In this post, the first of two related ones, I describe the closure of the space of Radon measures with zero total mass with the respect to the Kantorovich-Rubinstein norm. Somebody asked me what this was after some talk – I did not know the answer and couldn’t figure it out, so I asked this over on mathoverflow, got the right pointers, and here’s the post. A great deal of the post comes from the book Lipschitz Algebras by Nik Weaver (the very same Nik Weaver who answered over at MO) but beware that the notation is slightly different in this post (especially the usage of ${\mathrm{KR}}$).

1. The Kantorovich-Rubinstein norm on the space of Radon measures

Let ${(X,d)}$ be a metric space and denote by ${\mathfrak{M}(X)}$ the space of Radon measures on ${X}$. We will work with compact ${X}$ throughout this post, although in many places compactness is not needed. Following Bourbaki we say that the space ${\mathfrak{M}(X)}$ is defined as the dual of the space ${C(X)}$ of continuous functions on ${X}$, equipped with the supremum norm. The dual pairing of a measure ${\mu}$ and a continuous function ${f}$ is then

$\displaystyle \langle \mu,f\rangle = \int_X f{\mathrm d} \mu$

(and the integral can also be understood in terms of standard measure theory where a measure is defined by a set function of a ${\sigma}$-algebra). The norm of ${\mathfrak{M}(X)}$ is

$\displaystyle \|\mu\|_{\mathfrak{M}} = \sup\{\int_X f{\mathrm d}\mu\ :\ \|f\|_\infty\leq 1\}$

and is called variation norm or Radon norm (sometimes also total variation, not to be confused with the seminorm on the space of functions with bounded variation with a similar name). The variation norm captures some of the topological structure of ${X}$ but does not take into account the metric structure of ${X}$ (in fact, in can be defined on compact topological spaces ${X}$).

One can define further norms and semi-norms on ${\mathfrak{M}(X)}$ and subspaces thereof and two of them that are particularly suited to capture the underlying metric structure of ${X}$ are:

1. The dual Lipschitz-norm

$\displaystyle \|\mu\|_{\mathrm{Lip}_0}^* = \sup\{\int f{\mathrm d} \mu\ :\ \mathrm{Lip}(f)\leq 1\}$

gives finite values on the closed subspace ${\mathfrak{M}_0(X)}$ of measures with total mass equal to zero, i.e. ${\mu(X)=0}$.

2. The Kantorovich-Rubinstein norm

$\displaystyle \|\mu\|_{\mathrm{KR}} = \sup\{\int f{\mathrm d} \mu\ :\ \mathrm{Lip}(f)\leq 1,\ \|f\|_\infty\leq 1\}$

gives finite values for all Radon measures.

Actually, it is not totally trivial to show that these are indeed norm but also not too hard (you may look up this fact for ${\mathrm{Lip}_0^*}$ in Proposition 2.3.2 in the above mentioned book Lipschitz Algebras and the case of ${\mathrm{KR}}$ is basically similar).

Obviously ${\|\mu\|_{\mathrm{KR}}\leq \|\mu\|_{\mathfrak{M}}}$. The reversed inequality is not true and also does not even hold up to a constant if ${X}$ has a non-constant convergent sequence. This can be seen by observing that for ${x\neq y}$ one has ${\|\delta_x - \delta_y\|_{\mathfrak{M}} = 2}$ while ${\|\delta_x - \delta_y\|_{\mathrm{KR}} = \|\delta_x-\delta_y\|_{\mathrm{Lip}_0}^* \leq d(x,y)}$.

While ${(\mathfrak{M}(X),\|\cdot\|_{\mathfrak{M}})}$ is a Banach space (by its very definition as a dual space) it is not a priori clear if ${(\mathfrak{M}(X),\|\cdot\|_{\mathrm{KR}})}$ is also a Banach space. In fact, it is clear that it is not:

Theorem 1 The space ${(\mathfrak{M}(X),\|\cdot\|_{\mathrm{KR}})}$ is complete if and only if ${X}$ is finite, similarly for ${(\mathfrak{M}_0(X),\|\cdot\|_{\mathrm{Lip}_0}^*)}$.

