KU Math 800: Lecture 2/23/2026
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Recall from last time:
[theorem=Part 1] Given $\{a_n\}\subseteq \CC$ such that $|a_n| \to \infty$ as $n \to \infty$. Then there exists an entire $f:\CC \to \CC$ such that $f(a_n) = 0$. [/theorem]
We saw that such a function had the form
\[\begin{align*} f(z) = z^m \prod_{n=1}^\infty E_n\left(\frac{z}{a_n}\right) \end{align*}\]where $m$ is the number of zeroes in $\{a_n\}$ and
\[\begin{align*} E_0(z) &= 1 - z, \\ E_k(z) &= (1 - z)e^{z + \frac{z^2}{2} + \cdots + \frac{z^k}{k}}. \end{align*}\][theorem=Part 2] Any other such function is of the form
\[f(z)e^{g(z)}\]where $g(z)$ is entire. [/theorem]
The proof of the second part is as follows. Say $h:\CC \to \CC$ satisfies $h(a_n) = 0$, $h(z) \neq 0$, and $z \notin \{a_n\}$, then we have that
\[\frac{h}{f}\]is a nowhere vanishing entire function. So then we know that there exists entire $g$ for which
\[\frac{h}{f} = e^g\]by taking the logarithm.
This form might not be too ideal to work with, though, because $n$ becomes infinitely large and the canonical factors become unwieldy. So then we have a refinement due to Hadamard.
[theorem=(Hadamard’s Refinement)] Let $f$ be entire and have growth order $\rho_0$. That is,
\[|f(z)| \le A e^{B |z|^{\rho_0}}.\]Pick $k \in \ZZ$ such that $k \le \rho_0 < k+1$ (that is, $k$ is the floor $\rho_0$). If $a_1, \ldots$ are nonzero zeros of $f$, then
\[f(z) = e^{P(z)}z^m \prod_{n=1}^\infty E_k\left(\frac{z}{a_n}\right)\]where $P(z) \in \CC[z]$ is a polynomial of degree $\le k$ where $m$ is the order of the zero of $f$ at zero. [/theorem]
[lemma=%counter%]
\[|E_k(z)| \ge e^{-C|z|^{k+1}}\]if $|z| \le \frac{1}{2}$. [/lemma]
[lemma=%counter% (Corollary 5.4)] There exists of a sequence of real numbers $r_1, r_2, \ldots, r_m \to \infty$ such that
\[\left|\prod_{n=1}^\infty E_k\left(\frac{z}{a_n}\right)\right| \ge e^{-C|z|^s}\]if $|z| = r_m$. [/lemma]
Grant these 2 Lemmas, we first claim that
\[E(z) = z^m \prod_{n=1}^\infty E_k\left(\frac{z}{a_n}\right)\]converges. To do so, we need
\[|F_n(z) - 1| \le C_n\]where $\sum C_n < \infty$. If that happens to be true, then $\prod_{n} F_n(z)$ converges to a holomorphic function. Start by setting
\[F_n(z) = E_k\left(\frac{z}{a_n}\right)\]where $|a_n| \to \infty$ implies $|z/a_n| \le 1/2$ for $n \gg 0$. So by the Lemma,
\[|1 - F_n(z)| \le C\left|\frac{z}{a_n}\right|^{k+1}.\]By Theorem 2.1 of Chapter 5, we have that
\[\sum \left|\frac{1}{a_n}\right|^s < \infty\]if $s > \rho_0$. Take $s = k + 1$. So then $\prod F_n(z)$ converges.
Now we make a second claim that $f/E$ has no zeros. This would imply that there is a nowhere vanishing entire function $f/E = e^{g}$ with $g$ entire. So then $f = E e^{g}$. So then it suffices to show that $g$ is a polynomial of degree $\le k$. Since
\[\left|\frac{f(z)}{E(z)}\right| = \left|e^{g(z)}\right| = e^{\Re(g(z))}.\]Recall that $|f(z)| \le A e^{B|z|^{\rho_0}}$ and that
\[\left|E(z)\right| \ge e^{-C|z|^2}\]for $|z| = r_m$. So then
\[e^{\Re g(z)} \le D e^{C_0|z|^s}\]for $|z| = r_m$. Taking logarithms gives us
\[\Re g(z) \le C|z|^s\]for $|z| = r_m$.
