TITLE: If the image of an operator is closed, is the image of the powers of the operator also closed?
QUESTION [6 upvotes]: Say $T$ is a bounded linear operator in a normed space that maps to itself (Banach or Hilbert space is fine). If the image $\text{Im}(T)$ is closed, then is it true that $\text{Im}(T^n)$ is closed? If not, what is a counterexample?
I know there are some theorems for compact operators that make use of this, and it is true for compact, but I'm not sure if this is true in general.

REPLY [12 votes]: Let $H$ a separable Hilbert space with orthonormal basis $\{e_n\}$. Define $T$ as the linear operator induced by
$$
Te_n=\begin{cases}
e_{n+1},&\ n\ \text{ even } \\[0.3cm]
\frac1n\,e_n,&\ n\ \text{ odd} 
\end{cases}
$$
Let $M=\overline{\operatorname{span}}\{e_n,\ n\ \text{ odd}\}$. Then $\operatorname{Im} T=M$ is closed (note that $T(M)\subset M$ and $T(M^\perp)=M$, and that $M$ is closed by definition). But
$$
T^2e_n=\begin{cases}
\tfrac1{n+1}\,e_{n+1},&\ n\ \text{ even}\\[0.3cm]
\tfrac1{n^2}\,e_n,&\ n\ \text{ odd}
\end{cases}
$$
The operator $T^2$ is compact with infinite-dimensional range, so it has non-closed range.