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Spring 2014 - Problem 10

analytic continuation

Let $\alpha \in \br{0, 1} \setminus \Q$ and let $\set{a_n}_n \subseteq \ell^1\p{\N}$ with $a_n \neq 0$ for all $n \geq 1$ . Set $\D = \set{z \in \C \mid \abs{z} < 1}$ . Show that

f\p{z} = \sum_{n\geq1} \frac{a_n}{z - e^{i\alpha n}}, \quad z \in \D,

converges and defines a function that is analytic in $\D$ which does not admit an analytic continuation to any domain larger than $\D$ .

Solution.

It suffices to show that the partial sums are uniformly bounded on compact sets. Without loss of generality, suppose $K = \cl{B\p{0, R}}$ , where $R \in \p{0, 1}$ , so for $z \in K$ ,

\abs{f\p{z}} \leq \sum_{n=1}^\infty \frac{\abs{a_n}}{1 - \abs{z}} \leq \sum_{n=1}^\infty \frac{\abs{a_n}}{1 - R} < \infty,

since $\set{a_n}_n \in \ell^1\p{\N}$ . Thus, the sum converges locally uniformly, so $f$ defines a holomorphic function in the disk $\D$ . For the second claim, first notice that if $k \in \N$ , then

\p{z - e^{i\alpha k}}f\p{z} = a_k + \sum_{n \neq k} \frac{z - e^{i\alpha k}}{z - e^{i\alpha n}} a_n.

Thus, if $z = re^{i\alpha k}$ ,

\abs{\frac{z - e^{i\alpha k}}{z - e^{i\alpha n}}} \abs{a_n} \leq \frac{1 - r}{1 - r} \abs{a_n} = \abs{a_n}

for all $r$ . Hence, by dominated convergence,

\lim_{r\to1}\,\p{re^{i\alpha k} - e^{i\alpha k}}f\p{re^{i\alpha k}} = a_k \neq 0,

so $f\p{re^{i\alpha k}} \to \infty$ as $r \to 1$ . Since $\alpha$ is irrational, $\set{\alpha k \mid k \geq 1}$ is dense in the torus $\R/\Z$ . If $f$ had a holomorphic extension to some $e^{i\theta} \in \partial\D$ , then by density, any neighborhood of $e^{i\theta}$ contains a pole, which is impossible, so $f$ cannot be extended analytically past the disk.