Proof: Let ${X}$ be finite with ${n}$ elements, say. Then ${\mathfrak{M}(X)}$ is naturally identified with ${{\mathbb R}^n}$ and there all norm are equivalent, which shows completeness with all norms. Now let ${X}$ be infinite. Since ${X}$ is also compact, there are, for every ${\epsilon>0}$ two points ${x,y\in X}$ with ${d(x,y)<\epsilon}$. For these points we’ve already observed that ${\|\delta_x - \delta_y\|_{\mathrm{KR}} <\epsilon}$ while ${\|\delta_x - \delta_y\|_{\mathfrak{M}} = 2}$. This shows that the identity ${(\mathfrak{M}(X),\|\cdot\|_{\mathfrak{M}}) \rightarrow (\mathfrak{M}(X),\|\cdot\|_{\mathrm{KR}})}$ is a continuous linear bijection but does not have a continuous inverse. So the open mapping theorem shows that ${(\mathfrak{M}(X),\|\cdot\|_{\mathrm{KR}})}$ can not be complete. The argument for ${(\mathfrak{M}_0(X),\|\cdot\|_{\mathrm{Lip}_0}^*)}$ is the same. $\Box$

This raises two related questions:

1. What is the completion of the space of Radon measures ${\mathfrak{M}(X)}$ with respect to Kantorovich-Rubinstein norm ${\|\cdot\|_\mathrm{KR}}$?
2. What is the completion of the space of Radon measures with zero mass ${\mathfrak{M}_0(X)}$ with respect to the dual Lipschitz norm ${\|\cdot\|_{\mathrm{Lip}_0}^*}$?

In this post we’ll see the answer to the second question.

Note that the proof above indicates some important phenomenon: Diracs with opposite signs that are getting closer seem to be a particular thing here. Indeed, if ${x_k}$ and ${y_k}$ are two sequences in ${X}$ with ${x_k-y_k\rightarrow 0}$ then for

$\displaystyle \mu_k = \delta_{x_k} - \delta_{y_k}$

it holds that ${\|\mu_k\|_{\mathrm{KR}}\leq d(x_k,y_k)\rightarrow 0}$ (similarly for ${\|\cdot\|_{\mathrm{Lip}_0}^*}$) but ${\mu_k}$ is not converging to zero with respect to ${\|\cdot\|_{\mathfrak{M}}}$. It holds that ${\mu_k}$ does converge weakly to zero if ${x_k}$ (and hence ${y_k}$) does converge to something, but in the case that ${x_k}$ (and hence ${y_k}$) does not converge, ${\mu_k}$ does not even converge weakly.

We’ll come to a description of the completion in a minute but first we introduce several notions that will be used.

2. Spaces of Lipschitz functions

For a compact metric space ${(X,d)}$ we say that ${f:X\rightarrow{\mathbb R}}$ is Lipschitz continuous if it has finite Lipschitz-constant

$\displaystyle \mathrm{Lip}(f) = \sup_{x\neq y}\frac{|f(x)-f(y)|}{d(x,y)}.$

The theorem of Arzela-Ascoli shows that the space ${\mathrm{Lip}(X)}$ of all Lipschitz continuous functions is a Banach space, when equipped with the norm

$\displaystyle \|f\|_{\mathrm{Lip}} = \max(\|f\|_\infty,\mathrm{Lip}(f)).$

If we additionally mark a distinguished base point ${e\in X}$ we can also consider the space

$\displaystyle \mathrm{Lip}_0(X) = \{f\in \mathrm{Lip}(X)\ :\ f(e)=0\}.$

On this space the Lipschitz constant itself is a norm (and not only a semi-norm) and we denote it by

