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

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.

• Fix $y\in H$ with $\lVert y\rVert = 1$ and consider the map $x \mapsto \langle\,\cdot\,,y\rangle\cdot x$. – Daniel Fischer Oct 5 '14 at 16:34
• Sorry i dont understand your answer. how this map can help me to prove that (B(H))_1 is not SO-compact? – joker Oct 5 '14 at 18:29
• The map is an isometric embedding of $H$ into $B(H)$. What about the strong operator topology? – Daniel Fischer Oct 5 '14 at 18:39

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.