Proof of A Priori Second-Moment Bound for the State Fluctuation Process

Write $\mathbf{1}=\mathbf{1}_{\Omega_0}$, and let $b$, $g_s$, and $\Lambda$ be as in the statement. Since $\Omega_0$ has probability $1$, expectations are unchanged when their integrands are modified off $\Omega_0$, and we use this silently below. All applications of the Tonelli theorem below are on the product of $([0,T]$ with the trace Borel $\sigma$-algebra and restricted Lebesgue measure$)$ and $(\Omega,\mathcal{F},P)$; both factors are finite measure spaces (the first of total mass $T$ by the toolkit, the second a probability space), hence $\sigma$-finite.

Part (a). By the joint measurability lemma, each $\mathbf{1}\Sigma^\gamma_t$ and each $\mathbf{1}\alpha^j_t$ is product-measurable. Each map $(t,\omega)\mapsto S^\gamma_t$ or $A^j_t$ is product-measurable, being the composition of the measurable map $(t,\omega)\mapsto t$ (rectangle preimages) with a continuous, hence sequentially continuous, component of the trajectory pair, via measurability of sequentially continuous functions of measurable maps. Differences, scalar multiples, squares, and finite sums of product-measurable real maps are again product-measurable by the same composition lemma applied to the continuous arithmetic operations; hence $\mathbf{1}|\mathfrak{s}_t|^2$ and $\mathbf{1}|\mathfrak{a}_t|^2$ are product-measurable. For $\mathbf{1}(g^\gamma_t)^2$: replace $(\Sigma_t,\alpha_t)$ off $\Omega_0$ by a fixed point $(e_1,a_0)\in\Delta^l\times\mathcal{A}$ with $e_1=(1,0,\dots,0)$ and $a_0\in\mathcal{A}$, which is nonempty; the resulting componentwise-measurable map composed with the sequentially continuous $b^\gamma$ is product-measurable by the composition lemma (sequential continuity of $b^\gamma$ on $\Delta^l\times\mathcal{A}$ follows from the joint continuity clause of the transition-rate family together with continuity of the coordinate factors in its defining formula), while $b^\gamma(S_t,A_t)$ is continuous in $t$; hence $\mathbf{1}g^\gamma_t$ and $\mathbf{1}(g^\gamma_t)^2$ are product-measurable. By the Tonelli theorem, $t\mapsto\mathbb{E}[|\mathfrak{s}_t|^2]$, $t\mapsto\mathbb{E}[|\mathfrak{a}_t|^2]$, and $t\mapsto\mathbb{E}[(g^\gamma_t)^2]$ are measurable $[0,\infty]$-valued functions on $[0,T]$, so $\mathcal{A}_2$ is a well-defined Lebesgue integral in $[0,\infty]$. Every $\Sigma_t$ lies in the probability simplex (each agent occupies exactly one state, by the derived notation of the solution definition), and any $\Sigma\in\Delta^l$ has $|\Sigma|^2=\sum_\gamma(\Sigma^\gamma)^2\le\sum_\gamma\Sigma^\gamma=1$; likewise $|S_t|\le1$. Hence $|\mathfrak{s}_t|\le2\sqrt{N}$ and $\mathbb{E}[|\mathfrak{s}_t|^2]\le4N$; and $|b^\gamma|\le2(l-1)B$ (part (a) of the martingale decomposition theorem) gives $|g^\gamma_t|\le4\sqrt{N}(l-1)B$ and $\mathbb{E}[(g^\gamma_t)^2]\le16N(l-1)^2B^2$.

Part (b). Step 1 (integral representation of $\mathfrak{s}$). By part (b) of the martingale decomposition theorem, $M^\gamma_t=\Sigma^\gamma_t-\Sigma^\gamma_0-\int_{[0,t]}\mathbf{1}_{\Omega_0}\,b^\gamma(\Sigma_s,\alpha_s)\,ds$ defines a square-integrable martingale with $M^\gamma_0=0$, the integrals existing at every point of $\Omega$ by part (a) of that theorem; on $\Omega_0$ the indicator factor equals $1$. By condition 2 of the mean-field trajectory pair and the agreement of the Riemann and Lebesgue integrals for continuous integrands, $S^\gamma_t=S^\gamma_0+\int_{[0,t]}b^\gamma(S_s,A_s)\,ds$. Hence, almost surely, for all $t\in[0,T]$ and all $\gamma$,

$$\mathfrak{s}^\gamma_t=\mathfrak{s}^\gamma_0+\int_{[0,t]}g^\gamma_s\,ds+\sqrt{N}\,M^\gamma_t,$$

the integrals existing since $|g^\gamma_s|\le4\sqrt{N}(l-1)B$ pathwise.

