Author 
Message 
eugene science forum Guru
Joined: 24 Nov 2005
Posts: 331

Posted: Fri Jun 23, 2006 10:52 pm Post subject:
vectors from l_2 and their difference isn't in l_1+uncountable



Is it possible to choose uncountable set A \in l_2 of
elements with unit norm such that for any distinct
x = (x_1,x_2,...,x_n,...),y = (y_1,y_2,...,y_n,...),from
A the series sum_{n=1^infty} x_ny_n is divergent?
Thanks 

Back to top 


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

Posted: Sat Jun 24, 2006 12:24 pm Post subject:
Re: vectors from l_2 and their difference isn't in l_1+uncountable



On Fri, 23 Jun 2006 18:52:53 EDT, eugene <jane1806@mail.ru> wrote:
Quote:  Is it possible to choose uncountable set A \in l_2 of
elements with unit norm such that for any distinct
x = (x_1,x_2,...,x_n,...),y = (y_1,y_2,...,y_n,...),from
A the series sum_{n=1^infty} x_ny_n is divergent?

Yes.
Notation: If x is in l_2 I'm going to write
x = (x[1], x[2], ...)
instead of x = (x_1, ...), so that I can use the
notation x_1, x_2,... for a sequence of elements
of l_2.
Start by S_1, S_2, ..., a countable collection of
disjoint sets of natural numbers, with each S_n infinite.
For each n, choose a sequence x_n such that
x_n[j] = 0 for j not in S_n
x_n[j] > 0 for j in S_n
x_n_2 < 2^(n)
x_n_1 = infinity.
Now for each set E of positive integers, let
y_E = sum_{n in E} x_n,
and then let
z_E = y_E /y_E_2.
There are uncountably many z_E's, they all have
l_2 norm 1, and if E <> E' then there exists n in E
but not in E' (or the other way around), and that
shows that
z_E  z_E'_1 >= sum_{j in S_n} x_n[j] = infinity.
************************
David C. Ullrich 

Back to top 


The World Wide Wade science forum Guru
Joined: 24 Mar 2005
Posts: 790

Posted: Sat Jun 24, 2006 8:39 pm Post subject:
Re: vectors from l_2 and their difference isn't in l_1+uncountable



In article <05bq92hk6amru58bs5dnruii57d7shmf0k@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
Quote:  On Fri, 23 Jun 2006 18:52:53 EDT, eugene <jane1806@mail.ru> wrote:
Is it possible to choose uncountable set A \in l_2 of
elements with unit norm such that for any distinct
x = (x_1,x_2,...,x_n,...),y = (y_1,y_2,...,y_n,...),from
A the series sum_{n=1^infty} x_ny_n is divergent?
Yes.
Notation: If x is in l_2 I'm going to write
x = (x[1], x[2], ...)
instead of x = (x_1, ...), so that I can use the
notation x_1, x_2,... for a sequence of elements
of l_2.
Start by S_1, S_2, ..., a countable collection of
disjoint sets of natural numbers, with each S_n infinite.
For each n, choose a sequence x_n such that
x_n[j] = 0 for j not in S_n
x_n[j] > 0 for j in S_n
x_n_2 < 2^(n)
x_n_1 = infinity.
Now for each set E of positive integers, let
y_E = sum_{n in E} x_n,
and then let
z_E = y_E /y_E_2.
There are uncountably many z_E's, they all have
l_2 norm 1, and if E <> E' then there exists n in E
but not in E' (or the other way around), and that
shows that
z_E  z_E'_1 >= sum_{j in S_n} x_n[j] = infinity.

Nice and elementary.
Less elementary but perhaps quicker: identify L^2 = L^2[0,2Pi]
with l^2. For 0 < a < 2Pi, let f_a be the characteristic function
of (0,a), divided by sqrt(a). Then {f_a} is an uncountable set of
unit vectors in L^2. If a < b, then sum_(n=oo,oo) (f_b 
f_a)^(n) = oo, otherwise f_b  f_a would be continuous, which it
isn't. Here f^(n) is the nth Fourier coefficient of f. 

