# regularity of singular set in optimal transportation

Shibing Chen  Shibing Chen, School of Mathematical Sciences, University of
Science and Technology of China, Hefei, 230026, P.R. China.
<EMAIL_ADDRESS>and  Jiakun Liu  Jiakun Liu, School of Mathematics and
Applied Statistics, University of Wollongong, Wollongong, NSW 2522, AUSTRALIA
<EMAIL_ADDRESS>
###### Abstract.

We consider the optimal transport problem when the target consists of two
disjoint convex domains. This is a basic model where singularities of optimal
transport maps arise. When the domains are $C^{2}$, uniformly convex and the
densities are $C^{\alpha}$ smooth, we show that the singular set of the
optimal transport map is an $(n-1)$-dimensional $C^{2,\alpha}$ regular sub-
manifold of $\mathbb{R}^{n}.$

###### Key words and phrases:

Optimal transportation, Monge-Ampère equation, free boundary

###### 2000 Mathematics Subject Classification:

35J96, 35J25, 35B65.

The second were supported by ARC DP 200101084.

##  1\. Introduction

In [3], Caffarelli constructed an important example showing that optimal
transport maps may have singularity if the target domain is non-convex. In
that example, the source domain is the unit ball $B_{1}(0)$ in
$\mathbb{R}^{n},$ and the target is a union of two half balls
$B_{1}^{+}+e_{1}$ and $B_{1}^{-}-e_{1}$, where
$B_{1}^{+}=B_{1}(0)\cap\\{x_{1}>0\\}$, $B_{1}^{-}=B_{1}(0)\cap\\{x_{1}<0\\}$
and $e_{1}=(1,0,\cdots,0)\in\mathbb{R}^{n}$. Assuming constant densities, then
the potential function of the optimal transport map is
$\frac{1}{2}|x|^{2}+|x_{1}|,$ which has a singular set $\\{x_{1}=0\\}\cap
B_{1}.$ A natural question is what can we say about the singular set of the
optimal transport map, when the target consists of two disjoint general convex
domains associated with non-constant densities.

In this paper we investigate this problem. Consider an optimal transport
problem from the source domain $\Omega$ associated with density $f$ to the
target $\Omega^{*}=\Omega^{*}_{1}\cup\Omega^{*}_{2}$ associated with density
$g$, where $\Omega^{*}_{1},\Omega^{*}_{2}$ are two domains separated by a
hyperplane $\mathcal{H}$, and the densities satisfy

$\int_{\Omega}f(x)\,dx=\int_{\Omega^{*}}g(y)\,dy$

and $1/\lambda<f,g<\lambda$ for some positive constant $\lambda$. Denote by
$u$ (resp. $v$) the convex function solving $(\partial
u)_{\\#}f\chi_{\Omega}=g\chi_{\Omega^{*}}$ (resp. $(\partial
v)_{\\#}g\chi_{\Omega^{*}}=f\chi_{\Omega}$). It is known in [7, 17] that

  * $(i)$

The interior of $\Omega_{1}:=\partial v(\Omega^{*}_{1})$ and
$\Omega_{2}:=\partial v(\Omega^{*}_{2})$ are disjoint and separated by a
Lipschitz hypersurface.

  * $(ii)$

Suppose that $\Omega$, $\Omega^{*}_{1}$ and $\Omega^{*}_{2}$ are convex, then
the singular set $\mathcal{F}:=\partial\Omega_{1}\cap\Omega$ is $C^{1,\alpha}$
for some $\alpha\in(0,1)$.

The main result of this paper is the following theorem.

######  Theorem 1.1.

Assume that $\Omega$ is convex and $\partial\Omega^{*}_{i}\in C^{2}$ are
uniformly convex ($i=1,2$), and the densities $f,g\in C^{\alpha}$ for some
$\alpha\in(0,1)$. Then, the singular set $\mathcal{F}\in C^{2,\alpha}$.

Let $u$ be the potential function of the optimal transport problem from
$\Omega$ to $\Omega^{*}$. Denote by $u_{i}$ the restriction of $u$ on
$\Omega_{i}$, $i=1,2$. Note that $u_{1}=u_{2}$ on $\mathcal{F}$. A useful
observation in [7, 17] is that the unit normal of $\mathcal{F}$ at $x$ is
given by

(1.1)
$\nu_{\mathcal{F}}(x)=\frac{Du_{1}(x)-Du_{2}(x)}{|Du_{1}(x)-Du_{2}(x)|}.$

Hence, higher regularity of $u_{i}$ up to the singular set $\mathcal{F}$ will
automatically imply higher regularity of $\mathcal{F}.$ In [7, 17] the authors
established the $C^{1,\alpha}$ regularity of $u_{i},i=1,2$ up to the singular
set $\mathcal{F}.$

To prove the higher order regularity of $u_{i}$ up to $\mathcal{F},$ the key
is to establish the following obliqueness estimate:

######  Proposition 1.1.

Assume $0\in\mathcal{F},$ denote $y_{0}=Du_{1}(0)$ and
$\hat{y}_{0}=Du_{2}(0).$ Let $\nu,\nu^{*},\hat{\nu}^{*}$ be the unit inner
normals of $\Omega_{1},\Omega_{1}^{*},\Omega_{2}^{*}$ at $0,y_{0},\hat{y}_{0}$
respectively, then $\nu\cdot\nu^{*}>0$ and $-\nu\cdot\hat{\nu}^{*}>0.$

In [5] and [8], the obliqueness estimate was also the key estimate to the
global regularity of the optimal maps, convexity of domains plays crucial
roles in the proof. In [10], the authors studied the optimal partial transport
problem, they proved the obliqueness estimate at the free boundary point.
Although the free boundary there is in general non-convex, they overcame this
difficulty by exploiting the interior ball property of the free boundary
(which implies the semi-convexity of the free boundary).

In the current paper, to establish the obliqueness estimate, serious
difficulty presents due to the fact that we do not have any sort of convexity
of the singular set, neither convexity nor semi-convexity. Moreover, we need
to rule out three different cases:
I) $\nu\cdot\nu^{*}=0,\hat{\nu}\cdot\hat{\nu}^{*}=0,$ and
$\nu,\nu^{*},\hat{\nu}^{*}$ are not in the same plane;
II) $\nu\cdot\nu^{*}=0,\hat{\nu}\cdot\hat{\nu}^{*}=0,$ and
$\nu,\nu^{*},\hat{\nu}^{*}$ are in the same plane;
III) $\nu\cdot\nu^{*}=0,\hat{\nu}\cdot\hat{\nu}^{*}>0.$

New ideas are introduced to rule out Case I) and Case II). For Case II), we
use (1.1) and the disjointness of $\Omega_{1}^{*}$ and $\Omega_{2}^{*}$ to
show that the singular set will be flatter and flatter during the
normalisation process, which will contradict to the strict convexity estimate
of $u_{1}.$ For Case I), we design a new strategy to prove the splitting
property of the singular set and $\partial\Omega_{1}^{*}$ during the blow-up
process: 1, we use the assumption that the obliqueness fails to establish a
$C^{1,\frac{1}{2}-\epsilon}$ estimate of the singular set at $0;$ 2, in the
first blow-up, we manage to control the shape of the level set of $v$ near
$y_{0},$ so that $\mathcal{F}$ becomes $\omega\times\mathbb{R}^{n-2}$ for some
$2d$-convex set in the limit, and $\partial\Omega_{1}^{*}$ becomes a graph of
some non-negative quadratic polynomial in the limit; 3, in the second blow-up,
we use the techniques from [8] to show that both $\Omega_{1}$ and
$\Omega_{1}^{*}$ will become cylindrical shape as that in [8].

This paper is organised as follows. In §2 we recall some results from [5, 6,
8, 10] which will be used in subsequent sections. In §3 we establish the
obliqueness estimate. In §4 we prove the main theorem.

##  2\. Preliminaries

In the following we always assume $\Omega,\Omega_{1}^{*},\Omega_{2}^{*}$ are
$C^{2}$ and uniformly convex. The closures of $\Omega_{1}^{*}$ and
$\Omega_{2}^{*}$ are disjoint. $f,g$ satisfy $\lambda^{-1}<f,g<\lambda$ in
$\Omega,\Omega^{*}$, respectively, for a positive constant $\lambda.$ Let
$u,v$ be as in the introduction. Let $u_{i}:=u\big{|}_{\Omega_{i}},i=1,2.$ We
extend $u_{1},u_{2},v$ to $\mathbb{R}^{n}$ as follows.

