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?

  • 3
    $\begingroup$ I assume that any function $f(z)=a\sin(z)$ where $|a|\le1$ is also ok. $\endgroup$
    – PC1
    Commented Jun 28, 2022 at 23:42
  • 3
    $\begingroup$ 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. $\endgroup$
    – peek-a-boo
    Commented Jun 28, 2022 at 23:48
  • $\begingroup$ @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? $\endgroup$
    – Andres2003
    Commented Jun 29, 2022 at 0:33
  • $\begingroup$ @PC1 Sorry, I did missunderstand something. I edited my question. $\endgroup$
    – Andres2003
    Commented Jun 29, 2022 at 0:49
  • $\begingroup$ @Andres2003 it's Liouville, not Liouvielle. $\endgroup$
    – PC1
    Commented Jun 29, 2022 at 3:37

1 Answer 1


Hint: From $\sin z=0$ we get $f(z)=0$. Therefore $\frac{f(z)}{\sin z}$ is holomorphic.

  • $\begingroup$ $\sin z /z$ is not bounded. $\endgroup$ Commented Jun 28, 2022 at 23:48
  • $\begingroup$ @geetha290krm: Thanks for the heads up, improved the hint $\endgroup$
    – orangeskid
    Commented Jun 28, 2022 at 23:50

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