# A trace inequality between self-adjoint operators

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 $$\|A\|_p:=(Tr(|A|^p)^{1/p}.$$

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 $$\|A\|_p^p \leq \|B\|_p^p$$.
• 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 $\leq \tau(|B|^p)$. Apr 9 at 3:45
• I dont know how to obtain the last inequality. I know example where $-B\leq A\leq B$ does not imply that $|A|\leq B.$ Apr 10 at 19:33
• 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 $v$. Apr 11 at 4:08