if $int_{0}^{1}|f_n(x)-f(x)|dxto 0$ then $f_nto f$ uniformly on $[0,1]$: True/false












1












$begingroup$


Let ${f_n}$ be a sequence of continuous real valued functions on $[0, infty)$. Suppose $f_n(x)to f(x) ~~~forall xin [0,infty)$ and $f$ is integrable. Then



if $int_{0}^{1}|f_n(x)-f(x)|dxto 0$ then $f_nto f$ uniformly on $[0,1]$: True/false



Im not able to find a counter example ?



Pliz help me






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    @henningmakholm Your first example that you deleted is valid if you let $f(1)=1$ and $f(x)=0$ for all other $x$. Then, $f_n(x)=x^n$ is continuous, has $f(x)$ as its limit, $f(x)$ is integrable, and $||f_n-f||_1to 0$. Yet, $x^n$ fails to converge uniformly. So, it works.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:24


















1












$begingroup$


Let ${f_n}$ be a sequence of continuous real valued functions on $[0, infty)$. Suppose $f_n(x)to f(x) ~~~forall xin [0,infty)$ and $f$ is integrable. Then



if $int_{0}^{1}|f_n(x)-f(x)|dxto 0$ then $f_nto f$ uniformly on $[0,1]$: True/false



Im not able to find a counter example ?



Pliz help me






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    @henningmakholm Your first example that you deleted is valid if you let $f(1)=1$ and $f(x)=0$ for all other $x$. Then, $f_n(x)=x^n$ is continuous, has $f(x)$ as its limit, $f(x)$ is integrable, and $||f_n-f||_1to 0$. Yet, $x^n$ fails to converge uniformly. So, it works.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:24
















1












1








1


1



$begingroup$


Let ${f_n}$ be a sequence of continuous real valued functions on $[0, infty)$. Suppose $f_n(x)to f(x) ~~~forall xin [0,infty)$ and $f$ is integrable. Then



if $int_{0}^{1}|f_n(x)-f(x)|dxto 0$ then $f_nto f$ uniformly on $[0,1]$: True/false



Im not able to find a counter example ?



Pliz help me






real-analysis







share|cite|improve this question









$endgroup$




Let ${f_n}$ be a sequence of continuous real valued functions on $[0, infty)$. Suppose $f_n(x)to f(x) ~~~forall xin [0,infty)$ and $f$ is integrable. Then



if $int_{0}^{1}|f_n(x)-f(x)|dxto 0$ then $f_nto f$ uniformly on $[0,1]$: True/false



Im not able to find a counter example ?



Pliz help me






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 4:15









Messi fifaMessi fifa

53111




53111












  • $begingroup$
    @henningmakholm Your first example that you deleted is valid if you let $f(1)=1$ and $f(x)=0$ for all other $x$. Then, $f_n(x)=x^n$ is continuous, has $f(x)$ as its limit, $f(x)$ is integrable, and $||f_n-f||_1to 0$. Yet, $x^n$ fails to converge uniformly. So, it works.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:24




















  • $begingroup$
    @henningmakholm Your first example that you deleted is valid if you let $f(1)=1$ and $f(x)=0$ for all other $x$. Then, $f_n(x)=x^n$ is continuous, has $f(x)$ as its limit, $f(x)$ is integrable, and $||f_n-f||_1to 0$. Yet, $x^n$ fails to converge uniformly. So, it works.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:24


















$begingroup$
@henningmakholm Your first example that you deleted is valid if you let $f(1)=1$ and $f(x)=0$ for all other $x$. Then, $f_n(x)=x^n$ is continuous, has $f(x)$ as its limit, $f(x)$ is integrable, and $||f_n-f||_1to 0$. Yet, $x^n$ fails to converge uniformly. So, it works.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:24






$begingroup$
@henningmakholm Your first example that you deleted is valid if you let $f(1)=1$ and $f(x)=0$ for all other $x$. Then, $f_n(x)=x^n$ is continuous, has $f(x)$ as its limit, $f(x)$ is integrable, and $||f_n-f||_1to 0$. Yet, $x^n$ fails to converge uniformly. So, it works.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:24












