Uniqueness of optimal solutions for semi-discrete transport with p-norm cost functions

TL;DR

This paper investigates conditions under which semi-discrete optimal transport solutions can be characterized by shifts that partition the continuous region, providing examples of failures and identifying classes where partitioning always occurs.

Contribution

It identifies when shift-characterized solutions in semi-discrete transport always form partitions, expanding understanding of solution structures in optimal transport problems.

Findings

Examples where partitioning fails in semi-discrete transport.

A large class of problems where shift solutions always partition.

Insights into the structure of optimal solutions in semi-discrete transport.

Abstract

Semi-discrete transport can be characterized in terms of real-valued shifts. Often, but not always, the solution to the shift-characterized problem partitions the continuous region. This paper gives examples of when partitioning fails, and offers a large class of semi-discrete transport problems where the shift-characterized solution is always a partition.

Figures

Equations

$$\Pi(\mu,\,\nu):=\left\{\pi\in\mathcal{P}(X\times Y)\left|\begin{array}[]{c}\pi[A\times Y]=\mu[A],\,\pi[X\times B]=\nu[B]\ ,\\ \forall\text{ meas.\ }A\subseteq X,\,B\subseteq Y\end{array}\right.\right\},$$

$$P(\pi ):={\int }_{X\times Y}c(\mathbf{x},\mathbf{y})d\pi (\mathbf{x},\mathbf{y}).$$

$$P^{\ast }:=\underset{\pi \in \Pi (\mu ,\nu )}{\inf }P(\pi ),$$

$${\pi }^{\ast }:=\underset{\pi \in \Pi (\mu ,\nu )}{\mathrm{arginf}}P(\pi ).$$

$${\pi }^{\ast }(\mathbf{x},\mathbf{y})={\pi }_{T^{\ast }}^{\ast }(\mathbf{x},\mathbf{y}):=\mu (\mathbf{x})\chi [\mathbf{y}=T^{\ast }(\mathbf{x})],$$

$$\pi (\mathbf{x},\mathbf{y})={\pi }_T(\mathbf{x},\mathbf{y}):=\mu (\mathbf{x})\chi [\mathbf{y}=T(\mathbf{x})],$$

$$P(\pi ):={\int }_{X}c(\mathbf{x},T(\mathbf{x}))d\mu (\mathbf{x}).$$

$$P(\pi ):=\sum _{i=1}^n{\int }_{A_i}c(\mathbf{x},{\mathbf{y}}_i)d\mu (\mathbf{x}).$$

$$F(\mathbf{x}):=\underset{1\le i\le n}{\max }\{a_i-c(\mathbf{x},{\mathbf{y}}_i)\}.$$

$$A_i:=\{\mathbf{x}\in A\mid F(\mathbf{x})=a_i-c(\mathbf{x},{\mathbf{y}}_i)\}.$$

$$A_{ij}:=A_i\cap A_j.$$

$$B:=\bigcup _{1\le i<n}\bigcup _{i<j\le n}{A}_{ij}.$$

$$g_{ij}(\mathbf{x}):=c(\mathbf{x},{\mathbf{y}}_i)-c(\mathbf{x},{\mathbf{y}}_j).$$

$$\mathbf{x}\in {\mathring{A}}_i\text{ for some }i\in {\mathbb{N}}_n⟹T(\mathbf{x})={\mathbf{y}}_i.$$

$$\mu \left(\{\mathbf{x}\in A|c(\mathbf{x},{\mathbf{y}}_i)-c(\mathbf{x},{\mathbf{y}}_j)=k\}\right)=0\forall i,j\in {\mathbb{N}}_n,i\ne j,\forall k\in \mathbb{R}.$$

$$\mu \left(\left\{\mathbf{x}\in A|c(\mathbf{x},{\mathbf{y}}_2)-c(\mathbf{x},{\mathbf{y}}_1)=k\right\}\right)=\frac{1}{16}\text{ if }k\in \left\{-\frac{1}{2},\frac{1}{2}\right\}.$$

$$\nu ({\mathbf{y}}_1)\in \left(\frac{1}{16},\frac{15}{16}\right).$$

$$\mu \left(\left\{\mathbf{x}\in A|c(\mathbf{x},{\mathbf{y}}_2)-c(\mathbf{x},{\mathbf{y}}_1)=k\right\}\right)=\frac{1}{8}\text{ if }k\in \left\{-\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2}\right\}.$$

$$\nu ({\mathbf{y}}_1)\in \left(\frac{1}{16},\frac{7}{16}\right)\cup \left(\frac{9}{16},\frac{15}{16}\right).$$