Step 2 (three-term estimate). For real numbers, $(a_1+a_2+a_3)^2\le3(a_1^2+a_2^2+a_3^2)$, since $2a_pa_q\le a_p^2+a_q^2$. Applying this componentwise, summing over $\gamma$, and taking expectations,

$$\mathbb{E}\big[|\mathfrak{s}_t|^2\big]\le3\,\mathbb{E}\big[|\mathfrak{s}_0|^2\big]+3\sum_{\gamma=1}^l\mathbb{E}\Big[\Big(\int_{[0,t]}g^\gamma_s\,ds\Big)^2\Big]+3N\sum_{\gamma=1}^l\mathbb{E}\big[(M^\gamma_t)^2\big].$$

For $t>0$, the Cauchy-Schwarz inequality of the interval toolkit applied pathwise to the pair $(1,g^\gamma)$ on $[0,t]$ gives $(\int_{[0,t]}g^\gamma_s ds)^2\le t\int_{[0,t]}(g^\gamma_s)^2ds\le T\int_{[0,t]}(g^\gamma_s)^2ds$; for $t=0$ both sides vanish. Taking expectations and applying the Tonelli theorem to the nonnegative product-measurable (after $\mathbf{1}$-modification) integrand,

$$\sum_{\gamma}\mathbb{E}\Big[\Big(\int_{[0,t]}g^\gamma_s ds\Big)^2\Big]\le T\int_{[0,t]}\sum_{\gamma}\mathbb{E}\big[(g^\gamma_s)^2\big]\,ds .$$

By the covariation identity, part (c) of the martingale decomposition theorem, with $r=0$, $D=\Omega$, $\delta=\gamma$, and $M^\gamma_0=0$: $N\,\mathbb{E}[(M^\gamma_t)^2]=\mathbb{E}[\int_{[0,t]}\mathbf{1}_{\Omega_0}\,\Theta^{\gamma\gamma}(\Sigma_s,\alpha_s)ds]\le2(l-1)B\,t\le2(l-1)B\,T$, using the bound $|\Theta^{\gamma\gamma}|\le2(l-1)B$ from part (a) of that theorem; summing over $\gamma$ gives at most $2l(l-1)BT$. Combining the three estimates proves part (b).

Part (c). If $\mathcal{A}_2=\infty$ there is nothing to prove, so assume $\mathcal{A}_2<\infty$. By part (i) of the drift regularity lemma, $b$ agrees with the extended aggregate state drift $\bar{b}$ on $\Delta^l\times\mathcal{A}$, and by its part (ii) — applicable since $\mathcal{A}$ is convex by hypothesis and, at every point of $\Omega_0\times[0,T]$, both $(\Sigma_s,\alpha_s)$ and $(S_s,A_s)$ lie in $\Delta^l\times\mathcal{A}$ — $|b^\gamma(\Sigma_s,\alpha_s)-b^\gamma(S_s,A_s)|\le\sqrt{l+m}\,l(B+K)\,d\big((\Sigma_s,\alpha_s),(S_s,A_s)\big)$ at every such point, while $N\,d\big((\Sigma_s,\alpha_s),(S_s,A_s)\big)^2=|\mathfrak{s}_s|^2+|\mathfrak{a}_s|^2$. Hence, at every point of $\Omega_0\times[0,T]$,

$$\sum_{\gamma}(g^\gamma_s)^2\le l\,(l+m)\,l^2(B+K)^2\,\big(|\mathfrak{s}_s|^2+|\mathfrak{a}_s|^2\big)=\Lambda^2\big(|\mathfrak{s}_s|^2+|\mathfrak{a}_s|^2\big).$$

Let $u(t)=\mathbb{E}[|\mathfrak{s}_t|^2]$, measurable by part (a) and bounded by $4N$, and set $a=3\,\mathbb{E}[|\mathfrak{s}_0|^2]+6\,l(l-1)BT+3\,T\Lambda^2\mathcal{A}_2$. Part (b) together with the display above and $\int_{[0,t]}\mathbb{E}[|\mathfrak{a}_s|^2]ds\le\mathcal{A}_2$ yields $u(t)\le a+3T\Lambda^2\int_{[0,t]}u(s)\,ds$ for every $t\in[0,T]$. Define $w(t)=a+3T\Lambda^2\int_{[0,t]}u(s)\,ds$; then $u\le w$ on $[0,T]$, and $w$ is continuous by the absolute continuity of the Lebesgue integral applied to the bounded integrand $u$. By the monotonicity of the Lebesgue integral, $w(t)\le a+3T\Lambda^2\int_{[0,t]}w(s)\,ds$, and for the continuous function $w$ the Lebesgue integral agrees with the Riemann integral. Gronwall's lemma applied to $w$ with constants $a$ and $3T\Lambda^2$ gives $w(t)\le a\exp(3T\Lambda^2t)$, and therefore $u(t)\le a\exp(3T\Lambda^2t)$ for every $t\in[0,T]$, which is the claimed bound.