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?
Best Answer
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.