$$c(\mathbf{x},\mathbf{y}):={\left[\sum _{i=1}^d|x_i-y_i{|}^p\right]}^{1/p},\forall \mathbf{x}=(x_1,\dots ,x_d)\in X,\forall \mathbf{y}=(y_1,\dots ,y_d)\in Y.$$

$$g_{ij}({\mathbf{x}}_0)=c({\mathbf{x}}_0{\mathbf{y}}_i)-c({\mathbf{x}}_0{\mathbf{y}}_j)=a_i-a_j>c({\mathbf{y}}_i,{\mathbf{y}}_j),$$

$$(x_k-y_k^i)|x_k-y_k^i{|}^{p-2}{\left(c_i({\mathbf{x}}_0)\right)}^{1-p}=(x_k-y_k^j)|x_k-y_k^j{|}^{p-2}{\left(c_j({\mathbf{x}}_0)\right)}^{1-p}.$$

$$\frac{x_k-y_k^i}{c_i({\mathbf{x}}_0)}=\frac{x_k-y_k^j}{c_j({\mathbf{x}}_0)}\forall k\in {\mathbb{N}}_d.$$

$$|g_{ij}({\mathbf{x}}_0)|=|(x_k-y_k^i)-(x_k-y_k^j)|=|y_k^j-y_k^i|=c({\mathbf{y}}_i,{\mathbf{y}}_j).$$

$$\frac{(p-1)|x_k-y_k^i{|}^{p-2}}{(c_i({\mathbf{x}}_0){)}^{2p-1}}\left[(c_i({\mathbf{x}}_0){)}^p-|x_k-y_k^i{|}^p\right]=\frac{(p-1)|x_k-y_k^j{|}^{p-2}}{(c_j({\mathbf{x}}_0){)}^{2p-1}}\left[(c_j({\mathbf{x}}_0){)}^p-|x_k-y_k^j{|}^p\right],$$

$$\frac{p-1}{c_i({\mathbf{x}}_0)}{\left(\frac{|x_k-y_k^i|}{c_i({\mathbf{x}}_0)}\right)}^{p-2}\left[1-{\left(\frac{|x_k-y_k^i|}{c_i({\mathbf{x}}_0)}\right)}^p\right]=\frac{p-1}{c_j({\mathbf{x}}_0)}{\left(\frac{|x_k-y_k^j|}{c_j({\mathbf{x}}_0)}\right)}^{p-2}\left[1-{\left(\frac{|x_k-y_k^j|}{c_j({\mathbf{x}}_0)}\right)}^p\right].$$

$${\sigma }_k=\frac{|x_k-y_k^i|}{c_i({\mathbf{x}}_0)}=\frac{|x_k-y_k^j|}{c_j({\mathbf{x}}_0)}.$$

$$\begin{gathered} \frac{p-1}{c_i({\mathbf{x}}_0)}{\sigma }_k^{p-2}(1-{\sigma }_k^p) \\ =\frac{p-1}{c_j({\mathbf{x}}_0)}{\sigma }_k^{p-2}(1-{\sigma }_k^p). \end{gathered}$$

$${\sigma }_k=\frac{|x_k-y_k^i|}{c_i({\mathbf{x}}_0)}<1,$$

$$g_{ij}(\mathbf{x})=a_i-a_j⟺c(\mathbf{x},{\mathbf{y}}_i)-c(\mathbf{x},{\mathbf{y}}_j)=c({\mathbf{y}}_i,{\mathbf{y}}_j)⟺c(\mathbf{x},{\mathbf{y}}_i)=c(\mathbf{x},{\mathbf{y}}_j)+c({\mathbf{y}}_i,{\mathbf{y}}_j).$$

Full text

Acknowledgments. This material is based upon work supported by the National Science Foundation Graduate Research Fellowship Program under Grant No. DGE-1650044. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation.

1. Introduction

Optimal transport offers a way to measure the distance between two probability spaces, $X$ and $Y$. In the class of transport problems known as semi-discrete optimal transport, the probability distribution on $X$ is almost-everywhere continuous and the probability distribution on $Y$ is discrete, with $N$ points of positive measure. Given minimal assumptions, described below, the semi-discrete problem always has at least one solution that partitions $X$ into $N$ regions based on transport destination.

