I've found myself embarrassingly baffled by the following seemingly
simple problem:
Let (a_n) be an enumeration of all complex numbers having rational
real and imaginary parts, and modulus (strictly) greater than one,
i.e., all elements of Q[i] not contained in the closed unit disk.
Let (c_n) be a decreasing sequence of real numbers which converges
rapidly toward zero. (I'm deliberately being vague on what exactly I
mean by "rapidly", but I'll say more in a few lines.)
Let F(z) be the sum of c_n/(z-a_n) for all n, provided this series
converges. We let f(z)=F(z) for z in the open unit disk.
If c_n is chosen to decrease fast enough, then the series converges
uniformly on the closed unit disk. Then F is continuous on the closed
unit disk and its restriction f to the open unit disk is holomorphic.
At the very least, we shall assume (c_n) such that the series
converges uniformly on every compact set within the open unit disk
(so, again, f is holomorphic on the open unit disk).
The question is then: show (or refute) that f does not admin a
holomorphic extension to any larger (connected) open set (that is, the
unit circle is the natural boundary of holomorphy of f).
Intuitively the reason seems clear: we have created a "pole" at every
rational point outside the closed unit disk, so we cannot get a
holomorphic extension there (nor even a meromorphic extension, since
the poles would be dense). I have been unable, however, to make this
reasoning rigorous.
Note that if (c_n) decreases rapidly enough then f and all its
derivatives extend continuously to the closed unit disk. One might
hope to prove that the derivatives at a point on the unit circle
increase too rapidly, but I do not see why this should be.
It is infuriating that the series for F converges - at least when
(c_n) decreases rapidly enough - for many points outside the unit
disk: in fact, on the complement H of a (dense) set of Lebesgue
measure zero. It is probably easy to see that F on H, defined as the
sum of the converging series, cannot be extended to a holomorphic
function on any open set not contained in the open unit disk: however,
I see no reason why a holomorphic extension of f should indeed
coincide with the sum F of the series outside the unit disk, so this
line of attack seems doomed.
Did I miss a simple fact from complex analysis?
I would welcome any thoughts on this problem. Thanks in advance!
--
David A. Madore
(david.mad...@ens.fr,
http://www.eleves.ens.fr:8080/home/madore/ )