Operator topology

An operator topology specifies convergence of linear operators. Common choices include the norm, strong, and weak operator topologies.

On the bounded linear operators $B(X,Y)$ between normed vector spaces, the strong operator topology is generated by the seminorms $p_x(T)=\Vert Tx{\Vert }_Y$, one for every $x\in X$. Thus a net $T_i$ converges to $T$ precisely when $\Vert (T_i-T)x{\Vert }_Y\to 0$ for each fixed vector $x$. This defines a topology, including cases where sequences alone do not characterize it.

Convergence in operator norm implies convergence in the strong operator topology, since $\Vert (T_i-T)x\Vert \le \Vert T_i-T\Vert \Vert x\Vert$. The converse fails even for orthogonal projections. On the Hilbert space ${\ell }^2$, let $P_n$ retain the first $n$ coordinates. The tail of a square-summable sequence tends to zero, so $P_{n}x\to x$ for every $x$, but $\Vert P_n-I\Vert =1$. An increasing sequence of finite-dimensional approximations can therefore converge strongly without uniformly approximating every unit vector.

This topology on bounded operators is generated by the seminorms $(\Vert T\xi {\Vert }^2+\Vert T^{\ast }\xi {\Vert }^2{)}^{1/2}$. A net converges exactly when both it and its adjoint net converge in the strong operator topology. It is the topology used in the full-unit-ball version of the Kaplansky density theorem.

On $B(H)$ for a Hilbert space $H$, the weak operator topology is generated by the seminorms $T\mapsto |⟨Tx,y⟩|$, for all vectors $x,y\in H$. A net $T_i$ converges to $T$ exactly when every matrix coefficient $⟨(T_i-T)x,y⟩$ tends to zero. The strong operator topology implies the weak operator topology by the Cauchy-Schwarz inequality.

The reverse implication fails. For the unilateral shift $S$ on ${\ell }^2$, $S^n$ tends to zero in the weak operator topology. For finitely supported vectors $x,y$, the coefficient $⟨S^{n}x,y⟩$ eventually vanishes; approximation by finitely supported vectors and $\Vert S^n\Vert =1$ extend this to arbitrary $x,y$. But $\Vert S^{n}x\Vert =\Vert x\Vert$, so $S^n$ does not tend strongly to zero for nonzero $x$. Matrix coefficients can vanish while the total norm of a moving vector stays constant.

Let $H$ be a Hilbert space and let $E_0$ be the space of finite-rank operators on $H$, equipped with the trace norm. A rank-one operator $R_{x,y}z=⟨z,y⟩x$ gives the matrix coefficient $T\mapsto \mathrm{Tr}(T{R}_{x,y})=⟨Tx,y⟩$, with the inner product linear in its first argument. Trace-class duality identifies $E_0^{\ast }$ isometrically with $B(H)$. However, $E_0$ is generally incomplete: the usual Banach space predual is its completion, the trace-class operators ${\mathcal{S}}_1(H)$.

The weak operator topology tests only these matrix coefficients, equivalently all pairings with $E_0$. The weak-star topology supplied by the complete predual tests every trace-class operator. These two topologies agree on each operator-norm-bounded subset. Indeed, if $\Vert T_i\Vert ,\Vert T\Vert \le M$ and $A_N$ approximates $A$ in trace norm, thenFirst choose $N$ and then use matrix-coefficient convergence for $A_N$. This also explains why the operator unit ball is compact in the weak operator topology: apply the Banach-Alaoglu theorem to the complete predual and restrict to that ball. Completeness of a predual and agreement of topologies on bounded sets are distinct issues.