Let $\phi(x)\in C_0^{\infty}(\Bbb R)$ be a mollifier such that $\int\limits_{-\infty}^{+\infty}\phi(x)dx=1$, $\phi\ge0$ and $\phi(x)=0\ \forall |x|>1$
$\phi_\nu(x)=\nu\phi(\nu x)$ is the corresponding mollifying sequence.
Let $f\in C_0(a,b)$ be continuous of compact support i.e. $\overline{\{x\in\Bbb R\ :\ f(x)\ne0\}}\subset (a,b)$
We define $f_\nu(x):=\int\limits_{-\infty}^{+\infty}\phi_\nu(x-y)f(y)dy$
I have already shown that $f_\nu\in C^{\infty}(a,b) \ \forall\nu\ge1$ but I wasn't able to prove that for $\nu$ large enough $f_\nu\in C_0^{\infty}(a.b)$
The idea is to prove that there exists $\epsilon>0$ such that $\forall x\in (-\infty,a+\epsilon)\cup(b-\epsilon,\infty)$ we have $f(x)=0$
Best Answer
Since $\phi(x)=0$ for all $|x|>1$, you have that $\phi_\nu(x-y)=\nu \phi(\nu(x-y))=0$ whenever $|\nu(x-y)|>1$, that is, $|x-y|>\frac1\nu$. This means that in the integral you only see the interval $(x-\frac1\nu,x+\frac1\nu)$, so you can write $$f_\nu(x)=\int\limits_{x-\frac1\nu}^{x+\frac1\nu}\phi_\nu(x-y)f(y)dy.$$ So if the interval $(x-\frac1\nu,x+\frac1\nu)$ does not intersects $(a,b)$ then $f_\nu(x)=0$.