 Solution

Let $A=\mathrm{Leb}(\mathcal{C})>0$ and let $\mu$ be the uniform distribution on the body. The hit-and-run kernel density lower bound impliesFor the last range, $\mathcal{C}$ lies in a radius-$D$ disc about any of its points, so $A\le \pi D^2$ and $\epsilon \le 1/2$. This global Doeblin condition proves irreducibility with respect to $\mu$ and aperiodicity. The entire state space is a small set; $V=1$, $\lambda =1/2$, and $b=1/2$ satisfy the geometric drift condition.

In fact the conclusion is stronger. Write $P=\epsilon \Pi +(1-\epsilon )R$, where $\Pi (x,\cdot )=\mu$ and $R$ is the residual Markov kernel. Since $\mu P=\mu$, also $\mu R=\mu$. Each transition has a probability $\epsilon$ of resetting to stationarity, and after such a reset every subsequent distribution remains stationary. Equivalently,Thus the algorithm has uniform geometric ergodicity, using the full-dimensional and boundary conventions stated in part (a).