# On entire functions of finite order

An entire function $$f(z)$$ has said to have finite order if there exist positive constants $$c$$ and $$n$$ such that $$|f(z)|\le ce^{|z|^n} .$$ Prove that if such a function has only a finite number of zeros, then it must be of the form $$f(z)=p(z)e^{q(z)} ,$$ where $$p$$ and $$q$$ are polynomials.

My attempt:

I have tried considering the function $$g(z)\colon=f(z)/e^{z^n}$$ but we can not use the condition since $$|e^{z^n}|\le e^{|z|^n}$$ but $$|g(z)|\ge\frac{|f(z)|}{e^{|z|^n}}\le c.$$ Then I am stuck... I really need some hints to move on. Thank you.

This is an immediate consequence of the Hadamard factorization theorem. A direct proof goes as follows:

If $$f$$ has only finitely many zeros then $$f(z) = p(z) e^{h(z)}$$ for some polynomial $$p$$ and an entire function $$h$$. Then $$e^{\operatorname{Re}h(z)} = |e^{h(z)}| = \frac{|f(z)|}{|p(z)|} \le c e^{|z|^n} \\ \implies \operatorname{Re}h(z) \le |z|^n + \log(c)$$ for sufficiently large $$z$$. Now use Can the real part of an entire function be bounded above by a polynomial? to conclude that $$h$$ is in fact a polynomial.

• Could you please explain why $\frac{|f(z)|}{|p(z)|} \le c e^{|z|^n}$? I tried this problem before you answered and this is where I stuck. Thanks in advance! – Feng Shao Jul 12 at 8:47
• @FengShao: $|f(z)|\le ce^{|z|^n}$ is given. If $f$ has no zeros then we can choose $p(z) = 1$. Otherwise the degree of $p$ is at least one, so that $|p(z)| \to \infty$ for $z \to \infty$. So in all cases, $|p(z)| \ge 1$ for sufficiently large $z$. – Martin R Jul 12 at 8:50
• It's clear now. Thanks.(+1 – Feng Shao Jul 12 at 8:52
• Why do you have $f(z)=p(z)e^{h(z)}$ given $f(z)$ entire with finitely many zeros? – Bach Jul 12 at 9:16
• @Bach: If $a_1, \ldots, a_n$ are the zeros of $f$ then $f(z)/\prod(z-a_i)$ is an entire function without zeros, and that is necessarily of the form $e^h$, see for example math.stackexchange.com/q/267805/42969. – Martin R Jul 12 at 9:19

If $$f$$ has finitely zeros let $$p(z) = \prod_i (z-a_i)$$ then $$g(z) = f(z)/p(z)$$ is entire and has no zeros thus $$g(z) = e^{G(z)}$$ where $$G$$ is entire.

Then the main point is Borel–Carathéodory theorem : from $$|f(z)|\le ce^{|z|^n}$$ we know $$\Re(G(z)) \le C |z|^{n+1/2}$$ which implies $$|G(z)| \le A |z|^{n+1/2}$$ so that (Liouville theorem, Cauchy integral formula) $$G^{(k)}(0) = 0$$ for $$k \ge n+1$$ and $$G(z) = \sum_{k=0}^n \frac{G^{(k)}(0)}{k!} z^k, \qquad f(z) = p(z)e^{G(z)}$$

• Sure I wanted to make clear Hadamard is a corollary of Borel-Catheodory – reuns Jul 12 at 10:53