I'm going through Peter Walters's Introduction to Ergodic Theory (pg. 50) and I'm having trouble understanding the proof of Theorem 1.28, which says that for an endomorphism of a compact group strong mixing, weak mixing and ergodicity are equivalent.
In particular, the proof starts off with the assumption $G$ is abelian. It then says it suffices to show that if the endomorphism $A: G \to G$ is ergodic, then $A$ is strong-mixing, which is clear. However, why is the following true?
If $\lambda, \delta \in \widehat{G}$, then $(U_A^n \lambda, \delta) = 0$ eventually unless $\lambda = \delta \equiv 1$.
Here, for our purposes $U_A: L^2(m) \to L^2(m)$ is the induced operator defined by $U_A f = f \circ A$, where $f \in L^2(G, \mathscr{B}, m)$. It is known that this operator is linear and unitary, and that $U_A(fg) = U_A(f) U_A(g)$ and $U_A c = c$ when $c$ is constant.
$\widehat{G}$ is the character group of $G$, and we know that the members of $\widehat{G}$ form an orthonormal basis for $L^2(m)$, where $m$ is the Haar measure.
There is also an earlier theorem (Theorem 1.10), which says that if $G$ is a compact abelian group equipped with normalized Haar measure $m$ and $A: G \to G$ is a continuous surjective endomorphism of $G$, then $A$ is ergodic iff the trivial character $\lambda \equiv 1$ is the only $\lambda \in G$ that satisfies $U_A\lambda =\lambda \circ A^n = \lambda$ for some $n > 0$.
So basically we know that $\lambda, U_A \lambda, U^2_A \lambda, \dots$ are all distinct, but how does it follow that eventually $(U_A^n \lambda, \delta) = 0$? I get that this isn't true for $\lambda = \delta \equiv 1$, but why does it hold for all other choices of characters?