Let $f(z)$ be analytic in the whole plane, and suppose that $f(z)$ has a nonessential singularity at $\infty$, Prove that $f(z)$ reduces to a polynomial.
My Thoughts so far :
Since $\infty $ is not an essential singularity of $f$ one of the following can happen
1) $\lim_{z \rightarrow \infty} f(z) = \infty $ ($\infty$ is a pole of finite order)
or
2) $\lim_{z \rightarrow \infty} f(z) = a \in \mathbb{C} $ ($\infty$ is a removable singularity)
Because $f$ has no poles , 2) implies that $f(z)=c, $ and we are done.
But in case 1) should I try to somehow show that $f^{(n)}(a)$ vanishes for some $n \in \mathbb{N}$ and for all integers $k > n$ ? Cauchy's estimate seems not to be helpful .
On the other hand, we can say that the behavior of $g(z)=f(\frac{1}{z})$ around zero is the same as the behavior of $f(z)$ at $\infty$ and Because $\infty$ is a pole of finite order, Can I say that $g(z)=\frac{h(z)}{z^k}$, where $\lim_{z \rightarrow 0} h(z) \neq 0 \ \ \ $ AND ? $ \ \ \ \lim_{z \rightarrow 0} h(z) \neq \infty $ Hence
$f(z)=g(\frac{1}{z})=z^kh(\frac{1}{z})$
According to above , Can we conclude that $f(z)$ is a polynomial ?
Thank you in advance.
Best Answer
If $|f(z)|\to \infty$ as $|z|\to \infty$ then look at the function $g(z)=1/ f(1/z)$ which has a removable singularity at $0$ and $g(0)=0$. If $m$ is the order of the zero of $g$ at $0$ then there exists an entire function $h$ which doesn't vanish around $0$ such that $$g(z)=z^m h(z)\Rightarrow f(1/z)= \frac{1}{z^m} \frac{1}{h(z)}$$ for $z$ in a neighborhood of $0$. Since $h$ is non-zero around $0$, $1/h$ is holomorphic around $0$, so it is bounded in some ball $|z|\leq R$. Hence, $1/|h(z)| \leq C$ for $|z|\leq R$ and by the above $$|f(1/z)|\leq C/ |z|^m$$ for $|z|\leq R$. Equivalently $$|f(z)|\leq C|z|^m$$ for $|z|\geq R'=1/R$. Now using Cauchy estimates for the derivatives of entire functions you can prove that $f$ is a polynomial of degree at most $m$.