# Do distance functionals separate probability measures?

Let $$(\Omega,d)$$ be a compact metric space and $$\mathcal P(\Omega)$$ its space of Borel probability measures. Let $$D=\{ d_p\mid p\in\Omega\}$$ where $$d_p(x)=d(p,x)$$ be the set of all "distance functionals". As usual, we can think of $$D$$ acting on $$\mathcal P(\Omega)$$ (or vice versa) via integration i.e. $$\langle d_p,\mu\rangle = \int_\Omega d_p(x)\,\mathrm d\mu(x)$$.

# Title Question

Does $$D$$ acting on $$\mathcal P(\Omega)$$ via integration separate points?

Or equivalently,

If $$\mu,\nu \in \mathcal P(\Omega)$$ and $$\langle d_p,\mu\rangle = \langle d_p,\nu\rangle$$ for all $$p\in \Omega$$, then must $$\mu=\nu$$?

# Alternative Formulations

There are a few other ways to frame the question as well.

### Probabilistic Formulation

Rewriting all integrals as expectations the question becomes,

If $$\mathbb E_{X\sim\mu}[d_p(X)] = \mathbb E_{Y\sim\nu}[d_p(Y)]$$ for all $$p\in \Omega$$, then must $$\mu=\nu$$?

In other words, does knowing the expected distance to a point for all points determine the measure?

### Geometric Formulation

Recall that the 1-Wasserstein distance is given by $$W_1(\mu,\nu) = \inf_{\gamma\in\Gamma(\mu,\nu)} \int_{\Omega\times\Omega} d(x,y) \,\mathrm d\gamma(x,y)$$ where $$\Gamma(\mu,\nu)$$ is the set of couplings between $$\mu$$ and $$\nu$$ i.e. Borel probability measures on $$\Omega\times\Omega$$ with marginals $$\mu$$ and $$\nu$$ respectively. Since the product measure $$\delta_p\otimes\mu$$ is the unique coupling between a Dirac delta measure $$\delta_p$$ and $$\mu$$, we have that

$$W_1(\delta_p,\mu)=\int_{\Omega\times\Omega} d(x,y)\,\mathrm d(\delta_p\otimes\mu)(x,y)=\int_\Omega d(p,y)\,\mathrm d\mu(y)=\langle d_p,\mu\rangle$$

Now the question can be stated geometrically as

If $$W_1(\delta_p,\mu)=W_1(\delta_p,\nu)$$ for all $$p\in \Omega$$, then must $$\mu=\nu$$?

In other words, does knowing the $$W_1$$ distance to the extreme points of $$\mathcal P(\Omega)$$ completely determine the probability measure?

### Integral Transform Forumlation

Define the distance transform of $$\mu\in\mathcal P(\Omega)$$ as the function $$\phi_\mu:\Omega\to\mathbb R$$ given by $$\phi_\mu(p) = \int_\Omega d(p,x)\,\mathrm d\mu(x)$$. The question can now be restated as,

Is the distance transform injective on $$\mathcal P(\Omega)$$?

Moreover, by the geometric formulation we have $$\phi_\mu(p) = W_1(\delta_p,\mu)$$. We will use the weak-$$*$$ topology for $$\mathcal P(\Omega)$$ (which coincides with the $$W_1$$ topology). Since the map $$p\mapsto \delta_p$$ is an embedding of $$\Omega$$ into $$\mathcal P(\Omega)$$, it follows that $$\phi_\mu:\Omega\to\mathbb R$$ is continuous. Denote the distance transform by $$\Phi(\mu)=\phi_\mu$$. Since $$\mathcal P(\Omega)$$ is compact Hausdorff and $$C(\Omega)$$ is Hausdorff we can restate the question as

If $$\Phi:\mathcal P(\Omega)\to C(\Omega)$$ is continuous, is it an embedding?

# Final Thoughts

Are any of these equivalent statements true? I have unfortunately only been able to reformulate the question and have not identified any clear proof, though I wouldn't be surprised if there is an easy one I'm overlooking. The geometric formulation of the problem leads me to believe that $$D$$ does indeed separate points in $$\mathcal P(\Omega)$$. However, if the answer is affirmative then I feel the resulting nice properties of $$\Phi$$ would make it something that would be easy to look up. Any insight would be appreciated.

