# Example of an Highly Branching CD Space

Mattia Magnabosco

###### Abstract

In [3] Ketterer and Rajala showed an example of metric measure space,
satisfying the measure contraction property $\mathsf{MCP}(0,3)$, that has
different topological dimensions at different regions of the space. In this
article I propose a refinement of that example, which satisfies the
$\mathsf{CD}(0,\infty)$ condition, proving the non-constancy of topological
dimension for CD spaces. This example also shows that the weak curvature
dimension bound, in the sense of Lott-Sturm-Villani, is not sufficient to
deduce any reasonable non-branching condition. Moreover, it allows to answer
to some open question proposed by Schultz in [7], about strict curvature
dimension bounds and their stability with respect to the measured Gromov
Hausdorff convergence.

In their remarkable works Lott, Villani [4] and Sturm [10, 11] introduced a
weak notion of curvature dimension bounds, which strongly relies on the theory
of Optimal Transport. Inspired by some results that hold in the Riemannian
case, they defined a consistent notion of curvature dimension bound for metric
measure spaces, that is known as CD condition. The metric measure spaces
satisfying the CD condition are called CD spaces and enjoy some remarkable
analytic and geometric properties.

In this work I present an example of an highly branching CD space, that shows
how the weak curvature dimension bound is not sufficient to deduce any type of
non-branching condition. This example is a refinement of the one by Ketterer
and Rajala [3], in particular, the topological and metric structure of the
space will be essentially the same, while the reference measure will be more
complicated. The metric measure space considered by Ketterer and Rajala in [3]
satisfies the so called measure contraction property $\mathsf{MCP}(0,3)$, but,
as they pointed out in their work, it does not satisfy any CD condition. As I
will highlight in the last section, the modification I am going to present
allows to extend the some of the results in [3] to CD spaces. Moreover it
gives an answer to some questions proposed by Schultz in [7], regarding the
very strict CD condition and the stability of strict CD condition.

Before going on, I want to emphasize that the CD condition is much more
complicated to prove than the MCP condition. In fact there is much more
freedom in choosing the marginals, and consequently it is much more difficult
to avoid mass overlap in a Wasserstein geodesic. For this reason, not only
will the proof be longer than Ketterer and Rajala’s one, but it will also be
more complicated. In particular, I will prove the CD condition for the desired
metric measure space, by proving it for a sequence of metric measure spaces
that converges to it, and using the stability result by Villani (Theorem
1.15). I will also take great inspiration by the work of Rajala in [5], which
shows an efficient way to prove the CD condition in branching metric measure
spaces, through the application of Jacobi equation.

The main consequences of the example I will present in this article are the
following:

  * •

any reasonable non-branching condition does not hold in general for (weak) CD
spaces,

  * •

the existence of an optimal transport map between two absolutely continuous
marginals is not granted in (weak) CD spaces, without assuming a non-branching
condition,

  * •

the very strict CD condition studied by Schultz ([7],[8],[9]) is strictly
stronger than the weak one,

  * •

the constancy of topological dimension does not hold in general for (weak) CD
spaces,

  * •

the strict CD condition I will define in the last section is not stable with
respect of the measured Gromov Hausdorff convergence,

  * •

the strict CD condition is strictly stronger than the weak one.

Let me now briefly explain the structure of this work. In the first section I
recall some preliminary results regarding both the basis of the Optimal
Transport theory and CD spaces. In the second section I am going to introduce
the metric measure spaces which will be the subject of the rest of the
article. In section 3 I simply state and prove some algebraic lemma, which
will help me doing the subsequent computations. In the fourth section I
present some result by Rajala [5], related to Jacobi equation and how it can
be used to prove entropy convexity. This theory requires the existence of a
suitable midpoint selection map, which will be introduced in Section 5, where
I will also prove some of its properties. The sixth section contains the proof
of the main theorem, that puts together all the results proven so far. The
last section aims to draw all the conclusions listed before.

## 1 Preliminary Notions

This first section aims to collect all the preliminary results this work needs
in order to be self contained. In particular I am going to introduce the
Wasserstein space and the entropy functional on it, being then able to define
the notions of curvature dimension bound and CD space. Moreover I am going to
briefly discuss the relation between curvature dimension bound and non-
branching conditions, that is one of the main motivation for this work.
Finally I will define the measured Gromov Hausdorff convergence of metric
measure spaces, stating in the end the stability of curvature dimension bounds
with respect to this convergence.

### 1.1 The Wasserstein Space

Denote by $\mathscr{P}(X)$ the set of Borel probability measures on a Polish
metric space $(X,\mathsf{d})$. Given two measures $\mu,\nu\in\mathscr{P}(X)$
and a Borel cost function $c:X\times X\to[0,\infty]$, the Optimal Transport
problem asks to find minima and minimizers of the quantity

$\min\int_{X\times Y}c(x,y)\,\mathrm{d}\pi(x,y),$ (1)

where $\pi$ varies among all probability measures in $\mathscr{P}(X\times X)$
with first marginal equal to $\mu$ and second marginal equal to $\nu$. If the
cost function $c$ is lower semicontinuous, the minimum in (1) is attained. The
minimizers of this problem are called optimal transport plans and the set of
all of them will be denoted by $\operatorname{OptPlans}(\mu,\nu)$. An optimal
transport plan $\pi\in\operatorname{OptPlans}(\mu,\nu)$ is said to be induced
by a map if there exists a $\mu$-measurable map $T:X\to X$ so that
$\pi=(\operatorname{id},T)_{\\#}\mu$, such a map $T$ will be called optimal
transport map.

A fundamental approach in facing the Optimal Transport problem is the one of
$c$-duality, which allows to prove some very interesting and useful results.
Below I report only the most basic statement, which is the only result I will
need in this work.

######  Definition 1.1.

A set $\Gamma\subset X\times X$ is said to be $c$-cyclically monotone if

$\sum_{i=1}^{N}c\left(x_{i},y_{\sigma(i)}\right)\geq\sum_{i=1}^{N}c\left(x_{i},y_{i}\right)$

for every $N\geq 1$, every permutation $\sigma$ of $\\{1,\dots,N\\}$ and every
$(x_{i},y_{i})\in\Gamma$ for $i=1,\dots,N$.

######  Proposition 1.2.

Let $X$ be a Polish space and $c:X\times X\to[0,\infty]$ a lower
semicontinuous cost function. Then every optimal transport plan
$\pi\in\operatorname{OptPlans}(\mu,\nu)$ such that $\int
c\,\mathrm{d}\pi<\infty$ is concentrated in a $c$-cyclically monotone set.

From now on I am going to consider the Optimal Transport problem in the
special case in which the cost function is equal to the distance squared, that
is $c(x,y)=\mathsf{d}^{2}(x,y)$. In this context the minimization problem
induces the so called Wasserstein distance on the space $\mathscr{P}_{2}(X)$
of probabilities with finite second order moment, that is

$\mathscr{P}_{2}(X):=\left\\{\mu\in\mathscr{P}(X)\,:\,\int\mathsf{d}^{2}(x,x_{0})\,\mathrm{d}\mu(x)<\infty\text{
for one (and thus all) }x_{0}\in X\right\\}.$

######  Definition 1.3 (Wasserstein distance).

Given two measures $\mu,\nu\in\mathscr{P}_{2}(X)$ define their Wasserstein
distance $W_{2}(\mu,\nu)$ as

$W_{2}^{2}(\mu,\nu):=\min\left\\{\int
d^{2}(x,y)\,\mathrm{d}\pi(x,y)\,:\,\pi\in\Gamma(\mu,\nu)\right\\}.$

It is easy to realize that $W_{2}$ is actually a distance on
$\mathscr{P}_{2}(X)$, moreover $(\mathscr{P}_{2}(X),W_{2})$ is a Polish metric
space.

Let me now deal with the geodesic structure of $(\mathscr{P}_{2}(X),W_{2})$,
which, as the following statement shows, is heavily related to the one of the
base space $(X,\mathsf{d})$. First of all, notice that every measure
${\mbox{\boldmath$\pi$}}\in\mathscr{P}(C([0,1],X))$ induces a curve $[0,1]\ni
t\to\mu_{t}=(e_{t})_{\\#}{\mbox{\boldmath$\pi$}}\in\mathscr{P}(X)$, therefore
in the following I will consider measures in $\mathscr{P}(C([0,1],X))$ in
order to consider curves in the Wasserstein space.

######  Proposition 1.4.

If $(X,\mathsf{d})$ is a geodesic space than $(\mathscr{P}_{2}(X),W_{2})$ is
geodesic as well. In particular, given two measures
$\mu,\nu\in\mathscr{P}_{2}(X)$, the measure
${\mbox{\boldmath$\pi$}}\in\mathscr{P}(C([0,1],X))$ is a constant speed
Wassertein geodesic connecting $\mu$ and $\nu$ if and only if it is
concentrated in $\operatorname{Geo}(X)$ (that is the space of constant speed
geodesics in $(X,\mathsf{d})$) and
$(e_{0},e_{1})_{\\#}{\mbox{\boldmath$\pi$}}\in\operatorname{OptPlans}(\mu,\nu)$.
In this case it is said that $\pi$ is an optimal geodesic plan between $\mu$
and $\nu$ and this will be denoted as
${\mbox{\boldmath$\pi$}}\in\operatorname{OptGeo}(\mu,\nu)$.

### 1.2 Curvature Dimension Bounds

In this subsection I introduce the notions of curvature dimension bound and CD
space, pioneered by Lott and Villani [4] and Sturm [10, 11]. Their definition
relies on the notion of entropy functional. As it will be soon clear, the most
appropriate framework in which deal with the entropy functional, is the one of
metric measure spaces.

######  Definition 1.5.

A metric measure space is a triple $(X,\mathsf{d},\mathfrak{m})$, where
$(X,\mathsf{d})$ is a Polish metric space and $\mathfrak{m}$ is a non-negative
and non-null Borel measure on $X$, finite on bounded sets.

In this work I will only deal with compact metric measure spaces, and in
particular $\mathfrak{m}(X)<\infty$. Let me now properly define the entropy
functional.

######  Definition 1.6.

In a metric measure space $(X,\mathsf{d},\mathfrak{m})$, define the relative
entropy functional with respect to the reference measure $\mathfrak{m}$
$\operatorname{Ent}:\mathscr{P}_{2}(X)\to\mathbb{R}\cup\\{+\infty\\}$ as

$\operatorname{Ent}(\mu):=\begin{cases}\int\rho\log\rho\,\mathrm{d}\mathfrak{m}&\text{if
}\mu\ll\nu\text{ and }\mu=\rho\mathfrak{m}\\\
+\infty&\text{otherwise}\end{cases}.$

In this context of this work the entropy functional $\operatorname{Ent}$ is
lower semicontinuous with respect to the Wasserstein convergence, this is not
always true in the non-compact case, when it might happen that
$\mathfrak{m}(X)=+\infty$.
I can now give the definitions of CD condition and CD space.

######  Definition 1.7.

A metric measure space $(X,\mathsf{d},\mathfrak{m})$ is said to satisfy the
(weak) $\mathsf{CD}(K,\infty)$ condition and to be a (weak)
$\mathsf{CD}(K,\infty)$ space, if for every absolutely continuous measures
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$ there exists a Wasserstein geodesic
with constant speed $(\mu_{t})_{t\in[0,1]}\subset\mathscr{P}_{2}(X)$
connecting them, along which the relative entropy functional is $K$-convex,
that is

$\operatorname{Ent}(\mu_{t})\leq(1-t)\operatorname{Ent}(\mu_{0})+t\operatorname{Ent}(\mu_{1})-t(1-t)\frac{K}{2}W_{2}^{2}(\mu_{0},\mu_{1}),\qquad\text{for
every }t\in[0,1].$ (2)

Moreover $(X,\mathsf{d},\mathfrak{m})$ is said to satisfy a strong
$\mathsf{CD}(K,\infty)$ condition and to be a strong $\mathsf{CD}(K,\infty)$
space if, for every absolutely continuous measures
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$, the relative entropy functional is
$K$-convex along every Wasserstein geodesic with constant speed connecting
them.

Let me also state a very useful proposition, which provides a simple strategy
to prove the (weak) $\mathsf{CD}(K,\infty)$ condition. Its proof can be found
in [10], anyway I present a brief sketch of it, in order to be self contained.