Back to top 


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

Posted: Sun Jun 25, 2006 11:25 am Post subject:
Re: vectors from l_2 and their difference isn't in l_1+uncountable



On Sat, 24 Jun 2006 13:39:30 0700, The World Wide Wade
<waderameyxiii@comcast.remove13.net> wrote:
Quote:  In article <05bq92hk6amru58bs5dnruii57d7shmf0k@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
On Fri, 23 Jun 2006 18:52:53 EDT, eugene <jane1806@mail.ru> wrote:
Is it possible to choose uncountable set A \in l_2 of
elements with unit norm such that for any distinct
x = (x_1,x_2,...,x_n,...),y = (y_1,y_2,...,y_n,...),from
A the series sum_{n=1^infty} x_ny_n is divergent?
Yes.
Notation: If x is in l_2 I'm going to write
x = (x[1], x[2], ...)
instead of x = (x_1, ...), so that I can use the
notation x_1, x_2,... for a sequence of elements
of l_2.
Start by S_1, S_2, ..., a countable collection of
disjoint sets of natural numbers, with each S_n infinite.
For each n, choose a sequence x_n such that
x_n[j] = 0 for j not in S_n
x_n[j] > 0 for j in S_n
x_n_2 < 2^(n)
x_n_1 = infinity.
Now for each set E of positive integers, let
y_E = sum_{n in E} x_n,
and then let
z_E = y_E /y_E_2.
There are uncountably many z_E's, they all have
l_2 norm 1, and if E <> E' then there exists n in E
but not in E' (or the other way around), and that
shows that
z_E  z_E'_1 >= sum_{j in S_n} x_n[j] = infinity.
Nice and elementary.
Less elementary but perhaps quicker: identify L^2 = L^2[0,2Pi]
with l^2. For 0 < a < 2Pi, let f_a be the characteristic function
of (0,a), divided by sqrt(a). Then {f_a} is an uncountable set of
unit vectors in L^2. If a < b, then sum_(n=oo,oo) (f_b 
f_a)^(n) = oo, otherwise f_b  f_a would be continuous, which it
isn't. Here f^(n) is the nth Fourier coefficient of f.

Ah  that's nice too, and not all that nonelementary.
At the words "less elementary" I thought you were going to
invoke the existence of an uncountable set of sets of integers,
any two of which have finite intersection. The result follows
from that, but _that's_ not elementary.
************************
David C. Ullrich 

Back to top 


eugene science forum Guru
Joined: 24 Nov 2005
Posts: 331

Posted: Mon Jun 26, 2006 9:50 am Post subject:
Re: vectors from l_2 and their difference isn't in l_1+uncountable



Quote:  On Sat, 24 Jun 2006 13:39:30 0700, The World Wide
Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
05bq92hk6amru58bs5dnruii57d7shmf0k@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
On Fri, 23 Jun 2006 18:52:53 EDT, eugene
jane1806@mail.ru> wrote:
Is it possible to choose uncountable set A \in
l_2 of
elements with unit norm such that for any
distinct
x = (x_1,x_2,...,x_n,...),y =
(y_1,y_2,...,y_n,...),from
A the series sum_{n=1^infty} x_ny_n is
divergent?
Yes.
Notation: If x is in l_2 I'm going to write
x = (x[1], x[2], ...)
instead of x = (x_1, ...), so that I can use the
notation x_1, x_2,... for a sequence of elements
of l_2.
Start by S_1, S_2, ..., a countable collection of
disjoint sets of natural numbers, with each S_n
infinite.
For each n, choose a sequence x_n such that
x_n[j] = 0 for j not in S_n
x_n[j] > 0 for j in S_n
x_n_2 < 2^(n)
x_n_1 = infinity.
Now for each set E of positive integers, let
y_E = sum_{n in E} x_n,
and then let
z_E = y_E /y_E_2.
There are uncountably many z_E's, they all have
l_2 norm 1, and if E <> E' then there exists n in
E
but not in E' (or the other way around), and that
shows that
z_E  z_E'_1 >= sum_{j in S_n} x_n[j] =
infinity.
Nice and elementary.
Less elementary but perhaps quicker: identify L^2 =
L^2[0,2Pi]
with l^2. For 0 < a < 2Pi, let f_a be the
characteristic function
of (0,a), divided by sqrt(a). Then {f_a} is an
uncountable set of
unit vectors in L^2. If a < b, then sum_(n=oo,oo)
(f_b 
f_a)^(n) = oo, otherwise f_b  f_a would be
continuous, which it
isn't. Here f^(n) is the nth Fourier coefficient of
f.
Ah  that's nice too, and not all that
nonelementary.
At the words "less elementary" I thought you were
going to
invoke the existence of an uncountable set of sets of
integers,
any two of which have finite intersection. The result
follows
from that, but _that's_ not elementary.

