Differentiability of the convolution $\int_0^tf(t-s)g(s)\;ds$

Question

Given two continuously differentiable functions $f,g:[0,\infty)\to\mathbb{R}$. I want to know what we can tell about the differentiability of $$(f\ast g)(t)=\int_0^tf(t-s)g(s)\;ds$$ Especially, why we've always got $$\frac{d}{dt}(f\ast g)(t)=(f'\ast g)(t)=(f\ast g')(t)$$

Answer 1

Let $h(t)=(f\star g)(t)$ and using Leibniz integral rule we have $$\frac{d}{dt}(f\star g)(t)\equiv h'(t)=\int^{t}_{0}f'(t-s)g(s)\,ds+f(0)g(t)=(f'\star g)(t)+f(0)g(t)$$ On the other hand a change of variable $t-s=s'$ yields for the original convolution $$h(t)\equiv (f\star g)(t)=\int^{t}_{0}f(t-s)g(s)\,ds=\int^{t}_{0}f(s')g(t-s')\,ds'$$ So we have again under differentation that $$\frac{d}{dt}(f\star g)(t)\equiv h'(t)=\int^{t}_{0}f(s')g'(t-s')\,ds'+f(t)g(0)=(f\star g')(t)+f(t)g(0)$$ Under the assumption $f(0)=g(0)=0$ one gets the desired result.