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David C. Ullrich science forum Guru
Joined: 28 Apr 2005
Posts: 2250

Posted: Thu Jul 20, 2006 11:55 pm Post subject:
Re: Fourier Transform / Decay / Canot set



On 20 Jul 2006 13:51:42 0400, hrubin@odds.stat.purdue.edu (Herman
Rubin) wrote:
Quote:  In article <e9mvt4$46l$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual Cantor set
(middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f as x > infinity ?
Note that f(3x) = cos(x) f(x). In particular, f(3^n pi) = (1)^n f(pi)
(and that is nonzero) for every positive integer n, so there is no
decay.
But more can be said.
ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(nk) pi + t/3^k)/cos(3^(nk) pi))
= sum_{k=1}^infty ln(cos(t/3^k)  sin(t/3^k) tan(3^(nk) pi))
The terms for k <= n are of course ln(cos(t/3^k)), while all terms
are dominated by c/3^k for some c (depending on t), so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) > sum_{k=1}^infty ln(cos(t/3^k)) = ln(f(t))
i.e. (1)^n f(3^n pi + t) > f(t) f(pi) as n > infty
(and this should be uniform on, say, t in [1,1]).
Therefore...
Therefore not much. The numbers of the form 3^n*pi+t,
t in a bounded set, have asymptotic density 0.
I believe that there is a theorem which states that
if, in the definition of f, the "3" is replaced by
2 or smaller, the resulting characteristic function
corresponds to a density.

By "correspond to a density" mean that the measure
we're taking the Fourier transform of is absolutely
continuous? If so I don't _think_ it's quite that
simple (not sure).
Quote:  So f(x)*f(sqrt(3)*x) is
at least in L^2.
What I am really interested is what L^p(R) spaces is f in ?
... None except L^infty.
John Schutkeker's shoe is safe.
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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David C. Ullrich science forum Guru
Joined: 28 Apr 2005
Posts: 2250

Posted: Thu Jul 20, 2006 11:53 pm Post subject:
Re: Fourier Transform / Decay / Canot set



On Thu, 20 Jul 2006 01:59:27 EDT, don <ctcowan@sfu.ca> wrote:
Quote:  In article
13869322.1153345299269.JavaMail.jakarta@nitrogen.math
forum.org>,
don <ctcowan@sfu.ca> wrote:
let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for
real x.
This is the Fourier transform of a measure supported
on the usual Cantor set
(middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f
as x > infinity ?
[...]
thanks for the help.
New question:
How about if I use a fat Cantor set instead ?
Since the measure is "smoother " I would expect the transform to have better decay.

I had a comment on this this morning, decided to look up
something I might have been misremembering.
Even for Cantor sets of measure zero, obtained by using a
constant ratio of dissection, the decay is not a monotone
function of the "smoothness", ie of the dimension!
Bizarre true fact: Say you construct a Cantor set by repeatedly
removing the middle (12a)th part of the intervals, keeping
the left ath and the right ath. Then the Fourier transform
of that measure tends to 0 at infinity if and only if 1/a is
not a Pisot number. Pisot numbers are algebraic, so there's
only countably many of them.
In particular you can have a Cantor set that's definitely
thinner than the middlethirds set, so the measure is in
some sense less smooth, such that the Fourier transform
does tend to zero.
(And since this is a perfectly ordinary Cantor set
with a constant ratio and measure zero and everything,
if anything appearing worse to the naked eye than the
middle thirds set, I think that someone really _should_
eat his shoe...))
Quote:  Can I expect this to be in some L^p space ? (p < inf.)
thanks
craig

************************
David C. Ullrich 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Thu Jul 20, 2006 8:25 pm Post subject:
Re: Fourier Transform / Decay / Canot set



