Compact Linear Transformations Composed with Bounded Linear Operator

Theorem

Let $H, K$ be Hilbert spaces.

Let $T \in \map {B_0} {H, K}$ be a compact linear transformation.

Let $A \in \map B H, B \in \map B K$ be bounded linear operators.

Then the compositions $T A$ and $B T$ are also compact linear transformations.

Proof

Let $\sequence {h_n}_{n \in \N}$ is a bounded sequence in $H$.

That is, there exists a $M > 0$ such that:

$\forall n \in \N : \norm {h_n}_H \le M$

Then $\sequence {A h_n}_{n \in \N}$ is also bounded, since:

$\norm {A h_n}_H \le \norm {A}_{\map B H} \norm {h_n}_H \le \norm {A}_{\map B H} M$

As $T$ is compact, the sequence $\sequence {T A h_n}_{n \in \N}$ has a subsequence convergent in $K$.

Hence $T A$ is compact.

$\Box$

Let $\sequence {k_n}_{n \in \N}$ is a bounded sequence in $K$.

As $T$ is compact, the sequence $\sequence {T k_n}_{n \in \N}$ has a subsequence convergent in $H$.

Say $\sequence {T k_{n_r} }_{r \in \N}$ is convergent.

Recall that $B$ is continuous due to Continuity of Linear Transformation between Normed Vector Spaces.

Thus $\sequence {B T k_{n_r} }_{r \in \N}$ is convergent.

Hence $B T$ is compact.

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\text {II}.4.2 \text {(c)}$