Theorem:
Let $u(x)$ is locally integrable function on $R^n$, $\phi= e^{\frac{1}{|x|^2-1}}, when |x|<1;\phi= 0, when |x|\geq 1 $
$\alpha(x)=\frac{1}{C}\phi$ where $C=\int_{R^n}\phi(x)dx$
and $\alpha_{\epsilon}(x)=\frac{1}{{\epsilon}^n}\alpha(\frac{x}{\epsilon})$
then $u_\epsilon(x)=\int_{R^n}u(x-y)a_\epsilon(y)dy$ is a $C^\infty$ function.
The proof is so brief:
$a_\epsilon$ is a infinite continues differentiable function, so use the interchangeabilty of $\frac{d}{dx}$ and $\int$, we can have $u_\epsilon(x)=\int_{R^n}u(x-y)a_\epsilon(y)dy$ is a $C^\infty$ function.
I'm still don't know how to proof it. can you give me a step-by-step proof? thanks a lot!
Let $\epsilon >0$, we will try to prove that $(u * a_\epsilon)'(x)=u * a_\epsilon'(x).$
To begin, let $h>0$ :
$$\frac{u * a_\epsilon(x+h)-u * a_\epsilon(x)}h=\frac 1 h \int_{R^n}u(y)a_\epsilon(x+h-y)dy-\frac 1 h \int_{R^n}u(y)a_\epsilon(x-y)dy$$
By linearity of the integral :
$$\frac{u * a_\epsilon(x+h)-u * a_\epsilon(x)}h=\int_{\mathbb{R}^n}u(y) \frac 1 h\left[ a_\epsilon(x+h)- a_\epsilon(x) \right] dy.$$
Hence :
$$\lim _{ h \to 0 } \frac{u * a_\epsilon(x+h)-u * a_\epsilon(x)}h = \lim _{ h \to 0 } \int_{\mathbb{R}^n}u(y) \frac 1 h\left[ a_\epsilon(x+h-y)- a_\epsilon(x-y) \right] dy.$$
Let $x \in \mathbb{R}^n$, then : $\forall 0<h<1, \frac 1 h\left[ a_\epsilon(x+h-y)- a_\epsilon(x-y) \right]$ is compactly supported in some $K_{x,\epsilon}$. Hence :
$$\lim _{ h \to 0 } \int_{\mathbb{R}^n}u(y) \frac 1 h\left[ a_\epsilon(x+h-y)- a_\epsilon(x-y) \right]dy= \lim _{ h \to 0 } \int_{K_{x,\epsilon}}u(y) \frac 1 h\left[ a_\epsilon(x+h-y)- a_\epsilon(x-y) \right]dy.$$
Now, we have, by some famous inequality, since $a_\epsilon \in C^1$ and has its support in $K_{x,\epsilon}$ being compact :
$$\left| a_\epsilon(x+h-y)- a_\epsilon(x-y) \right| \le \| a_\epsilon'\|_\infty h.$$
In other words :
$$\left|u(y) \frac 1 h\left[ a_\epsilon(x+h-y)- a_\epsilon(x-y) \right] \right| \le \| a_\epsilon'\|_\infty\left|u(y)\right|.$$
Since $u$ is locally integrable : $\| a_\epsilon'\|_\infty\left|u(y)\right| \in L^1(K_{x,\epsilon})$, then we can, by dominated convergence theorem, exchange $\lim$ and $\int$ :
$$\lim _{ h \to 0 } \frac{u * a_\epsilon(x+h)-u * a_\epsilon(x)}h = \int_{\mathbb{R}^n}u(y) \lim _{ h \to 0 } \frac 1 h\left[ a_\epsilon(x+h-y)- a_\epsilon(x-y) \right] dy=\int_{\mathbb{R}^n}u(y) a_\epsilon'(x-y) dy.$$
The right term exists and is well defined, then, if $a_\epsilon \in C^1$ :
$$(u * a_\epsilon)'(x)=u * a_\epsilon'(x).$$
Since $a_\epsilon \in C^\infty$, we have, by induction, for all $n \in \mathbb{N}$ :
$$(u * a_\epsilon)^{(n)}(x)=u * a_\epsilon^{(n)}(x).$$
Since $x$ is arbitrary, $u * a_\epsilon \in C^\infty$.