= Solution
Put $F=\ker(I-U)$. For a <unitary operator>, $\ker(I-U^*)=F$, so
$$
H=F\oplus\overline{\operatorname{Ran}(I-U)}.
$$
This follows from the general identity $\operatorname{Ran}(A)^\perp=\ker A^*$. The averaging operators have $\|T_n\|\leq1$ and equal the identity on $F$. On a <vector> $(I-U)\eta$, telescoping gives
$$
T_n(I-U)\eta=\frac{I-U^n}{n}\eta,\qquad
\|T_n(I-U)\eta\|\leq\frac{2\|\eta\|}{n}\longrightarrow0.
$$
The uniform <norm> bound extends this convergence to the <closure> of the range. Thus $T_n$ tends to identity on $F$ and to zero on its <orthogonal complement>. Consequently
$$
\boxed{T_n\xi\longrightarrow P\xi\quad\text{for every }\xi\in H,}
$$
which is exactly convergence in the <strong operator topology>. This proves the <Von Neumann mean ergodic theorem> in the requested unitary case.
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