Reweighting probability measures by convex potentials, and contraction in transport distance

Question

Let $W: \mathbf{R}^d \to \mathbf{R}$ be a convex function such that $\int \exp(-W) = 1$, and define probability measures $\mu_y$ by

$$\mu_y (dx) = \exp( - W (x - y)) \,dx,$$

i.e. each $\mu_y$ is a translation of the measure $\mu_0$ in the direction $y$.

Now, let $V: \mathbf{R}^d \to \mathbf{R}$ be another convex function which is uniformly quadratically convex with parameter $m > 0$, i.e. $V''(x) \succeq m\cdot I_d$ for all $x$.

Define reweighted measures $\nu_y$ by

\begin{align} \nu_y (dx) &= \exp( - W (x - y) - V(x) + F(y)) \,dx \\ &= \mu_y (dx) \cdot \exp( - V(x) + F(y)), \end{align}

with $F(y)$ chosen so that $\nu_y$ integrates to $1$.

For some specific choices of $W$, it is possible to show that this reweighting operation is a contraction, in the sense that

\begin{align} d ( \nu_{y_1}, \nu_{y_2}) &\leqslant \kappa_{V, W} \cdot d ( \mu_{y_1}, \mu_{y_2}) \\ &= \kappa_{V, W} \cdot | y_2 - y_1 | \end{align}

with $\kappa_{V, W} < 1$, and $d$ some transport distance.

For a bit of intuition, one can imagine that $\kappa_{V, W}$ gets smaller as the strength of the reweighting operation grows, e.g. as $m$ increases. I am not making a rigorous claim to this effect.

My question is: is there a general result which would guarantee that, given a specific $(V, W)$, there exists a $\kappa_{V, W} < 1$ such that the earlier estimate holds? In the best case, I would also hope for quantitative estimates of $\kappa_{V, W}$.

I could believe that one might need to make further assumptions on $W$ as well (e.g. uniform convexity, smoothness), but I would ideally like to avoid this.

Answer 1

There is a simple sufficient condition: If $\nabla W$ is $L$-Lipschitz, then $y \mapsto \nu_y$ is $(L/m)$-Lipschitz with respect to the quadratic Wasserstein distance $d=\mathcal{W}_2$. You thus have a contraction if $L<m$, and this condition fits with your "bit of intuition." Though this may be a stronger assumption than you are willing to impose on $W$.

Proof: Identifying $\nu_y$ with its density, the convexity of $W$ and $m$-convexity of $V$ ensure that $(-\log\nu_y)$ is $m$-convex, for each $y$. By the Bakry-Emery criterion, $\nu_y$ satisfies the log-Sobolev inequality $$H(\mu\,|\,\nu_y) \le \frac{1}{2m}I(\mu\,|\,\nu_y),$$ for every probability measure $\mu$ on $\mathbb{R}^d$. Here $H(\mu\,|\,\nu_y) = \int \log \tfrac{d\mu}{d\nu_y}\,d\mu$ denotes the relative entropy (KL divergence) and $I(\mu\,|\,\nu_y) = \int |\nabla \log \tfrac{d\mu}{d\nu_y}|^2\,d\mu$ the relative Fisher information. By Otto-Villani, we also have the quadratic transport inequality $$\mathcal{W}_2^2(\mu,\nu_y) \le \frac{2}{m}H(\mu\,|\,\nu_y),$$ for all $\mu$. Combine these two inequalities to get $$\mathcal{W}_2^2(\mu,\nu_y) \le \frac{1}{m^2}I(\mu\,|\,\nu_y),$$ for all $\mu$. For any $y_1,y_2$, we thus find \begin{align*} \mathcal{W}_2^2(\nu_{y_1},\nu_{y_2}) &\le \frac{1}{m^2}I(\nu_{y_1}\,|\,\nu_{y_2}) \\ &= \frac{1}{m^2} \int |\nabla W(x-y_1) - \nabla W(x-y_2)|^2\,\nu_{y_1}(dx) \\ &\le \frac{L^2}{m^2} |y_1-y_2|^2. \end{align*}