If $|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$, then exists a subsequence which converges uniformly [duplicate]

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  • Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.

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Let $f_n:[0,1]rightarrow mathbbR$ be continuous functions such that:
$|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$ for every $nin N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.




I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.



What I have noted/done so far:



Since $|f'_n(x)|leq frac1sqrt x$, if $x,yin [0,1],xneq y$, then



$$|f_n(y)-f_n(x)|=|int_x^yf'_n(t)dt |leqint_x^y|f'_n(t)|dtleqint_x^yfrac1sqrt tdtleqint_0^1frac1sqrt tdt = K$$



but I can't use this to solve the problem. Any help? Thanks in advance







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marked as duplicate by Xander Henderson, Sil, Taroccoesbrocco, Theoretical Economist, Lord Shark the Unknown Aug 12 at 11:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















    up vote
    2
    down vote

    favorite
    1













    This question already has an answer here:



    • Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.

      1 answer




    Let $f_n:[0,1]rightarrow mathbbR$ be continuous functions such that:
    $|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$ for every $nin N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.




    I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.



    What I have noted/done so far:



    Since $|f'_n(x)|leq frac1sqrt x$, if $x,yin [0,1],xneq y$, then



    $$|f_n(y)-f_n(x)|=|int_x^yf'_n(t)dt |leqint_x^y|f'_n(t)|dtleqint_x^yfrac1sqrt tdtleqint_0^1frac1sqrt tdt = K$$



    but I can't use this to solve the problem. Any help? Thanks in advance







    share|cite|improve this question














    marked as duplicate by Xander Henderson, Sil, Taroccoesbrocco, Theoretical Economist, Lord Shark the Unknown Aug 12 at 11:03


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
















      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1






      This question already has an answer here:



      • Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.

        1 answer




      Let $f_n:[0,1]rightarrow mathbbR$ be continuous functions such that:
      $|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$ for every $nin N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.




      I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.



      What I have noted/done so far:



      Since $|f'_n(x)|leq frac1sqrt x$, if $x,yin [0,1],xneq y$, then



      $$|f_n(y)-f_n(x)|=|int_x^yf'_n(t)dt |leqint_x^y|f'_n(t)|dtleqint_x^yfrac1sqrt tdtleqint_0^1frac1sqrt tdt = K$$



      but I can't use this to solve the problem. Any help? Thanks in advance







      share|cite|improve this question















      This question already has an answer here:



      • Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.

        1 answer




      Let $f_n:[0,1]rightarrow mathbbR$ be continuous functions such that:
      $|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$ for every $nin N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.




      I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.



      What I have noted/done so far:



      Since $|f'_n(x)|leq frac1sqrt x$, if $x,yin [0,1],xneq y$, then



      $$|f_n(y)-f_n(x)|=|int_x^yf'_n(t)dt |leqint_x^y|f'_n(t)|dtleqint_x^yfrac1sqrt tdtleqint_0^1frac1sqrt tdt = K$$



      but I can't use this to solve the problem. Any help? Thanks in advance





      This question already has an answer here:



      • Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.

        1 answer









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 12 at 7:02









      Sil

      5,14121443




      5,14121443










      asked Jun 20 at 11:26









      math.h

      1,034516




      1,034516




      marked as duplicate by Xander Henderson, Sil, Taroccoesbrocco, Theoretical Economist, Lord Shark the Unknown Aug 12 at 11:03


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Xander Henderson, Sil, Taroccoesbrocco, Theoretical Economist, Lord Shark the Unknown Aug 12 at 11:03


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















          1 Answer
          1






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          oldest

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          up vote
          3
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          accepted










          Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
          $$
          |f_n(y) - f_n(x)| leq left|
          int_x^y frac1sqrts, ds
          right|
          = 2 left| sqrty - sqrtxright|
          leq 2 sqrt,
          $$
          so that the sequence $(f_n)$ is equicontinuous.



          On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
          $$
          |f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
          qquad forall xin [0,1].
          $$






          share|cite|improve this answer




















          • perfectly noted, thank you
            – math.h
            Jun 20 at 11:46

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
          $$
          |f_n(y) - f_n(x)| leq left|
          int_x^y frac1sqrts, ds
          right|
          = 2 left| sqrty - sqrtxright|
          leq 2 sqrt,
          $$
          so that the sequence $(f_n)$ is equicontinuous.



          On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
          $$
          |f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
          qquad forall xin [0,1].
          $$






          share|cite|improve this answer




















          • perfectly noted, thank you
            – math.h
            Jun 20 at 11:46














          up vote
          3
          down vote



          accepted










          Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
          $$
          |f_n(y) - f_n(x)| leq left|
          int_x^y frac1sqrts, ds
          right|
          = 2 left| sqrty - sqrtxright|
          leq 2 sqrt,
          $$
          so that the sequence $(f_n)$ is equicontinuous.



          On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
          $$
          |f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
          qquad forall xin [0,1].
          $$






          share|cite|improve this answer




















          • perfectly noted, thank you
            – math.h
            Jun 20 at 11:46












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
          $$
          |f_n(y) - f_n(x)| leq left|
          int_x^y frac1sqrts, ds
          right|
          = 2 left| sqrty - sqrtxright|
          leq 2 sqrt,
          $$
          so that the sequence $(f_n)$ is equicontinuous.



          On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
          $$
          |f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
          qquad forall xin [0,1].
          $$






          share|cite|improve this answer












          Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
          $$
          |f_n(y) - f_n(x)| leq left|
          int_x^y frac1sqrts, ds
          right|
          = 2 left| sqrty - sqrtxright|
          leq 2 sqrt,
          $$
          so that the sequence $(f_n)$ is equicontinuous.



          On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
          $$
          |f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
          qquad forall xin [0,1].
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jun 20 at 11:39









          Rigel

          9,61511319




          9,61511319











          • perfectly noted, thank you
            – math.h
            Jun 20 at 11:46
















          • perfectly noted, thank you
            – math.h
            Jun 20 at 11:46















          perfectly noted, thank you
          – math.h
          Jun 20 at 11:46




          perfectly noted, thank you
          – math.h
          Jun 20 at 11:46


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