Sliced and Radon Wasserstein Barycenters of Measures

Abstract

This article details two approaches to compute barycenters of measures using 1-D Wasserstein distances along radial projections of the input measures. The first method makes use of the Radon transform of the measures, and the second is the solution of a convex optimization problem over the space of measures. We show several properties of these barycenters and explain their relationship. We show numerical approximation schemes based on a discrete Radon transform and on the resolution of a non-convex optimization problem. We explore the respective merits and drawbacks of each approach on applications to two image processing problems: color transfer and texture mixing.

Appendices

Appendix 1: Proofs of Section 2

Proof of Proposition 1

From the definition (5), one verifies that

$$\begin{aligned} \text {W}_{\mathbb {R}^d}(\varphi _{s,u} \sharp \mu _1,\varphi _{s,u} \sharp \mu _2) = s \text {W}_{\mathbb {R}^d}(\mu _1,\mu _2), \end{aligned}$$

(49)

so that

$$\begin{aligned} \fancyscript{E}_{s,u}(\mu )&= \sum _{i \in I} \lambda _i \text {W}_{\mathbb {R}^d}( \varphi _{s,u} \sharp \mu _i, \mu )^2\\&= s^2 \sum _{i \in I} \lambda _i \text {W}_{\mathbb {R}^d}( \mu _i, \varphi _{s,u}^{-1} \sharp \mu )^2 = s^2 \fancyscript{E}_{1,0}(\tilde{\mu }), \end{aligned}$$

where we have introduced the following change of variable

$$\begin{aligned} \mu = \varphi _{s,u} \sharp \tilde{\mu } \quad \Longleftrightarrow \quad \tilde{\mu } = \varphi _{s,u}^{-1} \sharp \mu , \end{aligned}$$

(note that $\varphi _{s,u}^{-1} = \varphi _{s^{-1},-s^{-1}u}$). One thus has

$$\begin{aligned} \underset{ \mu }{{{\mathrm{argmin}}}}\; \fancyscript{E}_{s,u}(\mu )&= \varphi _{s,u} \sharp \underset{ \tilde{\mu }}{{{\mathrm{argmin}}}}\; \fancyscript{E}_{1,0}(\tilde{\mu }) \end{aligned}$$

which proves (7). Property (8) is proved similarly. Properties (9) and (11) directly follow from  (8). $\square$

Proof of Proposition 2

We aim at determining $(s^\star ,u^\star )$ such that

$$\begin{aligned} \mu ^\star \in \text {Bar}_{\mathbb {R}^d}^W(\mu _i,\lambda _i)_{i \in I} \quad \text {where} \quad \left\{ \begin{array}{l} \mu ^\star = \varphi ^\star \sharp \mu , \\ \mu _i = \varphi _i \sharp \mu , \end{array} \right. \end{aligned}$$

and where for simplicity we have denoted $\varphi _i = \varphi _{s_i,u_i}$ and $\varphi ^\star = \varphi _{s^\star ,u^\star }$. First, let us notice that

$$\begin{aligned} \varphi _{s,u}(x) = \nabla \left( \frac{s}{2}|| x+u/s ||^2 \right) , \end{aligned}$$

so that the set $\fancyscript{T}$ of maps of the form $\varphi _{s,u}$ is a subset of gradients of convex functions. This point is important since optimal maps between $\mu _i$ and $\mu ^\star$ are characterized as the gradient of convex functions that push forward $\mu _i$ onto $\mu ^\star$, see [32]. Following [1], we thus only need to show that

$$\begin{aligned}&\sum _{i \in I} \lambda _i T_i = \mathrm {Id}_{\mathbb {R}^d} \quad \text {where} \quad T_i = \varphi ^\star \circ \varphi _i^{-1} = \varphi _{ \tilde{s}_i, \tilde{u}_i }\\&\quad \text {where} \quad \left\{ \begin{array}{l} \tilde{s}_i = s^\star s_i^{-1} \\ \tilde{u}_i = u^{\star }-s^\star s_i^{-1} u_i \end{array} \right. \end{aligned}$$

since $T_i \sharp \mu _i = \mu ^\star$ and $T_i \in \fancyscript{T}$ is a gradient of a convex function. So that $\mu ^\star$ is a barycenter if and only if

$$\begin{aligned} \sum _{i \in I} \lambda _i T_i&= \sum _{i \in I} \lambda _i \varphi _{\tilde{s}_i, \tilde{u}_i} \\&= \varphi _{\sum _{i \in I} \lambda _i \tilde{s}_i, \sum _{i \in I} \lambda _i \tilde{u}_i } = \mathrm {Id}_{\mathbb {R}^d} = \varphi _{1,0}. \end{aligned}$$