Rüschendorf and Uckelmann developed a way to characterize semi-discrete transport in terms of a set of real-valued shifts. This shift characterization often results in a solution that partitions $X$ into $N$ regions. Unfortunately, the shift characterization does not always partition $X$. This important fact has not always been recognized or clearly expressed in the literature; see [5, 6, 7]. To remedy that ambiguity, this paper gives clear, specific examples where shift-characterized partitioning fails, and it offers a large class of problems where the shift characterization is guaranteed to partition $X$.

2. Background

2.1. General optimal transport: the Monge-Kantorovich and Monge

problems

Though this paper focuses on the semi-discrete problem, it is worth describing it in terms of the more general, Monge-Kantorovich transport problem.

Definition 2.1** (Monge-Kantorovich problem).**

Let $X,\,Y\subseteq\mathbb{R}^{d}$, let $\mu$ and $\nu$ be probability densities defined on $X$ and $Y$, and let $c(\mathbf{x},\,\mathbf{y}):X\times Y\to\mathbb{R}$ be a continuous measurable ground cost function. Define the set of transport plans

[TABLE]

where $\mathcal{P}(X\times Y)$ is the set of probability measures on the product space, and define the primal cost function $P:\Pi(\mu,\,\nu)\to\mathbb{R}$ as

[TABLE]

The Monge-Kantorovich problem is to find the optimal primal cost

[TABLE]

and an associated optimal transport plan

[TABLE]

Under the conditions given, an optimal transport plan, $\pi^{*}$, is guaranteed to exist. However, $\pi^{*}$ may not be unique, or even a.e.-unique. Furthermore, the existence of $\pi^{*}$, an optimal plan, does not ensure that $\pi^{*}$ is a map, or that an optimal map exists. Nonetheless, consider the form such an optimal map would take.

Definition 2.2** (Monge problem).**

In certain cases, there exists at least one solution to the semi-discrete Monge-Kantorovich problem that does not split transported masses. In other words, there exists some $\pi^{*}$ such that

[TABLE]

where $T^{*}:X\to Y$ is a measurable map called the optimal transport map. When such a $\pi^{*}$ exists, we say the solution also solves the Monge problem.

If the Monge problem has a solution, we can assume without loss of generality that every $\pi\in\Pi(\mu,\,\nu)$ satisfies

[TABLE]

for some measurable transport map $T:X\to Y$, and that the primal cost can be written

[TABLE]

2.2. Semi-discrete optimal transport and the

shift characterization

The semi-discrete optimal transport problem is the Monge-Kantorovich problem of Definition 2.1, with restrictions on $\mu$ and $\nu$:

1. (1)

Assume that $\mu$ satisfies the following:

1. (a)

$\mu$ is bounded. 2. (b)

$\mu$ is nonatomic. 3. (c)

$\mu$ is continuous except on a set of Lebesgue measure zero. 4. (d)

The support of $\mu$ is contained in the convex compact region $A\subseteq X$. 2. (2)

Assume $\nu$ has exactly $n\geq 2$ non-zero values, located at $\{\mathbf{y}_{i}\}_{i=1}^{n}\subseteq Y$.

Because $c$ is continuous and $\mu$ is nonatomic, at least one solution to the semi-discrete Monge-Kantorovich problem also satisfies the Monge problem, described in Definition 2.2; see [4]. Thus, by applying Equation 2.6, we can assume without loss of generality that any transport plan $\pi$ has an associated map $T$, and that $T$ partitions $A$ into $n$ sets $A_{i}$, where $A_{i}$ is the set of points in $A$ that are transported by $T$ to $\mathbf{y}_{i}$. Using this partitioning scheme in combination with Equation 2.7 allows us to rewrite the primal cost function for the semi-discrete problem as

[TABLE]

This idea of sets $A_{i}$ is central to describing the shift characterization of the semi-discrete optimal transport problem. The following definition is based on one given by Rüschendorf and Uckelmann in [5, 7].

Definition 2.3** (Shift characterization).**

Let $\{a_{i}\}_{i=1}^{n}$ be a set of $n$ finite values, referred to as shifts. Define

[TABLE]

For $i\in\mathbb{N}_{n}$, where $\mathbb{N}_{n}=\{1,\,\ldots,\,n\}$, let

[TABLE]

Note that $\cup_{i=1}^{n}A_{i}=A$. The problem of determining an optimal transport plan $\pi^{*}$ is equivalent to determining shifts $\{a_{i}\}_{i=1}^{n}$ such that for all $i\in\mathbb{N}_{n}$, the total mass transported from $A_{i}$ to $\mathbf{y}_{i}$ equals $\nu(\mathbf{y}_{i})$.

