4
$\begingroup$

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?

$\endgroup$

1 Answer 1

7
$\begingroup$

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 $q$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Feb 5, 2014 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 $\tau(qx) \leq -\varepsilon \tau(q)<0$. $\endgroup$ Feb 5, 2014 at 8:00

Your Answer

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

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