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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

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    $\begingroup$ 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 $\overline{D_{r_0}(0)} \subset D_R(z_0)$. $\endgroup$ Commented Dec 23, 2014 at 15:25
  • $\begingroup$ @DanielFischerSorry about my previous comment $\endgroup$
    – Heisenberg
    Commented Dec 23, 2014 at 15:33
  • $\begingroup$ @DanielFischer So I guess $R>|z_0|+r_0$ will do? $\endgroup$
    – Heisenberg
    Commented Dec 23, 2014 at 15:34
  • $\begingroup$ 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 $n\geqslant 3$. $\endgroup$ Commented Dec 23, 2014 at 15:36
  • $\begingroup$ 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 ? $\endgroup$
    – Heisenberg
    Commented Dec 23, 2014 at 15:40

1 Answer 1

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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 $|f^{(n)}(0)|$.

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  • $\begingroup$ Thanks for the answer. I too figured this out as mentioned in the comment above $\endgroup$
    – Heisenberg
    Commented Dec 23, 2014 at 15:48
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    $\begingroup$ Actually my reasoning is needlessly complicated. Let me repair it. Dire caffeine need... $\endgroup$
    – copper.hat
    Commented Dec 23, 2014 at 15:50
  • $\begingroup$ No worries idea noted $\endgroup$
    – Heisenberg
    Commented Dec 23, 2014 at 15:55
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    $\begingroup$ @copper.hat: you need to visit your local dealer, the Med, for a fix! Cheers! $\endgroup$ Commented Dec 23, 2014 at 16:07
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    $\begingroup$ @RobertLewis: That is a great idea (if I could stomach the holiday traffic). HH! $\endgroup$
    – copper.hat
    Commented Dec 23, 2014 at 16:09

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