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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.

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

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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.

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