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  "body": "\n## Definition\n\nLet $n \\ge 1$ and let $\\abs{R}$ be the [[Lebesgue measure|2Z2QVWZ262]] of a set $R \\subseteq \\R^n$.\nA *rectangle* here is a set $R = I_1 \\times \\cdots \\times I_n$, where each $I_k$ is a bounded interval of positive length.\nLet $f$ be a locally [[integrable|X7GTF3D98F]] function on $\\R^n$.\nThe *strong maximal function* of $f$ is the function $M_S f \\colon \\R^n \\to [0, \\infty]$ given by\n$$M_S f(x) = \\sup_{R \\ni x} \\frac{1}{\\abs{R}} \\int_R \\abs{f(y)} \\, dy,$$\nwhere the supremum is over all such rectangles $R$ that contain $x$.\n\n## Remarks\n\nEach ball of radius $r$ lies in a cube of side $2r$, which is a rectangle of comparable volume.\nSo $Mf \\le C_n M_S f$ for the [[Hardy–Littlewood maximal function|VX1A56P5KQ]] $Mf$, with $C_n$ depending only on $n$.\nLet $M_k$ be the one-dimensional maximal function in the $k$-th variable, with the other variables fixed.\nThen $M_S f \\le M_1 M_2 \\cdots M_n f$ almost everywhere, and the one-dimensional bounds give $\\norm{M_S f}_p \\le C \\norm{f}_p$ for $1 \u003c p \\le \\infty$.\nFor $n \\ge 2$ the operator $M_S$ is not of weak type $(1, 1)$.\nIt satisfies the weaker bound $\\abs{\\set{M_S f \u003e \\lambda}} \\le C \\int \\frac{\\abs{f}}{\\lambda} \\left( 1 + \\log^+ \\frac{\\abs{f}}{\\lambda} \\right)^{n - 1}$, due to Jessen, Marcinkiewicz and Zygmund.\n\nExample: let $n = 2$, $f = \\mathbf{1}_{[0, 1]^2}$ and $x = (a, 1/2)$ with $a \u003e 1$.\nThe rectangle $R = [0, a] \\times [0, 1]$ contains $x$ and has $\\frac{1}{\\abs{R}} \\int_R f = \\frac{1}{a}$.\nAny rectangle $R' = [s, t] \\times J$ containing $x$ and meeting the square has $s \\le 1 \u003c a \\le t$.\nIts average of $f$ is at most $(1 - \\max(s, 0))/(t - s)$, which is at most $1/a$.\nSo $M_S f(x) = 1/a$.\n\nNon-example: a square centred at $x$ and rotated by $45$ degrees is not a rectangle in the sense above.\nIts sides are not parallel to the axes.\nAllowing rectangles in all directions gives a different operator, which is unbounded on $L^p(\\R^2)$ for every $p \u003c \\infty$, because of Besicovitch sets.\n",
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