This in turn is equivalent to the relationships

$$\begin{aligned} \sum _{i \in I} \lambda _i \tilde{s}_i = 1 \quad \text {and} \quad \sum _{i \in I} \lambda _i \tilde{u}_i = 0, \end{aligned}$$

which corresponds to (14). $\square$

Proof of Proposition 3

The proof is done in [1] for $\mu = \mu _j$ for some $j \in I$, which is supposed to be absolutely continuous. It extends to an arbitrary measure $\mu$. $\square$

Proof of Corollary 1

When using $\mu$, the uniform and normalized measure on $[0,1]$, with the notation of Proposition (3), one has $T_i = C_{\mu _i}^+$. This is indeed a classical result for 1-D optimal transport, see for instance [1], Section 6.1. One then recognizes that formula (18) is the same as formula (15). $\square$

Proof of Proposition 4

One has

$$\begin{aligned} \nu ^\star&\in \underset{\nu \in \bar{\fancyscript{M}}_1^+(\varOmega ^d)}{{{\mathrm{argmin}}}}\; \sum _{i \in I} \lambda _i \text {W}_{\varOmega ^d}(\nu _i,\nu )^2 \\&= \underset{\nu \in \bar{\fancyscript{M}}_1^+(\varOmega ^d)}{{{\mathrm{argmin}}}}\; \int _{\mathbb {S}^{d-1}} \sum _{i \in I} \lambda _i \text {W}_{\mathbb {R}}( \nu _i^\theta , \nu ^\theta )^2 \mathrm {d}\theta . \end{aligned}$$

This is equivalent to the fact that for almost all $\theta \in \mathbb {S}^{d-1}$, one has

$$\begin{aligned} \nu ^{\star , \theta } \in \text {Bar}_{\mathbb {R}}^W(\nu _i^\theta ,\lambda _i)_{i \in I}. \end{aligned}$$

$\square$

Proof of Proposition 5

Proof of (20). Similarly to the proof of (7), the proof of (20) is obtained by using the following invariance of the Wasserstein distance on $\varOmega ^d$

$$\begin{aligned} \text {W}_{\varOmega ^d}( \psi _{s,u} \sharp \nu _1, \psi _{s,u} \sharp \nu _2 ) = s \text {W}_{\varOmega ^d}( \nu _1, \nu _2 ). \end{aligned}$$

(50)

Proof of (21). One has that $\nu ^\star \in \text {Bar}_{\varOmega ^d}^W( \psi _{s_i,u_i} \sharp \nu , \lambda _i )_{i \in I}$ is equivalent to

$$\begin{aligned} \text {for almost all}\, \theta \in \mathbb {S}^{d-1}, \,\, (\nu ^\star )^\theta \in \text {Bar}_{\mathbb {R}}^W( \varphi _{s_i,\langle u_i,\,\theta \rangle } \sharp \nu ^\theta , \lambda _i )_{i \in I}. \end{aligned}$$

Using the property of proposition 2 for $d=1$, one obtains that

$$\begin{aligned} \text {Bar}_{\mathbb {R}}^W( \varphi _{s_i,\langle u_i,\,\theta \rangle } \sharp \nu ^\theta , \lambda _i )_{i \in I} \;\ni \; \varphi _{s^\star ,\langle u^\star ,\,\theta \rangle } \sharp \nu ^\theta , \end{aligned}$$

which gives the desired result. $\square$

Appendix 2: Proofs of Section 3

Proof of Proposition 6

For all $g \in \fancyscript{C}_0(\varOmega ^d)$, one has

$$\begin{aligned}&\int _{\mathbb {S}^{d-1}}\int _{\mathbb {R}} g(t, \theta ) \mathrm {d}(R(\mu )^\theta )(t) \mathrm {d}\theta = \int _{\varOmega ^d}g(t, \theta )\mathrm {d}(R(\mu ))(t, \theta ) \\&\qquad = \int _{\mathbb {R}^d}(R^*g)(x)\mathrm {d}\mu (x) \\&\qquad = \int _{\mathbb {R}^d}\int _{\mathbb {S}^{d-1}} g(P_\theta (x),\theta ) \mathrm {d}\theta \mathrm {d}\mu (x)\\&\qquad = \int _{\mathbb {S}^{d-1}} \int _{\mathbb {R}} g(y,\theta ) \mathrm {d}(P_\theta \sharp \mu )(y) \mathrm {d}\theta . \end{aligned}$$

