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  "body": "\n## Definition\n\nLet $I \\subseteq \\R$ be an [[interval|0Q1D85B8G6]] and let $f \\colon I \\to \\R$.\nLet $A$ be an $n \\times n$ [[Hermitian matrix|0P17NGQQMQ]] with all eigenvalues in $I$.\nBy the [[spectral theorem|0DGZBQWZWN]], $A = U \\operatorname{diag}(\\lambda_1, \\dots, \\lambda_n) U^*$ with $U$ unitary.\nThe [[matrix function|W7MPTRPXYD]] $f(A)$ is $U \\operatorname{diag}(f(\\lambda_1), \\dots, f(\\lambda_n)) U^*$; it does not depend on the choice of $U$.\nFor Hermitian matrices $A$ and $B$ of the same size, $A \\le B$ in the [[Loewner order|9G9S92T1VJ]] means that $B - A$ is [[positive semidefinite|01MMVW8CKC]].\nThe function $f$ is *operator monotone* on $I$ if the following holds for every $n \\ge 1$.\nFor all $n \\times n$ Hermitian matrices $A$ and $B$ with eigenvalues in $I$, if $A \\le B$, then $f(A) \\le f(B)$.\n\n## Remarks\n\nFor $n = 1$ the condition says that $f$ is nondecreasing, so an operator monotone function is a [[monotone function|0A7KV2WB7A]].\nExample: $f(t) = -1/t$ on $(0, \\infty)$.\nIf $0 \u003c A \\le B$, then $B^{-1/2} A B^{-1/2} \\le I$, so its eigenvalues lie in $(0, 1]$, so $B^{1/2} A^{-1} B^{1/2} \\ge I$, which gives $A^{-1} \\ge B^{-1}$, that is, $-A^{-1} \\le -B^{-1}$.\nNon-example: $f(t) = t^2$ on $[0, \\infty)$.\nTake $A = \\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 2 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$, both positive semidefinite, with $B - A = \\operatorname{diag}(1, 0) \\ge 0$.\nThen $B^2 - A^2 = \\begin{pmatrix} 3 \u0026 1 \\\\ 1 \u0026 0 \\end{pmatrix}$ has determinant $-1$, so it has a negative eigenvalue and $A^2 \\le B^2$ fails.\nThe same definition with self-adjoint operators on a Hilbert space gives the same class of functions.\n\n## Explanations\n\n### L1 human\n\nThe definition asks that $f$ keep the order of Hermitian matrices of all sizes, not only the order of numbers.\nThe size-two example with $t^2$ shows that this is a real restriction.\n",
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