Closed unit ball of $B(H)$ is not compact in strong operator topology of $B(H)$.

Question

In operator theory we prove that closed unit ball of $B(H)$ is compact in weak operator topology and is closed in strong operator topology. But a book of operator theory states that closed unit ball of $B(H)$ is not compact in strong operator topology.

Is there any straightforward proof for this statement? thanks for your guidance.

Answer 1

It's not so much that the closed unit ball of $B(H)$ is never compact in the strong operator topology, but it is not compact in general. More precisely, it is compact if and only if $H$ is finite dimensional. For convenience, let $S$ denote the closed unit ball of $B(H)$.

• If $H$ is finite dimensional, then $B(H)$ is a finite dimensional normed space, so its closed unit ball is compact in the norm topology. Since the identity map $(S,\text{norm}) \to (S,\text{strong})$ is continuous, it follows that its image in compact as well.
• Conversely, suppose that $S$ is compact. Now the identity map $(S,\text{strong}) \to (S,\text{weak})$ is a continuous bijection from a compact space to a Hausdorff space, so it is a homeomorphism. We see that the strong and weak topologies coincide on $S$. Now suppose, for the sake of contradiction, that $H$ contains an orthonormal sequence $\{e_n\}_{n=1}^\infty$. We define a sequence $\{u_n\}_{n=1}^\infty$ in $S$ by $$ u_n(x) = \langle x, e_1\rangle e_n. $$ For all $x,y\in H$ we have $$ \lim_{n\to\infty} \langle u_n(x), y\rangle = \langle x,e_1\rangle\cdot\lim_{n\to\infty} \langle e_n, y\rangle = 0. $$ Thus, it follows that $\{u_n\}_{n=1}^\infty$ converges weakly to $0$. On the other hand, we have $$ \lim_{n\to\infty} ||u_n(x)|| = |\langle x, e_1\rangle|,\qquad\text{for all $x\in H$}, $$ so we see that the sequence $\{u_n\}_{n=1}^\infty$ does not converge strongly to $0$. This contradicts our earlier finding that the strong and weak topologies coincide on $S$. It follows that $H$ must be finite dimensional.