Positive-matrix comparison identity

Put $u:=\mathrm{tr}\mathrm{\Psi }$. The inverse formula in [F1] makes each ${\mathrm{\Phi }}_i^{-1}$ smooth on $\bar{\mathrm{\Omega }}$, so $u$ is smooth there. Differentiating ${\mathrm{\Phi }}_i{\mathrm{\Phi }}_i^{-1}=I_n$ entrywise and using the product rule gives ${\mathrm{\partial }}_a{\mathrm{\Phi }}_i^{-1}=-{\mathrm{\Phi }}_i^{-1}({\mathrm{\partial }}_a{\mathrm{\Phi }}_i){\mathrm{\Phi }}_i^{-1}=-J_{i,a}{\mathrm{\Phi }}_i^{-1}$. Since ${\mathrm{\partial }}_a{\mathrm{\Phi }}_i={\mathrm{\Phi }}_i{J}_{i,a}$, it follows that ${\mathrm{\partial }}_a\mathrm{\Psi }={\mathrm{\Phi }}_1\mathrm{\Delta }{J}_a{\mathrm{\Phi }}_0^{-1}$. The two divergence equations imply ${\sum }_a{\mathrm{\partial }}_a\mathrm{\Delta }{J}_a=0$.

By [F2], each positive definite Hermitian ${\mathrm{\Phi }}_i$ has a positive Hermitian square root ${\mathrm{\Phi }}_i^{1\mathrm{/}2}$. Its eigenvalues are the positive square roots of the positive eigenvalues of ${\mathrm{\Phi }}_i$, so ${\mathrm{\Phi }}_i^{1\mathrm{/}2}$ is invertible; define ${\mathrm{\Phi }}_i^{-1\mathrm{/}2}:=({\mathrm{\Phi }}_i^{1\mathrm{/}2}{)}^{-1}$. Differentiate the identity in step 1.1, sum over $a$, take traces, and discard the term $\mathrm{tr}({\mathrm{\Phi }}_1({\sum }_a{\mathrm{\partial }}_a\mathrm{\Delta }{J}_a){\mathrm{\Phi }}_0^{-1})=0$. This gives $\mathrm{\Delta }u={\sum }_a\mathrm{tr}({\mathrm{\Phi }}_1(J_{1,a}\mathrm{\Delta }{J}_a-\mathrm{\Delta }{J}_a{J}_{0,a}){\mathrm{\Phi }}_0^{-1})$. Since ${\mathrm{\Phi }}_i^{†}={\mathrm{\Phi }}_i$ pointwise and derivatives are taken componentwise, $({\mathrm{\partial }}_a{\mathrm{\Phi }}_i{)}^{†}={\mathrm{\partial }}_a{\mathrm{\Phi }}_i$. Thus $J_{i,a}^{†}={\mathrm{\Phi }}_i{J}_{i,a}{\mathrm{\Phi }}_i^{-1}$. Substituting this relation and cyclically permuting factors inside the trace rewrites each summand as $\mathrm{tr}((\mathrm{\Delta }{J}_a{)}^{†}{\mathrm{\Phi }}_1\mathrm{\Delta }{J}_a{\mathrm{\Phi }}_0^{-1})$. With $M_a:={\mathrm{\Phi }}_1^{1\mathrm{/}2}\mathrm{\Delta }{J}_a{\mathrm{\Phi }}_0^{-1\mathrm{/}2}$, this trace equals $\mathrm{tr}(M_a^{†}{M}_a)={\sum }_{i,j}\mathrm{\mid }(M_a{)}_{ij}{\mathrm{\mid }}^2\ge 0$. Hence $\mathrm{\Delta }u\ge 0$.

