TITLE: Is the expectation value Lipschitz for the Wasserstein metric?
QUESTION [6 upvotes]: Consider the space $M$ of Borel measures on the real unit interval $[0,1]$, equipped with the 1-Wasserstein metric $d_W$ (or "Earth mover's distance"). 
The expected value is then a map $M\to [0,1]$ given by:
$$
p \mapsto E_p := \int_{[0,1]} x \, dp(x)\;.
$$
Is it true that this map is Lipschitz? That is, can we find a constant $C$ such that for every $p,q\in M$, $|E_p-E_q|\le C\,d_W(p,q)$ ?

REPLY [2 votes]: The Kantorovich-Rubinstein formula states that for $(M,d)$ compact we have,
$$
W_1(\mu,\nu) =\!\!\!\sup_{\substack{f:M\to\mathbb R \\ \text{1-Lipschitz}}} \left| \int_M f\,d\mu - \int_M f\,d\nu \right|,
$$
Define $I_f(\mu):=\int_M f\,d\mu$ and immediately we see that if $f$ is 1-Lipschitz then,
$$
|I_f(\mu)-I_f(\nu)| \leq \!\!\!\sup_{\substack{f:M\to\mathbb R \\ \text{1-Lipschitz}}} |I_f(\mu)-I_f(\nu)| =  W_1(\mu,\nu)
$$
Thus $I_f$ is 1-Lipschitz w.r.t $W_1$. Moreover, if $f$ is $K$-Lipschitz then so is $I_f$ (simply let $g=\tfrac{1}{K}f$).
For your question, let $M=[0,1]$ and $f(x)=x$. Then $E_p=I_f(p)$ is 1-Lipschitz because $f$ is.