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  "body": "\n## Statement\n\nLet $\\mathbb{D} = \\set{z \\in \\C : \\abs{z} \u003c 1}$.\nLet $\\omega \\colon \\mathbb{D} \\to \\mathbb{D}$ be a [[holomorphic function|0NAHXR1FRH]] with $\\omega(0) = 0$, and let $F$ be a holomorphic function on $\\mathbb{D}$.\nThen for every $p$ with $0 \u003c p \u003c \\infty$ and every $r$ with $0 \u003c r \u003c 1$,\n$$\\int_0^{2\\pi} \\abs{F(\\omega(r e^{i\\theta}))}^p \\, d\\theta \\le \\int_0^{2\\pi} \\abs{F(r e^{i\\theta})}^p \\, d\\theta.$$\n\n## Remarks\n\nThe function $\\abs{F}^p$ is [[subharmonic|Z1MHMGFNFW]].\nLet $u$ be the harmonic function on the disc $\\abs{z} \u003c r$ with boundary values $\\abs{F}^p$ on $\\abs{z} = r$, given by the [[Poisson kernel|TCYSWCTYYM]]; then $\\abs{F}^p \\le u$ on $\\abs{z} \\le r$.\nBy the [[Schwarz lemma|0MB48XDBY3]], $\\abs{\\omega(z)} \\le \\abs{z}$, so $\\omega$ maps the closed disc of radius $r$ into itself and $\\abs{F \\circ \\omega}^p \\le u \\circ \\omega$ there.\nSince $u \\circ \\omega$ is harmonic, its mean over $\\abs{z} = r$ is $u(\\omega(0)) = u(0)$, which is the mean of $\\abs{F}^p$ over $\\abs{z} = r$.\nThe hypothesis $\\omega(0) = 0$ is needed.\nFor the constant map $\\omega \\equiv c$ with $0 \u003c c \u003c 1$ and $F(z) = z$, the left side is $2\\pi c^p$ and the right side is $2\\pi r^p$.\nSo the inequality fails for $r \u003c c$.\n\n## Explanations\n\n### L1 human\n\nComposing a function on the disc with a self-map of the disc that fixes the centre can only make the function smaller on average along circles.\nThe self-map pulls each circle inside a smaller region, and averages of nice functions over circles grow with the radius.\nSo the theorem compares the size of $F \\circ \\omega$ with the size of $F$ itself.\n\n### L2 human\n\nTaking the supremum over $r$ shows that the composition operator $F \\mapsto F \\circ \\omega$ is a contraction on each [[Hardy space|PNNNFJVF6W]] $H^p$ when $\\omega(0) = 0$.\nFor a general holomorphic self-map $\\omega$ one composes with a disc automorphism moving $\\omega(0)$ to $0$, which shows that every composition operator on $H^p$ is bounded.\nThe proof only uses that $\\abs{F}^p$ is subharmonic, so the same inequality holds for any subharmonic function in place of $\\abs{F}^p$.\nThis is the starting point of the theory of composition operators on spaces of analytic functions.\n",
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