$A = { x = { x_n } vert sum_{n=1}^{infty} vert x_{n+1} - x_n vert < infty }$
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Let $A = { x = { x_n } in mathbb{R} vert sum_{n=1}^{infty} vert x_{n+1} - x_n vert < infty }$ with norm:
$$ Vert x Vert = vert x_1 vert + sum_{n=1}^{infty} vert x_{n+1} - x_n vert $$
Now, this is norm and I proved it. I want to say that A is a Banach space, so I want to check if it is complete with this norm.
I pick $x_n$ Cauchy sequence in A.
Then I have ${x_{n,j}}_{j=1}^infty$. Fixing n, I have that ${x_j}$ is Cauchy in $mathbb{R}$, so it converges. I can then say that my limit candidate is as such:
$$ exists lim_{nto infty} x_{n,j} = x_j$$
I call $x = { x_j }_{j=1}^{infty}$. I want now to prove that:
$ Vert x_n - x Vert to 0 $ as n $to infty $ in the norm defined before.
I start writing:
$$ Vert x_n - x Vert = vert x_{n,1} - x_1 vert + sum_{n=1}^infty vert x_{n+1,j} + x_{j+1} - x_{n,j} - x_j vert $$
But now I am stuck, I do not know how to use epsilons to make it go to 0. I think that up to here it is all right though.
Any help?
functional-analysis
asked Nov 20 at 21:00
qcc101
458113
|
show 1 more comment
up vote
1
down vote
favorite
Let $A = { x = { x_n } in mathbb{R} vert sum_{n=1}^{infty} vert x_{n+1} - x_n vert < infty }$ with norm:
$$ Vert x Vert = vert x_1 vert + sum_{n=1}^{infty} vert x_{n+1} - x_n vert $$
Now, this is norm and I proved it. I want to say that A is a Banach space, so I want to check if it is complete with this norm.
I pick $x_n$ Cauchy sequence in A.
Then I have ${x_{n,j}}_{j=1}^infty$. Fixing n, I have that ${x_j}$ is Cauchy in $mathbb{R}$, so it converges. I can then say that my limit candidate is as such:
$$ exists lim_{nto infty} x_{n,j} = x_j$$
I call $x = { x_j }_{j=1}^{infty}$. I want now to prove that:
$ Vert x_n - x Vert to 0 $ as n $to infty $ in the norm defined before.
I start writing:
$$ Vert x_n - x Vert = vert x_{n,1} - x_1 vert + sum_{n=1}^infty vert x_{n+1,j} + x_{j+1} - x_{n,j} - x_j vert $$
But now I am stuck, I do not know how to use epsilons to make it go to 0. I think that up to here it is all right though.
Any help?
functional-analysis
asked Nov 20 at 21:00
qcc101
458113
I think there is a potential mixup in subscripts going on here. $x_n$ seems to represent both an element of $A$ and an element of a sequence in $A$.
– Paul
Nov 20 at 21:04
I added the subscript j when I want to say the j-th element of the sequence $x_n$. Did I miss it somewhere?
– qcc101
Nov 20 at 21:07
I was looking at the definition of the norm, which has it there.
– Paul
Nov 20 at 21:08
Yes, but then when I say let $x_n$ be a Cauchy sequence I mean another n. The one in the norm is just an example. I do not know if I have explained myself correctly.
– qcc101
Nov 20 at 21:11
Shouldn't the last sum be over $j$?
– Marco
Nov 20 at 21:38
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A = { x = { x_n } in mathbb{R} vert sum_{n=1}^{infty} vert x_{n+1} - x_n vert < infty }$ with norm:
$$ Vert x Vert = vert x_1 vert + sum_{n=1}^{infty} vert x_{n+1} - x_n vert $$
Now, this is norm and I proved it. I want to say that A is a Banach space, so I want to check if it is complete with this norm.
I pick $x_n$ Cauchy sequence in A.
Then I have ${x_{n,j}}_{j=1}^infty$. Fixing n, I have that ${x_j}$ is Cauchy in $mathbb{R}$, so it converges. I can then say that my limit candidate is as such:
$$ exists lim_{nto infty} x_{n,j} = x_j$$
I call $x = { x_j }_{j=1}^{infty}$. I want now to prove that:
$ Vert x_n - x Vert to 0 $ as n $to infty $ in the norm defined before.
I start writing:
$$ Vert x_n - x Vert = vert x_{n,1} - x_1 vert + sum_{n=1}^infty vert x_{n+1,j} + x_{j+1} - x_{n,j} - x_j vert $$
But now I am stuck, I do not know how to use epsilons to make it go to 0. I think that up to here it is all right though.
