This is the follow up to this post on the completion of the space of Radon measures with respect to transport norm. In the that post we have seen that $\displaystyle \mathrm{KR}_0(X)^*\cong \mathrm{Lip}_0(X),$

i.e. that the completion of the Radon measure with zero total mass with respect to he dual Lipschitz norm $\displaystyle \|\mu\|_{\mathrm{Lip}_0}^* = \sup\{\int f{\mathrm d}\mu\ :\ \mathrm{Lip}(f)\leq 1,\ f(e)=0\}$

where ${e\in X}$ is some base point in the metric space ${X}$.

Recall that on a compact metric space ${(X,d)}$ we have ${\mathfrak{M}(X)}$, the space of Radon measures on ${X}$. The Kantorovich-Rubinstein norm for measure is defined by $\displaystyle \|\mu\|_{\mathrm{KR}} = \sup\{\int f{\mathrm d} \mu\ :\ \mathrm{Lip}(f)\leq 1,\ \|f\|_\infty\leq 1\}.$

Theorem 1 For a compact metric space ${(X,d)}$ it holds that $\displaystyle \mathrm{KR}(X)^* \cong \mathrm{Lip}(X).$

Proof: We give an explicit identification for ${\mathrm{KR}(X)^*}$ as follows:

1. Define a Lipschitz function from an element of ${\mathrm{KR}(X)^*}$: For every ${\phi\in\mathrm{KR}(X)^*}$ and ${x\in X}$ we set $\displaystyle (T\phi)(x) = \phi(\delta_x).$

Now we check that by linearity and continuity of ${\phi}$ that for any ${x\in X}$ it holds that $\displaystyle |(T\phi)(x)| = |\phi(\delta_x)|\leq \|\phi\|\|\delta_x\|_{\mathrm{KR}} = \|\phi\|.$

This shows that ${T\phi:X\rightarrow {\mathbb R}}$ is a bounded function. Similarly for all ${x,y\in X}$ we have $\displaystyle |(T\phi)(x)-(T\phi)(y)| = |\phi(\delta_x-\delta_y)|\leq \|\phi\|\|\delta_x-\delta_y\|_{\mathrm{KR}}\leq \|\phi\|\min(2,d(x,y))\leq \|\phi\|d(x,y).$

This shows that ${(T\phi)}$ is actually Lipschitz continuous, and moreover, that ${T:\mathrm{KR}(X)^*\rightarrow\mathrm{Lip}(X)}$ is continuous with norm ${\|T\|\leq 1}$.

2. Define an element in ${\mathrm{KR}(X)^*}$ from a Lipschitz function: We just set for ${f\in\mathrm{Lip}(X)}$ and ${\mu\in\mathfrak{M}(X)}$ $\displaystyle (Sf)(\mu) = \int_Xf{\mathrm d}\mu.$

By the definition of the ${\mathrm{KR}}$-norm we have $\displaystyle |(Sf)(\mu)\leq \|f\|_{\mathrm{Lip}}\|\mu\|_{\mathrm{KR}},$

which shows that ${S(f)}$ can be extended to a continuous and linear functional ${S:\mathrm{KR}(X)\rightarrow{\mathbb R}}$, i.e. ${Sf\in\mathrm{KR}(X)^*}$, and we also have that ${\|S\|\leq 1}$.

3. Check that ${S}$ and ${T}$ invert each other: Finally we check that ${T}$ and ${S}$ are inverses of each other. We begin with ${TS}$: For ${x\in X}$ and ${f\in\mathrm{Lip}(X)}$ observe that $\displaystyle T(Sf)(x) = Sf(\delta_x) = \int_X f{\mathrm d}\delta_x = f(x)$

i.e. ${TS}$ is the identity on ${\mathrm{Lip}(X)}$. Conversely, for any ${\phi\in\mathrm{KR}(X)^*}$ and ${x\in X}$ we check $\displaystyle S(T\phi)(\delta_x) = \int_X T\phi{\mathrm d}\delta_x = (T\phi)(x_0) = \phi(\delta_x).$

By density of the Dirac measures in ${\mathrm{KR}(X)}$ we conclude that indeed ${ST\phi = \phi}$, i.e. ${ST}$ is the identity on ${\mathrm{KR}(X)^*}$. $\Box$