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In Hilbert space $H$, is it true that the composition of operators $ST$ and $TS$ of the bounded operator $S$ and the compact operator $T$ are compact?

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1 Answer 1

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Yes (and more generally in a Banach space). If $B$ is the closed unit ball, $\overline{TB}$ is compact, so $STB \subseteq S(\overline{TB})$ which is compact (in a Hausdorff space, a continuous image of a compact set is compact). Therefore $ST$ is a compact operator.

$SB \subseteq \|S\| B$, so $TSB \subseteq \|S\| TB \subseteq \|S\| \overline{TB}$ which is compact, so $TS$ is a compact operator.

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  • $\begingroup$ Thank you for your nice answer. $\endgroup$
    – Zbigniew
    Aug 11, 2019 at 15:37
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    $\begingroup$ I believe you do not need the space to be Hausdorff in order for the continuous image of a compact set to be compact $\endgroup$ Nov 29, 2020 at 18:28

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