# Hölder and Sobolev regularity of optimal transportation potentials with
rough measures

Pierre-Emmanuel Jabin  and  Antoine Mellet

###### Abstract.

We consider a Kantorovich potential associated to an optimal transportation
problem between measures that are not necessarily absolutely continuous with
respect to the Lebesgue measure, but are comparable to the Lebesgue measure
when restricted to balls with radius greater than some $\delta>0$. Our main
results extend the classical regularity theory of optimal transportation to
this framework. In particular, we establish both Hölder and Sobolev regularity
results for Kantorovich potentials up to some critical length scale depending
on $\delta$. Our assumptions are very natural in the context of the numerical
computation of optimal maps, which often involves approximating by sums of
Dirac masses some measures that are absolutely continuous with densities
bounded away from zero and infinity on their supports.

<EMAIL_ADDRESS>Pennsylvania State University, Department of Mathematics and
Huck Institutes, State College, PA 16802. Partially supported by NSF DMS Grant
161453, 1908739, 2049020 and NSF Grant RNMS (Ki-Net) 1107444.

<EMAIL_ADDRESS>University of Maryland, Department of Mathematics and CSCAMM,
College Park, MD 20742 USA. Partially supported by NSF Grant NSF Grant
DMS-2009236 and DMS-2307342.

##  1\. Introduction

###  1.1. Optimal transportation with general measure

The theory of optimal transportation, first introduced by Monge in the late
1800 (and later generalized by Kantorovich), has been extensively studied from
a theoretical point of view in the past few decades. It has also found
numerous applications to problems both theoretical and applied. Following the
work of Brenier [5], a major topic of mathematical research has been the study
of the regularity of the optimal transportation map between absolutely
continuous measures.

More recently, partly motivated by numerical applications, there has been a
lot of interest in optimal transportation problems between general measures.
Indeed, numerical schemes often involve measures that are sums of Dirac
masses. Solutions of optimal transportation problems between such singular
measures cannot be expected to be smooth in general. However it is natural to
expect these solutions to have some nice properties when the general measures,
possibly discrete, are close (in some sense) to some nice measures (see
Assumption 1). For example, if some measures are obtained by discretizing some
smooth densities at a scale $\delta$, they “look” smooth when seen only at
scales larger than $\delta$. One can then ask whether the solution of the
corresponding optimal transportation problem also looks smooth when seen at a
correspondingly large enough scale (whose relation to $\delta$ needs to be
identified).

This article introduces a general strategy to extend the classical regularity
theory for optimal transportation problem between continuous densities to such
a discrete setting. The main take away of our analysis is that the solutions
of this discrete optimal transportation problem have the usual Sololev
regularity when mollified at a scale $\delta^{1/2}$ (see Theorems 2.5 and 2.6)
- this scale is optimal in the sense that it is consistent with known
convergence rates of numerical schemes (see Section 1.3).

Let us now briefly recall the setting of optimal transportation with general
measures. Throughout the paper, we will focus exclusively on the classical
quadratic cost (we refer for example to [36] for a detailed presentation of
the topic in a much more general framework):

Given two convex bounded open sets $\Omega$ and $\mathcal{O}$ in
$\mathbb{R}^{n}$ ($n\geq 1$), we consider two Borel probability measures
$\mu$, $\nu$ such that $\mathrm{spt}\,\mu\subset\overline{\Omega}$ and
$\mathrm{spt}\,\nu\subset\overline{\mathcal{O}}$. In Kantorovich’s setting,
admissible transference plans are probability measures on
$\mathbb{R}^{n}\times\mathbb{R}^{n}$ with marginals $\mu$ and $\nu$. We denote
by $\Pi(\mu,\nu)$ the set of transference plans:

$\Pi(\mu,\nu):=\\{\pi\in\mathcal{P}(\mathbb{R}^{n}\times\mathbb{R}^{n})\,;\,\pi(A\times\mathbb{R}^{n})=\mu(A),\;\pi(\mathbb{R}^{n}\times
B)=\nu(B)\\}$

(where $A$ and $B$ are any measurable sets in $\mathbb{R}^{n}$). The classical
optimal transportation problem (with quadratic cost), is concerned with the
minimization problem

(1)
$C(\mu,\nu)=\inf_{\pi\in\Pi(\mu,\nu)}\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}|x-z|^{2}d\pi(x,z).$

The existence of a minimizer for (1) is given by the following classical
result (see for instance [35, Theorem 2.12]):

######  Theorem 1.1.

If $C(\mu,\nu)<\infty$, then there exists an optimal transference plan
$\pi\in\Pi(\mu,\nu)$ (i.e. a minimizer of (1)). Furthermore, any
$\pi\in\Pi(\mu,\nu)$ is optimal if and only if there is a convex, lower semi-
continuous, function $\psi:\mathbb{R}^{n}\to(-\infty,+\infty]$ such that $\pi$
is concentrated in the graph of $\partial\psi$, that is

(2)
$\mathrm{spt}\,(\pi)\subset\mathrm{Graph}(\partial\psi):=\left\\{(x,z)\in\mathbb{R}^{n}\times\mathbb{R}^{n}\;|\;z\in\partial\psi(x)\right\\}\mbox{.}$

Recall that for a convex function $\psi$, the subdifferential of $\psi$ at $x$
is given by

$\partial\psi(x):=\left\\{z\in\mathbb{R}^{n}\,;\,\psi(y)\geq\psi(x)+z\cdot(y-x)\mbox{
for all }y\in\mathbb{R}^{n}\right\\}.$

We will also use the notation $\partial\psi(A)=\cup_{x\in A}\partial\psi(x)$.

In this general framework, neither the optimal plan $\pi$ nor the Kantorovich
potential $\psi$ are unique. We also note that without changing the values of
$\psi$ in $\mathrm{spt}\,\mu$, we can assume (see Appendix C) that
$\psi(x)=\sup_{z\in\mathcal{O}}(x\cdot z-\phi(z))$ for some function
$\phi:\mathcal{O}\to(-\infty,+\infty]$. Since $\mathcal{O}$ is convex, we can
thus assume that

(3) $\partial\psi(\mathbb{R}^{n})\subset\overline{\mathcal{O}},$

and in fact we can even assume that $\partial\psi(\mathbb{R}^{n})$ is a subset
of the closed convex envelop of $\mathrm{spt}\,\nu$. Since $\mathcal{O}$ is
bounded, this implies that $\psi$ is Lipschitz continuous in $\mathbb{R}^{n}$.

There is a vast literature devoted to the properties of Kantorovich potentials
when the measures $\mu$ and $\nu$ are of the form

(4) $\mu=f(x)\,dx\quad\mbox{ and }\quad\nu=g(y)\,dy$

for some measurable functions $f$ and $g$ satisfying

(5) $\lambda\leq f(x)\leq\Lambda\qquad\forall x\in\Omega,\quad\lambda\leq
g(y)\leq\Lambda\qquad\forall y\in\mathcal{O}.$

Before stating our assumptions and main results, we will briefly review (in a
completely non-exhaustive way) some key aspects of the classical regularity
theory in that framework.

###  1.2. Regularity theory under assumptions (4)-(5)

Under assumptions (4)-(5), the Kantorovich potential is uniquely defined (up
to a constant) on $\Omega$. In fact, we have:

######  Theorem 1.2 (Brenier’s theorem [6]).

If $\mu$ does no give mass to small sets, then there is a unique optimal
transference plan $\pi$, given by $d\pi(x,y)=d\mu(x)\delta(y=\nabla\psi(x)$
where $\nabla\psi$ is the unique ($\mu$-a.e.) gradient of a convex function
satisfying $\nabla\psi\\#\mu=\nu$. Furthermore, we have

$\mathrm{spt}\,(\nu)=\overline{\nabla\psi(\mathrm{spt}\,(\mu)))}.$

In this framework, the so-called Brenier map $T=\nabla\psi$ is the unique
solution of the Monge problem and when the measures $\mu$ and $\nu$ satisfy
(4)-(5), the function $\psi$ solves the generalized Monge-Ampère equation

(6) $\det(D^{2}\psi(x))=h(x)\qquad\mbox{ with
}h(x):=\frac{\mu(x)}{\nu(\nabla\psi(x))}\qquad\mbox{ in }\Omega$

with second type boundary conditions (3). In particular, $\psi$ satisfies

(7) $\frac{\lambda}{\Lambda}\chi_{\Omega}\leq\det
D^{2}\psi\leq\frac{\Lambda}{\lambda}\chi_{\Omega}$

in a weak sense (the Brenier sense) - see for instance [13, 16]. When
$\mathcal{O}$ is convex, Caffarelli proved in [13] that $\psi$ is a strictly
convex solution of (7) in the following Alexandrov sense:

(8)
$\frac{\lambda}{\Lambda}|A\cap\Omega|\leq|\partial\psi(A)|\leq\frac{\Lambda}{\lambda}|A\cap\Omega|\qquad\mbox{for
any Borel set $A\subset\mathbb{R}^{n}$.}$

The regularity theory for the classical Monge-Ampère equation, developed by
Caffarelli [9, 10, 12, 13, 14], for strictly convex solutions of (8) then
implies that $\psi$ is $C^{1,\alpha}_{loc}$ (see also [21]). Further
regularity ($C^{2,\alpha}_{loc}(\Omega)$) for the solution of Monge-Ampère
equation (6) with Dirichlet boundary conditions where proved for example by
Caffarelli [8] when the right hand side $h$ is $C^{\alpha}_{loc}(\Omega)$.

Since further Hölder regularity is not expected under condition (5) alone, it
is natural to ask about the Sobolev regularity of $\psi$. In the framework of
Monge-Ampère equation (6) (with Dirichlet boundary condition), Caffarelli [8]
proved that the solution is in $W^{2,p}$ provided the right-hand side
satisfies $\|h-1\|_{L^{\infty}}\leq\eta$ for some small enough $\eta$
(depending on $p$). When $f$ and $g$ only satisfy condition (5), De Philippis
and Figalli [17] proved that $D^{2}\psi\in L\log L_{loc}$. This result was
improved by De Philippis, Figalli and Savin [15] who proved that $\psi\in
W^{2,1+\varepsilon}_{loc}$ for some small $\varepsilon>0$.

###  1.3. General measures and formal presentation of our results

In this paper, we do not make either assumption (4) or (5), instead requiring
only that $\mu$ and $\nu$ are comparable to the Lebesgue measure when
restricted to balls of radius greater than or equal to some fixed $\delta>0$
(see Assumption 1). A simple and very important example we have in mind is
that of measures which are sums of Dirac masses approximating some absolutely
continuous measures satisfying (4)-(5). Our goal is then to prove the
equivalent of the $C^{1,\alpha}_{loc}$, $W^{2,1}_{loc}$ and
$W^{2,1+\varepsilon}_{loc}$ regularity results mentioned above, in our
framework. Since we consider measures that may not be absolutely continuous
with respect to the Lebesgue measure, the Kantorovich potential $\psi$ (which
still exists but may not be unique) does not solve the Monge-Ampère equation
(either (7) or (8)) and none of the results discussed above apply. In fact,
Kantorovich potentials are not expected to be either strictly convex or
$C^{1}$: Even in the semi-discrete case (when $\nu$ is absolutely continuous
but $\mu$ is a sum of Dirac masses), $\psi$ is a piecewise affine function
(see [3] and Appendix C). In particular $\psi$ has both ‘flat parts’ and
‘corners’. To our knowledge, few quantitative estimates on the convex function
$\psi$ are known in this setting.

This discrete framework with $\mu$, $\nu$ sums of Dirac masses is important
for numerical computations of optimal maps between regular measures, as
described in Appendix A. There exists an extensive literature investigating
numerical schemes for optimal transportation. This is a challenging question,
both from the point of view of the strongly non-linear Monge-Ampère equation
or when starting from the Monge-Kantorovich formulation. In the later case for
example, it would be natural to discretize the starting and target measures by
sums of Dirac masses (which fits well with our framework) since that formally
reduces optimal transportation to a linear programming problem. But this
linear programming behave quadratically in the number of points in the
discretization, leading to highly non-trivial computational challenges. We
briefly refer to [1, 2], [28], [31, 32] or [33] for examples of various
numerical schemes. Several recent works have focused on proving convergence
rates for these numerical schemes: see for instance [3] for the semi-discrete
case and [29] for the fully discrete case, together with [27, 30].

Such convergence rates are not the goal of this paper. Instead, we develop a
regularity theory for the potential $\psi$, "up to a length scale depending on
$\delta$", which parallels the classical regularity theory developed under
assumptions (4)-(5). To roughly summarize the main results stated in the next
section, if we consider the optimal transportation problem corresponding to
some potential $\psi$ between measures $\mu$ and $\nu$ that are discretized at
some scale $\delta$, then we will prove that $\bar{K}_{r}\star\Delta\psi$
belongs to $L^{p}$ for some $p>1$ where $\bar{K}_{r}$ is a mollifying kernel
at scale $r$ with $\delta^{1/2}\lesssim r$. We expect that the precise uniform
regularity results that we derive here will prove useful for the analysis of
complex numerical schemes; and in particular multi-scale schemes.

We want to emphasize that an important aspect of our results is to obtain the
regularity of $\bar{K}_{r}\star\Delta\psi$ for the conjectured critical scale
$r\sim\delta^{1/2}$. Indeed, this rate is consistent with the known rates of
convergence of numerical schemes: If $\mu_{\delta}$ and $\nu_{\delta}$ are
discretizations at scale $\delta$ of some nice measures $\mu_{0}$ and
$\nu_{0}$, then the corresponding Kantorovich potentials $\psi_{\delta}$ and
$\psi_{0}$ are expected to satisfy $\psi_{\delta}-\psi_{0}=O(\delta^{1/2})$ in
some appropriate norms - see [29] where weighted $L^{2}$ error estimates on
the associated transport maps are derived (see also [3] in the semi-discrete
case). Using the regularity of $\psi_{0}$, it is possible, via some
interpolation, to derive some $L^{p}$ bounds on
$\bar{K}_{r}\star\Delta\psi_{\delta}$. However such an approach will lead to
regularity results at scale $r\geq\delta^{\alpha}$ with $\alpha<1/2$. In
addition, we note that the convergence results of [3, 29] requires some
$C^{2}$ regularity assumptions for the limiting potential $\psi_{0}$ while our
result is completely independent of the properties of $\psi_{0}$ (in fact it
does not even require the existence of $\psi_{0}$).

The results of the present paper should also be compared with the previous
work of the authors (with M. Molina) in [26]: Under similar assumptions on the
measures $\mu$ and $\nu$, in dimension $2$, the function $\psi$ is proved to
be strictly convex up to some discrete scale and approximately $C^{1}$ again
up to some discrete scale. The $C^{1,\alpha}$ and $W^{2,p}$ regularity results
proved here (in any dimension) are thus a significant improvement of these
results. On the other hand, we note that the results in [26] are local and do
not require any control on $\psi$ near the boundary of $\Omega$, while in the
present article we require some sort of boundary conditions (see a precise
discussion in the next section). This is in complete accord with what is known
for solutions of Monge-Ampère equation with a right-hand side bounded from
below and from above: strict convexity holds locally in dimension $2$
independently of the boundary condition but in higher dimensions, classical
counter-examples show that boundary conditions are necessary.

Finally, we mention the question of partial regularity to optimal
transportation between $C^{0,\alpha}$ densities which shows that the map
$T=\nabla\psi$ is $C^{1,\alpha}$ up to removing sets of measure $0$; see [18,
20] and [23] for a variational approach. This suggests that some notion of
approximate $C^{2,\alpha}$ estimate on $\psi$ (and not only $C^{1,\alpha}$ as
we prove here) may hold, though with stronger assumptions on the measures
$\mu$ and $\nu$ than what we are making here.

##  2\. Main assumptions and main results

###  2.1. The measures $\mu$ and $\nu$

In this paper, we assume that the measures $\mu$ and $\nu$ are comparable to
the Lebesgue measure up to a certain scale. Our main assumption can be written
as follows:

######  Assumption 1.

There exist $\lambda$, $\Lambda>0$ such that

(9) $\lambda|B_{r}|\leq\mu(B_{r})\leq\Lambda|B_{r}|,\quad\mbox{for all ball
$B_{r}\subset\Omega$ with $r\geq\delta$}$

and

(10) $\lambda|B_{r}|\leq\nu(B_{r})\leq\Lambda|B_{r}|\quad\mbox{for all ball
$B_{r}\subset\mathcal{O}$ with $r\geq\delta$.}$

We note that (9) is equivalent to assuming that the measure convolution
$\mu_{\delta}=\mu\star\frac{1}{|B_{\delta}|}\chi_{B_{\delta}}$ is absolutely
continuous with respect to the Lebesgue measure and has density $f_{\delta}$
satisfying

$\lambda\leq f_{\delta}\leq\Lambda\quad\mbox{ a.e. in
}\Omega_{\delta}=\\{x\in\Omega\,;\,d(x,\partial\Omega)>\delta\\}.$

We also point out that the use of balls in (9)-(10) is not crucial. Indeed, up
to multiplying $\delta$ by a constant depending only on the dimension, we
could formulate our assumption with ellipsoids whose smallest radius are
greater than $\delta$ or even with general convex sets with inner radius
greater than $\delta$ (see Proposition B.1)

Assumption 1 is natural in situation in which the measures $\mu$ and $\nu$ are
obtained by approximating measures satisfying (4)-(5). This construction,
which plays a key role in the numerical computation of optimal maps (see [3,
29]) is recalled in Appendix A.

We already mentioned that in this setting, it does not make sense to write the
Monge-Ampère equation and that many of the classical tools (e.g. comparison
principle) are thus no longer available to prove our result. Another important
observation, which is less fundamental and more technical, is the fact that
the renormalization of convex set provided by John’s Lemma 3.1 is extensively
used in the literature to prove regularity results. Unfortunately, Assumption
1 is not invariant under affine transformations (since the image of a ball
could be a very elongated ellipsoid). We will thus need to largely forgo the
use of such renormalization in this paper.

###  2.2. The potential $\psi$

Throughout the paper, $\psi:\mathbb{R}^{n}\to\mathbb{R}$ is a convex lower
semi-continuous function associated to an optimal transference plan $\pi$ as
in Theorem 1.1 and satisfying (3). In particular, $\psi$ satisfies

(11) $\mu(A)\leq\nu(\partial\psi(A))\quad\mbox{ for all Borel sets
$A\subset\Omega$}$

where $\partial\psi(A)=\cup_{x\in A}\partial\psi(x)$. Indeed, we can write

$\mu(A)=\int_{A}\int_{\mathbb{R}^{n}}d\pi=\int_{A}\int_{\partial\psi(A)}d\pi\leq\int_{\mathbb{R}^{n}}\int_{\partial\psi(A)}d\pi=\nu(\partial\psi(A)).$

This inequality might be strict in general but that we also have the
inequality

(12) $\mu(\partial\psi^{*}(B))\geq\nu(B)\quad\mbox{ for all Borel sets
$B\subset\mathcal{O}$}$

where $\psi^{*}$ denotes the Legendre transform, defined by

$\psi^{*}(z)=\sup_{x\in\mathbb{R}^{n}}\big{(}x\cdot z-\psi(x)\big{)}\qquad
z\in\mathbb{R}^{n}.$

Indeed, we recall that $\psi^{*}$ is also convex and lower semi-continuous
(and (3) implies that $\psi^{*}(z)=+\infty$ when
$z\notin\overline{\mathcal{O}}$) and that we have the following equivalences:

(13) $x\cdot
y=\psi(x)+\psi^{*}(y)\;\Longleftrightarrow\;y\in\partial\psi(x)\;\Longleftrightarrow\;x\in\partial\psi^{*}(y).$

In particular (2) is equivalent to
$\mathrm{spt}\,(\pi)\subset\left\\{(x,z)\in\mathbb{R}^{n}\times\mathbb{R}^{n}\;|\;x\in\partial\psi^{*}(z)\right\\}$
and we can prove (12) just as we proved (11).

These two properties (11) and (12), together with the boundary condition (3),
are actually the only properties of $\psi$ that we will use to prove our
result.

The relation between these inequalities and Theorem 1.1 was made clear by a
result that we proved in [26] and which we recall here for the reader’s
convenience:

######  Theorem 2.1.

Assume $\mu,\;\nu$ are two probability measures on $\mathbb{R}^{n}$ such that
$C(\mu,\nu)<\infty$ and let $\psi$ be a proper lower semi-continuous convex
function satisfying (11) and (12). Then there exists an optimal transference
plan $\pi\in\Pi(\mu,\nu)$ (minimizer of (1)) such that
$\mathrm{supp}(\pi)\subset\mathrm{Graph}(\partial\psi)$.

###  2.3. Sections of convex functions

As mentioned above, a fundamental difference between our work and most
previous work that deals with absolutely continuous measure $\mu$ and $\nu$,
is the fact that we cannot use Monge-Ampère equation and the comparison
principle usually associated to elliptic equations and which plays a key role
in the work of Caffarelli as well as in the more recent partial regularity
theory of [37, 19, 20]. Note that a variational approach (relying on optimal
transportation arguments rather than using some barriers for Monge-Ampère
equation) to the partial regularity theory of [20] was recently developed in
[23, 22].

