June 03, 2026
We isolate a layerwise refinement of the terminal testing-discrepancy step in Chen’s perturbed reverse-heat approach [1] to Talagrand’s convolution conjecture on the Boolean cube. Built on the joint-filtration martingale formulation of Chen’s coupling, and on Chen’s approximate monotonicity and conditional squared-score estimates being available in the joint-filtration form stated below, we prove the localized testing estimate \[D_E\le C_\tau\bigl(\mathcal{S}_E+\sqrt{\mathcal{S}_E\,\mathbb{P}(E)}\bigr), \qquad E\in\mathcal{F}_\theta,\] where \(D_E\) is the localized terminal testing discrepancy and \(\mathcal{S}_E\) is the stopped perturbative score energy. Applying this estimate to the layers \(G_r(\theta)=\{r\le R_\theta<r+1\}\) replaces the global Cauchy–Schwarz discrepancy cost by the layerwise cost \[O_\tau\left(\frac{\alpha}{\sqrt r}+\frac{\alpha^2}{r}\right) \mathbb{P}(G_r(\theta)), \qquad \alpha\simeq\log\log\eta.\] Under these imported joint-filtration inputs, combining the localized estimate with the time-smoothed anti-concentration profile yields the black-box consequence \[\mu\{P_\tau f>\eta\|f\|_1\} \le C_\tau \frac{\log\log\eta}{\eta\sqrt{\log\eta}}, \qquad \eta>e^3,\] for the Boolean heat semigroup. This makes a \((\log\log\eta)^{1/2}\) improvement over Chen’s result.
Talagrand’s convolution conjecture [2], [3] asks for the sharp dimension-free regularization of nonnegative \(L^1\) functions under convolution by a biased coin on the Boolean cube. Let \(\mu\) be the uniform probability measure on \(\{-1,1\}^n\), and let \((P_t)_{t\ge0}\) be the Boolean heat semigroup \[P_t f(x)=\mathbb{E}f(x\odot \xi_t), \qquad \mathbb{P}(\xi_t^{(i)}=1)=\frac{1+e^{-t}}{2}.\] Talagrand conjectured that, for every fixed \(\tau>0\), \[\mu\{P_\tau f>\eta\|f\|_1\} \le \frac{C_\tau}{\eta\sqrt{\log\eta}}, \qquad \eta>1,\] uniformly over the dimension and over \(f\ge0\). The conjectured order is known to be optimal up to the value of the constant \(C_\tau\).
The Gaussian analogue, with the Ornstein–Uhlenbeck semigroup replacing the Boolean heat semigroup, was initiated by Ball, Barthe, Bednorz, Oleszkiewicz and Wolff [4] and was resolved through the stochastic Föllmer-process approach of Eldan–Lee [5] and the refinement of Lehec [6]. In the Boolean setting, Chen [1] introduced a perturbed reverse-heat process and proved the first dimension-free improvement over Markov’s inequality. That is, Chen [1] obtained the bound as 1 \[\mu\{P_\tau f>\eta\|f\|_1\} \le C_\tau \frac{(\log\log\eta)^{3/2}}{\eta\sqrt{\log\eta}}.\]
The purpose of this note is to record a localization route that removes the extra factor \((\log\log\eta)^{1/2}\) within Chen’s framework, provided that the monotonicity and score-energy inputs being precisely in the form stated below. This note should be read as a conditional refinement of the reverse-heat framework, not as an independent proof of the imported monotonicity and score-energy estimates. The sole new ingredient is a localization of the terminal discrepancy cost with respect to the initial remaining gap \(R_\theta\); no new reverse process is introduced.
The key observation is elementary. Note that Chen [1] estimates the global discrepancy cost of a perturbation by first averaging over all values of the initial remaining gap and then applying Cauchy–Schwarz. This creates a factor \[\alpha\sqrt{\mathbb{E}\frac{\mathbf{1}_{\{R_\theta\ge \alpha\}}}{R_\theta+1}}, \qquad \alpha\simeq \log\log\eta.\] Because the time-smoothed profile estimate only gives \[\int \mathbb{E}\frac{\mathbf{1}_{\{R_\theta\ge \alpha\}}}{R_\theta+1}\,d\theta \lesssim \frac{\log\log\eta}{\log\eta},\] one obtains the exponent \(3/2\). Instead, we employ a layerwise refinement \[G_r(\theta) :=\{r\le R_\theta<r+1\}\] and apply the localized discrepancy estimate on each layer separately. For \(G_r\), the perturbation size is \(\alpha/r\), while the score energy is of order \(r\). As a result, the local discrepancy contribution is \[\lesssim \left(\frac{\alpha}{\sqrt r}+\frac{\alpha^2}{r}\right)\mathbb{P}(G_r(\theta)).\] After averaging \(\theta\) over a unit interval and using the time-smoothed profile estimate, the layer sum is \[\frac{\alpha}{L}\sum_{\alpha\le r\le L/2}r^{-1/2} +\frac{\alpha^2}{L}\sum_{\alpha\le r\le L/2}r^{-1} \lesssim \frac{\alpha}{\sqrt L}, \qquad L=\log\eta.\] This gives the announced conditional log–log bound under the Chen inputs stated below.
The new part of the argument is only the following localized terminal testing estimate. For every \(\mathcal{F}_\theta\)-measurable event \(E\), the discrepancy between the localized terminal sub-laws satisfies \[D_E\le C_\tau\left(\mathcal{S}_E+\sqrt{\mathcal{S}_E\,\mathbb{P}(E)}\right).\] All reverse-heat monotonicity and score-energy estimates are imported from Chen’s framework. We do not invoke Chen’s global total-variation estimate in its final global form. Instead, we reuse the bridge and Doob-transform calculation underlying that estimate, with the \(\mathcal{F}_\theta\)-measurable multiplier inserted before the final Cauchy–Schwarz step. This yields the localized terminal testing estimate for sub-probability laws displayed above.
We work under Chen’s reverse-heat framework. The perturbed coupling, approximate monotonicity estimate, and conditional score-energy estimate are used as imported ingredients. The time-smoothed profile estimate is standard in the same framework and is recalled below. The only new step is the localized accounting of the terminal testing discrepancy between the corresponding localized terminal sub-probability laws. We reproduce the bridge-gradient calculation only because the localization factor is inserted there. Constants denoted by \(C\) are universal, while constants denoted by \(C_\tau\) may depend on \(\tau\), but never on \(n,f,L,\eta,T\), or \(\theta\); both may change from line to line.
Let \(P_t=e^{t\mathcal{L}}\), where \[\mathcal{L}h(x)=\frac{1}{2}\sum_{i=1}^n\bigl(h(\sigma_i x)-h(x)\bigr).\] We write \(u_t=P_t f\), \(g_t=\log u_t\), and \[d\nu_t=u_t\,d\mu, \qquad \mathcal{A}_t(I)=\nu_t\{g_t\in I\}.\] By homogeneity we assume \(\|f\|_1=1\). The stochastic notation below is written for strictly positive \(f\), so that \(g_t\) and the reverse score are everywhere defined. If \(f\equiv0\), the theorem is trivial. For a general nonnegative \(f\not\equiv0\), apply the argument to \[F_\varepsilon=\frac{f+\varepsilon}{\|f+\varepsilon\|_1}.\] Then \(P_\tau F_\varepsilon(x)\to P_\tau f(x)/\|f\|_1\) for every \(x\). Since \(\{P_\tau f/\|f\|_1>\eta\}\subseteq\liminf_{\varepsilon\downarrow0} \{P_\tau F_\varepsilon>\eta\}\), Fatou’s lemma transfers the weak-type estimate from \(F_\varepsilon\) to \(f\).
Theorem 1 (Black-box consequence of Chen’s reverse-heat inputs). Assume the reverse-heat inputs 3 4 5 hold for every strictly positive normalized \(f\), every \(L\ge8\), every \(T>\tau+1\), and every \(\theta\in[T-\tau-1,T-\tau)\), with constants depending only on \(\tau\). Then, for every \(\tau>0\) there exists \(C_\tau<\infty\) such that, for every \(n\ge1\), every \(f:\{-1,1\}^n\to\mathbb{R}_+\), and every \(\eta>e^3\), \[\mu\{x:P_\tau f(x)>\eta\|f\|_1\} \le C_\tau \frac{\log\log\eta}{\eta\sqrt{\log\eta}}.\]
The theorem is a conditional implication. It becomes an unconditional improvement of Chen’s bound only after 3 4 5 are verified in the joint-filtration form stated here.
The argument is a black-box improvement inside Chen’s framework: it replaces the global discrepancy Cauchy–Schwarz step by a layerwise one. It should not be read as a proof of the conjectural \((\eta\sqrt{\log\eta})^{-1}\) bound; the remaining \(\log\log\eta\) factor would require information beyond the localized discrepancy bound and the time-smoothed profile used here. More specifically, we do not invoke Chen’s global total-variation estimate in its final global form. Instead, 6 keeps the bridge and Doob-transform calculation localized before the final Cauchy–Schwarz step.
