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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.

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    $\begingroup$ How do you make sense of $\sqrt{A}$ when $A$ is not symmetric $\ge 0$? $\endgroup$
    – YCor
    Commented Jul 31, 2019 at 14:33
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    $\begingroup$ @YemonChoi It should be true as stated. I have an idea how to get it (though it'll take me some time to check the details) but, if indeed true, it should be a textbook stuff, so even if I'm not mistaken, I'll wait a bit before sharing my home-made computations in case somebody has a good reference. $\endgroup$
    – fedja
    Commented Jul 31, 2019 at 18:18
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    $\begingroup$ @DavidRoberts: This arose in a quantum information context, in trying to prove that trace distance between density matrices being small implies fidelity of their canonical purifications is large. $\endgroup$
    – Pratik Rath
    Commented Aug 1, 2019 at 7:16
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    $\begingroup$ @fedja Thanks for the help. I found a reference where something equivalent to this is proved. Look at theorem 1 of arxiv.org/pdf/1207.1197.pdf $\endgroup$
    – Pratik Rath
    Commented Aug 1, 2019 at 7:18
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    $\begingroup$ PratikRath @fedja Indeed I am now extremely embarrassed for not recognising this result straightaway; not because I claim I could easily come up with the proof (unlike fedja) but because this result is mentioned and used a lot in areas related to my research. It is (assuming positivity etc as in YCor's edit) the Powers-Stormer inequality c.f. mathoverflow.net/questions/198610/… $\endgroup$
    – Yemon Choi
    Commented Aug 1, 2019 at 13:28

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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).

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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.

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  • $\begingroup$ Thanks. I had discovered this result here math.stackexchange.com/q/1934184. Unfortunately, I really needed the Schatten 2-norm to prove the result I was interested in. $\endgroup$
    – Pratik Rath
    Commented Aug 1, 2019 at 7:21

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