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.

-

marked as duplicate by Fundamental, Jonas Meyer, hardmath, Robert Soupe, Eric StuckyJan 26 at 4:45

Try another conjecture! What if $f(z)=g(z)=e^z$? –  Ted Shifrin Feb 27 '14 at 3:07

I claim that $f(z) = C g(z)$ for some constant $C$ with $|C|\leq1$.

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, a constant with absolute value $\leq1$.

-

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 $|e^z|>0$).

-
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 '14 at 3:18
But $f(-1)=-1 ,g(-1)=0$ in your example. –  gilliatt Feb 27 '14 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 '14 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 '14 at 3:45
Thank you for your help and counter example –  gilliatt Feb 27 '14 at 3:52