$\displaystyle \|f\|_{\mathrm{Lip}_0} = \mathrm{Lip}(f).$

In fact the spaces ${\mathrm{Lip}}$ and ${\mathrm{Lip}_0}$ are really closely related since in the case of metric spaces of finite diameter one space can be turned into the other and vice versa. The idea is to enhance a given metric space ${X}$ with an additional base point that we put in some way “right in the middle of ${X}$” or, maybe more accurately, “at equal distance of anything” as follows:

Theorem 2 For a compact metric space ${(X,d)}$ (with no special base point) and diameter ${\mathrm{diam}(X) = D}$ denote ${X^+ := X\cup\{e\}}$ for some new point ${e}$ and let ${d^+:X^+\times X^+\rightarrow{\mathbb R}}$ be equal to ${d}$ on ${X}$ and satisfy ${d^+(x,e)=D/2}$ for all ${x\in X}$. Then:

1. ${(X^+,d^+)}$ is a compact metric space with diameter ${\mathrm{diam}(X^+)=D}$.
2. For ${f\in\mathrm{Lip}(X)}$ define ${f^+\in \mathrm{Lip}_0(X^+)}$ by ${f^+(x) = f(x)}$ for ${x\in X}$ and ${f^+(e)=0}$. Then the mapping ${T:f\mapsto f^+}$ is a homeomorphism from ${\mathrm{Lip}(X)}$ to ${\mathrm{Lip}_0(X^+)}$.

Proof:

1. The only thing to check for ${d^+}$ to be a metric is that ${d^+}$ also fulfills the triangle inequality: but since for ${x,y\in X}$ it holds that ${d^+(x,e)+d^+(e,y) = D/2 + D/2 = D\geq d(x,y) = d^+(x,y)}$ this is fulfilled.
2. Let ${f\in \mathrm{Lip}(X)}$. Then

$\displaystyle \frac{|f^+(x)-f^+(e)|}{d^+(x,e)} = \frac{|f(p)-0|}{D/2}\leq \frac2D\|f\|_\infty \leq \frac2D\|f\|_{\mathrm{Lip}}$

and consequently

$\displaystyle \mathrm{Lip}(f^+) \leq\tfrac2D\|f\|_{\mathrm{Lip}}.$

On the other hand for all ${x\in X}$

$\displaystyle |f(x)| = \frac{D}{2}\frac{|f(x)-0|}{D/2}\leq \frac{D}2\mathrm{Lip}(f^+)$

which shows ${\|f\|_\infty\leq\tfrac{D}2\mathrm{Lip}(f^+)}$. Obviously we have ${\mathrm{Lip}(f)\leq \mathrm{Lip}(f^+)}$, which finishes the argument.

$\Box$

3. The Arens-Eells space

That spaces of Lipschitz functions should play a role for the ${\mathrm{KR}}$– and ${\mathrm{Lip}_0^*}$-norm seems pretty obvious—Lipschitz functions are used it the very definition of the norm. Now we come to some space for which the relation with the ${\mathrm{KR}}$– and ${\mathrm{Lip}_0^*}$-norm may not be that obvious but which is indeed very much related.

Definition 3 For a compact metric space ${(X,d)}$ we define a molecule as a function ${m:X\rightarrow{\mathbb R}}$ with finite support with ${\sum_{x\in X} m(x)=0}$.

Obviously, the set of molecules is a linear space.

Any molecule can be identified with a measure on ${X}$ as follows: For a molecule ${m}$ define a measure

$\displaystyle \mu_m = \sum_{x\in X} m(x)\delta_{x}.$

In other words: At every point ${x}$ of the support of ${m}$ we put a delta with weight ${m(x)}$.

Since the support of ${m}$ is finite, the sum is actually finite and in addition, this measure ${\mu_m}$ has zero total measure since

$\displaystyle \mu_m(X) = \sum_{x\in X}m(x)\delta_{x}(X) = \sum_{x\in X}m(x)=0.$

So one may say that the linear space of all molecules is naturally identified with the linear space of all finite linear combinations of deltas with zero total mass.

On the space of molecules we define the so-called Arens-Eells norm. To do so we introduce elementary molecules consisting having a two-element support with values ${\pm 1}$ on them, namely for ${x,y\in X}$

$\displaystyle m_{xy}(z) = \begin{cases} 1, & z=x\\ -1, & z=y\\ 0, & \text{else.} \end{cases}$

Note that any molecule ${m}$ can be written as finite linear combination of elementary molecules. (To do so you may order the support of some given ${m}$ as ${\{x_1,\dots x_n\}}$, start with ${m(x_1)m_{x_1 x_2}}$, then form ${m^1 = m - m(x_1)m_{x_1x_2}}$, observe that ${m^1}$ has a support of size ${n-1}$, and proceed iteratively as started). Importantly, the decomposition of a molecule into elementary molecules is not unique.

Definition 4 For a molecule ${m}$ on ${X}$ we set

$\displaystyle \|m\|_{\AE} = \inf\{\sum_{i=1}^n |a_i|d(x_i,y_i)\ :\ m = \sum_{i=1}^n a_i m_{x_iy_i}\}$

where the infimum is taken over all decomposition of ${m}$ into elementary molecules.

Actually, it is not clear from the start that ${\|\cdot\|_{\AE}}$ is indeed a norm. While homogeneity and the triangle inequality are more or less straight forward, definiteness is not clear from the beginning. We will see that in a minute, but it makes sense to first introduce the completion of the space of molecules with respect to the (semi-)norm ${\|\cdot\|_{\AE}}$ since its dual is indeed familiar. Since we do not know yet that ${\|\cdot\|_{\AE}}$ is a norm, we have to be careful with the completion:

Definition 5 The Arens-Eells space ${\AE(X)}$ on a compact metric space is the completion of the space of molecules with respect to the seminorm ${\|\cdot\|_{\AE}}$ modulo elements with zero norm.

The great thing is that the dual of ${\AE(X)}$ is a well known space:

Theorem 6 For a compact metric space ${(X,d)}$ with some base point ${e}$ it holds that

$\displaystyle \AE(X)^* \cong \mathrm{Lip}_0(X).$

On bounded sets, weak* convergence in ${\mathrm{Lip}_0(X)}$ agrees with pointwise convergence.

Proof: We represent elements of ${\AE(X)^*}$ as Lipschitz functions by the mapping ${T_1:\AE(X)^*\rightarrow \mathrm{Lip}_0(X)}$ be setting for ${\phi\in \AE(X)^*}$ and ${x\in X}$

$\displaystyle (T_1\phi)(x) = \phi(m_{xe})$

Note that ${(T_1\phi)(e) = \phi(m_{ee}) = 0}$. By the definition of ${\|\cdot\|_{\AE}}$ it holds that ${\|m_{xy}\|_{\AE}\leq d(x,y)}$ and hence,

$\displaystyle |(T_1\phi)(x)-(T_1\phi)(y)| = |\phi(m_{xe} - m_{ye})| = |\phi(m_{xy})|\leq \|\phi\|d(x,y)$

which shows that ${\mathrm{Lip}(T_1\phi)\leq \|\phi\|}$ and we conclude that ${T_1\phi\in\mathrm{Lip}_0(X)}$ and also that ${\|T_1\|\leq 1}$. Conversely, we now define a map that associates to every Lipschitz function vanishing at the base point an element in ${\AE(X)^*}$: Define ${T_2:\mathrm{Lip}_0(X)\rightarrow \AE(X)^*}$ for any molecule ${m}$ and any ${f\in \mathrm{Lip}_0(X)}$ by

