# Find all entire functions $f$ such that $|f(z)| \leq |\sin(z)|$

I am trying to solve the following exercise:

Find all entire functions $$f$$ such that $$|f(z)| \leq |\sin(z)|$$, $$\forall z \in \mathbb{C}$$

I think Liouville's Theorem is the way to go.

Liouville's Theorem states that:

Every bounded entire function must be constant.

Since $$\cos(z)=0$$ for $$z=\frac{2k+1}{2} \pi$$,

my answer would be that the only entire function is the zero function $$g\equiv 0$$.

Am I correct?

Edit: I got a little bit confused, because in $$\mathbb{R}$$, sin is bounded with $$|\sin(x)|<1$$. Because of this I thought that I only need to search constant functions f, such that $$|f(z)| \leq |\sin(z)|$$.

This is why I thought that the Zero-Function is the only option.

Considering the comments, $$f(z):= a \sin(z)$$ with $$|a| \leq 1$$ also fullfill the condition wanted.

How can I proof that these are all function?

• I assume that any function $f(z)=a\sin(z)$ where $|a|\le1$ is also ok.
– PC1
Commented Jun 28, 2022 at 23:42
• In general, if $f,g$ are entire functions such that for all $z\in\Bbb{C}$, $|f(z)|\leq |g(z)|$, then $f=ag$ for some constant $|a|\leq 1$ (if $g=0$, this is obvious, otherwise, $f/g$ is meromorphic and bounded, so has an entire extension by Riemann's theorem, and hence constant by Liouville). See Liouville's theorem on Wikipedia. Commented Jun 28, 2022 at 23:48
• @PC1 Why should f(z)= a sin(z), with |a|<1 be an answer. Wouldn't Liouvielle's Theorem suggest that the functions I am looking for are constant? Commented Jun 29, 2022 at 0:33
• @PC1 Sorry, I did missunderstand something. I edited my question. Commented Jun 29, 2022 at 0:49
• @Andres2003 it's Liouville, not Liouvielle.
– PC1
Commented Jun 29, 2022 at 3:37

Hint: From $$\sin z=0$$ we get $$f(z)=0$$. Therefore $$\frac{f(z)}{\sin z}$$ is holomorphic.
• $\sin z /z$ is not bounded. Commented Jun 28, 2022 at 23:48