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  "body": "\n## Definition\n\nLet $n \\ge 1$, and let $\\chi$, $\\psi$ and $\\Delta_j$, $j \\in \\Z$, be as in the [[Littlewood–Paley decomposition|D1PAT0K446]] on $\\R^n$.\nLet $f$ be in the [[Lp space|SASP14MA1Z]] $L^p(\\R^n)$ for some $1 \\le p \\le \\infty$.\nThe *Littlewood–Paley square function* of $f$ is the function $Sf \\colon \\R^n \\to [0, \\infty]$ given by\n$$Sf(x) = \\left( \\sum_{j \\in \\Z} \\abs{\\Delta_j f(x)}^2 \\right)^{1/2}.$$\n\n## Remarks\n\nThe operator $f \\mapsto Sf$ is not linear, but it is sublinear: $S(f + g) \\le Sf + Sg$, by the triangle inequality in $\\ell^2$.\nIt depends on the choice of $\\chi$, but the $L^p$ norms of square functions built from two choices are comparable for $1 \u003c p \u003c \\infty$.\n\nExample: assume moreover that $\\chi$ is radial and nonincreasing in $\\abs{\\xi}$, so that $\\psi \\ge 0$.\nFor each $\\xi \\ne 0$ at most two of the numbers $\\psi(2^{-j}\\xi)$ are nonzero, and they sum to $1$.\nSo $\\frac{1}{2} \\le \\sum_j \\psi(2^{-j}\\xi)^2 \\le 1$.\nBy Plancherel's theorem for the [[Fourier transform|DPVM34ADZV]], $\\norm{Sf}_2^2 = \\int \\sum_j \\psi(2^{-j}\\xi)^2 \\abs{\\hat f(\\xi)}^2 \\, d\\xi$, hence $\\frac{1}{2}\\norm{f}_2^2 \\le \\norm{Sf}_2^2 \\le \\norm{f}_2^2$ for $f \\in L^2(\\R^n)$.\n\nNon-example: the constant function $1$ is not in any $L^p(\\R^n)$ with $p \u003c \\infty$.\nAs a tempered distribution its Fourier transform is the point mass at $0$, and $\\psi(0) = 0$, so every $\\Delta_j 1$ vanishes and the formula gives $S1 = 0$ although $1 \\ne 0$.\nThis is why the comparison of $\\norm{Sf}_p$ and $\\norm{f}_p$ needs $p \u003c \\infty$.\n",
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