######  Proposition 1.8.

The metric measure space $(X,\mathsf{d},\mathfrak{m})$ is a
$\mathsf{CD}(K,\infty)$ space if for every pair of absolutely continuous
measures $\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$ there exists a midpoint
$\eta\in\mathscr{P}_{2}(X)$ of $\mu_{0}$ and $\mu_{1}$, absolutely continuous
with respect to $\mathfrak{m}$, satisfying

$\operatorname{Ent}(\eta)\leq\frac{1}{2}\operatorname{Ent}(\mu_{0})+\frac{1}{2}\operatorname{Ent}(\mu_{1})-\frac{K}{8}W_{2}^{2}(\mu_{0},\mu_{1}).$
(3)

######  Proof.

Given two absolutely continuous measures
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$, define $\mu_{1/2}$ as a midpoint of
$\mu_{0}$ and $\mu_{1}$ satisfying (3). Similarly define $\mu_{1/4}$ as a
midpoint of $\mu_{0}$ and $\mu_{1/2}$ and $\mu_{3/4}$ as a $\mu_{1/2}$ and
$\mu_{1}$, both satisfying (3). Proceeding in this way, it is possible to
define $\mu_{t}$ for all dyadic times
$t\in\big{\\{}\frac{k}{2^{h}}:h\in\mathbb{N}^{+},k=1,\dots,2^{h}-1\big{\\}}$.
An easy induction argument on $h$ shows that

$\operatorname{Ent}(\mu_{t})\leq(1-t)\operatorname{Ent}(\mu_{0})+t\operatorname{Ent}(\mu_{1})-t(1-t)\frac{K}{2}W_{2}^{2}(\mu_{0},\mu_{1}).$

for every dyadic time
$t\in\\{\frac{k}{2^{h}}:h\in\mathbb{N}^{+},k=1,\dots,2^{h}-1\\}$. Defining the
geodesic $(\mu_{t})_{t\in[0,1]}$ as the continuous extension, the lower
semicontinuity of the entropy ensures that, for every $t\in[0,1]$, the measure
$\mu_{t}$ satisfies the equation (2). ∎

In the last part of this subsection I want to present the relation between
curvature dimension bounds and non-branching conditions. The most important
result in this context was proven by Rajala and Sturm in [6]:

######  Theorem 1.9.

Every strong $\mathsf{CD}(K,\infty)$ metric measure space
$(X,\mathsf{d},\mathfrak{m})$ is essentially non-branching, that is for every
absolutely continuous measures $\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$, every
optimal geodesic plan connecting them is concentrated on a non-branching set
of geodesics.

The work of Rajala and Sturm was then generalized by Schultz [7] to the
context of very strict CD spaces.

######  Definition 1.10.

A metric measure space $(X,\mathsf{d},\mathfrak{m})$ is called a very strict
$\mathsf{CD}(K,\infty)$ space if for every absolutely continuous measures
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$ there exists an optimal geodesic plan
$\eta\in\operatorname{OptGeo}(\mu_{0},\mu_{1})$, so that the entropy
functional $\operatorname{Ent}$ satisfies the K-convexity inequality along
$(\operatorname{restr}_{t_{0}}^{t_{1}})_{\\#}(f\eta)$ for every
$t_{0}<t_{1}\in[0,1]$, and for all bounded Borel functions
$f:\operatorname{Geo}(X)\to\mathbb{R}^{+}$ with $\int f\,\mathrm{d}\eta=1$.

As the reader can easily notice, these spaces are not in general essentially
non-branching, but they satisfy a weaker condition that I will call weak
essentially non-branching.

######  Definition 1.11 (Weak Essentially Non-Branching).

A metric measure space $(X,\mathsf{d},\mathfrak{m})$ is said to be weakly
essentially non-branching if for every absolutely continuous measures
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$, there exists an optimal geodesic plan
connecting them, that is concentrated on a non-branching set of geodesics.

######  Theorem 1.12 (Schultz [7]).

Every very strict $\mathsf{CD}(K,\infty)$ space is weakly essentially non-
branching.

Notice that the very strict CD condition is intermediate between the weak and
the strong one. It is easy to find examples of very strict CD spaces, which
are not strong CD spaces, while it is not obvious if the very strict condition
is strictly stronger than the weak one. In this work I am going to present an
example of an highly branching (weak) CD space which is not very strict CD.

### 1.3 Measured Gromov Hausdorff Convergence and Stability of CD Spaces

In this subsection I introduce (following [12]) a notion of convergence for
metric measure spaces, that is called measured Gromov Hausdorff convergence.
Roughly speaking, it is the combination of Hausdorff topology for the metric
side, and weak topology for the measure side. In order to properly define the
measured Gromov Hausdorff convergence I have to preliminary introduce the
notion of $\varepsilon$-isometry.

######  Definition 1.13.

A measurable map
$f:(X,\mathsf{d},\mathfrak{m})\to(X^{\prime},\mathsf{d}^{\prime},\mathfrak{m}^{\prime})$
between two metric measure spaces is called an $\varepsilon$-isometry if

  1. 1.

it almost preserves the distances, that is:

$\left|\mathsf{d}\left(f(x),f\left(x^{\prime}\right)\right)-\mathsf{d}\left(x,x^{\prime}\right)\right|\leq\varepsilon\quad\text{for
every }x,x^{\prime}\in X,$

  2. 2.

it is almost surjective, that is:

$\forall y\in X^{\prime},\,\,\text{there exists }x\in X\text{ such that
}\,\mathsf{d}(f(x),y)\leq\varepsilon.$

######  Definition 1.14.

Let $(X_{k},\mathsf{d}_{k},\mathfrak{m}_{k})_{k\in\mathbb{N}}$ and
$(X,\mathsf{d},\mathfrak{m})$ be compact metric measure spaces. It is said
that the sequence $(X_{k},\mathsf{d}_{k},\mathfrak{m}_{k})_{k\in\mathbb{N}}$
converges to $(X,\mathsf{d},\mathfrak{m})$, in the measured Gromov Hausdorff
sense, if for every $k$ there exists a measurable $\varepsilon_{k}$-isometry
$f_{k}:X_{k}\to X$, where $\varepsilon_{k}\to 0$, such that

$(f_{k})_{\\#}\mathfrak{m}_{k}\rightharpoonup\mathfrak{m}\quad\text{as
}k\to\infty.$ (4)

The measured Gromov Hausdorff convergence can be in some sense metrized by the
$\mathbb{D}$ distance, introduced by Sturm in [11]. Moreover in [2] Gigli,
Mondino and Savaré showed that some different notion of convergence for
(pointed) metric measure spaces are equivalent to the (pointed) measured
Gromov Hausdorff convergence.
I end this subsection, stating the stability of the (weak) CD condition with
respect to the measured Gromov Hausdorff convergence.

######  Theorem 1.15.

Let $(X_{k},\mathsf{d}_{k},\mathfrak{m}_{k})_{k\in\mathbb{N}}$ be a sequence
of compact metric measure spaces converging in the measured Gromov Hausdorff
sense to a compact metric measure space $(X,\mathsf{d},\mathfrak{m})$. Given
$K\in\mathbb{R}$, if each
$(X_{k},\mathsf{d}_{k},\mathfrak{m}_{k})_{k\in\mathbb{N}}$ satisfies the weak
curvature dimension condition $\mathsf{CD}(K,\infty)$, then also
$(X,\mathsf{d},\mathfrak{m})$ satisfies $\mathsf{CD}(K,\infty)$.

## 2 Definition of the Metric Measure Spaces

$\varepsilon$$f_{k,\varepsilon}(x)$ Figure 1: The metric measure space
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$ with
$\varepsilon>0$. $f_{k,0}(x)$ Figure 2: The metric measure space
$(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$.

In this section I am going to introduce the metric measure spaces that will be
studied in the rest of this work. The definitions that follow are actually
involved and quite complicated, thus I invite the reader to look at Figure 1
and Figure 2, in order to better understand.

######  Definition 2.1.

Fix $\varepsilon,k\in\mathbb{R}$ such that $0\leq\varepsilon<k<\frac{1}{2}$
and let $\varphi:\mathbb{R}\to[0,1]$ be a continuous function such that
$\int\varphi=k$, $\varphi\leq 1$ on the whole $\mathbb{R}$ and $\varphi=0$
outside $[-k,k]$. Define the function
$f_{k,\varepsilon}:[-1,1]\to\mathbb{R}^{+}$ prescribing

$f_{k,\varepsilon}^{\prime\prime}(x)=\varphi(x),\qquad
f_{k,\varepsilon}^{\prime}(-1)=0,\qquad f_{k,\varepsilon}(-1)=\varepsilon.$

Consequently define the set

$X_{k,\varepsilon}=\\{(x,y)\in\mathbb{R}^{2}\,:\,x\in[-1,1]\text{ and }0\leq
y\leq f_{k,\varepsilon}(x)\\}.$

In the following I will use this notation:

$L=X_{k,0}\cap\big{(}\\{f_{k,0}=0\\}\times\mathbb{R}\big{)}\quad\text{and}\quad
C=X_{k,0}\cap\big{(}\\{f_{k,0}\neq 0\\}\times\mathbb{R}\big{)},$

for sake of simplicity, I will not explicit the $k$ dependence, because it
will be clear from the context.

######  Definition 2.2.

Given $\varepsilon,k\in\mathbb{R}$ with $0<\varepsilon<k<\frac{1}{2}$ and
$K\geq 1$, define the measure $\mathfrak{m}_{k,K,\varepsilon}$ on
$X_{k,\varepsilon}$ as

$\mathfrak{m}_{k,K,\varepsilon}=m_{k,K,\varepsilon}(x,y)\cdot\mathscr{L}^{2}|_{X_{k,\varepsilon}}:=\frac{1}{f_{k,\varepsilon}(x)}\exp\left(-K\left(\frac{y}{f_{k,\varepsilon}(x)}\right)^{2}\right)\mathscr{L}^{2}|_{X_{k,\varepsilon}}.$

While for every $0<k<\frac{1}{2}$ and $K\geq 1$, define the measure
$\mathfrak{m}_{k,K,0}$ on $X_{k,0}$ as

$\mathfrak{m}_{k,K,0}:={\left.\kern-1.2ptC_{K}\cdot\chi_{\\{f_{k,0}(x)=0\\}}\cdot\mathcal{H}^{1}\right|_{\\{y=0\\}}}+\chi_{\\{f_{k,0}(x)>0\\}}\frac{1}{f_{k,0}(x)}\exp\left(-K\left(\frac{y}{f_{k,0}(x)}\right)^{2}\right)\cdot\mathscr{L}^{2}|_{X_{k,0}},$

where

$C_{K}=\int_{0}^{1}e^{-Ky^{2}}\,\mathrm{d}y.$

Notice that a simple change of variable shows that

$(\mathtt{p}_{x})_{\\#}\mathfrak{m}_{k,K,0}=(\mathtt{p}_{x})_{\\#}\mathfrak{m}_{k,K,\varepsilon}=C_{K}\cdot\chi_{\\{-1\leq
x\leq 1\\}}\cdot\mathcal{H}^{1},$

for every suitable $k$, $K$ and $\varepsilon$. Moreover, since I have imposed
$\varepsilon<k$, it is easy to realize that $f_{k,\varepsilon}(x)<3k$ for
every $x\in[-1,1]$ (see Figure 1).

In the following I am going to prove that for suitable $k$ and $K$ the metric
measure space
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$ is a
$\mathsf{CD}(0,\infty)$ space for every $\varepsilon<k$. In particular, in the
next four sections I will consider the metric measure space
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$, but
I will avoid to indicate the parameters $k$, $K$ and $\varepsilon$ at the
subscript, in order to ease the notation. Later, in the last section, I am
going to prove the measured Gromov Hausdorff convergence of the spaces
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$ to
the space $(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$, as
$\varepsilon$ goes to $0$. Combining this with the stability result (Theorem
1.15), will follow that $(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$
is itself a $\mathsf{CD}(0,\infty)$ space.

## 3 Preliminary Algebraic Lemmas

In this section I simply state and prove four algebraic lemmas that will be
fundamental in the proof of the main theorem. In particular they will only
help in carrying on the computation, but they do not hide any particular or
sophisticated idea. For this reason I do not spend a lot of words on them and
I immediately go through the proofs.