Proposition 2 (Conditional fixed-time anti-concentration). Under the same reverse-heat inputs, for every \(\tau>0\) there exists \(C_\tau<\infty\) such that, for every \(L\ge8\), \[\mathcal{A}_\tau((L,L+1]) \le C_\tau \frac{\log L}{\sqrt L}.\]
Proof of 1 from 2. Put \(L=\log\eta\). For \(L\ge8\), \[\begin{align} \mu\{g_\tau>L\} &\le \sum_{k=0}^{\infty} \mu\{L+k<g_\tau\le L+k+1\} \\ &\le \sum_{k=0}^{\infty} e^{-(L+k)}\nu_\tau\{L+k<g_\tau\le L+k+1\} \\ &\le C_\tau e^{-L} \sum_{k=0}^{\infty}e^{-k}\frac{\log(L+k)}{\sqrt{L+k}} \le C_\tau e^{-L}\frac{\log L}{\sqrt L}. \end{align}\] Since \(e^L=\eta\), this proves the desired estimate for \(\eta>e^8\). The range \(e^3<\eta\le e^8\) is absorbed by Markov’s inequality after changing \(C_\tau\). ◻
Lemma 1 (Edge-ratio bound). Let \(f\ge0\) and \(f\not\equiv0\). For \(t>0\), \(x\in\{-1,1\}^n\), and \(i\in[n]\), \[\frac{1-e^{-t}}{1+e^{-t}} \le \frac{P_t f(\sigma_i x)}{P_t f(x)} \le \frac{1+e^{-t}}{1-e^{-t}}.\] Consequently, \[|g_t(\sigma_i x)-g_t(x)| \le \log\frac{1+e^{-t}}{1-e^{-t}}.\]
Proof. Using the multilinear extension, \(P_t f(x)=f(e^{-t}x)\). Fix \(z\in(-1,1)^n\). Since \(f\ge0\) on the cube, the multilinear function can be written in the \(i\)-th coordinate as \[f(z)=(1+z_i)A+(1-z_i)B,\] where \(A,B\ge0\). Replacing \(z_i\) by \(-z_i\) gives \[\frac{f(\sigma_i z)}{f(z)} = \frac{(1-z_i)A+(1+z_i)B}{(1+z_i)A+(1-z_i)B} \in \left[ \frac{1-|z_i|}{1+|z_i|}, \frac{1+|z_i|}{1-|z_i|} \right].\] Taking \(z=e^{-t}x\) proves the claim. ◻
Lemma 2 (Level-one inequality). Let \(h:\{-1,1\}^n\to\{0,1\}\), and let \(H\) be its multilinear extension. For every \(z\in(-1,1)^n\), \[\sum_{i=1}^n(1-z_i^2)(\partial_i H(z))^2 \le H(z)-H(z)^2.\]
Proof. This is the standard level-one inequality for biased Fourier analysis [7]. Let \(\mu_z\) be the product measure with coordinate means \(z_i\). The biased Fourier expansion gives \[(1-z_i^2)^{1/2}\partial_i H(z) = \widehat h_{\mu_z}(\{i\}).\] Summing over \(i\) and using Parseval gives the variance \(\operatorname{Var}_{\mu_z}(h)=H(z)-H(z)^2\). ◻
Lemma 3 (Time-smoothed anti-concentration). For every \(\ell>2\), \[\int_0^\infty \mathcal{A}_s((\ell,\ell+1])\,ds \le \frac{C}{\ell},\] where \(C\) is universal.
Proof. For \(s>0\), \(u_s=P_s f\) is strictly positive whenever \(f\not\equiv0\), since the Boolean heat kernel has full support; the endpoint \(s=0\) is irrelevant for the time integral. Thus all logarithms below are legitimate. Equivalently, one may first replace \(f\) by \((f+\varepsilon)/\|f+\varepsilon\|_1\) and then let \(\varepsilon\downarrow0\). Let \(E_s=\{x:\ell<g_s(x)\le \ell+1\}\). Choose a \(C^1\) cutoff \(\chi:\mathbb{R}\to[0,1]\), equal to \(1\) on \([0,1]\), supported on \((-1,2)\), and with \(\|\chi'\|_\infty\le C\). Define \[h_s(x)=u_s(x)\chi(g_s(x)-\ell)^2.\] Then \(h_s=u_s\) on \(E_s\), while \(h_s=0\) unless \(\ell-1<g_s<\ell+2\). Put \[m_s:=\mathcal{A}_s((\ell,\ell+1])=\int_{E_s}u_s\,d\mu, \qquad H_s:=\int h_s\,d\mu\le1.\] Since \(u_s>e^\ell\) on \(E_s\), one has \(\mu(E_s)\le e^{-\ell}m_s\). We use the entropy variational formula \[\operatorname{Ent}_\mu(h_s) = \sup_{\int e^\varphi d\mu\le1}\int h_s\varphi\,d\mu.\] Take \[\varphi_s=\frac{\ell}{2}\mathbf{1}_{E_s}-\log Z_s, \qquad Z_s=\int\exp\left(\frac{\ell}{2}\mathbf{1}_{E_s}\right)d\mu .\] Then \(\int e^{\varphi_s}d\mu=1\), and \[\log Z_s \le (e^{\ell/2}-1)\mu(E_s) \le e^{-\ell/2}m_s.\] Therefore \[\operatorname{Ent}_\mu(h_s) \ge \frac{\ell}{2}m_s-H_s\log Z_s \ge \left(\frac{\ell}{2}-e^{-\ell/2}\right)m_s,\] which, for \(\ell>2\), gives \[\operatorname{Ent}_\mu(h_s)\ge c\ell\,\mathcal{A}_s((\ell,\ell+1]).\] The Boolean log-Sobolev inequality [8] yields \[\operatorname{Ent}_\mu(h_s)\le C\,\mathcal{E}(\sqrt{h_s},\sqrt{h_s}),\] where \[\mathcal{E}(a,b)=-\int a\mathcal{L}b\,d\mu =\frac{1}{4}\int\sum_i\Delta_i a\,\Delta_i b\,d\mu.\] The discrete coarea estimate for the Lipschitz function \(\psi(r)=e^{r/2}\chi(r-\ell)\) gives \[\mathcal{E}(\sqrt{h_s},\sqrt{h_s}) \le C\int_{\ell-1}^{\ell+2} \mathcal{E}(\mathbf{1}_{\{u_s>e^v\}},u_s)\,dv.\] For completeness, this follows edge by edge. Put \(\Psi(A)=A^{1/2}\chi(\log A-\ell)\). If \(A\ge B>0\), then \[|\Psi(A)-\Psi(B)|^2 \le C(A-B)\int_{\ell-1}^{\ell+2} \mathbf{1}_{\{B<e^v<A\}}\,dv = C\int_{\ell-1}^{\ell+2} (\mathbf{1}_{\{A>e^v\}}-\mathbf{1}_{\{B>e^v\}})(A-B)\,dv.\] Applying this with \(A=u_s(x)\) and \(B=u_s(\sigma_i x)\), after orienting each edge so that \(A\ge B\), and summing over Boolean edges gives the displayed coarea bound. All identities involving level sets are first interpreted for a.e. level \(v\); the subsequent integration in \(v\) removes these exceptional levels. Finally, for \[F_s(v)=\int (u_s-e^v)_+\,d\mu,\] we have \[\frac{d}{ds}F_s(v) = -\mathcal{E}(\mathbf{1}_{\{u_s>e^v\}},u_s)\] for a.e. \(s\). Integrating in \(s\) and \(v\), and using \[\int_0^\infty \mathcal{E}(\mathbf{1}_{\{u_s>e^v\}},u_s)\,ds = F_0(v)-F_\infty(v) \le 1,\] then integrating over \(v\in[\ell-1,\ell+2]\), gives the result. ◻
Throughout [sec:reverse-heat] [sec:localized-discrepancy] and the proof of 2, we work first with \(f>0\) and \(\|f\|_1=1\). The general nonnegative case is obtained by the approximation explained in 1.
Fix \(\tau>0\), choose \(T>\tau+1\), and set \({T_{\circ}}=T-\tau\). The reverse heat process \((V_t)_{0\le t\le T}\) is the time reversal of the forward heat chain started from \(f\,d\mu\). Thus \(V_t\sim\nu_{T-t}\), and its generator is \[\widetilde{\mathcal{L}}_t h(x) = \frac{1}{2}\sum_{i=1}^n \frac{P_{T-t}f(\sigma_i x)}{P_{T-t}f(x)} \bigl(h(\sigma_i x)-h(x)\bigr).\] It is convenient to write \[S_i(t,x) = e^{-(T-t)} \frac{x_i\,\partial_i f(e^{-(T-t)}x)}{f(e^{-(T-t)}x)}.\] Then \[\frac{P_{T-t}f(\sigma_i x)}{P_{T-t}f(x)} = 1-2S_i(t,x),\] and the \(i\)-th reverse jump rate is \(1/2-S_i(t,x)\). The factor \(e^{-(T-t)}\) is part of the definition of the score. With this normalization, the edge-ratio bound gives \(1-2S_i(t,x)\in[C_\tau^{-1},C_\tau]\) for \(t\le{T_{\circ}}\).
