Let $M$ be a von Neumann algebra and $\tau$ a faithful (semi-finite?) normal trace on $M$; as is standard, the $L^p$-norm is defined as $||u||\_p=\tau(|u|^p)^{1/p}$. Let $\{u\_i\}\_{i=1}^\infty$ be a sequence of hermitian elements that converges to $u$ in the $L^p$-norm; i.e. $||u\_i-u||\_p\to 0$ as $i\to\infty$. If $u\_i\geq\psi$ for $i=1,2,\ldots$, does it hold that $u\geq\psi$? How can one see this? 



Does noncommutative Lp-convergence respect orderings?
Yes. Use two facts: the first is that $x \geq 0$ if and only if $\tau(xq) \geq 0$ for all finite projections $q$. The second fact is Hölder's inequality, which implies that $x \mapsto \tau(xq)$ is continuous on $L^p$ for all finite projections $q$.


