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  "body": "\n## Definition\n\nLet $(X, \\mu)$ and $(Y, \\nu)$ be [[measure spaces|P15VQS5DJ9]] and let $1 \\le p \\le \\infty$ and $1 \\le q \u003c \\infty$.\nLet $T$ be an operator that maps each $f$ in the [[Lp space|SASP14MA1Z]] $L^p(X, \\mu)$ to a [[measurable function|YYNCJSR0PJ]] $Tf$ on $Y$.\nThe operator $T$ is of *weak type $(p, q)$* if there is a constant $C$ such that\n$$\\nu(\\set{y \\in Y : \\abs{Tf(y)} \u003e \\lambda}) \\le \\left( \\frac{C \\norm{f}_{L^p}}{\\lambda} \\right)^q$$\nfor every $f \\in L^p(X, \\mu)$ and every $\\lambda \u003e 0$.\nOne also says that $T$ satisfies a weak-type $(p, q)$ estimate.\n\n## Remarks\n\nEquivalently, $\\norm{Tf}_{q, \\infty} \\le C \\norm{f}_p$, where $\\norm{\\cdot}_{q,\\infty}$ is the quasinorm of the [[weak Lp space|H4C888M5AZ]] $L^{q,\\infty}(\\nu)$.\nThe operator is of *strong type $(p, q)$* if $\\norm{Tf}_q \\le C \\norm{f}_p$.\nBy Chebyshev's inequality $\\nu(\\set{\\abs{g} \u003e \\lambda}) \\le \\lambda^{-q} \\norm{g}_q^q$, strong type $(p, q)$ implies weak type $(p, q)$ with the same constant.\nBy convention, weak type $(p, \\infty)$ means strong type $(p, \\infty)$.\n\nExample: let $X = (0, 1)$ and $Y = (0, \\infty)$ with Lebesgue measure, and let $Tf(y) = c_f / y$ with $c_f = \\int_0^1 f$.\nThen $\\set{y : \\abs{Tf(y)} \u003e \\lambda} = (0, \\abs{c_f}/\\lambda)$, of measure $\\abs{c_f}/\\lambda \\le \\norm{f}_1/\\lambda$.\nSo $T$ is of weak type $(1, 1)$ with $C = 1$, but not of strong type $(1, 1)$, because $\\int_0^\\infty dy/y = \\infty$ when $c_f \\neq 0$.\n\nNon-example: on the same spaces, with $Y = (0, 1)$, let $Tf(y) = c_f / y^2$.\nFor $f = 1$ one has $c_f = 1$ and $\\set{Tf \u003e \\lambda} = (0, \\lambda^{-1/2})$ for $\\lambda \u003e 1$.\nThe ratio $\\lambda^{-1/2} / (C/\\lambda) = \\lambda^{1/2}/C$ is unbounded as $\\lambda \\to \\infty$, so $T$ is not of weak type $(1, 1)$.\n",
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