Let \(L\ge8\), and set \[\alpha=\frac{1}{2}\log L+1.\] For a perturbation start time \(\theta\in[{T_{\circ}}-1,{T_{\circ}})\), define the remaining gap \[R_\theta=[L-g_{T-\theta}(V_\theta)]_+.\] The perturbation amplitude is the \(\mathcal{F}_\theta\)-measurable random variable \[\bar\delta=\frac{\alpha\,\mathbf{1}_{\{R_\theta\ge \alpha\}}}{R_\theta+1}.\] The perturbed process \(W\) starts from \(W_\theta=V_\theta\), is driven by the same Poisson clocks as \(V\), and has coordinate perturbation \(\delta_i(t)=\delta_i(t,V_{t-})\), where \[\delta_i(t,x) = \bar\delta\left[ \mathbf{1}_{\{S_i(t,x)>0\}} + \frac{1-2S_i(t,x)}{1-2\bar\delta S_i(t,x)} \mathbf{1}_{\{S_i(t,x)\le0\}} \right].\] The perturbation is stopped at \[{\sigma_\theta} = \inf\left\{t\in[\theta,{T_{\circ}}]: \max\{g_{T-t}(V_t)-\alpha,\;g_{T-t}(W_t)\}\ge L\right\} \wedge {T_{\circ}}.\] The process \(t\mapsto\mathbf{1}_{\{t\le{\sigma_\theta}\}}\) is left-continuous and adapted, hence predictable. All predictable statements are with respect to the natural filtration \[\mathcal{F}_t=\sigma(V_s:s\le t)\vee\sigma(W_s:\theta\le s\le t), \qquad t\in[\theta,{T_{\circ}}],\] completed in the usual way. Under conditioning on \(V_T=\zeta\), we use the same raw filtration on \([\theta,{T_{\circ}}]\), completed under the conditioned law. Conditioned on \(\mathcal{F}_\theta\), the joint process \((V_t,W_t)\) is a finite-state pure-jump process with predictable generator \[\bar{\mathcal{L}}_t^\delta h(x,y) = \bar{\mathcal{L}}_t^0h(x,y)+\mathbf{1}_{\{t\le{\sigma_\theta}\}}\mathcal{B}_t h(x,y),\] where \[\Delta_i^yh(x,y)=h(x,\sigma_i y)-h(x,y),\qquad \Delta_i^{xy}h(x,y)=h(\sigma_i x,\sigma_i y)-h(x,y),\] and, writing \(S_i=S_i(t,x)\), \(\delta_i=\delta_i(t,x)\), \[\begin{align} \label{eq:joint-generator} \bar{\mathcal{L}}_t^0h(x,y) &= \frac{1}{2}\sum_{i=1}^n(1-2S_i)\Delta_i^{xy}h(x,y), \notag\\ \mathcal{B}_t h(x,y) &= \sum_{i=1}^n \mathbf{1}_{\{S_i>0\}}\delta_iS_i\Delta_i^yh(x,y) + \sum_{i=1}^n \mathbf{1}_{\{S_i\le0\}}\delta_iS_i \Delta_i^yh(\sigma_i x,y). \end{align}\tag{1}\] The second line is a signed perturbation of the synchronized generator, not a generator by itself; the full operator \(\bar{\mathcal{L}}_t^\delta\) is the predictable generator of the coupled process. Throughout the sequel, \(T\) denotes the fixed terminal horizon in the reverse construction, while \({\sigma_\theta}\) denotes this stopping time. The perturbation part of the generator is always multiplied by \(\mathbf{1}_{\{t\le{\sigma_\theta}\}}\), so integrals of perturbative terms may be written over \([\theta,{\sigma_\theta}]\) or over \([\theta,{T_{\circ}}]\) with this indicator.
Input 3 (Joint-filtration martingale problem). We use Chen’s perturbed coupling in the following predictable generator formulation. For every strictly positive normalized \(f\), every \(L\ge8\), every \(T>\tau+1\), and every \(\theta\in[{T_{\circ}}-1,{T_{\circ}})\), the perturbed coupling described above is well-defined on \([\theta,{T_{\circ}}]\), has predictable generator 1 , and makes the \(V\)-coordinate a time-inhomogeneous Markov chain with generator \(\widetilde{\mathcal{L}}_t\) with respect to the joint filtration \((\mathcal{F}_t)\). Equivalently, for every bounded \(h=h(x)\), \[h(V_t)-h(V_\theta) -\int_\theta^t\widetilde{\mathcal{L}}_s h(V_s)\,ds\] is an \((\mathcal{F}_t)\)-martingale. In particular, for every bounded test function depending only on \(x\), the perturbation operator \(\mathcal{B}_t\) vanishes.
Verification from Chen’s PRM construction. This Markov statement is used below; no new reverse process is introduced. This is the only structural property of the coupled construction used below beyond the explicit predictable generator. The point needed below is not that the full pair \((V,W)\) is Markov in the ordinary state-space sense. Rather, in the Poisson random measure formulation, the additional information carried by \(W\) up to time \(t\) is generated by the same past Poisson increments and by predictable thinning decisions based on the history up to time \(t\). It does not reveal increments of the driving Poisson random measures after \(t\). Those future increments remain independent of \(\mathcal{F}_t\), and the \(V\)-coordinate jump intensities are functions only of \((t,V_{t-})\), namely \(\frac{1}{2}(1-2S_i(t,V_{t-}))\). Hence the \(V\)-coordinate has the same martingale problem with respect to \((\mathcal{F}_t)\) as in the original reverse-heat process. Equivalently, if \(P^V_{t,T}\) denotes the transition operator of the \(V\)-coordinate from \(t\) to \(T\), then for every bounded terminal test \(\Phi\), \[\mathbb{E}[\Phi(V_T)\mid\mathcal{F}_t] = P^V_{t,T}\Phi(V_t).\] This is the precise joint-filtration property used in 5. ◻
The next two quantitative inputs are reverse-heat estimates imported from Chen’s arXiv framework cited here [1]. They are stated in the notation above, with the normalized score \(S_i\), the stopping time \({\sigma_\theta}\), and the perturbation amplitude \(\bar\delta\). To make the comparison explicit, the main translation is \[\begin{array}{c|c} \text{Chen notation} & \text{notation in this note}\\ \hline \log \eta & L\\ \alpha=\frac{1}{2}\log\log\eta+1 & \alpha=\frac{1}{2}\log L+1\\ R_\theta=[\log\eta-\log P_{T-\theta}f(V_\theta)]_+ & R_\theta=[L-g_{T-\theta}(V_\theta)]_+\\ S_i(\rho_t\mathbf{V}_t) & S_i(t,V_t)\\ \text{Chen's stopping time } \boldsymbol{\mathfrak T} & \sigma_\theta\\ \kappa=(1+e^{-\tau})/(1-e^{-\tau}) & \text{absorbed into }C_\tau \end{array}\] The symbol \(\boldsymbol{\mathfrak T}\) in the table denotes Chen’s stopping time, not the terminal horizon \(T\) used here. Here the score row means explicitly \[S_i(t,V_t)= e^{-(T-t)} \frac{V_t^{(i)}\,\partial_i f(e^{-(T-t)}V_t)}{f(e^{-(T-t)}V_t)}.\] In Chen’s numbering, the coupling construction and generator formula correspond to the structural input in 3. 4 is the present-notation version of Chen’s approximate monotonicity estimate [1]. 5 is the present-notation version of the first conditional squared-score estimate in Chen [1], after translating Chen’s scaled score notation to \(S_i(t,V_t)\) and absorbing \(\kappa\) into \(C_\tau\). The proof below uses 5 only in this stated conditional form. If one works directly from Chen’s notation, this is the point where the translation of filtrations and stopping conventions must be checked. The time-smoothed profile in 3 is Chen’s Lemma 4, recalled above with proof. The paragraphs following the statements are sketches in the present notation, not independent reproofs of those imported martingale estimates. We do not claim that 3 is stated in Chen’s paper as a standalone proposition; it is the joint-filtration consequence of the predictable generator formulation needed for 5.
Input 4 (Approximate monotonicity). For the coupling above, \[\mathbb{P}\{g_\tau(V_{T_{\circ}})>L+1\} \ge \mathbb{P}\{g_\tau(W_{T_{\circ}})>L\} -\mathcal{A}_{T-\theta}((L-\alpha,L+\alpha]) -\frac{3}{\sqrt L}.\]
Explanation of imported input in the present notation. This is the approximate monotonicity estimate. We recall the argument. After conditioning on \(\mathcal{F}_\theta\), the perturbation size \(\bar\delta\) is frozen. In scaled variables \(\bar V_t=e^{-(T-t)}V_t\) and \(\bar W_t=e^{-(T-t)}W_t\), Itô’s formula for the jump process gives stochastic differential identities for \(\log f(\bar V_t)\) and \(\log f(\bar W_t)\). Conditionally on \(\mathcal{F}_\theta\), three exponential martingales give, with probability at least \(1-3/\sqrt L\), \[\begin{align} \log f(\bar V_{T_{\circ}}) &> \log f(\bar V_{{\sigma_\theta}})-\alpha+1,\\ \log f(\bar W_{{\sigma_\theta}})-\log f(\bar W_\theta) &< 2\sum_i\int_\theta^{{\sigma_\theta}}(1-\delta_i)w_iv_i\,dt+\alpha-1,\\ \log f(\bar W_{T_{\circ}})-\log f(\bar V_{T_{\circ}}) &< -2\sum_i\int_\theta^{{\sigma_\theta}} \delta_i\left(\mathbf{1}_{\{v_i>0\}}+ \frac{\mathbf{1}_{\{v_i\le0\}}}{1-2v_i}\right)w_iv_i\,dt+\alpha-1, \end{align}\] where \(v_i=S_i(t,V_{t-})\) and \(w_i=S_i(t,W_{t-})\). The special choice of \(\delta_i\) gives the exact algebraic identity \[\delta_i\left(\mathbf{1}_{\{v_i>0\}}+ \frac{\mathbf{1}_{\{v_i\le0\}}}{1-2v_i}\right) = \frac{\bar\delta}{1-\bar\delta}(1-\delta_i).\] Hence, on these three good events, \[\log f(\bar V_{T_{\circ}}) > \log f(\bar W_{T_{\circ}}) + \frac{\bar\delta}{1-\bar\delta} \bigl(\log f(\bar W_{{\sigma_\theta}})-\log f(\bar W_\theta)\bigr) -\frac{\alpha-1}{1-\bar\delta}.\] If \(g_\tau(W_{T_{\circ}})>L\), then either \(g_{T-{\sigma_\theta}}(V_{{\sigma_\theta}})\ge L+\alpha\), which immediately implies \(g_\tau(V_{T_{\circ}})>L+1\), or \(g_{T-{\sigma_\theta}}(W_{{\sigma_\theta}})\ge L\). In the second case, outside the initial band \((L-\alpha,L+\alpha]\), either the process starts already above \(L+\alpha\), reducing to the first case, or \(R_\theta\ge \alpha\). Then \(\bar\delta(R_\theta+1)=\alpha\), and the displayed lower bound again gives \(g_\tau(V_{T_{\circ}})>L+1\). The excluded initial band has probability \(\mathcal{A}_{T-\theta}((L-\alpha,L+\alpha])\). ◻
Input 5 (Conditional score energy). Conditionally on \(\mathcal{F}_\theta\), \[\mathbb{E}\left[ \int_\theta^{{\sigma_\theta}} \sum_{i=1}^n S_i(t,V_{t-})^2\,dt \,\middle|\,\mathcal{F}_\theta \right] \le C_\tau(R_\theta+\alpha+1).\]
Explanation of imported input in the present notation. Apply Itô’s formula to the process \[t\mapsto \log f(e^{-(T-t)}V_t).\] The drift contains \[S_i+\left(\frac{1}{2}-S_i\right)\log(1-2S_i).\] By the edge-ratio bound, \(1-2S_i\in[C_\tau^{-1},C_\tau]\) on \([0,{T_{\circ}}]\), and therefore \[S_i+\left(\frac{1}{2}-S_i\right)\log(1-2S_i) \ge c_\tau S_i^2.\] After integrating to \({\sigma_\theta}\), taking conditional expectation, and using the stopping definition \[g_{T-{\sigma_\theta}}(V_{{\sigma_\theta}}) \le L+\alpha+O_\tau(1),\] one obtains the stated bound. The \(O_\tau(1)\) term accounts for the possible one-jump overshoot at the hitting time. If the \(V\)-branch triggers, this is immediate from the threshold plus the edge-ratio bound. If the \(W\)-branch triggers first, then immediately before the triggering jump the \(V\)-branch is still below \(L+\alpha\), and any simultaneous \(V\)-jump again has \(O_\tau(1)\) overshoot by the edge-ratio bound. The stopping time \({\sigma_\theta}\) is a bounded stopping time for the enlarged joint filtration. Since 3 gives the \(V\)-martingale problem with respect to this enlarged filtration, the Itô identity may be stopped at \({\sigma_\theta}\), although \({\sigma_\theta}\) depends on the \(W\)-coordinate. ◻
The main estimate in this section is the point at which Chen’s global discrepancy argument is sharpened. It is the same Duhamel and bridge-gradient argument as in the reverse-heat method, but with an \(\mathcal{F}_\theta\)-measurable localization kept throughout the estimates.