(2.1) $\begin{split}&v(x)=\sup\\{L(x)\,:\,L\ \text{is affine},L\leq v\
\text{in}\ \Omega^{*},\ \text{and}\ DL\in\Omega\\},\\\
&u_{i}(x)=\sup\\{L(x)\,:\,L\ \text{is affine},L\leq u\ \text{in}\ \Omega_{i},\
\text{and}\ DL\in\Omega_{i}^{*}\\},i=1,2.\end{split}$

Then $v,u_{i},i=1,2$ satisfy the Monge-Ampere inequality in the sense of
Alexandrov

(2.2) ${\begin{split}C^{-1}\chi_{{}_{\Omega^{*}}}&\leq\mbox{det}\,D^{2}v\leq
C\chi_{{}_{\Omega^{*}}},\\\
C^{-1}\chi_{{}_{\Omega_{i}}}&\leq\mbox{det}\,D^{2}u_{i}\leq
C\chi_{{}_{\Omega_{i}}},i=1,2.\end{split}}$

It is shown in [7, 17] that $v\in C^{1}(\mathbb{R}^{n}),u_{i}\in
C^{1}(\mathbb{R}^{n}),i=1,2,$ and that $v$ (resp. $u$) is strictly convex in
$\Omega^{*}$ (resp. $\Omega$). We also need the following

######  Definition 2.1.

Let $y_{0}\in\overline{\Omega^{*}}$ and $h>0$ be a small constant. We denote
by

(2.3)
$S^{c}_{h}[v](y_{0}):=\left\\{y\in\mathbb{R}^{n}\,:\,v(y)<v(y_{0})+(y-y_{0})\cdot\bar{p}+h\right\\}$

the centred sub-level set of $v$ with height $h$, where
$\bar{p}\in\mathbb{R}^{n}$ is chosen such that the centre of mass of
$S^{c}_{h}[v](y_{0})$ is $y_{0}$. We denote by

(2.4)
$S_{h}[v](y_{0}):=\left\\{y\in\mathbb{R}^{n}\,:\,v(y)<\ell_{y_{0}}(y)+h\right\\}$

the sub-level set of $v$ with height $h$, where $\ell_{y_{0}}$ is a support
function of $v$ at $y_{0}$.

Assume $x_{0}\in\mathcal{F},$ $y_{0}=Du_{1}(x_{0})\in\partial\Omega_{1}^{*}.$
Denote $S_{h}^{c}[v]=S_{h}^{c}[v](y_{0})$. In the following we will call a
constant universal if it depends only on
$n,\Omega,\Omega^{*},\lambda,\text{dist}(x_{0},\partial\Omega).$ When $h>0$ is
sufficiently small, it is proved in [7, 17] that $S_{h}^{c}[v]$ decays
geometrically, namely, for any given $s_{1}<1,$ there exists a constant
$s_{0}<1$ independent of $h$ such that

$S^{c}_{\bar{s}h}[v]\subset s_{1}S^{c}_{h}[v]\quad\
\forall\,\bar{s}\in(0,s_{0}).$

Furthermore, we have the strict convexity estimate

(2.5) $v(y)\geq
v(y_{0})+Dv(y_{0})\cdot(y-y_{0})+C|y-y_{0}|^{1+\beta}\quad\forall\,y\in\overline{\Omega_{1}^{*}}\text{
near }y_{0}$

for some constant $\beta>1$, which in turn implies $u_{1}\in
C^{1,\alpha^{\prime}}$.

As in the partial transport case we still have the following uniform density
property and tangential $C^{1,1-\epsilon}$ regularity.

######  Lemma 2.1 (Uniform density).

$\frac{|S_{h}^{c}[v]\cap\Omega^{*}_{1}|}{|S_{h}^{c}[v]|}\geq\delta$ for some
constant $\delta>0$ depending only on $\Omega,\Omega^{*},\lambda,$ and
$\text{dist}(x_{0},\partial\Omega).$

A direct corollary of Lemma 2.1 is the volume estimate

(2.6)
$|S_{h}[v]\cap\Omega^{*}_{1}|\approx|S^{c}_{h}[v]\cap\Omega^{*}_{1}|\approx|S^{c}_{h}[v]|\approx
h.$

Moreover, if one of $S_{h}[v]\cap\Omega^{*}_{1}$ or $S^{c}_{h}[v]$ is
normalised, the other one will also be normalised simultaneously.

After a change of coordinates, we may assume that $\\{x_{1}=0\\}$ is the
tangent hyperplane of $\Omega_{1}^{*}$ at $y_{0}.$ Then we have

######  Lemma 2.2.

[Tangential $C^{1,1-\epsilon}$ for $v$] For any $\epsilon>0$ there exists a
constant $C_{\epsilon}$ such that
$B_{C_{\epsilon}h^{\frac{1}{2}+\epsilon}}(y_{0})\cap\\{x_{1}=0\\}\subset
S_{h}^{c}[v]$ for $h$ small.

######  Remark 2.1.

_As $v$ is strictly convex up to the boundary, we actually have an equivalency
relation between its sub-level sets $S_{h}[v](y_{0})$ and centred sections
$S^{c}_{h}[v](y_{0})$, that is for all small $h>0$,_

(2.7) $S^{c}_{b^{-1}h}[v](y_{0})\cap\Omega_{1}^{*}\subset
S_{h}[v](y_{0})\cap\Omega_{1}^{*}\subset
S^{c}_{bh}[v](y_{0})\cap\Omega_{1}^{*},$

_where $b\geq 1$ is a constant independent of $h$. For the proof of (2.7), we
refer the reader to [5] and [8, Lemma 2.2]. Then, by Lemma 2.2 we have that
$B_{C_{\epsilon}h^{\frac{1}{2}+\epsilon}}\cap\partial\Omega_{1}^{*}\subset
S_{h}[v]$ for $h$ small. _

##  3\. Obliqueness

For $x\in\mathcal{F}$, denote by $\nu=\nu(x),\hat{\nu}=\hat{\nu}(x)$ the unit
inner normals of $\partial\Omega_{1},\partial\Omega_{2}$ at $x$, respectively.
Since $\mathcal{F}\in C^{1,\alpha}$, one has $\nu=-\hat{\nu}$, and

$\displaystyle Du_{1}(x)$ $\displaystyle=:y\in\partial\Omega^{*}_{1};$
$\displaystyle Du_{2}(x)$ $\displaystyle=:\hat{y}\in\partial\Omega^{*}_{2},$

where $\partial u(x)=\\{t\hat{y}+(1-t)y:t\in[0,1]\\}$ is a segment. Denote by
$\nu^{*}=\nu^{*}(y),\hat{\nu}^{*}=\hat{\nu}^{*}(\hat{y})$ the unit inner
normals of $\partial\Omega^{*}_{1},\partial\Omega^{*}_{2}$ at $y,\hat{y}$,
respectively.

######  Proposition 3.1.

$\nu\cdot\nu^{*}>0$ and $\hat{\nu}\cdot\hat{\nu}^{*}>0$ hold simultaneously.

We shall prove the above obliqueness by contradiction. Suppose the obliqueness
conditions fails on $\partial\Omega^{*}_{1}$, namely
$\nu(x_{0})\cdot\nu^{*}(y_{0})=0$ at some
$x_{0}\in\mathcal{F},y_{0}=Du_{1}(x_{0})\in\partial\Omega^{*}_{1}$, then up to
a translation and rotation of coordinates, we may assume that $x_{0}=0$, and
$\nu=\nu(0)=e_{n}$, $\nu^{*}=\nu^{*}(y_{0})=e_{1}$.

Let $\hat{y}_{0}:=Du_{2}(0)$. Since $y_{0}=Du_{1}(0)$, from (1.1) one has
$e_{n}=\frac{y_{0}-\hat{y}_{0}}{|y_{0}-\hat{y}_{0}|}$. By subtracting a linear
function from $u$, we may assume that $y_{0}=le_{n}$ and
$\hat{y}_{0}=-\hat{l}e_{n}$ for some constants $l,\hat{l}>l_{0}$, where
$l_{0}>0$ is a universal constant.

Since $\mathcal{F}\in C^{1,\alpha}$, and $\partial\Omega^{*}_{1}\in C^{2}$ is
uniformly convex, there exist two functions $\rho\in C^{1,\alpha}$,
$\rho^{*}\in C^{2}$ such that
$\mathcal{F}=\\{x_{n}=\rho(x_{1},\cdots,x_{n-1})\\}\text{ locally near }0,$
and $\partial\Omega^{*}_{1}=\\{x_{1}=\rho^{*}(x_{2},\cdots,x_{n}-l)\\}\text{
locally near }y_{0},$ with $\rho^{*}$ uniformly convex and $C^{2}$ .

Recall that $v$ is convex and $Dv(y_{0})=0$. By adding a constant, we may
assume that $v(y_{0})=0$ and $v\geq 0$.

