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It seems to be a fairly common exercise to show that $$ 0\preceq A \preceq B \implies \sqrt A \preceq \sqrt B $$ for positive semi-definite matrices. On the other hand, in general, $$ 0\preceq A \preceq B \ \nRightarrow\ A^2 \preceq B^2. $$ (See e.g. https://math.stackexchange.com/a/510999/620957).

Along with being able to take inverses, this is more or less the usual simple results I've seen of this kind. But do they hint at a more general theorem behind the scenes? I'm imaging some result along the lines of

If $0\preceq A \preceq B$, and if $f$ is a [concave, increasing, positive, ...?] function, then $f(A)\preceq f(B)$.

I realize that this question is somewhat vague, so sorry about that, but is there a nice result to discovered here? My googling is being unsuccessful... Thanks in advance.

Milten
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    Such an $f$ is called operator-monotone, and these were classified by Lowner. See here – Brevan Ellefsen Oct 24 '23 at 20:12
  • Thank you, that's all I need then! I'll try to find a duplicate to close this question. – Milten Oct 24 '23 at 21:00
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    To other readers, https://doi.org/10.1155/2015/649839 seems to be a nice survey on related results. (I'll look further for a suitable duplicate later). – Milten Oct 24 '23 at 21:37
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    Bhatia's Matrix Analysis has a very nice chapter on operator-monotone and operator-convex functions for anyone interested. There is a condition that is a bit stronger than "concave" that is needed here, and that condition is "operator-concavity". For functions defined over $[0,\infty)$, the conditions of operator monotonicity and operator concavity are equivalent. – Ben Grossmann Oct 24 '23 at 22:57

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