2.3. Formalizing the shift-characterized partition

“Partitioning” $A$ is described in [7, 5] as $\mu(A_{i})=\nu(\mathbf{y}_{i})$. However, it is beneficial to describe the shift-characterized partition in more detail. Doing so requires a few additional definitions.

Definition 2.4** (Boundaries and boundary sets).**

For all $i,\,j\in\mathbb{N}_{n}$ such that $i\neq j$, let

[TABLE]

The boundary set is defined as

[TABLE]

For all $i,\,j\in\mathbb{N}_{n}$ such that $i\neq j$, define $g_{ij}:X\to\mathbb{R}$ as

[TABLE]

Definition 2.5** ($F$ $\mu$-partitions $A$).**

Let $F$ be as defined in Equation 2.9, and the sets $A_{i}$ as defined in Equation 2.10 for $i\in\mathbb{N}_{n}$. Then one says $F$ $\mu$-partitions the set $A$, or $F$ is called a $\mu$-partition, if

1. (1)

$\mu(A)<\infty$, 2. (2)

for all $i,\,j\in\mathbb{N}_{n}$, $i\neq j$, $\mu(A_{ij})=0$, 3. (3)

$\sum_{i=1}^{n}\mu(A_{i})=\mu(A)$, and 4. (4)

for all $i\in\mathbb{N}_{n}$, $\mu(A_{i})=\nu(\mathbf{y}_{i})>0$.

Definition 2.6** (Monge under the shift characterization).**

We say a transport plan $\pi$ is Monge under the shift characterization if $\pi$ has an associated transport map $T$, a function $F$, as described in Equation 2.9, and sets $\{A_{i}\}_{i=1}^{n}$, as described in Equation 2.10, such that for all $\mathbf{x}\in A$,

[TABLE]

In other words, $F$ $\mu$-partitions $A$ and $T$ agrees with $F$ on $A\setminus B$.

If $\mu(B)>0$ for the shifts $\{a_{i}\}_{i=1}^{n}$, no such transport plan $\pi$ can exist, and the transport problem itself can be said to be not Monge under the shift characterization. Conversely, if $\mu(B)=0$, then such a transport plan exists, and so the transport problem itself is said to be Monge under the shift characterization. In other words, $F$ $\mu$-partitions $A$ if and only if the transport problem is Monge under the shift characterization.

The following result, from [3], allows us to go further:

Theorem 2.7**.**

Suppose one has a semi-discrete transport problem, as described in Section 2.2. Let $F$ be as defined in Equation 2.9, and the sets $A_{i}$ as defined in Equation 2.10 for $i\in\mathbb{N}_{n}$. Then $F$ $\mu$-partitions $A$ if and only if $\mu(B)=0$.

Taken together, these statements provide a formal definition and condition for what it means for the shift-characterized solution to partition $A$:

The shift-characterized semi-discrete transport problem partitions $A$ — that is, $F$ $\mu$-partitions $A$ — if and only if the semi-discrete transport problem is Monge under the shift characterization, which is true if and only if $\mu(B)=0$.

2.4. Uniqueness of semi-discrete transport

solutions

Given the semi-discrete transport problem described in Section 2.2, Corollary 4 of [2] provides a sufficient condition for the existence of a Monge solution that is unique $\mu$-a.e.:

[TABLE]

If Equation 2.15 is satisfied, $\mu(B)=0$. Therefore, if Equation 2.15, then the transport problem is Monge under the shift characterization and the transport solution is unique $\mu$-a.e.

However, if a transport problem is Monge under the shift characterization, then it has a unique $\mu$-a.e. shift-characterized solution, whether or not Equation 2.15 is satisfied. This statement is formalized and proved in [3] as the following theorem:

Theorem 2.8** (The optimal transport map is unique $\mu$-a.e.).**

Given a semi-discrete transport problem, let $\pi^{*}$ and $\tilde{\pi}^{*}$ be optimal transport plans that are both Monge under the shift characterization. If $T$ is a transport map associated with $\pi^{*}$, and $\widetilde{T}$ a transport map associated with $\tilde{\pi}^{*}$, then $T=\widetilde{T}$ except on a set of $\mu$-measure zero.

3. Mathematical support

While Equation 2.15 implies $\mu(B)=0$, the converse is not true, as Section 3.1 shows. Next, Section 3.2 identifies a large class of problems where both conditions hold and the solution is always unique $\mu$-a.e.

3.1. Partitioning with the 1-norm

and $\bm{\infty}$-norm

Let $X=[0,\,1]^{2}$, $Y=\{\mathbf{y}_{1},\,\mathbf{y}_{2}\}$, and let $\mu$ be the continuous uniform distribution. This simple setup can be used to demonstrate failure to partition for both the uniform norm ($\infty$-norm) and the Manhattan norm ($1$-norm).

