The reference text for this question is: Pedersen, Analysis Now, GTM 118.

The $\sigma$-weak topology on $B(H)$ (the bounded linear operators on a Hilbert space $H$) is the weak$^*$-topology on $B(H)$ seen as the dual of the Banach algebra $B_1(H)$ ($T\in B_1(H)$ if $T$ is a compact operator such that its norm in $B_1(H)$ is $tr(|T|)<\infty$ - see Pedersen 4.6.10, 3.4.6, 3.4.12, 3.4.13).

The weak operator topology on $B(H)$ is the one generated by the semi norms $|(Tx,y)|$ as $x,y$ range in $H$ and $(\cdot,\cdot)$ is the inner product on $H$ - see Pedersen 4.6.1.

It is shown in Proposition 4.6.14 that the weak operator topology and the $\sigma$-weak topology overlap on the ball $B(H,n)$ of operators $T\in B(H)$ of norm at most $n$.

This gives that any $\sigma$-weakly continuous functional $\phi$ when restricted to $B(H,1)$ is continuous with respect to the weak operator topology.

Now propositions 4.6.11 and 4.6.4 give the following characterizations of continuous functionals in these topologies:

4.6.11: a functional $\phi:B(H)\to\mathbb{C}$ is $\sigma$-weakly continuous if and only if there exists sequences $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ of elements of $H$ such that $\sum_{n\in\mathbb{N}}||x_n||+\sum_{n\in\mathbb{N}}||y_n||<\infty$ and $\phi(S)=\sum_{n\in\mathbb{N}}(S(x_n),y_n)$.

4.6.4: a functional $\phi:B(H)\to\mathbb{C}$ is continuous with respect to the weak operator topology if and only if there exists sequences $(x_0,\dots, x_n)$ and $(y_0,\dots,y_n)$ of elements of $H$ such that $\phi(S)=\sum_{j=0,\dots, n}(S(x_j),y_j)$.

I can follow the proof of proposition 4.6.14, but I don't understand how this translates at the level of these different characterizations, i.e.:

Assume $\phi(S)=\sum_{n\in\mathbb{N}}(S(x_n),y_n)$ is $\sigma$-weakly continuous, what are the vectors $(z_0,\dots, z_n)$ and $(w_0,\dots,w_n)$ such that $\phi(S)=\sum_{i\leq n}(S(z_i),w_i)$ for all operators $S$ of norm at most 1?

How can it happen that a $\sigma$-weak continuous $\phi$ functional is such that $\phi\restriction B(H,n)=\psi_n$ with $\psi_n$ continuos for the weak operator topology but for all $n$ $\phi\neq\psi_n$?

Some examples would be illuminating.

Thanks in advance


1 Answer 1


For your first question, I think you are misunderstanding what the theorems say. When you restrict your $\sigma$-weakly continuous functional to the unit ball, you don't get a wot-continuous functional on the whole space: so 4.6.4 does not apply.

For your second question, here is an example: fix an orthonormal basis $\{e_n\}$ and let $$ \phi(S)=\sum_n\frac1{n^2}\langle Se_n,e_n\rangle. $$ This is clearly $\sigma$-weakly continuous by 4.6.11. And by 4.6.14 (I'm using your numbers, I don't have the book at hand) it is weakly continuous on bounded sets. But it is not weakly continuous: here is an example I posted a few days ago. I'm not sure if you are aware, but the continuity does not fail on sequences (a weakly convergent sequence is bounded) so we need nets.

Let $\mathcal F=\{F\subset H:\ F \text{ is a finite-dimensional subspace} \}$, ordered by inclusion. We construct a net of operators indexed by $\mathcal F$ as follows: $$ T_F=\frac1{\phi(P_{F^\perp})}\,P_{F^\perp}, $$ where $P_{F^\perp}$ is the orthogonal projection onto $F^\perp$. Then, for any $x\in H$, if we move far enough along the net we will have $x\in F$, so $T_Fx=0$. So $T_F\to0$ sot, and then $$ \langle T_Fx,y\rangle\to0 $$ for any $x,y$. That is, $T_F\to0$ weakly.

But, for any $F$, $$ \phi(T_F)=\frac{\phi(P_{F^\perp})}{\phi(P_{F^\perp})}=1. $$ So $\{\phi (T_F)\}$ does not converge to $\phi (0)=0$, showing that $\phi $ is not strongly/weakly continuous.

  • $\begingroup$ Thanks! useful answer, I got the first part of your reply, i.e. the pre image under $\phi$ of a bounded set which is wot-closed is wot-closed, but the same does not occur for unbounded wot closed sets. However I do not follow your counterexample for the second part. What is the relation between the net $T_F$ and $\phi$? what is $A$ in the definition of $T_F$? Thanks again in any case $\endgroup$ Nov 21, 2014 at 8:21
  • $\begingroup$ Sorry, I was using the wrong notation. Edited. $\endgroup$ Nov 21, 2014 at 12:53
  • $\begingroup$ Thanks, now everything is clear, I voted for your reply. $\endgroup$ Nov 22, 2014 at 12:04

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