Proof of Von Neumann Ergodic Theorem

Question

I am trying to understand the line that says: and hence $U_nf \rightarrow 0$ as $n \rightarrow \infty$. I understand everything up until that claim.

Answer 1

Since $U$ is an isometry, \begin{align*} ||g - U^n g|| &\leq ||g|| + ||U^n g|| \\ &= ||g|| + ||g|| \\ &= 2||g|| \text{.} \end{align*}

So $|| \sum_{i=0}^{n-1} U^n f || \leq 2||g||$ and when you divide through by $n$, making the left-hand side $U_n f$, the right-hand side goes to zero as $n \rightarrow \infty$.

Answer 2

Since $U$ is an isometry, $\|U\|=1$ and thus $$\|U_nf\|=\frac 1n\left\|g-U^ng\right\|\leq\frac2n\|g\|\to0$$ as $n\to\infty$.