I am currently reading a proof on characterizations of a compact operator for Hilbert spaces. Let $\mathscr{B}_1$ be the closed unit ball of the Hilbert space $H$. One of the equivalent statement involves: $T\in B(H)$ is the norm-limit of finite rank operators if and only if $T_{|\mathscr{B}_1}:\mathscr{B}_1\to H$ is continuous as a function from $\mathscr{B}_1$ endowed with weak topology to $H$ endowed with the norm topology.
The author then proceeds to prove the $\implies$ direction of the result basically by showing that $\{x_n\}$ converges weakly to $x$ in $\mathscr{B}_1$ implies that $\{Tx_n\}$ converges to $Tx$ in norm of $H.$ I know this is the sequential criterion for continuity, but this is only valid if $\mathscr{B}_1$ is a (weak) sequential space. Is it?
I've tried looking for other sources, but to no avail.
Best Answer
This is just from the definition of weak convergence. First, suppose $Tx = \phi(x) y$ is an operator of rank $1$, where $\phi \in H^*, y \in H$. Let $(x_\alpha)_\alpha$ be a net in $\mathscr{B}_1$ converging weakly to $x \in \mathscr{B}_1$. Then $$\|Tx - T x_\alpha \| = | \phi(x - x_\alpha) | \|y \| \to 0 .$$ We can then extend this fact to an operator of finite rank by finite additivity. Now if $S$ is an arbitrary compact operator, then if $T$ is a finite-rank operator that norm-approximates $S$, then \begin{align*} \| Sx - S x_\alpha \| & \leq \| Sx - Tx \| + \| Tx - T x_\alpha \| + \| T x_\alpha - S x_\alpha \| \\ & \leq \| S - T \| \|x \| + \| Tx - T x_\alpha \| + \| T - S \| \|x_\alpha \| \\ & \leq 2 \| S - T \| + \| Tx - T x_\alpha \| . \end{align*} Now fix $\epsilon > 0$, and choose $T$ finite-rank so $\| S - T \| < \epsilon / 3$. Then choose $\beta$ so $\alpha \geq \beta \Rightarrow \| Tx - T x_\alpha \| < \epsilon / 3$. Then if $\alpha \geq \beta$, then $\| S x - S x_\alpha \| < \epsilon$.
Therefore $S$ is continuous from the weak $\mathscr{B}_1$ to the normed $H$.