I have to prove that $w(t)$ is in $\mathcal{D}(R^n)$, with $$w(t):= C \begin{cases} e^{-\frac{1}{1-|x|^2}};& \quad |x|<1,\\ 0, &\quad |x| \geq 1, \end{cases} $$ where $C > 0$ is such that $\int_{\mathbb{R}^n}w(x)\;dx=1$.
I consider first the function $$f(t):= \begin{cases} e^{\frac{1}{t}}& \quad t<0,\\ 0, &\quad t \geq 0,\\ \end{cases} $$ and check that is continuous in $0$. $$\lim_{h \rightarrow 0^-} \frac{ e^{\frac{1}{0+h}} - 0}{h} = \lim_{h \rightarrow 0^+} \frac {0 -0}{h}$$
Now the function $w(x)$ behave like $f(t)$ in $x=-1$ and $x=1$. The support is compact, because $w(x)=0$ outside of $[-1,1]$ and being and exponential function $w(x) \in \mathcal{C}^\infty\; \forall \alpha$.
The question is: do I have to ckeck that $\partial^\alpha w(x)$ is continuous in $x=-1$ and $x=1$?