Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 22F / c / Solution

Fix $z_0\in S^1$, and let ${\delta }_{z_0}\in C(S^1{)}^{\ast }$ be the point evaluation functional ${\delta }_{z_0}(f)=f(z_0)$. Define orbit averages in the dual space byEach ${\mu }_n$ has norm at most one and satisfies ${\mu }_n(1_{S^1})=1$. The space $C(S^1)$ is separable, so part (a) gives a subsequence converging pointwise to some $\mu \in C(S^1{)}^{\ast }$. In particular,For every $f\in C(S^1)$, the Cesaro average telescopes at the endpoints:whose absolute value is at most $2\Vert f{\Vert }_{\infty }/n$. Passing to the convergent subsequence provesThus $\mu$ is an invariant functional of a composition operator.

It need not be unique when $T$ is not the identity operator. For the nonidentity homeomorphism $\tau (z)=\overline{z}$, both $1$ and $-1$ are fixed. Hence the two distinct point evaluationshave value one on $1_{S^1}$ and are invariant under $T$. Therefore nonidentity does not imply uniqueness.