How can i prove that this function is bounded in order to apply Liouville's theorem?

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?

• Don't use the hint, just use Cauchy's estimates, the tool to prove Liouville's theorem. – orole Jan 18 '18 at 1:32
• but how would i proced with that inequality using that integral? – Eduardo Silva Jan 18 '18 at 1:35
• No integral, the inequality at the end of that section. the inequality has $|f(z)|$ in it. And you are given a further inequality $|f(z)|\leq\sqrt{1+|z|}$. use transitivity. And then use that the maximum is when $|z|=r$. So the $|z|$ inside $\sqrt{1+|z|}$ is just $r$. – orole Jan 18 '18 at 1:36
• but then i would not need the fact that $f \leq \sqrt{1+|z|}$ – Eduardo Silva Jan 18 '18 at 1:39
• What are you talking about? In what I told you it is clearly used. – orole Jan 18 '18 at 1:44

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.