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Suppose $f$ and $g$ are entire functions with $|f(z)|\leq|g(z)|$ for all $z$.

How can we show that $f=cg$ for some complex constant $c$?

Thanks for any help :)

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You may want to check out math.stackexchange.com/questions/50421/… –  algebra_fan Jul 18 '11 at 8:51
    
Hint: Use Riemann's theorem of removable singularities. –  Hendrik Vogt Jul 18 '11 at 8:52
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@all: algebra_fan's proposed link is better than mine, so if you vote for closing this question, please use his. –  t.b. Jul 18 '11 at 8:58
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@Hendrik: yes, sure. now we have a good answer by Chandru, I see no reason for closure. –  t.b. Jul 18 '11 at 9:07
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1 Answer

up vote 6 down vote accepted

Assume $g(z) \neq 0$. Consider the quotient $\nu(z)=\displaystyle\small\frac{f(z)}{g(z)}$. Then the singularities of $\nu$ are isolated since the zeros of $g$ are isolated. Clearly $\nu(z)$ is bounded in each deleted neighborhood of each zero of $g$. By Riemann's theorem, $\nu$ extends, uniquely to an entire function and using continuity we have $|\nu(z)| \leq 1$ for all $z \in \mathbb{C}$. Now use Liouville's theorem.

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Actually its a neat (and very elementary) result from the fact that entire functions are analytic that $v$ can be extended to the roots of $g$. –  Listing Jul 18 '11 at 9:13
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@Listing: But that's exactly the usual proof of Riemann's theorem on removable singularities. –  t.b. Jul 18 '11 at 9:37
    
Oh ok I didn't know it under that name. I looked on wikipedia for Riemann's theorem and I think I came up with a wrong one. –  Listing Jul 18 '11 at 9:49
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@Listing: Well, that's hardly your fault :) Riemann's theorem is about as descriptive as Lebesgue's theorem or Banach's theorem. –  t.b. Jul 18 '11 at 9:52
    
In fact, I do not think it is referred to as "Riemann's theorem on removable singularities" in Walter Rudin's Real and Complex Analysis. Of course, this does not imply that the name "Riemann's theorem on removable singularities" is uncommon ... –  Amitesh Datta Jul 23 '11 at 13:55
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