######  Lemma 3.1.

There exists a constant $C$ such that for every $A\in\mathbb{R}^{+}$ and every
$\delta$ with $|\delta|<\frac{1}{32}$ it holds

$\frac{1}{2}\log(A)\leq\log\bigg{(}1+\bigg{(}\frac{1}{2}+\delta\bigg{)}(A-1)\bigg{)}+C\delta^{2}$
(5)

######  Proof.

Notice that for some $A\in\mathbb{R}^{+}$ the inequality (5) holds also
without the term $+C\delta^{2}$. In particular this is true if and only if

$\sqrt{A}\leq 1+\bigg{(}\frac{1}{2}+\delta\bigg{)}(A-1).$ (6)

Through elementary computation it is easy to show that inequality (6) holds
for every suitable $\delta$ if

$A\notin\bigg{[}1-\frac{8|\delta|}{(2\delta+1)^{2}},1+\frac{8|\delta|}{(2\delta+1)^{2}}\bigg{]}\subset[1-16|\delta|,1+16|\delta|].$

Therefore, in order to conclude the proof, it is sufficient to prove
inequality (5) for $A\in[1-16|\delta|,1+16|\delta|]$. In this case

$\displaystyle\log\bigg{(}1+\bigg{(}\frac{1}{2}+\delta\bigg{)}(A-1)\bigg{)}$
$\displaystyle=\log\bigg{(}1+\frac{1}{2}(A-1)+\delta(A-1)\bigg{)}$
$\displaystyle=\log\bigg{(}1+\frac{1}{2}(A-1)\bigg{)}+\int_{0}^{\delta(A-1)}\frac{1}{1+\frac{1}{2}(A-1)+t}\,\mathrm{d}t$
$\displaystyle\geq\log\bigg{(}1+\frac{1}{2}(A-1)\bigg{)}-\int_{0}^{|\delta(A-1)|}\frac{1}{1-8|\delta|-16\delta^{2}}\,\mathrm{d}t$
$\displaystyle\geq\log\bigg{(}1+\frac{1}{2}(A-1)\bigg{)}-2|\delta(A-1)|$
$\displaystyle\geq\frac{1}{2}\log(A)-32\delta^{2},$

where the last inequality follows from the concavity of the logarithm. ∎

######  Lemma 3.2.

Given $(x_{0},y_{0}),(x_{1},y_{1})\in X$, let $\gamma:[0,1]\to X$ be the
function:

$t\mapsto\bigg{(}(1-t)\frac{y_{0}}{f(x_{0})}+t\frac{y_{1}}{f(x_{1})}\bigg{)}f((1-t)x_{0}+tx_{1})$

then

  1. 1.

for every $t\in[0,1]$ it holds

$\left|\frac{\gamma^{\prime}(t)}{x_{1}-x_{0}}-\frac{y_{1}-y_{0}}{x_{1}-x_{0}}\right|\leq
3k,$

  2. 2.

if $k<\frac{1}{4}$, for every $t\in[0,1]$ it holds

$\left|\gamma^{\prime\prime}(t)\right|\leq(x_{1}-x_{0})^{2}\frac{2k}{f((1-t)x_{0}+tx_{1})}\left(\left|\frac{y_{1}-y_{0}}{x_{1}-x_{0}}\right|+2\right).$

######  Proof.

The first derivative of $\gamma$ is

$\gamma^{\prime}(t)=\bigg{(}\frac{y_{1}}{f(x_{1})}-\frac{y_{0}}{f(x_{0})}\bigg{)}f((1-t)x_{0}+tx_{1})+\bigg{(}(1-t)\frac{y_{0}}{f(x_{0})}+t\frac{y_{1}}{f(x_{1})}\bigg{)}f^{\prime}((1-t)x_{0}+tx_{1})(x_{1}-x_{0}).$

Therefore, denoting $x_{t}=(1-t)x_{0}+tx_{1}$, it holds

$\displaystyle\frac{\gamma^{\prime}(t)}{x_{1}-x_{0}}$
$\displaystyle=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}+\bigg{(}\frac{1}{f(x_{1})}-\frac{1}{f(x_{t})}\bigg{)}\frac{y_{1}}{x_{1}-x_{0}}f(x_{t})+\bigg{(}\frac{1}{f(x_{t})}-\frac{1}{f(x_{0})}\bigg{)}\frac{y_{0}}{x_{1}-x_{0}}f(x_{t})$
$\displaystyle\quad+\bigg{(}(1-t)\frac{y_{0}}{f(x_{0})}+t\frac{y_{1}}{f(x_{1})}\bigg{)}f^{\prime}(x_{t})$
$\displaystyle=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}+\frac{y_{1}}{f(x_{1})}\frac{f(x_{t})-f(x_{1})}{x_{1}-x_{0}}+\frac{y_{0}}{f(x_{0})}\frac{f(x_{0})-f(x_{t})}{x_{1}-x_{0}}$
$\displaystyle\quad+\bigg{(}(1-t)\frac{y_{0}}{f(x_{0})}+t\frac{y_{1}}{f(x_{1})}\bigg{)}f^{\prime}(x_{t}).$

On the other hand, it is possible to perform the following estimate

$\displaystyle\left|\frac{y_{1}}{f(x_{1})}\frac{f(x_{t})-f(x_{1})}{x_{1}-x_{0}}\right|$
$\displaystyle\leq\left|\frac{y_{1}}{f(x_{1})}\right|\left|\frac{f(x_{t})-f(x_{1})}{x_{1}-x_{0}}\right|\leq\left|\frac{f(x_{t})-f(x_{1})}{x_{1}-x_{0}}\right|\leq\sup
f^{\prime}\leq k,$

and the same calculation can be done for the symmetric term, thus

$\left|\frac{y_{0}}{f(x_{0})}\frac{f(x_{0})-f(x_{t})}{x_{1}-x_{0}}\right|\leq
k.$

Moreover, a similar procedure shows that

$\left|\bigg{(}(1-t)\frac{y_{0}}{f(x_{0})}+t\frac{y_{1}}{f(x_{1})}\bigg{)}f^{\prime}(x_{t})\right|\leq
k.$

Putting together all this estimates, it is possible to conclude that

$\left|\frac{\gamma^{\prime}(t)}{x_{1}-x_{0}}-\frac{y_{1}-y_{0}}{x_{1}-x_{0}}\right|\leq
3k.$

The second derivative of $\gamma$ is

$\displaystyle\gamma^{\prime\prime}(t)=2\bigg{(}\frac{y_{1}}{f(x_{1})}-\frac{y_{0}}{f(x_{0})}\bigg{)}f^{\prime}(x_{t})(x_{1}-x_{0})+\bigg{(}(1-t)\frac{y_{0}}{f(x_{0})}+t\frac{y_{1}}{f(x_{1})}\bigg{)}f^{\prime\prime}(x_{t})(x_{1}-x_{0})^{2}$

Consider the first term of the right hand side, through algebraic
manipulations similar to the ones performed in the first part of the proof, I
obtain

$\displaystyle 2\bigg{(}\frac{y_{1}}{f(x_{1})}-$
$\displaystyle\frac{y_{0}}{f(x_{0})}\bigg{)}f^{\prime}(x_{t})(x_{1}-x_{0})$
$\displaystyle=2\frac{f^{\prime}(x_{t})(x_{1}-x_{0})^{2}}{f(x_{t})}\bigg{[}\frac{y_{1}-y_{0}}{x_{1}-x_{0}}+\frac{y_{1}}{f(x_{1})}\frac{f(x_{t})-f(x_{1})}{x_{1}-x_{0}}+\frac{y_{0}}{f(x_{0})}\frac{f(x_{0})-f(x_{t})}{x_{1}-x_{0}}\bigg{]}.$

Using the same estimates as before, I get

$\displaystyle\left|2\bigg{(}\frac{y_{1}}{f(x_{1})}-\frac{y_{0}}{f(x_{0})}\bigg{)}f^{\prime}(x_{t})(x_{1}-x_{0})\right|$
$\displaystyle\leq
2(x_{1}-x_{0})^{2}\left|\frac{f^{\prime}(x_{t})}{f(x_{t})}\right|\bigg{(}\left|\frac{y_{1}-y_{0}}{x_{1}-x_{0}}\right|+2k\bigg{)}$
$\displaystyle\leq\frac{2k(x_{1}-x_{0})^{2}}{f(x_{t})}\bigg{(}\left|\frac{y_{1}-y_{0}}{x_{1}-x_{0}}\right|+2k\bigg{)}.$

On the other hand, also the second term of the right hand side can be easily
estimate:

$\left|\bigg{(}(1-t)\frac{y_{0}}{f(x_{0})}+t\frac{y_{1}}{f(x_{1})}\bigg{)}f^{\prime\prime}(x_{t})(x_{1}-x_{0})^{2}\right|\leq(x_{1}-x_{0})^{2}.$

Adding this last two inequalities, and using that $f(x_{t})\leq 3k$ I can
conclude

$\left|\gamma^{\prime\prime}(t)\right|\leq(x_{1}-x_{0})^{2}\bigg{[}1+\frac{2k}{f(x_{t})}\bigg{(}\left|\frac{y_{1}-y_{0}}{x_{1}-x_{0}}\right|+2k\bigg{)}\bigg{]}\leq(x_{1}-x_{0})^{2}\frac{2k}{f(x_{t})}\left(\left|\frac{y_{1}-y_{0}}{x_{1}-x_{0}}\right|+2\right),$

where the last inequality holds if $k<\frac{1}{4}$. ∎

######  Lemma 3.3.

Given a fixed a constant $H$, let $y:I=[x_{0},x_{1}]\to\mathbb{R}^{+}$ be a
$C^{2}$ function such that $y^{\prime}(x)\geq\frac{1}{4}$ and
$y^{\prime\prime}(x)\leq H\frac{k}{f(x)}$ for every $x\in I$. Then, for $k$
small enough, it holds

$\log\left(m\left(\frac{x_{0}+x_{1}}{2},y\left(\frac{x_{0}+x_{1}}{2}\right)\right)\right)\geq\log(m(x_{0},y(x_{0})))+\log(m(x_{1},y(x_{1})))+\frac{K}{128f(x_{1})^{2}}(x_{1}-x_{0})^{2}$

######  Proof.

Before going into the proof, I want to point out that the inequality I have to
prove is basically a $K$-convexity inequality. The strategy of the proof
consists in deducing the $K$-convexity from a second derivative estimate. So,
start take the first derivative:

$\displaystyle\frac{\partial}{\partial
x}\left(\log(f(x))+K\left(\frac{y(x)}{f(x)}\right)^{2}\right)=\frac{f^{\prime}(x)}{f(x)}+2K\frac{y(x)}{f(x)}\left(\frac{y^{\prime}(x)}{f(x)}-\frac{y(x)f^{\prime}(x)}{f(x)^{2}}\right)$

Taking another derivative, I obtain

$\displaystyle\frac{\partial^{2}}{\partial
x^{2}}\left(\log(f(x))+K\left(\frac{y(x)}{f(x)}\right)^{2}\right)$
$\displaystyle=\frac{f^{\prime\prime}(x)}{f(x)}-\frac{f^{\prime}(x)^{2}}{f(x)^{2}}+2K\left(\frac{y^{\prime}(x)}{f(x)}-\frac{y(x)f^{\prime}(x)}{f(x)^{2}}\right)^{2}$
$\displaystyle\,+2K\frac{y(x)}{f(x)}\left(\frac{y^{\prime\prime}(x)}{f(x)}-2\frac{y^{\prime}(x)f^{\prime}(x)}{f(x)^{2}}-\frac{y(x)f^{\prime\prime}(x)}{f(x)^{2}}+2\frac{y(x)f^{\prime}(x)^{2}}{f(x)^{3}}\right).$

Therefore, neglecting some positive terms, the following estimate holds

$\displaystyle\frac{\partial^{2}}{\partial
x^{2}}\left(\log(f(x))+K\left(\frac{y(x)}{f(x)}\right)^{2}\right)$
$\displaystyle\geq
2K\frac{y^{\prime}(x)^{2}}{f(x)^{2}}-\frac{f^{\prime}(x)^{2}}{f(x)^{2}}-8K\left|\frac{y(x)y^{\prime}(x)f^{\prime}(x)}{f(x)^{3}}\right|-2K\left|\frac{y(x)y^{\prime\prime}(x)}{f(x)^{2}}\right|$
$\displaystyle-2K\left|\frac{y(x)^{2}f^{\prime\prime}(x)}{f(x)^{3}}\right|.$

Noticing that $\left|\frac{y(x)}{f(x)}\right|\leq 1$,
$\left|f^{\prime}(x)\right|\leq k$ and $\left|f(x)\right|\leq 3k$, I deduce

$\displaystyle\frac{\partial^{2}}{\partial x^{2}}$
$\displaystyle\left(\log(f(x))+K\left(\frac{y(x)}{f(x)}\right)^{2}\right)$
$\displaystyle\qquad\qquad\qquad\qquad\geq\frac{2K}{f(x)^{2}}\bigg{(}y^{\prime}(x)^{2}-\frac{k^{2}}{2K}-4k\left|y^{\prime}(x)\right|-\left|f(x)\right|\left|y^{\prime\prime}(x)\right|-\left|f(x)\right|\left|f^{\prime\prime}(x)\right|\bigg{)}$
$\displaystyle\qquad\qquad\qquad\qquad\geq\frac{2K}{f(x)^{2}}\bigg{(}y^{\prime}(x)^{2}-k^{2}-4k\left|y^{\prime}(x)\right|-kH-3k\bigg{)}$
$\displaystyle\qquad\qquad\qquad\qquad\geq\frac{K}{16f(x)^{2}},$

where the last inequality holds for every suitably small $k$. The thesis
follows by making the uniform convexity explicit and noticing that
$f(x_{1})\geq f(x)$ for every $x\in I$. ∎

Performing the same computations of the previous proof, it possible to prove
the following corollary.

