[proofplan] We first check that every $1$-Lipschitz [test function](/page/Test%20Function) is integrable against both measures, so the dual expression is well-defined. The inequality from right to left follows directly from the defining Lipschitz estimate $f(y)-f(x)\le d(x,y)$, integrated against an arbitrary coupling. For the reverse inequality, we invoke the general [Kantorovich duality](/theorems/6799) theorem for non-negative lower semicontinuous costs on Polish spaces, applied to the cost $c(x,y)=d(x,y)$. The two-potential dual problem is then reduced to a single $1$-Lipschitz potential by taking the $d$-Lipschitz envelope of one potential. [/proofplan] [step:Verify that the Lipschitz test integrals are finite] Fix a point $x_0 \in X$. Let $f:X\to\mathbb R$ be $1$-Lipschitz. Since $f$ is Lipschitz, it is continuous and therefore $\mathcal{B}(X)$-measurable. For every $x\in X$, the Lipschitz condition gives \begin{align*} |f(x)| \le |f(x_0)| + d(x,x_0). \end{align*} Since $\mu$ and $\nu$ have finite first moments, this estimate implies \begin{align*} \int_X |f|\,d\mu < \infty \end{align*} and \begin{align*} \int_X |f|\,d\nu < \infty. \end{align*} Thus both integrals in the displayed supremum are finite [real numbers](/page/Real%20Numbers). The same first-moment assumption also implies that $W_1(\mu,\nu)<\infty$. Indeed, the product probability measure $\mu\otimes\nu$ belongs to $\Pi(\mu,\nu)$, and the triangle inequality gives \begin{align*} d(x,y) \le d(x,x_0)+d(y,x_0) \end{align*} for all $(x,y)\in X\times X$. Therefore Tonelli's theorem applies to the non-negative functions involved and yields \begin{align*} \int_{X\times X} d(x,y)\,d(\mu\otimes\nu)(x,y) \le \int_X d(x,x_0)\,d\mu(x)+\int_X d(y,x_0)\,d\nu(y) < \infty. \end{align*} [/step] [step:Integrate the pointwise Lipschitz bound against an arbitrary coupling] Let $\pi\in\Pi(\mu,\nu)$ be arbitrary, and let $f:X\to\mathbb R$ be $1$-Lipschitz. Since $\pi$ has first marginal $\mu$ and second marginal $\nu$, the marginal identities give \begin{align*} \int_{X\times X} f(y)\,d\pi(x,y)=\int_X f\,d\nu \end{align*} and \begin{align*} \int_{X\times X} f(x)\,d\pi(x,y)=\int_X f\,d\mu. \end{align*} The Lipschitz condition gives $f(y)-f(x)\le d(x,y)$ for every $(x,y)\in X\times X$. Integrating this pointwise inequality with respect to $\pi$ gives \begin{align*} \int_X f\,d\nu-\int_X f\,d\mu \le \int_{X\times X} d(x,y)\,d\pi(x,y). \end{align*} Taking the infimum over all $\pi\in\Pi(\mu,\nu)$ yields \begin{align*} \int_X f\,d\nu-\int_X f\,d\mu \le W_1(\mu,\nu). \end{align*} Finally, taking the supremum over all $1$-Lipschitz functions $f:X\to\mathbb R$ gives \begin{align*} \sup\left\{\int_X f\,d\nu-\int_X f\,d\mu : f:X\to\mathbb R \text{ is }1\text{-Lipschitz}\right\} \le W_1(\mu,\nu). \end{align*} [/step] [step:Apply Kantorovich duality to the metric cost] Define the cost function \begin{align*} c:X\times X\to[0,\infty),\qquad c(x,y):=d(x,y). \end{align*} Because $d$ is continuous on $X\times X$, the function $c$ is lower semicontinuous. Because $\mu$ and $\nu$ have finite first moments, the cost has finite value for at least one coupling, as shown using $\mu\otimes\nu$ in the first step. By the general Kantorovich duality theorem for non-negative lower semicontinuous costs on Polish spaces, applied to $c=d$ and to the Borel probability measures $\mu,\nu$ (citing a result not yet in the wiki: Kantorovich duality for lower semicontinuous costs on Polish spaces), we have \begin{align*} W_1(\mu,\nu)=\sup\left\{\int_X \varphi\,d\nu-\int_X \psi\,d\mu : \varphi(y)-\psi(x)\le d(x,y)\text{ for all }x,y\in X\right\}, \end{align*} where