# Are Strong- and weak Operator topologies on separable Hilbert spaces sequential?

If I am not mistaken, the norm operator topology should make the set of bounded operators into a sequential space, since the norm defines a metric. I was wondering if the Weak and Strong Operator topologies also turn the bounded operators into a sequential space, i.e. is a sequentially closed set in those topologies automatically closed? If that makes things easier, I am only interested in bounded operators on a separable Hilbert space.

Best Lev

I’ll first prove that the strong operator topology is not sequential on $$B(l^2)$$. We consider the following subset of $$B(l^2)$$:

$$A = \{np_{V^\perp}: V \subset l^2 \, \mathrm{finite \, dimensional \, subspace}, \, n \in \mathbb{N}_+, \, n \geq \dim(V)\}$$

It is not closed in the strong operator topology. Indeed, let $$\lambda$$ be the net of nontrivial finite dimensional subspaces of $$l^2$$, ordered by inclusion, then it is not hard to see that $$\lim_{V \in \lambda} \dim(V)p_{V^\perp} = 0$$ in the strong operator topology. Indeed, for any $$h \in l^2$$, as long as $$V \supset \mathrm{span}\{h\}$$, we have $$\dim(V)p_{V^\perp}(h) = 0$$. But $$0 \notin A$$.

However, I claim that $$A$$ is sequentially closed. Indeed, let $$(a_m) _{m \in \mathbb{N}} \subset A$$ be a sequence converging strongly to $$a \in B(l^2)$$. By uniform boundedness principle, there exists $$C > 0$$ s.t. $$\|a_m\| \leq C$$ for all $$m$$. But $$\|np_{V^\perp}\| = n$$, so this means all $$a_m$$ must be of the form $$a_m = n_mp_{V_m^\perp}$$ where $$\dim(V_m) \leq n_m \leq C$$. There are only finitely many choices of $$n_m$$, so by passing to a subsequence if necessary, we may assume there exists $$N > 0$$ s.t. $$n_m = N$$ for all $$m$$. Thus, $$a_m = Np_{V_m^\perp}$$ where $$\dim(V_m) \leq N$$.

Recall that the set of projections is strongly closed. As $$a_m \to a$$ strongly, we must have $$a = Np$$ for some projection $$p$$ and $$p_{V_m^\perp} \to p$$ strongly. Thus, $$p_{V_m} \to 1-p$$ strongly. I claim that $$\dim(1-p) \leq N$$. Indeed, assume to the contrary, then we may let $$\{h_1, \cdots, h_t\}$$ be a finite orthonormal set in the range of $$1-p$$ with $$t > N$$. The Gramian $$G_{kl} = \langle h_k, h_l \rangle = \langle (1-p)(h_k), (1-p)(h_l) \rangle$$ is then the identity matrix, so in particular $$\det(G) \neq 0$$. But $$p_{V_m} \to 1-p$$ strongly, so the Gramian $$G_m$$ of $$\{p_{V_m}(h_1), \cdots, p_{V_m}(h_t)\}$$, given by $$(G_m)_{kl} = \langle p_{V_m}(h_k), p_{V_m}(h_l) \rangle$$, converges to $$G$$. Hence, for large $$m$$, $$\det(G_m) \neq 0$$, so for large $$m$$, $$\{p_{V_m}(h_1), \cdots, p_{V_m}(h_t)\}$$ is linearly independent, whence $$t \leq \dim(V_m) \leq N < t$$, a contradiction.

But this means $$1-p = p_V$$ for some $$V \subset l^2$$ with $$\dim(V) \leq N$$, so $$a = Np = Np_{V^\perp} \in A$$. That is, $$A$$ is sequentially closed.

The proof that the weak operator topology on $$B(l^2)$$ is not sequential is somewhat similar. We change the definition of $$A$$ a bit:

$$A = \{n(1 - T): T \, \mathrm{is \, a \, finite \, rank \, operator}, \, n \in \mathbb{N}_+, \, n \geq \mathrm{rank}(T)\}$$

By the exact same reasoning, $$A$$ is not closed in the weak operator topology (in fact, not even closed in the strong operator topology). To show it’s sequentially closed, let $$(a_m) _{m \in \mathbb{N}} \subset A$$ be a sequence converging weakly to $$a \in B(l^2)$$. Again by uniform boundedness principle, $$(a_m)$$ is uniformly bounded. When $$T$$ is of finite rank, $$\|1 - T\| \geq 1$$, so again by the same reasoning, we may assume there exists $$N > 0$$ s.t. $$a_m = N(1-T_m)$$ where $$\mathrm{rank}(T_m) \leq N$$ for all $$m$$.

Write $$a$$ as $$a = N(1 - T)$$, then $$T_m \to T$$ weakly. Again, I claim that $$\mathrm{rank}(T) \leq N$$. Assume to the contrary, then we may let $$\{h_1, \cdots, h_t\}$$ be a finite orthonormal set in the range of $$T$$ with $$t > N$$. Let $$\xi_k$$ be chosen so that $$T(\xi_k) = h_k$$ for all $$1 \leq k \leq t$$. Let $$p$$ be the orthogonal projection onto $$\mathrm{span}\{h_1, \cdots, h_t\}$$. The Gramian $$G_{kl} = \langle h_k, h_l \rangle = \langle pT(\xi_k), pT(\xi_l) \rangle$$ is then the identity matrix, so in particular $$\det(G) \neq 0$$. But $$T_m \to T$$ weakly, so for each $$1 \leq k \leq t$$, $$T_m(\xi_k) \to T(\xi_k)$$ weakly. As $$p$$ is a finite rank projection, $$pT_m(\xi_k) \to pT(\xi_k)$$ in norm. Thus, the Gramian $$G_m$$ of $$\{pT_m(\xi_1), \cdots, pT_m(\xi_t)\}$$, given by $$(G_m)_{kl} = \langle pT_m(\xi_k), pT_m(\xi_l) \rangle$$, converges to $$G$$. Hence, for large $$m$$, $$\det(G_m) \neq 0$$, so for large $$m$$, $$\{pT_m(\xi_1), \cdots, pT_m(\xi_t)\}$$ is linearly independent, whence $$t \leq \mathrm{rank}(pT_m) \leq \mathrm{rank}(T_m) \leq N < t$$, a contradiction. So, again, $$a \in A$$ and $$A$$ is sequentially closed.

• Note that on uniformly bounded subsets of $B(H)$ where $H$ is separable, both the strong and the weak operator topologies are metrizable, so in particular sequential. Thus, the fact that $A$ in my answer does not have a uniform norm bound is necessary. Commented Mar 25 at 18:40