$2$-norm distance between square roots of matrices 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.
 A: I don't have the answer to your question, but I can give you the following:
$$\|\sqrt A-\sqrt B\|_\infty\le\sqrt{\|A-B\|_\infty\,}\,,$$
where the $\infty$-Schatten norm is nothing but the operator norm.
A proof for this claim can be found here.
A: As you yourself discovered by finding a paper of Audenaert: an upper bound of the form you require is provided by the Powers–Størmer inequality:

Theorem (Powers–Størmer, 1970, Lemma 4.1; link) Let $S$ and $T$ be positive Hilbert-Schmidt operators on a Hilbert space. Then $\Vert S - T \Vert_2^2 \leq \Vert S^2-T^2\Vert_1$, where $\Vert \quad\Vert_p$ denotes the Schatten $p$-norm.

The starting idea of the proof is to work in an ONB with respect to which $R=S-T$ is diagonal (which is possible by the spectral theorem for compact self-adjoint operators). One then observes that, putting $Q=S+T$, we have $S^2-T^2 = (RQ+QR)/2$ and then one exploits the fact that $Q\geq \pm R$.
More general inequalities are known, see the discussion in Section X.1 of Bhatia's Springer GTM book on Matrix Analysis (Springer GTM).
