Unilateral convolution

Let $f,g \colon \mathbb{R} \rightarrow \mathbb{R}$ be Lebesgue integrable on $\mathbb{R}$. The classical (i.e. time scale $\mathbb{T}=\mathbb{R}$) convolution of $f$ and $g$ is the function $f*g \colon \mathbb{R} \rightarrow \mathbb{R}$ given by $$(f*g)_{\mathbb{R}}(t)=\displaystyle\int_{\mathbb{R}} f(\tau)g(t-\tau) \mathrm{d}\tau.$$ The reason the convolution is of interest is because of the so-called convolution theorem for the classical Laplace transform: $$\mathscr{L}_{\mathbb{R}}\{f*g\}(z)=\mathscr{L}_{\mathbb{R}}\{f\}(z)\mathscr{L}_{\mathbb{R}}\{g\}(z).$$

Let $\mathbb{T}$ be any time scale and $f,g \colon \mathbb{T} \rightarrow \mathbb{C}$ be delta integrable on $\mathbb{T}$. We cannot simply use the definition of the convolution for time scales because an arbitrary time scale is not closed under addition and subtraction. The integrand $f(s)g(t-s) \mathrm{d}s$ will be written in the time scale convolution as $\hat{f}(t,\sigma(s))g(s) \Delta s$. Thus we define for $t \in \mathbb{T}$, $$(f*g)_{\mathbb{T}}(t;t_0)=\displaystyle\int_{t_0}^t \hat{f}(t,\sigma(s))g(s)\Delta s.$$

Properties

Theorem: (Convolution theorem) The following formula holds:

where $\mathscr{L}_{\mathbb{T}}$ denotes the Laplace transform.

Proof: █