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I'm studying for my final exam and came across this problem: Let f and g be entire analytic functions and |f(z)|<|g(z)| when |z|>1. Show that f/g is a rational function.

I'm not quite sure where to start at all, but it does seem like a rather interesting result that I wouldn't expect.

Jonathan
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    Note that g(z) is non-zero outside $|z|>1$. – Marra May 09 '13 at 01:24
  • Ah good point. So it is bounded entire outside the unit disk but what about inside? – Jonathan May 09 '13 at 01:27
  • You just have to show it's meromorphic on the whole extended complex plane. There are no singularities outside the unit disk, including at infinity. What kind of singularities can there be inside the disk? – GCD May 09 '13 at 02:08
  • There can't be any essential singularities inside the disk right? – Jonathan May 09 '13 at 02:09
  • so if f/g only has isolated singularities inside the disk we are done, because then f/g is a meromorphic function on the riemann sphere and every meromorphic function on the riemann sphere is a rational function. Right idea? – Jonathan May 09 '13 at 02:16
  • @Jonathan Yes, and the reason the singularities are isolated is that the zeroes of $g$ are isolated. (As of any holomorphic function that's not identically zero). ¶ You can write your own answer below and accept it. – 75064 May 09 '13 at 04:41
  • See http://math.stackexchange.com/questions/261081/proving-an-entire-function-is-a-polynomial – lhf May 10 '13 at 01:41

1 Answers1

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First, $g$ can only have a finite number of zeros say $a_1,\ldots, a_n$. More importantly $g(z)=(z-a_1)\cdots(z-a_n)\cdot h$ where $h$ is some entire nonzero function. But then $|\frac{f(z)}{h(z)}|<|(z-a_1)\cdots(z-a_n)|$ for every $|z|>1$. That is $\frac{f}{h}$ is bounded by a polynomial. Thus, $\frac{f}{h}$ is a polynomial, which shows both that $h$ must be a constant and that $f$ is a polynomial. Thus both $f$ and $g$ are polynomials.

$\hspace{1in}$

Theorem. Let $f$ be an entire function and $n\in\mathbb{N}$ such that $|f|\le |z|^n$. Then $f$ must be a polynomial.

Proof. Let $f(z)=\sum_0^\infty a_k z^k $, we will show that the $a_k$'s must be zero eventually; they will in fact be zero when $k>n$. We perform the following calculation ($n$ is fixed, $m$ is arbitrary):

$\displaystyle |a_{n+m}|=\left|\frac{f^{(n+m)}(0)}{(n+m)!}\right|=\lim_{R\to\infty} \left|\frac{1}{2\pi i} \int_{B(0,R)} \frac{f(z)}{z^{n+m+1}} dz\right|\le\lim_{R\to\infty} \frac{1}{2\pi} \underset{z\in B(0,R)}{max}\ \left|\frac{f(z)}{z^{n+m+1}} \right|\cdot 2\pi R \le\lim_{R\to\infty} \underset{z\in B(0,R)}{max}\frac{|z|^n}{|z|^{n+m+1}} \cdot R \le\lim_{R\to\infty} \frac{R^n}{R^{n+m+1}} \cdot R = 0$

Thus, $a_k=0$ for $k>n$ and so $f$ must be a polynomial.

For the general result notice that any polynomial $|p|\le M\cdot |z|^n$ for some $M\in\mathbb{R}$ and $n\in\mathbb{N}$.

Interestingly, using the real or imaginary version of cauchy integral formula for taylor coefficients, we can extend this result to the real or imaginary part of a function. That is if $f=u+vi$ and $|u|\le |p|$ for some polynomial $p$ then $f$ must a polynomial. We can even (shockingly) drop the absolute values.

Theorem (Markushevich - Volume 2 - Page 265) Let $f=u+vi$. Suppose that $u(z)\le|z|^n$ for some $n\in\mathbb{N}$, then $f$ must be a polynomial.

Bobby Ocean
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    Excuse me, I am not an expert in this area. Why must $f/h$ be a polynomial when it is bounded by a polynomial for $|z|>1$? – TCL May 10 '13 at 01:46
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    Sorry about that, I went ahead and appended the appropriate material. Let me know if that makes sense. – Bobby Ocean May 13 '13 at 09:02
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    $h$ has no reason whatsoever to be constant. Let for example $f(z) = e^z$, and $g(z) = 2e^z$. The only thing you need - and that holds - is that $\dfrac{f}{h}$ is a polynomial. – Daniel Fischer Dec 11 '13 at 14:57