If $|f(z)|\le |g(z^2)|$ for all $ z\in\mathbb{C} $, then $ f(z) =f(-z), \forall z\in\mathbb{C} $

Question

Consider two entire functions $f $ and $ g$. If $|f(z)|\le |g(z^2)|$ for all $ z\in\mathbb{C} $, show that $ f$ is even.

I'm having a hard time coming up with a way to solve this. This is a follow up question to this one: if $|f(z)|\le |g(z)|$ for all $ z\in\mathbb{C} $, then $ f=cg$ for some constant $c\in\mathbb{C} $. I managed to solve this one, but I can't find a possible link between the two problems.

Could someone give a hint, please?

Thanks.

Answer 1

You know that, for some $c\in\mathbb C$, you have$$(\forall z\in\mathbb C):f(z)=cg(z^2).$$Therefore, if $z\in\mathbb C$,$$f(-z)=cg\bigl((-z)^2\bigr)=cg(z^2)=f(z).$$