How the sup norm works to make a normed space complete












2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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
$$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    2












    2








    2





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    0












    $begingroup$

    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
    $$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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
      $$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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
        $$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$






        share|cite|improve this answer









        $endgroup$



        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
        $$|f_{n^2}-f_n|_sup ge (f_{n^2}-f_n)left(sqrt[n]{1-frac1n}right)= left(1-frac1nright)^n - left(1-frac1nright) xrightarrow{ntoinfty} frac1e - 1 ne 0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 13:49









        mechanodroidmechanodroid

        28.9k62648




        28.9k62648






























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