Left Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.

Let $T : X \to Y$ be a compact linear transformation.

Let $S : Y \to Z$ be a bounded linear transformation.

Then $S T : X \to Z$ is a compact linear transformation.

Proof

Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence.

Since $T$ is compact, there exists a subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ such that:

$T x_{n_j} \to y$

as $n \to \infty$ for some $y \in Y$.

Since $S$ is bounded, it is continuous from Continuity of Linear Transformations, and so:

$S T x_{n_j} \to S y$

So for each bounded sequence $\sequence {x_n}_{n \mathop \in \N}$, there exists a subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ such that $\sequence {S T x_n}_{n \mathop \in \N}$ converges.

$\blacksquare$