 Solution

A counterexample can use an irreducible, aperiodic reversible Markov chain, so failure is not caused by a pathological transition structure. Take two binary coordinates, enumerate the four states as $00,01,10,11$, and let $\pi$ be their uniform distribution. Use the Markov kernelIts rows sum to one and it is symmetric, proving detailed balance with $\pi$. Every entry is positive. Let the univariate parametric family be $\tilde{\mathcal{Q}}=\{\mathrm{Bernoulli}(r):0\le r\le 2/5\}$.

For a product initial probability distribution with parameters $r_1,r_2$, additivity of the Kullback-Leibler divergence givesEach summand has derivative $\log (r_i/(1-r_i))<0$ on $(0,2/5]$, so the unique optimum at $t=0$ has $r_1=r_2=2/5$. ThusOn the other hand the allowed initial probability distribution ${\delta }_{00}$, obtained with $r_1=r_2=0$, is sent by one step to $\pi$. Its Kullback-Leibler divergence is zero, the smallest possible value. ConsequentlyThe transition reduces Kullback-Leibler divergence for each input, but may reduce it by different amounts for different inputs. It therefore need not preserve the ordering of competing initial probability distributions, which is exactly why Markov chain variational inference must optimize the initial parameters for its chosen number of steps.