Let $f$ be a entire function such that $|f(z)| \leq\sqrt{1+|z|}$ for all $z \in \mathbb{C}$. Prove that $f$ is constant.
My try:
Well, the hint was saying to use Liouville's theorem so i need to prove that $f$ is bounded in order to apply that. I was thinking in use derivatives to do that but a got to nowhere.
Any more hint?
Actually, the hint should be "use the idea from the proof of Liouville's theorem", i.e. use Cauchy integral formula.
Since $f$ is entire, then it follows \begin{align} f'(z) = \frac{1}{2\pi i} \int_{|w|=R} \frac{f(w)}{(z- w)^2}\ dw. \end{align} In particular, by $ML$-estimate, we see that \begin{align} |f'(z)| \leq \frac{1}{2\pi} \int_{|w|=R} \frac{|f(w)|}{(R-|z|)^2}|dw| \leq \frac{1}{2\pi} \int_{|w|=R} \frac{\sqrt{1+R}}{(R-|z|)^2}\ |dw| \leq \frac{R\sqrt{1+R}}{(R-|z|)^2} \rightarrow 0 \end{align} as $R\rightarrow \infty$. This shows that $f'(z) \equiv 0$ which means $f(z)$ is constant.