Nonpositive asymptotic-variance change for the observable-gradient coupling

Let $\pi$ be the invariant measure, let $f\in L_0^2(\pi)$, and let $\phi\in L_0^2(\pi)$ be the corresponding solution of the overdamped Poisson equation. Let $\alpha_f$ be the matrix-valued coupling function defined from the normalised gradients of $f$, and let $\Gamma_{\alpha}$ be the associated coupling operator. Observable-gradient conjecture.

$$\delta\sigma_F^2(\Gamma_{\alpha})=\int_{\mathbb{R}^d\times\mathbb{R}^d}\left(\nabla\phi(y)\cdot\alpha_f(x,y)\nabla\phi(x)\right)e^{-(V(x)+V(y))}\,\mathrm{d}x\,\mathrm{d}y\leq 0.$$

The claim predicts that using the observable gradient as a surrogate for the Poisson-solution gradient improves, or at least does not worsen, the asymptotic variance for small perturbations. It is motivated by an integration-by-parts identity and numerical experiments, but the source gives no proof.