Lemma 4 (Boolean bridge algebra). Fix \(t\in[\theta,{T_{\circ}}]\) and condition on \(V_T=\zeta\). Put \[\gamma_t=e^{-({T_{\circ}}-t)},\qquad \beta=e^{-\tau},\qquad \rho_t=\gamma_t\beta=e^{-(T-t)}.\] Define \[\lambda_{t,i}^{\zeta}(x) = \frac{1-\rho_t x_i\zeta_i}{1+\rho_t x_i\zeta_i}.\] Let \[a_t=\frac{\gamma_t(1-\beta^2)}{1-\rho_t^2} =\frac{\sinh(T-{T_{\circ}})}{\sinh(T-t)},\qquad b_t=\frac{\beta(1-\gamma_t^2)}{1-\rho_t^2} =\frac{\sinh({T_{\circ}}-t)}{\sinh(T-t)},\qquad \omega_i=x_iy_i\zeta_i,\] and \[m_t^{[i]}(x,y,\zeta)=a_ty_i+b_t\omega_i.\] Equivalently, \[m_t^{[i]}(x,y,\zeta) = y_i\frac{\gamma_t+\beta x_i\zeta_i}{1+\rho_t x_i\zeta_i}.\] For a multilinear extension \(\phi\) of a \(\{0,1\}\)-valued function, set \[q_t^\zeta(x,y)=\phi(m_t(x,y,\zeta)).\] Here \(\Delta_i^y h(x,y)=h(x,\sigma_i y)-h(x,y)\) and \(\Delta_i^{xy}h(x,y)=h(\sigma_i x,\sigma_i y)-h(x,y)\). Then \[\begin{align} \label{eq:bridge-delta-y} \Delta_i^yq_t^\zeta(x,y) &=-2(a_ty_i+b_t\omega_i)\partial_i\phi(m_t(x,y,\zeta)),\\ \label{eq:bridge-delta-y-flipx} \Delta_i^yq_t^\zeta(\sigma_i x,y) &=-2(a_ty_i-b_t\omega_i)\partial_i\phi(m_t(x,y,\zeta)),\\ \label{eq:bridge-delta-xy} \Delta_i^{xy}q_t^\zeta(x,y) &=-2a_ty_i\partial_i\phi(m_t(x,y,\zeta)). \end{align}\] {#eq: sublabel=eq:eq:bridge-delta-y,eq:eq:bridge-delta-y-flipx,eq:eq:bridge-delta-xy} Moreover, \[\begin{align} \label{eq:bridge-b-control} \lambda_{t,i}^{\zeta}(x)b_t^2 \le \frac{e^{-2\tau}}{1-e^{-2\tau}} \bigl(1-m_t^{[i]}(x,y,\zeta)^2\bigr), \end{align}\qquad{(1)}\] and, under the Doob-transformed unperturbed generator \[\mathcal{L}_t^{0,\zeta}h(x,y) = \frac{1}{2}\sum_{i=1}^n \lambda_{t,i}^{\zeta}(x)\Delta_i^{xy}h(x,y),\] one has \[\begin{align} \label{eq:bridge-harmonic} (\partial_t+\mathcal{L}_t^{0,\zeta})q_t^\zeta&=0,\\ \label{eq:bridge-square} (\partial_t+\mathcal{L}_t^{0,\zeta})(q_t^\zeta)^2 &= \frac{1}{2}\sum_i\lambda_{t,i}^{\zeta} \bigl(\Delta_i^{xy}q_t^\zeta\bigr)^2 \notag\\ &= 2a_t^2\sum_i\lambda_{t,i}^{\zeta} |\partial_i\phi(m_t)|^2. \end{align}\] {#eq: sublabel=eq:eq:bridge-harmonic,eq:eq:bridge-square}
Proof. The one-dimensional bridge calculation gives \[\mathbb{E}[V_{T_{\circ}}^{(i)}\mid V_t^{(i)}=x_i,V_T^{(i)}=\zeta_i] = \frac{\gamma_t x_i+\beta\zeta_i}{1+\rho_t x_i\zeta_i}.\] Since under the unperturbed synchronized coupling the sign discrepancy \(x_iy_i\) is preserved, the conditional mean of \(W_{T_{\circ}}^{0,(i)}\) is \[x_iy_i \frac{\gamma_t x_i+\beta\zeta_i}{1+\rho_t x_i\zeta_i} = y_i\frac{\gamma_t+\beta x_i\zeta_i}{1+\rho_t x_i\zeta_i}.\] The identity \[\frac{A+\beta\varepsilon}{1+A\beta\varepsilon} = \frac{A(1-\beta^2)}{1-A^2\beta^2} + \varepsilon \frac{\beta(1-A^2)}{1-A^2\beta^2}, \qquad \varepsilon\in\{-1,1\},\] with \(A=\gamma_t\), proves \(m_t^{[i]}=a_ty_i+b_t\omega_i\). The formula for \(\lambda_{t,i}^\zeta\) follows from the heat bridge kernel \[K_t^\zeta(x)=2^{-n}\prod_{j=1}^n(1+\rho_t x_j\zeta_j),\] and hence \[\frac{K_t^\zeta(\sigma_i x)}{K_t^\zeta(x)} = \frac{1-\rho_t x_i\zeta_i}{1+\rho_t x_i\zeta_i}.\] This gives \(q_t^\zeta(x,y)=\mathbb{E}[\phi(W_{T_{\circ}}^0)\mid V_t=x,W_t^0=y,V_T=\zeta]\), hence the harmonicity ?? . The three difference identities ?? –?? follow by observing how \(m_t^{[i]}\) changes when one flips \(y_i\), or flips both \(x_i\) and \(y_i\), and using multilinearity of \(\phi\). In particular, \(\partial_i\phi\) does not depend on the \(i\)-th coordinate of its argument, so the derivative may be evaluated at the common point \(m_t(x,y,\zeta)\) in the display above even when the finite difference passes through \((\sigma_i x,y)\).
For the \(b_t\)-coefficient, with \(\varepsilon=x_i\zeta_i\), \[1-\bigl(m_t^{[i]}(x,y,\zeta)\bigr)^2 = \frac{(1-\gamma_t^2)(1-\beta^2)}{(1+\rho_t\varepsilon)^2}\] and \[b_t^2= \frac{\beta^2(1-\gamma_t^2)^2}{(1-\rho_t^2)^2}.\] Consequently, \[\frac{\lambda_{t,i}^{\zeta}(x)b_t^2}{1-(m_t^{[i]}(x,y,\zeta))^2} = \frac{\beta^2(1-\gamma_t^2)}{(1-\rho_t^2)(1-\beta^2)} \le \frac{\beta^2}{1-\beta^2} = \frac{e^{-2\tau}}{1-e^{-2\tau}}.\] Finally, the square identity is the standard carré-du-champ identity for the finite-state generator \(\mathcal{L}_t^{0,\zeta}\): \[(\partial_t+\mathcal{L}_t^{0,\zeta})(q_t^\zeta)^2 = 2q_t^\zeta(\partial_t+\mathcal{L}_t^{0,\zeta})q_t^\zeta + \frac{1}{2}\sum_i\lambda_{t,i}^{\zeta} (\Delta_i^{xy}q_t^\zeta)^2,\] and the first term vanishes by ?? . Plugging in ?? gives ?? . ◻
The following consequence of 3 is used repeatedly. Even though the joint coupling is path-dependent through \(\mathbf{1}_{\{t\le{\sigma_\theta}\}}\), the \(V\)-coordinate has the same predictable characteristics with respect to the enlarged filtration. Consequently, future terminal tests of \(V_T\) may be conditioned through the original reverse-heat transition once \(V_t\) is fixed.
