Proof of The Even-Split Equilibrium of the Ising Population Model is a Stationary Mean-Field Triple and is Optimal

Throughout write $S_{t}=\bigl(\tfrac12,\tfrac12\bigr)$, $A_{t}=(1,1)$ and $P_{t}=(0,0)$ for the equilibrium triple defined in the statement, and recall from The Regularised Entropic Rate Cost §sign-bounds that $\phi\ge0$ with $\phi(u)=0$ exactly for $u=1$, and from The Regularised Entropic Rate Cost §cost-function that $\phi(1)=0$ and $\phi'(1)=0$.

Claim 1. By claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §rates, $b(\Sigma,\alpha)=(\Sigma^{2}\alpha^{2}-\Sigma^{1}\alpha^{1})v$; at $(S_{t},A_{t})$ this is $(\tfrac12\cdot1-\tfrac12\cdot1)v=0$.

The map $S$ takes the value $\bigl(\tfrac12,\tfrac12\bigr)$, which lies in $\Delta^{2}$ because its entries are nonnegative and sum to $1$; the map $A$ takes the value $(1,1)$, which lies in $\mathcal{A}$ because $\underline{a}<1<\bar{a}$. Both are constant, hence have continuous components, which is the continuity clause of the trajectory-pair definition. For its dynamics clause, the map $s\mapsto b^{\gamma}(S_{s},A_{s})$ is constant with value $0$, hence continuous, and its Riemann integral over $[0,t]$ is $0$; therefore $S^{\gamma}_{0}+\int_{0}^{t}b^{\gamma}(S_{s},A_{s})\,ds=S^{\gamma}_{0}=S^{\gamma}_{t}$. So $(S,A)$ is a mean-field trajectory pair for $\beta$ with horizon $T$.

The same computation with the Lebesgue integral over the compact interval $[0,t]$, which is likewise $0$ for the constant integrand $0$, gives $S^{\gamma}_{t}=S^{\gamma}_{0}+\int_{[0,t]}b^{\gamma}(S_{s},A_{s})\,ds$. Together with $S$ taking values in $\Delta^{2}$ and $S_{0}=\bigl(\tfrac12,\tfrac12\bigr)$, these are exactly the three requirements of the standing hypothesis on $S^{*}$ in The Block Cascade of Anchored Good-Set Clocks: Adapted Good Sets, Matched Escape Bounds, and the Energy Ledger, read with $S^{*}=S$ and $x_{0}=S_{0}$.

Claim 2. At $(S_{t},A_{t})$ we have $x^{2}-x^{1}=0$, $x^{1}+x^{2}-1=0$ and $a^{1}=a^{2}=1$. Substituting into the formulas of claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §cost-extension gives

We verify the three clauses of the definition of a stationary co-state for the constant map $P$. Continuity: each component is constant, hence continuous. Co-state equation: the integrand is

since every $P^{\delta}_{s}$ vanishes and the first-order partial derivatives of $\bar{L}$ vanish at $(S_{s},A_{s})$ by the previous paragraph; it is therefore continuous, its Riemann integral over $[t,T]$ is $0$, and $-\partial_{\gamma}\bar{G}(S_{T})=0$ because $\bar{G}$ is constant by claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §cost-extension. Hence the right-hand side of the co-state equation is $0=P^{\gamma}_{t}$. Stationarity: for $j\in\{1,2\}$ the left-hand side is $\partial_{2+j}\bar{L}(S_{t},A_{t})=0$ and the right-hand side is $\sum_{\delta}\partial_{2+j}\bar{b}^{\delta}(S_{t},A_{t})P^{\delta}_{t}=0$. So $P$ is a stationary co-state and $(S,A,P)$ is a stationary mean-field triple. Finally $\sum_{\delta}|P^{\delta}_{t}|=0\le C_{P}$ with $C_{P}=0$.

Claim 3. By definition the mean-field Hamiltonian along the triple is $\mathcal{H}_{t}(a)=\bar{L}(S_{t},a)-\sum_{\delta}P^{\delta}_{t}\bar{b}^{\delta}(S_{t},a)$, and the second term vanishes because $P_{t}=0$. Substituting $x=S_{t}$ into the formula for $\bar{L}$ of clause The Ising Population Data §cost, and using $x^{2}-x^{1}=0$ and $x^{1}+x^{2}-1=0$ there, gives $\mathcal{H}_{t}(a)=\chi^{-1}\bigl(\tfrac12\phi(a^{1})+\tfrac12\phi(a^{2})\bigr)$, the displayed formula.

