Let $A$ and $B$ be self-adjoint operators on some Hilbert space and $B$ is postive. Suppose we have $-B\leq A\leq B$.Is it true then that $\|A\|_p\leq\|B\|_p$ where $\|.\|_p$ is the Schatten-$p$ norm defined as $\|A\|_p:=(Tr(|A|^p)^{1/p}.$
$\begingroup$
$\endgroup$
asked Apr 8, 2022 at 11:45
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Yes, this follows from the fact that $\|B\|_p^p \geq \sum |\langle Be_i, e_i\rangle|^p$ for any orthonormal basis $(e_i)$ (see here). If $(e_i)$ diagonalizes $A$ then we have $\|A\|_p^p = \sum |\langle A e_i, e_i\rangle|^p$, and also $|\langle Ae_i,e_i\rangle| \leq |\langle Be_i, e_i\rangle|$ for all $i$ because $-B \leq A \leq B$, and putting all that together yields $\|A\|_p^p \leq \|B\|_p^p$.
answered Apr 8, 2022 at 15:51
-
$\begingroup$ @ Nik. Very nice. Is it also true for any von Neumann algebra with a normal faithful semifinite trace? $\endgroup$ Apr 8, 2022 at 17:33
-
1$\begingroup$ Good question! Surely the answer is yes. Maybe something like: assuming $\|A\| < 1$, fix $n$ and for $-n \leq i < n$ let $p_i = P_{[i, i+1)}(A)$ (spectral projection), so that $\sum p_i = 1$ and each $p_i$ commutes with $A$. Then $\tau(|A|^p) = \tau((\sum p_i)|A|^p) = \sum \tau(p_i|A|^p) = \sum \tau(p_i|A|^p p_i) \leq \sum \tau(p_i|B|^p p_i)$, and maybe you can adapt the discrete case argument to get that this is $\leq \tau(|B|^p)$. $\endgroup$ Apr 9, 2022 at 3:45
-
$\begingroup$ I dont know how to obtain the last inequality. I know example where $-B\leq A\leq B$ does not imply that $|A|\leq B.$ $\endgroup$ Apr 10, 2022 at 19:33
-
1$\begingroup$ By the definition of $\leq$ the inequality $-B \leq A \leq B$ implies $-\langle Bv,v\rangle \leq \langle Av,v\rangle \leq \langle Bv,v\rangle$ for all $v$. $\endgroup$ Apr 11, 2022 at 4:08