Lemma 5 (Terminal conditioning under Chen’s coupling). Assume 3. For \(\zeta\in\{-1,1\}^n\) with \(\mathbb{P}(V_T=\zeta)>0\), set \(\mathbb{P}^\zeta=\mathbb{P}(\,\cdot\,\mid V_T=\zeta)\). For \(t\in[\theta,{T_{\circ}}]\), \[\mathbb{P}(V_T=\zeta\mid\mathcal{F}_t) = \mathbb{P}(V_T=\zeta\mid V_t) = H_t^\zeta(V_t),\] where \[H_t^\zeta(x) = \frac{K_t^\zeta(x)f(\zeta)}{P_{T-t}f(x)},\qquad K_t^\zeta(x)=2^{-n}\prod_{j=1}^n(1+\rho_t x_j\zeta_j), \qquad \rho_t=e^{-(T-t)}.\] Let \(r_{t,i}^\zeta(x)=H_t^\zeta(\sigma_i x)/H_t^\zeta(x)\). Under \(\mathbb{P}^\zeta\), the stopped perturbed process on \([\theta,{T_{\circ}}]\) has predictable generator \[\mathcal{L}_t^{0,\zeta}+\mathbf{1}_{\{t\le{\sigma_\theta}\}}\mathcal{B}_t^\zeta,\] where \[\mathcal{L}_t^{0,\zeta}h(x,y) = \frac{1}{2}\sum_i\lambda_{t,i}^{\zeta}(x)\Delta_i^{xy}h(x,y), \qquad \lambda_{t,i}^{\zeta}(x)= \frac{1-\rho_t x_i\zeta_i}{1+\rho_t x_i\zeta_i},\] and \[\mathcal{B}_t^\zeta h(x,y) = \sum_i\mathbf{1}_{\{S_i>0\}}\delta_iS_i\Delta_i^yh(x,y) + \sum_i\mathbf{1}_{\{S_i\le0\}}r_{t,i}^\zeta\delta_iS_i \Delta_i^yh(\sigma_i x,y).\] Here \(S_i=S_i(t,x)\) and \(\delta_i=\delta_i(t,x)\). Moreover, if \(E\in\mathcal{F}_\theta\) and \(M_t^\zeta\) is a bounded martingale on \([\theta,{T_{\circ}}]\) under \(\mathbb{P}^\zeta\) with respect to the conditioned filtration, then \(\mathbf{1}_E(M_t^\zeta-M_\theta^\zeta)\) is again a martingale; the same statement holds under \(\mathbb{P}\). As before, \(\mathcal{B}_t^\zeta\) is a signed perturbation; only the full operator \(\mathcal{L}_t^{0,\zeta}+\mathbf{1}_{\{t\le{\sigma_\theta}\}}\mathcal{B}_t^\zeta\) is the conditioned predictable generator.
Proof. We first work with the raw natural filtration; statements for the usual completions follow by modifying \(E\in\mathcal{F}_\theta\) on null sets under the corresponding measures. The same pathwise stopping time \({\sigma_\theta}\) is a stopping time for the conditioned raw filtration under each law \(\mathbb{P}^\zeta\). By 3, the \(V\)-coordinate is Markov with respect to the joint filtration. Hence \(V_T\) is conditionally independent of the joint past, including \(W_t\), once \(V_t\) is fixed. This gives the displayed conditional probability formula. The bridge weight \(H_t^\zeta\) is space-time harmonic for the \(V\)-generator, \[(\partial_t+\widetilde{\mathcal{L}}_t)H_t^\zeta=0.\] Since \(\mathcal{B}_t\) vanishes on functions of \(x\) alone, it is also harmonic for the full joint generator: \[(\partial_t+\bar{\mathcal{L}}_t^\delta)H_t^\zeta=0.\] For a test function \(h\), the time-inhomogeneous Doob transform is \[\mathcal{L}_t^{\delta,\zeta}h = \frac{1}{H_t^\zeta} (\partial_t+\bar{\mathcal{L}}_t^\delta)(H_t^\zeta h) - \frac{h}{H_t^\zeta} (\partial_t+\bar{\mathcal{L}}_t^\delta)H_t^\zeta.\] The second term is zero by the harmonicity just displayed. Thus the time-inhomogeneous Doob transform has no extra killing term; it only multiplies each transition changing \(x\) by the ratio \(H_t^\zeta(\sigma_i x)/H_t^\zeta(x)\). The Doob transform is applied to the full predictable generator; the displayed transform of \(\mathcal{B}_t\) is only the perturbative part of that transformed full generator. Transition by transition: the synchronized \(x,y\)-flip receives this ratio, a \(y\)-only perturbative flip receives none, and the \(S_i\le0\) term, evaluated after the \(x\)-flip, receives \(r_{t,i}^\zeta(x)\). For the synchronized part this gives \[r_{t,i}^\zeta(x)(1-2S_i(t,x)) = \frac{K_t^\zeta(\sigma_i x)}{K_t^\zeta(x)} = \lambda_{t,i}^\zeta(x),\] which is the stated \(\mathcal{L}_t^{0,\zeta}\). The \(S_i>0\) perturbative term flips only \(y\), so no bridge ratio appears. The \(S_i\le0\) term is evaluated after flipping \(x\), so it picks up exactly \(r_{t,i}^\zeta(x)\). The martingale assertion follows because \(E\in\mathcal{F}_\theta\) is fixed at the initial time of the identities used below; boundedness on the finite state space makes the stopped local martingales true martingales. ◻
The next estimate is the local version of the terminal testing-discrepancy calculation: the factor \(\mathbf{1}_E\) is inserted before the final Cauchy–Schwarz step. The point is that \(E\in\mathcal{F}_\theta\) is frozen at the perturbation start time. Therefore it may be inserted before the Duhamel identity, before the conditional bridge expansion, and before the Doob-transform energy estimate. No stopping-time localization after \(\theta\) is used.
Lemma 6 (Localized terminal testing discrepancy). Let \(E\in\mathcal{F}_\theta\). Define \[\mu_E^W(A)=\mathbb{P}(E,W_{T_{\circ}}\in A),\qquad \mu_E^V(A)=\mathbb{P}(E,V_{T_{\circ}}\in A).\] With the probability convention \[d_{\operatorname{TV}}(\mu,\nu)=\sup_{A\subseteq\{-1,1\}^n}|\mu(A)-\nu(A)|,\] define \[D_E = d_{\operatorname{TV}}(\mu_E^W,\mu_E^V) = \sup_{\phi:\{-1,1\}^n\to\{0,1\}} \left| \mathbb{E}\left[ \mathbf{1}_E\{\phi(W_{T_{\circ}})-\phi(V_{T_{\circ}})\} \right]\right|.\] Thus \(D_E\) is a localized terminal testing discrepancy between two sub-probability measures; when \(E=\Omega\), it reduces to the usual terminal testing discrepancy between the laws of \(W_{T_{\circ}}\) and \(V_{T_{\circ}}\). The two sub-probability measures have the same total mass \(\mathbb{P}(E)\), so this convention agrees with the usual indicator-testing normalization of total variation. Also define \[\mathcal{S}_E = \mathbb{E}\left[ \mathbf{1}_E \int_\theta^{{\sigma_\theta}} \bar\delta^2\sum_{i=1}^n S_i(t,V_{t-})^2\,dt \right].\] For \(\theta\in[{T_{\circ}}-1,{T_{\circ}})\), \[D_E \le C_\tau\left(\mathcal{S}_E+\sqrt{\mathcal{S}_E\,\mathbb{P}(E)}\right).\]
Proof. Fix \(\phi:\{-1,1\}^n\to\{0,1\}\). We prove the estimate for this \(\phi\) and then take the supremum. Throughout the proof, \(\phi\) is identified with its multilinear extension to \([-1,1]^n\).
Localized Duhamel identity. Let \((V_t,W_t^0)\) denote the unperturbed synchronized joint process, started from the same pair \((V_\theta,W_\theta)\), and let \[U_t(x,y)= \mathbb{E}\bigl[\phi(W^0_{T_{\circ}})\mid (V_t,W_t^0)=(x,y)\bigr].\] This function is deterministic once \(f,T,t\) are fixed; below we evaluate the same backward solution along the perturbed path. Then \(U_{T_{\circ}}(x,y)=\phi(y)\), and \(U_t\) solves the backward equation for the unperturbed joint generator. More explicitly, let \(\mathcal{L}_t^0\) be the synchronized unperturbed predictable generator and let \(\mathcal{L}_t^\delta=\mathcal{L}_t^0+\mathbf{1}_{\{t\le{\sigma_\theta}\}}\mathcal{B}_t\) be the predictable generator of the perturbed joint process. Since the state space is finite and \(t\le{T_{\circ}}<T\), each coordinate jump rate and each finite difference appearing below is bounded by a constant depending only on \(\tau\); for each fixed \(n\), the total rate is finite. Thus \[U_t(V_t,W_t)-U_\theta(V_\theta,W_\theta) -\int_\theta^t\mathbf{1}_{\{s\le{\sigma_\theta}\}} \mathcal{B}_s U_s(V_{s-},W_{s-})\,ds\] is a martingale. Since \(E\in\mathcal{F}_\theta\), multiplying this identity by \(\mathbf{1}_E\) preserves the martingale property. Also, on the unperturbed synchronized coupling, \(W_t^0=V_t\) when \(W_\theta^0=V_\theta\). Hence \[U_\theta(V_\theta,W_\theta) = \mathbb{E}[\phi(V_{T_{\circ}})\mid\mathcal{F}_\theta], \qquad U_{T_{\circ}}(V_{T_{\circ}},W_{T_{\circ}})=\phi(W_{T_{\circ}}),\] and, since \(E\in\mathcal{F}_\theta\), \[\mathbb{E}[\mathbf{1}_E U_{T_{\circ}}(V_{T_{\circ}},W_{T_{\circ}})] = \mathbb{E}[\mathbf{1}_E\phi(W_{T_{\circ}})], \qquad \mathbb{E}[\mathbf{1}_E U_\theta(V_\theta,W_\theta)] = \mathbb{E}[\mathbf{1}_E\phi(V_{T_{\circ}})].\] Therefore \[\begin{align} \label{eq:localized-duhamel} \mathbb{E}\left[\mathbf{1}_E(\phi(W_{T_{\circ}})-\phi(V_{T_{\circ}}))\right] = \mathbb{E}\left[ \mathbf{1}_E\int_\theta^{{\sigma_\theta}} \mathcal{B}_t U_t(V_{t-},W_{t-})\,dt \right], \end{align}\tag{2}\] where \(\mathcal{B}_t\) is the perturbation part of the predictable joint generator.
