Proof of The Fluctuation LQG Data of the Ising Equilibrium and Its Joint Coercivity

Throughout, $t\in[0,T]$ is fixed and we use $S_{t}=\bigl(\tfrac12,\tfrac12\bigr)$, $A_{t}=(1,1)$ and $P_{t}=(0,0)$, together with $\phi(1)=\phi'(1)=0$ from The Regularised Entropic Rate Cost §cost-function and $\phi''(1)=\varpi(1)=1$ from The Regularised Entropic Rate Cost §profile and The Regularised Entropic Rate Cost §cost-function, the value $\varpi(1)=1$ holding because $\underline{a}\le1\le\bar{a}$.

Claim 1. By definition $H_{ij}(t)=\partial_{j}\partial_{i}\bar{L}(S_{t},A_{t})-\sum_{\delta=1}^{2}P^{\delta}_{t}\,\partial_{j}\partial_{i}\bar{b}^{\delta}(S_{t},A_{t})$, and the second term vanishes because $P_{t}=0$; likewise $F_{\gamma\delta}=\partial_{\delta}\partial_{\gamma}\bar{G}(S_{T})=0$, all partial derivatives of the constant function $\bar{G}$ vanishing by claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §cost-extension. Substituting $x^{2}-x^{1}=0$, $x^{1}+x^{2}-1=0$, $x^{1}=x^{2}=\tfrac12$ and $a^{1}=a^{2}=1$ into the ten second-order partial derivatives of $\bar{L}$ listed in that same claim gives

These values are independent of $t$ and symmetric in $(i,j)$.

Claim 2. By claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §rate-extension the extended aggregate state drift is $\bar{b}^{1}(x,a)=x^{2}a^{2}-x^{1}a^{1}$ and $\bar{b}^{2}=-\bar{b}^{1}$ on $U\times V$. Differentiating the slice functions, which are affine in each coordinate, gives $\partial_{1}\bar{b}^{1}=-a^{1}$, $\partial_{2}\bar{b}^{1}=a^{2}$, $\partial_{3}\bar{b}^{1}=-x^{1}$, $\partial_{4}\bar{b}^{1}=x^{2}$, and the negatives of these for $\bar{b}^{2}$. At $(S_{t},A_{t})$ this yields

that is, $E_{t}=-vv^{\top}$ and $\mathsf{B}_{t}=-\tfrac12vv^{\top}$, since $vv^{\top}$ has entries $1,-1,-1,1$.

By claim 1 the coefficients $H_{ij}$ are symmetric, so $Q^{\gamma\delta}_{t}=\tfrac12H_{\gamma\delta}$, $V^{\gamma j}_{t}=H_{\gamma,2+j}$ and $R^{ij}_{t}=\tfrac12H_{2+i,2+j}$. Hence $V_{t}=0$ and $R_{t}=\tfrac12\cdot(2\chi)^{-1}I=(4\chi)^{-1}I$, while

Finally $\hat{F}=0$ because $F_{\gamma\delta}=0$.

Claim 3. The state matrix has entries $(\mathcal{E}_{t})_{\gamma\delta}=\partial_{\delta}\bar{b}^{\gamma}(S_{t},A_{t})$ and the control matrix $(\mathcal{B}_{t})_{\gamma j}=\partial_{2+j}\bar{b}^{\gamma}(S_{t},A_{t})$, the same numbers as in claim 2; so $\mathcal{E}_{t}=E_{t}=-vv^{\top}$ and $\mathcal{B}_{t}=\mathsf{B}_{t}=-\tfrac12vv^{\top}$.

The observation matrix has entries $(\tilde{\mathcal{E}}_{t})_{\upsilon\gamma}=\partial_{\gamma}\bar{\tilde{b}}^{\upsilon}(S_{t})$. By claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §observation-extension, $\bar{\tilde{b}}^{\upsilon}(x)=q\,x^{\upsilon}+q_{0}(x^{1}+x^{2})$, an affine function whose slice derivative in the coordinate $x^{\gamma}$ is $q\,\mathbf{1}_{\{\upsilon=\gamma\}}+q_{0}$; hence $\tilde{\mathcal{E}}_{t}=q\,I+q_{0}\,\mathsf{n}\mathsf{n}^{\top}$, the matrix $\mathsf{n}\mathsf{n}^{\top}$ having all entries equal to $1$.

