Help with $\lambda \in \Bbb C$ so that $f(z)= \lambda \cdot g(z)$ $\forall z \in \Bbb C$

Question

Let $f$ and $g$ be two entire functions, s.t. $|f(z)|\leq |g(z)|$ $\forall z \in \Bbb C$. I wanna show that it exists $\lambda \in \Bbb C$, with $|\lambda |\leq 1$ so that $f(z)= \lambda \cdot g(z)$ $\forall z \in \Bbb C$.

Let's observe $h:= \frac{f}{g}$, with $g$ not equal to zero. By assumption $\vert h \vert \leq 1$ so all the singularities of h, which are isolated by the identity principle, are removable by the Riemann extension theorem.

Why can we conclude this?

https://math.stackexchange.com/questions/613524/suppose-f-and-g-are-entire-functions-and-fz-leq-gz-for-all-z-i?rq=1 — Martin R, Jan 28 '21 at 13:16
What you added is correct. Which part exactly is unclear to you? — Martin R, Jan 28 '21 at 13:43
Why are all the singularities of $h$ isolated by the identity principle? Maybe if I understand this, I will also understand the next sentence. — Frederick Manfred, Jan 28 '21 at 13:45
The zeros of $g$ are isolated (unless $g$ is identially zero). — Martin R, Jan 28 '21 at 13:48
Alright but how now the Riemann extensions theorem comes into the game? Is it because the limit of $h(z)$ when $z$ tends to the isolated zero exists because $h$ is bounded? — Frederick Manfred, Jan 28 '21 at 13:57
Yes. $h$ is bounded so that all its singularities are removable. That is Riemann's theorem. — Martin R, Jan 28 '21 at 14:19

score 1 · Accepted Answer · answered Jan 28 '21 at 14:42

As pointed out in the comments, either $g$ vanishes identically (and in this case the conclusion is trivial, since both $g$ and $f$ are $0$) or it has isolated zeroes.

In the latter case, the quotient $h$ is in modulus $\le1$ and defines an holomorphic function on the whole plane minus a discrete set (the zeros of $g$). By Riemann thm, as pointed out, $h$ extends across these isolated singularities to an entire function which is then bounded. Then by Liouville theorem $h$ is constant, that is, $\exists \lambda\in\Bbb C$ such that $0<|\lambda|\le1$ such that $h\equiv\lambda$. Since $h=f/g$ you get the conclusion.

Help with $\lambda \in \Bbb C$ so that $f(z)= \lambda \cdot g(z)$ $\forall z \in \Bbb C$

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