The perturbation leaves the \(V\)-marginal unchanged: for every test function depending only on \(x\), the operator \(\mathcal{B}_t\) vanishes. Hence \(V_T\) and the bridge weights \(H_t^\zeta(V_t)\) appearing below are those of the original reverse heat process throughout the Duhamel computation.
Bridge expansion. We next recall the bridge representation used to estimate \(\mathcal{B}_t U_t\); the coefficients \(a_t,b_t,m_t,\lambda_{t,i}^\zeta\) are those of 4. For \(\zeta\in\{-1,1\}^n\), set \[K_t^\zeta(x)=2^{-n}\prod_{j=1}^n(1+\rho_t x_j\zeta_j), \qquad \rho_t=e^{-(T-t)}.\] Thus \(K_t^\zeta(x)\) is the forward heat probability of being at \(\zeta\) at time \(T\), starting from \(x\) at time \(t\). The actual reverse-process bridge weight is \[H_t^\zeta(x) = \mathbb{P}(V_T=\zeta\mid V_t=x) = \frac{K_t^\zeta(x)f(\zeta)}{P_{T-t}f(x)}, \qquad r_{t,i}^\zeta(x)=\frac{H_t^\zeta(\sigma_i x)}{H_t^\zeta(x)}.\] Since \(P_{T-t}f(\sigma_i x)/P_{T-t}f(x)=1-2S_i(t,x)\), the heat-kernel ratio gives the useful identity \[r_{t,i}^\zeta(x)(1-2S_i(t,x)) = \frac{K_t^\zeta(\sigma_i x)}{K_t^\zeta(x)} = \lambda_{t,i}^\zeta(x).\] Under the conditional law \(\mathbb{P}^\zeta=\mathbb{P}(\,\cdot\,\mid V_T=\zeta)\), the unperturbed synchronized generator is the Doob transform \[\mathcal{L}_t^{0,\zeta}h(x,y) = \frac{1}{2}\sum_{i=1}^n\lambda_{t,i}^\zeta(x)\Delta_i^{xy}h(x,y),\] where \[\lambda_{t,i}^\zeta(x) = \frac{1-\rho_t x_i\zeta_i}{1+\rho_t x_i\zeta_i}.\] For \(t\le{T_{\circ}}\), the edge-ratio bound implies \[C_\tau^{-1}\le \lambda_{t,i}^\zeta(x)\le C_\tau.\] The perturbation part under \(\mathbb{P}^\zeta\) is \[\mathcal{B}_t^\zeta h(x,y) = \sum_i\mathbf{1}_{\{S_i>0\}}\delta_iS_i\Delta_i^yh(x,y) + \sum_i\mathbf{1}_{\{S_i\le0\}}r_{t,i}^\zeta\delta_iS_i \Delta_i^yh(\sigma_i x,y),\] where \(S_i=S_i(t,x)\) and \(\delta_i=\delta_i(t,x)\). Moreover, \[\left|\delta_i \bigl(\mathbf{1}_{\{S_i>0\}}+r_{t,i}^\zeta\mathbf{1}_{\{S_i\le0\}}\bigr)\right|^2 \le C_\tau\,\bar\delta^2\,\lambda_{t,i}^\zeta(x).\] Indeed, if \(S_i>0\), then \(\delta_i=\bar\delta\) and \(\lambda_{t,i}^\zeta\ge C_\tau^{-1}\). If \(S_i\le0\), then \[\delta_i=\bar\delta\,\frac{1-2S_i}{1-2\bar\delta S_i}, \qquad r_{t,i}^\zeta(1-2S_i)=\lambda_{t,i}^\zeta,\] so \[r_{t,i}^\zeta\delta_i = \bar\delta\,\frac{\lambda_{t,i}^\zeta}{1-2\bar\delta S_i}.\] Since \(S_i\le0\), the denominator is at least \(1\), and \(\lambda_{t,i}^\zeta\le C_\tau\). This gives the displayed bound.
For \(t<{T_{\circ}}\), let \(m_t(x,y,\zeta)\in(-1,1)^n\) be the product mean of the Boolean heat bridge from \(t\) to \({T_{\circ}}\), conditioned on \(V_t=x\) and \(V_T=\zeta\), after the synchronized sign change \(x\odot y\). Write \[q_t^\zeta(x,y)=\phi(m_t(x,y,\zeta)).\] Equivalently, \(q_t^\zeta\) is the conditional expectation of the \(\{0,1\}\)-valued terminal test \(\phi(W_{T_{\circ}}^0)\) under the Boolean bridge, so \(0\le q_t^\zeta\le1\). The bridge formula in 4 gives \[\begin{align} \label{eq:bridge-derivatives-local} \Delta_i^yq_t^\zeta(x,y) &=-2(a_ty_i+b_t\omega_i)\partial_i\phi(m_t(x,y,\zeta)), \notag\\ \Delta_i^yq_t^\zeta(\sigma_i x,y) &=-2(a_ty_i-b_t\omega_i)\partial_i\phi(m_t(x,y,\zeta)), \end{align}\tag{3}\] where \(\omega_i=x_iy_i\zeta_i\). Consequently, \[|\Delta_i^yq_t^\zeta(x,y)|^2+ |\Delta_i^yq_t^\zeta(\sigma_i x,y)|^2 \le C(a_t^2+b_t^2)\,|\partial_i\phi(m_t(x,y,\zeta))|^2.\] By finite-state regular disintegration, \[\mathbb{E}[\cdot\mid\mathcal{F}_t] = \sum_{\zeta\in\{-1,1\}^n} \mathbb{P}(V_T=\zeta\mid\mathcal{F}_t) \mathbb{E}[\cdot\mid\mathcal{F}_t,V_T=\zeta].\] By 5, \(H_t^\zeta(V_t)\) is the conditional law of \(V_T\) even after conditioning on the joint past \(\mathcal{F}_t\). Therefore expanding \(U_t\) through \(V_T\) gives \[U_t(x,y)=\sum_{\zeta\in\{-1,1\}^n}H_t^\zeta(x)q_t^\zeta(x,y).\] Consequently the perturbative generator decomposes as \[\mathcal{B}_t U_t(x,y) = \sum_{\zeta\in\{-1,1\}^n} H_t^\zeta(x)\mathcal{B}_t^\zeta q_t^\zeta(x,y).\] Indeed, for each fixed \(\zeta\), \[\mathcal{B}_t(H_t^\zeta q_t^\zeta)(x,y) = H_t^\zeta(x)\mathcal{B}_t^\zeta q_t^\zeta(x,y).\] For the \(S_i>0\) part, \(\mathcal{B}_t\) flips only the \(y\)-coordinate, so the coefficient \(H_t^\zeta(x)\) is unchanged. For the \(S_i\le0\) part, \[\Delta_i^y(H_t^\zeta q_t^\zeta)(\sigma_i x,y) = H_t^\zeta(\sigma_i x)\Delta_i^yq_t^\zeta(\sigma_i x,y) = H_t^\zeta(x)r_{t,i}^\zeta(x) \Delta_i^yq_t^\zeta(\sigma_i x,y),\] which is exactly the factor appearing in \(\mathcal{B}_t^\zeta\).
Pointwise perturbation bound. Combining the last three displays gives \[|\mathcal{B}_t U_t(x,y)| \le C_\tau \sum_{\zeta}H_t^\zeta(x)\sum_i \bar\delta\left|S_i(t,x)\right|\,\lambda_{t,i}^\zeta(x)^{1/2} (a_t^2+b_t^2)^{1/2} |\partial_i\phi(m_t(x,y,\zeta))|.\] Applying weighted Cauchy–Schwarz over the product index \((\zeta,i)\), with weights \(H_t^\zeta(x)\), yields the pointwise estimate \[\begin{align} \label{eq:pointwise-AtU} |\mathcal{B}_t U_t(x,y)| \le C_\tau \left(\bar\delta^2\sum_iS_i(t,x)^2\right)^{1/2} \Gamma_t(x,y)^{1/2}, \end{align}\tag{4}\] where \[\Gamma_t(x,y)= \sum_{\zeta}H_t^\zeta(x) \sum_i\lambda_{t,i}^\zeta(x)(a_t^2+b_t^2) |\partial_i\phi(m_t(x,y,\zeta))|^2.\] From 2 , 4 , and Cauchy–Schwarz, \[\begin{align} \label{eq:localized-tv-cauchy} \left|\mathbb{E}\left[\mathbf{1}_E(\phi(W_{T_{\circ}})-\phi(V_{T_{\circ}}))\right]\right| \le C_\tau\,\mathcal{S}_E^{1/2} \left(\mathbb{E}\left[\mathbf{1}_E\int_\theta^{T_{\circ}} \Gamma_t(V_{t-},W_{t-})\,dt\right]\right)^{1/2}. \end{align}\tag{5}\]
Energy closure. It remains to localize the bridge-gradient energy. Define \[\begin{align} \Psi_a^E &= \mathbb{E}\left[\mathbf{1}_E\int_\theta^{T_{\circ}} a_t^2\sum_i\lambda_{t,i}^{V_T}(V_t) |\partial_i\phi(m_t(V_t,W_t,V_T))|^2\,dt\right],\\ \Psi_b^E &= \mathbb{E}\left[\mathbf{1}_E\int_\theta^{T_{\circ}} b_t^2\sum_i\lambda_{t,i}^{V_T}(V_t) |\partial_i\phi(m_t(V_t,W_t,V_T))|^2\,dt\right]. \end{align}\] In Lebesgue-time integrals we freely replace \(V_{t-},W_{t-}\) by \(V_t,W_t\), since the jump times are countable almost surely. Then the energy in 5 is bounded by \(C(\Psi_a^E+\Psi_b^E)\). For endpoint rigor, fix \(0<\varepsilon<{T_{\circ}}-\theta\) and let \(\Psi_{a,\varepsilon}^E,\Psi_{b,\varepsilon}^E\) denote the same quantities with the upper limit \({T_{\circ}}-\varepsilon\). We first prove the estimates below for these truncated energies, with constants independent of \(\varepsilon\). Monotone convergence then gives the displayed full-time bounds. To keep notation readable, the subscript \(\varepsilon\) is suppressed until the final limiting step.
