Show that $\int_{-\infty}^{\infty}f(x)\overline {g(x)}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat{f}(\mu)\overline{\hat{g}(\mu)}d\mu.$ Given: Show that if f(x) is defined as:
The Fourier transform $\hat(\mu)$ of a function $f(x)$ specified on $\mathbb R$ is often defined by the formula:
$$\hat{f}(\mu) = \int_{-\infty}^{\infty}e^{i\mu x}f(x)dx \quad for \;\mu \in \mathbb C$$
Also, $g(x)$ is just another function of the same form as $f(x)$. Then:
Question:
$$\int_{-\infty}^{\infty}f(x)\overline {g(x)}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat{f}(\mu)\overline{\hat{g}(\mu)}d\mu.$$
Intuition:
I'm trying to use the inner product space $\sum_{k=1}^4i^k\langle f+i^kg, f+i^kg\rangle = 4\langle f,g\rangle$ to solve this.
More precisely, use this Lemma:

(fixed intuition)
 A: 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
$$
         \int|\hat{f}+i^n\hat{g}|^2ds=\int|\widehat{f+i^n g}|^2ds=\int|f+i^ng|^2dx
$$
A: Substitute your expression for $\hat{f}$ into the RHS integral.  Now substitute the similar expression for $\bar{\hat{g}}$ as well, and remember that this is the complex conjugate.  For clarity, make the integration variable on $g$ equal to, for example $x^\prime$.  Now, you will have two exponential factors that can be combined into $e^{i\mu(x-x^\prime)}$.  Pull that factor out into the integral over $\mu$.  Do you know what
$\displaystyle
\int_{-\infty}^{\infty}d\mu e^{iu(x-x^\prime)} $
gives?