[lemma=5.5] Suppose $g$ is entire and $\Re g(z) \le Cr^s$ whenever $|z| = r$ for a sequence of positive real numbers $r \to \infty$. Then $g$ is a polynomial of degree $\le s$. [/lemma]
By this lemma, it follows that $g$ is a polynomial of degree $\le s$. In particular, $g$ is a polynomial of degree $\le k$ where $k \le \rho_0 < k+1$. And so we are done!
[proof=Proof of Lemma] Since $g$ is entire, we can expand it into a power series
\[g(z) = \sum_{n=0}^\infty a_n z^n.\]A fact from Ch. 3 is that
\[\frac{1}{2\pi}\int_0^{2\pi} g(re^{i\theta}) e^{-in\theta} \dd{\theta} = \begin{cases} a_n r^n & \text{ if } n \ge 0, \\ 0 & \text{ if } n < 0. \end{cases}\]Taking the complex conjugate gives us
\[\frac{1}{2\pi}\int_0^{2\pi} \overline{g(re^{i\theta})} e^{in\theta} \dd{\theta} = 0\]if $n > 0$. So then if $n > 0$, we have
\[\frac{1}{2\pi}\int_0^{2\pi } \left[g(re^{i\theta}) + \overline{g(re^{i\theta})}\right]e^{-in\theta} \dd{\theta} = a_n r^n.\]But we can also write $u(z) = \Re(g(z))$ and the integral just becomes
\[a_n r^n = \frac{1}{\pi}\int_{0}^{2\pi} u(re^{i\theta}) e^{-in\theta} \dd{\theta}\]where $n > 0$. Notive that if $n = 0$ and we take the real part, we obtain
\[2 \Re(a_0) = \frac{1}{\pi}\int_{0}^{2\pi} u(re^{i\theta}) \dd{\theta}.\]Recall that
\[\int_{0}^{2\pi} e^{-in\theta} \dd{\theta} = 0\]if $n \neq 0$. So then we have
\[a_n = \frac{1}{\pi r^n} \int_{0}^{2\pi} \left[u(re^{i\theta}) - Cr^s\right] e^{-in\theta} \dd{\theta}.\]So then
\[\begin{align*} |a_n| &\le \frac{1}{\pi r^n} \int_{0}^{2\pi} \left|u(re^{i\theta}) - Cr^s\right|\dd\theta \\ &= \frac{1}{\pi r^n} \int_{0}^{2\pi} (Cr^s - u(re^{i\theta})) \dd{\theta} \\ &\le 2Cr^{s-n} - \frac{1}{\pi r^n}\int_0^{2\pi} u(re^{i\theta}) \dd{\theta} \\ &= 2Cr^{s - n} - 2\Re(a_0) r^{-n} \end{align*}\]which goes to zero as $r\to \infty$ if $n > s$. Thus, $|a_n| = 0$ if $n > s$. So then $g(z)$ is a polynomial of degree $\le s$. [/proof]
[proof=Proof of Lemma 1] If $|z| \le \frac{1}{2}$.
\[\begin{align*} E_k(z) &= (1 - z)e^{z + \frac{z^2}{2} + \cdots + \frac{z^k}{k}} \\ &= e^{\log(1 - z) + z + \frac{z^2}{2} + \cdots + \frac{z^k}{k}} \\ &= e^{-\sum_{n=k+1}^\infty z^k/n} = e^w \end{align*}\]Claim: $|w| \le C \cdot |z|^{k+1}$ (proved last time). If $|z| \le \frac{1}{2}$.
Claim 2: $|e^{w}| \ge e^{-|w|}$. Let $w = a + ib$ and so s it true that
\[|e^w| = e^a \ge e^{-\sqrt{a^2 + b^2}}$?\]Yes. Can be shown by cases: $a = 0$ and $a > 0$. [/proof]