5
$\begingroup$

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($n=10$): enter image description here

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:enter image description here

Which would tend to indicate that my conjecture has at least a grain of truth in it.

$\endgroup$
  • $\begingroup$ Suggested by your recent computations for $\sum_{n=1}^\infty z^{n^2}$? mathoverflow.net/questions/272990 $\endgroup$ – Noam D. Elkies Jun 30 '17 at 0:00
  • 1
    $\begingroup$ @NoamD.Elkies Yes, see the edit... $\endgroup$ – Igor Rivin Jun 30 '17 at 1:24
  • 1
    $\begingroup$ An analytic function can be continuous and even smooth in the closed disk and have the circle as a natural boundary. $\endgroup$ – Alexandre Eremenko Jun 30 '17 at 5:43
14
$\begingroup$

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 $\sum_n 1/n^2$.

$\endgroup$
7
$\begingroup$

No, this analogue is false. For example function

$$f(z) = \sum\limits_{n = 1}^\infty \frac{z^{2^n}}{n^2}$$

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 $10$.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.