Prove that the limit exist II First question was here. I add one new condition.
Let  $T : H \rightarrow H$ is a linear continuous unitary ($T^*=T^{-1}$) operator, $H$ is a Hilbert space (not necessary). Suppose that


*

*$\forall h \in H \Rightarrow Th=h$.

*$\forall n \in \mathbb{N} \ || T_n ||\leq 1 $. NEW CONDITION

*$T_n$ - a sequence of linear operators $T_n \underset{n\rightarrow \infty}{\longrightarrow}  T$, $|| T_n - T||\rightarrow 0$. So the sequence $T_n$ tends to $T$ in the norm and we have $\forall h \in H \ \ || T_n h - T h|| \leq || T_n - T|| \ ||h||\rightarrow 0$.


Can we prove that there exists a limit of the sum $S_n= \frac{1}{n}\left( T_1 h + T_1 T_2 h + \dots + T_1 \dots T_n h \right)$ where $n \rightarrow \infty$?
 A: The limit does not necessarily exist.
Assume that $H$ is a complex Hilbert space and define $T_k=e^{i\alpha_k} Id$, where $(\alpha_k)$ is a sequence of real numbers tending to $0$. Then all $T_k$'s are unitary and $\Vert T_k-Id\Vert\to 0$. On the other hand, we have 
$$S_n=\frac1n\left(e^{i A_1}+\dots +e^{i A_n} \right) Id , $$
where $A_k=\alpha_1+\dots +\alpha_k$.
So one just needs to choose $(\alpha_k)$ in such a way that the sequence $(e^{iA_k})$ is not convergent in the Cesaro sense.
For example, one can proceed as follows (this is probably not the simplest way to do so). Let $(I_p)$ be a sequence of consecutive intervals of $\mathbb N$ and denote by $N_p$ the cardinality of the interval $I_p$. Put $\alpha_k=0$ if $k\in I_{p}$ for some odd $p$, and $\alpha_k=\frac{2\pi}{N_p}$ if $k\in I_p$ for some even $p$. Then $e^{iA_k}=1$ if $k\in I_p$ for some odd $p$, and the $e^{iA_k}$, $k\in I_p$ enumerate the $N_p$ roots of $1$ is $p$ is even (starting with $e^{i{2\pi}/{N_p}}$ and ending with $1$). Hence 
$$ \sum_{k\in I_p} e^{iA_k}=\left\{ \begin{matrix} N_p&p\;{\rm odd}\cr
0&p\; {\rm even} \end{matrix}\right.$$
It follows that if the sequence $(N_p)$ is very fast increasing (say $N_1+\dots N_p=o(N_{p+1})$) and if we put $n_p=N_1+\dots +N_p$, then $\frac{1}{n_{2l}}\sum_{k=1}^{n_{2l}} e^{iA_k}\to 0$ as $l\to\infty$, whereas $\frac{1}{n_{2l+1}}\sum_{k=1}^{n_{2l+1}} e^{iA_k}\to 1$.
