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Let $f$ and $g$ be entire functions with $\lvert f\rvert\leq\lvert g\rvert$. Show that then $$ f=cg $$ for a constant $c\in\mathbb{C}$.

Can I just do it like this:

$\lvert f(z)\rvert\leq\sup\limits_{z\in\mathbb{C}}\lvert g(z)\rvert$, so f is limited and therefore constant (Liouville), i.e. $f(z)=c~\forall~z\in\mathbb{C}$.

  1. $c=0$: Then $0=f(z)=0\cdot g(z)$

  2. $c\neq 0$: Then $g(z)\neq 0$ and therefore $\frac{f(z)}{g(z)}$ is defined. This is an entire function because $f$ and $g$ are entire function and it is $$ \left\lvert\frac{f(z)}{g(z)}\right\rvert=\frac{\lvert f(z)\rvert}{\lvert g(z)\rvert}\leq 1, $$ so $\frac{f(z)}{g(z)}$ is a bounded entire function and therefore constant, lets say for a $d\in\mathbb{C}$ $$ \frac{f(z)}{g(z)}=d~\forall~z\in\mathbb{C}\Leftrightarrow f(z)=d\cdot g(z). $$

This is my proof. Is it okay or does one need a special sentence for this proof?

  • In your first step, why does $|f(z)| \leq \sup_{z\in \mathbb{C}} |g(z)|$ allow you to conclude that $f$ is bounded? I don't see why $\sup_{z\in \mathbb{C}} |g(z)|$ has to be finite; if that were the case, then $g$ would also be bounded and the problem would be trivial. – Alex Wertheim Jul 15 '13 at 17:17
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    "Can I just do it like this:" Not quite. $\sup\limits_{z \in \mathbb{C}} \lvert g(z)\rvert$ is usually $\infty$. Distinguish two cases, a) $g \equiv 0$, then evidently $f = 3\cdot g$, and b) $g\not\equiv 0$. Then what can you say about the quotient $f/g$? – Daniel Fischer Jul 15 '13 at 17:17
  • Hm. Can you say me then how to show that? –  Jul 15 '13 at 17:19
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    I saw the question several times: for example

    http://math.stackexchange.com/questions/52121/property-of-entire-functions

     – ziang chen Jul 15 '13 at 17:19
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    See this answer too. – Ayman Hourieh Jul 15 '13 at 17:28
  • Why do you think $\sup |g(z)|<\infty$? – Thomas Andrews Jul 15 '13 at 17:30

1 Answers1

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As you've seen in the comments, your method of proof isn't valid. What you can do, though, is to consider two separate cases: $g\equiv 0$ and $g\not\equiv0$. The first case is obvious, so I'll leave that to you. In the second case, show that $f$ has zeros at the same places as $g$ with at least the same order. If you can show that, then $f/g$ is an entire function and $|f/g|\leq1$, so by Liouville's theorem is constant. By the open mapping theorem, then $f/g$ is constant, so $f=cg$ as desired.

Clayton
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  • I guess one can show that with Riemanns theorem on removable singularities: $h=f/g$ is 1. an entire function and 2. limited in the neighbourhood of every singularity (i.e. the zeros of g), so all singularities are removable. So there is an extension $\tilde{h}$ of $h$. Furtermore $\tilde{h}$ is limited. So Liouville says, that $\tilde{h}$ is constant, i.e. $\tilde{h}(z)=c\Leftrightarrow f(z)=c\cdot g(z)$ for all $z\in\mathbb{C}$. –  Jul 15 '13 at 20:32