On the other hand, we will rely heavily on another tool that has proved
extremely useful in the study of Monge-Ampère equation: The sections of convex
functions. We recall that given $x_{0}\in\Omega$,
$p_{0}\in\partial\psi(x_{0})$ and $t\geq 0$, the section centered at $x_{0}$
with height $t$ is the convex set:

$S(x_{0},p_{0},t):=\\{x\in\mathbb{R}^{n}\,;\,\psi(x)\leq\psi(x_{0})+p_{0}\cdot(x-x_{0})+t\\}.$

Sections associated to solutions of Monge-Ampère equation have good properties
and play a key role in most regularity results, see for instance [11], [25],
[24, Chapter 3]. An important contribution of this paper, and a key aspect of
our proofs, will be to show that these classical properties still hold under
Assumption 1, provided $t\geq C_{0}\delta$ for some constant $C_{0}$ depending
only on the dimension $n$ and on the diameters of $\Omega$ and $\mathcal{O}$.
These properties are summarized in Proposition 3.5 and proved in Section 7. In
order to avoid the use of renormalization of convex sets (which, as noted
earlier, is not compatible with Assumption 1), we will rely on the notion of
polar body of a convex set to prove several of our results.

A final remark is that our regularity results will be interior regularity
results: they hold in a subset $\Omega^{\prime}\subset\subset\Omega$ such that
there exists $\rho$ such that

(14) $S(x,p,\rho)\subset\Omega\quad\forall x\in\Omega^{\prime}\quad\mbox{for
some }p\in\partial\psi(x).$

The existence of such a $\rho>0$ for a given subset $\Omega^{\prime}$ is
classical when the measures satisfy (4)-(5) (and $\mathcal{O}$ is convex), but
it is not obvious in more general framework. We will show in Appendix A that
when $\mu$ and $\nu$ are obtained as discrete approximation of regular
measures, then $\rho$ satisfying (14) exists provided $\delta$ is small enough
for any $\Omega^{\prime}\subset\subset\Omega$. Technically speaking (14) then
holds for all $p\in\partial\psi(x)$ though we do not need this stronger
version for our analysis.

The fact that some sort of condition is needed near the boundary of $\Omega$
is however well-understood. In the case $\delta=0$, the regularity of the
solutions to Monge-Ampère equation depends on the boundary conditions. For
example in dimension $3$ or more, local solutions to Monge-Ampère equation
with a right-hand side bounded from below and from above may fail to be
strictly convex (dimension $n=2$ is a special case for which we refer in
particular to the discussion in [26]). Because of our setting with discrete
measure, straightforward boundary conditions would not make much sense and are
replaced by (14) which is a natural extension: To violate (14), one would in
particular need to have very elongated sections for small $\rho$ that touch
the boundary of $\Omega$.

###  2.4. Main results

Our first result concerns the $C^{1,\alpha}_{loc}$ regularity of $\psi$ up to
some length scale depending on $\delta$. Forzani and Maldonado [21] proved
that the engulfing property of sections (see Proposition 3.5-(iii)) is
equivalent to the $C^{1,\alpha}$ regularity of the potential. Using similar
ideas, we will show the $C^{1,\alpha}$ regularity up to a scale depending on
$\delta$. In the theorem below, the constants $C_{0}$, $\theta$ and $\beta$
are the constants appearing in Proposition 3.5 and in (23):

######  Theorem 2.2 ($C^{1,\alpha}$ regularity).

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (3),
(11) and (12) with $\mu$ and $\nu$ satisfying (9) and (10) for some
$\delta>0$. Let $\Omega^{\prime}\subset\subset\Omega$ be such that there
exists $\rho>C_{0}\delta$ such that (14) holds. For all
$\Omega^{\prime\prime}\subset\subset\Omega^{\prime}$ there exists a constant
$C$ depending on
$\mathrm{dist}(\partial\Omega^{\prime},\partial\Omega^{\prime\prime})$,
$\rho$, $\partial\psi(\Omega)$, $\Omega$, $\lambda$, $\Lambda$ and $n$ such
that

$0\leq\psi(x)-\psi(y)-q\cdot(x-y)\leq\overline{C}|x-y|^{1+\frac{1}{\theta}}\qquad\forall
q\in\partial\psi(y)$

for all $x,y\in\Omega^{\prime\prime}$ satisfying

$|x-y|\geq C\delta^{1/\beta}.$

Next, we turn our attention to the Sobolev regularity of $\psi$. When
$\delta=0$ (that is when (4)-(5) holds), De Philippis and Figalli [17] proved
the following result:

######  Theorem 2.3 ([17]).

Let $\Omega\subset\mathbb{R}^{n}$ be bounded convex domain and let
$u:\overline{\Omega}\to\mathbb{R}$ be an Alexandrov solution of

(15) $\begin{cases}\det D^{2}u=h&\mbox{ in }\Omega\\\ u=0&\mbox{ on
}\partial\Omega\end{cases}$

with $0<\lambda\leq h(x)\leq\Lambda$ for all $x\in\Omega$. Then for any
$\Omega^{\prime}\subset\subset\Omega$ and $k\in\mathbb{N}$, there exists a
constant $C$ depending on $k$, $n$, $\lambda$, $\Lambda$, $\Omega$ and
$\Omega^{\prime}$ such that

(16) $\int_{\Omega^{\prime}}\|D^{2}u\|\log^{k}(2+\|D^{2}u\|)\,dx\leq C.$

The function $u$ in Theorem 2.3 satisfies

(17) $\lambda|B|\leq|\partial u(B)|\leq\Lambda|B|\qquad\mbox{ for all Borel
set $B\subset\Omega$.}$

These conditions are also satisfied by our function $\psi$ when Assumption 1
holds with $\delta=0$ if we assume that $|\partial\mathcal{O}|=0$ and that
$\psi$ is $C^{1}$ and strictly convex. Indeed, when $\mu$ and $\nu$ satisfy
(9)-(10) with $\delta=0$, then (11), (12) imply

$\lambda|A|\leq|\partial\psi(A)|\qquad\forall
A\subset\Omega,\qquad|B|\leq\Lambda|\partial\psi^{*}(B)|\qquad\forall
B\subset\mathcal{O}.$

Since $\partial\psi(A)\subset\overline{\mathcal{O}}$ for all $A\subset\Omega$
and if $|\partial\mathcal{O}|=0$, the second inequality gives

$|\partial\psi(A)|\leq\Lambda|\partial\psi^{*}(\partial\psi(A))|.$

When $\psi$ is strictly convex, we have $\partial\psi^{*}(\partial\psi(A))=A$
so we recover (17).

Conditions (9)-(10) (with $\delta=0$) are actually enough to carry out an
adaptation of the arguments of [17] and derive (16). In fact, we prove:

######  Theorem 2.4 ($W^{2,1}_{loc}$ regularity when $\delta=0$).

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (3),
(11), (12), with $\mu$ and $\nu$ satisfying (9) and (10) with $\delta=0$.
Assume moreover that $\Delta\psi\in L^{1}(\Omega)$ and let
$\Omega^{\prime}\subset\subset\Omega$ such that there exists $\rho>0$ such
that (14) holds. There exists a constant $C$ depending on $\Omega$,
$\Omega^{\prime}$, $\mathcal{O}$, $n$, $\lambda$, $\Lambda$ such that

(18) $\int_{\Omega^{\prime}}\Delta\psi\log(2+\Delta\psi)\,dx\leq C.$

Since $\psi$ is a convex function, we have
$\frac{1}{n}\Delta\psi\leq\|D^{2}\psi\|\leq\Delta\psi$ and so (18) is
equivalent to the result of [17]. We will nevertheless give a detailed proof
of Theorem 2.4. The overall arguments are the same as in [17], although some
minor details are different (for example we avoid the use of the maximal
function and renormalization of convex sets). Since Theorem 2.4 can also be
obtained by taking the limit $r\to 0$ in the next result, this proof is not
really necessary, but it allows the reader to see the main ideas of our proof
in a simpler framework.

The main goal of the paper is to derive Sobolev regularity results when
$\delta>0$. Since the potential $\psi$ might be piecewise affine, the measure
$\Delta\psi$ may be singular and we cannot expect (16) to hold in that case.
Instead, we will derive a similar bound for an approximation of the Laplacian
which measures the mass of the second derivative up to a scale $r$. More
precisely, we fix a radially symmetric kernel $K:\mathbb{R}^{n}\to[0,\infty]$,
supported in $B_{1}$ and such that

$\int K(x)\,dx=1,\qquad\int_{B_{1}}|x|^{2}K(x)\,dx>0.$

We then define for any $r>0$

$\Delta_{r}\psi(x):=\frac{m_{0}}{r^{2}}\int
K_{r}(y)[\psi(x+y)-\psi(x)]\,dy,\qquad
K_{r}(x)=\frac{1}{r^{n}}K\left(\frac{x}{r}\right)$

with $m_{0}=\frac{2}{\int_{\mathbb{R}^{n}}y_{1}^{2}K(y)\,dy}$. We have in
particular that $\Delta_{r}\psi(x)\geq 0$ as long as $\psi$ is convex and we
will see that $\lim_{r\to 0}\Delta_{r}\psi=\Delta\psi$ when $\psi$ is $C^{2}$.
We shall derive estimates on $\Delta_{r}\psi$ for any choice of $K$ satisfying
the assumptions listed above. Some natural choices of $K$ include
$K=\frac{1}{|B_{1}|}\chi_{B_{1}}$ for which we have

$\Delta_{r}\psi(x):=\frac{1}{r^{2}}\left[\fint_{B_{r}(x)}\psi(y)\,dy-\psi(x)\right],$

and $K=\frac{1}{\mathcal{H}^{n-1}(\partial B_{1})}\mathcal{H}^{n-1}|_{\partial
B_{1}}$ which leads to

$\Delta_{r}\psi(x):=\frac{1}{r^{2}}\left[\fint_{\partial
B_{r}(x)}\psi(y)\,dy-\psi(x)\right].$

With these notations, the second main result of this paper is the following
theorem:

######  Theorem 2.5 ($W^{2,1}_{loc}$ regularity when $\delta>0$).

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (11),
(12), with $\mu$ and $\nu$ satisfying (9) and (10) with $\delta>0$. Let
$\Omega^{\prime}\subset\subset\Omega$ be such that there exists $\rho>0$ such
that (14) holds. There exist a constant $C$ depending on $\Omega$, $\rho$,
$\mathcal{O}$, $n$, $\lambda$, $\Lambda$ such that

$\int_{\Omega^{\prime}}\Delta_{r}\psi\log(2+\Delta_{r}\psi)\,dx\leq
C\qquad\forall r>\delta^{\frac{1}{2}}.$

Since the constant $C$ does not depend on $r$, we see that when $\delta=0$, we
can take the limit $r\to 0$ and recover the result of Theorem 2.4. For that
reason, we will not be very careful with the a priori regularity of $\psi$ in
the proof of Theorem 2.4: A priori, $\Delta\psi$ is a Radon measure and some
of the computations in Section 5.2 should involve a decomposition of that
measure into a absolutely continuous part and a singular part. For simplicity,
we just assume that $\Delta\psi$ is locally integrable. This is not an issue
with the proof of Theorem 2.5 since $\Delta_{r}\psi$ is a Lipschitz function
for all $r>0$. Our proof of Theorem 2.5 thus also provides a proof of Theorem
2.4 which does not require any a priori knowledge on the structure of
$\Delta\psi$.

Following [15], we can then improve the argument to show:

######  Theorem 2.6 ($W^{2,1+\varepsilon}_{loc}$ regularity when $\delta>0$).

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (11),
(12), with $\mu$ and $\nu$ satisfying (9) and (10)) with $\delta>0$. Let
$\Omega^{\prime}\subset\subset\Omega$ be such that there exists $\rho>0$ such
that (14) holds. For any $x_{0}\in\Omega^{\prime}$ and $R$ such that
$B(x_{0},2R)\subset\Omega^{\prime}$, there exists $p>1$, $C$ (depending on
$R$, $\Omega$, $\rho$, $n$, $\lambda$, $\Lambda$, $L(\partial\psi(\Omega))$)
such that

$\int_{B(x_{0},R)}\,(\Delta_{r}\psi)^{p}\,dx\leq C\qquad\forall
r>\delta^{\frac{1}{2}}.$

We point out that we can also write $\Delta_{r}\psi(x)$ as
$\Delta(G_{r}\star\psi)$ or $\bar{K}_{r}\star\Delta\psi$. So that our results
can be interpreted as a priori Sobolev norm on a regularization
$G_{r}\star\psi$ of $\psi$, instead of a $L^{p}$ bound on the discrete
Laplacian $\Delta_{r}$ of $\psi$. A simple computation leads to the relation
$K_{r}=\delta+\Delta G_{r}$ and $G_{r}=\Gamma-\Gamma\star K_{r}$ (where
$\Gamma$ denotes the fundamental solution $-\Delta\Gamma=\delta$). For example
we can check that the kernel $K=\mathcal{H}^{n-1}|_{\partial B_{1}}$
corresponds to $G(x)=c_{n}\left(\frac{1}{|x|^{n-2}}-1\right)_{+}$ for some
normalization constant $c_{n}$.

###  2.5. What is new in our method

We emphasize that the results stated above rely on some key improvements with
respect to the existing strategies:

  * •

Precise properties of sections. A key point both for the $C^{1,\alpha}$ and
$W^{2,p}$ regularity in the continuous case is the so-called engulfing
property of sections $S(x,p,t)$ (see Proposition 3.5 later for a precise
formulation). This in particular allows to use Vitali’s covering lemma on
sections. Unfortunately, the engulfing property does not hold for sections of
a generic convex function and instead relied on the function $\psi$ solving
the Monge-Ampère equation. An important technical contribution of this paper
is thus to extend the engulfing property to our discrete setting, based on
optimal transportation instead of Monge-Ampère and up to the precise critical
discrete scale $t\sim\delta$.

  * •

“Fattening” small sets. The classical proof of $W^{2,p}$ regularity in the
case $\delta=0$ relies on the identification in each section $S(x,p,t)$ of a
critical set $\Sigma$ over which $\Delta\psi$ is large enough and with
$\Sigma=\partial\psi^{*}(V)$ for $|V|$ large enough as well. The proof of
Theorem 2.5 requires enlarging those sets $\Sigma$ and $V$ so that
$\Delta_{r}\psi$ is large on $\Sigma$ (which requires $\Delta\psi$ to be large
on a neighborhood of $\Sigma$) and $\Sigma$ and $V$ to be large enough to
apply Assumption 1. The way to enlarge those sets is delicate if one wants to
keep some notion of duality between them $\Sigma=\partial\psi^{*}(V)$ and be
compatible with the critical scale $r\sim\delta^{1/2}$.

Outline of the paper: In the next section, we recall several facts about
convex sets and sections of convex functions that will be used throughout the
rest of the paper. The main result of that section is Proposition 3.5, which
summarizes the properties of the sections of $\psi$. These properties are
classical when $\delta=0$, and generalizing them to the case $\delta>0$ is an
important contribution of this paper. The proof of Proposition 3.5 is detailed
in Section 7. The proof of Theorem 2.4 is presented in Section 5. It relies on
ideas introduced in [17] but it has been slightly modified to avoid the use of
the maximal function and the use of renormalizations. We present it here
mainly because it allows us to present the main ideas and estimates in the
simpler framework $\delta=0$ before presenting the more complicated proof of
Theorem 2.5 in Section 6.

##  3\. Preliminaries

###  3.1. Notations

The proofs will make extensive use of the notion of inner and outer radii of a
convex set, defined as follows:

$\displaystyle\ell(K)$ $\displaystyle:=\mbox{ the radius of the largest ball
contained in $K$ (inner radius)}$ $\displaystyle L(K)$ $\displaystyle:=\mbox{
the radius of the smallest ball containing $K$ (outer radius).}$

Throughout the paper, $c$ and $C$ denote some constants, which may change
value from line to line, that depend (unless otherwise indicated) only on the
dimension $n$, the sets $\Omega$, $\mathcal{O}$ and the constant $\lambda$,
$\Lambda$. When comparing quantities, we will use the notation $a\sim b$ to
indicate that there are constant $c_{1}$, $c_{2}$ depending only on the
dimension $n$ such that

$c_{1}a\leq b\leq c_{2}a.$

The norm of a matrix $A$ is defined by $\|A\|=\sup_{|x|=1}|Ax|$. Given a
positive definite matrix $A$, we will use the notation

$|x|_{A}:=\sqrt{x^{*}Ax}\qquad x\in\mathbb{R}^{n}.$

###  3.2. John’s Lemma

We recall the following classical lemma:

######  Lemma 3.1.

[John’s Lemma] Let $K$ be a bounded convex set with non-empty interior. There
exists an affine transformation $T(x)=Nx+b$ such that

$B_{1}(0)\subset T(K)\subset B_{n}(0).$

Up to a translation of $K$, we can always assume that $b=0$. We then have

(19) $\\{x\in\mathbb{R}^{n}\,;\,x^{*}Ax\leq 1\\}\subset
K\subset\\{x\in\mathbb{R}^{n}\,;\,x^{*}Ax\leq n^{2}\\}\qquad\mbox{ with
}A=N^{*}N.$

We denote by $\lambda_{1}\geq\cdots\geq\lambda_{n}>0$ the eigenvalues of the
positive symmetric matrix $A=N^{*}N$. The set $x^{*}Ax=1$ is an ellipsoid with
radii $\frac{1}{\sqrt{\lambda_{i}}}$, $i=1,\dots,n$. In particular, (19)
implies that the inner and outer radii of $K$ satisfy
$\ell(K)\sim\frac{1}{\sqrt{\lambda_{1}}}$ and
$L(K)\sim\frac{1}{\sqrt{\lambda_{n}}}$. More precisely, we have

$\frac{1}{\sqrt{\lambda_{1}}}\leq\ell(K)\leq\frac{n}{\sqrt{\lambda_{1}}}\quad\mbox{and}\qquad\frac{1}{\sqrt{\lambda_{n}}}\leq
L(K)\leq\frac{n}{\sqrt{\lambda_{n}}}.$

We also have $\|N^{*}\|\|N\|=\|A\|\sim\mathrm{Tr}(A)$ (the first equality can
be proved using the polar decomposition of matrices) and

(20) $\frac{1}{\ell(K)^{2}}\leq\mathrm{Tr}(A)\leq\frac{n^{3}}{\ell(K)^{2}}.$

###  3.3. Revisiting Assumption 1

We stated Assumption 1 for balls where it is easily understood. We will
however need to compare $\mu(A)$ or $\nu(A)$ with $|A|$ for more general sets,
including some convex sets and sections. The following proposition allows us
to use the bounds (9)-(10) in such cases.

######  Proposition 3.2.

Let $\mu$ and $\nu$ satisfy Assumption 1. Then there exists $\lambda^{\prime}$
and $\Lambda^{\prime}$ and $\beta$ such that

(21)
$\lambda^{\prime}|A_{\delta}|\leq\mu(A_{\delta}),\;\nu(A_{\delta})\leq\Lambda^{\prime}|A_{\delta}|,$

for any Borel set $A$ and with $A_{\delta}=\cup_{x\in
A}(S_{\delta}(x))\subset\Omega$ where $\\{S_{\delta}(x)\\}_{x\in A}$ is a
family of convex sets satisfying $x\in S_{\delta}$,
$\ell(S_{\delta})\geq\beta\delta$, and for which Vitali’s covering lemma (with
constant $C_{*}$) can be applied. Furthermore, the constant
$\lambda^{\prime}$, $\Lambda^{\prime}$ and $\beta$ only depend on $\lambda$,
$\Lambda$ and $C_{*}$. In particular, (21) holds if $A$ is a convex set
satisfying $\ell(A)\geq\beta\delta$.

The proof of this proposition can be found in Appendix B.