Since $\phi\ge0$, we have $\mathcal{H}_{t}(a)\ge0$ for every $a$, and $\mathcal{H}_{t}(A_{t})=\chi^{-1}\phi(1)=0$. If $a\in\mathcal{A}$ satisfies $\mathcal{H}_{t}(a)=0$ then $\phi(a^{1})=\phi(a^{2})=0$, both summands being nonnegative, hence $a^{1}=a^{2}=1$ and $a=A_{t}$. Therefore $\mathcal{H}_{t}(a)>0=\mathcal{H}_{t}(A_{t})$ for every $a\in\mathcal{A}$ with $a\neq A_{t}$, which is hypothesis (U).

Claim 4. Fix a real $\theta>0$ and let $x\in\Delta^{2}$ and $\xi\in\mathcal{U}_{\mathcal{A}}$, with an admissible representative also written $\xi$, so that $\xi(t)\in\mathcal{A}$ for every $t\in[0,\theta]$. By the definition of the mean-field cost of a control from an initial state, $F^{[\theta]}(x,\xi)$ is the generalized mean-field cost of the pair $(S(x,\xi),\xi)$, namely $\int_{[0,\theta]}L\bigl(S_{t}(x,\xi),\xi(t)\bigr)\,dt+G\bigl(S_{\theta}(x,\xi)\bigr)$. The flow takes values in $\Delta^{2}$, so the integrand is nonnegative by claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §cost-data, and $G=0$; by monotonicity of the integral, Linearity and Monotonicity of the Lebesgue Integral, we get $F^{[\theta]}(x,\xi)\ge0$. Hence $0$ is a lower bound of the set whose infimum is $J^{*[\theta]}_{x}$, and $J^{*[\theta]}_{x}\ge0$.

Now let $x=\bigl(\tfrac12,\tfrac12\bigr)$ and let $\xi^{\circ}$ be the constant map on $[0,\theta]$ with value $(1,1)$. It is square integrable and takes values in $\mathcal{A}$, so its class lies in $\mathcal{U}_{\mathcal{A}}$, and it is an admissible representative of that class. The constant map $t\mapsto\bigl(\tfrac12,\tfrac12\bigr)$ on $[0,\theta]$, paired with $\xi^{\circ}$, is a generalized mean-field trajectory pair with value $x$ at $t=0$: its components are constant, hence absolutely continuous with vanishing derivative, and the drift $b\bigl((\tfrac12,\tfrac12),(1,1)\bigr)$ vanishes by claim 1, so the integral equation holds exactly as in claim 1. By the uniqueness in claim 1 of Existence and Uniqueness of the Generalized Mean-Field Trajectory for a Measurable Control, this constant map is the mean-field flow $S(x,\xi^{\circ})$. Consequently

since $L\bigl((\tfrac12,\tfrac12),(1,1)\bigr)=0$ by claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §cost-data. So the infimum $J^{*[\theta]}_{x}$, which is at least $0$ and at most $F^{[\theta]}(x,\xi^{\circ})=0$, equals $0$, and the class of $\xi^{\circ}$ lies in $\mathcal{M}^{*[\theta]}_{x}$ by the definition of that set. Taking $\theta=T$ and noting that $A$ is exactly the constant map with value $(1,1)$ on $[0,T]$ and that $S_{0}=\bigl(\tfrac12,\tfrac12\bigr)$ gives $[A]\in\mathcal{M}^{*}_{S_{0}}$.

Claim 5. Let $t_{0}\in[0,T)$, put $T^{\sharp}=T-t_{0}>0$, and let $x\in\Delta^{2}$ with $|x-S_{t_{0}}|\le1$. Since $S_{t_{0}}=\bigl(\tfrac12,\tfrac12\bigr)$, claim 4 applied with $\theta=T^{\sharp}$ gives $J^{*[T^{\sharp}]}_{S_{t_{0}}}=0$ and $J^{*[T^{\sharp}]}_{x}\ge0$. As $P_{t_{0}}=0$, the right-hand side of the inequality of (TG) with $C_{tg}=0$ equals $J^{*[T^{\sharp}]}_{S_{t_{0}}}-0-0=0$, and the inequality reads $J^{*[T^{\sharp}]}_{x}\ge0$, which holds. So (TG) holds with $\varepsilon_{tg}=1$ and $C_{tg}=0$.