# Behaviour at natural boundary

Suppose I have a holomorphic function $f$ in a domain $\Omega$ with natural boundary $\partial \Omega.$ Let $p \in \partial \Omega.$ Is it true that there is some analogue of Picard's little theorem - that is, by choosing an appropriate sequence $x_1, \dotsc, x_n, \dotsc \in \Omega$ converging to $p$ we can have the limit of $f(x_i)$ be (almost) any complex number?

EDIT As divined by Noam Elkies, this was inspired by the recent discussion of $\sum_{i=0}^n z^{i^2},$ and the fact that its zeros seem to cluster near its natural boundary. Indeed, consider the function (a slight variant of that occurring in both answers) $$\sum_{i=0}^n \frac{z^{2^i-1}}{(i+1)^2}.$$

Here is the graph of its zeros($n=10$): This would tend to imply that you don't have to work very hard to tend to zero when approaching the boundary, and I assume that zero is not a particularly exceptional value. Indeed, if you plot the $1$s of the function (preimages of the value $1$), you get an identical plot: Which would tend to indicate that my conjecture has at least a grain of truth in it.

• Suggested by your recent computations for $\sum_{n=1}^\infty z^{n^2}$? mathoverflow.net/questions/272990 Jun 30 '17 at 0:00
• @NoamD.Elkies Yes, see the edit... Jun 30 '17 at 1:24
• An analytic function can be continuous and even smooth in the closed disk and have the circle as a natural boundary. Jun 30 '17 at 5:43

No. Consider $$f(z) = \sum_{n=1}^\infty \frac{z^{2^n}}{n^2}$$ By the Ostrowski-Hadamard gap theorem, the natural boundary is the unit circle. But the series converges absolutely on the unit circle, and the limit is always $f(p)$, which is bounded in absolute value by $\sum_n 1/n^2$.
$$f(z) = \sum\limits_{n = 1}^\infty \frac{z^{2^n}}{n^2}$$
can not be analytically continued beyond $\mathbb{D}$ but $|f|$ is bounded by $10$ in $\mathbb{D}$ so modulus of any limit of $f(x_i)$ as well can not be greater then $10$.
There is, indeed, a grain of truth in your conjecture. One possible formalization of it is as follows. Suppose that there is a sequence of analytic in the unit disk $\mathbb D$ functions $f_n$ such that $f_n$ converge pointwise on some set $E\subset \mathbb D$ having an accumulation point inside $\mathbb D$. If $f_n$ omit two fixed different values in $\mathbb D$, then $f_n$ converge uniformly inside $\mathbb D$. This is just the usual mumbo-jumbo about normal families (Montel's theorem, to be exact) plus the uniqueness theorem (any limit of a subsequence has prescribed values on $E$). If you replace $\mathbb D$ by a disk around $p$ and take $f_n$ to be Taylor polynomials of $f$, say, you'll see that either $f$ can be analytically extended to a neighborhood of $p$, or you have the effect you observed for $a$-points of $f_n$ with almost every $a\in \mathbb C$. Alas, more often than not, the zeroes on your picture will lie outside $\Omega$, so you cannot extract too much information about $f$ itself from that picture.