Home > Complex questions > Complex #26: polynomial properties

## Complex #26: polynomial properties

Problem #26: If $f(z)$ is an entire function that is not a polynomial, prove that, given arbitrary $C > 0$, $R > 0$, and integer $m > 0$, there exists a $z$ such that  $|z|> R$ and $|f(z)| > C|z|^m$.

The way this is worded makes my head hurt, so let’s formulate it equivalently as Not B implies Not A instead of A implies B.
Reformulation: Let $f(z)$ be an entire function. Suppose that there exists a $C>0$$R > 0$, and an integer $m>0$  such that $|f(z)|\le C|z|^m$ whenever $|z|>R$. Prove that $f$ is a polynomial.

Proof: By the given and a limiting process, $|f(z)|\le CR^m$ for $|z|=R$. By the maximum modulus principle, the same fact is true for $|z|\le R$. By the Cauchy-Lagrange inequalities if $|z|=\frac{R}{2}$ , we have $|f^{(m)}(z)|\le \frac{m! M}{(R/2)^m}$, where $M$ is the maximum value of $|f(w)|$ for $|w-z|=\frac{R}{2}$. Since this circle is completely contained in the disk $\{ w:|w|\le R\}$, $M\le CR^m$, and so we have $|f^{(m)}(z)|\le (m!) 2^m C$ for $|z|=\frac{R}{2}$ and thus for $|z|\le \frac{R}{2}$. By Liouville’s theorem, $f^{(m)}(z)$ is a constant, and thus $f$ is a polynomial.