Does the set of all $C^infty$ functions having compact support an ideal in $C(BbbR)$?

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Question. Let $C(BbbR)$ denote the ring of real-valued continuous functions on $BbbR$, with pointwise addition and multiplication. Which of the following form an ideal in this ring?



  1. The set of all $C^infty$ functions having compact support.

  2. $C_c(BbbR)$

  3. The set of all continuous functions which vanish at infinity.



My Solution.



  1. True.(Follows from: $Support(f+g) subset support(f) cup support(g)$ also $Support(fg) subset support(f)$ and $support(f):=cl f(x) neq 0$)


  2. False. (Take, $f(x)=frac11+x^2 in C_0(BbbR)$ and $r(x)=1+x^2in C(BbbR)$, then $rf$ doesn't belong to $C_0(BbbR)$).


But I cannot find a counter example in (1) as I did in (3) ...



Suppose I choose $f(x)=1$ in $[-1,1]$ and take $r(x)=|x|$. Then $rf(x)=|x|$ and $rf$ doesn't belong to $C_0(BbbR)$. But the problem is if I define $f=0$ outside $[-1,1]$ it wouldn't be smooth...



So I think I have to find another counter example. Can any one please help me to find an counter example here?







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    Question. Let $C(BbbR)$ denote the ring of real-valued continuous functions on $BbbR$, with pointwise addition and multiplication. Which of the following form an ideal in this ring?



    1. The set of all $C^infty$ functions having compact support.

    2. $C_c(BbbR)$

    3. The set of all continuous functions which vanish at infinity.



    My Solution.



    1. True.(Follows from: $Support(f+g) subset support(f) cup support(g)$ also $Support(fg) subset support(f)$ and $support(f):=cl f(x) neq 0$)


    2. False. (Take, $f(x)=frac11+x^2 in C_0(BbbR)$ and $r(x)=1+x^2in C(BbbR)$, then $rf$ doesn't belong to $C_0(BbbR)$).


    But I cannot find a counter example in (1) as I did in (3) ...



    Suppose I choose $f(x)=1$ in $[-1,1]$ and take $r(x)=|x|$. Then $rf(x)=|x|$ and $rf$ doesn't belong to $C_0(BbbR)$. But the problem is if I define $f=0$ outside $[-1,1]$ it wouldn't be smooth...



    So I think I have to find another counter example. Can any one please help me to find an counter example here?







    share|cite|improve this question






















      up vote
      4
      down vote

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      up vote
      4
      down vote

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      Question. Let $C(BbbR)$ denote the ring of real-valued continuous functions on $BbbR$, with pointwise addition and multiplication. Which of the following form an ideal in this ring?



      1. The set of all $C^infty$ functions having compact support.

      2. $C_c(BbbR)$

      3. The set of all continuous functions which vanish at infinity.



      My Solution.



      1. True.(Follows from: $Support(f+g) subset support(f) cup support(g)$ also $Support(fg) subset support(f)$ and $support(f):=cl f(x) neq 0$)


      2. False. (Take, $f(x)=frac11+x^2 in C_0(BbbR)$ and $r(x)=1+x^2in C(BbbR)$, then $rf$ doesn't belong to $C_0(BbbR)$).


      But I cannot find a counter example in (1) as I did in (3) ...



      Suppose I choose $f(x)=1$ in $[-1,1]$ and take $r(x)=|x|$. Then $rf(x)=|x|$ and $rf$ doesn't belong to $C_0(BbbR)$. But the problem is if I define $f=0$ outside $[-1,1]$ it wouldn't be smooth...



      So I think I have to find another counter example. Can any one please help me to find an counter example here?







      share|cite|improve this question













      Question. Let $C(BbbR)$ denote the ring of real-valued continuous functions on $BbbR$, with pointwise addition and multiplication. Which of the following form an ideal in this ring?



      1. The set of all $C^infty$ functions having compact support.

      2. $C_c(BbbR)$

      3. The set of all continuous functions which vanish at infinity.



      My Solution.



      1. True.(Follows from: $Support(f+g) subset support(f) cup support(g)$ also $Support(fg) subset support(f)$ and $support(f):=cl f(x) neq 0$)


      2. False. (Take, $f(x)=frac11+x^2 in C_0(BbbR)$ and $r(x)=1+x^2in C(BbbR)$, then $rf$ doesn't belong to $C_0(BbbR)$).


      But I cannot find a counter example in (1) as I did in (3) ...



      Suppose I choose $f(x)=1$ in $[-1,1]$ and take $r(x)=|x|$. Then $rf(x)=|x|$ and $rf$ doesn't belong to $C_0(BbbR)$. But the problem is if I define $f=0$ outside $[-1,1]$ it wouldn't be smooth...



      So I think I have to find another counter example. Can any one please help me to find an counter example here?









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      asked Aug 29 at 9:07









      Indrajit Ghosh

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          There exists a $C^infty$ function $f$ such that $f(x)=1$ for all $x in (-1,1)$ and $f(x)=0$ for $|x| >2$. [Construction of such functions using $e^-1/x$ is standard]. If you multiply this by $|x|$ you will go out of $C^infty$.






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            up vote
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            accepted










            There exists a $C^infty$ function $f$ such that $f(x)=1$ for all $x in (-1,1)$ and $f(x)=0$ for $|x| >2$. [Construction of such functions using $e^-1/x$ is standard]. If you multiply this by $|x|$ you will go out of $C^infty$.






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              up vote
              4
              down vote



              accepted










              There exists a $C^infty$ function $f$ such that $f(x)=1$ for all $x in (-1,1)$ and $f(x)=0$ for $|x| >2$. [Construction of such functions using $e^-1/x$ is standard]. If you multiply this by $|x|$ you will go out of $C^infty$.






              share|cite|improve this answer
























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                There exists a $C^infty$ function $f$ such that $f(x)=1$ for all $x in (-1,1)$ and $f(x)=0$ for $|x| >2$. [Construction of such functions using $e^-1/x$ is standard]. If you multiply this by $|x|$ you will go out of $C^infty$.






                share|cite|improve this answer














                There exists a $C^infty$ function $f$ such that $f(x)=1$ for all $x in (-1,1)$ and $f(x)=0$ for $|x| >2$. [Construction of such functions using $e^-1/x$ is standard]. If you multiply this by $|x|$ you will go out of $C^infty$.







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                edited Aug 29 at 9:20

























                answered Aug 29 at 9:13









                Kavi Rama Murthy

                25k31334




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