In article <e9ofre$32s0@odds.stat.purdue.edu>,
Herman Rubin <hrubin@odds.stat.purdue.edu> wrote:
Quote:  In article <e9mvt4$46l$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual Cantor set
(middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f as x > infinity ?
Note that f(3x) = cos(x) f(x). In particular, f(3^n pi) = (1)^n f(pi)
(and that is nonzero) for every positive integer n, so there is no
decay.
But more can be said.
ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(nk) pi + t/3^k)/cos(3^(nk) pi))
= sum_{k=1}^infty ln(cos(t/3^k)  sin(t/3^k) tan(3^(nk) pi))
The terms for k <= n are of course ln(cos(t/3^k)), while all terms
are dominated by c/3^k for some c (depending on t), so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) > sum_{k=1}^infty ln(cos(t/3^k)) = ln(f(t))
i.e. (1)^n f(3^n pi + t) > f(t) f(pi) as n > infty
(and this should be uniform on, say, t in [1,1]).
Therefore...
Therefore not much. The numbers of the form 3^n*pi+t,
t in a bounded set, have asymptotic density 0.

Asymptotic density doesn't matter here. The L^p norm requires the
integral of f^p, not the asymptotic average of f^p, to be
finite. So if f is bounded away from 0 on, say, the intervals
(3^n pi  1, 3^n pi + 1), then it can't be in L^p for any
p < infinity.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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Herman Rubin science forum Guru
Joined: 25 Mar 2005
Posts: 730

Posted: Thu Jul 20, 2006 5:51 pm Post subject:
Re: Fourier Transform / Decay / Canot set



In article <e9mvt4$46l$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
Quote:  In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual Cantor set
(middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f as x > infinity ?
Note that f(3x) = cos(x) f(x). In particular, f(3^n pi) = (1)^n f(pi)
(and that is nonzero) for every positive integer n, so there is no
decay.
But more can be said.
ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(nk) pi + t/3^k)/cos(3^(nk) pi))
= sum_{k=1}^infty ln(cos(t/3^k)  sin(t/3^k) tan(3^(nk) pi))
The terms for k <= n are of course ln(cos(t/3^k)), while all terms
are dominated by c/3^k for some c (depending on t), so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) > sum_{k=1}^infty ln(cos(t/3^k)) = ln(f(t))
i.e. (1)^n f(3^n pi + t) > f(t) f(pi) as n > infty
(and this should be uniform on, say, t in [1,1]).
Therefore...

Therefore not much. The numbers of the form 3^n*pi+t,
t in a bounded set, have asymptotic density 0.
I believe that there is a theorem which states that
if, in the definition of f, the "3" is replaced by
2 or smaller, the resulting characteristic function
corresponds to a density. So f(x)*f(sqrt(3)*x) is
at least in L^2.
Quote:  What I am really interested is what L^p(R) spaces is f in ?
... None except L^infty.
John Schutkeker's shoe is safe.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada


This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558 

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Herman Rubin science forum Guru
Joined: 25 Mar 2005
Posts: 730

Posted: Thu Jul 20, 2006 5:41 pm Post subject:
Re: Fourier Transform / Decay / Canot set



In article <Xns9805CD7BEA1A7lkajehoriuasldfjknak@207.115.17.102>,
John Schutkeker <jschutkeker@sbcglobal.net.nospam> wrote:
Quote:  don <ctcowan@sfu.ca> wrote in
news:13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org:
let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual
Cantor set (middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f as x > infinity
?
Any comments or suggestion would be greatly appreciated
If you get any kind of decay in f, as x > infinity, I'll eat my shoe.
Taking the Fourier transform of the Cantor set is asking for trouble.

I have already commented about it not approaching 0.
However, and I have not checked this, I believe that the
measure of the set of reals between 0 and T for which
f(x) > c is o(T) for all c > 0.
One reason for believing this is if one takes
g(x) = product_{k=1}^infinity cos(x 4^{k} ),
then g(x)*g(2x) is the characteristic function of a
uniform distribution.

This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558 

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Herman Rubin science forum Guru
Joined: 25 Mar 2005
Posts: 730

Posted: Thu Jul 20, 2006 5:19 pm Post subject:
Re: Fourier Transform / Decay / Canot set



In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
Quote:  let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual Cantor set (middle third) but translated so symmetric about 0.

