Does this sequence of function converges uniformly?












0












$begingroup$


I was practicing for my final and stumbled on this problem.



Let $f_n$:$[0,1]$ $mapsto$ $R$ such that



$f(x)_n$ =
begin{cases}
nx, & 0leq text{ $x$ } lt1/n\
1, & 1/nleqtext{ $x$ }leq 1
end{cases}



I could find its piecewise limit, which is



$f(x)$ =
begin{cases}
0, & text{ $x$ =0} \
1, & text{ $otherwise$ }
end{cases}



I was thinking about using the $sup$$|f(x)_n-f(x)|<epsilon$



then I realized there were more than one interval of both $f_n$ and $f$.



So I used $x$ to indicate the supremum of each interval



for $x=0$



Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$



Since both function yield $0$



for $1/nleqtext{ $x$ }leq 1$



Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$



Since both function yield $1$



Now I am stuck here



for $0lt text{ $x$ } lt1/n$



What should i do with this interval?



And does $f(x)_n$ converges uniformly?






real-analysis uniform-convergence







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I was practicing for my final and stumbled on this problem.



    Let $f_n$:$[0,1]$ $mapsto$ $R$ such that



    $f(x)_n$ =
    begin{cases}
    nx, & 0leq text{ $x$ } lt1/n\
    1, & 1/nleqtext{ $x$ }leq 1
    end{cases}



    I could find its piecewise limit, which is



    $f(x)$ =
    begin{cases}
    0, & text{ $x$ =0} \
    1, & text{ $otherwise$ }
    end{cases}



    I was thinking about using the $sup$$|f(x)_n-f(x)|<epsilon$



    then I realized there were more than one interval of both $f_n$ and $f$.



    So I used $x$ to indicate the supremum of each interval



    for $x=0$



    Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$



    Since both function yield $0$



    for $1/nleqtext{ $x$ }leq 1$



    Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$



    Since both function yield $1$



    Now I am stuck here



    for $0lt text{ $x$ } lt1/n$



    What should i do with this interval?



    And does $f(x)_n$ converges uniformly?






    real-analysis uniform-convergence







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I was practicing for my final and stumbled on this problem.



      Let $f_n$:$[0,1]$ $mapsto$ $R$ such that



      $f(x)_n$ =
      begin{cases}
      nx, & 0leq text{ $x$ } lt1/n\
      1, & 1/nleqtext{ $x$ }leq 1
      end{cases}



      I could find its piecewise limit, which is



      $f(x)$ =
      begin{cases}
      0, & text{ $x$ =0} \
      1, & text{ $otherwise$ }
      end{cases}



      I was thinking about using the $sup$$|f(x)_n-f(x)|<epsilon$



      then I realized there were more than one interval of both $f_n$ and $f$.



      So I used $x$ to indicate the supremum of each interval



      for $x=0$



      Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$



      Since both function yield $0$



      for $1/nleqtext{ $x$ }leq 1$



      Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$



      Since both function yield $1$



      Now I am stuck here



      for $0lt text{ $x$ } lt1/n$



      What should i do with this interval?



      And does $f(x)_n$ converges uniformly?






      real-analysis uniform-convergence







      share|cite|improve this question









      $endgroup$




      I was practicing for my final and stumbled on this problem.



      Let $f_n$:$[0,1]$ $mapsto$ $R$ such that



      $f(x)_n$ =
      begin{cases}
      nx, & 0leq text{ $x$ } lt1/n\
      1, & 1/nleqtext{ $x$ }leq 1
      end{cases}



      I could find its piecewise limit, which is



      $f(x)$ =
      begin{cases}
      0, & text{ $x$ =0} \
      1, & text{ $otherwise$ }
      end{cases}



      I was thinking about using the $sup$$|f(x)_n-f(x)|<epsilon$



      then I realized there were more than one interval of both $f_n$ and $f$.



      So I used $x$ to indicate the supremum of each interval



      for $x=0$



      Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$



      Since both function yield $0$



      for $1/nleqtext{ $x$ }leq 1$



      Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$



      Since both function yield $1$



      Now I am stuck here



      for $0lt text{ $x$ } lt1/n$



      What should i do with this interval?



