Phragmén–Lindelöf principle for subharmonic functions

Remarks

Proof idea: pick $\beta$ with $\max(\alpha, 0) < \beta < 1$ and put $w(z) = \operatorname{Re} \cosh(\beta z) = \cosh(\beta x) \cos(\beta y)$ for $z = x + iy$. The function $w$ is harmonic, and on $\overline{S}$ it satisfies $w \ge \cos(\beta\pi/2) \cosh(\beta x) > 0$. For $\varepsilon > 0$, the subharmonic function $u - \varepsilon w$ tends to $-\infty$ as $\abs{\operatorname{Re} z} \to \infty$, because $\beta > \alpha$. The maximum principle for subharmonic functions on long rectangles gives $u \le \varepsilon w$, and letting $\varepsilon \to 0$ gives $u \le 0$.

The bound on $\alpha$ is sharp. The 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 $\alpha = 1$. But $u > 0$ on $S$. For holomorphic $f$, applying the theorem to $\log\abs{f}$ gives the Phragmén–Lindelöf principle for strips, and conformal maps carry it to sectors.

Going deeper

For a student

This is the maximum principle for unbounded domains, with a growth condition replacing boundedness of the domain. The threshold growth is the growth of the positive harmonic function $\operatorname{Re} e^{z}$, which vanishes on the boundary. The proof compares $u$ with the harmonic function $\operatorname{Re}\cosh(\beta z)$, which grows more slowly than $\operatorname{Re} e^{z}$ but faster than $u$. The 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.