Let $(X, d, m)$ be an ergodic random walk on a metric space, with invariant distribution $\nu$. Suppose that the coarse Ricci curvature of $X$ is at least $\kappa > 0$ and that the average diffusion constant $\sigma$ verifies $\sigma < \infty$. Suppose that $\nu$ is reversible. For $f \in L^2(X, \nu)$, let the averaging operator $M$ be $$ Mf(x) := \int_X f(y) \, \text{d}m_x(y). $$ I am trying to prove that $$ \operatorname{Var}_\nu(Mf) \leq (1 - \kappa)^2 \operatorname{Var}_\nu(f), $$ inequality used in the proof of corollary 31 of Yann Ollivier's paper Ricci curvature of Markov chains on metric spaces.
I know that $$ \operatorname{Var}_\nu (f) = \int_X \operatorname{Var}_{m_x} (f) \, \text{d}\nu(x) + \operatorname{Var}_\nu(Mf), $$ and since the coarse Ricci curvature of $X$ is $\kappa$, for every $k$-Lipschitz function $f : X \mapsto R$, the function $Mf$ is $k(1 − \kappa)$-Lipschitz. Moreover, Lipschitz functions are dense in $L^2(X, \nu)$. Finally, I know that the spectral radius of the averaging operator $M$ acting on $L^2(X, \nu)/\{ \text{const} \}$ is at most $1 - \kappa$. However, I am still unsure how to justify that $$ \operatorname{Var}_\nu(Mf) \leq (1 - \kappa)^2 \operatorname{Var}_\nu(f). $$