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