# Topological Modules/Orthogonal projection

Theorem (Von Neumann ergodic theorem):

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

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

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

${\displaystyle \left\|{\frac {1}{n}}\sum _{k=1}^{n}U^{k}-P_{W}\right\|\leq {\frac {2}{n}}}$

is a valid estimate for the convergence rate.

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

{\displaystyle {\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

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

we obtain

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

{\displaystyle {\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 ${\displaystyle m\geq 1}$ this is the case; the gaps are closed using that

{\displaystyle {\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 ${\displaystyle m\to \infty }$ 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 ${\displaystyle U^{m}}$ to it does not change it by a large amount. ${\displaystyle \Box }$