We need to consider three cases:
I) $\hat{\nu}\cdot\hat{\nu}^{*}=0,$ and $\nu,\nu^{*},\hat{\nu}^{*}$ are not in
the same plane;
II) $\hat{\nu}\cdot\hat{\nu}^{*}=0,$ and $\nu,\nu^{*},\hat{\nu}^{*}$ are in
the same plane;
III) $\hat{\nu}\cdot\hat{\nu}^{*}>0.$

###  3.1. Case I): $\hat{\nu}\cdot\hat{\nu}^{*}=0,$ and
$\nu,\nu^{*},\hat{\nu}^{*}$ are not in the same plane.

Up to an affine transform we may assume $\Omega_{1}^{*}\subset\\{x_{1}\geq
0\\}$ and $\Omega_{2}^{*}\subset\\{x_{2}\geq 0\\}.$ We may assume

$\partial\Omega^{*}_{2}=\\{x_{2}=\rho_{2}^{*}(x_{1},x_{3},\cdots,x_{n}+\hat{l})\\}\text{
locally near }\hat{y}_{0},$

with $\rho_{2}^{*}$ uniformly convex and $C^{2}.$

Let

$d_{e}:=\sup\\{|x\cdot e|:x\in S_{h}[v]\cap\Omega_{1}^{*}\\},$

for any unit vector $e.$ Denote $d_{e_{i}}$ by $d_{i}$ for short. Let
$p_{e}\in\overline{S_{h}[v]\cap\Omega_{1}^{*}}$ be the point such that
$p_{e}\cdot e=d_{e}.$ When $e=e_{1},$ we denote $p_{e_{1}}$ by $p$ for short.

######  Lemma 3.1.

$|d_{1}|\leq C_{\epsilon}h^{\frac{2}{3}-\epsilon},$ $d_{e}^{2}\leq
C_{\epsilon}d_{1}$ for any unit vector
$e\in\text{span}\\{e_{2},\cdots,e_{n}\\},$ where $\epsilon$ can be as small as
we want.

###### Proof.

Let $q$ be the intersection of segment $op$ and $\partial\Omega_{1}^{*}.$ By
convexity we have that $q\in S_{h}[v].$ Suppose $p_{1}\geq
h^{\frac{2}{3}-4n\epsilon},$ then since $S_{h}[v]\cap\Omega_{1}^{*}$ converges
to the point $y_{0}$ we have that $\frac{q_{n}}{p_{n}}\geq\frac{1}{2}$
provided $h$ is small enough. Hence $q_{1}\geq\frac{q_{n}}{p_{n}}p_{1}\geq
Ch^{\frac{2}{3}-4n\epsilon}.$

Denote $e_{q}:=\frac{q-y_{0}}{|q-y_{0}|}.$ Denote by $D$ the planar region in
$\text{span}\\{e_{q},e_{1}\\},$ enclosed by
$\partial\Omega_{1}^{*}\cap\text{span}\\{e_{q},e_{1}\\}$ and the segment
$y_{0}q.$ By the $C^{2}$ regularity and uniform convexity of
$\partial\Omega_{1}^{*}$ we have that $\mathcal{H}^{2}(D)\geq
Cd_{1}^{\frac{3}{2}}.$ Let $\tilde{e}_{2},\cdots,\tilde{e}_{n-1}$ be an
orthonormal basis of the orthogonal complement of
$\text{span}\\{e_{1},e_{p}\\}$ in $\mathbb{R}^{n}.$ By the tangential
$C^{1,1-\epsilon}$ estimate of $v$ we have that
$y_{0}+h^{\frac{1}{2}+\epsilon}\tilde{e}_{i}\in S^{c}_{bh}[v],i=2,\cdots,n-1.$
Let $G$ be the convex envelope of $D$ and the points
$y_{0}+h^{\frac{1}{2}+\epsilon}\tilde{e}_{i},i=2,\cdots,n-1.$ By convexity we
have $G\subset S^{c}_{bh}[v].$ Hence

$h^{(\frac{1}{2}+\epsilon)(n-2)}\mathcal{H}^{2}(D)\leq|G|\leq|S^{c}_{bh}[v]|\approx
h^{\frac{n}{2}},$

which implies that $|d_{1}|\leq C_{\epsilon}h^{\frac{2}{3}-\epsilon}.$

Fix any unit vector $e\in\text{span}\\{e_{2},\cdots,e_{n}\\},$ by the uniform
convexity of $\partial\Omega_{1}^{*}$ we have that $d_{1}\geq p_{e}\cdot
e_{1}\geq C(p_{e}\cdot e)^{2},$ which implies $d_{e}^{2}\leq
C_{\epsilon}d_{1}.$ ∎

The above lemma has a very useful corollary.

######  Corollary 3.1.

$v(x)\geq C_{\epsilon}|x-y_{0}|^{3+\epsilon}$ for all $x\in\Omega_{1}^{*}.$

###### Proof.

By Lemma 3.1 we have that $v(x)\geq h$ for any $x\in\Omega_{1}^{*}$ satisfying
$|x|\geq C_{\epsilon}h^{\frac{1}{3}-\epsilon}.$ This implies that $v(x)\geq
C_{\epsilon}|x-y_{0}|^{3+\epsilon}$ for all $x\in\Omega_{1}^{*},$ as desired.
∎

######  Remark 3.1.

Note that by the same reasoning, we have that $v(x)\geq
C_{\epsilon}|x-\hat{y}_{0}|^{3+\epsilon}$ for all $x\in\Omega_{2}^{*}.$

######  Lemma 3.2.

$|\rho(x^{\prime})|\leq C_{\epsilon}|x^{\prime}|^{\frac{3}{2}-\epsilon}.$

###### Proof.

For any $x\in\mathbb{R}^{n},$ we have

$\displaystyle 0\leq u_{1}(x)-lx_{n}$ $\displaystyle=$
$\displaystyle\sup_{y\in\Omega_{1}^{*}}x\cdot(y-y_{0})-v(y)$
$\displaystyle\leq$
$\displaystyle\sup_{y\in\Omega_{1}^{*}}x\cdot(y-y_{0})-C_{\epsilon}|y-y_{0}|^{3+\epsilon}$
$\displaystyle\leq$ $\displaystyle\sup_{y\in\mathbb{R}^{n}}x\cdot
y-C_{\epsilon}|y|^{3+\epsilon}$ $\displaystyle\leq$ $\displaystyle
C_{\epsilon}|x|^{\frac{3}{2}-\epsilon}.$

Similarly, $0\leq u_{2}(x)+\hat{l}x_{n}\leq
C_{\epsilon}|x|^{\frac{3}{2}-\epsilon}.$ By convexity, we have that

$|D(u_{1}(x)-lx_{n})|\leq C_{\epsilon}|x|^{\frac{1}{2}-\epsilon},\
|D(u_{2}(x)+\hat{l}x_{n})|\leq C_{\epsilon}|x|^{\frac{1}{2}-\epsilon}.$

Since $\mathcal{F}$ is $C^{1,\alpha},$ we have that $|\rho(x^{\prime})|\leq
C|x^{\prime}|^{1+\alpha},$ it follows that

(3.1) $|D_{x^{\prime}}u_{i}(x)|\leq
C_{\epsilon}|x^{\prime}|^{\frac{1}{2}-\epsilon},\ i=1,2,$

for $x\in\mathcal{F}.$

For any $x\in\mathcal{F},$ we have that the unit inner normal of $\Omega_{1}$
at $x$ is given by

$\nu(x)=\frac{(-D_{x^{\prime}}\rho(x^{\prime}),1)}{\sqrt{|D_{x^{\prime}}\rho(x^{\prime})|^{2}+1}}=\frac{\big{(}D_{x^{\prime}}(u_{1}(x)-D_{x^{\prime}}u_{2}(x)),\partial_{x_{n}}(u_{1}(x)-u_{2}(x)\big{)}}{|Du_{1}(x)-Du_{2}(x)|}.$

By (3.1) we have that $|D_{x^{\prime}}\rho(x^{\prime})|\leq
C_{\epsilon}|x^{\prime}|^{\frac{1}{2}-\epsilon}.$ It follows that
$|\rho(x^{\prime})|\leq C_{\epsilon}|x^{\prime}|^{\frac{3}{2}-\epsilon}$ as
wanted. ∎

######  Lemma 3.3.

There exists $t_{0}>0$ small such that

(3.2) $\rho(t,0,\cdots,0)>0,\ \text{provided}\ 0<|t|<t_{0}.$

###### Proof.

Suppose not, then there exits $t\neq 0$ with $|t|$ as small as we want such
that $te_{1}\in\overline{\Omega_{1}}.$ Let
$q:=Du_{1}(te_{1})\in\overline{\Omega_{1}^{*}}.$ Observe that $Dv(q)=te_{1}.$
If $t<0,$ then on one hand by the uniform convexity of $\Omega_{1}^{*}$ we
have $(Dv(q)-0)\cdot(q-y_{0})<0,$ contradicting to the convexity of $v.$ Hence
$t>0.$ Let $h=v(q).$ By Lemma 3.1 we have that $q_{1}\leq
C_{\epsilon}h^{\frac{2}{3}-\epsilon}.$ Since $Dv(q)=te_{1},$ we have that
$e_{1}$ is the unit out normal of $S_{h}[v]$ at $q.$ It follows that