Thanks for you solution. Meanwhile, i thought about this solution you meant with this result that there exist an uncountable collection of sets of N any two of which have finite intersection: here is the thread where the proofs where given:
http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1403177&messageID=4841466#4841466
> David C. Ullrich 

Back to top 


eugene science forum Guru
Joined: 24 Nov 2005
Posts: 331

Posted: Mon Jun 26, 2006 9:53 am Post subject:
Re: vectors from l_2 and their difference isn't in l_1+uncountable



Quote:  In article
05bq92hk6amru58bs5dnruii57d7shmf0k@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
On Fri, 23 Jun 2006 18:52:53 EDT, eugene
jane1806@mail.ru> wrote:
Is it possible to choose uncountable set A \in l_2
of
elements with unit norm such that for any distinct
x = (x_1,x_2,...,x_n,...),y =
(y_1,y_2,...,y_n,...),from
A the series sum_{n=1^infty} x_ny_n is
divergent?
Yes.
Notation: If x is in l_2 I'm going to write
x = (x[1], x[2], ...)
instead of x = (x_1, ...), so that I can use the
notation x_1, x_2,... for a sequence of elements
of l_2.
Start by S_1, S_2, ..., a countable collection of
disjoint sets of natural numbers, with each S_n
infinite.
For each n, choose a sequence x_n such that
x_n[j] = 0 for j not in S_n
x_n[j] > 0 for j in S_n
x_n_2 < 2^(n)
x_n_1 = infinity.
Now for each set E of positive integers, let
y_E = sum_{n in E} x_n,
and then let
z_E = y_E /y_E_2.
There are uncountably many z_E's, they all have
l_2 norm 1, and if E <> E' then there exists n in E
but not in E' (or the other way around), and that
shows that
z_E  z_E'_1 >= sum_{j in S_n} x_n[j] =
infinity.
Nice and elementary.
Less elementary but perhaps quicker: identify L^2 =
L^2[0,2Pi]
with l^2. For 0 < a < 2Pi, let f_a be the
characteristic function
of (0,a), divided by sqrt(a). Then {f_a} is an
uncountable set of
unit vectors in L^2. If a < b, then sum_(n=oo,oo)
(f_b 
f_a)^(n) = oo, otherwise f_b  f_a would be
continuous, which it
isn't. Here f^(n) is the nth Fourier coefficient of
f.

Thanks a lot nice idea. But i;m a bit vaguous whether i completely understand your solution:
In the proof you probably used this fact:

If f is in L_2[0, 2Pi] and
sum_[n=infty,  infty] c_n < + infty, then
f is continious. There c_n is the n'th Fourier coefficient of f.

Did you mean it? Is it always true. I'm probably missing something obvious, but why is it true ?
Thanks 

Back to top 


eugene science forum Guru
Joined: 24 Nov 2005
Posts: 331

Posted: Mon Jun 26, 2006 10:16 am Post subject:
Re: vectors from l_2 and their difference isn't in l_1+uncountable



Quote:  In article
05bq92hk6amru58bs5dnruii57d7shmf0k@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu
wrote:
On Fri, 23 Jun 2006 18:52:53 EDT, eugene
jane1806@mail.ru> wrote:
Is it possible to choose uncountable set A \in
l_2
of
elements with unit norm such that for any
distinct
x = (x_1,x_2,...,x_n,...),y =
(y_1,y_2,...,y_n,...),from
A the series sum_{n=1^infty} x_ny_n is
divergent?
Yes.
Notation: If x is in l_2 I'm going to write
x = (x[1], x[2], ...)
instead of x = (x_1, ...), so that I can use the
notation x_1, x_2,... for a sequence of elements
of l_2.
Start by S_1, S_2, ..., a countable collection
of
disjoint sets of natural numbers, with each S_n
infinite.
For each n, choose a sequence x_n such that
x_n[j] = 0 for j not in S_n
x_n[j] > 0 for j in S_n
x_n_2 < 2^(n)
x_n_1 = infinity.
Now for each set E of positive integers, let
y_E = sum_{n in E} x_n,
and then let
z_E = y_E /y_E_2.
There are uncountably many z_E's, they all have
l_2 norm 1, and if E <> E' then there exists n in
E
but not in E' (or the other way around), and
that
shows that
z_E  z_E'_1 >= sum_{j in S_n} x_n[j] =
infinity.
Nice and elementary.
Less elementary but perhaps quicker: identify L^2
=
L^2[0,2Pi]
with l^2. For 0 < a < 2Pi, let f_a be the
characteristic function
of (0,a), divided by sqrt(a). Then {f_a} is an
uncountable set of
unit vectors in L^2. If a < b, then sum_(n=oo,oo)
(f_b 
f_a)^(n) = oo, otherwise f_b  f_a would be
continuous, which it
isn't. Here f^(n) is the nth Fourier coefficient
of
f.
Thanks a lot nice idea. But i;m a bit vaguous
whether i completely understand your solution:
In the proof you probably used this fact:

