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  "body": "\n## Definition\n\nLet $n \\ge 2$, let $B = \\set{x \\in \\R^n : \\abs{x} \u003c 1}$ be the open unit ball, and let $S = \\partial B$ be the unit sphere.\nLet $\\sigma$ be the rotation-invariant Borel probability measure on $S$.\nLet $1 \\le p \u003c \\infty$.\nThe *harmonic Hardy space* $h^p(B)$ is the set of [[harmonic functions|0DEB0JMN0W]] $u \\colon B \\to \\R$ such that\n$$\\norm{u}_{h^p} = \\sup_{0 \u003c r \u003c 1} \\left( \\int_S \\abs{u(r\\zeta)}^p \\, d\\sigma(\\zeta) \\right)^{1/p} \u003c \\infty.$$\nThe space $h^\\infty(B)$ is the set of bounded harmonic functions on $B$, with $\\norm{u}_{h^\\infty} = \\sup_B \\abs{u}$.\n\n## Remarks\n\nEach $h^p$ with $1 \\le p \\le \\infty$ is a Banach space for the norm above.\nWrite $P(x, \\zeta) = (1 - \\abs{x}^2) / \\abs{x - \\zeta}^n$ for the [[Poisson kernel|TCYSWCTYYM]] of $B$.\nFor $1 \u003c p \\le \\infty$, $u \\in h^p$ if and only if $u(x) = \\int_S P(x, \\zeta) f(\\zeta)\\, d\\sigma(\\zeta)$ for some $f$ in the [[Lp space|SASP14MA1Z]] $L^p(\\sigma)$, and then $\\norm{u}_{h^p} = \\norm{f}_{L^p}$.\nFor $p = 1$, $u \\in h^1$ if and only if $u$ is the Poisson integral of a finite signed Borel measure on $S$.\nExample: fix $\\eta \\in S$ and let $u = P(\\cdot, \\eta)$.\nIt is positive and harmonic, and by the mean value property its mean over each sphere $\\abs{x} = r$ is $u(0) = 1$, so $u \\in h^1$ with norm $1$.\nIt is not in $h^p$ for $p \u003e 1$, since it is the Poisson integral of the point mass at $\\eta$, which has no $L^p$ density.\nNon-example: in the plane, let $u = \\operatorname{Re}\\left( \\frac{1 + z}{1 - z} \\right)^2$ on the unit disc.\nFor $e^{i\\theta} \\ne 1$, $u(r e^{i\\theta}) \\to -\\cot^2(\\theta / 2)$ as $r \\to 1$, and this function is not integrable on the circle.\nBy Fatou's lemma the means of $\\abs{u}$ over circles are unbounded, so $u \\notin h^1$.\nIn the plane, if $1 \u003c p \u003c \\infty$ and $f = u + iv$ is holomorphic with $v(0) = 0$, then $u \\in h^p$ if and only if $f$ is in the holomorphic [[Hardy space|PNNNFJVF6W]] $H^p$, by the [[M. Riesz theorem|PJFH51Z48M]].\nThis fails for $p = 1$ and for $p = \\infty$.\n",
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