Let $A$ and $B$ be selfadjoint 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 $\A\_p:=(Tr(A^p)^{1/p}.$
1 Answer
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 $\A\_p^p \leq \B\_p^p$.

$\begingroup$ @ Nik. Very nice. Is it also true for any von Neumann algebra with a normal faithful semifinite trace? $\endgroup$ Apr 8 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 $\tau(A^p) = \tau((\sum p_i)A^p) = \sum \tau(p_iA^p) = \sum \tau(p_iA^p p_i) \leq \sum \tau(p_iB^p p_i)$, and maybe you can adapt the discrete case argument to get that this is $\leq \tau(B^p)$. $\endgroup$ Apr 9 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 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 at 4:08