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



A trace inequality between self-adjoint operators
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](https://math.stackexchange.com/questions/2269975/is-this-inequality-on-schatten-p-norm-and-diagonal-elements-true)). 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 $\|A\|\_p^p \leq \|B\|\_p^p$.


