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Given $f$ and $g$ be entire functions and: $$|(f(z)| \leq |g(z)|$$ can you show me how to prove that $$f(z)=cg(z)$$ , where c is a (complex) constant? I have been able to prove this for the case $g(z) \neq 0$ by dividing $f$ into $g$: $$ \frac{|f|}{|g|} \leq 1 \rightarrow f(z)=cg(z) \text{by Liouville's theorem} $$ However, I don't know how to prove this result for the more general case in which $g(z)$ is arbitrary.

A Slow Learner
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  • Another one (for entire functions) here: https://math.stackexchange.com/questions/1313298/there-exists-a-constact-lambda-s-t-f-lambda-g. – Martin R Mar 07 '18 at 06:40

2 Answers2

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The zeros of $g$ must be discrete, if not, $g=0$ and hence the result follows. We conclude that $f/g$ has only removable singularities, an entire function $h$ coincides with $f/g$ except for removable singularities, and $h$ is bounded, so $h=c$, and the result follows.

user284331
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  • Could you show me how to prove it? – A Slow Learner Mar 07 '18 at 06:04
  • Actually you can look at the "duplicate" link. – user284331 Mar 07 '18 at 06:05
  • I did. The link talked about meromorphic functions, and I just started learning complex analysis and have only learned about entire functions so far. Also, the link assumes without proving for the case $f$ and $g$ being entire. – A Slow Learner Mar 07 '18 at 06:11
  • Find a ball such that $g$ has no zero inside, then $f=cg$ on this ball. And we know that if entire function that coincide with a connected open set must coincide throughout. – user284331 Mar 07 '18 at 06:13
  • @TheOblivious: Entire functions are just meromorphic functions in $\Bbb C$ without poles, so the same proof can be used here, with $P_f$ being an empty set. – Martin R Mar 07 '18 at 06:28
  • @user284331: The essential point (that $f/g$ has only removable singularities and therefore can be extended to an entire function) is missing in your answer. There is no Liouville theorem for functions holomorphic in a ball. – Martin R Mar 07 '18 at 06:30
  • Ah... yes, you are right, thanks for reminding. – user284331 Mar 07 '18 at 06:31
  • I looked at the second link, too. Could you explain to me what it means by saying that the singularities are "removable"? Is there a theorem related to that statement? – A Slow Learner Mar 07 '18 at 06:54
  • @TheOblivious: The first Google hit for "removable singularity" is https://en.wikipedia.org/wiki/Removable_singularity, with Riemann's theorem ... – Martin R Mar 07 '18 at 07:12
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If you know some basic facts about order of zeros of an analytic function (and the fact that if f is analytic at all points except a and if f is bounded near a then it is actually analytic in a neighborhood of a) you can argue as follows: we first show that if g has a zero of order n at a point a then f has a zero of order m at a with $m \geq n$. Indeed, $g(a)=0$ implies $f(a)=0$; let m be the order of zero of f at a. Then we can write $f(z)= (z-a)^{m}f_1 (z)$ and $g(z)= (z-a)^{m}g_1 (z)$ in a disc around a where $f_1,g_1$ are analytic in the disc and do not vanish at a. If $m<n$ divide the inequality $|(z-a)^{m}f_1 (z)| \leq |(z-a)^{n}g_1 (z)|$ by $(z-a)^{m}$ and let $z \to a$ to get a contradiction. Thus $m \geq n$. From this it follows $\frac f g$ can be defined in an obvious way at a whenver $g(a)=0$. This makes $\frac f g$ a bounded entire function, hence a constant by Liouville's Theorem.

Kavi Rama Murthy
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