######  Corollary 3.4.

Given a fixed a constant $H$, let $y:I=[x_{0},x_{1}]\to\mathbb{R}^{+}$ be a
$C^{2}$ function such that $y^{\prime}(x)\geq\frac{1}{4}$ and
$y^{\prime\prime}(x)\leq H\frac{k}{f(x)}$ for every $x\in I$. Then, for $k$
small enough, it holds

$K\left(\frac{y\left(\frac{x_{0}+x_{1}}{2}\right)}{f\left(\frac{x_{0}+x_{1}}{2}\right)}\right)^{2}\leq\frac{K}{2}\left(\frac{y(x_{0})}{f(x_{0})}\right)^{2}+\frac{K}{2}\left(\frac{y(x_{1})}{f(x_{1})}\right)^{2}-\frac{K}{128f(x_{1})^{2}}(x_{1}-x_{0})^{2}.$

######  Lemma 3.5.

Let $x\in[-1,1]$ and $\delta>0$ such that $x-\delta,x+\delta\in[-1,1]$, then

$\left|\frac{f(x)}{f(x-\delta)}+\frac{f(x)}{f(x+\delta)}-2\right|\leq\frac{[2k^{2}+f(x)]\delta^{2}}{f(x-\delta)f(x+\delta)}.$

######  Proof.

Denote $I_{1}=\int_{x}^{x-\delta}f^{\prime}(t)\,\mathrm{d}t$ and
$I_{2}=\int_{x}^{x+\delta}f^{\prime}(t)\,\mathrm{d}t$, then

$\displaystyle\left|\frac{f(x)}{f(x-\delta)}+\frac{f(x)}{f(x+\delta)}-2\right|$
$\displaystyle=\left|\frac{f(x)}{f(x)+I_{1}}+\frac{f(x)}{f(x)+I_{2}}-2\right|$
$\displaystyle=\left|\frac{-2I_{1}I_{2}-f(x)(I_{1}+I_{2})}{(f(x)+I_{1})(f(x)+I_{2})}\right|$
$\displaystyle\leq\left|\frac{2I_{1}I_{2}}{(f(x)+I_{1})(f(x)+I_{2})}\right|+\left|\frac{f(x)(I_{1}+I_{2})}{(f(x)+I_{1})(f(x)+I_{2})}\right|.$

But the following estimates hold

$\left|I_{1}+I_{2}\right|=\left|\int_{0}^{\delta}f^{\prime}(x+t)-f^{\prime}(x-\delta+t)\,\mathrm{d}t\right|\leq\int_{0}^{\delta}\left|f^{\prime}(x+t)-f^{\prime}(x-\delta+t)\right|\,\mathrm{d}t\leq\delta^{2}\sup\left|f^{\prime\prime}\right|\leq\delta^{2}$

and

$\left|I_{1}\right|,\left|I_{2}\right|\leq k\delta.$

Using this last two estimates I conclude

$\left|\frac{f(x)}{f(x-\delta)}+\frac{f(x)}{f(x+\delta)}-2\right|\leq\frac{[2k^{2}+f(x)]\delta^{2}}{(f(x)+I_{1})(f(x)+I_{2})}$

∎

## 4 How to Prove Convexity of the Entropy

In this section I prove an important result (Proposition 4.2) that will be
fundamental in the following, in order to prove the CD condition. This results
relies on the possibility to compute the density of a pushforward measure,
through Jacobi equation. For example, consider two measures
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}(\mathbb{R}^{2})$ which are absolutely
continuous with respect to the Lebesgue measure $\mathscr{L}^{2}$, with
density $\rho_{0}$ and $\rho_{1}$ respectively. Suppose there exists a smooth
one-to-one map $T:\mathbb{R}^{2}\to\mathbb{R}^{2}$ such that
$T_{\\#}\mu_{0}=\mu_{1}$, then it is well known that

$\rho_{1}(T(x,y))J_{T}(x,y)=\rho_{0}(x,y),$ (7)

for $\mu_{0}$-almost every $(x,y)$. As shown in [1] the assumptions on the map
$T$ can be relaxed and it is sufficient to require $T$ to be approximately
differentiable and injective outside a $\mu_{0}$-null set. However, in this
work I am going to deal with transport map which are not necessarily
approximately differentiable, but have another rigidity property. Therefore,
the version of Jacobi equation I will need is the following, which is an easy
consequence of Proposition 2.1 in [5].

######  Proposition 4.1.

Let $\mu_{0},\mu_{1}\in\mathscr{P}_{2}(\mathbb{R}^{2})$ be absolutely
continuous with respect to the Lebesgue measure $\mathscr{L}^{2}$. Assume that
there exists a map $T=(T_{1},T_{2})$ which is injective outside a
$\mu_{0}$-null set, such that $T_{\\#}\mu_{0}=\mu_{1}$. Suppose also that
$T_{1}$ locally does not depend on the $y$ coordinate and it is increasing in
$x$, while $T(x,y)$ is increasing in $y$ for every fixed $x$. Then the Jacobi
equation (7) is satisfied with $J_{T}=\frac{\partial T_{1}}{\partial
x}\frac{\partial T_{2}}{\partial y}$.

Now that I have shown that Jacobi equation can be adapted to the context of
this work, I am going to explain how it can be useful in proving convexity of
the entropy functional. The following proposition will be an important element
in the main proof of this article.

######  Proposition 4.2.

Let $\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$ be absolutely continuous measure
and let $T:X\to X$ be an optimal transport map between $\mu_{0}$ and
$\mu_{1}$, in particular $T_{\\#}\mu_{0}=\mu_{1}$. Consider a midpoint
$\mu_{1/2}$ of $\mu_{0}$ and $\mu_{1}$, assume that
$\mu_{1/2}=[M\circ(\operatorname{id},T)]_{\\#}\mu_{0}$ where the map
$M:X\times X\to X$ is a midpoint selection. Suppose also that the maps $T$ and
$M\circ(\operatorname{id},T):X\to X$ are injective outside a set of measure
zero and they satisfy the Jacobian equation, with suitable Jacobian functions
$J_{T}$ and $J_{M\circ(\operatorname{id},T)}$. If

$\log\left(m\big{(}M((x,y),T(x,y))\big{)}J_{M\circ(\operatorname{id},T)}(x,y)\right)\geq\frac{1}{2}\log\left(m(T(x,y))J_{T}(x,y)\right)+\frac{1}{2}\log(m(x,y))$

for $\mu_{0}$ almost every $(x,y)$, then

$\operatorname{Ent}(\mu_{1/2})\leq\frac{1}{2}\operatorname{Ent}(\mu_{0})+\frac{1}{2}\operatorname{Ent}(\mu_{1}).$
(8)

######  Proof.

Suppose $\mu_{0}=\rho_{0}\mathfrak{m}=\tilde{\rho}_{0}\mathscr{L}^{2}$,
$\mu_{1}=\rho_{1}\mathfrak{m}=\tilde{\rho}_{1}\mathscr{L}^{2}$ and
$\mu_{1/2}=\rho\mathfrak{m}=\tilde{\rho}_{1}\mathscr{L}^{2}$. It easy to
realize that, in order to prove (8), it is sufficient to prove that

$\log\big{(}\rho_{1/2}\big{(}M((x,y),T(x,y))\big{)}\big{)}\leq\frac{1}{2}\log\big{(}\rho_{1}(T(x,y))\big{)}+\frac{1}{2}\log(\rho_{0}(x,y)),$
(9)

for $\mu_{0}$-almost every $(x,y)$. On the other hand, the validity of Jacobi
equation ensures that

$\tilde{\rho}_{1}(T(x,y))J_{T}(x,y)=\tilde{\rho}_{0}(x,y),$

for $\mu_{0}$-almost every $(x,y)$, and thus that

$\rho_{1}(T(x,y))m(T(x,y))J_{T}(x,y)=\rho_{0}(x,y)m(x,y).$

for $\mu_{0}$-almost every $(x,y)$. Analogously, it is possible to infer that

$\rho_{1/2}\big{(}M((x,y),T(x,y))\big{)}m\big{(}M((x,y),T(x,y))\big{)}J_{M\circ(\operatorname{id},T)}(x,y)=\rho_{0}(x,y)m(x,y).$

for $\mu_{0}$-almost every $(x,y)$. Therefore (9) is equivalent to

$\log\left(\frac{\rho_{0}(x,y)m(x,y)}{m\big{(}M((x,y),T(x,y))\big{)}J_{M\circ(\operatorname{id},T)}(x,y)}\right)\leq\frac{1}{2}\log\left(\frac{\rho_{0}(x,y)m(x,y)}{m(T(x,y))J_{T}(x,y)}\right)+\frac{1}{2}\log(\rho_{0}(x,y)).$

Some easy algebraic computations show that this last equation is equivalent to

$\log\left(m\big{(}M((x,y),T(x,y))\big{)}J_{M\circ(\operatorname{id},T)}(x,y)\right)\geq\frac{1}{2}\log\left(m(T(x,y))J_{T}(x,y)\right)+\frac{1}{2}\log(m(x,y)),$

concluding the proof. ∎

Notice that the result of this last proposition gains importance if seen in
relation with Proposition 1.8. In fact, Proposition 4.2 provides a strategy to
prove entropy convexity in a suitable midpoint, which is sufficient to deduce
CD condition, according to Proposition 1.8.

I conclude the section with a simple corollary of Proposition 4.2, which will
be useful in the last section of this work. This result is a straightforward
consequence of the previous proof, and takes full advantage of the fact that
Jacobi equation allows to prove entropy convexity pointwise as well as
globally.

######  Corollary 4.3.

Under the same assumptions of Proposition 4.2, it holds that

$\operatorname{Ent}\big{(}[M\circ(\operatorname{id},T)]_{\\#}(f\mu_{0})\big{)}\leq\frac{1}{2}\operatorname{Ent}(f\mu_{0})+\frac{1}{2}\operatorname{Ent}(T_{\\#}(f\mu_{0})),$

for every bounded measurable function $f:X\to\mathbb{R}^{+}$, with $\int
f\,\mathrm{d}\mu_{0}=1$.

## 5 Definition of the Midpoint

As previously pointed out, in order to prove CD condition I am going to prove
entropy convexity in a suitable midpoint of any pair of absolutely continuous
measures. Notice that in an highly branching metric measure space the choice
of a midpoint can be done with great freedom. This is because, in general,
both the optimal transport map and the geodesic interpolation are not unique,
and thus they must be selected in a clever way. In this section, for any pair
of absolutely continuous measures, I define a suitable midpoint and in the
following section I will show that it satisfies the convexity of the entropy.
This midpoint selection is actually quite complicated but it does the job, in
particular I believe there is no way to obtain a considerably simpler one.

Before going into the details, I introduce the following subsets of
$\mathbb{R}^{2}\times\mathbb{R}^{2}$ that will play an important role in the
definition of the midpoint.

######  Definition 5.1.

Define the sets
$V,D,H,H_{0},H_{\frac{1}{2}},H_{1}\subset\mathbb{R}^{2}\times\mathbb{R}^{2}$
as:

$V:=\left\\{\left(\left(x_{0},y_{0}\right),\left(x_{1},y_{1}\right)\right)\in\mathbb{R}^{2}\times\mathbb{R}^{2}:\left|x_{0}-x_{1}\right|<\left|y_{0}-y_{1}\right|\right\\},$
$D:=\left\\{\left(\left(x_{0},y_{0}\right),\left(x_{1},y_{1}\right)\right)\in\mathbb{R}^{2}\times\mathbb{R}^{2}:\left|x_{0}-x_{1}\right|=\left|y_{0}-y_{1}\right|\right\\},$
$H:=\left\\{\left(\left(x_{0},y_{0}\right),\left(x_{1},y_{1}\right)\right)\in\mathbb{R}^{2}\times\mathbb{R}^{2}:\left|x_{0}-x_{1}\right|>\left|y_{0}-y_{1}\right|\right\\}=H_{0}\cup
H_{1},$

where

$H_{0}:=\left\\{\left(\left(x_{0},y_{0}\right),\left(x_{1},y_{1}\right)\right)\in\mathbb{R}^{2}\times\mathbb{R}^{2}:\frac{1}{2}\left|x_{0}-x_{1}\right|\geq\left|y_{0}-y_{1}\right|\right\\},$
$H_{1}:=\left\\{\left(\left(x_{0},y_{0}\right),\left(x_{1},y_{1}\right)\right)\in\mathbb{R}^{2}\times\mathbb{R}^{2}:\left|x_{0}-x_{1}\right|>\left|y_{0}-y_{1}\right|>\frac{1}{2}|x_{0}-x_{1}|\right\\}.$

The first step in the selection of the midpoint consists in choosing a
suitable optimal transport map between two given absolutely continuous
measures. To this aim I use a nice result by Rajala [5], who was able to show
the existence of an optimal transport map with different nice properties. The
idea behind his work is to use consecutive minimization, in order to select a
particular optimal transport plan. The result of Rajala can be summarized in
the following statement.