Quote:  The question is what kind of decay can I expect in f as x > infinity ?

If one considers f(3^k * 2pi), this is the same for all
k and is not zero.
Quote:  What I am really interested is what L^p(R) spaces is f in ?

Certainly not for p <= 2. As f is bounded, if it is in
any L^p, it is in all higher ones, and if it were in L^2,
the measure of which it is a transform would have to have
a density.
Quote:  Any comments or suggestion would be greatly appreciated


This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Thu Jul 20, 2006 5:04 pm Post subject:
Re: Fourier Transform / Decay / Canot set



don wrote:
Quote:  New question:
How about if I use a fat Cantor set instead ?
Since the measure is "smoother " I would expect the transform to have better decay.
Can I expect this to be in some L^p space ? (p < inf.)

If I understand you correctly, you're now taking the Fourier transform
of a
constant times the indicator function I_C of your "fat Cantor set".
Since I_C
is in L^1 and L^2, by the HausdorffYoung theorem its Fourier transform
is
in L^p for 2 <= p <= infinity. But actually more can be said. Suppose
your
"fat Cantor set" is obtained by starting with an interval C_0 of length
s_0 = 1
and removing open subintervals; at the n'th stage you remove 2^{n1}
middle
subintervals, each of length r_n < s_{n1}, from C_{n1}, and what
remains,
C_n, is the union of 2^n subintervals, each of length s_n, where
2 s_n + r_n = s_{n1}.
Thus the Lebesgue measure of C_n is
m(C_n) = 2^n s_n = (1  r_n/s_{n1}) m(C_{n1}).
The intersection of all C_n is C, with
m(C) = product_{n=1}^infty (1  r_n/s_{n1})
and for it to be "fat" you want this infinite product to be nonzero,
which is
the case iff sum_{n=1}^infty r_n/s_{n1} < infty. The simplest way is
to take r_n/s_{n1} = w^n for some constant w, 0 < w < 1.
Now the Fourier transform of the indicator function of an interval, say
(a,b) is
g_(a,b)(x) = (exp(iax)  exp(ibx)) i/x
which is in all L^p for p > 1. Noting that g_(a,b){(x) <= min(ba,
2/x),
g_(a,b)_p <= c (ba)^(11/p) where c is a constant (depending only
on p).
Now since C_n \ C_{n1} consists of 2^{n1} intervals of length r_n,
the Fourier
transform of its indicator function has L^p norm at most
2^{n1} c r_n^(11/p). The L^p norm of the Fourier transform of I_C is
then
bounded by
c + sum_{n=1}^infty 2^(n1) c r_n^(11/p).
Now if r_n/s_{n1} = w^n we have r_n < w^n/2^(n1), so this L^p norm is
bounded by c (1 + sum_{n=1}^infty 2^((n1)/p) w^(n(11/p)))
which converges if w < 2^(1/(1p)). So in this case, the Fourier
transform
is in L^p at least for infinity >= p > (ln(w)  ln(2))/ln(w) > 1.
However, if instead of making w constant we make r_n decrease rapidly
enough that (r_n/s_{n1})^(1/n) > 0+, we should get the Fourier
transform
in L^p for all p in (1,infty].
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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David C. Ullrich science forum Guru
Joined: 28 Apr 2005
Posts: 2250

Posted: Thu Jul 20, 2006 10:53 am Post subject:
Re: Fourier Transform / Decay / Canot set



On Wed, 19 Jul 2006 17:41:09 EDT, don <ctcowan@sfu.ca> wrote:
Quote:  let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual Cantor set (middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f as x > infinity ?

None. For example if N is a positive integer then
f(3^N 2 pi) = f(2 pi) <> 0.
Quote:  What I am really interested is what L^p(R) spaces is f in ?