First consider \(\Psi_b^E\). The bridge algebra ?? gives \[\frac{\lambda_{t,i}^{\zeta}(x)b_t^2}{1-m_t^{[i]}(x,y,\zeta)^2} \le C_\tau.\] Using 2 at the point \(m_t(x,y,\zeta)\), \[\sum_i(1-m_t^{[i]}(x,y,\zeta)^2) |\partial_i\phi(m_t(x,y,\zeta))|^2 \le \frac{1}{4}.\] Hence the integrand defining \(\Psi_b^E\) is at most \(C_\tau\), and since \({T_{\circ}}-\theta\le1\), \[\begin{align} \label{eq:local-psi-b} \Psi_b^E\le C_\tau\mathbb{P}(E). \end{align}\tag{6}\]
We now estimate \(\Psi_a^E\). Fix \(\zeta\) and work under \(\mathbb{P}^\zeta\). By ?? , \[(\partial_t+\mathcal{L}_t^{0,\zeta})(q_t^\zeta)^2 = \frac{1}{2}\sum_i\lambda_{t,i}^{\zeta} \bigl(\Delta_i^{xy}q_t^\zeta\bigr)^2 = 2a_t^2\sum_i\lambda_{t,i}^\zeta |\partial_i\phi(m_t)|^2.\] Because \(E\in\mathcal{F}_\theta\), multiplying Itô’s formula for \((q_t^\zeta(V_t,W_t))^2\) by \(\mathbf{1}_E\) is legitimate by 5. More explicitly, applying Itô’s formula under \(\mathbb{P}^\zeta\), averaging over \(\zeta=V_T\), and using the conditioned generator from 5, gives \[\begin{align} 2\Psi_{a,\varepsilon}^E &= \mathbb{E}\left[ \mathbf{1}_E (q_{{T_{\circ}}-\varepsilon}^{V_T})^2(V_{{T_{\circ}}-\varepsilon},W_{{T_{\circ}}-\varepsilon}) \right] - \mathbb{E}\left[ \mathbf{1}_E (q_{\theta}^{V_T})^2(V_\theta,W_\theta) \right] \\ &\quad - \mathbb{E}\left[ \mathbf{1}_E\int_\theta^{{T_{\circ}}-\varepsilon} \mathbf{1}_{\{t\le{\sigma_\theta}\}} \mathcal{B}_t^{V_T}(q_t^{V_T})^2(V_{t-},W_{t-})\,dt \right]. \end{align}\] Since \(0\le q_t^\zeta\le1\), this yields \[\Psi_{a,\varepsilon}^E \le C\mathbb{P}(E) + C\,\mathbb{E}\left[\mathbf{1}_E\int_\theta^{{T_{\circ}}-\varepsilon} \mathbf{1}_{\{t\le{\sigma_\theta}\}} |\mathcal{B}_t^{V_T}(q_t^{V_T})^2(V_{t-},W_{t-})|\,dt\right].\] Since \(0\le q_t^\zeta\le1\), \[|\Delta_i^y(q_t^\zeta)^2(x,y)| \le2|\Delta_i^yq_t^\zeta(x,y)|,\qquad |\Delta_i^y(q_t^\zeta)^2(\sigma_i x,y)| \le2|\Delta_i^yq_t^\zeta(\sigma_i x,y)|.\] Using the displayed formula for \(\mathcal{B}_t^\zeta\), the estimate \[\left|\delta_i \bigl(\mathbf{1}_{\{S_i>0\}}+r_{t,i}^\zeta\mathbf{1}_{\{S_i\le0\}}\bigr)\right|^2 \le C_\tau\bar\delta^2\lambda_{t,i}^\zeta,\] and Cauchy–Schwarz in the coordinate \(i\), we get \[|\mathcal{B}_t^\zeta(q_t^\zeta)^2| \le C_\tau \left(\bar\delta^2\sum_iS_i(t,V_{t-})^2\right)^{1/2} \left(\Gamma_{a,t}^\zeta+\Gamma_{b,t}^\zeta\right)^{1/2},\] where \[\Gamma_{a,t}^\zeta = a_t^2\sum_i\lambda_{t,i}^{\zeta} |\partial_i\phi(m_t(x,y,\zeta))|^2,\qquad \Gamma_{b,t}^\zeta = b_t^2\sum_i\lambda_{t,i}^{\zeta} |\partial_i\phi(m_t(x,y,\zeta))|^2.\] After evaluating at \((V_{t-},W_{t-},V_T)\), multiplying by \(\mathbf{1}_E\mathbf{1}_{\{t\le{\sigma_\theta}\}}\), integrating in time, and applying Cauchy–Schwarz, the last display gives \[\Psi_a^E \le C_\tau\mathbb{P}(E) + C_\tau\mathcal{S}_E^{1/2}(\Psi_a^E+\Psi_b^E)^{1/2}.\] Set \(Y=\Psi_a^E+\Psi_b^E\). Combining the last display with 6 gives \[Y\le C_\tau\mathbb{P}(E)+C_\tau\mathcal{S}_E^{1/2}Y^{1/2}.\] Using \(uv\le \frac{1}{2}v^2+C_\tau u^2\), with \(u=\mathcal{S}_E^{1/2}\) and \(v=Y^{1/2}\), we obtain \[Y\le C_\tau\bigl(\mathbb{P}(E)+\mathcal{S}_E\bigr).\] Restoring the suppressed truncation parameter, this estimate is uniform for \(Y_\varepsilon=\Psi_{a,\varepsilon}^E+\Psi_{b,\varepsilon}^E\). Letting \(\varepsilon\downarrow0\) and using monotone convergence gives the same bound for the full energies \(\Psi_a^E+\Psi_b^E\). In particular, \[\Psi_a^E \le C_\tau\bigl(\mathbb{P}(E)+\mathcal{S}_E\bigr).\] Plugging the bound on \(Y=\Psi_a^E+\Psi_b^E\) into 5 gives \[\left|\mathbb{E}\left[\mathbf{1}_E(\phi(W_{T_{\circ}})-\phi(V_{T_{\circ}}))\right]\right| \le C_\tau\mathcal{S}_E^{1/2}\bigl(\mathbb{P}(E)+\mathcal{S}_E\bigr)^{1/2} \le C_\tau\left(\sqrt{\mathcal{S}_E\mathbb{P}(E)}+\mathcal{S}_E\right).\] Taking the supremum over \(\phi\) proves the lemma. ◻
Corollary 1 (Layered discrepancy bound). Let \[G_r(\theta)=\{r\le R_\theta<r+1\}, \qquad r\ge \alpha/2.\] Then, for \(r\ge \alpha/2\), \[D_{G_r(\theta)} \le C_\tau \left( \frac{\alpha}{\sqrt r}+\frac{\alpha^2}{r} \right)\mathbb{P}(G_r(\theta)).\] More generally, the same estimate holds with \(G_r(\theta)\) replaced by any \(E\in\mathcal{F}_\theta\) such that \(E\subset G_r(\theta)\). Moreover, for \(G_{\ge L/2}(\theta)=\{R_\theta\ge L/2\}\), \[D_{G_{\ge L/2}(\theta)} \le C_\tau\left(\frac{\alpha}{\sqrt L}+\frac{\alpha^2}{L}\right).\]
Proof. Let \(E\in\mathcal{F}_\theta\) with \(E\subset G_r(\theta)\). The active part is \(E\cap\{R_\theta\ge\alpha\}\). On this active part, \[\bar\delta^2(R_\theta+\alpha+1) = \frac{\alpha^2(R_\theta+\alpha+1)}{(R_\theta+1)^2} \le C\frac{\alpha^2}{r}, \qquad r\ge\alpha/2,\] while outside it \(\bar\delta=0\). By 5, \[\mathcal{S}_E \le C_\tau\frac{\alpha^2}{r}\mathbb{P}(E).\] Plugging this into 6 gives the subset estimate, and taking \(E=G_r(\theta)\) gives the first displayed bound. On \(E=G_{\ge L/2}(\theta)\), since \(L\ge8\) gives \(\alpha\le L/2\), the perturbation is active and \[\bar\delta^2(R_\theta+\alpha+1) = \frac{\alpha^2(R_\theta+\alpha+1)}{(R_\theta+1)^2} \le \frac{C \alpha^2}{R_\theta+1} \le \frac{C \alpha^2}{L}.\] Together with 5 this gives \(\mathcal{S}_E\le C_\tau \alpha^2\mathbb{P}(E)/L\), and the second estimate follows from 6 and \(\mathbb{P}(E)\le1\). No layer-profile estimate is needed for this tail layer: it appears only once in the final decomposition, and the crude bound \(\mathbb{P}(E)\le1\) already gives \(O_\tau(\alpha/\sqrt L)\). ◻
