We recall that if $f\_1\in L^p(\mathbb R)$ and if $f\_2\in L^q(\mathbb R)$ where  

$1 \lt p \lt \infty$ and $\frac 1p+\frac1q=1$ then the function $f\_1\ast f\_2(x)=\int\_{\mathbb R} f\_1(x-y) f\_2(y)dy$ is a continuous function in $x$. Now if we take $f$ to be a $L^\infty$ function on $\mathbb R$ and $\mu$ an element of the dual space of $L^\infty(\mathbb R)$ which is a finitely additive measure, is $f\ast \mu(x)=\int\_{\mathbb R} f(x-y)d\mu(y)$ a continuous function? 



Convolution with an element in the dual space
This should be a comment on Yulia Kuznetsova's answer, but I didn't have the points and it's gone now anyway:


Can't you take $\mu$ as corresponding to a functional $f \mapsto f(0)$ if $f$ is an $L^\infty$ function continuous at $0$, extended to all of $L^\infty(\mathbb{R})$ by Hahn-Banach? Then if you take e.g. a step function $g = \mathbb{1}\_{(0,\infty)}$ which is $0$ to the left of $0$ and $1$ to the right, $g \star \mu$ should be $0$ a.e. on $(-\infty,0)$ and $1$ a.e. on $(0,\infty)$, and so wouldn't have a continuous representative. What am I missing?


