Proof of Sharp Drift Linearization Error under a Control-Affine Extension

Throughout, fix $s\in[0,T]$. For conclusion 2 also fix $\omega\in\Omega_0$ and abbreviate $\Sigma=\Sigma_s(\omega)$, $a=\alpha_s(\omega)$, $S=S_s$, $A=A_s$; then $\Sigma,S\in\Delta^l$ (the empirical state measure and the mean-field trajectory take values in the probability simplex) and $a,A\in\mathcal{A}$ by hypothesis (A2). For conclusion 1, instead let $a\in\mathcal{A}$ be arbitrary and keep $S=S_s$, $A=A_s$.

Step 1 (Slice maps and their partial derivatives). By clause (i) of the drift regularity lemma, each $\bar{b}^\delta$ is a $C^1$ map on $U\times\mathbb{R}^m$ and each $\partial_i\bar{b}^\delta$ is again a $C^1$ map there. Fix $\delta\in\{1,\dots,l\}$ and define the slice maps

For every $v\in\mathbb{R}^m$ and $j\in\{1,\dots,m\}$, the difference quotients defining the partial derivative $\partial_j\varphi(v)$ and those defining $\partial_{l+j}\bar{b}^\delta(S,v)$ coincide, because moving the $j$-th coordinate of $v$ is the same as moving the $(l+j)$-th coordinate of $(S,v)$; hence $\partial_j\varphi(v)$ exists and equals $\partial_{l+j}\bar{b}^\delta(S,v)$. Its continuity in $v$ follows from the continuity of $\partial_{l+j}\bar{b}^\delta$: the first block of coordinates being fixed, the Euclidean distance satisfies $d((S,v),(S,v'))=d(v,v')$, so any $\varepsilon$--$\rho$ modulus of continuity for $\partial_{l+j}\bar{b}^\delta$ at $(S,v)$ is one for the slice at $v$ (we write $\rho$ for the radius, the letter $\delta$ being in use as a component index). Thus $\varphi$ is a $C^1$ map on the open set $\mathbb{R}^m$. The same argument applied to the $C^1$ maps $\partial_{l+j}\bar{b}^\delta$ and $\partial_\gamma\bar{b}^\delta$ shows: each $\partial_j\varphi$ is again $C^1$ on $\mathbb{R}^m$ with $\partial_k\partial_j\varphi(v)=\partial_{l+k}\partial_{l+j}\bar{b}^\delta(S,v)$; each $\psi_\gamma$ is $C^1$ on $\mathbb{R}^m$ with $\partial_j\psi_\gamma(v)=\partial_{l+j}\partial_\gamma\bar{b}^\delta(S,v)$; and $\chi$ is $C^1$ on the open set $U$ with $\partial_\gamma\chi(\Sigma')=\partial_\gamma\bar{b}^\delta(\Sigma',a)$, each $\partial_\gamma\chi$ again $C^1$ with $\partial_{\gamma'}\partial_\gamma\chi(\Sigma')=\partial_{\gamma'}\partial_\gamma\bar{b}^\delta(\Sigma',a)$.

Step 2 (Vanishing control curvature). Let $v\in\mathcal{A}$ and $j,k\in\{1,\dots,m\}$. In the second-derivative formula of clause (i) of the drift regularity lemma, evaluated at $x=(S,v)$ with derivative indices $l+j$ and $l+k$, all four Kronecker factors of the formula vanish: with the formula's two derivative indices set to $l+j$ and $l+k$, each factor pairs one of these against a state index --- the summation index $\sigma$ or the drift's component index (written $\gamma$ in the formula, $\delta$ here) --- and the state indices lie in $\{1,\dots,l\}$ while $l+j,l+k>l$. Hence

by hypothesis (A3), since $(S,v)\in\Delta^l\times\mathcal{A}$.

Step 3 (Conclusion 1: exact control linearity). Apply part (iii) of the multivariate Taylor lemma to $f=\varphi$ on the open set $\mathbb{R}^m$, with base point $A$, endpoint $a$, and $h=a-A$. The segment $\{A+\tau(a-A):\tau\in[0,1]\}$ lies in $\mathcal{A}$, by hypothesis (A1) and the definition of a convex subset. By Step 1, $\varphi$ is $C^1$ with each $\partial_j\varphi$ again $C^1$; by Step 2, $\partial_k\partial_j\varphi$ vanishes at every point of the segment and at $A$, so the Taylor lemma applies with $\bar{\varepsilon}=0$ and its second-order term vanishes. Hence

that is, $\bar{b}^\delta(S,a)-\bar{b}^\delta(S,A)=\sum_{j=1}^{m}\partial_{l+j}\bar{b}^\delta(S,A)\,(a-A)^j=\big(\mathsf{B}_s(a-A)\big)^\delta$, with the matrix-vector product. Since $\bar{b}$ agrees with $b$ on $\Delta^l\times\mathbb{R}^m$ (clause (i) of the drift regularity lemma), this is conclusion 1, the index $\delta$ being arbitrary.

