Suppose\begin{align*}a_n &\to \delta_0 \text{ as a distribution}, \\f_n &\stackrel{\ast}{\rightharpoonup} f \text{ in $L^\infty$} \\f_n \ast a_n &\stackrel{\ast}{\rightharpoonup} g \text{ in $L^\infty$}\end{align*}Can we conclude that $f =g$ a.e.?
This would be true obviously if $f_n$ converges strongly instead of weak-star.