Proof of convolution I would like to know how I could prove the following convolution:
$$
D (f*g) =D f* g =f* Dg
$$
 A: \begin{align}
(f'*g)(x) & = \int_{-\infty}^\infty f'(u)g(x-u)\,du \\[12pt]
& = \int_{-\infty}^\infty \underbrace{g(x-u)}_{s} \,  \underbrace{f'(u)\,du}_{dt} = \int s\,dt \\[12pt]
& = st - \int t\,ds \\[12pt]
& = \left.\phantom{\frac{}{}}f(u)g(x-u)\right|_{u\to-\infty}^{u\to\infty} - \int_{-\infty}^\infty f(u) g'(x-u) \, (-du).
\end{align}
The minus sign in $(-du)$ comes from the chain rule applied to $\dfrac{d}{du}g(x-u)$, giving us $ds=g'(x-u)\,(-du)$.
The last integral is $(f*g')(x)$.  The identity $f'*g=f*g'$ therefore holds when the expression before the integral is $0$.
A: You need some smoothness assumptions. Then you can interchange differentiation (with respect to $t$ below) with integration to get the result.
$$(f * g) (t) = \int_{\mathbb{R}} f(x)g(t-x) \, dx = \int_{\mathbb{R}} f(t-x)g(x) \, dx.$$
Differentiating gives
$$(f * g)' (t) = \int_{\mathbb{R}} f(x)g'(t-x)\,dx = \int_{\mathbb{R}} f'(t-x)g(x)\,dx,$$ which is the desired result.
A: I found a nice proof using Shift-Invariant Linear Systems:
Every Linear System that is Shift-Invariant is in-fact a Convolution.
Let $T$ be a system such that $g(x) = T\{f(x)\}$
Linear System Requirements:
Homogeneity: $T\{af(x)\} = aT\{f(x)\}$
Additivity: $T\{f(x) + h(x)\} = T\{f(x)\} + T\{h(x)\}$
Shift-Invariant Requirements:
$T\{f(x-x_0)\} = g(x-x_0)$
Clearly $\frac{d}{dx}$ satisfies these conditions.
Therefore if $T$ is the derivative operation, we get that $T\{f(x)\} =  \frac{d}{dx}f(x) = f(x) * T\{\delta(x)\}$ where $\delta(x)$ is Dirac's Delta function. This is the system's response to an impulse signal.
If we do this for both sides we get:
$\frac{d}{dx}(f(x))*g(x) \stackrel{?}{=} f(x)*\frac{d}{dx}(g(x))$
$ f(x) * T\{\delta(x)\}*g(x) \stackrel{?}{=} f(x)* g(x) * T\{\delta(x)\}$
Due to the commutative convolution operation, we can safely remove $T\{\delta(x)\}$ from the equation, and find that:
$\frac{d}{dx}(f(x))*g(x) = f(x)*\frac{d}{dx}(g(x))$, or 
$ f'(x)*g(x) = f(x)*g'(x)$
Note: This proof is also valid for any expression with the same amount of derivatives on both sides.
