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