that $\int_{-\infty}^{\infty}f(x)\overline {g(x)}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat{f}(\mu)\overline{\hat{g}(\mu)}d\mu.$

This is the polarization identity. It's use is to reconstruct the inner product from knowledge of the norm. Presumably you are trying to go from $$ \int_{-\infty}^{\infty}|\hat{f}|^2ds = \int_{-\infty}^{\infty}|f|^2dx $$ to $$ \int_{-\infty}^{\infty}\hat{f}\overline{\hat{g}}ds = \int_{-\infty}^{\infty}f\overline{g}dx $$ That's straightforward application of the polarization identity because $$ f\,\overline{g}=\frac{1}{4}\sum_{n=0}^{3}i^n|f+i^ng|^2 \\ \hat{f}\overline{\hat{g}}=\frac{1}{4}\sum_{n=0}^{3}i^n|\hat{f}+i^n\hat{g}|^2 $$ and $$ \int|\hat{f}+i^n\hat{g}|^2ds=\int|\widehat{f+i^n g}|^2ds=\int|f+i^ng|^2dx $$