Unilateral convolution
For $t \in \mathbb{T}$, the convolution on a time scale is defined by the formula $$(f*g)_{\mathbb{T}}(t,s)=\displaystyle\int_{s}^t \hat{f}(t,\sigma(\xi))g(\xi)\Delta \xi,$$ where $\hat{f}$ denotes the solution of the shifting problem. The classic definition of the convolution using a shift in the integrand is not appropriate for time scales, since a time scale is not closed under addition and subtraction.
Properties
Covolution theorem for unilateral Laplace transform
Theorem: The convolution is associative: $$(f*g)*h = f*(g*h).$$
Proof: █
Theorem: If $f$ is $\Delta$-differentiable, then $$(f*g)^{\Delta}=f^{\Delta}*g+f(t_0)g.$$ If $g$ is $\Delta$-differentiable, then $$(f*g)^{\Delta}=f*g^{\Delta}+fg(t_0).$$
Proof: █
Theorem: Suppose that $\hat{f}$ has partial $\Delta$-derivatives of all orders. Then $$\dfrac{\partial^k \hat{f}}{\Delta^k t} (t,t)=f^{\Delta^k}(t_0).$$
Proof: █
Theorem: (Convolution theorem) The following formula holds: $$\mathscr{L}_{\mathbb{T}}\{f*g\}(z)=\mathscr{L}\{f\}(z) \mathscr{L}\{g\}(z),$$ where $\mathscr{L}$ denotes the Laplace transform and $f*g$ denotes the convolution.
Proof: █
Theorem: Define $u_a(t)= \left\{\begin{array}{ll} 0 &; t < a \\ 1 &; t \geq a \end{array} \right..$ Then $$\mathscr{L}_{\mathbb{T}}\{u_s \hat{f}(\cdot,s) \}(z) = e_{\ominus z}(s,t_0)\mathscr{L}_{\mathbb{T}}\{f\}(z).$$
Proof: █