(3.3) $S_{h}[v]\subset\\{x\in\mathbb{R}^{n}:x_{1}\leq
C_{\epsilon}h^{\frac{2}{3}-\epsilon}\\}.$

By tangential $C^{1,1-\epsilon}$ of $v$ at $\hat{y}_{0},$ we have that
$p:=\hat{y}_{0}+C_{\epsilon}h^{\frac{1}{2}+\epsilon}e_{1}\subset
S^{c}_{\frac{1}{2}b^{-1}h}[v](\hat{y}_{0})\subset S_{\frac{h}{2}}[v],$ for $h$
small. Hence $p\in S_{h}[v]$ with
$p_{1}=C_{\epsilon}h^{\frac{1}{2}+\epsilon}\gg
C_{\epsilon}h^{\frac{2}{3}-\epsilon}$ when $h$ is sufficiently small, this
contradicts to (3.3) provided $h$ is small enough, guaranteed by taking $t$
sufficiently small initially. Therefore, (3.2) holds for small $t_{0}.$ ∎

First blow up. Up to a translation of coordinates we may assume $y_{0}=0.$
Suppose $S_{h}^{c}[v]\sim E$ for some ellipsoid centered at $0.$ Then,
$E\cap\\{x_{1}=0\\}$ is an $(n-1)$-dimensional ellipsoid with principal
directions $\bar{e}_{2},\cdots,\bar{e}_{n}.$ Note that
$\text{span}\\{\bar{e}_{2},\cdots,\bar{e}_{n}\\}=\text{span}\\{e_{2},\cdots,e_{n}\\}.$
Now, we can write

$E=\\{x=x_{1}e_{1}+\sum_{i=2}^{n}\bar{x}_{i}\bar{e}_{i}:\frac{x_{1}^{2}}{a_{1}^{2}}+\sum_{i=2}^{n}\frac{(\bar{x}_{i}-k_{i}x_{1})^{2}}{a_{i}^{2}}\leq
1\\}.$

By Lemma 3.1 and the tangential $C^{1,1-\epsilon}$ estimate of $v$ at $y_{0},$
we have that

$0<a_{1}<C_{\epsilon}h^{\frac{2}{3}-\epsilon},\ \text{and}\
C_{\epsilon}h^{\frac{1}{2}+\epsilon}<a_{i}<C_{\epsilon}h^{\frac{1}{3}-\epsilon}\
\text{for}\ i=2,\cdots,n.$

Let $T_{1},T_{2}$ be the affine transformations as following.

$T_{1}:x=x_{1}e_{1}+\sum_{i=2}^{n}\bar{x}_{i}\bar{e}_{i}\mapsto
z=x_{1}e_{1}+\sum_{i=2}^{n}(\bar{x}_{i}-k_{i}x_{1})\bar{e}_{i};$
$T_{2}:z=z_{1}e_{1}+\sum_{i=2}^{n}\bar{z}_{i}\bar{e}_{i}\mapsto
y=\frac{z_{1}}{a_{1}}e_{1}+\sum_{i=2}^{n}\frac{\bar{z}_{i}}{a_{i}}\bar{e}_{i}.$

Then $T(E)=B_{1},$ where $T=T_{2}T_{1}.$ Hence $TS_{h}^{c}[v]\sim B_{1}.$ Let
$v_{h}(x):=\frac{1}{h}v(T^{-1}x).$ Then $Dv_{h}(x)=T^{*}Dv(T^{-1}x),$ where
$T^{*}=\frac{(T^{t})^{-1}}{h}=\frac{1}{h}(T_{2}^{t})^{-1}(T_{1}^{t})^{-1}.$ A
direct computation shows that

$(T^{t}_{1})^{-1}:x=x_{1}e_{1}+\sum_{i=2}^{n}\bar{x}_{i}\bar{e}_{i}\mapsto
z=(x_{1}+\sum_{i=2}^{n}k_{i}\bar{x}_{i})e_{1}+\sum_{i=2}^{n}\bar{x}_{i}\bar{e}_{i};$
$(T^{t}_{2})^{-1}:z=z_{1}e_{1}+\sum_{i=2}^{n}\bar{z}_{i}\bar{e}_{i}\mapsto
y=a_{1}z_{1}e_{1}+\sum_{i=2}^{n}a_{i}\bar{z}_{i}\bar{e}_{i}.$

Let $\Omega_{h1}^{*}:=T\Omega^{*}_{1}$, then we have the following estimate.

######  Lemma 3.4.

There exist constants $0<\alpha<1,\beta>1$ such that

(3.4) $0\leq v_{h}(x)\leq C|x|^{1+\alpha}\ \text{for any}\ x\in B_{r_{0}}$

and

(3.5) $v_{h}(x)\geq\frac{1}{C}|x|^{1+\beta}\ \text{for any}\ x\in
B_{r_{0}}\cap\Omega_{h1}^{*},$

where $C,r_{0}>0$ are constants independent of $h.$

###### Proof.

Note that $v_{h}$ satisfies
$\frac{1}{C}\chi_{\Omega^{*}_{h1}}\leq\mbox{det}D^{2}v_{h}<C\chi_{\Omega^{*}_{h1}}$
in $B_{r_{0}}$ for some universal constant $C,r_{0}.$ The estimate (3.5)
follows from the localised version of Caffarelli’s strict convexity estimate
of optimal transport maps [3]. By localised version of [1, Theorem 2.1], we
have $0\leq v_{h}(x)\leq C|x|^{1+\alpha}\ \text{for any}\ x\in
B_{r_{0}}\cap\Omega^{*}_{h1}$ for some $C>0$ independent of $k.$ Combine this
with the uniform density property we have that
$B_{\tilde{h}^{\frac{1}{1+\alpha}}}\cap\Omega^{*}_{h1}\subset
S^{c}_{\tilde{h}}[v_{h}]\cap\Omega^{*}_{h1}$ for $\tilde{h}\leq h_{0},$ where
$h_{0}>0$ is independent of $k.$ Hence there exists some constant $C$
independent of $h,$ such that
$B_{\frac{1}{C}\tilde{h}^{\frac{1}{1+\alpha}}}\subset S_{h}^{c}[v_{h}]$ for
$h\leq h_{0}.$ The estimate (3.4) easily follows from the above discussion. ∎

Passing to a subsequence we may assume that $v_{h}\rightarrow v_{\infty}$
locally uniformly for some convex function $v_{\infty}$ defined on
$\mathbb{R}^{n},$ and that $T\Omega_{1}^{*}$ converges to some convex set
$\Omega_{1,\infty}^{*}$ locally uniformly in Hausdorff distance. Denote by
$\Omega_{1,\infty}$ the interior of $Dv_{\infty}(\mathbb{R}^{n}).$ Since $v$
is a convex function defined on the entire $\mathbb{R}^{n},$ it is well-known
that $\Omega_{1,\infty}$ is a convex set. A direct consequence of Lemma 3.7 is
the following

######  Corollary 3.2.

There exist constants $0<\alpha<1,\beta>1$ such that

(3.6) $0\leq v_{\infty}(x)\leq C|x|^{1+\alpha}\ \text{for any}\ x\in B_{1}$

and

(3.7) $v_{\infty}(x)\geq\frac{1}{C}|x|^{1+\beta}\ \text{for any}\ x\in
B_{1}\cap\Omega_{1,\infty}^{*},$

where $C>0$ is a constant independent of $h.$

######  Lemma 3.5.

Up to an affine transform we have:
1) $\Omega_{1,\infty}^{*}=\\{x\in\mathbb{R}^{n}:x_{1}\geq P(x)\\}$ for some
non-negative homogeneous quadratic polynomial satisfying $P(0)=0,DP(0)=0,$
2) $\Omega_{1,\infty}=\\{x\in\mathbb{R}^{n}:x_{n}\geq\rho_{\infty}(x_{1})\\}$
for some convex function $\rho_{\infty}$ when $|x_{1}|$ is small, moreover,
$\rho_{\infty}(0)=0,\rho_{\infty}\geq 0.$

###### Proof.

Fix any unit vector $e\in\text{span}\\{e_{2},\cdots,e_{n-1}\\}.$ Let $t_{h}$
be the positive number such that $|\frac{1}{h}(T_{2}^{t})^{-1}(t_{h}e)|=1,$ we
have that

