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Suppose $f : \mathbb{D}\backslash \{0\} \rightarrow \mathbb{C} $ is holomorphic. Prove that the family $\{f_n\}$ defined by $f_n(z) = f(\frac{z}{2^n}), z \in \mathbb{D}\backslash \{0\}$ is normal in $\mathbb{D}\backslash \{0\}$ if and only if f has a removable singularity at $z = 0$.

I already had something for the first direction: If the family is normal, then we have that for an holomorphic function in the punctured disc, that when $f$ is bounded on the punctured disc, $0$ will be a removable singularity. Now because we now that it is uniformly bounded when the family is normal, this condition is satified and we have what we want.

However the other way is harder to start. I wanted to do something with the fact that exept for $0$ it is analytic, but I can not see how I can prove that it is uniformly bounded and equicontinuous. (The 2 conditions I want to check to prove that $\{f_n\}$ is a normal family.

Can anybody help me with this?

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If $(f_n)$ is normal then it is uniformly bounded on the circle of radius $1/2$, say $$ |f_n(z)| \le M \text{ for } |z| = \frac 12 \, . $$ It follows that $|f(z)| \le M$ on all radii $|z|=1/2^{n+1}$. Using the maximum modulus principle one can conclude that in fact $|f(z)| \le M$ for $0 < |z| \le 1/2$. It follows that the singularity at the origin is removable.

Conversely, if $f$ has a removable singularity at $z=0$ then it is bounded on each disk $\{ |z| \le r \}$. It follows that $(f_n)$ is locally uniformly bounded, and according to Montel's theorem, that is equivalent to $(f_n)$ being normal.

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