Mean ergodic theorem different assumption 
Show that the mean ergodic theorem still holds if we replace the assumption that $T$ is an isometry by the assumption that $T$ is contraction, that is $||T f|| ≤ ||f||$ for all $f ∈ H$ where $H$ is Hilbert space?

Why the theorem still holds? what is the relation between isometry  and contraction?
And how to show that $T $ is contraction iff $T ^∗ $ (adjoint operator of $T$) is a contraction?
Any help please.
Thanks.
 A: I knew I remembered this problem. I have it saved on my computer from exactly 4 years ago!
Chapter 6 Exercise 25: Note that all we needed for the proof of the mean ergodic Theorem was to write $f= f_0+f_1$ for $f_0 \in S$ and $f_1 \in \overline{S_1}$. If $T$ is a contraction instead of an isometry, $A_nf_0$ is still equal to $f_0 = P(f)$ and $||A_n(f_1-f_1')|| \le \frac{1}{n} \sum_{k=0}^{n-1} ||T^k(f_1-f_1')|| \le ||f_1-f_1'||$. Thus, all we need to show is $S^\perp \subseteq \overline{S_1}$. It thus suffices to show $\overline{S_1}^\perp \subseteq S$. Observe that $f \in \overline{S_1}^\perp$ iff $\langle f,g-TG \rangle = 0$ $\forall g \in H \iff f = T^*f \iff f \in S_*$. Therefore, it suffices to show if $T$ is a contraction, then $S_* \subseteq S$.
\vspace{2mm}
Suppose $T^*f = f$. Then, $||Tf||\cdot||f|| \le ||f||^2 = \langle f, f \rangle = \langle f,T^*f \rangle = \langle Tf,f\rangle$ so we have equality in Cauchy-Schwarz implying $Tf = cf$ for some $c$. Then, $c\langle f,f\rangle = \langle cf,f \rangle = \langle Tf,f\rangle = \langle f,T^*f \rangle = \langle f,f\rangle \implies c = 1$, as desired.
