Let $M$ be a von Neumann algebra and $\tau$ a faithful (semifinite?) 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. $u_iu_p\to 0$ 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?
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 $q$.

$\begingroup$ Thanks! The proof of Hölder's inequality one can find in any book, but the other statement: $\tau(xq)\geq 0$ (for all finite projections) iff $x\geq 0$, do you have a reference for it, or is it trivial to prove? $\endgroup$ – Joakim Arnlind Feb 5 '14 at 6:46

$\begingroup$ If $x \geq 0$ you can write $x = y^* y$, so that $\tau(xq) = \tau((yq)^*(yq)) \geq 0$. Conversely, if the spectral projection $1_{(\infty,\varepsilon)}(x)$ is nonzero, you can pick a finite projection $q$ that is smaller and hence $\tau(qx) \leq \varepsilon \tau(q)<0$. $\endgroup$ – Mikael de la Salle Feb 5 '14 at 8:00