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Let $\mu,\nu$ be probability measures on $\mathbb{R}^d$ with finite $p$-th moment ($p\in [1,\infty)$) and define the set of couplings by $\mathcal{C}(\mu,\nu)$ i.e. the set of probability measures on $\mathbb{R}^d\times \mathbb{R}^d$ so that the projection onto each component is $\mu,\nu$.

Define an integral functional $I_{p}(\pi,\omega):\mathcal{C}(\mu,\nu)\times \mathbb{S}^{d-1}\to \mathbb{R}$ by

$\displaystyle \begin{equation} I_p(\omega, \pi) =\left(\int_{\mathbb{R}^n\times \mathbb{R}^n}|\langle \omega ,x-y\rangle |^p\pi(dx\times dy)\right)^{1/p} \end{equation}\tag*{}$

where $\langle x,y \rangle$ denotes the standard Euclidean inner product on $\mathbb{R}^n$.

I am interested in proving or disproving that:

$\displaystyle \inf_{\pi\in \mathcal{C}(\mu,\nu)}\int_{\mathbb{S}^{d-1}}I_p(\omega,\pi)^pd\sigma_{d-1}=\int_{\mathbb{S}^{d-1}}\inf_{\pi\in \mathcal{C}(\mu,\nu)}I_p(\omega,\pi)^pd\sigma_{d-1}\tag*{}$

Here, $\sigma_{n-1}$ is the standard volume measure on $\mathbb{S}^{d-1}$.

Background. This is related to sliced Wasserstein distance. The RHS is almost the definition of the sliced W distance (aside from $1/p$ power).

What I tried:

Following this MSE post, I tried to construct a sequence $\pi_n$ such that $I_p(\pi_n, \cdot) \to \textrm{inf}_{\pi}I_p(\pi,\cdot)$ uniformly in $\omega$. However, I was unable to do so. On the way to prove it, I have observed the following facts.

(1) $I_p$ is continuous assuming we give $\mathcal{C}(\mu,\nu)$ the topology of weak convergence. (The proof is a bit long, so I am omitting it here. The Cauchy-Schwartz $|\langle \omega, x-y\rangle | \leq \|x-y\|$ is heavily in use.)

(2) Since $\mathcal{C}(\mu,\nu)$ and $\mathbb{S}^{d-1}$ are compact, $I_p$ is uniformly continuous, and the family $\{I_p(\cdot,\omega)|\omega \in \mathbb{S}^{d-1}\}$ is uniformly equicontinuous.

(3) I wanted to use the above facts to find $\pi_n$, but this won't work. For example, $f(x,y)=|x-y|$ on $[0,1]\times [0,1]$ satisfies the same properties as above, but $\inf_{x}\int_{0}^{1}|x-y|dy\neq \int_{0}^{1}\inf_{x}|x-y|dy$.

(4) If we can find the uniform minimizer $\pi^*$ (i.e. for any $\omega$ and $\pi$, $I_p(\pi^*, \omega)\leq I_p(\pi,\omega)$), then its trivial since we can take $\pi_n=\pi^*$, but I think this is too much to ask for.

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1 Answer 1

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$\newcommand\om\omega$This is of course not true in general (and almost never true).

Indeed, $I:=I_p$ is a continuous function, at least when $\mu$ and $\nu$ are compactly supported. Then the equality in question is $L=R$, where $$L:=\inf_\pi\int d\om\,I(\om,\pi)=\min_\pi\int d\om\,I(\om,\pi) =\int d\om\,J(\om),$$ where $J(\om):=I(\om,\pi_*)$ for some coupling $\pi_*$, and $$R:=\int d\om\,K(\om),$$ where $K(\om):=\min_\pi I(\om,\pi)$. The functions $J$ and $K$ are continuous, and $J\ge K$. So, $$L=R\iff J=K. \tag{1}\label{1}$$

More specifically, let $\mu(\{d\})=2/3=1-\mu(\{b\})$, and $\nu(\{a\})=2/3=1-\mu(\{c\})$, where $$(a,b,c,d):=((0,0),(1,0),(1,1),(0,1)),$$ the quadruple of the vertices of the unit square. Then (assuming $L=R$ and recalling \eqref{1}) we have the following:

  • For $\om_1:=(1,0)$ one has $I(\om_1,\pi_1)=0<I(\om_1,\pi)$ for all $\pi\ne\pi_1$, where $\pi_1(\{(d,a)\})=2/3=1-\pi_1(\{(b,c)\})$. (The equality $I(\om_1,\pi_1)=0$ holds because $\om_1$ only "cares" about the first coordinate of a point, whereas the "transportation plan" $\pi_1$ does not change the first coordinate: $d$ and $a$ have the same first coordinate, and $b$ and $c$ have the same first coordinate.) Therefore and because $K\ge0$, we have $I(\om_1,\pi_1)=K(\om_1)=J(\om_1)=I(\om_1,\pi_*)$. So, $\pi_*=\pi_1$.
  • On the other hand, for $\om_2:=(0,1)$ and $\pi_2$ such that $\pi_2(\{(d,a)\})=\pi_2(\{(d,c)\})=\pi_2(\{(b,a)\})=1/3$ we have $J(\om_2)=I(\om_2,\pi_*)=I(\om_2,\pi_1)=1$ and hence $K(\om_2)\le I(\om_2,\pi_2)=(1/3)^{1/p}<1=J(\om_2)=K(\om_2)$, a contradiction. $\quad\Box$
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