---
tag: VX7BFH22GX
kind: definition
name: Strong operator topology
msc: [47L05, 46L10, 46A03]
defines: KHJ2G2R5G7
uses: [HM2HHE3J51, BMDG4JQPEH, ZVFXQ7S6FY, 4XM1VN1C1N]
sources:
  - url: https://en.wikipedia.org/wiki/Strong_operator_topology
---

## Definition

Let $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $B(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$.
The *strong operator topology* on $B(H)$ is the coarsest topology such that each map $B(H) \to H$, $T \mapsto T x$, with $x \in H$, is continuous.
Here $H$ carries its norm topology.

## Remarks

Equivalently, it is the [[topology generated by the seminorms|ZVFXQ7S6FY]] $T \mapsto \|T x\|$ 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|4XM1VN1C1N]].
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.