Theorem about mean value of iterations of isometries Let $\mathcal{H}$ be a Hilbert space and let $T\in \mathcal{L}(\mathcal{H})$ be an isometry, and for every integer n let
$$A_n=\frac{1}{n} \sum_{j=0}^{n-1}T^j.$$
Then $A_nv$ converges to $P_Sv$ as n diverges, for every $v\in \mathcal{H}$, where $P_S$ is the orthogonal projection on the space $S:=$Ker$(\mathbb{I} - T)$, i.e. of vectors which are $T$-invariant.

I am not able to make any real progress on proving this statement, any hints?

My only observations:
I saw that if $v\in S$ then, by definition, $Tv=\mathbb{I}v=v$, so that
$$A_nv=\frac{1}{n} \sum_{j=0}^{n-1}T^jv = \frac{1}{n} \sum_{j=0}^{n-1}v=\frac{n}{n}v=v$$
for all $n\in \mathbb{N}$, so that $A_nv \to v=P_Sv$.
The struggle starts when I consider $v\not \in S$.
 A: Consider the subspace $\mathrm{Im}(I - T)$. We can check that $\mathrm{Im}(I - T)^\perp = \mathrm{Ker}(I - T) = \mathrm{Im}(P_S) = S$. Therefore $\mathcal{H} = \overline{\mathrm{Im}(I - T)} \oplus S$ (where overline is the closure). You checked that it holds for $v \in S$.
Let $v \in \mathrm{Im}(I - T)$. Then $v = w - Tv$ for some $w \in \mathcal{H}$. We can check that $P_S v = 0$. Then
$$
\frac{1}{n} \sum^{n-1}_{j=0} T^j v = \frac{1}{n} \sum^{n-1}_{j=0} (T^j w - T^{j+1} w) = \frac{1}{n} (w - T^nw) .
$$
(We have a telescoping series.) Since $T$ is an isometry, we can apply the triangle inequality to show that
$$
\frac{1}{n} \sum^{n-1}_{j=0} T^j v \to 0 , \quad n \to \infty .
$$
Since $P_S v = 0$, the claim is also true for $v \in \mathrm{Im}(I - T)$.
To finish the argument, we have to check that this holds in the closure as well. To do this, for $v \in \overline{\mathrm{Im}(I - T)}$, we can take a sequence $v_n \in \mathrm{Im}(I - T)$ such that $v_n \to v$ as $n \to \infty$, and use what we have just shown (again, using the fact that $T$ is an isometry).
