Transport maps, non-branching sets of geodesics and measure rigidity

TL;DR

This paper explores the connection between transport maps, non-branching geodesic sets, and measure rigidity in metric measure spaces, extending previous results and establishing new conditions for measure uniqueness and absolute continuity.

Contribution

It generalizes existing results by linking qualitative non-degenericity to the existence of unique transport maps in essentially non-branching spaces and proves a new measure rigidity theorem.

Findings

Qualitative non-degenericity implies unique transport maps in essentially non-branching spaces.

Any two essentially non-branching, qualitatively non-degenerate measures are mutually absolutely continuous.

Results apply to spaces with generalized finite dimensional Ricci curvature bounds.

Abstract

In this paper we investigate the relationship between a general existence of transport maps of optimal couplings with absolutely continuous first marginal and the property of the background measure called essentially non-branching introduced by Rajala-Sturm (Calc.Var.PDE 2014). In particular, it is shown that the qualitative non-degenericity condition introduced by Cavalletti-Huesmann (Ann. Inst. H. Poincar\'e Anal. Non Lin\`eaire 2015) implies that any essentially non-branching metric measure space has a unique transport maps whenever initial measure is absolutely continuous. This generalizes a recently obtained result by Cavalletti-Mondino (Commun. Contemp. Math. 2017) on essentially non-branching spaces with the measure contraction condition $\mathsf{MCP}(K,N)$. In the end we prove a measure rigidity result showing that any two essentially non-branching, qualitatively non-degenerate measures on a fixed metric spaces must be mutually absolutely continuous. This result was obtained under stronger conditions by Cavalletti-Mondino (Adv.Math. 2016). It applies, in particular, to metric measure spaces with generalized finite dimensional Ricci curvature bounded from below.

Equations

$$\pi \mapsto \int c(x,y)d\pi (x,y)$$

$${\phi }_{\Gamma }(x)=\{\begin{array}{ll}{\sup }_{(x,y)\in \Gamma }d(x,y)-{\inf }_{(x,y)\in \Gamma }d(x,y)& x\in p_1(\Gamma )\\ 0& \text{otherwise}.\end{array}$$

$$\underset{(x,y_1)\in K\times T_1(K)}{\sup }d(x,y_1)+\delta \le \underset{(x,y_2)\in K\times T_2(K)}{\inf }d(x,y_2).$$

$${\mathrm{graph}}_{\Omega }T:=\{(x,T(x)|x\in \Omega \}$$

$$T(x)\ne S(x^{\prime })\text{for all }x,x^{\prime }\in {\bar{B}}_r(x_0)\cap K_1.$$

$${\Gamma }^{+}$$

$${\Gamma }^{-}$$

$$d(x,T^{-}(x^{\prime }))+\delta \le d(x,T^{+}(x^{\prime \prime }))\text{for all }x,x^{\prime },x^{\prime \prime }\in {\bar{B}}_r(x_0)\cap K_3$$

$$\sigma (C\cap S^{-1}(B))={\int }_C{\sigma }_y(B)d\varpi (y).$$

$$\sigma =\int {\sigma }_{y}d\varpi (y).$$

$$\pi (K\times A_1),\pi (K\times A_2)>0.$$

$${\left(A_1\right)}_{ϵ}\cap {\left(A_2\right)}_{ϵ}=\varnothing$$

$$\underset{(x,y_1)\in K\times A_1}{\sup }d(x,y)+\delta \le \underset{(x,y_1)\in K\times A_1}{\inf }d(x,y).$$

$$\mu_{x}=\mu_{x}\big{|}_{A_{1}}+\mu_{x}\big{|}_{A_{2}}+\mu_{x}\big{|}_{M\backslash(A_{1}\cup A_{2})}$$

$$\pi (K\times (T_i(K){)}_{ϵ})>0.$$

$$0$$

$$\le \pi ((B_{ϵ}(x)\cap K)\times B_{ϵ}(T(x)))$$

$$\le \pi (K\times B_{ϵ}(T(x)))\le \pi (K\times (T_i(K){)}_{ϵ}).$$

$$(T_1(K){)}_{2ϵ}\cap (T_2(K){)}_{2ϵ}=\varnothing .$$

$${\mu }_x(A_1),{\mu }_x(A_2)\in (\delta ,1-\delta )\text{for all }x\in K^{\prime }.$$

$${\pi }_1$$

$${\pi }_2$$

$$\sigma =\int {\delta }_{T(y)}d\varpi (y).$$

$$\sigma_{x}\big{|}_{A_{i}}\neq 0,\;x\in K,i=1,2.$$

$$d({\gamma }_t,{\gamma }_s)=|t-s|d({\gamma }_0,{\gamma }_1)\text{for }t,s\in [0,1]$$

$$({\mathsf{restr}}_{s,t}\gamma )(r)={\gamma }_{s+(t-s)r}.$$

$${\mathcal{P}}_p(M)=\left\{\mu \in \mathcal{P}(M)|\int d(x,x_0{)}^{p}d\mu (x)\right\}$$

$$W_p({\mu }_0,{\mu }_1)={\left(\underset{\pi \in \Pi ({\mu }_0,{\mu }_1)}{\inf }\int d(x,y{)}^{p}d\pi (x,y)\right)}^{\frac{1}{p}}$$

$$W_p({\mu }_0,{\mu }_1)={\left(\int d(x,y{)}^{p}d{\pi }_{\mathrm{opt}}(x,y)\right)}^{\frac{1}{p}}.$$

$$\sigma =\int {\sigma }_{x,y}d\pi (x,y)$$

$${\sigma }_f={\lambda }^{-1}\int {\sigma }_{x,y}f(x,y)d\pi (x,y)$$

$${\mu }_0^f$$

Full text

In the end we prove a measure rigidity result showing that any two essentially non-branching, qualitatively non-degenerate measures on a fixed metric spaces must be mutually absolutely continuous. This result was obtained under stronger conditions by Cavalletti–Mondino (Adv.Math. 2016). It applies, in particular, to metric measure spaces with generalized finite dimensional Ricci curvature bounded from below.

Fachbereich Mathematik, Universität Tübingen, Germany

e-mail: martin.kell@math.uni-tuebingen.de

1. 1 Introduction
2. 2 Preliminaries
3. 3 Spaces with good transport behavior
4. 4 Existence of absolutely continuous interpolations for non-degenerate measures
5. 5 Existence of transport maps
6. 6 Measure rigidity
7. A Proof of Theorem 2.11 and Corollary 2.12

1. Introduction

In the theory of optimal transport one of the first questions one asks is whether an optimal coupling $\pi_{\operatorname{opt}}$ that minimizes the functional

[TABLE]

among all $\pi$ with fixed marginals $(p_{1})_{*}\pi=\mu$ and $(p_{2})_{*}\pi=\nu$ can be written as a coupling induced by a transport map, i.e. whether there is a measurable map $T:M\to M$ such that $\pi=(\operatorname{id}\times T)_{*}\mu$. On sufficiently nice spaces this result can be deduced from Rademacher’s Theorem, the (weak) differential structure and the exponential map whenever $\mu$ is absolutely continuous, see Brenier [Bre91] in the Euclidean setting, McCann [McC01] on Riemannian manifolds, Ambrosio–Rigot [AR04] and Figalli–Rigot [FR10] on sub-class of sub-Riemannian manifolds and also Bertrant [Ber08] on Alexandrov spaces.

A first proof for non-smooth non-branching spaces with generalized Ricci curvature bounded from below was obtained by Gigli [Gig12]. He showed that the non-existence of a transport map would imply that two disjoint parts of a Wasserstein geodesic whose initial densities with respect to the background measure $\mathsf{m}$ are bounded would overlap at intermediate times. This, however, cannot happen for non-branching spaces (compare Lemma 2.9 below).

Consequently, Gigli’s idea was adapted to essentially non-branching spaces with generalized Ricci curvature bounded from below (see [RS14, GRS16, CM17]). Here essentially non-branching is a weak version of the non-overlapping property described above, i.e. it prohibits that initially disjoint parts of a Wasserstein geodesic overlap at intermediate times whenever the initial and final measures are absolutely continuous.

Both Gigli–Rajala–Sturm [GRS16] and Cavalletti–Mondino [CM17] had to prove that there are absolutely continuous interpolations along which a corresponding interpolation inequality holds.

Rather than using density bounds of intermediate measures, Cavalletti–Huesmann [CH15] showed that if not too much mass is lost in a uniform way when transported towards a fixed point (compare Definition 5.1), then, together with the non-overlapping property implied by non-branching property, the non-existence of transport maps would yield a contradiction.

Whereas the interpolation inequality implied density bounds, Cavalletti–Huesmann’s approach only relied on an “easier to measure” quantity. In this paper we want to combine this approach with the one of Cavalletti–Mondino [CM17]. The difficulty is that an arbitrary interpolation might only “see” part of the interpolation points. We avoid this by proving the following:

• –

at finitely many fixed times $\{t_{i}\}_{i=1}^{n}$ there are absolutely continuous interpolations $\mu_{t_{n}}\ll\mathsf{m}$ (see first part of Theorem 4.10 and Corollary 4.11)

• –

at a fixed time $t$ the interpolation $\mu_{t}$ “sees” $\mathsf{m}$-almost every possible interpolation point $\Gamma_{t}$ (see second part of Theorem 4.10)

Whereas the first part gives us the non-overlapping if the space is essentially non-branching, the second part makes sure that the set-wise interpolation $\Gamma_{t}$ does not contain a set of positive $\mathsf{m}$-measure that is not seen by the interpolation $\mu_{t}$.

The remaining parts follow along the line of Cavalletti–Huesmann [CH15], more precisely, if the background measure is qualitatively non-degenerate, i.e. there is a function $f:(0,1)\to(0,\infty)$ with $\limsup_{t\to 0}f(t)>\frac{1}{2}$ such that for all Borel set $A$ and all $x\in X$ it holds $\mathsf{m}(A_{t,x})\geq f(t)\mathsf{m}(A)$ where $A_{t,x}=\{\gamma_{t}\,|\,\gamma_{0}\in A,\gamma_{1}=x\}$, then

• –

every optimal coupling between $\mu_{0}\ll\mathsf{m}$ and $\mu_{1}=\sum\lambda_{i}\delta_{x_{i}}$ is induced by a transport map (Lemma 5.4 and Corollary 5.6)

• –

any $c_{p}$-cyclically monotone set $\Gamma$ satisfies a qualitative non-degenericity condition, i.e. $\mathsf{m}(\Gamma_{t})\geq f(t)\mathsf{m}(\Gamma_{0})$ (Lemma 5.7)

• –

every optimal coupling between $\mu_{0}\ll\mathsf{m}$ and $\mu_{1}$ is induced by a transport map (Theorem 5.8).

As it turns out the general existence of transport maps implies not only uniqueness of the optimal coupling, but also unique geodesics between almost all points coupled via the optimal transport, see Lemma 3.4. This in return gives uniqueness of the interpolation measures. Furthermore, only assuming a priori the existence of transport maps from an absolutely continuous initial measure implies that the space must be essentially non-branching (Proposition 3.6). We call spaces with such an a priori existence of transport maps spaces having good transport behavior $\mathsf{(GTB)}_{p}$, see Definition 3.1.

We may summarize one of the results of this note as follows.

Theorem** (see Proposition 3.6 and Theorem 5.8).**

Let $(M,d,\mathsf{m})$ be a metric measure space and assume $\mathsf{m}$ is (uniformly) qualitatively non-degenerate then the following properties are equivalent:

1. (i)

$(M,d,\mathsf{m})$* is $p$-essentially non-branching, i.e. any $p$-optimal dynamical coupling $\pi\in\mathcal{P}(\mathsf{Geo}_{[0,1]}(M,d))$ with $(e_{0})_{*}\sigma,(e_{1})_{*}\sigma\ll\mathsf{m}$ is concentrated on a set of non-branching geodesics.* 2. (ii)

for every $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0}\ll\mathsf{m}$ there is a unique $p$-optimal dynamical coupling $\sigma\in\mathcal{P}_{p}(\mathsf{Geo}_{[0,1]}(M,d))$ between $\mu_{0}$ and $\mu_{1}$ and for this coupling $\sigma$ the $p$-optimal coupling $(e_{0},e_{1})_{*}\sigma$ is induced by a transport map and every interpolation $\mu_{t}=(e_{t})_{*}\sigma$, $t\in[0,1)$, is absolutely continuous, i.e. $\mu_{t}\ll\mathsf{m}$.

It is well-known that on smooth spaces, there is an abundance of measures satisfying all but the last property of the second statement in the theorem above, in particular, they have the good transport behavior $\mathsf{(GTB)}_{p}$.

If interpolations We call the last property of the second statement in the theorem above strong interpolation property $\mathsf{(sIP)}_{p}$. Note that two reference measures $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ have the strong interpolation property and $\mu_{0}\ll\mathsf{m}_{1}$ and $\mu_{1}\ll\mathsf{m}_{2}$ then the interpolation $(e_{t})_{*}\sigma$ must be absolutely continuous with respect to both $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ implying $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ cannot be mutually singular. This property can be used to obtain the following measure rigidity theorem.

Theorem** (Measure Rigidity).**

Any two measures on a complete separable metric space $(M,d)$ which are $p$-essentially non-branching and qualitatively non-degenerate must be mutually absolutely continuous.

Under a more involved inversion property and a stronger qualitative non-degenericity with $\limsup_{t\to 0}f(t)=1$ such a statement was obtained by Cavalletti–Mondino [CM16].

By [RS14, AGS14] the statement applies in particular to spaces with finite dimensional Ricci curvature bounded from below.

Corollary**.**

If both $(M,d,\mathsf{m}_{1})$ and $(M,d,\mathsf{m}_{2})$ are $\mathsf{RCD}^{*}(K,N)$-spaces with $N\in[1,\infty)$ then $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ must be mutually absolutely continuous.

A slightly different kind of measure rigidity of the background measure $\mathsf{m}$ for metric measure spaces satisfying the $\mathsf{RCD}(K,N)$-condition with $N\in[1,\infty)$ was obtain independently by Gigli–Pasqualetto [GP16] and by Mondino and the author [KM17] using the weak converse of Rademacher’s theorem proven in [DPR16].

The section dealing with the measure rigidity theorem (Section 6) only relies on few properties which are restated in the beginning of that section and can be read independently of the rest of this note. We also present a proof of the rigidity theorem that relies on a bounded density property and applies to strong $\mathsf{CD}_{p}(K,\infty)$-spaces with the strong interpolation property, see Theorem 6.5.

2. Preliminaries

Throughout this paper we always assume that $(M,d,\mathsf{m})$ is a* geodesic metric measure space*, i.e. $(M,d)$ is a complete separable geodesic metric space and $\mathsf{m}$ is measure on $M$ which is finite on bounded sets. It is proper if every closed bounded set $B\subset M$ is compact.

Selection Dichotomy and Disintegration Theorem

We present two technical results which help to classify couplings that are induced by a transport map. The first can be obtained by combining the Measurable Selection Theorem and Lusin’s Theorem. This form of the selection dichotomy was used by Cavalletti–Huesmann [CH15] to select $p$-optimal couplings that overlap at the initial measure.

In the following for a set $\Gamma\subset M\times M$ we let $\Gamma(x)=\{y\in M\,|\,(x,y)\in\Gamma\}$. We say a map $T:M\to M$ is a selection of $\Gamma$ is $(x,T(x))\in\Gamma$ for all $x\in p_{1}(\Gamma)$.

Theorem 2.1** (Selection Dichotomy of Sets).**

Assume $\mu$ is a probability measure on $M$ and $\Gamma\subset M\times M$ a Borel set with $\mu(p_{1}(\Gamma))=1$. Then exactly one of the following holds:

1. (i)

For $\mu$-almost all $x\in M$ the set $\Gamma(x)$ contains exactly one element. Furthermore, if $\pi\in\mathcal{P}(M\times M)$ with $\operatorname{supp}\pi\subset\Gamma$ and $(p_{1})_{*}\pi=\mu$ then $\pi=(\operatorname{id}\times T)_{*}\mu$ for a $\mu$-measurable selection $T:M\to M$ of $\Gamma$. In particular, $T$ is unique up to $\mu$-measure zero and $\pi$ is unique among all measure $\tilde{\pi}\in\mathcal{P}(M\times M)$ with support in $\Gamma$ and first marginal $\mu$. 2. (ii)

There are a compact set $K\subset\operatorname{supp}\mu$ with $\mu(K)>0$ and two $\mu$-measurable selections $T_{1},T_{2}:M\to M$ of $\Gamma$ which are continuous when restricted to $K$ and $T_{1}(K)\cap T_{2}(K)=\varnothing$. Furthermore, if the function

[TABLE]

is positive on a set of positive $\mu$-measure then $K$ can be chosen such that for some $\delta>0$

[TABLE]

Proof.

It is easy to see that the conditions are mutually exclusive. Indeed, the measures $(\operatorname{id}\times T_{1})_{*}\mu$ and $(\operatorname{id}\times T_{2})_{*}\mu$ are distinct and supported on $\Gamma$.

By the Measurable Selection Theorem there is a $\mu$-measurable map $T$ such that $(x,T(x))\in\Gamma$ for all $x\in p_{1}(\Gamma)$. Using Lusin’s Theorem one can show that there is a Borel set $\Omega\subset M$ of full $\mu$-measure such that

[TABLE]

is a Borel subset of $M\times M$. Thus $\Gamma^{{}^{\prime}}=\Gamma\backslash\operatorname{graph}_{\Omega}T$ is a Borel set and there is a $\mu$-measurable selection $S:M\to M$ with $(x,S(x))\in\Gamma^{{}^{\prime}}$ for all $x\in p_{1}(\Gamma^{{}^{\prime}})$. Since $p_{1}(\Gamma^{{}^{\prime}})$ is $\mu$-measurable, we can redefine $S$ outside of $p_{1}(\Gamma^{{}^{\prime}})$ and assume $T(x)=S(x)$ for all $x\in M\backslash p_{1}(\Gamma^{{}^{\prime}})$.

If $\mu(p_{1}(\Gamma^{{}^{\prime}}))=0$ then $\mu(p_{1}(\Gamma^{{}^{\prime}})\cap\Omega)=0$ and thus $S(x)=T(x)$ for $\mu$-almost all $x\in M$. In particular, the first case holds.

Otherwise, by Lusin’s Theorem there is a compact set $K_{1}\subset\operatorname{supp}\mu\cap p_{1}(\Gamma^{\prime})\cap\Omega$ with $\mu(K_{1})>0$ such that $T$ and $S$ are continuous on $K_{1}$ and $S(x)\neq T(x)$ for $x\in K_{1}$. Thus for sufficiently small $r>0$ and a fixed $x_{0}\in K_{1}$, it holds

[TABLE]

In case $\mu(\{\varphi_{\Gamma}=0\})=0$ we can choose $K=\bar{B}_{r}(x_{0})\cap K_{1}$, $T_{1}=T$ and $T_{2}=S$ and conclude.