3 Answers
3






active

oldest

votes


















2












$begingroup$

Here is an example where $f_n(x)$ does not converge at any point! Let $I_n=[frac {i-1} {2^{n}},frac i {2^{n}})$ for $1leq i leq 2^{n}$ and $n geq 1$. Arrange the characteristic functions of these intervals in a sequence by first listing those with $n=1$, then the one's with $n=2$ etc. Then $int_0^{1} |f_n(x)-0|, dx to 0$ but the functions do not converge even at one point.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks u sir @Kavi ram
    $endgroup$
    – Messi fifa
    Nov 30 '18 at 21:59



















2












$begingroup$

False .. $x^n$ is not uniformly convergent on $[0,1]$. Here the limit function is $f(x) = 0$ when $x in [0,1)$ and $f(1) = 1$



But $int_{0}^{1}|x^n-f(x)|dxto 0$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    (+1) Clear and concise.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:25










  • $begingroup$
    Except that $x^n$ does not converge pointwise on $[0,infty)$ as the question (rather pointlessly, to be sure) requires.
    $endgroup$
    – Henning Makholm
    Nov 30 '18 at 4:27





















1












$begingroup$

Let $g(x)=xe^{-x}$, and then consider $f(x)=0$ and $f_n(x)=g(nx)$.






share|cite|improve this answer









$endgroup$













    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Here is an example where $f_n(x)$ does not converge at any point! Let $I_n=[frac {i-1} {2^{n}},frac i {2^{n}})$ for $1leq i leq 2^{n}$ and $n geq 1$. Arrange the characteristic functions of these intervals in a sequence by first listing those with $n=1$, then the one's with $n=2$ etc. Then $int_0^{1} |f_n(x)-0|, dx to 0$ but the functions do not converge even at one point.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks u sir @Kavi ram
      $endgroup$
      – Messi fifa
      Nov 30 '18 at 21:59
















    2












    $begingroup$

    Here is an example where $f_n(x)$ does not converge at any point! Let $I_n=[frac {i-1} {2^{n}},frac i {2^{n}})$ for $1leq i leq 2^{n}$ and $n geq 1$. Arrange the characteristic functions of these intervals in a sequence by first listing those with $n=1$, then the one's with $n=2$ etc. Then $int_0^{1} |f_n(x)-0|, dx to 0$ but the functions do not converge even at one point.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks u sir @Kavi ram
      $endgroup$
      – Messi fifa
      Nov 30 '18 at 21:59














    2












    2








    2





    $begingroup$

    Here is an example where $f_n(x)$ does not converge at any point! Let $I_n=[frac {i-1} {2^{n}},frac i {2^{n}})$ for $1leq i leq 2^{n}$ and $n geq 1$. Arrange the characteristic functions of these intervals in a sequence by first listing those with $n=1$, then the one's with $n=2$ etc. Then $int_0^{1} |f_n(x)-0|, dx to 0$ but the functions do not converge even at one point.






    share|cite|improve this answer









    $endgroup$



    Here is an example where $f_n(x)$ does not converge at any point! Let $I_n=[frac {i-1} {2^{n}},frac i {2^{n}})$ for $1leq i leq 2^{n}$ and $n geq 1$. Arrange the characteristic functions of these intervals in a sequence by first listing those with $n=1$, then the one's with $n=2$ etc. Then $int_0^{1} |f_n(x)-0|, dx to 0$ but the functions do not converge even at one point.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 30 '18 at 6:08









    Kavi Rama MurthyKavi Rama Murthy

    59k42161




    59k42161












    • $begingroup$
      thanks u sir @Kavi ram
      $endgroup$
      – Messi fifa
      Nov 30 '18 at 21:59


















    • $begingroup$
      thanks u sir @Kavi ram
      $endgroup$
      – Messi fifa
      Nov 30 '18 at 21:59
















    $begingroup$
    thanks u sir @Kavi ram
    $endgroup$
    – Messi fifa
    Nov 30 '18 at 21:59




