How to derive through a convolution?

Question

Let $f(t) = \alpha e^{-\beta t}$, where $\alpha, \beta$ are constants

Let $g(t) = y(t)$

Then the resulting convolution $f\ast g$ is:

$$f \ast g = \int_0^t \alpha e^{-\beta (t-\tau)} y(\tau) d\tau$$

• Does anyone know how one would take the derivative of this expression?

• In general, are there rules for taking derivative of $f \ast g$, for some given $f,g$?

Idea: using fundamental theorem of calculus, treat the integrand as a single function $h(t)$

$$\int_0^t \alpha e^{-\beta (t-\tau)} y(\tau) d\tau = \int_0^t h(\tau) d\tau$$

Answer 1

It suffices that one of the two functions is in $L^1$ and the other is in $L^{\infty} \cap C^1$ and its derivative is in $L^{\infty}$. More generally:

If $f \in L^1(\Bbb R^d)$ and $g \in C^k(\Bbb R^d)$, such that $D^{\alpha}g \in L^{\infty}(\Bbb R^d)$ for all $|\alpha| \le k$, then $f \star g \in C^k(\Bbb R^d)$ and: $D^{\alpha}(f \star g) = f \star D^{\alpha} g$.

Where $\alpha = (\alpha_1,\ldots, \alpha_d) \in \Bbb N^d$, $|\alpha| = \alpha_1 + \cdots + \alpha_d$, and:

$$D^{\alpha} g := \frac{\partial^{|\alpha|} g}{\partial x_1 ^{\alpha_1} \cdots \partial x_d ^{\alpha_d}}$$

This can be proved by induction on $k$. To prove the case $k=1$, we differentiate under the integral sign.

Answer 2

If $f$ and $g$ are $\mathcal{C}^1$ and $f',g'$ are both integrable, then: $$(f\ast g)'=f'\ast g=f\ast g'.$$ Basically, whenever $f'\ast g$ and $f\ast g'$ are well-defined, one has the equality above.

Answer 3

To avoid using concepts maybe unfamiliar, but also maintain rigor, I'll introduce some lemmas.

Let $f: [a,b]\to \mathbb{R}$ be a function. We say that $f$ is uniformly differentiable if for every $\epsilon > 0$ there exists a $\delta > 0$ so that $$ 0 < |x-y| < \delta \qquad\Rightarrow\qquad \left|\frac{f(y)-f(x)}{y-x} - f'(x)\right| < \epsilon. $$ for all $x,y\in[a,b]$.

Lemma 1: The function $f$ is uniformly differentiable if and only if it is continuously differentiable on $[a,b]$.

proof: Suppose that $f$ is continuously differentiable. Let $\epsilon>0$. Since $f^{\prime}$ is continuous on a compact set of $\mathbb{R}$, implying that $f^{\prime}$ is uniformly continuous. Then there exists $\delta$ such that $|x-y|<\delta$ implies that $|f^{\prime}(x)-f^{\prime}(y)|<\epsilon$ for all $x,y \in [a,b]$.

Let $x,y \in [a,b]$ such that $|x-y|<\delta$. By the mean value theorem, there exists $c \in [a,b]$ such that $f^{\prime}(c)=\frac{f(x)-f(y)}{x-y}$. It immediately follows that $|x-c|<\delta$, meaning that $|f^{\prime}(x)-\frac{f(x)-f(y)}{x-y}|=|f^{\prime}(x)-f^{\prime}(x)|<\epsilon$.

