If an entire function grows slower than a polynomial, then it is a polynomial!

Question

I was investigating the following corollary to Liouville's Theorem in Complex Analysis: if $f(z)$ is entire and $\lim_{z\rightarrow \infty}z^{-n}f(z)=0$, then $f(z)$ is a polynomial in $z$ of degree at most $n-1$.

My proof uses the complex L'Hospital Rule and Induction on $n$. The base case $n=0$ is obvious. To illustrate the idea: for the case $n=2$ Suppose that $f(z)$ is not bounded, so that $$\lim_{z\rightarrow \infty}{f(z)}=\infty.$$ Then by using L'Hospital, we have $$\lim_{z\rightarrow \infty}{\frac{f(z)}{z^2}}=\lim_{z\rightarrow \infty}{\frac{f'(z)}{2z}}.$$ If $f'(z)$ is not bounded, then we have further $$\lim_{z\rightarrow \infty}{\frac{f(z)}{z^2}}=\lim_{z\rightarrow \infty}{\frac{f'(z)}{2z}}=\lim_{z\rightarrow \infty}{\frac{f''(z)}{2}}=0.$$ Thus, by Liouville's Theorem, we know that $f''(z)$ is equal to some constant, and since it approaches 0 at infinity, is simply 0. Then $f'(z)=A$ (technically this actually leads to a contradiction though, since then $f'(z)$ is bounded, contrary to assumption) for some complex number $A$, and thus $f(z)=Az+B$ for complex numbers $A$ and $B$. To complete the case analysis, if either $f(z)$ or $f'(z)$ were bounded, we could have applied the Liouville's Theorem to see that they were both constants, and thus $f(z)$ would be a polynomial in $z$ of degree at most 1.

Now suppose for the induction hypothesis that $\lim_{z\rightarrow \infty}{\frac{f(z)}{z^n}}=0$, we know that $f(z)$ is a polynomial in $z$ of degree at most $n-1$, and consider the case $n+1$. Then if $f(z)$ is not bounded, we know that $$\lim_{z\rightarrow \infty}{\frac{f(z)}{z^{n+1}}}=\lim_{z\rightarrow \infty}{\frac{f'(z)}{nz^n}}=0$$ so that by our induction hypothesis we know that $f'(z)$ is a polynomial in $z$ of degree at most $n-1$, and thus $f(z)$ is a polynomial in $z$ of degree at most $n$.

Is this proof correct? My main concerns are the use of L'Hospital's Rule, for which there seems to be little information about its validity for complex valued functions. Searching around I could find the following link, but it deals with complex-valued functions of a real variable, rather than the more general complex-valued functions of a complex variable.

Answer 1

This exercise is solved using Cauchy's Integral Formula: $$ f^{(m)}(0)=\frac{m!}{2\pi i}\int_{|z-z_0|}\frac{f(z)\,dz}{z^{m+1}} $$ and hence, for $m\ge n$: \begin{align} \lvert f^{(m)}(0)\rvert&=\frac{m!}{2\pi}\left|\int_{|z-z_0|=R}\frac{f(z)\,dz}{z^{m+1}}\right| \le \frac{m!}{2\pi}\cdot \frac{2\pi R}{R^{m+1}}\max_{|z|=R}\lvert f(z)\rvert =m!\frac{\max_{|z|=R}\lvert f(z)\rvert}{R^m}\\&= \frac{m!}{R^{m-n}}\frac{\max_{|z|=R}\lvert f(z)\rvert}{R^n}. \end{align} By hypothesis $$ \lim_{R\to\infty}\frac{\max_{|z|=R}\lvert f(z)\rvert}{R^n}=0, $$ and thus $f^{(m)}(0)=0$, for all $m\ge 0$, which means that the power series of $f$ at $z=0$ is a finite sum missing all the terms higher than the $(n\!-\!1)$-th. Hence $f$ is polynomial of degree at most $n-1$.

Answer 2

Let $f$ be an entire function and suppose that there exist constants $M\geq0$, $R>0$ and an integer $n\geq1$ such that $$|f(z)|\leq M|z|^n,\quad\text{for all $z$ with $|z|>R$}.$$ Show that $f$ is a polynomial with degree less than or equal to $n$.
Due to Cauchy's inequality $|f^{(k)}(0)|\leq k!M(\rho)\rho^{-k}$, where $M(\rho) = \sup_{|z|=\rho}|f(z)|$.

As $|f(z)|\leq M|z|^n$ then $$M(\rho)\leq M\rho^n\quad\text{for}~\rho>R.$$
Therefore $$|f^{(k)}(0)|\leq k!M\rho^n\rho^{-k} = k!M\rho^{n-k}\to0,\quad \rho\to\infty,\quad \text{for $k>n$}.$$ As $f\in\mathcal{H}(\mathbb{C})$ then $f(z) = \sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!}z^k$, and because of what was calculated before, the sum is finite, that is $$f(z) = \sum_{k=0}^{n}\frac{f^{k}(0)}{k!}z^k,$$ which implies that $f$ is a polynomial with degree less than or equal to $n$.