Subtracting the <expected value> of $f$ does not alter either side, so suppose $\pi(f)=0$. Since reversibility makes $K$ a <self-adjoint operator> and stationarity makes it a contraction,
$$
\mathcal E_{K^2}(f)
=\lVert f\rVert_{L^2(\pi)}^2-\lVert Kf\rVert_{L^2(\pi)}^2.
$$
The <Discrete-time Poincaré inequality for a Markov kernel> is therefore equivalent to
$$
\lVert Kf\rVert_2^2
\leq\left(1-\frac1C\right)\lVert f\rVert_2^2.
$$
Applying this inequality successively to $f,Kf,\ldots,K^{t-1}f$ yields
$$
\operatorname{Var}_\pi(K^tf)
=\lVert K^tf\rVert_2^2
\leq\left(1-\frac1C\right)^t
\lVert f\rVert_2^2
=\left(1-\frac1C\right)^t\operatorname{Var}_\pi(f).
$$
Conversely, the asserted variance contraction with $t=1$ rearranges to the Poincaré inequality. Hence the two statements are equivalent.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 216 / 3 / d / Solution

{c}

For a reversible kernel $K$, the inequality $\operatorname{Var}_\pi(f)\leq C\mathcal E_{K^2}(f)$ is equivalent to geometric contraction of variance under iteration of $K$.


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 Incoming links: Discrete-time Poincaré inequality for a Markov kernel