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Since Lipschitz functions are continuous, the feasible class here is smaller, giving $m_L\le m_d$. To prove the reverse inequality, take any continuous feasible pair $f,g$ and perform Lipschitz regularization of transport potentials:The minimum exists by compactness. For every $x,x^{\prime }$ the triangle inequality gives $h(x)\le d(x,x^{\prime })+h(x^{\prime })$; interchanging $x,x^{\prime }$ proves $|h(x)-h(x^{\prime })|\le d(x,x^{\prime })$. Thus $h$ has Lipschitz constant at most $1$.

The original constraint yields $f(x)\le h(x)$, and evaluating the minimum at $y=x$ gives $h(x)\le -g(x)$. Moreover, $h(x)-h(y)\le d(x,y)$, so $(h,-h)$ is a feasible Lipschitz pair withTaking the supremum over the original continuous pairs proves $m_L(P,Q)=m_d(P,Q)=W(P,Q)$. Notice that this is an exact regularization, with no density or approximation assumption about the initial pair.

Every function $h$ with Lipschitz constant at most $1$ yields a feasible pair $(h,-h)$, and replacing $h$ by $-h$ reverses the objective. HenceConversely, the Lipschitz regularization of transport potentials in part iii dominates every feasible pair by $(h,-h)$ with $\Vert h{\Vert }_L\le 1$, so $m_L\le \gamma$. Together with the preceding parts this proves the Kantorovich–Rubinstein theorem here in all four requested forms:The seminorm convention is essential. If instead $\Vert h{\Vert }_L$ meant $\Vert h{\Vert }_{\infty }+\mathrm{Lip}(h)$, the equality would fail: on a two-point space at distance $2$, with point masses at the two different points, $W=2$ whereas the latter unit-ball supremum is $1$. Indeed, an oscillation $a$ needs supremum norm at least $a/2$ and Lipschitz constant $a/2$, so the sum norm bounds $a$ by $1$, attained by values $-1/2,1/2$. This distinguishes the Lipschitz seminorm used above from a full bounded-Lipschitz norm.