When is the mollification of an $L^p$ function in $L^p\cap L^{\infty}$?

Question

Let $f\in L^p(\mathbb{R}^3)$ and $f_{\varepsilon}=f\ast \varphi_{\varepsilon}$ its convolution with the standard mollifier $\varphi_{\varepsilon}$. Then it is well-known that $f_{\varepsilon}\in L^p(\mathbb{R}^n)$. I would like to find (weak) conditions on $f$ to ensure that $f_{\varepsilon}\in L^p\cap L^{\infty}$ at infinity, e.g. $f_{\varepsilon}\in L^p\cap L^{\infty}(B_R(0)^C)$ for some $R>0$. I would be grateful for any suggestions!
Since yesterday I know that in general $C^{\infty}\cap L^p(B_R(0)^C)$ is no subspace of $L^q$ for $q>p$ on any $[M,\infty)$.

Answer 1

Hint:

Young's theorem: Suppose $p,q,r\in[1,\infty]$, If $\frac1r=\frac1p+\frac1q-1$ and $f\in L_p$ and $g\in L_q$ then, $f*g$ exits a.s., $f*g\in L_r$, and $$\|f*g\|_r\leq\|f\|_p\|g\|_q$$ Furthermore, if $\frac1p+\frac1q=1$ ($r=\infty$) then $f*g\in C_0$ (continuous vanishing at infinity).

There are many textbooks that present this result, for example Jone, F. Lebeguse integration in Euclidean space, or Folland, G, Real Analysis.

Finally, notice that the typicall mollifier is in any $L_p$ space vy virtue of being compactly supported and continuous (in fact smooth).