if $f$ is in weak $L^p$ and $\phi$ is $C_0^{1}$ then $f \ast \phi$ is in weak $L^p$

Question

Okay, so I'd like to know if what I wrote in the title is true. Suppose that $f \in L^{p,\infty}(\mathbb{R}^n)$ (weak $L^p$ space) and $\phi \in C_0^1(\mathbb{R}^n)$ [or even $C_0^{\infty}$ if it matters, but I think not], where $C_0^1$ denotes continuously differentiable functions with compact support. is it true that $$ f \ast \phi \in L^{p,\infty}(\mathbb{R}^n)$$ ? if so, I'd also like to know how we can estimate this thing's weak norm - hopefully it'd be something that depends only on $f$'s weak norm and $||\phi||_{\infty}$. Is is true? If so does this result have a name?

Answer 1

Certainly you cannot control the convolution by $\|\phi\|_\infty$. Take $f(x) = |x|^{-p/n}$ and let $\phi$ be a smooth function such that $\phi=1$ on the ball $B(0,R)$, and $0\le \phi\le 1$ everywhere. Then for $|x|\gg R$ we have $$f*\phi(x) \approx |B(0,R)| |x|^{-p/n}$$ where $ |B(0,R)|$ is the Lebesgue measure of the ball. So, the weak $L^p$ norm gets multiplied by factor of about $ |B(0,R)|$.

The smoothness and even continuity of $\phi$ play no role: the relevant quantity is its $L^1$ norm. You need an analogue of Young's inequality for weak $L^p$ spaces. Note that $p=1$ does not work: weak $L^1$ does not even guarantee that $f*\phi$ is finite a.e. For $p>1$, Young-type inequalities are available: see theorem 3.3 here and Theorem 2.2 in Convolution Inequalities in Lorentz Spaces by Nursultanov and Tikhonov.