It is well known that "genuinely different" entire functions cannot dominate each other. More precisely, let $f$ and $g$ be entire functions in $\mathbb{C}$ satisfying $|f(z)|\le |g(z)|$ for all $z \in \mathbb{C}.$ Then $f=Cg$ for some $C\in \mathbb{C}.$
I'd like to know the result for meromorphic functions similar to the above result for entire functions.
Please let me know if you have any comment about this question. Thanks in advance!
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The exact same result holds for meromorphic functions: if $f$ and $g$ are meromorphic on $\mathbb{C}$ and $|f(z)|\leq |g(z)|$ for all $z$, then $f(z)=Cg(z)$ for some constant $C$. The proof is also basically the same: consider the function $h(z)=g(z)/f(z)$. This function $h$ is meromorphic, and $|h(z)|\leq 1$ for any $z$ except possibly the poles or zeroes of $g$. But the poles and zeroes of $g$ are isolated, and so by continuity $|h(z)|\leq 1$ everywhere and in particular $h$ has no poles. Since $h$ is an entire bounded function, it is constant.
If $f$ and $g$ are meromorphic in $\mathbb C$ with $|f(z)| \le |g(z)|$, then $f(z)/g(z)$ is meromorphic and bounded. Thus its singularities are all removable; after removing them, you have a bounded entire function which Liouville says is constant.