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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_n-y_n| is divergent?

Thanks
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_n-y_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.

 Quote: Thanks

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

David C. Ullrich
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 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_n-y_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.
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

 Quote: In article <05bq92hk6amru58bs5dnruii57d7shmf0k@4ax.com>, David C. Ullrich wrote: On Fri, 23 Jun 2006 18:52:53 EDT, eugene 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_n-y_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 non-elementary.

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
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 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_n-y_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 non-elementary. 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:

> David C. Ullrich
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 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_n-y_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
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 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_n-y_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).
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 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_n-y_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 non-elementary. 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

(or something similar) with no >> signs. See the first words in
automatically for you. It will help if you figure out how to do
this properly.

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