Let $\mu$ be a compactly supported distribution on $\mathbb R$ such that $\mu$ is positive, finite measure and its distributional derivative $\mu'$ is also a measure. Define a measure $\mu_\epsilon$ by $\int f(x)d\mu_{\epsilon}(x)=\int f(\epsilon x)d\mu(x)$ for continuous f.
Show that if $a\in C^1(\mathbb R)$(i.e a is continuously differentiable and $a$, $a'$ are both bounded) and $f\in L^2(\mathbb R)$ is continuous, then $||a\cdot(f'*\mu_\epsilon)-(af')*\mu_\epsilon||_{L^2}\leq C||f||_{L^2}$ with $C$ independent of $f,\epsilon$. Show that the left hand side tends to zero as $\epsilon\to0$.
I have trouble to understand this question. Since $f$ may not be differentiable, so $f'$ is the distributional derivative, and why is the convolution well-defined? Can someone help me to start?