# Is the closed unit ball of a Hilbert space a weak sequential space?

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.

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 $$H$$.
• Does the sequential continuity, as what you have shown imply that $T$ then is continuous wrt to the appropriate topologies? Edit: I am not really asking how to show sequential continuity, but I am asking how showing it would imply actual continuity of $T$. Jul 17, 2020 at 16:28
• @Kurome Are you familiar with nets? A net is a generalization of a sequence such that a function $f : X \to Y$ between any topological spaces is continuous iff for every net $(x_\alpha)_\alpha$ in $X$ converging to a point $x \in X$, the net $(f(x_\alpha))_\alpha$ in $Y$ converges to $f(x)$. I claim you could replace all the sequences $(x_n)$ in my answer with nets $(x_\alpha)$ and things would work out just the same.