TITLE: Upper bound total variation by Wasserstein distance for continuous distance
QUESTION [5 upvotes]: I am reading the survey of the relationships between metrics of distributions (see https://arxiv.org/pdf/math/0209021.pdf for the paper).
The general results show that for general distributions, we cannot upper bound the total variation by Wasserstein distance. Many answers in MO give the same intuitive counterexample: consider a discrete distribution and a continuous distribution.
However, assuming the two underlying distributions are both continuous (for example consider the simpliest when they are both Gaussian mixtures with zeros mean or just Gaussians with zero mean), can we find such an upper bound? 
Any related suggestions or comments are welcomed!

REPLY [4 votes]: If both probability measures have smooth densities, the total variation can be bounded by Wasserstein (even by a weaker metric Levy-Prokhorov), see this paper. 
More precisely, see Lemma 5.1 and proof of Theorem 2.1. The basic idea is given below.
If both $p$ and $q$ are smooth densities, $d_V(p, p_\gamma)$ and $d_V(q, q_\gamma)$ are sufficiently small for every small gamma, where $d_V$ is the total variation and $p_\gamma$ is a convolution of p and the uniform density as defined in the paper.
Note that $d_V(p, q) < d_V(p, p_\gamma) + d_V(p_\gamma, q_\gamma) + d_v(q_\gamma, q)$ by the triangle inequality.
Lemma 5.1 guarantees that if $p$ and $q$ are close in Levy-Prokhorov metric, then $d_V(p_\gamma, q_\gamma)$ is also small provided that $\gamma$ is much bigger than the Levy-Prochorov distance.