Linear Transformation Compact iff Adjoint Compact

Theorem
Let $H, K$ be Hilbert spaces.

Let $T: H \to K$ be a linear transformation.

Then $T$ is compact its adjoint $T^*$ is.

Proof
First suppose that $T$ is compact.

From the construction of the adjoint, $T^\ast$ is bounded.

From Right Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact, $T T^\ast$ is compact.

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

Then there exists a convergent subsequence of $\sequence {T T^\ast x_n}_{n \mathop \in \N}$, say $\sequence {T T^\ast x_{n_j} }_{j \mathop \in \N}$.

We show that $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ converges.

Pick $M > 0$ such that:


 * $\norm {x_n}_K \le M$

for each $n \in \N$.

We have:

Since $\sequence {\map {T T^\ast} {x_{n_j} } }_{j \mathop \in \N}$ is convergent, it is Cauchy from Convergent Sequence is Cauchy Sequence.

Let $\epsilon > 0$.

Pick $N \in \N$ so that:


 * $\ds \norm {\map {T T^\ast} {x_{n_j} - x_{n_k} } }_K < \frac {\epsilon^2} {2 M}$

for $n, m \ge N$.

Then for $n, m \ge N$ we have:


 * $\norm {\map {T^\ast} {x_{n_j} - x_{n_k} } }_K^2 < \epsilon^2$

so:


 * $\norm {\map {T^\ast} {x_{n_j} - x_{n_k} } }_K < \epsilon$

So $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ is Cauchy.

Since $H$ is Hilbert, $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ is convergent.

So for each bounded sequence $\sequence {x_n}_{n \mathop \in \N}$ in $K$, $\sequence {T^\ast x_n}_{n \mathop \in \N}$ has a convergent subsequence.

So $T^\ast$ is compact.

So for a bounded linear transformation $T : H \to K$, if $T$ is compact then $T^\ast$ is compact.

So for a bounded linear transformation $T : H \to K$, if $T^\ast$ is compact then $T^{\ast \ast}$ is compact.

From Adjoint is Involutive, we therefore have that that if $T^\ast$ is compact then $T$ is compact.

So $T$ is compact its adjoint $T^*$ is.