If f is in L_2[0, 2Pi] and
sum_[n=infty,  infty] c_n < + infty, then
f is continious. There c_n is the n'th Fourier
coefficient of f.

Did you mean it? Is it always true. I'm probably
missing something obvious, but why is it true ?
Thanks

Oops, sorry fot the stupid question. It is really obvious:
If sum_[n=infty,  infty] c_n < + infty, then the Fourier series of f converges uniformly to some continious g which Fourier coefficients are the same as of f and thus f = g ( L_2[0, 2Pi] is complete space). 

Back to top 


The World Wide Wade science forum Guru
Joined: 24 Mar 2005
Posts: 790

Posted: Mon Jun 26, 2006 6:28 pm Post subject:
Re: vectors from l_2 and their difference isn't in l_1+uncountable



In article
<4406831.1151315431453.JavaMail.jakarta@nitrogen.mathforum.org>,
eugene <jane1806@mail.ru> wrote:
Quote:  On Sat, 24 Jun 2006 13:39:30 0700, The World Wide
Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
05bq92hk6amru58bs5dnruii57d7shmf0k@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
On Fri, 23 Jun 2006 18:52:53 EDT, eugene
jane1806@mail.ru> wrote:
Is it possible to choose uncountable set A \in
l_2 of
elements with unit norm such that for any
distinct
x = (x_1,x_2,...,x_n,...),y =
(y_1,y_2,...,y_n,...),from
A the series sum_{n=1^infty} x_ny_n is
divergent?
Yes.
Notation: If x is in l_2 I'm going to write
x = (x[1], x[2], ...)
instead of x = (x_1, ...), so that I can use the
notation x_1, x_2,... for a sequence of elements
of l_2.
Start by S_1, S_2, ..., a countable collection of
disjoint sets of natural numbers, with each S_n
infinite.
For each n, choose a sequence x_n such that
x_n[j] = 0 for j not in S_n
x_n[j] > 0 for j in S_n
x_n_2 < 2^(n)
x_n_1 = infinity.
Now for each set E of positive integers, let
y_E = sum_{n in E} x_n,
and then let
z_E = y_E /y_E_2.
There are uncountably many z_E's, they all have
l_2 norm 1, and if E <> E' then there exists n in
E
but not in E' (or the other way around), and that
shows that
z_E  z_E'_1 >= sum_{j in S_n} x_n[j] =
infinity.
Nice and elementary.
Less elementary but perhaps quicker: identify L^2 =
L^2[0,2Pi]
with l^2. For 0 < a < 2Pi, let f_a be the
characteristic function
of (0,a), divided by sqrt(a). Then {f_a} is an
uncountable set of
unit vectors in L^2. If a < b, then sum_(n=oo,oo)
(f_b 
f_a)^(n) = oo, otherwise f_b  f_a would be
continuous, which it
isn't. Here f^(n) is the nth Fourier coefficient of
f.
Ah  that's nice too, and not all that
nonelementary.
At the words "less elementary" I thought you were
going to
invoke the existence of an uncountable set of sets of
integers,
any two of which have finite intersection. The result
follows
from that, but _that's_ not elementary.
Thanks for you solution. Meanwhile, i thought about this solution you meant
with this result that there exist an uncountable collection of sets of N any
two of which have finite intersection: here is the thread where the proofs
where given:
http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1403177&messageID=4841
466#4841466
David C. Ullrich

Eugene, if you reply to me, the first words should be
In article
<waderameyxiii4589C3.13392924062006@comcast.dca.giganews.com>,
The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote:
(or something similar) with no >> signs. See the first words in
my reply here for example. Most newsreaders will do this
automatically for you. It will help if you figure out how to do
this properly. 

Back to top 


Google


Back to top 



The time now is Thu Jan 24, 2019 12:52 pm  All times are GMT