Update: In light of George Lowther's elegant 4-point counter-example and Pietro Majer's affirmative answer for $$\Omega=[0,1]$$, it would be interesting to better understand what factors determine whether the underlying metric space yields an affirmative answer.

George's counter-example can be extended to counter-examples where $$\Omega$$ is a sphere (with intrinsic metric). Thus, requiring $$\Omega$$ to be positive-dimensional, a manifold, connected, path-connected, simply-connected, etc, will not make the issue go away. On the other hand, Pietro suspects that the answer is again affirmative in the case when $$\Omega$$ is a compact convex subset of Euclidean space.

• I don't know the answer, but one can reduce the problem to the case that $\Omega$ is finite, since probability measures with finite support are weak*-dense. – Michael Greinecker Aug 17 at 8:24
• Isn't this a consequence of the Riesz representation theorem and the Stone-Weierstraß approximation theorem? – Jochen Glueck Aug 17 at 8:45
• But Stone-Weierstraß only implies that the algebra generated by the distance functionals is dense. One needs the weak$^*$ density of the generated subspace. – Jochen Wengenroth Aug 17 at 9:15
• @JochenWengenroth: You're right, of course. I confused a few arguments and thus thought this would not be a problem; but of course, it actually is a problem... – Jochen Glueck Aug 17 at 9:29

No. Suppose that $$\Omega$$ consists of four points arranged in a square, where adjacent points have distance 1 between them and opposite points have distance 2. Specifically, if the points are labeled A,B,C,D then \begin{align} & d(A,C)=d(B,D)=2,\\ & d(A,B)=d(B,C)=d(C,D)=d(D,A)=1. \end{align} For example, A,B,C,D could be equally spaced around a circle, using the internal circle metric.

There are precisely two probability measures assigning probability 1/2 to each of two opposite points and probability zero to the remaining two points. \begin{align} & \mu(\{A\})=\mu(\{C\}) = 1/2,\ \mu(\{B,D\})=0,\\ & \nu(\{B\})=\nu(\{D\})=1/2,\ \nu(\{A,C\})=0. \end{align} You can check that these two measures give the same integral for all $$`$$distance functions'. The average distance from every point is equal to 1 under both of these.

• Elegant counter-example.This idea can be pushed a bit further as well though it was perhaps implicit: The empirical measure of a pair of antipodal points on the unit circle $\delta_p / 2+\delta_{-p} / 2$ produces an expected intrinsic distance of $\pi/2$ to any point on the circle. Thus, any two "empirical measure of antipodal points" cannot be separated by distance functionals. This shows the obstruction isn't simply due to finiteness of $\Omega$ and that the obstruction can't be avoided by requiring the underlying space to be nonzero dimensional, a manifold, connected, path-connected, etc. – Christian Bueno Aug 18 at 8:29
• Actually, this idea using antipodal points works for spheres as well. Thus being simply-connected is not a sufficient condition to ensure an affirmative answer either. – Christian Bueno Aug 18 at 8:48

On the positive side, the answer is affirmative if $$\Omega$$ is the unit interval $$[0,1]$$ with its standard distance. In this case $$\phi_\mu$$ is a convex $$1$$-Lipschitz function (in fact, it is also defined for all $$p\in\mathbb{R}$$, with $$\phi'(p)=\mathrm{sgn} p$$ for $$p\notin[0,1]$$), with left and right derivatives $$\phi_-'(p)=\mu[0,p)-\mu[p,1]= 2\mu[0,p)-1$$ $$\phi_+'(p)=\mu[0,p] -\mu (p,1]= 2\mu[0,p] -1=1-2\mu(p,1]$$ so that $$\mu$$ is determined on all intervals, hence on all Borel subsets.

Conversely, note that any convex function $$\phi$$ as above
may be written in the form $$\phi(p)=\int_{[0,1]}|t-p|dm(t)$$ for some Borel probability measure $$m$$ on $$[0,1]$$. This because $$g:= \frac{1}{2}\big(1-\phi_+'\big)$$ is a nonnegative bounded cadlag function, so there is a Borel probability function $$m$$ such that $$g(p)=m(p,1]$$, whence $$\phi(p)=\int_{[0,1]}|t-p|dm(t)$$ follows easily from the above relations.

I'd guess the answer is also affirmative for $$\Omega$$ a convex compact set of $$\mathbb{R}^n$$ with the Euclidean distance.