Sum of unitary operators converges to projection operator So let $\mathcal{H}$ be a hilbert space and $U$ is a unitary operator on $\mathcal{H}$. Let $I=\{v\in\mathcal{H}:U(v)=v\}$. Show that $\frac{1}{N}\sum_{n=1}^NU^n(v)\rightarrow Pv$ where $P:\mathcal{H}\rightarrow\mathcal{H}$ is the projection operator onto $I$. I don't know how to proceed with this problem. Can someone suggest hints as to how to do this.

if $v \in I$ the result is trivial. Let $v \in I^{\perp}$ since the restriction of $Id - U$ on $I^{\perp}$ is bijective (Indeed it is injective : if $x \in I^{\perp}$, $(Id-U)(x) = 0 \Rightarrow x \in I \Rightarrow x = 0$ it is also surjective because if $z \in Im(Id-U)$ then there is $y \in H$ such that $z = (Id -U)(y)$. Since $(Id-U)(P(y)) = 0$ setting $x = (Id-P)(y)$ then $(Id-U)(x) = z$), Let $v_n = \frac{1}{N}\sum_{n=1}^{N}U^n(v) \in I^\perp$ and $w_n = (I-U)(v_n)$ then $w_n = \frac{1}{N}\sum_{n=1}^{N}\left(U^n(v) - U^{n+1}(v)\right) = \frac{1}{N}(U(v) - U^{N+1}(v)) \rightarrow 0$ because $||U|| \le 1$. Setting $V$ the inverse of the restriction of $Id - U$ on $I^\perp$, we have $v_n = V(w_n) \rightarrow 0$. The result is also verified for $v \in I^{\perp}$. Using $H = I + I ^{\perp}$ and the linearity of $P$ we have the result for all $v$. 
