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Fall 2015 - Problem 11

Phragmén-Lindelöf, subharmonic functions

Let $\Omega = \set{\p{x, y} \in \R^2 \mid x > 0,\ y > 0}$ and let $u$ be subharmonic in $\Omega$ , continuous in $\cl{\Omega}$ , such that

u\p{x, y} \leq \abs{x + iy},

for large $\p{x, y} \in \Omega$ . Assume that

u\p{x, 0} \leq ax, \quad u\p{0, y} \leq by, \quad x, y \geq 0,

for some $a, b > 0$ . Show that

u\p{x, y} \leq ax + by, \quad \p{x, y} \in \Omega.

Solution.

Observe that

u\p{x, y} - \p{ax + by} \leq u\p{x, y} \leq \abs{x + iy}

for large $\p{x, y} \in \Omega$ , by assumption. Recall that $\phi\p{z} = \abs{z}^k$ is a Phragmén-Lindelöf function for $\Omega$ , where $0 < k < 2$ . In particular, $k = 1$ works. Thus, because $u\p{x, y} - \p{ax + by}$ is subharmonic ( $u$ is subharmonic and $ax + by$ is harmonic), it attains its maximum on the boundary.

On $y = 0$ , our assumption gives $u\p{x, 0} - ax \leq 0$ . Similarly, if $x = 0$ , we have $u\p{0, y} - by \leq 0$ as well. Hence, $u\p{x, y} - \p{ax + by} \leq 0$ on all of $\Omega$ , i.e.,

u\p{x, y} \leq ax + by

on all of $\Omega$ .

For completeness, we will prove that $\phi\p{z} = \abs{z}^k$ for $0 < k < \frac{\pi}{\beta - \alpha}$ is Phragmén-Lindelöf function (PL function) for the sector $\Omega = \set{\alpha < \Arg{z} < \beta}$ .

Let $0 < k < k_1 < \frac{\pi}{\beta - \alpha}$ and $\gamma = \alpha + \frac{\beta - \alpha}{2} = \frac{\alpha + \beta}{2}$ (this is the rotation needed to center the sector about the positive real axis). Set

\begin{aligned} \psi\p{z} = \Re\p{\p{e^{-i\gamma}z}^{k_1}} &= \Re\p{e^{k_1\Log\p{e^{-i\gamma}z}}} \\ &= \Re\p{e^{k_1\log\,\abs{z} + ik_1\p{\Arg\p{z}-\gamma}}} \\ &= \abs{z}^{k_1}\cos\p{k_1\p{\Arg\p{z} - \gamma}}, \end{aligned}

which is harmonic as the real part of an analytic function. Notice also that

-\frac{\beta - \alpha}{2} = \alpha - \gamma < \Arg\p{z} - \gamma < \beta - \gamma = \frac{\beta - \alpha}{2}.

Hence,

k_1\p{\Arg\p{z} - \gamma} \in \br{-k_1\p{\frac{\beta - \alpha}{2}}, k_1\p{\frac{\beta - \alpha}{2}}} \subseteq \p{-\frac{\pi}{2}, \frac{\pi}{2}},

i.e., $k$ is chosen so that the cosine term in $\psi$ is bounded strictly above $0$ . Thus, for $z$ large,

\frac{\phi\p{z}}{\psi\p{z}} \leq C\abs{z}^{k-k_1} \xrightarrow{z\to\infty} 0.

Let $u$ be subharmonic on $\Omega$ and continuous on $\cl{\Omega}$ . Suppose further that $u\p{z} \leq M$ on $\partial\Omega$ and satisfies $u\p{z} \leq C\phi\p{z}$ for large $z$ and some $C > 0$ .

Let $\epsilon > 0$ and set $u_\epsilon\p{z} = u\p{z} - \epsilon\psi\p{z}$ . This is subharmonic since $u$ is and because $\psi$ is harmonic. Then

u_\epsilon\p{z} \leq C\phi\p{z} - \epsilon\psi\p{z} = -\psi\p{z}\p{\epsilon - C\frac{\phi\p{z}}{\psi\p{z}}} \xrightarrow{z\to\infty} -\infty.

In particular, there exists $R > 0$ so that for large $z$ , $u_\epsilon\p{z} \leq M$ . Since $B\p{0, R} \cap \Omega$ is a bounded set, we may apply the maximum principle on $u_\epsilon$ to see $u_\epsilon\p{z} \leq M$ on $\cl{B\p{0, R}} \cap \Omega$ . Hence, $u_\epsilon \leq M$ on all of $\Omega$ . Sending $\epsilon \to 0$ , we see that $u \leq M$ on all of $\Omega$ , which completes the proof.