If $U$ is a linear isometry on $H$, then $\frac1n\left\|(1-U^n)x\right\|_H\xrightarrow{n\to\infty}0$

Question

Let $H$ be a $\mathbb R$-Hilbert space and $U$ be a linear isometry on $H$. Trivial question: How can we show that $$\frac1n\left\|(1-U^n)x\right\|_H\xrightarrow{n\to\infty}0\tag1$$ for all $x\in H$? Can't figure out the trick for the moment. Maybe writing $1-U^n=1-U^{n-1}+U^{n-1}-U^n$ so that $\left\|(1-U^n)x\right\|_H\le\left\|(1-U^{n-1})x\right\|_H+\left\|(1-U)x\right\|_H$?

Answer 1

Hint - use the operator norm. The norm of an isometry is $1$, and so you get (using the triangle inequality):

$$\frac{1}{n}||1-U^n||\leq \frac{1}{n}(||1||+||U||^n)=\frac{2}{n}$$