3.1.1. The uniform norm

Let $\mathbf{y}_{1}=(\nicefrac{{1}}{{4}},\,\nicefrac{{1}}{{2}})$ and $\mathbf{y}_{2}=(\nicefrac{{3}}{{4}},\,\nicefrac{{1}}{{2}})$, and let $c:\mathbb{R}^{2}\times\mathbb{R}^{2}\to\mathbb{R}$ be the uniform norm ($\infty$-norm): $c(\mathbf{x},\,\mathbf{y})=\max_{i\in\{1,\,2\}}\,\lvert x_{i}-y_{i}\rvert$ for all $\mathbf{x}=(x_{1},\,x_{2})\in X$, $\mathbf{y}=(y_{1},\,y_{2})\in Y$. Consider two examples:

1. (1)

If $\nu(\mathbf{y}_{1})=\nicefrac{{1}}{{32}}$, then $\nu(\mathbf{y}_{2})=\nicefrac{{31}}{{32}}$. In this case, $\mu(B)=\nicefrac{{1}}{{16}}$, and the shift-characterized solution fails to partition $A$. See Figure 1(a). 2. (2)

However, if $\nu(\mathbf{y}_{1})=\nicefrac{{1}}{{8}}$, then $\nu(\mathbf{y}_{2})=\nicefrac{{7}}{{8}}$, $\mu(B)=0$ and the shift-characterized solution does partition $A$. See Figure 1(b).

Even though one of the problems illustrated in Figure 1 results in a partition, Equation 2.15 fails in both cases:

[TABLE]

In general, for this choice of $X$, $Y$, $\mu$, and $c$, the shift-characterized solution partitions $A$ if and only if

[TABLE]

Thus, when $c$ is the uniform norm, one can have $\mu(B)=0$, giving a shift-characterized partition of $A$ that is unique $\mu$-a.e., whether or not Equation 2.15 is satsified.

3.1.2. The Manhattan norm

Now let $\mathbf{y}_{1}=(\nicefrac{{1}}{{4}},\,\nicefrac{{1}}{{4}})$ and $\mathbf{y}_{2}=(\nicefrac{{3}}{{4}},\,\nicefrac{{3}}{{4}})$. Let $c:\mathbb{R}^{2}\times\mathbb{R}^{2}\to\mathbb{R}$ be the Manhattan norm (1-norm): $c(\mathbf{x},\,\mathbf{y})=\lvert x_{1}-y_{1}\rvert+\lvert x_{2}-y_{2}\rvert$ for all $\mathbf{x}=(x_{1},\,x_{2})\in X$, $\mathbf{y}=(y_{1},\,y_{2})\in Y$. Consider three examples:

1. (1)

If $\nu(\mathbf{y}_{1})=\nicefrac{{1}}{{2}}$, then $\nu(\mathbf{y}_{2})=\nicefrac{{1}}{{2}}$. In this case, $\mu(B)=\nicefrac{{1}}{{8}}$, and the shift-characterized solution fails to partition $A$. See Figure 2(a) for an illustration. 2. (2)

If $\nu(\mathbf{y}_{1})=\nicefrac{{1}}{{32}}$, then $\nu(\mathbf{y}_{2})=\nicefrac{{31}}{{32}}$ and $\mu(B)=\nicefrac{{1}}{{8}}$, so the shift-characterized solution again fails to partition $A$. This is shown in Figure 2(b). 3. (3)

However, if $\nu(\mathbf{y}_{1})=\nicefrac{{1}}{{4}}$, then $\nu(\mathbf{y}_{2})=\nicefrac{{3}}{{4}}$. In this case $\mu(B)=0$ and the shift-characterized solution does partition $A$. See Figure 2(c).

Once again, Equation 2.15 fails in all the Figure 2 cases:

[TABLE]

In fact, for this choice of $X$, $Y$, $\mu$, and $c$, the shift-characterized solution partitions $A$ if and only if

[TABLE]

Thus, as Figure 2 illustrates, when $c$ is the 1-norm, one can have $\mu(B)=0$, giving a shift-characterized partition of $A$ that is unique $\mu$-a.e., whether or not Equation 2.15 is satsified.

Figure 2(a) is worth special consideration, because it is not simply a non-partitioning shift-characterized transport solution: it also constitutes a failed Voronoi diagram. One can see a similar example in Figure 37 of [1], offered as part of a discussion on methods for resolving lack of partitioning and uniqueness for certain Voronoi diagrams.

