Proposition 23 ([Critical conditional-copy uniqueness]). For the invariant conditional coupling with common innovations at θ=b^-1/2 , the site-field discrepancy satisfies V(t)=0 for every t≥0 .

Work on the fixed horizon [0,T] used above. Inserting the noise and drift bounds and into the variance identity, and using n_rθ^ 2r =db^ r-1 b^ -r =d/b for r≥1 , gives V'(t)≤C( V(t)+∑_ r≥1 Q_r(t)+∑_ r≥1 L_r(t) +√V(t)∑_ r≥1 √Q_r(t)) for almost every t . The sums in Lemma 22 ([The noise and drift path sums]) are finite at the positive times with V(t)>0 where we now use this inequality. Insert those three bounds into. Use S≤CV , and hence √VS≤C V , for all terms except Sω^ -2 . The bounds V≤H_T^2 imply V^ 3/2 ≤H_TV and V^2≤H_T^2V . Finally ( I)^p≤C_p I (I≥1, 0<p≤2), since x^p e^ -x is bounded on x≥0 . As ω^ -1 ≤ ω^ -3/2 , all the terms just treated are at most C_TV(I+ω^ -3/2 ). This identifies the worst drift contribution explicitly: √VS\,ω^ -3/2 ≤CVω^ -3/2 . For 0<t_*,T , we have ω=√t . Lemma 21 ([Small-time log-factor estimate]) gives the remaining estimate Sω^ -2 ≤C_TV(t^ -3/4 +(I)/(t^2(1/t))). Together with I≥1 , the preceding bounds prove V'(t)≤k_T(t)V(t) (1+(K_T)/(V(t))) almost every t∈(0,T] with V(t)>0 , where, after increasing C_T , we may take k_T(t)= C_T(1+t^ -3/4 +(1)/(t^2(1/t))),, 0<t_*,T,; [2mm] C_T,, t_*<t≤T. For the second branch, ω(t)≥√t_*>0 because t_*≤1 ; there we use S≤CV also for Sω^ -2 . Thus that branch is bounded on the whole later interval. If T≤t_* there is no later interval. The small-time expression is used only for t≤t_*≤e^ -1 , and it is integrable there: ∫_0^ t_* t^ -3/4 \, dt=4t_*^ 1/4 , ∫_0^ t_* ( dt)/(t^2(1/t)) =(1)/((1/t_*)). In particular k_T∈L^1(0,T) . We finish with an Osgood-type comparison~, proving the needed absolutely continuous form directly. Suppose that V is positive somewhere on [0,T] . Choose such a time t , and let a<t be the left endpoint of its component in the positivity set of V . Continuity and V(0)=0 give V(a)=0 . On any closed interval [u,t] with a<u<t , continuity bounds V away from zero. The function z↦(1+(K_T/z)) has bounded derivative on the range of V on that interval. Its composition with the absolutely continuous function V is therefore absolutely continuous, and the chain rule and give ( I)'=-(V')/(VI)≥-k_T everywhere on [u,t]. Integrating yields I(u)≤ I(t)+∫_u^t k_T(s)\, ds ≤ I(t)+∫_a^t k_T(s)\, ds<∞. The last bound is independent of u . But V(u)→V(a)=0 as u a , so I(u)→∞ and I(u)→∞ , a contradiction. Hence V vanishes on [0,T] . Since the horizon was arbitrary, it vanishes at every time.