Suppose two square real matrices $A$ and $B$ are close in the Schatten 1-norm, i.e. $\|A-B\|_1=\varepsilon$. Can this be used to put a bound on the Schatten 2-norm distance between their square roots. Namely, is there something of the form $\|\sqrt{A}-\sqrt{B}\|_2\leq f(\varepsilon)$? It is important that $f(\varepsilon)$ be independent of the dimension of the matrices. One can assume that these are symmetric, positive definite matrices.

I have a proof for the above statement when $A$ and $B$ are taken to be simultaneously diagonal. However, I was wondering if there is a more general proof.

Powers-Stormer inequalityc.f. mathoverflow.net/questions/198610/… $\endgroup$ – Yemon Choi Aug 1 '19 at 13:283more comments