Mixing
Let $c = { x = {x_k}_{k=1}^{infty} in l^infty vert exists lim_{k to infty} x_k in mathbb{C} }$
Let $c = { x = {x_k}_{k=1}^{infty} in l^infty vert exists lim_{k to infty} x_k in mathbb{C} }$.
Let $x_n in c$, with $x_n to x = {x_k}$ with the sup norm.
I want to prove that $ x in c$.
So I am a bit stuck here, I want to use the fact that $x_n$ belongs to c, so it has a limit that I call $l_n$. But from now on I do not know how to keep going. Any help?
functional-analysis
asked Nov 20 '18 at 16:09
qcc101
458113
add a comment |
Let $c = { x = {x_k}_{k=1}^{infty} in l^infty vert exists lim_{k to infty} x_k in mathbb{C} }$.
Let $x_n in c$, with $x_n to x = {x_k}$ with the sup norm.
I want to prove that $ x in c$.
So I am a bit stuck here, I want to use the fact that $x_n$ belongs to c, so it has a limit that I call $l_n$. But from now on I do not know how to keep going. Any help?
functional-analysis
asked Nov 20 '18 at 16:09
qcc101
458113
4
You should improve your notation: in your question, $x_k$ is sometimes a real number and sometimes an element in $c$. Maybe you should use $x^{k}$ for one or the other, which would enable notation such as $x_n^{k}$.
– supinf
Nov 20 '18 at 16:13
add a comment |
Let $c = { x = {x_k}_{k=1}^{infty} in l^infty vert exists lim_{k to infty} x_k in mathbb{C} }$.
Let $x_n in c$, with $x_n to x = {x_k}$ with the sup norm.
I want to prove that $ x in c$.
So I am a bit stuck here, I want to use the fact that $x_n$ belongs to c, so it has a limit that I call $l_n$. But from now on I do not know how to keep going. Any help?
functional-analysis
asked Nov 20 '18 at 16:09
qcc101
458113
Let $c = { x = {x_k}_{k=1}^{infty} in l^infty vert exists lim_{k to infty} x_k in mathbb{C} }$.
Let $x_n in c$, with $x_n to x = {x_k}$ with the sup norm.
I want to prove that $ x in c$.
So I am a bit stuck here, I want to use the fact that $x_n$ belongs to c, so it has a limit that I call $l_n$. But from now on I do not know how to keep going. Any help?
functional-analysis
functional-analysis
asked Nov 20 '18 at 16:09
qcc101
458113
asked Nov 20 '18 at 16:09
qcc101
458113
asked Nov 20 '18 at 16:09
qcc101
458113
asked Nov 20 '18 at 16:09
qcc101
458113
asked Nov 20 '18 at 16:09
qcc101
458113
458113
4
You should improve your notation: in your question, $x_k$ is sometimes a real number and sometimes an element in $c$. Maybe you should use $x^{k}$ for one or the other, which would enable notation such as $x_n^{k}$.
– supinf
Nov 20 '18 at 16:13
add a comment |
4
You should improve your notation: in your question, $x_k$ is sometimes a real number and sometimes an element in $c$. Maybe you should use $x^{k}$ for one or the other, which would enable notation such as $x_n^{k}$.
– supinf
Nov 20 '18 at 16:13
4
4
You should improve your notation: in your question, $x_k$ is sometimes a real number and sometimes an element in $c$. Maybe you should use $x^{k}$ for one or the other, which would enable notation such as $x_n^{k}$.
– supinf
Nov 20 '18 at 16:13
You should improve your notation: in your question, $x_k$ is sometimes a real number and sometimes an element in $c$. Maybe you should use $x^{k}$ for one or the other, which would enable notation such as $x_n^{k}$.
– supinf
Nov 20 '18 at 16:13
add a comment |
2 Answers
2
active
oldest
votes
As said in the comments, you should improve your notation. It will help.
Let $(x_n)subset c$ with $x_n=(x_n^1,x_n^2,dots)$ for any $n$.
Let $x=(b_1,b_2,dots)$. It suffices to prove that $(b_k)$ is a Cauchy sequence.
