Let $f$ be a continuous but nowhere differentiable function. Is $f$ convolved with mollifier, a smooth function?
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The key is that when you (as you say in the comments) get the two scenarios:
$$ f \star (D g) = D(f \star g) = (D f) \star g $$
then you get to choose which!
So if $D f$ doesn't make sense, then you can ignore it and choose to use the identity $D(f \star g) = f \star (D g)$.
Taking this to the extreme, you get the - bizarre, in my opinion - result that if $p$ is a polynomial, then $f \star p$ is always a polynomial.