SOT and WOT Dual

Question

Let $H$ be an infinite dimensional Hilbert space and let $e_{n}$ be an orthonormal basis of $H$. Let $\phi$ be a linear functional defined on $B(H)$ as follows. $ϕ(A)=\underset{n}∑(\frac{1}{2})^n⟨Ae_{n},e_n⟩$ for A ∈ B(H). Clearly $\phi$ is SOT continuous. Also ϕ is faithful. But we know that no wot continuous linear functional on B(H) is faithful when H is infinite dimensional. In particular $\phi$ is not WOT continuous. This is a contradiction since WOT dual and SOT dual are same. Please help me in finding the error?

Answer 1

The functional $\phi$ is not SOT-continuous. I will use a tweak of the net used in this answer.

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: let $B=\sum_n2^{-n}\,\langle \,\cdot\, e_n,e_n\rangle$, so $\phi=\operatorname{Tr}(B\,\cdot\,)$, and let $$ T_F=\frac1{\operatorname{Tr}(BP_{F^\perp})}\,P_{F^\perp}, $$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 $T_Fx=0$, and then $T_F\to0$ in the sot topology.

On the other hand, $$ \phi(T_F)=\frac{\operatorname{Tr}(BP_{F^\perp})}{\operatorname{Tr}(BP_{F^\perp})}=1. $$