###  3.4. Sections of a convex function

The notion of sections of a convex function $\psi:\mathbb{R}^{n}\to\mathbb{R}$
plays a central role in many regularity results of the theory of optimal
transportation, including in the approach that we present here. Given
$x_{0}\in\Omega$, $p_{0}\in\partial\psi(x_{0})$ and $t\geq 0$, we recall that
the section centered at $x_{0}$ with height $t$ is defined as the convex set:

$S(x_{0},p_{0},t):=\\{x\in\mathbb{R}^{n}\,;\,\psi(x)\leq\psi(x_{0})+p_{0}\cdot(x-x_{0})+t\\}.$

When $\partial\psi(x_{0})$ reduces to one point $\\{\nabla\psi(x_{0})\\}$, we
will simply write $S(x_{0},t)$ for $S(x_{0},\nabla\psi(x_{0}),t)$.
Furthermore, given $\gamma>0$ we denote by $\gamma S(x_{0},p_{0},t)$ the
dilation of $S(x_{0},p_{0},t)$ with respect to the center of mass
$x^{*}_{0}$111We could also use dilations with respect to $x_{0}$, as in [17].
The properties of section listed in Proposition 3.5 are valid with dilation
with respect to any point in $S$.:

$\gamma S(x_{0},p_{0},t):=\left\\{x_{0}^{*}+\gamma(x-x_{0}^{*})\,;\,x\in
S(x_{0},p_{0},t)\right\\}.$

The following simple result will be useful to control the inner radius of
sections:

######  Proposition 3.3.

Let $\Omega$ be a bounded subset of $\mathbb{R}^{n}$ and
$\psi:\Omega\to\mathbb{R}$ be a convex function such that
$\partial\psi(\Omega)$ is also a bounded subset of $\mathbb{R}^{n}$. For any
section $S=S(x_{0},p_{0},t)\subset\Omega$, we have

$\ell(S)\geq\frac{t}{L(\partial\psi(\Omega))},\qquad\ell(\partial\psi(S))\geq\frac{t}{L(\Omega)}.$

The scaling of these bounds is not optimal in the framework of optimal
transportation, but is the best one can hope for in the case of general convex
functions; in particular the first inequality is an equality when
$\psi(x)=L|x|$.

###### Proof.

After replacing the function $\psi$ with
$\overline{\psi}(x)=\psi(x_{0}+x)-p_{0}\cdot x$, we can assume that $x_{0}=0$
and $p_{0}=0$. We then have
$|\nabla\overline{\psi}(x)|=|\nabla\psi(x_{0}+x)-p_{0}|\leq
L(\partial\psi(\Omega))$ a.e. $x\in S$. This implies in particular that
$\bar{\psi}(x)\leq\bar{\psi}(0)+L(\partial\psi(\Omega))|x|$ and so
$B_{t/L(\partial\psi(\Omega))}(0)\subset S(0,0,t)$. The first inequality
follows.

To prove the second inequality, we note that if $q\in\mathbb{R}^{n}$ is such
that $|q|<\frac{t}{L(\Omega)}$, then the function
$u(x)=\overline{\psi}(x)-\overline{\psi}(0)-q\cdot x$ satisfies $u(0)=0$ and
$u(x)=t-q\cdot x\geq t-|q||L(\Omega)|>0$ for $x\in\partial S$. It follows that
$u$ has a minimum in the interior of $S$. At such a point we have
$0\in\partial u(x)$ and so
$q\in\partial\overline{\psi}(x)\subset\partial\overline{\psi}(S)$. We deduce
that $B_{\frac{t}{L(\Omega)}}(0)\subset\partial\bar{\psi}(S)$ which implies
the second inequality ∎

We also recall the classical Alexandrov estimate (see for example [25]):

######  Proposition 3.4.

Let $\psi:\Omega\to\mathbb{R}$ be a convex function and assume that
$0\in\Omega$, $0\in\partial\psi(0)$ and $t\geq 0$ is such that
$S:=S(0,0,t)\subset\Omega$. Then

$|\psi(y)-t|^{n}\leq c_{n}L(S)^{n-1}\mathrm{dist}(y,\partial
S)|\partial\psi(S)|$

for some constant $c_{n}$ depending only on the dimension.

Finally, we summarize in the following proposition the key properties
satisfied by the sections of convex potentials appearing in the context of
optimal transportation (that is under our assumptions (11)-(12) and (9)-(10)).

######  Proposition 3.5.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying
(11)-(12) for measures $\mu$ and $\nu$ satisfying Assumption 1. There exists a
constant $C_{0}$ depending only on $n$, $L(\Omega)$ and
$L(\partial\psi(\Omega))$ such that when

$t\geq C_{0}\delta$

we have

(i) There exists $C=C(n,\lambda,\Lambda)$ such that
$C^{-1}t^{n}\leq|S(x_{0},p_{0},t)|^{2}\leq Ct^{n}$ for all sections
$S(x_{0},p_{0},t)\subset\Omega$.

(ii) There exists $M=M(n,\lambda,\Lambda)$ such that $2S(x_{0},p_{0},t)\subset
S(x_{0},p_{0},Mt)\subset MS(x_{0},p_{0},t)$ for all sections such that
$S(x_{0},p_{0},Mt)\subset\Omega$.

(iii) There exists $\theta=\theta(n,\lambda,\Lambda)>1$ such that if
$S(x_{0},p_{0},2t)\subset\Omega$, then $S(x_{0},p_{0},t)\subset S(y,q,\theta
t)$ for all $y\in S(x_{0},p_{0},t)$ and $q\in\partial\psi(y)$.

These properties are classical when $\delta=0$ (see for instance [11], [25],
[24, Chapter 3]), but proving that they hold when $\delta>0$ as long as $t\geq
C_{0}\delta$ is an important contribution of this paper. Complete proofs for
the case $\delta>0$ are presented in Section 7: Property (i) will be proved in
Proposition 7.5. Since $\omega_{n}\ell(S)^{n}\leq|S|\leq\omega_{n}L(S)^{n}$,
it implies

(22) $\ell(S)\leq Ct^{1/2},\qquad L(S)\geq C^{-1}t^{1/2}$

for some constant $C=C(n,\lambda,\Lambda)$. Property (ii) follows from
Proposition 7.6; note that the second inclusion $S(x_{0},p_{0},Mt)\subset
MS(x_{0},p_{0},t)$ holds for any convex function. Finally property (iii)
follows from Proposition 7.9.

We also make the following important observation: by iterating Proposition
3.5-(ii), we find $2^{k}S(x_{0},p_{0},t)\subset S(x_{0},p_{0},M^{k}t)$ that is

(23) $KS(x_{0},p_{0},t)\subset S(x_{0},p_{0},K^{\beta}t)\qquad\forall t\geq
C_{0}\delta\quad\mbox{ if }\quad S(x_{0},p_{0},K^{\beta}t)\subset\Omega$

with $\beta=\frac{\ln M}{\ln 2}$.

Properties (ii) and (iii) above are required to use Vitali’s covering lemma
with sections rather than Euclidean balls (see [34, Section 1.1]):

######  Corollary 3.6 (Vitali’s covering Lemma).

Under the assumptions of Proposition 3.5, there exist constants
$C_{0}(n,L(\Omega),L(\partial\psi(\Omega))$ and $C_{*}(n,\lambda,\Lambda)$
such that given a collection of sections $S_{i}=S(x_{i},t_{i})\subset\Omega$
with $t_{i}\geq C_{0}\delta$ there exists a countable subcollection
$S_{i_{j}}$ such that the $S_{i_{j}}$ are pairwise disjoint and satisfy

$\bigcup_{i}S_{i}\subset\bigcup_{j}C_{*}S_{i_{j}}.$

##  4\. $C^{1,\alpha}$ regularity up to scale $\delta$

It was shown by Forzani and Maldonado [21] that the engulfing property of
sections is actually equivalent to the $C^{1,\alpha}$ regularity of the
potential. Using this idea, we can prove an approximate $C^{1,\alpha}$
regularity at some scale depending on $\delta$. The following theorem implies
Theorem 2.2:

######  Theorem 4.1.

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded convex domain and
$\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (11), (12),
with $\mu$ and $\nu$ satisfying (9) and (10) with $\delta>0$. Let
$\Omega^{\prime}\subset\subset\Omega$ be such that there exists
$\rho>C_{0}\delta$ such that (14) holds. Given $y\in\Omega^{\prime}$, there
exist a constant $\overline{C}$ depending on
$\mathrm{dist}(y,\partial\Omega^{\prime})$, $\rho$, $\partial\psi(\Omega)$,
$\Omega$, $\lambda$, $\Lambda$ and $n$ such that

$0\leq\psi(x)-\psi(y)-q\cdot(x-y)\leq\overline{C}|x-y|^{1+\frac{1}{\theta}}$

for all $q\in\partial\psi(y)$ and $x\in\Omega^{\prime}$ such that

(24)
$C\delta^{1/\beta}\leq|x-y|\leq\min\left(\mathrm{dist}(y,\partial\Omega^{\prime}),\frac{\rho}{2L(\partial\psi(\Omega))}\right).$

We emphasize that the $C^{1,\alpha}$ regularity of solutions to Monge-Ampère
in dimension $n\geq 3$ cannot be derived purely locally but requires
appropriate boundary conditions. In Theorem 4.1, the role of the boundary
condition is played by the assumption that $S(x,p,\rho)\subset\Omega$ for all
$x\in\Omega^{\prime}$ (condition (14)).

To prove Theorem 4.1 we start with the following observation.

######  Lemma 4.2.

There exists $C$ and $\beta$ depending on $\Omega$, $\rho$, $\mathcal{O}$,
$n$, $\lambda$, $\Lambda$ such that for all $x\in\Omega^{\prime}$ and
$p\in\partial\psi(x)$ so that (14) holds we have

$L(S(x,p,t))\leq Ct^{\frac{1}{\beta}}\qquad\mbox{ for all }\
{t\in[C_{0}\delta,\rho]}$

###### Proof.

The dilation property of sections (23) together with condition (14) imply

$S(x,p,t)\subset\left(\frac{t}{\rho}\right)^{\frac{1}{\beta}}S(x,p,\rho)\subset\left(\frac{t}{\rho}\right)^{\frac{1}{\beta}}\Omega$

and so
$L(S(x,p,t))\leq\frac{L(\Omega)}{\rho^{\frac{1}{\beta}}}t^{\frac{1}{\beta}}$.
∎

###### Proof of Theorem 4.1.

Given $x,y\in\Omega^{\prime}$ we fix $p\in\partial\psi(x)$ such that (14)
holds and choose any $q\in\partial\psi(y)$. We then take $t$ to be the
smallest positive number such that $y\in S(x,p,t)$. We then have

(25) $t=\psi(y)-\psi(x)-p\cdot(y-x),$

and $L(S(x,p,t))\geq|y-x|$.

If $t\leq C_{0}\delta$, then $|x-y|\leq L(S(x,p,t))\leq
L(S(x,p,C_{0}\delta))$, and Lemma 4.2 gives $L(S(x,p,C_{0}\delta))\leq
C(C_{0}\delta)^{\frac{1}{\beta}}$. We then have $|x-y|\leq
C(C_{0}\delta)^{\frac{1}{\beta}}$ which contradicts (24) (for an appropriate
choice of the constant $C$. We must thus have $t\geq C_{0}\delta$.

Proposition 3.5-(iii) (the engulfing property of sections) then gives
$S(x,t)\subset S(y,q,\theta t)$ provided $2t\leq\rho$ (so that (14) guarantees
that $S(x,p,2t)\subset\Omega$). This implies

(26) $\psi(x)\leq\psi(y)+q\cdot(x-y)+\theta t.$

In view of the definition of $t$, the condition $2t\leq\rho$ is satisfied as
long as $y\in S(x,\rho/2)$. Since $B_{\rho/2L(\partial\psi(\Omega))}(x)\subset
S(x,\rho/2)$ (see the proof of Proposition 3.3), this holds as long as
$|x-y|\leq\frac{\rho}{2L(\partial\psi(\Omega))}$.

Combining (25) and (26), we deduce

(27) $\frac{1+\theta}{\theta}[\psi(x)-\psi(y)-q\cdot(x-y)]\leq(p-q)\cdot(x-y)$

which holds for all $x,y\in\Omega^{\prime}$ such that
$C\delta^{1/\beta}\leq|x-y|\leq\frac{\rho}{2L(\partial\psi(\Omega))}$.

Let now $v$ be a vector satisfying $|v|=1$ and define

$f(s)=\psi(y+sv)-\psi(y)-q\cdot sv.$

Notice that $\psi$ is convex in $s\in\mathbb{R}$. It is hence Lipschitz
regular and differentiable almost everywhere. Hence for $a.e.\;s$ such that

$C\delta^{1/\beta}\leq s\leq
s_{0}=\min\left(\mathrm{dist}(y,\partial\Omega^{\prime}),\frac{\rho}{2L(\partial\psi(\Omega))}\right)$

inequality (27) implies

$\frac{1+\theta}{\theta}f(s)\leq sf^{\prime}(s).$

Define $g(s)=s^{-\frac{1+\theta}{\theta}}f(s)$ which is also Lipschitz on the
corresponding interval of $s$ (as it is away from $s=0$) and notice that
$g^{\prime}(s)\geq 0$ for $a.e.\;s$ so that $g$ is monotone increasing. Thus

$0\leq f(s)\leq s^{\frac{1+\theta}{\theta}}g(s_{0})\qquad\forall
s\in[C\delta^{1/\beta},s_{0}],$

which gives by taking $x=y+sv$

$0\leq\psi(x)-\psi(y)-q\cdot(x-y)\leq|x-y|^{1+\frac{1}{\theta}}g(s_{0})$

for all $x$ such that $C\delta^{1/\beta}\leq|x-y|\leq s_{0}$.

It remains to see that

$g(s_{0})=s_{0}^{-\frac{1+\theta}{\theta}}f(s_{0})\leq\min\left(\mathrm{dist}(y,\partial\Omega^{\prime}),\frac{\rho}{2L(\partial\psi(\Omega))}\right)^{-\frac{1+\theta}{\theta}}L(\partial\psi(\Omega))\,L(\Omega)$

to get the result.

∎

##  5\. The case $\delta=0$: Proof of Theorem 2.4

In this section we present the strategy for proving the Sobolev regularity of
$\psi$ in the simpler case $\delta=0$. The general argument relies on the
ideas introduced in [17] with only slight modifications. Among the noteworthy
modifications, we will not rely on the maximal function (this does not
fundamentally change the arguments but makes the proof more self contained).
We will also avoid the use of renormalization of the sections, something that
will be very important when we extend the proof to the case $\delta>0$ in the
next section.

###  5.1. Preliminaries

We consider a convex function $\psi:\mathbb{R}^{n}\to\mathbb{R}$ and a section
$S(x_{0},p_{0},t)$ with $x_{0}\in\mathbb{R}^{n}$ and
$p_{0}\in\partial\psi(x_{0})$. Since we can always replace $\psi$ with
$\psi(x)-\psi(x_{0})-p\cdot(x-x_{0})$, we can assume in this subsection,
without loss of generality, that $\psi(x_{0})=0$ and $p_{0}=0$. We recall (see
Lemma 3.1) that up to a translation, we can also assume that there is a
positive definite symmetric matrix $A$ such that

(28) $\\{x\in\mathbb{R}^{n}\,;\,|x|_{A}\leq 1\\}\subset
S(x_{0},0,t)\subset\\{x\in\mathbb{R}^{n}\,;\,|x|_{A}\leq n\\}$

and we stress out the fact that we cannot assume that $x_{0}=0$ since we are
already assuming that the ellipsoid in (28) are centered at $0$.

The following two lemmas, which will play a crucial role in the proof, are
reminiscent of Lemma 3.2 and Lemma 3.3 in [17]:

######  Lemma 5.1.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function and $A$ a positive
definite symmetric matrix such that $S=S(x_{0},0,t)$ satisfies (28). There
exists a constant $C$ depending only on the dimension $n$ such that

$\fint_{S}\Delta\psi(x)\,dx\leq C\mathrm{Tr}(A)\sup_{\\{|x|_{A}\leq
2n\\}}\psi$

where $\fint f(x)\,dx=\frac{1}{|S|}\int_{S}f(x)\,dx$.

######  Lemma 5.2.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function and $A$ a positive
definite symmetric matrix such that $S=S(x_{0},0,t)$ satisfies (28). There
exists a constant $c$ depending only on the dimension $n$ and a set
$V\subset\partial\psi(S)$ such that

(29) $|V|\geq c\frac{t^{n}}{|S|},\qquad\partial\psi^{*}(V)\subset S$

and

(30) $\Delta\psi(y_{0})\geq\frac{t}{4n^{2}}\mathrm{Tr}(A)\qquad\forall
y_{0}\in\Sigma:=\partial\psi^{*}(V)\subset S.$

Furthermore, for all $q\in V$ and $y_{0}\in\partial\psi^{*}(q)$ we have

(31)
$\psi(y)-\psi(y_{0})\geq\frac{t}{8n^{2}}|y-y_{0}|_{A}^{2}+q\cdot(y-y_{0})\qquad\forall
y\in\\{x\in\mathbb{R}^{n}\,;\,|x|_{A}\leq 2n\\}.$

Note that when (31) holds, the convexity of $\psi$ implies that
$\partial\psi^{*}(q)$ cannot contain any other point besides $y_{0}$, that is

$\partial\psi^{*}(q)=\\{y_{0}\\}.$

On the other hand it is not necessarily true that (31) will hold for any
another $q^{\prime}\in\partial\psi(y_{0})$. This means that we may have
$V\varsubsetneq\partial\psi(\Sigma)$ (even though
$\Sigma=\partial\psi^{*}(V)$). When $\psi$ is $C^{1}$ and strictly convex (as
is the case in [17]) this distinction is not relevant (since
$V=\partial\psi(\Sigma)$). But it constitutes an important difficulty in
extending the proof to the case $\delta>0$. Since we will not be able to
assume that $\psi\in C^{1}$ in that case it will be very important that (29)
provides a lower bound on the measure of $V$ rather than the measure of
$\partial\psi(\Sigma)$.

An inequality similar to (30) is used in [17], and this inequality is enough
to prove our main result when $\delta=0$. Inequality (31), which carries
additional nonlocal information, will be needed to extend the result to the
case $\delta>0$.

###### Proof of Lemma 5.1.

We consider the polynomial $p(x)=2-\frac{x^{*}Ax}{n^{2}}$ which, in view of
(28), satisfies $p(x)\geq 1$ for $x\in S$ and
$\\{p>0\\}=\\{x\in\mathbb{R}^{n}\,;\,|x|_{A}\leq\sqrt{2}n\\}.$ Let
$\overline{\psi}:=\psi-\sup_{\\{p>0\\}}\psi$. Since $\psi$ is convex,
$\Delta\psi=\Delta\overline{\psi}\geq 0$ and a couple of integrations by parts
give

$\displaystyle\int_{S}\Delta\psi\,dx\leq\int_{\\{p>0\\}}p\Delta\overline{\psi}\,dx$
$\displaystyle=\int_{\partial\\{p>0\\}}p\nabla\overline{\psi}\cdot
n+\int_{\partial\\{p>0\\}}\overline{\psi}(-\nabla p\cdot
n)+\int_{\\{p>0\\}}\overline{\psi}\Delta p.$

The first term vanishes since $p=0$ on ${\partial\\{p>0\\}}$ while the second
term is non-positive since $-\nabla p\cdot n\geq 0$ on ${\partial\\{p>0\\}}$
and $\overline{\psi}\leq 0$ in $\\{p>0\\}$. We deduce:

$\displaystyle\int_{S}\Delta\psi\,dx\leq\int_{\\{p>0\\}}\overline{\psi}\Delta
p=\frac{2}{n^{2}}\mathrm{Tr}(A)\int_{\\{p>0\\}}(-\overline{\psi})$
$\displaystyle\leq\frac{2}{n^{2}}\mathrm{Tr}(A)\int_{\\{p>0\\}}[-\inf_{x\in{\\{p>0\\}}}\overline{\psi}]$
$\displaystyle\leq\frac{2}{n^{2}}\mathrm{Tr}(A)|\\{p>0\\}|\,\sup_{\\{p>0\\}}\psi.$

The result follows since $|\\{p>0\\}|=(\sqrt{2}n)^{n}|\\{|x|_{A}\leq 1\\}|\leq
C|S|$. ∎

###### Proof of Lemma 5.2.

The main idea stems from the proof of Lemma 3.3 in [17]: We denote
$U=\\{x\in\mathbb{R}^{n}\,;\,|x|_{A}\leq 2n\\}$ and set
$p(x)=\frac{t}{2}(\frac{1}{4n^{2}}x^{*}Ax-1)$. We have $-\frac{t}{2}\leq p\leq
0$ in $U$, so that the function $w(x):=\psi(x)-t-p(x)$ satisfies
$w(x)\geq\psi(x)-t>0$ in $U\setminus S$ and
$w(x_{0})\leq\psi(x_{0})-t+\frac{t}{2}=-\frac{t}{2}$. We then consider
$\overline{w}$ the convex hull of $w$ in $U$, defined by

$\overline{w}(x)=\sup\\{\ell(x)\,;\,\ell\mbox{ affine function s.t. }\ell\leq
w\mbox{ in }U,\;\ell\leq 0\mbox{ on }\partial U\\},$

and the contact set

$\Sigma=\\{x\in U\,;\,\overline{w}(x)=w(x)\\}.$

Since $w>0$ in $U\setminus S$ and $\overline{w}\leq 0$, we have $\Sigma\subset
S$. Given $y_{0}\in\Sigma$ and $q_{0}\in\partial\overline{w}(y_{0})$, since
$\overline{w}$ is convex, we have

$w(x)\geq\overline{w}(x)\geq\overline{w}(y_{0})+q_{0}\cdot(x-y_{0})=w(y_{0})+q_{0}\cdot(x-y_{0})\qquad\forall
x\in U.$

The definition of $w$ and the fact that $p$ is a quadratic polynomial then
imply

$\displaystyle\psi(x)-\psi(y_{0})$ $\displaystyle\geq
p(x)-p(y_{0})+q_{0}\cdot(x-y_{0})$ $\displaystyle\geq\nabla
p(y_{0})\cdot(x-y_{0})+\frac{1}{2}(x-y_{0})^{*}D^{2}p(y_{0})(x-y_{0})+q_{0}\cdot(x-y_{0})$
$\displaystyle\geq\frac{t}{8n^{2}}|x-y_{0}|^{2}_{A}+(q_{0}+\nabla
p(y_{0}))\cdot(x-y_{0})\qquad\forall x\in U.$

We now denote

(32) $V=\\{q_{0}+\nabla p(y_{0})\,|\,y_{0}\in\Sigma,\quad
q_{0}\in\partial\overline{w}(y_{0})\\}.$

The inequality above implies in particular that $q:=q_{0}+\nabla
p(y_{0})\in\partial\psi(y_{0})$ so that $V\subset\partial\psi(\Sigma)$. As
noted earlier, it also implies that $\partial\psi^{*}(q)=\\{y_{0}\\}$.

