Strong operator topology

Remarks

Equivalently, it is the topology generated by the seminorms $T \mapsto \|T x\|$ for $x \in H$. A net $(T_i)$ converges to $T$ in this topology if and only if $\|T_i x - T x\| \to 0$ for every $x \in H$.

Example: on $H = \ell^2(\N)$ with standard basis $(e_k)$, let $P_n$ be the orthogonal projection onto the span of $e_0, \dots, e_{n-1}$. For every $x$, $\|x - P_n x\|^2 = \sum_{k \ge n} \abs{x_k}^2 \to 0$, so $P_n \to I$ strongly. But $\|I - P_n\| = 1$ for all $n$, so there is no convergence in norm.

Non-example: let $S$ be the right shift on $\ell^2(\N)$. The powers $S^n$ tend to $0$ in the weak operator topology. If $S^n$ converged strongly to some $T$, then $T$ would also be the weak limit, so $T = 0$; but $\|S^n x\| = \|x\|$ does not tend to $0$ for $x \ne 0$. So $(S^n)$ has no strong limit, while the adjoint powers $(S^*)^n$, which delete the first $n$ entries, do tend to $0$ strongly. In particular $T \mapsto T^*$ is not continuous in the strong operator topology.

The strong operator topology is finer than the weak operator topology and coarser than the norm topology. A convex subset of $B(H)$ has the same closure in both the weak and the strong operator topology.