Suppose $f$ and $g$ are entire functions, and $|f(z)|≤|g(z)||f(z)|≤|g(z)|$ for all $∈ℂ$; what conclusion can you draw? 
Suppose $f$ and $g$ are entire functions, and $|f(z)|≤|g(z)||f(z)|≤|g(z)|$ for all $∈ℂ$; what conclusion can you draw?

This is the second exercise from the tenth chapter of Walter Rudin's real and complex analysis. I understand that if $f$ and $g$ are entire functions, then that means that they are holomorphic on the whole complex plane. Furthermore, I understand that if they are holomorphic then there exists $f'(z_0)$ s.t. we get the limit definition for $f'(z_0)$. what I am having trouble with is figuring out exactly what $f(z)$ and $g(z)$ are. I am not looking for an answer, just really looking for a hint to get started on the problem. thank you!
 A: First, if $g$ is $0$ then $f$ has to be $0$ as well.
If $g$ is not identically $0$ then it is not $0$ on a dense set.
But then, from $|f(z)||g(z)| \leq |g(z)|$ one gets that $|f(z)| \leq 1$ on a dense set so everywhere. Hence $f$ is constant since entire bounded function is constant.
Here, one can add then if $f$ (which is constant) is not identically $0$ then since
$|f(z)| \leq |g(z)||f(z)|$ we see that $|g(z)|\geq 1$, hence has no roots. In particular it can be lifted, i.e. $g(z) = e^{h(z)}$ for some entire function $h(z)$.
Taking all these considerations together we see that $f(z) = c_1$ and $g(z) = c_2e^{h(z)}$ for some constants $c_1,c_2$.
A: Hopefully my hint was sufficient for OP, now I will propose my answer:
Let's start by noting that if $f$ is the zero function, then there is no condition on $g$.
Assume now that $f$ is nonzero.
Let $z \in \mathbb{C}$. Then:

*

*If $g(z) = 0$, we have $f(z) = 0$;

*If $g(z) \neq 0$, we obtain, by dividing the second part of the given inequality by $|g(z)|$: $|f(z)| \leq 1$.

Thus, $f$ is entire and bounded, and so by Liouville $f$ is constant equal to, say $\alpha \in \mathbb{C}$, and $\alpha$ is nonzero by assumption on $f$. Thus, by the left side inequality, $|g(z)| \geq |\alpha| > 0$ for all $z$.
Therefore, $\frac{1}{g}$ is well-defined and entire due to $g$ being entire and having no roots, with, for all $z$: $\left|\frac{1}{g(z)}\right| \leq \left|\frac{1}{\alpha}\right|$.
Hence, by Liouville again, $\frac{1}{g}$ and thus $g$, is constant, and $g$ is equal to a nonzero constant since $|g|\geq |\alpha| > 0$ nonzero.
We can hereby conclude that $f$ is always a constant, and that $g$ is a nonzero constant if $f$ is nonzero, with no conditions on $g$ if $f$ is the zero function.
