x/n uniformly convergent?












0












$begingroup$


Let $f_n(x) : mathbb{R} rightarrow mathbb{R}$ be defined as $dfrac{x}{n}$. Is $f_n$ uniformly convergent?



With "$(a_n)$ is limited and $(b_n)$ converges to 0 $Rightarrow (a_nb_n)$ converges also to 0" i would guess that $xcdot dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?






analysis uniform-convergence







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$endgroup$












  • $begingroup$
    You need to specify an interval for the values of $x$.
    $endgroup$
    – GaC
    Jan 30 '17 at 13:50
















0












$begingroup$


Let $f_n(x) : mathbb{R} rightarrow mathbb{R}$ be defined as $dfrac{x}{n}$. Is $f_n$ uniformly convergent?



With "$(a_n)$ is limited and $(b_n)$ converges to 0 $Rightarrow (a_nb_n)$ converges also to 0" i would guess that $xcdot dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?






analysis uniform-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    You need to specify an interval for the values of $x$.
    $endgroup$
    – GaC
    Jan 30 '17 at 13:50














0












0








0





$begingroup$


Let $f_n(x) : mathbb{R} rightarrow mathbb{R}$ be defined as $dfrac{x}{n}$. Is $f_n$ uniformly convergent?



With "$(a_n)$ is limited and $(b_n)$ converges to 0 $Rightarrow (a_nb_n)$ converges also to 0" i would guess that $xcdot dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?






analysis uniform-convergence







share|cite|improve this question











$endgroup$




Let $f_n(x) : mathbb{R} rightarrow mathbb{R}$ be defined as $dfrac{x}{n}$. Is $f_n$ uniformly convergent?



With "$(a_n)$ is limited and $(b_n)$ converges to 0 $Rightarrow (a_nb_n)$ converges also to 0" i would guess that $xcdot dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?






analysis uniform-convergence




analysis uniform-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 '17 at 13:58







Arjihad

















asked Jan 30 '17 at 13:43









ArjihadArjihad

390112




390112












  • $begingroup$
    You need to specify an interval for the values of $x$.
    $endgroup$
    – GaC
    Jan 30 '17 at 13:50


















  • $begingroup$
    You need to specify an interval for the values of $x$.
    $endgroup$
    – GaC
    Jan 30 '17 at 13:50
















$begingroup$
You need to specify an interval for the values of $x$.
$endgroup$
– GaC
Jan 30 '17 at 13:50




$begingroup$
You need to specify an interval for the values of $x$.
$endgroup$
– GaC
Jan 30 '17 at 13:50










2 Answers
2






active

oldest

votes


















0












$begingroup$

It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is $mathbb{R}$ limited?
    $endgroup$
    – Arjihad
    Jan 30 '17 at 13:59










  • $begingroup$
    of course not (if by limited you mean bounded)
    $endgroup$
    – b00n heT
    Jan 30 '17 at 15:01



















0












$begingroup$

If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.



If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.






share|cite|improve this answer









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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Is $mathbb{R}$ limited?
      $endgroup$
      – Arjihad
      Jan 30 '17 at 13:59










    • $begingroup$
      of course not (if by limited you mean bounded)
      $endgroup$
      – b00n heT
      Jan 30 '17 at 15:01
















    0












    $begingroup$

    It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Is $mathbb{R}$ limited?
      $endgroup$
      – Arjihad
      Jan 30 '17 at 13:59










    • $begingroup$
      of course not (if by limited you mean bounded)
      $endgroup$
      – b00n heT
      Jan 30 '17 at 15:01














    0












    0








    0





    $begingroup$

    It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.






    share|cite|improve this answer









    $endgroup$



    It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 30 '17 at 13:50









    b00n heTb00n heT

    10.4k12235




    10.4k12235












    • $begingroup$
      Is $mathbb{R}$ limited?
      $endgroup$
      – Arjihad
      Jan 30 '17 at 13:59










    • $begingroup$
      of course not (if by limited you mean bounded)
      $endgroup$
      – b00n heT
      Jan 30 '17 at 15:01


















    • $begingroup$
      Is $mathbb{R}$ limited?
      $endgroup$
      – Arjihad
      Jan 30 '17 at 13:59










    • $begingroup$
      of course not (if by limited you mean bounded)
      $endgroup$
      – b00n heT
      Jan 30 '17 at 15:01
















    $begingroup$
    Is $mathbb{R}$ limited?
    $endgroup$
    – Arjihad
    Jan 30 '17 at 13:59




    $begingroup$
    Is $mathbb{R}$ limited?
    $endgroup$
    – Arjihad
    Jan 30 '17 at 13:59












    $begingroup$
    of course not (if by limited you mean bounded)
    $endgroup$
    – b00n heT
    Jan 30 '17 at 15:01




    $begingroup$
    of course not (if by limited you mean bounded)
    $endgroup$
    – b00n heT
    Jan 30 '17 at 15:01











    0












    $begingroup$

    If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.



    If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.



      If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.



        If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.






        share|cite|improve this answer









        $endgroup$



        If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.



        If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 '17 at 13:57









        Nicolas PelletierNicolas Pelletier

        18116




        18116






























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