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Show that there is an analytic function $f$ on unit open disc $D$ which is not analytic on any connected open set $G$ which properly contains $D$.

My attempt : I know Weierstrass factorization theorem for a region $G$ then I can find a sequence $\{a_n\}$ of points which lies in unit circle and construct an analytic function $g$ which only have zeros on that points $\{a_n\}$. Finally if I consider $f=1/g$ then $f$ satisfy the require properties. Am I right or is there is any other issues or nice examples?

Any help/hint in this regards would be highly appreciated. Thanks in advance!

David C. Ullrich
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2 Answers2

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No, your solution is not right. The points $a_n$ would have to be dense in the circle or it doesn't work, and if the $a_n$ are dense in the circle then $g=0$.

Instead take $(a_n)$ contained in the open disk, with no limit point in the open disk, such that every point of the circle is a limit point. If $(a_n)$ is the zero set of $f$ then $f$ cannot extend to a larger connected open set, since if it did the zero set would have a limit point in the domain.

David C. Ullrich
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Classic example: For $|z|<1,$ let

$$f(z) = \sum_{n=0}^{\infty}z^{n!}.$$

Then $f$ is holomorphic in the open unit disc.

For any $p,q\in \mathbb N,$ you can verify that

$$\lim_{r\to 1^-} f(re^{2\pi ip/q}) = \infty.$$

Since the set of all such $e^{2\pi ip/q}$ is dense on the unit circle, $f$ is unbounded in any neighborhood of any point on the circle. Thus $f$ cannot be extended analytically to any larger open connected set.

zhw.
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