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  "body": "\n## Statement\n\nLet $a \u003c b$ be real numbers, possibly $a = -\\infty$ or $b = \\infty$, and let $f \\colon (a, b) \\to \\R$.\nLet $\\mathbb{H} = \\set{z \\in \\C : \\operatorname{Im} z \u003e 0}$ be the upper half-plane.\nThen $f$ is [[operator monotone|N155BSJ2DN]] on $(a, b)$ if and only if there is a holomorphic function $F$ on $\\C \\setminus \\bigl( (-\\infty, a] \\cup [b, \\infty) \\bigr)$ such that:\n\n1. $F(t) = f(t)$ for all $t \\in (a, b)$;\n2. $F(\\bar z) = \\overline{F(z)}$ for all $z$ in the domain of $F$;\n3. $\\operatorname{Im} F(z) \\ge 0$ for all $z \\in \\mathbb{H}$.\n\n## Remarks\n\nSo an operator monotone function is real analytic, and it extends to a holomorphic map from $\\mathbb{H}$ into the closed upper half-plane, a so-called Pick function.\nExample: for $0 \u003c p \u003c 1$, the function $t^p$ on $(0, \\infty)$ extends to $z^p$ with the principal branch.\nIt maps $\\mathbb{H}$ into the sector of arguments between $0$ and $p\\pi$, which lies in $\\mathbb{H}$, so $t^p$ is operator monotone.\nIn the same way $\\log t$ is operator monotone on $(0, \\infty)$, since $\\log z$ maps $\\mathbb{H}$ into the strip $0 \u003c \\operatorname{Im} w \u003c \\pi$.\nIn contrast, $z^2$ maps $-1 + i \\in \\mathbb{H}$ to $-2i$, so $t^2$ is not operator monotone on $(0, \\infty)$.\nOne route to the proof goes through divided differences.\nA function $f$ of class $C^1$ is operator monotone if and only if the following holds for all points $t_1, \\dots, t_n$ of $(a, b)$.\nThe matrix with entries $(f(t_i) - f(t_j)) / (t_i - t_j)$ off the diagonal and $f'(t_i)$ on the diagonal is [[positive semidefinite|01MMVW8CKC]].\n\n## Explanations\n\n### L1 human\n\nKeeping the order of Hermitian matrices of every size is a very strong demand on a function.\nLöwner's theorem says that exactly the functions that extend to the complex upper half-plane and send it into itself meet this demand.\nThis turns a question about matrices into a question about holomorphic functions.\n\n### L2 human\n\nThe theorem gives integral representations of operator monotone functions, through the Nevanlinna representation of Pick functions.\nIt explains at once why $t^p$ is operator monotone on $(0, \\infty)$ exactly for $0 \\le p \\le 1$.\nOperator monotone functions appear in matrix inequalities, in the theory of operator means and in quantum information.\n",
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