I'm trying to prove the Extended Liouville's Theorem:
Let $f$ be an entire function. Assume that for some $k \in \mathbb{N}$, and sufficiently large $|z|$, we have that $|f(z)| \leq A + B |z|^k$. Prove that $f$ is a polynomal of degree at most $k$.
Proof: The case $k = 0$ is the original Liouville Theorem. Assume the claim is true for some $k \in \mathbb{N}$. We prove it for $k + 1 \in \mathbb{N}$. Define the function,
\begin{align}
g(z) =
\begin{cases}
\frac{f(z) – f(0)}{z} \quad z \neq 0, \\
f'(0) \; \; \; \; \quad z = 0.
\end{cases}
\end{align}
Noting that $0 \leq |f(0)| \leq A$, we have that, for $z \neq 0$,
\begin{align}
|g(z)| & = \frac{|f(z) – f(0)|}{|z|}, \\
& \leq \frac{|f(z)| + |f(0)|}{|z|}, \\
& \leq \frac{A + B|z|^{k+1} + A}{|z|}, \\
& = \frac{2A + B|z|^{k+1}}{|z|}, \\
& = \frac{2A}{|z|} + B |z|^k, \\
& \leq \frac{2A}{M} + B |z|^k, \\
& \equiv D + B |z|^k, \\
\end{align}
where the last inequality follows since the theorem is stated for $|z| \geq M$ for some $M \in \mathbb{R}$. Considering the compact domain bounded by a closed and bounded circle of radius $M$, we have that, since $g(z)$ is entire, and hence continuous, we have that,
\begin{align}
g(z) \leq N \quad for \quad |z| \leq M.
\end{align}
Since $g(z)$ is bounded such that is satisfies the statement of theorem for some $k \in \mathbb{N}$, we have that $g(z)$ is a polynomial of degree at most $k$. A simple rearrangement of the given piecewise function now allows us to argue that $f(z)$ is a polynomial of degree at most $k + 1$.
Does this proof make sense?
Best Answer
Yes, your proof is fine, but you should write
\begin{align} |g(z) |\leq N \quad for \quad |z| \leq M. \end{align}