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The definition of Wasserstein $p$-distance between two measures $\mu$ and $\nu$ on a Polish space $X$ is given by $$ W_p(\mu, \nu)^p = \inf_{\gamma \in \Pi(\mu, \nu)} \int_{X \times X} d(x, y)^p\; d\gamma, $$ where $\gamma$ is a measure on $X \times X$ with marginals $\mu$ and $\nu$. It is also well-known that for the special case $p = 1$, the Monge-Kantorovich-Rubinstein duality gives the following alternative definition: $$ W_1(\mu, \nu) = \sup_{\|f\|_{Lip} \leq 1} \int_X f \; d(\mu - \nu). $$ These notes seem to suggest (as do many other resources online) that an extension of the latter characterization to larger values of $p$ is not possible.

With that in the background, here is my question: via Theorem 1.3 in Villani's book (also see here), there is a Monge-Kantorovich-Rubinstein duality which works for general cost functions $c(x, y)$. Using this theorem with the special cost function $c(x, y) = d(x, y)^p$, and $g(x) = -f(y)$, it seems that we can at least have the one-sided estimate $$ W_p(\mu, \nu)^p \geq \sup_{\|f\|_{H_p} \leq 1} \int_X f \; d(\mu - \nu), $$ where $\|.\|_{H_p}$ denotes the Hölder norm $\|f\|_{H_p} = \sup_{x, y \in X} \frac{|f(x) - f(y)|}{d(x, y)^{1/p}}$.

Is this correct, or am I missing something? Any insight will be gratefully appreciated.

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    $\begingroup$ Ok, this is really simple: for any $p$-Holder $f$, using the fact that $\mu, \nu$ are marginals, we have that $\int d(x, y)^p\; d\gamma \geq \int (f(x) - f(y))\; d\gamma = \int f\; d(\mu - \nu)$, and things follow from the first definition itself. $\endgroup$
    – SMS
    Commented Mar 15, 2021 at 12:11
  • $\begingroup$ Is there something off about $p$ and $1/p$ in this question, especially in your definition of Hölder norm? First a typo: I believe you are taking $g(y) = - f(y)$. Then you would need $f(x) - f(y) \le c(x,y) = d(x,y)^p$. For $p > 1$, if we take $X = Y$ to be some intervals with $d$ being Euclidean norm, all such functions $f$ are constant function according to the Wikipedia page of Hölder condition. Then RHS becomes 0 and what you got is $W_p \ge 0$ which is uninteresting. $\endgroup$
    – Shuyao Li
    Commented Sep 10, 2023 at 15:00

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