Hadamard gap theorem

Remarks

Proof idea: suppose that $f$ extends holomorphically across an arc of the circle, and rotate so that the arc contains $1$. A standard proof composes $f$ with the polynomial $w \mapsto (w^m + w^{m+1})/2$, for an integer $m$ chosen from $\lambda$. This map sends the closed unit disc into the unit disc except at $1$, so the composed series would converge on a disc of radius greater than $1$. Because of the gaps, the expansion of $f$ composed with this map in powers of $w$ splits into disjoint blocks, one for each term $a_k z^{n_k}$. So the series of $f$ would converge at some real point greater than $1$, a contradiction. Example: $\sum_{n \ge 0} z^{2^n}$ and $\sum_{n \ge 1} z^{n!}$ have the unit circle as natural boundary. In the second series, $n_{k+1}/n_k = k + 1 \ge 2$. The gap condition cannot just be dropped: $\sum_{n \ge 0} z^n = \frac{1}{1 - z}$ has radius $1$ but extends across every point except $1$. The Fabry gap theorem reaches the same conclusion under the weaker condition $n_k / k \to \infty$.

Going deeper

For a student

The theorem is the classical source of functions with a natural boundary, called lacunary functions. It shows that the obstruction to analytic continuation can be spread over the whole circle of convergence. It is a special case of the Fabry gap theorem, and its proof is closely related to the phenomenon of overconvergence studied by Ostrowski.