0
$\begingroup$

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 $f(x)=0$

$\endgroup$
1
$\begingroup$

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 $f_\nu(x)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.