Proof of Mean Deviation Bound for the Aggregate Fluctuation Covariance along a Mean-Field Trajectory Pair

Throughout, adopt the notation of the statement, and write $x_s=(S_s,A_s)$ and $y_s=(\Sigma_s,\alpha_s)$ as points of $\mathbb{R}^{l+m}$ under the coordinate identification of the extension definition, so that $y_s-x_s=z_s/\sqrt{N}$ pointwise. Products with an infinite factor are read with the convention $0\cdot\infty=0$. If $c_\Theta=0$ then $B=K=0$, the rate bound forces $\beta\equiv0$, hence $\Theta\equiv0$ by its entry formulas, every left-hand side of the statement vanishes, and all three clauses hold; assume $c_\Theta>0$ below.

Step 1: pointwise bound (clause (a)). Fix $\omega\in\Omega$, $s\in[0,T]$, and $\gamma,\delta\in\{1,\dots,l\}$. By the definition of the controlled $N$-agent dynamics, the state processes take values in $\{1,\dots,l\}$ at every point, so the empirical state measure $\Sigma_s(\omega)$ lies in the probability simplex $\Delta^l$; and $S_s\in\Delta^l$, the mean-field trajectory pair having $S:[0,T]\to\Delta^l$. Hence $x_s$ and $y_s$ lie in $\Delta^l\times\mathbb{R}^m$, and so does every point of the segment $\{x_s+\tau(y_s-x_s):\tau\in[0,1]\}$: the convex combination $(1-\tau)S_s+\tau\Sigma_s(\omega)$ of two points of the simplex has nonnegative entries with sum $(1-\tau)+\tau=1$, and the last $m$ coordinates are unconstrained. Fix an ordered pair $(\sigma,\gamma')$ with $\sigma\neq\gamma'$. By clause 1 of the extension definition, $\bar{\beta}(\sigma,\gamma',\cdot,\cdot)$ agrees with $\beta(\sigma,\gamma',\cdot,\cdot)$ on $\Delta^l\times\mathbb{R}^m$; by clause 2 it is a $C^1$ map on the open set $U\times\mathbb{R}^m\supseteq\Delta^l\times\mathbb{R}^m$; and by clause 3, $|\partial_i\bar{\beta}(\sigma,\gamma',x)|\le K$ for all $i$ and all $x\in U\times\mathbb{R}^m$, in particular on the segment. Part (i) of the Taylor lemma, with $n=l+m$ and $M_1=K$, therefore gives

Moreover $|\Sigma^\sigma_s-S^\sigma_s|\le|y_s-x_s|$, a single coordinate of the difference being at most its Euclidean norm, and, by the transition-rate family definition, $0\le\beta(\sigma,\gamma',\cdot,\cdot)\le B$, while $0\le S^\sigma_s\le1$ on the simplex. Hence, for each ordered pair $(\sigma,\gamma')$ with $\sigma\neq\gamma'$,

the first step by the triangle inequality (the Euclidean distance is a metric, applied in $\mathbb{R}$) after adding and subtracting $S^\sigma_s\,\beta(\sigma,\gamma',y_s)$. By the entry formulas of the aggregate fluctuation covariance, the entry $\Theta^{\gamma\gamma}$ is a sum of $2(l-1)$ terms of this form (the pairs $(\sigma,\gamma)$ and $(\gamma,\sigma)$ for $\sigma\neq\gamma$), and an off-diagonal entry $\Theta^{\gamma\delta}$ ($\gamma\neq\delta$) is, up to sign, a sum of $2$ such terms; since $l\ge2$, in either case the triangle inequality gives

which is clause (a).

Step 2: mean bound (clause (b)). Fix $s\in[0,T]$ and $\gamma,\delta$. The map $\Theta^{\gamma\delta}(\Sigma_s,\alpha_s)$ is a random variable: the entry $\Theta^{\gamma\delta}$ is a finite sum of products of coordinate maps with members of the rate family, hence sequentially continuous on $\Delta^l\times\mathbb{R}^m$ by the joint-continuity clause of the transition-rate family definition, and measurability of sequentially continuous functions of measurable Euclidean maps applies to the components of $(\Sigma_s,\alpha_s)$, which are random variables by the controlled-dynamics definition (as in the proof of the martingale decomposition). By clause (a) of the weighted second-moment evolution lemma, applied with the constant matrix family $Z_t=0$ (whose densities $\dot{z}^{\gamma\delta}\equiv0$ are continuous), $\mathbb{E}[\Theta^{\gamma\delta}(\Sigma_s,\alpha_s)]$ is finite, and it is bounded and measurable as a function of $s$. Subtracting the constant $\Theta^{\gamma\delta}(S_s,A_s)$, whose expectation is itself, and using linearity of the expectation and the bound $|\mathbb{E}[X]|\le\mathbb{E}[|X|]$ from the integrable-case clause there,

