# Strong operator topology

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Strong operator topology

In functional analysis, a branch of mathematics, the strong operator topology, often abbreviated SOT, is the weakest topology on the set of bounded operators on a Hilbert space (or, more generally, on a Banach space) such that the evaluation map sending an operator "T" to the real number $|Tx|$ is continuous for each vector "x" in the Hilbert space.

The SOT is stronger than the weak operator topology and weaker than the norm topology.

The SOT lacks some of the nicer properties that the weak operator topology has, but being stronger, things are sometimes easier to prove in this topology. It is more natural too, since it is simply the topology of pointwise convergence for an operator.

As an example of this lack of nicer properties, let us mention that the involution map is not continuous in this topology: fix an orthonormal basis $\left\{e_n:ninmathbb\left\{N\right\}\right\}$ of a Hilbert space and consider the unilateral shift $S$ given by

:$S\left(e_n\right)=e_\left\{n+1\right\}.$

Then the adjoint $S^*$ is given by

:$S^*\left(e_\left\{n\right\}\right)=e_\left\{n-1\right\}, nge1, S^*\left(e_0\right)=0.$

The sequence $\left\{S^n\right\}$ satisfies

:$|S^n\left(x\right)|=|x|$,

for every vector $x$, but

:$lim_\left\{n ightarrowinfty\right\}\left(S^*\right)^n=0$

in the SOT topology. This means that the adjoint operation is not SOT-continuous.

On the other hand, the SOT topology provides the natural language for the generalization of the spectral theorem to infinite dimensions. In this generalization (due to John von Neumann), the sum of multiples of projection is replaced by an integral over a projection-valued measure. The required notion of convergence is then that of the SOT topology. The SOT topology also provides the framework for the measurable functional calculus, just as the norm topology does for the continuous functional calculus.

The linear functionals on the set of bounded operators on a Hilbert space which are continuous in the SOT are precisely those which are continuous in the WOT. 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.

ee also

*Topologies on the set of operators on a Hilbert space

References

*cite book |last=Rudin |first=Walter |title=Functional Analysis |year=1991 |month=January |publisher=McGraw-Hill Science/Engineering/Math |id=ISBN 0-07-054236-8
*cite book |last=Pedersen |first=Gert |title=Analysis Now |year=1989 |publisher=Springer |id=ISBN 0-387-96788-5

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