Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 / 5

Assume the invariant set $C$ is nonempty, as required for any fixed-point assertion. For $x\in C$, put $r(x)={\sup }_{c\in C}\Vert x-c\Vert$ and $r_{\ast }={\inf }_{x\in C}r(x)$. Boundedness makes these finite. The parallelogram identity gives, for $x,y\in C$,The midpoint lies in $C$, so a minimizing sequence $x_n$ satisfiesSince $C$ is closed in a Hilbert space, the sequence converges to $x_{\ast }\in C$. The function $r$ is 1-Lipschitz, hence $r(x_{\ast })=r_{\ast }$. The same midpoint inequality proves uniqueness of this minimizing center.

An affine isometry from the group maps $C$ onto itself, since its inverse also preserves $C$. Therefore it preserves $r$ and sends its unique minimizer to another minimizer. Uniqueness givesThis proof applies to a closed bounded convex subset, the substantive meaning of the printed “convex subspace”; if it is literally a bounded vector subspace, it is just the zero subspace.

The topology needed here is the weak-star topology $\sigma (E^{\ast },E)$, meaning pointwise convergence on $E$. With the literal weak topology $\sigma (E^{\ast },E^{\ast \ast })$, the claimed compactness is false. For example, $E=c_0$ is separable and $E^{\ast }={\ell }^1$. Its unit vectors $e_n$ have no weak cluster point: every coordinate functional forces a cluster point to have all coordinates zero, while the bounded functional $x\mapsto {\sum }_j{x}_j$ has value one throughout. Thus the dual unit ball is not weakly compact. We prove the intended weak-star assertion in full.

Choose a countable dense set $(x_k)$ in the unit ball of $E$. On $E_1^{\ast }$ defineEach term is at most $2\cdot 2^{-k}$, so the series converges. The triangle inequality is immediate, and $d=0$ forces equality on the dense set and hence on all of $E$. It is therefore a metric. Convergence in this metric is equivalent to convergence on every $x_k$: one direction follows from any individual term, and the other by bounding the series tail uniformly. Because $\Vert \varphi \Vert ,\Vert \psi \Vert \le 1$, approximation of any unit vector by $x_k$ then gives convergence on all of $E$. The same finite-tail and approximation argument for neighborhoods proves equality with the weak-star topology, not just agreement of its convergent sequences.

Given a sequence $({\varphi }_n)$ in this ball, diagonal selection gives a subsequence converging at every $x_k$, since each coordinate lies in a compact disk. Uniform norm bounds then make it pointwise Cauchy at every $x\in E$. Define $\varphi (x)$ as the limit. Linearity passes to the limit, and $|\varphi (x)|\le \Vert x\Vert$, so $\varphi \in E_1^{\ast }$. The subsequence converges to it in the metric. Thus this metric space is sequentially compact, hence compact, proving

Use the same intended weak-star topology as in (b), and assume $C\ne \varnothing$. The induced operator is $T=S^{\ast }$, so $T\varphi (x)=\varphi (Sx)$; it is weak-star continuous because evaluation after applying $T$ is still evaluation at a vector of $E$.

Choose $\varphi \in C$ and form the averagesInvariance and convexity put every ${\varphi }_n$ in $C$. Weak-star closedness and (b) make $C$ compact, so some subsequence converges to $\psi \in C$. Telescoping givesbecause both orbit points lie in the dual unit ball. Weak-star continuity now gives $T\psi -\psi =0$. Hence $\boxed{T\text{ has a fixed point in }C.}$ This is the weak-star fixed point theorem for an adjoint operator; it does not require $T$ to be a contraction on the entire dual space.

If “weakly closed” is interpreted literally as closed for $\sigma (E^{\ast },E^{\ast \ast })$, the result can fail. Take $E=c_0$, let $S$ be the left shift and $T=S^{\ast }$ the right shift on ${\ell }^1$. The setis a nonempty weakly closed convex subset of the unit ball: its defining coordinate inequalities and sum equality are weakly closed. It is invariant under the right shift. But a fixed vector of that shift has first coordinate zero and then every coordinate zero, so none belongs to $C$. This supplies a counterexample to the literal topology and explains why the weak-star qualification is essential.

Put $F=\ker (I-U)$. For a unitary operator, $\ker (I-U^{\ast })=F$, soThis follows from the general identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$. The averaging operators have $\Vert T_n\Vert \le 1$ and equal the identity on $F$. On a vector $(I-U)\eta$, telescoping givesThe uniform norm bound extends this convergence to the closure of the range. Thus $T_n$ tends to identity on $F$ and to zero on its orthogonal complement. Consequentlywhich is exactly convergence in the strong operator topology. This proves the Von Neumann mean ergodic theorem in the requested unitary case.