Finding holomorphic function such that $|f|$ tends to $0$ as $|z|$ tends to $\infty$

Question

I am trying to answer this by Liouville's theorem but I am not sure if my solution is rigorous. I am given that $f$ is holomorphic on $\mathbb{C}$ so if I could show that $|f|$ was bounded then I would be done (since $f$ would be constant and so $f(z)=0$). But it is the bounded part that I am unsure about. My solution is the following:

Assume that $|f|$ is not bounded. That is, there exists some $z_0\in\mathbb{C}$ such that $|f(z)|\rightarrow \infty$ as $z\rightarrow z_0$. But if this were the case, then $f$ would not be continuous since $|f(z)|\rightarrow0$ as $|z|\rightarrow\infty$.

Is this ok? I feel as though the implication that $f$ would not be continuous should be better explained but I don't really know how to do that.

Answer 1

By the very definition of "$|f|\to 0$ as $|z|\to \infty$", for every $\epsilon>0$, there exists $M$ such $\bigl||f(z)|-0\bigr|<\epsilon$ for all $z$ with $|z|>M$. In particular, this works for $\epsilon=42$. Hence for such $M$, we know that $|f(z)|<42$ for all $|z|>M$; and on the compact disk of radius $M$, the continuous function $|f|$ is bounded as well.

Answer 2

You are given $|f(z)| \to 0$ as $|z| \to \infty.$ Thus, there exists $R > 0$ such that $|f(z)| < 43$ for every $|z| > R$.
On the other hand, the set $K = \{z : |z| \le R\}$ is compact and thus $M = \displaystyle\sup_K|f|$ exists.

Now, $|f|$ is bounded by $\max\{M, 43\}.$

As $f$ is entire, it must be constant which gives $f \equiv 0.$

Answer 3

You are interpreting the "not bounded" part wrong. $f$ being not bounded means:

You can't find $M$ s.t. for all $z$, $|f(z)|<M$.

which is equivalent to

For any $M$, there exists $z$, s.t. $|f(z)|\geq M$

Try to derive a contradiction from this.

Answer 4

The other answers are the standard ones, but there is a way to argue exactly as you have, using the Riemann sphere, a one point compactification of $\mathbb{C}$, denoted $\hat{\mathbb{C}}=\mathbb{C}\cup \infty$.

Since $f\to 0$ as $z\to \infty$, $f$ extends to a holomorphic, and, in particular a continuous map on $\hat{\mathbb{C}}$, (the derivative at infinity is defined as the limit $zf(1/z)$ as $z\to 0$, in this case it's zero). Namely, $$ \hat{f}(z)=\begin{cases} f(z),&z\in \mathbb{C}\\ 0,&z=\infty\end{cases} $$ Note that from here we can conclude immediately from the extreme value theorem.

But, if you're wedded to your strategy, then suppose $f$ is unbounded. Find a sequence of $z_n$ such that $|f(z_n)|>n$. These accumulate somewhere by compactness. That somewhere can't be $\infty$ by assumption, so $z_n\to z\in \mathbb{C}$. Now you can conclude from what you have wrote.