If $\mu(\{\varphi_{\operatorname{supp}\pi}>0\})>0$ then there are $\epsilon>0$ and a compact set $K_{2}\subset\operatorname{supp}\mu$ with $\mu(K_{2})>0$ and $\varphi_{\operatorname{supp}\pi}(x)>\epsilon$ for all $x\in K_{2}$. Note also that the sets

[TABLE]

are non-empty Borel subsets of $\Gamma$ with $p_{1}(\Gamma^{+})=p_{1}(\Gamma^{-})=K_{2}$. Thus there are two $\mu$-measurable selection $T^{+}$ and $T^{-}$ with $(x,T^{\pm}(x))\in\Gamma^{\pm}$ for all $x\in K_{2}$. As above we may assume that $T^{\pm}$ agree with $T$ outside of $K_{2}$.

Choose another compact $K_{3}\subset K_{2}$ such that the maps $T^{\pm}$ are continuous on $K_{3}$. In particular, for some $x_{0}\in K_{3}$ and sufficiently small $r>0$ it holds

[TABLE]

To conclude observe that the compact set $K=\bar{B}_{r}(x_{0})\cap K_{3}$ and the maps $T_{1}=T^{-}$ and $T_{2}=T^{+}$ satisfy the last part of the second statement. ∎

Note that in general the second possibility of the Selection Dichotomy above does not say anything about the relationship of the measures $(\operatorname{id}\times T_{i})_{*}\mu$ and a fixed measure $\pi\in\mathcal{P}(M\times M)$ with $(p_{1})_{*}\pi=\mu$ and $\operatorname{supp}\pi\subset\Gamma$. More precisely, in general, $(T_{i})_{*}\mu$ might be singular with respect to $(p_{2})_{*}\pi$, or more generally, it is possible that $(T_{i})_{*}\mu\bot\mathsf{m}$ even if $(p_{2})_{*}\pi\ll\mathsf{m}$.

The following lemma is a more general version of the Selection Dichotomy and can be extracted from Gigli’s work [Gig12, Proof of Theorem 3.3]. It shows that any measure $\pi$, regarded as a generalized transport map $\int\delta_{x}\otimes\mu_{x}d\mu(x)$, is either already induced by a transport map, i.e. $\mu_{x}=\delta_{T(x)}$, or can be decomposed into (at least) two partial transport with target transport on a compact set $K$ of positive $\mu$-measure. We give a simpler proof relying on the Selection Dichotomy for Sets.

First, recall the the statement of the Disintegration Theorem. Let $(X,d)$ and $(Y,d)$ be two complete separable metric spaces. Denote the Borel $\sigma$-algebra of $X$ and $Y$ by $\mathcal{B}(X)$ and resp. $\mathcal{B}(Y)$.

Definition 2.2** (Disintegration over $S$).**

Let $\sigma$ a probability measure on $X$, $S:X\to Y$ a Borel map and $\varpi=S_{*}\sigma$. An assignment $\sigma:\mathcal{B}(X)\times Y\to[0,1]$, denoted $(B,y)\mapsto\sigma_{y}(B)$, is called *a disintegration of $\sigma$ over $S$ *if

1. (1)

$\sigma_{y}(\cdot)$ is a probability measure on $X$ for all $y\in Y$ 2. (2)

$y\mapsto\mu_{y}(B)$ is $\varpi$-measurable for all Borel sets $B\in\mathcal{B}(X)$. 3. (3)

$\mu_{y}(S^{-1}(y))=1$ for all $y\in Y$

and it holds

[TABLE]

Regarding the assignment $y\mapsto\sigma_{y}=\sigma_{y}(\cdot)$ as a map from $Y$ to $\mathcal{P}(X)$ we abbreviate this as

[TABLE]

Remark*.*

(1) A disintegration of $\sigma$ over $S$ as above is usually called a disintegration of $\sigma$ which strongly consistent with $S$.

(2) If $\pi\in\mathcal{P}(M\times M)$ for some complete separable metric space $M$ with $\mu=(p_{1})_{*}\pi$, then any disintegration $\pi=\int\pi_{x}d\mu(x)$ must satisfy $\operatorname{supp}\pi_{x}\subset\{x\}\times M$ and hence $\pi_{x}=\delta_{x}\otimes\mu_{x}$ for a measure $\mu_{x}\in\mathcal{P}(M)$.

The following theorem can be deduced from the general Disintegration Theorem [Fre06, Section 452].

Lemma 2.3** (Disintegration Theorem).**

For every $\sigma\in\mathcal{P}(X)$ and every Borel map $S:X\to Y$ there exists a unique disintegration $\sigma_{\cdot}(\cdot)$ of $\sigma$ over $S$ and for any other disintegration $\tilde{\sigma}_{\cdot}(\cdot)$ of $\sigma$ over $S$ it holds $\sigma_{y}(\cdot)=\tilde{\sigma}_{y}(\cdot)$ for $S_{*}\sigma$-almost all $y\in Y$.

The theorem allows us to say that up to a $\mu$-null set $\sigma=\int\sigma_{x}d\mu$ is the disintegration of $\sigma$ over $S$.

Theorem 2.4** (Selection Dichotomy for Measures).**

Let $\pi$ be a probability measure on $M\times M$ and $\mu=(p_{1})_{*}\pi$. Then exactly one of the following holds:

1. (i)

There is a $\mu$-measurable map $T:M\to M$ such that $\pi(\operatorname{graph}T)=1$. 2. (ii)

There are compact set $K\subset M$ and two closed bounded sets $A_{1},A_{2}\subset M$ with $A_{1}\cap A_{2}=\varnothing$ and

[TABLE]

Furthermore, there are two measure $\pi_{1},\pi_{2}\in\mathcal{P}(M\times M)$ with \frac{1}{\mu(K)}\mu\big{|}_{K}=(p_{1})_{*}\pi_{1}=(p_{1})_{*}\pi_{2} and $\pi_{1},\pi_{2}\ll\pi$.

Remark*.*

(1) The construction shows that for some $\epsilon>0$ it holds

[TABLE]

where $A_{\epsilon}=\bigcup_{x\in A}B_{\epsilon}(x)$ for a set $A\subset M$. Furthermore, as above it is possible to choose $K$, $A_{1}$ and $A_{2}$ such that for some $\delta>0$

[TABLE]

(2) If $\pi=\int\delta_{x}\otimes\mu_{x}d\mu(x)$ is the disintegration over $p_{1}$ and $\pi$ is not induced by a map then for $\mu$-almost all $x\in K$ the measure $\mu_{x}$ is not a delta measure. Indeed, for $\mu$-almost all $x\in K$ it holds

[TABLE]

and the choice of $K$ shows that \mu_{x}\big{|}_{A_{1}} and \mu_{x}\big{|}_{A_{2}} are non-trivial for $\mu$-almost all $x\in K$.

Proof.

We apply Theorem 2.1 to $\Gamma=\operatorname{supp}\pi$. If the first option of the dichotomy holds then $\pi=(\operatorname{id}\times T)_{*}\mu$ and thus $\pi(\operatorname{graph}T)=1$.

Otherwise let $K$, $T_{1}$ and $T_{2}$ as in the second possibility of Theorem 2.1. We may restrict $K$ further and assume \operatorname{supp}(\mu\big{|}_{K})=K.

We claim that for all $\epsilon>0$ and $i=1,2$ it holds

[TABLE]

Indeed, note that \pi\big{|}_{K}=\pi(\cdot\cap(K\times M))\neq\mathbf{0}, so that (x,T_{1}(x)),(x,T_{2}(x))\in\operatorname{supp}(\pi\big{|}_{K})=\operatorname{supp}\pi\cap(K\times M). Because $B_{\epsilon}^{M\times M}(x,y)\subset B_{\epsilon}(x)\times B_{\epsilon}(y)$ for $i=1,2$ it holds

[TABLE]

Since $T_{1}$ and $T_{2}$ are continuous on $K$ and $K$ is compact there is an $\epsilon>0$ such that

[TABLE]

Choosing $A_{1}=\operatorname{cl}(T_{1}(K))_{\epsilon}$ and $A_{1}=\operatorname{cl}(T_{2}(K))_{\epsilon}$ gives first part of the claim.

To obtain the second part, note that there are a $\delta>0$ and compact $K^{\prime}$ of positive $\mu$-measure such that

[TABLE]

Restricting $K^{\prime}$ again, assume K^{\prime}=\operatorname{supp}(\mu\big{|}_{K^{\prime}}) and define two non-trivial measures $\pi_{1},\pi_{2}\in\mathcal{P}(M\times M)$ as follows

[TABLE]

It is easy to see that $\pi_{1},\pi_{2}\ll\pi$ and \frac{1}{\mu(K^{\prime})}\mu\big{|}_{K^{\prime}}=(p_{1})_{*}\pi_{1}=(p_{1})_{*}\pi_{2} which proves the claim. ∎

For completeness we present the following more general form of the Selection Dichotomy.

Corollary 2.5** (General Selection Dichotomy).**

Assume $(X,d)$ and $(Y,d)$ are complete separable metric spaces. Let $\sigma$ be a measure on $X$ and $S:X\to Y$ a Borel map. Then exactly one of the following holds:

1. (i)

There is a measurable map $T:Y\to X$ such that $S(T(y))=y$ and $T_{*}\varpi=\sigma$ where $\varpi=S_{*}\sigma$. In particular, the disintegration of $\sigma$ via $S$ is given by

[TABLE] 2. (ii)

There are a compact set $K\subset X$ and two closed bounded sets $A_{1},A_{2}\subset X$ with $A_{1}\cap A_{2}=\varnothing$ and

[TABLE]

In particular, for $\varpi$-almost all $x\in K$ the measures $\sigma_{x}$ are not delta measures.

Proof.

Just note that the proofs above did not rely on the product structure of $M\times M$ and that $p_{1}$ is a projection. Thus replace $M\times M$ by $X$ and $p_{1}$ by $S$ we can follow the proofs above line by line. ∎

Wasserstein spaces on geodesic spaces

Let $(X,d)$ be a complete, separable metric space. A map $\gamma:[0,1]\to X$ satisfying

[TABLE]

is called a geodesic connecting $\gamma_{0}$ and $\gamma_{1}$. Note that our terminology implies that any geodesic is the curve of minimal length between its endpoints. Denote by $\mathsf{Geo}_{[0,1]}(X,d)$ the set of geodesics. On $\mathsf{Geo}_{[0,1]}(X,d)$ there are natural evaluation maps $e_{t}:\mathsf{Geo}_{[0,1]}(M,d)\to M$, $t\in[0,1]$, defined by $e_{t}:\gamma\mapsto\gamma_{t}$. Denote the length of a geodesic $\gamma$ by $\ell(\gamma):=d(\gamma_{0},\gamma_{1})$ and define a restriction map $\mathsf{restr}_{s,t}:\mathsf{Geo}_{[0,1]}(M,d)\to\mathsf{Geo}_{[0,1]}(M,d)$ for all $0\leq s,t\leq 1$ by

[TABLE]

We say the metric space $(X,d)$ is a *geodesic metric space if between each *$x,y\in X$ there is a geodesic connecting $x$ and $y$, i.e. $(e_{0},e_{1})(\mathsf{Geo}_{[0,1]}(X,d))=X\times X$.

In the following we introduce the main concepts used from the theory of optimal transport. For a comprehensive introduction we refer the reader to Villani’s book [Vil08]. Recall that $(M,d)$ is a complete separable geodesic metric space. Let $\mathcal{P}(M)$ be the set of probability measures on $M$ and for a fixed $x_{0}\in M$ let

[TABLE]

the space of probability measures with finite $p$-th moment. On $\mathcal{P}_{p}(M)$ we define the $p$-Wasserstein metric $W_{p}$ as follows

[TABLE]

where $\Pi(\mu_{0},\mu_{1})$ is the set of $\pi\in\mathcal{P}(M\times M)$ with $(p_{1})_{*}\pi=\mu_{0}$ and $(p_{2})_{*}\pi=\mu_{1}$. This defines a complete metric on $\mathcal{P}_{p}(M)$ with a topology which is strictly stronger than the subspace topology induced by $\mathcal{P}_{p}(M)\subset\mathcal{P}(M)$ unless $(M,d)$ is bounded. We call the convergence induced by the subspace topology weak convergence.

One can show that for each $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ there is a $\pi_{\operatorname{opt}}\in\Pi(\mu_{0},\mu_{1})$ such that

[TABLE]

In this case we say $\pi_{\operatorname{opt}}$ is a $p$-optimal coupling. Let $\mathsf{Opt}_{p}(\mu_{0},\mu_{1})$ denote the set of all $p$-optimal couplings between $\mu_{0}$ and $\mu_{1}$. A general measure $\pi\in\mathcal{P}(M\times M)$ is said to be $p$-optimal if it is a $p$-optimal coupling between $(e_{0})_{*}\pi$ and $(e_{1})_{*}\pi$.

Since $(M,d)$ is geodesic it is possible to show that $(\mathcal{P}_{p}(M),W_{p})$ is geodesic as well. Just note that $(x,y)\mapsto(e_{0},e_{1})^{-1}(x,y)\subset\mathsf{Geo}_{[0,1]}(M,d)$ is a measurable closed-valued map and any selection $\mathcal{T}$ will lift a coupling $\pi$ to a dynamical coupling $\sigma=\mathcal{T}_{*}\pi\in\mathcal{P}(\mathsf{Geo}_{[0,1]}(M,d))$ between two measure $\mu_{0}$ and $\mu_{1}$. If $\pi$ is $p$-optimal than we say $\sigma\in\mathcal{P}(\mathsf{Geo}_{[0,1]}(M,d))$ is a $p$-optimal dynamical coupling. Now one may readily verify that $t\mapsto(e_{t})_{*}\sigma$ is a geodesic connecting $(e_{0})_{*}\sigma$ and $(e_{1})_{*}\sigma$. Denote the set of of $p$-optimal dynamical couplings between $\mu_{0}$ and $\mu_{1}$ by $\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$.

Also note each geodesic $t\mapsto\mu_{t}$ in $\mathcal{P}_{p}(M)$ is induced by a (unique) measure $\sigma\in\mathcal{P}(\mathsf{Geo}_{[0,1]}(M,d))$ such that $(e_{t})_{*}\sigma=\mu_{t}$. In this case it is easy to see that $(e_{t},e_{s})_{*}\sigma$ is a $p$-optimal coupling between $\mu_{t}$ and $\mu_{s}$.

Recall that disintegrating a dynamical coupling $\sigma$ over $(e_{0},e_{1}):\mathsf{Geo}_{[0,1]}(M,d)\to M\times M$ shows that

[TABLE]

where $\pi=(e_{0},e_{1})_{*}\sigma$ and $(x,y)\mapsto\sigma_{x,y}$ is a measurable assignment of dynamical couplings between $\delta_{x}$ and $\delta_{y}$. Similarly, we can disintegrate $\sigma$ over $e_{0}$ to obtain $\sigma=\int\sigma_{x_{0}}d\mu_{0}(x_{0})$ such that for $\mu_{0}$-almost all $x_{0}\in M$ the probability measure $\sigma_{x_{0}}$ is a dynamical coupling between $\delta_{x_{0}}$ and a probability measure $\mu_{x_{0}}$ with $\pi=\int\delta_{x_{0}}\otimes\mu_{x_{0}}d\mu(x_{0})$. Furthermore, if $\sigma$ is $p$-optimal then $\sigma_{x_{0}}$ is $p$-optimal for $\mu_{0}$-almost all $x_{0}\in M$.

The following is the well-known restriction property of optimal couplings, see [Vil08]. Compare the following notation also the the concept push-forward via a plan, see e.g. [AGS14, Definition 2.1].

Lemma 2.6**.**

Assume $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ and $\sigma$ is a $p$-optimal dynamical coupling between $\mu_{0}$ and $\mu_{1}$. Then for $f:M\times M\to[0,\infty)$ with $\lambda=\int fd\pi\in(0,\infty)$ the measure

[TABLE]

is a $p$-optimal coupling between $\mu_{0}^{f}$ and $\mu_{1}^{f}$ where

[TABLE]

Furthermore, if, in addition $f\leq 1$,$\int fd\pi\in(0,1)$ and $\tilde{\sigma}_{f}$ is another $p$-optimal dynamical coupling between $\mu_{0}^{f}$ and $\mu_{1}^{f}$ then

[TABLE]

is also a $p$-optimal coupling between $\mu_{0}$ and $\mu_{1}$.

Remark*.*

(1) If $f$ is a function depending only on the first coordinate then $\mu_{0}^{f}=\lambda^{-1}f\mu_{0}$.

(2) If $\Gamma\subset M\times M$ is Borel set with $\sigma(\Gamma)>0$ then we write \sigma_{\Gamma}=\sigma_{\chi_{\Gamma}}=\frac{1}{\pi(\Gamma)}\sigma\big{|}_{\hat{\Gamma}} where $\hat{\Gamma}=(e_{0},e_{1})^{-1}(\Gamma)$.

Proof.

The first part follows from the restriction property of optimal transport and the second from linearity of the cost functional

[TABLE]

and the fact that $\tilde{\sigma}$ is still a dynamical coupling between $\mu_{0}$ and $\mu_{1}$. ∎

It is easy to see that whenever there are two distinct $p$-optimal coupling $\pi_{1}$ and $\pi_{2}$ between $\mu_{0}$ and $\mu_{1}$ then there are at least two distinct $p$-optimal dynamical coupling $\sigma_{1}$ and $\sigma_{2}$ between $\mu_{0}$ and $\mu_{1}$. By convexity this would actually give a continuum of $p$-optimal (dynamical) couplings. If, however, the dynamical coupling is unique we get the following for restrictions of the endpoints.

Corollary 2.7**.**

Let $\mu_{0}$ and $\mu_{1}$ be two measures in $\mathcal{P}_{p}(M)$. If there is a unique $p$-optimal dynamical coupling $\sigma$ between $\mu_{0}$ and $\mu_{1}$ then for any $f\in L^{\infty}(\pi)$ the $p$-optimal dynamical coupling $\sigma_{f}$ is unique between $\mu_{0}^{f}$ and $\mu_{1}^{f}$.

Proof.

Note that $\sigma_{f}=\sigma_{cf}$ for all $c>0$. Thus if $f\in L^{\infty}(\pi)$ then it is possible to replace $f$ by $\frac{1}{\|f\|_{\infty}}f$ and assume without loss of generality $f\leq 1$ and $\int fd\mathsf{m}\in(0,1)$. Thus the dynamical coupling $\tilde{\sigma}:=\lambda\tilde{\sigma}_{f}+(1-\lambda)\sigma_{1-f}$ is $p$-optimal between $\mu_{0}$ and $\mu_{1}$ whenever $\tilde{\sigma}_{f}$ is $p$-optimal between $\mu_{0}^{f}$ and $\mu_{1}^{f}$. Furthermore, if $\tilde{\sigma}_{f}$ is distinct from $\sigma_{f}$ then $\tilde{\sigma}$ is also distinct from $\sigma$ proving the claim. ∎

Non-branching geodesics

Given a set $\Gamma\subset M\times M$ we frequently use the following abbreviations

[TABLE]

and for $t,s\in[0,1]$

[TABLE]

Thus $\hat{\Gamma}$ is the set of geodesics with endpoints $(x,y)\in\Gamma$ and $\Gamma_{t}$ is the set of $t$-midpoints, where $z$ is a $t$-midpoint of $x$ and $y$ if $\gamma_{t}=z$ for a geodesic connecting $x$ and $y$.

