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Fall 2009 - Problem 12

Jensen's formula, meromorphic functions

Let $f$ be a non-constant meromorphic function in the complex plane. Assume that if $f$ has a pole at the point $z \in \C$ , then $z$ is of the form $n\pi$ with integer $n \in \Z$ . Assume that for all non-real $z$ we have the estimate

\abs{f\p{z}} \leq \p{1 + \abs{\Im{z}}^{-1}}e^{-\Im{z}}.

Prove that for every integer $n \in \Z$ , $f$ has a pole at the point $n\pi$ .

Solution.

Suppose $f$ has a pole at $n\pi$ and let $\epsilon > 0$ . Then for $z = n\pi + i\epsilon$ ,

\begin{aligned} \abs{\p{z - n\pi}f\p{z}} &\leq \p{\abs{z - n\pi} + \frac{\abs{z - n\pi}}{\abs{\Im{z}}}}e^{-\Im{z}} \\ &= \p{\epsilon + 1}e^{-\epsilon}. \end{aligned}

Thus, $\p{z - n\pi}f\p{z}$ is bounded as $\epsilon \to 0$ , so $f$ must have a simple pole at $n\pi$ . Indeed, if $f$ had a pole with higher order, then $\abs{\p{z - n\pi}f\p{z}}$ must diverge to $\infty$ from any path.

Let $g\p{z} = f\p{z}\sin{z}$ . Because $f$ had a simple pole at each $n\pi$ and $\sin{z}$ has a simple zero at each $n\pi$ , we have by the factor theorem that $\sin{z} = \p{z - n\pi}h\p{z}$ , where $h$ is holomorphic near $n\pi$ . Thus, $g\p{z} = \p{z - n\pi}f\p{z}h\p{z}$ , so $g$ is holomorphic at $n\pi$ . In particular, $g$ is entire.

Also notice that by definition,

\abs{\sin{z}} \leq \abs{\frac{e^{iz} - e^{-iz}}{2i}} \leq \frac{e^{\abs{z}} + e^{\abs{z}}}{2} = e^{\abs{z}},

so we have the bound

\abs{g\p{z}} \leq 1 + \frac{1}{\abs{\Im{z}}}.

We will show that $g$ vanishes nowhere. By Jensen's formula, if $g$ has zero anywhere, then

\frac{1}{2\pi} \int_0^{2\pi} \log\abs{g\p{Re^{i\theta}}} \,\diff\theta \geq A\log{R}

for $A$ large enough, so it suffices to prove that this integral tends to $0$ . By our previous estimate,

\begin{aligned} \frac{1}{2\pi} \int_0^{2\pi} \log\abs{g\p{Re^{i\theta}}} \,\diff\theta &\leq \frac{1}{2\pi} \int_0^{2\pi} \log\abs{1 + \frac{1}{R\abs{\sin\theta}}} \,\diff\theta \\ &= \frac{2}{\pi} \int_0^{\pi/2} \log\abs{1 + \frac{1}{R\sin\theta}} \,\diff\theta. \end{aligned}

The equality comes from periodicity of $\sin\theta$ . Write $\theta = \frac{t\pi}{2}$ for $t \in \br{0, 1}$ so that by concavity of $\sin\theta$ ,

\sin\theta \geq t\sin\frac{\pi}{2} = t = \frac{2\theta}{\pi}.

Since $t \in \br{0, 1}$ this inequality holds for $\theta \in \br{0, \frac{\pi}{2}}$ and combining this with the change of variables $u = \frac{1}{\theta}$ to get

\begin{aligned} \frac{1}{2\pi} \int_0^{2\pi} \log\abs{g\p{Re^{i\theta}}} \,\diff\theta &\leq \frac{2}{\pi} \int_0^{\pi/2} \log\abs{1 + \frac{\pi}{2R\theta}} \,\diff\theta \\ &\leq \frac{2}{\pi} \int_{2/\pi}^\infty \frac{1}{u^2} \log\abs{1 + \frac{u\pi}{2R}} \,\diff{u}. \end{aligned}

Finally, notice that for $u$ large enough,

\frac{1}{u^2} \log\abs{1 + \frac{u\pi}{2R}} \leq \frac{u^{1/2}}{u^2} = \frac{1}{u^{3/2}} \in L^1\p{\pco{\frac{2}{\pi}, \infty}}.

Thus, by dominated convergence,

\begin{aligned} \lim_{R\to\infty} \frac{1}{2\pi} \int_0^{2\pi} \log\abs{g\p{Re^{i\theta}}} \,\diff\theta &\leq \lim_{R\to\infty} \frac{2}{\pi} \int_{2/\pi}^\infty \frac{1}{u^2} \log\abs{1 + \frac{u\pi}{2R}} \,\diff{u} \\ &= \frac{2}{\pi} \int_{2/\pi}^\infty \frac{1}{u^2} \log{1} \,\diff{u} \\ &= 0. \end{aligned}

Hence, $g$ has no zeroes. Because $g$ is entire, it follows from Liouville's theorem applied to $\frac{1}{g}$ that $g$ is constant, i.e.,

f\p{z}\sin{z} = C \implies f\p{z} = \frac{C}{\sin{z}}.

As a result, $f$ has a simple pole at $n\pi$ for every $n \in \Z$ , as desired.