A sticky general-state chain can converge without a geometric rate

For any $A$ with $\lambda (A)>0$, $P(x,A)\ge \sqrt{x}\lambda (A)>0$, so [F1] holds. For $x<1$, the kernel has a positive self-loop of mass $1-\sqrt{x}$; since $\pi (\{1\})=0$, a cyclic decomposition with positive $\pi$-mass classes would contain a $\pi$-positive set of states with positive self-loop probability, contradicting deterministic movement to a disjoint next class. Thus the chain is aperiodic.

To realize refreshes, use the marked state space $\mathsf{X}\times \{0,1\}$ and the probability kernel $K((x,b),A\times \{1\})=\sqrt{x}\lambda (A)$, $K((x,b),A\times \{0\})=(1-\sqrt{x})1_A(x)$. Every measurable set is a disjoint union of these two slice forms, so [F7] verifies countable additivity and measurable evaluations. Its first-coordinate transition is $P$. Apply [F8] with initial state $(x,0)$. Direct multiplication of the transition probabilities gives probability $(1-\sqrt{x}{)}^{t-1}\sqrt{x}\lambda (A)$ that the first marked refresh occurs at time $t\ge 1$ with new state in $A$. Summing over $t$ shows that the waiting time is finite almost surely and the new state has law $\lambda$. The same finite-history calculation, iterated for any finite number of marked refreshes, gives independent $\lambda$ refresh states and almost-surely finite waiting times. A countable intersection then gives infinitely many refreshes almost surely. For $\lambda (A)>0$, the probability that the first $N$ refresh draws all miss $A$ is $(1-\lambda (A){)}^N\to 0$. Hence every positive-$\lambda$ set is hit almost surely from every start, proving [F3].

The one-step law under $\pi$ has a holding part with density $(1-\sqrt{x})\mathrm{/}(2\sqrt{x})$ and a refresh part of total mass $\int \sqrt{x}\pi (dx)=1\mathrm{/}2$, distributed as $\lambda$. For every measurable $A$, $\int P(x,A)\pi (dx)={\int }_A\frac{1-\sqrt{x}}{2\sqrt{x}}dx+\frac{1}{2}\lambda (A)={\int }_A\frac{dx}{2\sqrt{x}}=\pi (A)\mathrm{.}$ Thus $\pi$ is invariant. By [F6], convergence holds from every state.

Fix any $r\in (0,1)$ and set $B_r=(0,(1-r{)}^2)$. Since $\pi (B_r)=1-r>0$, this set meets every full-$\pi$-measure set on which a proposed geometric bound with rate $r$ holds and its multiplier is finite. For each $x\in B_r$, a refresh draw has no atom at $x$, so $P^n(x,\{x\})=(1-\sqrt{x}{)}^n$, while $\pi (\{x\})=0$. Hence $\mathrm{\parallel }{P}^n(x,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}\ge (1-\sqrt{x}{)}^n$. Choose $x$ in the intersection just described. Then $1-\sqrt{x}>r$, so no finite $M(x)$ can satisfy $(1-\sqrt{x}{)}^n\le M(x)r^n$ for every $n\ge 1$. As this holds for every $r\in (0,1)$, the chain is not geometrically ergodic, despite convergence from every start.