Mollifier and totally bounded function

Question

I have set L which is a subset of $L^p (\mathbb R^N)$. I mollify it to another set $L_\sigma $ using mollifier $\phi_\sigma$ $$ L_\sigma = \{ \phi_\sigma * f \mid f \in L \}$$ where $$ \phi_\sigma * f = \int_{\mathbb R^N} \phi_\sigma (s,t) f(s) \,ds$$ then I restrict the The function to $\bar {B_R(0)}$ closed ball having radius 0. $$ \phi_\sigma * f = \int_{\bar {B_R(0)}} \phi_\sigma (s,t) f(s) \,ds$$ Have I done the procedure correctly that I define and I want to know how can I show that The set $L_\sigma $ after restriction is totally bounded

Answer 1

Usually we take $\phi \in C^\infty_c(\Bbb{R}^N)$ and $\ast$ the convolution and $\phi_\sigma=\sigma^{-N}\phi(./\sigma)$ so that $$\|f\ast \phi_\sigma\|_p\le \|\phi\|_1 \|f\|_p$$

Proof: use that $$f\ast \phi_\sigma= \lim_{K\to \infty} 2^{-KN} \sum_{n\in \Bbb{Z}^N} \phi_\sigma(n/2^K) f(.- n/2^K)$$ $$\|f\ast \phi_\sigma\|_p\le \lim_{K\to \infty} 2^{-KN} \sum_{n\in \Bbb{Z}^N} \|\phi_\sigma(n/2^K) f(.- n/2^K)\|_p$$ $$ =\lim_{K\to \infty} 2^{-KN} \sum_{n\in \Bbb{Z}^N} |\phi_\sigma(n/2^K)| \|f\|_p= \|\phi_\sigma\|_1\|f\|_p= \|\phi\|_1\|f\|_p$$