Talk:Strong operator topology

The definition is wrong.

The article page says "... 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."

So defined topology is not translation invariant, nor locally convex, nor even Hausdorff, nor satisfies most claims of the article. For example it is not Hausdorff since there is no open set distinguishing the right shift S and its power $S^{2}$. (The zero operator is the only one separated by its neighborhoods from the rest of the space.)

Please, correct the article (provisional correction done by adding 'locally convex' to include the translational invariance ad hock). Remember that no information is much better than wrong information.

Copy the definition correctly from the source, usually no word (such as 'locally convex' for example) can be ommited.

(Note that also the example turned affected by correcting the definition.)

90.180.192.165 (talk) 16:44, 2 April 2010 (UTC)

I corrected the definition and added an equivalent one.

Kappaenne (talk) 11:04, 16 September 2014 (UTC)

The example about * removed

I removed the text (example) below from the article, since it is affected by correcting the definition. First to be said, the example claims to show $*$ is not continuous, for which it (silently) argues (or it seems to) that $S^{n}(x)$ converges in SOT to identity (while the adjoints do not), supporting it by formula

$\|S^{n}(x)\|=\|x\|$.

After the correction of definition, we would need that

$\|S^{n}(x)-x\|$

converge to 0 which is not true, e.g. for $x=e_{1}$.

Well, it looks like the example could be reversed:

$\lim _{n\rightarrow \infty }(S^{*})^{n}=0$

in the SOT topology while its adjoints $((S^{*})^{n})*=S^{n}$ do not converge to $0*=0$.

Please would you check that and reformulate the example?

——————— 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 $\{e_{n}:n\in \mathbb {N} \}$ of a Hilbert space and consider the unilateral shift $S$ given by

$S(e_{n})=e_{n+1}.$

Then the adjoint $S^{*}$ is given by

$S^{*}(e_{n})=e_{n-1},n\geq 1,\ \ S^{*}(e_{0})=0.$

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

$\|S^{n}(x)\|=\|x\|$,

for every vector $x$, but

$\lim _{n\rightarrow \infty }(S^{*})^{n}=0$

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

SOT as an acronym should be a disambiguation page, not a redirect.

There are other uses for SOT as an acronym. e.g. Small Outline Transistor. Sound on tape. DFH (talk) 18:22, 25 May 2010 (UTC)