When $f,g \in L^1(R)$ and $g$ is differentiable with $g' \in L^1(R)$, I want to show that $$(f*g)'=f*g'$$ holds. I guess that the proof should be something like this. $$\begin{align} (f*g)'(x) &= \lim\limits_{h \to 0} \frac{1}{h} \{(f*g)(x+h)-(f*g)(x)\}\\ &= \lim\limits_{h \to 0} \frac{1}{h} \{\int_R f(y)g(x+h-y)dy-\int_R f(y)g(x-y)dy\}\\ &= \lim\limits_{h \to 0} \int_R f(y)\frac{g(x+h-y)-g(x-y)}{h}dy\\ (*) &= \int_R f(y)\lim\limits_{h \to 0}\frac{g(x+h-y)-g(x-y)}{h}dy\\ &= \int_R f(y)g'(x-y)dy\\ &= (f*g')(x)\\ \end{align}$$
However, in order to take the limit inside the integral, we need to show that there exists $\delta>0$ and $h \in L^1(R)$ s.t. $$|f(y)\frac{g(x+h-y)-g(x-y)}{h}|\leq h(y)$$ holds.
Is the above condition sufficient to find the appropriate function $h$? We have $g'\in L^1(R)$ so my guess is that $h$ should be expressed as a function of $g'$. I can create a pointwise bound on the difference quotient of $g$, but without further assumption I can't seem to find the uniform bound on the difference quotient. Any help, comment or counterexample is appreciated.
Thanks!