Proof of Injection Weights from a Bounded Profile Along Mean-Field Clocks: Label-Rate Form of the Fluctuation Covariance, Weight Bounds, the Prior Quadratic Form, and the Step-Function Injection

Throughout, $c=(\sigma,\gamma)$ denotes a label, sums over $c$ run over $\mathcal{L}$, and we fix a label $c$ whenever a single clock is discussed. We first record some preliminaries.

(P1) Vectors. $|v_c|=\sqrt{2}$: $v_c=\delta_\gamma-\delta_\sigma$ with $\gamma\neq\sigma$ has two coordinates equal to $\pm1$ and the others $0$, so $|v_c|^{2}=2$ by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. Hence $|v_c\cdot y|\le\sqrt{2}\,|y|$ for $y\in\mathbb{R}^l$ by The Cauchy-Schwarz Inequality on $\mathbb{R}^n$; in particular $|v_c\cdot\lambda_s|\le\sqrt{2}\,\Lambda$ for all $s$. For vectors $y_1,\dots,y_k\in\mathbb{R}^l$ and reals $a_1,\dots,a_k$, $|\sum_ia_iy_i|\le\sum_i|a_i|\,|y_i|$ by induction from claims 6 and 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. Also the coordinates of $v_c=\delta_\gamma-\delta_\sigma$ are $v_c^{i}=\mathbf{1}\{i=\gamma\}-\mathbf{1}\{i=\sigma\}$, by the difference of points in Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, and hence $v_c\cdot y=y^{\gamma}-y^{\sigma}$ by the dot product of that definition; the dot product is bilinear by its defining formula, so in particular $y\cdot(v_c(v_c\cdot y))=(v_c\cdot y)^{2}$.

(P2) Integrals. Every interval contained in $[0,T]$ is a Borel set by Borel Sigma-Algebra on the Real Line and, being a subset of $[0,T]$, belongs to $\mathcal{B}_{[0,T]}=\{S\cap[0,T]:S\in\mathcal{B}(\mathbb{R})\}$ (claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval, which also makes $([0,T],\mathcal{B}_{[0,T]},\mathrm{Leb})$ a measure space); $\mathrm{Leb}([0,t])=t$ and $\mathrm{Leb}(\{u\})=\mathrm{Leb}([u,u])=0$ for $0\le t\le T$ and $u\in[0,T]$ by claim 4 of Existence of Lebesgue Measure on the Real Line (with $a=b=u$ for the singleton). If $f$ is bounded and measurable, $|f|\le M$, and $J\in\mathcal{B}_{[0,T]}$, then $\mathbf{1}_{J}f$ is measurable (claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions) and integrable, since $\int_{[0,T]}|\mathbf{1}_{J}f|\,d\mathrm{Leb}\le\int_{[0,T]}M\mathbf{1}_{J}\,d\mathrm{Leb}=M\,\mathrm{Leb}(J)\le M\,\mathrm{Leb}([0,T])=MT<\infty$ by monotonicity (applied also to $\mathbf{1}_{J}\le\mathbf{1}_{[0,T]}$), the integral of the simple function $M\mathbf{1}_{J}$ being $M\,\mathrm{Leb}(J)$ by Simple Function and Its Integral and the agreement of the two notions of integral for nonnegative simple functions stated in Lebesgue Integral of a Nonnegative Measurable Function. For nonnegative bounded measurable $f$ the integral of a nonnegative function and the integral of an integrable function agree, since $f^{+}=f$ and $f^{-}=0$ in Integrable Function and the Lebesgue Integral; we use both notions interchangeably for such $f$. Moreover $|\int_{J}f(s)\,ds|\le\int_{J}|f(s)|\,ds\le M\,\mathrm{Leb}(J)$ (monotonicity applied to $\pm\mathbf{1}_{J}f\le\mathbf{1}_{J}|f|$). Linearity of the integral gives additivity: if $J,J'\in\mathcal{B}_{[0,T]}$ are disjoint then $\mathbf{1}_{J\cup J'}=\mathbf{1}_{J}+\mathbf{1}_{J'}$ and $\int_{J\cup J'}f=\int_{J}f+\int_{J'}f$, and similarly for finitely many pairwise disjoint sets. For maps $g$ into $\mathbb{R}^l$ with bounded measurable components, $|\int_{[0,T]}g(s)\,ds|\le\int_{[0,T]}|g(s)|\,ds$ by Norm Bound for a Vector-Valued Lebesgue Integral over a Compact Interval.

(P3) The clocks. Fix $c$. For $0\le t\le t'\le T$ one has $0\le\mathbf{1}_{[0,t]}\phi_c\le\mathbf{1}_{[0,t']}\phi_c\le B$ pointwise, so by monotonicity $0\le\bar{\mathsf{C}}^{c}_t\le\bar{\mathsf{C}}^{c}_{t'}$ and $\bar{\mathsf{C}}^{c}_{t'}\le NB\,\mathrm{Leb}([0,t'])\le NBT\le R$. For $0\le s_1\le s_2\le T$, additivity with $[0,s_2]=[0,s_1]\cup(s_1,s_2]$ and $\mathrm{Leb}(\{s_1\})=0$ give

$$N\int_{[s_1,s_2]}\phi_c(s)\,ds=N\int_{(s_1,s_2]}\phi_c(s)\,ds=\bar{\mathsf{C}}^{c}_{s_2}-\bar{\mathsf{C}}^{c}_{s_1}.$$

For reals $r_1\ler_2$ the set $\{s\in[0,T]:r_1\le\bar{\mathsf{C}}^{c}_s\ler_2\}$ is an interval: if $s_1\le s\le s_2$ with $s_1,s_2$ in the set, then $r_1\le\bar{\mathsf{C}}^{c}_{s_1}\le\bar{\mathsf{C}}^{c}_s\le\bar{\mathsf{C}}^{c}_{s_2}\ler_2$. The same argument shows that the preimage under $s\mapsto\bar{\mathsf{C}}^{c}_s$ of any interval is an interval, hence belongs to $\mathcal{B}_{[0,T]}$ by (P2).

(M) Increment bound. Let $J\in\mathcal{B}_{[0,T]}$ and let $r_1\ler_2$ be reals with $\bar{\mathsf{C}}^{c}_s\in[r_1,r_2]$ for every $s\in J$. Then $N\int_{J}\phi_c(s)\,ds\ler_2-r_1$. Indeed, $J\subseteq J'=\{s\in[0,T]:r_1\le\bar{\mathsf{C}}^{c}_s\ler_2\}$, an interval in $\mathcal{B}_{[0,T]}$ by (P3), and $\mathbf{1}_{J}\phi_c\le\mathbf{1}_{J'}\phi_c$, so it suffices to bound $N\int_{J'}\phi_c$. If $J'$ is empty the integral is $0$. Otherwise $J'$ is a nonempty subset of $[0,T]$, so it has a least upper bound $v$ and (applying the same to $\{-s:s\in J'\}$) a greatest lower bound $u$, both in $[0,T]$, by the completeness of the real numbers; thus $J'\subseteq[u,v]$ and, $J'$ being an interval, $(u,v)\subseteq J'$ (for $u<s<v$ there are points of $J'$ below and above $s$). For $s_1\le s_2$ in $J'$, (P3) gives $N\int_{[s_1,s_2]}\phi_c=\bar{\mathsf{C}}^{c}_{s_2}-\bar{\mathsf{C}}^{c}_{s_1}\ler_2-r_1$. If $u=v$ then $J'\subseteq\{u\}$ and $N\int_{J'}\phi_c\le NB\,\mathrm{Leb}(\{u\})=0$. If $u<v$, put $s^{n}_1=u+(v-u)/(n+1)$ and $s^{n}_2=v-(v-u)/(n+1)$ for natural numbers $n\ge1$; then $u<s^{n}_1\le s^{n}_2<v$, so $[s^{n}_1,s^{n}_2]\subseteq J'$ and $N\int_{[s^{n}_1,s^{n}_2]}\phi_c\ler_2-r_1$. The functions $f_n=\mathbf{1}_{[s^{n}_1,s^{n}_2]}\phi_c$ ($n\in\mathbb{N}$) are nonnegative and measurable on the measure space $([0,T],\mathcal{B}_{[0,T]},\mathrm{Leb})$, nondecreasing in $n$ (the intervals increase), and $\sup_nf_n=\mathbf{1}_{(u,v)}\phi_c$ pointwise (every $s\in(u,v)$ lies in $[s^{n}_1,s^{n}_2]$ for all large $n$), so by the monotone convergence theorem $N\int_{(u,v)}\phi_c=\lim_nN\int f_n\,d\mathrm{Leb}\ler_2-r_1$, limits preserving non-strict inequalities by claim 1 of Order Properties of Limits of Real Sequences. Finally $J'\subseteq(u,v)\cup\{u\}\cup\{v\}$, and the two singletons contribute at most $NB\,\mathrm{Leb}(\{u\})+NB\,\mathrm{Leb}(\{v\})=0$ by (P2) and monotonicity, so $N\int_{J'}\phi_c\le N\int_{(u,v)}\phi_c\ler_2-r_1$.

Step 1 (Claim 1). Fix $\Sigma\in\Delta^l$, $\alpha\in\mathcal{A}$ and $y\in\mathbb{R}^l$, and abbreviate $\beta(\sigma,\gamma)=\beta(\sigma,\gamma,\Sigma,\alpha)$. By (P1), the $\gamma$-th coordinate of $\sum_cv_c(v_c\cdot y)\Sigma^{\sigma}\beta(\sigma,\gamma')$ (labels written $c=(\sigma,\gamma')$) equals

$$\sum_{\sigma\neq\gamma}(y^{\gamma}-y^{\sigma})\,\Sigma^{\sigma}\beta(\sigma,\gamma)\ -\ \sum_{\gamma'\neq\gamma}(y^{\gamma'}-y^{\gamma})\,\Sigma^{\gamma}\beta(\gamma,\gamma')=y^{\gamma}\sum_{\delta\neq\gamma}\bigl(\Sigma^{\delta}\beta(\delta,\gamma)+\Sigma^{\gamma}\beta(\gamma,\delta)\bigr)+\sum_{\delta\neq\gamma}y^{\delta}\bigl(-\Sigma^{\gamma}\beta(\gamma,\delta)-\Sigma^{\delta}\beta(\delta,\gamma)\bigr),$$

the first sum collecting the labels with $\gamma'=\gamma$ (contributing $+1$ to $v_c^{\gamma}$) and the second those with $\sigma=\gamma$ (contributing $-1$); the right-hand side is $\Theta^{\gamma\gamma}(\Sigma,\alpha)y^{\gamma}+\sum_{\delta\neq\gamma}\Theta^{\gamma\delta}(\Sigma,\alpha)y^{\delta}$ by the entries of Aggregate Fluctuation Covariance, which is the $\gamma$-th coordinate of $\Theta(\Sigma,\alpha)y$ by the definition of the matrix-vector product. This proves the first identity; taking the dot product with $y$ and using $y\cdot(v_c(v_c\cdot y))=(v_c\cdot y)^{2}$ from (P1) termwise gives the second.

Step 2 (Claim 2). Fix $c$. Monotonicity and the bounds on $\bar{\mathsf{C}}^{c}$ are (P3). Since $\bar{\mathsf{C}}^{c}_s\ge0$, $\bar{J}_{c,1}$ is the preimage of $[0,b^{c}_1]$ and $\bar{J}_{c,j}$ that of $I_{c,j}$, so each time cell is an interval in $\mathcal{B}_{[0,T]}$ by (P3). The sets $[0,b^{c}_1]$ and $I_{c,j}$ ($2\le j\le J_c$) are pairwise disjoint with union $[0,R]$, and $\bar{\mathsf{C}}^{c}_s\in[0,R]$ for every $s$, so the time cells are pairwise disjoint with union $[0,T]$. The bound $N\int_{\bar{J}_{c,j}}\phi_c\le\mu_{c,j}$ is (M) with $[r_1,r_2]=[b^{c}_{j-1},b^{c}_j]$ (for $j=1$, $[0,b^{c}_1]$). The map $s\mapsto v_c\cdot\lambda_s=\sum_iv_c^{i}\lambda^{i}_s$ is measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and bounded by $\sqrt{2}\Lambda$ (P1); products with $\phi_c$, with itself and with indicators of time cells are bounded and measurable by claims 1 and 3 of that lemma; so all integrands are bounded and measurable and the integrals exist by (P2). By (P2), (P1) and (M),

$$|w_{c,j}|\le\frac{N}{\mathsf{m}}\int_{\bar{J}_{c,j}}|v_c\cdot\lambda_s|\,\phi_c(s)\,ds\le\frac{\sqrt{2}\Lambda}{\mathsf{m}}\,N\int_{\bar{J}_{c,j}}\phi_c(s)\,ds\le\frac{\sqrt{2}\,\Lambda\,\mu_{c,j}}{\mathsf{m}},$$

and summing, $\sum_q|w_q|\le\frac{\sqrt{2}\Lambda}{\mathsf{m}}\sum_c\sum_{j=1}^{J_c}\mu_{c,j}=\frac{\sqrt{2}\Lambda}{\mathsf{m}}\sum_c(b^{c}_{J_c}-b^{c}_0)=\frac{\sqrt{2}\,\Lambda\,l(l-1)R}{\mathsf{m}}$. For $\bar{F}_t$, the integrand $g_s=\mathbf{1}_{[0,t]}(s)\sum_cv_c(v_c\cdot\lambda_s)\phi_c(s)$ satisfies $|g_s|\le\sum_c\sqrt{2}\cdot\sqrt{2}\Lambda B=2\Lambda B\,l(l-1)$ by (P1), so $|\bar{F}_t|\le2\Lambda B\,l(l-1)\,\mathrm{Leb}([0,T])$ by (P2). The integrand of $\mathcal{P}$ is nonnegative and at most $\sum_c2\Lambda^{2}B=2\Lambda^{2}B\,l(l-1)$, whence the bounds on $\mathcal{P}$.

Step 3 (Claim 3). Fix $q=(c,j)\in\mathsf{L}$. Apply claim 1 of Weighted Cauchy-Schwarz Inequality on a Measure Space and the Symmetrised Score Functional: Bounds and Averaging on the measure space $([0,T],\mathcal{B}_{[0,T]},\mathrm{Leb})$ (claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval) to $\beta_{\mathrm{CS}}=\mathbf{1}_{\bar{J}_q}\phi_c$, which is measurable with values in $[0,\infty)$ and $\int\beta_{\mathrm{CS}}\,d\mathrm{Leb}\le BT<\infty$, and to $\alpha_{\mathrm{CS}}=\mathbf{1}_{\bar{J}_q}(v_c\cdot\lambda)\phi_c$, which is measurable (Step 2) and vanishes wherever $\beta_{\mathrm{CS}}$ does. The function $q_{\mathrm{CS}}$ of that claim (equal to $\alpha_{\mathrm{CS}}^{2}/\beta_{\mathrm{CS}}$ where $\beta_{\mathrm{CS}}>0$ and to $0$ elsewhere) is $\mathbf{1}_{\bar{J}_q}(v_c\cdot\lambda)^{2}\phi_c$ at every point: where $\beta_{\mathrm{CS}}>0$ this is $\alpha_{\mathrm{CS}}^{2}/\beta_{\mathrm{CS}}$, and where $\beta_{\mathrm{CS}}=0$ both vanish. Its integral is at most $2\Lambda^{2}BT<\infty$. The claim yields

$$\Bigl(\frac{\mathsf{m}}{N}w_q\Bigr)^{2}=\Bigl(\int_{\bar{J}_q}(v_c\cdot\lambda_s)\phi_c(s)\,ds\Bigr)^{2}\le\Bigl(\int_{\bar{J}_q}\phi_c(s)\,ds\Bigr)\Bigl(\int_{\bar{J}_q}(v_c\cdot\lambda_s)^{2}\phi_c(s)\,ds\Bigr)\le\frac{\mu_q}{N}\int_{\bar{J}_q}(v_c\cdot\lambda_s)^{2}\phi_c(s)\,ds ,$$

the last step by claim 2. Multiplying by $N^{2}\mathsf{h}_q/\mathsf{m}^{2}$ and using $\mu_q\mathsf{h}_q\le\kappa\mathsf{m}^{2}$,

$$w_q^{2}\mathsf{h}_q\le\frac{N\,\mu_q\,\mathsf{h}_q}{\mathsf{m}^{2}}\int_{\bar{J}_q}(v_c\cdot\lambda_s)^{2}\phi_c(s)\,ds\le\kappa N\int_{\bar{J}_q}(v_c\cdot\lambda_s)^{2}\phi_c(s)\,ds .$$

Summing over $j$ for fixed $c$ and using additivity over the partition of $[0,T]$ into time cells (claim 2), then summing over $c$, gives $\sum_qw_q^{2}\mathsf{h}_q\le\kappa N\sum_c\int_{[0,T]}(v_c\cdot\lambda_s)^{2}\phi_c(s)\,ds=\kappa N\mathcal{P}$ by linearity.

Step 4 (Claim 4). Fix $t\in[0,T]$, the family $x=(x_c)_c$, and a label $c$; write $\xi=x_c$ and $\bar{x}=\bar{\mathsf{C}}^{c}_t\in[0,R]$, drop the superscript $c$ from $b^{c}_j$, and put

$$\mathsf{S}_c(z)=\sum_{j=1}^{J_c}w_{c,j}\,n_{c,j}(z),\qquad \mathsf{W}_c(z)=\mathsf{m}\sum_{j:\,1\le j\le J_c,\ b_j\le z}w_{c,j}\qquad(z\in[0,R]).$$

(i) Straddling cell. For $j$ with $b_j\le\xi$ one has $n_{c,j}(\xi)=\mathsf{m}$, and for $j$ with $b_{j-1}\ge\xi$ one has $n_{c,j}(\xi)=0$; the remaining indices satisfy $b_{j-1}<\xi<b_j$, i.e. $\xi$ lies in the open interval $(b_{j-1},b_j)$, and since these open intervals are pairwise disjoint there is at most one such $j$. Hence $\mathsf{S}_c(\xi)-\mathsf{W}_c(\xi)$ is either $0$ or $w_{c,j}n_{c,j}(\xi)$ for that single $j$, so $|\mathsf{S}_c(\xi)-\mathsf{W}_c(\xi)|\le\mathsf{m}\max_j|w_{c,j}|\le\sqrt{2}\Lambda\mu_{\max}$ by claim 2.

(ii) Realized versus mean-field clock. We claim $|\mathsf{W}_c(\xi)-\mathsf{W}_c(\bar{x})|\le\sqrt{2}\Lambda(|\xi-\bar{x}|+\mu_{\max})$. By symmetry of the assertion in $\xi$ and $\bar{x}$ we may assume $\xi\le\bar{x}$; then $\mathsf{W}_c(\bar{x})-\mathsf{W}_c(\xi)=\mathsf{m}\sum_{j:\,\xi<b_j\le\bar{x}}w_{c,j}$. If no $j$ satisfies $\xi<b_j\le\bar{x}$ the difference is $0$. Otherwise, since $b_1<\dots<b_{J_c}$, the indices $j$ with $\xi<b_j\le\bar{x}$ form a set of consecutive integers $\{j_1,\dots,j_2\}$, and $\sum_{j=j_1}^{j_2}\mu_{c,j}=b_{j_2}-b_{j_1-1}\le\bar{x}-(b_{j_1}-\mu_{c,j_1})\le\bar{x}-\xi+\mu_{\max}$, using $b_{j_2}\le\bar{x}$ and $b_{j_1}>\xi$. By claim 2, $\mathsf{m}\sum_{j=j_1}^{j_2}|w_{c,j}|\le\sqrt{2}\Lambda\sum_{j=j_1}^{j_2}\mu_{c,j}\le\sqrt{2}\Lambda(\bar{x}-\xi+\mu_{\max})$.

(iii) Mean-field clock versus profile injection. We claim $\bigl|\mathsf{W}_c(\bar{x})-N\int_{[0,t]}(v_c\cdot\lambda_s)\phi_c(s)\,ds\bigr|\le\sqrt{2}\Lambda\mu_{\max}$. Let $j^{*}$ be the largest $j\in\{0,\dots,J_c\}$ with $b_j\le\bar{x}$ (it exists since $b_0=0\le\bar{x}$). If $j^{*}=0$ then $\mathsf{W}_c(\bar{x})=0$ and $\bar{x}<b_1$; for $s\in[0,t]$ one has $\bar{\mathsf{C}}^{c}_s\in[0,\bar{x}]$ by (P3), so by (P1), (P2) and (M), $N\bigl|\int_{[0,t]}(v_c\cdot\lambda_s)\phi_c(s)\,ds\bigr|\le\sqrt{2}\Lambda\,N\int_{[0,t]}\phi_c(s)\,ds\le\sqrt{2}\Lambda\bar{x}\le\sqrt{2}\Lambda b_1=\sqrt{2}\Lambda\mu_{c,1}\le\sqrt{2}\Lambda\mu_{\max}$. If $j^{*}\ge1$, then by the definition of the weights and additivity over the pairwise disjoint time cells,

$$\mathsf{W}_c(\bar{x})=\mathsf{m}\sum_{j=1}^{j^{*}}w_{c,j}=N\int_{\bar{J}^{*}}(v_c\cdot\lambda_s)\phi_c(s)\,ds,\qquad \bar{J}^{*}=\bigcup_{j=1}^{j^{*}}\bar{J}_{c,j}=\{s\in[0,T]:\bar{\mathsf{C}}^{c}_s\le b_{j^{*}}\},$$

the last equality because $[0,b_1]\cup I_{c,2}\cup\dots\cup I_{c,j^{*}}=[0,b_{j^{*}}]$ and $\bar{\mathsf{C}}^{c}\ge0$. Since $|\mathbf{1}_{\bar{J}^{*}}-\mathbf{1}_{[0,t]}|=\mathbf{1}_{[0,t]\setminus\bar{J}^{*}}+\mathbf{1}_{\bar{J}^{*}\setminus[0,t]}$, linearity, (P2) and (P1) give

$$\Bigl|\mathsf{W}_c(\bar{x})-N\int_{[0,t]}(v_c\cdot\lambda_s)\phi_c(s)\,ds\Bigr|\le\sqrt{2}\Lambda\Bigl(N\int_{[0,t]\setminus\bar{J}^{*}}\phi_c(s)\,ds+N\int_{\bar{J}^{*}\setminus[0,t]}\phi_c(s)\,ds\Bigr),$$

both sets belonging to $\mathcal{B}_{[0,T]}$. On $[0,t]\setminus\bar{J}^{*}$ one has $b_{j^{*}}<\bar{\mathsf{C}}^{c}_s\le\bar{\mathsf{C}}^{c}_t=\bar{x}$ by (P3), so (M) with $[r_1,r_2]=[b_{j^{*}},\bar{x}]$ bounds the first term by $\bar{x}-b_{j^{*}}$; if $j^{*}<J_c$ this is less than $b_{j^{*}+1}-b_{j^{*}}\le\mu_{\max}$ by maximality of $j^{*}$, and if $j^{*}=J_c$ then $R=b_{J_c}\le\bar{x}\le R$ forces $\bar{x}-b_{j^{*}}=0$. On $\bar{J}^{*}\setminus[0,t]$ one has $s>t$, hence $\bar{x}=\bar{\mathsf{C}}^{c}_t\le\bar{\mathsf{C}}^{c}_s\le b_{j^{*}}\le\bar{x}$, so $\bar{\mathsf{C}}^{c}_s=\bar{x}$ there and (M) with $[r_1,r_2]=[\bar{x},\bar{x}]$ shows the second term is $0$. Thus the claim of (iii) holds in all cases.

Assembly. By (i), (ii), (iii) (recall $\xi=x_c$) and the triangle inequality for absolute values,

$$\Bigl|\frac{1}{N}\mathsf{S}_c(x_c)-\int_{[0,t]}(v_c\cdot\lambda_s)\phi_c(s)\,ds\Bigr|\le\frac{\sqrt{2}\Lambda}{N}\bigl(\mu_{\max}+|x_c-\bar{\mathsf{C}}^{c}_t|+\mu_{\max}+\mu_{\max}\bigr)=\frac{\sqrt{2}\Lambda}{N}\bigl(|x_c-\bar{\mathsf{C}}^{c}_t|+3\mu_{\max}\bigr).$$

By componentwise linearity of the integral, $\bar{F}_t=\sum_cv_c\int_{[0,t]}(v_c\cdot\lambda_s)\phi_c(s)\,ds$, so

$$\frac{1}{N}\sum_cv_c\,\mathsf{S}_c(x_c)-\bar{F}_t=\sum_cv_c\Bigl(\frac{1}{N}\mathsf{S}_c(x_c)-\int_{[0,t]}(v_c\cdot\lambda_s)\phi_c(s)\,ds\Bigr),$$

and the triangle inequality of (P1) with $|v_c|=\sqrt{2}$ gives the bound $\frac{2\Lambda}{N}\sum_c(|x_c-\bar{\mathsf{C}}^{c}_t|+3\mu_{\max})$ of claim 4.