$\displaystyle (T_2f)(m) = \sum_{x\in X} f(x)m(x).$

Note that for a decomposed molecule ${m = \sum_{i=1}^n a_i m_{x_iy_i}}$ we have

$\displaystyle \begin{array}{rcl} |(T_2 f)(m)| & =&|(T_2 f)(\sum_{i=1}^n a_i m_{x_iy_i})\\ & \leq & \sum_{i=1}^n |a_i| |(T_2 f)(m_{x_iy_i})|\\ & = &\sum_{i=1}^n |a_i| |f(x_i) - f(y_i)|\\ & \leq & \mathrm{Lip}(f) \sum_{i=1}^n |a_i| d(x_i,y_i). \end{array}$

Taking the infimum over all decompositions, we conclude ${|(T_2 f)(m)|\leq \mathrm{Lip}(f)\|m\|_{\AE}}$ showing that ${T_2 f\in \AE(X)^*}$ and ${\|T_2f\|\leq\mathrm{Lip}(f)}$, i.e, ${\|T_2\|\leq 1}$. Now let’s look at ${T_1T_2}$: For ${x\in X}$ and ${f\in\mathrm{Lip}_0(X)}$ it holds that ${T_1(T_2 f)(x) = (T_2 f)(m_{xe}) = f(x) - f(e) = f(x)}$, i.e. ${T_1T_2 = I}$. Also for ${\phi\in \AE(X)^*}$ and a molecule ${m}$: ${T_2(T_1\phi)(m) = \sum_{x\in X}(T_1\phi)(x)m(x) = \sum_{x\in X}\phi(m_{xe})m(x) = \phi(\sum_{x\in X}m(x)m_{xe})= \phi(m)}$. So ${T_2}$ and ${T_1}$ are indeed inverses and hence provide identifications of ${\AE(X)^*}$ and ${\mathrm{Lip}_0(X)}$. Now consider a sequence of function ${f_n\in \mathrm{Lip}_0(X)}$ that converges weakly* to ${f}$. This says that (in the notation above) for any molecule ${m}$ it holds that ${(T_2f_n)(m)\rightarrow (T_2 f)(m)}$. Particularly for the molecules ${m_{xe}}$ we have

$\displaystyle (T_2 f_n)(m_{xe}) = \sum_{y\in X} f_n(y)m_{xe}(y) = f_n(y).$

It follows that ${f_n(x)\rightarrow f(x)}$ for all ${x}$. This shows that weak* convergence implies pointwise convergence. If, conversely, ${f_n}$ converges pointwise, we conclude that ${(T_2 f_n)(m_{xe}) \rightarrow (T_2 f)(m_{xe})}$ for all ${m_{xe}}$. By definition, these are dense in ${\AE(X)}$ and since we assume that the ${f_n}$ are bounded by assumption, this implies weak* convergence. $\Box$

The above proof also shows that ${\|\cdot\|_{\AE}}$ is indeed a norm and moreover, that it can be written as

$\displaystyle \|m\|_{\AE} = \sup\{\sum_{x\in X} f(x)m(x)\ :\ f(e)=0,\ \mathrm{Lip}(f)\leq 1\}$

(also, by Arzela-Ascoli, there is a Lipschitz function ${f}$ with ${\mathrm{Lip}(f)\leq 1}$ and ${f(e)=0}$ where the supremum is attained, but we will not need this). This shows the following fact:

Corollary 7 For any molecule ${m}$ on a compact metric space ${X}$ with base point ${e}$ and the corresponding measure ${\mu_m}$ it follows that

$\displaystyle \|m\|_{\AE} = \|\mu_m\|_{\mathrm{KR}}.$

4. The completion of ${\mathfrak{M}_0(X)}$ under the dual Lipschitz norm

Now we come to the answer of the second question posed in the beginning. We define by ${\mathrm{KR}_0(X)}$ the completion of ${\mathfrak{M}_0(X)}$ (the space of Radon measures with zero total mass) with respect to ${\|\cdot\|_{\mathrm{Lip}_0}^*}$ and aim at a characterization of this space.