$\square$

Proof of Lemma 1

Proof of (26): For all $g \in \fancyscript{C}_0(\varOmega ^d)$, one has

$$\begin{aligned} \int _{\mathbb {R}^d} g \mathrm {d}[ R (\varphi _{s,u} \sharp \mu )]&= \int _{\mathbb {R}^d} R^*(g) \mathrm {d}[ \varphi _{s,u} \sharp \mu ] \\&= \int _{\mathbb {R}^d} \int _{\mathbb {S}^{d-1}} g(\langle sx+u,\,\theta \rangle ,\theta ) \mathrm {d}\theta \mathrm {d}\mu (x) \\&= \!\int _{\mathbb {R}^d}\! \int _{\mathbb {S}^{d-1}} \!(g \!\circ \!\psi _{s,u})(\langle x,\,\theta \rangle ,\theta ) \mathrm {d}\theta \mathrm {d}\mu (x) \\&= \int _{\mathbb {R}^d} (g \circ \psi _{s,u}) \mathrm {d}[R(\mu )] \\&= \int _{\mathbb {R}^d} g \mathrm {d}[ \psi _{s,u} \sharp R(\mu )]. \end{aligned}$$

Proof of (27): First we notice, using (22), that

$$\begin{aligned}&R( f \circ \varphi _{s,u} )(t,\theta ) = \int _{{\mathbb R}^{d-1} } f\left( s(t\theta + U_\theta \gamma ) + u \right) \mathrm {d}\gamma \\&\qquad = \int _{{\mathbb R}^{d-1} } f\left( st\theta + U_\theta s\gamma + \langle u,\,\theta \rangle \theta + U_\theta (U_\theta )^{T}u \right) \mathrm {d}\gamma \\&\qquad = \int _{{\mathbb R}^{d-1} } f\left( (st+\langle u,\,\theta \rangle )\theta + U_\theta (s\gamma + (U_\theta )^{T}u) \right) \mathrm {d}\gamma \\&\qquad = s^{1-d} \int _{{\mathbb R}^{d-1} } f\left( \psi _{s,u} (t,\theta ) \theta + U_\theta \gamma ' \right) \mathrm {d}\gamma ' \end{aligned}$$

which proves

$$\begin{aligned} R( f \circ \varphi _{s,u} ) = s^{1-d} R(f) \circ \psi _{s,u}. \end{aligned}$$

(51)

We write $H = (R^*R)^{-1}$ the filtering operator with kernel $h^+$. One has, for smooth functions $f \in \fancyscript{S}(\mathbb {R}^d)$, denoting $\fancyscript{F}(f)=\hat{f}$,

$$\begin{aligned} \fancyscript{F}( H(f \circ \varphi _{s,u}) )&= c^{-1}|| \omega ||^{1-d} \hat{f}(s\omega ) e^{-\mathrm {i}\langle \omega ,\,u\rangle }, \\ \fancyscript{F}( H(f) \circ \varphi _{s,u} )&= c^{-1}|| s\omega ||^{1-d} \hat{f}(s\omega ) e^{-\mathrm {i}\langle \omega ,\,u\rangle }, \end{aligned}$$

and hence

$$\begin{aligned} H(f) \circ \varphi _{s,u} = s^{1-d} H(f \circ \varphi _{s,u}). \end{aligned}$$

(52)

This shows, using (51) and (52) that for all $f \in \fancyscript{D}(\mathbb {R}^d)$,

$$\begin{aligned} \int _{\mathbb {R}^d} f \mathrm {d}[ R^+( \psi _{s,u} \sharp \nu ) ]&= \int _{\mathbb {R}^d} (RHf) \circ \psi _{s,u} \mathrm {d}\nu \\&= s^{d-1} \int _{\mathbb {R}^d} R(H(f) \circ \varphi _{s,u}) \mathrm {d}\nu \\&= \int _{\mathbb {R}^d} RH(f \circ \varphi _{s,u}) \mathrm {d}\nu \\&= \int _{\mathbb {R}^d} f \mathrm {d}[ \varphi _{s,u} \sharp R^+(\nu ) ]. \end{aligned}$$

