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Spring 2011 - Problem 9

Cantor set, Painlevé's theorem

Let $E \subseteq \br{0, 1}$ denote the Cantor 'middle thirds' set; namely, the set $E = \set{\sum_{i\geq1} b_i3^{-i} \mid b_i = 0, 2}$ . Embedding $\br{0, 1}$ naturally into $\C$ , we regard $E$ as a subset of $\C$ . Suppose $\func{f}{\C \setminus E}{\C}$ is holomorphic and (uniformly) bounded. Show that $f$ is constant.

Solution.

Notice that $E$ has Lebesgue measure zero, i.e.,

m\p{E} = \inf\,\set{\sum_{n=1}^\infty \abs{b_n - a_n} \st E \subseteq \bigcup_{n=1}^\infty \p{a_n, b_n}},

by outer regularity of the Lebesgue measure.

Let $\epsilon > 0$ . Then there exists a family $\set{B\p{x_n, r_n}}_n$ of balls in $\C$ such that $E \subseteq \bigcup_{n=1}^\infty B\p{x_n, r_n}$ and $\sum_{n=1}^\infty r_n < \epsilon$ . By removing balls which are disjoint from $E$ , we see that such a family is contained in the ball $B\p{0, 1 + \epsilon}$ . By compactness, there exists $N \in \N$ such that $E \subseteq \bigcup_{n=1}^N B\p{x_n, r_n}$ . Let

\gamma_\epsilon = \partial\p{\bigcup_{n=1}^N B\p{x_n, r_n}}

and orient it counter-clockwise. Let $B = B\p{0, 2}$ so that $\gamma_\epsilon \subseteq B$ for all $\epsilon \in \p{0, 1}$ , which gives us the following annular decomposition of $f$ on $B \setminus \bigcup_{n=1}^N B\p{x_n, r_n}$ :

f\p{z} = \frac{1}{2\pi i} \int_{\partial B} \frac{f\p{\zeta}}{\zeta - z} \,\diff\zeta - \frac{1}{2\pi i} \int_{\gamma_\epsilon} \frac{f\p{\zeta}}{\zeta - z} \,\diff\zeta.

Thus, if $z \in B \setminus E$ , then $d\p{z, E} > 0$ . If we shrink $\epsilon$ to, say, smaller than a quarter of this distance, then $d\p{z, \gamma_\epsilon} \geq \frac{d\p{z, E}}{2} \eqqcolon \delta > 0$ since we required that all the balls above intersect $E$ . Hence,

\begin{aligned} \abs{f\p{z} - \frac{1}{2\pi i} \int_{\partial B} \frac{f\p{\zeta}}{\zeta - z} \,\diff\zeta} &\leq \frac{1}{2\pi} \int_{\gamma_\epsilon} \abs{\frac{f\p{\zeta}}{\zeta - z}} \,\diff\abs{\zeta} \\ &\leq \frac{1}{2\pi} \frac{\norm{f}_{L^\infty}}{\delta} \abs{\gamma_\epsilon} \\ &\leq \frac{1}{2\pi} \frac{\norm{f}_{L^\infty}}{\delta} \sum_{n=1}^N 2\pi r_n \\ &= \frac{\norm{f}_{L^\infty}}{\delta}\epsilon \xrightarrow{\epsilon\to0} 0. \end{aligned}

In other words, if $g\p{z} = \frac{1}{2\pi i} \int_{\partial B} \frac{f\p{\zeta}}{\zeta - z} \,\diff\zeta$ , then $g$ is holomorphic on $B\p{0, 2}$ and $f$ agrees with $g$ on $B\p{0, 2} \setminus E$ . Hence, $f$ extends to an entire function, which completes the proof.