# On every simply connected domain, there exists a holomorphic function with no analytic continuation.

I am working on a question that requires me to prove that on every simply connected open subset of $\mathbb{C}$, there exists a holomorphic function that cannot be extended to a holomorphic function on a larger connected open set.

I know an example of such a holomorphic function on the open unit disc: $$f:z\mapsto\sum_{n=1}^\infty z^{n!}.$$ I tried to combine this with the Riemann mapping theorem.

Let $\Omega$ be a simply connected open subset of $\mathbb{C}$ and one may assume that $\Omega$ is not the entire complex plane. By the Riemann mapping theorem, there exists a conformal equivalence $$\phi:\Omega\rightarrow D$$where $D$ is the open unit disc. Then my guess is that $f\circ\phi$ has no analytic continuation, but then I had to link the boundaries of $\Omega$ and of $D$ and I don't know what is going on between $\partial\Omega$ and $\partial D$.

Could anyone offer any idea? Many thanks!

• If $\partial \Omega$ is a Jordan curve then boundary maps to boundary.
– A.Γ.
Jul 31 '15 at 2:06
• @A.G.homeomorphisms map boundaries to boundaries. Jul 31 '15 at 5:06
• @Gary.That is not true. For example, the unit disc and the right half plane are conformally equivalent, but a circle is not homeomorphic to a straight line (by compactness). Jul 31 '15 at 5:32
• Better example: A disk and a slit disk. Aug 2 '15 at 23:37

I don't think this has much to do with simple connectivity. Let $U\subset \mathbb {C}$ be any open set with non-empty boundary. Claim: There exists $f\in H(U)$ such that

$$\sup_{D(a,r)\cap U} |f|=\infty$$

for every $a \in \partial U$ and $r>0.$ This $f$ cannot be be extended analytically to any larger open set containing a point in $\partial U.$

Proof: Let $\{a_1,a_2, \dots \}$ be a countable dense subset of $\partial U.$ The idea is to define

$$f(z) = \sum_{n=1}^{\infty} \frac{c_n}{z-a_n}$$

for appropriate positive constants $c_n.$

To do this, we choose pairwise disjoint countably infinite sets $E_1, E_2,\dots \subset U$ such that for each $n,$ the only limit point of $E_n$ in $\mathbb {C}$ is $a_n.$ (For each $n,$ $E_n$ is just a sequence of distinct points in $U$ whose limit is $a_n.$ You can choose the $E_n$'s inductively.)

Now there are compact sets $K_1,K_2, \dots \subset U$ such that

$$K_1\subset \text {int}(K_2)\subset K_2 \subset \text {int}(K_3) \subset \dots$$

such that $U= \cup K_n.$ Choose $c_1$ such that $c_1/|z-a_1| < 1/2$ on $K_1.$ If $c_1,\dots c_n$ have been chosen, we choose $c_{n+1}$ such that

$$\frac{c_{n+1}}{|z-a_{n+1}|} < \frac{1}{2^{n+1}},\ \ z\in K_{n+1}\cup E_1 \cup \cdots E_n.$$

Then the series $\sum c_n/(z-a_n)$ converges uniformly on each $K_n,$ hence defines a holomorphic function $f$ on $U.$ Let $a\in \partial U, r > 0.$ Then $D(a,r)$ contains some $a_n,$ hence contains the tail-end of $E_n.$ Thus as $z\to a_n$ within $E_n,$ we have

$$|f(z)| \ge \left|\sum_{k=1}^{n}\frac{c_n}{z-a_k}\right| - \sum_{k=n+1}^{\infty}\frac{1}{2^{k+1}}\to \infty.$$

This completes the proof.

HINT:

The function $\sum_{n\ge 0}z ^{n!}$ can be seen to be un-extendable for an intrinsic reason: there exists a sequence $(a_n)$ of points in the disk such that $|f(a_n)| \to \infty$, and every open subset of the disk that is not relatively compact contains a point $a_n$ ( so in fact infinitely many).

Try to find such a sequence $(a_n)$.

$\bf{Added}$ If $0 < z < 1$ and $z = r e^{2\pi \frac{k}{N}}$ then for all $n \ge N$ we have $z^{n!} = |z|^{n!}$. If you fix the argument $2 \pi\frac{k}{N}$ but increase the absolute value towards $1$ you see that $|f(z)| \to \infty$. Just choose $N$ points with arguments $0$, $2\pi \frac{1}{N}$, $\ldots$ , $2 \pi \frac{N-1}{N}$. such that say $f(\cdot)$ is in absolute value $> N$.

• Thank you for the hint! But I cannot quite follow it. A subset $A$ of the unit disc not being relatively compact means that the closure of $A$ in the unit disc is not compact, so $A$ has a limit point on the unit circle. But if $A$ only has one limit point on the unit circle, why should the value of $f$ be unbounded on $A$? Jul 31 '15 at 6:20
• @Chuwei Zhang : $A$ will contain infinitely many points $a_n$ Jul 31 '15 at 6:29
• I still cannot work out what the $a_n$ should be. Could you tell me how to construct them? Jul 31 '15 at 17:11
• @Chuwei Zhang: I sketched a sequence that may work. I don't have yet a complete argument. Aug 1 '15 at 4:13
• Thanks for the added information! I have thought about what you suggested but I couldn't get it to work. Using the notation in your solution, as $N$ increases, $r$ needs to get closer and closer to $1$ in order that $|f(re^{2i\pi k/N})|>N$. So I can take a sequence $b_n$ in the disc that approaches the unit circle such that $b_n$ is not in the closure of $\{a_m:m\in\mathbb{N}\}$. Then there is a neighborhood of $\{b_n:n\in\mathbb{N}\}$, which is open and not relatively compact (since $|b_n|\rightarrow 1$), and which does not contain any $a_n$. Aug 1 '15 at 22:26

Another proof of the claim of zhw is the following:

Given an open set $$A$$ and a set $$B\subset A$$ without cluster points in $$A$$, a slight generalization of Weierstrass factorization theorem (see here, or go to Rudin's "Real and Complex analysis" for a proof) ensure the existence of a function $$f\in \mathcal{H}(A):\{z\in A|f(z)=0\}=B$$.

Thanks to this theorem, we can easily prove the claim:

Choose as $$B$$ a sequence that accumulates to every element of $$\partial A$$: the associated $$f$$ cannot be analytically extended to a larger open set, as the identity principle would imply $$f\equiv 0$$.

• Doesn't this only account for one boundary point? Feb 26 at 14:05
• @Chappers Indeed. Edited, thanks. Feb 26 at 15:37