Orbital integral for matrix coefficients I am currently aiming at estimating orbital integrals. Maybe surprizingly, I hope for some help in the compact case (ramified places), in proving the usual formula

$$O_\gamma(f) = \int_G f(x^{-1}\gamma x)dx = \Theta_\pi(\gamma)  dim(\pi)^{-1}$$

where $f$ is a suitably normalized matrix coefficient for $\pi$. There are general proofs of this fact (for instance in Arthur or Kottwitz, using Peter-Weyl theorem or properties of supercuspidal representations), yet I would rather find an easiest way to do so in the compact case.
For instance, it is quite easy to prove 
$$O_\gamma(f) = \Theta_\pi (f) \Theta_\pi (\gamma)$$
so is there any way to see that $\Theta_\pi(f) = dim(\pi)^{-1}$ (if true) ?
Thank you in advance for any clue or reference.
 A: You said in comments that you are OK with using orthogonality of matrix coefficients, but let's suppose you changed your mind.  Let $\langle\cdot, \cdot\rangle$ be a $G$-invariant pairing on the space $V$ of $\pi$, and let $\mathrm dg$ be the Haar measure on $G$ normalised to give it total mass $1$.  What follows is all very standard, so I assume it's not what you want, but let's have something out here so that you can clarify where it falls short.
Note that, for each $b, w \in V$, we have that $v \otimes v' \mapsto {\displaystyle\int} \langle g b, v'\rangle\cdot\overline{\langle g w, v\rangle}\mathrm dg$ is an element of $\operatorname{Hom}_G(V \otimes V, \mathbb C)$, hence, by Schur's lemma, a scalar multiple of $\langle\cdot, \cdot\rangle$; and that the map sending $b \otimes w$ to this scalar is itself is an element of $\operatorname{Hom}_G(V \otimes V, \mathbb C)$, hence again a scalar multiple of $\langle\cdot, \cdot\rangle$.  Call this scalar $c$.  (Note the illustration of Yemon Choi's claim that Schur's lemma lies at the bottom of everything!)
Now
$$
\Theta_\pi(g) = \sum_{b \in B} \langle g b, b\rangle,
$$
where $B$ is an orthonormal basis of the space $V$ of $\pi$; so, if we write $f_{v, w}$ for the matrix coefficient $g \mapsto \langle v, g w\rangle$, then
\begin{align*}
\Theta_\pi(f_{v, w})
& {}= \sum_{b \in B} \int \langle g b, b\rangle\cdot\langle v, g w\rangle\mathrm dg \\
& {}= \sum_{b \in B} \int \langle g b, b\rangle\cdot\overline{\langle g w, v\rangle}\mathrm dg \\
& {}= \sum_{b \in B} c\langle b, w\rangle\cdot\overline{\langle b, v\rangle} \\
& {}= c\langle v, w\rangle = c f_{v, w}(1).
\end{align*}
On the other hand,
\begin{align*}
f_{v, w}(1)
& {}= \int \langle g v, g w\rangle\mathrm dg \\
& {}= \int \sum_{b \in B} \langle g v, b\rangle\cdot\overline{\langle g w, b\rangle}\mathrm dg \\
& {}= \sum_{b \in B} c\langle v, w\rangle\cdot\langle b, b\rangle \\
& {}= c\cdot\lvert B\rvert\cdot f_{v, w}(1) = c\cdot\dim(\pi)f_{v, w}(1).
\end{align*}
Since this is true for all $v, w \in V$, we have that $c = \dim(\pi)^{-1}$.