3.2. Partitioning with $\bm{p}$-norms when

$\bm{p\in(1,\,\infty)}$-norm

Given the semi-discrete transport assumptions already stated, let $c:\mathbb{R}^{d}\times\mathbb{R}^{d}\to\mathbb{R}$ be a $p$-norm with $p\in(1,\,\infty)$:

[TABLE]

Then the semi-discrete transport problem is always Monge under the shift characterization.

This assertion will be shown in two steps:

1. (1)

If $g_{ij}$, defined in Equation 2.13, is equal to the constant value $a_{i}-a_{j}$ in some neighborhood of $\mathbf{x}_{0}\in A_{ij}$, then $\lvert a_{i}-a_{j}\rvert=c(\mathbf{y}_{i},\,\mathbf{y}_{j})$. [Theorem 3.1] 2. (2)

It follows from Step (1) that $\mu(B)>0$ implies the existence of a ball of positive radius whose points are all collinear with both $\mathbf{y}_{i}$ and $\mathbf{y}_{j}$. [Theorem 3.2]

Because of the contradiction inherent in Step (2), $\mu(B)=0$, and so Theorem 3.2 concludes that the problem must be Monge under the shift characterization.

Theorem 3.1**.**

Let $c$ be a $p$-norm with $p\in(1,\,\infty)$, and $\mathbf{x}_{0}\in A_{ij}$ for some $i,\,j\in\mathbb{N}_{n}$, $i\neq j$. If $g_{ij}(\mathbf{x})=a_{i}-a_{j}$ for all $\mathbf{x}$ in a neighborhood of $\mathbf{x}_{0}$, then $\lvert a_{i}-a_{j}\rvert=c(\mathbf{y}_{i},\,\mathbf{y}_{j})$.

Proof.

Let $c$ be a $p$-norm with $p\in(1,\,\infty)$, $\mathbf{x}_{0}\in A_{ij}$, and $g_{ij}(\mathbf{x})=a_{i}-a_{j}$ for all $\mathbf{x}$ in some neighborhood of $\mathbf{x}_{0}$. Suppose to the contrary, however, that $\lvert a_{i}-a_{j}\rvert\neq c(\mathbf{y}_{i},\,\mathbf{y}_{j})$.

Say $\lvert a_{i}-a_{j}\rvert>c(\mathbf{y}_{i},\,\mathbf{y}_{j})$, and assume without loss of generality that $\lvert a_{i}-a_{j}\rvert=a_{i}-a_{j}$. Then

[TABLE]

which implies $c(\mathbf{x}_{0}\,\mathbf{y}_{i})>c(\mathbf{x}_{0}\,\mathbf{y}_{j})+c(\mathbf{y}_{i},\,\mathbf{y}_{j})$. This is a violation of the triangle inequality. Therefore, it must be the case that $\lvert a_{i}-a_{j}\rvert<c(\mathbf{y}_{i},\,\mathbf{y}_{j})$.

For all $k\in\mathbb{N}_{n}$, define $c_{k}(\mathbf{x}):=c(\mathbf{x},\,\mathbf{y}_{k})$. Because $\lvert a_{i}-a_{j}\rvert<c(\mathbf{y}_{i},\,\mathbf{y}_{j})$, $\mathbf{x}_{0}\neq\mathbf{y}_{i}$ and $\mathbf{x}_{0}\neq\mathbf{y}_{j}$. Hence, $c_{i}(\mathbf{x}_{0})>0$ and $c_{j}(\mathbf{x}_{0})>0$.

Because $g_{ij}$ is constant in a neighborhood of $\mathbf{x}_{0}$, $\nabla g_{ij}(\mathbf{x}_{0})=\nabla c_{i}(\mathbf{x}_{0})-\nabla c_{j}(\mathbf{x}_{0})=0$, which implies $\nabla c_{i}(\mathbf{x}_{0})=\nabla c_{j}(\mathbf{x}_{0})$. Hence, each of the first-order partial derivatives of $c_{i}$ and $c_{j}$ are equal at $\mathbf{x}_{0}$.