Proof of (28): the proof is similar to the one of (26). $\square$

Proof of Proposition 8

Using Lemma 1, one has

$$\begin{aligned} \text {Bar}_{\mathbb {R}^d}^R(\varphi _{s,u} \sharp \mu _i,\lambda _i)_{i \in I}&= R^+ \text {Bar}_{\varOmega ^d}^W(R(\varphi _{s,u}\sharp \mu _i), \lambda _i)_{i \in I}\\&= R^+ \text {Bar}_{\varOmega ^d}^W(\psi _{s,u}\sharp (R(\mu _i)), \lambda _i)_{i \in I}\\&= R^+ \psi _{s,u} \sharp \text {Bar}_{\varOmega ^d}^W(R(\mu _i), \lambda _i)_{i \in I}\\&= \varphi _{s,u} \sharp R^+ \text {Bar}_{\varOmega ^d}^W(R(\mu _i), \lambda _i)_{i \in I}\\&= \varphi _{s,u} \sharp \text {Bar}_{\mathbb {R}^d}^R(\mu _i,\lambda _i)_{i \in I}. \end{aligned}$$

which proves (7) for $\text {Bar}_{\mathbb {R}^d}^R$. Property (8) for $\text {Bar}_{\mathbb {R}^d}^R$ is proved similarly using (28). $\square$

Proof of Proposition 9

One has

$$\begin{aligned} \text {Bar}_{\mathbb {R}^d}^R(\varphi _{s_i,u_i} \sharp \mu ,\lambda _i)_{i \in I}&= R^+ \text {Bar}_{\varOmega ^d}^W(R(\varphi _{s_i,u_i}\sharp \mu ), \lambda _i)_{i \in I}\\&= R^+ \text {Bar}_{\varOmega ^d}^W( \psi _{s_i,u_i}\sharp R(\mu ), \lambda _i)_{i \in I}\\&= R^+ \psi _{s^\star ,u^\star } \sharp \text {Bar}_{\varOmega ^d}^W(R(\mu ), \lambda _i)_{i \in I}\\&= \varphi _{s^\star ,u^\star } \sharp R^+ \text {Bar}_{\varOmega ^d}^W(R(\mu ), \lambda _i)_{i \in I}\\&= \varphi _{s^\star ,u^\star } \sharp \text {Bar}_{\mathbb {R}^d}^R(\mu ,\lambda _i)_{i \in I}, \end{aligned}$$

which proves (13) for $\text {Bar}_{\mathbb {R}^d}^R$. $\square$

Appendix 3: Proof of Section 4

Proof of Proposition 10

Property (34) is a re-statement of property (19). Property (35) corresponds to the change of variable $\nu = R\mu \in \text { Im}(R)$ in (32), which is a bijection thanks to the injectivity of $R$, see proposition 7. $\square$

Proof of Proposition 11

The proof is the same as Proposition 1, replacing the invariance (49) by

$$\begin{aligned} \text {SW}_{\mathbb {R}^d}(\varphi _{s,u} \sharp \mu _1,\varphi _{s,u} \sharp \mu _2)&= \!\text {W}_{\varOmega ^d}( \!R( \varphi _{s,u} \sharp \mu _1 ), R( \varphi _{s,u} \sharp \mu _2 ) ) \\&= \text {W}_{\varOmega ^d}( \psi _{s,u}\! \sharp \!R( \mu _1 ), \psi _{s,u} \sharp R( \mu _1 ) ) \\&= \text {W}_{\varOmega ^d}( R( \mu _1 ), R( \mu _1 ) ) \\&= \text {SW}_{\mathbb {R}^d}( \mu _1, \mu _2), \end{aligned}$$

where we have used the invariance (50) of the Wasserstein distance on $\varOmega ^d$. $\square$

Proof of Proposition 12

One has,

$$\begin{aligned} \forall \,\theta \in \mathbb {S}^{d-1}, \quad P_\theta \sharp \varphi _{s,u} \sharp \mu = \varphi _{s,\langle u,\,\theta \rangle } \sharp P_\theta \sharp \mu . \end{aligned}$$