For a set $A\subset M$ and $x\in M$ we also use the abbreviation

[TABLE]

Let $\mathsf{L}$ be a subset of geodesics, then we denote by $\mathsf{L}^{-1}$ the set of reversed geodesics, i.e.

[TABLE]

Similarly, let $\Gamma^{-1}=\{(y,x)\,|\,(x,y)\in\Gamma\}$. It is easy to see that $(\Gamma^{-1})^{\wedge}=(\hat{\Gamma})^{-1}$.

Definition 2.8** (non-branching set).**

A set of geodesics $\mathsf{L}\subset\operatorname{Geo}_{[0,1]}(M,d)$ is non-branching to the right if for all $\gamma,\eta\in\mathsf{L}$ with $\mathsf{restr}_{0,t}\gamma=\mathsf{restr}_{0,t}\eta$ it holds $\gamma\equiv\eta$. Similarly, $\mathsf{L}$ is non-branching to the left if $\mathsf{L}^{-1}$ is non-branching to the right. Furthermore, $\mathsf{L}$ is non-branching if it is both non-branching to the left and to the right.

Remark*.*

Non-branching to the left is the same as Rajala–Sturm’s non-branching condition [RS14, Section 2.2]. This lack of symmetry is irrelevant in their study as essentially non-branching is a symmetric condition when one changes initial and final points (see below).

The following is well-known and follows from strict convexity of $r\mapsto r^{p}$ and the triangle inequality [CH15, Kel15].

Lemma 2.9**.**

Assume for $\gamma,\eta\in\mathsf{Geo}_{[0,1]}(M,d)$ it holds

[TABLE]

Then $\gamma_{t}=\eta_{t}$ for some $t\in(0,1)$ implies $\ell(\gamma)=\ell(\eta)$ and if $(M,d)$ is, in addition, non-branching then $\gamma\equiv\eta$.

Finally recall some properties of Wasserstein geodesics on essential non-branching spaces. We collect the result which can be deduced from [RS14, CM16]. The reader may consult the appendix for a proof of the theorem and its corollary.

Definition 2.10** ($p$-essentially non-branching).**

A metric measure space $(M,d,\mathsf{m})$ is $p$-essentially non-branching if for all $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0},\mu_{1}\ll\mathsf{m}$, any optimal dynamical coupling $\sigma\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ is concentrated on a set of non-branching geodesics, i.e. there is a measurable set $\mathsf{L}\subset\mathsf{Geo}_{[0,1]}(M,d)$ such $\sigma(\mathsf{L})=1$. For brevity we say a measure $\mathsf{m}$ is $p$-essentially non-branching if $(M,d,\mathsf{m})$ is a metric measure space which is $p$-essentially non-branching.

Remark*.*

The property essentially non-branching introduced in [RS14] is equivalent the property $2$-essentially non-branching.

Theorem 2.11**.**

Assume $(M,d,\mathsf{m})$ is a $p$-essentially geodesic measure space. Then for every $p$-optimal dynamical coupling $\sigma\in\mathcal{P}_{p}(M)$ with $(e_{0})_{*}\sigma,(e_{1})_{*}\sigma\ll\mathsf{m}$ the following holds:

1. (i)

For each $t\in(0,1)$ there is a Borel map $\mathsf{T}_{t}:M\to\mathsf{Geo}_{[0,1]}(M,d)$ such that the disintegration of $\sigma$ over $e_{t}$

[TABLE]

where $\mu_{t}=(e_{t})_{*}\sigma$. 2. (ii)

For each $t\in(0,1)$ there is a measurable set of geodesics $\mathsf{L}\subset\mathsf{Geo}_{[0,1]}(M,d)$ with $\sigma(\mathsf{L})=1$ and whenever $\gamma_{t}=\eta_{t}$ for $\gamma,\eta\in\mathsf{L}$ then $\gamma\equiv\eta$. 3. (iii)

For all disjoint Borel sets $\Gamma^{(1)},\Gamma^{(2)}\subset M\times M$ of positive $(e_{0},e_{1})_{*}\sigma$-measure the $t$-midpoints of the restricted geodesics $s\mapsto(e_{s})_{*}\sigma_{\Gamma^{(i)}}$, $i=1,2$, are mutually singular, i.e.

[TABLE]

Corollary 2.12**.**

Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching. Then for each geodesic $t\mapsto\mu_{t}$ in $\mathcal{P}_{p}(M)$ connecting $\mu_{0},\mu_{1}\ll\mathsf{m}$ and $t\in(0,1)$ the geodesics $s\mapsto\mu_{st}$ and $s\mapsto\mu_{t+s(1-t)}$ are the unique geodesics connecting $\mu_{0}$ and $\mu_{t}$ and respectively $\mu_{t}$ and $\mu_{1}$. Furthermore, the (unique) $p$-optimal couplings of $(\mu_{t},\mu_{0})$ and $(\mu_{t},\mu_{1})$ are induced by transport maps $T_{t,0},T_{t,1}:M\to M$.

3. Spaces with good transport behavior

In this section we study spaces where the existence of transport maps is a priori assumed whenever the initial measure is absolutely continuous. It turns out that such spaces are already $p$-essentially non-branching.

Definition 3.1** (Good transport behavior).**

A metric measure space $(M,\mathsf{d},\mathsf{m})$ has good transport behavior $\mathsf{(GTB)}_{p}$ for $p\in(1,\infty)$, if for all $\mu,\nu\in\mathcal{P}_{p}(M)$ with $\mu\ll\mathsf{m}$ any optimal transport plan between $\mu$ and $\nu$ is induced by a map.

Remark*.*

The condition was used in a recent work by F. Galaz-García, A. Mondino, G. Sosa and the author [GGKMS17] in order to study the orbit structure of groups acting isometrically on metric measure spaces with $\mathsf{(GTB)}_{p}$. In particular, it can be used to exclude isometries with too large fixed point set, see also [Sos16, Lemma 4.1].

Proposition 3.2**.**

The following spaces have $\mathsf{(GTB)}_{p}$:

1. (i)

Essentially non-branching $\mathsf{MCP}(K,N)$-spaces for $p=2$, $K\in\mathbb{R}$, and $N\in[1,\infty)$. In particular, this includes, essentially non-branching $\mathsf{CD}^{*}(K^{\prime},N^{\prime})$-spaces, essentially non-branching \text{\mathsf{CD}}(K,N)-spaces, and $\mathsf{RCD^{*}(K^{\prime},N^{\prime})}$-spaces, see **[GRS16, CM17]**. 2. (ii)

Non-branching, qualitatively non-degenerate spaces for all $p\in(1,\infty)$, see **[CH15]** and Definition 5.1 below. 3. (iii)

Any (local) doubling measure $\mu$ on $(\mathbb{R}^{n},\|\cdot\|_{\operatorname{Euclid}})$ or more generally on a Riemannian manifold, see **[GM96]**.

The last example shows that there is an abundance of spaces with good transport behavior. However, we will show that the existence of transport maps prevents too much branching and excludes therefore normed spaces whose norm is not strictly convex. Note that the main theorem of this note extends the list above to $p$-essentially non-branching, qualitatively non-degenerate spaces, see Theorem 5.8.

The first two lemmas were proved in a slightly different form in [GGKMS17]. Recall that for $\Gamma\subset M\times M$ and $s,t\in[0,1]$ we define $\Gamma_{s,t}:=(e_{s},e_{t})\left((e_{0},e_{1})^{-1}\Gamma\right)$.

Lemma 3.3**.**

Let $\Gamma\subset M\times M$ be a $c_{p}$-cyclically monotone set. Then for any $s,t\in[0,1]$ the set $\Gamma_{s,t}$ is $c_{p}$-cyclically monotone.

Proof.

Choose $(x_{s}^{i},x_{t}^{i})\in\Gamma_{s,t}$, $i=1,\ldots,n$, and note that there are geodesics $\gamma^{(i)}\in\hat{\Gamma}$, $i=1,\ldots,n$, with $(\gamma_{s}^{(i)},\gamma_{t}^{(i)})=(x_{s}^{(i)},x_{t}^{(i)}).$ By assumption

[TABLE]

is $c_{p}$-cyclically monotone and hence

[TABLE]

is a $p$-optimal dynamical coupling. Observe that

[TABLE]

is $c_{p}$-cyclically monotone because $(e_{s},e_{t})_{*}\sigma$ is $p$-optimal. Since $\bigcup_{n=1}^{n}\{(\gamma_{0}^{(i)},\gamma_{1}^{(i)})\}\subset\Gamma_{s,t}$ this shows that $\Gamma_{s,t}$ is $c_{p}$-cylically monotone. ∎

Recall that for a subset $\Gamma\subset M\times M$ we define for $x\in M$

[TABLE]

Lemma 3.4** ([GGKMS17, Lemma 4.5]).**

A metric measure space $(M,\mathsf{d},\mathsf{m})$ has $\mathsf{(GTB)}_{p}$ if and only if for every $c_{p}$-cyclically monotone $\Gamma$, the set $\Gamma(x)$ contains at most one point for $\mathsf{m}$-almost all $x\in M$.

In particular, if $(M,\mathsf{d},\mathsf{m})$ has $\mathsf{(GTB)}_{p}$ then for any closed $c_{p}$-cyclically monotone set $\Gamma$ and $\mathsf{m}$-almost all $x\in M$ there exists a unique geodesic connecting $x$ and $\Gamma(x)$, whenever the set $\Gamma(x)$ is non-empty.

Remark*.*

The lemma applies in particular to the $c_{p}$-superdifferential $\partial^{c_{p}}\varphi$ of $c_{p}$-concave functions $\varphi$.

Proof.

The first part follows from the Selection Dichotomy for Sets (Theorem 2.1) and the fact that the support $\Gamma=\operatorname{supp}\pi$ of a $p$-optimal coupling $\pi$ is $c_{p}$-cyclically monotone. Indeed, the second possibility of the Selection Dichotomy applied to $\Gamma$ and $\mu=(p_{1})_{*}\pi$ would imply that there is a compact set $K\subset p_{1}(\Gamma)$ and two maps $T_{1},T_{2}:K\to M$ with $T_{1}(x)\neq T_{2}(x)$ for $x\in K$ and $\{T_{1}(x),T_{2}(x)\}\subset\Gamma(x)$ for $x\in p_{1}(\Gamma)$ and that for \mu_{0}=\frac{1}{\mu(K)}\mu\big{|}_{K} the following coupling

[TABLE]

is $p$-optimal and not induced by a transport map. Therefore, either of the condition implies that $\mu_{0}$ cannot be absolutely continuous with respect $\mathsf{m}$.

To prove the last statement suppose $(M,\mathsf{d},\mathsf{m})$ has $\mathsf{(GTB)}_{p}$ and observe that by the previous lemma whenever $\gamma$ is a geodesic connecting $x$ and $y\in\Gamma(x)$ then $\gamma_{t}\in\Gamma_{0,t}(x)$.

Let $D_{t}=\{x\in M\,|\,\Gamma_{0,t}(x)\neq\varnothing\}$ and note that $D_{t}\subset D_{t^{\prime}}$ whenever $0\leq t^{\prime}\leq t\leq 1$. Let $(t_{n})_{n\in\mathbb{N}}$ be dense in $(0,1]$ with $t_{1}=1$ and choose a measurable set $\Omega_{n}\subset D_{1}$ of full $\mathsf{m}$-measure in $D_{1}$ such that $\Gamma_{0,t_{n}}(x)$ is single-valued for all $x\in\Omega_{n}$. Then $\Omega=\cap_{n\in\mathbb{N}}\Omega_{n}$ also has full $\mathsf{m}$-measure in $D_{1}$. Let $\gamma$ and $\eta$ be two geodesics connecting $x\in\Omega$ and $y\in\Gamma_{0,1}(x)$. If $\gamma$ and $\eta$ were distinct then there is an open interval $I\subset(0,1)$ such that $\gamma_{s}\neq\eta_{s}$ for all $s\in I$. In particular, there is an $n>0$ such that $t_{n}\in I$. Hence $\gamma_{t_{n}}\neq\eta_{t_{n}}$ and $\Gamma_{0,t_{n}}(x)$ is not single-valued. However, this is a contradiction as $x\in\Omega\subset\Omega_{n}$ implies that $\Gamma_{0,t_{n}}(x)$ is single-valued. ∎

Lemma 3.5**.**

Let $\varphi$ be a $c_{p}$-concave function and $(x_{0},x_{1}),(y_{0},y_{1})\in\partial^{c_{p}}\varphi$ be such that for some $t_{0}\in(0,1)$ it holds that $x_{t_{0}}=y_{t_{0}}$, where $x_{t}$ and $y_{t}$ are $t$-midpoints of $(x_{0},x_{1})$ and $(y_{0},y_{1})$ respectively. Then $(x_{0},y_{1}),(y_{0},x_{1})\in\partial^{c_{p}}\varphi$.

Proof.

Choose geodesics $s\mapsto x_{s}$ and $s\mapsto y_{s}$ between $x_{0}$ and $x_{1}$ and resp. $y_{0}$ and $y_{1}$ and define

[TABLE]

Note that $(\varphi,\varphi^{c_{p}})$ is a dual solution for the measures $\mu_{0}$ and $\mu_{1}$.

We write $\varphi_{t}=t^{p-1}\varphi$ and note that the function $\varphi_{t}$ is $c$-concave and $(\varphi_{t},\varphi_{t}^{c_{p}})$ a dual solution for the measure $\mu_{0}$ and $\mu_{t}$ ([Kel15, 2.9 and Remark after 2.1]). Denote the $c_{p}$-duals of $\varphi$ and $\varphi_{t}$ by $\psi$ and $\psi_{t}$ respectively.

Since $\partial^{c_{p}}\varphi$ is $c_{p}$-cyclically monotone, Lemma 2.9 shows that

[TABLE]

Furthermore, $(x_{0},x_{t}),(y_{0},y_{t})\in\partial^{c_{p}}\varphi_{t}$ by the choice of geodesics $s\mapsto x_{s}$ and $s\mapsto y_{s}$. Hence

[TABLE]

For $t=1$ we obtain

[TABLE]

and because $x_{t_{0}}=y_{t_{0}}$ for some $t_{0}\in(0,1)$, we also have

[TABLE]

implying

[TABLE]

Therefore,

[TABLE]

which shows $(x_{0},y_{1})\in\partial^{c_{p}}\varphi$. Similarly, it holds $(y_{0},x_{1})\in\partial^{c_{p}}\varphi$. ∎

The following is a direct application of the last lemma.

Proposition 3.6**.**

Assume $(M,\mathsf{d},\mathsf{m})$ has $\mathsf{(GTB)}_{p}$ and let $\mu_{0},\mu_{1}\in\mathcal{P}^{p}(M)$ with $\mu_{0}\ll\mathsf{m}$. Then there is a unique $p$-optimal dynamical plan $\sigma\in\mathrm{OptGeo}_{p}(\mu_{0},\mu_{1})$ and a measurable set $\mathsf{L}$ with $\sigma(\mathsf{L})=1$ which is non-branching to the right. In particular, a metric measure space with $\mathsf{(GTB)}_{p}$ is $p$-essentially non-branching.

Proof.

Let $\mu_{0},\mu_{1},$ and $\pi$ be as above and $T$ be a $p$-optimal transport map between $\mu_{0}$ and $\mu_{1}$. Assume $\mathcal{T}:M\times M\to\mathsf{Geo}(M,d)$ is a measurable selection such that $\mathcal{T}(x,y)_{0}=x$ and $\mathcal{T}(x,y)_{0}=y$ and define a measurable map $\mathcal{S}:M\to\mathsf{Geo}(M,d)$ by

[TABLE]

Note if $\varphi$ is a dual solution then $\operatorname{supp}\pi\subset\partial^{c_{p}}\varphi$.

By Lemma 3.4 there a Borel set $A$ of full $\mu_{0}$-measure such that

[TABLE]

and for all $x\in A$ the geodesic $\mathcal{S}(x)$ is the unique geodesic connecting $x$ and $T(x)$. This implies immediately that the dynamical coupling $\sigma=\mathcal{S}_{*}\mu_{0}$ is the unique $p$-optimal dynamical coupling between $\mu_{0}$ and $\mu_{1}$.

It suffices to show that $\mathsf{L}=\mathcal{S}(A)$ is non-branching to the right. For this let $\gamma,\eta\in\mathsf{L}$ be two geodesics with $\ell(\gamma)=\ell(\eta)$ with $\gamma_{t}=\eta_{t}$ for some $t\in(0,1)$. Lemma 3.5 implies that $(\gamma_{0},\eta_{1})$ and $(\eta_{0},\gamma_{1})$ are both in $\partial^{c_{p}}\varphi$. However, since $\gamma_{0},\eta_{0}\in A$ this means $\gamma_{1}=\eta_{1}=T(\gamma_{0})$. If we define now

[TABLE]

then $\tilde{\gamma}_{s}$ is also a geodesic connecting $\gamma_{0}$ and $\gamma_{1}$. The choice of $A$ yields $\tilde{\gamma}\equiv\gamma$. Thus $\gamma_{s}=\eta_{s}$ for $s\in[t,1]$ implying that $\mathsf{L}$ is non-branching to the right. ∎

Remark*.*

The conclusion in the first part of Proposition 3.6 above is stronger than the ordinary $p$-essentially non-branching property as it takes into account arbitrary final measures rather than just absolutely continuous ones.

Corollary 3.7**.**

Assume $(M,d,\mathsf{m})$ has good transport behavior $\mathsf{(GTB)}_{p}$ and is strongly non-degenerate $\mathsf{(sND)}_{p}$ (see Definition 4.4 below). Then for any $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0}\ll\mathsf{m}$ there is a unique $p$-optimal dynamical coupling $\sigma$ between $\mu_{0}$ and $\mu_{1}$ and this coupling is concentrated on a set of non-branching geodesics. Furthermore, $(e_{t})_{*}\sigma\ll\mathsf{m}$ for all $t\in(0,1)$. In particular, $\mathsf{m}$ has the strong interpolation property $\mathsf{(sIP)}_{p}$.

Example of essentially non-branching spaces with bad geometric behavior

In this section we construct a measure on the tripod that is essentially non-branching and for any two absolutely continuous measures there is a unique transport map. However, the obvious branching in the tripod shows that there is no measure that makes the tripod into a space with good transport behavior.

Definition 3.8**.**

A metric measure space $(M,d,\mathsf{m})$ has the weak good transport behavior $\mathsf{(GTB)}_{w,p}$ for $p\in(1,\infty)$, if for all $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0},\mu_{1}\ll\mathsf{m}$ any optimal transport plan between $\mu_{0}$ and $\mu_{1}$ is induced by a map.

Let $(\mathsf{T},d)$ be the tripod, i.e. $\mathsf{T}$ is obtained by gluing together three intervals $I_{i}=[0_{i},1_{i}]$, $i=1,2,3$ at $\mathbf{0}=0_{1}=0_{2}=0_{3}$ and $d$ is the corresponding length metric. Denote by $T_{i}$ the natural inclusions $[0,1]\to I_{i}\subset\mathsf{T}$.