$t_{h}\leq\frac{h}{\min_{2\leq i\leq n}a_{i}}\leq
C_{\epsilon}h^{\frac{1}{2}-\epsilon}.$

For any $t<h^{-\epsilon}t_{h},$ denote
$x^{t}:=te=(0,x^{t}_{2},\cdots,x^{t}_{n-1},0).$ It is straightforward to check
that $|T^{*}(h^{-\epsilon}t_{h}e)|\rightarrow\infty,$ as $h\rightarrow 0.$

Now $q^{t}:=x^{t}+\rho(0,x^{t}_{2},\cdots,x^{t}_{n-1})e_{n}\in\mathcal{F}.$
Let $z_{i}=e_{n}\cdot\bar{e}_{i},$ we have that
$e_{n}=\sum_{i=2}^{n}z_{i}\bar{e}_{i}.$ Then

$T^{*}(\rho(0,x^{t}_{2},\cdots,x^{t}_{n-1})e_{n})=\frac{1}{h}\rho(0,x^{t}_{2},\cdots,x^{t}_{n-1})\left(a_{1}(\sum_{i=2}^{n}k_{i}z_{i})e_{1}+\sum_{i=2}^{n}a_{i}z_{i}\bar{e}_{i}\right).$

Since $\rho(0,x^{t}_{2},\cdots,x^{t}_{n-1})\leq
C_{\epsilon}t^{\frac{3}{2}-\epsilon},$ $|t|\leq
C_{\epsilon}h^{\frac{1}{2}-2\epsilon},$ $|z_{i}|\leq 1,$ and $|a_{i}|\leq
C_{\epsilon}h^{\frac{1}{3}-\epsilon}$ for $i=2,\cdots,n,$ we have

$|\frac{1}{h}\rho(0,x^{t}_{2},\cdots,x^{t}_{n-1})\sum_{i=2}^{n}a_{i}z_{i}\bar{e}_{i}|\leq
C_{\epsilon}h^{\frac{1}{12}-10\epsilon}$

provided $h$ is small.

Let $\tilde{d}_{e}:=\\{|x\cdot e|:x\in S_{h}^{c}[v]\\}.$ By uniform density,
Remark 2.1 and Lemma 3.1 we have

(3.8) $\tilde{d}_{e}^{2}\leq Ca_{1}\ \text{for any}\
e\in\text{span}\\{e_{2},\cdots,e_{n}\\}.$

It follows that

(3.9) $|k_{i}|\leq C\frac{a_{1}^{\frac{1}{2}}}{a_{1}}=Ca_{1}^{-\frac{1}{2}}.$

Note that $|a_{1}|\leq C_{\epsilon}h^{\frac{2}{3}-\epsilon}.$ Hence

$\displaystyle|\frac{1}{h}\rho(0,x^{t}_{2},\cdots,x^{t}_{n-1})a_{1}(\sum_{i=2}^{n}k_{i}z_{i})e_{1}|\leq
C_{\epsilon}\frac{1}{h}\left(C_{\epsilon}h^{\frac{1}{2}-2\epsilon}\right)^{\frac{3}{2}-\epsilon}\left(C_{\epsilon}h^{\frac{2}{3}-\epsilon}\right)^{\frac{1}{2}}\leq
C_{\epsilon}h^{\frac{1}{12}-10\epsilon}.$

Hence

(3.10) $|T^{*}(q^{t})-T^{*}(x^{t})|\leq
2C_{\epsilon}h^{\frac{1}{12}-10\epsilon}\rightarrow 0$

as $h\rightarrow 0,$ provided initially we choose $\epsilon$ small enough. Up
to a subsequence we may assume $T^{*}(\text{span}\\{e_{2},\cdots,e_{n-1}\\})$
converges to $H,$ an $(n-2)$-dimensional subspace of $\mathbb{R}^{n}.$ Then,
by (3.10) we may assume $H\subset\partial\Omega_{1,\infty}.$

Now, let $u_{\infty}(x):=\sup\\{x\cdot y-v_{\infty}(y):y\in
B_{1}(0)\cap\Omega^{*}_{1,\infty}\\}.$ By Corollary (3.2) we have that

(3.11) $u_{\infty}(x)\geq C|x|^{\beta^{\prime}}\ \text{in}\
B_{r}\cap\overline{\Omega_{1,\infty}}.$

for some small positive number $r$ and $\beta^{\prime}>1.$ First, we observe
that

(3.12) $e_{1}\notin H,$

since otherwise $u_{\infty}$ is not strictly convex in
$B_{1}\cap\overline{\Omega_{1,\infty}}.$ Note that $\frac{Te_{n}}{|Te_{n}|}$
is the unit inner normal of $T^{*}\Omega_{1}.$ Up to a subsequence of $h$ we
may assume $\frac{Te_{n}}{|Te_{n}|}$ converges to a unit vector $e_{\infty}.$
We have that $e_{\infty}$ is orthogonal to $H$ and $e_{1}.$ By Lemma 3.3 we
have that $\\{te_{1}:t\in\mathbb{R}\\}$ can not pass the interior of
$\Omega_{1,\infty}.$ Hence

(3.13) $\Omega_{1,\infty}\subset\\{x:x\cdot e_{\infty}\geq 0\\}.$

Recall that
$\partial\Omega^{*}_{1}=\\{x:x_{1}=\rho^{*}(x_{2},\cdots,x_{n})\\}$ for some
$C^{2}$ uniformly convex function $\rho^{*}$ near $0$ satisfying
$\rho^{*}(0)=0,D\rho^{*}(0)=0.$ Denote
$x^{\prime\prime}=(x_{2},\cdots,x_{n}).$ Then there exits a positive
homogeneous quadratic polynomial $P^{\prime}$ such that
$\rho^{*}(x^{\prime\prime})=P^{\prime}(x^{\prime\prime})+o(|P^{\prime}(x^{\prime\prime})|).$
Extend $P^{\prime}$ to $\mathbb{R}^{n}$ such that
$\tilde{P}(x_{1},x^{\prime\prime})=P^{\prime}(x^{\prime\prime}).$ Hence

$\partial\Omega^{*}_{1}=\\{x:x\cdot e_{1}=\tilde{P}(x)+o(|\tilde{P}(x)|)\\}.$

By a straightforward computation, we have

(3.14) $\partial(T_{h}\Omega_{1}^{*})=\big{\\{}x\,:\,\langle
x,e_{1}\rangle=\tilde{P}_{h}(x)+o(\tilde{P}_{h})\big{\\}}\ \ \text{ near }0,$

where
$\tilde{P}_{h}(x)=\frac{1}{|(T_{h}^{t})^{-1}e_{1}|}\tilde{P}(T_{h}^{-1}x)\geq
0$, and

(3.15) $B_{1}(0)\cap T_{h}\Omega_{1}^{*}\subset\Big{\\{}x\,:\,\langle
x,e_{1}\rangle\geq\frac{1}{2}\tilde{P}_{h}(x)\Big{\\}}\ \ \text{ for $h>0$
small}.$

Then, by the uniform density property of $v$ near $0,$ we have that the
coefficients of $\tilde{P}_{h}$ is bounded by some constant independent of
$h.$ Hence, by passing to a subsequence we may assume that $\tilde{P}_{h}$
converges to a non-negative homogeneous quadratic polynomial $P.$ Hence
$\Omega_{1,\infty}^{*}=\\{x\in\mathbb{R}^{n}:x_{1}\geq P(x)\\}.$

Finally, since $e_{1}$ is the unit inner normal of $\Omega^{*}_{1,\infty},$ by
(3.12) and (3.13), up to an affine transformation we can achieve the desired
property as stated in the statement of the lemma. ∎

In the following, for short of notations we denote $v_{\infty},u_{\infty}$ by
$\tilde{v},\tilde{u}$ respectively. Let
$U:=Dv_{\infty}(B_{1}(0)\cap\Omega^{*}_{1,\infty}),$ and let
$V:=B_{1}(0)\cap\Omega^{*}_{1,\infty}.$ By Corollary 3.2 we have that
$B_{r_{0}}\cap\Omega_{1,\infty}\subset U$ for some small $r_{0}>0.$ Now,

