Let $H$ be a Hilbert space and $U\in\mathcal{L}(H)$ an unitary operator. Let $P$ be the orthogonal projection whose image is $\ker(U-I)$. Let, for $n\geq 1$, $$S_n=\frac{I+U+\ldots+U^n}{n+1}.$$
I would like to show that, for all $u\in H$, $S_nu\to Pu$ when $n\to\infty$, but I don't really know how to do it. All I have got is that $H=\ker(U-I)\oplus\overline{\text{Im}(U-I)}$, but I'm not sure that it is really useful to answer my question.
Could anybody help me?
Thanks in advance.
On
I assume that $\lim S_n(x)$ exists. (Consider $f(x)=-x$ in $\mathbb{R}$.
It is enough to show that $S_n(x)\in \ker(U-I)$.
We have $(U-I)S_n=-{I\over{n+1}}$, we deduce that $\lim_n(U-I)S_n(x)=(U-I)\lim_n(S_n(x))=0$ and $\lim_nS_n\in\ker(U-I)$.
On
This comment is too long for the comment section. The other answers make sense but if I use the Functional Calculus, I am getting an actual formula for $S.$ So something is not right, but I don't know what it is.
We have an isometric $∗$-isomorphism $C^∗(U)\cong C(\sigma(U))$ that sends $p(U)$ to $p(z)$ for any finite polynomial in $I$ and $U$. Now $\sigma (U)\subseteq \mathbb T$ and since
$p_n(z)=\frac{1}{n+1}\sum_{k=0}^nz^n=\begin{cases} 1\quad z=1\\\frac{1-z^{n+1}}{(1+n)(1-z)}\quad z\neq 1\end{cases}$ we have that $\lim p_n(z)=\begin{cases} 1\quad z=1\\0\quad z\neq 1\end{cases}\quad $
so if $p_n\to S\in C^∗(U)$ then $p_n$ has to converge to a continuous function on $\sigma(U).$ Therefore, if $1\in \sigma(U)$ then $S=I$ and if $1\notin \sigma(U)$ then $S=0.$
If $y \in\operatorname{Ker}(U-I)$ then $Uy=y$ so $U^{n}y=y$ for all $n$ and $S_n y=\frac {y+\cdots+y} {n+1}=y$ for all $n$ So $S_n y \to y= Py$.
If $y \in\operatorname{Im}(U-I)$ then $y=Ux-x$ for some $x$. Now $S_ny$ simplifies to $\frac {U^{n+1}x-x} {n+1}$ after cancelling some terms. Hence $S_n y \to 0=Py$. If $y \in \overline {\operatorname{Im}(U-I)}$ and $\epsilon >0$ then there exists $z \in \operatorname{Im}(U-I)$ such that $\|y-z\|<\epsilon$. Use triangle inequality and the fact that $\|U^{i} \|\leq 1$ for all $i$ to finish the proof.