Ergodic theorem

Proof. Let $𝒩$ be the range of the linear transformation $1-U$ . If $x=y-Uy$ is in $𝒩$ , then \begin{align} V_{n} x &= \frac{1}{n}(y-U y+U y-U^{2} y+\cdots+U^{n-1} y-U^{n} y) \\ &= \frac{1}{n}(y-U^{n} y), \end{align}so that \begin{align} \|V_{n} x\| &= \frac{1}{n}\|y-U^{n} y\|\\ &\leq \frac{1}{n}\big(\|y\|+\|U^{n} y\|\big) \\ &= \frac{2}{n}\|y\|. \end{align}This implies that $V_{n}x$ converges to zero when $x$ is in $𝒩$ .

On the other hand, if $x$ is in $ℳ$ , that is, $Ux=x$ , then $V_{n}x=x$ , so that in this case $V_{n}x$ certainly converges to $x$ .

We shall complete the proof by showing that ${𝒩}^{⟂}=ℳ$ . (This will imply that every vector is a sum of two vectors for which $(V_n)$ converges, so that $(V_n)$ converges everywhere. What we have already proved about the limit of $(V_n)$ in $ℳ$ and in $𝒩$ shows that $(V_{n}x)$ always converges to the projection of $x$ in $ℳ$ .) To show that ${𝒩}^{⟂}=ℳ$ , we observe that $x$ is in the orthogonal complement of $𝒩$ if and only if $(x,y-Uy)=0$ for all $y$ . This in turn implies that \begin{align} 0 &= (x, y-U y)\\ &= (x, y)-(x, U y)\\ &= (x, y)-(U^{*} x, y) \\ &= (x-U^{*} x, y), \end{align}that is, that $x-U^{\ast }x=x-U^{-1}x$ is orthogonal to every vector $y$ , so that $x-U^{-1}x=0$ , $x=U^{-1}x$ , or $Ux=x$ . Reading the last computation from right to left shows that this necessary condition is also sufficient; we need only to recall the definition of $ℳ$ to see that $ℳ={𝒩}^{⟂}$ . ◻