=\int_{\tilde Y}L_tv\,w\,d\mu^\varphi
=\int_{\tilde Y}T_tv\,w\,d\mu^\varphi$, so that $$\hat\rho(s)=\int_0^\infty e^{-st}\rho(t)\,dt
=\int_0^\infty e^{-st}\int_{\tilde Y} T_tv\, w\,d\mu^\varphi\,dt
=\int_{\tilde Y}\hat T(s)v\, w\,d\mu^\varphi,$$ as required.

\[thm-renew\] $\hat T(s)\hat R(s)=\hat T(s)-\hat U(s)$ for $s\in{{\mathbb H}}$.

Compute that $$\begin{aligned}
& \int_{\tilde Y}\hat T(s)\hat R(s)v\,w\,d\tilde\mu  = 
 \int_{\tilde Y} \int_0^\infty e^{-st}L_t\tilde R(e^{-s\tilde\varphi}v)\,w\,dt\,d\tilde\mu
 =\int_0^\infty e^{-st}\int_{\tilde Y} e^{-s\tilde\varphi}v\,w\circ f_t\circ \tilde F\,d\tilde\mu\,dt \\
& \qquad = \int_{\tilde Y} \int_0^\infty e^{-s(t+\tilde\varphi)} v\, w\circ f_{t+\tilde\varphi}\,dt\,d\tilde\mu
 = \int_{\tilde Y} \int_{\tilde\varphi}^\infty e^{-st} v\,w\circ f_t\,dt\,d\tilde\mu \\ & \qquad
 = \int_{\tilde Y} \int_0^\infty e^{-st} v\,w\circ f_t\,dt\,d\tilde\mu
 - \int_{\tilde Y} \int_0^{\tilde\varphi} e^{-st} v\,w\circ f_t\,dt\,d\tilde\mu \\ & \qquad
 = \int_{\tilde Y} \hat T(s)v\,w\,d\tilde\mu
 - \int_{\tilde Y} \int_0^\infty 1_{\{\tilde\varphi>t\}} e^{-st} v\,w\circ f_t\,dt\,d\tilde\mu 
 = \int_{\tilde Y} \hat T(s)v\,w\,d\tilde\mu
 - \int_{\tilde Y} \hat U(s)v\,w\,d\tilde\mu\end{aligned}$$ so $\hat T\hat R=\hat T-\hat U$ as required.

\[rmk-LT\] Throughout the paper, we write $\hat a(s)$ to denote a function that is analytic on ${{\mathbb H}}$, with inverse Laplace transform $a(t)$.

We note that $\hat R_0(s)$ has the formal inverse Laplace transform $R_0(t)v=R(\delta_{\varphi}(t)v)$, where $\delta_x$ is the $\delta$-measure at $x$, but this is not used explicitly in the paper.

For future reference, we record the following formula for $U_t$.

\[prop-U\] Suppose that $v\in L^1(\tilde Y)$. Then $$(U_tv)(y,u)=\begin{cases}
v(y,u-t)1_{[t,1]}(u), & 0\le t\le 1 \\
(\tilde Rv_t)(y,u), & t>1
\end{cases},$$ where $v_t(y,u)=1_{\{t<\varphi(y)<t+1-u\}}v(y,u-t+\varphi(y))$.