######  Proposition 5.2.

Given two measures $\mu_{0},\mu_{1}\in\mathscr{P}(\mathbb{R}^{2})$ which are
absolutely continuous with respect to the Lebesgue measure $\mathscr{L}^{2}$,
there exists a measurable optimal transport map $T=(T_{1},T_{2})$, injective
outside a $\mu_{0}$-null set, satisfying $T_{\\#}\mu_{0}=\mu_{1}$, with some
nice rigidity properties. In particular the optimal transport plan
$(\operatorname{id},T)_{\\#}\mu_{0}$ is concentrated in a set $\Gamma\subset
X\times X$, such that for all
$\left(x,y_{1}\right),\left(x,y_{2}\right),\left(x_{1},y\right),\left(x_{2},y\right)\in\left\\{(x,y)\in
X:((x,y),T(x,y))\in\Gamma\right\\}$ it holds that

$\text{if }y_{1}\neq y_{2}\text{ and
}T_{1}\left(x,y_{1}\right)=T_{1}\left(x,y_{2}\right),\text{ then
}\frac{T_{2}\left(x,y_{1}\right)-T_{2}\left(x,y_{2}\right)}{y_{1}-y_{2}}\geq
0$

and

$\text{if }x_{1}\neq x_{2}\text{ and
}T_{2}\left(x_{1},y\right)=T_{2}\left(x_{2},y\right),\text{ then
}\frac{T_{1}\left(x_{1},y\right)-T_{1}\left(x_{2},y\right)}{x_{1}-x_{2}}\geq
0.$

Moreover for $\mu_{0}$-almost every $(x,y)$ I have that

$\begin{array}[]{l}T_{1}\text{ is locally constant in }y,\text{ if
}((x,y),T(x,y))\in H\text{ and }\\\ T_{2}\text{ is locally constant in
}x,\text{ if }((x,y),T(x,y))\in V\end{array}.$

Combining this two properties with some monotonicity properties one can deduce
that the function $T_{1}(x,y)$ is increasing in $x$ for every fixed $y$ and
the function $T_{2}(x,y)$ is increasing in $y$ for every fixed $x$, as a
consequence for $\mu_{0}$-almost every $(x,y)$ it holds

$\frac{\partial T_{1}}{\partial x}\geq 0\text{ and }\frac{\partial
T_{2}}{\partial y}\geq 0,\text{ if }((x,y),T(x,y))\in H\cup V.$

Finally let me point out that, since
$(\operatorname{id},T)_{\\#}\mu_{0}\in\operatorname{OptPlans}(\mu_{0},\mu_{1})$
the usual cyclical monotonicity holds, thus I can assume that for every
$\left(z_{1},w_{1}\right),\left(z_{2},w_{2}\right)\in\Gamma$ it holds

$\mathsf{d}_{\infty}^{2}(z_{1},w_{1})+\mathsf{d}_{\infty}^{2}(z_{2},w_{2})\leq\mathsf{d}_{\infty}^{2}(z_{1},w_{2})+\mathsf{d}_{\infty}^{2}(z_{2},w_{1}).$

Now fix two measures $\mu_{0},\mu_{1}\in\mathscr{P}(X)$ absolutely continuous
with respect to the reference measure $\mathfrak{m}$, and thus also with
respect to the Lebesgue measure $\mathscr{L}^{2}$. Call $T$ the optimal
transport map between $\mu_{0}$ and $\mu_{1}$, that satisfies the requirements
of Proposition 5.2. Moreover denote by $\Gamma$ the full
$(\operatorname{id},T)_{\\#}\mu_{0}$-measure set with all the monotonicity
property stated in Proposition 5.2. In order to identify a midpoint of
$\mu_{0}$ and $\mu_{1}$, I need to choose a proper midpoint interpolation,
that is a measurable map $M:X\times X\to X$ such that

$\mathsf{d}_{\infty}(M(z,w),z)=\mathsf{d}_{\infty}(M(z,w),w)=\frac{1}{2}\mathsf{d}_{\infty}(z,w)\quad\text{for
every }(z,w)\in X\times X,$ (10)

the desired midpoint will be
$M_{\\#}\big{(}(\operatorname{id},T)_{\\#}\mu_{0}\big{)}=[M\circ(\operatorname{id},T)]_{\\#}\mu_{0}$.

Let me now define the midpoint interpolation map $M$ that I will use for the
proof of the main theorem. This definition is actually quite involved, in fact
the map $M$ is defined in different ways on the sets $V$, $D$, $H_{0}$ and
$H_{1}$. In particular the precise definition is the following:

  * •

If $\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}\in V\cup D$

$M\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}:=\left(\frac{x_{0}+x_{1}}{2},\frac{y_{0}+y_{1}}{2}\right)$

  * •

If $\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}\in H_{0}$,

$M\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}=\left(\frac{x_{0}+x_{1}}{2},\frac{1}{2}\bigg{(}\frac{y_{0}}{f(x_{0})}+\frac{y_{1}}{f(x_{1})}\bigg{)}f\left(\frac{x_{0}+x_{1}}{2}\right)\right).$

  * •

If $\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}\in H_{1}$, with $x_{0}<x_{1}$
and $y_{0}<y_{1}$, introduce the quantity

$\tilde{y}(x_{0},x_{1},y_{0})=\frac{1}{2}\bigg{(}\frac{y_{0}}{f(x)}+\frac{y_{0}+\frac{x_{1}-x_{0}}{2}}{f(x_{1})}\bigg{)}f\left(\frac{x_{1}+x_{0}}{2}\right)-y_{0},$

and consequently define

$\displaystyle M$ $\displaystyle\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}$
$\displaystyle\qquad\qquad=\left(\frac{x_{0}+x_{1}}{2},y_{0}+\tilde{y}(x_{0},x_{1},y_{0})+\left(\frac{x_{1}-x_{0}}{2}-\tilde{y}(x_{0},x_{1},y_{0})\right)\left(2\frac{y_{1}-y_{0}}{x_{1}-x_{0}}-1\right)\right).$

In the other cases $M$ can be defined analogously, but I prefer not to
explicitly do it, in order to avoid unnecessary complications in this
definition and in the following. In particular every proof from now on will be
done using only the definition above, implying it can be easily adapted to the
other cases.

With such a complex definition is not completely obvious that the map $M$
satisfies condition (10), but this can be easily proven.

######  Proposition 5.3.

The map $M$ actually defines a midpoint interpolation.

######  Proof.

This is completely obvious if $\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}\in
V\cup D$. While when $\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}\in H_{0}$ the
statement is an easy consequence of Lemma 3.2, provided that $k$ is
sufficiently small, and of the fact that

$0\leq\frac{1}{2}\bigg{(}\frac{y_{0}}{f(x_{0})}+\frac{y_{1}}{f(x_{1})}\bigg{)}f\left(\frac{x_{0}+x_{1}}{2}\right)\leq
f\left(\frac{x_{0}+x_{1}}{2}\right).$

On the other hand if $\big{(}(x_{0},y_{0}),(x_{1},y_{1})\big{)}\in H_{1}$ it
is sufficient to notice that the point

$(\bar{x},\bar{y}):=\left(\frac{x_{0}+x_{1}}{2},y_{0}+\tilde{y}(x_{0},x_{1},y_{0})+\left(\frac{x_{1}-x_{0}}{2}-\tilde{y}(x_{0},x_{1},y_{0})\right)\left(2\frac{y_{1}-y_{0}}{x_{1}-x_{0}}-1\right)\right)$

lies on a convex combination of a 45 degree line and a curve
$\gamma:[x_{0},x_{1}]\to X$, with

$\frac{1}{2}-3k\leq\gamma^{\prime}(x)\leq\frac{1}{2}+3k\quad\text{ for every
}x\in[x_{0},x_{1}],$

according to Lemma 3.2. As a consequence, for a suitably small $k$, this
ensures both that $(\bar{x},\bar{y})\in X$ and that $(\bar{x},\bar{y})$ is a
$\mathsf{d}_{\infty}$-midpoint of $(x_{0},y_{0})$ and $(x_{1},y_{1})$. ∎

In order to efficiently apply Proposition 4.2, I need to compute the Jacobian
of the map $M\circ(\operatorname{id},T)$. Observe that the way $M$ is defined,
combined with the properties of $T$, ensures that
$M\circ(\operatorname{id},T)$ satisfies all the rigidity assumptions of
Proposition 4.1. Therefore, proving the following result, will allow to use
Proposition 4.1 to compute $J_{M\circ(\operatorname{id},T)}$.

######  Proposition 5.4.

The map $M\circ(\operatorname{id},T)$ is injective outside a $\mu_{0}$-null
set.

######  Proof.

First of all notice that it is sufficient to prove the injectivity of $M$ on
$\Gamma$, because $\Gamma$ has full
$(\operatorname{id},T)_{\\#}\mu_{0}$-measure. Thus suppose by contradiction
that there exist

$((x_{1},y_{1}),(x_{1}^{\prime},y_{1}^{\prime}))\neq((x_{2},y_{2}),(x_{2}^{\prime},y_{2}^{\prime}))\in\Gamma$

such that

$M((x_{1},y_{1}),(x_{1}^{\prime},y_{1}^{\prime}))=M((x_{2},y_{2}),(x_{2}^{\prime},y_{2}^{\prime})).$

Following the proof of Lemma 3.7 in [5], I can limit myself to consider the
case when

$((x_{1},y_{1}),(x_{1}^{\prime},y_{1}^{\prime}))\neq((x_{2},y_{2}),(x_{2}^{\prime},y_{2}^{\prime}))\in\Gamma\cap
H.$

In this case, one can easily realize that the cyclical monotonicity properties
of $\Gamma$ imply that $x_{1}=x_{2}$ and $x_{1}^{\prime}=x_{2}^{\prime}$. So,
if
$((x_{1},y_{1}),(x_{1}^{\prime},y_{1}^{\prime})),((x_{2},y_{2}),(x_{2}^{\prime},y_{2}^{\prime}))\in\Gamma\cap
H_{0}$, thesis simply follows from the definition of $M$ and from the
monotonicity of $T_{2}$. While if
$((x_{1},y_{1}),(x_{1}^{\prime},y_{1}^{\prime})),((x_{2},y_{2}),(x_{2}^{\prime},y_{2}^{\prime}))\in\Gamma\cap
H_{1}$ the statement is a consequence of the monotonicity property of $T_{2}$
associated with the fact that the quantity

$y+\tilde{y}(x,x^{\prime},y)+\left(\frac{x^{\prime}-x}{2}-\tilde{y}(x,x^{\prime},y)\right)\left(2\frac{y^{\prime}-y}{x^{\prime}-x}-1\right)$

is locally increasing in $y$ and $y^{\prime}$, when
$((x,y),(x^{\prime},y^{\prime}))\in H_{1}$ (with $x<x^{\prime}$ and
$y<y^{\prime}$). The first monotonicity is not straightforward, therefore I am
going to prove it. First of all notice that, according to Lemma 3.5 and since
$f\leq 3k$, it holds that

$2\left|\frac{\partial}{\partial
y}\tilde{y}(x,x^{\prime},y)\right|=\left|\frac{f\big{(}\frac{x+x^{\prime}}{2}\big{)}}{f(x)}+\frac{f\big{(}\frac{x+x^{\prime}}{2}\big{)}}{f(x^{\prime})}-2\right|\leq\frac{[2k^{2}+3k]\big{(}\frac{x^{\prime}-x}{2}\big{)}^{2}}{f(x)f(x^{\prime})}.$

Moreover, the geometry of the set $X$ allows to deduce that

$\frac{x^{\prime}-x}{2}\leq y+\frac{x^{\prime}-x}{2}\leq f(x^{\prime})\leq
f(x)+k(x^{\prime}-x),$

and consequently $\big{(}1-2k\big{)}\frac{x^{\prime}-x}{2}\leq f(x)\leq
f(x^{\prime})$. On the other hand, observe that Lemma 3.2 guarantees that

$\left(\frac{x^{\prime}-x}{2}-\tilde{y}(x,x^{\prime},y)\right)\leq\left(\frac{1}{2}+3k\right)\frac{x^{\prime}-x}{2}$

Those estimates allow to conclude that

$\displaystyle\frac{\partial}{\partial y}\bigg{[}y+\tilde{y}(x,x^{\prime},y)$
$\displaystyle+\left(\frac{x^{\prime}-x}{2}-\tilde{y}(x,x^{\prime},y)\right)\left(2\frac{y^{\prime}-y}{x^{\prime}-x}-1\right)\bigg{]}$
$\displaystyle\geq\frac{1}{2}-3k-\left|\frac{\partial}{\partial
y}\tilde{y}(x,x^{\prime},y)\right|\left(2-2\frac{y^{\prime}-y}{x^{\prime}-x}\right)$
$\displaystyle\geq\frac{1}{2}-3k-\frac{[2k^{2}+3k]}{(1-2k)^{2}}>0,$

for $k$ sufficiently small. The case when
$((x_{1},y_{1}),(x_{1}^{\prime},y_{1}^{\prime}))\in\Gamma\cap
H_{1},((x_{2},y_{2})$ and $(x_{2}^{\prime},y_{2}^{\prime}))\in\Gamma\cap
H_{0}$ can be treated analogously. ∎

## 6 Proof of CD Condition

In the previous sections I have introduced all I need to prove that the metric
measure space $(X,\mathsf{d}_{\infty},\mathfrak{m})$ satisfies the
$\mathsf{CD}(0,\infty)$ condition. Let me now go into the details of the
proof.

######  Theorem 6.1.

For suitable $k$ and $K$, the metric measure space
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$ is a
$\mathsf{CD}(0,\infty)$ space, for every $0<\varepsilon<k$.

