Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A.$












0












$begingroup$


Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A$



begin{equation*}
f_{n}(x) = g(nx),quad g(x)=
begin{cases}
x & text{if }0 leq x < 1/2,\
1-x & text{if } 1/2 leq x leq 1,\
0 & text{if } x > 1.
end{cases}
end{equation*}



I know that the first step is to find the limit of $f_{n}(x),$ which I do not know how, could anyone help me in doing so?



Also the book did not wrote what is the set $A$, can I assume that it is $mathbb{R}$ and start my solution?






calculus sequences-and-series uniform-convergence







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A$



    begin{equation*}
    f_{n}(x) = g(nx),quad g(x)=
    begin{cases}
    x & text{if }0 leq x < 1/2,\
    1-x & text{if } 1/2 leq x leq 1,\
    0 & text{if } x > 1.
    end{cases}
    end{equation*}



    I know that the first step is to find the limit of $f_{n}(x),$ which I do not know how, could anyone help me in doing so?



    Also the book did not wrote what is the set $A$, can I assume that it is $mathbb{R}$ and start my solution?






    calculus sequences-and-series uniform-convergence







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A$



      begin{equation*}
      f_{n}(x) = g(nx),quad g(x)=
      begin{cases}
      x & text{if }0 leq x < 1/2,\
      1-x & text{if } 1/2 leq x leq 1,\
      0 & text{if } x > 1.
      end{cases}
      end{equation*}



      I know that the first step is to find the limit of $f_{n}(x),$ which I do not know how, could anyone help me in doing so?



      Also the book did not wrote what is the set $A$, can I assume that it is $mathbb{R}$ and start my solution?






      calculus sequences-and-series uniform-convergence







      share|cite|improve this question











      $endgroup$




      Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A$



      begin{equation*}
      f_{n}(x) = g(nx),quad g(x)=
      begin{cases}
      x & text{if }0 leq x < 1/2,\
      1-x & text{if } 1/2 leq x leq 1,\
      0 & text{if } x > 1.
      end{cases}
      end{equation*}



      I know that the first step is to find the limit of $f_{n}(x),$ which I do not know how, could anyone help me in doing so?



      Also the book did not wrote what is the set $A$, can I assume that it is $mathbb{R}$ and start my solution?






      calculus sequences-and-series uniform-convergence




      calculus sequences-and-series uniform-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 10:00









      user549397

      1,5101418




      1,5101418










      asked Jan 21 at 0:35









      hopefullyhopefully

      247114




      247114






















          1 Answer
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          1












          $begingroup$

          $f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is the value of this $epsilon$?
            $endgroup$
            – hopefully
            Jan 21 at 0:50










          • $begingroup$
            what is the intuition behind this solution?
            $endgroup$
            – hopefully
            Jan 21 at 0:51










          • $begingroup$
            what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
            $endgroup$
            – hopefully
            Jan 21 at 0:52






          • 1




            $begingroup$
            @hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:14






          • 1




            $begingroup$
            @hopefully Yes, you are right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 22 at 0:12











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          $f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is the value of this $epsilon$?
            $endgroup$
            – hopefully
            Jan 21 at 0:50










          • $begingroup$
            what is the intuition behind this solution?
            $endgroup$
            – hopefully
            Jan 21 at 0:51










          • $begingroup$
            what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
            $endgroup$
            – hopefully
            Jan 21 at 0:52






          • 1




            $begingroup$
            @hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:14






          • 1




            $begingroup$
            @hopefully Yes, you are right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 22 at 0:12
















          1












          $begingroup$

          $f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is the value of this $epsilon$?
            $endgroup$
            – hopefully
            Jan 21 at 0:50










          • $begingroup$
            what is the intuition behind this solution?
            $endgroup$
            – hopefully
            Jan 21 at 0:51










          • $begingroup$
            what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
            $endgroup$
            – hopefully
            Jan 21 at 0:52






          • 1




            $begingroup$
            @hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:14






          • 1




            $begingroup$
            @hopefully Yes, you are right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 22 at 0:12














          1












          1








          1





          $begingroup$

          $f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].






          share|cite|improve this answer











          $endgroup$



          $f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 21 at 9:50

























          answered Jan 21 at 0:40









          Kavi Rama MurthyKavi Rama Murthy

          61.9k42262




          61.9k42262












          • $begingroup$
            what is the value of this $epsilon$?
            $endgroup$
            – hopefully
            Jan 21 at 0:50










          • $begingroup$
            what is the intuition behind this solution?
            $endgroup$
            – hopefully
            Jan 21 at 0:51










          • $begingroup$
            what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
            $endgroup$
            – hopefully
            Jan 21 at 0:52






          • 1




            $begingroup$
            @hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:14






          • 1




            $begingroup$
            @hopefully Yes, you are right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 22 at 0:12


















          • $begingroup$
            what is the value of this $epsilon$?
            $endgroup$
            – hopefully
            Jan 21 at 0:50










          • $begingroup$
            what is the intuition behind this solution?
            $endgroup$
            – hopefully
            Jan 21 at 0:51










          • $begingroup$
            what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
            $endgroup$
            – hopefully
            Jan 21 at 0:52






          • 1




            $begingroup$
            @hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:14






          • 1




            $begingroup$
            @hopefully Yes, you are right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 22 at 0:12
















          $begingroup$
          what is the value of this $epsilon$?
          $endgroup$
          – hopefully
          Jan 21 at 0:50




          $begingroup$
          what is the value of this $epsilon$?
          $endgroup$
          – hopefully
          Jan 21 at 0:50












          $begingroup$
          what is the intuition behind this solution?
          $endgroup$
          – hopefully
          Jan 21 at 0:51




          $begingroup$
          what is the intuition behind this solution?
          $endgroup$
          – hopefully
          Jan 21 at 0:51












          $begingroup$
          what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
          $endgroup$
          – hopefully
          Jan 21 at 0:52




          $begingroup$
          what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
          $endgroup$
          – hopefully
          Jan 21 at 0:52




          1




          1




          $begingroup$
          @hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
          $endgroup$
          – Kavi Rama Murthy
          Jan 21 at 23:14




          $begingroup$
          @hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
          $endgroup$
          – Kavi Rama Murthy
          Jan 21 at 23:14




          1




          1




          $begingroup$
          @hopefully Yes, you are right.
          $endgroup$
          – Kavi Rama Murthy
          Jan 22 at 0:12




          $begingroup$
          @hopefully Yes, you are right.
          $endgroup$
          – Kavi Rama Murthy
          Jan 22 at 0:12


















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