Assume $\mathbf{x}_{0}=(x_{1},\,\ldots,\,x_{d})$, $\mathbf{y}_{i}=(y^{i}_{1},\,\ldots,\,y^{i}_{d})$, and $\mathbf{y}_{j}=(y^{j}_{1},\,\ldots,\,y^{j}_{d})$. Then the equality of the $k$-th partial derivatives, $\nabla_{\!x_{k}}c_{i}(\mathbf{x}_{0})=\nabla_{\!x_{k}}c_{j}(\mathbf{x}_{0})$, gives

[TABLE]

Thus, $x_{k}-y^{i}_{k}$ and $x_{k}-y^{j}_{k}$ have the same sign or are both zero. Because $p>1$, $p-1>0$. Hence, taking the $(p-1)$-th root of both sides,

[TABLE]

As a consequence of Equation 3.2, $x_{k}-y^{i}_{k}=0$ if and only if $x_{k}-y^{j}_{k}=0$. Hence, $x_{k}=y^{i}_{k}$ if and only if $x_{k}=y^{j}_{k}$.

Let $K$ be the total number of $k$-th directional components satisfying $x_{k}\neq y^{i}_{k}$. Consider three cases: $K=0$, $K=1$, and $K>1$.

$\bm{K=0}$**: **

Then $x_{k}=y^{i}_{k}=y^{j}_{k}$ for all $k\in\mathbb{N}_{d}$, in which case $\mathbf{y}_{i}=\mathbf{y}_{j}$. Since the semi-discrete transport problem requires distinct non-zero points in $Y$, it must be the case that $i=j$, contradicting the initial assumption that $i\neq j$. Hence, $K\geq 1$.

$\bm{K=1}$**: **

There exists exactly one $k$ such that the components are not equal. Since $x_{k}-y^{i}_{k}$ and $x_{k}-y^{j}_{k}$ have the same sign,

[TABLE]

This contradicts the assumption that $\lvert a_{i}-a_{j}\rvert<c(\mathbf{y}_{i},\,\mathbf{y}_{j})$, and hence $K>1$.

$\bm{K>1}$**: **

Because $g_{ij}$ is constant in some neighborhood of $\mathbf{x}_{0}$, it must also be the case that $\nabla^{2}g_{ij}(\mathbf{x}_{0})=0$. Hence, $\nabla^{2}c_{i}(\mathbf{x}_{0})=\nabla^{2}c_{j}(\mathbf{x}_{0})$, so each of the second-order partial derivatives of $c_{i}$ and $c_{j}$ are equal at $\mathbf{x}_{0}$. The equality of the second-order partial derivatives taken with respect to $x_{k}$ gives

[TABLE]

which can be rewritten as

[TABLE]

Applying Equation 3.2, define

[TABLE]

Then Equation 3.4 can be rewritten as

[TABLE]

By assumption, for all $k\in\mathbb{N}_{d}$, $x_{k}-y^{i}_{k}\neq 0$ and $x_{k}-y^{j}_{k}\neq 0$. Hence, $\sigma_{k}>0$.

Since $d>1$, and $\lvert x_{k}-y^{i}_{k}\rvert>0$ for all $k\in\mathbb{N}_{d}$, it must be that $\lvert x_{k}-y^{i}_{k}\rvert<c_{i}(\mathbf{x}_{0})$ for all $k\in\mathbb{N}_{d}$. Therefore,

[TABLE]

which implies $1-\sigma_{k}^{p}>0$. Therefore, $(p-1)\,\sigma_{k}^{p-2}(1-\sigma_{k}^{p})>0$, and Equation 3.5 simplifies to $\frac{1}{c_{i}(\mathbf{x}_{0})}=\frac{1}{c_{j}(\mathbf{x}_{0})}$.

Thus, $c_{i}(\mathbf{x}_{0})=c_{j}(\mathbf{x}_{0})$. Combining this with Equation 3.2 implies $y^{i}_{k}=y^{j}_{k}$ for all $k\in\mathbb{N}_{d}$, and so $\mathbf{y}_{i}=\mathbf{y}_{j}$. Since $\mathbf{y}_{i}=\mathbf{y}_{j}$, and the semi-discrete transport problem requires distinct non-zero points in $Y$, it must be the case that $i=j$, contradicting the initial assumption that $i\neq j$. Thus, $K\ngtr 1$.

All choices of $K$ lead to contradictions. Hence, if $c$ is a $p$-norm for some $p\in(1,\,\infty)$, $\mathbf{x}_{0}\in A_{ij}$, $i\neq j$, and $g_{ij}(\mathbf{x})=a_{i}-a_{j}$ for all $\mathbf{x}$ in some neighborhood of $\mathbf{x}_{0}\in A_{ij}$, then it must be the case that $\lvert a_{i}-a_{j}\rvert=c(\mathbf{y}_{i},\,\mathbf{y}_{j})$. ∎

Theorem 3.2**.**

If $c$ is a $p$-norm for some $p\in(1,\,\infty)$, then the semi-discrete transport problem is Monge under the shift characterization.

