Kantorovich–Rubinstein duality on compact metric spaces — Examples

By [F1], $X=[0,1]$ is a compact metric space, so [F2] applies. The point $(0,1)$ is Borel by [F5], and [F3] makes ${\delta }_{(0,1)}$ a Borel probability measure. Its coordinate marginals are ${\delta }_0$ and ${\delta }_1$, so it is admissible; [F3] gives its cost as $d(0,1)=1$. Thus $W_1^X(\mu ,\nu )\le 1$.

Let $h$ be any feasible real-valued function, so it is 1-Lipschitz and hence continuous and Borel measurable. Its parts $h^{+}=\max \{h,0\}$ and $h^{-}=\max \{-h,0\}$ are nonnegative measurable functions by [F4]. Applying [F3] to each part, for $c\in \{0,1\}$ we get $\int h^{+}d{\delta }_c=h^{+}(c)<\mathrm{\infty }$ and $\int h^{-}d{\delta }_c=h^{-}(c)<\mathrm{\infty }$. Thus [F4] makes $h$ integrable against both endpoint masses and gives $\int hd{\delta }_c=h^{+}(c)-h^{-}(c)=h(c)$. In particular, $f(x)=-x$ is feasible and has integrals $f(0)=0$ and $f(1)=-1$, so its objective is $1$. Every feasible $g$ has objective $g(0)-g(1)\le \mathrm{\mid }g(0)-g(1)\mathrm{\mid }\le 1$, so the dual supremum is exactly $1$.

For any admissible coupling $\gamma$, the marginal conditions give $\gamma (\{0\}\times X)=1$ and $\gamma (X\times \{1\})=1$. The complement of their intersection has measure zero: each of the two complements is null, and finite subadditivity follows from the measure axioms [F6]. Their intersection is $\{(0,1)\}$. For any Borel $E$, if $(0,1)\in E$ then $\gamma (E)=1$, while if $(0,1)\mathrm{\notin }E$ then $E$ lies in the null complement of that singleton, so $\gamma (E)=0$; hence $\gamma ={\delta }_{(0,1)}$. Therefore every admissible cost is $1$, and step 1.1 gives $W_1^X(\mu ,\nu )=1$.

Steps 1.2 and 2.1 compute the primal and dual values as $1$, proving the example's claimed equality.

By [F1], $X\times X$ consists of four Borel singletons. Given any admissible Borel coupling $\gamma$, define ${\pi }_{uv}:=\gamma (\{(u,v)\})$ for $u,v\in \{a,b\}$. Finite additivity [F5] makes $\pi$ a nonnegative matrix with total sum $1$; its row and column sums are the masses of the marginals, namely $(p,1-p)$ and $(q,1-q)$ by [F2]. Thus $p-q={\pi }_{ab}-{\pi }_{ba}$. By [F4], its cost is $D({\pi }_{ab}+{\pi }_{ba})\ge D\mathrm{\mid }p-q\mathrm{\mid }$, so [F3] gives $W_1^X(\mu ,\nu )\ge D\mathrm{\mid }p-q\mathrm{\mid }$.

Every real-valued function $f$ on the finite discrete set is Borel measurable, and $f^{+}$ and $f^{-}$ are nonnegative simple functions. For either $\lambda =\mu$ or $\lambda =\nu$, apply the nonnegative finite-sum formula [F4] separately to $f^{+}$ and $f^{-}$ to get $\int f^{\pm }d\lambda =f^{\pm }(a)\lambda (\{a\})+f^{\pm }(b)\lambda (\{b\})<\mathrm{\infty }$. Subtracting these two finite quantities by the signed-integral definition in [F4] proves $\int fd\lambda =f(a)\lambda (\{a\})+f(b)\lambda (\{b\})$. Using the atom masses [F2], the objective for a feasible $f$ with $f(a)=0$ is therefore $(p-q)(f(a)-f(b))$. Since $\mathrm{\mid }f(a)-f(b)\mathrm{\mid }\le D$, the objective is at most $D\mathrm{\mid }p-q\mathrm{\mid }$. If $p\ge q$, choose $f(b)=-D$; if $p<q$, choose $f(b)=D$. Each choice is feasible and attains $D\mathrm{\mid }p-q\mathrm{\mid }$, including $p=q$.

If $p\ge q$, the matrix $(q,p-q,0,1-p)$ is nonnegative, has the required row sums $(p,1-p)$ and column sums $(q,1-q)$, and costs $D(p-q)$. If $p<q$, the matrix $(p,0,q-p,1-q)$ is nonnegative, has the same required marginals, and costs $D(q-p)$. Either matrix defines a Borel probability on the four-point set by assigning each singleton its matrix entry; countable additivity follows because only finitely many disjoint subsets can be nonempty, and [F5] gives finite additivity. Its row and column sums give the required marginals, and [F4] gives its stated cost. Thus [F3] yields $W_1^X(\mu ,\nu )\le D\mathrm{\mid }p-q\mathrm{\mid }$, and step 1.1 gives equality.

Steps 1.1 and 2.1 give the primal value $D\mathrm{\mid }p-q\mathrm{\mid }$, and step 1.2 gives the same dual supremum, including $p=q$ and endpoints $p,q\in \{0,1\}$.