Thus, for an arbitrary $\tilde{\mu } \in \fancyscript{M}_1^+(\mathbb {R}^d)$, one has

$$\begin{aligned}&\sum _{i\in I} \lambda _i \text {W}_{\mathbb {R}}( P_\theta \sharp (\varphi _{s_i,u_i} \sharp \mu ), P_\theta \sharp \tilde{\mu } )^2\\&\qquad = \sum _{i\in I} \lambda _i \text {W}_{\mathbb {R}}( \varphi _{s_i,\langle u_i,\,\theta \rangle } \sharp (P_\theta \sharp \mu ), P_\theta \sharp \tilde{\mu } )^2 \\&\qquad \geqslant \sum _{i\in I} \lambda _i \text {W}_{\mathbb {R}}( \varphi _{s_i,\langle u_i,\,\theta \rangle } \sharp (P_\theta \sharp \mu ), \varphi _{s^\star ,\langle u^\star ,\,\theta \rangle } \sharp (P_\theta \sharp \mu ) )^2 \\&\qquad = \sum _{i\in I} \lambda _i \text {W}_{\mathbb {R}}( P_\theta \sharp (\varphi _{s_i,u_i} \sharp \mu ), P_\theta \sharp (\varphi _{s^\star ,u^\star } \sharp \mu ) )^2 \end{aligned}$$

where the inequality comes from the properties of 1-D Wasserstein barycenters. Integrating the resulting inequality with respect to $\theta \in \mathbb {S}^{d-1}$ gives

$$\begin{aligned}&\sum _i \lambda _i \text {SW}_{\mathbb {R}^d}( \varphi _{s_i,u_i} \sharp \mu , \tilde{\mu } )^2 \\&\quad \geqslant \sum _i \lambda _i \text {SW}_{\mathbb {R}^d}( \varphi _{s_i,u_i} \sharp \mu , \varphi _{s^\star ,u^\star } \sharp \mu )^2. \end{aligned}$$

This inequality is an equality if and only for almost all $\theta \in \mathbb {S}^{d-1}$, one has

$$\begin{aligned} P_\theta \sharp \tilde{\mu } = P_\theta \sharp (\varphi _{s^\star ,u^\star } \sharp \mu ) \end{aligned}$$

so that, using Proposition (7), this corresponds to $\tilde{\mu } = \varphi _{s^\star ,u^\star } \sharp \mu$. Since the measure $\tilde{\mu }$ is arbitrary, this gives the desired result. This proves (13) in the case $\text {Bar}_{\mathbb {R}^d}^S$. $\square$

Appendix 4: Proof of Theorem 1

Notations. Without loss of generality, for a fixed $Y \in \mathbb {R}^{d \times N}$, we study the smoothness of

$$\begin{aligned}&\forall \,X \in \mathbb {R}^{d\times N}, \quad \fancyscript{E}(X) = \frac{1}{2} \text {SW}_{\mathbb {R}^d}(\mu _X,\mu _Y)^2\\&\quad = \int _{\mathbb {S}^{d-1}} \fancyscript{E}_\theta (X) \mathrm {d}x\\&\quad \text {where} \quad \fancyscript{E}_\theta (X) = \frac{1}{2} \fancyscript{W}(X_\theta ,Y_\theta )^2. \end{aligned}$$

We have used, for $x, y \in \mathbb {R}^N$, the shorthand notation

$$\begin{aligned} \fancyscript{W}(x,y) = \text {W}_{\mathbb {R}}(\mu _x,\mu _y). \end{aligned}$$

The result of Theorem 1 then follows by summations of such functionals.

We define $\mathbb {U}(N,d)$ to be vectors of $\mathbb {R}^{d \times N}$ with distinct entries:

$$\begin{aligned}&\mathbb {U}(N,d) \nonumber \\&\quad = \left\{ W = (W_1, \ldots , W_N) \in {\mathbb R}^{d \times N} \;;\; \forall \,i \not =j,\, X_i \not = X_j \right\} . \end{aligned}$$

The hypothesis is that $X \in \mathbb {U}(N,d)$. One has

$$\begin{aligned} \fancyscript{E}_\theta (X) = \frac{1}{2} || X_\theta - Y_\theta \circ \sigma _\theta ||^2 \quad \text {where} \quad \sigma _\theta = \sigma _X^\theta \circ (\sigma _Y^{\theta })^{-1} \end{aligned}$$

is a permutation depending on both $X$ and $Y$. Note that the permutation involved are not necessarily unique, and are assumed to be arbitrary valid sorting permutations.

