Stationary finite quantum semigroups decrease relative entropy

On $M_d(\mathbb{C})$ use the max-entry norm $\mathrm{\parallel }X{\mathrm{\parallel }}_{\max }:={\max }_{i,j}\mathrm{\mid }{X}_{ij}\mathrm{\mid }$. The space is finite-dimensional and Banach by [F3]. In coordinates, $(\mathcal{L}X{)}_{ij}={\sum }_{k,l}{a}_{ij,kl}{X}_{kl}$ for finitely many coefficients, so $\mathrm{\parallel }\mathcal{L}X{\mathrm{\parallel }}_{\max }\le C\mathrm{\parallel }X{\mathrm{\parallel }}_{\max }$ with $C:={\max }_{i,j}{\sum }_{k,l}\mathrm{\mid }{a}_{ij,kl}\mathrm{\mid }<\mathrm{\infty }$; thus $\mathcal{L}$ is bounded. Hence [F4] gives operator-norm convergence of the exponential series, its compact-uniform convergence, and the law $T_u{T}_v=T_{u+v}$. For $0\le s\le t$, put $u=t-s$ and $v=s$. The map $T_{t-s}$ is a finite channel by the hypothesis and [F1], and fixes $\sigma$ by stationarity. Apply [A1] to $X=T_s\rho$ and $Y=\sigma$; these are positive semidefinite, so $D(T_t\rho \mathrm{\parallel }\sigma )\le D(T_s\rho \mathrm{\parallel }\sigma )$. This proves global monotonicity, including singular initial densities, because $\sigma >0$ makes every displayed relative entropy finite.

For $x>0$ and $R>0$, direct integration gives ${\int }_0^R((1+s{)}^{-1}-(x+s{)}^{-1})ds=\log x+\log (1+R)-\log (x+R)$, which tends to $\log x$ as $R\to \mathrm{\infty }$. Applying this scalar identity to each eigenvalue using [F2] yields $\log A={\int }_0^{\mathrm{\infty }}((1+s{)}^{-1}I-(A+sI{)}^{-1})ds$ for positive definite $A$. Let $u\mapsto A(u)$ be a $C^1$ path of positive-definite matrices on an open interval, or on a one-sided interval at an endpoint, around the parameter under consideration, and write $X=\dot{A}(u)$. Differentiating $(A+sI)(A+sI{)}^{-1}=I$ gives $\frac{d}{du}(A+sI{)}^{-1}=-(A+sI{)}^{-1}X(A+sI{)}^{-1}$, so $\frac{d}{du}\log A={\int }_0^{\mathrm{\infty }}(A+sI{)}^{-1}X(A+sI{)}^{-1}ds$. At a fixed parameter $u_0$, let ${\lambda }_0>0$ be the least eigenvalue of $A(u_0)$. By continuity, on a neighborhood $\mathrm{\parallel }A(u)-A(u_0)\mathrm{\parallel }<{\lambda }_0\mathrm{/}2$, so $A(u)\ge ({\lambda }_0\mathrm{/}2)I$ there; since $A^{\mathrm{\prime }}$ is locally bounded, the derivative integrand is uniformly bounded by $C({\lambda }_0\mathrm{/}2+s{)}^{-2}$ nearby for some finite $C$, an integrable function. The product rule gives $\frac{d}{du}\mathrm{Tr}(A\log A)=\mathrm{Tr}(X\log A)+{\int }_0^{\mathrm{\infty }}\mathrm{Tr}(A(A+sI{)}^{-1}X(A+sI{)}^{-1})ds$. Diagonalize $A$ with eigenvalues ${\lambda }_j>0$; the integral is ${\sum }_j{X}_{jj}{\int }_0^{\mathrm{\infty }}{\lambda }_j({\lambda }_j+s{)}^{-2}ds={\sum }_j{X}_{jj}=\mathrm{Tr}X$. Hence $\frac{d}{du}\mathrm{Tr}(A\log A)=\mathrm{Tr}(X(\log A+I))$.

For $f(t):=D(T_t\rho \mathrm{\parallel }\sigma )$, every quotient $(f(t+h)-f(t))\mathrm{/}h$ with $h>0$ is nonpositive by step 1.1. Thus $D^{+}f(t):={\mathrm{lim\hspace{0.17em}sup}}_{h\downarrow 0}(f(t+h)-f(t))\mathrm{/}h\le 0$ as an extended-real upper right Dini derivative, without inserting $\log {\rho }_t$ when ${\rho }_t$ is singular.

If ${\rho }_t>0$, continuity gives a positive-definite neighborhood (right-sided at $t=0$). By [F4], $T_t$ is differentiable in operator norm and ${\dot{\rho }}_t=\mathcal{L}({\rho }_t)$. Trace preservation implies $\mathrm{Tr}{\dot{\rho }}_t=0$, by differentiating $\mathrm{Tr}(T_t\rho )=\mathrm{Tr}\rho$. Apply step 1.2 to $A={\rho }_t$ and differentiate the linear term $-\mathrm{Tr}({\rho }_t\log \sigma )$ to obtain the displayed derivative formula. Its sign is nonpositive because the same function is nonincreasing by step 1.1.