# Proving that mollified function has compact support for $\nu$ large enough

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 $$f(x)=0$$

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 $$f_\nu(x)=0$$.