Prove $f$ entire and with a pole at infinity to be polynomial

Question

The question has been asked before, but I can't seem to wrap my head around it. I guess I'm missing something.

There is a certain part where I'm experiencing problems, and I can't seem to find an answer to that specific part in the other question(s).

If $f$ is entire and $\lim_{z\to \infty} f(z)= \infty$ then $f$ is polynomial.

Proof:

Since $f$ is entire it can be expanded as a Taylor-series.

$$f(z) =\sum_{n=0}^{+\infty} a_n z^n$$

To prove $f$ would be polynomial I would have to prove $(\exists N)(\forall n > N)(a_n = 0)$

Define:

$$g(z) = f\left( \frac{1}{z}\right) = \sum_{n=0}^{+\infty} a_n\left(\frac{1}{z}\right)^n$$

Since $0$ is a pole (see below) of $g(z)$ the expansion above would be it's corresponding Laurent-expansion. And by definition of a pole $(\exists N)(\forall n >N)(a_n = 0)$.

Q.E.D.

However

Why is $0$ a pole of $g(z)$, why does $\lim_{z \to \infty} f(z) = \infty$ imply $0$ to be a pole of $g(z)$? I know $\lim_{z \to \infty} f(z) = \infty$ implies $\infty$ to be a pole of $f(z)$, but then...?

Answer 1

Well, $g(0)=f\left(\frac10\right)=f(\infty)=\infty$.

Answer 2

Note that when $z\rightarrow \infty, w=\frac{1}{z}\rightarrow 0$. And since $f(z)\rightarrow \infty$ as $z\rightarrow \infty$, there is no essential singularity. So $g(w)$ must have a pole of finite order at $\infty$.

Answer 3

To show that 0 is a pole for f(1/z), try this:

Define $h(z)=\frac{1}{f(\frac{1}{z})}$, then $h(z)$ is analytic for $f(\frac{1}{z})$ non-zero. And note that $\lim_{z\to 0} |h(z)|=0 \implies lim_{z\to 0}h(z)=0$.

As the limit exists, we can take $h^*$ to be an analytic extension. Knowing that $h^*(z)$ is not identically zero, gives us that zeros are separated, so we can find a ball of radius $R>0$ such that $h^*(z)$ is analytic on $B_R(0)$.

Thus $h^*$ has a Taylor expansion centered about 0 on that region. As the first coefficient of that is 0 (Looking at the limit), but not all coefficients are non-zero or $h^*$ would be identically zero.

We have that $h^*(z)=z^N*\sum_{n=N}^\infty a_n z^{n-N}=z^N*k(z)$, where k is analytic and doesn't vanish on the disk.

$$lim_{z\to 0}Z^{N+1}f\left(\frac{1}{z}\right)=lim_{z\to 0} \frac{z}{k(z)}=0$$

Therefore $f(1/z)$ has a pole of order N at 0 $\implies$ f(z) has a pole of order N at infinity. Given this, dietervdf proof follows easily, but proving that the singularity at infinity is a pole and not an essential singularity is, I think, non-trivial.

Answer 4

For the purposes of the proof, let $|f(z)| \geq M$ for $|z| > R$. Assume $f$ is not a polynomial and is entire. Then $f$ can only have an essential singularity at $\infty$ (let's convince ourselves of this first ). It follows that

$$g(w) = f\left(\frac{1}{w}\right)$$

has an essential singularity at $w = 0$. By the Casorati–Weierstrass theorem, for all $c \in \mathbb{C}$, there exists a nonzero sequence $w_n$ converging to $0$ such that $g(w_n) = c$ as $n \to \infty$.

Now, observe that $|g(w)| \geq M$ in $|w| < R$. Setting $c = \frac{1}{2} M$ gives us a contradiction, since $g(w_n)$ never achieves this value (roughly speaking).