- $c$ is bounded, i.e., $\sup_{x, y\in X} c(x, y) < C$ for some $C \in \mathbb R$.
Then $\|f\|_{\mathrm{Lib}} \le 1$ implies $f$ is bounded, and thus $f \in L_1 (|\mu- \nu|)$ and $(-f, f) \in \Phi_{c}$. By Kantorovich duality, it suffices to prove
$$
\sup _{\Phi_{c}} \mathbb J(\varphi, \psi) = \sup \left \{ \int_X f d (\mu - \nu) \,\middle\vert\, f \in L_1 (|\mu- \nu|) , \|f\|_{\mathrm{Lib}} \le 1 \right \}.
$$
If $f \in L_1 (\mu)$, then
$$
f^c:Y \to \mathbb R \cup \{\pm \infty\}, y \mapsto \inf_{x \in X} [c(x, y) - f(x)]
$$
is such that $\|f^c\|_{\mathrm{Lib}} \le 1$ and $f^{cc} = -f^c$. By this lemma, we have
$$
\sup _{\Phi_{c}} \mathbb J(\varphi, \psi) = \sup _{f \in L_1 (\mu)} \mathbb J(f^{cc}, f^c) = \sup _{f \in L_1 (\mu)} \mathbb J(-f^{c}, f^c) \le \sup _{\|f\|_{\mathrm{Lib}} \le 1} \mathbb J(-f, f) \le \sup _{\Phi_{c}} \mathbb J(\varphi, \psi).
$$
- $c$ is unbounded.
We have
$$
\sup \left \{ \int_X f d (\mu - \nu) \,\middle\vert\, f \in L_1 (|\mu- \nu|) , \|f\|_{\mathrm{Lib}} \le 1 \right \} \le \sup _{\Phi_{c}} \mathbb J(\varphi, \psi).
$$
It remains to prove
$$
\sup \left \{ \int_X f d (\mu - \nu) \,\middle\vert\, f \in L_1 (|\mu- \nu|) , \|f\|_{\mathrm{Lib}} \le 1 \right \} \ge \sup _{\Phi_{c}} \mathbb J(\varphi, \psi).
$$
Let $c_n := \min\{n, c\}$. Then $c$ is a bounded metric on $X$ such that $c_n \nearrow c$. By this result, we have
$$
\lim_n \inf_{\pi \in \Pi(\mu, \nu)} \int c_n d\pi = \inf_{\pi \in \Pi(\mu, \nu)} \int c d\pi = \sup _{\Phi_{c}} \mathbb J(\varphi, \psi).
$$
On the other hand,
$$
\begin{align}
\inf_{\pi \in \Pi (\mu, \nu)} \int_{X \times Y} c_n d \pi &= \sup \left \{ \int_X f d (\mu - \nu) \,\middle\vert\, f \in L_1 (|\mu- \nu|) , \|f\|_{\mathrm{Lib}, c_n} \le 1 \right \} \\
&\le \sup \left \{ \int_X f d (\mu - \nu) \,\middle\vert\, f \in L_1 (|\mu- \nu|) , \|f\|_{\mathrm{Lib}} \le 1 \right \}.
\end{align}
$$
This completes the proof.
We have
$$
W^p_1 (\mu, \nu) \le W^p_p (\mu, \nu) \le (\mathrm{diam} X)^p W_1 (\mu, \nu) \quad \forall p \in [1, +\infty).
$$
So it suffices to prove for $p=1$. Let
$$
\mathrm{Lib} (f) := \sup_{x\neq y} \frac{|f(x) - f(y)|}{|x-y|}.
$$
- Assume $W_1 (\mu_m, \mu) \to 0$.
By Kantorovich-Rubinstein theorem, we have
$$
W_1 (\mu_m, \nu) = \sup \left \{\int_X f \mathrm d (\mu_m-\nu) \,\middle\vert\, f \in L_1 (|\mu_m-\nu|) , \mathrm{Lib} (f) \le 1\right \}.
$$
Let $f$ be bounded Lipschitz-continuous such that $\mathrm{Lib} (f) > 0$. Then $\frac{f}{\mathrm{Lib} (f)}$ is $1$-Lipschitz. Then
$$
\int_X \frac{f}{\mathrm{Lib} (f)} \mathrm d (\mu_m-\nu) \to 0.
$$
So
$$
\int_X f \mathrm d (\mu_m-\nu) \to 0,
$$
The result then follows from Portmanteau theorem.
