# Showing that a family is normal iff a singularity is non-essential

I'm just learning about normal families and am having a little trouble. I was told that we have an analytic function $$f:B_\mathbb{C}(0,1)\backslash\{0\}\to\mathbb{C}$$. Then to show that $$f_n(z):=f(z/n)$$ is normal iff $$0$$ is a non-essential singularity.

I've attempted to show the spherical derivative is unbounded provided the singularity was essential, but I've ran into trouble. I believe I can show $$f'(z/n)$$ is unbounded, but I'm not sure how to relate that to the spherical derivative in this case. I don't know of a lot of other techniques to show a family is normal. Any hints would be appreciated. I don't want a solution, just a hint or two to help me get through this. Thank you.

• What does B(0,1) mean in this context? Commented Apr 28, 2020 at 3:04
• @memerson the complex open ball of radius $1$ centered at $0.$ I thought that notation was standard. I guess maybe I should have written $B_\mathbb{C}(0,1)$? Sorry for the confusion. Commented Apr 28, 2020 at 3:11
• For the implication essential$\rightarrow$ non-normal, it may be more useful to use directly the definition of normality: remember that, if the singularity is essential, the function $f$ will send every punctured neighbourhood of the origin in a dense subset of the riemann sphere (casorati-weierstrass theorem). For the other implication (non-essential $\rightarrow$ normal) the approach you are following now is quite effective (if the singularity is a pole of order $n$, how does the spherical derivative behave?)
– user515010
Commented Apr 28, 2020 at 8:35

Essential$$\rightarrow$$ non-normal: If the family is normal we can extract from the sequence $$f_n:=f(\frac{z}{n})$$ a convergent subsequence $$f_{n_k}$$. The limit of this sequence (which we will write as $$\hat{f}$$) is either an analytic function or the identically infinite function: we shall show that both options lead to a contradiction.
• Suppose that $$\hat{f}$$ is analytic on $$\mathbb{D}-\{0\}$$. In particular, let $$A$$ be the annulus of inner radius $$\frac{1}{3}$$, outer radius $$\frac 12$$. Then $$\hat{f}(A)$$ is bounded, and this implies that $$f_{n_k}(A)$$ is bounded for $$k$$ large enough. By definition of $$f_{n_k}$$, we have that $$f(A/{n_k})$$ is bounded, and by the Cauchy formula for Laurent series $$|a_{-n}|\le r^n\underset{|z|=r}\max |f|$$ Choosing $$r$$ as the radius of a circle contained in $$A/n_k$$ and letting $$k\to \infty$$ we obtain $$a_{-n}=0$$, i.e. $$f$$ is analytic in $$\mathbb{D}$$, a contradiction.
• Suppose instead that $$\hat{f}=\infty$$. Defining $$g=\frac{1}{f}$$ bring us back to the previous situation, as $$g$$ is a function (analytic if we restrict to a small enough punctured neighbourhood of $$0$$, because since $$f_{n_k}\to \infty$$, $$f(\frac{z}{N})$$ must be nonzero in $$\mathbb{D}-\{0\}$$ and thus $$f$$ must be non-zero in $$\mathbb{D}/N-\{0\}$$) with an essential singularity in $$0$$ such that $$\lim g_{n_k}=\lim \frac{1}{f_{n_k}}=0$$ is an analytic function.
Non-essential$$\rightarrow$$ normal: if the singularity is removable, this is easy. If the singularity is a pole of order $$n\ge 1$$, then the derivative has a pole of order $$n+1$$ and we have
$$\frac{|f'_m(z)|}{1+|f_m(z)|^2}=\frac{|f'(z/m)|/m}{1+|f(z/m)|^2}=\frac{1}{m}\frac{O(1/|z/m|^{n+1})}{O(1/|z/m|^{2n})}=|z|^{n-1}O\left(m^{-n}\right)=O(m^{-n})$$