It is known that $L^1(\mathbb{R}) \ast f$ is dense in $L^1(\mathbb{R})$ for some $f\in L^1(\mathbb{R})$. So for such $f$ the closure of $L^1(\mathbb{R}) \ast f$ in the $L^1$ norm is $L^1(\mathbb{R})$. But apparently

(1)$\quad\quad L^1(\mathbb{R}) \ast f \neq L^1(\mathbb{R})$ *for every*
$f\in L^1(\mathbb{R})$.

Is there a simple proof of (1)?