Prove the inequality
$$d_{H}(\mathrm{spt}(\mu),\mathrm{spt}(\nu))\leq W_{\infty}(\mu,\nu)$$
where $d_H$ denotes the Hausdorff distance between the supports of the measures $\mu$ and $\nu$, and $W_\infty$ is the Wasserstein infinity distance between $\mu$ and $\nu$. I believe this is obvious because min max > max min, but this seems to be different. Any guidance on how to tackle this inequality would be greatly appreciated. Please let me know if I need to provide any additional context.
EDIT: answer 2 below is completely false, as pointed out by the OP. However this is such a typical example of wishful thinking that I believe it is worth leaving for the posterity. (I'll record it below as False Answer 2.) The right line of reasoning goes the other way around (take quasi-optimizers in the Hausdorff distance instead of taking a quasi-optimal plan).
Preliminaries: I'll consider as usual the case of a Polish space $(X,d)$ with cost function $c(x,y)=d(x,y)$, and I'll also write $Y=X$ for notational convenience. By definition one has $$ W_{\infty}(\mu,\nu)=\inf\limits_{\pi} \|f_d\|_{L^\infty(\pi)}, $$ where the infimum is taken along admissible plans $\pi\in\Pi(\mu,\nu)\subset \mathcal P(X\times Y)$ with first and second marginals $\mu,\nu$, respectively, and $f_d$ is just a short notation for $f_d(x,y)=d(x,y)$, the distance function as a function of two variables $(x,y)\in X\times Y$.
Correct answer 1
We only consider the case when $D= d_H(\mathrm{spt}\mu,\mathrm{spt}\nu)>0$ (otherwise the statement is trivial). In words, the proof goes as follows: by definition there are points $x\in\mathrm{spt}\mu$ or $y\in\mathrm{spt}\nu$ which must be at distance at least $D$ from the other support. Hence any admissible transportation plan must eventually ship some (positive amount of) mass from $\mathrm{spt}\mu$ to $\mathrm{spt}\nu$ over a distance at least $D$. This immediately gives the result, since the infinity-Wasserstein distance then just corresponds to taking the best such plan.
More rigorously, recall that by definition the Hausdorff distance is $$ d_H(\mathrm{spt}\mu,\mathrm{spt}\nu)=\max\Bigg\{ \sup\limits_{x\in \mathrm{spt}\mu} d(x,\mathrm{spt}\nu) , \sup\limits_{y\in \mathrm{spt}\nu} d(y,\mathrm{spt}\mu) \Bigg\}. $$ Let us assume that the maximum is $$ D=d_H(\mathrm{spt}\mu,\mathrm{spt}\nu)=\sup\limits_{x\in \mathrm{spt}\mu} d(x,\mathrm{spt}\nu) $$ (the other case is completely symmetric). Picking a maximizing sequence $x_n\in \mathrm{spt}\mu$, we have $$ D_n:=d(x_n,\mathrm{spt}\nu)\to D. $$ Since $x_n\in \mathrm{spt}\mu$, by definition of the support of a measure, we have $\mu(B_\epsilon(x))>0$ for all $\epsilon >0$, as small as desired. Taking in particular $\epsilon_n =\frac 1n D_n\to 0$ we see by triangular inequality that any point $x\in B_\epsilon(x_n)$ is at distance at least $(1-\frac 1n)D_n$ from the other support, \begin{equation} d(x,y)\geq (1-\frac 1n)D_n,\qquad \forall x\in B_{\epsilon_n}(x_n)\text{ and }\forall\,y\in \mathrm{spt}\nu. \tag{1} \end{equation} This strongly suggests that the $L^\infty$ norm of $f_d=d(\cdot,\cdot)$ should be at least $(1-\frac 1n)D_n\approx D$, but some extra care is needed to check that the set $B_{\epsilon_n}(x_n)\times\mathrm{spt}\nu $ is not negligible. To this end, take any admissible plan $\pi\in \Pi(\mu,\nu)$, and recall that any such plan actually satisfies $\mathrm{spt}\pi\subset \mathrm{spt}\mu\times \mathrm{spt}\nu$. By definition of the marginals we have $$ \pi(B_{\epsilon_n}(x_n)\times \mathrm{spt}\nu) = \pi(B_{\epsilon_n}(x_n)\times Y) = \mu(B_{\epsilon_n}(x_n))>0. $$ As a consequence, and from (1), $$ \|f_d\|_{L^\infty(\pi)}\geq (1-\frac 1n)D_n. $$ Taking $n\to\infty$ we see that $$ \|f_d\|_{L^\infty(\pi)}\geq D=d_H(\mathrm{spt}\mu,\mathrm{spt}\nu) \qquad \forall\,\pi\in\Pi(\mu,\nu) $$ and the conclusion follows.
False answer 2
Let $\pi_n$ be minimizing sequence. By definition of the essential supremum, for any such fixed $n$ there exists $(x_n,y_n)\in \mathrm{spt}(\pi_n)$ such that $$ d(x_n,y_n)=f_d(x_n,y_n)\geq \|f_d\|_{L^\infty(\pi_n)}-\frac 1n. $$ Since also $d(x_n,y_n)\leq \|f_d\|_{L^\infty(\pi_n)}$, and because $\pi_n$ is a minimizing sequence, we conclude that $$ d(x_n,y_n)\rightarrow W_\infty(\mu,\nu) $$ Recall now that, for any admissible plan $\pi\in \Pi(\mu,\nu)$, one has $\mathrm{spt}(\pi)\subset\mathrm{spt}(\mu)\times \mathrm{spt}(\nu)$. The previous inequality immediately gives, by definition of the Hausdorff distance, $$ d_H(\mathrm{spt}(\mu),\mathrm{spt}(\nu)) \leq d(x_n,y_n) $$ this is blatantly false!!! and taking $n\to\infty$ finally gives the result.
Call $A,B$ the two supports, and $h$ the Hausdorff distance. Given a transport plan $\gamma$ we need to infer $h(A,B)\leq\sup^\gamma d(x,y)$, that is, for any $x\in A$, that $\inf_{y\in B} d(x,y)\leq m$ for all $m\in\mathbb R$ such that $\gamma(\{(x,y):d(x,y)>m\})=0$. This is equivalent to $\exists y_n\in B : \limsup d(x,y_n)\leq m$. Assume it is false. By contradiction, $\forall y_n\in B$ we get $\limsup d(x,y_n)> m$ which implies $d(x,y)>m$ for all $y\in B$. So $0=\gamma(\{x\}\times B)=\nu(B)$, contradiction.
