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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 $$ d_{TV}(\mu_1\ast\gamma,\mu_2\ast\gamma) \leq \left(\sup_{x\neq y} \frac{d_{TV}(\gamma_x,\gamma_y)}{\|x-y\|} \right)\cdot d_W(\mu_1,\mu_2)? $$

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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.

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  • $\begingroup$ Lots of steps missing here. First note that $\delta_x * \gamma = \gamma_x$ for any transition kernel $\gamma$. Now, $$ \begin{split} \|\mu_1 * \gamma - \mu_2 * \gamma\|_{TV} &= \|(\mu_1-\mu_2)*\gamma\|_{TV} = \|\int ((\delta_x-\delta_y)dM(x,y)) * \gamma\|_{TV}\\ & = \|\int((\delta_x-\delta_y)* \gamma) dM(x,y) \|_{TV} = \|\int(\delta_x * \gamma -\delta_y * \gamma)dM(x,y) \|_{TV}\\ & = \|\int(\gamma_x-\gamma_y)dM(x,y)\|_{TV}\\ &\le \int \|\gamma_x-\gamma_y\|_{TV}dM(x,y) \le K_\gamma \int c(x,y) dM(x,y), \end{split} $$ where the last bust one inequality is the triangle inequality. $\endgroup$
    – dohmatob
    Jan 19, 2020 at 21:51
  • $\begingroup$ It seems one can get another bound: $\|\mu_1 * \gamma - \mu_2 * \gamma\|_W \le (\underset{x,y,\;x \ne y}{\sup}\;\frac{\|\gamma_x - \gamma_y\|_W}{c(x,y)})\|\mu_1-\mu_2\|_W$, for any Polish space $(X,c)$ and transition kernel $\gamma$ thereupon. $\endgroup$
    – dohmatob
    Jan 19, 2020 at 21:57

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