For all $q\in V$, we can write $q=q_{0}+\nabla p(y_{0})$ with $y_{0}\in\Sigma$
(and $q_{0}\in\partial\overline{w}(y_{0})$) and the computation above shows
that (31) then holds (and (31) implies (30)). We saw above that this also
implies that $\partial\psi^{*}(q)=\\{y_{0}\\}$ and so this proves that (31)
actually holds for all $q\in V$ and $y_{0}\in\partial\psi^{*}(q)$ (as claimed
in the lemma). This also implies that

$\partial\psi^{*}(V)\subset\Sigma.$

The fact that $\Sigma\subset\partial\psi^{*}(V)$ is easy to show since for all
$y_{0}\in\Sigma$ (and $q_{0}\in\partial\overline{w}(y_{0})$) the vector
$q=q_{0}+\nabla p(y_{0})$ satisfies $q\in V$ and
$y_{0}\in\partial\psi^{*}(q)$. We thus have

$\Sigma=\partial\psi^{*}(V).$

So the only thing left to show to complete the proof of Lemma 5.2 is the lower
bound on $|V|$: Using the convexity of $\overline{w}$ together with the
smoothness of $p$, (32) implies:

$V=\partial(\overline{w}+p)(\Sigma)$

and so (since $D^{2}\overline{w}+D^{2}p\geq D^{2}\overline{w}\geq 0$)

$|V|=|\partial(\overline{w}+p)(\Sigma)|=\int_{\Sigma}\det(D^{2}\overline{w}+D^{2}p)\,dx\geq\int_{\Sigma}\det(D^{2}\overline{w})\,dx=|\partial\overline{w}(\Sigma)|.$

Furthermore, the definition of the convex hull gives
$\partial\overline{w}(\Sigma)=\partial\overline{w}(U)$, hence

$|V|\geq|\partial\overline{w}(U)|.$

Since $\overline{w}=0$ on $\partial U$ and $\overline{w}(x_{0})\leq-t/2$, $U$
is a section of $\overline{w}$ with some center $\bar{x}_{0}\in U$ and height
$\overline{t}\geq\frac{t}{2}$. In particular, (64) implies

$|\partial\overline{w}(U)|\,|U|\geq\frac{\omega_{n}^{2}}{(2n)^{n}}\left(\frac{t}{2}\right)^{n}$

(we can also use Alexandrov’s estimate, Proposition 3.4) and so

$|V|\geq|\partial\overline{w}(U)|\geq ct^{n}\frac{1}{|U|}$

and we conclude our proof by using the fact that $|U|=(2n)^{n}|\\{|x|_{A}\leq
1\\}|\leq(2n)^{n}|S|$. ∎

Lemmas 5.1 and 5.2 hold for any convex function and do not make use of the
optimal transportation assumptions. When $\psi$ is an optimal transportation
potential satisfying our assumptions, these lemmas yield:

######  Proposition 5.3.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (3),
(11), (12), with $\mu$ and $\nu$ satisfying (9) and (10) with $\delta=0$.
There exists some constants $c$ and $C_{1}$ depending on $n$, $\lambda$ and
$\Lambda$ such that the following holds: Given $x_{0}\in\Omega$,
$p_{0}\in\partial\psi(x_{0})$, $t$ such that
$S(x_{0},p_{0},C_{1}t)\subset\Omega$ and setting

$\alpha:=\fint_{S(x_{0},p_{0},t)}\Delta\psi(x)\,dx,$

there exists $\Sigma\subset S(x_{0},p_{0},t)$ such that

$\Delta\psi(x)\geq c\alpha\quad\forall x\in\Sigma,\quad\mbox{ and
}\quad|\Sigma|\geq c|S(x_{0},p_{0},t)|.$

###### Proof.

Up to subtracting a linear function to $\psi$, we can assume that

$p_{0}=0,\quad\psi(x_{0})=0$

so that $0\leq\psi(x)\leq t$ in $S(x_{0},0,t)$ and $\psi(x)=t$ on $\partial
S(x_{0},0,t)$. Up to a translation, Lemma 3.1 implies the existence of a
matrix $A$ such that (28) holds. Lemma 5.1 then implies

$\fint_{S(x_{0},0,t)}\Delta\psi(x)\,dx\leq C\mathrm{Tr}(A)\sup_{\\{|x|_{A}\leq
2n\\}}\psi$

Using (23), we find $\\{|x|_{A}\leq 2n\\}=2n\\{|x|_{A}\leq 1\\}\subset
2nS(x_{0},0,t)\subset S(x_{0},0,C_{1}t)$ with $C_{1}=(2n)^{\beta}$ provided
$S(x_{0},0,C_{1}t)\subset\Omega$. It follows that $\sup_{\\{|x|_{A}\leq
2n\\}}\psi\leq C_{1}t$ and so

(33) $\alpha=\fint_{S(x_{0},0,t)}\Delta\psi(x)\,dx\leq C\mathrm{Tr}(A)t.$

for some $C$ depending only on $n$, $\lambda$ and $\Lambda$.

We now use Lemma 5.2 to deduce the existence of a set
$\Sigma=\partial\psi^{*}(V)\subset S(x_{0},0,t)$ such that

$\Delta\psi(x)\geq\frac{t}{4n^{2}}\mathrm{Tr}(A)\quad\mbox{ a.e. in
}\Sigma,\qquad|V|\geq c\frac{t^{n}}{|S|}.$

Together with (33), this implies

$\Delta\psi(x)\geq c\alpha\quad\mbox{ a.e. in }\Sigma,$

for some $c$ depending only on $n$, $\lambda$ and $\Lambda$. Using the lower
bound on $|V|$ together with (12) (note that
$V\subset\partial\psi(S)\subset\mathcal{O}$ since $\psi$ satisfies (3)) and
Assumption 1, we get

$c\lambda\frac{t^{n}}{|S|}\leq\lambda|V|\leq\nu(V)\leq\mu(\partial\psi^{*}(V))=\mu(\Sigma)\leq\Lambda|\Sigma|.$

The volume inequality of Proposition 3.5 (i) give $\frac{t^{n}}{|S|}\geq
c|S|$, hence

$|\Sigma|\geq c|S|$

for some constant $c$ depending on $n$, $\lambda$ and $\Lambda$. ∎

###  5.2. A reversed Chebychev’s inequality

The heart of the proof of Theorem 2.4 is the following reversed Chebychev’s
inequality (which is equivalent to (3.15) in [17]):

######  Theorem 5.4.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (3),
(11), (12), with $\mu$ and $\nu$ satisfying (9) and (10) with $\delta=0$.
Assume moreover that $\Delta\psi\in L^{1}_{loc}$. Let
$\Omega^{\prime}\subset\subset\Omega$ be such that there exists $\rho>0$ such
that (14) holds. Then there exists $\alpha_{0}$, $c$ and $C$ depending on
$\rho$, $n$, $\lambda$, $\Lambda$, $\Omega$ and $\partial\psi(\Omega)$ such
that

(34)
$\int_{\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)\geq\alpha\\}}\Delta\psi(x)\,dx\leq
C\alpha|\\{x\in\Omega\,|\,\Delta\psi(x)\geq c\alpha\\}|\qquad\mbox{ for all
$\alpha>\alpha_{0}$}.$

Let us first show that this theorem implies Theorem 2.4 (the proof is similar
to the proof in [17]. We recall it here for the sake of completeness).

###### Proof of Theorem 2.4.

For $\bar{c}\geq\max\\{2,\alpha_{0}\\}$ (so that $\log(2+\gamma)\leq
2\log\gamma$ for $\gamma\geq\bar{c}$), we can write

$\displaystyle\int_{\Omega^{\prime}}\Delta\psi(x)\log(2+\Delta\psi(x))\,dx$
$\displaystyle\leq\bar{c}\log(2+\bar{c})|\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)\leq\bar{c}\\}|$
$\displaystyle\qquad\qquad+2\int_{\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)\geq\bar{c}\\}}\Delta\psi(x)\log(\Delta\psi(x))\,dx$
$\displaystyle\leq\bar{c}\log(2+\bar{c})|\Omega|+2\int_{\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)\geq\bar{c}\\}}\Delta\psi(x)\int_{1}^{\Delta\psi(x)}\frac{1}{\gamma}\,d\gamma\,dx$
$\displaystyle\leq\bar{c}\log(2+\bar{c})|\Omega|+2\log\bar{c}\int_{\Omega^{\prime}}\Delta\psi(x)\,dx$
$\displaystyle\qquad+2\int_{\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)\geq\bar{c}\\}}\Delta\psi(x)\int_{\bar{c}}^{\Delta\psi(x)}\frac{1}{\gamma}\,d\gamma\,dx$

Now, by Fubini’s theorem, the last term can be written as

$\int_{\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)\geq\bar{c}\\}}\Delta\psi(x)\int_{\bar{c}}^{\Delta\psi(x)}\frac{1}{\gamma}\,d\gamma\,dx=\int_{\bar{c}}^{\infty}\frac{1}{\gamma}\int_{\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)\geq\gamma\\}\\}}\Delta\psi(x)\,dx\,d\gamma$

and we can use Theorem 5.4: Since $\gamma>\bar{c}>\alpha_{0}$, we get

$\displaystyle\int_{\bar{c}}^{\infty}\frac{1}{\gamma}\int_{\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)\geq\gamma\\}\\}}\Delta\psi(x)\,dx\,d\gamma$
$\displaystyle\leq
C\int_{\bar{c}}^{\infty}|\\{x\in\Omega\,|\,\Delta\psi(x)\geq
c\gamma\\}|\,d\gamma$
$\displaystyle\leq\frac{C}{c}\int_{c\bar{c}}^{\infty}|\\{x\in\Omega\,|\,\Delta\psi(x)\geq\gamma\\}|\,d\gamma$
$\displaystyle\leq\frac{C}{c}\int_{\Omega}\Delta\psi(x)\,dx$

Putting things together, we proved that for some constant $C$ we have

$\int_{\Omega^{\prime}}\Delta\psi(x)\log(2+\Delta\psi(x))\,dx\leq
C+C\int_{\Omega}\Delta\psi(x)\,dx$

and we conclude using the fact that

$\int_{\Omega}\Delta\psi(x)\,dx\leq\int_{\Omega}\Delta\psi\,dx=\int_{\partial\Omega}\nabla\psi\cdot
n\,d\mathcal{H}^{n-1}(x)\leq\mathcal{H}^{n-1}(\partial\Omega)L(\partial\psi(\Omega)).$

∎

The proof of Theorem 5.4 is where we explicitly use the $L^{1}$ regularity of
$\Delta\psi$. More precisely we use the fact that the map
$t\mapsto\fint_{S(x,t)}\Delta\psi(x)\,dx$ is continuous and satisfies

$\lim_{t\to 0}\fint_{S(x,t)}\Delta\psi(x)\,dx=\Delta\psi(x)\ a.e.$

As remarked in the introduction, it should be possible to remove this
assumption by correctly decomposing $\Delta\psi$ into a regular and a singular
part in the argument. Since this more general case is thoroughly covered by
our theorem in the case $\delta>0$, we stay at the more elementary level where
$\Delta\psi$ is absolutely continuous w.r.t. the Lebesgue measure.

###### Proof of Theorem 5.4.

We fix some $\varepsilon>0$ (to be chosen later) and take $\alpha_{0}$ such
that

$\fint_{S(x,t)}\Delta\psi(x)\,dx\leq\alpha_{0}\qquad\forall
x\in\Omega^{\prime},\;\forall t\in[\varepsilon\rho,\rho].$

The existence of such an $\alpha_{0}$ follows from the fact that for
$t\in[\varepsilon\rho,\rho]$ we have

$\fint_{S(x,t)}\Delta\psi(x)\,dx\leq\frac{1}{|S(x,t)|}\int_{\Omega}\Delta\psi(x)\,dx\leq\frac{1}{|S(x,\varepsilon\rho)|}\int_{\Omega}\Delta\psi\,dx\leq\frac{\mathcal{H}^{n-1}(\partial\Omega)L(\partial\psi(\Omega))}{|S(x,\varepsilon\rho)|},$

where $|S(x,\varepsilon\rho)|\geq c(\varepsilon\rho)^{n/2}$ (see Proposition
3.5 (i)).

We now take $\alpha>\alpha_{0}$ and denote

$E_{\alpha}=\\{x\in\Omega^{\prime}\,|\,\Delta\psi(x)>\alpha\\}.$

For a.e. $x\in E_{\alpha}$ we have $\fint_{S(x,t)}\Delta\psi(x)\,dx<\alpha$
for all $t\in[\varepsilon\rho,\rho]$ (by the choice of $\alpha$) and

$\lim_{t\to 0}\fint_{S(x,t)}\Delta\psi(x)\,dx=\Delta\psi(x)>\alpha$

by the Lebesgue differentiation theorem.

We can thus define

$t(x)=\sup\left\\{t\leq\rho\,;\,\fint_{S(x,t)}\Delta\psi(x)\,dx=\alpha\right\\}$

which satisfies in particular $t(x)\leq\varepsilon\rho$ and

(35) $\fint_{S(x,t)}\Delta\psi(x)\,dx\leq\alpha\quad\mbox{ for all }t(x)\leq
t\leq\rho,\qquad\fint_{S(x,t(x))}\Delta\psi(x)\,dx=\alpha.$

Since $E_{\alpha}\subset\bigcup_{x\in E_{\alpha}}S(x,t(x))$, we can use
Vitali’s covering lemma (see Corollary 3.6) to extract a countable family
$\\{x_{k}\\}_{k\in\mathbb{N}}$ such that the sections
$S_{k}=S(x_{k},t(x_{k}))$ are pairwise disjoint and
$E_{\alpha}\subset\bigcup_{k}C_{*}S_{k}$ with a $C_{*}$ depending only on $n$,
$\lambda$ and $\Lambda$.

Using (23), we get $C_{*}S_{k}\subset S(x_{k},C_{**}t_{k})$ with
$C_{**}=C_{*}^{\beta}$, provided $S(x_{k},C_{**}t_{k})\subset\Omega$, which
holds if $C_{**}t_{k}\leq\rho$ (since $x_{k}\in\Omega^{\prime}$). Since we
know that $t_{k}\leq\varepsilon\rho$, this last condition can be enforced by
choosing $\varepsilon:=\frac{1}{C_{**}}$.

Note that we then also have $t_{k}\leq C_{**}t_{k}\leq\rho$. The first
inequality in (35) thus implies

$\int_{C_{*}S_{k}}\Delta\psi(x)\,dx\leq\int_{S(x_{k},C_{**}t_{k})}\Delta\psi(x)\,dx\leq\alpha|S(x_{k},C_{**}t_{k})|.$

Proposition 3.5 (i) gives $|S(x_{k},C_{**}t_{k})|\leq Ct_{k}^{\frac{n}{2}}\leq
C|S(x_{k},t_{k})|$ hence

$\int_{C_{*}S_{k}}\Delta\psi(x)\,dx\leq C\alpha|S_{k}|.$

We are ready to conclude: First, we write (since
$E_{\alpha}\subset\bigcup_{k}C_{*}S_{k}$)

$\displaystyle\int_{E_{\alpha}}\Delta\psi(x)\,dx$
$\displaystyle\leq\sum_{k}\int_{C_{*}S_{k}}\Delta\psi(x)\,dx\leq
C\alpha\sum_{k}|S_{k}|.$

Then, we apply Proposition 5.3: We note that $S(x_{k},C_{1}t_{k})\subset
S(x_{k},C_{1}\varepsilon\rho)\subset\Omega$ (up to taking a smaller
$\varepsilon$ so that $\varepsilon\leq 1/C_{1}$) and
$\fint_{S_{k}}\Delta\psi(x)\,dx=\alpha$ (see (35)). Proposition 5.3 thus
provides the existence of a subset $\Sigma_{k}\subset S_{k}$ with
$|\Sigma_{k}|\geq c|S_{k}|$ and $\Delta\psi\geq c\alpha$ a.e. in $\Sigma_{k}$.
We can then write:

$\displaystyle\int_{E_{\alpha}}\Delta\psi(x)\,dx$ $\displaystyle\leq
C\alpha\sum_{k}|\Sigma_{k}|$ $\displaystyle\leq C\alpha|\cup_{k}\Sigma_{k}|$
$\displaystyle\leq C\alpha|\\{x\in\Omega\,|\,\Delta\psi(x)\geq c\alpha\\}|$

where the second inequality follows from the fact that the $S_{k}$, and
therefore the $\Sigma_{k}$, are disjoint. ∎

##  6\. The case $\delta>0$: Proof of Theorem 2.5

We now turn to the proof of our main result when $\delta>0$. We recall that we
fix a kernel $K:\mathbb{R}^{n}\to[0,\infty]$ radially symmetric, supported in
the ball $B_{1}(0)$ and such that $\int_{B_{1}}|x|^{2}K(x)\,dx>0$ and we
define for any $r>0$

$\Delta_{r}\psi(x):=\frac{m_{0}}{r^{2}}\int
K_{r}(y)[\psi(x+y)-\psi(x)]\,dy,\qquad
K_{r}(x)=\frac{1}{r^{n}}K\left(\frac{x}{r}\right)$

with $m_{0}=\frac{2}{\int_{\mathbb{R}^{n}}y_{1}^{2}K(y)\,dy}$.

We note that

  1. (1)

For any $p\in\partial\psi(x)$, the radial symmetry of $K$ allows us to write

(36) $\Delta_{r}\psi(x)=\frac{m_{0}}{r^{2}}\int
K_{r}(y)[\psi(x+y)-\psi(x)-p\cdot y]\,dy$

This formula makes the definition of $\Delta_{r}$ useful when working with
sections of the function $\psi$. It also implies that $\Delta_{r}\psi(x)\geq
0$ for all $x\in\mathbb{R}^{n}$ for any convex function.

  2. (2)

If $p(x)$ is an homogeneous polynomial of degree $2$, that is $p(x)=x^{*}Ax$
for some symmetric matrix $A$, then we have (using the radial symmetry of $K$
again):

(37) $\Delta_{r}p(x)=\frac{m_{0}}{r^{2}}\sum_{i,j}\int
K_{r}(y)A_{ij}y_{i}y_{j}\,dy=\frac{m_{0}}{r^{2}}\sum_{i}A_{ii}\int
K_{r}(y)y_{i}^{2}\,dy=2\mathrm{Tr}A=\Delta p(x).$

The choice of the normalization constant $m_{0}$ is naturally critical here.

  3. (3)

Together with Taylor’s formula, (37) also implies that if $\psi$ is $C^{2}$
then

$\Delta_{r}\psi(x)=\frac{m_{0}}{r^{2}}\sum_{i,j}\int
K_{r}(y)\frac{1}{2}\partial_{ij}\psi(x)y_{i}y_{j}\,dy+o(1)=\Delta\psi(x)+o(1).$

###  6.1. Preliminary results

Our first task it to generalize Lemmas 5.1 and 5.2 using $\Delta_{r}$ instead
of $\Delta$. The generalization of Lemma 5.1 is relatively straightforward
(see Lemma 6.1 below), and the main issue with Lemma 5.2 will be to fatten the
set $\Sigma$ so that it is big enough to allow us to use the estimates
(9)-(10) when $\delta>0$ (see Propositions 6.4 and 6.6).

