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)?



Proof of $L^1(\mathbb{R}) \ast f \neq L^1(\mathbb{R})$
I'm guessing $\*$ means convolution (since this is a math forum) and not pointwise multiplication (since this is not a computer forum). 


Some steps to try ... suppose $L^1 \* f = L^1$


$\widehat{f} \ne 0$ a.e.


There is $g \in L^1$ so that $g \* f = f$


$ \widehat{g} \widehat{f} = \widehat{f}$


$\widehat{g} = 1$ identically


$\widehat{g} \notin C\_0$


$g \notin L^1$


