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Spring 2012 - Problem 12

Jensen's formula

Show that the only entire function $f\p{z}$ obeying both

\abs{f'\p{z}} \leq e^{\abs{z}} \quad\text{and}\quad f\p{\frac{n}{\sqrt{1 + \abs{n}}}} = 0 \quad\text{for all}\quad n \in \Z

is the zero function. Here $'$ denotes differentiation.

Solution.

Suppose there exists such a function $f$ which is not the zero function. First, by the fundamental theorem of calculus,

f\p{z} = \int_0^z f'\p{z} \,\diff{z} + f\p{0} \implies \abs{f\p{z}} \leq \abs{z}e^{\abs{z}},

since $f\p{0} = 0$ . Since $e^x \geq x$ , we see $\abs{f\p{z}} \leq e^{2\abs{z}}$ .

Since $f$ is not identically zero, there exists $n_0 \in \N$ so that $g\p{z} \coloneqq z^{-n_0}f\p{z}$ satisfies $g\p{0} \neq 0$ . The zeroes of $f$ cannot accumulate, so the zeroes are countable, so let $\set{a_n}_n$ be an enumeration of the zeroes of $f$ , which are precisely the zeroes of $g$ .

First, notice that if $n$ is large enough, say, $n \geq N$ , then $\abs{\frac{n}{\sqrt{1 + \abs{n}}}} \leq \sqrt{n}$ . Thus, for any $M \geq N$ ,

\begin{aligned} \sum_{n=N}^M \log{\abs{\frac{n}{\sqrt{1 + \abs{n}}}}} &\leq \sum_{n=N}^M \frac{1}{2}\log{n} \\ &\leq \frac{1}{2} \int_N^M \log{x} \,\diff{x} \\ &\leq \frac{1}{2} \int_1^M \log{x} \,\diff{x} \\ &\leq \frac{1}{2}\p{M\log{M} - M + 1}, \end{aligned}

where the second inequality comes from considering lower Riemann sums of $\log{x}$ on $\br{N, M}$ .

Let $R > 0$ be such that $g\p{z} \neq 0$ on $\abs{z} = R$ . Then if $\abs{a} < R$ , we have $\log{\abs{\frac{a}{R}}} < 0$ , and so by Jensen's formula,

\begin{aligned} \log{\abs{g\p{0}}} &= \frac{1}{2\pi} \int_0^{2\pi} \log{\abs{g\p{Re^{i\theta}}}} \,\diff\theta + \sum_{\abs{a_n} < R} \log{\abs{\frac{a_n}{R}}} \\ &\leq \log{\abs{R^{-n_0}e^{2R}}} + \sum_{\abs{a_n} < R; n \geq N} \log{\abs{\frac{a_n}{R}}} \\ &\leq 2R - n_0\log{R} + \sum_{\frac{n}{\sqrt{1+\abs{n}}} < R; n \geq N} \log{\frac{\abs{n}}{R\sqrt{1 + \abs{n}}}} \\ &\leq 2R - n_0\log{R} + \sum_{\sqrt{n} < R; n \geq N} \log{\frac{\abs{n}}{R\sqrt{1 + \abs{n}}}} \\ &= 2R - n_0\log{R} - \sum_{\sqrt{n} < R} \log{R} + \sum_{\sqrt{n} < R; n \geq N} \log{\frac{\abs{n}}{\sqrt{1 + \abs{n}}}} \\ &\leq 2R - n_0\log{R} - \floor{R^2} \log{R} + \frac{1}{2}\p{R\log{R} - R + 1}. \end{aligned}

Since $g$ only has countably many zeroes, it follows that we may send $R \to \infty$ . But because the dominating term is $-\floor{R^2}\log{R}$ , we see $\log{\abs{g\p{0}}} \leq -\infty$ , which is impossible as $g\p{0} \neq 0$ . Thus, $f$ must have been identically zero to begin with, which completes the proof.