It is well known that the convolution of a $L^1$ function and a Schwartz function is also in $L^1$, by Young's inequality for convolution. Let $f\in L^1(\mathbb{R}^n)$ and $\phi\in S(\mathbb{R}^n)$, where $\phi$ is nonnegative, radial, and radially decreasing. Set $\phi_\epsilon(x)=\epsilon^{-n}\phi(x/\epsilon)$. My question is, whether the supremum of this convolution $${\rm sup}_{\epsilon>0}|f\ast \phi_\epsilon|\in L^1(\mathbb{R}^n)\ ?$$

**Remark:** We denote the Hardy-Littlewood maximal function of $f$ by $Mf$. Recall that $$ {\rm sup}_{\epsilon>0}|f\ast \phi_\epsilon|(x)\le Mf(x)\int_{\mathbb{R}^n} \phi dx.$$ The Hardy-Littlewood maximal theorem tells us that if $f\in L^p(\mathbb{R}^n)$, $1<p\le \infty$, then $ Mf\in L^p(\mathbb{R}^n),$ while if $f\in L^1(\mathbb{R}^n)$, then $Mf\notin L^1(\mathbb{R}^n)$, whenever $f\ne 0$ on some positive measure set. Therefore,
we get $$f\in L^p(\mathbb{R}^n),\ 1<p\le \infty\Rightarrow{\rm sup}_{\epsilon>0}|f\ast \phi_\epsilon|\in L^p(\mathbb{R}^n).$$ However, we don't know whether it still holds for $p=1$.