Given a convex function $\psi:\mathbb{R}^{n}\to\mathbb{R}$, we fix
$x_{0}\in\mathbb{R}^{n}$ and $p_{0}\in\partial\psi(x_{0})$. Since we can
always replace $\psi$ with $\psi(x)-\psi(x_{0})-p\cdot(x-x_{0})$, we can
assume, without loss of generality, that $\psi(x_{0})=0$ and $p_{0}=0$. We
recall (see Lemma 3.1) that up to a translation, we can also assume that there
is a positive definite symmetric matrix $A$ such that

(38) $\\{x\in\mathbb{R}^{n}\,;\,|x|_{A}\leq 1\\}\subset
S(x_{0},0,t)\subset\\{x\in\mathbb{R}^{n}\,;\,|x|_{A}\leq n\\}.$

We start with the following generalization of Lemma 5.1:

######  Lemma 6.1.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function and $A$ be a
positive definite symmetric matrix $A$ such that the section $S=S(x_{0},0,t)$
satisfies (38). There exists a constant $C$ depending only on $n$ such that

(39) $\fint_{S}\Delta_{r}\psi(x)\,dx\leq C\mathrm{Tr}(A)\sup_{\\{|x|_{A}\leq
2n\\}+B_{r}}\psi.$

In particular, if $r\leq\ell(S)$, then

$\fint_{S}\Delta_{r}\psi(x)\,dx\leq C\mathrm{Tr}(A)\sup_{\\{|x|_{A}\leq
3n\\}}\psi.$

###### Proof.

We introduce the polynomial $p(x)=2-\frac{x^{*}Ax}{n^{2}}$ which satisfies
(see (38))

$\\{p>0\\}=\\{x\in\mathbb{R}^{n}\,;\,x^{*}Ax<2n^{2}\\},$

and $p\geq 1$ in $S$.

We denote $U=\\{p>0\\}+B_{r}$ and $\overline{\psi}:=\psi-\sup_{U}\psi$. Since
$\Delta_{r}\psi=\Delta_{r}\overline{\psi}$, a simple change of variable gives

$\displaystyle\int_{S}\Delta_{r}\psi\,dx\leq\int_{\\{p>0\\}}p(x)\Delta_{r}\overline{\psi}(x)\,dx$
$\displaystyle=\int_{\\{p>0\\}}\overline{\psi}(x)\Delta_{r}p(x)\,dx$
$\displaystyle\quad+\frac{m_{0}}{r^{2}}\int_{\\{p>0\\}^{c}}\int_{\\{p>0\\}}K_{r}(x-y)p(y)\overline{\psi}(x)\,dy\,dx$
$\displaystyle\quad-\frac{m_{0}}{r^{2}}\int_{\\{p>0\\}}\int_{\\{p>0\\}^{c}}K_{r}(x-y)p(y)\overline{\psi}(x)\,dy\,dx.$

Since $\mathrm{Supp}\,K_{r}\subset B_{r}$, we can write the last two terms as

$\frac{m_{0}}{r^{2}}\int_{U\setminus\\{p>0\\}}\int_{\\{p>0\\}}K_{r}(x-y)p(y)\overline{\psi}(x)\,dy\,dx-\frac{m_{0}}{r^{2}}\int_{\\{p>0\\}}\int_{U\setminus\\{p>0\\}}K_{r}(x-y)p(y)\overline{\psi}(x)\,dy\,dx$

which is negative since $\overline{\psi}(x)\leq 0$ in $U$, $p(y)>0$ in
$\\{p>0\\}$ and $p(y)<0$ in $U\setminus\\{p>0\\}$.

Using (37), we deduce

$\displaystyle\int_{S}\Delta_{r}\psi\,dx$
$\displaystyle\leq\int_{\\{p>0\\}}\overline{\psi}(x)\Delta_{r}p(x)\,dx$
$\displaystyle=-\frac{2}{n^{2}}\mathrm{Tr}(A)\int_{\\{p>0\\}}\overline{\psi}(x)\,dx$
$\displaystyle\leq-\frac{2}{n^{2}}\mathrm{Tr}(A)|\\{p>0\\}|\inf_{{\\{p>0\\}}}\overline{\psi}.$

Since
$\inf_{{\\{p>0\\}}}\overline{\psi}=\inf_{{\\{p>0\\}}}\psi-\sup_{U}\psi\geq-\sup_{U}\psi$
(recall that $\psi\geq 0$ in $\mathbb{R}^{n}$) and $|\\{p>0\\}|\leq C|S|$, we
deduce

$\int_{S}\Delta_{r}\psi\,dx\leq C\mathrm{Tr}(A)|S|\sup_{U}\psi$

for some constant $C$ depending only on $n$. The result follows. ∎

On the other hand, Lemma 5.2 would easily imply

######  Lemma 6.2.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function and $A$ be a
positive definite symmetric matrix $A$ such that the section $S=S(x_{0},0,t)$
satisfies (38). There exists a constant $c$ depending only on the dimension, a
subset $V\subset\partial\psi(S)$ such that $\partial\psi^{*}(V)\subset S$ and

(40) $|V|\geq c\frac{t^{n}}{|S|}$

and for all $r\leq\ell(S)$ we have

(41) $\Delta_{r}\psi(y_{0})\geq\frac{t}{4n^{2}}\,\mathrm{Tr}(A)\qquad\forall
y_{0}\in\Sigma:=\partial\psi^{*}(V).$

###### Proof.

Lemma 5.2 implies the existence of the set $V\subset\partial\psi(S)$
satisfying $\partial\psi^{*}(V)\subset S$ and (40) and such that for all
$y_{0}\in\Sigma$, we have

(42)
$\psi(y)-\psi(y_{0})\geq\frac{t}{8n^{2}}|y-y_{0}|_{A}^{2}+q\cdot(y-y_{0})\qquad\forall
y\in\\{x\in\mathbb{R}^{n}\,;\,|x|_{A}\leq 2n\\}$

for some $q\in\partial\psi(y_{0})$.

If $r\leq\ell(S)$ then for any $y_{0}\in\Sigma\subset S$ we have
$B_{r}(y_{0})\subset 2S\subset\\{|x|_{A}\leq 2n\\}$. Since $K_{r}$ is
supported in $B_{r}$, we can multiply (42) by
$\frac{m_{0}}{r^{2}}K_{r}(y_{0}-y)$ and integrate with respect to $y$ to get
(using the radial symmetry of $K_{r}$ and (37)):

$\displaystyle\Delta_{r}\psi(y_{0})$ $\displaystyle\geq\frac{m_{0}}{r^{2}}\int
K_{r}(y_{0}-y)\left[\frac{t}{8n^{2}}|y-y_{0}|_{A}^{2}+q\cdot(y-y_{0})\right]\,dy$
$\displaystyle\geq\frac{t}{8n^{2}}\frac{m_{0}}{r^{2}}\int K_{r}(y)y^{*}Ay\,dy$
$\displaystyle\geq\frac{t}{4n^{2}}\,\mathrm{Tr}(A).$

∎

This result is however not enough for our purpose because (40) does not yield
any control on $|\Sigma|$ (we can have $|\Sigma|=0$, even if $|V|>0$). We thus
need to "fatten" the set $\Sigma$ by showing that (41) holds on a slightly
larger set. This will be done by considering the set

$\Sigma_{h}=\bigcup_{q_{0}\in
V,y_{0}\in\partial\psi^{*}(q_{0})}S(y_{0},q_{0},h)$

for some appropriate $h$.

We emphasize that the choice of $\Sigma_{h}$ is delicate. There could a priori
be many ways of enlarging $\Sigma$, for example by simply taking
$\Sigma+B_{h}$. However we need several precise properties on $\Sigma_{h}$
(see Prop. 6.4 later) that we are not able to prove on $\Sigma+B_{h}$. Roughly
speaking, we need to preserve some equivalent of the duality that we have
between $V$ and $\Sigma$. The definition above of $\Sigma_{h}$ will naturally
lead to a set $V_{h}$ that properly enlarges $V$ and maintains the inclusion
$\partial\psi^{*}(V_{h})\subset\Sigma_{h}$.

First, we need to prove the following lemma which will allow us to extend the
inequality (41) to the larger set $\Sigma_{h}$:

######  Lemma 6.3.

For all $q_{0}\in V$ and $y_{0}\in\partial\psi^{*}(q_{0})$, if $x$ is such
that

$B_{r}(x)\subset\\{x\,;\,|x|_{A}\leq 2n\\},$

then

$\Delta_{r}\psi(x)\geq\frac{t}{4n^{2}}\mathrm{Tr}(A)-\frac{m_{0}}{r^{2}}(q-q_{0})\cdot(x-y_{0})\qquad\forall
q\in\partial\psi(x).$

###### Proof.

We fix $q_{0}\in V$ and $y_{0}\in\partial\psi^{*}(q_{0})$ (we recall that
$\partial\psi^{*}(q_{0})=\\{y_{0}\\}$ for $q_{0}\in V$). If $x$ is such that
$B_{r}(x)\subset\\{x\,;\,|x|_{A}\leq 2n\\}$, then (31) applied to $x+y$
implies that

$\psi(x+y)-\psi(y_{0})\geq\frac{t}{8n^{2}}|x+y-y_{0}|_{A}^{2}+q_{0}\cdot(x+y-y_{0})\qquad\forall
y\in B_{r}(0).$

Given $q\in\partial\psi(x)$, we have $\psi(y_{0})\geq\psi(x)+q\cdot(y_{0}-x)$
and so

$\displaystyle\psi(x+y)-\psi(x)$
$\displaystyle\geq\frac{t}{8n^{2}}|x+y-y_{0}|_{A}^{2}+q_{0}\cdot(x+y-y_{0})+\psi(y_{0})-\psi(x)$
$\displaystyle\geq\frac{t}{8n^{2}}|x+y-y_{0}|_{A}^{2}+q_{0}\cdot(x+y-y_{0})+q\cdot(y_{0}-x)$
$\displaystyle\geq\frac{t}{8n^{2}}|x+y-y_{0}|_{A}^{2}+q_{0}\cdot
y-(q-q_{0})\cdot(x-y_{0})\qquad\forall y\in B_{r}(0).$

We deduce

$\displaystyle\Delta_{r}\psi(x)$ $\displaystyle=\frac{m_{0}}{r^{2}}\int
K_{r}(y)[\psi(x+y)-\psi(x)]\,dy$
$\displaystyle\geq\frac{t}{8n^{2}}\frac{m_{0}}{r^{2}}\int
K_{r}(y)\left[|x-y_{0}|_{A}^{2}+|y|_{A}^{2}\right]\,dy-\frac{m_{0}}{r^{2}}(q-q_{0})\cdot(x-y_{0})$
$\displaystyle\geq\frac{t}{4n^{2}}\mathrm{Tr}(A)+\frac{m_{0}}{r^{2}}\frac{t}{8n^{2}}|x-y_{0}|_{A}^{2}-\frac{m_{0}}{r^{2}}(q-q_{0})\cdot(x-y_{0}),$

where we used the symmetry of $K_{r}$ and the normalization $\int
K_{r}(y)\,dy=1$. ∎

The sets $V_{h}$ and $\Sigma_{h}$: For $h>0$ and $q_{0}\in V$, we introduce
the set

$\displaystyle W_{h}(q_{0})$
$\displaystyle:=q_{0}+\frac{h}{2}(S(y_{0},q_{0},h)-y_{0})^{\circ}$
$\displaystyle=\\{q\in\mathbb{R}^{n}\,;\,(q-q_{0})\cdot(x-y_{0})\leq\frac{h}{2},\;\forall
x\in S(y_{0},q_{0},h)\\}$

where $\\{y_{0}\\}=\partial\psi^{*}(q_{0})$. We note that the polar body
$(S(y_{0},q_{0},h)-y_{0})^{\circ}$ of a convex set (see definition (58)) is
convex and contains $0$. So $W_{h}(q_{0})$ is expected to be a small convex
set around $q_{0}$. We then introduce the sets

$V_{h}=\bigcup_{q_{0}\in V}W_{h}(q_{0})$

and

$\Sigma_{h}=\bigcup_{y_{0}\in\Sigma}S(y_{0},q_{0},h).$

The properties of these sets are summarized in the proposition below.

######  Proposition 6.4.

The following properties hold for some $k$ depending only on $n$

$\partial\psi^{*}(W_{h}(q_{0}))\subset S(y_{0},q_{0},h)$ and so
$\partial\psi^{*}(V_{h})\subset\Sigma_{h}$

$\ell(S(y_{0},q_{0},h))\geq\frac{h}{L(\partial\psi(\Omega))}$ and
$\ell(W_{h}(q_{0}))\geq\frac{h}{4nL(\Omega)}$

If $t\geq C_{0}\delta$, $h\leq Mt$ and $S(y_{0},q_{0},2M^{k}t)\subset\Omega$
for all $y_{0}\in\Sigma$, then

$\Sigma_{h}\subset\theta M^{k}S(x_{0},0,t)$

For all $x\in S(y_{0},q_{0},h)$ and $q\in\partial\psi(x)$ we have

$(x-y_{0})\cdot(q-q_{0})\leq M^{k}h$

where the constants $C_{0}$, $M$ and $\theta$ are the constant appearing in
Proposition 3.5.

Before proving this proposition, we state the following lemma which follows
from Proposition 7.4, but for which we also give a short proof below.

######  Lemma 6.5.

Let $\psi$ be a convex function and $S=S(x_{0},p_{0},t)$ be a section of
$\psi$. If $x\in S$ is such that $2x-x_{0}\in S$ (which means that
$x\in\frac{1}{2}S$ where the contraction is done with respect to $x_{0}$) then

$(x-x_{0})\cdot(p-p_{0})\leq t\qquad\forall p\in\partial\psi(x)$

and

$x_{0}\in S(x,p,t).$

###### Proof.

Without loss of generality, we can assume that $x_{0}=0$, $p_{0}=0$ and
$\psi(x_{0})=0$. We are then assuming $x\in S$ and $2x\in S$, which gives
$\psi(2x)\leq t$. The convexity of $\psi$ implies

$t\geq\psi(2x)\geq\psi(x)+p\cdot x\geq p\cdot x\qquad\forall
p\in\partial\psi(x)$

which is the first inequality in the lemma. We can then write

$\psi(0)\leq\psi(x)\leq\psi(x)-p\cdot x+t$

which implies $0\in S(x,p,t)$ and concludes the proof. ∎

###### Proof of Proposition 6.4.

The first property is an immediate consequence of the definition of
$W_{h}(q_{0})$ since Proposition 7.4 (see (62)) implies

$\partial\psi^{*}(W_{h}(q_{0}))\subset S(y_{0},q_{0},h).$

Next, we note that Proposition 3.3 gives
$\ell(S(y_{0},q_{0},h))\geq\frac{h}{L(\partial\psi(\Omega))}$ which is the
first inequality in (ii) and Lemma 7.2 yields

$\ell(W_{h}(q_{0}))=\frac{h}{2}\ell((S(y_{0},q_{0},h)-y_{0})^{\circ})\geq\frac{h}{4nL(S(y_{0},q_{0},h))}\geq\frac{h}{4nL(\Omega)}$

which is the second inequality in (ii).

Lemma 6.5 together with the scaling and engulfing property of sections implies
(iii). Recall that the section $S(x_{0},0,t)$ is included in the ellipsoid
$\\{(x-x_{0}^{*})^{*}\,A\,(x-x_{0}^{*})\leq n^{2}\\}$ for $x_{0}^{*}$ the
center of mass. Since the ellipsoid
$\\{(x-x_{0}^{*})^{*}\,A\,(x-x_{0}^{*})\leq 1\\}$ is included in the section,
this implies that $2x_{0}^{*}-x_{0}\in n\,S(x_{0},0,t)$. Since $t\geq
C_{0}\,\delta$ the scaling property of sections, Proposition 3.5-(ii), gives
that $S(x_{0},0,t)\subset\frac{1}{3n}S(x_{0},0,M^{k}t)$ for some $k$ depending
only on $n$. Hence if $y_{0}\in S(x_{0},0,t)$,

$2y_{0}-x_{0}=x_{0}^{*}+2(y_{0}-x_{0}^{*})-(x_{0}-x_{0}^{*})=x_{0}^{*}+3\,\left(\frac{2}{3}\,y_{0}+\frac{1}{3}\,(2\,x_{0}^{*}-x_{0})-x_{0}^{*}\right)\in
3\,n\,S(x_{0},0,t).$

Hence we satisfy the assumption of Lemma 6.5 for $S=S(x_{0},0,M^{k}\,t)$
implying that $x_{0}\in S(y_{0},q_{0},M^{k}\,t)$. The engulfing property of
sections, namely Proposition 3.5-(iii), implies

$y_{0}\in S(y_{0},q_{0},M^{k}t)\subset S(x_{0},0,\theta M^{k}t).$

Since this holds for all $y_{0}\in\Sigma$ as $\Sigma\subset S(x_{0},0,t)$, it
follows that as long as $h\leq Mt$ we have

$\Sigma_{h}\subset S(x_{0},0,\theta
M^{k}t)\subset\theta\,M^{k}\,S(x_{0},0,t).$

Finally, if $x\in S(y_{0},q_{0},h)$ then the scaling property of sections
again gives $x\in\frac{1}{3n}\,S(y_{0},q_{0},M^{k}h)$ and so Lemma 6.5 implies
(iv). ∎

The choice of $h$: For any $x\in\Sigma_{h}$, we have $x\in S(y_{0},q_{0},t)$
for some $q_{0}\in V$, $y_{0}\in\Sigma$. So Lemma 6.3 together with
Proposition 6.4 implies that for some constant $C$

$\Delta_{r}\psi(x)\geq\frac{t}{4n^{2}}\mathrm{Tr}(A)-Cm_{0}\frac{h}{r^{2}}\qquad\forall
x\in\Sigma_{h}.$

We thus define

$h_{0}:=\frac{t}{8n^{2}Cm_{0}}r^{2}\mathrm{Tr}(A),\qquad\widetilde{\Sigma}:=\Sigma_{h_{0}}$

so that we have:

$\Delta_{r}\psi(x)\geq\frac{t}{8n^{2}}\,\mathrm{Tr}(A)\qquad\forall
x\in\widetilde{\Sigma}$

which is the most important property of $\widetilde{\Sigma}$. In addition,
unlike $\Sigma$, the set $\widetilde{\Sigma}$ is "large" enough for us to use
Assumption 1 to control its size. Indeed, we recall that

(43)
$\widetilde{\Sigma}=\bigcup_{y_{0}\in\Sigma}S(y_{0},q_{0},h_{0}),\quad\tilde{V}=\bigcup_{q_{0}\in
V}W_{h_{0}}(q_{0}),$

and since $\mathrm{Tr}(A)\geq\frac{1}{\ell(S)^{2}}$, we have $h_{0}\geq
c\frac{t}{\ell(S)^{2}}r^{2}.$ In particular, Proposition 6.4 (ii) implies that
if

$r\geq\delta^{1/2}\quad\mbox{ and }\quad\frac{t}{\ell(S)^{2}}\geq C$

then

(44) $\ell(S(y_{0},q_{0},h_{0}))\geq\delta\quad\mbox{ and
}\quad\ell(W_{h_{0}}(q_{0}))\geq\delta.$

We can now prove following proposition, which generalizes Proposition 5.3 to
the case $\delta>0$:

######  Proposition 6.6.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (3),
(11), (12), with $\mu$ and $\nu$ satisfying (9) and (10) with $\delta>0$.
There exists some constants $c$, $C_{1}$, $C_{2}$ depending on $n$, $\lambda$,
$\Lambda$, $\Omega$ and $\partial\psi(\Omega)$ such that the following holds:
Given $x_{0}\in\Omega$, $p_{0}\in\partial\psi(x_{0})$, $t$ such that
$S(x_{0},p_{0},C_{1}t)\subset\Omega$ and if

$t\geq C_{0}\delta\quad\mbox{ and }\quad\ell(S)\geq r\geq\delta^{1/2}$

and

(45) $\frac{t}{\ell(S)^{2}}\geq C_{2}$

then there exists a set $\widetilde{\Sigma}\subset CS(x_{0},p_{0},t)$ such
that

$|\widetilde{\Sigma}|\geq c|S|$

and

$\Delta_{r}\psi(x)\geq c\frac{t}{\ell(S)^{2}}\quad\mbox{ a.e. in
$\widetilde{\Sigma}$}.$

The assumptions require a short discussion: The condition $t\geq C_{0}\delta$
is necessary to use the standard properties of section Proposition 3.5. The
condition $r<\ell(S)$ is necessary to control the nonlocal quantity
$\Delta_{r}\psi$. The condition $r\geq\delta^{1/2}$ will then be used to
ensure that the sets $\tilde{\Sigma}$ and $\tilde{V}$ are "fat" enough to be
seen by the (possibly discrete) measures $\mu$ and $\nu$. Finally, we note
that (22) implies $\frac{t}{\ell(S)^{2}}\geq c$ for some constant $c$, but
(45) requires this quantity to be arbitrarily large. Instead of (45), we could
actually assume that $r\geq C_{3}\delta^{1/2}$ for some large constant
$C_{3}$. However, we will only use this proposition for sections for which
$\frac{t}{\ell(S)^{2}}$ is very large (see (52)), so this condition will not
be an issue.

