Finite cutoff differs from the unbounded pivot model by a controlled return error

At time zero, ${\eta }_0-{\nu }_0$ is the restriction of ${\eta }_0$ to $(R,\mathrm{\infty })$. Since every admissible bounded-Lipschitz test has supremum norm at most $1$, $d_{BL}({\eta }_0,{\nu }_0)\le {\eta }_0((R,\mathrm{\infty }))\le M_1\mathrm{/}R$ by [F4].

At any finite-cutoff state, the fragmentation gains and all parent-loss terms in $F_{\mathrm{\Delta }}$ agree with those of the no-cutoff field. The only omitted retained-measure gain is $P_{\mathrm{\Delta }}(x+y)$ for parent pairs with $x+y>R$. Their ordered event rate is at most $\frac{K_{\ast }}{2}{\iint }_{x+y>R}{\nu }_t(dx){\nu }_t(dy)\le K_{\ast }{M}_0({\nu }_t){\nu }_t((R\mathrm{/}2,\mathrm{\infty }))\mathrm{.}$ The finite mass bound gives $M_0({\nu }_t)\le M_1\mathrm{/}a$, while [F4] gives ${\nu }_t((R\mathrm{/}2,\mathrm{\infty }))\le 2M_1\mathrm{/}R$. Each omitted gain pairs with a bounded test by at most its event rate, so $\mathrm{\parallel }{F}_R({\nu }_t)-F_{\mathrm{\Delta }}({\nu }_t){\mathrm{\parallel }}_{BL}\le 2K_{\ast }{M}_1^2\mathrm{/}(aR)$.

Write $F_R$ for the retained-measure field of the finite algorithm [F1], and $F_{\mathrm{\Delta }}$ for the no-cutoff field. Subtract the equations for ${\nu }_t$ and ${\eta }_t$, add and subtract $F_{\mathrm{\Delta }}({\nu }_t)$, and test against any admissible bounded-Lipschitz function. By [F3] the vector-field difference contributes at most $C_T{d}_{BL}({\nu }_t,{\eta }_t)$; step 1.2 contributes at most $r:=2K_{\ast }{M}_1^2\mathrm{/}(aR)$. Taking the supremum and integrating yields $d(t)\le d(0)+C_T{\int }_0^{t}d(s)ds+\frac{2K_{\ast }{M}_1^{2}t}{aR}\mathrm{.}$ Let $g(t)=d(0)+rt+C_T{\int }_0^{t}d(s)ds$. Then $d\le g$ and $g^{\mathrm{\prime }}\le r+C_{T}g$ almost everywhere. If $C_T>0$, multiplying by $e^{-C_{T}t}$ and integrating gives $g(t)\le d(0)e^{C_{T}t}+r(e^{C_{T}t}-1)\mathrm{/}{C}_T\le e^{C_{T}t}(d(0)+rt)$; if $C_T=0$, $g(t)=d(0)+rt$. Substituting $d(0)\le M_1\mathrm{/}R$ from step 1.1 proves the bound.