Weak operator topology

Remarks

Equivalently, it is the topology generated by the seminorms $T \mapsto \abs{\langle T x, y \rangle}$ for $x, y \in H$. A net $(T_i)$ converges to $T$ in this topology if and only if $\langle T_i x, y \rangle \to \langle T x, y \rangle$ for all $x, y \in H$.

Example: on $H = \ell^2(\N)$, let $S$ be the right shift, $S(x_0, x_1, \dots) = (0, x_0, x_1, \dots)$. Then $\langle S^n x, y \rangle = \sum_{k \ge 0} x_k \overline{y_{k+n}}$, and by the Cauchy–Schwarz inequality its modulus is at most $\|x\| \, (\sum_{k \ge n} \abs{y_k}^2)^{1/2}$, which tends to $0$. So $S^n \to 0$ in the weak operator topology, while $\|S^n\| = 1$ for all $n$.

Non-example: let $\lambda \in \C$ with $\abs{\lambda} = 1$ and $\lambda \ne 1$, and let $T_n = \lambda^n I$. For $x = y \ne 0$, the numbers $\langle T_n x, x \rangle = \lambda^n \|x\|^2$ do not converge, so $(T_n)$ has no limit in this topology.

The weak operator topology is coarser than the strong operator topology, which is coarser than the norm topology. The closed unit ball of $B(H)$ is compact in the weak operator topology. Multiplication is separately continuous but not jointly continuous when $H$ is infinite-dimensional, and $T \mapsto T^*$ is continuous.