Can we dispense with the assumption of compact support in the definition of mollifier and still retain Lp approximation of the identity? explicitly:
Let $\phi\in L_1(\mathbb{R}),\ f\in L_p(\mathbb{R})$ and denote: $\forall\epsilon>0:\phi_\epsilon=\frac{\phi(x\epsilon^{-1})}{\epsilon}$
If we dont assume phi is compactly supported, does it still follow that:
$f\star \phi_\epsilon\overset{L_p}\rightarrow f$
(where ($f\star g)(x)=\int f(y)g(x-y)$ is the convolution)
Best Answer
Yes. $\|f*\phi_{\epsilon} -f*\psi_{\epsilon}\|_p \leq \|f\|_p \|\phi_{\epsilon} -\psi_{\epsilon} \|_1$ and we can find $\psi$ with compact support such that $\|\phi_{\epsilon} -\psi_{\epsilon} \|_1=\|\phi-\psi\|_1$ is as small as we want.