the supremum is taken over admissible real-valued Borel functions $\varphi,\psi:X\to\mathbb R$ for which the displayed integrals are well-defined. [guided] We now pass from the transport problem to its two-potential dual problem. The cost function used in optimal transport is \begin{align*} c:X\times X\to[0,\infty),\qquad c(x,y):=d(x,y). \end{align*} The general Kantorovich duality theorem for Polish spaces requires the cost to be non-negative and lower semicontinuous. Non-negativity follows from the definition of a metric, and lower semicontinuity follows because $d:X\times X\to\mathbb R$ is continuous. The theorem also requires the primal value not to be forced into the indeterminate form $+\infty-\infty$; here the primal value is finite because the product coupling $\mu\otimes\nu$ has finite cost: \begin{align*} \int_{X\times X} d(x,y)\,d(\mu\otimes\nu)(x,y) \le \int_X d(x,x_0)\,d\mu(x)+\int_X d(y,x_0)\,d\nu(y) < \infty. \end{align*} Applying the general Kantorovich duality theorem to this cost gives the two-potential formula \begin{align*} W_1(\mu,\nu)=\sup\left\{\int_X \varphi\,d\nu-\int_X \psi\,d\mu : \varphi(y)-\psi(x)\le d(x,y)\text{ for all }x,y\in X\right\}. \end{align*} Here $\varphi:X\to\mathbb R$ and $\psi:X\to\mathbb R$ are Borel potentials satisfying the pointwise admissibility constraint. The rest of the proof is therefore not about transport plans anymore; it is about showing that this two-function admissibility condition is exactly captured by one $1$-Lipschitz function. [/guided] [/step] [step:Compress every admissible pair into one Lipschitz potential] Let $\varphi,\psi:X\to\mathbb R$ be an admissible pair in the Kantorovich dual problem, so that \begin{align*} \varphi(y)-\psi(x)\le d(x,y) \end{align*} for all $x,y\in X$. Define the envelope function \begin{align*} f:X\to\mathbb R,\qquad f(y):=\inf_{x\in X}\{\psi(x)+d(x,y)\}. \end{align*} For every $y\in X$, the admissibility inequality gives $\varphi(y)\le \psi(x)+d(x,y)$ for every $x\in X$, hence \begin{align*} \varphi(y)\le f(y). \end{align*} Taking $x=y$ in the definition of $f$ gives \begin{align*} f(y)\le \psi(y). \end{align*} We claim that $f$ is $1$-Lipschitz. Let $y,z\in X$. For every $x\in X$, the triangle inequality gives \begin{align*} \psi(x)+d(x,y)\le \psi(x)+d(x,z)+d(z,y). \end{align*} Taking the infimum over $x\in X$ gives \begin{align*} f(y)\le f(z)+d(y,z). \end{align*} Interchanging $y$ and $z$ gives \begin{align*} f(z)\le f(y)+d(y,z). \end{align*} Therefore $|f(y)-f(z)|\le d(y,z)$, so $f$ is $1$-Lipschitz. Since $\varphi\le f\le \psi$, we obtain \begin{align*} \int_X \varphi\,d\nu-\int_X \psi\,d\mu \le \int_X f\,d\nu-\int_X f\,d\mu. \end{align*} Thus every admissible two-potential value is bounded above by a value achieved by a single $1$-Lipschitz potential. Consequently, \begin{align*} W_1(\mu,\nu)\le \sup\left\{\int_X f\,d\nu-\int_X f\,d\mu : f:X\to\mathbb R \text{ is }1\text{-Lipschitz}\right\}. \end{align*} [/step] [step:Conclude equality of the primal and Lipschitz dual values] The second step proved \begin{align*} \sup\left\{\int_X f\,d\nu-\int_X f\,d\mu : f:X\to\mathbb R \text{ is }1\text{-Lipschitz}\right\} \le W_1(\mu,\nu). \end{align*} The preceding step proved the reverse inequality \begin{align*} W_1(\mu,\nu)\le \sup\left\{\int_X f\,d\nu-\int_X f\,d\mu : f:X\to\mathbb R \text{ is }1\text{-Lipschitz}\right\}. \end{align*} Combining the two inequalities gives \begin{align*} W_1(\mu,\nu)=\sup\left\{\int_X f\,d\nu-\int_X f\,d\mu : f:X\to\mathbb R \text{ is }1\text{-Lipschitz}\right\}. \end{align*} This is the desired Kantorovich-Rubinstein duality formula. [/step]