Signed Dobrushin stability with bounded force defects

If $M=0$, the zero-mass condition and equality of total variation masses make ${\omega }_t={\nu }_t=0$, so all conclusions hold. Suppose $M>0$. Fix the prescribed couplings ${\pi }_0^{+},{\pi }_0^{-}$. Transport each by the corresponding two characteristic flows to obtain couplings ${\pi }_t^{+},{\pi }_t^{-}$. Let $D(t)={\sum }_{s\in \{+,-\}}\int \mathrm{\mid }{X}_t^{\omega }(x,s)-X_t^{\nu }(y,s)\mathrm{\mid }d{\pi }_0^s(x,y)\mathrm{.}$ The initial-position identities give $D(0)=D_0$. Their marginals are the Jordan parts at time $t$ by the sign-preservation assumption, so [F3] gives $C_{\pm }({\omega }_t,{\nu }_t)\le D(t)$. Integrable path suprema and [F6] make $D$ continuous.

For a coupled pair, subtract the two label equations. Decompose each signed interaction into its positive and negative Jordan-label integrals and apply [F2]. The interaction difference is at most $L_{\epsilon }M\mathrm{\mid }{X}^{\omega }-X^{\nu }\mathrm{\mid }+L_{\epsilon }D(t)$: the first term uses total variation mass $M$, and the inner second-coordinate differences integrate to the two transported coupling costs whose sum is $D(t)$. The defects add at most ${\eta }_{\omega }+{\eta }_{\nu }$. Integrating this pointwise estimate against the outer couplings, whose total mass is $M$, yields $D(t)\le D_0+2L_{\epsilon }M{\int }_0^{t}D(s)ds+M({\eta }_{\omega }+{\eta }_{\nu })t\mathrm{.}$ The last term and $D_0$ are nondecreasing, so [F4] gives the displayed exponential bound.

Apply [F1] and step 1.1 to obtain the bounded-Lipschitz estimate. By [F7], the initial coupling sets are nonempty and their infimum is finite; taking the infimum of the step 2.1 bound gives the stated result with $D_0=C_{\pm }({\omega }_0,{\nu }_0)$. No infimum is assumed to be attained.