How the sup norm works to make a normed space complete
$begingroup$
Let $(x_n)_{nin N}$ be the sequence of continuous bounded functions $x_n=x^n$ on the unit interval $[0,1]$ and $C([0,1])$ be the space of continuous bounded real-valued functions on $[0,1].$ It is well known that this Vector space is complete in respect to the sup norm. But we also know that $(x_n)=x^n$ is converging towards a discontinuous function which is thus not element of $C[0,1]$.
I am wondering in what sense does the sup norm resolves this problem, i.e. how the limit of the sequence $x_n=x^n$ become an element of $C([0,1], {leftlVert cdot rightrVert}_{sup}) ,,?$ Many thanks.
real-analysis functional-analysis
asked Dec 31 '18 at 21:57
user249018user249018
440138
$endgroup$
add a comment |
$begingroup$
Let $(x_n)_{nin N}$ be the sequence of continuous bounded functions $x_n=x^n$ on the unit interval $[0,1]$ and $C([0,1])$ be the space of continuous bounded real-valued functions on $[0,1].$ It is well known that this Vector space is complete in respect to the sup norm. But we also know that $(x_n)=x^n$ is converging towards a discontinuous function which is thus not element of $C[0,1]$.
I am wondering in what sense does the sup norm resolves this problem, i.e. how the limit of the sequence $x_n=x^n$ become an element of $C([0,1], {leftlVert cdot rightrVert}_{sup}) ,,?$ Many thanks.
real-analysis functional-analysis
asked Dec 31 '18 at 21:57
user249018user249018
440138
$endgroup$
4
$begingroup$
$x^n$ is not a Cauchy sequence with respect to this norm is it?
$endgroup$
– Yanko
Dec 31 '18 at 22:02
$begingroup$
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
$endgroup$
– user249018
Dec 31 '18 at 22:51
1
$begingroup$
Instead of speaking of "the limit of $x^n,, $" bear in mind that "the function $x^n,$" is actually the set ${(x,x^n):xin [0,1[}, $ and there may be more than one kind of limit of a sequence of infinite sets. A point-wise limit is not the same thing as a uniform limit. Convergence in the $sup$ norm means uniform convergence only..... $x^n$ does not converge uniformly.
$endgroup$
– DanielWainfleet
Jan 1 at 4:49
$begingroup$
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
$endgroup$
– DanielWainfleet
Jan 1 at 4:56
add a comment |
$begingroup$
Let $(x_n)_{nin N}$ be the sequence of continuous bounded functions $x_n=x^n$ on the unit interval $[0,1]$ and $C([0,1])$ be the space of continuous bounded real-valued functions on $[0,1].$ It is well known that this Vector space is complete in respect to the sup norm. But we also know that $(x_n)=x^n$ is converging towards a discontinuous function which is thus not element of $C[0,1]$.
I am wondering in what sense does the sup norm resolves this problem, i.e. how the limit of the sequence $x_n=x^n$ become an element of $C([0,1], {leftlVert cdot rightrVert}_{sup}) ,,?$ Many thanks.
real-analysis functional-analysis
asked Dec 31 '18 at 21:57
user249018user249018
440138
$endgroup$
Let $(x_n)_{nin N}$ be the sequence of continuous bounded functions $x_n=x^n$ on the unit interval $[0,1]$ and $C([0,1])$ be the space of continuous bounded real-valued functions on $[0,1].$ It is well known that this Vector space is complete in respect to the sup norm. But we also know that $(x_n)=x^n$ is converging towards a discontinuous function which is thus not element of $C[0,1]$.
I am wondering in what sense does the sup norm resolves this problem, i.e. how the limit of the sequence $x_n=x^n$ become an element of $C([0,1], {leftlVert cdot rightrVert}_{sup}) ,,?$ Many thanks.
real-analysis functional-analysis
real-analysis functional-analysis
asked Dec 31 '18 at 21:57
user249018user249018
440138
asked Dec 31 '18 at 21:57
user249018user249018
440138
asked Dec 31 '18 at 21:57
user249018user249018
440138
asked Dec 31 '18 at 21:57
user249018user249018
440138
asked Dec 31 '18 at 21:57
user249018user249018
440138
440138
4
$begingroup$
$x^n$ is not a Cauchy sequence with respect to this norm is it?
