Proof of Quadratic Growth of the Mean-Field Hamiltonian in the Control along a Stationary Mean-Field Triple

Throughout, $t\in[0,T]$ is arbitrary, $x_t=(S_t,A_t)\in\Delta^l\times\mathcal{A}$, and points of $\mathbb{R}^{l}\times\mathbb{R}^{m}$ are written $(\Sigma,\alpha)$ and identified with points of $\mathbb{R}^{l+m}$ as in the extension definitions. We use three facts about the data. (F1) $|\partial_j\partial_i\bar{L}(z)|\le K_c$ for all $i,j\in\{1,\dots,l+m\}$ and $z\in U_c\times\mathbb{R}^m$ (clause 3 of the cost extension definition), and $|\partial_j\partial_i\bar{b}^\delta(z)|\le3\,l\,K$ for all $i,j$, all $\delta$, and all $z\in\Delta^l\times V$ (part (iii) of the regularity of the extended aggregate state drift). (F2) For every real $\varepsilon>0$ there is a real $\eta>0$ such that $|\partial_j\partial_i\bar{L}(z)-\partial_j\partial_i\bar{L}(z')|\le\varepsilon$ whenever $z,z'\in U_c\times\mathbb{R}^m$ satisfy $d(z,z')\le\eta$, and $|\partial_j\partial_i\bar{b}^\delta(z)-\partial_j\partial_i\bar{b}^\delta(z')|\le\varepsilon$ for all $i,j$ and all $\delta$ whenever $z,z'\in\Delta^l\times V$ satisfy $d(z,z')\le\eta$: take $\eta$ to be the smaller of the two moduli supplied for this $\varepsilon$ by clause 4 of the cost extension definition and by part (iii) of the same regularity lemma. (The letter $\delta$ is reserved for state indices throughout this proof.) (F3) For $a\in\mathcal{A}$ the segment $\{(S_t,A_t+\tau(a-A_t)):\tau\in[0,1]\}$ from $x_t$ to $(S_t,a)$ lies in $\{S_t\}\times\mathcal{A}$, because $\mathcal{A}$ is convex (under (A) in (d), and by the standing assumption of (c) there); hence it lies both in $U_c\times\mathbb{R}^m$ and in $\Delta^l\times V\subseteq U\times V$, since $\Delta^l\subset U_c$, $\Delta^l\subset U$, and $\mathcal{A}\subseteq V$ by the extension definitions. The Euclidean distance between $x_t$ and $(S_t,a)$ is $|a-A_t|$, and the difference vector $h=(S_t,a)-x_t$ has components $h_i=0$ for $i\le l$ and $h_{l+j}=a^j-A^j_t$ for $j\le m$.

Proof of (a). The map $(t,a)\mapsto(S_t,a)$ from $[0,T]\times V$ to $\mathbb{R}^{l+m}$ is continuous at every point: each component $t\mapsto S^\gamma_t$ is continuous on $[0,T]$ by clause 1 of the trajectory-pair definition, metric continuity and Euclidean continuity agreeing for real-valued maps by claim 1 of the continuity agreement lemma, the coordinate maps $(t,a)\mapsto a^j$ are continuous, and a map into $\mathbb{R}^{l+m}$ with continuous components is continuous since $d\big((\Sigma,\alpha),(\Sigma',\alpha')\big)\le\sum_{\gamma}|\Sigma^\gamma-\Sigma'^\gamma|+\sum_j|\alpha^j-\alpha'^j|$ by claim 1 of the componentwise estimates. The functions $\bar{L}$ and $\bar{b}^\delta$ are continuous at every point of their open domains (clause 1 of the $C^k$ definition, via clause 2 of the cost extension definition and part (i) of the regularity lemma), so $(t,a)\mapsto\bar{L}(S_t,a)$ and $(t,a)\mapsto\bar{b}^\delta(S_t,a)$ are continuous by continuity of compositions. Each $(t,a)\mapsto P^\delta_t$ is continuous, being the composition of the continuous projection $(t,a)\mapsto t$ with $t\mapsto P^\delta_t$, continuous by clause 1 of the co-state definition and the same agreement lemma (the same remark applies to $(t,a)\mapsto S^\gamma_t$ above). Finite sums and products of real-valued continuous functions are continuous by continuity of sums and products, applied pointwise with the agreement lemma. This proves the continuity of $(t,a)\mapsto\mathcal{H}_t(a)$. For $(t,a)\mapsto\mathcal{H}_t(A_t)$, note that $t\mapsto(S_t,A_t)$ is continuous into $\mathbb{R}^{l+m}$ by the same componentwise argument (clause 1 of the trajectory-pair definition also covering $A$), and argue identically.

Proof of (b). The first identity is clause 3 of the co-state definition, rewritten by subtracting its right-hand side from both sides. For the second, by the definition of $R_t$ and a relabeling of the summation indices,

$$h\cdot R_th=\tfrac{1}{4}\sum_{i,j=1}^{m}H_{l+i,l+j}(t)h^ih^j+\tfrac{1}{4}\sum_{i,j=1}^{m}H_{l+j,l+i}(t)h^ih^j=\tfrac{1}{2}\sum_{i,j=1}^{m}H_{l+i,l+j}(t)h^ih^j .$$

A combined Taylor estimate. Fix $t$ and $a\in\mathcal{A}$, let $h$ be as in (F3), and let $\bar{\varepsilon}_L,\bar{\varepsilon}_b\ge0$ be real numbers such that $|\partial_j\partial_i\bar{L}(z)-\partial_j\partial_i\bar{L}(x_t)|\le\bar{\varepsilon}_L$ and $|\partial_j\partial_i\bar{b}^\delta(z)-\partial_j\partial_i\bar{b}^\delta(x_t)|\le\bar{\varepsilon}_b$ for every point $z$ of the segment of (F3), all $i,j$, and all $\delta$. Part (iii) of the multivariate Taylor expansion applies to $f=\bar{L}$, which is of class $C^2$ on the open set $U_c\times\mathbb{R}^m$ (open by clause 2 of the cost extension definition), with $n=l+m$, $x=x_t$, $y=(S_t,a)$, and $\bar{\varepsilon}=\bar{\varepsilon}_L$; and to $f=\bar{b}^\delta$ on the open set $U\times V$, of class $C^2$ by part (i) of the regularity lemma, with $\bar{\varepsilon}=\bar{\varepsilon}_b$. Multiplying the $\bar{b}^\delta$ estimates by $-P^\delta_t$, adding them to the $\bar{L}$ estimate, and using the triangle inequality together with $\sum_\delta|P^\delta_t|\le C_P$, we obtain, since only the components $h_{l+j}$ of $h$ are nonzero,

$$\Big|\mathcal{H}_t(a)-\mathcal{H}_t(A_t)-\sum_{j=1}^{m}\Big(\partial_{l+j}\bar{L}(x_t)-\sum_{\delta}P^\delta_t\partial_{l+j}\bar{b}^\delta(x_t)\Big)h_{l+j}-\tfrac{1}{2}\sum_{i,j=1}^{m}H_{l+i,l+j}(t)\,h_{l+i}h_{l+j}\Big|\le\tfrac{1}{2}(l+m)\big(\bar{\varepsilon}_L+C_P\bar{\varepsilon}_b\big)|a-A_t|^2,$$

where we used that, by the definition of the fluctuation Hessian coefficients, $H_{l+i,l+j}(t)=\partial_{l+j}\partial_{l+i}\bar{L}(x_t)-\sum_\delta P^\delta_t\partial_{l+j}\partial_{l+i}\bar{b}^\delta(x_t)$. By (b), applied to the vector $\hat{h}=(h_{l+1},\dots,h_{l+m})=a-A_t\in\mathbb{R}^m$ (so that $h_{l+i}=\hat{h}^i$), the first-order sum vanishes and the second-order sum equals $\hat{h}\cdot R_t\hat{h}=(a-A_t)\cdot R_t(a-A_t)$, so

$$\Big|\mathcal{H}_t(a)-\mathcal{H}_t(A_t)-(a-A_t)\cdot R_t(a-A_t)\Big|\le\tfrac{1}{2}(l+m)\big(\bar{\varepsilon}_L+C_P\bar{\varepsilon}_b\big)|a-A_t|^2 .\qquad(\ast)$$

Proof of (c). Apply part (ii) of the same Taylor lemma instead of part (iii), to $\bar{L}$ with the second-derivative bound of that lemma taken to be $K_c$ and to each $\bar{b}^\delta$ with that bound taken to be $3lK$ (the symbol $M_2$ is reserved for the constant $K_c+3lKC_P$ of the statement), the bounds (F1) holding on the segment by (F3), whose convexity requirement is the standing assumption of (c). Combining as above and using the vanishing of the first-order sum from (b),

$$\big|\mathcal{H}_t(a)-\mathcal{H}_t(A_t)\big|\le\tfrac{1}{2}(l+m)\big(K_c+C_P\cdot3lK\big)|a-A_t|^2=C_{\mathcal{H}}|a-A_t|^2 .$$

Proof of (d). Suppose, for a contradiction, that no real $r_0>0$ has the stated property. Then for every natural number $n$ the number $r_0=1/n$ fails, so we may choose $t_n\in[0,T]$ and $a_n\in\mathcal{A}$ with

$$\mathcal{H}_{t_n}(a_n)-\mathcal{H}_{t_n}(A_{t_n})<\tfrac{1}{n}\,|a_n-A_{t_n}|^2 .\qquad(\ast\ast)$$

In particular $a_n\neq A_{t_n}$, since for $a_n=A_{t_n}$ both sides of $(\ast\ast)$ vanish.

Extraction of a convergent subsequence. The interval $[0,T]$ is sequentially compact in the real line by sequential compactness of closed intervals, and $\mathcal{A}$ is sequentially compact in $\mathbb{R}^m$ with the Euclidean distance by (A) and the corollary that compact subsets of metric spaces are sequentially compact. By claim 2 of the product-metric lemma, $[0,T]\times\mathcal{A}$ is sequentially compact for the product metric, so the sequence $((t_n,a_n))_{n}$ has a subsequence $((t_{n_k},a_{n_k}))_{k}$ converging in the product metric to some $(t_\ast,a_\ast)\in[0,T]\times\mathcal{A}$ (the letters $t$ and $a$ of the preamble are not reused for this limit point), and by claim 1 of the same lemma $t_{n_k}\to t_\ast$ in $\mathbb{R}$ and $a_{n_k}\to a_\ast$ in $\mathbb{R}^m$. Consequently $(t_{n_k},a_{n_k})\to(t_\ast,a_\ast)$ in the Euclidean distance of $\mathbb{R}^{1+m}$, by claim 1 of the componentwise estimates, which bounds a Euclidean distance by the sum of the coordinate distances and each coordinate distance by the Euclidean distance. Since $A$ has continuous components, likewise $(t_{n_k},A_{t_{n_k}})\to(t_\ast,A_{t_\ast})$ in $\mathbb{R}^{1+m}$. A map continuous at a point in the Euclidean sense preserves limits of sequences converging to that point: given $\epsilon'>0$, choose $\eta'>0$ from the definition, then an index beyond which the sequence is within $\eta'$ of the point. Hence, by (a) and the continuity of $t\mapsto A_t$,

$$\mathcal{H}_{t_{n_k}}(a_{n_k})\to\mathcal{H}_{t_\ast}(a_\ast),\qquad \mathcal{H}_{t_{n_k}}(A_{t_{n_k}})\to\mathcal{H}_{t_\ast}(A_{t_\ast}),\qquad |a_{n_k}-A_{t_{n_k}}|\to|a_\ast-A_{t_\ast}| ,$$

the last by continuity of the norm. Moreover $\mathcal{A}$ is bounded by boundedness of compact subsets of Euclidean space: it lies in some ball of centre $c$ and radius $\varrho$, so $|a'-a''|\le|a'-c|+|c-a''|\le2\varrho=:D$ for all $a',a''\in\mathcal{A}$, and the right-hand side of $(\ast\ast)$ is at most $D^2/n\to0$.

Case 1: $a_\ast\neq A_{t_\ast}$. Passing to the limit along the subsequence in $(\ast\ast)$, non-strict inequalities being preserved under limits, gives $\mathcal{H}_{t_\ast}(a_\ast)-\mathcal{H}_{t_\ast}(A_{t_\ast})\le0$ with $a_\ast\in\mathcal{A}$ and $a_\ast\neq A_{t_\ast}$, contradicting (U).

Case 2: $a_\ast=A_{t_\ast}$. Then $h_k:=a_{n_k}-A_{t_{n_k}}$ satisfies $|h_k|\to0$, by the third displayed limit above together with $a_\ast=A_{t_\ast}$, and $h_k\neq0$ for every $k$. Fix a real $\varepsilon>0$ with $\tfrac{1}{2}(l+m)(1+C_P)\varepsilon\le r/2$, and let $\eta>0$ be furnished by (F2) for this $\varepsilon$. Choose $k$ so large that $|h_k|\le\eta$ and $1/n_k<r/2$. Every point $z$ of the segment from $x_{t_{n_k}}$ to $(S_{t_{n_k}},a_{n_k})$ satisfies $d(z,x_{t_{n_k}})\le|h_k|\le\eta$ and lies in $\Delta^l\times V$ and in $U_c\times\mathbb{R}^m$ by (F3), so (F2) shows that $\bar{\varepsilon}_L=\bar{\varepsilon}_b=\varepsilon$ are admissible in $(\ast)$ at $t=t_{n_k}$ and $a=a_{n_k}$. By $(\ast)$ and (H1),

$$\mathcal{H}_{t_{n_k}}(a_{n_k})-\mathcal{H}_{t_{n_k}}(A_{t_{n_k}})\ \ge\ h_k\cdot R_{t_{n_k}}h_k-\tfrac{1}{2}(l+m)(1+C_P)\varepsilon\,|h_k|^2\ \ge\ r|h_k|^2-\tfrac{r}{2}|h_k|^2=\tfrac{r}{2}|h_k|^2\ >\ \tfrac{1}{n_k}|h_k|^2 ,$$

the last inequality because $|h_k|^2>0$ and $1/n_k<r/2$. This contradicts $(\ast\ast)$ at $n=n_k$.

Both cases being contradictory, some real $r_0>0$ has the stated property, which proves (d). $\blacksquare$