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For a Hilbert space $\mathcal{H}$ with a space of bounded operators $\mathcal{B}(\mathcal{H}),$ the strong topology on $\mathcal{B}(\mathcal{H})$ is defined so that the net $T_{\alpha}$ converges to $T$ if and only if $(T - T_{\alpha}) x$ converges to zero for every $x \in \mathcal{H}.$ The ultraweak topology is defined so that the net $T_{\alpha}$ converges to $T$ if and only if $\operatorname{tr}((T-T_{\alpha}) A)$ converges to zero for every trace class operator $A$.

I have found many sources claiming that the ultraweak topology and strong topology are incomparable on infinite-dimensional Hilbert spaces; this would mean that there exist nets that converge in the ultraweak topology but not in the strong topology, and conversely that there exist nets that converge in the strong topology but not the ultraweak topology. This is claimed, for example, in Dixmier's book on von Neumann algebras, and in the current version of the Wikipedia page on operator topologies. I have been unable, however, to find a proof.

Can anyone provide a proof or a good resource for understanding the incomparability of these topologies?

user_35
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1 Answers1

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Consider a net of projections $\{P_j\}\subset B(H)$. Suppose that $P_j\to T$ in the strong operator topology. We have $$ \langle T^*x,y\rangle=\langle x,Ty\rangle=\lim_j\langle x,P_jy\rangle=\lim_j P_jx,y\rangle=\langle Tx,y\rangle. $$ Then $T^*=T$. Using that $\|P_j\|\leq1$ for all $j$,
\begin{align} \|(T^2-T)x\|&\leq\|T^2x-P_jTx\|+\|P_jTx-P_j^2x\|+\|P_j^2x-Tx\|\\[0.3cm] &\leq \|(T-P_j)Tx\|+\|P_j\|\,\|Tx-P_jx\|+\|P_jx-Tx\|\\[0.3cm] &\leq \|(T-P_j)Tx\|+2\|(T-P_j)x\|\to0. \end{align} So $T^2=T$. We have shown that a strong limit of projections is a projection.

Meanwhile, given any $T\in B(H)$ with $\|T\|\leq1$, there exists a net $\{P_j\}$ of projections with $P_j\to T$ in the weak operator topology (proof here). The ultraweak topology agrees with the weak operator topology on bounded sets. If in particular we take $T$ to not be a projection, we are getting a net of projections that converges ultraweakly but does not converge strongly.

On the other direction, here is an example of a net of operators that converges strongly but not ultraweakly. Such a net is of course unbounded, as on bounded sets strong convergence implies weak convergence.

Martin Argerami
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  • Good answer, thanks! For anyone who looks at this in the future, here's a slightly more explicit version of the "net that converges strongly but not ultraweakly." Let $A$ be any trace-class operator, let $C$ be the directed set of finite-dimensional subspaces ordered by inclusion. We define a net by setting $T_F = P_{F^{\perp}} / \operatorname{tr}(A P_{F^{\perp}})$ for $F \in C.$ This net converges strongly to the zero operator, but we have $\operatorname{tr}(A T_F) = 1,$ so it does not converge ultraweakly to zero. – user_35 Nov 30 '22 at 14:35
  • I should clarify that in my comment above, we need $\operatorname{tr}(A P_{F^{\perp}})$ to be nonzero for all $P_{F^{\perp}},$ which requires that $A$ is infinite-rank and $\mathcal{H}$ is separable. If $\mathcal{H}$ is non-separable, we can restrict to the directed set of finite-dimensional subspaces of the support of $A$. – user_35 Nov 30 '22 at 14:41