Mixing
How the sup norm works to make a normed space complete
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
393127
|
show 1 more comment
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
393127
4
$x^n$ is not a Cauchy sequence with respect to this norm is it?
– Yanko
Dec 31 '18 at 22:02
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
– user249018
Dec 31 '18 at 22:51
1
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.
– DanielWainfleet
Jan 1 at 4:49
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
– DanielWainfleet
Jan 1 at 4:56
which is the nice thing about Cauchy sequences...we may avoid invoking a limit
– Matt A Pelto
Jan 1 at 4:58
|
show 1 more comment
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
393127
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
393127
asked Dec 31 '18 at 21:57
user249018user249018
393127
asked Dec 31 '18 at 21:57
user249018user249018
393127
asked Dec 31 '18 at 21:57
user249018user249018
393127
asked Dec 31 '18 at 21:57
user249018user249018
393127
393127
4
$x^n$ is not a Cauchy sequence with respect to this norm is it?
– Yanko
Dec 31 '18 at 22:02
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
– user249018
Dec 31 '18 at 22:51
1
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.
– DanielWainfleet
Jan 1 at 4:49
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
– DanielWainfleet
Jan 1 at 4:56
which is the nice thing about Cauchy sequences...we may avoid invoking a limit
– Matt A Pelto
Jan 1 at 4:58
|
show 1 more comment
4
$x^n$ is not a Cauchy sequence with respect to this norm is it?
– Yanko
Dec 31 '18 at 22:02
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
– user249018
Dec 31 '18 at 22:51
1
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.
– DanielWainfleet
Jan 1 at 4:49
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
– DanielWainfleet
Jan 1 at 4:56
which is the nice thing about Cauchy sequences...we may avoid invoking a limit
– Matt A Pelto
Jan 1 at 4:58
4
4
$x^n$ is not a Cauchy sequence with respect to this norm is it?
– Yanko
Dec 31 '18 at 22:02
$x^n$ is not a Cauchy sequence with respect to this norm is it?
– Yanko
Dec 31 '18 at 22:02
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
– user249018
Dec 31 '18 at 22:51
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
– user249018
Dec 31 '18 at 22:51
1
1
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.
– DanielWainfleet
Jan 1 at 4:49
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.
– DanielWainfleet
Jan 1 at 4:49
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
– DanielWainfleet
Jan 1 at 4:56
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
– DanielWainfleet
Jan 1 at 4:56
which is the nice thing about Cauchy sequences...we may avoid invoking a limit
– Matt A Pelto
Jan 1 at 4:58
which is the nice thing about Cauchy sequences...we may avoid invoking a limit
– Matt A Pelto
Jan 1 at 4:58
|
show 1 more comment
3 Answers
3
active
oldest
votes
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
21.5k41351
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 ?
– user249018
Dec 31 '18 at 22:10
2
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
– Foobaz John
Dec 31 '18 at 22:18
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?
– user249018
Dec 31 '18 at 22:28
add a comment |
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
27k62446
add a comment |
The sequence of functions ${f_n}_{n=1}^infty$ given by $f_n(x):=x^n$ is not uniformly Cauchy on $[0,1]$, or rather the sequence is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$. To see this we first make two simple observations:
$|x^n-x^m|=|x^n||1-x^{m-n}|$, and
the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$.
So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$ by definition.
By saying a sequence ${x_n}_{n=1}^infty$ is not Cauchy in a metric space $(X,d)$ we consider the negation of what it means for ${x_n}_{n=1}^infty$ to be a Cauchy sequence in $(X,d)$:
$exists varepsilon in mathbb R_{>0}: forall N in mathbb N: exists m, n in mathbb N: left(m, n ge N land d left({x_n, x_m}right) geq varepsilon right)$
Let $(X,d)$ be a metric space. If ${x_n}_{n=1}^infty$ is a convergent sequence in $(X,d)$, then ${x_n}_{n=1}^infty$ is a Cauchy sequence in $(X,d)$. Considering the contrapositive of this proposition and notwithstanding pointwise convergence (a weaker property than uniform convergence), we have thus already shown that ${f_n}_{n=1}^infty$ is not a convergent sequence in $left(mathcal C[0,1], |cdot|_{sup}right)$.