Example**.**

There is continuum of measures $\mathsf{m}$ of full support on $\mathsf{T}$ such that $(\mathsf{T},d,\mathsf{m})$ is $p$-essentially non-branching and has the weak good transport behavior $\mathsf{(GTB)}_{w,p}$ for all $p\in(1,\infty)$.

Sketch of the construction.

Let $\nu_{0},\nu_{1},\nu_{2}$ three non-atomic probability measures on $[0,1]$ with full support and $\Omega_{i}$ three disjoint sets such that $\mu_{i}(\Omega_{j})=\delta_{ij}$. Define a measure $\mathsf{m}$ on $\mathsf{T}$ by

[TABLE]

and a set $\Omega=\cup T_{i}(\Omega_{i}).$ Note that $\mathsf{m}(M\backslash\Omega)=0.$

If $x,y\in\Omega$ satisfy $d(\mathbf{0},x)=d(\mathbf{0},y)$ then $x,y\in\Omega_{i}$ for exactly one $i=1,2,3$ and $x=y$. Thus since branching can only happen at $\mathbf{0}$, any two geodesics with endpoints in a $c_{p}$-cyclically monotone set $\Gamma\subset\Omega\times\Omega$ which intersect at a point $t\in(0,1)$ must be equal. In particular, any such $\Gamma$ is already non-branching. Note that whenever $\mu_{0},\mu_{1}\ll\mathsf{m}$ and $\pi$ is a $p$-optimal coupling then

[TABLE]

so that $\pi$ is concentrated on the non-branching set $\operatorname{supp}\pi\cap(\Omega\times\Omega)$.

To obtain transport maps it is sufficient to assume \mu_{0}\ll\mathsf{m}\big{|}_{I_{i}}. In that case let $S_{i}:\mathsf{T}\to[-1,1]$ be the map that collapses $I_{j}$ and $I_{k}$ for $i\neq k,j$ where we assume $I_{i}$ corresponds to $[-1,0]$. Note also that $S_{i}$ restricted to $\Omega$ is invertible hence $(S_{i})_{*}\mathsf{m}$ is a non-atomic measure. This can be used to show that the (unique) $p$-optimal transport map between $(S_{i})_{*}\mu_{0}$ and $(S_{i})_{*}\mu_{1}$ can be pulled back to a $p$-optimal transport map.

This construction works more general for all cost function $h(d(\cdot,\cdot))$ with $h$ strictly convex and increasing. ∎

4. Existence of absolutely continuous interpolations for non-degenerate

measures

In this section we prove the existence of absolutely continuous interpolation measures if the initial measure is absolutely continuous and the background measure satisfies certain non-degenericity conditions. In order to avoid proving very similar results for final measures supported on finite sets and then on general sets, we generalize the construction to optimal couplings concentrated on so called non-degenerate sets.

Non-degenerate measures and sets

The following condition was introduced in [CM16] and is based on stronger variant called qualitative non-degenericty (see blow) introduced earlier in [CH15].

Definition 4.1** (non-degenerate measure).**

A metric measure space $(M,d,\mathsf{m})$ is called non-degenerate if for all Borel sets $A$ with $\mathsf{m}(A)>0$ it holds $\mathsf{m}(A_{t,x})>0$ for $t\in(0,1)$.

Our main goal is to prove existence of absolutely continuous interpolations. We formalize the general a priori existence by the following condition.

Definition 4.2** (interpolation property).**

A metric measure space $(M,d,\mathsf{m})$ is said to have the interpolation property $\mathsf{(IP)}_{p}$ for some $p\in(1,\infty)$ if for all $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0}\ll\mathsf{m}$, all $p$-optimal couplings $\pi\in\mathsf{Opt}_{p}(\mu_{0},\mu_{1})$ and all $t\in(0,1)$ there is a $p$-optimal dynamical coupling $\sigma$ between $\mu_{0}$ and $\mu_{1}$ with $(e_{t})_{*}\sigma\ll\mathsf{m}$.

It has the strong interpolation property $\mathsf{(sIP)}_{p}$ for all $p$-optimal dynamical coupling $\sigma$ between $\mu_{0}$ and $\mu_{1}$ it holds $(e_{t})_{*}\sigma\ll\mathsf{m}$.

In order to show that the interpolation property $\mathsf{(IP)}_{p}$ holds we will study the supports of optimal couplings and need a non-degenericity condition of sets $\Gamma\subset M\times M$.

The following notation will be used: Given a set $A$ and $s\in[0,1]$ define the set $\Gamma^{A,s}$ by

[TABLE]

i.e. we throw out all endpoints which cannot be reached via geodesics having an $s$-midpoint in $A$. As above $\Gamma_{t}^{A,s}$ equals $e_{t}(\Gamma^{A,s})$ whenever $t\in[0,1]$. Observe that if $A$ is analytic then $\Gamma_{t}^{A,s}$ is analytic for all $s,t\in[0,1]$. Also in case $s=t=0$ this simplifies to $\Gamma_{0}^{A,0}=\Gamma_{0}\cap A$.

Definition 4.3** (non-degenerate set).**

An Borel set $\Gamma\subset M\times M$ is non-degenerate (with respect to $\mathsf{m}$) if for all Borel sets $A$ with $\mathsf{m}(\Gamma_{0}\cap A)>0$ it holds $\mathsf{m}(\Gamma_{t}^{A,0})>0$ whenever $t\in(0,1)$.

It is easy to see that $B\times\{x\}$ is non-degenerate for all $x\in M$ and all Borel sets $B\subset M$ whenever $\mathsf{m}$ is non-degenerate.

Definition 4.4** (strong non-degenerate measure).**

A metric measure space $(M,d,\mathsf{m})$ is *strongly non-degenerate $\mathsf{(sND)}_{p}$ *for some $p\in(1,\infty)$ if every $c_{p}$-cyclically monotone Borel set $\Gamma$ is non-degenerate.

Remark*.*

By abuse of notation we say $\mathsf{m}$ is (strongly) non-degenerate or has the (strong) interpolation property if $(M,d,\mathsf{m})$ is (resp. has) the corresponding property.

It is easy to see that any strongly non-degenerate measure $\mathsf{m}$ is also non-degenerate. Furthermore, a measure with strong interpolation property $\mathsf{(sIP)}_{p}$ is necessarily strongly non-degenerate $\mathsf{(sND)}_{p}$. The converse is true if the $p$-optimal dynamical coupling $\sigma$ is unique. In general this is wrong as can be seen by the metric measure space $(\mathbb{R}^{n},\|\cdot-\cdot\|_{\infty},\lambda^{n})$ which has many non-absolutely continuous interpolations from the Lebesgue measures restricted to the unit ball to the delta measure at the origin, see also Remark after the proof of Lemma 5.7.

However, the interpolation property is sufficient to show that the space is strong non-degenerate. Via the existence of absolutely continuous interpolations in the next section one can show that both properties are actually equivalent.

Lemma 4.5**.**

Assume $(M,d,\mathsf{m})$ is a metric measure space having the interpolation property $\mathsf{(IP)}_{p}$. Then $(M,d,\mathsf{m})$ is strongly non-degenerate $\mathsf{(sND)}_{p}$.

Proof.

Let $A$ be a Borel set and $\Gamma$ be a $c_{p}$-cyclically monotone Borel set with $\mathsf{m}(p_{1}(\Gamma)\cap A)>0$. Without loss of generality $A\subset p_{1}(\Gamma)=\Gamma_{0}$. Let \mu_{0}=\frac{1}{\mathsf{m}(A)}\mathsf{m}\big{|}_{A} and choose a measurable selection $T$ of $\Gamma\cap(A\times M)$. Then $\pi=(\operatorname{id}\times T)_{*}\mu_{0}$ is a $p$-optimal coupling. Let $\sigma$ be given by the interpolation property. Then $(e_{t})_{*}\sigma(\Gamma_{t}^{A,0})=1$ and $(e_{t})_{*}\sigma\ll\mathsf{m}$ implying $\mathsf{m}(\Gamma_{t}^{A,0})>0$. Because $A$ and $\Gamma$ are arbitrary we conclude that $(M,d,\mathsf{m})$ ist strongly non-degenerate $\mathsf{(sND)}_{p}$. ∎

The GKS-Construction

In this section we construct an absolutely continuous interpolation $\mu_{t}$ given a $p$-optimal coupling $\pi$ which is concentrated (in a consistent way) on a non-degenerate set $\Gamma$ and its first marginal $(p_{1})_{*}\pi$ is absolutely continuous. Furthermore, we find a Borel set $\tilde{\Gamma}\subset\Gamma$ of full $\pi$-measure, such that $\mu_{t}$ “sees” all points in $\tilde{\Gamma}_{t}$ of positive $\mathsf{m}$-measure. The last property turns out to be crucial in order to apply the idea of Cavalletti–Huesmann [CH15] in the setting of essentially non-branching spaces, see proof of Lemma 5.4 and Theorem 5.8.

The proof of Theorem 4.10 below is based on the following generalized form of the Lebesgue decomposition which can be found in [Rud08, Section 9.4]. One part of the result was proven by Glicksberg and the other by König and Sievers owing the name GKS-Decomposition, see [Rud08, 9.4.1].

Lemma 4.6** (GKS-Decomposition [Rud08, 9.4.4]).**

Let $(M,d)$ be a locally compact complete separable metric space and $\mathcal{B}\subset\mathcal{P}(M)$ be a weakly compact and linearly convex subset of probability measures. Then every non-negative finite measure $\tilde{\mathsf{m}}$ has a unique decomposition

[TABLE]

such that $\tilde{\mathsf{m}}_{a}\ll\mu$ for some $\mu\in\mathcal{B}$ and there is a Borel set $F$ which is a countable union of closed subsets such that $\mathsf{m}_{s}$ is concentrated on $F$ and, in addition, $F$ is $\mathcal{B}$-null, i.e. it holds $\tilde{\mathsf{m}}_{s}(M\backslash F)=0$ and $\nu(F)=0$ for all $\nu\in\mathcal{B}$.

Remark*.*

(1) Linearly convex of $\mathcal{B}$ means that whenever $\mu,\nu\in\mathcal{B}$ then also $(1-\lambda)\mu+\lambda\nu\in\mathcal{B}$ for all $\lambda\in[0,1]$.

(2) The lemma is usually stated for compact Hausdorff spaces. However, one can embed $M$ into the one-point-compactification $M^{*}=\{*\}\cup M$ such that $\mathcal{B}$ is still compact in $\mathcal{P}(M^{*})$. Note that $M^{*}$ is a compact Hausdorff space. Since each of the involved measures gives zero measure to the set $\{*\}$, we see that the lemma also holds for general locally compact Hausdorff spaces. In particular, it holds for proper metric spaces.

(3) Recall that $\mathsf{m}$ is a locally bounded measure if $(M,d,\mathsf{m})$ is a proper metric measure space. In that case there is a continuous function $\varphi:[0,\infty)\to(0,1]$ such that $\tilde{\mathsf{m}}=\varphi(d(x_{0},\cdot))\mathsf{m}$ is a finite measure. Then the unique decomposition of $\mathsf{m}$ with respect to $\mathcal{B}$ is given by $\mathsf{m}=(\varphi(d(x_{0},\cdot))^{-1}\tilde{\mathsf{m}}_{a}+(\varphi(d(x_{0},\cdot))^{-1}\tilde{\mathsf{m}}_{s}$.

Before stating the main theorem of this section we need the following technical lemmas.

Lemma 4.7**.**

If $A$ is an analytic set and $\pi$ is a coupling concentrated on $\Gamma^{A,t}$ then there is a dynamical coupling $\sigma$ concentrated on $e_{t}^{-1}(A)\cap\hat{\Gamma}$. In particular, $(e_{t})_{*}\sigma(A)=1$.

Proof.

Since $A$ is analytic, the set

[TABLE]

is also analytic. Thus by von Neumann’s Measurable Selection Theorem there is a selection $S:\Gamma\to\mathsf{Geo}_{[0,1]}(M,d)$ such that $(x,y,S(x,y))\in\Lambda$ for all $(x,y)\in\Gamma$. In particular, $S(x,y)_{t}\in A$ . To conclude just observe that $\sigma=S_{*}\pi$ is concentrated on $e_{t}^{-1}(A)\cap\hat{\Gamma}$. ∎

Lemma 4.8**.**

Assume $(M,d)$ is a proper geodesic space. Then for all measures $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ and every $p$-optimal coupling $\pi$ the following set of $t$-midpoints

[TABLE]

is linearly convex, compact in $\mathcal{P}_{p}(M)$ and weakly compact in $\mathcal{P}(M)$.

Remark*.*

Similar arguments also show that for finitely many $\{s_{1},\ldots,s_{n}\}\subset[0,1]$ and a measure $\boldsymbol{\mathcal{\pi}}\in\mathcal{P}(M^{n})$ the set

[TABLE]

is linearly convex, compact in $\mathcal{P}_{p}(M)$ and weakly compact in $\mathcal{P}(M)$.

Note that the result shows that the GKS-Decomposition can be applied to the set $\mathcal{B}$. Its proof is given at the end of this section.

In order to make the main theorem more readable we introduce the following condition. It won’t be used anywhere else but here.

Definition 4.9**.**

A coupling $\pi\in\mathcal{P}(M\times M)$ is strongly consistent if for all $\tilde{\pi}\ll\pi$ and every measurable set $\Gamma^{\prime}$ with $\tilde{\pi}(\Gamma^{\prime})=1$ there is a non-degenerate, $c_{p}$-cyclically monotone Borel set $\Gamma\subset\Gamma^{\prime}$ with $\tilde{\pi}(\Gamma)=1$.

Note that whenever $\pi$ is strongly consistent then any coupling $\pi^{\prime}$ with $\pi^{\prime}\ll\pi$ is strongly consistent as well.

Theorem 4.10** (GKS-Construction).**

Let $(M,d,\mathsf{m})$ be a proper metric measure space. Assume $\pi$ is a strongly consistent, $p$-optimal coupling between $\mu_{0}\ll\mathsf{m}$ and $\mu_{1}$. Then for every $t\in(0,1)$ there is a $p$-optimal dynamical coupling $\sigma$ such that $(e_{0},e_{1})_{*}\sigma=\pi$ and

[TABLE]

Furthermore, $\mu_{t}$ is maximal in the following sense: Let $\Gamma$ be a Borel set of full $\pi$-measure and

[TABLE]

be the the Lebesgue decomposition of \mathsf{m}\big{|}_{\Gamma_{t}} with respect to $\mu_{t}$ where $F\subset\Gamma_{t}$ is a Borel with $\mu_{t}(F)=0$. Then $\pi$ is concentrated on a Borel set $\tilde{\Gamma}\subset\Gamma\backslash\Gamma^{F,t}$ and it holds

[TABLE]

Remark*.*

Absolute continuity and maximality imply \mathsf{m}\big{|}_{\tilde{\Gamma}_{t}}\ll\mu_{t}\ll\mathsf{m}\big{|}_{\tilde{\Gamma}_{t}} as $\pi$ is concentrated on $\tilde{\Gamma}$.

Corollary 4.11**.**

Suppose $\mu_{t}\ll\mathsf{m}$ and $\tilde{\Gamma}$ are constructed from $\pi$ as above. If there is a strongly consistent coupling $\pi_{t,1}$ of $\mu_{t}$ and $\mu_{1}$ then for each $s\in(t,1)$ there is a $p$-optimal dynamical coupling $\hat{\sigma}$ such that $(e_{0},e_{1})_{*}\hat{\sigma}=\pi$, $(e_{t})_{*}\hat{\sigma}=\mu_{t}$ and $(e_{s})_{*}\hat{\sigma}\ll\mathsf{m}$.

Proof of the theorem.

Let $\mathcal{B}$ be defined as in Lemma 4.8 above. We split the proof into two steps.

Step $\mathbf{1}$: There is a $\mu_{t}\in\mathcal{B}$ which is maximal in the sense of the theorem and $\rho_{t}\neq 0$ where $\mu_{t}=\rho_{t}\mathsf{m}+\mu_{t}^{s}$ is the Lebesgue decomposition of $\mu_{t}$ with respect to $\mathsf{m}$.

Let $\Gamma\subset\operatorname{supp}\pi$ be a non-degenerate, $c_{p}$-cyclically monotone Borel set with $\pi(\Gamma)=1$. Since $\mathcal{B}$ is weakly compact and linearly convex we can apply the GKS-Decomposition to \mathsf{m}\big{|}_{\Gamma_{t}} and obtain a measure $\mu_{t}\in\mathcal{B}$ such that

[TABLE]

and there is a Borel set $F\subset\Gamma_{t}$ such that $\tilde{\mu}_{t}(F)=0$ for all $\tilde{\mu}_{t}\in\mathcal{B}$ and $\mathsf{m}_{s}(M\backslash F)=0$. Thus \mathsf{m}_{s}=\mathsf{m}\big{|}_{F} and $g(x)>0$ for $\mu_{t}$-almost all $x\in M$.

We claim that $\pi$ is concentrated on $\Gamma\backslash\Gamma^{F,t}$. Assume, by contradiction, that

[TABLE]

Then for $f=\chi_{\Gamma^{F,t}}$ the coupling$\sigma_{f}$ is a $p$-optimal dynamical coupling between $(e_{0})_{*}\sigma$ and $(e_{1})_{*}\sigma$ and $\pi_{F}=(e_{0},e_{1})_{*}\sigma$ is concentrated on $\Gamma^{F,t}$. Note that $\pi=\lambda\pi_{F}+(1-\lambda)\check{\pi}$ where $\check{\pi}=(e_{0},e_{1})_{*}\sigma_{1-f}$ for some $p$-optimal dynamical coupling $\sigma$ with $(e_{0},e_{1})_{*}\sigma=\pi$ and $(e_{t})_{*}\sigma=\mu_{t}$.

By Lemma 4.7 there is a $p$-optimal dynamical coupling $\tilde{\sigma}_{f}$ induced by $\pi_{F}$ which concentrated on $\hat{\Gamma}^{F,t}$ such that $(e_{t})_{*}\sigma(F)=1$. However, by Lemma 2.6 the dynamical coupling

[TABLE]

is also $p$-optimal with

[TABLE]

Thus $\tilde{\mu}_{t}=(e_{t})_{*}\tilde{\sigma}\in\mathcal{B}$ with

[TABLE]

contradicting the properties of GKS-Decomposition. Hence $\pi(\Gamma\backslash\Gamma^{F,t})=1$.

Let

[TABLE]

be the Lebesgue decomposition of $\mu_{t}$ with respect to $\mathsf{m}$ with $\mu_{t}^{s}\perp\mathsf{m}$. Note that $\rho_{t}(x)>0$ for $\mathsf{m}$-almost all $x\in A_{t}=(\Gamma\backslash\Gamma^{F,t})_{t}$. By assumption $\pi$ is concentrated on a non-degenerate $c_{p}$-cyclically monotone Borel set $\tilde{\Gamma}\subset\Gamma\backslash\Gamma^{F,t}$. Since $\mu_{0}(\tilde{\Gamma}_{0})=1$ and $\mu_{0}\ll\mathsf{m}$ it holds $\mathsf{m}(\tilde{\Gamma}_{0})>0$ and thus $\mathsf{m}(\tilde{\Gamma}_{t})>0$ by non-degenericity of $\tilde{\Gamma}$ implying $\rho_{t}\neq 0$, i.e. the absolutely continuous part of $\mu_{t}$ is non-trivial. Finally observe that $\rho_{t}(x)>0$ for $\mathsf{m}$-almost all $x\in A_{t}$ shows that $\mathsf{m}(A_{t}\backslash\tilde{\Gamma}_{t})=0$ hence \mathsf{m}\big{|}_{\tilde{\Gamma}_{t}}\ll\mu_{t} yields maximality of $\mu_{t}$.

