A locally bounded solution $Z(t)$ to the renewal equation $ Z(t)=z(t)+\int^t_0Z(t-y)F(dy)$ is of the form $Z(t)=U*z(t)=\int^t_0z(t-u)U(du)$, where $U(t)$ is the renewal function (expected value of the number of renewal times until t).
If $F(x)=1-e^{-ax}$, for $x>0$, is the distribution of the interarrival times of a renewal process, we know that the renewal function is $U(t)=1+at$.
In Resnick's Adventures of Stochastic Processes, page 203, the author states that in this context $$U*z(t)=z(t)+a\int^t_0z(t-y)dy$$ Why is this?
Since, $U'(t)=a$, I would think that $U*z(t)=\int^t_0z(t-u)U(du)=a\int^t_0z(t-u) du$. Therefore substituting in the renewal equation one would get $U*z(t)=z(t)+F*U*z(t)$, but I have no idea how to obtain the equation in the book.
Any help would be appreciated.
$$ \begin{align*} Z &= z + Z*F = z + F*Z = z+F*(z+F*Z) \\ &= z+F*z + F_{2}*Z\\ &\vdots\\ &= z + z*\sum_{n=1}^{\infty}F_{n}\\ &= z+z*m_{F} \end{align*}$$
where $m_{F}(t) = \mathbb{E}[N(t)]$ where $N(t)$ is a poisson process with rate $a$ (because the interarrival times are exponential). Therefore, $m_{F}(t) = at$ and finally,
$$ Z(t) = z(t) + (z*at)(t) = z(t) + a \int_{0}^{t}z(t-y)dy $$
P.S. Please comment if something is not clear.