Suppose that $\mu\_1$ and $\mu\_2$ are two distributions defined on $\mathbb{R}^n$ and $\gamma$ is a symmetric distribution (around $0$) on $\mathbb{R}^n$ with compact support. Let $\gamma\_x$ denote the resulting distribution by translating the centre of $\gamma$ from $0$ to $x$, $d\_{TV}(\cdot,\cdot)$ denote the total variation distance and $d\_W(\cdot,\cdot)$ 1-Wasserstein distance. 


**Question:** Does it hold that
$$
d\_{TV}(\mu\_1\ast\gamma,\mu\_2\ast\gamma) \leq \left(\sup\_{x\neq y} \frac{d\_{TV}(\gamma\_x,\gamma\_y)}{\|x-y\|} \right)\cdot d\_W(\mu\_1,\mu\_2)?
$$



1-wasserstein distance v.s. total variation distance
Yes.


I presume that your "1-Wasserstein" distance is what is otherwise called the transportation metric.


Let $M$ be any measure with marginals $\mu\_1$ and $\mu\_2$, so that $\mu\_1=\int \delta\_x\,dM(x,y)$ and $\mu\_2=\int\delta\_y\,dM(x,y)$, whence $\mu\_1-\mu\_2=\int (\delta\_x-\delta\_y)\,dM(x,y)$, which, after taking convolution with $\gamma$ and passing to the total variation norm, yields
$$
\| \mu\_1\*\gamma-\mu\_2\*\gamma \| \le \int \| \gamma\_x-\gamma\_y \|\,dM(x,y) \le K \int \|x-y\|\,dM(x,y) \;,
$$
where $K$ is the $\sup$ from your question (none of the additional conditions on $\gamma$ are required). Taking for $M$ a measure which realizes the transportation distance between $\mu\_1$ and $\mu\_2$, one gets the claim.


