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  "body": "\n## Definition\n\nLet $\\mathbb{D} = \\set{z \\in \\C : \\abs{z} \u003c 1}$ be the [[unit disc|8G5ABXRK85]] and let $0 \u003c p \u003c \\infty$.\nThe *Hardy space* $H^p$ is the set of [[holomorphic functions|0NAHXR1FRH]] $f \\colon \\mathbb{D} \\to \\C$ such that\n$$\\norm{f}_{H^p} = \\sup_{0 \u003c r \u003c 1} \\left( \\frac{1}{2\\pi}\\int_0^{2\\pi} \\abs{f(r e^{i\\theta})}^p \\, d\\theta \\right)^{1/p} \u003c \\infty.$$\nThe space $H^\\infty$ is the set of bounded holomorphic functions on $\\mathbb{D}$, with $\\norm{f}_{H^\\infty} = \\sup_{z \\in \\mathbb{D}} \\abs{f(z)}$.\n\n## Remarks\n\nThe integral means increase with $r$, because $\\abs{f}^p$ is [[subharmonic|Z1MHMGFNFW]], so the supremum is a limit as $r \\to 1$.\nFor $f(z) = \\sum_{n \\ge 0} a_n z^n$, a [[power series|0D6KXBHKC0]] on $\\mathbb{D}$, Parseval's identity gives $\\frac{1}{2\\pi}\\int_0^{2\\pi} \\abs{f(re^{i\\theta})}^2 d\\theta = \\sum \\abs{a_n}^2 r^{2n}$, so $\\norm{f}_{H^2}^2 = \\sum_{n \\ge 0} \\abs{a_n}^2$.\nExample: $f(z) = \\log(1 - z) = -\\sum_{n \\ge 1} z^n/n$ has $\\sum 1/n^2 \u003c \\infty$, so $f \\in H^2$.\nNon-example: $g(z) = 1/(1 - z) = \\sum_{n \\ge 0} z^n$ has all coefficients $1$, so $g \\notin H^2$; in fact $g \\in H^p$ exactly for $p \u003c 1$.\nFor $1 \\le p \\le \\infty$, $H^p$ is a [[Banach space|06GZ956657]]; for $0 \u003c p \u003c 1$ it is a complete metric space but not a normed space.\nThere are also Hardy spaces of the upper half-plane and of domains in $\\C^n$ and $\\R^n$.\n",
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