Editing Weak operator topology (section)

==Relationship with other topologies on ''B''(''H'')==

The WOT is the weakest among all common [[Topologies on the set of operators on a Hilbert space|topologies on <math>B(H)</math>]], the bounded operators on a Hilbert space <math>H</math>.

===Strong operator topology===

The [[strong operator topology]], or SOT, on <math>B(H)</math> is the topology of pointwise convergence. Because the inner product is a continuous function, the SOT is stronger than WOT. The following example shows that this inclusion is strict. Let <math>H = \ell^2(\mathbb N)</math> and consider the sequence <math>\{T^n\}</math> of right shifts.  An application of Cauchy-Schwarz shows that <math>T^n \to 0</math> in WOT. But clearly <math>T^n</math> does not converge to <math>0</math> in SOT.

The [[linear functional]]s on the set of bounded operators on a Hilbert space that are continuous in the [[strong operator topology]] are precisely those that are continuous in the WOT (actually, the WOT is the weakest operator topology that leaves continuous all strongly continuous linear functionals on the set <math>B(H)</math> of bounded operators on the Hilbert space ''H''). Because of this fact, the closure of a [[convex set]] of operators in the WOT is the same as the closure of that set in the SOT.

It follows from the [[polarization identity]] that a net <math>\{T_\alpha\}</math> converges to <math>0</math> in SOT if and only if <math>T_\alpha^* T_\alpha \to 0</math> in WOT.

===Weak-star operator topology===

The predual of ''B''(''H'') is the [[trace class]] operators C<sub>1</sub>(''H''), and it generates the w*-topology on&nbsp;''B''(''H''), called the [[weak-star operator topology]] or σ-weak topology. The weak-operator and σ-weak topologies agree on norm-bounded sets in&nbsp;''B''(''H'').

A net {''T<sub>&alpha;</sub>''} ⊂ ''B''(''H'') converges to ''T'' in WOT if and only Tr(''T<sub>&alpha;</sub>F'') converges to Tr(''TF'') for all [[finite-rank operator]] ''F''. Since every finite-rank operator is trace-class, this implies that WOT is weaker than the σ-weak topology. To see why the claim is true, recall that every finite-rank operator ''F'' is a finite sum

:<math> F = \sum_{i=1}^n \lambda_i u_i v_i^*.</math>

So {''T<sub>&alpha;</sub>''} converges to ''T'' in WOT means

:<math> \text{Tr} \left ( T_{\alpha} F \right )  =  \sum_{i=1}^n \lambda_i v_i^* \left ( T_{\alpha} u_i \right ) \longrightarrow \sum_{i=1}^n \lambda_i v_i^* \left ( T u_i \right ) = \text{Tr} (TF).</math>

Extending slightly, one can say that the weak-operator and σ-weak topologies agree on norm-bounded sets in ''B''(''H''): Every trace-class operator is of the form

:<math> S = \sum_i \lambda_i u_i v_i^*,</math>

where the series <math>\sum\nolimits_i \lambda_i</math> converges. Suppose <math>\sup\nolimits_{\alpha} \|T_{\alpha} \| = k < \infty,</math> and <math>T_{\alpha} \to T</math> in WOT. For every trace-class ''S'',

:<math> \text{Tr} \left ( T_{\alpha} S \right )  =  \sum_i \lambda_i v_i^* \left ( T_{\alpha} u_i \right ) \longrightarrow \sum_i \lambda_i v_i^* \left ( T u_i \right ) = \text{Tr} (TS),</math>

by invoking, for instance, the [[dominated convergence theorem]].

Therefore every norm-bounded closed set is compact in WOT, by the [[Banach–Alaoglu theorem]].