Given a normal operator $T\in\mathcal{L}(H)$ with $||T||\leq 1$, let $M\subseteq H$ the closed subspace defined by $M=\{x\in H|T(x)=x\}$ and $P$ the orthogonal projection on $M$. Prove that for all $x\in H$, we have:
$$\lim_{n\to\infty}\left|\left|P(x)-\dfrac{x+T(x)+T^2(x)+...+T^n(x)}{n+1}\right|\right|=0$$
The proof goes as follows. Let $Q$ be the orthogonal projection on $M^\perp$. By linearity, it's very easy to see that it's enough to show the result when $x\in M^\perp$. Fixed $x\in M^\perp$, we need to demonstrate the following:
$$\lim_{n\to\infty}\left|\left|\dfrac{x+T(x)+T^2(x)+...+T^n(x)}{n+1}\right|\right|=0$$
Since $T$ is a normal operator, using the Spectral Theorem, there exists a unique spectral measure $E$ defined on $\mathcal{B}_{\sigma(T)}$ such that:
$$T=\int_{\sigma(T)}{z\,dE(z)}.$$
It's clear that if $T$ is normal, then, for all $n\in\mathbb{N}$, $T^n$ is normal and:
$$T^n=\int_{\sigma(T)}{z^n\,dE(z)}.$$
Let $f_n:\sigma(T)\rightarrow\mathbb{C}$ given by:
$$f_n(z)=\dfrac{1+z+z^2+...+z^n}{n+1}.$$
Therefore:
$$\dfrac{x+T(x)+T^2(x)+...+T^n(x)}{n+1}=\left(\int_{\sigma(T)}{f_n(z)\,dE(z)}\right)(x).$$
Let $S_n=\int_{\sigma(T)}{f_n(z)\,dE(z)}$. Then:
$$S_n^*\circ S_n=\int_{\sigma(T)}{|f_n(z)|^2\,dE(z)}$$
Furthermore, we know that for all $y\in H$, it is verified:
$$\langle S_n(x),y\rangle=\int_{\sigma(T)}{f_n(z)\,dE_{x,y}(z)}$$
Taking $y=S_n(x)$, we get:
$$||S_n(x)||^2=\langle S_n(x),S_n(x)\rangle=\langle S_n^*(S_n(x)),x\rangle=\int_{\sigma(T)}{|f_n(z)|^2\,dE_{x,x}(z)},$$
where $E_{x,x}$ is a finite positive measure. It's very easy to check that $f_n\stackrel{n\to\infty}{\rightarrow}\chi_{\{1\}}$ pointwise on $\sigma(T)$ and $|f_n|\leq 1$ on $\sigma(T)$. Since $\chi_{\{1\}},1\in\mathcal{L}^1_{E_{x,x}}(\sigma(T))$, we can use the Dominated Convergence Theorem and we have:
$$\lim_{n\to\infty}{||S_n(x)||^2}=E_{x,x}(\sigma(T)\cap\{1\})=\langle E(\sigma(T)\cap\{1\})(x),x\rangle.$$
If $1\notin\sigma(T)$, clearly $E_{x,x}(\sigma(T)\cap\{1\})=E_{x,x}(\varnothing)=0$. Suppose $1\in\sigma(T)$. I have to prove that $E_{x,x}(\{1\})=0$. How can I prove this?