# Does the set of all $C^\infty$ functions having compact support an ideal in $C(\Bbb{R})$?

Question. Let $C(\Bbb{R})$ denote the ring of real-valued continuous functions on $\Bbb{R}$, 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. $C_c(\Bbb{R})$
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 $support(f):=cl\{x | f(x) \neq 0\}$)

2. False. (Take, $f(x)=\frac{1}{1+x^2} \in C_0(\Bbb{R})$ and $r(x)=1+x^2\in C(\Bbb{R})$, then $rf$ doesn't belong to $C_0(\Bbb{R})$).

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 $C_0(\Bbb{R})$. 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?

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 $C^{\infty}$.