We have that $|x_n-x|_inftyto0$, therefore $displaystyle{lim_{ntoinfty}sup_{kinmathbb{N}}|x_n^k-b_k|=0}$. Let $varepsilon>0$; then there exists $n_0inmathbb{N}$ such that for all $ngeq n_0$ and for all $kinmathbb{N}$ it is $|x_n^k-b_k|<varepsilon$. Now since $x_{n_0}in c$ it is a Cauchy sequence therefore for this $varepsilon$ there exists $Ngeq 0$ such that for all $m,ngeq N$ it is $|x_{n_0}^n-x_{n_0}^m|<varepsilon$.
Now we have for $m,ngeq N$: $|b_n-b_m|leq|b_n-x_{n_0}^n|+|x_{n_0}^n-x_{n_0}^m|+|x_{n_0}^m-b_m|<3varepsilon$.
Since for a random $varepsilon>0$ we have found an integer $N$ such that for all $m,n$ greater than $N$ it is $|b_n-b_m|<3varepsilon$, by the definition of a Cauchy sequence we have that $(b_n)$ is Cauchy.
answered Nov 20 '18 at 17:11
JustDroppedIn
1,764420
Good. I added a different style of proof.
– DanielWainfleet
Nov 20 '18 at 18:46
Why is it enough to prove that the sequence is Cauchy? Moreover, why $x_{n_0}$ is Cauchy?
– qcc101
Nov 20 '18 at 19:59
@qcc101 because a sequence of $mathbb{C}$ is convergent if-f it is Cauchy; $x_{n_0}in c$ and $c$ is by definition the space of convergent sequences.
– JustDroppedIn
Nov 20 '18 at 20:01
add a comment |
With the notation of the Answer given by JustDroppedIn.
Since you want to show that $c$ is closed in $l^{infty},$ you can prove that $l^{infty}setminus c$ is open, as follows:
Let $x in l^{infty}setminus c.$ Then $x$ is not a Cauchy sequence, so there exists $r>0,$ and functions $f:Bbb Nto Bbb N,, g:Bbb Nto Bbb N,$ both strictly increasing, such that $$forall kin Bbb N,(,|x^{f(k)}-x^{g(k)}|>r,).$$ Consider the open ball $B(x,r/3)={yin l^{infty}: |y-x|<r/3}.$ If $yin B(x,r/3)$ then for all $kin Bbb N$ we have $$|y^{f(k)}-y^{g(k)}|=$$ $$=|(y^{f(k)}-x^{f(k)})+(x^{f(k)}-x^{g(k)})+(x^{g(k)}-y^{g(k)})|geq$$ $$geq -|y^{f(k)}-x^{f(k)}|+|x^{f(k)}-x^{g(k)}|-|x^{g(k)}-y^{g(k)}|geq$$ $$geq -|y-x|+|x^{f(k)}-x^{g(k)}|-|x-y|>$$ $$>-r/3+r-r/3=r/3.$$ Since ${f(k):jin Bbb N}$ and ${g(k):kin Bbb N}$ are infinite sets, this implies that any $yin B(x,r/3)$ is not a Cauchy sequence. So $B(x,r/3) cap c =emptyset.$
The idea is that you cannot uniformly approximate a non-convergent sequence $x$ to an arbitrary degree by a convergent sequence because of the "$r$". For a bounded non-convergent sequence $x$ we can take $0<r<(lim sup_{jto infty} x^j)-(lim inf_{jto infty} x^j).$
answered Nov 20 '18 at 18:41
DanielWainfleet
34.1k31647
add a comment |
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2 Answers
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As said in the comments, you should improve your notation. It will help.
Let $(x_n)subset c$ with $x_n=(x_n^1,x_n^2,dots)$ for any $n$.
Let $x=(b_1,b_2,dots)$. It suffices to prove that $(b_k)$ is a Cauchy sequence.
