Directed sets of positive elements in noncommutative $\mathrm L^p$ spaces

Question

Let $\tau$ be a faithful normal semifinite trace on a von Neumann algebra $\mathcal M$.

If $1<p<\infty$ and $E$ is a nonempty subset of $\mathrm L^p(\mathcal M,\tau)_+$ such that

• for every $x\in E$ and $y\in E$, there is a $z\in E$ such that $x\leq z$ and $y\leq z$,
• $\sup_{x\in E}\|x\|_{\mathrm L^p(\mathcal M,\tau)}<\infty$,

does $E$ necessarily have a supremum in $\mathrm L^p(\mathcal M,\tau)_+$? And if it does, does the filter of sections of $E$ converge to $\sup E$ in $\mathrm L^p(\mathcal M,\tau)$ (i.e. for every $\varepsilon>0$, is there a $x\in E$ such that $\mathopen\|\sup E-y\|_{\mathrm L^p(\mathcal M,\tau)}\leq\varepsilon$ whenever $y\in E$ satisfies $y\geq x$)?

This is true in classical $\mathrm L^p$ spaces over a measure space and can be shown using, among other things, the fact that $a^p+b^p\leq (a+b)^p$ for every nonnegative real numbers $a$ and $b$. But this inequality, I believe, does not generalize to positive operators.

Answer 1

Yes, the net $E$ is Cauchy. There are several ways of proving it (the first that came to my mind was using uniform convexity of $L_p(\mathcal{M},\tau)$). If one wants to stay close to your suggestion, one can use the inequality $\|a\|_p^p + \|b\|_p^p \leq \|a+b\|_p^p$ for every nonnegative $a,b \in L_p(\mathcal{M},\tau)$. The validity of this inequality illustrates a general fact: many inequalities for real numbers are no longer true for general operators, but stay true under the trace.

Proof of the inequality: it can be rewritten as $\|aa^*\|_p^p + \|b b^*\|_p^p \leq \| aa^* +b b^*\|_p^p$ for every $a,b \in L_{2p}(\mathcal M,\tau)$, or equivalently $$ \|a\|_{2p}^{2p} + \|b\|_{2p}^{2p} \leq \big\| \begin{pmatrix} a &b \end{pmatrix}\big\|_{2p}^{2p}.$$ By interpolation, it is enough to check this inequality for $p=1$ (where it is an equality) and for $p=\infty$ (where it becomes the obvious inequality $\max(\|a\|_\infty,\|b\|_{\infty}) \leq \big\| \begin{pmatrix} a &b \end{pmatrix}\big\|_{\infty}$.

Answer 2

In addition to Mikael de la Salle's answer, here's another argument which relies less on the non-commutative $L^p$-structure but rather on the general theory of ordered Banach spaces.

Let $X$ be a Banach space ordered by a closed convex cone $X_+$.

Theorem 1. If the space $X$ is reflexive and the cone $X_+$ is normal - i.e., there exists $M \ge 1$ such that $0 \le x \le y$ implies $\|x\| \le M \|y\|$ - then every increasing norm bounded net in $X$ converges in norm.

Proof. Since the cone $X_+$ is normal, a classical result on ordered Banach spaces says that the dual cone $X'_+$ (that consists of all functionals $x' \in X'$ which map $X_+$ to $[0,\infty)$) satisfies $X'_+ - X'_+ = X'$ (i.e., the dual cone is generating). This together with the reflexivity of $X$ implies that every norm bounded increasing net in $X$ is weakly convergent.

But again due to the normality of $X_+$, every increasing and weakly convergent net in $X$ is automatically norm convergent. (This is a version of Dini's theorem and can be proved by precisely the same argument as Dini's classical theorem on monotone convergence of continuous functions.) $\square$

Theorem 1 covers the case of non-commutative $L^p$-spaces for $p \in (1,\infty)$.

Theorem 2. Assume that the norm on $X$ is additive on $X_+$, meaning that $\|x+y\| = \|x\| + \|y\|$ for all $x,y \in X_+$. Then every increasing norm bounded net in $X$ is norm convergent.

Proof. Since the norm is additive on $X_+$ one can easily show that every increasing norm bounded net in $X$ is Cauchy. $\square$

Theorem 2 covers non-commutative $L^1$-spaces (and the proof is, of course, a version of the argument given in Mikael de la Salle's answer).