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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $u_1, \\dots, u_n$ be [[complex numbers|09FTAET5Z4]].\nThen\n$$\\Big| \\sum_{\\substack{1 \\le r, s \\le n \\\\ r \\ne s}} \\frac{u_r \\overline{u_s}}{r - s} \\Big| \\le \\pi \\sum_{r=1}^{n} \\abs{u_r}^2.$$\n\n## Remarks\n\nProof idea: the function $\\varphi(t) = i(\\pi - t)$ on $(0, 2\\pi)$ has Fourier coefficients $1/k$ for $k \\ne 0$ and $0$ for $k = 0$, and $\\abs{\\varphi} \\le \\pi$.\nSo the double sum equals $\\frac{1}{2\\pi} \\int_0^{2\\pi} \\varphi(t) \\, \\abs{P(t)}^2 \\, dt$ for the trigonometric polynomial $P(t) = \\sum_r u_r e^{-i r t}$.\nBounding $\\abs{\\varphi}$ by $\\pi$ and using [[Parseval's identity|047RFCNVSK]] $\\frac{1}{2\\pi} \\int_0^{2\\pi} \\abs{P}^2 = \\sum_r \\abs{u_r}^2$ gives the result.\n\nThe constant $\\pi$ cannot be replaced by a smaller one that works for all $n$.\n\nA closely related inequality, also called Hilbert's inequality, says that for nonnegative $a_m, b_n$,\n$$\\sum_{m=1}^{\\infty} \\sum_{n=1}^{\\infty} \\frac{a_m b_n}{m + n} \\le \\pi \\Big( \\sum_{m=1}^{\\infty} a_m^2 \\Big)^{1/2} \\Big( \\sum_{n=1}^{\\infty} b_n^2 \\Big)^{1/2}.$$\n\n## Explanations\n\n### L1 human\n\nHilbert's inequality controls a double sum in which each pair of terms is weighted by $1/(r - s)$.\nThese weights do not shrink fast, and they change sign, so it is not obvious that the sum is small.\nThe inequality says it is never more than $\\pi$ times the total squared size of the numbers.\n\n### L2 human\n\nThe matrix with entries $1/(r - s)$ for $r \\ne s$ is a discrete version of the Hilbert transform.\nThe inequality says that this matrix has operator norm at most $\\pi$ on $\\C^n$ for every $n$, uniformly in $n$.\nThe proof is a model of a common technique: write a matrix as the Fourier coefficients of a bounded function, then use Parseval's identity.\n",
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