Now suppose that $f$ is uniformly differentiable. Let $\epsilon>0$. Then there exists $\delta>0$ such that $|x-y|<\delta$ implies that $|f^{\prime}(x)-\frac{f(x)-f(y)}{x-y}|<\frac{\epsilon}{2}$ for all $x,y \in [a,b]$. By the mean value theorem, there exists $c \in (x,y)$ such that $f^{\prime}(c)=\frac{f(x)-f(y)}{x-y}$. Note that $|x-c|<\delta$. Then

\begin{align*} |f^{\prime}(x)-f^{\prime}(y)| &=|f^{\prime}(x)-f^{\prime}(c)+f^{\prime}(c)-f^{\prime}(y)|\\ &\leq |f^{\prime}(x)-f^{\prime}(c)|+|f^{\prime}(c)-f^{\prime}(y)|\\ &<\epsilon/2+\epsilon/2=\epsilon \end{align*} thus concluding our proof.

Lemma 2: Let $f: \mathbb{R} \to \mathbb{R}$ and $g:[a,b] \to \mathbb{R}$ be continuous functions. Then $f*g$ is continuous.

proof: Let $\epsilon>0$. Since $g:[a,b] \to \mathbb{R}$ is continuous on a non-degenerate closed bounded interval, there exists $M \in \mathbb{R}$ such that $|g(t)| \leq M$ for all $t \in [a,b]$. Let $x \in \mathbb{R}$. Then define interval $[x-r,x+r]$. Then let $I=[x-r-a,x+r-b]$. It follows that $x-t \in I$ for all $t \in [a,b]$. It is also clear that Since $f|_{I}$ is uniformly continuous, there exists $\delta>0$ such that $|x-y|< \delta$ implies that $|f(x)-f(y)|<\epsilon/(m(b-a))$. Let $x,y \in I$ such that $|x-y|<\delta$. Note that $|(x-t)-(y-t)|<\delta$. Then \begin{align*} |f*g(x)-f*g(y)| &=\left|\int_{a}^{b} (f(x-t)-f(y-t))\cdot g(t) dt\right|\\ &\leq \int_{a}^{b}\left|(f(x-t)-f(y-t))\right|\cdot |g(t)| dt\\ &\leq M \int_{a}^{b} \frac{\epsilon}{M(b-a)} dt\\ &=\frac{M \epsilon (b-a)}{M(b-a)}=\epsilon. \end{align*}

Proposition 3: Let $f: \mathbb{R} \to \mathbb{R}$ and $g:[a,b] \to \mathbb{R}$ be continuous functions. Show that if $f^{\prime}$ is continuously differentiable, then $f*g$ is differentiable, where $(f*g)^{\prime}(x)=f^{\prime}*g(x)$.

proof: Let $\epsilon>0$. Since $g:[a,b] \to \mathbb{R}$ is continuous on a non-degenerate closed bounded interval, there exists $M \in \mathbb{R}$ such that $|g(t)| \leq M$ for all $t \in [a,b]$. Then define interval $[x-r,x+r]$. Then let $I=[x-r-a,x+r-b]$. It follows that $x-t \in I$ for all $t \in [a,b]$.

Since $f^{\prime}$ is continuously differentiable, by lemma 1, it is uniformly differentiable on $I$, implying that there exists $\delta$ so that $x,y \in I$ and $|x-y|<\delta$ imply that $|f^{\prime}(x)-\frac{f(x)-f(y)}{x-y}|<\frac{\epsilon}{M(b-a)}$. Then, we have that

\begin{align*}|f*g(x)-f*g(y)| &=\left|\int_{a}^{b}\frac{f(x-t)-f(y-t)}{x-y}\cdot g(t)dt-\int_{a}^{b}f^{\prime}(x-t)g(t)dt\right|\\ &=\left|\int_{a}^{b}(\frac{f(x-t)-f(y-t)}{x-y}-f^{\prime}(x-t))\cdot g(t)dt\right|\\ &\leq \int_{a}^{b}\left|\left(\frac{f(x-t)-f(y-t)}{x-y}-f^{\prime}(x-t)\right)\right|\cdot |g(t)|dt\\ &\leq M \int_{a}^{b} \frac{\epsilon}{M(b-a)} dt\\ &=\frac{M \epsilon (b-a)}{M(b-a)}=\epsilon. \end{align*}