A family of infinitely differentiable functions with compact support Hi everyone: Let $ B(r) $ designate the ball of radius $ r>0 $ and center $ O $ in $ \mathbb{R}^{N} $ and $ N\geq2 $. we know that for all $ \varepsilon>0 $
there exist an infinitely differentiable function $ \varphi_\varepsilon $ with  compact support in $ B(1+\varepsilon) $
and that is equal to $1$ on $ B(1) $. My question is: can we then let $ \varepsilon $ approach zero?! If yes, what do we get at the limit?
Thanks for your help.
 A: Pick some $\epsilon>0$. Since $\phi_\epsilon(x) = 1$ for $\|x\|<1$ and $\phi_\epsilon$ is continuous, we see that $\phi_\epsilon(x) = 1$ for $\|x\| \le 1$.
Hence $\lim_{\epsilon \downarrow 0} \phi_\epsilon(x) = 1$.
Now suppose $\|x\|>1$, then choose $\delta={1 \over 2} (\|x\|-1)$. Then if $\epsilon>0$
satisfies $\epsilon<\delta$, we see that the support of
$\phi_\epsilon$ is contained in $B(0,1+\epsilon) \subset B(0,{1 \over 2} (\|x\|+1))$, and hence $\phi_\epsilon(x) = 0$.
Hence $\lim_{\epsilon \downarrow 0} \phi_\epsilon(x) = 0$.
This can be written as $\lim_{\epsilon \downarrow 0} \phi_\epsilon(x) = 1_{\overline{B(0,1)}}(x)$.
A: If I understand the intent of the question, it means: Is there some function equal to $1$ at every point in the ball and equal to $0$ everywhere else and that is infinitely differentiable?
The answer is no because differentiable functions are continuous.  If a function is $0$ everywhere outside the ball and $1$ everwhere inside the ball, then it is discontinuous on the boundary of the ball.