$endgroup$
– Yanko
Dec 31 '18 at 22:02
$begingroup$
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
$endgroup$
– user249018
Dec 31 '18 at 22:51
1
$begingroup$
Instead of speaking of "the limit of $x^n,, $" bear in mind that "the function $x^n,$" is actually the set ${(x,x^n):xin [0,1[}, $ and there may be more than one kind of limit of a sequence of infinite sets. A point-wise limit is not the same thing as a uniform limit. Convergence in the $sup$ norm means uniform convergence only..... $x^n$ does not converge uniformly.
$endgroup$
– DanielWainfleet
Jan 1 at 4:49
$begingroup$
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
$endgroup$
– DanielWainfleet
Jan 1 at 4:56
add a comment |
4
$begingroup$
$x^n$ is not a Cauchy sequence with respect to this norm is it?
$endgroup$
– Yanko
Dec 31 '18 at 22:02
$begingroup$
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
$endgroup$
– user249018
Dec 31 '18 at 22:51
1
$begingroup$
Instead of speaking of "the limit of $x^n,, $" bear in mind that "the function $x^n,$" is actually the set ${(x,x^n):xin [0,1[}, $ and there may be more than one kind of limit of a sequence of infinite sets. A point-wise limit is not the same thing as a uniform limit. Convergence in the $sup$ norm means uniform convergence only..... $x^n$ does not converge uniformly.
$endgroup$
– DanielWainfleet
Jan 1 at 4:49
$begingroup$
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
$endgroup$
– DanielWainfleet
Jan 1 at 4:56
4
4
$begingroup$
$x^n$ is not a Cauchy sequence with respect to this norm is it?
$endgroup$
– Yanko
Dec 31 '18 at 22:02
$begingroup$
$x^n$ is not a Cauchy sequence with respect to this norm is it?
$endgroup$
– Yanko
Dec 31 '18 at 22:02
$begingroup$
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
$endgroup$
– user249018
Dec 31 '18 at 22:51
$begingroup$
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
$endgroup$
– user249018
Dec 31 '18 at 22:51
1
1
$begingroup$
Instead of speaking of "the limit of $x^n,, $" bear in mind that "the function $x^n,$" is actually the set ${(x,x^n):xin [0,1[}, $ and there may be more than one kind of limit of a sequence of infinite sets. A point-wise limit is not the same thing as a uniform limit. Convergence in the $sup$ norm means uniform convergence only..... $x^n$ does not converge uniformly.
$endgroup$
– DanielWainfleet
Jan 1 at 4:49
$begingroup$
Instead of speaking of "the limit of $x^n,, $" bear in mind that "the function $x^n,$" is actually the set ${(x,x^n):xin [0,1[}, $ and there may be more than one kind of limit of a sequence of infinite sets. A point-wise limit is not the same thing as a uniform limit. Convergence in the $sup$ norm means uniform convergence only..... $x^n$ does not converge uniformly.
$endgroup$
– DanielWainfleet
Jan 1 at 4:49
$begingroup$
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
$endgroup$
– DanielWainfleet
Jan 1 at 4:56
$begingroup$
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
$endgroup$
– DanielWainfleet
Jan 1 at 4:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The pointwise limit of the sequence is not become an element of $C([0,1])$. This poses no contradiction to the fact that $C([0,1])$ is complete. Indeed, the sequence of functions given by $f_{n}(x)=x^n$ is not uniformly Cauchy.
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
23k41552
$endgroup$
$begingroup$
Thanks. But I still dont get it. The limit of $f_n(x)=x^n$ is not in $C([0,1])$: Why it does not pose a contradiction ? Would you please comment on that ?