###### Proof.

In view of (43) and (44), Assumption 1 and Proposition 3.2 imply

$\mu(\widetilde{\Sigma})=\mu(\widetilde{\Sigma}\cap\Omega)\leq\Lambda|\widetilde{\Sigma}|\quad\mbox{
and
}\quad\nu(\widetilde{V}\cap\mathcal{O})\geq\lambda|\widetilde{V}\cap\mathcal{O}|$

so that (12) and Proposition 6.4 (i) (which gives
$\partial\psi^{*}(\widetilde{V})\subset\widetilde{\Sigma}$) imply

$\lambda|\widetilde{V}\cap\mathcal{O}|\leq\nu(\widetilde{V}\cap\mathcal{O})\leq\mu(\partial\psi^{*}(\widetilde{V}\cap\mathcal{O}))\leq\mu(\partial\psi^{*}(\widetilde{V}))\leq\mu(\widetilde{\Sigma})\leq\Lambda|\widetilde{\Sigma}|.$

Using (40), and the fact that $V\subset\partial\psi(S)\subset\mathcal{O}$, we
deduce

$c\frac{t^{n}}{|S|}\leq|V|\leq|\widetilde{V}\cap\mathcal{O}|\leq\frac{\Lambda}{\lambda}|\widetilde{\Sigma}|.$

Finally, Proposition 3.5 (i) gives $t^{n}\geq c|S|^{2}$ which yields
$|\widetilde{\Sigma}|\geq c|S|$ for some constant $c$. ∎

###  6.2. A reversed Chebychev’s inequality when $\delta>0$

As in the case $\delta=0$, Theorem 2.5 follows from the following reversed
Chebychev’s inequality:

######  Theorem 6.7.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (11),
(12), with $\mu$ and $\nu$ satisfying (9) and (10)) with $\delta>0$. Let
$\Omega^{\prime}\subset\subset\Omega$ be such that there exists $\rho>0$ such
that (14) holds. Then there exists $\alpha_{0}$, $r_{0}$, $c$ and $C$
(depending on $\Omega$, $\rho$, $n$, $\lambda$, $\Lambda$,
$L(\partial\psi(\Omega))$) such that whenever

$\alpha>\alpha_{0},\qquad r<r_{0},\qquad\delta\leq r^{2},$

the following holds:

(46)
$\int_{\\{x\in\Omega^{\prime}\,|\,\Delta_{r}\psi\geq\alpha\\}}\Delta_{r}\psi\,dx\leq
C\alpha|\\{x\in\Omega\,|\,\Delta_{r}\psi\geq c\alpha\\}|.$

Before proving this result, we show that it implies our main Theorem 2.5:

###### Proof of Theorem 2.5.

Since $|\Delta_{r}\psi|\leq\frac{C}{r_{0}^{2}}\sup\psi$ when $r\geq r_{0}$, we
only need to prove the result when $r<r_{0}$. The proof is identical to the
proof of Theorem 2.4: Using Theorem 6.7 and Fubini’s theorem we can show that
there exists a constant $C$ such that

$\int_{\Omega^{\prime}}\Delta_{r}\psi(x)\log(2+\Delta_{r}\psi(x))\,dx\leq
C+C\int_{\Omega^{\prime\prime}}\Delta_{r}\psi(x)\,dx,$

where $\Omega\subset\subset\Omega^{\prime\prime}\subset\subset\Omega$ and we
can further have that $\Omega^{\prime\prime}+B(0,r)\subset\Omega$ by taking
$r_{0}$ small enough with respect to $\rho$.

We can now observe that

$\Delta_{r}\psi(x)=L_{r}\star\Delta\psi(x),$

where $\Delta\psi$ is a non-negative Radon measure and $L_{r}$ is a smooth
convolution kernel, bounded in $L^{1}$ and obtained through averages of
$K_{r}$.

Consequently,

$\int_{\Omega^{\prime\prime}}\Delta_{r}\psi(x)\,dx=\int_{\Omega^{\prime\prime}}L_{r}\star\Delta\psi(x)\,dx\leq\int_{\Omega}\Delta\psi\leq
C.$

∎

We now turn to the proof of Theorem 6.7.

###### Proof of Theorem 6.7.

The proof follows the same idea as that of Theorem 5.4 but requires several
adjustments: Given the set

$E_{\alpha}=\\{x\in\Omega^{\prime}\,|\,\Delta_{r}\psi(x)>\alpha\\},$

we need to identify, for each $x\in E_{\alpha}$ the value $t(x)$ of $t$ with
the following properties:

  1. (1)

We should have $t(x)\geq C_{0}\delta$ so that we are only working with
sections for which the properties listed in Proposition 3.5 are valid.

  2. (2)

We need to have $\fint_{S(x,t)}\Delta_{r}\psi\,dy\leq C\alpha$ for all
$t\in[t(x),\rho]$

  3. (3)

We need to have $\frac{t(x)}{\ell(S(x,t(x))^{2}}\geq c\alpha$ in order to use
Proposition 6.6.

  4. (4)

The particular choice of $q\in\partial\psi(x)$ to define the section is not
important except for the fact that we need to take $q$ such that (14) holds.

Lemma 6.1 can be used to show that $\fint_{S(x,q,t)}\Delta_{r}\psi\,dy\leq
C\frac{t}{\ell(S(x,q,t(x))^{2}}$ if $t\geq C_{0}\delta$ and
$r\leq\ell(S(x,q,t(x))$. This allows us to replace items 2 and 3 with the
condition

$\frac{t}{\ell(S(x,q,t(x))^{2}}\leq C\alpha\quad\forall
t\in[t(x),\rho],\qquad\frac{t(x)}{\ell(S(x,q,t(x))^{2}}\geq c\alpha.$

Making sure that $t(x)\geq C_{0}\delta$ is more delicate, but we note that if
$B_{r}(x)\subset S(x,q,t)$, then the definition of $\Delta_{r}$ implies
$\Delta_{r}\psi(x)\leq m_{0}\frac{t}{r^{2}}$. For $x\in E_{\alpha}$, we thus
have $t\geq\frac{\alpha}{m_{0}}r^{2}\geq\frac{\alpha_{0}}{m_{0}}\delta$ and we
get the desired inequality if $\alpha_{0}$ is large enough. It turns that
requiring that $B_{r}(x)\subset S(x,q,t(x))$ is too strong so we will require

$B_{r}(x)\subset S(x,q,Mt(x))$

where $M$ is the constant given in Proposition 3.5 such that $2S(x,q,t)\subset
S(x,q,Mt)$.

All these considerations lead us to introduce for all $x\in E_{\alpha}$,

(47)
$t(x)=\sup\left\\{t\leq\rho\,\Big{|}\,\max\left\\{\frac{t}{\ell(S(x,q,t))^{2}},\frac{r\alpha}{d(x,\partial
S(x,q,Mt))}\right\\}\geq\alpha\right\\}.$

Given $\varepsilon=\varepsilon_{1}\varepsilon_{2}>0$ (to be chosen later) we
claim that if $\alpha$ is large enough and $r$ is small enough, then we must
have $t(x)\leq\varepsilon\rho$. Indeed, we recall (see Proposition 3.3) that
$\ell(S(x,q,t))\geq\frac{t}{L(\partial\psi(\Omega))}$ and so

$\frac{t}{\ell(S(x,q,t))^{2}}\leq\frac{L(\partial\psi(\Omega))^{2}}{t}\leq\alpha_{0}:=\frac{L(\partial\psi(\Omega))^{2}}{\varepsilon\rho}\qquad\forall
x\in\Omega^{\prime},\;\forall t\in[\varepsilon\rho,\rho]$

while Alexandrov’s estimate (Proposition 3.4) implies

$d(x,\partial S(x,q,Mt))\geq
r_{0}:=\frac{(\varepsilon\rho)^{n}}{|\partial\psi(\Omega)|L(\Omega)^{n-1}}\qquad\forall
x\in\Omega^{\prime},\;\forall t\in[\varepsilon\rho,\rho].$

From now on, we assume that $r<r_{0}$ and $\alpha>\alpha_{0}$. We then have,
for all $x\in\Omega^{\prime}$,

$\max\left\\{\frac{t}{\ell(S(x,q,t))^{2}},\frac{r\alpha}{d(x,\partial
S(x,q,Mt))}\right\\}<\alpha\qquad\forall t\in[\varepsilon\rho,\rho]$

and so $t(x)\leq\varepsilon\rho$.

The definition of $t(x)$, (47) implies

(48) $\frac{t}{\ell(S(x,q,t))^{2}}\leq\alpha\qquad\forall x\in
E_{\alpha},\;\forall t\in[t(x),\rho]$

and

(49) $d(x,\partial S(x,q,Mt))\geq r\qquad\qquad\forall x\in
E_{\alpha},\;\forall t\in[t(x),\rho].$

The first condition was one of the required conditions for $t(x)$ and the
second condition can be used to show that $t(x)\geq C_{0}\delta$. Indeed, for
$x\in E_{\alpha}$, condition (49) implies that $B_{r}(x)\subset S(x,q,Mt)$ and
so

$\displaystyle\alpha<\Delta_{r}\psi(x)$
$\displaystyle=\frac{m_{0}}{r^{2}}\int_{B_{r}(x)}K_{r}(y-x)[\psi(y)-\psi(x)-q\cdot(y-x)]\,dy$
(50) $\displaystyle\leq\frac{m_{0}}{r^{2}}Mt\int
K_{r}(y)\,dy=\frac{m_{0}M}{r^{2}}t\qquad\forall t\in[t(x),\rho]$

and so (with $t=t(x)$)

(51) $t(x)\geq\frac{\alpha}{m_{0}M}r^{2}\geq\frac{\alpha_{0}}{m_{0}M}\delta.$

We can always take a larger $\alpha_{0}$ so that
$\frac{\alpha_{0}}{m_{0}M}>C_{0}$ (which we assume in what follows).

Finally, we can prove:

######  Lemma 6.8.

Let $t(x)$ be defined by (47). Then

(52) $\frac{t(x)}{\ell(S(x,q,t(x)))^{2}}\geq c\alpha,\qquad\forall x\in
E_{\alpha}$

and

(53) $\fint_{S(x,q,t)}\Delta_{r}\psi(y)\,dy\leq C\alpha\qquad\forall x\in
E_{\alpha},\quad t\in[t(x),\varepsilon_{1}\rho].$

###### Proof of Lemma 6.8.

Observe that for a fixed $x$ and $q\in\partial\psi(x)$, the map $t\to
S(x,q,t)$ is continuous thanks to the convexity of $\psi$. Indeed considering
a sequence $t_{n}\to t$ with for example $t_{n}<t$, we first trivially have
that $\bigcup_{n}S(x,q,t_{n})\subset S(x,q,t)$ and thus
$\overline{\bigcup_{n}S(x,q,t_{n})}\subset S(x,q,t)$. On the other hand if
there exists $y\in S(x,q,t)\setminus\overline{\bigcup_{n}S(x,q,t_{n})}$ then
necessarily $q\in\partial\psi(y)$. As $q\in\partial\psi(x)$, by the convexity
of $\psi$, this is possible only if $t=\psi(x)$.

The definition (47) then implies that

$\max\left\\{\frac{t(x)}{\ell(S(x,q,t(x)))^{2}},\frac{r\alpha}{d(x,\partial
S(x,q,Mt(x)))}\right\\}=\alpha.$

In particular we either have $\frac{t(x)}{\ell(S(x,q,t(x)))^{2}}=\alpha$ (and
(52) is true) or $\frac{t(x)}{\ell(S(x,q,t(x)))^{2}}<\alpha$ and $d(x,\partial
S(x,q,Mt(x)))=r$. In that last case we go back to (50) which implies

$\alpha\leq m_{0}M\frac{t(x)}{r^{2}}=m_{0}M\frac{t(x)}{d(x,\partial
S(x,q,Mt))^{2}}.$

In order to conclude, we note that (here we write $\ell=\ell(S(x,q,t(x)))$)
$B_{\ell}(0)\subset S(x,q,t(x))$ and so (using the convexity of $S$)
$B_{\ell}(x)\subset 2S(x,q,t(x))\subset S(x,q,Mt(x))$ (using Proposition 3.5
and (51)). It follows that $\ell(S(x,q,t(x))\leq Cd(x,\partial S(x,q,Mt))$ and
the inequality above implies

$\alpha\leq m_{0}M\frac{t(x)}{\ell(S(x,q,t(x)))^{2}}$

which is (52).

To prove inequality (53), we take $t\in[t(x),\varepsilon_{1}\rho]$ and $x\in
E_{\alpha}$. Lemma 6.1, together with the fact that
$\mathrm{Tr}(A)\leq\frac{n}{\ell(S)^{2}}$ and $\ell(S(x,q,t))>r$ implies

$\fint_{S(x,q,t)}\Delta_{r}\psi(y)\,dy\leq\frac{C}{\ell(S(x,q,t))^{2}}\sup_{y\in
3nS(x,q,t)}[\psi(y)-\psi(x)-q\cdot(y-x)].$

Condition (51) implies that the scaling property of section (23) can be used
to prove that there exists a constant $C_{1}=(3n)^{\beta}$ such that
$3nS(x,q,t)\subset S(x,q,C_{1}t)$ provided $S(x,q,C_{1}t)\subset\Omega$ which
holds if we choose $\varepsilon_{1}=\frac{1}{C_{1}}$ (since
$t\leq\varepsilon_{1}\rho$). We deduce

$\fint_{S(x,q,t)}\Delta_{r}\psi(y)\,dy\leq\frac{Ct}{\ell(S(x,q,t))^{2}}.$

Inequality (48) then implies (53). ∎

The rest of the proof is similar to the case $\delta=0$: We have
$E_{\alpha}\subset\bigcup_{x\in E_{\alpha}}S(x,q,t(x))$ and since $t(x)\geq
C\delta$, we can use Vitali’s covering lemma (see Corollary 3.6) to extract a
countable family $x_{k}$ such that the sections
$S_{k}=S(x_{k},q_{k},t(x_{k}))$ are pairwise disjoint and
$E_{\alpha}\subset\bigcup_{k}C_{*}S_{k}$.

Successive applications of Proposition 7.6 gives $C_{*}S_{k}\subset
S(x_{k},q_{k},C_{**}t_{k})$ for some constant $C_{**}$ provided
$C_{**}t_{k}\leq\rho.$ Furthermore, in order to be able to use Inequality
(53), we will require that $C_{**}t_{k}\leq\varepsilon_{1}\rho$. We thus
choose

$\varepsilon_{2}:=\frac{1}{C_{**}}$

The condition $t_{k}\leq\varepsilon\rho$ with
$\varepsilon=\varepsilon_{1}\varepsilon_{2}$ then imply that
$C_{**}t_{k}\in[t_{k},\varepsilon_{1}\rho]$. Inequality (53) thus implies

$\int_{C_{*}S_{k}}\Delta_{r}\psi(x)\,dx\leq\int_{S(x_{k},q_{k},C_{**}t_{k})}\Delta_{r}\psi(x)\,dx\leq\alpha|S(x_{k},q_{k},C_{**}t_{k})|$

and Proposition 3.5 (i) gives $|S(x_{k},q_{k},C_{**}t_{k})|\leq
C_{n}|S(x_{k},q_{k},t_{k})|$. We deduce that

$\int_{C_{*}S_{k}}\Delta_{r}\psi(x)\,dx\leq C\alpha|S_{k}|.$

We are ready to conclude: First, we write

$\displaystyle\int_{E_{\alpha}}\Delta_{r}\psi(x)\,dx$
$\displaystyle\leq\sum_{k}\int_{C_{*}S_{k}}\Delta_{r}\psi(x)\,dx\leq
C\alpha\sum_{k}|S_{k}|.$

We then use Proposition 6.6: We have $\frac{t_{k}}{\ell(S_{k})^{2}}\geq
c\alpha$ (see (52)), so there is a subset $\Sigma_{k}\subset S_{k}$ such that

$|\Sigma_{k}|\geq c|S_{k}|,\qquad\Delta_{r}\psi(x)\geq
c\frac{t_{k}}{\ell(S_{k})^{2}}\geq c\alpha\quad\forall x\in\Sigma_{k},$

and since the $S_{k}$, and therefore the $\Sigma_{k}$, are disjoint, we have
that

$\displaystyle\int_{E_{\alpha}}\Delta_{r}\psi(x)\,dx$ $\displaystyle\leq
C\alpha\sum_{k}|\Sigma_{k}|$ $\displaystyle\leq
C\alpha|\\{x\in\Omega\,|\,\Delta_{r}\psi(x)\geq c\alpha\\}|$

which is the desired inequality. ∎

###  6.3. Sobolev regularity

A key step in deriving Sobolev $W^{2,p}$ estimates on $\psi$ up to scale $r$
consists in localizing the reversed Chebychev’s inequality of Theorem 6.7.

######  Theorem 6.9.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (11),
(12), with $\mu$ and $\nu$ satisfying (9) and (10)) with $\delta>0$. Let
$\Omega^{\prime}\subset\subset\Omega$ be such that there exists $\rho>0$ such
that (14) holds. Then there exists $\alpha_{0}$, $r_{0}$, $c$ and $C$
(depending on $\Omega$, $\rho$, $n$, $\lambda$, $\Lambda$,
$L(\partial\psi(\Omega))$) such that whenever

$\alpha>\alpha_{0},\qquad
r<r_{0},\quad\alpha<\frac{c}{r^{1/n}},\qquad\delta\leq r^{2},$

the following holds for any ball $B(x_{0},R)\subset\Omega^{\prime}$

(54) $\int_{\\{x\in
B(x_{0},R)\,|\,\Delta_{r}\psi\geq\alpha\\}}\Delta_{r}\psi\,dx\leq
C\alpha|\\{x\in B(x_{0},R+C\,\alpha^{-1/\beta})\,|\,\Delta_{r}\psi\geq
c\alpha\\}|.$

###### Proof.

The proof follows the exact same lines as the proof of Theorem 6.7 by covering
the set $\\{x\in B(x_{0},R)\,|\,\Delta_{r}\psi\geq\alpha\\}$ with sections
$S(x,t(x))$ where $x\in B(x_{0},R)$. This leads to

$\int_{\\{x\in
B(x_{0},R)\,|\,\Delta_{r}\psi\geq\alpha\\}}\Delta_{r}\psi\,dx\leq
C\alpha|\\{x\in B(x_{0},R+R^{\prime})\,|\,\Delta_{r}\psi\geq c\alpha\\}|,$

where $R^{\prime}=\sup_{x\in B(x_{0},R)}L(S(x,t(x))$.

The only difference with Theorem 6.7 therefore lies in bounding precisely
$R^{\prime}$. We go back to the definition (47) of $t(x)$ and use first again
Proposition 3.3 which gives
$\ell(S(x,q,t))\geq\frac{t}{L(\partial\psi(\Omega))}$, so that for some
constant $C$ we have

$\frac{t}{\ell(S(x,q,t))^{2}}\leq\frac{L(\partial\psi(\Omega))^{2}}{t}<\alpha,\quad\forall
t>\frac{C}{\alpha}.$

Similarly using Proposition 3.4, we have that

$d(x,\partial
S(x,q,Mt))\geq\frac{t^{n}}{|\partial\psi(\Omega)|\,L(\Omega)^{n-1}}>r,\quad\forall
t>C\,r^{1/n}.$

Since we assumed that $\alpha<\frac{c}{r^{1/n}}$, this last inequality holds
in particular for all $t>\frac{C}{\alpha}$. Taken together, the two
inequalities and (47) imply that $t(x)\leq\frac{C}{\alpha}$.

To conclude, it is now enough to use Lemma 4.2 which shows that

$L(S(x,p,t))\leq C\,t(x)^{1/\beta}\leq\frac{C}{\alpha^{1/\beta}}.$

Since this bound holds for all $x\in B(x_{0},R)$, we find $R^{\prime}\leq
C\alpha^{-1/\beta}$, which concludes the proof. ∎

Using Theorem 6.9, we can now prove the Sobolev regularity Theorem 2.6

###### Proof of Theorem 2.6.

The proof mostly follows the arguments presented in [17] with only minor
adaptations. Since $|\Delta_{r}\psi|\leq\frac{C}{r_{0}^{2}}\sup\psi$ when
$r\geq r_{0}$, we only need to prove the result when $r<r_{0}$.

For some $\gamma>1$ to be chosen later, we define $\alpha_{k}$ by induction
with $\alpha_{k+1}=\gamma\,\alpha_{k}$, $\alpha_{1}$ large enough, and
$R_{k+1}=R_{k}-C/\alpha_{k}^{1/\beta}$ with $R_{1}=2\,R$.