We have that $|x_n-x|_inftyto0$, therefore $displaystyle{lim_{ntoinfty}sup_{kinmathbb{N}}|x_n^k-b_k|=0}$. Let $varepsilon>0$; then there exists $n_0inmathbb{N}$ such that for all $ngeq n_0$ and for all $kinmathbb{N}$ it is $|x_n^k-b_k|<varepsilon$. Now since $x_{n_0}in c$ it is a Cauchy sequence therefore for this $varepsilon$ there exists $Ngeq 0$ such that for all $m,ngeq N$ it is $|x_{n_0}^n-x_{n_0}^m|<varepsilon$.
Now we have for $m,ngeq N$: $|b_n-b_m|leq|b_n-x_{n_0}^n|+|x_{n_0}^n-x_{n_0}^m|+|x_{n_0}^m-b_m|<3varepsilon$.
Since for a random $varepsilon>0$ we have found an integer $N$ such that for all $m,n$ greater than $N$ it is $|b_n-b_m|<3varepsilon$, by the definition of a Cauchy sequence we have that $(b_n)$ is Cauchy.
answered Nov 20 '18 at 17:11
JustDroppedIn
1,764420
Good. I added a different style of proof.
– DanielWainfleet
Nov 20 '18 at 18:46
Why is it enough to prove that the sequence is Cauchy? Moreover, why $x_{n_0}$ is Cauchy?
– qcc101
Nov 20 '18 at 19:59
@qcc101 because a sequence of $mathbb{C}$ is convergent if-f it is Cauchy; $x_{n_0}in c$ and $c$ is by definition the space of convergent sequences.
– JustDroppedIn
Nov 20 '18 at 20:01
add a comment |
As said in the comments, you should improve your notation. It will help.
Let $(x_n)subset c$ with $x_n=(x_n^1,x_n^2,dots)$ for any $n$.
Let $x=(b_1,b_2,dots)$. It suffices to prove that $(b_k)$ is a Cauchy sequence.
We have that $|x_n-x|_inftyto0$, therefore $displaystyle{lim_{ntoinfty}sup_{kinmathbb{N}}|x_n^k-b_k|=0}$. Let $varepsilon>0$; then there exists $n_0inmathbb{N}$ such that for all $ngeq n_0$ and for all $kinmathbb{N}$ it is $|x_n^k-b_k|<varepsilon$. Now since $x_{n_0}in c$ it is a Cauchy sequence therefore for this $varepsilon$ there exists $Ngeq 0$ such that for all $m,ngeq N$ it is $|x_{n_0}^n-x_{n_0}^m|<varepsilon$.
Now we have for $m,ngeq N$: $|b_n-b_m|leq|b_n-x_{n_0}^n|+|x_{n_0}^n-x_{n_0}^m|+|x_{n_0}^m-b_m|<3varepsilon$.
Since for a random $varepsilon>0$ we have found an integer $N$ such that for all $m,n$ greater than $N$ it is $|b_n-b_m|<3varepsilon$, by the definition of a Cauchy sequence we have that $(b_n)$ is Cauchy.
answered Nov 20 '18 at 17:11
JustDroppedIn
1,764420
Good. I added a different style of proof.
– DanielWainfleet
Nov 20 '18 at 18:46
Why is it enough to prove that the sequence is Cauchy? Moreover, why $x_{n_0}$ is Cauchy?
– qcc101
Nov 20 '18 at 19:59
@qcc101 because a sequence of $mathbb{C}$ is convergent if-f it is Cauchy; $x_{n_0}in c$ and $c$ is by definition the space of convergent sequences.
– JustDroppedIn
Nov 20 '18 at 20:01
add a comment |
As said in the comments, you should improve your notation. It will help.
Let $(x_n)subset c$ with $x_n=(x_n^1,x_n^2,dots)$ for any $n$.
Let $x=(b_1,b_2,dots)$. It suffices to prove that $(b_k)$ is a Cauchy sequence.
We have that $|x_n-x|_inftyto0$, therefore $displaystyle{lim_{ntoinfty}sup_{kinmathbb{N}}|x_n^k-b_k|=0}$. Let $varepsilon>0$; then there exists $n_0inmathbb{N}$ such that for all $ngeq n_0$ and for all $kinmathbb{N}$ it is $|x_n^k-b_k|<varepsilon$. Now since $x_{n_0}in c$ it is a Cauchy sequence therefore for this $varepsilon$ there exists $Ngeq 0$ such that for all $m,ngeq N$ it is $|x_{n_0}^n-x_{n_0}^m|<varepsilon$.
