## 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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