Proof.

Assume the contrary is true. Then $\mu(B)>0$, so $\mu(A_{ij})>0$ for some $i,\,j\in\mathbb{N}_{n}$, $i\neq j$. Because $\mu$ is nonatomic, there exist $\mathbf{x}_{0}\in A_{ij}$ and $\epsilon>0$ such that the ball $\mathcal{B}_{\epsilon}(\mathbf{x}_{0})$, defined with respect to the Euclidean space $\mathbb{R}^{d}$, satisfies $\mathcal{B}_{\epsilon}(\mathbf{x}_{0})\subseteq A_{ij}$ and $\mu(\mathcal{B}_{\epsilon}(\mathbf{x}_{0}))>0$. By Theorem 3.1, $\lvert a_{i}-a_{j}\rvert=c(\mathbf{y}_{i},\,\mathbf{y}_{j})$. Assume without loss of generality that $\lvert a_{i}-a_{j}\rvert=a_{i}-a_{j}$.

Let $\mathbf{x}\in\mathcal{B}_{\epsilon}(\mathbf{x}_{0})$. Since $\mathbf{x}\in A_{ij}$,

[TABLE]

Because $c$ is a $p$-norm and $p\in(1,\,\infty)$, Minkowski’s inequality implies that $\mathbf{x}$, $\mathbf{y}_{i}$, and $\mathbf{y}_{j}$ are all collinear. The choice of $\mathbf{x}$ was nonspecific, and therefore every point in the ball $\mathcal{B}_{\epsilon}(\mathbf{x}_{0})$ must be collinear with the points $\mathbf{y}_{i}$ and $\mathbf{y}_{j}$.

Of course, this is impossible, and so $\mu(A_{ij})=0$ for all $i,\,j\in\mathbb{N}_{n}$, $i\neq j$. Therefore, $\mu(B)=0$. From this final contradiction, it is clear that the semi-discrete transport problem must be Monge under the shift characterization. ∎

Corollary 3.3**.**

If the semi-discrete transport problem is defined as given in Section 2.2, and $c$ is a $p$-norm for some $p\in(1,\,\infty)$, then Equation 2.15 is satisfied, and the optimal transport solution is unique $\mu$-a.e.

Proof.

Suppose $c$ is a $p$-norm for some $p\in(1,\,\infty)$, and assume the semi-discrete transport problem is characterized by shifts as given in Definition 2.3. By the triangle inequality, for all $\mathbf{x}\in X$, $i,\,j\in\mathbb{N}_{n}$, $i\neq j$, $c(\mathbf{x},\,\mathbf{y}_{i})\leq c(\mathbf{x},\,\mathbf{y}_{j})+c(\mathbf{y}_{i},\,\mathbf{y}_{j})$. Hence, one consequence of the triangle inequality is that $g_{ij}(\mathbf{x})\leq c(\mathbf{y}_{i},\,\mathbf{y}_{j})$ for all $\mathbf{x}\in A$, $i,\,j\in\mathbb{N}_{n}$ such that $i\neq j$. Therefore, $\{\,\mathbf{x}\in A\;|\;g_{ij}(\mathbf{x})=k\,\}=\varnothing$ if $k<-c(\mathbf{y}_{i},\,\mathbf{y}_{j})$ or $k>-c(\mathbf{y}_{i},\,\mathbf{y}_{j})$. This implies

[TABLE]

By Theorem 3.2, $\mu(B)=0$. Thus, for any $i,\,j\in\mathbb{N}_{n}$, $i\neq j$, $\mu(A_{ij})=0$. Since the problem assumes nothing about the probability density $\nu$, it must be the case that

[TABLE]

Therefore, $\mu\left(\{\,\mathbf{x}\in A\;|\;g_{ij}(\mathbf{x})=k\,\}\right)=0$ for all $i,\,j\in\mathbb{N}_{n}$, $i\neq j$, and for all $k\in\mathbb{R}$, and uniqueness follows from Corollary 4 of [2]. ∎

4. Conclusions

This paper resolves issues of partitioning and uniqueness for semi-discrete transport problems using a large class of ground cost functions: the $p$-norms. If the cost function is a $p$-norm with $p\in(1,\,\infty)$, the above arguments ensure that $\mu$-a.e. unique solutions exist for semi-discrete transport problems. As the examples show, if the cost function is a $p$-norm with $p=1$ or $p=\infty$, the solution may or may not constitute a $\mu$-a.e. unique partition of the continuous space.