Now we have for $m,ngeq N$: $|b_n-b_m|leq|b_n-x_{n_0}^n|+|x_{n_0}^n-x_{n_0}^m|+|x_{n_0}^m-b_m|<3varepsilon$.
Since for a random $varepsilon>0$ we have found an integer $N$ such that for all $m,n$ greater than $N$ it is $|b_n-b_m|<3varepsilon$, by the definition of a Cauchy sequence we have that $(b_n)$ is Cauchy.
answered Nov 20 '18 at 17:11
JustDroppedIn
1,764420
As said in the comments, you should improve your notation. It will help.
Let $(x_n)subset c$ with $x_n=(x_n^1,x_n^2,dots)$ for any $n$.
Let $x=(b_1,b_2,dots)$. It suffices to prove that $(b_k)$ is a Cauchy sequence.
We have that $|x_n-x|_inftyto0$, therefore $displaystyle{lim_{ntoinfty}sup_{kinmathbb{N}}|x_n^k-b_k|=0}$. Let $varepsilon>0$; then there exists $n_0inmathbb{N}$ such that for all $ngeq n_0$ and for all $kinmathbb{N}$ it is $|x_n^k-b_k|<varepsilon$. Now since $x_{n_0}in c$ it is a Cauchy sequence therefore for this $varepsilon$ there exists $Ngeq 0$ such that for all $m,ngeq N$ it is $|x_{n_0}^n-x_{n_0}^m|<varepsilon$.
Now we have for $m,ngeq N$: $|b_n-b_m|leq|b_n-x_{n_0}^n|+|x_{n_0}^n-x_{n_0}^m|+|x_{n_0}^m-b_m|<3varepsilon$.
Since for a random $varepsilon>0$ we have found an integer $N$ such that for all $m,n$ greater than $N$ it is $|b_n-b_m|<3varepsilon$, by the definition of a Cauchy sequence we have that $(b_n)$ is Cauchy.
answered Nov 20 '18 at 17:11
JustDroppedIn
1,764420
answered Nov 20 '18 at 17:11
JustDroppedIn
1,764420
answered Nov 20 '18 at 17:11
JustDroppedIn
1,764420
answered Nov 20 '18 at 17:11
JustDroppedIn
1,764420
1,764420
Good. I added a different style of proof.
– DanielWainfleet
Nov 20 '18 at 18:46
Why is it enough to prove that the sequence is Cauchy? Moreover, why $x_{n_0}$ is Cauchy?
– qcc101
Nov 20 '18 at 19:59
@qcc101 because a sequence of $mathbb{C}$ is convergent if-f it is Cauchy; $x_{n_0}in c$ and $c$ is by definition the space of convergent sequences.
– JustDroppedIn
Nov 20 '18 at 20:01
add a comment |
Good. I added a different style of proof.
– DanielWainfleet
Nov 20 '18 at 18:46
Why is it enough to prove that the sequence is Cauchy? Moreover, why $x_{n_0}$ is Cauchy?
– qcc101
Nov 20 '18 at 19:59
@qcc101 because a sequence of $mathbb{C}$ is convergent if-f it is Cauchy; $x_{n_0}in c$ and $c$ is by definition the space of convergent sequences.
– JustDroppedIn
Nov 20 '18 at 20:01
Good. I added a different style of proof.
– DanielWainfleet
Nov 20 '18 at 18:46
Good. I added a different style of proof.
– DanielWainfleet
Nov 20 '18 at 18:46
Why is it enough to prove that the sequence is Cauchy? Moreover, why $x_{n_0}$ is Cauchy?
– qcc101
Nov 20 '18 at 19:59
Why is it enough to prove that the sequence is Cauchy? Moreover, why $x_{n_0}$ is Cauchy?
– qcc101
Nov 20 '18 at 19:59
@qcc101 because a sequence of $mathbb{C}$ is convergent if-f it is Cauchy; $x_{n_0}in c$ and $c$ is by definition the space of convergent sequences.
