# Holomorphic extension to a perfect, nowhere dense set.

Suppose $f$ is holomorphic and uniformly bounded by $M$ on $\mathbb{C}\setminus E$, where $E$ is perfect and nowhere dense? Can $f$ be extended to a holomorphic function on $\mathbb{C}$?

Riemann's theorem on removable singularities let's you remove one singularity $\{a\}$ at a time, provided $f$ is holomorphic and bounded on $\mathbb{C}\setminus\{a\}$. I suspect that the proof may be tailored in this case, since $E$ is nowhere dense, but would like your help on this. Furthermore, as is usual in cases of "I want to do something uncountably many times, but only have a method for doing it once", it seems natural to invoke Zorn's Lemma, so I tried:

Let $\mathfrak{F}$ be the collection of all pairs $\{(F,\Omega)\}$ satisfying:

1. $\Omega$ is an open subset of $\mathbb{C}$.
2. $\mathbb{C}\setminus E\subseteq \Omega$
3. $F: \Omega\to \mathbb{C}$ is uniformly bounded by $M$ and is holomorphic.
4. $F\mid_{C\setminus E}=f$

Since $f\in \mathfrak{F}$, it's nonempty. So we may define a partial order on $\mathfrak{F}$ by declaring that $(F,\Omega)\le (G,\Omega')$ if:

• $\Omega\subseteq \Omega'$
• $G\mid_{\Omega}=F$

Now if $\{(F_{\alpha},\Omega_{\alpha})\}$ is a chain in $(\mathfrak{F},\le)$, we set $O=\bigcup_{\alpha\in A}F_{\alpha}$ and $F(x)=f_{\alpha}(x)$ for all $x\in \Omega_{\alpha}$. It is clear that $O$ is open, does not meet $E$, and that $F$ is well-defined, uniformly bounded by $M$, and $F\mid_{C\setminus E}=f$. It remains to show that $F$ is holomorphic. So here's my try: pick $a\in O$. Find $\epsilon>0$ so that $B(a,\epsilon)\subseteq O$ and $\overline{(B(a,\frac{\epsilon}{2}))}\subseteq O$. Then the closed ball is compact and, being covered by $\{\Omega_{\alpha}\}$, admits a finite subcover. Since they lie in a chain, there exists some $\alpha^*$ for which $a\in \Omega_{\alpha^*}$. But $F=F_{\alpha^*}$ on $\Omega_{\alpha^*}$, and so taking difference quotients shows that $F$ is holomorphic at $a$. So $F\in \mathfrak{F}$, and hence $(F,O)$ is clearly an upper bound (if this all worked) on the chain.

By Zorn's Lemma, there exists some maximal element $(\mathcal{F},\mathcal{O})$. I claim that $\mathcal{O}=\mathbb{C}$. It is clear that there must be some $a\in E$ for which $a\not\in E$. But now we run through the proof of Riemann's Removable Singularity Theorem and find an extension of $F$ to...what!? $E$ has no isolated points, so I'm out of luck. I can't extend $F$ in a way to contradict its maximality.

Does anyone have an idea of how to fix this, or do this in general? The question I'm actually interested in knowing is if one embeds the Cantor Set into $\mathbb{C}$ and finds a holomorphic function $F$ that is uniformly bounded on the complement of the Cantor set, can one extend it to all of $\mathbb{C}$.

Thanks!

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My guess is no, and maybe $$F(z)=\int_{\Gamma} \frac{f(w)}{w-z}dw$$ for appropriate $\Gamma, f$ could give a counterexample. (Notice that $F$ is holomorphic on the complement of the perfect nowhere dense set $\Gamma$). My hope is that something along the lines of the jump condition for the double layer potential can be achieved. – Jose27 Sep 19 '12 at 7:26

## 2 Answers

As mentioned by LVK's answer, this is related to so-called analytic capacity. You're asking if perfect and nowhere dense sets are removable for bounded holomorphic functions. Compact sets that are removable for bounded holomorphic functions are exactly those of zero analytic capacity.

Now, to answer your question, let $E$ be a compact subset of the plane and suppose that $E$ is contained in the real line $\mathbb{R}$. A theorem of Pommerenke says that $E$ is removable if and only if it's one-dimensional Lebesgue measure is zero.

Hence, just take for $E$ your favorite Cantor set with positive measure, and it will be non-removable. One can even give an explicit example of a bounded holomorphic function in $\mathbb{C} \setminus E$ which is not constant :

Define $$f(z):= \int_{E} \frac{dt}{z-t}.$$ Then $f$ is holomorphic in $\mathbb{C} \setminus E$. Furthermore, it is non-constant since $f(\infty)=0$ and $\lim_{z \rightarrow \infty} zf(z)= m(E) > 0$. However, it is easy to show that $\operatorname{Im} f(z)$ is bounded in $\mathbb{C} \setminus E$. Hence, $$g(z):= e^{i f(z)}$$ is holomorphic and bounded in $\mathbb{C} \setminus E$, and yet non-constant.

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It depends on the metric structure of the set; the topological assumptions you made are not enough. In particular, if the Hausdorff dimension of your set is greater than one, it is not removable. If the Hausdorff dimension is less than 1, it is removable. The story is quite complicated when the dimension is equal to one: see this question and the Wikipedia article on analytic capacity linked there.

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