Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 22F (Linear Analysis)

Choose a countable dense subset $\{x_1,x_2,\dots \}$ of the normed vector space $X$. For each fixed $j$, the scalar sequence $(f_n(x_j))$ is bounded becauseRepeated use of the Bolzano-Weierstrass theorem followed by a diagonal subsequence argument produces a subsequence $(f_n{)}_{n\in \Lambda }$ for which $f_n(x_j)$ converges for every $j$.

Let $D$ be the linear span of the $x_j$. On every $x\in D$, defineFinite linear combinations show that $f:D\to \mathbb{R}$ is linear, and passage to the limit in $|f_n(x)|\le \Vert x\Vert$ gives $|f(x)|\le \Vert x\Vert$. Thus $f$ is a bounded linear functional on the dense linear subspace $D$ and extends uniquely to an element of the dual space $X^{\ast }$ with $\Vert f\Vert \le 1$.

For arbitrary $x\in X$ and $y\in D$,First choose $y$ close to $x$, then let $n\in \Lambda$ tend to infinity. This provesThis is the Sequential Banach-Alaoglu theorem for a separable predual.

The fixed-point space $I=\ker (U-I)$ is closed. If $y\in I$ and $w=Ux-x\in W$, then unitarity givesso $I\perp W$.

Conversely, $z\in W^{\perp }$ exactly whenfor every $x\in H$. Thus $U^{\ast }z=z$, equivalently $Uz=z$, and $z\in I$. Hence $W^{\perp }=I$, so the double orthogonal complement identity givesThe orthogonal decomposition by a closed subspace now yields

Define the Cesaro averageFor $y\in I$, $A_{n}y=y$. For a generator $w=Uz-z$ of $W$, the sum telescopes:Moreover $\Vert A_n\Vert \le 1$, so approximation extends this convergence from $W$ to $\overline{W}$. Writing $x=Px+(I-P)x$ according to the orthogonal decomposition gives the Von Neumann mean ergodic theorem

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.