– JustDroppedIn
Nov 20 '18 at 20:01
@qcc101 because a sequence of $mathbb{C}$ is convergent if-f it is Cauchy; $x_{n_0}in c$ and $c$ is by definition the space of convergent sequences.
– JustDroppedIn
Nov 20 '18 at 20:01
add a comment |
With the notation of the Answer given by JustDroppedIn.
Since you want to show that $c$ is closed in $l^{infty},$ you can prove that $l^{infty}setminus c$ is open, as follows:
Let $x in l^{infty}setminus c.$ Then $x$ is not a Cauchy sequence, so there exists $r>0,$ and functions $f:Bbb Nto Bbb N,, g:Bbb Nto Bbb N,$ both strictly increasing, such that $$forall kin Bbb N,(,|x^{f(k)}-x^{g(k)}|>r,).$$ Consider the open ball $B(x,r/3)={yin l^{infty}: |y-x|<r/3}.$ If $yin B(x,r/3)$ then for all $kin Bbb N$ we have $$|y^{f(k)}-y^{g(k)}|=$$ $$=|(y^{f(k)}-x^{f(k)})+(x^{f(k)}-x^{g(k)})+(x^{g(k)}-y^{g(k)})|geq$$ $$geq -|y^{f(k)}-x^{f(k)}|+|x^{f(k)}-x^{g(k)}|-|x^{g(k)}-y^{g(k)}|geq$$ $$geq -|y-x|+|x^{f(k)}-x^{g(k)}|-|x-y|>$$ $$>-r/3+r-r/3=r/3.$$ Since ${f(k):jin Bbb N}$ and ${g(k):kin Bbb N}$ are infinite sets, this implies that any $yin B(x,r/3)$ is not a Cauchy sequence. So $B(x,r/3) cap c =emptyset.$
The idea is that you cannot uniformly approximate a non-convergent sequence $x$ to an arbitrary degree by a convergent sequence because of the "$r$". For a bounded non-convergent sequence $x$ we can take $0<r<(lim sup_{jto infty} x^j)-(lim inf_{jto infty} x^j).$
answered Nov 20 '18 at 18:41
DanielWainfleet
34.1k31647
add a comment |
With the notation of the Answer given by JustDroppedIn.
Since you want to show that $c$ is closed in $l^{infty},$ you can prove that $l^{infty}setminus c$ is open, as follows:
Let $x in l^{infty}setminus c.$ Then $x$ is not a Cauchy sequence, so there exists $r>0,$ and functions $f:Bbb Nto Bbb N,, g:Bbb Nto Bbb N,$ both strictly increasing, such that $$forall kin Bbb N,(,|x^{f(k)}-x^{g(k)}|>r,).$$ Consider the open ball $B(x,r/3)={yin l^{infty}: |y-x|<r/3}.$ If $yin B(x,r/3)$ then for all $kin Bbb N$ we have $$|y^{f(k)}-y^{g(k)}|=$$ $$=|(y^{f(k)}-x^{f(k)})+(x^{f(k)}-x^{g(k)})+(x^{g(k)}-y^{g(k)})|geq$$ $$geq -|y^{f(k)}-x^{f(k)}|+|x^{f(k)}-x^{g(k)}|-|x^{g(k)}-y^{g(k)}|geq$$ $$geq -|y-x|+|x^{f(k)}-x^{g(k)}|-|x-y|>$$ $$>-r/3+r-r/3=r/3.$$ Since ${f(k):jin Bbb N}$ and ${g(k):kin Bbb N}$ are infinite sets, this implies that any $yin B(x,r/3)$ is not a Cauchy sequence. So $B(x,r/3) cap c =emptyset.$
The idea is that you cannot uniformly approximate a non-convergent sequence $x$ to an arbitrary degree by a convergent sequence because of the "$r$". For a bounded non-convergent sequence $x$ we can take $0<r<(lim sup_{jto infty} x^j)-(lim inf_{jto infty} x^j).$
answered Nov 20 '18 at 18:41
DanielWainfleet
34.1k31647
add a comment |
With the notation of the Answer given by JustDroppedIn.
