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.

  • $\begingroup$ What does B(0,1) mean in this context? $\endgroup$
    – memerson
    Commented Apr 28, 2020 at 3:04
  • $\begingroup$ @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. $\endgroup$ Commented Apr 28, 2020 at 3:11
  • $\begingroup$ 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?) $\endgroup$
    – user515010
    Commented Apr 28, 2020 at 8:35

1 Answer 1


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


The conclusion easily follows from this and Marty's theorem.

Note: these results do not depend much on the domain: the claim still follows if you substitute the punctured disc with an open set containing a circle encircling the origin.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .