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

Write $\mathbf{1}=\mathbf{1}_{\Omega_0}$, and let $b$, $g_s$, $\Lambda$, $c_M$, and $\mathcal{A}_4$ 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. The proof transposes the published proof of the a priori second-moment bound from second to fourth moments; all applications of the Tonelli theorem are on the product of $[0,T]$ (trace Borel $\sigma$-algebra, restricted Lebesgue measure) and $(\Omega,\mathcal{F},P)$, both finite measure spaces by the interval toolkit, hence $\sigma$-finite.

Part (a). By the joint measurability of the state and control together with measurability of sequentially continuous functions of measurable maps, as recorded in part (a) of the second-moment lemma, the maps $\mathbf{1}|\mathfrak{s}_t|^{2}$, $\mathbf{1}|\mathfrak{a}_t|^{2}$, and $\mathbf{1}(g^\gamma_t)^{2}$ (each $\gamma$) are measurable for the product $\sigma$-algebra. 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 $\mathbf{1}|\mathfrak{s}_t|^{4}=(\mathbf{1}|\mathfrak{s}_t|^{2})^{2}$, $\mathbf{1}|\mathfrak{a}_t|^{4}=(\mathbf{1}|\mathfrak{a}_t|^{2})^{2}$, and $\mathbf{1}|g_t|^{4}=(\sum_\gamma\mathbf{1}(g^\gamma_t)^{2})^{2}$ are product-measurable. By the Tonelli theorem, $t\mapsto\mathbb{E}[|\mathfrak{s}_t|^{4}]$, $t\mapsto\mathbb{E}[|\mathfrak{a}_t|^{4}]$, and $t\mapsto\mathbb{E}[|g_t|^{4}]$ are measurable $[0,\infty]$-valued functions on $[0,T]$, and $\mathcal{A}_4$ is a well-defined Lebesgue integral with value in $[0,\infty]$. For the bounds: every $\Sigma_t$ and every $S_t$ lies in the probability simplex (each agent occupies exactly one state, by the derived notation of the solution definition, and condition 1 of the trajectory pair), and any $\Sigma\in\Delta^{l}$ has $|\Sigma|^{2}=\sum_\gamma(\Sigma^\gamma)^{2}\le\sum_\gamma\Sigma^\gamma=1$; hence $|\mathfrak{s}_t|\le2\sqrt{N}$ and $\mathbb{E}[|\mathfrak{s}_t|^{4}]\le16N^{2}$. Likewise $|b^\gamma|\le2(l-1)B$ by part (a) of the martingale decomposition theorem, so $|g^\gamma_t|\le4\sqrt{N}(l-1)B$, $|g_t|^{2}\le16\,l\,N(l-1)^{2}B^{2}$, and $\mathbb{E}[|g_t|^{4}]\le256\,l^{2}N^{2}(l-1)^{4}B^{4}$.

Part (b). Step 1 (integral representation). Exactly as in Step 1 of the published proof of the second-moment lemma: by part (b) of the martingale decomposition theorem, condition 2 of the mean-field trajectory pair, and the agreement of the Riemann and Lebesgue integrals for continuous integrands, almost surely, for all $t\in[0,T]$ and all $\gamma$,

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

Step 2 (quartic three-term estimate). For real numbers, $(a_1+a_2+a_3)^{2}\le3(a_1^{2}+a_2^{2}+a_3^{2})$ (expand and use $2a_pa_q\le a_p^{2}+a_q^{2}$); applied componentwise and summed over $\gamma$, this gives $|x+y+z|^{2}\le3(|x|^{2}+|y|^{2}+|z|^{2})$ for vectors $x,y,z\in\mathbb{R}^{l}$, and a second application to the three nonnegative reals $|x|^{2},|y|^{2},|z|^{2}$ gives

Applying this to Step 1 and taking expectations (each term is nonnegative, so the expectation splits in $[0,\infty]$),

