If $X$ is a Banach space and $T : X \to X$ is a continuous linear operator with the property that $T^{n}X$ equals $T^{n+1}X$ for some $n \ge 1$, does it follow that $T^{n}X$ is a closed subspace?
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4$\begingroup$ No. Suppose $X=X_0\oplus X_1\oplus X_2$ and $T=A+B$, where $A$ is an isomorphism from $X_0$ onto $X_0\oplus X_1$ and $B$ is arbitrary from $X_1$ into $X_2$. Then $T$ has a "pseudo-inverse" $S:=A^{-1}$ that satisfies $TS=I_{X_0\oplus X_1}$ and $T^2S=T$. The range of $T$ need not be closed because of $B$. $\endgroup$– Narutaka OZAWACommented Jul 20, 2021 at 4:19
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$\begingroup$ That looks correct to me. Thanks! $\endgroup$– Andy HammerlindlCommented Jul 20, 2021 at 6:00
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$\begingroup$ BTW, for a positive direction, one could "forbid" existence of an isomorphism between $X_0$ and $X_0\oplus X_1$. For example, if $X$ is a Hilbert space and $T$ belongs to a finite von Neumann algebra, then the answer to your problem is (probably) positive. $\endgroup$– Narutaka OZAWACommented Jul 20, 2021 at 6:40
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