Convergence of a Series of Operators Does the following sequence of operators converge:
$$
\frac{1}{n}\sum_{k=0}^{n-1} T^k
$$
where $T:X\to X$ for some Banach space $X$ and $\|T\|= 1$? Intuitively the answer should be yes, but I am struggling showing that the sequence is Cauchy. Any hints are appreciated.
 A: In the case when $(T-I)$ is invertible, the answer is easy:
$$
\frac{1}{n}\sum_{k=0}^{n-1} T^k=(T-I)^{-1}(T-I)\frac{1}{n}\sum_{k=0}^{n-1}T^k=(T-I)^{-1}\frac{1}{n}(T^n-I)\to 0,\ n\to\infty,
$$
since the norm of $T^n-I$ is $\leq 2$, you get the norm convergence.
If $(T-I)^{-1}$ does not exist, there is a counteraxemple. 
Lemma. Let $f\in C[0,1]$. Then the norm of the multiplication operator $T_f$ on the Hilbert space $L^2[0,1]$ defined by $T_f\varphi=\varphi$ is equal to $||f||_\infty=\sup_{t\in[0,1]}|f(t)|$. (The proof is easy.)
Consider the multiplication operator $T_f$ on the Hilbert space $L^2[0,1]$, where $f(x)=x$. Then $||T||=1$. The operator $$\frac{1}{n}\sum_{k=0}^{n-1} T^k$$ is again a multiplication operator $T_{g_n}$ where $g_n(x)=\frac{1}{n}\sum_{k=0}^{n-1} x^k$. Since $||g_n||=1,$ we have $||\frac{1}{n}\sum_{k=0}^{n-1} T^k||=1$, i.e. the sequence does not converge to $0$.
On the other hand, by dominated convergence theorem we have $||T_{g_n}\varphi||\to 0$ for every $\varphi\in L^2[0,1]$. It means that the norm-convergence of $T_{g_n}$ could be only against $0$.
