Continuous function as pointwise limit but not as uniform limit of a sequence of continuous functions on...












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I recentaly find an article where it is said that there is a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ that converges pointqise almost everywhere to zero function , but not converges uniformly.



My question is the following :--



Let $f:[0,1]rightarrow Bbb R$ be continuous , does there always exists a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ such that $f_nrightarrow f$ pointwise, but not uniformly on $[0,1]$.






real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence







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    $begingroup$


    I recentaly find an article where it is said that there is a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ that converges pointqise almost everywhere to zero function , but not converges uniformly.



    My question is the following :--



    Let $f:[0,1]rightarrow Bbb R$ be continuous , does there always exists a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ such that $f_nrightarrow f$ pointwise, but not uniformly on $[0,1]$.






    real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      I recentaly find an article where it is said that there is a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ that converges pointqise almost everywhere to zero function , but not converges uniformly.



      My question is the following :--



      Let $f:[0,1]rightarrow Bbb R$ be continuous , does there always exists a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ such that $f_nrightarrow f$ pointwise, but not uniformly on $[0,1]$.






      real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence







      share|cite|improve this question









      $endgroup$




      I recentaly find an article where it is said that there is a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ that converges pointqise almost everywhere to zero function , but not converges uniformly.



      My question is the following :--



      Let $f:[0,1]rightarrow Bbb R$ be continuous , does there always exists a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ such that $f_nrightarrow f$ pointwise, but not uniformly on $[0,1]$.






      real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence




      real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 12 '18 at 13:30









      MathloverMathlover

      1447




      1447






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          Yes, there is always such a function.



          Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.



          For an example of such $g_n$, you can take the following:
          $$
          g_n(x) = cases{
          nx & if $x< frac1n$\
          2-nx & if $frac1n leq x < frac2n$\
          0& otherwise}
          $$
          The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
            $endgroup$
            – 5xum
            Dec 12 '18 at 13:33










          • $begingroup$
            @5xum I gues it's pretty standardized how one makes the "trivial" examples.
            $endgroup$
            – Arthur
            Dec 12 '18 at 13:38



















          3












          $begingroup$

          Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).



          Let $f$ be a continuous function.



          Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            Yes, there is always such a function.



            Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.



            For an example of such $g_n$, you can take the following:
            $$
            g_n(x) = cases{
            nx & if $x< frac1n$\
            2-nx & if $frac1n leq x < frac2n$\
            0& otherwise}
            $$
            The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
              $endgroup$
              – 5xum
              Dec 12 '18 at 13:33










            • $begingroup$
              @5xum I gues it's pretty standardized how one makes the "trivial" examples.
              $endgroup$
              – Arthur
              Dec 12 '18 at 13:38
















            5












            $begingroup$

            Yes, there is always such a function.



            Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.



            For an example of such $g_n$, you can take the following:
            $$
            g_n(x) = cases{
            nx & if $x< frac1n$\
            2-nx & if $frac1n leq x < frac2n$\
            0& otherwise}
            $$
            The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
              $endgroup$
              – 5xum
              Dec 12 '18 at 13:33










            • $begingroup$
              @5xum I gues it's pretty standardized how one makes the "trivial" examples.
              $endgroup$
              – Arthur
              Dec 12 '18 at 13:38














            5












            5








            5





            $begingroup$

            Yes, there is always such a function.



            Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.



            For an example of such $g_n$, you can take the following:
            $$
            g_n(x) = cases{
            nx & if $x< frac1n$\
            2-nx & if $frac1n leq x < frac2n$\
            0& otherwise}
            $$
            The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.






            share|cite|improve this answer











            $endgroup$



            Yes, there is always such a function.



            Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.



            For an example of such $g_n$, you can take the following:
            $$
            g_n(x) = cases{
            nx & if $x< frac1n$\
            2-nx & if $frac1n leq x < frac2n$\
            0& otherwise}
            $$
            The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 12 '18 at 13:36

























            answered Dec 12 '18 at 13:31









            ArthurArthur

            116k7116198




            116k7116198








            • 1




              $begingroup$
              Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
              $endgroup$
              – 5xum
              Dec 12 '18 at 13:33










            • $begingroup$
              @5xum I gues it's pretty standardized how one makes the "trivial" examples.
              $endgroup$
              – Arthur
              Dec 12 '18 at 13:38














            • 1




              $begingroup$
              Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
              $endgroup$
              – 5xum
              Dec 12 '18 at 13:33










            • $begingroup$
              @5xum I gues it's pretty standardized how one makes the "trivial" examples.
              $endgroup$
              – Arthur
              Dec 12 '18 at 13:38








            1




            1




            $begingroup$
            Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
            $endgroup$
            – 5xum
            Dec 12 '18 at 13:33




            $begingroup$
            Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
            $endgroup$
            – 5xum
            Dec 12 '18 at 13:33












            $begingroup$
            @5xum I gues it's pretty standardized how one makes the "trivial" examples.
            $endgroup$
            – Arthur
            Dec 12 '18 at 13:38




            $begingroup$
            @5xum I gues it's pretty standardized how one makes the "trivial" examples.
            $endgroup$
            – Arthur
            Dec 12 '18 at 13:38











            3












            $begingroup$

            Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).



            Let $f$ be a continuous function.



            Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).



              Let $f$ be a continuous function.



              Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).



                Let $f$ be a continuous function.



                Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.






                share|cite|improve this answer









                $endgroup$



                Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).



                Let $f$ be a continuous function.



                Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 12 '18 at 13:32









                5xum5xum

                91.1k394161




                91.1k394161






























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