Step 4 (State move). Apply part (ii) of the Taylor lemma to $f=\chi$ on $U$, with base point $S$, endpoint $\Sigma$, and $h=\Delta\Sigma:=\Sigma-S$. For $\tau\in[0,1]$ the point $S+\tau\Delta\Sigma$ has components $(1-\tau)S^\gamma+\tau\Sigma^\gamma\ge0$ whose sum is $(1-\tau)+\tau=1$, so the segment lies in the probability simplex, and $\Delta^l\subseteq U$ by the extension definition. By Step 1 and clause (iii) of the drift regularity lemma, $|\partial_{\gamma'}\partial_\gamma\chi(z)|=|\partial_{\gamma'}\partial_\gamma\bar{b}^\delta(z,a)|\le3\,l\,K$ at every point $z$ of the segment, the argument $(z,a)$ lying in $\Delta^l\times\mathbb{R}^m$. Hence, with $n=l$ and $M_2=3lK$,

Step 5 (Derivative transport). For each $\gamma$, apply part (i) of the Taylor lemma to $f=\psi_\gamma$ on $\mathbb{R}^m$, with base point $A$ and endpoint $a$; the segment lies in $\mathcal{A}$ as in Step 3. By Step 1 and clause (iii) of the drift regularity lemma, $|\partial_j\psi_\gamma(z)|=|\partial_{l+j}\partial_\gamma\bar{b}^\delta(S,z)|\le3\,l\,K$ at every point $z$ of the segment. Hence, with $n=m$ and $M_1=3lK$, and writing $\Delta\alpha:=a-A$,

Step 6 (Assembly and conclusion 2). At the fixed $(\omega,s)\in\Omega_0\times[0,T]$, using the agreement of $b$ and $\bar{b}$ on $\Delta^l\times\mathbb{R}^m$ once more,

Dividing the defining identities $\mathfrak{s}_s=\sqrt{N}(\Sigma_s-S_s)$ and $\mathfrak{a}_s=\sqrt{N}(\alpha_s-A_s)$ of the fluctuation processes by $\sqrt{N}$ gives $\Delta\Sigma=N^{-1/2}\mathfrak{s}_s(\omega)$ and $\Delta\alpha=N^{-1/2}\mathfrak{a}_s(\omega)$ componentwise. By Step 3 the second bracket equals $\sum_j\mathsf{B}^{\delta j}_s\Delta\alpha^j=N^{-1/2}(\mathsf{B}_s\mathfrak{a}_s)^\delta$. Hence, from the definition of $e_s$ and $E^{\delta\gamma}_s=\partial_\gamma\bar{b}^\delta(S,A)$,

By Steps 4 and 5 together with $\sum_\gamma|\Delta\Sigma^\gamma|\le l\max_\gamma|\Delta\Sigma^\gamma|\le l\,|\Delta\Sigma|$ (claim 1 of the componentwise toolkit) and the triangle inequality for the absolute value (claim 5 of its properties lemma, applied finitely many times across the sums),

using $\tfrac{3}{2}\le3\sqrt{m}$ for $m\ge1$. By absolute homogeneity of the norm (claim 5 of the norm properties lemma), $|\Delta\Sigma|=N^{-1/2}|\mathfrak{s}_s|$ and $|\Delta\alpha|=N^{-1/2}|\mathfrak{a}_s|$, so $\sqrt{N}\,|\Delta\Sigma|^2=N^{-1/2}|\mathfrak{s}_s|^2$ and $\sqrt{N}\,|\Delta\Sigma||\Delta\alpha|=N^{-1/2}|\mathfrak{s}_s||\mathfrak{a}_s|$, giving

Finally, by claim 1 of the componentwise toolkit, $|e_s|\le\sum_{\delta=1}^{l}|e^\delta_s|\le l\cdot3l^2\sqrt{m}KN^{-1/2}\big(|\mathfrak{s}_s|^2+|\mathfrak{s}_s||\mathfrak{a}_s|\big)=c_a\,N^{-1/2}\big(|\mathfrak{s}_s|^2+|\mathfrak{s}_s||\mathfrak{a}_s|\big)$, which is conclusion 2. $\blacksquare$