As long as $\alpha_{k}\leq\frac{C}{r^{1/n}}$ and provided that $\gamma$ was
chosen large enough so that $R_{k}\geq R$ for all $k$, we can apply Theorem
6.9 to have that

$\int_{\\{x\in
B(x_{0},R_{k+1})\,|\,\Delta_{r}\psi\geq\alpha_{k}/c\\}}\Delta_{r}\psi\,dx\leq\frac{C}{c}\alpha_{k}|\\{x\in
B(x_{0},R_{k})\,|\,\Delta_{r}\psi\geq\alpha_{k}\\}|.$

As a consequence

$\begin{split}\int_{\\{x\in
B(x_{0},R_{k+1})\,|\,\Delta_{r}\psi\geq\alpha_{k+1}\\}}\Delta_{r}\psi\,dx\leq&\frac{C}{c}\alpha_{k}|\\{x\in
B(x_{0},R_{k})\,|\,\alpha_{k}\leq\Delta_{r}\psi\leq\alpha_{k+1}\\}|\\\
&+\frac{C}{c\,\gamma}\,\alpha_{k+1}\,|\\{x\in
B(x_{0},R_{k})\,|\,\Delta_{r}\psi\geq\alpha_{k+1}\\}|,\end{split}$

or, taking $\gamma\geq 2C/c$,

$\begin{split}\int_{\\{x\in
B(x_{0},R_{k+1})\,|\,\Delta_{r}\psi\geq\alpha_{k+1}\\}}\Delta_{r}\psi\,dx\leq&\frac{C}{c}\,\int_{\\{x\in
B(x_{0},R_{k})\,|\,\alpha_{k}\leq\Delta_{r}\psi\leq\alpha_{k+1}\\}}\Delta_{r}\psi\,dx\\\
&+\frac{1}{2}\,\int_{\\{x\in
B(x_{0},R_{k}))\,|\,\Delta_{r}\psi\geq\alpha_{k+1}\\}}\Delta_{r}\psi\,dx.\end{split}$

We deduce:

$\begin{split}\int_{\\{x\in
B(x_{0},R_{k+1})\,|\,\Delta_{r}\psi\geq\alpha_{k+1}\\}}\Delta_{r}\psi\,dx\leq&\frac{2C}{c}\,\int_{\\{x\in
B(x_{0},R_{k})\,|\,\alpha_{k}\leq\Delta_{r}\psi\leq\alpha_{k+1}\\}}\Delta_{r}\psi\,dx\\\
&+\,\int_{\\{x\in B(x_{0},R_{k})\setminus
B(x_{0},R_{k+1})\,|\,\Delta_{r}\psi\geq\alpha_{k+1}\\}}\Delta_{r}\psi\,dx.\end{split}$

Defining correspondingly $D_{k}=\\{x\in
B(x_{0},R_{k})\,|\,\Delta_{r}\psi\geq\alpha_{k}\\}$, this is equivalent, for a
different constant $C$, to

$\int_{D_{k+1}}\Delta_{r}\psi\,dx\leq C\,\int_{D_{k}\setminus
D_{k+1}}\Delta_{r}\psi\,dx,$

or

$(1+C)\,\int_{D_{k+1}}\Delta_{r}\psi\,dx\leq
C\,\int_{D_{k}}\Delta_{r}\psi\,dx.$

This implies that

$\int_{D_{k+1}}\Delta_{r}\psi\,dx\leq\tau\,\int_{D_{k}}\Delta_{r}\psi\,dx,$

for $\tau=C/(1+C)<1$.

By induction, and since $B(x_{0},R)\subset B(x_{0},R_{k})$, we can immediately
deduce that

(55) $\int_{\\{x\in
B(x_{0},R)\,|\,\Delta_{r}\psi\geq\alpha_{k}\\}}\Delta_{r}\psi\,dx\leq\tau^{k}\int_{\Omega}\Delta_{r}\psi\,dx\leq\frac{C}{\alpha_{k}^{\theta}},$

as long as $\alpha_{k}\leq\frac{C}{r^{1/n}}$ and for some $\theta>0$ since
$\alpha_{k}=\gamma^{k}\,\alpha_{1}$ (so that we can take
$\theta=-\log\tau/\log\gamma$).

Denote $k_{0}$ the last index $k$ s.t. $\alpha_{k_{0}}\leq\frac{C}{r^{1/n}}$
and decompose

$\begin{split}\int_{B(x_{0},R)}(\Delta_{r}\psi)^{p}\,dx=\int_{B(x_{0},R)\cap\\{\Delta_{r}\psi\leq\alpha_{k_{0}}\\}}(\Delta_{r}\psi)^{p}\,dx+\int_{B(x_{0},R)\cap\\{\Delta_{r}\psi>\alpha_{k_{0}}\\}}(\Delta_{r}\psi)^{p}\,dx.\end{split}$

Because $\Delta_{r}\psi(x)\leq\frac{C}{r^{2}}$, the last term is in fact easy
to handle,

$\begin{split}\int_{B(x_{0},R)\cap\\{\Delta_{r}\psi>\alpha_{k_{0}}\\}}(\Delta_{r}\psi)^{p}\,dx&\leq\frac{C}{r^{2(p-1)}}\,\int_{B(x_{0},R)\cap\\{\Delta_{r}\psi>\alpha_{k_{0}}\\}}\Delta_{r}\psi\,dx\\\
&\leq\frac{C}{r^{2(p-1)}}\,\frac{1}{\alpha_{k_{0}}^{\theta}}\leq
C\,\frac{r^{\theta/n}}{r^{2(p-1)}}\leq C,\end{split}$

provided that $p$ is close enough to $1$ so that $\theta/n\geq 2(p-1)$.

We now recall that for any measurable function $f(x)\geq 0$ and $p>1$,

$\int_{B(x_{0},R)}f^{p}\,dx=(p-1)\,\int_{0}^{\infty}\alpha^{p-2}\,\int_{\\{x\in
B(x_{0},R)\,|\,f(x)\geq\alpha\\}}f\,dx.$

Hence

$\begin{split}\int_{B(x_{0},R)\cap\\{\Delta_{r}\psi\leq\alpha_{k_{0}}\\}}(\Delta_{r}\psi)^{p}\,dx&=(p-1)\,\int_{0}^{\alpha_{k_{0}}}\alpha^{p-2}\,\int_{\\{x\in
B(x_{0},R)\,|\,\Delta_{r}\psi\geq\alpha\\}}\Delta_{r}\psi\,dx\\\ &\leq
C+C\,\sum_{k=1}^{k_{0}}\alpha_{k}^{p-1}\,\int_{\\{x\in
B(x_{0},R)\,|\,\Delta_{r}\psi\geq\alpha_{k}\\}}\Delta_{r}\psi\,dx.\end{split}$

By using again (55), we finally conclude that

$\int_{B(x_{0},R)\cap\\{\Delta_{r}\psi\leq\alpha_{k_{0}}\\}}(\Delta_{r}\psi)^{p}\,dx\leq
C+C\,\sum_{k=1}^{k_{0}}\gamma^{k(p-1-\theta)}\leq C,$

as long as $p$ is again selected close enough to $1$ s.t. $p-1-\theta<0$. ∎

##  7\. Properties of sections

In this section, we establish some important properties of the sections of a
convex function that satisfies inequalities (11)-(12) for measures $\mu$ and
$\nu$ such that Assumption 1 (inequalities (9)-(10)) holds. Namely we
establish some bounds on the volume of the sections (Proposition 7.5), a
scaling property of sections (Proposition 7.6) and the engulfing property of
sections (Proposition 7.9). These three properties are classical when
$\delta=0$ (see for instance [11], [25], [24, Chapter 3]) and play an
important role in the regularity theory for Monge-Ampère equation. When
$\delta>0$, it is clear that they can only hold for large enough sections. The
main interest of this section is thus to clearly identify the condition on $t$
(depending on $\delta$) for these conditions to hold.

An important tool in the proof of these results when $\delta=0$ is the
renormalization of convex sets provided by John’s lemma. However, this
renormalization procedure is not compatible with Assumption 1 since it changes
the length scales differently depending on the direction. For this reason, our
proofs will avoid the use of renormalization whenever Assumption 1 must be
used. They will make extensive use of the notion of polar body of a convex
set, a notion well suited to optimal transportation. To our knowledge this
approach is new even in the case $\delta=0$.

Throughout this section, we will assume that $\mu$ and $\nu$ satisfy the
following condition, which is equivalent to Assumption 1 (see Proposition
B.1):

######  Assumption 2.

There exist $\lambda$, $\Lambda>0$ such that

(56) $\lambda|K|\leq\mu(K)\leq\Lambda|K|,\quad\mbox{for all convex set
$K\subset\Omega$ with $\ell(K)\geq\delta$}$

and

(57) $\lambda|K|\leq\nu(K)\leq\Lambda|K|,\quad\mbox{for all convex set
$K\subset\mathcal{O}$ with $\ell(K)\geq\delta$.}$

###  7.1. Polar bodies of a convex set

Several proofs in this section will make use of the notion of polar body of a
convex set, which turns out to be a convenient tool to relate the section $S$
of a convex function $\psi$ and its image $\partial\psi(S)$. For a convex set
$K$ containing the origin, we recall that the polar body is defined by

(58) $K^{\circ}=\\{z\in\mathbb{R}^{n}\,;\,z\cdot x\leq 1,\quad\forall x\in
K\\}.$

We recall the following basic facts about polar bodies:

  1. (1)

If $A\subset B$ then $B^{\circ}\subset A^{\circ}$

  2. (2)

$B_{r}(0)^{\circ}=B_{1/r}(0)$

  3. (3)

The volume product $|K|\,|K^{\circ}|$ is unchanged by linear (but not affine)
transformations: Indeed,

$z\in(L(K))^{\circ}\Leftrightarrow z\cdot L(x)\leq 1\;\forall x\in
K\Leftrightarrow L^{*}(z)\in K^{\circ}$

and so $|L(K)|\,|(L(K))^{\circ}|=|\det
L||K|\,|\det(L^{*})^{-1}||K^{\circ}|=|K|\,|K^{\circ}|$.

There is an abundant literature devoted to the study of the polar body and in
particular to the derivation of optimal lower and upper bounds for the volume
product $|K|\,|K^{\circ}|$. We will use the following classical result:

######  Proposition 7.1.

Let $K$ be a bounded convex subset of $\mathbb{R}^{n}$ with non-empty
interior. We have:

(59) $\inf_{x_{0}\in K}|K|\,|(K-x_{0})^{\circ}|\leq n^{n}\omega_{n}^{2}$

(the unique point $x_{0}$ where the infimum in (59) is attained is called the
Santaló point of $K$) and

(60) $|K|\,|K^{\circ}|\geq\frac{\omega_{n}^{2}}{(2n)^{n}}$

where $\omega_{n}$ denotes the volume of the $n$-dimensional unit ball.

These inequalities are classical and we give a short proof below for the sake
of completeness. Note that the constants in (59) and (60) are not optimal. We
refer for example to Bourgain-Milman [4] for discussion about optimal
constants.

###### Proof.

When $K$ is a convex body that is symmetric with respect to the origin, John’s
Lemma 3.1 implies that there is a linear transformation $L$ such that
$B_{1}(0)\subset L(K)\subset B_{n}(0)$. We then have
$B_{1/n}(0)\subset(L(K))^{\circ}\subset B_{1}(0)$ and so

(61)
$\frac{\omega_{n}^{2}}{n^{n}}\leq|L(K)|\,|(L(K))^{\circ}|=|K|\,|K^{\circ}|\leq
n^{n}\omega_{n}^{2}.$

When $K$ is not symmetric with respect to the origin, John’s Lemma gives
$B_{1}(a)\subset L(K)\subset B_{n}(a)$ for some $a=L(y)$ with $y\in K$. The
same computation as above, applied to $K-y$ thus gives

$\frac{\omega_{n}^{2}}{n^{n}}\leq|K|\,|(K-y)^{\circ}|\leq n^{n}\omega_{n}^{2}$

which gives in particular (59).

We note that if $0\notin K$, then $|K^{\circ}|$ is infinite and (60) holds. So
we can assume that $0\in K$ and thus $0\in L(K)\subset B_{n}(a)$. It follows
that $L(K)\subset B_{2n}(0)$ and so $L(K)^{\circ}\supset B_{1/2n}(0)$.
Proceeding as above, we deduce

$|K|\,|K^{\circ}|=|L(K)|\,|(L(K))^{\circ}|\geq|B_{1}||B_{1/2n}|=\frac{\omega_{n}^{2}}{(2n)^{n}}$

which proves (60). ∎

Finally, we can get the following bounds on the inner and outer radii of $K$
and $K^{\circ}$.

######  Lemma 7.2.

Let $K$ be a bounded convex set with non-empty interior such that $0\in K$.
Then

$\ell(K)L(K^{\circ})\geq\frac{1}{2n}\qquad\ell(K^{\circ})L(K)\geq\frac{1}{2n}$

###### Proof.

John’s Lemma implies the existence of an ellipsoid $E$ such that

$E\subset K\subset nE,$

where we note that $nE$ denotes a dilation of $E$ about its center, which is
not necessarily $0$. In particular, we have $\ell(K)\geq\ell(E)$ and
$(nE)^{\circ}\subset K^{\circ}$.

Next, we note that the polar dual of a centered ellipsoid
$E_{0}=\left\\{\sum_{i=1}^{n}\left(\frac{x_{i}}{a_{i}}\right)^{2}\leq
1\right\\}$ is the ellipsoid
$E_{0}^{\circ}=\left\\{\sum_{i=1}^{n}\left(a_{i}u_{i}\right)^{2}\leq
1\right\\}$. In particular $\ell(E_{0}^{\circ})L(E_{0})=1$ and
$L(E_{0}^{\circ})\ell(E_{0})=1$ for such an ellipsoid.

The ellipsoid $F:=nE$ might not be centered, but we have $0\in F$, so there is
a centered ellipsoid $F_{0}$, twice as large as $F$, such that $F\subset
F_{0}$ and $F_{0}^{\circ}\subset F^{\circ}$. We have $L(F^{\circ})\geq
L(F_{0}^{\circ})=\frac{1}{\ell(F_{0})}=\frac{1}{2\ell(F)}$ and
$L(F)=\frac{1}{2}L(F_{0})=\frac{1}{2\ell(F_{0}^{\circ})}\geq\frac{1}{2\ell(F^{\circ})}$.
We thus have:

$\ell(nE)L((nE)^{\circ})\geq\frac{1}{2},\qquad
L(nE)\ell((nE)^{\circ})\geq\frac{1}{2}.$

Finally, we can write

$\ell(K)\geq\ell(E)=\frac{1}{n}\ell(nE)\geq\frac{1}{2nL((nE)^{\circ})}\geq\frac{1}{2nL(K^{\circ})}$

which gives the first inequality in Lemma 7.2. $K^{\circ}$ is a convex set
containing zero and $(K^{\circ})^{\circ}=K$ so the second inequality follows
by applying the first inequality to $K^{\circ}$. ∎

This lemma readily implies.

######  Corollary 7.3.

Let $\Omega$ be a bounded subset of $\mathbb{R}^{n}$ and
$\psi:\Omega\to\mathbb{R}$ convex. For any section $S\subset\Omega$, we have

$\ell((S-y)^{\circ})\geq\frac{1}{2nL(\Omega)}\qquad\forall y\in S.$

###### Proof.

The result follows from Lemma 7.2 since we have

$\frac{1}{2n}\leq\ell((S-y)^{\circ})L(S-y)\leq\ell((S-y)^{\circ})L(\Omega).$

∎

The usefulness of polar bodies in the context of optimal transportation is
made clear by the next proposition. We recall that for a convex function
$\psi:\mathbb{R}^{n}\to\mathbb{R}$ and given $x_{0}\in\mathbb{R}^{n}$,
$p_{0}\in\partial\psi(x_{0})$ and $t\geq 0$, the section centered at $x_{0}$
with height $t$ is the convex set

$S(x_{0},p_{0},t):=\\{x\in\mathbb{R}^{n}\,;\,\psi(x)\leq\psi(x_{0})+p_{0}\cdot(x-x_{0})+t\\}.$

######  Proposition 7.4.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function. For all
$x_{0}\in\mathbb{R}^{n}$, $p_{0}\in\partial\psi(x_{0})$ and $t\geq 0$ such
that $S(x_{0},p_{0},t)$ is bounded, we have

(62) $p_{0}+t(S-x_{0})^{\circ}\subset\partial\psi(S)\quad\mbox{ and
}\quad\partial\psi^{*}(p_{0}+\frac{t}{2}(S-x_{0})^{\circ})\subset S$

and

(63) $\partial\psi\left(\frac{1}{2}(S+y_{0})\right)\subset
p_{0}+2t(S-y_{0})^{\circ}\qquad\forall y_{0}\in S.$

When $x_{0}=0$ and $p_{0}=0$, (62) can be written simply as
$\partial\psi^{*}(\frac{t}{2}S^{\circ})\subset S$ and
$tS^{\circ}\subset\partial\psi(S)$. The proof will make it clear that we also
have $\partial\psi^{*}((1-\eta)tS^{\circ})\subset S$ for all $\eta>0$ and even
$\partial\psi^{*}(tS^{\circ})\subset S$ if $\psi$ is strictly convex. This
simple relation between the sections of $\psi$ and their dual body, combined
with the volume product inequality (59) readily imply in particular the
estimate

(64) $|\partial\psi(S)|\,|S|\geq\frac{\omega_{n}^{2}}{(2n)^{n}}t^{n}.$

###### Proof of Proposition 7.4.

Up to replacing $\psi$ with the function
$\overline{\psi}(x)=\psi(x_{0}+x)-p_{0}\cdot x-\psi(x_{0})$, it is enough to
prove the result when $x_{0}=0$, $p_{0}=0$ and $\psi(x_{0})=0$. We then have
$\bar{S}=S-x_{0}$, $\partial\overline{\psi}(x)=\partial\psi(x_{0}+x)-p_{0}$
and $\partial\overline{\psi}^{*}(z)=\partial\psi^{*}(p_{0}+z)-x_{0}$.

Let $z\in\mathbb{R}^{n}$ be such that $z\cdot x\leq t$ for all
$x\in\overline{S}$ (that is $z\in t\overline{S}^{\circ}$). The function
$u(x)=\overline{\psi}(x)-z\cdot x$ is convex and satisfies

$u(0)=0,\qquad u(x)\geq 0\quad\mbox{ for }x\in\partial\overline{S},$

using the fact that $\overline{\psi}(x)=t$ and $z\cdot x\leq t$ for all
$x\in\partial\overline{S}$.

In particular, there is at least one $y\in\overline{S}$ such that $u$ has a
minimum at $y$, and for such a point we have $0\in\partial u(y)$ and so
$z\in\partial\overline{\psi}(y)$. This implies
$t\overline{S}^{\circ}\subset\partial\overline{\psi}(\overline{S})$ which is
the first inclusion in (62).

Next, we note that if $z\in\mathbb{R}^{n}$ is such that $z\cdot
x\leq\frac{t}{2}$ for all $x\in\overline{S}$ (that is
$z\in\frac{t}{2}\overline{S}^{\circ}$), then the same function $u$ as above
satisfies $u(x)>0$ for $x\in\partial\overline{S}$. In particular, any minimum
$y\in\mathbb{R}^{n}$ of $u$ must lie $S$. Since $u$ is convex, the set of
points such that $0\in\partial u$ coincides with the set of points where $u$
is minimum. So we can claim that for all $z\in\frac{t}{2}S^{\circ}$, we have
$z\in\partial\psi(x)\Rightarrow x\in S$. This implies
$\partial\psi^{*}(\frac{t}{2}S^{\circ})\subset S$.

In order to prove (63), we recall that for any $y_{0}\in\overline{S}$, we have

$(\overline{S}-y_{0})^{\circ}=\\{z\in\mathbb{R}^{n}\,;\,z\cdot(y-y_{0})\leq
1\,\mbox{ for all }y\in\overline{S}\\}.$

We now fix $\bar{x}\in\frac{1}{2}(\overline{S}+y_{0})$. Then
$z\in\partial\overline{\psi}(\bar{x})$ implies
$\overline{\psi}(\bar{x})+z\cdot(x-\bar{x})\leq\overline{\psi}(x)$ for all
$x\in\mathbb{R}^{n}$ and so

(65) $z\cdot(x-\bar{x})\leq t\quad\mbox{ for all }x\in\overline{S}.$

In particular, if we take $x=2\bar{x}-y_{0}$ in (65) (which we can do since
$2\bar{x}-y_{0}\in\overline{S}$ with our choice of $\bar{x}$), we get that
$z\cdot(\bar{x}-y_{0})\leq t$. Using (65) again, we deduce:

$z\cdot(x-y_{0})=z\cdot(x-\bar{x})+z\cdot(\bar{x}-y_{0})\leq 2t\quad\mbox{ for
all }x\in\overline{S}.$

We proved that $\partial\overline{\psi}(\bar{x})\subset
2t(\overline{S}-y_{0})^{\circ}$ for all
$\bar{x}\in\frac{1}{2}(\overline{S}+y_{0})$, which implies (63). ∎

###  7.2. Volume of sections: Proof of Proposition 3.5-(i)

We now prove the following proposition, which implies Proposition 3.5-(i):

######  Proposition 7.5.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (3),
(11)-(12) for measures $\mu$ and $\nu$ satisfying Assumption 1. Then there
exist two constants $c$ and $C$ depending only on $\lambda$, $\Lambda$ and $n$
such that if

(66) $t\geq\max\\{2L(\partial\psi(\Omega)),2nL(\Omega)\\}\delta$

then

(67) $c|S(x_{0},p_{0},t)|^{2}\leq t^{n}\leq C|S(x_{0},p_{0},t)|^{2}$

for all sections with $x_{0}\in\Omega$, $p_{0}\in\partial\psi(x_{0})$ and
$t\geq 0$ such that $S(x_{0},p_{0},t)\subset\Omega$.

