Modulus of of continuity of a convolution operator with respect to Wasserstein metric

Question

For a (discrete) measure $G$ on some reasonable metric space $\Theta$, consider the map $G \mapsto f_G$ defined as $$ f_G := f*G(dx) := \int f(dx|\theta) G(d\theta) $$ for some nice kernel function $f(dx|\theta)$ on $X\times \Theta$ where $X$ is some other reasonable metric space. We think of $f_G$ as a measure on $X$, which is a smoothed version of $G$. For simplicity one can think of $X = \mathbb{R}^d$ and $f(dx|\theta) = f(x|\theta) dx$ for some density $f(x|\theta)$ with respect to Lebesgue measure. We can even restrict to the case $f(x|\theta) = f(x-\theta)$ assuming $X =\Theta = \mathbb{R}^d$.

Let us equip both the spaces of measures on $\Theta$ and $X$ with some Wasserstein distance, say $W_2$. The question is how to obtain a bound of the form $$ \alpha(W_2(G,G')) \le W_2(f_G,f_{G'}), \quad \forall G,G' $$ for some $\alpha : [0,\infty) \to [0,\infty)$. In other words, we are interested in the modulus of continuity of the "inverse" of $G \mapsto f_{G}$. One can even consider the simplest case where $G = \delta_{\theta}$ and $G' = \delta_{\theta'}$ so that $W_2(G,G') = d_{\Theta}(\theta,\theta')$.

Answer 1

It is intuitively clear that for the simplest case $G = \delta_\theta$ and $G' = \delta_{\theta'}$ in $\mathbb{R}^n$ we have $W_2(G,G') = W_2(f_G,f_{G'})$ for all $f$. The less trivial direction can be seen for instance by considering the orthogonal projection $p$ of a transport $\sigma$ between $f_G$ and $f_{G'}$ to the line $l$ passing through $\theta$ and $\theta'$. Since the orthogonal projection does not increase distances and since on the line the monotone rearrangement is the optimal way to tranport, we get $$ W_2^2(f_G,f_{G'}) = \inf_\sigma\int|x-y|^2\,d\sigma(x,y) \ge \inf_{\sigma}\int|p(x)-p(y)|^2\,d\sigma(x,y) \ge W_2^2(G,{G'}). $$

However, in general it is harder to give useful bounds:

Example Let $X = \Theta = \mathbb{R}$ and $f(x) = \frac12\chi_{[-1,1]}(x)$. Take $n \in \mathbb{N}$ and define $$ G = \sum_{i=1}^n \frac{1}{n}\delta_{2i} $$ and $$ G' = \frac{1}{2n}\delta_{1} + \frac{1}{2n}\delta_{2n+1} + \sum_{i=1}^{n-1} \frac{1}{n}\delta_{2i+1}. $$

Then $W_2(G,G') = 1$, but by looking at the monotone rearrangement for the diffused measures $f_G$ and $f_{G'}$ we see that $$ W_2(f_G,f_{G'}) = \sqrt{\frac1n\int_0^1t^2\,dt} = \frac{1}{\sqrt{3n}}. $$ Thus the modulus $\alpha$ has the constraint $\alpha(1) = 0$ and consequently $\alpha(t) = 0$ for all $t \in (0,1]$.