######  Proof.

Let $\mu_{0},\mu_{1}\in\mathscr{P}(X)$ be absolutely continuous with respect
to the reference measure $\mathfrak{m}$, then, according to Proposition 1.8,
it is sufficient to prove that

$\operatorname{Ent}\big{(}[M\circ(\operatorname{id},T)]_{\\#}\mu_{0}\big{)}\leq\frac{1}{2}\operatorname{Ent}(\mu_{0})+\frac{1}{2}\operatorname{Ent}(\mu_{1}).$

Given Proposition 4.2, it is enough check the validity of

$\log\left(m\big{(}M((x,y),T(x,y))\big{)}J_{M\circ(\operatorname{id},T)}(x,y)\right)\geq\frac{1}{2}\log\left(m(T(x,y))J_{T}(x,y)\right)+\frac{1}{2}\log(m(x,y))$
(11)

for $\mu_{0}$-almost every $(x,y)$. For $\mu_{0}$-almost every $(x,y)\in V\cup
D$ this can be done following [5]. Thus I will treat the other cases and
applying Lemma 3.3.
Notice that, for $\mu_{0}$-almost every $(x,y)$ such that $((x,y),T(x,y))\in
H_{0}\cap\Gamma$, I have

$M\circ(\operatorname{id},T)(x,y)=\left(\frac{x+T_{1}}{2},\frac{1}{2}\bigg{(}\frac{y}{f(x)}+\frac{T_{2}}{f(T_{1})}\bigg{)}f\left(\frac{x+T_{1}}{2}\right)\right).$

Then, according to what I did in previous sections, it is possible to apply
Proposition 4.1 and deduce that

$J_{M\circ(\operatorname{id},T)}(x,y)=\frac{1}{2}\left(1+\frac{\partial
T_{1}}{\partial
x}\right)\frac{1}{2}\bigg{(}\frac{1}{f(x)}+\frac{\frac{\partial
T_{2}}{\partial y}}{f(T_{1})}\bigg{)}f\left(\frac{x+T_{1}}{2}\right),$

for $\mu_{0}$-almost every $(x,y)$ such that $((x,y),T(x,y))\in
H_{0}\cap\Gamma$. Furthermore it holds that

$m\big{(}M((x,y),T(x,y))\big{)}=f\left(\frac{x+T_{1}}{2}\right)^{-1}\exp\left(\frac{-K}{4}\bigg{(}\frac{y}{f(x)}+\frac{T_{2}}{f(T_{1})}\bigg{)}^{2}\right),$

thus, putting together this last two relations, I obtain

$\displaystyle\log\left(m\big{(}M((x,y),T(x,y))\big{)}J_{M\circ(\operatorname{id},T)}(x,y)\right)$
$\displaystyle=\log\Bigg{(}\frac{1}{2}\left(1+\frac{\partial T_{1}}{\partial
x}\right)\frac{1}{2}\bigg{(}\frac{1}{f(x)}+\frac{\frac{\partial
T_{2}}{\partial y}}{f(T_{1})}\bigg{)}$
$\displaystyle\qquad\exp\Bigg{(}\frac{-K}{4}\bigg{(}\frac{y}{f(x)}+\frac{T_{2}}{f(T_{1})}\bigg{)}^{2}\Bigg{)}\Bigg{)}$
$\displaystyle=\log\left(\frac{1}{2}\left(1+\frac{\partial T_{1}}{\partial
x}\right)\right)+\log\left(\frac{1}{2}\bigg{(}\frac{1}{f(x)}+\frac{\frac{\partial
T_{2}}{\partial y}}{f(T_{1})}\bigg{)}\right)$
$\displaystyle\quad-K\bigg{(}\frac{1}{2}\bigg{(}\frac{y}{f(x)}+\frac{T_{2}}{f(T_{1})}\bigg{)}\bigg{)}^{2}$

On the other hand it holds

$\log(m(x,y))=\log\left(\frac{1}{f(x)}\exp\left(-K\bigg{(}\frac{y}{f(x)}\bigg{)}^{2}\right)\right)=\log(1)+\log\left(\frac{1}{f(x)}\right)-K\bigg{(}\frac{y}{f(x)}\bigg{)}^{2}$

and, applying once again Proposition 4.1, this time to the map $T$, also

$\displaystyle\log\left(m(T(x,y))J_{T}(x,y)\right)$
$\displaystyle=\log\left(\frac{\partial T_{1}}{\partial x}\frac{\partial
T_{2}}{\partial
y}\frac{1}{f(T_{1})}\exp\left(-K\bigg{(}\frac{T_{2}}{f(T_{1})}\bigg{)}^{2}\right)\right)$
$\displaystyle=\log\left(\frac{\partial T_{1}}{\partial
x}\right)+\log\left(\frac{\frac{\partial T_{2}}{\partial
y}}{f(T_{1})}\right)-K\bigg{(}\frac{T_{2}}{f(T_{1})}\bigg{)}^{2},$

for $\mu_{0}$-almost every $(x,y)$ such that $((x,y),T(x,y))\in
H_{0}\cap\Gamma$. Putting together this last three equations, inequality (11)
follows from the concavity of the functions $\log$ and $-Kx^{2}$.
Passing now to the last case, for $\mu_{0}$-almost every $(x,y)$ such that
$((x,y),T(x,y))\in H_{1}\cap\Gamma$ (with $x<T_{1}(x,y)$ and $y<T_{2}(x,y)$) I
have

$\displaystyle(S_{1},S_{2})(x,y)$
$\displaystyle:=M\circ(\operatorname{id},T)(x,y)$
$\displaystyle\quad=\left(\frac{x+T_{1}}{2},y+\tilde{y}(x,T_{1},y)+\left(\frac{T_{1}-x}{2}-\tilde{y}(x,T_{1},y)\right)\left(2\frac{T_{2}-y}{T_{1}-x}-1\right)\right).$

Reasoning as before, Proposition 4.1 ensures that, for $\mu_{0}$-almost every
$(x,y)$ such that $((x,y),T(x,y))\in H_{1}\cap\Gamma$ (with $x<T_{1}(x,y)$ and
$y<T_{2}(x,y)$),

$J_{M\circ(\operatorname{id},T)}(x,y)=\frac{\partial S_{1}}{\partial
x}\frac{\partial S_{2}}{\partial y},$

and in particular it holds that

$\frac{\partial S_{1}}{\partial x}=\frac{1}{2}\left(1+\frac{\partial
T_{1}}{\partial x}\right),$

and

$\displaystyle\frac{\partial S_{2}}{\partial y}$
$\displaystyle=1+\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)\left(2-2\frac{T_{2}-y}{T_{1}-x}\right)+2\left(\frac{T_{1}-x}{2}-\tilde{y}(x,T_{1},y)\right)\frac{\frac{\partial
T_{2}}{\partial y}-1}{T_{1}-x}$ $\displaystyle=1+\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)\left(2-2\frac{T_{2}-y}{T_{1}-x}\right)+\left(\frac{\partial
T_{2}}{\partial
y}-1\right)\left(1-\frac{\tilde{y}(x,T_{1},y)}{\frac{T_{1}-x}{2}}\right)$
$\displaystyle=1+\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)\left(2-2\frac{T_{2}-y}{T_{1}-x}\right)+\left(\frac{\partial
T_{2}}{\partial
y}-1\right)\left(\frac{1}{2}-\frac{\tilde{y}(x,T_{1},y)-\frac{T_{1}-x}{4}}{\frac{T_{1}-x}{2}}\right).$

I can now consider the explicit value of $\tilde{y}(x,T_{1},y)$ and notice
that

$\displaystyle\tilde{y}(x,T_{1},y)-\frac{x+T_{1}}{4}$
$\displaystyle=\frac{1}{2}\left[\bigg{(}\frac{y}{f(x)}+\frac{y+\frac{T_{1}-x}{2}}{f(T_{1})}\bigg{)}f\left(\frac{T_{1}+x}{2}\right)-2y-\frac{T_{1}-x}{2}\right]$
$\displaystyle=\frac{1}{2}\left[y\left(\frac{f\left(\frac{T_{1}+x}{2}\right)}{f(x)}+\frac{f\left(\frac{T_{1}+x}{2}\right)}{f(T_{1})}-2\right)+\frac{T_{1}-x}{2}\left(\frac{f\left(\frac{T_{1}+x}{2}\right)}{f(T_{1})}-1\right)\right].$

Moreover, I can easily obtain that

$\left|\frac{f\left(\frac{T_{1}+x}{2}\right)}{f(T_{1})}-1\right|=\left|\frac{f\left(\frac{T_{1}+x}{2}\right)-f(T_{1})}{f(T_{1})}\right|\leq\sup
f^{\prime}\cdot\frac{\frac{T_{1}-x}{2}}{f(T_{1})}=k\cdot\frac{\frac{T_{1}-x}{2}}{f(T_{1})},$

thus applying Lemma 3.5 and noticing that $\frac{y}{f(x)}\leq 1$, I can
conclude that

$\displaystyle\left|\tilde{y}(x,T_{1},y)-\frac{T_{1}-x}{4}\right|\leq\frac{\left[2k^{2}+k+f\left(\frac{T_{1}+x}{2}\right)\right]\left(\frac{T_{1}-x}{2}\right)^{2}}{f(T_{1})}\leq\frac{[2k^{2}+4k]\left(\frac{T_{1}-x}{2}\right)^{2}}{f(T_{1})}$

where in the last inequality I used that $f(x)\leq 3k$ for every $x\in[-1,1]$.
Therefore it is possible to infer that

$\left|\frac{\tilde{y}(x,T_{1},y)-\frac{T_{1}-x}{4}}{\frac{T_{1}-x}{2}}\right|\leq\frac{[2k^{2}+4k]\frac{T_{1}-x}{2}}{f(T_{1})}\leq\frac{1}{64}\frac{\frac{T_{1}-x}{2}}{f(T_{1})}\leq\frac{\frac{T_{1}-x}{2}}{f(T_{1})},$
(12)

for a sufficiently small $k$. Now suppose that

$\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)=\frac{1}{2}\frac{f\big{(}\frac{x+T_{1}}{2}\big{)}}{f(x)}+\frac{1}{2}\frac{f\big{(}\frac{x+T_{1}}{2}\big{)}}{f(T_{1})}-1>0,$

then, after noticing that the geometry of the set $X$ ensures that
$\frac{T_{1}-x}{2}\leq f(T_{1})$, it is possible to apply Lemma 3.1 and obtain
that

$\displaystyle\log\left(\frac{\partial S_{2}}{\partial y}\right)$
$\displaystyle\geq\log\left(1+\left(\frac{\partial T_{2}}{\partial
y}-1\right)\left(\frac{1}{2}-\frac{\tilde{y}(x,T_{1},y)-\frac{T_{1}-x}{4}}{\frac{T_{1}-x}{2}}\right)\right)$
(13) $\displaystyle\geq\log(1)+\log\left(\frac{\partial T_{2}}{\partial
y}\right)-C\frac{\left(\frac{T_{1}-x}{2}\right)^{2}}{f(T_{1})^{2}}.$

On the other hand, it is easy to realize that the point
$M\circ(\operatorname{id},T)(x,y)$ lies on a curve, which is a convex
combination of a 45 degree line and of the curve

$t\mapsto\bigg{(}(1-t)\frac{y}{f(x)}+t\,\frac{y+\frac{T_{1}-x}{2}}{f(T_{1})}\bigg{)}f((1-t)x+tT_{1}).$

Therefore, up to take a suitably small $k$, Lemma 3.2 allows to apply Lemma
3.3, and obtain

$\log\left(m\left(M\circ(\operatorname{id},T)(x,y)\right)\right)\geq\log(m(x,y))+\log(m(T(x,y)))+\frac{K}{128f(T_{1})^{2}}(T_{1}-x)^{2}.$
(14)

Inequality (11) then follows as before, putting together (13) and (14) and
taking $K$ sufficiently large.
Suppose instead that

$\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)=\frac{1}{2}\frac{f\big{(}\frac{x+T_{1}}{2}\big{)}}{f(x)}+\frac{1}{2}\frac{f\big{(}\frac{x+T_{1}}{2}\big{)}}{f(T_{1})}-1<0,$

then notice that

$\displaystyle\log\left(f\bigg{(}\frac{T_{1}+x}{2}\bigg{)}\right)-\frac{1}{2}\big{(}\log(f(x))+\log(f(T_{1}))\big{)}=\frac{1}{2}\log\left(\frac{f\left(\frac{T_{1}+x}{2}\right)}{f(x)}\cdot\frac{f\left(\frac{T_{1}+x}{2}\right)}{f(T_{1})}\right)$
(15)
$\displaystyle\leq\log\left(\frac{1}{2}\frac{f\left(\frac{T_{1}+x}{2}\right)}{f(x)}+\frac{1}{2}\frac{f\left(\frac{T_{1}+x}{2}\right)}{f(T_{1})}\right)=\log\left(1+\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)\right)$

Moreover, according to the estimates done in the proof of Proposition 5.4 and
to (12), it is easy to realize that, for $k$ small enough

$\left(\frac{1}{2}-\frac{\tilde{y}(x,T_{1},y)-\frac{T_{1}-x}{4}}{\frac{T_{1}-x}{2}}\right)\left(1+\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)\right)^{-1}=\frac{1}{2}+\tilde{\delta}$

for some $\tilde{\delta}$ such that

$\big{|}\tilde{\delta}\big{|}\leq\frac{1}{32}\frac{\frac{T_{1}-x}{2}}{f(T_{1})}.$

Consequently I can infer that

$\displaystyle\log\left(\frac{\partial S_{2}}{\partial y}\right)$
$\displaystyle\geq\log\left(1+\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)+\left(\frac{\partial T_{2}}{\partial
y}-1\right)\left(\frac{1}{2}-\frac{\tilde{y}(x,T_{1},y)-\frac{T_{1}-x}{4}}{\frac{T_{1}-x}{2}}\right)\right)$
(16) $\displaystyle=\log\left(1+\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)\right)+\log\left(1+\left(\frac{\partial T_{2}}{\partial
y}-1\right)\left(\frac{1}{2}+\tilde{\delta}\right)\right)$
$\displaystyle\geq\log\left(1+\frac{\partial}{\partial
y}\tilde{y}(x,T_{1},y)\right)+\log(1)+\log\left(\frac{\partial T_{2}}{\partial
y}\right)-C\frac{\left(\frac{T_{1}-x}{2}\right)^{2}}{f(T_{1})^{2}}$

where the last passage follows from Lemma 3.1. Finally it is possible to prove
(11), putting together (15) with (16), applying Corollary 3.4 and taking $K$
sufficiently large. ∎

As I did in section 4, I exploit the local nature of Jacobi equation to
improve the last result. The following result is an easy consequence of
Theorem 6.1 and Corollary 4.3, and it will be useful in the end of this work.