Step $\mathbf{2}$: Given $\mu_{t}=\rho_{t}\mathsf{m}+\mu_{t}^{s}$ and $\tilde{\Gamma}$ as in Step 1, there is a $\mu_{t}^{*}\in\mathcal{B}$ with $\mu_{t}^{*}=\rho_{t}^{*}\mathsf{m}$ and $\rho_{t}\leq\rho_{t}^{*}$, and $\mu_{t}^{*}$ is maximal in the sense of the theorem.

We define a partial ordering on subsets of $\mathcal{B}$ and show that maximal elements exist and are absolutely continuous: For $\rho\in L_{\geq 0}^{1}(\mathsf{m})$ with $\int\rho d\mathsf{m}\in[0,1]$ set

[TABLE]

where $\mu=\rho\mathsf{m}+\mu^{s}$ is the Lebesgue decomposition of $\mu$ with respect to $\mathsf{m}$. Note that we identify two $L^{1}(\mathsf{m})$-functions which agree $\mathsf{m}$-almost everywhere.

Let

[TABLE]

and write

[TABLE]

This is a partial ordering of $\mathbb{K}$. Also note that

[TABLE]

is a partition of $\mathcal{B}$. Hence the partial order $\succ$ induces one on this partition.

Assume $\mu\in\mathcal{B}_{\rho}$ is not absolutely continuous with respect to $\mathsf{m}$. Then the decomposition $\mu=\rho\mathsf{m}+\mu^{s}$ induces a decomposition of $\pi$ as follows

[TABLE]

where $\pi_{a}$ and $\pi_{s}$ are uniquely defined measures in $\mathcal{P}(M\times M)$ with

[TABLE]

and

[TABLE]

Let

[TABLE]

Since $\pi_{s}\ll\pi$, we see that $\pi_{s}$ is a strongly consistent $p$-optimal coupling between $\tilde{\mu}_{0}\ll\mathsf{m}$ and $\tilde{\mu}_{1}$. Thus Step $1$ above is applicable to $(\tilde{\mu}_{0},\tilde{\mu}_{1})$ and there is a $t$-midpoint $\tilde{\mu}_{t}$

[TABLE]

with $\tilde{\rho}\neq 0$ such that

[TABLE]

is still in $\mathcal{B}$. Hence $\rho+\lambda\tilde{\rho}\in\mathbb{K}$ and $\rho+\lambda\tilde{\rho}\succ\rho$. In particular, any $\rho\in\mathbb{K}$ with $\int\rho d\mathsf{m}\neq 1$ is not maximal with respect the partial order $\succ$. Also note that any element $\rho\in\mathbb{K}$ satisfying $\int\rho d\mathsf{m}=1$ is automatically maximal. To finish the proof it suffices to show that there are maximal elements above any $\rho\in\mathbb{K}$.

For this we want to apply Zorn’s Lemma: Let $\{\rho_{i}\}_{i\in I}$ be a totally ordered chain where $I$ is a totally ordered index set. Then choose $\mu_{i}\in\mathcal{B}_{\rho_{i}}$ and observe by compactness of $\mathcal{B}$ there is a subnet $I^{\prime}\subset I$ such that $\lim_{i\in I^{\prime}}\mu_{i}=\mu\in\mathcal{B}$. Then the net $(\rho_{i})_{i\in I^{\prime}}$ is an increasing family of non-negative $L^{1}(\mathsf{m})$-function so that by monotone convergence there is an $\rho\in L^{1}(\mathsf{m})$ such that

[TABLE]

and $\int\rho d\mathsf{m}\in[0,1]$. Since

[TABLE]

whenever $i\leq j$, it holds $\rho_{i}\mathsf{m}\leq\mu$ and thus $\rho\mathsf{m}\leq\mu$ where $\mu\leq\nu$ means $\mu(A)\leq\nu(A)$ for all Borel sets $A$. In particular, the Lebesgue decomposition of $\mu$ is given by

[TABLE]

for some $\hat{\rho}\in\mathbb{K}$ with $\hat{\rho}\succeq\rho\succeq\rho_{i}$, i.e. the chain $\{\rho_{i}\}_{i\in I}$ has a maximal element in $\mathbb{K}$. Therefore, Zorn’s Lemma applied to $(\mathbb{K},\succeq)$ gives the existence of at least one maximal element $\rho_{t}^{*}\in\mathbb{K}$ with $\rho_{t}^{*}\geq\rho_{t}$. Choosing $\mu_{t}^{*}=\rho_{t}^{*}\mathsf{m}$ gives a measure satisfying the statements of the theorem.

Finally, the properties of $\mu_{t}$ and $\tilde{\Gamma}$ imply \mathsf{m}\big{|}_{\tilde{\Gamma}_{t}}\ll\rho_{t}\mathsf{m}\ll\mu_{t}^{*}\ll\mathsf{m}\big{|}_{\tilde{\Gamma}_{t}}. Thus for any other Borel set $\Gamma$ with $\pi(\Gamma)=1$ there is a Borel set $\Gamma^{{}^{\prime}}\subset\Gamma\cap\tilde{\Gamma}$ with $\pi(\Gamma^{{}^{\prime}})=1$. But then \mathsf{m}\big{|}_{\tilde{\Gamma}_{t}}\ll\mu_{t}^{*}\ll\mathsf{m}\big{|}_{\Gamma_{t}^{{}^{\prime}}} implying $\mathsf{m}(\tilde{\Gamma}_{t}\backslash\Gamma_{t}^{{}^{\prime}})=0$. This yields immediately \mathsf{m}\big{|}_{\bar{\Gamma}_{t}}\ll\mu_{t}^{*} and thus maximality of $\mu_{t}^{*}$. ∎

In order to apply the theorem we need to prove that we find $p$-optimal couplings satisfying the assumptions of the theorem.

Lemma 4.12**.**

Assume $(M,d,\mathsf{m})$ is non-degenerate. Then any $c_{p}$-cyclically monotone set $\Gamma$ such that $\Gamma_{1}=p_{2}(\Gamma)$ is finite is non-degenerate. In particular, if $\pi$ is a $p$-optimal coupling with $(p_{1})_{*}\pi\ll\mathsf{m}$ and $(p_{2})_{*}\pi=\sum_{i=1}^{n}a_{i}\delta_{x_{i}}$ then any measurable set $\Gamma$ of full $\pi$-measures contains a non-degenerate, $c_{p}$-cyclically monotone Borel set $\tilde{\Gamma}\subset\operatorname{supp}\pi\cap\Gamma$ of full $\pi$-measure.

Proof.

Observe

[TABLE]

for measurable subsets $B^{i}\subset M$. Since $\Gamma_{1}=\mathsf{m}(\cup_{i=1}^{n}B^{i})$ the condition $\mathsf{m}(\Gamma_{0}^{A})>0$ implies there is at least one $i\in\{1,\ldots,n\}$ with $\mathsf{m}(B^{i}\cap A)>0$. But then

[TABLE]

showing that $\Gamma$ is non-degenerate.

For the last statement note that if $\pi(\Gamma^{\prime})=1$ then $\pi(\Gamma^{\prime}\cap\operatorname{supp}\pi)=1$ so that there is a Borel set $\tilde{\Gamma}\subset\Gamma^{\prime}\cap\operatorname{supp}\pi$ with $\pi(\tilde{\Gamma})=1$. Since

[TABLE]

is finite, the result follows. ∎

The same argument also holds more general if the background measure is strongly non-degenerate.

Lemma 4.13**.**

If $(M,d,\mathsf{m})$ is strongly non-degenerate $\mathsf{(sND)}_{p}$ then for any $p$-optimal coupling $\pi$ with $(p_{1})_{*}\pi\ll\mathsf{m}$ the following holds: Whenever $\pi(\Gamma^{\prime})=1$ for a $\pi$-measurable set $\Gamma^{\prime}$ then there is a $c_{p}$-cyclically monotone, non-degenerate Borel subset $\Gamma\subset\Gamma^{\prime}$ of full $\pi$-measure.

Proof.

Just note that $\pi$ is concentrated on $\Gamma^{\prime}\cap\operatorname{supp}\pi$ which is measurable, non-degenerate and $c_{p}$-cyclically monotone. Hence there is a Borel subset $\Gamma\subset\Gamma^{\prime}\cap\operatorname{supp}\pi$ with $\pi(\Gamma)=1$. By strong non-degenericity $\Gamma$ is non-degenerate. ∎

The two lemmas allow us to apply Theorem 4.10 and Corollary 4.11.

Corollary 4.14**.**

Let $(M,d,\mathsf{m})$ be a proper metric measure space and $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0}\ll\mathsf{m}$. Assume either $(M,d,\mathsf{m})$ is non-degenerate and $\mu_{1}=\sum_{i=1}^{n}\lambda_{i}\delta_{x_{i}}$ or that $(M,d,\mathsf{m})$ is strongly non-degenerate $\mathsf{(sND)}_{p}$. Then for any $0<t<s<1$ there is a $p$-optimal dynamical coupling $\sigma$ between $\mu_{0}$ and $\mu_{1}$ with $(e_{t})_{*}\sigma,(e_{s})_{*}\sigma\ll\mathsf{m}$ and $(e_{t})_{*}\sigma$ is maximal in the sense of Theorem 4.10. In particular, $(M,d,\mathsf{m})$ is strongly non-degenerate $\mathsf{(sND)}_{p}$ if and only if it has the interpolation property $\mathsf{(IP)}_{p}$.

Proof.

For strongly non-degenerate measures the result follows immediately from the previous lemma.

For the case of $\mathsf{m}$ being non-degenerate and $\mu_{1}=\sum_{i=1}^{n}\lambda_{i}\delta_{x_{i}}$ just observe that for any $p$-optimal coupling $\pi_{t,1}$ between $\mu_{t}\ll\mathsf{m}$ and $\mu_{1}$ is strongly consistent. Indeed, if $\tilde{\pi}\ll\pi_{t,1}$ then the set $p_{2}(\operatorname{supp}\tilde{\pi}\cap\Gamma^{\prime})$ is finite hence contains a $c_{p}$-cylically monontone Borel set $\Gamma$ with $\tilde{\pi}(\Gamma)=1$. ∎

Proof of Lemma 4.8.

Since $(M,d)$ is geodesic $\mathcal{B}$ is non-empty and closed in $\mathcal{P}_{p}(M).$ Furthermore, properness of $M$ together with $\mathcal{B}$ being bounded implies that $\mathcal{B}$ is weakly precompact. If $\mu_{t}^{0}$ and $\mu_{t}^{1}$ are measures in $\mathcal{B}$ then there are two $p$-optimal dynamical couplings $\sigma^{0}$ and $\sigma^{1}$ such that $(e_{t})_{*}\sigma^{i}=\mu_{t}^{i}$ and $(e_{0},e_{1})\sigma^{i}$, $i=0,1$.

It holds

[TABLE]

for

[TABLE]

so that

[TABLE]

and similarly $W_{p}^{p}(\mu_{0},\mu_{t}^{\lambda})\leq(1-t)^{p}W_{p}^{p}(\mu_{0},\mu_{1})$. Thus $W_{p}(\mu_{0},\mu_{t}^{\lambda})+W_{p}(\mu_{t}^{\lambda},\mu_{1})\leq W_{p}(\mu_{0},\mu_{1})$ which implies that $\mu_{t}^{\lambda}$ is a $t$-midpoint. Since $(e_{0},e_{1})_{*}\sigma^{\lambda}=\pi$ we have $\mu_{t}^{\lambda}\in\mathcal{B}$ implying linear convexity of $\mathcal{B}$.

Now let $(\mu_{t}^{n})_{n\in\mathbb{N}}$ be a sequence in $\mathcal{B}$. By weak compactness we can assuming after picking a subsequence and relabeling that $(\mu_{t}^{n})_{n\in\mathbb{N}}$ converges weakly to some $\mu\in\mathcal{P}(M)$. Since

[TABLE]

and

[TABLE]

we see that $\mu$ is a $t$-midpoint as well. Hence $W_{p}(\mu_{0},\mu_{t}^{n})\to W_{p}(\mu_{0},\mu)$ so that the sequence $(\mu_{t}^{n})_{n\in\mathbb{N}}$ also converges in the $p$-th moment. This shows that $\mu_{t}^{n}\to\mu$ in $\mathcal{P}_{p}(M)$ (see [Vil08, Definition 6.8]). Thus any sequence in $\mathcal{B}$ has a subsequence converging in $\mathcal{P}_{p}(M)$. In particular, $\mathcal{B}$ is compact in $\mathcal{P}_{p}(M)$. ∎

5. Existence of transport maps

In this section we want to prove the existence of transport maps using a combined approach of [CH15] and [CM17].

Qualitatively non-degenerate measures

Non-degenericity and the GKS-Construction in the previous section imply that there are absolutely continuous interpolations between $\mu_{0}\ll\mathsf{m}$ and $\mu_{1}=\sum\lambda_{i}\delta_{x_{i}}$. However, the non-degenericity condition is too weak to use approximation arguments for general $\mu_{1}$. For this we need the following uniform variant which was introduced by Cavalletti–Huesmann [CH15] and represents a weak form of the measure contraction condition $\mathsf{MCP}(K,N)$, see e.g. [Stu06a, CM17] and references therein.

Definition 5.1**.**

The measure $\mathsf{m}$ is said to be qualitatively non-degenerate if for all $R>0$ and $x_{0}\in M$ there is a function $f_{R,x_{0}}:(0,1)\to(0,\infty)$ with

[TABLE]

such that for every measurable $A\subset B_{R}(x_{0})$ and all $x\in B_{R}(x_{0})$ and $t\in(0,1)$ it holds

[TABLE]

Corollary 5.2**.**

Any qualitatively non-degenerate measure is non-degenerate.

The following proposition shows that qualitatvely non-degenerate spaces are proper and make it possible to use GKS-Construction of the previous section.

Proposition 5.3**.**

A qualitatively non-degenerate measure $\mathsf{m}$ is locally doubling, i.e. for each $R>0$ and $x_{0}\in M$ there is a constant $C_{R,x_{0}}>0$ such that

[TABLE]

whenever $B_{2r}(x)\subset B_{R}(x_{0})$. In particular, $(M,d)$ is a proper metric space.

Proof.

Just note that $B_{r}(x)\subset(B_{2r}(x))_{\frac{1}{2},x}$ for all $x\in M$ and $r>0$. Thus qualitative non-degenericity implies for $B_{2r}(x)\subset B_{R}(x_{0})$

[TABLE]

Finally, properness follows from $\mathsf{m}$ being locally doubling (see e.g. [Hei01]). ∎

Lemma 5.4**.**

Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching for some $p\in(1,\infty)$ and $\mathsf{m}$ is qualitatively non-degenerate. If for a Borel set $A$, the set $A\times\{x,y\}$ is $c_{p^{\prime}}$-cyclically monotone for $x\neq y\in M$ and $p^{\prime}\in(1,\infty)$ then $\mathsf{m}(A)=0$.

Remark*.*

One may replace the qualitative non-degenericity by the following pointwise variant

[TABLE]

This condition is, however, too weak to do approximations of general $c_{p}$-cyclically monotone sets as in Lemma 5.7.

Proof.

By inner regularity we can assume $A$ is compact and $\{x,y\}\cup A\subset B_{R}(x_{0})$ for some $R>0$. By compactness of $A$ we find $t$ close to [math] and $s$ close to $1$ such that $f_{R,x_{0}}(t)\geq\frac{1}{2}+\epsilon$,

[TABLE]

and

[TABLE]

for some $\epsilon,\delta>0$.

Decompose $A$ into two Borel sets

[TABLE]

and

[TABLE]

It suffices to show that the claim is true for the cases $A=A^{eq}$ and $A=A^{ne}$.

First assume $A=A^{ne}$ and observe that by $c_{p^{\prime}}$-cyclic monotonicity and the fact that $d(z,x)\neq d(z,y)$ for all $z\in A$ it holds

[TABLE]

Hence

[TABLE]

which implies that $\mathsf{m}(A)=0$.

For the case $A=A^{eq}$ assume by contradiction $\mathsf{m}(A)>0$. Set

[TABLE]

and observe that $A=A^{eq}$ implies that $A\times\{x,y\}$ is $c_{p"}$-cyclically monotone for all $p"\in[1,\infty)$. In particular, $A\times\{x,y\}$ is $c_{p}$-cyclically monotone.

Apply Corollary 4.14 to $(\mu_{0},\delta_{x})$ and $(\mu_{0},\delta_{y})$ to get two dynamical couplings $\sigma^{x}$ and $\sigma^{y}$ whose interpolations at times $s$ and $t$ are absolutely continuous. The choice of $s$ shows that $\mu_{s}^{x}=(e_{s})_{*}\sigma^{x}$ and $\mu_{s}^{y}=(e_{s})_{*}\sigma^{y}$ have disjoint support. Furthermore, the measures $\mu_{t}^{x}=(e_{t})_{*}\sigma^{x}$ and $\mu_{t}^{y}=(e_{t})_{*}\sigma^{y}$ are maximal with respect to $\tilde{A}\times\{x\}$ and resp. $\tilde{A}\times\{y\}$ for some $\tilde{A}\subset A$ with $\mathsf{m}(A\backslash\tilde{A})=0$. Since the set $\tilde{A}\times\{x,y\}$ is still $c_{p}$-cyclically monotone, the dynamical coupling $\frac{1}{2}(\sigma^{x}+\sigma^{y})$ is $p$-optimal between $\mu_{0}$ and $\frac{1}{2}(\delta_{x}+\delta_{y})$. Because $\mu_{s}=\frac{1}{2}(\mu_{s}^{x}+\mu_{s}^{y})$ is a decomposition into mutually singular measures, Theorem 2.11 shows

[TABLE]

By maximality of $\mu_{t}^{x}$ and $\mu_{t}^{y}$ it holds \mathsf{m}\big{|}_{\tilde{A}_{t,x}}\ll\mu_{t}^{x} and \mathsf{m}\big{|}_{\tilde{A}_{t,y}}\ll\mu_{t}^{y} so that $\mathsf{m}(\tilde{A}_{t,x}\cap\tilde{A}_{t,y})=0$. In particular, since $\mathsf{m}$ is qualitatively non-degenerate

[TABLE]

Combining those facts we obtain

[TABLE]

which is a contradiction. This shows that $\mathsf{m}(A)=0$. ∎

Corollary 5.5**.**

Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching and $\mathsf{m}$ qualitatively non-degenerate. If for some $p^{\prime}\in(1,\infty)$ the set $\Gamma$ is a $c_{p^{\prime}}$-cyclically monotone set in $B_{R}(x_{0})\times B_{R}(x_{0})$ and $\Gamma_{1}$ is finite then

[TABLE]

Proof.