(3.16) ${\begin{split}\mbox{det}\,D^{2}\tilde{u}&=c_{0}\quad\ \text{ in }\
U,\\\ D\tilde{u}(U)&=V\end{split}}$

for some positive constant $c_{0}.$ We summarize the properties of
$\tilde{u},\tilde{v}$ as follows:
1) Since $\partial U$ and $\partial V$ is convex near $0,$ by the results in
[5] we have that $S_{h}^{c}[\tilde{u}],S_{h}^{c}[\tilde{v}]$ decay
geometrically, $\tilde{u}$ (resp. $\tilde{v}$) is $C^{1,\alpha}$ and strictly
convex in $(B_{r_{1}}\cap\overline{U})$ (resp. $(B_{r_{1}}\cap\overline{V})$
for some positive $r_{1}>0$
2) Since the uniform density property is affine invariant, it implies that
$\tilde{v}$ also satisfies the uniform density property, namely,
$\frac{|S_{h}^{c}[\tilde{v}]\cap V|}{|S_{h}^{c}[\tilde{v}]|}\geq\delta$ for
some constant $\delta>0$ independent of $h.$ By duality, it also implies that
$\tilde{u}$ satisfies the uniform density property.
3) The uniform density property implies that $S^{c}_{h}[{\tilde{u}}]$ and
$S^{c}_{h}[\tilde{v}]$ are conjugate for small $h$ in the sense that if
$AS^{c}_{h}[{\tilde{u}}]\sim B_{\frac{h}{2}}$ for some affine transform with
$\mbox{det}A=1,$ then $(A^{t})^{-1}S^{c}_{h}[\tilde{v}]\sim B_{\frac{h}{2}}.$
4) Since $\partial U$ is flat in $e_{2},\cdots,e_{n-1}$ directions, by [5,
Corollary 1.1] we have that $\tilde{u}$ is $C^{1,1}$ in these directions.
Hence $B_{Ch^{\frac{1}{2}}}\cap\\{x_{n}=0\\}\subset S_{h}^{c}[\tilde{u}]$ for
$h$ small. Then, by duality we have that

(3.17) $\sup\\{|x\cdot e|:x\in S_{h}^{c}[\tilde{v}]\\}\leq Ch^{\frac{1}{2}},$

for $e\in\text{span}\\{e_{2},\cdots,e_{n-1}\\}$.
5) Since $\partial V$ is smooth, by the result in [8, Lemma 3.1] we also have
that $\tilde{v}$ is $C^{1,1-\epsilon}$ tangentially, namely

(3.18) $B_{C_{\epsilon}h^{\frac{1}{2}+\epsilon}}\cap\\{x_{1}=0\\}\subset
S_{h}^{c}[\tilde{v}].$

Second blow up. Suppose $S_{h}^{c}[\tilde{v}]\sim E$ for some ellipsoid
centered at $0.$ Then, $E\cap\\{x_{1}=0\\}$ is an $(n-1)$-dimensional
ellipsoid with principal directions $\bar{e}_{2},\cdots,\bar{e}_{n}.$ Note
that
$\text{span}\\{\bar{e}_{2},\cdots,\bar{e}_{n}\\}=\text{span}\\{e_{2},\cdots,e_{n}\\}.$
Now, we can write

$E=\\{x=x_{1}e_{1}+\sum_{i=2}^{n}\bar{x}_{i}\bar{e}_{i}:\frac{x_{1}^{2}}{a_{1}^{2}}+\sum_{i=2}^{n}\frac{(\bar{x}_{i}-k_{i}x_{1})^{2}}{a_{i}^{2}}\leq
1\\}.$

By (3.17) and (3.18) we have that

(3.19) $c_{\epsilon}h^{\frac{1}{2}+\epsilon}\leq a_{i}\leq
C_{\epsilon}h^{\frac{1}{2}-\epsilon},\ \text{for}\ i=2,\cdots,n-1,\ \
a_{n}\geq c_{\epsilon}h^{\frac{1}{2}+\epsilon}.$

By the $C^{1,\delta}$ regularity of $\tilde{v}$ in $B_{r_{0}}\cap V,$ we have
that

(3.20) $B_{Ch^{\frac{1}{1+\delta}}}\cap V\subset S_{h}^{c}[\tilde{v}].$

Let $A_{h}$ be the affine transformation such that $A_{h}E=B_{1}.$ We may
write $A_{h}=A_{h2}A_{h1},$ where

$A_{h1}:x=x_{1}e_{1}+\sum_{i=2}^{n}\bar{x}_{i}\bar{e}_{i}\mapsto
z=x_{1}e_{1}+\sum_{i=2}^{n}(\bar{x}_{i}-k_{i}x_{1})\bar{e}_{i};$
$A_{h2}:z=z_{1}e_{1}+\sum_{i=2}^{n}\bar{z}_{i}\bar{e}_{i}\mapsto
y=\frac{z_{1}}{a_{1}}e_{1}+\sum_{i=2}^{n}\frac{\bar{z}_{i}}{a_{i}}\bar{e}_{i}.$

Now, for any $x=(x_{1},x^{\prime\prime})\subset\\{x_{n}=0\\}\cap\partial V$
with $|x^{\prime\prime}|=h^{\frac{1}{2}-5\epsilon},$ we have that
$x_{1}=\tilde{\rho}(x^{\prime\prime})\leq Ch^{1-10\epsilon}.$ Hence, by (3.20)
we have that

$\text{dist}\left(A_{h}x,A_{h}\text{span}\\{e_{2},\cdots,e_{n-1}\\}\right)\leq
C_{\epsilon}h^{1-10\epsilon-\frac{1}{1+\delta}}\rightarrow 0$

as $h\rightarrow 0$ provided $\epsilon$ is chosen small enough. Note that for
such $x$ we also have

$|A_{h}x|\geq C_{\epsilon}h^{\frac{1}{2}-5\epsilon-\frac{1}{2}+\epsilon}\geq
C_{\epsilon}h^{-4\epsilon}\rightarrow\infty$

as $h\rightarrow 0$ provided $\epsilon$ is chosen small enough. Therefore, by
convexity, up to a subsequence $A_{h}(\\{x_{n}=0\\}\cap\partial V)$ locally
uniformly converges to $\tilde{H}^{*},$ which is an $(n-2)$-dimensional
subspace of $\mathbb{R}^{n}.$ Hence, by convexity, up to a subsequence,
$A_{h}\partial V$ locally uniformly converges to a smooth convex set in the
form of $\tilde{\Omega}^{*}=\omega^{*}\times\tilde{H}^{*},$ where $\omega^{*}$
is a two dimensional smooth convex set. It is easy to see that
$(A_{h}^{t})^{-1}(U\cap B_{r_{0}})$ locally uniformly converges to a convex
set $\tilde{\Omega}=\omega\times\tilde{H}$ for some $(n-2)$-dimensional
subspace of $\mathbb{R}^{n}.$ Note that up to a subsequence,
$(A_{h}^{t})^{-1}e_{1},(A_{h}e_{n})$ converges to $\tilde{e}^{*},\tilde{e}$
respectively up to a subsequence, moreover,

$\tilde{\Omega}\subset\\{x:x\cdot\tilde{e}>0\\}\ \text{and}\
\tilde{\Omega}^{*}\subset\\{x:x\cdot\tilde{e}^{*}>0\\}.$

Moreover, by performing another affine transform we may assume that

$\tilde{\Omega}=\\{x\in\mathbb{R}^{n}:x_{n}\geq\tilde{\rho}(x_{1})\\},$

and that

$\tilde{\Omega}^{*}=\\{x\in\mathbb{R}^{n}:x_{1}\geq\tilde{\rho}^{*}(x_{n})\\},$

where $\tilde{\rho}$ is a convex function defined in a neighborhood of $0$
with $\tilde{\rho}(0)=0$ and $\tilde{\rho}\geq 0,$ and $\tilde{\rho}^{*}$ is a
smooth convex function defined in a neighborhood of $0$ with
$\tilde{\rho}^{*}(0)=0$ and $\tilde{\rho}^{*}\geq 0.$

Finally, contradiction can be made by following the argument in [10].