For $X \in \mathbb {R}^{N \times d}$ and $\varepsilon >0$ we introduce

$$\begin{aligned}&\Theta _\varepsilon (X)\\&\quad = \left\{ \theta \in \mathbb {S}^{d-1} \;;\; \forall \,|| \delta ||_{\mathbb {R}^{N \times d}} \leqslant \varepsilon , \quad X_\theta + \delta _\theta \in \mathbb {U}(N,1) \right\} . \end{aligned}$$

This is the set of directions for which any perturbation of $X$ of amplitude smaller than $\varepsilon$ has a projection with disjoint points.

Overview of the proof. In the following, we thus aim at proving that $\fancyscript{E}$ is $C^1$, that

$$\begin{aligned}&\tilde{\nabla } \fancyscript{E}(X) = \int _{\mathbb {S}^{d-1}} \tilde{\nabla } \fancyscript{E}_\theta (X) \mathrm {d}\theta \\&\quad \quad \text {where} \quad \tilde{\nabla } \fancyscript{E}_\theta (X) = (X_\theta - Y_\theta \circ \sigma _\theta ) \theta \end{aligned}$$

is indeed equal to $\nabla \fancyscript{E}(X)$, and that this gradient is Lipschitz continuous.

The general strategy of the proof is to split the integration between the directions $\theta \in \Theta _\varepsilon (X)$, for which we can locally assume that the permutations $\sigma _\theta$ are constant (see Lemma 2), which in turn defines a smooth quadratic energy, and the remaining directions in $\Theta _\varepsilon (X)^c$, which are shown to have a negligible contribution to the energy and to the derivative (see Lemma 3).

Preparatory results. The following lemma shows that if $\theta \in \Theta _\varepsilon (X)$ the permutations $\sigma _X^\theta$ are stable to small perturbations of $X$.

Lemma 2

Let $X \in \mathbb {U}(N,d)$. For all $\theta \in \Theta _\varepsilon (X)$, for all $\delta$ with $|| \delta ||_{\mathbb {R}^{N \times d}} \leqslant \varepsilon$, the permutation $\sigma _{X+\delta }^\theta$ that sorts $( \langle X_i+\delta _i,\,\theta \rangle )_i$ is uniquely defined and satisfies $\sigma _{X+\delta }^\theta = \sigma _X^\theta$.

Proof

If one has $\sigma _{X+\delta }^\theta \ne \sigma _X^\theta$, then necessarily there exists some $t \in [0,1]$ such that $\sigma _{X+t\delta }^\theta$ is not uniquely defined, which is equivalent to $X_\theta +t\delta _\theta$ not being in $\mathbb {U}(N,1)$. Since $|| t \delta ||_{\mathbb {R}^{N \times d}} \leqslant \varepsilon$, this shows that $\theta \notin \Theta _\varepsilon (X)$. $\square$

In order to prove Theorem 1, we need the following lemma.

Lemma 3

For $X \in \mathbb {U}(N, d)$, one has

$$\begin{aligned} \text {Vol}(\Theta _\varepsilon (X)^c) = \int _{\Theta _\varepsilon (X)^c} \mathrm {d}\theta = O( \varepsilon ). \end{aligned}$$

(53)

Proof

One has $X_\theta + \delta _\theta \notin \mathbb {U}(N,1)$ if and only there exists a pair of points $u=X_i+\delta _i$ and $v=X_j+\delta _j$ with $i \ne j$ such that

$$\begin{aligned}&\theta \in A(u,v)\\&\quad \quad \text {where} \quad A(u,v) = \left\{ \xi \in \mathbb {S}^{d-1} \;;\; \langle \xi ,\,u-v\rangle =0 \right\} \end{aligned}$$

Note that $A(u,v)$ is a great circle of the sphere $\mathbb {S}^{d-1}$.

One can thus covers $\Theta _\varepsilon (X)^c$ using the union of all such circles $A(u,v)$, which shows

$$\begin{aligned}&\Theta _\varepsilon (X)^c \subset \bigcup _{i \ne j} A_\varepsilon (X_i,X_j) \quad \text {where} \quad A_\varepsilon (x,y) \\&\quad = \bigcup _{ {\begin{matrix} || u-x || \leqslant \varepsilon \\ || v-y || \leqslant \varepsilon \end{matrix}} } A(u,v) \end{aligned}$$

