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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 

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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 

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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. 

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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 

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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 

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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 

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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). 

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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. 

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