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.



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