$endgroup$
– user249018
Dec 31 '18 at 22:10
2
$begingroup$
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
$endgroup$
– Foobaz John
Dec 31 '18 at 22:18
$begingroup$
So in general the pointwise limit of a sequence of functions must not have a limit in the vector space which is complete with respect to a norm. Right ? Does then the pointwise limit is used at all in normed spaces or must one always use the norm when working with limits?
$endgroup$
– user249018
Dec 31 '18 at 22:28
add a comment |
$begingroup$
Your sequence $(f_n)_n$ is not Cauchy with respect to $|cdot|_{sup}$, not even on $[0,1)$.
Indeed, for any $n inmathbb{N}$ we have
$$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$
answered Jan 1 at 13:49
mechanodroidmechanodroid
28.9k62648
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
The pointwise limit of the sequence is not become an element of $C([0,1])$. This poses no contradiction to the fact that $C([0,1])$ is complete. Indeed, the sequence of functions given by $f_{n}(x)=x^n$ is not uniformly Cauchy.
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
23k41552
$endgroup$
$begingroup$
Thanks. But I still dont get it. The limit of $f_n(x)=x^n$ is not in $C([0,1])$: Why it does not pose a contradiction ? Would you please comment on that ?
$endgroup$
– user249018
Dec 31 '18 at 22:10
2
$begingroup$
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
$endgroup$
– Foobaz John
Dec 31 '18 at 22:18
$begingroup$
So in general the pointwise limit of a sequence of functions must not have a limit in the vector space which is complete with respect to a norm. Right ? Does then the pointwise limit is used at all in normed spaces or must one always use the norm when working with limits?
$endgroup$
– user249018
Dec 31 '18 at 22:28
add a comment |
$begingroup$
The pointwise limit of the sequence is not become an element of $C([0,1])$. This poses no contradiction to the fact that $C([0,1])$ is complete. Indeed, the sequence of functions given by $f_{n}(x)=x^n$ is not uniformly Cauchy.
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
23k41552
$endgroup$
$begingroup$
Thanks. But I still dont get it. The limit of $f_n(x)=x^n$ is not in $C([0,1])$: Why it does not pose a contradiction ? Would you please comment on that ?
$endgroup$
– user249018
Dec 31 '18 at 22:10
2
$begingroup$
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
$endgroup$
– Foobaz John
Dec 31 '18 at 22:18
$begingroup$
So in general the pointwise limit of a sequence of functions must not have a limit in the vector space which is complete with respect to a norm. Right ? Does then the pointwise limit is used at all in normed spaces or must one always use the norm when working with limits?
$endgroup$
– user249018
Dec 31 '18 at 22:28
add a comment |
$begingroup$
The pointwise limit of the sequence is not become an element of $C([0,1])$. This poses no contradiction to the fact that $C([0,1])$ is complete. Indeed, the sequence of functions given by $f_{n}(x)=x^n$ is not uniformly Cauchy.
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
23k41552
$endgroup$
The pointwise limit of the sequence is not become an element of $C([0,1])$. This poses no contradiction to the fact that $C([0,1])$ is complete. Indeed, the sequence of functions given by $f_{n}(x)=x^n$ is not uniformly Cauchy.
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
23k41552
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
23k41552
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
23k41552
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
23k41552
23k41552
$begingroup$
Thanks. But I still dont get it. The limit of $f_n(x)=x^n$ is not in $C([0,1])$: Why it does not pose a contradiction ? Would you please comment on that ?
$endgroup$
– user249018
Dec 31 '18 at 22:10
2
$begingroup$
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
$endgroup$
– Foobaz John
Dec 31 '18 at 22:18
$begingroup$
So in general the pointwise limit of a sequence of functions must not have a limit in the vector space which is complete with respect to a norm. Right ? Does then the pointwise limit is used at all in normed spaces or must one always use the norm when working with limits?
$endgroup$
– user249018
Dec 31 '18 at 22:28
add a comment |
$begingroup$
Thanks. But I still dont get it. The limit of $f_n(x)=x^n$ is not in $C([0,1])$: Why it does not pose a contradiction ? Would you please comment on that ?
