smoothness of $f(x) = \dfrac{((1-\|x\|_2)_+)^2}{1+\|x\|_2}$

Question

Consider $f(x) = \dfrac{((1-\|x\|_2)_+)^2}{1+\|x\|_2}$, where $(x)_+ = \max\{0,x\}$. Is $f$ infinite smooth on $\mathbb{R}^d$?

Answer 1

Your function is $$f(x)=\frac{\max\big(0,1-|x|\big)^2}{1+|x|} \\ =\begin{cases}\frac{(1-|x|)^2}{1+|x|} & x\in\mathbb B(0,1) \\ 0 & x\notin \mathbb B(0,1)\end{cases}$$

This function is not $C^\infty$. Consider $d=1$, $$f(x)=\begin{cases}\frac{(1-|x|)^2}{1+|x|} & |x|<1 \\ 0 & |x|\geq 1\end{cases}$$

Show (exercise)

$$f'(x)=\begin{cases}\text{undefined} & x=0 \\ -\operatorname{sgn}(x)\frac{(3+|x|)(1-|x|)}{(1+|x|)^2} & |x|<1 ~\text{and}~x\neq 0 \\ 0 & x\geq 1\end{cases}$$

It is obvious in particular that $\lim_{x\to 0+}f'(x) \neq \lim_{x\to 0^-}f'(x)$. This is easily to generalize to higher dimensions. Therefore $f$ is not $C^1$ in $\mathbb R^d$.

However, $f$ is $C^\infty$ on the domain $\mathbb R^d \setminus \{0\}$.