# Proving $f$ is a polynomial of degree at most $2$.

Suppose $f$ is entire and that $|f(z)|\leq|z|^2$ for all $|z|>r_0$. I need to prove that $f$ is a polynomial of degree at most $2$. This is my approach.

Let $z_0$ be such that $|z_0|>r_0$. Choose $R>0$ such that $R={|z_0|-r_0\over 2}$. Then $D_{R(z_0)}$ is outside the disk $D_{r_0}(0)$. So if we apply the Cauchy estimate for each $z$ such that $|z-z_0|=R$ $$f^{(n)}(z_0)\leq\frac{n!}{R^n}Max_{|z-z_0|=R}|z|^2\leq \frac{n!}{R^n}(R+z_0)^2$$ since $$|z|-|z_0|\leq|z-z_0|=R$$ So for $n\geq 3$ as $R\rightarrow \infty$ , $f^{(n)}(z_0)\rightarrow 0$, so $f^{(n)}=0$ which implies that $f$ is a polynomial of degree 2. Is my answer correct? THanks

• You have on the one hand arranged it so that $D_R(z_0)$ is disjoint from $D_{r_0}(0)$, and on the other, you want to let $R\to\infty$, you cannot have both at the same time. Being able to let $R\to\infty$ is the important thing here (and having the estimate $\lvert f(z)\rvert \leqslant \lvert z\rvert^2$ for the $z$ on the boundary). You can get both if you choose $R$ so large that $\overline{D_{r_0}(0)} \subset D_R(z_0)$. Commented Dec 23, 2014 at 15:25
• @DanielFischerSorry about my previous comment Commented Dec 23, 2014 at 15:33
• @DanielFischer So I guess $R>|z_0|+r_0$ will do? Commented Dec 23, 2014 at 15:34
• Yes, that will do. Another way to show it is to look at $f^{(n)}(0)$ and use the Cauchy estimates to find that $f^{(n)}(0) = 0$ for $n\geqslant 3$. Commented Dec 23, 2014 at 15:36
• So just pick an $R>r_0$ and then $f^{(n)}(0)\leq \frac{n!}{R^n}R^2$ so for $n\geq 3$ $f^{(n)}(0)=0$ as tends to infinity ? Commented Dec 23, 2014 at 15:40

Suppose $R > r_0$, the maximum modulus theorem gives $M= \sup_{|z|\le R} |f(z)| = \sup_{|z| = R} |f(z)| \le R^2$.
Now estimate $|f^{(n)}(0)|$.