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Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.

Consider the bounded operators: $$\mathcal{B}(\mathcal{H},\mathcal{K}):=\{T:\mathcal{H}\to\mathcal{K}:\|T\|<\infty\}$$

Regard the linear functionals: $$l_{(\varphi,\psi)}T:=\langle T\varphi,\psi\rangle_\mathcal{K}:\quad|l_{(\varphi,\psi)}T|\leq\|\varphi\|_\mathcal{H}\|\psi\|_\mathcal{K}\cdot\|T\|$$

Do these exhaust its topological dual: $$\mathcal{B}(\mathcal{H},\mathcal{K})'=\overline{\langle\{l_{(\varphi,\psi)}:(\varphi,\psi)\in\mathcal{H}\times\mathcal{K}\}\rangle}$$ (I'm wondering about this already for some time.)

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  • $\begingroup$ In finite dimension spaces, these are only the rank 1 matrices. So no, the dual space is larger. $\endgroup$ – user251257 Mar 17 '16 at 6:21
  • $\begingroup$ @user251257: Linear functionals are always rank one. What do you mean exactly by matrices? $\endgroup$ – C-Star-W-Star Mar 17 '16 at 6:23
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    $\begingroup$ If this was the case, the set would be closed under addition, which it isn't. However, the closed linear span of them is the entire dual. $\endgroup$ – Friedrich Philipp Mar 17 '16 at 6:36
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    $\begingroup$ Consider the annihilator. If the span of your set was not dense, there would be some $T\in\mathcal B(\mathcal H,\mathcal K)$ such that $l_{(\varphi,\psi)}T = 0$ for all $\varphi,\psi$. But this is of course only possible for $T = 0$. Note moreover that you can reduce your set to all $\varphi,\psi$ from orthonormal bases. $\endgroup$ – Friedrich Philipp Mar 17 '16 at 6:41
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    $\begingroup$ @FriedrichPhilipp: Your argument is incorrect, I think. For example, consider the space $L^\infty = X$. Then $L^1 \lneq X^\ast$, but still $f =0$ if $\phi(f)=0$ for all $\phi \in L^1$ and $f\in X$. $\endgroup$ – PhoemueX Mar 17 '16 at 7:33

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