Weak operator topology

In functional analysis, the weak operator topology, often abbreviated WOT, is the weakest topology on the set of bounded operators on a Hilbert space H {displaystyle H} , such that the functional sending an operator T {displaystyle T} to the complex number ⟨ T x , y ⟩ {displaystyle langle Tx,y angle } is continuous for any vectors x {displaystyle x} and y {displaystyle y} in the Hilbert space. In functional analysis, the weak operator topology, often abbreviated WOT, is the weakest topology on the set of bounded operators on a Hilbert space H {displaystyle H} , such that the functional sending an operator T {displaystyle T} to the complex number ⟨ T x , y ⟩ {displaystyle langle Tx,y angle } is continuous for any vectors x {displaystyle x} and y {displaystyle y} in the Hilbert space. Explicitly, for an operator T {displaystyle T} there is base of neighborhoods of the following type: choose a finite number of vectors x i {displaystyle x_{i}} , continuous functionals y i {displaystyle y_{i}} , and positive real constants ε i {displaystyle varepsilon _{i}} indexed by the same finite set I {displaystyle I} . An operator S {displaystyle S} lies in the neighborhood if and only if | y i ( T ( x i ) − S ( x i ) ) | < ε i {displaystyle |y_{i}(T(x_{i})-S(x_{i}))|<varepsilon _{i}} for all i ∈ I {displaystyle iin I} . Equivalently, a net T i ⊂ B ( H ) {displaystyle T_{i}subset B(H)} of bounded operators converges to T ∈ B ( H ) {displaystyle Tin B(H)} in WOT if for all y ∈ H ∗ {displaystyle yin H^{*}} and x ∈ H {displaystyle xin H} , the net y ( T i x ) {displaystyle y(T_{i}x)} converges to y ( T x ) {displaystyle y(Tx)} . The WOT is the weakest among all common topologies on B ( H ) {displaystyle B(H)} , the bounded operators on a Hilbert space H {displaystyle H} . The strong operator topology, or SOT, on B ( H ) {displaystyle B(H)} 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 H = ℓ 2 ( N ) {displaystyle H=ell ^{2}(mathbb {N} )} and consider the sequence { T n } {displaystyle {T^{n}}} of unilateral shifts. An application of Cauchy-Schwarz shows that T n → 0 {displaystyle T^{n} o 0} in WOT. But clearly T n {displaystyle T^{n}} does not converge to 0 {displaystyle 0} in SOT. 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 B ( H ) {displaystyle B(H)} 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 { T α } {displaystyle {T_{alpha }}} converges to 0 {displaystyle 0} in SOT if and only if T α ∗ T α → 0 {displaystyle T_{alpha }^{*}T_{alpha } o 0} in WOT. 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