I'm reading through Diaconis' notes Group Representations in Probability and Statistics, and working on the following exercise, where $\hat{P}$ denotes the Fourier transform of $P$:
Recall that the uniform distribution is defined by $U(s) = \frac{1}{|G|}$ where $|G|$ is the order of a group $G$. Then at the trivial representation $\hat{U}(\rho) = 1$ and at any other non-trivial irreducible representation $\hat{U}(\rho) = 0$.
I think I have proof by contradiction, but I'm a little unsure about it and would love some feedback. Also, if there is a direct proof I would love to hear about it.
Proof: First, let $\rho$ be the trivial representation. Then $$\hat{U}(\rho) = \sum_s U(s) \rho(s) = \sum_s \frac{1}{|G|} = 1.$$
Next, let $\rho$ be a nontrivial irreducible representation carried by $V$. First, note that $\operatorname{im}\hat{U}(\rho)$ is an invariant subspace of $V$: $$\rho(t)\hat{U}(\rho)(v) = \rho(t)\left( \frac{1}{|G|} \sum_s \rho(s) \right) v = \left( \frac{1}{|G|} \sum_s \rho(t) \rho(s) \right) v = \hat{U}(\rho)(v)$$ for all $v \in V$. Thus, $\operatorname{im}\hat{U}(\rho) = 0$ or $V$. Suppose the latter for the sake of contradiction. Then $\hat{U}(\rho)$ is an isomorphism. Let $v \in V$ and $s \in G$ such that $\rho(s)v \neq v$. Such a pair of $v, s$ exists because otherwise, the span of $v$ would be an invariant subspace corresponding to the trivial representation. Then $$ \hat{U}(\rho)(\rho(s) v) = \hat{U}(\rho) v$$ by translation, contradicting the injectivity of $\hat{U}(\rho)$. Thus, $\operatorname{im}\hat{U}(\rho) = 0$, so $\hat{U}(\rho) = 0$. $\square$
Thanks!