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  "body": "\n## Definition\n\nLet $(X, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]] and let $f$ be a real or complex [[measurable function|YYNCJSR0PJ]] on $X$.\nIts [[distribution function|HQVXKEMN5N]] is $d_f(s) = \\mu(\\set{x : \\abs{f(x)} \u003e s})$ for $s \\ge 0$.\nIts [[decreasing rearrangement|BWS042NC2P]] is $f^*(t) = \\inf \\set{s \\ge 0 : d_f(s) \\le t}$ for $t \\ge 0$.\nLet $0 \u003c p \u003c \\infty$ and $0 \u003c q \\le \\infty$, and put\n$$\\norm{f}_{p,q} = \\left( \\int_0^\\infty \\left( t^{1/p} f^*(t) \\right)^q \\frac{dt}{t} \\right)^{1/q} \\text{ if } q \u003c \\infty, \\qquad \\norm{f}_{p,\\infty} = \\sup_{t \u003e 0} t^{1/p} f^*(t).$$\nThe *Lorentz space* $L^{p,q}(\\mu)$ is the set of measurable $f$ with $\\norm{f}_{p,q} \u003c \\infty$, modulo equality [[almost everywhere|RARTMA0ZEC]].\n\n## Remarks\n\nExample: for $q = p$, $\\norm{f}_{p,p}^p = \\int_0^\\infty f^*(t)^p \\, dt = \\int_X \\abs{f}^p \\, d\\mu$, because $f^*$ and $\\abs{f}$ have the same distribution function.\nSo $L^{p,p}(\\mu) = L^p(\\mu)$ with equal norms.\n\nNon-example: on $(0, \\infty)$ with Lebesgue measure, $f(x) = x^{-1/p}$ is not in $L^{p,q}$ for $q \u003c \\infty$.\nHere $f^* = f$, so $t^{1/p} f^*(t) = 1$ and the integral $\\int_0^\\infty dt / t$ diverges.\nIt does lie in $L^{p,\\infty}$, with $\\norm{f}_{p,\\infty} = 1$.\n\nThe space $L^{p,\\infty}$ is the weak $L^p$ space, since $\\sup_t t^{1/p} f^*(t) = \\sup_s s \\, d_f(s)^{1/p}$.\nFor fixed $p$, the spaces increase with $q$: $L^{p,q_1} \\subseteq L^{p,q_2}$ when $q_1 \u003c q_2$.\nIn general $\\norm{\\cdot}_{p,q}$ is only a quasinorm.\n",
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