$endgroup$
– user249018
Dec 31 '18 at 22:10
2
$begingroup$
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
$endgroup$
– Foobaz John
Dec 31 '18 at 22:18
$begingroup$
So in general the pointwise limit of a sequence of functions must not have a limit in the vector space which is complete with respect to a norm. Right ? Does then the pointwise limit is used at all in normed spaces or must one always use the norm when working with limits?
$endgroup$
– user249018
Dec 31 '18 at 22:28
$begingroup$
Thanks. But I still dont get it. The limit of $f_n(x)=x^n$ is not in $C([0,1])$: Why it does not pose a contradiction ? Would you please comment on that ?
$endgroup$
– user249018
Dec 31 '18 at 22:10
$begingroup$
Thanks. But I still dont get it. The limit of $f_n(x)=x^n$ is not in $C([0,1])$: Why it does not pose a contradiction ? Would you please comment on that ?
$endgroup$
– user249018
Dec 31 '18 at 22:10
2
2
$begingroup$
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
$endgroup$
– Foobaz John
Dec 31 '18 at 22:18
$begingroup$
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
$endgroup$
– Foobaz John
Dec 31 '18 at 22:18
$begingroup$
So in general the pointwise limit of a sequence of functions must not have a limit in the vector space which is complete with respect to a norm. Right ? Does then the pointwise limit is used at all in normed spaces or must one always use the norm when working with limits?
$endgroup$
– user249018
Dec 31 '18 at 22:28
$begingroup$
So in general the pointwise limit of a sequence of functions must not have a limit in the vector space which is complete with respect to a norm. Right ? Does then the pointwise limit is used at all in normed spaces or must one always use the norm when working with limits?
$endgroup$
– user249018
Dec 31 '18 at 22:28
add a comment |
$begingroup$
Your sequence $(f_n)_n$ is not Cauchy with respect to $|cdot|_{sup}$, not even on $[0,1)$.
Indeed, for any $n inmathbb{N}$ we have
$$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$
answered Jan 1 at 13:49
mechanodroidmechanodroid
28.9k62648
$endgroup$
add a comment |
$begingroup$
Your sequence $(f_n)_n$ is not Cauchy with respect to $|cdot|_{sup}$, not even on $[0,1)$.
Indeed, for any $n inmathbb{N}$ we have
$$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$
answered Jan 1 at 13:49
mechanodroidmechanodroid
28.9k62648
$endgroup$
add a comment |
$begingroup$
Your sequence $(f_n)_n$ is not Cauchy with respect to $|cdot|_{sup}$, not even on $[0,1)$.
Indeed, for any $n inmathbb{N}$ we have
$$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$
answered Jan 1 at 13:49
mechanodroidmechanodroid
28.9k62648
$endgroup$
Your sequence $(f_n)_n$ is not Cauchy with respect to $|cdot|_{sup}$, not even on $[0,1)$.
Indeed, for any $n inmathbb{N}$ we have
$$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$
answered Jan 1 at 13:49
mechanodroidmechanodroid
28.9k62648
answered Jan 1 at 13:49
mechanodroidmechanodroid
28.9k62648
answered Jan 1 at 13:49
mechanodroidmechanodroid
28.9k62648
answered Jan 1 at 13:49
mechanodroidmechanodroid
28.9k62648
28.9k62648
add a comment |
add a comment |
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4
$begingroup$
$x^n$ is not a Cauchy sequence with respect to this norm is it?
$endgroup$
– Yanko
Dec 31 '18 at 22:02
$begingroup$
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
$endgroup$
– user249018
Dec 31 '18 at 22:51
1
$begingroup$
Instead of speaking of "the limit of $x^n,, $" bear in mind that "the function $x^n,$" is actually the set ${(x,x^n):xin [0,1[}, $ and there may be more than one kind of limit of a sequence of infinite sets. A point-wise limit is not the same thing as a uniform limit. Convergence in the $sup$ norm means uniform convergence only..... $x^n$ does not converge uniformly.
$endgroup$
– DanielWainfleet
Jan 1 at 4:49
$begingroup$
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
$endgroup$
– DanielWainfleet
Jan 1 at 4:56