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Suppose $f(z)$ and $g(z)$ are entire functions and that $f(z)$ is not constant. If $|f(z)| < |g(z)|$ for all $z \in \mathbb C, $ prove that $f(z)$ can not be a polynomial.

I was thinking what I could do was using the fact $|f(z)| < |g(z)|$ , I can argue $ \frac {|f(z)|} {|g(z)|}< 1$ if $g$ not equal $0$. And, I use the Louiville's Theorem to conclude $ \frac {|f(z)|} {|g(z)|}$ is constant. Then I don't know where to go with that. I think I am not going in the right direction. Please help.

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There is something missing since taking f(z)=g(z)=z meets the condition but f(z) is a polynomial. –  Ittay Weiss Dec 30 '12 at 4:10
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Possibly relevant to whatever the intended question is: If $f$ and $g$ are entire functions such that $|f(z)|\leq|g(z)|$ for all $z\in\mathbb C$, then there is a constant $c$ such that $f=cg$. –  Jonas Meyer Dec 30 '12 at 4:16
    
oops, there is one thing missing, that equal sign should be removed, let me edit that. Sorry about that. –  Deepak Dec 30 '12 at 4:24

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up vote 4 down vote accepted

You are almost done.

Since $\frac{f(z)}{g(z)}=C$, you get that $f(z)=C g(z)$.

Now, if $f$ is a polynomial, since it is not constant it has some toot $z_0$. Then $0=C g(z_0)$ which implies that $g(z_0)=0$, contradiction.

P.S. You actually get something stronger: You can prove that $f(z)$ has no zeroes. The funny part is that the stronger version probably makes the problem easier, since it guides you towards the last step....

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$|f(z)/g(z)| < 1$ where $g(z) \ne 0$, but the singularities of $f/g$ at zeros of $g$ are removable (because $f/g$ is bounded in a deleted neighbourhood). After removing the singularities, you have a bounded entire function. –  Robert Israel Dec 30 '12 at 9:59
    
In fact it must be that $\,g(z)\neq 0\,\,,\,\forall\,z\in\Bbb C\,$, otherwise the condition $\,|f(z)|<|g(z)|\,$ isn't fulfilled... –  DonAntonio Dec 30 '12 at 12:00
    
@BrettFrankel Since $|f(z)| < |g(z)|$ it is impossible for $g$ to be zero. Note that with $\leq$, one can still prove that $\frac{f}{g}$ is entire, BUT then the rest of the problem is not true ;) –  N. S. Dec 30 '12 at 17:08

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