If the image of an operator is closed, is the image of the powers of the operator also closed?

Solution 1:

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.