if $\displaystyle\lim_{z\to\infty} f(z)=0$ then f is bounded, for a holomorphic function $f$

Question

I would like to ask you for some help, I have a doubt that I have not been able to solve yet...

let $f:\mathbb{C}\rightarrow\mathbb{C}$ a holomorphic function. if $\displaystyle\lim_{z\to\infty} f(z)=0$ then $f$ is bounded

I don't know yet if this is true, however this is what I have been trying to solve it.

Suppose that $f$ is not bounded, then \begin{align*} \forall r>0 : \exists z_r\in\mathbb{C} : f(z_r)\not\in B(0,r) \end{align*} iff
\begin{align*} \forall r>0 : r\leq|f(z_r)| \end{align*} then \begin{align*} \infty=\lim_{r\rightarrow\infty}r\leq\lim_{r\rightarrow\infty}|f(z_r)| \end{align*}

This would be complete provided that $|z_r|\rightarrow\infty$ However, I believe that what I have written is not necessarily true.

If someone could give me some guidance it would be great.

Answer 1

Since $\lim_{z\to \infty}|f(z)|=0$, there exists $R>0$ such that $|z|>R\implies |f(z)|<1$

$|f|$ is continuous on compact set $\overline{ B(0, R)}$ so attains a maximum $M\ge0$.

$|f|$ is bounded by $\max(M, 1)$.

Answer 2

This is essentially Koro's argument but with some details.

$\lim_{z\rightarrow \infty}f(z)=0$ means that for any $\epsilon>0$, there exists and $M$ such that for all $|z|>M$, $|f(z)|<\epsilon$. So, fix, say, $\epsilon=1$. Then, there exists an $M$ such that for all $|z|>M$, $|f(z)|<1$. But what about when $|z|\leq M$? Well, in which case, we are looking at the closed ball $B^c(0,M)$. Since closed balls are compact ( in $\mathbb{C}$), and that $f$ is continuous, then so is $|f|$. Hence, $|f|$ attains its maximum on $B^c(0,M)$. In other words, there exists a $L$ such that if $|z|\leq M$ then $|f(z)|\leq L$. Hence the maximum of $L$ and 1 gives you the desired bound