    $begingroup$
    thanks u sir @Kavi ram
    $endgroup$
    – Messi fifa
    Nov 30 '18 at 21:59











    2












    $begingroup$

    False .. $x^n$ is not uniformly convergent on $[0,1]$. Here the limit function is $f(x) = 0$ when $x in [0,1)$ and $f(1) = 1$



    But $int_{0}^{1}|x^n-f(x)|dxto 0$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      (+1) Clear and concise.
      $endgroup$
      – Mark Viola
      Nov 30 '18 at 4:25










    • $begingroup$
      Except that $x^n$ does not converge pointwise on $[0,infty)$ as the question (rather pointlessly, to be sure) requires.
      $endgroup$
      – Henning Makholm
      Nov 30 '18 at 4:27


















    2












    $begingroup$

    False .. $x^n$ is not uniformly convergent on $[0,1]$. Here the limit function is $f(x) = 0$ when $x in [0,1)$ and $f(1) = 1$



    But $int_{0}^{1}|x^n-f(x)|dxto 0$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      (+1) Clear and concise.
      $endgroup$
      – Mark Viola
      Nov 30 '18 at 4:25










    • $begingroup$
      Except that $x^n$ does not converge pointwise on $[0,infty)$ as the question (rather pointlessly, to be sure) requires.
      $endgroup$
      – Henning Makholm
      Nov 30 '18 at 4:27
















    2












    2








    2





    $begingroup$

    False .. $x^n$ is not uniformly convergent on $[0,1]$. Here the limit function is $f(x) = 0$ when $x in [0,1)$ and $f(1) = 1$



    But $int_{0}^{1}|x^n-f(x)|dxto 0$






    share|cite|improve this answer









    $endgroup$



    False .. $x^n$ is not uniformly convergent on $[0,1]$. Here the limit function is $f(x) = 0$ when $x in [0,1)$ and $f(1) = 1$



    But $int_{0}^{1}|x^n-f(x)|dxto 0$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 30 '18 at 4:24









    cmicmi

    1,116312




    1,116312








    • 1




      $begingroup$
      (+1) Clear and concise.
      $endgroup$
      – Mark Viola
      Nov 30 '18 at 4:25










    • $begingroup$
      Except that $x^n$ does not converge pointwise on $[0,infty)$ as the question (rather pointlessly, to be sure) requires.
      $endgroup$
      – Henning Makholm
      Nov 30 '18 at 4:27
















    • 1




      $begingroup$
      (+1) Clear and concise.
      $endgroup$
      – Mark Viola
      Nov 30 '18 at 4:25










    • $begingroup$
      Except that $x^n$ does not converge pointwise on $[0,infty)$ as the question (rather pointlessly, to be sure) requires.
      $endgroup$
      – Henning Makholm
      Nov 30 '18 at 4:27










    1




    1




    $begingroup$
    (+1) Clear and concise.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:25




    $begingroup$
    (+1) Clear and concise.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:25












    $begingroup$
    Except that $x^n$ does not converge pointwise on $[0,infty)$ as the question (rather pointlessly, to be sure) requires.
    $endgroup$
    – Henning Makholm
    Nov 30 '18 at 4:27






    $begingroup$
    Except that $x^n$ does not converge pointwise on $[0,infty)$ as the question (rather pointlessly, to be sure) requires.
    $endgroup$
    – Henning Makholm
    Nov 30 '18 at 4:27













    1












    $begingroup$

    Let $g(x)=xe^{-x}$, and then consider $f(x)=0$ and $f_n(x)=g(nx)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $g(x)=xe^{-x}$, and then consider $f(x)=0$ and $f_n(x)=g(nx)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $g(x)=xe^{-x}$, and then consider $f(x)=0$ and $f_n(x)=g(nx)$.






        share|cite|improve this answer









        $endgroup$



        Let $g(x)=xe^{-x}$, and then consider $f(x)=0$ and $f_n(x)=g(nx)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 4:18









        Henning MakholmHenning Makholm

        240k17305541




        240k17305541






























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