# 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

$\lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}U^{k}=P_{W}$ ,

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

$\left\|{\frac {1}{n}}\sum _{k=1}^{n}U^{k}-P_{W}\right\|\leq {\frac {2}{n}}$ is a valid estimate for the convergence rate.

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

{\begin{aligned}\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-P_{W}(x),z\rangle &=\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-P_{W}(x),z\rangle +{\frac {n}{n}}\langle P_{W}(x)-x,z\rangle \\&=\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-x,z\rangle \\&={\frac {1}{n}}\sum _{k=1}^{n}\left(\langle U^{k}x,z\rangle -\langle x,z\rangle \right)\\&{\overset {z\in W}{=}}{\frac {1}{n}}\sum _{k=1}^{n}\left(\overbrace {\langle U^{k}x,U^{k}z\rangle } ^{=\langle x,z\rangle }-\langle x,z\rangle \right)=0.\end{aligned}} Further, if we set

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

we obtain

{\begin{aligned}\|Uy_{n}-y_{n}\|^{2}&=\langle y_{n}-Uy_{n},y_{n}-Uy_{n}\rangle \\&={\frac {1}{n^{2}}}\langle Ux-U^{n+1}x,Ux-U^{n+1}x\rangle \\&={\frac {1}{n^{2}}}\left(\langle Ux,Ux\rangle -\langle U^{n+1}x,Ux\rangle -\langle Ux,U^{n+1}x\rangle +\langle U^{n+1}x,U^{n+1}x\rangle \right)\\&{\overset {\text{Cauchy‒Schwarz}}{\leq }}{\frac {4\|x\|^{2}}{n^{2}}}.\end{aligned}} 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{aligned}\|y_{n}-y_{mn}\|&=\left\|y_{n}-\sum _{k=1}^{m}{\frac {1}{m}}U^{nk}y_{n}\right\|\\&\leq \sum _{k=1}^{m}{\frac {1}{m}}\|y_{n}-U^{nk}y_{n}\|\\&\leq {\frac {2\|x\|}{n}}\end{aligned}} for $m\geq 1$ this is the case; the gaps are closed using that

{\begin{aligned}\|y_{n}-y_{m}\|&=\left\|{\frac {m-n}{mn}}\sum _{k=1}^{n}U^{k}x-{\frac {1}{m}}\sum _{k=n+1}^{m}U^{k}x\right\|\\&\leq {\frac {m-n}{mn}}\|x\|+{\frac {m-n}{m}}\|x\|.\end{aligned}} Taking $m\to \infty$ 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. $\Box$ 