$A = { x = { x_n } vert sum_{n=1}^{infty} vert x_{n+1} - x_n vert < infty }$











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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







share|cite|improve this question






















  • 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















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







share|cite|improve this question






















  • 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













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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










1 Answer
1






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oldest

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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 $N to infty$].






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    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 $N to infty$].






    share|cite|improve this answer

























      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 $N to infty$].






      share|cite|improve this answer























        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 $N to infty$].






        share|cite|improve this answer












        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 $N to infty$].







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 at 23:58









        Kavi Rama Murthy

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