The Hermitian matrix $C:={\mathrm{\Phi }}_0^{-1\mathrm{/}2}{\mathrm{\Phi }}_1{\mathrm{\Phi }}_0^{-1\mathrm{/}2}$ is positive definite: for $v\mathrm{\ne }0$, $v^{†}Cv=({\mathrm{\Phi }}_0^{-1\mathrm{/}2}v{)}^{†}{\mathrm{\Phi }}_1({\mathrm{\Phi }}_0^{-1\mathrm{/}2}v)>0$. Let ${\mu }_0,\dots ,{\mu }_{n-1}>0$ be the eigenvalues of ${\mathrm{\Phi }}_0$. In an orthonormal eigenbasis, determinant multiplicativity and $\det (U^{†}U)=\det I_n=1$ for the unitary change-of-basis matrix $U$ give $\det {\mathrm{\Phi }}_0={\prod }_j{\mu }_j=1$; hence $\det ({\mathrm{\Phi }}_0^{-1\mathrm{/}2})={\prod }_j{\mu }_j^{-1\mathrm{/}2}=1$. Therefore determinant multiplicativity gives $\det C=\det ({\mathrm{\Phi }}_0^{-1\mathrm{/}2}{)}^2\det {\mathrm{\Phi }}_1=1$. For the positive eigenvalues ${\lambda }_0,\dots ,{\lambda }_{n-1}$ of $C$, orthonormal diagonalization gives $\det C={\prod }_j{\lambda }_j$ and $\mathrm{tr}C={\sum }_j{\lambda }_j$. Cyclicity of trace gives $\mathrm{tr}C=\mathrm{tr}({\mathrm{\Phi }}_1{\mathrm{\Phi }}_0^{-1})$, and [F4] gives $1=({\prod }_j{\lambda }_j{)}^{1\mathrm{/}n}\le \frac{1}{n}{\sum }_j{\lambda }_j=\mathrm{tr}C\mathrm{/}n$, so $u=\mathrm{tr}C-n\ge 0$. Equality $u=0$ forces equality in AM-GM, hence all ${\lambda }_j=1$ and $C=I_n$; multiplying by ${\mathrm{\Phi }}_0^{1\mathrm{/}2}$ on both sides gives ${\mathrm{\Phi }}_1={\mathrm{\Phi }}_0$. Therefore $u=0$ holds exactly when ${\mathrm{\Phi }}_1={\mathrm{\Phi }}_0$.

Suppose ${\mathrm{\Phi }}_1={\mathrm{\Phi }}_0$ on $\mathrm{\partial }\mathrm{\Omega }$, so $u=0$ there. Since $\mathrm{\Omega }$ is nonempty and bounded, $\bar{\mathrm{\Omega }}$ is nonempty, closed and bounded, hence compact by [F6]; choose $R>0$ with $\bar{\mathrm{\Omega }}\subseteq \{x:\mathrm{\mid }x\mathrm{\mid }\le R\}$. For each $ϵ>0$, the continuous function $u_{ϵ}(x):=u(x)+ϵ\mathrm{\mid }x{\mathrm{\mid }}^2$ attains a maximum on $\bar{\mathrm{\Omega }}$ by [F6]. Since $\mathrm{\Delta }{u}_{ϵ}=\mathrm{\Delta }u+2dϵ>0$, this maximum cannot occur at an interior point: at any interior local maximum $x$, each symmetric second difference $[u_{ϵ}(x+h{e}_a)+u_{ϵ}(x-h{e}_a)-2u_{ϵ}(x)]\mathrm{/}{h}^2$ is nonpositive for sufficiently small $h$, so taking $h\to 0$ gives ${\mathrm{\partial }}_a^2{u}_{ϵ}(x)\le 0$ for every $a$ and hence $\mathrm{\Delta }{u}_{ϵ}(x)\le 0$, a contradiction. The maximum is therefore on the boundary, where $u_{ϵ}\le ϵR^2$. Thus $u(x)\le ϵR^2$ for every $x\in \mathrm{\Omega }$; letting $ϵ\downarrow 0$ gives $u\le 0$.

Step 3.1 gives $u\ge 0$, whereas step 3.2 gives $u\le 0$. Hence $u=0$ on $\mathrm{\Omega }$ and, by continuity, on $\bar{\mathrm{\Omega }}$. The equality case in step 3.1 then yields ${\mathrm{\Phi }}_1={\mathrm{\Phi }}_0$ throughout $\bar{\mathrm{\Omega }}$.