Geometric $L^2$ decay of the transition operator on mean-zero functions

Geometric decay of the transition operator on mean-zero $L^2$ functions. If the $N$-step kernel is uniformly close to $\pi$ in total variation — precisely, $\sup_x\|P^N(x,\cdot)-\pi\| \le \rho$ with $4\rho \le 1/4$ — then for every square-integrable $h$ with $\mathbb E_\pi h = 0$ and every $j$,

What this gives. Uniform ergodicity supplies such an $N$ (take $Rt^N$ small), so $\|P^n h\|_{L^2(\pi)} \le 2^{-\lfloor n/N\rfloor}\|h\|_{L^2(\pi)}$: the transition operator decays geometrically on the mean-zero subspace of $L^2(\pi)$, with a rate depending only on the chain. Two consequences drive the Markov chain central limit theorem for square-integrable observables:

• Summable covariances. Under the stationary chain, $\mathbb E[h(X_0)h(X_n)] = \int h\,(P^nh)\,d\pi$, so by Cauchy–Schwarz $|\mathbb E[h(X_0)h(X_n)]| \le \|h\|_2\|P^nh\|_2$, and $\sum_n \|P^nh\|_2 \le 2N\|h\|_2 < \infty$. Hence $\operatorname{Var}(\sum_{k<n}h(X_k)) = O(n\|h\|_2^2)$ with a constant depending only on $P$ — the estimate that controls the truncation error when passing from bounded to $L^2$ observables.
• The Poisson equation in $L^2$. The Neumann series $\sum_{n\ge0}P^nh$ converges in $L^2(\pi)$ and solves $\hat h - P\hat h = h$.

Proof. Induction on $j$, with the function generalized so that the induction hypothesis can be applied to $P^Nh$ rather than to $h$. Three ingredients recur:

1. Invariance for all powers: $P^n\circ\pi = \pi$, by induction from $P\circ\pi=\pi$.
2. Jensen: $(P^m u)^2 \le P^m(u^2)$ pointwise almost everywhere, from the nonnegativity of $\int (u - P^mu(x))^2\,dP^m(x,\cdot)$. This gives both integrability of $(P^mu)^2$ and, with invariance, that $P^mu$ still has $\pi$-mean zero — so the hypotheses of the contraction estimate are preserved along the induction.
3. Semigroup: $P^{(j+1)N} = P^N\circ P^{jN}$, so $P^{(j+1)N}u = P^{jN}(P^Nu)$ almost everywhere (Fubini for a composed kernel, applied at those $x$ where $u$ is integrable — a full-measure set by invariance).

Then $\|P^{(j+1)N}u\|_2^2 = \|P^{jN}(P^Nu)\|_2^2 \le 4^{-j}\|P^Nu\|_2^2 \le 4^{-j}\cdot\tfrac14\|u\|_2^2$, the last step being one application of the contraction estimate.