In functional analysis, the weak operator topology, often abbreviated WOT,[1] is the weakest topology on the set of bounded operators on a Hilbert space
there is base of neighborhoods of the following type: choose a finite number of vectors
The WOT is the weakest among all common topologies on
, the bounded operators on a Hilbert space
The strong operator topology, or SOT, on
Because the inner product is a continuous function, the SOT is stronger than WOT.
The linear functionals 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
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.
The predual of B(H) is the trace class operators C1(H), and it generates the w*-topology on 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 B(H).
A net {Tα} ⊂ B(H) converges to T in WOT if and only Tr(TαF) 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 So {Tα} converges to T in WOT means 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 where the series
For every trace-class S, by invoking, for instance, the dominated convergence theorem.
Therefore every norm-bounded closed set is compact in WOT, by the Banach–Alaoglu theorem.
The adjoint operation T → T*, as an immediate consequence of its definition, is continuous in WOT.
Multiplication is not jointly continuous in WOT: again let
Appealing to Cauchy-Schwarz, one has that both Tn and T*n converges to 0 in WOT.
(Because WOT coincides with the σ-weak topology on bounded sets, multiplication is not jointly continuous in the σ-weak topology.)
However, a weaker claim can be made: multiplication is separately continuous in WOT.
We can extend the definitions of SOT and WOT to the more general setting where X and Y are normed spaces and
is the space of bounded linear operators of the form
The resulting family of seminorms generates the weak operator topology on
is a locally convex topological vector space when endowed with the WOT.
Thus, a topological base for the SOT is given by open neighborhoods of the form where as before
For instance, "strong convergence" for vectors in a normed space sometimes refers to norm-convergence, which is very often distinct from (and stronger than) than SOT-convergence when the normed space in question is
The weak topology on a normed space
is the coarsest topology that makes the linear functionals in
And while the WOT is formally weaker than the SOT, the SOT is weaker than the operator norm topology.
is a formally weaker topology than the SOT, but they nevertheless share some important properties.