Strong convergence of an "averaging" operator Let $X$ be an Hilbert space and $S:X \rightarrow X$ be a bounded linear operator with $||S||=1 
$
Define $$T_n= \frac{1}{n} \sum_{r=0}^{n-1} S^r$$
I want to show it converges strongly to some bounded operator $T$ and to identify it.
I have showed that $T_n x$ has a limit in $X$ for $x\in \ker(I-S) + \operatorname{ran}(I-S)$
and that $||T_n||$ is uniformly bounded.
Hence I just need to show that $\ker(I-S) + \operatorname{ran}(I-S)$ is a dense subspace of $X$
If I can prove $\ker(I-S^*)=\ker(I-S)$ then I could conclude how can I prove this?
I also need to identify the limit: as Joel noted in the comments, by Mean Ergodic Theorem, the limit is a projection on $X$ onto $\ker (I-S)$ there should a direct way of proving this though, without appealing to the Mean Ergodic Theorem.
After help from T.A.E. we concluded that the only remaining issue is to prove that, for this $S$, $\ker(I-S^*)=\ker(I-S)$.
Just for reference (and to understand how the problem develops) it is from a past exam question: https://www.maths.ox.ac.uk/system/files/attachments/b04b-12.pdf 
question 1.
 A: Let $\mathcal{D}(T)$ be the set of vectors $x$ for which the limit $\lim_{n}T_{n}x$ exists, and let $Tx$ denote that limit. It's not hard to verify that $\mathcal{D}(T)$ is a linear space and that $T$ is linear on its domain. $T$ is bounded on its domain with $\|Tx\|\le \|x\|$ for all $x\in\mathcal{D}(T)$ because $\|T_{n}\| \le 1$ for all $n$.
The domain $\mathcal{D}(T)$ is closed. To see this, let $x\in\mathcal{D}(T)^{c}$. Because $x\in\mathcal{D}(T)^{c}$, then, for every $\epsilon > 0$, there exists $x_{\epsilon} \in \mathcal{D}(T)$ such that $\|x-x_{\epsilon}\| < \epsilon/3$; for this fixed $\epsilon$ we may then choose $N$ such that $\|T_{n}x_{\epsilon}-T_{m}x_{\epsilon}\|<\epsilon/3$ whenever $n,m \ge N$. It follows that, for $n,m \ge N$,
$$
\begin{align}
    \|T_{n}x-T_{m}x\| & \le\|T_{n}(x-x_{\epsilon})\|
            +\|T_{n}x_{\epsilon}-T_{m}x_{\epsilon}\|+\|T_{m}(x-x_{\epsilon})\| \\
      & < \|x-x_{\epsilon}\|+\epsilon/3+\|x-x_{\epsilon}\| < \epsilon.
\end{align}
$$
Consequently $\{ T_{n}x\}$ converges for $x \in \mathcal{D}(T)^{c}$, which proves that $\mathcal{D}(T)=\mathcal{D}(T)^{c}$.
It is easy to verify that $(I-S)T_{n}=T_{n}(I-S)=\frac{1}{n}(S^{n}-I)$, which converges to $0$ in $\mathcal{L}(X)$. If $x\in\mathcal{D}(T)$, then $(I-S)x\in\mathcal{D}(T)$ and $(I-S)Tx=T(I-S)x=0$. So $S$ maps $\mathcal{D}(T)$ into itself, and
$$
                                  STx=TSx = Tx,\;\;\; x \in \mathcal{D}(T).
$$
Using this, $T_{n}T=T$ on $\mathcal{D}(T)$, which further shows that $T$ maps $\mathcal{D}(T)$ into itself, with
$$
                                   T^{2}x=Tx,\;\;\; x \in \mathcal{D}(T).
$$
So $T$ is a projection on its domain. $T=0$ on $\mathcal{R}(I-S)^{c}$; and $T=1$ on $\mathcal{N}(I-S)$ because $T_{n}x=x$ for $x\in\mathcal{N}(I-S)$.
If the range of $I-S$ is dense in $X$, then $\mathcal{D}(T)=X$. I don't see how to take the domain issue any further than
$$
                 (\mathcal{N}(I-S)+ \mathcal{R}(I-S))^{c}\subseteq\mathcal{D}(T).
$$
New Addition: If $S^{\star}x=x$ and $x\ne 0$, then
$$
        (Tx,x) = \lim_{n}\frac{1}{n}\sum_{r=0}^{n-1}(x,(S^{\star})^{r}x)=(x,x).
$$
By the Cauchy-Schwarz inequality
$$
          (x,x)=(Tx,x)=|(Tx,x)|\le \|Tx\|\|x\| \le \|x\|^{2}.
$$
So you have equality for the Cauchy-Schwarz inequality. Since $Tx \ne 0$ is implied, then there exists $\lambda$ such that $Tx=\lambda x$ because $\{ Tx,x\}$ must be a linearly-dependent set of vectors. Then $(x,x)=(Tx,x)=\lambda(x,x)$ gives $\lambda=1$, which implies that $T=I$ on $\mathcal{N}(I-S^{\star})=\mathcal{R}(I-S)^{\perp}$. We already knew that $T=0$ on $\mathcal{R}(I-S)^{c}$. So $T$ is determined on
$$
        X=\mathcal{R}(I-S)^{c}\oplus\mathcal{R}(I-S)^{\perp}=\mathcal{R}(I-S)^{c}\oplus\mathcal{N}(I-S^{\star}).
$$
It appears that $T$ must be the orthogonal projection onto $\mathcal{N}(I-S^{\star})$. 
I think you can apply the same analysis again after replacing replacing $S$ by $S^{\star}$ and obtaining a corresponding $T'$. It would follow that $T'=T^{\star}$ is the orthogonal projection onto $\mathcal{N}(I-S^{\star\star})=\mathcal{N}(I-S)$.
However $T^{\star}=T$ is an orthogonal projection, which means that $\mathcal{N}(I-S)=\mathcal{N}(I-S^{\star})$. I'm assuming $\mathcal{D}(T')=X$ again. That's a point that apparently can be proved according to the problem statement in your PDF.
