Convolution derivative, $(f*g)'=f*g'$

90 Views Asked by At

Let $f,g \in L^1$ and $g$ is differentiable and $g' \in L^1$, then $(f*g)'=f*g'$

PROOF: For this proof I am asked to use the dominated convergence theorem

It can be quickly arrived at that

$$\begin{align} (f*g)'(x) &= \lim\limits_{h \to 0} \int_\mathbb{R} f(t)\frac{g(x+h-t)-g(x-t)}{h}dt\\ \end{align}$$

For the following, it is necessary to interchange the limit and the integral and it is here where the theorem mentioned above is used.

To do so, it is sufficient to find a $k\in L^1$ such that $|f(t)\frac{g(x+h-t)-g(x-t)}{h}|\leq k(t)$

However, I am having trouble finding such a function $k$