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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?

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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?

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  • $\begingroup$ Thanks. I meant ultrastrong and ultraweak respectively, but the question I linked used $\sigma$ terminology. Did anybody study these topologies for operators on Banach spaces? $\endgroup$
    – Conifold
    Commented Jun 1, 2014 at 21:02
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    $\begingroup$ The weak and strong operator topologie are uused sporadically in the context of Banach spaces but the ultraweak and ultrastrong have no direct analogies. As substitutes one can consider the finest locally convex topologies which agree with the WOT and the SOT on the unit ball. These have the advantage of being complete and, under suitable conditions (involving approximation properties), will have the appropriate dual spaces. To my knowledge, this has never been written up. $\endgroup$
    – couperin
    Commented Jun 2, 2014 at 5:06
  • $\begingroup$ In Ryan's book on tensor products it is mentioned that Banach predual of $\mathfrak{B}(X)$ is the projective tensor product $X\otimes_\pi X^*$ when $X$ is reflexive. Will ultraweak under your definition be the coarsest compatible with this duality? Will ultrastrong have the same dual? Sorry, I am used to metric spaces and have no feeling for what the finest locally convex topology that agrees on the unit ball might look like. Is there a way to see what the seminorms are? $\endgroup$
    – Conifold
    Commented Jun 2, 2014 at 22:16
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    $\begingroup$ Hilbert dpace has three properties which are relevant here: it is reflexive, it is self dual and it has the strongest possible approximation properties. All of these are relevant here. Thus the first topology on $L(E,F)$ I mentioned will be complete only if $F$ is reflexive, not in general as I sloppily claimed. The crucial property of the topologies I mentioned is that their duals are the completions of those of the operstors with wOT and SOT. To complete the analysis, one has to identity the latter with a space of finite rank operators and the completions with the trace class operators. $\endgroup$
    – couperin
    Commented Jun 3, 2014 at 5:00
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    $\begingroup$ I don't seem to be able to edit the above so forgive the typos. Also didn't have space for the final comment which is. As I said, nobody has finshed this analysis in detail to my knowledge so that I can't quote a theorem which nails precisely the conditions on $E$ and $F$ which ensure that the duals can be identified with a space of trace class operators. $\endgroup$
    – couperin
    Commented Jun 3, 2014 at 5:07
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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 $B(H)$.

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