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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

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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.

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    $\begingroup$ 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. $\endgroup$
    – David Gao
    Commented Mar 25 at 18:40

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