######  Corollary 6.2.

Given two absolutely continuous measures $\mu_{0},\mu_{1}\in\mathscr{P}(X)$,
assume that there exists a map $T$ such that $T_{\\#}\mu_{0}=\mu_{1}$,
satisfying all the properties of Proposition 5.2. Then, calling $M$ the
midpoint selection presented in section 5, it holds

$\displaystyle\operatorname{Ent}\big{(}[M\circ(\operatorname{id},T)]_{\\#}(f\mu_{0})\big{)}\leq\frac{1}{2}\operatorname{Ent}(f\mu_{0})+\frac{1}{2}\operatorname{Ent}(T_{\\#}(f\mu_{0})),$

for every bounded measurable function $f:X\to\mathbb{R}^{+}$ with $\int
f\,\mathrm{d}\mu_{0}=1$.

## 7 Conclusions

In this last section I conclude all the work done up to now. In particular I
am going to show why this example is relevant, asking also to some open
question related to strict curvature dimension bounds. First of all let me
prove the most important result, which had already been anticipated in
previous sections.

######  Theorem 7.1.

For suitable $k$ and $K$ the metric measure space
$(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$ is a
$\mathsf{CD}(0,\infty)$ space.

######  Proof.

I am going to prove that, for every sequence of positive real numbers
$(\varepsilon_{n})_{n\in\mathbb{N}}$ converging to zero, the sequence of
metric measure spaces
$(X_{k,\varepsilon_{n}},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon_{n}})$
measured Gromov Hausdorff converges to
$(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$. According to Theorem
1.15 and Theorem 6.1 this is sufficient to conclude the proof, up to choose
suitable $k$ and $K$.

Define the function $f_{n}:X_{k,\varepsilon_{n}}\to X_{k,0}$ as

$f_{n}(x,y)=\bigg{(}x,y\cdot\frac{f_{k,0}(x)}{f_{k,\varepsilon_{n}}(x)}\bigg{)},$

it is immediate to notice that its image is actually $X_{k,0}$. Moreover its
is easy to prove that

$(f_{n})_{\\#}\mathfrak{m}_{k,K,\varepsilon_{n}}=\mathfrak{m}_{k,K,0}\quad\text{and}\quad\big{|}\mathsf{d}_{\infty}\big{(}f_{n}(x_{1},y_{1}),f_{n}(x_{2},y_{2})\big{)}-\mathsf{d}_{\infty}\big{(}(x_{1},y_{1}),(x_{2},y_{2})\big{)}\big{|}\leq
2\varepsilon_{n},$

and this shows the desired measured Gromov-Hausdorff convergence. ∎

First of all notice that the space
$(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$ has different topological
dimensions at different regions of the space. In particular, this shows the
non-constancy of topological dimension also for CD spaces, extending one of
the results by Ketterer and Rajala [3]. Furthermore, the space
$(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$ is not a very strict
$\mathsf{CD}(K,\infty)$ space for every $K\in\mathbb{R}$, in fact it is not
weakly essentially non-branching (see Theorem 1.12). In order to see this, it
is sufficient to consider an absolutely continuous measure $\mu_{0}$ supported
on $L$ and an absolutely continuous measure $\mu_{1}$ supported on $C$, and
subsequently notice that every $\eta\in\operatorname{OptGeo}(\mu_{0},\mu_{1})$
is supported in branching geodesics. It is then possible to conclude that the
weak CD condition is not sufficient to ensure any type of non-branching
condition. Observe also that every
$\eta\in\operatorname{OptGeo}(\mu_{0},\mu_{1})$ is not induced by a map,
consequently the (weak) CD condition is not sufficient to ensure the existence
of an optimal transport map, between two absolutely continuous marginals.
Finally notice that the space
$(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$ is an example of (weak)
CD space which is not a very strict CD space, and this shows that this two
notions of curvature dimension bounds are actually different.

For the last part of this section I need to introduce another type of
curvature bounds, called strict CD condition, which is stronger than the weak
CD condition, but is weaker than the very strict one.

######  Definition 7.2.

A metric measure space $(X,\mathsf{d},\mathfrak{m})$ is called a strict
$\mathsf{CD}(K,\infty)$ space if for every absolutely continuous measures
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}(X)$ there exists an optimal geodesic plan
$\eta\in\operatorname{OptGeo}(\mu_{0},\mu_{1})$, so that the entropy
functional $\operatorname{Ent}$ satisfies the K-convexity inequality along
$f\eta$ for every bounded measurable function
$f:\operatorname{Geo}(X)\to\mathbb{R}^{+}$ with $\int f\,\mathrm{d}\eta=1$.

I am now going to prove that, for suitable constants, the spaces
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$ with
$0<\varepsilon<k$ are strict CD spaces, while their measured Gromov Hausdorff
limit as $\varepsilon\to 0$
$(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$ is not. As a consequence
the following proposition holds.

######  Proposition 7.3.

The strict CD condition is not stable under measured Gromov Hausdorff
convergence.

Before going into the details of the proofs, I want to make some
clarifications. The fact that the spaces
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$ are
strict CD is a consequence of Corollary 6.2 and of an iteration argument. On
the other hand the space $(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$
cannot be a strict CD space, because of its particular topological structure I
have already highlighted.

######  Proposition 7.4.

For suitable $k$ and $K$ and every $0<\varepsilon<k$ the metric measure space
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$ is a
strict $\mathsf{CD}(0,\infty)$ space.

######  Proof.

For every $n\in\mathbb{N}$ I am going to define a measurable map
$G_{n}:X\to\operatorname{Geo}(X)$ by induction. In particular I introduce
$G_{0}:X\to\operatorname{Geo}(X)$ as any measurable map such that
$(e_{0},e_{1})\circ G_{0}(x)=(\operatorname{id},T)$ $\mu_{0}$-almost
everywhere, consequently
$(G_{0})_{\\#}\mu_{0}\in\operatorname{OptGeo}(\mu_{0},\mu_{1})$. Given
$G_{n}:X\to\operatorname{Geo}(X)$, define $G_{n+1}:X\to\operatorname{Geo}(X)$
by imposing that:

  1. 1.

$e_{r}\circ G_{n+1}=e_{r}\circ G_{n}$ for every
$r\in\big{\\{}\frac{k}{2^{n}},k=0,\dots,2^{n}\big{\\}}$

  2. 2.

$e_{\frac{2k+1}{2^{n+1}}}\circ G_{n+1}=M\big{(}e_{\frac{k}{2^{n}}}\circ
G_{n},e_{\frac{k+1}{2^{n}}}\circ G_{n}\big{)}$ $\mu_{0}$-almost everywhere,
where $M$ is the midpoint map defined as in Section 5.

Notice that, if the optimal transport map that induces
$\big{(}e_{\frac{k}{2^{n}}}\circ G_{n},e_{\frac{k+1}{2^{n}}}\circ
G_{n}\big{)}_{\\#}\mu_{0}$ satisfies all the properties of Proposition 5.2,
then also the maps that induce $\big{(}e_{\frac{2k}{2^{n+1}}}\circ
G_{n+1},e_{\frac{2k+1}{2^{n+1}}}\circ G_{n+1}\big{)}_{\\#}\mu_{0}$ and
$\big{(}e_{\frac{2k+1}{2^{n+1}}}\circ G_{n+1},e_{\frac{2k+2}{2^{n+1}}}\circ
G_{n+1}\big{)}_{\\#}\mu_{0}$ satisfy all the properties of Proposition 5.2.
The reader can easily realize that this is a quite straightforward consequence
of the definition of map $M$ and of its properties highlighted in Section 6.
This observation shows that the inductive procedure I have introduced can be
done in accordance with the previous section, moreover it is possible to apply
Corollary 6.2 and obtain that

$\begin{split}\operatorname{Ent}\big{(}(e_{s})_{\\#}[(G_{n+1})_{\\#}(f\mu_{0})]\big{)}&\leq\frac{1}{2}\operatorname{Ent}\big{(}(e_{r})_{\\#}[(G_{n+1})_{\\#}(f\mu_{0})]\big{)}+\frac{1}{2}\operatorname{Ent}\big{(}(e_{t})_{\\#}[(G_{n+1})_{\\#}(f\mu_{0})]\big{)}\\\
&\quad-\frac{K}{8}W_{2}^{2}\big{(}(e_{r})_{\\#}[(G_{n+1})_{\\#}(f\mu_{0})],(e_{t})_{\\#}[(G_{n+1})_{\\#}(f\mu_{0})]\big{)},\end{split}$
(17)

where $r=\frac{2k}{2^{n+1}}$, $s=\frac{2k+1}{2^{n+1}}$,
$t=\frac{2k+2}{2^{n+1}}$ and $f$ is any bounded measurable function with $\int
f\,\mathrm{d}\mu_{0}=1$. Notice that, in order to infer (17), I also used that
the map $e_{r}\circ G_{n}$ is injective outside a $\mu_{0}$-null set, as a
consequence of Proposition 5.4.
An inductive argument allows to conclude that for every $n\in\mathbb{N}$ it
holds

$\operatorname{Ent}\big{(}(e_{r})_{\\#}[(G_{n})_{\\#}(f\mu_{0})]\big{)}\leq(1-r)\operatorname{Ent}(f\mu_{0})+r\operatorname{Ent}(T_{\\#}(f\mu_{0}))-\frac{K}{2}r(1-r)W_{2}^{2}(f\mu_{0},T_{\\#}(f\mu_{0})),$

for every $r\in\big{\\{}\frac{k}{2^{n}},k=0,\dots,2^{n}\big{\\}}$, and every
bounded measurable function $f$ with $\int f\,\mathrm{d}\mu_{0}=1$. In fact,
this is completely obvious for $n=0$, and assuming it true for an $n$ it is
possible to easily deduce it for $n+1$, using (17).
It is now easy to notice that the first property in the definition of
$G_{n+1}$ given $G_{n}$, ensures the existence of a measurable map
$G:X\to\operatorname{Geo}(X)$, such that $G_{n}\to G$ uniformly. Furthermore
it is obvious that

$(e_{r})_{\\#}[(G_{n})_{\\#}(f\mu_{0})]=(e_{r})_{\\#}[G_{\\#}(f\mu_{0})]$

for every $r\in\big{\\{}\frac{k}{2^{n}},k=0,\dots,2^{n}\big{\\}}$, and every
bounded measurable function $f$ with $\int f\,\mathrm{d}\mu_{0}=1$.
Consequently it holds that

$\operatorname{Ent}\big{(}(e_{r})_{\\#}[G_{\\#}(f\mu_{0})]\big{)}\leq(1-r)\operatorname{Ent}((f\mu_{0}))+r\operatorname{Ent}(T_{\\#}(f\mu_{0}))-\frac{K}{2}r(1-r)W_{2}^{2}(f\mu_{0},T_{\\#}(f\mu_{0})),$

for every dyadic time $r$ and every suitable function $f$. Then the lower
semicontinuity of $\operatorname{Ent}$ allows to infer that the $K$-convexity
inequality of the entropy is satisfied along $G_{\\#}(f\mu_{0})$ for every
suitable function $f$. Finally I can conclude by observing that every optimal
geodesic plan of the type $F\cdot G_{\\#}\mu_{0}$, for a measurable function
$F:\operatorname{Geo}(X)\to\mathbb{R}^{+}$ with $\int
F\,\mathrm{d}(G_{\\#}\mu_{0})=1$, can be written as $G_{\\#}(f\mu_{0})$ for a
suitable measurable $f$ with $\int f\,\mathrm{d}\mu_{0}=1$, since $G$ is
clearly injective. ∎

######  Proposition 7.5.

For every $k$ and $K$ the metric measure space
$(X_{k,0},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,0})$ is not a strict
$\mathsf{CD}(0,\infty)$ space.

$\mu_{0}$$\mu_{1}$$\mu_{s}$ Figure 3: A representation of the geodesic
$(\mu_{t})_{t\in[0,1]}$.

