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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.

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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.

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  • $\begingroup$ Nice! +1 I was thinking along the lines of shift operators, but couldn't make them to work. $\endgroup$
    – Jose27
    Dec 5, 2021 at 20:08
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    $\begingroup$ My thinking was to have $T=K+N$, with $K$ compact and $N$ nilpotent with closed range. $\endgroup$ Dec 5, 2021 at 20:34

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