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In Stratila and Zsido, as well as some other sources, the ultraweak topology on $B(H)$ is taken to be the smallest topology for which every element in the closure of the span in $B(H)$ of the elements $\omega_{\xi, \eta}$ where $\omega_{\xi, \eta}(x)=<\xi, x\eta>$ is continuous. Another popular definition is found in Dixmier Chapter 3, or just as well here: p.2 of this notes

A third popular definition can be found here.

Can someone help me establish the three are equivalent? Thanks.

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A linear functional of the form $x \mapsto \sum_{i = 1}^\infty \langle \xi_i, x \eta_i \rangle$ where $\sum_{i = 1}^\infty \| \xi_i \|^2 < \infty$ and $\sum_{i = 1}^\infty \| \eta_i \|^2 < \infty$ can be written more succinctly as $x \mapsto {\text Tr}(Tx)$ where $T$ is the trace class operator $T\zeta = \sum_{i = 1}^\infty \langle \xi_i, \zeta \rangle \eta_i$. (or something close to this formula)

Since the trace class operators from the predual of $B(H)$ it follows that the second and third definitions are equivalent.

Since every such linear functional above can be approximated by a finite sum (or equivalently since the finite rank operators are dense (in trace norm) in the trace class operators) it follows that the first and second/third definitions are equivalent.

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My by-now-usual knee-jerk comment would be to recommend that one ask oneself what "function" a given topology provides. The raw definitions have been distilled to a point that they do not confess their own purpose(s). –  paul garrett Jul 14 '12 at 0:40
    
@Jesse Petersen everything I know about trace class operators relies on testing the $T$ in question against orthonormal sets, which inconveniently the $\xi$s and $\eta$s are not. As such, I'm having difficulty even seeing that your suggested $T$ is trace class. Can you please let me know how to do this? It occurred to me to use Gram Schmidt orthogonalization tricks but I don't think that preserves the square summability. –  Jeff Jul 15 '12 at 8:06
    
@Jeff: No problem. The first thing you need is that the trace norm is indeed a norm, and the trace class operators form a Banach space with respect to this norm. You also need to know that the definition if the trace is independent of the orthonormal basis you use. You should be able to find proofs of these facts in any book which introduces the trace class operators. Once you have these facts then check that for vectors $\xi$ and $\eta$ the operator $T_{\xi, \eta}\zeta = \langle \xi, \zeta \rangle \eta$ has trace norm $\| T_{\xi, \eta} \|_{\text Tr} = \| \xi \| \| \eta \|$. –  Jesse Peterson Jul 15 '12 at 17:09
    
Since the trace norm satisfies the triangle inequality it then follows that for finite sums we have $\| \sum_i T_{\xi_i, \eta_i} \|_{\text Tr} \leq \sum_i \| \xi_i \| \| \eta_i \| \leq (\sum_i \| \xi_i \|^2 )^{1/2} (\sum_i \| \eta_i \|^2 )^{1/2}$. Since the trace class operators form a Banach space it follows that this holds also for infinite sums as long as $\sum_i \| \xi_i \|^2 < \infty$, and $\sum_i \| \eta_i \|^2 < \infty$. –  Jesse Peterson Jul 15 '12 at 17:13
    
@JessePeterson I agree with what you say, but now I also struggle to see that the setup you have shows $Tr(Tx)$ is a linear functional with the desired form. Furthermore, one has to see that every trace class operator corresponds to a linear functional which is among those that are responsible for inducing the ultraweak topology according to other definitions. –  Jeff Jul 16 '12 at 5:28

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