Since you want to show that $c$ is closed in $l^{infty},$ you can prove that $l^{infty}setminus c$ is open, as follows:
Let $x in l^{infty}setminus c.$ Then $x$ is not a Cauchy sequence, so there exists $r>0,$ and functions $f:Bbb Nto Bbb N,, g:Bbb Nto Bbb N,$ both strictly increasing, such that $$forall kin Bbb N,(,|x^{f(k)}-x^{g(k)}|>r,).$$ Consider the open ball $B(x,r/3)={yin l^{infty}: |y-x|<r/3}.$ If $yin B(x,r/3)$ then for all $kin Bbb N$ we have $$|y^{f(k)}-y^{g(k)}|=$$ $$=|(y^{f(k)}-x^{f(k)})+(x^{f(k)}-x^{g(k)})+(x^{g(k)}-y^{g(k)})|geq$$ $$geq -|y^{f(k)}-x^{f(k)}|+|x^{f(k)}-x^{g(k)}|-|x^{g(k)}-y^{g(k)}|geq$$ $$geq -|y-x|+|x^{f(k)}-x^{g(k)}|-|x-y|>$$ $$>-r/3+r-r/3=r/3.$$ Since ${f(k):jin Bbb N}$ and ${g(k):kin Bbb N}$ are infinite sets, this implies that any $yin B(x,r/3)$ is not a Cauchy sequence. So $B(x,r/3) cap c =emptyset.$
The idea is that you cannot uniformly approximate a non-convergent sequence $x$ to an arbitrary degree by a convergent sequence because of the "$r$". For a bounded non-convergent sequence $x$ we can take $0<r<(lim sup_{jto infty} x^j)-(lim inf_{jto infty} x^j).$
answered Nov 20 '18 at 18:41
DanielWainfleet
34.1k31647
With the notation of the Answer given by JustDroppedIn.
Since you want to show that $c$ is closed in $l^{infty},$ you can prove that $l^{infty}setminus c$ is open, as follows:
Let $x in l^{infty}setminus c.$ Then $x$ is not a Cauchy sequence, so there exists $r>0,$ and functions $f:Bbb Nto Bbb N,, g:Bbb Nto Bbb N,$ both strictly increasing, such that $$forall kin Bbb N,(,|x^{f(k)}-x^{g(k)}|>r,).$$ Consider the open ball $B(x,r/3)={yin l^{infty}: |y-x|<r/3}.$ If $yin B(x,r/3)$ then for all $kin Bbb N$ we have $$|y^{f(k)}-y^{g(k)}|=$$ $$=|(y^{f(k)}-x^{f(k)})+(x^{f(k)}-x^{g(k)})+(x^{g(k)}-y^{g(k)})|geq$$ $$geq -|y^{f(k)}-x^{f(k)}|+|x^{f(k)}-x^{g(k)}|-|x^{g(k)}-y^{g(k)}|geq$$ $$geq -|y-x|+|x^{f(k)}-x^{g(k)}|-|x-y|>$$ $$>-r/3+r-r/3=r/3.$$ Since ${f(k):jin Bbb N}$ and ${g(k):kin Bbb N}$ are infinite sets, this implies that any $yin B(x,r/3)$ is not a Cauchy sequence. So $B(x,r/3) cap c =emptyset.$
The idea is that you cannot uniformly approximate a non-convergent sequence $x$ to an arbitrary degree by a convergent sequence because of the "$r$". For a bounded non-convergent sequence $x$ we can take $0<r<(lim sup_{jto infty} x^j)-(lim inf_{jto infty} x^j).$
answered Nov 20 '18 at 18:41
DanielWainfleet
34.1k31647
answered Nov 20 '18 at 18:41
DanielWainfleet
34.1k31647
answered Nov 20 '18 at 18:41
DanielWainfleet
34.1k31647
answered Nov 20 '18 at 18:41
DanielWainfleet
34.1k31647
34.1k31647
add a comment |
add a comment |
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You should improve your notation: in your question, $x_k$ is sometimes a real number and sometimes an element in $c$. Maybe you should use $x^{k}$ for one or the other, which would enable notation such as $x_n^{k}$.
– supinf
Nov 20 '18 at 16:13