I want to solve the following exercise from Ahlfors' complex analysis text. Below is my attempt followed by questions I have regarding it:

If a function element is defined by a power series inside its circle of convergence, supposed to be of finite radius, prove that at least one radius is a singular path for the global analytic function which it determines. ("A power series has at least one singular point on its circle of convergence.")

Here is my attempt:

Let $f$ be defined and analytic in the disk $\Delta(z_0;R)$ where $R<\infty$ is the radius of convergence around the point $z_0$. Let $\mathbf{f}$ be the global analytic function determined by the function element $(f,\Delta(z_0;R))$, and let $\mathfrak{S}_0(\mathbf{f})$ be its Riemann surface (which is a connected component of the sheaf of germs of analytic functions).

For $\theta \in[0,2 \pi)$ denote by $\gamma_\theta:[0,R] \to \mathbb C$ the radius of the circle of convergence parallel to $e^{i \theta}$ (i.e. $\gamma_\theta(t)=z_0+t e^{i \theta}$), and by $\overline{\gamma_\theta}:[0,R] \to \mathfrak{S_0}(\mathbf{f})$ the unique analytic continuation of $\mathbf{f}$ along $\gamma$ (if it exists).

By way of contradiction suppose that none of the radii are a singular path. We can extend $f$ to a (univalent) function $F$ which is analytic in a larger concentric disk, and coincides with $f$ on the original disk $\Delta(z_0;R)$. Clearly, this should be done using analytic continuation along all radii.

I would like to define $F$ the following way:

  • For $z \in \Delta(z_0;R)$ set trivially $F(z)=f(z)$

  • For any germ $\overline{\gamma_\theta}(R)$, consider its power series expansion at $z_0+R e^{i \theta}$. For all points $z$ in the series' disk of convergence, which are not in $\Delta(z_0;R)$, set $F(z)$ to be the value of the series at $z$.

The second part seems to be ambiguous, since several different germs may define $F$ at the same point $z \notin \Delta(z_0;R)$. We will now prove that there is no ambiguity.

Let $z \notin \Delta(z_0;R)$ be a "possible ambiguous point" for $F$. Thus there exist germs $\overline{\gamma_{\theta_1}}(R),\overline{\gamma_{\theta_2}}(R)$ with $\theta_1 \neq \theta_2$ and with power series developments $F_1,F_2$ which converge in the disks $\Delta_1,\Delta_2$ centered at $z_0+Re^{i \theta_1},z_0+Re^{i \theta_2}$ respectively, where both disks contain $z$.

Ahlfors has previously proven the equivalence between analytic continuation along arcs and along chains of function elements. Using his construction we can form a chain $\{(f_k,\Omega_k \}_{k=1}^N$ of function elements which follow the radius $\gamma_{\theta_1}$. Combining this with the fact that the analytic continuation along an arc is unique, we find that $$f_{N-1} \big|_{\Delta(z_0;R) \cap \Omega_{N-1}} \equiv f,$$ thus the identity theorem for analytic functions gives $$F_1 \equiv f $$ on $\Delta_1 \cap \Delta(z_0;R)$. Similarly, we find that $$F_2 \equiv f $$ on $\Delta_2 \cap \Delta(z_0;R).$

Now I would like to say that the intersection $\Delta_1 \cap \Delta_2 \cap \Delta(z_0;R)$ is nonempty (by some geometric argument), and from there applying the identity theorem one last time will yield $F_1(z)=F_2(z)$.

Hence, $F$ is indeed a univalent analytic function which extends $f$ to an open neighborhood of $\overline{\Delta(z_0;R)}$. By elementary topology, this open neighborhood contains a larger disk $\Delta(z_0,R+\epsilon)$ concentric with $\Delta(z_0,R)$, contradicting the maximality of the radius of convergence.


  1. Overall, is my solution correct?

  2. More specifically, how would one go about proving the "geometric argument" that $\Delta(z_0,R) \cap \Delta_1 \cap \Delta_2 \neq \emptyset$?



1 Answer 1


The proof is correct, but I don't think you need to go back to chains. Analytic continuation along each radius gives you an analytic germ at each boundary point. This germ is defined by some function element $(f,\Omega)$ where $\Omega$ is a small disk centered at that boundary point.

Geometric argument: if two disks $\Delta_1,\Delta_2$ have nonempty intersection, then the line segment connecting their centers lies in the intersection. In the above setup, this line segment is a chord of $\Delta(z_0;R)$; thus, all three disks have a point in common.

Termonological remark: in complex analysis, univalent traditionally means "injective", not "single-valued".

  • $\begingroup$ how is injective ("one-to-one") not single-valued? $\endgroup$
    – postmortes
    Dec 17, 2013 at 9:19
  • $\begingroup$ @postmortes See en.wikipedia.org/wiki/Single-valued_function $\endgroup$ Dec 17, 2013 at 14:27
  • $\begingroup$ Oh ok, so all injective functions are single-valued, but single-valued functions are not necessarily injective. Thanks for the link (even if it is wikipedia) :) $\endgroup$
    – postmortes
    Dec 17, 2013 at 15:08

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