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I found myself curious whether there existed a complex function which is analytic on the interior of the unit disc, but such that there is no extension of the function to a holomorphic function on a strictly larger connected open set. This was because all the typical examples I could think of with a finite radius of convergence of the Taylor series had natural analytic extensions near all but finitely many points of the boundary of the region of convergence.

The example I came up with was: $$f(z) := \sum_{n=0}^\infty \frac{z^{2^n}}{2^n}.$$ This function has radius of convergence 1, and at every point of $|z| = 1$ it converges to a function whose imaginary part has $\Im f(e^{i \theta})$ equal to the Weierstrass function. Therefore, the restriction to the unit circle is not differentiable (as a function on the real $C^\infty$ manifold $S^1$) at any point, which makes it impossible for any extension of $f$ to be holomorphic at any point on the unit circle (since Abel's theorem implies such an extension wouldn't have a pole at such a point).

My question is: is this a valid example, or is there something I'm missing? And also, is there a more natural example of such a function in terms of the usual examples of analytic functions?

Daniel Schepler
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    Look up lacunary series – Robert Israel Jun 18 '19 at 20:14
  • If you want "closed form" examples, try Jacobi theta functions (as functions of the nome). Thus $\theta_3(0,q) = 1 + 2 \sum_{n=1}^\infty q^{n^2}$. The Fabry gap theorem implies that this has the unit circle as natural boundary. – Robert Israel Jun 18 '19 at 20:24
  • Whoops, I actually meant "meromorphic" in place of "holomorphic". Given the existing answer, might be a bit late to correct that as it would change the question a bit, though certainly not by much. – Daniel Schepler Jun 18 '19 at 20:54
  • I mean obviously, if $f$ had a meromorphic extension to a neighborhood of $e^{i \theta_0}$ then that would be an analytic extension to a neighborhood of $e^{i (\theta_0 + \Delta \theta)}$ for $\Delta \theta \ne 0$ sufficiently small. – Daniel Schepler Jun 18 '19 at 21:01

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This is a well-known phenomenon. If you have Rudin's Real and Complex analysis, on p. 320 he gives the related example $f(z) = \sum_{n=0}^{\infty} z^{2^n}$, which as he points out is unbounded on every radius of the unit disk ending at $e^{2\pi i k / 2^n}$ for $k$ and $n$ positive integers. Thus $f(z)$ can't be extended to an analytic function on a neighborhood of any such $e^{2\pi i k / 2^n}$. Consequently, $f(z)$ can't be extended to any larger connected open set.

Zarrax
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