then $x$ belongs to an interval of constancy for $F_\mu$. This in tandem with our hypothesis ${\operatorname{supp}}(\nu) \subseteq {\operatorname{supp}}(\mu)$, implies that the countable union of these intervals have $\nu$-measure zero.

\[lem: isomorphism\] The mapping $\varphi: \S^{*}(\nu,\mu) \to \S^{*}(\nu\circ {F}^{-1}_{\mu} ,\Lambda))$ defined by $\varphi(f) \coloneqq f \circ \check{F}^{-1}_{\mu}$, is an isometric isomorphism with inverse $\varphi^{-1}(f) \coloneqq f \circ F_{\mu}$, $*\in \{s,w\}$.

This is a consequence of together with the push-forward formula for measures in the strong case and the definition of $C_{\nu,\mu}$ in the weak case.

The following theorem is the main observation in this section needed for all subsequent corollaries.

\[prop:laplace\_backward\] For $\gamma \in [0,{\pi/2}]^{2}$, we have that $\Delta_{\nu \circ F_{\mu}^{-1}, \Lambda }\circ\varphi=\varphi\circ \Delta_{\nu,\mu }$ and $\varphi ( \mathcal{D}^*_{\gamma}(\Delta_{\nu,\mu}))=\mathcal{D}^*_{\gamma}(\Delta_{\nu \circ F_{\mu}^{-1},\Lambda})$.

If $f \in  \mathcal{D}^*_{\gamma}(\Delta_{\nu,\mu})$, then, for all $x \in  {\operatorname{supp}}(\mu)$, $$\begin{aligned}
f(x)=a+bF_{\mu}(x)+
\int \mathds{1}_{[0, x]}(y)(F_{\mu}(x)-F_{\mu}(y) ) \Delta_{\nu, \mu } f(y) \;{\mathrm{d}}\nu(y),\end{aligned}$$ where $a=f(0)$ and $b= \nabla_{\mu} f(0)$. Using and replacing $x$ with $\check{F}_{\mu}^{-1}(x)$ gives $$\begin{aligned}
    f(\check{F}_{\mu}^{-1}(x))
    &=a+bF_{\mu}(\check{F}_{\mu}^{-1}(x))+
\int \mathds{1}_{[0, \check{F}_{\mu}^{-1}(x)]}(y)(F_{\mu}(\check{F}_{\mu}^{-1}(x))-F_{\mu}(y) ) \Delta_{\nu, \mu }(f )(y)\;{\mathrm{d}}\nu(y) \\
&=a+b x+
\int \mathds{1}_{[0, \check{F}_{\mu}^{-1}(x)]}(y)(x-F_{\mu}(y) ) \Delta_{\nu, \mu }(f )(\check{F}_{\mu}^{-1}(F_{\mu}(y)))\;{\mathrm{d}}\nu(y) \\
&=a+bx+
\int \mathds{1}_{[0, x]}(F_{\mu}(y))(x-F_{\mu}(y) ) \Delta_{\nu, \mu }(f )(\check{F}_{\mu}^{-1}(F_{\mu}(y)))\;{\mathrm{d}}\nu(y)  \\
&=a+bx+
\int \mathds{1}_{[0, x]}(y)(x-y ) \Delta_{\nu, \mu }(f ) \circ \check{F}_{\mu}^{-1}(y) \;{\mathrm{d}}(\nu 