Derivative of convolution

Solution 1:

Using this thread, and the fact that if $f_1$ and $f_2$ are two integrable functions, $\mathcal F(f\star g)=\mathcal F(f)\cdot\mathcal F(g)$, we have $$\mathcal F\left(\frac d{dx}(f\star g)\right)(x)=ix\mathcal F\left((f\star g)\right)(x)=ix \mathcal F(f)(x)\cdot \mathcal F(g)(x),$$ and $$\mathcal F\left(\left(\frac d{dx}f\right)\star g\right)(x)=\left(\mathcal F\left(\frac d{dx}f\right)\right)\cdot\left(\mathcal F(g)(x)\right)=ix \mathcal F(f)(x)\cdot \mathcal F(g)(x).$$ We conclude by uniqueness of Fourier transform.

Solution 2:

Definition: $$h(x)=f*g(x)=\int_A f(x-t)g(t)dt$$ where A is a support of function $q()$, i.e. $A=\{t:q(t)\ne 0\}$

Let's calculate derivative:

$$\frac {dh}{dx}=\underset{dx\rightarrow0}{\lim} \frac {(\int_A f(x+dx-t)g(t)dt-\int_A f(x-t)g(t)dt)}{dx}=\underset{dx\rightarrow0}{\lim}(\int_A \frac{(f(x+dx-t)-f(x-t))}{dx}g(t)dt)$$

If we assume that there exists some integrable function $q(t)$, such that for $t$ almost everywhere $$ \left| \frac{(f(x+dx-t)-f(x-t))}{dx} \right| < q(t), \forall dx>0 $$

I.e. $$ \mu\{t: \left| \frac{(f(x+dx-t)-f(x-t))}{dx} \right| \ge q(t)\}=0,\forall dx >0 $$

then by the Lebesgue dominated convergence theorem we can push the limit inside integral.

$$\frac {dh}{dx}=\frac{d}{dx}(f*g(x))=\int_A f'(x-t)g(t)dt=f'*g$$

Under assumption that: $\int_A q(t)dt$ is bounded above. One situation is when A is a compact set and $f,g$ are continuous function in the set A with a finite number of dicontinuities.