Continuous linear functionals in strong operator and $\sigma$-strong topologies It was mentioned in the comments to https://math.stackexchange.com/questions/517369/comparison-of-strong-operator-and-weak-topologies-on-bh that continuous linear functionals on $\mathfrak{B}(\mathbb{H})$ are the same in the strong operator topology as in the weak operator topology. It doesn't seem obvious and I couldn't find it proved. Can someone give an idea of a proof or a reference to it? 
Does the same apply to $\sigma$-strong and $\sigma$-weak (weak*) topologies, can continuous linear functionals be identified with trace class operators in both cases?
 A: The duals under the strong and the weak topologies can be identified with the finite rank operators, under the ultrastrong and the ultraweak with the trace class operators.  This can be found in the classic "von Neumann algebras" by Dixmier.
Can you make precise what you mean by the $\sigma$-strong and $\sigma$-weak topologies?
A: The $\sigma$-case is just a trivial extension of the following argument. Take a strongly continuous functional $\phi$ on $B(H)$. Then by definition $|f(A)|\leq p(A)$ for some strong seminorm $p$. But $p$ is of the form $\Vert{\pi(A)z}\Vert$ for some vector $z$ in the algebraic direct sum of countably many copies of $H$, and where $\pi$ is the diagonal $*$-homomorphism $A\mapsto A\oplus A\oplus \cdots$. The linear functional $\psi$ that sends $\pi(A)z$ to $\phi(A)$ is clearly bounded by $p$ and therefore it can be extended to the closure of its domain, which is then a Hilbert space with the relative inner product structure coming from $H$. By Riesz representation theorem there exists $z'\in H$ such that $\psi(\pi(A)z) = (z',\pi(A)z)$, which is a weakly continuous linear functional by definition. Hence $f(A) = (z',\pi(A)z)$.
For the $\sigma$-case, start with a $\sigma$-strongly continuous functional, find $z$ in the Hilbert-space direct sum $H^{\oplus\infty}$. Repeat all the above steps to find a vector $z'\in H^{\oplus\infty}$ that gives you a $\sigma$-weakly continuous functional on $B(H)$.
