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.

  • 2
    $\begingroup$ The elements of $\tilde M$ are in general unbounded. When you talk about strong convergence, do you mean strong resolvent convergence? $\endgroup$
    – MaoWao
    Feb 20 at 13:57
  • $\begingroup$ @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. $\endgroup$
    – John
    Feb 20 at 14:08

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – John
    Feb 21 at 0:56
  • 1
    $\begingroup$ I have added that the general argument why strong convergence of a sequence in $M$ implies convergence in measure. $\endgroup$ Feb 21 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.