Bounded-kernel measure solutions are stable in bounded-Lipschitz discrepancy

Fix an admissible test $\phi$ in [F2]. The coagulation integrand $H_{\phi }(x,y)=\frac{1}{2}K(x,y)[\phi (x+y)-\phi (x)-\phi (y)]$ has a uniform supremum bound and uniform Lipschitz bounds in each variable: the bracket is bounded by $3$ and changes by at most $2\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }$ when only $x$ changes, so the product rule for differences gives a bound depending only on $K_{\ast },L_K$.

The function $J_{\phi }(x)=\beta (x)[\int \phi d{B}_x-\phi (x)]$ is bounded by ${\beta }_{\ast }(D+1)$ and Lipschitz with constant at most $L_{\beta }(D+1)+{\beta }_{\ast }(L_B+1)$, by the assumed boundedness and $d_{BL}(B_x,B_y)\le L_B\mathrm{\mid }x-y\mathrm{\mid }$. Thus its fragmentation derivative difference is bounded by a constant times $d_{BL}({\mu }_t,{\nu }_t)$.

Write $\mu \otimes \mu -\nu \otimes \nu =(\mu -\nu )\otimes \mu +\nu \otimes (\mu -\nu )$. For each fixed value of the other variable, the section of $H_{\phi }$ is a bounded-Lipschitz function with the constants from step 1.1. Scaling that section into the unit test class [F2] and integrating the other variable gives a coagulation derivative difference bounded by $C_K(M_0({\mu }_t)+M_0({\nu }_t))d_{BL}({\mu }_t,{\nu }_t)$.

Integrating the weak equations, steps 2.1 and 1.2 give for every admissible $\phi$ the bound $\mathrm{\mid }⟨\phi ,{\mu }_t-{\nu }_t⟩\mathrm{\mid }\le d_{BL}({\mu }_0,{\nu }_0)+C_T{\int }_0^t{d}_{BL}({\mu }_s,{\nu }_s)ds$, where [F3] makes $C_T$ uniform on $[0,T]$. Taking the supremum gives $d(t)\le d(0)+C_T{\int }_0^{t}d(s)ds$. Put $g(t)=d(0)+C_T{\int }_0^{t}d(s)ds$; then $d\le g$ and $g^{\mathrm{\prime }}\le C_{T}g$ almost everywhere, so $(e^{-C_{T}t}g(t){)}^{\mathrm{\prime }}\le 0$ and $d(t)\le g(t)\le e^{C_{T}t}d(0)$.