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  "body": "\n## Statement\n\nLet $0 \u003c r_1 \u003c r_2$ and let $f$ be [[holomorphic|0DG0Z72CQD]] on an open set containing the closed annulus $\\set{z : r_1 \\le \\abs{z} \\le r_2}$.\nFor $r_1 \\le r \\le r_2$ let $M(r) = \\max_{\\abs{z} = r} \\abs{f(z)}$.\nThen for each $r$ with $r_1 \\le r \\le r_2$,\n$$\\log M(r) \\le \\frac{\\log(r_2 / r)}{\\log(r_2 / r_1)} \\log M(r_1) + \\frac{\\log(r / r_1)}{\\log(r_2 / r_1)} \\log M(r_2).$$\nEquivalently, $\\log M(r)$ is a [[convex function|07Q7CNM5WF]] of $\\log r$.\n\n## Remarks\n\nProof idea: choose a real $\\alpha$ such that $r_1^{\\alpha} M(r_1) = r_2^{\\alpha} M(r_2)$.\nFor rational $\\alpha = p/q$ with $q \u003e 0$, the [[maximum modulus principle|048F2P90WR]] for $z^p f(z)^q$ on the annulus gives $r^{\\alpha} M(r) \\le r_1^{\\alpha} M(r_1)$.\nA limit over rational $\\alpha$, or the subharmonicity of $\\log \\abs{f(z)} + \\alpha \\log \\abs{z}$, handles irrational $\\alpha$, and the bound rearranges to the inequality.\nExample: for $f(z) = z^n$, $\\log M(r) = n \\log r$ is linear in $\\log r$, so equality holds.\nIf $M(r) = 0$ for some $r$, then $f$ vanishes on a circle, so $f \\equiv 0$ on the annulus and the inequality is trivial.\n\n## Explanations\n\n### L1 human\n\nTake three circles around $0$, one inside another.\nThe theorem bounds how large $\\abs{f}$ can be on the middle circle by its sizes on the inner and outer circles.\nPlot the logarithm of the maximum against the logarithm of the radius.\nThe result is a curve that bends upward, so between two radii it never rises above the straight line joining its values there.\n\n### L2 human\n\nThe three-circle theorem is a quantitative form of the maximum modulus principle.\nIt is the circle version of the [[Hadamard three-lines theorem|WHS58R43N0]]; the map $z \\mapsto e^{z}$ carries a vertical strip onto an annulus.\nIt is used in the theory of entire functions to study the growth of $M(r)$, and it extends to subharmonic functions.\n",
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