Proof of von Neumann’s Ergodic Theorem.
$\def\ker{\operatorname{Ker}}\DeclareMathOperator{\im}{Im}$Let $X$ be a Hilbert space and let $U\colon X → X$ be a unitary operator.
(a) Show that $\ker(I-U)=\ker(I-U^*)$;
(b) Show that $X=\overline{\im(I-U)} ⊕ \ker(I-U)$;
(c) Show that $\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x=x$ if $x ∈ \ker(I-U)$ and $\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x=0$ if $x ∈ \overline{\im(I-U)}$;
(d) Deduce that, for each $x ∈ X$,
\[\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x=P x,\]
where $P$ is the orthogonal projection onto $\ker(I-U)$.
(a)
$\ker(I-U)=\ker(UU^*-U)=\ker(U(U^*-I))⊃\ker(I-U^*)$.
$\ker(I-U^*)=\ker(U^*U-U^*)=\ker(U^*(U-I))⊃\ker(I-U)$.
(b)
By 1.5 for any $T∈B(X)$, $\overline{\im T}=(\ker T^*)^⟂$.
Applying $X=\overline{\im T}⊕\ker T^*$ to $T=I-U$
\[X=\overline{\im(I-U)} ⊕ \ker(I-U^*)\overset{\text{(a)}}=\overline{\im(I-U)} ⊕ \ker(I-U)\]
(c)
If $x ∈ \ker(I-U)$, $x=U^nx$, so $\frac{1}{N} \sum_{n=1}^{N-1} U^n x=\frac{N-1}Nx→x$.
If $x ∈ \im(I-U)$, let $x=y-Uy$. Since $U$ is isometric, $‖y‖=‖Uy‖=‖U^{n+1}y‖$,
\[\frac{1}{N} \sum_{n=1}^{N-1} U^n x=\frac{1}{N} \sum_{n=1}^{N-1}U^ny-U^{n+1}y=\frac{Uy-U^{n+1}y}N→0\]
Additionally we show this holds for $\overline{\im(I-U)}$, consider $x∈\overline{\im(I-U)}$ there exist $\tilde{x} ∈\im(I-U)$ s.t. $\|x-\tilde{x}\|<\frac{ϵ}{2}$ and large $N$ s.t. $\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n \tilde{x}\right\|<\frac{ϵ}{2}$
\begin{align*}\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n x\right\|&=\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n((x-\tilde{x})+\tilde{x})\right\|\\&≤\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n(x-\tilde{x})\right\|+\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n \tilde{x}\right\| \\&<\left(\frac{1}{N} \sum_{n=1}^{N-1}\left\|U^n\right\|\right)\frac{ϵ}{2}+\frac{ϵ}{2}\\&=\frac{N-1}N\frac{ϵ}{2}+\frac{ϵ}{2}<ϵ . \text { Using }\|U^n\|=1\end{align*}
so $\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x=0$ for all $x ∈ \overline{\im(I-U)}$.
Remark.(theorem 3.5) If $‖T_n‖≤M$ for all $n$, and $T_n→0$ for all $x$ in a dense set $D⊆X$, then $T_n→0$ for all $x$ in $X$.
Not true if dropping the condition $‖T_n‖≤M$ for all $n$, example: $X=l^2,T_n((x_k))=nx_n,D=c_{00},T_n|_{c_{00}}→0$ but $T_n((\frac1k))=1$
(d)
Using (b) and $Px∈ \ker(I-U),x-Px∈\ker(I-U)^⟂=\overline{\im(I-U)}$.
Using (c)
\begin{align*}\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x&=\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n(Px+(x-Px))\\&=P x+0=Px.\end{align*}