1
$\begingroup$

Let $\mathcal{H}$ be a Hilbert space. I would like to show:

A $\phi : \mathcal{B(H)}\to\mathbb{C}$ linear functional is continous in the strong operator-topology iff is continous in the weak operator-topology.

The SOT $\Rightarrow$ WOT implication is clear, but the reverse is not.

$\endgroup$
8
  • 2
    $\begingroup$ The question now makes sense, but now you have it the wrong way around. The trivial direction is "WOT$\implies$ SOT" $\endgroup$
    – Simon
    Nov 27, 2016 at 12:53
  • 1
    $\begingroup$ $\phi$ is WOT-continous iff for every WOT-convergent sequence $A_n\to A\in B(H)$ we have $\phi(A_n)\to\phi(A)$. If you replace "WOT" with "SOT" in that statement, you make the assumption stronger, so the statement becomes weaker. $\endgroup$
    – Simon
    Nov 27, 2016 at 12:59
  • $\begingroup$ Yes, I had realised and deleted my comment. $\endgroup$
    – s.harp
    Nov 27, 2016 at 13:00
  • 2
    $\begingroup$ Also, since WOT and SOT are not metrizisable, it is not enough to check simply that the implication holds on the level of sequential continuity. $\endgroup$
    – s.harp
    Nov 27, 2016 at 14:40
  • 1
    $\begingroup$ just replace "sequence" with "net". Also: I would be interested in the "SOT$\implies$ WOT" proof, as that is the non-trivial one. $\endgroup$
    – Simon
    Nov 27, 2016 at 15:34

1 Answer 1

1
$\begingroup$

$\phi$ is WOT-continous iff for every WOT-convergent net $A_n\to A\in\mathcal{B(H)}$ we have $\phi(A_n)\to\phi(A)$. If you replace "WOT" with "SOT" in that statement, you make the assumption stronger, so the statement becomes weaker.

$\endgroup$
3
  • 1
    $\begingroup$ just replace "sequence" with "net", and it works. $\endgroup$
    – Simon
    Nov 27, 2016 at 15:32
  • $\begingroup$ right, sometimes I wonder whether I think at all $\endgroup$
    – s.harp
    Nov 27, 2016 at 16:03
  • $\begingroup$ @s.harp well, I wasn't thinking too much either. I though "hey, $\mathbb{C}$ is metric, so sequences are enough"... which is not true. $\endgroup$
    – Simon
    Nov 27, 2016 at 16:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .