1. ## Complex Analysis

Suppose that $\displaystyle f$ is an entire functionand that there are positive constants $\displaystyle A$ and $\displaystyle m$ with $\displaystyle |f(z)| \leq A|z|^{m}$ if $\displaystyle |z| \geq R_{0}$. Show that $\displaystyle f$ is a polynomial of degree $\displaystyle m$ or less.

2. Originally Posted by chiph588@
Suppose that $\displaystyle f$ is an entire functionand that there are positive constants $\displaystyle A$ and $\displaystyle m$ with $\displaystyle |f(z)| \leq A|z|^{m}$ if $\displaystyle |z| \geq R_{0}$. Show that $\displaystyle f$ is a polynomial of degree $\displaystyle m$ or less.
isn't this just a slight twist on the (extended) Liouville's theorem?

3. Originally Posted by chiph588@
Suppose that $\displaystyle f$ is an entire functionand that there are positive constants $\displaystyle A$ and $\displaystyle m$ with $\displaystyle |f(z)| \leq A|z|^{m}$ if $\displaystyle |z| \geq R_{0}$. Show that $\displaystyle f$ is a polynomial of degree $\displaystyle m$ or less.
Let $\displaystyle R>R_0$.

If $\displaystyle f$ is an entire function then $\displaystyle f(z) = \sum_{n=0}^{\infty}a_n z^n$ and $\displaystyle a_n = \frac{1}{2\pi i}\oint \limits_{|z|=R} \frac{f(z)}{z^{n+1}} dz$.

If $\displaystyle |f| \leq M$ on $\displaystyle |z|=R$ then $\displaystyle \left| a_n \right| = \left| \frac{1}{2\pi i}\oint \limits_{|z|=R} \frac{f(z)}{z^{n+1}}dz \right| \leq \frac{M}{R^n}$

By hypothesis $\displaystyle |f(z)| \leq A |z|^m$ for $\displaystyle |z| \geq R_0$. Therefore, if $\displaystyle R > R_0$ then on $\displaystyle |z| = R$ we have $\displaystyle |f(z)| \leq A |z|^m = A\cdot R^m$

Thus, $\displaystyle |a_n| \leq \frac{A \cdot R^m}{R^n}$.
Thus, if $\displaystyle n>m$ then by making $\displaystyle R$ sufficiently large we have $\displaystyle |a_n| = 0 \implies a_n = 0$.

Thus, $\displaystyle f$ is a polynomial of degree at most $\displaystyle m$.

4. That's the proof I used, does that prove it for the disk $\displaystyle |z| < R_{0}$ also?