Note: We say that $(X,d)$ is complete if and only if the converse of this aforementioned proposition is true.
The pointwise limit $f(x):=lim_{nto infty} f_n(x)$ is definitely not in $mathcal C[0,1]$ and this poses no contradiction to the completeness of $left(mathcal C[0,1], |cdot|_{sup}right)$. Perhaps it is also worth noting that when we talk about the sequence of functions ${f_n}_{n=1}^infty$ converging to $f$ pointwise it is in the context of the topology of pointwise convergence on $mathbb{R}^{[0,1]}$ which is a significantly different context from that of the topology of uniform convergence on $mathcal C [0,1]$.
answered Jan 1 at 4:32
Matt A PeltoMatt A Pelto
2,397620
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
21.5k41351
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 ?
– user249018
Dec 31 '18 at 22:10
2
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
– Foobaz John
Dec 31 '18 at 22:18
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?
– user249018
Dec 31 '18 at 22:28
add a comment |
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
21.5k41351
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 ?
– user249018
Dec 31 '18 at 22:10
2
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
– Foobaz John
Dec 31 '18 at 22:18
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?
– user249018
Dec 31 '18 at 22:28
add a comment |
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
21.5k41351
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
21.5k41351
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
21.5k41351
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
21.5k41351
answered Dec 31 '18 at 22:02
Foobaz JohnFoobaz John
21.5k41351
21.5k41351
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 ?
– user249018
Dec 31 '18 at 22:10
2
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
– Foobaz John
Dec 31 '18 at 22:18
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?
– user249018
Dec 31 '18 at 22:28
add a comment |
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 ?
– user249018
Dec 31 '18 at 22:10
2
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
– Foobaz John
Dec 31 '18 at 22:18
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?
– user249018
Dec 31 '18 at 22:28
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 ?
– user249018
Dec 31 '18 at 22:10
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 ?
– user249018
Dec 31 '18 at 22:10
2
2
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
– Foobaz John
Dec 31 '18 at 22:18
Because the sequence $(f_n)$ is not Cauchy in $C([0,1])$ to begin with.
– Foobaz John
Dec 31 '18 at 22:18
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?
– user249018
Dec 31 '18 at 22:28
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?
– user249018
Dec 31 '18 at 22:28
add a comment |
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
27k62446
add a comment |
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
27k62446
add a comment |
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
27k62446
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
27k62446
answered Jan 1 at 13:49
mechanodroidmechanodroid
27k62446
answered Jan 1 at 13:49
mechanodroidmechanodroid
27k62446
answered Jan 1 at 13:49
mechanodroidmechanodroid
27k62446
27k62446
add a comment |
add a comment |
The sequence of functions ${f_n}_{n=1}^infty$ given by $f_n(x):=x^n$ is not uniformly Cauchy on $[0,1]$, or rather the sequence is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$. To see this we first make two simple observations:
$|x^n-x^m|=|x^n||1-x^{m-n}|$, and
the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$.
So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$ by definition.
By saying a sequence ${x_n}_{n=1}^infty$ is not Cauchy in a metric space $(X,d)$ we consider the negation of what it means for ${x_n}_{n=1}^infty$ to be a Cauchy sequence in $(X,d)$:
$exists varepsilon in mathbb R_{>0}: forall N in mathbb N: exists m, n in mathbb N: left(m, n ge N land d left({x_n, x_m}right) geq varepsilon right)$
Let $(X,d)$ be a metric space. If ${x_n}_{n=1}^infty$ is a convergent sequence in $(X,d)$, then ${x_n}_{n=1}^infty$ is a Cauchy sequence in $(X,d)$. Considering the contrapositive of this proposition and notwithstanding pointwise convergence (a weaker property than uniform convergence), we have thus already shown that ${f_n}_{n=1}^infty$ is not a convergent sequence in $left(mathcal C[0,1], |cdot|_{sup}right)$.
