By comparison with the geometric progression, the lacunary series defines a holomorphic function on the open unit disk . We will show that has no analytic continuation to a domain including unit circle .
We begin by showing that has singularities at each of the roots of units, for any . We proceed by induction. For , observe that
so that . Suppose we have shown that
and therefore has singularities at the roots of unity. Observe that
so that, for any , with and ,
Lemma 1. The subset is dense in .
Proof. This is immediate from the continuity of the map .
To complete the proof, suppose the function defined by the Lacunary series above has an analytic extension to a domain , for some . Then is holomorphic in some open neighborhood , for sufficiently small, and therefore continuous and bounded on , which is impossible by Lemma 1 and our work above.
We also see from our work that fails to have radial limits at any , with for some . Using a theorem of Zygmund, we will show in a later post that fails to radial limits almost everywhere.