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Let $X$ be a Banach space, the strong operator topology on the space of bounded linear operators $\mathcal{B}(X)$ is defined by the family of continuous semi-norms $A\to\|Ax\|$, $x\in X$. What is the dual space to $\mathcal{B}(X)$ in this topology? Can it be identified with something recognizable at least when $X$ is Hilbert or reflexive? For the weak operator topology I think that the dual space can be identified with the space of finite rank operators. I'll appreciate any references.

I am also interested in similar questions for the ultraweak and the ultrastrong topologies.

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  • $\begingroup$ the dual in ultraweak topology is nuclear operators $\endgroup$
    – Norbert
    Commented May 29, 2014 at 22:58
  • $\begingroup$ Is it the same for ultrastrong when $X$ is Hilbert? $\endgroup$
    – Conifold
    Commented May 30, 2014 at 1:30
  • $\begingroup$ I don't know... $\endgroup$
    – Norbert
    Commented May 30, 2014 at 9:26

1 Answer 1

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The answer is in part I of Linear Operators by Dunford and Schwartz, Theorem VI.1.4. For any Banach spaces $X$ and $Y$ linear functionals on $\mathcal{B}(X,Y)$ continuous in strong and weak operator topologies are the same. It follows from the proof that they are of the form $F(A)=\sum_{i=1}^n\varphi_i(Ax_i)$ for $x_i\in X$ and $\varphi_i\in Y^*$. In other words, the dual space can be identified with the algebraic tensor product $X\otimes Y^*$, or equivalently the space of finite rank operators $\mathcal{F}(Y,X)$.

For the ultraweak and the ultrastrong topologies on $\mathcal{B}(H)$, where $H$ is a Hilbert space, the answer is in Von Neumann Algebras by Dixmier, Lemma I.3.2. Again, the dual spaces coincide and can be identified with the projective tensor product $H\otimes_\pi H$, or the space of the trace class operators $\mathcal{L}_1(H)$. It also happens to coincide with the norm closure of $\mathcal{F}(H)$ in the Banach dual $\mathcal{B}(H)^*$, and is a Banach predual $\mathcal{B}(H)_*$ of $\mathcal{B}(H)$. Moreover, the ultraweak topology on $\mathcal{B}(H)$ coincides with the weak* topology generated by this predual, they are both $\sigma\big(\mathcal{B}(H),H\otimes_\pi H\big)$.

It doesn't appear that the ultraweak or the ultrastrong topologies were studied much beyond Hilbert spaces, I couldn't even find any established definitions for them. Since the weak operator topology coincides with $\sigma\big(\mathcal{B}(X,Y), X\otimes Y^*\big)$, by analogy to Hilbert spaces a natural candidate for the ultraweak is $\sigma\big(\mathcal{B}(X,Y), X\otimes_\pi Y^*\big)$. However, according to Introduction to Tensor Products of Banach Spaces by Ryan, Section 2.2 a predual to $\mathcal{B}(X,Y)$ is $X\otimes_\pi Y_*$, where $Y_*$ is a predual to $Y$. So the ultraweak so defined does not coincide with the weak* unless $Y$ is reflexive. And even if $Y$ is reflexive it is not clear what a natural candidate for the ultrastrong is or how the duals are related.

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  • $\begingroup$ I think this a partially related to absense of approximation property for general Banach spaces. $\endgroup$
    – Norbert
    Commented Jun 11, 2014 at 20:11
  • $\begingroup$ @Norbert The role of approximation properties is somewhat mysterious to me. To my surprise finite rank operators are strong operator dense in the space of bounded ones without any approximation properties math.stackexchange.com/questions/535645/… $\endgroup$
    – Conifold
    Commented Jun 12, 2014 at 23:58
  • $\begingroup$ I'm mean ultraweak topology $\endgroup$
    – Norbert
    Commented Jun 13, 2014 at 6:24

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