I am trying to prove the following:
Let $G$ be a finite group and let $\pi : G \to U(\mathcal{H})$ and $\rho : G \to U(\mathcal{K})$ be irreducible unitary representations. If $\pi$ is weakly contained in $\rho$, then $\pi$ is contained in $\rho$.
I am allowed to use the following fact:
If $\pi$ and $\rho$ are inequivalent irreducible representations of $G$, then $tr \circ \pi$ and $tr \circ \rho$ are orthogonal functions in $\ell^2(G)$.
So, arguing contrapositively, if $\pi$ is not contained in $\rho$, then there is no isometry $V : \mathcal{H} \to \mathcal{K}$ such that $V \pi ( \cdot ) V^* = \rho (\cdot)$. In particular, it means that there is no unitary operator with the stated property, so $\pi$ and $\rho$ are unitarily inequivalent representations. Hence, $tr \circ \pi$ and $tr \circ \rho$ are orthogonal functions in $\ell^2(G)$. But I'm having trouble seeing how this implies that $\pi$ is not weakly contained in $\rho$.