Theorem 8 It holds that ${\AE(X)\cong \mathrm{KR}_0(X)}$ isometrically.

Proof: We first show that any measure ${\mu}$ on ${X}$ with zero total mass can be approximated in the ${\mathrm{Lip}_0^*}$-norm by measures ${\mu_m}$ defined my molecules ${m}$. So let ${\epsilon>0}$ and ${\mu\in \mathrm{KR}_0(X)}$. We set ${\epsilon' = \epsilon/\|\mu\|_{\mathfrak{M}}}$. Now we cover ${X}$ by an ${\epsilon'}$-net, that is, we take ${x_1,\dots,x_n}$ such that balls ${B_k}$ around ${x_k}$ with radius ${\epsilon'}$ cover ${X}$. From this cover we define by ${V_1=U_1}$, ${V_k = U_k\setminus(U_1\cup\cdots\cup U_{k-1})}$ a disjoint cover of ${X}$ (i.e. a partition). Now we define a molecule ${m}$ by

$\displaystyle m(x) = \left\{ \begin{array}{ll} \mu(V_k), & \text{if}\ x=x_k\\ 0, & \text{else.} \end{array} \right.$

Clearly ${m}$ is a molecule (its support is the ${n}$ point ${x_k}$ and the sum ${\sum_{x}m(x)}$ equals ${\mu(X)=0}$). To see that ${\mu_m}$ approximates ${\mu}$ in the ${\mathrm{Lip}_0^*}$-norm we start as follows: For any function ${f}$ with ${\mathrm{Lip}(f)\leq 1}$ we see

$\displaystyle \begin{array}{rcl} |\int_{V_k}f{\mathrm d}(\mu-\mu_m)| &=& |\int_{V_k} f{\mathrm d}\mu - \int_{V_k}f{\mathrm d}\mu_m|\\ &=& |\int_{V_k} f{\mathrm d}\mu - f(x_k)\mu(V_k)|\\ &=& |\int_{V_k} (f-f(x_k)){\mathrm d}\mu|\\ &\leq& |\int_{V_k} \mathrm{Lip}(f)\epsilon'{\mathrm d}\mu| = \epsilon'\|\mu\|_{\mathfrak{M}(V_k)}. \end{array}$

Summing this over ${k}$ we get for every ${f}$ with ${\mathrm{Lip}(f)\leq 1}$

$\displaystyle |\int_X f{\mathrm d}(\mu-\mu_m)|\leq \epsilon'\|\mu\|_{\mathfrak{M}} = \epsilon.$

Taking the supremum over these ${f}$ gives ${\|\mu-\mu_m\|_{\mathrm{Lip}_0}^*\leq \epsilon}$ as desired. Corollary 7 shows that ${\|m|\|_{\AE}=\|\mu_m\|_{\mathrm{Lip}_0}^*}$ and from the above we see that these measures are dense in ${\mathrm{KR}_0(X)}$. Hence, the correspondence ${m\mapsto\mu_m}$ is an isometry on dense subsets of ${\AE(X)}$ and ${\mathrm{KR}_0(X)}$ which proves the assertion. $\Box$

Finally, we come to the announced description of the space ${\mathrm{KR}_0(X)}$ (defined as completion of ${(\mathfrak{M}_0(X),\|\cdot\|_{\mathrm{Lip}_0}^*)}$): It is equal to the Arens-Eells space and hence, its dual space is the space of Lipschitz functions rooted on a base point.

Corollary 9 For a compact metric space ${(X,d)}$ with basepoint it holds that

$\displaystyle \mathrm{KR}_0(X)^*\cong\mathrm{Lip}_0(X).$

To conclude this post, I just mention that the techniques used above are related to the Kuratowski embedding and the related Kuratowski–Wojdysławski embedding that embed metric spaces into certain Banach spaces.