      And does $f(x)_n$ converges uniformly?






      real-analysis uniform-convergence




      real-analysis uniform-convergence






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 '18 at 17:21









      AtchanaphongAtchanaphong

      1




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          2 Answers
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          active

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          0












          $begingroup$

          If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh! I totally forgot about that! Thank you, sir!
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:31










          • $begingroup$
            @Atchanaphong You're welcome!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 17:33










          • $begingroup$
            One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:43












          • $begingroup$
            @Atchanaphong Yup!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 19:11



















          -1












          $begingroup$

          The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. I totally forgot about this since I was focusing on the supremum.
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:33











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes









          0












          $begingroup$

          If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh! I totally forgot about that! Thank you, sir!
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:31










          • $begingroup$
            @Atchanaphong You're welcome!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 17:33










          • $begingroup$
            One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:43












          • $begingroup$
            @Atchanaphong Yup!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 19:11
















          0












          $begingroup$

          If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh! I totally forgot about that! Thank you, sir!
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:31










          • $begingroup$
            @Atchanaphong You're welcome!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 17:33










          • $begingroup$
            One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:43












          • $begingroup$
            @Atchanaphong Yup!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 19:11














          0












          0








          0





          $begingroup$

          If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.






          share|cite|improve this answer









          $endgroup$



          If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 17:27









          GenericMathematicianGenericMathematician

          863




          863












          • $begingroup$
            Oh! I totally forgot about that! Thank you, sir!
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:31










          • $begingroup$
            @Atchanaphong You're welcome!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 17:33










          • $begingroup$
            One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:43












          • $begingroup$
            @Atchanaphong Yup!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 19:11


















          • $begingroup$
            Oh! I totally forgot about that! Thank you, sir!
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:31










          • $begingroup$
            @Atchanaphong You're welcome!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 17:33










          • $begingroup$
            One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:43












          • $begingroup$
            @Atchanaphong Yup!
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 19:11
















          $begingroup$
          Oh! I totally forgot about that! Thank you, sir!
          $endgroup$
          – Atchanaphong
          Dec 9 '18 at 17:31




          $begingroup$
          Oh! I totally forgot about that! Thank you, sir!
          $endgroup$
          – Atchanaphong
          Dec 9 '18 at 17:31












          $begingroup$
          @Atchanaphong You're welcome!
          $endgroup$
          – GenericMathematician
          Dec 9 '18 at 17:33




          $begingroup$
          @Atchanaphong You're welcome!
          $endgroup$
          – GenericMathematician
          Dec 9 '18 at 17:33












          $begingroup$
          One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
          $endgroup$
          – Atchanaphong
          Dec 9 '18 at 17:43






          $begingroup$
          One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
          $endgroup$
          – Atchanaphong
          Dec 9 '18 at 17:43














          $begingroup$
          @Atchanaphong Yup!
          $endgroup$
          – GenericMathematician
          Dec 9 '18 at 19:11




          $begingroup$
          @Atchanaphong Yup!
          $endgroup$
          – GenericMathematician
          Dec 9 '18 at 19:11











          -1












          $begingroup$

          The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. I totally forgot about this since I was focusing on the supremum.
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:33
















          -1












          $begingroup$

          The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. I totally forgot about this since I was focusing on the supremum.
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:33














          -1












          -1








          -1





          $begingroup$

          The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.






          share|cite|improve this answer









          $endgroup$



          The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 17:24









          José Carlos SantosJosé Carlos Santos

          162k22128232




          162k22128232












          • $begingroup$
            Thank you. I totally forgot about this since I was focusing on the supremum.
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:33


















          • $begingroup$
            Thank you. I totally forgot about this since I was focusing on the supremum.
            $endgroup$
            – Atchanaphong
            Dec 9 '18 at 17:33
















          $begingroup$
          Thank you. I totally forgot about this since I was focusing on the supremum.
          $endgroup$
          – Atchanaphong
          Dec 9 '18 at 17:33




          $begingroup$
          Thank you. I totally forgot about this since I was focusing on the supremum.
          $endgroup$
          – Atchanaphong
          Dec 9 '18 at 17:33


















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