Sum of a infinite number of continuous functions on a set may not be continuous.

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I was asked to give an example of series of continuous functions whose limit is discontinuous .



I gave the following example: $f_n(x) = x^n - x^(n-1)$ . I thought any sequence of continuous functions which is convergent to a discontinuous function can be made by that way a series of
continuous function whose limit is discontinuous.
Am I correct?



Can anyone give me suggestion on this?






real-analysis sequences-and-series continuity discontinuous-functions







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  • Yes, $lim a_n =sum_n b_n$ where $b_1=a_1,b_k=a_k-a_k-1$ fro $k geq 2$. So any limit is an infinite sum and if $a_n$'s are continuous functions so are $b_n$'s.
    – Kavi Rama Murthy
    Aug 30 at 5:50














up vote
0
down vote

favorite












I was asked to give an example of series of continuous functions whose limit is discontinuous .



I gave the following example: $f_n(x) = x^n - x^(n-1)$ . I thought any sequence of continuous functions which is convergent to a discontinuous function can be made by that way a series of
continuous function whose limit is discontinuous.
Am I correct?



Can anyone give me suggestion on this?






real-analysis sequences-and-series continuity discontinuous-functions







share|cite|improve this question





















  • Yes, $lim a_n =sum_n b_n$ where $b_1=a_1,b_k=a_k-a_k-1$ fro $k geq 2$. So any limit is an infinite sum and if $a_n$'s are continuous functions so are $b_n$'s.
    – Kavi Rama Murthy
    Aug 30 at 5:50












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I was asked to give an example of series of continuous functions whose limit is discontinuous .



I gave the following example: $f_n(x) = x^n - x^(n-1)$ . I thought any sequence of continuous functions which is convergent to a discontinuous function can be made by that way a series of
continuous function whose limit is discontinuous.
Am I correct?



Can anyone give me suggestion on this?






real-analysis sequences-and-series continuity discontinuous-functions







share|cite|improve this question













I was asked to give an example of series of continuous functions whose limit is discontinuous .



I gave the following example: $f_n(x) = x^n - x^(n-1)$ . I thought any sequence of continuous functions which is convergent to a discontinuous function can be made by that way a series of
continuous function whose limit is discontinuous.
Am I correct?



Can anyone give me suggestion on this?






real-analysis sequences-and-series continuity discontinuous-functions




real-analysis sequences-and-series continuity discontinuous-functions






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asked Aug 30 at 5:39









cmi

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  • Yes, $lim a_n =sum_n b_n$ where $b_1=a_1,b_k=a_k-a_k-1$ fro $k geq 2$. So any limit is an infinite sum and if $a_n$'s are continuous functions so are $b_n$'s.
    – Kavi Rama Murthy
    Aug 30 at 5:50
















  • Yes, $lim a_n =sum_n b_n$ where $b_1=a_1,b_k=a_k-a_k-1$ fro $k geq 2$. So any limit is an infinite sum and if $a_n$'s are continuous functions so are $b_n$'s.
    – Kavi Rama Murthy
    Aug 30 at 5:50















Yes, $lim a_n =sum_n b_n$ where $b_1=a_1,b_k=a_k-a_k-1$ fro $k geq 2$. So any limit is an infinite sum and if $a_n$'s are continuous functions so are $b_n$'s.
– Kavi Rama Murthy
Aug 30 at 5:50




Yes, $lim a_n =sum_n b_n$ where $b_1=a_1,b_k=a_k-a_k-1$ fro $k geq 2$. So any limit is an infinite sum and if $a_n$'s are continuous functions so are $b_n$'s.
– Kavi Rama Murthy
Aug 30 at 5:50










1 Answer
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Yes! But here's the standard one:



Consider $$f_n(x)=fracx^2(1+x^2)^n$$
where $x in BbbR$ and $n=0,1,2,...$



Then $$f(x)=sum_n=0^infty f_n(x)= sum_n=0^inftyfracx^2(1+x^2)^n= begincases
0&x= 0\
1+x^2 &xneq0\
endcases
$$






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    1 Answer
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    1 Answer
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    up vote
    0
    down vote













    Yes! But here's the standard one:



    Consider $$f_n(x)=fracx^2(1+x^2)^n$$
    where $x in BbbR$ and $n=0,1,2,...$



    Then $$f(x)=sum_n=0^infty f_n(x)= sum_n=0^inftyfracx^2(1+x^2)^n= begincases
    0&x= 0\
    1+x^2 &xneq0\
    endcases
    $$






    share|cite|improve this answer


























      up vote
      0
      down vote













      Yes! But here's the standard one:



      Consider $$f_n(x)=fracx^2(1+x^2)^n$$
      where $x in BbbR$ and $n=0,1,2,...$



      Then $$f(x)=sum_n=0^infty f_n(x)= sum_n=0^inftyfracx^2(1+x^2)^n= begincases
      0&x= 0\
      1+x^2 &xneq0\
      endcases
      $$






      share|cite|improve this answer
























        up vote
        0
        down vote










        up vote
        0
        down vote









        Yes! But here's the standard one:



        Consider $$f_n(x)=fracx^2(1+x^2)^n$$
        where $x in BbbR$ and $n=0,1,2,...$



        Then $$f(x)=sum_n=0^infty f_n(x)= sum_n=0^inftyfracx^2(1+x^2)^n= begincases
        0&x= 0\
        1+x^2 &xneq0\
        endcases
        $$






        share|cite|improve this answer














        Yes! But here's the standard one:



        Consider $$f_n(x)=fracx^2(1+x^2)^n$$
        where $x in BbbR$ and $n=0,1,2,...$



        Then $$f(x)=sum_n=0^infty f_n(x)= sum_n=0^inftyfracx^2(1+x^2)^n= begincases
        0&x= 0\
        1+x^2 &xneq0\
        endcases
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 30 at 6:17

























        answered Aug 30 at 6:06









        Chinnapparaj R

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