maximal domain of holomorphy of a series

12 views
Skip to first unread message

David Madore

unread,
Mar 26, 2005, 10:00:06 AM3/26/05
to
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....@ens.fr,
http://www.eleves.ens.fr:8080/home/madore/ )

David C. Ullrich

unread,
Mar 27, 2005, 11:00:14 AM3/27/05
to
On Sat, 26 Mar 2005 15:00:06 +0000 (UTC), david....@ens.fr (David
Madore) wrote:

>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).

I would not be surprised if this was actually false. I would not
be surprised if it was true but harder to prove than you'd expect.

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.)

Sketch of a hint: The partial sums are going to have
a _subsequence_ S_n, constructed as follows: If S_n
has n terms then

S_{n+1} = S_n + R_n,

where R_n has 2n + 1 terms. For each term c/(z-p)
that occurs in S_n, R_n has two terms of the form

(-c/2)/(z-q) + (-c/2)/(z-r)

where q and r are very close to p, so close that
c/(z-p) + (-c/2)/(z-q) + (-c/2)/(z-r) is as small
as you want on the closed unit disk. R_n also
has one "extra" term inserted to make sure all the
a_n appear eventually: the extra term is c/(z-a),
where a is the first a_n that has not appeared yet,
and c is very small.

That gives a sum as above such that a subsequence
of the partial sums tends to zero uniformly on
the closed disk. If we'd taken each term in S_n
to be almost cancelled by just one term in R_n
then that would be all we could say, but since
each term in S_n is actually almost cancelled
by two terms in R_n, each with a smaller
coefficient, it's not hard to show that the
sum itself actually equals zero on the closed
disk.

So if you replace c_n > 0 by c_n <> 0 it's
simply false.

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)

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...

>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 C. Ullrich

Robert Israel

unread,
Mar 27, 2005, 4:30:06 PM3/27/05
to
In article <d26lae$cim$1...@news.ks.uiuc.edu>,

David C. Ullrich <ull...@math.okstate.edu> wrote:
>On Sat, 26 Mar 2005 15:00:06 +0000 (UTC), david....@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". It also makes your claim of convergence a bit
suspect. Also note that David said the c_n should be not only
positive but decreasing, which makes it harder. 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.

>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)

>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?

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

David C. Ullrich

unread,
Mar 28, 2005, 9:00:14 AM3/28/05
to

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 isr...@math.ubc.ca
>Department of Mathematics http://www.math.ubc.ca/~israel
>University of British Columbia Vancouver, BC, Canada


************************

David C. Ullrich

David C. Ullrich

unread,
Mar 29, 2005, 9:30:04 AM3/29/05
to
On Sun, 27 Mar 2005 21:30:06 +0000 (UTC), isr...@math.ubc.ca (Robert
Israel) wrote:

>suspect.[...]

I replied to most of your comments yesterday; a little later
I realized that I need to make a correction and include a
little more information, or my claim of convergence is indeed
very suspect. This was all based on my recollection of something
analogous that I wrote out in exruciating detail some years ago,
and there's an aspect of it I was forgetting.

I got convergence for a subsequence of the partial sums the
other day, and then stated without any justification that it
followed that the series itself converged. Thinking about
why that is I need to make a change. I said

>> Sketch of a hint: The partial sums are going to have
>> a _subsequence_ S_n, constructed as follows: If S_n
>> has n terms then
>>
>> S_{n+1} = S_n + R_n,
>>
>> where R_n has 2n + 1 terms. For each term c/(z-p)
>> that occurs in S_n, R_n has two terms of the form
>>
>> (-c/2)/(z-q) + (-c/2)/(z-r)
>>
>> where q and r are very close to p, so close that
>> c/(z-p) + (-c/2)/(z-q) + (-c/2)/(z-r) is as small
>> as you want on the closed unit disk. R_n also
>> has one "extra" term inserted to make sure all the
>> a_n appear eventually: the extra term is c/(z-a),
>> where a is the first a_n that has not appeared yet,
>> and c is very small.

Instead:

**************


The partial sums are going to have
a _subsequence_ S_n, constructed as follows: If S_n

has N terms then

S_{n+1} = S_n + R_n,

where R_n has N*2^n + 1 terms. For each term c/(z-p)
that occurs in S_n, R_n has 2^n terms of the form

(-c/2^n)/(z-q)

where q runs over 2^n values very close to p, so
close that c/(z-p) + the sum of the "corresponding"
terms in R_n is as small as you want on the closed
unit disk.

Now to get from convergence of the partial sums to
convergence you need to enumerate these terms in
an order that might not be the one you expect.
Say p_1, ... p_N are the p's that appear in S_n.
Say q_j,k are the q's in R_n, where each q_j,k
is one of the terms corresponding to p_j. Order
the terms in R_n like so:

q_1,1, q_2,1, ... q_N,1; q_1,2, q_2,2, ... .

R_n also has one "extra" term inserted to make sure all the
a_n appear eventually: the extra term is c/(z-a),
where a is the first a_n that has not appeared yet,
and c is very small.

***********

The idea is that when you're trying to show that
a partial sum of R_n is small, the partial sum
consists of some of the constant-k "blocks"
followed by a "tail" of fewer than N terms.
You can use the previous estimate on S_n to
show that the sum of each block is smaller
than you'd sort of expect, then estimate the
tail by looking at the size of the coefficients.

The details are somewhat tedious - if I can find
a preprint of the other thing I'll put it on a
web page and post a link to it. The argument
there should suffice to show that the construction
above does what I claim.

(Why I don't have the preprint:

Long ago a student asked me how to show that if
x_j was a sequence of distinct reals, c_j was
real, and sum c_j delta_x_j = 0 in the sense of
distributions then all the c_j = 0. The answer
is no, this doesn't follow. I had to write out
the proof in extreme detail to convince him that
"no" was the correct answer.

He wanted to publish this - I predicted that
nobody would be interested in publishing it because
although it was a little intricate it really wasn't
deep or all that surprising. So it was he who typed
it up - sure enough, a few journals turned it down.
I'll try to find him and ask whether he still has
the TeX.)

************************

David C. Ullrich

Reply all
Reply to author
Forward
0 new messages