A Holomorphic Function on the Unit Disk with No Analytic Continuation

By comparison with the geometric progression, the lacunary series \sum_{k=0}^{\infty}z^{2^{k}} defines a holomorphic function f on the open unit disk \mathbb{D}. We will show that f has no analytic continuation to a domain including unit circle \partial\mathbb{D}.

We begin by showing that f has singularities at each of the {2^{n}}th roots of units, for any n \in \mathbb{Z}^{\geq 0}. We proceed by induction. For n = 0, observe that

\displaystyle \liminf_{r \rightarrow 1^{-}}f(r) \geq \liminf_{r \rightarrow 1^{-}} \sum_{k=0}^{n}r^{2^{k}} = n, \indent \forall n \in \mathbb{Z}^{\geq 0}

so that \liminf_{r \rightarrow 1^{-}} f(r) = \infty. Suppose we have shown that

\displaystyle \liminf_{r \rightarrow 1^{-}}\left|f(re^{2\pi i\frac{j}{2^{n}}})\right| = \infty, \indent \forall j = 1,\cdots,2^{n}

and therefore f has singularities at the 2^{n}th roots of unity. Observe that

\displaystyle f(z^{2^{n+1}}) = f(z^{2^{n}}) - z^{2^{n}} \Longrightarrow \left|f(z^{2^{n}})\right| \geq \left|f(z^{2^{n+1}})\right| - \left|z^{2^{n}}\right|

so that, for any z = r^{\frac{1}{2^{n+1}}}e^{2\pi i \frac{j}{2^{n+1}}}, with 0 \leq r < 1 and j \in \left\{1,\cdots,2^{n+1}\right\},

\displaystyle \liminf_{r \rightarrow 1^{-}}\left|f(r^{\frac{1}{2}}e^{j\pi i})\right| \geq \liminf_{r \rightarrow 1^{-}}\left|f(r)\right| - r^{\frac{1}{2}} = \infty

Lemma 1. The subset \left\{z \in \partial\mathbb{D}: z^{2^{n}} = 1, n \in \mathbb{Z}^{\geq 0}\right\} is dense in \partial\mathbb{D}.

Proof. This is immediate from the continuity of the map \theta \mapsto e^{i\theta}.

To complete the proof, suppose the function f defined by the Lacunary series above has an analytic extension to a domain \Omega \supset D(0;1) \cup \left\{z_{0}\right\}, for some z_{0} \in \partial \mathbb{D}. Then f is holomorphic in some open neighborhood D(z_{0};r), for r > 0 sufficiently small, and therefore continuous and bounded on \overline{D}(z_{0};2^{-1}r), which is impossible by Lemma 1 and our work above.

We also see from our work that f fails to have radial limits at any z \in \partial\mathbb{D}, with z^{2^{n}} = 1 for some n \in \mathbb{Z}^{\geq 1}. Using a theorem of Zygmund, we will show in a later post that f fails to radial limits almost everywhere.

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