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Let $f$ be an entire function such that $ |f(z)| \to \infty$ as $|z| \to \infty$. Prove that $f$ is a polynomial.

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2 Answers 2

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Consider the holomorphic function on $\mathbb{C}^*$ $$ g(z)=f\left(\frac{1}{z}\right). $$ So $$ \lim_{z\rightarrow 0}\;|g(z)|=+\infty. $$ Obviously, $0$ is not a removable singularity.

Now by the Casorati-Weierstrass theorem, $0$ is not an essential singularity either. Indeed, take $r>0$ such that $|g(z)|\geq 1$ for all $0<|z|<r$. Then $g(z\;;\;0<|z|<r\})$ is certainly not dense in $\mathbb{C}$.

Edit: As pointed out by @mrf, it is actually easy to see that this implies $0$ is a pole for $g$. It suffices to use Riemann's theorem on removable singularities adequately. Indeed $h(z)=1/g(z)$ is holomorphic and bounded in some $\{z\;;\;0<|z|<r\}$ with limit $0$ at $0$, so $0$ is a removable singularity of $h$ and a zero of the holomorphic extension of $h$. Factor $h(z)=z^nk(z)$ with $k$ holomorphic and $k(0)\neq 0$. The result follows. And this is indeed more elementary than Casorati-Weierstrass. So keep the Casorati-Weierstrass argument for weaker assumptions such as the ones I mention below.

So $0$ must be a pole. Hence the Laurent series of $g$ at $0$ is of the form $$ g(z)=\frac{a_n}{z^n}+\ldots+\frac{a_1}{z}+a_0. $$ Note that there are no positive powers of $z$ because this comes from the power series of $f$.

The result follows clearly by going back to $f$.

Edit: Note the same proof works if we simply assume that $f$ is entire and there exist $\alpha>0$ and $M>0$ such that $$ |f(z)|\geq \alpha\qquad \forall |z|\geq M. $$

Of course, there could be even weaker assumptions like there exists $M>0$ such that $f(\{z\;;\; |z|>M\})$ is not dense in $\mathbb{C}$ and the proof would still work.

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This is a good answer +1, but is it possible to do with more elementary tools than Casorati-Weierstrass? –  user58512 Feb 23 '13 at 13:02
    
@user58512 Thanks. Maybe, since the assumption is much stronger than what is really needed to apply Casorati-Weierstrass. Let's wait and see. –  1015 Feb 23 '13 at 13:17
    
Since $|g|\to \infty$ as $z\to 0$, we see that $0$ is a pole directly (without Weierstrass-Casorati). –  mrf Feb 23 '13 at 13:22
    
@Seirios Thanks for the edit, you made me understand that it was possible to do this. After six months...it was high time. –  1015 Feb 23 '13 at 13:23
    
Often this property is used as the very definition of pole, but even with other plausible definitions, it's straight-forward to see that an isolated singularity is a pole iff $|f| \to\infty$ at the point in question. –  mrf Feb 23 '13 at 13:28

Hint: The set of zeroes of $f$ is bounded and discrete, hence finite and this allows us to find a polynomial $p$ such that $f(z)=p(z)g(z)$ for some entire function $g\colon \mathbb C\to\mathbb C^\times$. What scenarios can happen with $g(z)$ as $z\to\infty$?

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3  
If a set of points is bounded and discrete, can you then conclude is finite? Any sequence that converges is bounded and discrete, and infinite, like $\sum_0^\infty \frac{1}{2^k}$ –  MyUserIsThis Feb 23 '13 at 11:31
    
@GitGud Yes thanks. –  MyUserIsThis Feb 23 '13 at 11:33
1  
@MyUserIsThis: Some people take discrete to mean closed and discrete for which the statement that boundedness implies finiteness is true. The set of zeroes of $f$ is clearly closed. –  Martin Feb 23 '13 at 13:33
    
Hm, upon rereading I think that whoever upvoted me here should be ashamed :9 –  Hagen von Eitzen Feb 23 '13 at 15:25

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