Square-root total-variation bound for $L^2$ test functions

A square-root total-variation bound for square-integrable test functions. For probability measures $\mu,\nu$ and a measurable $h$ with $h^2$ integrable against both,

where $\|\mu-\nu\| = \sup_A|\mu(A)-\nu(A)|$.

Why the square root, and why this is the right inequality. For a bounded $h$ the elementary bound $|\int h\,d\mu - \int h\,d\nu| \le 2\|h\|_\infty\|\mu-\nu\|$ is linear in the total variation, but it is useless when $h$ is merely square-integrable. Replacing $\|h\|_\infty$ by $\|h\|_{L^2}$ costs a square root — and that is exactly the trade one needs, because $\sqrt{\|\mu-\nu\|}$ still decays geometrically when $\|\mu-\nu\|$ does.

The application: $L^2$ contraction for uniformly ergodic chains. Let $P$ be uniformly ergodic with invariant law $\pi$, so $\sup_x\|P^N(x,\cdot)-\pi\| \le \rho$ with $\rho$ as small as we like. For $h\in L^2(\pi)$ with $\int h\,d\pi = 0$, applying the inequality with $\mu = P^N(x,\cdot)$ and $\nu=\pi$ and then squaring and integrating in $x$ against $\pi$ — using the invariance $\int\!\!\int h^2\,dP^N(x,\cdot)\,d\pi(x) = \int h^2 d\pi$ — gives

Choosing $N$ with $\rho < 1/4$ makes $P^N$ a strict contraction on the mean-zero subspace of $L^2(\pi)$, which is what makes the Neumann series $\hat g = \sum_{n\ge0}P^n(f-\pi f)$ converge and solves the Poisson equation $\hat g - P\hat g = f - \pi f$ for every square-integrable $f$. Without this step the martingale approximation underlying the Markov chain CLT is available only for bounded $f$.

Proof. Let $A$ be a Hahn set for $\mu-\nu$, so $\nu \le \mu$ on subsets of $A$ and $\mu \le \nu$ on subsets of $A^c$. Then $\rho_1 := (\mu-\nu)|_A$ and $\rho_2 := (\nu-\mu)|_{A^c}$ are positive measures with

and $\rho_1 \le \mu$, $\rho_2 \le \nu$, with masses $\rho_1(X) = \mu(A)-\nu(A) \le \|\mu-\nu\|$ and $\rho_2(X) = \nu(A^c)-\mu(A^c) \le \|\mu-\nu\|$. Cauchy–Schwarz against a finite measure $\rho$ — obtained from the nonnegativity of $\int(|h|-\lambda)^2 d\rho$ at $\lambda = \int|h|\,d\rho\,/\,\rho(X)$ — gives $|\int h\,d\rho| \le \sqrt{\rho(X)}\,\sqrt{\int h^2 d\rho}$. Applying this to $\rho_1$ and $\rho_2$, and using monotonicity of the second moment in the measure, yields the two terms of the bound.