Any help?
functional-analysis
asked Nov 20 at 21:00
qcc101
458113
Let $A = { x = { x_n } in mathbb{R} vert sum_{n=1}^{infty} vert x_{n+1} - x_n vert < infty }$ with norm:
$$ Vert x Vert = vert x_1 vert + sum_{n=1}^{infty} vert x_{n+1} - x_n vert $$
Now, this is norm and I proved it. I want to say that A is a Banach space, so I want to check if it is complete with this norm.
I pick $x_n$ Cauchy sequence in A.
Then I have ${x_{n,j}}_{j=1}^infty$. Fixing n, I have that ${x_j}$ is Cauchy in $mathbb{R}$, so it converges. I can then say that my limit candidate is as such:
$$ exists lim_{nto infty} x_{n,j} = x_j$$
I call $x = { x_j }_{j=1}^{infty}$. I want now to prove that:
$ Vert x_n - x Vert to 0 $ as n $to infty $ in the norm defined before.
I start writing:
$$ Vert x_n - x Vert = vert x_{n,1} - x_1 vert + sum_{n=1}^infty vert x_{n+1,j} + x_{j+1} - x_{n,j} - x_j vert $$
But now I am stuck, I do not know how to use epsilons to make it go to 0. I think that up to here it is all right though.
Any help?
functional-analysis
functional-analysis
asked Nov 20 at 21:00
qcc101
458113
asked Nov 20 at 21:00
qcc101
458113
asked Nov 20 at 21:00
qcc101
458113
asked Nov 20 at 21:00
qcc101
458113
asked Nov 20 at 21:00
qcc101
458113
458113
I think there is a potential mixup in subscripts going on here. $x_n$ seems to represent both an element of $A$ and an element of a sequence in $A$.
– Paul
Nov 20 at 21:04
I added the subscript j when I want to say the j-th element of the sequence $x_n$. Did I miss it somewhere?
– qcc101
Nov 20 at 21:07
I was looking at the definition of the norm, which has it there.
– Paul
Nov 20 at 21:08
Yes, but then when I say let $x_n$ be a Cauchy sequence I mean another n. The one in the norm is just an example. I do not know if I have explained myself correctly.
– qcc101
Nov 20 at 21:11
Shouldn't the last sum be over $j$?
– Marco
Nov 20 at 21:38
|
show 1 more comment
I think there is a potential mixup in subscripts going on here. $x_n$ seems to represent both an element of $A$ and an element of a sequence in $A$.
– Paul
Nov 20 at 21:04
I added the subscript j when I want to say the j-th element of the sequence $x_n$. Did I miss it somewhere?
– qcc101
Nov 20 at 21:07
I was looking at the definition of the norm, which has it there.
– Paul
Nov 20 at 21:08
Yes, but then when I say let $x_n$ be a Cauchy sequence I mean another n. The one in the norm is just an example. I do not know if I have explained myself correctly.
– qcc101
Nov 20 at 21:11
Shouldn't the last sum be over $j$?
– Marco
Nov 20 at 21:38
I think there is a potential mixup in subscripts going on here. $x_n$ seems to represent both an element of $A$ and an element of a sequence in $A$.
– Paul
Nov 20 at 21:04
I think there is a potential mixup in subscripts going on here. $x_n$ seems to represent both an element of $A$ and an element of a sequence in $A$.
– Paul
Nov 20 at 21:04
I added the subscript j when I want to say the j-th element of the sequence $x_n$. Did I miss it somewhere?
– qcc101
Nov 20 at 21:07
I added the subscript j when I want to say the j-th element of the sequence $x_n$. Did I miss it somewhere?
– qcc101
Nov 20 at 21:07
I was looking at the definition of the norm, which has it there.
– Paul
Nov 20 at 21:08
I was looking at the definition of the norm, which has it there.
– Paul
Nov 20 at 21:08
Yes, but then when I say let $x_n$ be a Cauchy sequence I mean another n. The one in the norm is just an example. I do not know if I have explained myself correctly.
– qcc101
Nov 20 at 21:11
Yes, but then when I say let $x_n$ be a Cauchy sequence I mean another n. The one in the norm is just an example. I do not know if I have explained myself correctly.
– qcc101
Nov 20 at 21:11
Shouldn't the last sum be over $j$?
– Marco
Nov 20 at 21:38
Shouldn't the last sum be over $j$?