Proof of 2. Fix \(L\ge8\), and put \(\alpha=\frac{1}{2}\log L+1\). For a fixed \(\theta\in[{T_{\circ}}-1,{T_{\circ}})\), first compare the two tails at level \(L\). Since \(\bar\delta=0\) on \(\{R_\theta<\alpha\}\), the perturbation term is identically zero after \(\theta\) on this event and the two synchronized processes coincide up to \({T_{\circ}}\). Hence, for \(\varphi=\mathbf{1}_{\{g_\tau>L\}}\), \[\mathbb{P}\{g_\tau(V_{T_{\circ}})>L\} \le \mathbb{P}\{g_\tau(W_{T_{\circ}})>L\}+D_{\{R_\theta\ge \alpha\}}.\] Since \[\mathcal{A}_\tau((L,L+1]) = \mathbb{P}\{g_\tau(V_{T_{\circ}})>L\} - \mathbb{P}\{g_\tau(V_{T_{\circ}})>L+1\},\] combining the last display with 4 gives the bootstrap inequality in one line: \[\begin{align} \mathcal{A}_\tau((L,L+1]) &\le \mathbb{P}\{g_\tau(W_{T_{\circ}})>L\} +D_{\{R_\theta\ge \alpha\}} - \mathbb{P}\{g_\tau(V_{T_{\circ}})>L+1\} \\ &\le D_{\{R_\theta\ge \alpha\}} +\mathcal{A}_{T-\theta}((L-\alpha,L+\alpha]) +\frac{3}{\sqrt L}. \end{align}\] Decompose \[\{R_\theta\ge \alpha\} = \left(\bigcup_{r=\lfloor \alpha\rfloor}^{\lfloor L/2\rfloor} E_r(\theta)\right) \cup G_{\ge L/2}(\theta), \qquad E_r(\theta)=G_r(\theta)\cap\{\alpha\le R_\theta<L/2\}.\] This union is disjoint. The intersection in the definition of \(E_r(\theta)\) only trims the boundary pieces near \(R_\theta=\alpha\) and \(R_\theta=L/2\); the subset version of 1 is included precisely for this use. For any disjoint family \((F_j)\subset\mathcal{F}_\theta\), the definition of \(D_E\) and the triangle inequality give \[D_{\bigcup_jF_j} \le \sum_j D_{F_j}.\] Indeed, this holds for each fixed test function \(\phi\), and then one takes the supremum over \(\phi\). By 1, \[D_{\{R_\theta\ge \alpha\}} \le C_\tau \sum_{r=\lfloor \alpha\rfloor}^{\lfloor L/2\rfloor} \left(\frac{\alpha}{\sqrt r}+\frac{\alpha^2}{r}\right)\mathbb{P}(E_r(\theta)) + C_\tau\left(\frac{\alpha}{\sqrt L}+\frac{\alpha^2}{L}\right).\] Averaging \(\theta\) over \([{T_{\circ}}-1,{T_{\circ}}]\), and writing \(s=T-\theta\), we have \(s\in[\tau,\tau+1]\). For \(\lfloor \alpha\rfloor\le r\le L/2\), the identity below follows from \(R_\theta=[L-g_{T-\theta}(V_\theta)]_+\): on \(G_r(\theta)\), one has \(L-r-1<g_{T-\theta}(V_\theta)\le L-r\). Hence \[\int_{{T_{\circ}}-1}^{{T_{\circ}}}\mathbb{P}(E_r(\theta))\,d\theta \le \int_{{T_{\circ}}-1}^{{T_{\circ}}}\mathbb{P}(G_r(\theta))\,d\theta = \int_\tau^{\tau+1} \mathcal{A}_s((L-r-1,L-r])\,ds \le \frac{C}{L-r-1} \le \frac{C}{L}\] by 3, since \(L-r-1\ge L/2-1\ge L/4\) for \(L\ge8\). Therefore \[\begin{align} \int_{{T_{\circ}}-1}^{{T_{\circ}}}D_{\{R_\theta\ge \alpha\}}\,d\theta &\le C_\tau \frac{\alpha}{L}\sum_{r=\lfloor \alpha\rfloor}^{\lfloor L/2\rfloor}r^{-1/2} + C_\tau \frac{\alpha^2}{L}\sum_{r=\lfloor \alpha\rfloor}^{\lfloor L/2\rfloor}r^{-1} + C_\tau\frac{\alpha}{\sqrt L} \\ &\le C_\tau\frac{\alpha}{\sqrt L}. \end{align}\] Here the first sum is \(O(\sqrt L)\), while the second is \(O(\log L)\); since \(\alpha=\frac{1}{2}\log L+1\), the term \[\frac{\alpha^2}{L}\sum_{r=\lfloor \alpha\rfloor}^{\lfloor L/2\rfloor}r^{-1} \le C\frac{\alpha^2\log L}{L} = C\frac{\alpha}{\sqrt L}\cdot\frac{\alpha\log L}{\sqrt L} \le C\frac{\alpha}{\sqrt L}.\] The last inequality holds because, for \(L\) larger than an absolute constant, \(\alpha\log L\le C\sqrt L\), while the remaining compact range \(8\le L\le L_0\) is absorbed into the constant. The initial-band term is bounded by covering \((L-\alpha,L+\alpha]\) with the half-open unit intervals \((m,m+1]\) which intersect it; there are at most \(2\alpha+2\) of them. Since \(L\ge8\), every such interval has \(m\ge L-\alpha-1>2\), so these unit intervals lie in the range of 3. Thus \[\int_{{T_{\circ}}-1}^{{T_{\circ}}} \mathcal{A}_{T-\theta}((L-\alpha,L+\alpha])\,d\theta \le \frac{C \alpha}{L-\alpha} \le C\frac{\alpha}{L} \le C\frac{\alpha}{\sqrt L}.\] Also \(3/\sqrt L\le3\alpha/\sqrt L\), since \(\alpha\ge1\). Thus the averaged bootstrap inequality gives \[\mathcal{A}_\tau((L,L+1]) \le C_\tau\frac{\alpha}{\sqrt L} \le C_\tau\frac{\log L}{\sqrt L}.\] ◻
Remark 6 (Where the exponent \(3/2\) disappears). The global discrepancy estimate controls the perturbation by \[\alpha\sqrt{\mathbb{E}\frac{\mathbf{1}_{\{R_\theta\ge \alpha\}}}{R_\theta+1}}.\] After time averaging, the expectation is bounded by \((\log L)/L\), which gives \[\alpha\left(\frac{\log L}{L}\right)^{1/2} \sim \frac{(\log L)^{3/2}}{\sqrt L}.\] The localized lemma replaces this by the layer sum \[\sum_r\frac{\alpha}{\sqrt r}\mathbb{P}(G_r),\] and the same time-smoothed profile estimate then gives \[\frac{\alpha}{L}\sum_{r\le L/2}r^{-1/2}\lesssim \frac{\alpha}{\sqrt L}.\]
Remark 7 (Scope of the method, in black-box form). The previous calculation also explains why this particular route should not be expected to remove the remaining \(\log\log\eta\) factor. This is only a limitation of the present black-box estimates, not a lower bound for Talagrand’s conjecture.
Put \(L=\log\eta\) and \(\alpha\simeq\log L\). After averaging the start time \(\theta\in[{T_{\circ}}-1,{T_{\circ}}]\), the proof uses only the layer estimate \[\int_{{T_{\circ}}-1}^{{T_{\circ}}}D_{G_r(\theta)}\,d\theta \lesssim \left(\frac{\alpha}{\sqrt r}+\frac{\alpha^2}{r}\right)\overline{p}_r, \qquad \overline{p}_r:= \int_{{T_{\circ}}-1}^{{T_{\circ}}}\mathbb{P}(G_r(\theta))\,d\theta,\] and the time-smoothed profile consequence \[\overline{p}_r \lesssim \frac{1}{L-r-1}, \qquad \alpha\le r\le L/2.\] On the middle range \(L/4\le r\le L/2\), this gives only \(\overline{p}_r\lesssim L^{-1}\). The abstract layer profile \[\overline{p}_r=\frac{c}{L},\qquad L/4\le r\le L/2, \qquad \overline{p}_r=0\quad\text{otherwise},\] with \(c>0\) small, is therefore fully compatible with the profile information used in the argument. If, in addition, the localized Cauchy–Schwarz estimates are saturated on these layers, the contribution allowed by the method is \[\sum_{L/4\le r\le L/2} \frac{\alpha}{\sqrt r}\frac{c}{L} \asymp \frac{\alpha}{L}\sum_{L/4\le r\le L/2}r^{-1/2} \asymp \frac{\alpha}{\sqrt L}.\] Since \(\alpha\simeq\log L\), this is exactly the fixed-time \((\log L)/\sqrt L\) scale, equivalently the weak-type factor \[\frac{\log\log\eta}{\eta\sqrt{\log\eta}}.\] Thus any argument using only the localized discrepancy bound together with the time-smoothed profile estimate cannot, by itself, prove the conjectural \((\eta\sqrt{\log\eta})^{-1}\) bound. Removing the final log–log factor would require additional structure, such as a stronger occupation profile, layer interactions or cancellation, or a perturbative monotonicity mechanism whose cost is not linear in \(\alpha\simeq\log L\).
Remark 8 (Status of the imported inputs). The stochastic coupling, approximate monotonicity estimate, and conditional score-energy bound are imported from Chen’s reverse-heat framework. The localization is compatible with that calculation because \(E\in\mathcal{F}_\theta\) acts as a frozen multiplier in the Duhamel identity, the conditional bridge decomposition, and the Doob-transform energy estimate. The only part of Chen’s terminal testing-discrepancy calculation modified here is the insertion of the factor \(\mathbf{1}_E\) before the final Cauchy–Schwarz step.