Note: We say that $(X,d)$ is complete if and only if the converse of this aforementioned proposition is true.
The pointwise limit $f(x):=lim_{nto infty} f_n(x)$ is definitely not in $mathcal C[0,1]$ and this poses no contradiction to the completeness of $left(mathcal C[0,1], |cdot|_{sup}right)$. Perhaps it is also worth noting that when we talk about the sequence of functions ${f_n}_{n=1}^infty$ converging to $f$ pointwise it is in the context of the topology of pointwise convergence on $mathbb{R}^{[0,1]}$ which is a significantly different context from that of the topology of uniform convergence on $mathcal C [0,1]$.
answered Jan 1 at 4:32
Matt A PeltoMatt A Pelto
2,397620
add a comment |
The sequence of functions ${f_n}_{n=1}^infty$ given by $f_n(x):=x^n$ is not uniformly Cauchy on $[0,1]$, or rather the sequence is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$. To see this we first make two simple observations:
$|x^n-x^m|=|x^n||1-x^{m-n}|$, and
the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$.
So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$ by definition.
By saying a sequence ${x_n}_{n=1}^infty$ is not Cauchy in a metric space $(X,d)$ we consider the negation of what it means for ${x_n}_{n=1}^infty$ to be a Cauchy sequence in $(X,d)$:
$exists varepsilon in mathbb R_{>0}: forall N in mathbb N: exists m, n in mathbb N: left(m, n ge N land d left({x_n, x_m}right) geq varepsilon right)$
Let $(X,d)$ be a metric space. If ${x_n}_{n=1}^infty$ is a convergent sequence in $(X,d)$, then ${x_n}_{n=1}^infty$ is a Cauchy sequence in $(X,d)$. Considering the contrapositive of this proposition and notwithstanding pointwise convergence (a weaker property than uniform convergence), we have thus already shown that ${f_n}_{n=1}^infty$ is not a convergent sequence in $left(mathcal C[0,1], |cdot|_{sup}right)$.
Note: We say that $(X,d)$ is complete if and only if the converse of this aforementioned proposition is true.
The pointwise limit $f(x):=lim_{nto infty} f_n(x)$ is definitely not in $mathcal C[0,1]$ and this poses no contradiction to the completeness of $left(mathcal C[0,1], |cdot|_{sup}right)$. Perhaps it is also worth noting that when we talk about the sequence of functions ${f_n}_{n=1}^infty$ converging to $f$ pointwise it is in the context of the topology of pointwise convergence on $mathbb{R}^{[0,1]}$ which is a significantly different context from that of the topology of uniform convergence on $mathcal C [0,1]$.
answered Jan 1 at 4:32
Matt A PeltoMatt A Pelto
2,397620
add a comment |
The sequence of functions ${f_n}_{n=1}^infty$ given by $f_n(x):=x^n$ is not uniformly Cauchy on $[0,1]$, or rather the sequence is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$. To see this we first make two simple observations:
$|x^n-x^m|=|x^n||1-x^{m-n}|$, and
the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$.
So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$ by definition.
By saying a sequence ${x_n}_{n=1}^infty$ is not Cauchy in a metric space $(X,d)$ we consider the negation of what it means for ${x_n}_{n=1}^infty$ to be a Cauchy sequence in $(X,d)$:
$exists varepsilon in mathbb R_{>0}: forall N in mathbb N: exists m, n in mathbb N: left(m, n ge N land d left({x_n, x_m}right) geq varepsilon right)$
Let $(X,d)$ be a metric space. If ${x_n}_{n=1}^infty$ is a convergent sequence in $(X,d)$, then ${x_n}_{n=1}^infty$ is a Cauchy sequence in $(X,d)$. Considering the contrapositive of this proposition and notwithstanding pointwise convergence (a weaker property than uniform convergence), we have thus already shown that ${f_n}_{n=1}^infty$ is not a convergent sequence in $left(mathcal C[0,1], |cdot|_{sup}right)$.
