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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.

Peter Minford
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$|x|^{2}=x_1^{2}+x_2^{2}+...+x_n^{2}$ 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 $0$.

Kavi Rama Murthy
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  • Thank you. Okay why it is $C^{\infty}$ I got. But for the compact support I see it right that it is because, we have for $|x| < r$ that $\Phi_r(x) \neq 0$ and for $|x| \geq r$ that $\Phi_r(x) = 0$. So we have that $\Phi_r(x)$ has as compact support the $\overline{B}(0,r)$? – Peter Minford Apr 19 '20 at 12:03
  • @PeterMinford Yes that is correct. But to say that the function has compact support you don't have to find the exact support. If the function vanishes outside some compact set then the function has compact support. – Kavi Rama Murthy Apr 19 '20 at 12:09
  • This answer is a bit short. Maybe you should quickly address $x=0$ and maybe even the fact that $\phi\in\mathcal C^\infty(\mathbb R)$ – Maximilian Janisch Apr 19 '20 at 16:01
  • @MaximilianJanisch Thank you. I have added some details for differentiability at $0$. – Kavi Rama Murthy Apr 19 '20 at 23:16
  • Thank you, now maybe you can help me to understand why $r^n = \int_0^rs^n\phi_r^\prime(s)\mathrm{d}s$? Somehow nobody can help me – Peter Minford Apr 20 '20 at 07:03