# Prove that if $f$ is entire and $|f(z)| \leq |z|^2+12$ then $f$ is a polynomial of deg $\leq 2$

[Solution Verification]

Prove that if $f$ is entire and $|f(z)| \leq |z|^2+12$ for all $z \in \mathbb C$ then $f$ is a polynomial of degree $\leq 2$

So here's my proof, and there is a problematic thing I would like to address:

Since $f(z)$ is entire, we can develop a power series around $z=0$, so that $f(z)=\Sigma_{n=0}^\infty a_nz^n$, with $a_n= \frac {f^{(n)}(0)}{n!}$. It will suffice to prove that $f^{(n)}=0$ for all $n \geq 3$

Let there be a circle $C_R$ with radius $R$ centered at $z=0$. So by the general Cauchy Integral Formula we get that:

$$f^{(n)}(0)=\frac{n!}{2\pi i}\int_{C_R}\frac{f(z)}{z^{n+1}}dz$$

So we know that $$|f^{(n)}(0)|=|\frac{n!}{2\pi i}\int_{C_R}\frac{f(z)}{z^{n+1}}dz| \leq \frac{n!}{2\pi} \cdot Length(C_R) \cdot \max _{z \in C_R} \frac{1}{|z^{n+1}|} \cdot \max _{z \in C_R}|f(z)|=\\ =\frac{n!}{2\pi} \cdot 2 \pi R \cdot \frac {1}{R^{n+1}} \cdot \max _{z \in C_R}|f(z)|=n! \frac {|z|^2+12}{R^n} \leq n!\frac {R^2+12}{R^n}$$

So when $R \rightarrow \infty$, for $n \geq 3$ we get $|f^{(n)}(0)| \leq 0$ hence $f^{(n)}(0)=0$ for $n \geq 3$, hence QED.

My problem: Is this statement correct? $$\max _{z \in C_R}|f(z)|=|z|^2+12$$ As this is not a constant, it seems like $f$ does not have a maximum in a circle $C_R$, so I'm not sure I can use it.

Thank you!

• please edit your post to show the exponent $2$ in the statement. Also your last statement should be $\max _{z \in C_R}|f(z)|\le R^2+12.$
– zhw.
Commented Jun 21, 2015 at 16:29

1. You should be estimating $|a_n|=\lvert \frac{f^{(n)}(0)}{n!} \rvert$ rather then $|f^{(n)}(0)|$; the factorial that you have makes your limits infinite.
2. Having deduced that $f(z)=az^2+bz+c$ we can further conclude that $a=0$, since proper quadratics grow faster than linear functions.
• As for $1$, I see. So just evaluate normally, and before setting $R \rightarrow \infty$, divide both sides with $n!$? And as for my problem, is it OK to set $\max _{z \in C_R}|f(z)|=|z|^2+12$?
• @Reyo there is no equality there. You should use $\leq$ instead. Commented Jun 21, 2015 at 12:28