# Continuity of the extension of a tracial state with respect to the strong operator topology

Problem: Let $$M\subseteq B(H)$$ be a finite von Neumann algebra with a faithful tracial state $$\tau$$. Let $$\widetilde{M}$$ be the $$\tau$$-measurable operators on $$M$$ (recalled below). Extend the trace $$\tau$$ on $$\widetilde{M}_+$$ by $$\tau(a):=\int_0^\infty\lambda\tau(e_\lambda)$$ where $$a=\int_0^\infty\lambda\,de_\lambda$$ is the spectral decomposition (See Equation 4.6 of Hiwi). If $$\{x_n\}_{n\in\mathbb{N}}$$ be a sequence of positive elements from $$M$$, which converges strongly to an element $$x$$ of $$\widetilde{M}_+$$, then can we say that $$\{\tau(x_n)\}_{n\in\mathbb{N}}$$ converges to $$\tau (x)$$?

Or, can we at least say that if $$\{x_n\}_{n\in\mathbb{N}}$$ be a sequence of positive elements from $$M$$, which converges strongly to an element $$x$$ of $$\widetilde{M}_+$$, then $$\{x_n\}_{n\in\mathbb{N}}$$ converges to $$x$$ in the measure topology?

(A positive answer of any one of the above two questions would be sufficient for me.)

I got stuck with this problem while reading $$\tau$$-measurable operators from the book 'Lectures on Selected Topics in von Neumann Algebras' by Hiwi. Here I recall the definition of $$\tau$$-measurable operator.

Definition 1: For each $$\epsilon,\delta>0$$, define $$\mathscr{O}(\epsilon,\delta)=\{m\text{ affiliated to } M:eH\subseteq \mathcal{D}(m),\,\|me\|\leq \epsilon \text{ and }\tau(1-e)\leq\delta \text{ for some } e\in Proj(M)\}.$$ Let $$m$$ be a densely defined closed operator such that $$m$$ is affiliated to $$M$$. We say that $$m$$ is $$\tau$$-measurable if for any $$\delta >0$$, there exists an $$\epsilon >0$$ such that $$m\in\mathscr{O}(\epsilon,\delta)$$. We denote by $$\widetilde{M}$$ the set of such $$\tau$$-measurable operators.

Theorem 2: (Theorem 4.12 of Hiwi) The $$\widetilde{M}$$ is a complete metrizable Hausdorff topological *-algebra with $$\{\widetilde{M}\cap\mathscr{O}(\epsilon,\delta):\epsilon,\delta >0\}$$ as a neighborhood basis of $$0$$. Moreover, $$M$$ is dense in $$\widetilde{M}$$.

Thanks in advance for any help or suggestion.

• The elements of $\tilde M$ are in general unbounded. When you talk about strong convergence, do you mean strong resolvent convergence? Feb 20 at 13:57
• @MaoWao By strong convergence of a sequence $\{x_n\}_{n\in\mathbb{N}}$ in $M_+$ to $x\in\widetilde{M}_+$, I mean that for each $\xi\in\mathcal{D}(x)$, $\{x_n(\xi)\}_{n\in\mathbb{N}}$ converges to $x(\xi)$ in $H$. To make sense, I have deleted the first part of the question.
– John
Feb 20 at 14:08

With the specific definition of strong convergence in the comment (namely, a sequence $$x_n \in M_+$$ is said to converge strongly to $$x \in \widetilde{M}_+$$ if and only if $$x_n \xi \to x \xi$$ for all $$\xi \in D(x)$$), both properties indeed hold.

Take such a sequence $$x_n \in M_+$$ converging strongly to $$x \in \widetilde{M}_+$$. We prove that $$\tau(x_n) \to \tau(x)$$ and $$x_n - x \to 0$$ in measure.

First assume that $$\tau(x)=+\infty$$. Choose $$\kappa > 0$$. Since $$\tau(x) = +\infty$$, we can choose a spectral projection $$p = e_{[0,\lambda]}(x)$$ such that $$\tau(xp) > \kappa + 1$$. By assumption, $$x_n p \to x p$$ strongly in the usual sense. By the uniform boundedness principle, the sequence $$x_n p$$ is bounded in operator norm and $$\tau(x_n p) \to \tau(x p)$$. We can thus take $$n_0$$ such that $$|\tau(x_n p) - \tau(xp)| < 1$$ for all $$n \geq n_0$$. Thus, $$\tau(x_n p) > \kappa$$ for all $$n \geq n_0$$. Since $$\tau(x_n) \geq \tau(x_n^{1/2} p x_n^{1/2}) = \tau(x_n p) > \kappa \; ,$$ we conclude that $$\tau(x_n) > \kappa$$ for all $$n \geq n_0$$. Thus, $$\tau(x_n) \to + \infty$$.

Next assume that $$\tau(x) < +\infty$$. We then have a well-defined normal positive functional $$\omega$$ on $$M$$ satisfying $$\omega(a) = \tau((1+x)^{1/2} a (1+x)^{1/2}) \quad\text{for all a \in M.}$$ By assumption $$x_n (1+x)^{-1} \to x (1+x)^{-1}$$ strongly in the usual sense. Again by uniform boundedness, $$\omega(x_n (1+x)^{-1}) \to \omega(x (1+x)^{-1}) \; .$$ This precisely says that $$\tau(x_n) \to \tau(x)$$.

To prove that $$x_n - x \to 0$$ in measure, first note the following standard result: if $$y_n$$ is a sequence in $$M$$ such that $$y_n \to 0$$ strongly, then $$y_n \to 0$$ in measure. Indeed, given $$y \in M$$ and $$\delta > 0$$, consider the spectral projection $$e = e_{[0,\delta]}(y^* y)$$. Since $$y^* y \geq \delta (1-e)$$, we find that $$\delta \, \tau(1-e) \leq \tau(y^* y) \quad\text{and}\quad \|y e \| \leq \delta \; .$$ Writing $$\|y\|_2 = \sqrt{\tau(y^* y)}$$, taking $$\delta = \|y\|_2$$ and using the notation $$\mathcal{O}(\cdot,\cdot)$$ for the basic neighborhoods of $$0$$ in the measure topology as in the question, it follows that $$y \in \mathcal{O}(\|y\|_2,\|y\|_2) \quad\text{for all y \in M.}$$ When $$y_n \to 0$$ strongly, also $$\|y_n\|_2 \to 0$$ and thus $$y_n \to 0$$ in measure.

We now return to the sequence $$x_n - x$$. Fix $$\varepsilon > 0$$. Take a spectral projection $$p$$ of $$x$$ such that $$xp$$ is bounded and $$\tau(1-p) < \varepsilon/2$$. Since $$(x_n - x) p \to 0$$ strongly in the usual sense, also $$(x_n - x)p \to 0$$ in measure. So, for all $$n$$ large enough, we have that $$(x_n - x)p \in \mathcal{O}(\varepsilon,\varepsilon/2)$$. It follows that $$x_n - x \in \mathcal{O}(\varepsilon,\varepsilon)$$ for all $$n$$ large enough. Thus, $$x_n \to x$$ in measure.

• In the last paragraph, could you please elaborate how to show that $(x_n-x)p\rightarrow 0$ strongly implies $(x_n-x)p\rightarrow 0$ in measure?
– John
Feb 21 at 0:56
• I have added that the general argument why strong convergence of a sequence in $M$ implies convergence in measure. Feb 21 at 7:40