Let $(X,d)$ be a separable metric space with associated Borel $\sigma$-algebra $\mathcal{B}(X)$ and the set of Borel probability measures $\mathcal{P}(X)$. For $\mu,\mu'\in\mathcal{P}(X)$ Wasserstein-1 distance is defined as:
$$W_1(\mu, \mu') = \inf_{\gamma \in \Gamma(\mu, \mu')} \int_{X\times X} d(x_1,x_2)d\gamma(x1,x2),$$
where $\Gamma(\mu, \mu')$ is the set of all couplings of $\mu$ and $\mu'$ (i.e., a joint probability distribution with marginals $\mu$ and $\mu'$).
It can be shown that $W_1$ is a metric and that for $\mu, \mu', \nu \in \mathcal{P}(X)$ the triangle inequality holds: $$W_1(\mu,\mu')\leq W_1(\mu,\nu) + W_1(\nu,\mu')$$
The questions which are unclear to me are:
- Is the "induced" space $(\mathcal{P}(X), W_1)$ a standard metric space?
- Does the reverse triangle inequality holds for $W_1$?
Concretely, since we are operating on the space of probability measures, I am unsure if all the properties of the usual metric spaces hold.
If they do, the reverse triangle inequality is an immediate consequence. Nevertheless, if it is not the usual metric space, I am interested in if for $\mu, \mu', \nu \in \mathcal{P}(X)$ the following holds: $$W_1(\mu,\mu')\geq |W_1(\mu,\nu) - W_1(\nu,\mu')|$$
As was suggested, property (2) holds whenever the "regular" triangle inequality holds. So yes.
Start with the regular inequality one and rearrange in two ways: $$ d(x, z) \le d(x, y) + d(y, z) \iff d(x, y) \ge d(x, y) - d(y, z) $$ $$ d(y, z) \le d(y, x) + d(x, z) \iff d(x, y) \ge d(y, z) - d(x, y) $$
Note that we used symmetry, i.e., $d(x,y) = d(y,x)$. However, the two inequalities on the RHS are equivalent to your "reverse" triangle inequality.