Let $\{x_{i}\}_{i=1}^{n}=\Gamma_{1}$ with $x_{i}\neq x_{j}$ whenever $i\neq j$ and set $\Gamma^{i}=\Gamma\cap(M\times\{x_{i}\})$. Then the previous theorem shows

[TABLE]

for $i\neq j$. Thus there are disjoint sets $A^{i}\subset\Gamma^{i}$, $i=1,\ldots,n$, such that

[TABLE]

Similarly one may replace $A^{i}$ by a possibly smaller set which has full $\mathsf{m}$-measure in $A^{i}$ satisfying the condition above and, in addition, it holds

[TABLE]

for $i\neq j$. Setting

[TABLE]

we conclude

[TABLE]

∎

A similar argument also shows the following. As the result is not used below we leave the proof to the interested reader.

Corollary 5.6**.**

Assume $(M,d)$ is $p$-essentially non-branching and $\mathsf{m}$ is qualitatively non-degenerate. Then for any $p^{\prime}\in(1,\infty)$ and any $p^{\prime}$-optimal coupling $\pi$ with $(p_{1})_{*}\pi\ll\mathsf{m}$ and $(p_{2})_{*}\pi=\sum a_{i}\delta_{x_{i}}$ is induced by a transport map.

Via an approximation argument of Cavalletti–Huesmann [CH15, Proposition 4.3] qualitative non-degenericity implies strong non-degneraticity.

Lemma 5.7**.**

Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching and $\mathsf{m}$ is qualitatively non-degenerate. Then for any $c_{p^{\prime}}$-cyclically monotone Borel set $\Gamma$ in $B_{R}(x_{0})\times B_{R}(x_{0})$ it holds

[TABLE]

In particular, $(M,d,\mathsf{m})$ is strongly non-degenerate $\mathsf{(sND)}_{p^{\prime}}$ for all $p^{\prime}\in(1,\infty)$.

Proof.

For compact $\Gamma$ the argument is a in [CH15, Proposition 4.3]. For completeness, we present the argument: Let $(\Gamma^{(n)})_{n\in\mathbb{N}}$ be a sequence of $c_{p}$-cyclically monotone sets such that $\Gamma_{0}^{(n)}=\Gamma_{0}$ and $\Gamma_{1}^{(n)}\subset\Gamma_{1}$ is finite. More precisely, choose a countable dense sequence $y_{n}\in\Gamma_{1}$ and define

[TABLE]

and

[TABLE]

From the definition of $E_{i}^{(n)}$ it follows that $\Gamma^{(n)}$ is $c_{p^{\prime}}$-cyclically monotone. Furthermore, compactness of $\Gamma$ shows that for all $\epsilon>0$ there is an $N_{\epsilon}$ such that for all $n\geq N_{\epsilon}$ it holds

[TABLE]

This yields immediately the result for compact $\Gamma$ as follows

[TABLE]

For arbitrary $c_{p}$-monotone Borel sets $\Gamma$ in $B_{R}(x_{0})\times B_{R}(x_{0})$ we can use the Measurable Selection Theorem and Lusin’s Theorem to reduce the result to compact set. If $\mathsf{m}(\Gamma_{0})=0$ then there is nothing to prove. So assume $\mathsf{m}(\Gamma_{0})>0$. Choose a measurable selection $T$ of $\Gamma$. As $\mathsf{m}$ is locally bounded and $\Gamma_{0}\subset B_{R}(x_{0})$ we obtain by Lusin’s Theorem a family of compact sets $K^{1}\subset K^{2}\subset\ldots\subset\Gamma_{0}$ with $\mathsf{m}(K^{i})\to\mathsf{m}(\Gamma_{0})$ such that $T$ restricted to $K^{i}$ is continuous. Define $\Gamma^{i}=\operatorname{graph}_{K^{i}}T\subset\Gamma$ and note that $\Gamma^{i}$ is compact with $\Gamma_{0}^{i}=K^{i}$ so that

[TABLE]

It remains to show that $\mathsf{m}$ is strongly non-degenerate $\mathsf{(sND)}_{p}$. First observe

[TABLE]

where $\Gamma^{R}=\Gamma\cap(B_{R}(x_{0})\times B_{R}(x_{0}))$. As $\Gamma^{R}$ is bounded we have

[TABLE]

Assume now $\mathsf{m}(\Gamma_{0})>0$ then $\mathsf{m}(\Gamma_{0}^{R})\in(0,\infty)$ for all large $R>0$ so that

[TABLE]

Non-degenericity of $\Gamma$ follows by observing that for all Borel sets $A$ the set $\Gamma^{A,0}=\Gamma\cap(A\times M)$ is still a $c_{p}$-cyclically monotone Borel set. Thus $\mathsf{m}(\Gamma_{0}^{A,0})>0$ implies $\mathsf{m}(\Gamma_{t}^{A,0})>0$. ∎

Remark*.*

The proof of the results above relies only on the qualitative non-degenericity of $\mathsf{m}$ and that

[TABLE]

for all $x\neq y$. In particular, it holds for $(\mathbb{R}^{n},\|\cdot-\cdot\|_{\infty},\lambda^{n})$ which is highly branching.

By combining the previous lemma and the idea of the proof of Lemma 5.4 we obtain the main theorem of this section.

Theorem 5.8**.**

Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching and $\mathsf{m}$ is qualitatively non-degenerate. Then any $p$-optimal coupling $\pi\in\mathcal{P}_{p}(M\times M)$ with $(p_{1})_{*}\pi\ll\mathsf{m}$ is induced by a transport map. In particular, any such space has good transport behavior $\mathsf{(GTB)}_{p}$.

Combined with Proposition 3.6 and the existence of absolutely continuous interpolations (Corollary 4.14) we get the following two corollaries.

Corollary 5.9**.**

If $(M,d,\mathsf{m})$ is $p$-essentially non-branching and $\mathsf{m}$ qualitatively non-degenerate then between any two measure $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0}\ll\mathsf{m}$ there is a unique $p$-optimal dynamical coupling $\sigma$ and this coupling satisfies $(e_{t})_{*}\sigma\ll\mathsf{m}$ for all $t\in[0,1)$. In particular, it has the strong interpolation property $\mathsf{(sIP)}_{p}$.

Corollary 5.10**.**

Assume $\mathsf{m}$ is qualitatively non-degenerate. Then $\mathsf{m}$ is $p$-essentially non-branching if any only if it has good transport behavior $\mathsf{(GTB)}_{p}$ .

Proof of the Theorem.

Note by Proposition 5.3, $(M,d)$ is proper so that we can apply the GKS-Construction of the previous section.

Let $\Gamma=\operatorname{supp}\pi$ and note that $\mathsf{m}(p_{1}(\Gamma))=\mathsf{m}(\operatorname{supp}((p_{1})_{*}\pi))>0$. It suffices to show that $\Gamma(x)$ is single-valued for $\mathsf{m}$-almost all $x\in M$. This holds, if for all $R>0$, $\Gamma^{R}(x)$ is single-valued for $\mathsf{m}$-almost all $x\in M$ where $\Gamma^{R}=\Gamma\cap(\bar{B}_{R}(x_{0})\times\bar{B}_{r}(x_{0}))$. Note that large $R>0$ it holds $\mathsf{m}(p_{1}(\Gamma^{R}))\in(0,\infty)$.

Assume, by contradiction, that for some $R>0$ there is a Borel set $A$ with $\mathsf{m}(A)>0$ and the set $\Gamma^{R}(x)$ is non-empty and not single-valued for all $x\in A$. Then by the Selection Dichotomy of Sets (Theorem 2.1), there is a compact set $K\subset A$ of positive $\mathsf{m}$-measure, and two continuous maps $T_{1},T_{2}:M\to M$ with $T_{1}(K)\cap T_{2}(K)=\varnothing$ and

[TABLE]

Restricting $K$ further, we can also assume \operatorname{supp}(\mathsf{m}\big{|}_{K})=K\subset\bar{B}_{R}(x_{0}).

Define now \mu_{0}=\frac{1}{\mathsf{m}(K)}\mathsf{m}\big{|}_{K}, $\pi_{i}=(\operatorname{id}\times T_{i})_{*}\mu_{0}$ and $\mu_{1}^{i}=(p_{2})_{*}\pi_{i}$ for $i=1,2$. Let $\Gamma^{(i)}=\operatorname{supp}\pi_{i}$, $i=1,2$, and note that $\Gamma_{0}^{(i)}=K$ and both $\Gamma^{(1)}$ and $\Gamma^{(2)}$ are compact and $c_{p}$-cyclically monotone.

Choose $\delta>0$, $t$ close to [math] and $s$ close to $1$ such that $f_{R,x_{0}}(t)\geq\frac{1}{2}+\epsilon$,

[TABLE]

and

[TABLE]

Corollary 4.14 applied to $(\mu_{0},\mu_{1}^{1})$ and $(\mu_{0},\mu_{1}^{2})$ gives two $p$-optimal dynamical couplings $\sigma^{(1)}$ and $\sigma^{(2)}$ such that $(e_{s})_{*}\sigma^{(1)}$ and $(e_{s})_{*}\sigma^{(2)}$ are absolutely continuous with respect to $\mathsf{m}$ and have disjoint support. The choice of $\Gamma^{(1)}$ and $\Gamma^{(2)}$ implies that $\frac{1}{2}(\pi_{1}+\pi_{2})$ is supported on $\Gamma^{(1)}\cup\Gamma^{(2)}\subset\operatorname{supp}\pi$. Hence $(\mathsf{restr}_{0,s})_{*}\frac{1}{2}(\sigma^{(1)}+\sigma^{(2)})$ is a $p$-optimal dynamical coupling between $\mu_{0}$ and $\frac{1}{2}(\mu_{s}^{1}+\mu_{s}^{2})$ so that Theorem 2.11 shows

[TABLE]

Maximality at time $t$ shows that for $i=1,2$ there are measurable subsets $\tilde{\Gamma}^{(i)}\subset\Gamma^{(i)}$ with $\mathsf{m}(K\backslash\tilde{\Gamma}_{0}^{(i)})=0$ and \mathsf{m}\big{|}_{\tilde{\Gamma}_{t}^{(i)}}\ll\mu_{t}^{(i)}\ll\mathsf{m}\big{|}_{\tilde{\Gamma}_{t}^{(i)}}. Since $\mu_{t}^{(1)}\bot\mu_{t}^{(2)}$ we must have $\mathsf{m}(\tilde{\Gamma}_{t}^{(1}\cap\tilde{\Gamma}_{t}^{(2)})=0$. In combination with Lemma 5.7 this yields

[TABLE]

This, however, leads to the following contradiction

[TABLE]

Thus we have proved that the $\Gamma^{R}(x)$ at most single-valued for $\mathsf{m}$-almost all $x\in M$ proving that $\pi$ is induced by a transport map. ∎

The proof relies heavily on the $p$-essentially non-branching property of dynamical couplings between absolutely continuous measures. In contrast to the case of a discrete target measures we cannot show that that general $p^{\prime}$-optimal couplings with absolutely continuous first marginals are induced by transport maps. Nevertheless, $p$-essentially non-branching and the idea of Lemma 5.4 still exclude a too general behavior of the support of $p^{\prime}$-optimal couplings.

Theorem 5.11**.**

Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching for some $p\in(1,\infty)$, $\mathsf{m}$ is qualitatively non-degenerate and $p^{\prime}\in(1,\infty)$. Then for any $p^{\prime}$-optimal $\pi\in\mathcal{P}(M\times M)$ with $(p_{1})_{*}\pi\ll\mathsf{m}$ and for $\mu_{0}$-almost every $x\in M$ it holds

[TABLE]

Corollary 5.12**.**

The $c_{p^{\prime}}$-superdifferential $\partial^{c_{p^{\prime}}}\varphi$ of a $c_{p^{\prime}}$-concave function $\varphi$ satisfies for $\mathsf{m}$-almost every $x\in M$

[TABLE]

Remark*.*

The property $p$-essentially non-branching is used only to show that $c_{p^{\prime}}$-cyclically monotone sets are non-degenerate. As mentioned above, this holds if we replace $p$-essentially non-branching by the assumption

[TABLE]

for all $y\neq z$,

Proof.

If the claim was false then $\pi$ is not induced by a transport map and as above we get a compact set $K$ of positive $\mathsf{m}$-measure and measurable selections $T_{1}$ and $T_{2}$ as above which, in addition, satisfy

[TABLE]

Let \mu_{0}=\frac{1}{\mathsf{m}(K)}\mathsf{m}\big{|}_{K} and for $i=1,2$ define $\pi_{i}=(\operatorname{id}\times T_{i})_{*}\tilde{\mu}_{0}$ and $\Gamma^{(i)}=\operatorname{supp}\pi_{i}$. Again $\Gamma^{(i)}$ is $c_{p}$-cyclically monotone, but satisfies, in addition, the following

[TABLE]

Choosing $\epsilon$, $\delta$ and $t$ as in the previous proof, we arrive at the following contradiction

[TABLE]

∎

Density bounds of qualitatively non-degenerate measures

In [CM17] Cavalletti–Mondino showed that the measure contraction property $\mathsf{MCP}(K,N)$ implies the existence of an absolutely continuous interpolations with controlled $L^{\infty}$-bounds on their density. This was then used to prove the general existence of transport maps.

Definition 5.13**.**

The measure $\mathsf{m}$ has bounded density property if for all $R>0$ and $x_{0}\in M$ there is a function $g_{R,x_{0}}:(0,1)\to(0,\infty]$ with

[TABLE]

such that for some $p\in(1,\infty)$ and for every $\mu_{0}=f_{0}\mathsf{m}\in\mathcal{P}_{p}(M)$ with $\|f_{0}\|_{\infty}<\infty$, $\operatorname{supp}\mu_{0}\subset B_{R}(x_{0})$ and $x\in B_{R}(x_{0})$ there is a geodesic $t\mapsto\mu_{t}=f_{t}\mathsf{m}$ between $\mu_{0}$ and $\delta_{x}$ in $\mathcal{P}_{p}(M)$ such that

[TABLE]

Remark*.*

It is easy to see that the definition does not depend on $p\in(1,\infty)$.

First observe that the bounded density property is stronger than qualitative non-degenericity.

Lemma 5.14**.**

Every measure $\mathsf{m}$ with bounded density property is qualitatively non-degenerate.

Proof.

Let \mu_{0}=\frac{1}{\mathsf{m}(A_{0})}\mathsf{m}\big{|}_{A_{0}} and note that

[TABLE]

and

[TABLE]

we obtain

[TABLE]

Choosing $f_{R,x_{0}}=g_{R,x_{0}}^{-1}$ we obtain the result. ∎

The bounded density property was proven to hold for spaces with curvature dimension condition $\mathsf{CD}(K,N)$ by Rajala [Raj12b, Theorem 4.2] and later for the measure contraction property $\mathsf{MCP}(K,N)$ by Cavalletti–Mondino [CM17, Theorem 3.1]. Assuming $\mathsf{m}$ is $p$-essentially non-branching, the following result implies that the bounded density property is equivalent to qualitative non-degenericity.

Proposition 5.15**.**

Assume $(M,d,\mathsf{m})$ be $p$-essentially non-branching and $\mathsf{m}$ is qualitatively non-degenerate and $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0}=f_{0}\mathsf{m}$ and $\operatorname{supp}\mu_{0},\operatorname{supp}\mu_{1}\subset B_{R}(x_{0})$. Then for the unique $p$-optimal dynamical coupling $\sigma\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ it holds

[TABLE]

where $(e_{t})_{*}\sigma=f_{t}\mathsf{m}$. In particular, it holds

[TABLE]

so that $\mathsf{m}$ has the bounded density property.

Corollary 5.16**.**

In a $p$-essentially non-branching metric measure space $(M,d,\mathsf{m})$ the following are equivalent:

• •

The measure $\mathsf{m}$ is qualitatively non-degenerate.

• •

The measure $\mathsf{m}$ has the bounded density property.

Proof of the proposition.

We first assume $f_{0}\equiv\frac{1}{\mathsf{m}(A_{0})}$. If the claim was wrong then there is a compact set $\mathsf{L}\subset\mathsf{Geo}(M,d)$ with $\sigma(\mathsf{L})>0$ such that

[TABLE]

In particular, by restricting $\sigma$ to $\mathsf{L}$ we see that $\tilde{A}_{0}=e_{0}(\mathsf{L})\subset A_{0}$ it holds and

[TABLE]

where $\tilde{f}_{t}\mathsf{m}=(e_{t})_{*}\sigma_{\mathsf{L}}$. The qualitative non-degenericity yields

[TABLE]

Note that we always have

[TABLE]

This, however, leads to the following contradiction

[TABLE]

For general $\mu_{0}$, observe that

[TABLE]

is a geodesic in $\mathcal{P}_{p}(M)$ such that $\tilde{\mu}_{0}$ has constant density, i.e. \tilde{\mu}_{0}=\frac{1}{\mathsf{m}(\{f_{0}>0\})}\mathsf{m}\big{|}_{\{f_{0}>0\}}. Furthermore, $\tilde{f}_{t}$ satisfies

[TABLE]

so that

[TABLE]

which proves the claim. ∎

Recall that the $\mathsf{MCP}(0,N)$-condition holds if for all $\mu_{0}=\rho_{0}\mathsf{m}\in\mathcal{P}_{2}(M)$ and all $x\in M$ there is geodesic $t\mapsto\mu_{t}=\rho_{t}\mathsf{m}+\mu_{t}^{s}$ between $\mu_{0}$ and $\delta_{x}$ such that

[TABLE]

Cavalletti–Mondino showed that $\mathsf{MCP}(0,N)$-spaces have the bounded density property with $g_{R,x_{0}}(t)=(1-t)^{-N}$, see [CM17, Theorem 3.1]. Thus we obtain the following equivalent characterization of essentially non-branching $\mathsf{MCP}(0,N)$-spaces.

Corollary 5.17**.**

A $p$-essentially non-branching metric measure space satisfies the measure contraction property $\mathsf{MCP}(0,N)$ if and only if it is qualitatively non-degenerate with $f_{R,x_{0}}(t)=(1-t)^{N}$, i.e. $\mathsf{m}(A_{t,x})\geq(1-t)^{N}\mathsf{m}(A)$ for all $x\in M$ and all Borel set $A\subset M$ of finite $\mathsf{m}$-measure.

There are similar versions for the general measure contraction property $\mathsf{MCP}(K,N)$, $K\in\mathbb{R}$ and $N\in[1,\infty)$. This actually shows that one can regard the measure contraction property as a directional version of Bishop–Gromov volume comparison condition which for $K=0$ says that $\mathsf{m}(B_{r}(x))\geq(1-t)^{N}\mathsf{m}(B_{(1-t)r}(x))$.

Remark* (Removing essentially non-branching I).*

Using a construction of good geodesics as in Rajala [Raj12b] and Cavalletti–Mondino [CM17] combined with the GKS-Construction (Theorem 4.10) it might be possible to obtain the equivalence without assuming that measure is essentially non-branching.We leave the details to a future work.

