H
T
\langleTx,y\rangle
x
y
Explicitly, for an operator
T
xi
yi
\varepsiloni
I
S
|yi(T(xi)-S(xi))|<\varepsiloni
i\inI
Ti\subseteqB(H)
T\inB(H)
y\inH*
x\inH
y(Tix)
y(Tx)
The WOT is the weakest among all common topologies on B(H)
H
The strong operator topology, or SOT, on
B(H)
H=\ell2(N)
\{Tn\}
Tn\to0
Tn
0
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)
It follows from the polarization identity that a net
\{T\alpha\}
0
* | |
T | |
\alpha |
T\alpha\to0
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 ⊂ 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
F=
n | |
\sum | |
i=1 |
λiui
*. | |
v | |
i |
So converges to T in WOT means
Tr\left(T\alphaF\right)=
n | |
\sum | |
i=1 |
λi
* | |
v | |
i |
\left(T\alphaui\right)\longrightarrow
n | |
\sum | |
i=1 |
λi
* | |
v | |
i |
\left(Tui\right)=Tr(TF).
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
S=\sumiλiui
*, | |
v | |
i |
where the series
\sum\nolimitsiλi
\sup\nolimits\alpha\|T\alpha\|=k<infty,
T\alpha\toT
Tr\left(T\alphaS\right)=\sumiλi
* | |
v | |
i |
\left(T\alphaui\right)\longrightarrow\sumiλi
* | |
v | |
i |
\left(Tui\right)=Tr(TS),
by invoking, for instance, the dominated convergence theorem.
Therefore every norm-bounded 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
T
n
However, a weaker claim can be made: multiplication is separately continuous in WOT. If a net Ti → T in WOT, then STi → ST and TiS → TS in WOT.
We can extend the definitions of SOT and WOT to the more general setting where X and Y are normed spaces and
B(X,Y)
T:X\toY
x\inX
y*\inY*
\| ⋅ \| | |
x,y* |
B(X,Y)
\|T\| | |
x,y* |
=|y*(Tx)|
B(X,Y)
B(X,Y)
N(T,F,Λ,\epsilon):=\left\{S\inB(X,Y):\left|y*((S-T)x)\right|<\epsilon,x\inF,y*\inΛ\right\},
where
T\inB(X,Y),F\subseteqX
Λ\subseteqY*
\epsilon>0
B(X,Y)
The strong operator topology on
B(X,Y)
\| ⋅ \|x,x\inX,
\|T\|x=\|Tx\|
N(T,F,\epsilon):=\{S\inB(X,Y):\|(S-T)x\|<\epsilon,x\inF\},
where as before
T\inB(X,Y),F\subseteqX
\epsilon>0.
The different terminology for the various topologies on
B(X,Y)
B(X,Y)
X
X*
B(X,Y)
X
In general, the following inclusions hold:
\{WOT-opensetsinB(X,Y)\}\subseteq\{SOT-opensetsinB(X,Y)\}\subseteq\{operator-norm-opensetsinB(X,Y)\},
and these inclusions may or may not be strict depending on the choices of
X
Y
The WOT on
B(X,Y)
(B(X,Y),SOT)*=(B(X,Y),WOT)*.
Consequently, if
S\subseteqB(X,Y)
\overline{S}SOT=\overline{S}WOT,
in other words, SOT-closure and WOT-closure coincide for convex sets.