Let $H$ be a Hilbert space and $B(H)$ denotes the space of all continuous linear operators on $H$. I am looking for a class/example of bounded linear functionals $B(H)\to \mathbb C$ which cannot be reduced to the type $$T\mapsto\sum_{i=1}^\infty k_i\left<Tx_i,y_i\right>$$ for some fixed $x_i$'s and $y_i$'s in $H$ and $k_i$'s in $\mathbb C$. Do we have any explicit examples?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
So, you are asking about non-normal functionals on $B(H)$. This is very similar to the question of what are the functionals on $\ell_\infty$ that are not in $\ell_1$?
Fix an ultrafilter $U$ on $\mathbb N$ and let $(e_n)$ be an ONB for $H$. Define
$$\langle \phi, T\rangle = \lim_{n,U} \langle Te_n, e_n\rangle\quad (T\in B(H)).$$
Then this functional is not normal.
answered Sep 29, 2020 at 21:39
-
$\begingroup$ Thank you for answering. Can you suggest some literature where I could find these results along with proofs? $\endgroup$– NewBSep 29, 2020 at 22:14
-
2$\begingroup$ @NewB, I'd go for Takesaki, Theory of operator algebras, vol. 1. $\endgroup$ Sep 30, 2020 at 6:06
-
$\begingroup$ As far as I could understand, you gave a recipe for linear functionals on B(H) which are not coming through the natural map $J:H^*\otimes^\gamma H\to B(H)^*$, since projective tensor product $H^*\otimes^\gamma H$ is nothing but trace class operators, the predual of $B(H)$. Does this work for any Banach space projective tensor product? I mean, can we get linear functionals on $(X\otimes^\gamma Y)^{*}$ which are not coming through J map, through the same recipe? $\endgroup$– NewBSep 30, 2020 at 15:47
-
$\begingroup$ Also, do you mean non-principle ultrafilter? Because for principle ultrafilter generated by $k$, the functional defined above is $T\mapsto \left<Te_k,e_k\right>$, which is again normal. Am I missing here something? $\endgroup$– NewBOct 1, 2020 at 20:24
-