The state noise covariance is $\Theta^{\star}_{t}=\Theta(S_{t},A_{t})$, which by claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §covariance equals $\bigl(\tfrac12\cdot1+\tfrac12\cdot1\bigr)vv^{\top}=vv^{\top}$. The observation noise covariance has entries $\mathbf{1}_{\{\upsilon=\upsilon'\}}\tilde{b}^{\upsilon}(S_{t})$, and $\tilde{b}^{\upsilon}(S_{t})=q\cdot\tfrac12+q_{0}$ by claim The Ising Population Model Instantiates the Data of the Fluctuation Theory §observations; so $\tilde{\Theta}^{\star}_{t}=\bigl(\tfrac{q}{2}+q_{0}\bigr)I$. The terminal matrix is $F^{\star}=0$ by claim 1, and the Hessian blocks are read off from claim 1: $H^{SS}_{t}$ has entries $H_{\gamma\delta}$, which is $2Q_{t}$ by claim 2; $H^{SA}_{t}$ and $H^{AS}_{t}$ have entries $H_{\gamma,2+j}$ and $H_{2+j,\gamma}$, all zero; and $H^{AA}_{t}$ has entries $H_{2+j,2+k}=(2\chi)^{-1}\mathbf{1}_{\{j=k\}}$.

Claim 4. Both $R_{t}=(4\chi)^{-1}I$ and $\tilde{\Theta}^{\star}_{t}=cI$ with $c=\tfrac{q}{2}+q_{0}>0$ are symmetric, and for $z\neq0$ we have $z\cdot R_{t}z=(4\chi)^{-1}|z|^{2}>0$ and $z\cdot\tilde{\Theta}^{\star}_{t}z=c|z|^{2}>0$; so both are positive definite and hence invertible by Invertibility of Symmetric Positive Definite Matrices. Their inverses are $4\chi I$ and $c^{-1}I$, since the products with $R_{t}$ and $\tilde{\Theta}^{\star}_{t}$ are $I$.

Write $N=\mathsf{n}\mathsf{n}^{\top}$. Then $N^{2}=\mathsf{n}(\mathsf{n}^{\top}\mathsf{n})\mathsf{n}^{\top}=2N$, because $\mathsf{n}^{\top}\mathsf{n}=2$. The matrix $\tilde{\mathcal{E}}_{t}=qI+q_{0}N$ is symmetric, so

which is the displayed formula. Since $\mathsf{n}^{\top}v=1\cdot1+1\cdot(-1)=0$, we get $N\,vv^{\top}=\mathsf{n}(\mathsf{n}^{\top}v)v^{\top}=0$ and likewise $vv^{\top}N=0$; hence $\tilde{D}_{t}vv^{\top}=c^{-1}q^{2}vv^{\top}=\tilde{d}\,vv^{\top}$ and $vv^{\top}\tilde{D}_{t}=\tilde{d}\,vv^{\top}$. Finally $\tilde{d}=q^{2}/c>0$ because $q>0$, and multiplying numerator and denominator by $2$ gives $\tilde{d}=2q^{2}/(q+2q_{0})$.

Claim 5. Using the values of claim 1 and the symmetry of $H$,

On the other hand, expanding the squares,

which proves the displayed identity.

Put $s=w^{1}-w^{2}$ and $n=w^{1}+w^{2}$, so that $s^{2}+n^{2}=2\bigl((w^{1})^{2}+(w^{2})^{2}\bigr)$. Then

so the whole expression is at least $c_{J}\bigl((w^{1})^{2}+(w^{2})^{2}+(w^{3})^{2}+(w^{4})^{2}\bigr)=c_{J}|w|^{2}$ with $c_{J}=\min\{\psi,2\mu,(4\chi)^{-1}\}$, a positive number because $\psi>0$, $\mu>0$ and $\chi>0$. Since $F_{\gamma\delta}=0$, the terminal quadratic form is identically $0$, hence nonnegative; the two displayed requirements of (JC) are therefore met.

For (H1), $a\cdot R_{t}a=(4\chi)^{-1}|a|^{2}$ for every $a\in\mathbb{R}^{2}$ and every $t$, so the hypothesis holds with $r=(4\chi)^{-1}$, and a fortiori with the smaller constant $r=c_{J}$.