Why is it in $C_c^\infty(\mathbb{R}^n)$? I try to understand the following prove of an old problem:
https://math.stackexchange.com/a/61101/774621
There the following function is define: 
$\displaystyle\phi_r(s)=\left\{\begin{array}{cl}e^{s^2/(s^2-r^2)}&\text{for }0\le s<r\\0&\text{for }s\ge r\end{array}\right.\hspace{.25in}$ then $\Phi_r(x)=\phi_r(|x|)$ is in $C_c^\infty(\mathbb{R}^n)$
Now I don't get first how $\Phi_r(x)$ is defined and why it is in $C_c^\infty(\mathbb{R}^n)$. If I see it right, it should have a compact support in the closed ball with radius r in $\Bbb R^n$. But I am not sure. Thanks for your help. 

$|x|^{2}=x_1^{2}+x_2^{2}+...+x_n^{2}$ is infinitely differentiable. Composition of $C^{\infty}$ functions is $C^{\infty}$. Hence $\Phi_r$ is a $C^{\infty}$ function. It has compact support because  $\Phi_r(x)=0$ for $|x|\geq r$.
For showing differentiability at $0$ note that $\phi_r(s)=e^{-1/u}$ where $u=(r/s)^{2}-1$. As is well known The function $e^{-1/u}$ for $u>0$ and $0$ for $u=0$ has derivatives of all orders and the derivatives are all $0$. 