The proof of Proposition 7.5 relies on Proposition 7.4 and on the lower and
upper bound on the volume product proved in Proposition 7.1.

###### Proof.

We use (62) with (12) to get:

(68)
$\nu\left(p_{0}+\frac{t}{2}(S-x_{0})^{\circ}\right)\leq\mu\left(\partial\psi^{*}\left(p_{0}+\frac{t}{2}(S-x_{0})^{\circ}\right)\right)\leq\mu(S).$

Proposition 3.3 and condition (66) imply $\ell(S)\geq\delta$. Since $S$ is
convex and $S\subset\Omega$, (56) gives

$\mu(S)\leq\Lambda|S|.$

Similarly, Corollary 7.3 and condition (66) imply
$\ell(p_{0}+t(S-x_{0})^{\circ})=t\ell((S-x_{0})^{\circ})\geq\delta$, and (62)
implies $p_{0}+t(S-x_{0})^{\circ}\subset\partial\psi(S)\subset\mathcal{O}$. So
(57) gives

$\nu\left(p_{0}+\frac{t}{2}(S-x_{0})^{\circ}\right)\geq\lambda\left|\frac{t}{2}(S-x_{0})^{\circ}\right|=\lambda\frac{t^{n}}{2^{n}}\left|(S-x_{0})^{\circ}\right|.$

Going back to (68), we proved that

$t^{n}|(S-x_{0})^{\circ}|\leq\frac{2^{n}\Lambda}{\lambda}|S|.$

Using the lower bound on the volume product (60), we deduce

$|S|^{2}\geq ct^{n}|(S-x_{0})^{\circ}||S-x_{0}|\geq ct^{n},$

for a constant $c$ that depends only on $n$, $\lambda$ and $\Lambda$. This is
the second inequality in (67).

To prove the first inequality, we use (63) with (11) to write (for any $y\in
S$):

$\mu\left(\frac{1}{2}(S+y)\right)\leq\nu\left(\partial\psi\left(\frac{1}{2}(S+y)\right)\right)\leq\nu\left((p_{0}+2t(S-y)^{\circ}\right)\cap\mathcal{O}).$

We note that $\frac{1}{2}(S+y)\subset S\subset\Omega$ and Proposition 3.3
together with condition (66) imply
$\ell(\frac{1}{2}(S+y))=\frac{1}{2}\ell(S)\geq\delta$. Also, (62) implies that
$p+t(S-y)^{\circ}\subset(p_{0}+2t(S-y)^{\circ})\cap\mathcal{O}$, and so
Corollary 7.3 together with condition (66) implies
$\ell((p_{0}+2t(S-y)^{\circ})\cap\mathcal{O})\geq\delta$.

We can thus use Assumption 2 and get

$\frac{\lambda}{2^{n}}|S|=\lambda\left|\frac{1}{2}(S+y)\right|\leq\Lambda\left|p+2t(S-y)^{\circ}\right|=\Lambda
2^{n}t^{n}\left|(S-y)^{\circ}\right|$

and so

$|S|^{2}\leq Ct^{n}|(S-y)^{\circ}||S|$

for a constant $C$ depending only on $n$, $\lambda$ and $\Lambda$. The result
follows by taking the infimum over $y$ and using the upper bound on the volume
product (59). ∎

###  7.3. Scaling properties of sections: Proof of Proposition 3.5-(ii)

For any convex function $\psi$ and any $\tau\in(0,1)$ we have

(69) $\tau S(x_{0},p_{0},t)\subset S(x_{0},p_{0},\tau t).$

The main result of this section is the following proposition which provides
control in the other direction when $\psi$ is associated to our optimal
transportation problem:

######  Proposition 7.6.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function satisfying (3),
(11)-(12) for measures $\mu$ and $\nu$ such that (9)-(10) holds. For any fixed
$\tau\in(0,\ 1)$, there exists $\gamma\in(0,1)$, depending on $\lambda$,
$\Lambda$ and $n$ such that if $t$ satisfies (66) and

(70) $t\geq 8nL(\Omega)\delta$

then

(71) $S(x_{0},p_{0},\tau t)\subset\gamma S(x_{0},p_{0},t)$

for all $x_{0}\in\Omega$, $p_{0}\in\partial\psi(x_{0})$ and $t\geq 0$ such
that $S(x_{0},p_{0},t)\subset\Omega$.

First, we prove the following technical lemma for convex sets:

######  Lemma 7.7.

Let $K$ be a compact convex set with nonempty interior such that $0\in K$. For
all $\gamma\in(0,1)$ and $y\in K\setminus\gamma K$ we have:

$|K||(K-y)^{\circ}|\geq\frac{c}{1-\gamma}$

for some constant $c$ depending only on the dimension.

Note that $|(K-y)^{\circ}|\to\infty$ when $y\to\partial K$ since
$(K-y)^{\circ}$ is unbounded when $y\in\partial K$. So this lemma simply
quantifies how large the volume product will be when $y$ is close to $\partial
K$ (which corresponds to $\gamma\to 1$).

###### Proof.

John’s lemma implies that there exists a linear transform
$L:\mathbb{R}^{n}\to\mathbb{R}^{n}$ such that $B_{1}(b)\subset L(K)\subset
B_{n}(b)$ for some $b\in\mathbb{R}^{n}$ (we do not take $b=0$ as translations
have a strong effect on the polar body). We note that $L(K-y)=L(K)-L(y)$ and
$L(y)\in L(K)\setminus\gamma L(K)$. Since the product volume
$|K^{\circ}|\,|K|$ is invariant under linear transformation (although not
under affine transformation), it is enough to prove the result under the
condition

(72) $B_{1}(b)\subset K\subset B_{n}(b).$

We now claim that

(73) $|(K-y)^{\circ}|\geq\frac{c}{\mathrm{dist}(y,\partial
K)L(K)^{n-1}}\geq\frac{1}{1-\gamma}\frac{c}{L(K)^{n}}.$

To prove (73), we take $y_{0}\in\partial K$ such that
$|y_{0}-y|=\mathrm{dist}(y,\partial K)=:\eta$. Up to a rotation, we can always
assume that the vector $y_{0}-y$ is in the direction $e_{n}$ (that is
$y_{0}-y=\eta\,e_{n}$) and we denote $x=(x^{\prime},x_{n})$ a vector in
$\mathbb{R}^{n}$. By convexity of $K$ we have $(x-y_{0})\cdot(y-y_{0})\geq 0$
for all $x\in K$, and so

$x_{n}-y_{n}\leq\eta\qquad\forall x\in K.$

Given $z$ such that $0<z_{n}<1/\eta$ and
$|z^{\prime}|\leq\frac{1-z_{n}\,\eta}{L(K)}$ we then have, for all $x\in K$:

$z\cdot(x-y)=z^{\prime}\cdot(x^{\prime}-y^{\prime})+z_{n}(x_{n}-y_{n})\leq|z^{\prime}|L(K)+z_{n}\eta\leq
1.$

We deduce that

$\left\\{z\in\mathbb{R}^{n}\,;\,0<z_{n}<1/\eta,\;|z^{\prime}|\leq\frac{1-z_{n}\eta}{L(K)}\right\\}\subset(K-y)^{\circ}.$

Since the cone in the left hand side has volume $\frac{C}{\eta\,(L(K))^{n-1}}$
we get the first inequality in (73).

To obtain the second inequality, we use the fact that $\frac{y}{\gamma}\notin
K$ to get

$\mathrm{dist}(y,\partial
K)\leq\min\left\\{L(K),\left|\frac{y}{\gamma}-y\right|\right\\}\leq
L(K)\min\left\\{1,\frac{1-\gamma}{\gamma}\right\\}\leq 2L(K)(1-\gamma)$

where we used the fact that $0\in K$ and so $|y|\leq L(K)$, which concludes
the proof of (73).

Finally, (73) implies

$|(K-y)^{\circ}||K|\geq\frac{1}{(1-\gamma)}\frac{c_{n}|K|}{L(K)^{n}},$

and under condition (72) we have $\frac{|K|}{L(K)^{n}}\geq\frac{|B_{1}|}{2n}$
so the result follows. ∎

Next, we prove the following refinement of (62).

######  Lemma 7.8.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function such that
$0\in\partial\psi(0)$. For $t>0$ and $y\in S:=S(0,0,t)$, there holds:

$\partial\psi^{*}\left(\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}\right)\subset
S.$

###### Proof.

We can assume that $\psi(y)<t$ (otherwise the result is trivial). Let
$z\in\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}$, that is

$z\cdot(x-y)\leq\frac{1}{2}(t-\psi(y))\mbox{ for all }x\in S.$

The function $u(x)=\psi(x)-\psi(y)-z\cdot(x-y)$ hence satisfies that $u(x)>0$
for all $x\in\partial S$. Since $u(y)=0$ with $y\in S$ and $u$ is convex, it
follows that all minima of $u$ are in the interior of $S$. Furthermore again
since $u$ is convex, the set of points $x_{0}$ where $0\in\partial u(x_{0})$
coincides with the set of points where $u$ has a minimum. We deduce
$\\{x_{0}\,;\,0\in\partial u(x_{0})\\}\subset S$. Since $0\in\partial
u(x_{0})\Leftrightarrow z\in\partial\psi(x_{0})\Leftrightarrow
x_{0}\in\partial\psi^{*}(z)$. This implies $\partial\psi^{*}(z)\subset S$ and
the result follows. ∎

###### Proof of Proposition 7.6.

Up to replacing $\psi$ with the function
$\overline{\psi}(x)=\psi(x_{0}+x)-p_{0}\cdot x$, we see that it is enough to
prove the result when $x_{0}=0$ and $p_{0}=0$.

We recall that Lemma 7.8 implies

$\partial\psi^{*}\left(\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}\right)\subset
S,$

and so (12) gives

(74)
$\nu\left(\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}\right)\leq\mu\left(\partial\psi^{*}\left(\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}\right)\right)\leq\mu(S).$

We now want to use Assumption 2 to conclude, but this requires $\ell(S)$ and
$\ell\left(\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}\right)$ to be large
enough. For the first one, we use Proposition 3.3 with condition (66) to get
$\ell(S)\geq\delta$. For the second one, we note that Corollary 7.3 implies

(75)
$\ell\left(\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}\right)=\frac{1}{2}(t-\psi(y))\ell((S-y)^{\circ})\geq\frac{1}{4nL(\Omega)}(t-\psi(y)),$

and we need to distinguish two cases.

Case 1: If $\frac{1}{4nL(\Omega)}(t-\psi(y))\leq\delta$, then (70) implies
$t-\psi(y)\leq 4nL(\Omega)\delta\leq\frac{1}{2}t$ and so

$\psi(y)\geq\frac{1}{2}t.$

Case 2: If $\frac{1}{4nL(\Omega)}(t-\psi(y))\geq\delta$, then (75) gives
$\ell(\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ})\geq\delta$ and so we can
use Assumption 2 in (74) to get

$\displaystyle\lambda\left|\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}\right|\leq\nu\left(\frac{1}{2}\left(t-\psi(y)\right)(S-y)^{\circ}\right)\leq\mu(S)\leq\Lambda|S|.$

Proposition 7.5 implies

$\frac{\lambda}{2^{n}}\left(t-\psi(y)\right)^{n}|(S-y)^{\circ}||S|\leq\Lambda|S|^{2}\leq\Lambda
Ct^{n}$

and so (using Lemma 7.7) we deduce

$(t-\psi(y))^{n}\leq C|S|^{2}(1-\gamma)\leq C(1-\gamma)t^{n}\,,\qquad\forall
y\in S\setminus\gamma S$

that is

$\psi(y)\geq\left(1-\left(C(1-\gamma)\right)^{1/n}\right)t\,,\qquad\forall
y\in S\setminus\gamma S.$

If we take $\gamma=1-\frac{1}{C2^{n}}$, then we find

$\psi(y)\geq\frac{1}{2}t\qquad\forall y\in S\setminus\gamma S.$

We can now conclude: Combining both cases, we proved that $y\notin
S(\frac{1}{2}t)$ for all $y\in S\setminus\gamma S$ that is $S\setminus\gamma
S\subset S\setminus S(\frac{1}{2}t)$ which is equivalent to
$S(\frac{1}{2}t)\subset\gamma S$. ∎

###  7.4. Engulfing property of sections: Proof of Proposition 3.5-(iii)

We are now able to show that the scaling property implies the engulfing
property of sections through the following proposition.

######  Proposition 7.9.

Let $\psi:\mathbb{R}^{n}\to\mathbb{R}$ be a convex function,
$p_{0}\in\partial\psi(x_{0})$ and assume that there exists $\gamma\in(0,1)$
such that

(76) $\gamma^{-1}S(x_{0},p_{0},t)\subset S(x_{0},p_{0},2t)\subset\Omega.$

Then there exists $\theta>1$ depending only on $n$ and $\gamma$ such that if
$y\in S(x_{0},p_{0},t)$ then $S(x_{0},p_{0},t)\subset S(y,q,\theta t)$ for all
$q\in\partial\psi(y)$.

Proposition 7.6 implies that there exists $\gamma\in(0,1)$ such that if

$t\geq\max\left\\{2nL(\partial\psi(\Omega)),8nL(\Omega)\right\\}\delta\qquad\mbox{
and }\qquad S(x_{0},p_{0},2t)\subset\Omega$

then

$\gamma^{-1}S(x_{0},p_{0},t)\subset S(x_{0},p_{0},2t).$

So the assumption of Proposition 7.9 are satisfied in such a case and
Proposition 3.5-(iii) follows.

###### Proof.

We fix $y\in S(x_{0},p_{0},t)$. Using John’s lemma, we can assume (up to a
translation) that $S(x_{0},p_{0},t)$ is almost symmetric in the sense that

(77) $-S\subset nS.$

We want to prove that $\psi(x)\leq\psi(y)+q\cdot(x-y)+\theta t$ for all $x\in
S(x_{0},p_{0},t)$. Since $p_{0}\in\partial\psi(x_{0})$, we have

$\displaystyle\psi(y)+q\cdot(x-y)+\theta t$
$\displaystyle\geq\psi(x_{0})+p_{0}\cdot(y-x_{0})+q\cdot(x-y)+\theta t$
$\displaystyle\geq\psi(x_{0})+p_{0}\cdot(x-x_{0})+(q-p_{0})\cdot(x-y)+\theta
t$ $\displaystyle\geq\psi(x)-t+(q-p_{0})\cdot(x-y)+\theta t$

for all $x\in S(x_{0},p_{0},t)$. So the result will follow if we can show that
$(q-p_{0})\cdot(x-y)\geq-(\theta-1)t$ for all $x\in S(x_{0},p_{0},t)$.
Equivalently, we need to show that

(78) $(q-p_{0})\cdot(y-x)\leq Ct\quad\mbox{ for all }x\in S(x_{0},p_{0},t).$

for some constant $C$.

The fact that $q\in\partial\psi(y)$ and the definition of $S(x_{0},p_{0},2t)$
implies

$\psi(y)+q\cdot(x-y)\leq\psi(x)\leq\psi(x_{0})+p_{0}\cdot(x-x_{0})+2t\qquad\forall
x\in S(x_{0},p_{0},2t)$

while the fact that $p_{0}\in\partial\psi(x_{0})$ gives

$\psi(x_{0})+p_{0}\cdot(y-x_{0})\leq\psi(y).$

Adding these two inequality yields

(79) $(q-p_{0})\cdot(x-y)\leq 2t\quad\mbox{ for all }x\in S(x_{0},p_{0},2t).$

In order to get (78), we take $x\in S(x_{0},p_{)},t)$ and define
$\bar{x}=y+\delta(y-x)=(1+\delta)y-\delta x$. Condition (77) implies
$\bar{x}\in(1+\delta+\delta n)S(x_{0},p_{0},t)$ so, by choosing $\delta$ such
that $1+\delta(1+n)=\gamma^{-1}$, we get (using (76)):

$\bar{x}=y+\delta(y-x)\subset\gamma^{-1}S(x_{0},p_{0},t)\subset
S(x_{0},p_{0},2t)$

We can thus take $x=\bar{x}$ in (79) and deduce (78) with
$C=\frac{2}{\delta}=\frac{2\gamma(1+n)}{1-\gamma}$.

∎

##  Appendix A Approximation of regular measures by discrete measures

In this section, we assume that $\mu=f\,dx$ and $\nu=g\,dy$ are absolutely
continuous measures with density $f$ and $g$ that are bounded above and below.
In particular, there exists an optimal transport map $T$ which is the gradient
of a convex function $\psi:\Omega\to\mathbb{R}^{n}$ (see Theorem 1.2).

Given $\delta>0$, we can construct approximations $\mu_{\delta}$ and
$\nu_{\delta}$ which satisfies the assumptions of Theorem 2.5. The
construction is classical and recalled here briefly: We take a set of points
$x_{i}\in\Omega$, $i=1,\dots,n$ such that
$\Omega\subset\cup_{i=1}^{N}B_{\delta}(x_{i})$ and consider the associated
Voronoi tessellations

$V_{i}=\\{x\in\Omega\,;\,|x-x_{i}|=\min_{1\leq j\leq N}|x-x_{j}|\\}.$

We then set

$\mu_{\delta}:=\sum_{i=1}^{N}f_{i}\delta_{x_{i}},\qquad f_{i}:=\mu(V_{i}).$

We recall that $V_{i}\subset B_{\delta}(x_{i})$ for all $i$ and that
$W_{2}(\mu,\mu_{\delta})\leq\delta.$ Consequently, $\mu_{\delta}$ satisfies
(9). We proceed similarly with $\nu$, with a set of points
$\\{y_{j}\\}_{j=1}^{M}$ and $\nu_{\delta}:=\sum_{j=1}^{M}g_{i}\delta_{y_{j}}$
and obtain that $\nu_{\delta}$ satisfies (10) so that Assumption 1 holds.

The (discrete) optimal transport plan between $\mu_{\delta}$ and
$\nu_{\delta}$ has the form

$\gamma_{\delta}=\sum_{i=1}^{N}\sum_{j=1}^{M}\gamma_{\delta,ij}\delta_{x_{i}}\delta_{y_{j}}$

and if $\psi_{\delta}$ is an associated Kantorovich potential, we know that
$\gamma_{\delta}$ must be supported in the subdifferential of $\psi_{\delta}$,
that is

$\gamma_{\delta,ij}>0\Rightarrow y_{j}\in\partial\psi_{\delta}(x_{i}).$

As a consequence, for any Borel set $A\subset\Omega$, we have that

$\mu_{\delta}(A)=\sum_{i,\;x_{i}\in A}f_{i}=\sum_{i,\;x_{i}\in
A}\sum_{j,\;y_{j}\in\partial\psi(x_{i})}\gamma_{\delta,i,j}\leq\sum_{j,\;y_{j}\in\partial\psi(A)}\sum_{i}\gamma_{\delta,i,j}=\nu(\partial\psi_{\delta}(A)),$

since $y_{j}\in\partial\psi(x_{i})$ and $x_{i}\in A$ implies that
$y_{j}\in\partial\psi_{\delta}(A)$. This proves that
$\mu_{\delta},\;\nu_{\delta}$ and $\psi_{\delta}$ satisfy (11). Since
$\psi^{*}_{\delta}$ is a Kantorovich potential between $\nu_{\delta}$ and
$\mu_{\delta}$, the same argument also yields (12).

We have hence verified most of assumptions of Theorem 2.5. We only need to
show that we can find $\Omega^{\prime}\subset\subset\Omega$, and $\rho$ s.t.
$S(x,p,\rho)\in\Omega$ for every $x\in\Omega$. This is in fact non-trivial but
we can fortunately rely on some of the existing literature:

First, we recall one of the main results of [29] which can be written as
follows with our notations.

######  Theorem A.1.

If $-\psi(x)$ is $\lambda$-convex (in that case, the triple $(\mu,\nu,\psi)$
is said to be $\lambda$-regular), then there exists a constant $C$ such that

$\sum_{i=1}^{N}f_{i}\mathrm{dist}\,(\nabla\psi(x_{i}),\partial\psi_{\delta}(x_{i}))^{2}\leq
C\lambda\delta.$

We observe that the classical regularity results by Caffarelli, [11, 7], give
that $\psi$ and $\psi^{*}$ are $C^{2,\alpha}$ when $\mu$ and $\nu$ are
$C^{0,\alpha}$, strictly positive on their support and supported on $C^{2}$,
uniformly convex domains. In particular, $(\mu,\nu,\psi)$ is $\lambda$-regular
for some $\lambda$ under such conditions and Theorem A.1 applies.

We recall that $\psi$ is $C^{1}$, so $\partial\psi(x)=\\{\nabla\psi(x)\\}$ for
all $x$ and $\mathrm{dist}\,(\nabla\psi(x_{i}),\partial\psi_{\delta}(x_{i}))$
denotes the distance between the point $\nabla\psi(x_{i})$ and the set
$\partial\psi_{\delta}(x_{i})$ (in fact the result of [29] is slightly more
precise since it controls the distance between $\partial\psi_{\delta}(x_{i})$