# Operator on a Banach space

Let $T$ be a continuous operator on a Banach space $V$. Assume there exist $T$-stable finite-dimensional subspaces $V_i$ such that $\bigoplus_{i=1}^\infty V_i$ is dense in $V$, on $V_i$ the operator $T$ has only one eigenvalue $\lambda_i$. One has $|\lambda_i|<1$ for each $i$ and the $\lambda_i$ tend to zero.

My question is: Is it true that $T^nv$ tends to zero for every $v\in V$?

• Eric Wofsey has already answered your question very neatly, but I'd like to add the remark (implicit in his calculations) that the moral of the story is as follows: the spectral radius formula tells us that if the spectral radius is strictly less than one then the operator is power bounded, but it doesn't suffice to get control on how big the powers can get – Yemon Choi Jun 15 '15 at 0:34
• (Unless you have other information, such as normality, or control on the size of the Jordan blocks) – Yemon Choi Jun 15 '15 at 0:34

This is not necessarily true, and in fact you can get a counterexample where each $\lambda_i=0$. Consider $V=c_0$ and let $N_i=\sum_{j=0}^i j$ be the $i$th triangular number. Define $T:c_0\to c_0$ by $$T(x)_k=\begin{cases} 0 & \text{if }k=N_i\text{ for some }i \\ 2x_{k-1} & \text{otherwise}\end{cases}$$
That is, $T$ is twice the right shift, except that the $N_i$th coordinates of $T(x)$ are declared to be $0$ for each $i$. Clearly $T$ is bounded, with $\|T\|=2$.
If $V_i$ is the space of sequences supported on $[N_i,N_{i+1})$, then each $V_i$ is $T$-stable and $T$ is nilpotent on $V_i$. As $\bigoplus V_i$ is dense in $c_0$, we conclude that $T$ satisfies your hypotheses with $\lambda_i=0$ for all $i$. Now let $x\in c_0$ be the sequence such that $x_{N_i}=1/i$ for each $i$ and all other coordinates are $0$. Then for any $n$, $T^n(x)$ will have $(N_i+n)$th coordinate $2^n/i$ whenever $i\geq n$. In particular, $\|T^n(x)\|\to\infty$.