Showing that a family is not normal

Suppose $$f : mathbb{D}setminus {0} rightarrow mathbb{C}$$ is analytic and $$z=0$$ is an essential singularity of $$f$$. Show that the family $${f_n}$$ defined by
$$f_n(z) = f left( frac{z}{2^n} right), quad z in mathbb{D}setminus {0}$$
is not normal in $$mathbb{D}setminus {0}$$.

My attempt (inspired on this): Assume that $${f_n}$$ is a normal family in $$mathbb{D}setminus {0}$$. So there exists from the sequence $$(f_n)_n$$ a convergent subsequence $$(f_n)_{n_k}$$ with limit $$hat{f}$$. Because $$f$$ is analytic, it follows that $$hat{f}$$ is analytic on $$mathbb{D}setminus {0}$$. We take an annulus $$A$$ of inner radius $$1/3$$ and outer radius $$1/2$$. Then we have that $$hat{f}(A)$$ is bounded and because $$f_{n_k} rightarrow f$$ as $$k rightarrow infty$$, we have that $$f_{n_k} = f(frac{A}{2^{n_k}})$$ is bounded for $$k$$ large enough. By the Cauchy formula for Laurent series $$|a_{-j}| leq r^j max_{|z| = r}{|f|}.$$
We choose $$r$$ as the radius of the circle contained in $$A/n_k$$ and letting $$krightarrow infty$$. We then have to show that $$|a_{-j}|=0$$ for every $$j leq -1$$ such that we conclude that $$f$$ is an analytic function on $$mathbb{D}$$ which gives us a contradiction.

But in the last part I got stuck. Because if $$k rightarrow infty$$ then $$r rightarrow 0$$ and because $$f$$ has an essential singularity in $$z=0$$: $$max_{|z| = r}{|f|} rightarrow infty$$. So we have something of the form $$0 cdot infty$$ which is not defined.

Someone who can point out where I went wrong and help me out? Thanks!

You have shown that $$max_{|z| = r_k}{|f|}$$ is uniformly bounded for $$r_k = frac{1}{2 cdot 2^{n_k}}$$ therefore choosing these radii in $$|a_{-j}| leq r^j max_{|z| = r}{|f|}.$$ implies that $$a_{-j} = 0$$ for $$j ge 1$$.

Alternatively one can argue as follows: $$hat f$$ is bounded on the circle $$|z| = 1/2$$ and $$f_{n_k} to hat f$$ uniformly on that circle. It follows that $$|f_{n_k}(z)| le M text{ for } |z| = frac 12$$ with some $$M> 0$$ and sufficiently large $$k$$. Translated back to $$f$$ this means that $$|f(z)| le M text{ for } |z| = frac{1}{2 cdot 2^{n_k}} , .$$ So $$|f|$$ is bounded by $$M$$ on those concentric circles. Using the maximum modulus principle it follows that the same estimate holds in the annuli between those circles: $$|f(z)| le M text{ for } frac{1}{2 cdot 2^{n_{k+1}}} le |z| le frac{1}{2 cdot 2^{n_k}} , .$$ and that implies that $$f$$ is bounded in a neighborhood of $$z=0$$.

Using Riemann's theorem it follows that $$f$$ has a removable singularity at $$z=0$$.

Correct answer by Martin R on January 1, 2021

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