Note that the geodesic distance $d$ on the sphere $\mathbb {S}^{d-1}$ between two circles is equal to the angle between the normal to the planes of the circles

$$\begin{aligned} d(A(u,v),A(x,y))&= \text {Angle}(u-v,x-y) \\&= \text {Angle}(x-y + \varepsilon w,x-y) \end{aligned}$$

where $|| w ||\leqslant 2$. As $\varepsilon \rightarrow 0$, after some computations, one has the following asymptotic decay of the angle

$$\begin{aligned} \text {Angle}(x-y + \varepsilon w,x-y) = O(\varepsilon /|| x-y ||) \end{aligned}$$

and thus $d(A(u,v),A(x,y)) \leqslant C \varepsilon$ for some constant $C$. This proves that $\forall \,u,v$, one has

$$\begin{aligned} \left\{ \begin{array}{l} || u-x || \leqslant \varepsilon \\ || v-y || \leqslant \varepsilon \end{array} \right. \quad \Longrightarrow \quad A(u,v) \subset B_{C\varepsilon }(x,y) \end{aligned}$$

for some constant $C>0$, where

$$\begin{aligned} B_{\varepsilon }(x,y) = \left\{ \xi \in \mathbb {S}^{d-1} \;;\; d(\xi ,A(x,y)) \leqslant \varepsilon \right\} \end{aligned}$$

One thus has

$$\begin{aligned} A_{\varepsilon }(x,y) \subset B_{C\varepsilon }(x,y). \end{aligned}$$

The volume of the spherical band $B_{C\varepsilon }(x,y)$ of width $C\varepsilon$ is proportional to $\varepsilon$, and thus Vol$(A_\varepsilon (x,y)) = O(\varepsilon )$. Since $\Theta _\varepsilon (X)^c$ is a finite union of such sets, one obtains the result. $\square$

Proof of continuity. For each $\theta$, the function $\fancyscript{E}_\theta$ is continuous as a minimum of continuous functions. The function $\fancyscript{E}$ being an integral of $\fancyscript{E}_\theta$ on a compact set $\mathbb {S}^{d-1}$, it is thus continuous.

Proof of differentiability. Let $\delta \in \mathbb {R}^{N \times d}$ and $\varepsilon = || \delta ||_{\mathbb {R}^{N \times d}}$. The definition of the Wasserstein distance reads

$$\begin{aligned} \fancyscript{W}((X+\delta )_\theta ,Y_\theta )^2 = || (X_\theta + \delta _\theta ) \circ \sigma _{X+\delta }^\theta - Y_\theta \circ \sigma _Y^\theta ||^2. \end{aligned}$$

For all $\theta \in \Theta _\varepsilon (X)$, thanks to Lemma 2, $\sigma _{X+\delta }^\theta = \sigma _{X}^\theta$. One can thus compute the variation of the 1-D Wasserstein distance with respect to $\delta$ as

$$\begin{aligned}&\fancyscript{W}((X+\delta )_\theta ,Y_\theta )^2 = || X_\theta +\delta _\theta - Y_\theta \circ \sigma _\theta ||^2 \end{aligned}$$

(54)

$$\begin{aligned}&\quad = \fancyscript{W}(X_\theta ,Y_\theta )^2 + \langle \tilde{\nabla } \fancyscript{E}_\theta (X) ,\, \delta \rangle _{\mathbb {R}^{N \times d}} + || \delta _\theta ||^2. \end{aligned}$$

(55)

Note that the fact that $\sigma _Y^{\theta }$ might not be uniquely defined has no impact on the value of (55). One thus has

$$\begin{aligned}&\fancyscript{E}(X+\delta )-\fancyscript{E}(X) - \langle \tilde{\nabla } \fancyscript{E}(X),\,\delta \rangle _{\mathbb {R}^{N \times d}}\\&\quad = A(\delta ) + B(\delta ) + O(|| \delta ||_{\mathbb {R}^{N \times d}}^2) \end{aligned}$$

where

$$\begin{aligned}&A(\delta ) = \int _{\Theta _\varepsilon (X)^c} \left( \fancyscript{W}(X_\theta +\delta _\theta ,Y_\theta )^2 - \fancyscript{W}(X_\theta ,Y_\theta )^2 \right) \mathrm {d}\theta \\&\quad \text {and} \quad B(\delta ) = -\int _{\Theta _\varepsilon (X)^c} \langle \tilde{\nabla } \fancyscript{E}_\theta (X) ,\, \delta \rangle _{\mathbb {R}^{N \times d}} \mathrm {d}\theta \end{aligned}$$

Note that in the expression of $B(\delta )$ the permutation $\sigma _\theta$ involved in $\tilde{\nabla } \fancyscript{E}_\theta (X)$ is not necessary unique, and can be chosen arbitrarily.