Along the lines of [Raj12a] we also obtain local versions of the Poincaré inequality with constant

[TABLE]

i.e. for all Lipschitz functions $f:M\to\mathbb{R}$ and $B_{r}(x)\subset B_{R}(x_{0})$ it holds

[TABLE]

where

[TABLE]

and

[TABLE]

Remark* (Removing essentially non-branching II).*

Similar to Lemma 5.7 it is possible to show that under the condition

[TABLE]

for all $x\neq y$ the bounded density property holds between every $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ whenever $\operatorname{supp}\mu_{0},\operatorname{supp}\mu_{1}\subset B_{R}(x_{0})$ and the function $g_{R,x_{0}}$ is upper semi-continuous in $(0,1)$. In particular, if the density bounds are sufficiently nice then a local doubling condition and local Poincaré inequality holds.

We quickly sketch the argment: Note first that Lemma 5.4 holds for those spaces so that one obtains for $\mu_{1}=\sum_{i=1}^{n}\lambda_{i}\delta_{x_{i}}$ a geodesic in $\mathcal{P}_{p}(M)$ between $\mu_{0}$ and $\mu_{1}$ which has uniform density bounds only depending on the density of $\mu_{0}$. Now let $\mu_{1}^{n}\rightharpoonup\mu_{1}$. At a fixed time $t\in(0,1)$ there is a $\mu_{t}^{n}=\rho_{t}^{n}\mathsf{m}$ with $\int\rho_{t}^{n}\mathsf{m}=1$ and $\|\rho_{t}^{n}\|_{\infty}\leq g_{R,x_{0}}(t)\|\rho_{0}\|_{\infty}$ implying that $(\rho_{t}^{n})_{n\in\mathbb{N}}$ is precompact in $L^{1}(\mathsf{m})$. Hence up to extracting a subsequence $\rho_{t}^{n}\to\rho_{t}$ in $L^{1}(\mathsf{m})$, $\|\rho_{t}\|_{\infty}\leq C_{t}\|\rho_{0}\|_{\infty}$ and $\mu_{t}=\rho_{t}\mathsf{m}$ being a $t$-midpoint of $\mu_{0}$ and $\mu_{1}$. The same argument then gives a geodesic $t\mapsto\mu_{t}$ which is absolutely continuous with uniform density bounds at all points $t\in\mathbb{Q}\cap(0,1)$. By upper semi-continuity of $g_{R,x_{0}}$ and the again same argument this time applied to $\mu_{t_{n}}\rightharpoonup\mu_{t}$ with $t_{n}\in\mathbb{Q}\cap(0,1)$ and $t_{n}\to t\in(0,1)$ shows that $\mu_{t}$ is absolute continuous with uniform density bound, compare also with [CM17, Proof of Theorem 4.1].

Generalizations to $N=\infty$

As it turns out the idea of the proof of existence of transport maps can be easily generalized to a more general situation. Compare the results of this section with [Gig12, Theorem 3.3(ii)] where non-branching spaces were treated. Recall that the $\mathsf{CD}_{p}(K,\infty)$-condition (see [LV09, Stu06b, Kel15]) requires that for $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{i}\ll\mathsf{m}$ there is a geodesic $t\mapsto\mu_{t}\ll\mathsf{m}$ such that

[TABLE]

where $f_{t}$ is the density of $\mu_{t}$.

If we choose \mu_{0}=\frac{1}{\mathsf{m}(A_{0})}\mathsf{m}\big{|}_{A_{0}} and apply Jensen’s inequality on the left-hand side, then it holds

[TABLE]

where $\Gamma_{t}=\operatorname{supp}\mu_{t}$. Thus

[TABLE]

whenever $\int f_{1}\log f_{1}d\mathsf{m}<\infty$ and $A_{0}$ is compact.

Remark*.*

We can replace the $\mathsf{CD}_{p}(K,\infty)$-condition by the $\mathsf{CD}_{p}^{*}(K,N)$-condition with $N<0$ as defined by Ohta in [Oht16]. Indeed, following the proof of [Oht16, Theorem 4.8] gives a stronger variant of the Brunn–Minkowski inequality (replace $A_{t}$ by $\Gamma_{t}$) which for $K=0$ and $r=-\frac{1}{N}>0$ says

[TABLE]

implying again $\lim_{t\to 0}\mathsf{m}(\Gamma_{t})=\mathsf{m}(A_{0})$.

Lemma 5.18**.**

Let $A$ be a bounded Borel set and $\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{1}\ll\mathsf{m}$. Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching and satisfies the $\mathsf{CD}_{p}(K,\infty)$-condition. If the geodesic connecting \mu_{0}=\frac{1}{\mathsf{m}(A)}\mathsf{m}\big{|}_{A} and $\mu_{1}$ is unique then the (unique) $p$-optimal coupling $\pi$ of $\mu_{0}$ and $\mu_{1}$ is induced by a transport map.

Remark*.*

Strictly speaking the assumptions imply that the strong $\mathsf{CD}_{p}(K,\infty)$-condition holds between $\mu_{0}$ and $\mu_{1}$ thus the argument of the proof of [RS14, Corollary 1.4] can be used. For completeness we present the arguments based on the ideas above.

Proof.

Assume by contradiction that the claim is false for \mu_{0}=\frac{1}{\mathsf{m}(A)}\mathsf{m}\big{|}_{A} and $\mu_{1}\ll\mathsf{m}$. Then by the Selection Dichotomy (Theorem 2.4) there are a compact set $K\subset A$ and two disjoint bounded closed set $A_{1}$ and $A_{2}$ such that $\pi(K\times A_{1})>0$ and

[TABLE]

where \pi_{i}=\frac{1}{\pi(K\times A_{i})}\pi\big{|}_{K\times A_{i}} and $i=1,2$. Denote the density of $\mu_{1}^{i}$ by $f_{1}^{i}$ and note that for large $n\in\mathbb{N}$

[TABLE]

Thus we may restrict $K$ further (and obtain new $A_{i}$, $\pi_{i}$, and $\mu_{1}^{i}$) and assume that for $i=1,2$, $f_{1}^{i}(y)$ is bounded by $n$ for $\mu_{1}^{i}$-almost all $y\in M$. Since each $A_{1}^{i}$, $i=1,2$, is bounded we obtain

[TABLE]

Corollary 2.7 implies that the $p$-optimal dynamical coupling between $\mu_{0}$ and $\mu_{1}^{i}$ is still unique so that the interpolation inequality implies

[TABLE]

where $\Gamma^{(i)}=\operatorname{supp}\pi_{i}$ is the the support of the unique $p$-optimal dynamical coupling of $\mu_{0}$ and $\mu_{0}^{i}$. In particular, it holds

[TABLE]

Also note that $\Gamma^{(1)}\cup\Gamma^{(2)}$ is $c_{p}$-cyclically monotone, so that $(M,d,\mathsf{m})$ being $p$-essential non-branching shows for sequence $t_{n}\to 0$ we may replace $\Gamma^{(1)}$ and $\Gamma^{(2)}$ by smaller sets $\tilde{\Gamma}^{(1)}\subset\Gamma^{(1)}$ and $\tilde{\Gamma}^{(2)}\subset\Gamma^{(2)}$ such that $\pi_{1}(\tilde{\Gamma}^{(1)})=\pi_{2}(\tilde{\Gamma}^{(2)})=1$ and $\mathsf{m}(\Gamma_{t_{n}}^{(1)}\cap\Gamma_{t_{n}}^{(2)})=0$ for all large $n\in\mathbb{N}$. Note that still $\lim_{t\to 0}\mathsf{m}(\tilde{\Gamma}_{t}^{(i)})=\mathsf{m}(K)$. But then

[TABLE]

which is a contradiction. ∎

Theorem 5.19**.**

Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching and satisfies the $\mathsf{CD}_{p}(K,\infty)$-condition. If $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ such that there is a $p$-optimal dynamical coupling $\sigma\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ with $\mu_{0},(e_{t_{0}})_{*}\sigma\ll\mathsf{m}$ for some $t_{0}\in M$ then the $p$-optimal coupling $(e_{0},e_{1})_{*}\sigma$ is induced by a transport map $T$.

Proof.

We reduce the general case to the lemma above. Let $t\mapsto\mu_{t}$ be a geodesic between $\mu_{0}$ and $\mu_{1}$. By Corollary 2.12 we see that for fixed $t\in(0,1)$, $s\mapsto\mu_{st}$ is the unique geodesic connecting $\mu_{0}$ and $\mu_{t}$ and there is a $p$-optimal transport map $T_{t,1}$ between $\mu_{t}$ and $\mu_{1}$. Hence, it suffices to show that there is a transport map from $\mu_{0}$ to $\mu_{t}$. Since $s\mapsto\mu_{st}$ is unique between its endpoint it suffices to show the claim for $\mu_{0}$ and $\mu_{1}$ connected by a unique geodesic $t\mapsto\mu_{t}$.

Let $\sigma$ be the unique $p$-optimal dynamical coupling induced by $t\mapsto\mu_{t}$ and $\pi=(e_{0},e_{1})_{*}\pi$. Denote the densities of $\mu_{0}$ and $\mu_{1}$ with respect to $\mathsf{m}$ by $f_{0}$ and $f_{1}$ respectively. Since

[TABLE]

where $C_{n}=(\{f_{0}\geq\frac{1}{n}\}\cap B_{n}(x_{0}))\times M$ for a fixed $x_{0}\in M$, it suffices to show that the claim holds for $\pi$ restricted to $C_{n}$. Note by Corollary 2.7 the geodesic connecting the marginals of \frac{1}{\pi(C_{n})}\pi\big{|}_{C_{n}} is still unique.

Thus we can assume $\mu_{0}=f_{0}\mathsf{m}$ has bounded support with density $f_{0}$ bounded below by an $\epsilon>0$ on a set $A$ of full $\mu_{0}$-measure and $\sigma$ is the unique $p$-optimal dynamical coupling between $\mu_{0}$ and $\mu_{1}$. Now for

[TABLE]

Corollary 2.7 shows that $\sigma_{f}$ is still unique between \mu_{0}^{f}=\frac{1}{\mathsf{m}(A)}\mathsf{m}\big{|}_{A} and $\mu_{1}^{f}$. It is easy to see that $(e_{0},e_{1})_{*}\sigma_{f}$ is induced by a transport map if and only if $(e_{0},e_{1})_{*}\sigma$ is induced by a transport map. We conclude by noticing that $A=\operatorname{supp}\mu_{0}^{f}$ and $\mu_{1}^{f}$ satisfy the conditions of the previous lemma and hence $(e_{0},e_{1})_{*}\sigma_{f}$ is induced by a $p$-optimal transport map. ∎

Since by Rajala–Sturm [RS14] strong $\mathsf{CD}_{p}(K,\infty)$-spaces are $p$-essentially non-branching, we recover their result on the existence of transport maps [RS14, Corollary 1.4].

Corollary 5.20**.**

If $(M,d\mathsf{m})$ satisfies the strong $\mathsf{CD}_{p}(K,\infty)$-condition between every $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0},\mu_{1}\ll\mathsf{m}$ there is a $p$-optimal dynamical coupling $\sigma$ and $(e_{0},e_{1})_{*}\sigma$ is induced by a transport map.

If, instead, we know that between an absolutely continuous initial measure and an arbitrary measure every interpolation is absolutely continuousthen we can show general existence of transport maps.

Corollary 5.21**.**

Assume $(M,d\mathsf{m})$ has the strong interpolation property $\mathsf{(sIP)}_{p}$. Then $(M,d,\mathsf{m})$ is $p$-essentially non-branching and satisfies the (weak) $\mathsf{CD}_{p}(K,\infty)$-condition if and only if it satisfies the strong $\mathsf{CD}_{p}(K,\infty)$-condition. Furthermore, if either of the cases hold then $(M,d,\mathsf{m})$ has good transport behavior $\mathsf{(GTB)}_{p}$ as well.

In the more general setting there could be more than one geodesic connecting two absolutely continuous measure. However, it is possible to show that the absolutely continuous part of a geodesic connecting measures with finite entropy is just a restricting of a unique absolutely geodesic given by the interpolation inequality. Hence this geodesic is unique.

Corollary 5.22**.**

Assume $(M,d,\mathsf{m})$ is $p$-essentially non-branching and satisfies the $\mathsf{CD}_{p}(K,\infty)$-condition. Then for all $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0},\mu_{1}\ll\mathsf{m}$ and $\int f_{0}\log f_{0}d\mathsf{m},\int f_{1}\log f_{1}d\mathsf{m}<\infty$ there is a unique $p$-optimal dynamical coupling $\sigma\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ along which the $\mathsf{CD}_{p}(K,\infty)$-interpolation inequality holds and for this dynamical coupling the $p$-optimal coupling $(e_{0},e_{1})_{*}\sigma$ is induced by a transport map.

Furthermore, for any other $p$-optimal dynamical coupling $\tilde{\sigma}\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ such that for some $t\in(0,1)$ the $t$-midpoint $\mu_{t}=\rho_{t}\mathsf{m}+\mathsf{m}$ with $\rho_{t}\not\equiv 0$, there are a function $f:M\to[0,\infty)$ which is positive $\mathsf{m}$-almost everywhere on $\{\rho_{t}>0\}$ and a Borel set $A_{t}\subset M$ such that

[TABLE]

where $\tilde{f}(\gamma)=\chi_{A_{t}}(\gamma_{t})$, i.e. the absolutely continuous part of $\tilde{\sigma}$ is obtained via a restriction of $\sigma$.

Proof.

Let $\sigma$ be a $p$-optimal dynamical coupling along which the $\mathsf{CD}_{p}(K,\infty)$-interpolation inequality holds and $\tilde{\sigma}$ be another $p$-optimal dynamical coupling such that at time $t\in(0,1)$ the interpolation $\tilde{\mu}_{t}=(e_{t})_{*}\tilde{\sigma}$ has density with respect to $\mathsf{m}$, i.e.

[TABLE]

with $\tilde{\mu}_{t}^{s}\bot\mathsf{m}$, $\tilde{\mu}_{t}^{a}\ll\mathsf{m}$ and $\tilde{\mu}_{t}^{a}(M)>0$ . Let $A_{t}$ be a Borel set such that $\tilde{\mu}_{t}^{a}(M\backslash A_{t})=0$ and $\tilde{\mu}_{t}^{s}(A_{t})=0$.

Set $\tilde{f}(\gamma)=\chi_{A_{t}}(\gamma_{t})$, then by the theorem above the $p$-optimal coupling $(e_{0},e_{t_{0}})_{*}\tilde{\sigma}_{\tilde{f}}$ between $\tilde{\mu}_{0}^{\tilde{f}}$ and $\tilde{\mu}_{0}^{\tilde{f}}$ which is induced by a transport map $\tilde{T}$. Similarly, $(e_{0},e_{t})_{*}\sigma$ is induced by a transport map $T$.

We claim that $T=\tilde{T}$ on $A_{0}=\tilde{T}^{-1}(A_{t})$. Indeed, this would imply the result because \tilde{\mu}_{0}^{\tilde{f}}\leq\mu_{0}\big{|}_{A_{0}}.

The claim follows by observing that between $\hat{\mu}_{0}=\frac{1}{2}(\mu_{0}+\tilde{\mu}_{0}^{\tilde{f}})$ and $\hat{\mu}_{0}=\frac{1}{2}(\mu_{t}+\tilde{\mu}_{t}^{a})$ there is a unique $p$-optimal coupling which is induced by a transport map. ∎

Remark*.*

The corollary allows us to do localization so that we can show the following: between any two measure $\mu_{0}=f_{0}\mathsf{m}$ and $\mu_{1}=f_{1}\mathsf{m}$ with finite entropy there is an interpolation $\mu_{t}=f_{t}d\mathsf{m}$ such that

[TABLE]

This can be used to obtain density bounds of $f_{t}$ and thus a weak Poincaré inequality.

6. Measure rigidity

In this section we prove the measure rigidity theorem, i.e. we will show that two qualitatively non-degenerate measures on a $p$-essentially non-branching space are mutually absolutely continuous.

For convenience of the reader we recall the main properties of $p$-essentially non-branching, qualitatively non-degenerate measures $\mathsf{m}$.

Theorem**.**

Let $(M,d,\mathsf{m})$ be $p$-essentially non-branching and $\mathsf{m}$ be qualitatively non-degenerate. Then for every $\mu_{0},\mu_{1}\in\mathcal{P}_{p}(M)$ with $\mu_{0}\ll\mathsf{m}$ there is a unique $p$-optimal dynamical coupling $\sigma\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ and this dynamical coupling satisfies $(e_{t})_{*}\sigma\ll\mathsf{m}$ for $t\in[0,1)$.

Define

[TABLE]

to be the set of $t$-midpoints of geodesics starting at $x$. This set is analytic and hence measurable. Furthermore, for $0<t\leq s\leq 1$ it holds

[TABLE]

Thus the following set

[TABLE]

is also analytic and measurable. As an abbreviation we also write $R_{(0,t]}(x)=R_{t}(x)$. Finally define the set of strict $t$-midpoints by

[TABLE]

Lemma 6.1**.**

Assume $\mathsf{m}$ is non-degenerate. Then $\mathsf{m}(R_{t}^{*}(x))=0$.

Proof.

First note that

[TABLE]

is a non-decreasing function and finite for fixed $R>0$ so that for some set $\Omega\subset[0,1]$ whose complement is at most countable and it holds

[TABLE]

Assume by contradiction $\Omega\neq(0,1]$. Then there is a $t\notin\Omega$ with $\mathsf{m}(R_{t}^{*}(x)\cap B_{R}(x))>0$. It always holds

[TABLE]

Also note that every $t\in(0,1]\backslash\Omega$ there is an $s\in(0,1)$ with $st\in\Omega$. In combination with the non-degenercity of $\mathsf{m}$ this leads to the following contradiction

[TABLE]

∎

Proposition 6.2**.**

Assume $\mathsf{m}$ is a non-degenerate measure. For every measurable $A\subset M$ of finite measure and every $\epsilon>0$ there is a compact subset $K\subset A$ with $\mathsf{m}(A\backslash K)<\epsilon$ and $t\in(0,1)$ such that

[TABLE]

Furthermore, there is a $\mathsf{m}$-measurable map $T:K\to M$ such that

[TABLE]

In particular, there is a dynamical coupling $\sigma$ which is $p$-optimal for all $p\in[1,\infty)$ such that $(e_{0})_{*}\sigma=\delta_{x}$, (e_{t})_{*}\sigma=\frac{1}{\mathsf{m}(K)}\mathsf{m}\big{|}_{K} and $(e_{1})_{*}\sigma=T_{*}((e_{t})_{*}\sigma)$.

Proof.

Just note that because $\mathsf{m}(R_{1}^{*}(x))=0$ and $t\mapsto R_{t}(x)$ is monotone we have

[TABLE]

Now for all $\epsilon>0$ there is a $t\in(0,1)$ close to $1$ and a compact set $K\subset A\cap R_{t}(x)$ such that $\mathsf{m}(A\backslash K)<\epsilon$. Since $K\subset R_{t}(x)$ there is a measurable map $T:K\to M$ such that

[TABLE]

proving the first part of the claim. For the second just note that \mu_{t}=\frac{1}{\mathsf{m}(K)}\mathsf{m}\big{|}_{K} is a $t$-midpoint of $\delta_{x}$ and $T_{*}\mu_{t}$. ∎

Theorem 6.3**.**

Assume $\mathsf{m}_{i}$, $i=1,2$, are both $p$-essentially non-branching and qualitatively non-degenerate measures on $(M,d)$. Then $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ are mutually absolutely continuous.