– Marco
Nov 20 at 21:38
|
show 1 more comment
1 Answer
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votes
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2
down vote
accepted
You have mixed up the notations. First verify that $|x_{n,j}-x_{m,j}| leq |x_n-x_m|$ to show that $x_j=lim x_{n,j}$ actually exists for each $j$. Now let $epsilon >0$ and choose $k$ such that $|x_{n,1}-x_{m,1}|+sum_{j=1}^{infty} |x_{n,j+1}-x_{n,j}-x_{m,j+1}+x_{m,j}| <epsilon$ for $n,m geq k$. In this inequality let $m to infty$ and you get $|x_n-x| leq epsilon$ for $n geq k$ where $x=(x_1,x_2,cdots)$. [ In taking limit as $ m to infty$ yuo can take finite sums, say $j$ from $1$ to $N$, take the limit w.r.t. $m$ and then let $N to infty$].
answered Nov 20 at 23:58
Kavi Rama Murthy
47.1k31854
add a comment |
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1 Answer
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1 Answer
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active
oldest
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up vote
2
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accepted
You have mixed up the notations. First verify that $|x_{n,j}-x_{m,j}| leq |x_n-x_m|$ to show that $x_j=lim x_{n,j}$ actually exists for each $j$. Now let $epsilon >0$ and choose $k$ such that $|x_{n,1}-x_{m,1}|+sum_{j=1}^{infty} |x_{n,j+1}-x_{n,j}-x_{m,j+1}+x_{m,j}| <epsilon$ for $n,m geq k$. In this inequality let $m to infty$ and you get $|x_n-x| leq epsilon$ for $n geq k$ where $x=(x_1,x_2,cdots)$. [ In taking limit as $ m to infty$ yuo can take finite sums, say $j$ from $1$ to $N$, take the limit w.r.t. $m$ and then let $N to infty$].
answered Nov 20 at 23:58
Kavi Rama Murthy
47.1k31854
add a comment |
up vote
2
down vote
accepted
You have mixed up the notations. First verify that $|x_{n,j}-x_{m,j}| leq |x_n-x_m|$ to show that $x_j=lim x_{n,j}$ actually exists for each $j$. Now let $epsilon >0$ and choose $k$ such that $|x_{n,1}-x_{m,1}|+sum_{j=1}^{infty} |x_{n,j+1}-x_{n,j}-x_{m,j+1}+x_{m,j}| <epsilon$ for $n,m geq k$. In this inequality let $m to infty$ and you get $|x_n-x| leq epsilon$ for $n geq k$ where $x=(x_1,x_2,cdots)$. [ In taking limit as $ m to infty$ yuo can take finite sums, say $j$ from $1$ to $N$, take the limit w.r.t. $m$ and then let $N to infty$].
answered Nov 20 at 23:58
Kavi Rama Murthy
47.1k31854
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have mixed up the notations. First verify that $|x_{n,j}-x_{m,j}| leq |x_n-x_m|$ to show that $x_j=lim x_{n,j}$ actually exists for each $j$. Now let $epsilon >0$ and choose $k$ such that $|x_{n,1}-x_{m,1}|+sum_{j=1}^{infty} |x_{n,j+1}-x_{n,j}-x_{m,j+1}+x_{m,j}| <epsilon$ for $n,m geq k$. In this inequality let $m to infty$ and you get $|x_n-x| leq epsilon$ for $n geq k$ where $x=(x_1,x_2,cdots)$. [ In taking limit as $ m to infty$ yuo can take finite sums, say $j$ from $1$ to $N$, take the limit w.r.t. $m$ and then let $N to infty$].
answered Nov 20 at 23:58
Kavi Rama Murthy
47.1k31854
You have mixed up the notations. First verify that $|x_{n,j}-x_{m,j}| leq |x_n-x_m|$ to show that $x_j=lim x_{n,j}$ actually exists for each $j$. Now let $epsilon >0$ and choose $k$ such that $|x_{n,1}-x_{m,1}|+sum_{j=1}^{infty} |x_{n,j+1}-x_{n,j}-x_{m,j+1}+x_{m,j}| <epsilon$ for $n,m geq k$. In this inequality let $m to infty$ and you get $|x_n-x| leq epsilon$ for $n geq k$ where $x=(x_1,x_2,cdots)$. [ In taking limit as $ m to infty$ yuo can take finite sums, say $j$ from $1$ to $N$, take the limit w.r.t. $m$ and then let $N to infty$].
answered Nov 20 at 23:58
Kavi Rama Murthy
47.1k31854
answered Nov 20 at 23:58
Kavi Rama Murthy
47.1k31854
answered Nov 20 at 23:58
Kavi Rama Murthy
47.1k31854
answered Nov 20 at 23:58
Kavi Rama Murthy
47.1k31854
47.1k31854
add a comment |
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I think there is a potential mixup in subscripts going on here. $x_n$ seems to represent both an element of $A$ and an element of a sequence in $A$.
– Paul
Nov 20 at 21:04
I added the subscript j when I want to say the j-th element of the sequence $x_n$. Did I miss it somewhere?
– qcc101
Nov 20 at 21:07
I was looking at the definition of the norm, which has it there.
– Paul
Nov 20 at 21:08
Yes, but then when I say let $x_n$ be a Cauchy sequence I mean another n. The one in the norm is just an example. I do not know if I have explained myself correctly.
– qcc101
Nov 20 at 21:11
Shouldn't the last sum be over $j$?
– Marco
Nov 20 at 21:38