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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?

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    $\begingroup$ @MartinR I have edited the question. $\endgroup$
    – user82261
    Jul 5, 2018 at 7:17
  • $\begingroup$ Yes, your proof makes perfect sense. $\endgroup$ Jul 5, 2018 at 7:22

1 Answer 1

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Yes, your proof is fine, but you should write

\begin{align} |g(z) |\leq N \quad for \quad |z| \leq M. \end{align}

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