######  Proof.

In this proof I denote with $\mathfrak{m}$ the measure $\mathfrak{m}_{k,K,0}$,
in order to simplify the notation.
For every $t\in[0,1]$ define the measure

$\mu_{t}=\frac{1}{\mathfrak{m}\big{(}[-\frac{1}{2}+t,-\frac{1}{4}+t]\times\mathbb{R}\big{)}}\cdot{\left.\kern-1.2pt\mathfrak{m}\right|_{[-\frac{1}{2}+t,-\frac{1}{4}+t]\times\mathbb{R}}}=\frac{4}{C_{K}}\cdot{\left.\kern-1.2pt\mathfrak{m}\right|_{[-\frac{1}{2}+t,-\frac{1}{4}+t]\times\mathbb{R}}},$

see Figure 3 in order to visualize it. It is easy to realize that
$(\mu_{t})_{t\in[0,1]}$ is the unique geodesic connecting $\mu_{0}$ and
$\mu_{1}$, along which the entropy functional is convex. Moreover let
$\eta\in\mathscr{P}\big{(}C([0,1]),X\big{)}$ such that
$(e_{t})_{\\#}\eta=\mu_{t}$ for every $t\in[0,1]$, I am going to prove that
the entropy functional is not convex along $F\eta$, for a suitable bounded
measurable function $F:\operatorname{Geo}(X)\to\mathbb{R}^{+}$ with $\int
F\,\mathrm{d}\eta=1$. Before going on, let me point out that every Wasserstein
geodesic in $\mathscr{P}\big{(}C([0,1]),X\big{)}$ which connects a measure on
$L$ to a measure on $C$ (and thus $\eta$ in particular), consists only of
“horizontal” transports. Therefore the only useful coordinate, in order to
evaluate the distance $\mathsf{d}_{\infty}$, will be the $x$ coordinate. As a
consequence every such optimal geodesic plan (and $\eta$ in particular) will
only depend on the $x$ coordinate. Some of the considerations I will do in
this proof actually follows from this observation.
Now, define the set

$A:=\\{(x,y)\in\mathbb{R}^{2}:f_{k,0}(x)>0\text{ and }0\leq 2y\leq
f_{k,0}(x)\\}\subset X_{k,0},$

and the quantity

$C^{\prime}_{K}=\int_{0}^{\frac{1}{2}}e^{-Ky^{2}}\,\mathrm{d}y.$

Then fix a time $\bar{t}$ such that $\mu_{\bar{t}}$ is concentrated in $C$,
consider the map $\tilde{F}:\operatorname{Geo}(X)\to\mathbb{R}^{+}$ defined as
$\tilde{F}:=\frac{C_{K}}{C^{\prime}_{K}}\cdot\chi_{A}\circ e_{\bar{t}}$ and
call $\tilde{\mu}_{1}=(e_{1})_{\\#}(\tilde{F}\eta)$. Notice that
$\int\tilde{F}\,\mathrm{d}\eta=1$, thus $\tilde{\mu}_{1}$ is a probability
measure and it is absolutely continuous with respect to $\mathfrak{m}$, with
density $\tilde{\rho}_{1}$ bounded above by $\frac{4}{C^{\prime}_{K}}$, as a
consequence
$\mathfrak{m}(\\{\tilde{\rho}_{1}>0\\})\geq\frac{C^{\prime}_{K}}{4}$. Now,
suppose that
$\mathfrak{m}(\\{\tilde{\rho}_{1}>0\\})=\frac{C^{\prime}_{K}}{4}$, then
$\tilde{\rho}_{1}\equiv\frac{4}{C^{\prime}_{K}}$ on $\\{\tilde{\rho}_{1}>0\\}$
and therefore

$\operatorname{Ent}(\tilde{\mu}_{1})=\log\left(\frac{4}{C^{\prime}_{K}}\right)=\operatorname{Ent}\big{(}(e_{\bar{t}})_{\\#}(\tilde{F}\eta)\big{)}.$

On the other hand

$\operatorname{Ent}\big{(}(e_{0})_{\\#}(\tilde{F}\eta)\big{)}=\operatorname{Ent}(\mu_{0})=\log\left(\frac{4}{C_{K}}\right)<\log\left(\frac{4}{C^{\prime}_{K}}\right),$

and consequently the entropy functional is not convex along $\tilde{F}\eta$.
Otherwise, suppose that
$\mathfrak{m}(\\{\tilde{\rho}_{1}>0\\})>\frac{C^{\prime}_{K}}{4}$, call
$S:=\\{\tilde{\rho}_{1}>0\\}$ and define the set
$S_{x}:=\\{(x^{\prime},y^{\prime})\in S:x^{\prime}=x\\}$, for every
$x\in[-1,1]$. Now consider
$\mathfrak{m}_{1}:=(\mathtt{p}_{1})_{\\#}\mathfrak{m}$ and denote by
$(\mathfrak{m}_{x})_{x\in[-1,1]}\subset\mathscr{P}(\mathbb{R})$ the
disintegration of $\mathfrak{m}$ with respect to the projection map
$\mathtt{p}_{1}$. Notice that, since $\eta$ depends only on the $x$
coordinate, then
$(\mathtt{p}_{1})_{\\#}\tilde{\mu}_{1}=\frac{4}{C_{K}}\cdot{\left.\kern-1.2pt\mathfrak{m}_{1}\right|_{\left[\frac{1}{2},\frac{3}{4}\right]}}$.
Moreover, since the density $\tilde{\rho}_{1}$ is bounded above by
$\frac{4}{C^{\prime}_{K}}$, it holds that
$\mathfrak{m}_{x}(S_{x})\geq\frac{C^{\prime}_{K}}{C_{K}}$, for
$\mathfrak{m}_{1}$-almost every $x\in\left[\frac{1}{2},\frac{3}{4}\right]$.
Furthermore, the assumption on $S$, that is
$\mathfrak{m}(S)>\frac{C^{\prime}_{K}}{4}$, ensures that
$\mathfrak{m}_{x}(S_{x})>\frac{C^{\prime}_{K}}{C_{K}}$ for a
${\left.\kern-1.2pt\mathfrak{m}_{1}\right|_{\left[\frac{1}{2},\frac{3}{4}\right]}}$-positive
set of $x$, therefore

$\int_{\frac{1}{2}}^{\frac{3}{4}}\log\left(\mathfrak{m}_{x}(S_{x})\right)\,\mathrm{d}\mathfrak{m}_{1}(x)>\frac{C_{K}}{4}\log\left(\frac{C^{\prime}_{K}}{C_{K}}\right).$
(18)

On the other hand, for every positive constant $c>0$ define the set
$S^{c}:=\\{\tilde{\rho}_{1}>c\\}$ and call
$S^{c}_{x}:=\\{(x^{\prime},y^{\prime})\in S^{c}:x^{\prime}=x\\}$ for every
$x\in[-1,1]$. Notice that, for every sufficiently small constant $c$, since
$\tilde{\rho}_{1}$ is bounded and
$(\mathtt{p}_{1})_{\\#}\tilde{\mu}_{1}=\frac{4}{C_{K}}\cdot{\left.\kern-1.2pt\mathfrak{m}_{1}\right|_{\left[\frac{1}{2},\frac{3}{4}\right]}}$,
the quantity $\mathfrak{m}_{x}(S_{x})$ is uniformly bounded from below for
$\mathfrak{m}_{1}$-almost every $x\in\left[\frac{1}{2},\frac{3}{4}\right]$.
Consequently, it is possible to apply the monotone convergence theorem and
deduce that there exists a constant $\bar{c}>0$ such that

$\int_{\frac{1}{2}}^{\frac{3}{4}}\log\left(\mathfrak{m}_{x}(S^{\bar{c}}_{x})\right)\,\mathrm{d}\mathfrak{m}_{1}(x)>\frac{C_{K}}{4}\log\left(\frac{C^{\prime}_{K}}{C_{K}}\right).$

Now, define the measurable map $F:\operatorname{Geo}(X)\to\mathbb{R}^{+}$

$F:=\frac{4}{C_{K}}\tilde{F}\cdot\left(\frac{1}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\frac{\chi_{S^{\bar{c}}}(x,y)}{\tilde{\rho}_{1}(x,y)}\circ
e_{1}\right).$

I have already noticed that the quantity $\mathfrak{m}_{x}(S^{\bar{c}}_{x})$
is uniformly bounded from below for $\mathfrak{m}_{1}$-almost every
$x\in\left[\frac{1}{2},\frac{3}{4}\right]$, thus $F$ is well defined and
bounded, moreover it holds that

$\displaystyle\int F\,\mathrm{d}\eta$
$\displaystyle=\int\frac{4}{C_{K}}\cdot\left(\frac{1}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\frac{\chi_{S^{\bar{c}}}(x,y)}{\tilde{\rho}_{1}(x,y)}\circ
e_{1}\right)\,\mathrm{d}\tilde{F}\eta=\frac{4}{C_{K}}\int\frac{1}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\frac{\chi_{S^{\bar{c}}}(x,y)}{\tilde{\rho}_{1}(x,y)}\,\mathrm{d}\tilde{\mu}_{1}(x,y)$
$\displaystyle=\frac{4}{C_{K}}\int\frac{\chi_{S^{\bar{c}}}(x,y)}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\,\mathrm{d}\mathfrak{m}(x,y)=\frac{4}{C_{K}}\int_{\frac{1}{2}}^{\frac{3}{4}}\int\frac{\chi_{S^{\bar{c}}}(x,y)}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\,\mathrm{d}\mathfrak{m}_{x}(y)\,\mathrm{d}\mathfrak{m}_{1}(x)=\frac{4}{C_{K}}\int_{\frac{1}{2}}^{\frac{3}{4}}\,\mathrm{d}\mathfrak{m}_{1}(x)=1.$

In particular, observe that a computation similar to this last one shows that

$(e_{1})_{\\#}(F\eta)=\frac{4}{C_{K}}\frac{\chi_{S^{\bar{c}}}(x,y)}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\cdot\mathfrak{m},$

thus it is possible to estimate its entropy:

$\displaystyle\operatorname{Ent}\big{(}(e_{1})_{\\#}(F\eta)\big{)}$
$\displaystyle=\int\frac{4}{C_{K}}\frac{\chi_{S^{\bar{c}}}(x,y)}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\log\left(\frac{4}{C_{K}}\frac{\chi_{S^{\bar{c}}}(x,y)}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\right)\,\mathrm{d}\mathfrak{m}$
$\displaystyle=\int_{\frac{1}{2}}^{\frac{3}{4}}\int\frac{4}{C_{K}}\frac{\chi_{S^{\bar{c}}}(x,y)}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\log\left(\frac{4}{C_{K}}\frac{\chi_{S^{\bar{c}}}(x,y)}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\right)\,\mathrm{d}\mathfrak{m}_{x}(y)\,\mathrm{d}\mathfrak{m}_{1}(x)$
$\displaystyle=\int_{\frac{1}{2}}^{\frac{3}{4}}\int_{S^{\bar{c}}_{x}}\frac{4}{C_{K}}\frac{1}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\log\left(\frac{4}{C_{K}}\frac{1}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\right)\,\mathrm{d}\mathfrak{m}_{x}(y)\,\mathrm{d}\mathfrak{m}_{1}(x)$
$\displaystyle=\frac{4}{C_{K}}\int_{\frac{1}{2}}^{\frac{3}{4}}\log\left(\frac{4}{C_{K}}\frac{1}{\mathfrak{m}_{x}(S^{\bar{c}}_{x})}\right)\,\mathrm{d}\mathfrak{m}_{1}(x)$
$\displaystyle=\log\left(\frac{4}{C_{K}}\right)+\frac{4}{C_{K}}\int_{\frac{1}{2}}^{\frac{3}{4}}-\log\left(\mathfrak{m}_{x}(S^{\bar{c}}_{x})\right)\,\mathrm{d}\mathfrak{m}_{1}(x)<\log\left(\frac{4}{C^{\prime}_{K}}\right),$

where the last inequality follows from (18). On the other hand
$(e_{\bar{t}})_{\\#}(F\eta)\ll(e_{\bar{t}})_{\\#}(\tilde{F}\eta)$ and
consequently Jensen’s inequality ensures that

$\operatorname{Ent}\big{(}(e_{\bar{t}})_{\\#}(F\eta)\big{)}\geq\log\left(\frac{4}{C^{\prime}_{K}}\right),$

Furthermore, it is easy to realize that

$(\mathtt{p}_{1})_{\\#}\big{[}(e_{1})_{\\#}(F\eta)\big{]}=\frac{4}{C_{K}}\cdot{\left.\kern-1.2pt\mathfrak{m}_{1}\right|_{\left[\frac{1}{2},\frac{3}{4}\right]}}$

and thus, as before, I have

$\operatorname{Ent}\big{(}(e_{0})_{\\#}(F\eta)\big{)}=\operatorname{Ent}(\mu_{0})=\log\left(\frac{4}{C_{K}}\right)<\log\left(\frac{4}{C^{\prime}_{K}}\right).$

Putting together this last three inequalities it is easy to realize that the
entropy functional is not convex along $F\eta$. ∎

Notice that this last result shows in particular that the strict CD condition
and the weak one are two actually different notions. Moreover, the combination
of Proposition 7.4 and 7.5 obviously yields Proposition 7.3, according to the
proof of Theorem 7.1. On the other hand, observe that this work does not allow
to disprove the stability of the very strict CD condition. In fact the proof
of Proposition 7.4 heavily relies on an approximation argument, thus it is
impossible to modify it, in order to prove the very strict CD condition for
the spaces
$(X_{k,\varepsilon},\mathsf{d}_{\infty},\mathfrak{m}_{k,K,\varepsilon})$.
However, in my opinion, this example leaves little hope for the very strict CD
condition to be stable.

Aknowlegments : This article contains part of the work I did for my master
thesis, that was supervised by Luigi Ambrosio and Karl-Theodor Sturm.

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