Pointwise vs Uniform Convergence












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$begingroup$


Let {$f_n$ } be a sequence of functions defined by $f (x) =frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that




  1. {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$

  2. {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.


I have proved that $f_n$ converges uniformly to $f(x) = 0$.



However, I've realized that {$f′_n$ } = $frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?






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    0












    $begingroup$


    Let {$f_n$ } be a sequence of functions defined by $f (x) =frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that




    1. {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$

    2. {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.


    I have proved that $f_n$ converges uniformly to $f(x) = 0$.



    However, I've realized that {$f′_n$ } = $frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?






    real-analysis







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let {$f_n$ } be a sequence of functions defined by $f (x) =frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that




      1. {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$

      2. {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.


      I have proved that $f_n$ converges uniformly to $f(x) = 0$.



      However, I've realized that {$f′_n$ } = $frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?






      real-analysis







      share|cite|improve this question











      $endgroup$




      Let {$f_n$ } be a sequence of functions defined by $f (x) =frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that




      1. {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$

      2. {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.


      I have proved that $f_n$ converges uniformly to $f(x) = 0$.



      However, I've realized that {$f′_n$ } = $frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?






      real-analysis




      real-analysis






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      share|cite|improve this question













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      edited Dec 1 '18 at 19:31









      Mostafa Ayaz

      15.3k3939




      15.3k3939










      asked Dec 1 '18 at 19:23









      Ahmad LamaaAhmad Lamaa

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      1326






















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          $begingroup$

          The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
          $$g(x) = left{ array{0 &text{ if } xin (0,1] \
          1 & text{ if } x=0 }right. $$



          (Note that $f^prime_n(0) = 1$ for every $n$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's correct. I have overlooked the case where x=0. Thanks!
            $endgroup$
            – Ahmad Lamaa
            Dec 1 '18 at 19:50











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          1 Answer
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          active

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          $begingroup$

          The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
          $$g(x) = left{ array{0 &text{ if } xin (0,1] \
          1 & text{ if } x=0 }right. $$



          (Note that $f^prime_n(0) = 1$ for every $n$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's correct. I have overlooked the case where x=0. Thanks!
            $endgroup$
            – Ahmad Lamaa
            Dec 1 '18 at 19:50
















          2












          $begingroup$

          The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
          $$g(x) = left{ array{0 &text{ if } xin (0,1] \
          1 & text{ if } x=0 }right. $$



          (Note that $f^prime_n(0) = 1$ for every $n$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's correct. I have overlooked the case where x=0. Thanks!
            $endgroup$
            – Ahmad Lamaa
            Dec 1 '18 at 19:50














          2












          2








          2





          $begingroup$

          The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
          $$g(x) = left{ array{0 &text{ if } xin (0,1] \
          1 & text{ if } x=0 }right. $$



          (Note that $f^prime_n(0) = 1$ for every $n$).






          share|cite|improve this answer











          $endgroup$



          The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to
          $$g(x) = left{ array{0 &text{ if } xin (0,1] \
          1 & text{ if } x=0 }right. $$



          (Note that $f^prime_n(0) = 1$ for every $n$).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 20:02









          amWhy

          192k28225439




          192k28225439










          answered Dec 1 '18 at 19:46









          ThomasThomas

          16.7k21631




          16.7k21631












          • $begingroup$
            That's correct. I have overlooked the case where x=0. Thanks!
            $endgroup$
            – Ahmad Lamaa
            Dec 1 '18 at 19:50


















          • $begingroup$
            That's correct. I have overlooked the case where x=0. Thanks!
            $endgroup$
            – Ahmad Lamaa
            Dec 1 '18 at 19:50
















          $begingroup$
          That's correct. I have overlooked the case where x=0. Thanks!
          $endgroup$
          – Ahmad Lamaa
          Dec 1 '18 at 19:50




          $begingroup$
          That's correct. I have overlooked the case where x=0. Thanks!
          $endgroup$
          – Ahmad Lamaa
          Dec 1 '18 at 19:50


















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