In Jacques Faurat's book, Analysis on Lie Groups. An introduction, on page 105, he states the following theorem.
Theorem 6.3.3. Let $\pi$ be an irreducible unitary $\mathbb{C}$-linear representation of a compact group $G$ on a complex Euclidean vector space $\mathcal{H}$ with dimension $d_{\pi}$.
Then, for $u,v\in\mathcal{H}$,
$$\int_{G}\vert\langle{\pi(g)u,v}\rangle\vert^{2}\, \mu(dg) = \frac{1}{d_{\pi}}\|{u}\|^{2}\|v\|^{2},$$
and, by polarisation, for $u,v, u^{\prime}, v^{\prime}\in\mathcal{H}$,
$$\int_{G}\langle{\pi(g)u,v}\rangle \overline{\langle{\pi(g)u^{\prime},v^{\prime}}\rangle} \, \mu(dg) = \frac{1}{d_{\pi}}\langle{u,u^{\prime}}\rangle\overline{\langle{v,v^{\prime}}\rangle}$$
In the book, Jacques Faraut gives a very clear proof of the first equality, and one could say that the second one is left as an exercise for the reader, I have been trying to give a proof using the polarization identity for the inner product of $L^2( G)$, but I get nothing. Could someone give me some other suggestion, or some reference where I can find some outline of the proof of the second equality?
We shall use the polarization identity for inner spaces over the complex field, that is $$\langle x,y \rangle=\frac{1}{4} (\|x+y\|^2−\|x−y\|^2+\rm{i} \|x+iy\|^2−\rm{i} \|x−iy\|^2).$$ Thus, we can write \begin{align} \langle u,u' \rangle \overline{\langle v, v' \rangle} = \frac{1}{8}&(\|u+u'\|^2−\|u−u'\|^2+\rm{i} \|u+iu'\|^2−\rm{i} \|u−iu'\| ) \\ & (\|v+v'\|^2−\|v−v'\|^2+\rm{i} \|v+iv'\|^2−\rm{i} \|v−iv'\|). \end{align} Now, expand the product, use the first equation on each summand and reverse the previous steps.