Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f$ and $g$ are entire functions,and $|f(z)|\leq|g(z)|$ for every z.What conclusion can you draw?

My conjecture : $f$ and $g$ are constant but I don't know how to deal with it.

I will appreciate your help

share|improve this question
    
Try another conjecture! What if $f(z)=g(z)=e^z$? –  Ted Shifrin Feb 27 at 3:07

2 Answers 2

up vote 7 down vote accepted

I claim that $f(z) = C g(z)$ for some constant $C$. Note that if we let $h(z) = f(z)/g(z)$ then $h$ is analytic except at the isolated zeros of $g$. But since $|h(z)| \leq 1$ for all $z$, every isolated singularity of $h$ is removable. Hence, $h$ is entire and bounded, so $h$ is a constant.

share|improve this answer

I am not sure if this is what you are looking for, but you can apply Rouche's theorem in your case. Since the inequality holds over $\mathbb{C}$, we can conclude that $g$ and $f+g$ have the same number of zeros over $\mathbb{C}$.

Your conjecture doesn't hold simply by taking $f=g$ as a previous user mentioned. Even if you change the $\leq$ to $<$ to avoid this trivial example, you can still take $f(z)=0$ and $g(z)=e^z$ as a counter example (Since $|e^z|>0$).

share|improve this answer
    
Also, by $g$ and $f+g$ having the same number of zeros, I mean either both have infinitely many or both have finitely many with the same cardinality. –  Sergio Da Silva Feb 27 at 3:18
1  
But $f(-1)=-1 ,g(-1)=0$ in your example. –  gilliatt Feb 27 at 3:23
    
You are right, my apologies. Still, as Ted Shifrin mentioned, $f=g$ satisfies your conditions where $f$ is some non-constant entire function. –  Sergio Da Silva Feb 27 at 3:41
    
Less trivially, how about $f(z)=0$ and $g(z)=e^z$, and changing the $\leq$ to $<$. –  Sergio Da Silva Feb 27 at 3:45
    
Thank you for your help and counter example –  gilliatt Feb 27 at 3:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.