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 $v\in V$?



Operator on a Banach space
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, $\|T^n(x)\|\to\infty$.


