4

If $f(z)$ is an entire function and $|f(z)|\ge1$ for all $z$ with $|z|\ge \pi$ then show that $f$ is a polynomial.

I tried to apply Lioville's theorem on $f$. For $|z|\le \pi$ , $|f(z)|\le k$ for positive constant. But it does not help.

I've also tried with Taylor's series expansion as ,

$$f(z)=\sum_{n=0}^{\infty}a_nz^n$$where, $$a_n=\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)}{z^n}\, \mathrm{d}z$$

Then I wanted to find that $a_n=0$ for $n>p$ for some $p$ , but I failed to do so.

Empty
  • 13,012
  • 3
    Just an idea: The zeros of the function are in the disk $D(0,\pi)$, so there are a finite number of them. So you can write $f(z)=p(z)e^{g(z)}$ where $p$ is a polynomial whose zeros are in that disk, and $g$ is entire. – ajotatxe Jul 18 '15 at 11:11
  • Then how I proceed ? – Empty Jul 18 '15 at 11:36

1 Answers1

4

Hint 1: What type of singularity can $f$ have at $\infty$?

Hint 2 (based in ajotatxe's comment): Let $a_1,\dots,a_n$ be the zeroes of $f$, which are all in $\{|z|<\pi\}$, and consider $g(z)=\prod_{i=1}^n(z-a_i)/f(z)$.

Julián Aguirre
  • 76,354
  • (1) How you can factor $f$ ? (2) How you confirm that $f$ will be a polynomial ? (3) $|z|<\pi$ is NOT compact. So how you can say that $f$ has finite number of zeros ? – Empty Jul 18 '15 at 14:17
  • 1
    (2) $f$ does not have an essential singularity at $\infty$. This implies it is a polynomial. (3) $f$ has a finite number of zeroes on any compact set, in particular in ${|z|\le\pi}$, hence in ${|z|<\pi}$. – Julián Aguirre Jul 18 '15 at 15:43
  • 1
    Because the image of ${|z|>1}$ is not dense in $\mathbb C$. – Julián Aguirre Jul 19 '15 at 18:31
  • By the Casorati-Weierstrass theorem. – Julián Aguirre Jul 21 '15 at 09:57
  • Ok...Your Hint:1 is clear now..I have another question about Hint:2. Here, $|g(z)|\le \prod_{i=1}^n |z-a_i|$ for $|z|\ge \pi$. Then how I show that $g$ is bounded in $\mathbb C$ so that $g$ will be constant and then $f$ will be a polynomial ? – Empty Jul 21 '15 at 15:12