Dual operator of a compact operator

Question

Why is the dual operator $A^{\ast}$ of a compact operator $A:X \rightarrow Y$, where $X,Y$ are two Banach spaces again compact?

Answer 1

Let $A : X \rightarrow Y$ be a compact operator and $A^* : Y^* \rightarrow X^*$ its dual. The trick is to note that $$\Vert A^* y^* \Vert = \sup_{\Vert x \Vert \leq 1} \Vert y^* A x \Vert \leq \sup_{y \in \mathrm{cl}(A(B))} \Vert y^*(y) \Vert$$ where $B$ is the closed unit ball in $X$. Let $(y_n^*)_{n \in \mathbf{N}} \subset Y^*$ be a bounded sequence, say by $c > 0$. The sequence also lies in $C(\mathrm{cl} (A(B)), \mathbf{R})$. We view $\mathrm{cl}(A(B))$ as a metric space (the metric just given by the restriction of the metric on $Y$ to $\mathrm{cl}(A(B))$).

We now use Arzela-Ascoli to show that the sequence $(y_n^*)_{n \in \mathbf{N}}$ is precompact in $C(\mathrm{cl} (A(B)), \mathbf{R})$. Since $$\Vert y_n^*(y_1) - y_n^*(y_2) \Vert \leq \Vert y_n^* \Vert \cdot \Vert y_1 - y_2 \Vert \leq c \cdot \Vert y_1 - y_2 \Vert$$ all $y_n^*$ are Lipschitz with the same constant, hence equicontinuous.

The sequence is of course pointwise bounded, as $\Vert y_n^*(y) \Vert \leq c \cdot \Vert y \Vert$ for all $n \in \mathbf{N}$. It is thus pointwise compact, as $\mathrm{cl}(\{ y_n^*(y) : n \in \mathbf{N} \}) \subset \mathbf{R}$ is bounded and closed for any $y \in \mathrm{cl}(A(B))$.

By Arzela-Ascoli the sequence is precompact in $C(\mathrm{cl} (A(B)), \mathbf{R})$, hence has a Cauchy-subsequence $(y_{n_k}^*)_{k \in \mathbf{R}}$ w.r.t. $\Vert \cdot \Vert_\infty$ the supremum-norm on $\mathrm{cl}(A(B))$. This gives that $(A^* y_{n_k}^*)_{k \in \mathbf{N}}$ is a Cauchy-subsequence in $X^*$ by the very first estimate that we did.

Hence $A^*$ is compact.