One has,

$$\begin{aligned} |\langle \tilde{\nabla } \fancyscript{E}_\theta (X) ,\, \delta \rangle _{\mathbb {R}^{N \times d}}| \leqslant || X-Y\circ \sigma ^\theta ||_{\mathbb {R}^{N \times d}} || \delta ||_{\mathbb {R}^{N \times d}} \end{aligned}$$

which implies, using Lemma 3

$$\begin{aligned}&|B(\delta )| \leqslant O( \text {Vol}(\Theta _\varepsilon (X)^c) || \delta ||_{\mathbb {R}^{N \times d}} )\nonumber \\&\quad = O(|| \delta ||_{\mathbb {R}^{N \times d}}^2) = o(|| \delta ||_{\mathbb {R}^{N \times d}}). \end{aligned}$$

(56)

Since $(\theta ,X) \mapsto \fancyscript{E}_\theta (X)$ is continuous and defined on a compact set, it is uniformly continuous, and thus

$$\begin{aligned} |\fancyscript{W}(X_\theta +\delta _\theta ,Y_\theta )^2 - \fancyscript{W}(X_\theta ,Y_\theta )^2| \leqslant C(\delta ) \end{aligned}$$

where $C(\delta ) \rightarrow 0$ where $\delta \rightarrow 0$. This shows that

$$\begin{aligned} |A(\delta )| \leqslant \text {Vol}(\Theta _\varepsilon (X)^c) C(\delta ) = o(|| \delta ||_{\mathbb {R}^{N \times d}}). \end{aligned}$$

(57)

Putting together (56) and (57) leads to

$$\begin{aligned} |\fancyscript{E}(X+\delta )-\fancyscript{E}(X) - \langle \tilde{\nabla } \fancyscript{E}(X),\,\delta \rangle | = o(|| \delta ||_{\mathbb {R}^{N \times d}}) \end{aligned}$$

which shows that $\fancyscript{E}$ is differentiable with $\nabla \fancyscript{E}= \tilde{\nabla } \fancyscript{E}$.

Proof of Lipschitzianity of the gradient. For all $\theta \in \varTheta _0(X)$, $\nabla \fancyscript{E}_\theta (X)$ is continuous and uniformly bounded, and thus $\nabla \fancyscript{E}$ is continuous. One has, for $\delta \in \mathbb {R}^{N \times d}$, and denoting $\varepsilon =|| \delta ||$,

$$\begin{aligned}&\nabla \fancyscript{E}(X+\delta ) - \nabla \fancyscript{E}(X) = M( \varTheta _\varepsilon (X) ) + M( \varTheta _\varepsilon (X)^c )\\&\quad \text {where} \quad M(U) = \int _U ( \nabla \fancyscript{E}_\theta (X+\delta ) - \nabla \fancyscript{E}_\theta (X) ) \mathrm {d}\theta . \end{aligned}$$

One has

$$\begin{aligned} M( \varTheta _\varepsilon (X) ) = \int _{ \varTheta _\varepsilon (X) } \delta _\theta \theta \mathrm {d}\theta \end{aligned}$$

whereas

$$\begin{aligned} M( \varTheta _\varepsilon (X)^c )&= \int _{ \varTheta _\varepsilon (X)^c } \delta _\theta \theta \mathrm {d}\theta \\&+ \int _{ \varTheta _\varepsilon (X)^c } ( Y \circ \tilde{\sigma }_\theta - Y \circ \sigma _\theta ) \theta \mathrm {d}\theta \end{aligned}$$

where $\tilde{\sigma }_\theta = \sigma _{Y_\theta } \circ \,\, \sigma _{X_\theta + \delta _\theta }^{-1}$. Using Lemma (3), one has for some constant $C>0$, $\text {Vol}(\Theta _\varepsilon (X)^c) \leqslant C || \delta ||_{\mathbb {R}^{N \times d}}$ and hence

$$\begin{aligned}&|| \nabla \fancyscript{E}(X+\delta ) - \nabla \fancyscript{E}(X) ||_{\mathbb {R}^{N \times d}} \\&\quad \leqslant (1 + 2 C || Y ||_{\mathbb {R}^{N \times d}})|| \delta ||_{\mathbb {R}^{N \times d}} \end{aligned}$$

which shows that $\nabla \fancyscript{E}$ is $(1 + 2 C || Y ||_{\mathbb {R}^{N \times d}})$-Lipschitz continuous.