Derivative of convolution

Question

Assume that $f(x),g(x)$ are positive and are in $L^1$. Moreover, they are differentiable and their derivative is integrable. Let $h(x)=f(x)*g(x)$, the convolution of $f$ and $g$. Does the derivative of $h(x)$ exist? If yes, how can we prove that $$ \frac{d}{dx}(f(x)*g(x)) = \left(\frac{d}{dx}f(x)\right)*g(x)$$

Thanks

Answer 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.

Answer 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.

Answer 3

Note that, if $ f\in L_1(R)$ then it is Fourier transformable. Since,

$$ \left|\int_{-\infty}^{\infty} f(x) e^{-ixw}\right| \leq \int_{-\infty}^{\infty} |f(x)| < \infty$$.

To prove that the convolution of two $L_{1}(R)$ functions is again an $L_{1}(R)$ function, let

$$ h(x) = \int f(t) g(x-t) dt $$

$$ \int |h(x)|dx \leq \int\int |f(t)| |g(x-t)| dt dx = \int |f(t)|\int |g(x-t)|dxdt = \int |f(t)| ||g||_1 dt = ||f||_1 ||g||_1 \Rightarrow h \in L_1(R)\,.$$

The change of the order of integration is justified by Fubini's theorem. So, you can use the Fourier technique as in Davide's answer.

Answer 4

In the case there your functions are only defined on $[0,\infty)$, you get the following.

Given $f, g: [0,\infty) \to \mathbb{R}^n$, define the convolution $$ (f*g)(t) = \int_{\tau = 0}^t f(t)g(t-\tau)d\tau = \int_{\tau = 0}^t f(t-\tau)g(t)d\tau . $$

Let $f, g \in L_2([0,\infty))$ be differentiable, and suppose $\dot{g}\in L_2([0,t_0])$ for any $t_0>0.$ Setting $p = (f*g),$ we have $$\dot{p}(t) = (f*\dot{g})(t) + f(t) g(0). $$

To prove this, write \begin{align*} \dot{p}(t) & = \lim_{h\to 0} h^{-1}\left[ \int_0^{t+h} f(u) g(t+h-u) du - \int_0^t f(u) g(t-u)du\right] \\ & = \lim_{h\to 0 } \int_0^{t+h} f(u) \frac{g(t+h-u)-g(t-u)}{h} du + \int_t^{t+h} \frac{f(u) g(t-u)}{h} du\\ & = (f*\dot{g})(t) + f(t)g(0), \end{align*} where the last equality follows for the first integral by writing the integrand with the mean value theorem as $f(u) \chi_{[0, t+h]}(u) \dot{g}(\xi)$ for $\xi \in (t-u, t+h-u)$ or $(t-u, t+h-u)$ and applying Lebesgue Dominated Convergence Theorem, and for the second integral by the Fundamental Theorem of Calculus.

Answer 5

Using Leibniz integral rule: $$\frac{d}{dx}\left(f*g(x)\right) = \int_{-\infty}^{+\infty}{\frac{\partial}{\partial x}(g(t)f(x-t))dt} = \int_{-\infty}^{+\infty}g(t){\frac{\partial}{\partial x}f(x-t)dt}=g*\frac{\partial}{\partial x}f$$