Consider a continuous-time, irreducible Markov chain $X_t$ on a finite state space $E$. Assume the jump rates are $R(x,y)$ for $x,y\in E$, the generator is $L$, i.e for any function $f$ on E, $$Lf(x)=\sum_{y}R(x,y)[f(y)-f(x)].$$ Assume also the chain is reversible w.r.t. the measure $\pi$. Suppose the chain starts from an initial distribution $\mu$. Denote by $\mu P_t$ the distribution of chain at time $t$. Define the density $f_t=\frac{\mu P_t}{\pi}$. The corresponding entropy is given by $H(f_t)=\sum_{x\in E} f_t(x)\log f_t(x)\pi(x)$.
It is well know that $H(f_t)$ is decreasing in time $t$. More precisely, we have $$H(f_0)-H(f_t)=\int_0^t-\sum_xf_s(x)(L\log f_s)(x)\pi(x)ds.$$ Lets call the term inside the integral at the right hand side of the equation $E(f_t)$. Note that $E(f_t)$ is positive because it is bounded from below by $$D(\sqrt{f_s})=\sum_{x,y}R(x,y)[\sqrt{f_s(y)}-\sqrt{f_s(x)}]^2\pi(x),$$ by the inequality $a[\log b-\log a]\leq 2\sqrt{a}[\sqrt{b}-\sqrt{a}]$ for any $a,b\geq 0$.
$f_t$ eventually goes to the constant $1$, so $D(\sqrt{f_t})$ goes to $0$. My question is: is the true that $D(\sqrt{f_t})$ is decreasing along the time? What about $E(f_t)$?