Proof.

Assume first there are two mutually singular measure $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ satisfying the assumptions. Then we immediately arrive at a contradiction: For measures $\mu_{0}\ll\mathsf{m}_{1}$ and $\mu_{1}\ll\mathsf{m}_{2}$ the interpolations must be absolutely continuous with respect to both $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ which is not possible.

For general measures $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ there is a maximal decomposition

[TABLE]

for non-trivial mutually singular measures $\mathsf{m}$, $\mathsf{m}_{1}^{s}$ and $\mathsf{m}_{2}^{s}$ and a non-negative $L_{\operatorname{loc}}^{1}(\mathsf{m})$-function $f$ which is positive $\mathsf{m}$-almost everywhere.

Assume by contradiction $\mathsf{m}_{1}^{s}\not\equiv 0$ and let $A$ be a bounded set with $\mathsf{m}_{1}^{s}(A)>0$ and $\mathsf{m}_{2}(A)=0$. We claim that $\mathsf{m}_{2}(A_{t,x})=0$ for all $x$ and $t\in(0,1)$. Indeed, if this was not the case then for some $x\in M$ and $t\in(0,1)$ there is a compact $K\subset A$ and $\sigma$ as in the previous proposition such that $\mathsf{m}_{2}(K_{s,x})>0$ and $\mathsf{m}_{1}^{s}(K_{s,x})=0$ for $s\in(0,1)$. In that case it holds (e_{t})_{*}\sigma=\frac{1}{\mathsf{m}_{1}^{s}(K)}\mathsf{m}_{1}^{s}\big{|}_{K}, $(e_{st})_{*}\sigma\bot\mathsf{m}_{1}^{s}$ and $(e_{s,t})_{*}\sigma\ll\mathsf{m}_{2}$.

However, by the strong interpolation property $\mathsf{(sIP)}_{p}$ between the measures $(e_{st})_{*}\sigma$ and $(e_{1})_{*}\sigma$ this would imply that $(e_{t})_{*}\sigma\ll\mathsf{m}_{2}$ which is a contradiction as $\mathsf{m}_{2}$ and $\mathsf{m}_{1}^{s}$ are mutually singular.

This shows that

[TABLE]

Because $A$ is arbitrary, we see that $\mathsf{m}_{1}^{s}$ is qualitatively non-degenerate and $p$-essentially non-branching.

We arrive at a contradiction by observing that $\mathsf{m}_{1}^{s}$ and $\mathsf{m}_{2}$ are mutually singular. ∎

The following technical lemma can be extracted from the work of Cavalletti–Huesmann [CH14]. For convenience of the reader we include its short proof.

Lemma 6.4** (Self-Intersection Lemma).**

Assume $t\mapsto\mu_{t}=f_{t}\mathsf{m}$ is a geodesic in $\mathcal{P}_{p}(M)$ such that \mu_{0}=\frac{1}{\mathsf{m}(K)}\mathsf{m}\big{|}_{K} for some compact set $K$, $\operatorname{supp}\mu_{1}$ having bounded support and it holds $C:=\sup_{t\in[0,\delta]}\|f_{t}\|_{\infty}<\infty$ for $\delta\in(0,1)$ then there is a $t_{0}\in(0,\delta)$ such that for all $t\in[0,t_{0})$ it holds $\mu_{t}(K)>0$. In particular, $\mu_{t}$ and $\mu_{0}$ cannot be mutually singular.

Proof.

Assume this is not the case then there is a sequence $t_{n}\to 0$ such that $\mu_{t_{n}}\bot\mu_{0}$. In particular, there are Borel sets $A_{0}\subset K$ and $A_{n}\subset\operatorname{supp}\mu_{t_{n}}$ with $A_{n}\cap A_{0}=\varnothing$ and $\mu_{t_{n}}(A_{n})=\mu_{0}(A_{0})=1$. Note that this shows $\mathsf{m}(A_{0})=\mathsf{m}(K)$.

Since the support of $\mu_{0}$ and $\mu_{t}$ are bounded for all $\epsilon>0$ there is a $t_{\epsilon}\in(0,\delta)$ such that all $t_{n}\leq t_{\epsilon}$

[TABLE]

Also note that

[TABLE]

But then

[TABLE]

which is a contradiction. ∎

Theorem 6.5**.**

Assume $(M,d,\mathsf{m}_{1})$ and $(M,d,\mathsf{m}_{2})$ have the strong interpolation property $\mathsf{(sIP)}_{p}$ and satisfy the $\mathsf{CD}_{p}(K,\infty)$-condition. Then $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ are mutually absolutely continuous.

Remark*.*

The proof below also works in the setting of $p$-essentially non-branching, qualitatively non-degenerate measures. Indeed, by Proposition 5.15 the density $f_{t}$ of the $t$-interpolation $\mu_{t}$ is uniformly bounded by the density of $f_{0}$ if $t$ is close to $1$, compare this also to [Raj12b, Theorem 4.2] and [CM17, Theorem 4.1].

However, the strong interpolation property $\mathsf{(sIP)}_{p}$, which follows from qualitative non-degenericity, is essential in order to combine the singular part of one of the measures with the bounded density property. It is unclear whether without this property there could be more than one $\mathsf{CD}_{p}(K,\infty)$-measure.

Proof.

As above two such measures $\mathsf{m}_{1}$ and $\mathsf{m}_{2}$ cannot be mutually singular and must be of the form

[TABLE]

for non-trivial mutually singular measures $\mathsf{m}$, $\mathsf{m}_{1}^{s}$ and $\mathsf{m}_{2}^{s}$ and a non-negative $L_{\operatorname{loc}}^{1}(\mathsf{m})$-function $f$ which is positive $\mathsf{m}$-almost everywhere.

Assume by contradiction $\mathsf{m}_{1}^{s}\not\equiv 0$ and let $A$ and $B$ be compact sets with $\mathsf{m}_{1}^{s}(A)>0$, $\mathsf{m}_{2}(A)=0$, $\mathsf{m}_{1}^{s}(B)=0$, $\mathsf{m}_{2}^{s}(B)=0$ and $\mathsf{m}(B)>0$. Let \mu_{0}=\frac{1}{\mathsf{m}_{1}^{s}(A)}\mathsf{m}_{1}^{s}\big{|}_{A} and \mu_{1}=\frac{1}{\mathsf{m}(B)}\mathsf{m}\big{|}_{B}. Then there is a unique geodesic $t\mapsto\mu_{t}$ connecting $\mu_{0}$ and $\mu_{1}$ which is absolutely continuous with respect to $\mathsf{m}_{1}$.

Also note by the strong interpolation property for $\mathsf{m}_{2}$, $\mu_{1}\ll\mathsf{m}_{2}$ implies $\mu_{t}\ll\mathsf{m}_{2}$ for $t\in(0,1)$. In particular, $\mu_{t}(A)=0$ for all $t\in(0,1]$. Now the $\mathsf{CD}_{p}(K,\infty)$-condition implies

[TABLE]

where $\mu_{t}=f_{t}\mathsf{m}$.

We arrive at a contradiction by observing that $\mu_{t}(A)>0$ by the Self-Intersection-Lemma above. ∎

Appendix A Proof of Theorem 2.11 and Corollary 2.12

Before we prove the theorem we need the following technical lemmas.

Lemma A.1**.**

Let $\sigma$ be a $p$-optimal dynamical coupling $\mu_{0}$ and $\mu_{1}$ such that $(e_{t})_{*}\sigma=\delta_{x_{t}}$ for some $t\in(0,1)$ and $x_{t}\in M$ then $\mu_{0}\otimes\mu_{1}$ is a $p$-optimal coupling and $d(\cdot,\cdot)$ is constant on $\operatorname{supp}\mu_{0}\times\operatorname{supp}\mu_{1}$. In particular, if $\mu_{0}\otimes\mu_{1}$ is not a delta measure then there is a $p$-optimal dynamical coupling $\tilde{\sigma}$ such that $\tilde{\sigma}(\mathsf{L})<1$ for all measurable non-branching $\mathsf{L}\subset\mathsf{Geo}_{[0,1]}(M,d)$ of non-branching geodesics.

Proof.

First note that the trivial coupling $\mu_{0}\otimes\mu_{1}$ is a $p$-optimal coupling of $\mu_{0}$ and $\mu_{1}$ for all $p\in[1,\infty)$. Indeed, the assumptions imply that for each $\gamma,\eta\in\operatorname{supp}\sigma$ it holds $\gamma_{t}=\eta_{t}=x_{t}$ and hence $\ell(\gamma)=\ell(\eta)$. But then $\ell(\mathsf{restr}_{0,t}\gamma)=W_{p}(\mu_{0},\delta_{x_{t}})$ and $\ell(\mathsf{restr}_{t,1}\gamma)=W_{p}(\delta_{x_{t}},\mu_{1})$ and thus for all $x_{0}\in\operatorname{supp}\mu_{0}$ and $x_{1}\in\operatorname{supp}\mu_{1}$

[TABLE]

implying that $\operatorname{supp}\mu_{0}\times\operatorname{supp}\mu_{1}$ is $c_{p}$-cyclically monotone.

Let $T_{0,t}:M\to\mathsf{Geo}_{[0,1]}(M,d)$ be a measurable map such that $T_{0,t}(x_{0})$ is a geodesic between $x_{0}$ and $x_{t}$. Then $\sigma_{0,t}:=(T_{0,t})_{*}\mu_{0}$ is a $p$-optimal dynamical coupling between $\mu_{0}$ and $\delta_{x_{t}}$. Similarly, let $T_{t,1}:M\to\mathsf{Geo}_{[0,1]}(M,d)$ be a Borel map such that $T_{t,1}(x_{1})$ is a geodesic between $x_{t}$ and $x_{1}$. Note that for each $x_{0}\in\operatorname{supp}\mu_{0}$ and $x_{1}\in\operatorname{supp}\mu_{1}$

[TABLE]

is a geodesic between $x_{0}$ and $x_{1}$. Since $T_{0,t}$ and $T_{t,1}$ are Borel maps, so is $T_{0,1}^{x_{t}}:(x_{0},x_{1})\mapsto\gamma^{x_{0},x_{1}}$. In particular, $\tilde{\sigma}=(T_{0,1}^{x_{t}})_{*}\mu_{0}\otimes\mu_{1}$ is a $p$-optimal dynamical coupling of $\mu_{0}$ and $\mu_{1}$.

If $\mu_{0}\otimes\mu_{1}$ is not a delta measure then either $\mu_{0}$ or $\mu_{1}$ (or both) is not a delta measure. Assume $\mu_{1}$ is not a delta measure then for each set $\mathsf{L}$ with $\tilde{\sigma}(\Gamma)=1$ and for $\mu_{0}$-almost all $x_{0}\in e_{0}(\Gamma)$ there are at least two distinct geodesics $\gamma,\eta\in\Gamma$ with

[TABLE]

In particular, $\mu_{0}\otimes\mu_{1}$ is not concentrated on a non-branching set. ∎

Remark*.*

Assume $x\mapsto(\mu_{0}^{x},\mu_{1}^{x})$ is a measurable map such that $\delta_{x}$ is the $t$-midpoint of $\mu_{0}^{x}$ and $\mu_{1}^{x}$. Then $x\mapsto(T_{0,1}^{x})_{*}(\mu_{0}^{x}\otimes\mu_{1}^{x})$ is also measurable.

Lemma A.2**.**

Let $\mu_{0}$ and $\mu_{1}$ be probability measures such that any $p$-optimal dynamical coupling between $\mu_{0}$ and $\mu_{1}$ is concentrated on a set of non-branching geodesics. Then for any $t$-midpoint $\mu_{t}$ of $\mu_{0}$ and $\mu_{1}$, any $p$-optimal dynamical coupling between $\mu$ and $\mu_{t}$ is concentrated on a set of non-branching geodesics.

Proof.

It is easy to see that any $p$-optimal dynamical coupling $\sigma\in\mathsf{OptGeo}_{p}(\mu,\mu_{t})$ is obtained by restricting a $p$-optimal dynamical coupling $\tilde{\sigma}\in\mathsf{OptGeo}_{p}(\mu,\nu)$, i.e.

[TABLE]

and hence

[TABLE]

Furthermore, if $\mathsf{L}$ is non-branching and measurable then $\mathsf{restr}_{0,t}(\mathsf{L})$ is also non-branching and measurable. In particular, choosing $\mathsf{L}$ such that $\tilde{\sigma}(L)=1$ implies $\sigma(\mathsf{restr}_{0,t}(\mathsf{L}))=1$. ∎

Proposition A.3**.**

Let $\mu_{0}$ and $\mu_{1}$ be probability measures such that any $p$-optimal dynamical coupling between $\mu_{0}$ and $\mu_{1}$ is concentrated on a set of non-branching geodesics. Then for any $p$-optimal dynamical coupling $\sigma\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ and any $t\in(0,1)$ and $s\in[0,1]$ the $p$-optimal coupling $(e_{t},e_{s})_{*}\sigma$ are induced by a transport map.

Proof.

Let $\mu_{s}=(e_{s})_{*}\sigma$, $\pi_{t,0}=(e_{t},e_{0})_{*}\sigma$ and $\pi_{t,1}=(e_{t},e_{1})_{*}\sigma$. It suffices to show that $\pi_{t,0}$ is induced by a transport map $T$, i.e. $(\operatorname{id}\times T)_{*}\mu_{t}=\pi_{t,0}$. By disintegrating $\pi_{t,0}$ and $\pi_{t,1}$ over $(e_{t},e_{0})$ and resp. $(e_{t},e_{1})$ we get

[TABLE]

Let $\sigma=\int\sigma_{x}d\mu_{t}(x)$ denote the disintegration of $\sigma$ over $e_{t}$ and define a new dynamical coupling

[TABLE]

where $T_{0,1}^{x}$ is defined as in the proof of the previous lemma. Note that $(x_{0},x_{1},x_{t})\mapsto T_{0}^{x_{t}}(x_{0},x_{1})$ is measurable on $(e_{0},e_{1},e_{t})(\mathsf{Geo}_{[0,1]}(M,d))$ and hence $x\mapsto(T_{0,1}^{x})_{*}(\mu_{x}\otimes\nu_{x})$ measurable on $\operatorname{supp}\mu_{t}$.

We claim $\tilde{\sigma}$ is $p$-optimal. Indeed, by the previous lemma it holds

[TABLE]

If $\pi_{0,t}$ is not induced by a transport map then there is a Borel set $B\subset M$ of positive $\mu_{t}$-measure such that for all $x\in B$ the measure $\mu_{x}\otimes\nu_{x}$ is not a delta measure. This, however, implies that for all $x\in B$ the measure $\tilde{\sigma}_{x}=(T_{0,1}^{x})_{*}(\mu_{x}\otimes\nu_{x})$ is not concentrated on a set of non-branching geodesics. The assumption shows that there is a non-branching measurable set $\mathsf{L}\subset\mathsf{Geo}_{[0,1]}(M,d)$ with $\tilde{\sigma}(\mathsf{L})=1$. But this is a contradiction since $\sigma_{x}(\mathsf{L})<1$ for $\mu_{t}$-almost all $x\in B$ implies

[TABLE]

∎

Corollary A.4**.**

Let $\mu_{0}$ and $\mu_{1}$ be probability measures such that any $p$-optimal dynamical coupling between $\mu_{0}$ and $\mu_{1}$ is concentrated on a set of non-branching geodesics. Then for any $p$-optimal dynamical coupling $\sigma\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ and any $t\in(0,1)$ there is a Borel map $\mathsf{T}_{t}:M\to\mathsf{Geo}_{[0,1]}(M,d)$ such that

[TABLE]

where $\mu_{t}=(e_{t})_{*}\sigma$. In particular, whenever $\tilde{\sigma}\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ with $\mu_{t}=(e_{t})_{*}\tilde{\sigma}$ then $\sigma\equiv\tilde{\sigma}$.

Proof.

Let $\sigma=\int\sigma_{x}d\mu_{t}(x)$ be the disintegration of $\sigma$ over $e_{t}$. The proof above shows that $\sigma$ is unique among all dynamical couplings $\tilde{\sigma}\in\mathsf{OptGeo}_{p}(\mu_{0},\mu_{1})$ with $\mu_{t}=(e_{t})_{*}\tilde{\sigma}$. For fixed $s\in[0,1]$ there is a transport map $T_{t,s}$ such that

[TABLE]

In particular, there is a Borel set $\Omega_{s}\subset M$ with $\mu_{t}(\Omega_{s})=1$ and

[TABLE]

for all $x\in\Omega_{s}$. Let $(s_{n})_{n\in\mathbb{N}}$ be dense in $(0,1)$ and note $\mu_{t}(\Omega)=1$ where $\Omega=\cap_{n\in\mathbb{N}}\Omega_{s_{n}}$. Define $\gamma_{s}^{x}=T_{t,s}(x)$and observe that $\gamma^{x}\in\mathsf{Geo}_{[0,1]}(M,d)$ and

[TABLE]

This shows that $\sigma_{x}=\delta_{\gamma^{x}}$ on $\Omega$ and thus

[TABLE]

where $\mathsf{T}_{t}:M\to\mathsf{Geo}_{[0,1]}(M,d)$ is any measurable map with $\mathsf{T}_{t}(x)=\gamma^{x}$ on $\Omega$. ∎

Proof of Theorem 2.11.

By the $p$-essentially non-branching property we see that the second statement follows directly from the previous corollary. Furthermore, for $\Omega$ as in the previous proof we can choose

[TABLE]

Then $\sigma$ is concentrated on $\mathsf{L}$ and whenever $\gamma,\eta\in\mathsf{L}$ with $\gamma_{t}=\eta_{t}$ then $\gamma_{t},\eta_{t}\in\Omega$ so that $\mathsf{T}_{t}(\gamma_{t})\equiv\gamma\equiv\eta$. ∎

Proof of Corollary 2.12.

Assume $t\mapsto\tilde{\mu}_{s}^{0,t}$ and $t\mapsto\tilde{\mu}_{s}^{t,1}$ is a geodesic connecting $\mu_{0}$ and $\mu_{t}$ and resp. $\mu_{t}$ and $\mu_{1}$. Then

[TABLE]

is also a geodesic connecting $\mu_{0}$ and $\mu_{1}$. Denote the induced $p$-optimal dynamical coupling by $\hat{\sigma}$ and note that $\mathring{\sigma}=\frac{1}{2}\left(\sigma+\hat{\sigma}\right)$ is also a $p$-optimal dynamical coupling between $(e_{0})_{*}\sigma$ and $(e_{1})_{*}\sigma$. Thus it holds

[TABLE]

where $\mathsf{T}_{t}$, $\tilde{\mathsf{T}}_{t}$ and $\mathring{\mathsf{T}}_{t}$ are the maps given in the theorem. This, however, shows that the three maps agree $\mu_{t}$-almost everywhere implying $\hat{\sigma}=\sigma$ and thus $\mu_{s}=\hat{\mu}_{s}$for $s\in[0,1]$. In particular, $t\mapsto\mu_{ts}$ and $t\mapsto\mu_{s+(1-t)}$ are the unique geodesic between $\mu_{0}$ and $\mu_{t}$ and resp. $\mu_{t}$ and $\mu_{1}$. ∎