# $U \subset \mathbb{C}$ open, how do I construct $f$ holomorphic, such that there is no continuation of $f$ on any neighbourhood of $z \in \partial U$?

Let $U \subseteq \mathbb{C}$ be open. I want to construct a holomorphic function $f: U \to \mathbb{C}$, such that for all $z \in \partial U$ and for all $\varepsilon > 0$, there is no holomorphic continuation $\tilde{f}: U \cup B_\varepsilon(z) \to \mathbb{C}$ of $f$. For simply connected $U$, I could just find a diffeomorphism to the unit circle and there, the geometric series would be an example of such a function, if I am not mistaken. But what am I supposed to do if $U$ isn't simply connected? The Laurent-series comes to mind and also is the current topic of our complex-analysis lecture, however what would I do if those holes of $U$ were not "circle-shaped"?

Thanks for any help in advance.

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The geometric series has an analytic continuation at all points on the unit circle except $z = 1$, given by $\tilde{f}(z) = \frac1{1-z}$. – Lukas Geyer Dec 6 '12 at 20:43
@LukasGeyer: I was not aware of that. I thought since the geometric series diverges on the boundary of its convergence radius, it can't be analytically continued. Then now I am even more helpless than I was before. – Huy Dec 6 '12 at 20:50
As a hint: You can find solutions of the form $\sum_k \frac{a_k}{z-z_k}$, where $(z_k)$ is dense in the boundary of $U$, and $f(z) \to \infty$ as $z\to z_k$. – Lukas Geyer Dec 6 '12 at 20:56

The idea is to choose poles of first order $\frac{a_n}{z-z_n}$ at countable many $z_n \in \partial U$ such that they are dense in $\partial U$ and then to add these poles to a holomorphic function on $U$.

Let $z_n$ be a sequence of pairwise distinct complex numbers dense in $\partial U$.

Proposition 1: There exist two sequences $a_n, u_n$ of complex numbers such that

• for all $n \in \mathbb{N}$: $u_n \in U$ and $|u_n-z_n| < \frac{1}{n}$,
• for all $n \in \mathbb{N}$: $|a_n|\leq \frac{1}{n^2}$ and
• for all $k,n \in \mathbb{N}$ with $k \leq n$: $|f_n(u_k)| > k$,

where we define the $f_n$ through $$f_n(z) := \sum_{k=1}^n \frac{a_k}{z-z_k}.$$

Proposition 2: These $f_n$ converge locally uniformly to a holomorphic function $f$ on $U$, and $f$ is singular at each point in $\partial U$.

Proof of Proposition 1: We prove by induction. $z_1$ is given and we choose $a_1 = 1$. Choose further $u_1 \in U$ close enough to $z_0$ (since $z_0$ lies on the boundary of $U$ we have elements in $U$ arbitrarily close to $z_0$), such that $|f_1(u_1)| = |\frac{1}{u_1-z_1}| > 1$ and $|u_1-z_1|<1$. This was the case $n=1$.

Assume the proposition holds for the elements of the sequences up to $n-1$. Again we have $z_n$ given and choose $a_n$ such that for all $k \leq n-1$: $$|f_{n-1}(u_k)| > k \implies |f_n(u_k)| > k$$

For example, $$|a_n| < \min_{k=1, \dots, n-1} (|f_{n-1}(u_k)|-k)|u_k-z_n|$$ suffices. We can also ensure $|a_n| \leq \frac{1}{n^2}$. We have left to choose $u_n$: Choose it close enough to $z_n$ such that $|f_n(u_n)| > n$ and $|u_n-z_n| < \frac{1}{n}$ holds. This proves the proposition.

Proof of Proposition 2: Let $u \in U$ and $r > 0$ such that $B_{2r}(u) \subset U$. The sequence of holomorphic functions $f_n$ converges uniformly on $B_r(u)$, because each $v \in B_r(u)$ has distance further than $r$ to all $z_n \in \partial U$ and therefore, for all $n < m$: $$|f_n(v) - f_m(v)| \leq \sum_{k=n+1}^m \frac{|a_k|}{r} \leq\frac{1}{r} \sum_{k=n+1}^m \frac{1}{k^2}$$ This shows locally uniform convergence of $f_n$ to a holomorphic function $f$.

Further, if $f$ wasn't singular in $z \in \partial U$, then $f$ would be bounded near $z$. But in every neighbourhood of $z$ there are infinitely many $z_n$ and thus also infinitely many $u_n$. Now $|f(u_n)| \geq n$ contradicts boundedness (the condition $|f_n(u_k)| > k$ becomes $|f(u_n)| \geq n$ due to uniform convergence). Q.E.D.

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