If $T:X\to Y$ is a compact linear operator on Banach spaces $X$ and $Y$and $S:Y\to Z$ is a bounded linear operator where $Z$ is a Banach space. Prove $ST$ is compact.

Here's my proof, let $x_n \in X$ be a sequence such that $\| x_n\| \leq 1$ , then since $T$ is compact then $Tx_n$ has a convergent subsequence say $Tx_{n_k} \to y$, Then since $S$ is bounded we get $STx_{n_k} \to Sy$ which shows that $ST$ has a convergent subsequence proving that $ST$ is compact.

Is my proof correct?

  • $\begingroup$ Makes sense to me. Notation-wise, I think we usually write $S \circ T$ or $S(T(\cdot))$ instead of $ST$ $\endgroup$
    – gt6989b
    Mar 31, 2022 at 22:16
  • $\begingroup$ @gt6989b $ST$ is a common notation for composition of linear operators. Sometimes we even want to think of that as multiplication. (a standard example of a Banach algebra is the algebra of bounded linear operators of some Banach space) $\endgroup$
    – Mark
    Mar 31, 2022 at 22:21

1 Answer 1


Alternate proof: Let $A$ be any bounded subset of $X$. Then $Cl_Y(T[A])$ is compact in $Y$. Now $S$ is continuous because $S$ is bounded, and the continuous image of a compact set is compact,

so $S[Cl_Y(T[A])]$ is compact in $Z.$

In a normed linear space (& in any Hausdorff space), compact subsets are closed. So $$S[Cl_Y(T[A])]=Cl_Z(S[Cl_Y(T[A])]).$$ We now have $$Cl_Z((ST)[A])=Cl_Z(S[T[A]])\subseteq$$ $$\subseteq Cl_Z(S[Cl_Y(T[A])])=$$ $$=S[Cl_Y(T[A])].$$ So $Cl_Z((ST)[A])$ is a closed subset of the compact set $S[Cl_Y(T[A])],$ so $Cl_Z((ST)[A])$ is compact.

The proposer's proof is also correct.


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