The conditions under which a function family becomes a normal family - complex analysis

Let $$f$$ be holomorphic function on $$\{z\in\mathbb{C}: 0<|z|<1\}$$ and define $$\{f_n\}_{n\in\mathbb{N}}$$ as $$f_n(z)=f\left(\dfrac{z}{n}\right)$$. Under this condition, what are the necessary and sufficient conditions for $$\{f_n\}$$ to be a normal family?

I think the answer is that "$$f$$ is bounded around $$z=0$$." I have proved that this is a sufficient condition (use Montel's theorem), but I could not show that this is necessary (i.e. when $$\{f_n\}$$ is a normal family, then $$f$$ is bounded around $$z=0$$). Moreover, I know that the condition is equivalent to "$$f$$ is holomorphically extendable over $$z=0$$." (Riemann's theorem on removable singularities)

Could you teach me how to show that?

Thank you.

• Which Montel' theorem do you use? I think it only requires locally uniformly bounded inside the region.
– PPP
Commented Jun 17, 2023 at 9:09

If $$\{ f_n \}$$ is a normal family then $$|f_n|$$ is uniformly bounded by some constant $$M$$ on the compact set $$\{ 1/3 \le |z| \le 2/3 \}$$. Use this to conclude that $$|f|$$ is bounded by $$M$$ on each annulus $$\{ 1/(3n) \le |z| \le 2/(3n) \}$$ and consequently (since these annuli overlap), on $$0 < |z| \le 2/3 \}$$. So $$|f|$$ is bounded by $$M$$ in a neighborhood of the origin, and therefore has a removable singularity at the origin.
Conversely, if $$|f| \le M$$ in some neighborhood $$\{ 0 < |z| < \epsilon \}$$ of the origin then $$|f_n(z)| \le M$$ for $$0 < |z| < 1$$ if $$n > 1/\epsilon$$, so that $$\{ f_n \}$$ is uniformly bounded and therefore a normal family.