Ibragimov's covariance inequality: $|\\operatorname{cov}(U,V)| \\le 2\\sqrt{\\varphi(n)}\\,\\sigma_U \\sigma_V$

Let $(Y_i)_{i \ge 0}$ be a sequence of random variables on a probability space $(\Omega, \mathcal F, P)$, let $k, n \ge 0$, and let

be the corresponding past and future $\sigma$-algebras. Let $\varphi(n)$ denote the uniform (φ-) mixing coefficient of the sequence at lag $n$,

Then for all square-integrable $U$ measurable with respect to $\mathcal A$ and $V$ measurable with respect to $\mathcal B$,

This is Ibragimov's covariance inequality in the case $p = q = 2$: it is the pairwise form of the Cogburn–Ibragimov comparison $\rho(n) \le 2\sqrt{\varphi(n)}$ between the maximal-correlation and uniform-mixing coefficients, and taking the supremum over all admissible pairs $(U,V)$ turns this inequality into that comparison.

The classical argument interpolates between two endpoint estimates for the centered conditional-expectation operator $S g = E[g \mid \mathcal A] - E g$ acting on $\mathcal B$-measurable variables, namely $\|S\|_{L^\infty \to L^\infty} \le 2\varphi(\mathcal A, \mathcal B)$ and $\|S\|_{L^1 \to L^1} \le 2\varphi(\mathcal B, \mathcal A) \le 2$; the geometric mean of the two bounds gives $\|S\|_{L^2 \to L^2} \le 2\sqrt{\varphi(\mathcal A, \mathcal B)}$, and Cauchy–Schwarz then yields the displayed covariance bound. An interpolation-free route to the same $L^2$ bound uses that $T = S^{*}S$ is positive and self-adjoint, so that $\langle T g, g\rangle^{2^m} \le \langle T^{2^m} g, g\rangle$ by iterated Cauchy–Schwarz, while $\|T^{2^m} g\|_2^2 \le \|T^{2^m} g\|_\infty \|T^{2^m} g\|_1$ controls the right-hand side by the endpoint norms; letting $m \to \infty$ gives $\|T\| \le 4\varphi$.