###  3.2. Case II): $\hat{\nu}\cdot\hat{\nu}^{*}=0,$ and
$\nu,\nu^{*},\hat{\nu}^{*}$ are in the same plane.

Denote

$d_{h}:=\sup\\{x\cdot e_{1}:x\in S_{h}[v]\cap\Omega_{1}^{*}\\}$

and

$\hat{d}_{h}:=\sup\\{|x\cdot e_{1}|:x\in S_{h}[v]\cap\Omega_{2}^{*}\\}.$

Then either $d_{h}\geq\hat{d}_{h}$ or $d_{h}<\hat{d}_{h}$

Then there exists a sequence $h_{k}\rightarrow 0$ such that either
$d_{k}:=d_{h_{k}}\geq\hat{d}_{h_{k}}$ holds for all $k$ or the opposite
inequality holds for all $k.$ Suppose the former holds for all $k.$ Let

$d_{k,e}:=\sup\\{|x\cdot e|:x\in S_{h_{k}}[v]\cap\Omega_{1}^{*}\\},\ \
\hat{d}_{k,e}:=\sup\\{|x\cdot e|:x\in S_{h_{k}}[v]\cap\Omega_{2}^{*}\\},$

for any unit vector $e\in\text{span}\\{e_{2},\cdots,e_{n}\\}.$ By the same
proof of Lemma 3.1, we have

(3.21) $d_{k,e}\leq C_{\epsilon}h^{\frac{1}{3}-\epsilon}\ \text{and}\
\hat{d}_{k,e}\leq C_{\epsilon}h^{\frac{1}{3}-\epsilon},\forall
e\in\text{span}\\{e_{2},\cdots,e_{n}\\}.$

Suppose $S_{h_{k}}^{c}[v](y_{0})\sim E+\\{y_{0}\\}$ for some ellipsoid $E$
centered at $0.$ Then, $E\cap\\{x_{1}=0\\}$ is an $(n-1)$-dimensional
ellipsoid with principal directions $\bar{e}_{2},\cdots,\bar{e}_{n}.$ Note
that
$\text{span}\\{\bar{e}_{2},\cdots,\bar{e}_{n}\\}=\text{span}\\{e_{2},\cdots,e_{n}\\}.$
Similarly to that in the last section, we can write

$E=\\{x=x_{1}e_{1}+\sum_{i=2}^{n}\bar{x}_{i}\bar{e}_{i}:\frac{x_{1}^{2}}{a_{1}^{2}}+\sum_{i=2}^{n}\frac{(\bar{x}_{i}-k_{i}x_{1})^{2}}{a_{i}^{2}}\leq
1\\}.$

By Lemma 3.1 and the tangential $C^{1,1-\epsilon}$ estimate of $v$ at $y_{0},$
we have that

(3.22) $0<a_{1}<C_{\epsilon}h_{k}^{\frac{2}{3}-\epsilon},\ \text{and}\
C_{\epsilon}h_{k}^{\frac{1}{2}+\epsilon}<a_{i}<C_{\epsilon}h^{\frac{1}{3}-\epsilon}\
\text{for}\ i=2,\cdots,n.$

By Lemma 3.1 (apply the lemma to $v$ near $\hat{y}_{0}$) and the tangential
$C^{1,1-\epsilon}$ estimate of $v$ at $\hat{y}_{0},$ we have that

(3.23) $\hat{d}_{k,e}\leq C_{\epsilon}h^{\frac{1}{3}-\epsilon},\ \forall
e\in\text{span}\\{e_{2},\cdots,e_{n}\\}.$

By the same reason of (3.9), we have

(3.24) $|k_{i}|\leq C\frac{a_{1}^{\frac{1}{2}}}{a_{1}}=Ca_{1}^{-\frac{1}{2}}.$

Note that by our assumption we have

(3.25) $\hat{d}_{k}\leq d_{k}\approx a_{1}.$

Let $T_{k1},T_{k2}$ be the affine transformations as following.

$T_{k1}:x=x_{1}e_{1}+\sum_{i=2}^{n}\bar{x}_{i}\bar{e}_{i}\mapsto
z=x_{1}e_{1}+\sum_{i=2}^{n}(\bar{x}_{i}-k_{i}x_{1})\bar{e}_{i};$
$T_{k2}:z=z_{1}e_{1}+\sum_{i=2}^{n}\bar{z}_{i}\bar{e}_{i}\mapsto
y=\frac{z_{1}}{a_{1}}e_{1}+\sum_{i=2}^{n}\frac{\bar{z}_{i}}{a_{i}}\bar{e}_{i}.$

Then $T_{k}(E)=B_{1},$ where $T_{k}=T_{k2}T_{k1}.$ Hence
$T_{k}S_{h_{k}}^{c}[v]\sim B_{1}.$ Now $B_{\frac{1}{C}}(T_{k}y_{0})\cap
T_{k}\Omega_{1}^{*}\subset T_{k}(S_{h}[v]\cap\Omega_{1}^{*})\subset
B_{C}(T_{k}y_{0}).$ Denote by $\Omega^{*}_{ki}=T_{k}(\Omega^{*}_{i}),i=1,2.$
Denote $z_{k}=T_{k}y_{0}\in\partial\Omega^{*}_{k1}.$ Let

$v_{k}(\cdot)=\frac{1}{h_{k}}v(T_{k}^{-1}\cdot);\quad
u_{k}(\cdot)=\frac{1}{h_{k}}u(T_{k}^{t}\cdot).$

Denote $\Omega_{ki}=\frac{1}{h_{k}}(T_{k}^{t})^{-1}\Omega_{i},i=1,2,$ and
$\mathcal{F}_{k}=\frac{1}{h_{k}}(T_{k}^{t})^{-1}\mathcal{F}.$ Now, up to a
rotation of coordinates (changing $T_{k}e_{n}$ to $e_{n}$ direction), we may
assume $z_{k}=l_{k}e_{n},$ and $e_{1},-e_{1}$ are the unit inner normals of
$\partial\Omega^{*}_{ki},\ k=1,2$ at $z_{k},-z_{k}$ respectively. Note that
$l_{k}\rightarrow\infty$ as $k\rightarrow\infty.$ For any
$x\in\mathcal{F}_{k},$ we have that $\nu_{k}(x),$ the unit normal of
$\mathcal{F}_{k}$ at $x$ is given by
$\frac{Du_{k1}(x)-Du_{k2}(x)}{|Du_{k1}(x)-Du_{k2}(x)|},$ where $u_{ki}$ is the
restriction of $u_{k}$ on $\Omega_{ki}.$ Then we have that $\nu_{k}(x)\cdot
e_{n}>0$ for any $x\in B_{1}(0)\cap\mathcal{F}_{k},$ hence
$B_{1}(0)\cap\mathcal{F}_{k}$ is the graph of some function $\rho_{k}$ defined
on $B_{1}(0)\cap\\{x_{n}=0\\}.$

Then, by the assumption $d_{k}:=d_{h_{k}}\geq\hat{d}_{k}:=\hat{d}_{h_{k}}$
holds for all $k$ we have that the width of $S_{1}[v_{k}]\cap\Omega^{*}_{ki},\
i=1,2$ in $e_{1}$ direction are bounded by a universal constant $C.$ We will
need the following lemmas.

######  Lemma 3.6.

$\frac{\text{dist}(S_{1}[v_{k}]\cap\Omega^{*}_{k1},S_{1}[v_{k}]\cap\Omega^{*}_{k2})}{l_{k}}\rightarrow
1$ as $k\rightarrow\infty.$

###### Proof.

By the definition of $T_{k1}$ and (3.22)-(3.25), a direct computation shows
that the width of $T_{k1}(S_{h_{k}}\cap\Omega_{2}^{*})$ in $e_{i}$ direction
is bounded by $C_{\epsilon}h_{k}^{\frac{1}{3}-\epsilon}$ for $i=2,\cdots,n.$
Then, by (3.22) and the definition of $T_{k2},T_{k},$ we have that the width
of $S_{1}[v_{k}]\cap\Omega^{*}_{k2}=T_{k}(S_{h_{k}}\cap\Omega_{2}^{*})$ in
$e_{i}$ direction is bounded by

(3.26)
$\frac{C_{\epsilon}h_{k}^{\frac{1}{3}-\epsilon}}{\min\\{a_{i};i=2,\cdots,n\\}}\leq
C_{\epsilon}h_{k}^{-\frac{1}{6}-2\epsilon}.\text{for}\ i=2,\cdots,n.$

By (3.25) we also have that the width of $S_{1}[v_{k}]\cap\Omega^{*}_{k2}$ in
$e_{1}$ direction is bounded by some constant $C$ independent of $k.$ On the
other hand, by (3.23) we have that $l_{k}\geq
C_{\epsilon}h_{k}^{-\frac{1}{3}-\epsilon}.$ By the above discussion it is easy
to see that
$\frac{\text{dist}(S_{1}[v_{k}]\cap\Omega^{*}_{k1},S_{1}[v_{k}]\cap\Omega^{*}_{k2})}{l_{k}}\rightarrow
1$ as $k\rightarrow\infty.$

∎

######  Lemma 3.7.

There exist constants $0<\alpha<1,\beta>1$ such that

(3.27) $0\leq v_{k}(x)\leq C|x-z_{k}|^{1+\alpha}\ \text{for any}\ x\in
B_{1}(z_{k})$

and

(3.28) $v_{k}(x)\geq\frac{1}{C}|x-z_{k}|^{1+\beta}\ \text{for any}\ x\in
B_{1}(z_{k})\cap\Omega_{k1}^{*},$

where $C>0$ is a constant independent of $k.$

###### Proof.

The proof is exactly same to the proof of Lemma 3.4. ∎

######  Lemma 3.8.

There exists a universal constant $N,$ such that

(3.29) $S_{\frac{1}{N}}[u_{k1}-l_{k}x_{n}](0)\cap\mathcal{F}_{k}\subset
Dv_{k}(S_{1}[v_{k}]\cap\Omega^{*}_{k1}),$

and that

(3.30) $S_{\frac{1}{N}}[u_{k2}+l_{k}x_{n}](0)\cap\mathcal{F}_{k}\subset
Dv_{k}(S_{1}[v_{k}]\cap\Omega^{*}_{k2})$