the last step by clause (a) and monotonicity, $|z_s|$ being a random variable — a continuous function of the components of $\mathfrak{s}_s$ and $\mathfrak{a}_s$, which are random variables by the controlled-dynamics definition and the trajectory pair. If $\mathbb{E}[|\mathfrak{s}_s|^2]+\mathbb{E}[|\mathfrak{a}_s|^2]=\infty$, the right-hand side of clause (b) is $\infty$ ($c_\Theta>0$) and the clause is trivial. Otherwise $|z_s|$ is square-integrable, since $\mathbb{E}[|z_s|^2]=\mathbb{E}[|\mathfrak{s}_s|^2]+\mathbb{E}[|\mathfrak{a}_s|^2]$ by the pointwise identity $|z_s|^2=|\mathfrak{s}_s|^2+|\mathfrak{a}_s|^2$ and additivity; the constant $1$ is square-integrable with $\mathbb{E}[1^2]=1$, so the mean-square Cauchy--Schwarz inequality applied to the pair $(|z_s|,1)$ gives $\mathbb{E}[|z_s|]=\mathbb{E}[|z_s|\cdot1]\le\big(\mathbb{E}[|z_s|^2]\big)^{1/2}\cdot1$, the left side being nonnegative so the absolute value there is immaterial. This completes clause (b).

Step 3: integrated bound (clause (c)). The integrand $s\mapsto\big|\mathbb{E}[\Theta^{\gamma\delta}(\Sigma_s,\alpha_s)]-\Theta^{\gamma\delta}(S_s,A_s)\big|$ is measurable, by measurability of sequentially continuous functions of measurable maps applied to the continuous function $(u,v)\mapsto|u-v|$ and the two measurable functions of $s$: the bounded measurable function of Step 2, and $s\mapsto\Theta^{\gamma\delta}(S_s,A_s)$, which is continuous (as recorded in clause (c) of the statement) and hence measurable and bounded. Being measurable and bounded on $[0,T]$, the integrand has a defined, finite Lebesgue integral, by monotonicity against a constant and claim 1 of the toolkit ($\lambda([0,T])=T$). If $\mathcal{A}=\infty$ the right-hand side of clause (c) is $\infty$ ($c_\Theta>0$) and the claim is trivial; assume $\mathcal{A}<\infty$, so that $\mathcal{S}+\mathcal{A}<\infty$ by $\mathcal{S}\le4NT$. Set $\phi(s)=\big(\mathbb{E}[|\mathfrak{s}_s|^2]+\mathbb{E}[|\mathfrak{a}_s|^2]\big)^{1/2}\in[0,\infty]$; the map $s\mapsto\mathbb{E}[|\mathfrak{s}_s|^2]+\mathbb{E}[|\mathfrak{a}_s|^2]$ is a measurable $[0,\infty]$-valued function by clause (a) of the a priori second-moment bound and additivity, and $\phi$ is measurable: for real $a\ge0$, $\{\phi>a\}=\{\mathbb{E}[|\mathfrak{s}_\cdot|^2]+\mathbb{E}[|\mathfrak{a}_\cdot|^2]>a^2\}$ is measurable, for $a<0$ the set $\{\phi>a\}$ is all of $[0,T]$, and the sets $(a,\infty]$ over real $a$ generate the $\sigma$-algebra of $[0,\infty]$ used for $[0,\infty]$-valued measurability. Moreover $\int_{[0,T]}\phi^2\,ds=\mathcal{S}+\mathcal{A}<\infty$ by additivity for nonnegative measurable integrands. Let $D=\{s:\phi(s)<\infty\}$ and $\phi_0=\phi\,\mathbf{1}_{D}$, where $\mathbf{1}_D$ is the function equal to $1$ on $D$ and $0$ off $D$; then $\phi_0$ is measurable and real-valued, and $D$ is co-null: for every natural number $n$, $\mathcal{S}+\mathcal{A}\ge n^2\,\lambda([0,T]\setminus D)$ by monotonicity and the integral of a simple function ($\lambda$ the trace Lebesgue measure), forcing $\lambda([0,T]\setminus D)=0$. By claim 6 of the interval toolkit applied to the nonnegative measurable functions $\phi$ and $\phi^2$ with the co-null set $D$,

since $\phi_0=\phi\mathbf{1}_D$ and $\phi_0^2=\phi^2\mathbf{1}_D$ pointwise under the convention $0\cdot\infty=0$. By clause (b), the integrand of clause (c) is at most $\tfrac{c_\Theta}{\sqrt{N}}\,\phi(s)$ at every $s$, so by monotonicity and linearity (the constant $\tfrac{c_\Theta}{\sqrt{N}}$ passing out of the integral), and then claim 4 (Cauchy--Schwarz) of the interval toolkit applied to the pair $(1,\phi_0)$ on $[0,T]$ (both measurable with finite integrals of their squares, $T>0$), whose squared form $\big(\int_{[0,T]}\phi_0\,ds\big)^2\le T\int_{[0,T]}\phi_0^2\,ds$ passes to square roots because the nonnegative square root is nondecreasing,

where the first inequality also uses that $\phi$ and $\phi_0$ have equal integrals. This is clause (c).\ $\square$