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Mar 26, 2005, 10:00:06 AM3/26/05

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I've found myself embarrassingly baffled by the following seemingly

simple problem:

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/ )

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:

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

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:

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

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

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:

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

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