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  "body": "\n## Statement\n\nLet $S = \\set{z \\in \\C : \\abs{\\operatorname{Im} z} \u003c \\pi/2}$ be the horizontal strip of width $\\pi$, and let $u$ be [[subharmonic|Z1MHMGFNFW]] on $S$.\nAssume that $\\limsup_{z \\to \\zeta,\\, z \\in S} u(z) \\le 0$ for every point $\\zeta$ of the boundary of $S$.\nAssume also that there are constants $A \u003e 0$ and $\\alpha \u003c 1$ with $u(z) \\le A \\exp(\\alpha \\abs{\\operatorname{Re} z})$ for all $z \\in S$.\nThen $u \\le 0$ on $S$.\n\n## Remarks\n\nProof idea: pick $\\beta$ with $\\max(\\alpha, 0) \u003c \\beta \u003c 1$ and put $w(z) = \\operatorname{Re} \\cosh(\\beta z) = \\cosh(\\beta x) \\cos(\\beta y)$ for $z = x + iy$.\nThe function $w$ is [[harmonic|0DEB0JMN0W]], and on $\\overline{S}$ it satisfies $w \\ge \\cos(\\beta\\pi/2) \\cosh(\\beta x) \u003e 0$.\nFor $\\varepsilon \u003e 0$, the subharmonic function $u - \\varepsilon w$ tends to $-\\infty$ as $\\abs{\\operatorname{Re} z} \\to \\infty$, because $\\beta \u003e \\alpha$.\nThe [[maximum principle for subharmonic functions|9XJ5FBQD7M]] on long rectangles gives $u \\le \\varepsilon w$, and letting $\\varepsilon \\to 0$ gives $u \\le 0$.\n\nThe bound on $\\alpha$ is sharp.\nThe function $u(z) = \\operatorname{Re} e^{z} = e^{x}\\cos y$ is harmonic on $S$, vanishes on the boundary lines $y = \\pm\\pi/2$, and satisfies $u \\le e^{\\abs{\\operatorname{Re} z}}$, which is the case $\\alpha = 1$.\nBut $u \u003e 0$ on $S$.\nFor holomorphic $f$, applying the theorem to $\\log\\abs{f}$ gives the [[Phragmén–Lindelöf principle|FYR6V2NH6C]] for strips, and conformal maps carry it to sectors.\n\n## Explanations\n\n### L1 human\n\nOn a bounded region, a subharmonic function that is at most $0$ at the edge is at most $0$ inside.\nOn an unbounded strip this can fail, because the function may grow without limit along the strip.\nThe principle says it still holds as long as the function does not grow too fast: slower than $\\exp(\\abs{\\operatorname{Re} z})$, by a definite margin.\n\n### L2 human\n\nThis is the maximum principle for unbounded domains, with a growth condition replacing boundedness of the domain.\nThe threshold growth is the growth of the positive harmonic function $\\operatorname{Re} e^{z}$, which vanishes on the boundary.\nThe proof compares $u$ with the harmonic function $\\operatorname{Re}\\cosh(\\beta z)$, which grows more slowly than $\\operatorname{Re} e^{z}$ but faster than $u$.\nThe subharmonic form is more flexible than the holomorphic one, since it applies to $\\log\\abs{f}$, $\\abs{f}^p$ and to maxima of such functions.\n",
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