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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.

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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.

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    $\begingroup$ 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! $\endgroup$ – Feng Shao Jul 12 at 8:47
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    $\begingroup$ @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 $z$. $\endgroup$ – Martin R Jul 12 at 8:50
  • $\begingroup$ It's clear now. Thanks.(+1 $\endgroup$ – Feng Shao Jul 12 at 8:52
  • $\begingroup$ Why do you have $f(z)=p(z)e^{h(z)}$ given $f(z)$ entire with finitely many zeros? $\endgroup$ – Bach Jul 12 at 9:16
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    $\begingroup$ @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. $\endgroup$ – Martin R Jul 12 at 9:19
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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 $$G(z) = \sum_{k=0}^n \frac{G^{(k)}(0)}{k!} z^k, \qquad f(z) = p(z)e^{G(z)} $$

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

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