Re: maximal domain of holomorphy of a series
From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 03/28/05
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Date: Mon, 28 Mar 2005 14:00:14 +0000 (UTC)
On Sun, 27 Mar 2005 21:30:06 +0000 (UTC), israel@math.ubc.ca (Robert
Israel) wrote:
>In article <d26lae$cim$1@news.ks.uiuc.edu>,
>David C. Ullrich <ullrich@math.okstate.edu> wrote:
>>On Sat, 26 Mar 2005 15:00:06 +0000 (UTC), david.madore@ens.fr (David
>>Madore) wrote:
>
>>>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).
>
>>I can't quite give a counterexample off the top of my head, but
>>only because you assume that c_n > 0. If you say instead just
>>that the c_n should be non-zero reals then it's not too hard
>>to show that there exists a sequence (c_n) which tends to 0
>>fast enough that the series converges uniformly on the closed
>>disk, and such that the sum is identically 0 on the closed
>>disk! (The example has sum |c_n| infinite; I wouldn't be
>>surprised if sum|c_n| finite, or maybe a weighted sum,
>>with weights depending on a_n, finite, would suffice.)
>
>If sum_n |c_n| is infinite I wouldn't call that "tends to 0
>fast enough".
Well, it's not very fast perhaps, but it does allow the series
to converge uniformly on the closed unit disk - the point to the
construction is to show that "so fast that the series converges
uniformly on the closed disk" is not fast enough.
>It also makes your claim of convergence a bit
>suspect.
Hmm. Ok, the series probably does not converge absolutely
on the closed disk, so we have to take the terms in the
proper order, which means we may need to use a particular
enumeration of the rational points. Choosing a particular
enumeration is perfectly legal if we want a counterexample
to "If (a_n) is an enumeration then...".
>Also note that David said the c_n should be not only
>positive but decreasing, which makes it harder.
In this section I _said_ that I wasn't even getting
positive c_n, just real c_n. Hence it's not a counterexample
to the original question, but it does, or so it seems to
me, say that if the answer to the original question is
yes then the proof may be harder than one might have
expected. This was all I actually claimed.
If you look at the construction you see I do get |c_n|
non-increasing, hence a trivial modification would give
|c_n| strictly decreasing.
>Maybe you can
>do it if you have the freedom to choose the enumeration as
>well as the c_n. Well, you can certainly do it in a trivial
>way if you enumerate each point twice, since then you can
>have each even-numbered term cancel the preceding odd-numbered
>term.
In the construction above we use each rational point
exactly once.
>>Oh. I just realized that a person could give a
>>similar counterexample with c_n > 0. In the example
>>above we use the fact that 1/(z-p) - 1/(z-q)
>>is small on the unit disk if q is close to p.
>>Instead use the fact that if n is a positive
>>integer and "sum_w" means sum over all the
>>n-th roots of unity then
>
>> sum_w 1/(z-w*a)
That really should have been sum_w 1/(z-w*a)/n.
>>is small on the unit disk (if a is fixed and n is
>>large enough).
>
>> You have to jiggle the w*a a little
>>to get rational points, and writing down the
>>argument will be a little more complicated...
>
>It's not obvious to me how this will work
>if you want all the c_n > 0. Yes, you can get
>sum_w 1/(z-w*a) = n z^(n-1)/(z^n-a^n) to be small,
>but once you have it how do you "cancel" it with more
>terms?
I decided later that perhaps I should cancel the
part about getting c_n > 0. The idea _was_ that
a term 1/(z-a) at one stage is almost cancelled
by something like sum'_w 1/(z-w*a) at the next
stage, where sum' means we omit the term w = 1.
The problem is that it's really just sum_w 1/(z-w*a)/n
that's small on the closed disk, and here when
we want to get that cancellation we're not at
liberty to divide by that extra n.
Could be it works nonetheless, if in fact
sum_w 1/(z-w*a) is small on the closed disk, which
actually seems likely. But proving that would
involve some estimates, while the fact that
sum_w 1/(z-w*a)/n is small on the disk is just
because these are riemann sums for a certain
integral.
So I retract my claim about being able to get
c_n > 0 (although I'd be surprised if it were
not possible.)
>Robert Israel israel@math.ubc.ca
>Department of Mathematics http://www.math.ubc.ca/~israel
>University of British Columbia Vancouver, BC, Canada
************************
David C. Ullrich
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