# Prove composition of bounded and compact operators is compact.

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?

• Makes sense to me. Notation-wise, I think we usually write $S \circ T$ or $S(T(\cdot))$ instead of $ST$ Mar 31, 2022 at 22:16
• @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)
– Mark
Mar 31, 2022 at 22:21

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.