Note: We say that $(X,d)$ is complete if and only if the converse of this aforementioned proposition is true.
The pointwise limit $f(x):=lim_{nto infty} f_n(x)$ is definitely not in $mathcal C[0,1]$ and this poses no contradiction to the completeness of $left(mathcal C[0,1], |cdot|_{sup}right)$. Perhaps it is also worth noting that when we talk about the sequence of functions ${f_n}_{n=1}^infty$ converging to $f$ pointwise it is in the context of the topology of pointwise convergence on $mathbb{R}^{[0,1]}$ which is a significantly different context from that of the topology of uniform convergence on $mathcal C [0,1]$.
answered Jan 1 at 4:32
Matt A PeltoMatt A Pelto
2,397620
The sequence of functions ${f_n}_{n=1}^infty$ given by $f_n(x):=x^n$ is not uniformly Cauchy on $[0,1]$, or rather the sequence is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$. To see this we first make two simple observations:
$|x^n-x^m|=|x^n||1-x^{m-n}|$, and
the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$.
So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $left(mathcal C[0,1], |cdot|_{sup}right)$ by definition.
By saying a sequence ${x_n}_{n=1}^infty$ is not Cauchy in a metric space $(X,d)$ we consider the negation of what it means for ${x_n}_{n=1}^infty$ to be a Cauchy sequence in $(X,d)$:
$exists varepsilon in mathbb R_{>0}: forall N in mathbb N: exists m, n in mathbb N: left(m, n ge N land d left({x_n, x_m}right) geq varepsilon right)$
Let $(X,d)$ be a metric space. If ${x_n}_{n=1}^infty$ is a convergent sequence in $(X,d)$, then ${x_n}_{n=1}^infty$ is a Cauchy sequence in $(X,d)$. Considering the contrapositive of this proposition and notwithstanding pointwise convergence (a weaker property than uniform convergence), we have thus already shown that ${f_n}_{n=1}^infty$ is not a convergent sequence in $left(mathcal C[0,1], |cdot|_{sup}right)$.
Note: We say that $(X,d)$ is complete if and only if the converse of this aforementioned proposition is true.
The pointwise limit $f(x):=lim_{nto infty} f_n(x)$ is definitely not in $mathcal C[0,1]$ and this poses no contradiction to the completeness of $left(mathcal C[0,1], |cdot|_{sup}right)$. Perhaps it is also worth noting that when we talk about the sequence of functions ${f_n}_{n=1}^infty$ converging to $f$ pointwise it is in the context of the topology of pointwise convergence on $mathbb{R}^{[0,1]}$ which is a significantly different context from that of the topology of uniform convergence on $mathcal C [0,1]$.
answered Jan 1 at 4:32
Matt A PeltoMatt A Pelto
2,397620
edited Jan 1 at 10:11
answered Jan 1 at 4:32
Matt A PeltoMatt A Pelto
2,397620
answered Jan 1 at 4:32
Matt A PeltoMatt A Pelto
2,397620
answered Jan 1 at 4:32
Matt A PeltoMatt A Pelto
2,397620
2,397620
add a comment |
add a comment |
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$x^n$ is not a Cauchy sequence with respect to this norm is it?
– Yanko
Dec 31 '18 at 22:02
@Yanko. How one would characterise the pointwise limit of $f_n(x)=x^n$ as seen from $C([0,1],{leftlVert cdot rightrVert}_{sup}) )$ ?
– user249018
Dec 31 '18 at 22:51
1
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.
– DanielWainfleet
Jan 1 at 4:49
In $C[0,1]$ the set ${f_n: nin Bbb N}$ (where $f_n(x)=x^n$ ) is an infinite closed discrete subspace.
– DanielWainfleet
Jan 1 at 4:56
which is the nice thing about Cauchy sequences...we may avoid invoking a limit
– Matt A Pelto
Jan 1 at 4:58