I have seen some proofs of the fact that if $f,g$ are entire functions such that $|f(z)| \le |g(z)|, \forall z \in \mathbb C$ then $f(z)=cg(z) , \forall z \in \mathbb C$ for some $c \in \mathbb C$ . In all the proofs, it is proceeded like : Assume w.l.o.g. , $g$ is not identically zero . $|f(z)/g(z)| \le 1 $ for all $z$ for which $g(z)\ne 0$ ; now the set of zeroes of $g$ is isolated , and as $f/g$ is bounded in a neighbourhood of each such zero , each zero of $g$ is a removable singularity of $f/g$ , then it can be extended to a function holomorphic on $\mathbb C$ which is bounded .
Now I have one question with this proof . I agree that in some neighbourhood of each zero of $g$ , $f/g$ is bounded . So individually one by one we can holomorphically extend $f/g$ to incorporate each zero of $g$ . But $g$ can have infinitely many zeroes . How can we then extend $f/g$ holomorphically to whole complex plane ? The step by step procedure won't work . Is there any other way or any thing obvious I am missing ? Please help . Thanks in advance
I am not really sure about the meaning of your expression “step by step procedure”. Let $Z=\{z\in\mathbb{C}\,|\,g(z)=0\}$. Define$$\begin{array}{rccc}h\colon&\mathbb C&\longrightarrow&\mathbb C\\&z&\mapsto&\begin{cases}\frac{f(z)}{g(z)}&\text{ if }z\notin Z\\\lim_{w\to z}\frac{f(w)}{g(w)}&\text{ if }z\in Z.\end{cases}\end{array}$$This definition makes sense because the singularities of $\frac fg$ are removable. And it is easy to prove that $h$ is analytic. Since it is a bounded entire function, it is constant.