Topological Modules/Orthogonal projection

Theorem (Von Neumann ergodic theorem):

Let $H$ be Hilbert space, and let $U:H\to H$ be a unitary operator. Further, let the orthogonal projection onto the space $W:=\{x\in H|Ux=x\}$ be given by $P_W:H\to H$. Then

$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{k=1}^n{U}^k=P_W$,

where the limit is taken with respect to the operator norm on $B(H)$, the space of bounded operators on $H$. Moreover, the inequality

$\Vert \frac{1}{n}\sum _{k=1}^n{U}^k-P_W\Vert \le \frac{2}{n}$

is a valid estimate for the convergence rate.

Proof: Suppose first that $x\in H$ and $z\in W$. Then

$\begin{array}{cc}⟨\frac{1}{n}\sum _{k=1}^n{U}^{k}x-P_W(x),z⟩& =⟨\frac{1}{n}\sum _{k=1}^n{U}^{k}x-P_W(x),z⟩+\frac{n}{n}⟨P_W(x)-x,z⟩\\ & =⟨\frac{1}{n}\sum _{k=1}^n{U}^{k}x-x,z⟩\\ & =\frac{1}{n}\sum _{k=1}^n(⟨U^{k}x,z⟩-⟨x,z⟩)\\ & \stackrel{z\in W}{=}\frac{1}{n}\sum _{k=1}^n(\stackrel{=⟨x,z⟩}{\overbrace{⟨U^{k}x,U^{k}z⟩}}-⟨x,z⟩)=0.\end{array}$

Further, if we set

$y_n:=\frac{1}{n}\sum _{k=1}^n{U}^{k}x$,

we obtain

$\begin{array}{cc}\Vert U{y}_n-y_n{\Vert }^2& =⟨y_n-U{y}_n,y_n-U{y}_n⟩\\ & =\frac{1}{n^2}⟨Ux-U^{n+1}x,Ux-U^{n+1}x⟩\\ & =\frac{1}{n^2}(⟨Ux,Ux⟩-⟨U^{n+1}x,Ux⟩-⟨Ux,U^{n+1}x⟩+⟨U^{n+1}x,U^{n+1}x⟩)\\ & \stackrel{\text{Cauchy‒Schwarz}}{\le }\frac{4\Vert x{\Vert }^2}{n^2}.\end{array}$

If now the sequence $(y_n{)}_{n\in \mathbb{N}}$ is convergent, we see that its limit is indeed contained within $W$. From the respective former consideration, we may hence infer that the sequence $(y_n{)}_{n\in \mathbb{N}}$ does in fact converge to $P_W(x)$. We are thus reduced to proving the convergence of the sequence in operator norm. Since $H$ is Hilbert space, proving that ${\left(\frac{1}{n}\sum _{k=1}^n{U}^k\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence will be sufficient. But since

$\begin{array}{cc}\Vert y_n-y_{mn}\Vert & =\Vert y_n-\sum _{k=1}^m\frac{1}{m}{U}^{nk}{y}_n\Vert \\ & \le \sum _{k=1}^m\frac{1}{m}\Vert y_n-U^{nk}{y}_n\Vert \\ & \le \frac{2\Vert x\Vert }{n}\end{array}$

for $m\ge 1$ this is the case; the gaps are closed using that

$\begin{array}{cc}\Vert y_n-y_m\Vert & =\Vert \frac{m-n}{mn}\sum _{k=1}^n{U}^{k}x-\frac{1}{m}\sum _{k=n+1}^m{U}^{k}x\Vert \\ & \le \frac{m-n}{mn}\Vert x\Vert +\frac{m-n}{m}\Vert x\Vert .\end{array}$

Taking $m\to \mathrm{\infty }$ in the next to last computation yields the desired rate of convergence. These computations also reveal the underlying cause of convergence: The sequence becomes more and more uniform, since applying $U^m$ to it does not change it by a large amount. $◻$