Finite Rank Operator is Compact

Theorem

Let $H, K$ be Hilbert spaces.

Let $T \in \map {B_{00} } {H, K}$ be a bounded finite rank operator.

Then $T \in \map {B_0} {H, K}$, that is, $T$ is compact.

Proof

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Let $A$ be a bounded set in $H$.

Since $T$ is bounded, the image $T \sqbrk A$ is a bounded set by definition.

Since $T$ is finite rank operator, its range $\Rng T$ is finite dimensional.

The closure of a bounded set in $\Rng T$ is compact, by Heine-Borel Theorem.

So the closure of $T \sqbrk A$ in $\Rng T$ is compact.

Hence, $T$ is a compact linear operator.

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\text {II}.4$ Exercise $2$