For the middle term, fix $t>0$ (for $t=0$ it vanishes). Pathwise on $\Omega_0$, the Cauchy--Schwarz inequality of the interval toolkit applied to the pair $(1,g^\gamma)$ on $[0,t]$ gives $(\int_{[0,t]}g^\gamma_s\,ds)^{2}\le t\int_{[0,t]}(g^\gamma_s)^{2}ds$; summing over $\gamma$, $|\int_{[0,t]}g_s\,ds|^{2}\le t\int_{[0,t]}|g_s|^{2}ds$, and squaring together with a second application of the toolkit Cauchy--Schwarz inequality to the pair $(1,|g|^{2})$ (the path $s\mapsto|g_s|^{2}$ is measurable on $[0,t]$, being a finite sum of squares of the measurable paths $s\mapsto g^\gamma_s$, and bounded, so all integrals exist),

Taking expectations and applying the Tonelli theorem to the nonnegative product-measurable (after $\mathbf{1}$-modification) integrand, $\mathbb{E}[|\int_{[0,t]}g_s\,ds|^{4}]\le T^{3}\int_{[0,t]}\mathbb{E}[|g_s|^{4}]\,ds$. For the last term, part (c) of the aggregate counter moment lemma gives $\mathbb{E}[|\sqrt{N}M_t|^{4}]\le c_M\,(Bt+(Bt)^{2})$; we use the weaker $N$-free form of that bound, the sharper variant with $Bt/N$ in place of $Bt$ not being needed for an $N$-uniform estimate.

It remains to convert the $g$-integral. By part (i) of the drift regularity lemma, $b$ agrees with the extended aggregate state drift on $\Delta^{l}\times\mathbb{R}^{m}$, and by its part (ii), $|b^\gamma(\Sigma_s,\alpha_s)-b^\gamma(S_s,A_s)|\le\sqrt{l+m}\,l(B+K)\,d((\Sigma_s,\alpha_s),(S_s,A_s))$, while $N\,d((\Sigma_s,\alpha_s),(S_s,A_s))^{2}=|\mathfrak{s}_s|^{2}+|\mathfrak{a}_s|^{2}$. Hence, on $\Omega_0$,

and squaring, with $(u+v)^{2}\le2(u^{2}+v^{2})$ for nonnegative reals, $|g_s|^{4}\le2\,\Lambda^{4}\,(|\mathfrak{s}_s|^{4}+|\mathfrak{a}_s|^{4})$. Taking expectations and integrating (monotonicity and linearity of the Lebesgue integral for measurable $[0,\infty]$-valued integrands),

with both sides valued in $[0,\infty]$. Combining this with the three-term display and the counter bound proves part (b).

Part (c). If $\mathcal{A}_4=\infty$ the right-hand side is $+\infty$ and there is nothing to prove, so assume $\mathcal{A}_4<\infty$. Let $u(t)=\mathbb{E}[|\mathfrak{s}_t|^{4}]$, measurable by part (a) and bounded by $16N^{2}$, and set

Part (b), the monotonicity $Bt+(Bt)^{2}\le BT+(BT)^{2}$, and $\int_{[0,t]}\mathbb{E}[|\mathfrak{a}_s|^{4}]\,ds\le\mathcal{A}_4$ yield $u(t)\le a+54\,T^{3}\Lambda^{4}\int_{[0,t]}u(s)\,ds$ for every $t\in[0,T]$. Define $w(t)=a+54\,T^{3}\Lambda^{4}\int_{[0,t]}u(s)\,ds$; 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, $w(t)\le a+54\,T^{3}\Lambda^{4}\int_{[0,t]}w(s)\,ds$, and for the continuous $w$ the Lebesgue integral agrees with the Riemann integral. Gronwall's lemma applied to $w$ with constants $a$ and $54\,T^{3}\Lambda^{4}$ gives $w(t)\le a\exp(54\,T^{3}\Lambda^{4}t)$, and therefore $u(t)\le a\exp(54\,T^{3}\Lambda^{4}t)$ for every $t\in[0,T]$, which is the claimed bound, with the exponential function. In particular, if $\mathcal{A}_4<\infty$ then $\sup\{\mathbb{E}[|\mathfrak{s}_t|^{4}]:t\in[0,T]\}\le a\exp(54\,T^{3}\Lambda^{4}T)<\infty$, and among the quantities entering $a$ and the exponential rate, only $\mathbb{E}[|\mathfrak{s}_0|^{4}]$ and $\mathcal{A}_4$ depend on $N$ (the constants $B$, $T$, $l$, $m$, $K$, $c_M$, and $\Lambda$ do not). $\blacksquare$