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  "body": "\n## Definition\n\nLet $n \\ge 1$ and let $L^1(\\R^n)$ be the [[Lp space|SASP14MA1Z]] of integrable functions for Lebesgue measure.\nFor $j = 1, \\dots, n$ let $R_j$ be the $j$-th [[Riesz transform|FRNDFJ9WQ5]], which maps each $f \\in L^1(\\R^n)$ to a [[tempered distribution|AES2T5JPFP]] $R_j f$.\nThe *real Hardy space* $H^1(\\R^n)$ is the set of $f \\in L^1(\\R^n)$ such that each $R_j f$, $j = 1, \\dots, n$, is given by a function in $L^1(\\R^n)$.\nIt is a Banach space with the norm\n$$\\norm{f}_{H^1} = \\norm{f}_{L^1} + \\sum_{j=1}^n \\norm{R_j f}_{L^1}.$$\n\n## Remarks\n\nFor $n = 1$ the only Riesz transform is the [[Hilbert transform|NSCEBXM7ZM]] $H$, so $H^1(\\R) = \\set{f \\in L^1(\\R) : Hf \\in L^1(\\R)}$.\nEquivalent definitions use maximal functions or atoms.\nFix a Schwartz function $\\varphi$ with $\\int \\varphi = 1$ and let $\\varphi_t(x) = t^{-n}\\varphi(x/t)$.\nThen $f \\in H^1(\\R^n)$ exactly when $f$ is a tempered distribution with $\\sup_{t \u003e 0} \\abs{f * \\varphi_t} \\in L^1(\\R^n)$, and the two norms are comparable.\nEvery $f \\in H^1(\\R^n)$ has $\\int f = 0$.\n\nExample: $n = 1$ and $a = \\mathbf{1}_{[0, 1]} - \\mathbf{1}_{[-1, 0]}$.\nFor $\\abs{x} \u003e 2$, using $\\int a = 0$, $Ha(x) = \\frac{1}{\\pi} \\int a(y) \\left( \\frac{1}{x - y} - \\frac{1}{x} \\right) dy$, which is bounded by $C/x^2$.\nNear the points $-1$, $0$ and $1$, $Ha$ has only logarithmic singularities, which are integrable.\nSo $Ha \\in L^1(\\R)$ and $a \\in H^1(\\R)$.\n\nNon-example: $f = \\mathbf{1}_{[0, 1]}$ on $\\R$ is integrable but not in $H^1(\\R)$.\nIts Hilbert transform is $Hf(x) = \\frac{1}{\\pi} \\log \\abs{\\frac{x}{x - 1}}$, which behaves like $\\frac{1}{\\pi x}$ as $\\abs{x} \\to \\infty$, so it is not integrable.\nThis agrees with the fact that $\\int f = 1 \\ne 0$.\n",
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