How to take derivatives of a convolution when the kernel's derivative is in the distribution sense?

Question

I came need to take the derivative of the following convolution: $$ \int_{-\infty}^\infty \operatorname{sgn}(x-y)e^{-|x-y|}f(y) \, dy $$

However, the derivative of the kernel only exists in the sense of distributions, i.e. $$ -\frac{d}{dx}\operatorname{sgn}(x-y)e^{-|x-y|}=2\delta(x-y)e^{-|x-y|}-e^{-|x-y|} $$

My question is: According to this post, one cannot directly take the derivatives under the integral sign. So for my situation here, how am I supposed to do the differentiation?

Answer 1

HINT:

Split the integral as $$\int_{-\infty}^{\infty} \text{sgn} (x-y)e^{-|x-y|}f(y) \, dy = \int_{-\infty}^x e^{-(x-y)}f(y)\,dy-\int_x^{\infty} e^{(x-y)}f(y) \, dy$$ and use Leibnitz's Rule for differentiating under an integral.

SPOILER ALERT: SCROLL OVER SHADED AREA TO SEE ANSWER

$$\begin{align}\frac{d}{dx}\int_{-\infty}^{\infty} \text{sgn} (x-y)e^{-|x-y|}f(y) \, dy&=f(x)-\int_{-\infty}^x e^{-(x-y)}f(y)\,dy+f(x)-\int_x^{\infty} e^{(x-y)}f(y) \, dy\\\\&=2f(x)-\int_{-\infty}^{\infty} e^{-|x-y|}f(y) \, dy\end{align}$$

Answer 2

Let $g(x) = \operatorname{sgn}(x)e^{-|x|}$ and let $\delta$ be Dirac's delta "function". The problem is to find $(g*f)'(x)$.

$$ (g*f)'(x) = \underbrace{(\delta'*(g*f))(x) = ((\delta'*f)*g)(x)} = (f'*g)(x). \tag 1 $$

The thing $\underbrace{\text{in the middle}}$ has no meaning outside the context of some theory in which we can speak of $\delta'$. But the equality $$(g*f)'(x)=(f'*g)(x) \tag 2$$ is meaningful, and so can be true or false, outside such a context. If the whole chain of equalities $(1)$ is true in the Dirac context, then can $(2)$ be false in the context in which it can be understood without knowing of Dirac?

Here I'd say I'm rusty in that stuff and would crack open a book and remind myself of which hypotheses are needed for which conclusions to hold and ask whether those are satisfied.

That's how I would initially approach the problem.