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I want to determine all entire functions $f$ such that $|f(z)| \leq |\sin(z)|$. I searched around on MathSEx and I found the following question from which I tried to get inspired but I think it differs substantially from my question: Characterizing nonconstant entire functions with modulus 1 on the unit circle

Here's what I tried. It is not complete I am missing some cases. First if $|f(z)| = |\sin(z)|$ then let $h(z)= \frac{f(z)}{\sin(z)}$. Then $h$ is analytic on $\mathbb{C}\setminus \left\{n \pi,n\in\mathbb Z\right\}$, and we also have that $|h|=1$ on the unit disc $D$. So $\exists c$ such that $h(z)=c$ with $|c|=1$. $h$ is continuous at all $n \pi$, so $h(n\pi)=c.\sin(n\pi)$; so we have that $\exists c$ such that $|c|=1$ and $f(z)=c.\sin(z)$.

Now I need to take care of the case $|f(z)| < |\sin(z)|$ and I don't know what to do. Should I try to apply the Schwarz lemma to this case since we would get $|h|<1$?

Any help or solution is welcomed.

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marked as duplicate by Behaviour, Daniel Fischer Oct 29 at 19:09

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Well, $f(z)/sin z$ is analytic everywhere by the Riemann removable singularities theorem, and it is also bounded, so... –  Akhil Mathew Jul 8 '11 at 22:56
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+1 Nice question: you researched the site, provided motivation/context, developed your thoughts on the question, showing work, clearly stated your question...It's a "model" question (which I wish more posted questions would follow)! –  amWhy Jul 8 '11 at 22:59
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Here's the generalization to an entire function dominated by an arbitrary entire function: en.wikipedia.org/wiki/… –  joriki Jul 8 '11 at 23:01

2 Answers 2

up vote 8 down vote accepted

If $|f(z)|\leq|\sin z|$ for all $z$, then the quotient $h(z)=\frac{f(z)}{\sin z}$ is analytic at the complement of the zero set of $\sin z$, as you noted, but it is also analytic at the zeroes (or, rather, has removable singularities there)

Indeed, your inequality implies that whenever $\sin z_0$ is zero, then $f(z_0)$ is zero also: this and a little work will show that $h$ is bounded in a neighborhood of $z_0$.

We thus conclude that $h$ is entire and bounded by $1$. You can probably take it from here.

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So $h$ is constant by Liouville. Thx. –  user786 Jul 8 '11 at 22:59

The function $ h(z) $ must obey $ |h(z)| \le 1 $ on all of $ \mathbb{C} $, which means that $ h(z) $ is a bounded function. By Liouville's theorem, any bounded entire function must be constant, hence $ f(z) = \alpha \sin(z) $ for all $ |\alpha| \le 1 $ characterizes the set of functions you want.

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Notice that the OP's problem was in dealing with the points where $h$ is not a priori defined. You more or less glossed over this point in your first sentence :) –  Mariano Suárez-Alvarez Jul 8 '11 at 23:03
    
Shhhhhhhhhhh... –  anon Jul 9 '11 at 10:24

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