###### Proof.

Since $u_{k1}-l_{k}x_{n}$ is the Legendre transform of $v_{k}(\cdot-z_{k}),$
using Lemma 3.4 it is straightforward to check that

(3.31) $\frac{1}{C}|x|^{1+\beta^{\prime}}\leq u_{k1}-l_{k}x_{n}\leq
C|x|^{1+\alpha^{\prime}}$

in $B_{r}(0)\cap\Omega_{k1}$ for some constants
$\alpha^{\prime}\in(0,1),\beta>1,C>0,r>0$ independent of $k.$ Then, by
convexity, the desired estimate 3.29 follows from (3.31). The proof of (3.30)
is similar, indeed Lemma 3.4 holds for $v_{k}$ near $\hat{z}_{k}:=-z_{k}$ if
we normalize $S_{h_{k}}[v]\cap\Omega^{*}_{2}$ instead of
$S_{h_{k}}[v]\cap\Omega^{*}_{1}$ before Lemma 3.4. Note that (3.29) and (3.30)
are affine invariant. ∎

By (3.31) for sufficiently large $M,$ we have that

(3.32) $u_{k1}-l_{k}x_{n}\leq\frac{1}{4N}$

provided $x\in B_{\frac{1}{M}}(0)\cap\Omega_{k1}.$ By choosing $M$ large we
may also assume $B_{\frac{1}{M}}(0)\cap\Omega_{k1}\subset
Dv_{k}(S_{1}[v_{k}]\cap\Omega^{*}_{k1}).$ Recall that
$B_{1}(0)\cap\mathcal{F}_{k}$ is the graph of some function $\rho_{k}$ defined
in a neighborhood of $0$ inside $B_{1}(0)\cap\\{x_{n}=0\\}.$ Let
$p_{t}=\big{(}-t,0,\cdots,0,\rho_{k}(-t,0\cdots,0)\big{)}$ be a point on
$\mathcal{F}_{k}\cap\text{span}\\{e_{1},e_{n}\\}.$ We have that the width of
$S_{1}[v_{k}]\cap\Omega^{*}_{ki},\ i=1,2$ in $e_{1}$ direction are bounded by
a universal constant $C,$ and
$\frac{\text{dist}(S_{1}[v_{k}]\cap\Omega^{*}_{k1},S_{1}[v_{k}]\cap\Omega^{*}_{k2})}{l_{k}}\rightarrow
1$ as $k\rightarrow\infty.$ Then, if $p_{t}\in B_{\frac{1}{M}}(0)\cap
Dv_{k}(S_{1}[v_{k}]\cap\Omega^{*}_{k2}),$ we have

(3.33) $|\nu_{k}(p_{t})\cdot
e_{1}|=|\frac{Du_{k1}(p_{t})-Du_{k2}(p_{t})}{|Du_{k1}(p_{t})-Du_{k2}(p_{t})|}\cdot
e_{1}|\leq\frac{C}{l_{k}}$

for some universal constant $C$ provided $k$ is large enough.

By (3.33) we have that

(3.34) $|D_{1}\rho_{k}(p_{t})|\leq\frac{C}{l_{k}}$

for $k$ large.

Now, let

$s_{k0}:=\sup\\{s:p_{t}\in B_{\frac{1}{2M}}(0)\cap
Dv_{k}(S_{1}[v_{k}]\cap\Omega^{*}_{k2})\ \text{for}\ 0\leq t\leq s\\}.$

We will establish the following estimate on $s_{k0}.$

######  Lemma 3.9.

(3.35) $s_{k0}\geq\min\\{\frac{1}{2M},\frac{1}{4NC}\\}$

provided $k$ is sufficiently large.

###### Proof.

For simplicity of notation we denote $s_{k0}$ by $s_{0}.$ Suppose (3.35)
fails. Then by the definition of $s_{k0}$ we have that

(3.36) $u_{k2}(p_{s_{0}})+l_{k}p_{s_{0}}\cdot e_{n}\geq\frac{1}{N}.$

Then by 3.32 we have

(3.37) $u_{k1}(p_{s_{0}})\leq\frac{1}{4N}+l_{k}p_{s_{0}}\cdot
e_{n}\leq\frac{1}{4N}+l_{k}\frac{C}{l_{k}}s_{0}\leq\frac{1}{2N}.$

Since $u_{k1}=u_{k2}$ on $\mathcal{F}_{k},$ we have

$\displaystyle u_{k2}(p_{s_{0}})+l_{k}p_{s_{0}}\cdot e_{n}$ $\displaystyle=$
$\displaystyle u_{k1}(p_{s_{0}})+l_{k}p_{s_{0}}\cdot e_{n}$
$\displaystyle\leq$ $\displaystyle\frac{1}{2N}+\frac{1}{4N}<\frac{1}{N},$

which contradicts to (3.36). ∎

Let $t_{0}:=\min\\{\frac{1}{2M},\frac{1}{4NC}\\},$ by (3.34) and Lemma 3.9 we
have $\rho_{k}(-t_{0}e_{1})=p_{t_{0}}\cdot
e_{n}\leq\frac{C}{l_{k}}t_{0}\rightarrow 0$ as $k\rightarrow\infty.$ Now, let
$\tilde{u}_{k}(x):=\sup\\{x\cdot y-v_{k}(y+z_{k}):y+z_{k}\in
S_{1}[v_{k}]\cap\Omega^{*}_{k1}\\}.$ Then, $\tilde{u}_{k}=u_{k1}-l_{k}x_{n}$
on $B_{\frac{1}{2M}}(0)\cap\Omega_{k1}.$ Now, by the assumption that
$\Omega^{*}_{k1}\subset\\{x_{1}\geq 0\\}$ we have $D_{1}\tilde{u}_{k}\geq 0.$
Therefore

$\displaystyle u_{k1}(p_{t_{0}})-l_{k}p_{t_{0}}\cdot e_{n}$ $\displaystyle=$
$\displaystyle\tilde{u}_{k}(p_{t_{0}})$ $\displaystyle\leq$
$\displaystyle\tilde{u}_{k}(-t_{0}e_{1})+Cp_{t_{0}}\cdot e_{n}$
$\displaystyle\leq$ $\displaystyle Cp_{t_{0}}\cdot
e_{n}\leq\frac{C}{l_{k}}t_{0}\rightarrow 0$

as $k\rightarrow\infty,$ which contradicts to (3.31) for $k$ large enough.

###  3.3. Case III): $\hat{\nu}\cdot\hat{\nu}^{*}>0.$

In this case the arguments for the optimal partial transport problem([10]) for
$v$ restricted to $\Omega^{*}_{2},$ namely, $v$ is $C^{2,\alpha}$ regular near
$\hat{y}_{0}.$ Note that in Case I), the conditions
$\hat{\nu}\cdot\hat{\nu}^{*}=0$ and $\nu,\nu^{*},\hat{\nu}^{*}$ are not in the
same plane are only used in the proof of Lemma 3.2 and Lemma 3.3.

For the proof of Lemma 3.2 we only used the property that $v(x)\geq
C_{\epsilon}|x-\hat{y}_{0}|^{3+\epsilon}$ for all $x\in\Omega_{2}^{*}.$ In the
current case we have that

$v(x)=\frac{1}{2}D^{2}v(\hat{y}_{0})(x-\hat{y}_{0})\cdot(x-\hat{y}_{0})+o(|x-\hat{y}_{0}|^{2})$

for all $x\in\Omega_{2}^{*},$ where $D^{2}v(\hat{y}_{0})$ is positive
definite. Hence $v(x)\geq C|x-\hat{y}_{0}|^{2}$ for some positive constant
$C,$ which is also enough for the proof of Lemma 3.2.

For the proof of Lemma 3.3, up to an affine transformation we may assume that
$\nu$ is parallel to $\hat{\nu}^{*}.$ Since $v$ is $C^{2,\alpha}$ regular near
$\hat{y}_{0}$ with $v(\hat{y}_{0})=0$ and $Dv(\hat{y}_{0})=0,$ it is easy to
find a point $p\in S_{h}[v]$ such that $p_{1}\approx h^{\frac{1}{2}},$ which
contradicts to (3.3). Thus Lemma 3.3 can still be established for the current
case. Therefore we can repeat the remaining proof in Case I) to rule out the
current case.

##  4\. Proof of Theorem 1.1

Once we have proved Proposition 3.1, the same arguments for the optimal
partial transport problem ([10]) works.

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