11
$\begingroup$

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 :)

  • 1
    You may want to check out http://math.stackexchange.com/questions/50421/characterize-entire-functions-f-such-that-fz-leq-sinz2011-07-18
  • 0
    Hint: Use Riemann's theorem of removable singularities.2011-07-18
  • 3
    possible duplicate of [If $|f(z)|\lt a|q(z)|$ for some $a\gt 0$, then $f=bq$ for some $b\in \mathbb C$](http://math.stackexchange.com/questions/19536/if-fz-lt-aqz-for-some-a-gt-0-then-f-bq-for-some-b-in-mathbb-c)2011-07-18
  • 0
    @all: algebra_fan's proposed link is better than mine, so if you vote for closing this question, please use his.2011-07-18
  • 0
    To expand on @Hendrik's hint, observe that $f/g$ is bounded (if $g \neq 0$).2011-07-18
  • 0
    @Theo: It's not really a duplicate of either of those questions. This very question is the most general since $g$ is arbitrary here, so you need to use in addition that the zeros of $g$ are isolated; this part is easier for the other two questions.2011-07-18
  • 1
    @Hendrik: yes, sure. now we have a good answer by Chandru, I see no reason for closure.2011-07-18
  • 0
    thanks for the redirects/hints! In the future I'll try to search for more similar questions first.2011-07-18
  • 0
    @RHP, re your "Part 2": I think it's better if you ask a _new_ question.2011-07-18
  • 0
    ok, thanks for the advice2011-07-18

1 Answers 1

12

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.

  • 0
    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 $g$.2011-07-18
  • 1
    @Listing: But that's exactly the usual proof of Riemann's theorem on removable singularities.2011-07-18
  • 0
    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.2011-07-18
  • 3
    @Listing: Well, that's hardly your fault :) Riemann's theorem is about as descriptive as Lebesgue's theorem or Banach's theorem.2011-07-18
  • 0
    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 ...2011-07-23