Proof of convolution

Question

I would like to know how I could prove the following convolution: $$ D (f*g) =D f* g =f* Dg $$

Answer 1

\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 $0$.

Answer 2

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.

Answer 3

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.