Relationship between convolution for functions and for measures?

Question

Is the definition of the convolution of two measures in any way analogous to the definition of convolution of two functions? I know almost nothing about measure theory and have trouble making sense of the measure theory definition of convolution, so I have no idea if research into properties of convolution in that domain (e.g. finding convolution roots) maps back into the domain of convolution of functions (with applications to signal processing, computer vision, etc.). Are the two concepts related or independent but just sharing a name?

Answer 1

Consider the convolution $\mu * \nu$ of finite signed Borel measures $\mu$ and $\nu$ on $\mathbb{R}^d$ characterized by the condition

$$ \int_{\mathbb{R}^d} \varphi(x) \, (\mu * \nu)(\mathrm{d}x) = \iint_{\mathbb{R}^d\times\mathbb{R}^d} \varphi(x+y)\, \mu(\mathrm{d}x)\nu(\mathrm{d}y), \qquad \forall \varphi \in C(\mathbb{R}^d).$$

If $f$ and $g$ are integrable functions on $\mathbb{R}^d$, then the convolution of functions $f*g$ is related to the convolution of measures via the relation

$$ (f*g) \, \mathrm{d}x = (f \, \mathrm{d}x)*(g \, \mathrm{d}x), $$

where $\mathrm{d}x$ refers to the Lebesgue measure on $\mathbb{R}^d$. In this sense, convolution of measures can be regarded as a generalization of convolution of functions. Indeed, let $\mu(\mathrm{d}x) = f(x) \, \mathrm{d}x$ and $\nu(\mathrm{d}x) = g(x) \, \mathrm{d}x$. Then

\begin{align*} \int_{\mathbb{R}^d} \varphi(x) \, (\mu * \nu)(\mathrm{d}x) &= \iint_{\mathbb{R}^d\times\mathbb{R}^d} \varphi(x+y)\, \mu(\mathrm{d}x)\nu(\mathrm{d}y) \\ &= \int_{\mathbb{R}^d}\int_{\mathbb{R}^d} \varphi(x+y) f(x)g(y) \, \mathrm{d}x\mathrm{d}y \tag{by Fubini} \\ &= \int_{\mathbb{R}^d}\int_{\mathbb{R}^d} \varphi(x') f(x'-y)g(y) \, \mathrm{d}x'\mathrm{d}y \tag{$x'=x+y$} \\ &= \int_{\mathbb{R}^d} \varphi(x') \biggl( \int_{\mathbb{R}^d}f(x'-y)g(y) \, \mathrm{d}y \biggr) \, \mathrm{d}x' \tag{by Fubini}\\ &= \int_{\mathbb{R}^d} \varphi(x') (f * g)(x') \, \mathrm{d}x' \end{align*}

This shows that $\mu * \nu$ is absolutely continuous w.r.t. the Lebesgue measure, and the corresponding Radon–Nikodym derivative is precisely the convolution $f * g$.