- Assume $\mu_m \overset{\ast}{\rightharpoonup} \mu$.
Let $m_k$ be a subsequence such that
$$
\lim_k W_1 (\mu_{m_k}, \mu) = \limsup_m W_1 (\mu_m, \mu).
$$
Fix $a \in X$. Let $g_k$ be a Lipschitz funtion such that $\mathrm{Lib} (g_k) \le 1$ and
$$
W_1 (\mu_{m_k}, \mu) \le \int_X g_k \mathrm d (\mu_{m_k} - \mu) + \frac{1}{k} = \int_X (g_k - g_k (a)) \mathrm d (\mu_{m_k} - \mu) + \frac{1}{k} .
$$
It follows from $(\mu_{m_k} - \mu) (X) = 0$ that
$$
\int_X g_k (a) \mathrm d (\mu_{m_k} - \mu) = 0 \quad \forall k.
$$
Let $g'_k := g_k - g_k (a)$. Then $\mathrm{Lib} (g'_k) \le 1$ and $(g'_k)_k$ is bounded by $\mathrm{diam} X$. By Arzelà –Ascoli theorem, there is $g$ and a subsequence (re-labelled) such that $g'_k \to g$ in $\|\cdot\|_\infty$. Clearly, $\mathrm{Lib} (g) \le 1$. Finally,
$$
\begin{align}
\int_X g'_k \mathrm d (\mu_{m_k} - \mu) &= \int_X (g'_k-g) \mathrm d (\mu_{m_k} - \mu) + \int_X g \mathrm d (\mu_{m_k} - \mu) \\
&= \int_X (g'_k-g) \mathrm d (\mu_{m_k} - \mu)^+ - \int_X (g'_k-g) \mathrm d (\mu_{m_k} - \mu)^- + \int_X g \mathrm d (\mu_{m_k} - \mu) \\
&\le 2 \|g'_k-g\|_\infty (\mu_{m_k} - \mu)^+ (X) + \int_X g \mathrm d (\mu_{m_k} - \mu) \\
&\le 4 \|g'_k-g\|_\infty + \int_X g \mathrm d (\mu_{m_k} - \mu).
\end{align}
$$
The result then follows by taking the limit $k \to \infty$.
Best Answer
Let $\mu, \nu, \omega \in \mathcal P_p (X)$. Let $\pi_1 \in \Pi (\mu, \nu)$ and $\pi_2 \in \Pi (\mu, \nu)$ be optimal, i.e., $$ W^p_p (\mu, \nu) := \int_{X \times Y} |x-y|^p \mathrm d \pi_1 (x, y) \quad \text{and} \quad W^p_p (\nu, \omega) := \int_{Y \times Z} |y-z|^p \mathrm d \pi_2 (y, z). $$
Let $P^{X \times Y}$ and $P^{Y \times Z}$ be the projection maps from $X \times Y \times Z$ to $X \times Y$ and $Y \times Z$ respectively. Let $\gamma \in \mathcal P(X \times Y \times Z)$ such that $P^{X \times Y}_\sharp \gamma = \pi_1$ and $P^{Y \times Z}_\sharp \gamma = \pi_2$. Here $P^{X \times Y}_\sharp \gamma$ is the push-forward of $\gamma$ by $P^{X \times Y}$. Such $\gamma$ does exists by gluing lemma. Let $\pi = P^{X \times Z}_\sharp \gamma$. It's easy to prove that $\pi \in \Pi(\mu, \omega)$. Finally, $$ \begin{align} W_p(\mu, \omega) &\le \left [ \int_{X \times Z} |x-z|^p \mathrm d \pi (x, z) \right ]^{1/p} \\ &= \left [ \int_{X \times Y \times Z} |(x-y) - (y-z)|^p \mathrm d \gamma(x,y,z) \right ]^{1/p} \\ &\le \left [ \int_{X \times Y} |x-y|^p \mathrm d \pi_1 (x, y) \right ]^{1/p} + \left [ \int_{Y \times Z} |y-z|^p \mathrm d \pi_2 (y, z) \right ]^{1/p} \\ &= W_p(\mu, \nu) + W_p(\nu, \omega). \end{align} $$
This completes the proof.