None except for p = infinity. This follows from the above, plus
the fact that f' is bounded (either differentiate the formula
above or note that f' is also the Fourier transform of a
certain (complex) measure.)
Quote:  Any comments or suggestion would be greatly appreciated
craig

************************
David C. Ullrich 

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don11154 science forum beginner
Joined: 05 Jul 2005
Posts: 39

Posted: Thu Jul 20, 2006 5:59 am Post subject:
Re: Fourier Transform / Decay / Canot set



Quote:  In article
13869322.1153345299269.JavaMail.jakarta@nitrogen.math
forum.org>,
don <ctcowan@sfu.ca> wrote:
let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for
real x.
This is the Fourier transform of a measure supported
on the usual Cantor set
(middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f
as x > infinity ?
Note that f(3x) = cos(x) f(x). In particular, f(3^n
pi) = (1)^n f(pi)
(and that is nonzero) for every positive integer n,
so there is no
decay.
But more can be said.
ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(nk) pi +
+ t/3^k)/cos(3^(nk) pi))
= sum_{k=1}^infty ln(cos(t/3^k)  sin(t/3^k)
k) tan(3^(nk) pi))
The terms for k <= n are of course ln(cos(t/3^k)),
while all terms
are dominated by c/3^k for some c (depending on t),
so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) > sum_{k=1}^infty
ln(cos(t/3^k)) = ln(f(t))
i.e. (1)^n f(3^n pi + t) > f(t) f(pi) as n > infty
(and this should be uniform on, say, t in [1,1]).
Therefore...
What I am really interested is what L^p(R) spaces
is f in ?
... None except L^infty.
John Schutkeker's shoe is safe.
Robert Israel
israel@math.ubc.ca
Department of Mathematics
http://www.math.ubc.ca/~israel
University of British Columbia Vancouver,
BC, Canada

thanks for the help.
New question:
How about if I use a fat Cantor set instead ?
Since the measure is "smoother " I would expect the transform to have better decay.
Can I expect this to be in some L^p space ? (p < inf.)
thanks
craig 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Thu Jul 20, 2006 4:13 am Post subject:
Re: Fourier Transform / Decay / Canot set



In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
Quote:  let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual Cantor set
(middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f as x > infinity ?

Note that f(3x) = cos(x) f(x). In particular, f(3^n pi) = (1)^n f(pi)
(and that is nonzero) for every positive integer n, so there is no
decay.
But more can be said.
ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(nk) pi + t/3^k)/cos(3^(nk) pi))
= sum_{k=1}^infty ln(cos(t/3^k)  sin(t/3^k) tan(3^(nk) pi))
The terms for k <= n are of course ln(cos(t/3^k)), while all terms
are dominated by c/3^k for some c (depending on t), so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) > sum_{k=1}^infty ln(cos(t/3^k)) = ln(f(t))
i.e. (1)^n f(3^n pi + t) > f(t) f(pi) as n > infty
(and this should be uniform on, say, t in [1,1]).
Therefore...
Quote:  What I am really interested is what L^p(R) spaces is f in ?

.... None except L^infty.
John Schutkeker's shoe is safe.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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John Schutkeker science forum Guru Wannabe
Joined: 30 May 2005
Posts: 172

Posted: Thu Jul 20, 2006 12:11 am Post subject:
Re: Fourier Transform / Decay / Canot set



don <ctcowan@sfu.ca> wrote in
news:13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org:
Quote:  let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual
Cantor set (middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f as x > infinity
?
Any comments or suggestion would be greatly appreciated

If you get any kind of decay in f, as x > infinity, I'll eat my shoe.
Taking the Fourier transform of the Cantor set is asking for trouble. 

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don11154 science forum beginner
Joined: 05 Jul 2005
Posts: 39

Posted: Wed Jul 19, 2006 9:41 pm Post subject:
Fourier Transform / Decay / Canot set



let
f(x) = product_{k=1}^infinity cos(x 3^{k} ) for real x.
This is the Fourier transform of a measure supported on the usual Cantor set (middle third) but translated so symmetric about 0.
The question is what kind of decay can I expect in f as x > infinity ?
What I am really interested is what L^p(R) spaces is f in ?
Any comments or suggestion would be greatly appreciated
craig 

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