$C^{1,\alpha}$-estimates for the near field refractor

TL;DR

This paper proves local $C^{1,eta}$ regularity estimates for near field refractors in geometric optics, under certain structural target assumptions, without requiring smooth densities.

Contribution

It introduces new $C^{1,eta}$ regularity results for near field refractors with minimal assumptions on densities and target structure.

Findings

Established local $C^{1,eta}$ estimates for near field refractors

Achieved regularity results without smoothness assumptions on densities

Provided structural conditions on the target for regularity

Abstract

We establish local $C^{1,\alpha}$ estimates for one source near field refractors under structural assumptions on the target, and with no assumptions on the smoothness of the densities.

Equations

$$\frac{X}{|X|}\cdot \frac{Y-X}{|Y-X|}\ge \kappa ;$$

$$\rho (x,Y,b)=\frac{b-{\kappa }^{2}x\cdot Y-\sqrt{(b-{\kappa }^{2}x\cdot Y{)}^2-(1-{\kappa }^2)(b^2-{\kappa }^2|Y{|}^2)}}{1-{\kappa }^2}.$$

$$h(x,Y,X_0)=\rho (x,Y,b),$$

$${\Gamma }_{c_1{c}_2}=\left\{rx:x\in \Omega ,c_1\le r\le c_2\right\}.$$

$$\beta (\lambda )=\frac{-x_0\cdot \xi -\sqrt{(x_0\cdot \xi {)}^2-\left(1-{\kappa }^2\right)|\xi {|}^2}}{\kappa |\xi {|}^2},$$

$$[\bar{m},\hat{m}{]}_{x_0}=\left\{m(\lambda )=\frac{1}{\kappa }{x}_0+\beta (\lambda )\left(m_{\lambda }-\frac{1}{\kappa }{x}_0\right),\lambda \in [0,1]\right\}.$$

$$m=\frac{1}{\kappa }{x}_0+\bar{\beta }\left(\bar{m}-\frac{1}{\kappa }{x}_0\right)+\hat{\beta }\left(\hat{m}-\frac{1}{\kappa }{x}_0\right)$$

$${\mathcal{C}}_{\Omega }=\bigcap _{X\in {\Gamma }_{c_1{c}_2}}{C}_X.$$

$$\Sigma =\{X+s_X(m)m:m\in E(X)\},$$

$$\frac{1}{s_{X_0}(m(\lambda ))}+\frac{\mu }{|X_0|}\ge \bar{\beta }(\lambda )\left(\frac{1}{s_{X_0}(\bar{m})}+\frac{\mu }{|X_0|}\right)+\hat{\beta }(\lambda )\left(\frac{1}{s_{X_0}(\hat{m})}+\frac{\mu }{|X_0|}\right)$$

$$[\bar{Y},\hat{Y}{]}_{X_0}=\left\{Y(\lambda )=X_0+s_{X_0}(m(\lambda ))m(\lambda ):\lambda \in [0,1]\right\},$$

$$H^{n-1}\left(N_{\mu }\left(\left\{[\bar{Y},\hat{Y}{]}_{X_0}:\frac{1}{4}\le \lambda \le \frac{3}{4}\right\}\right)\cap \Sigma \right)\ge C{\mu }^{n-2}|\bar{Y}-\hat{Y}|,$$

$$s_X(m)\ge c_1/C(\kappa ),$$

$$|\hat{m}-\bar{m}|\le 2\min \left\{\frac{1}{|\bar{Y}-X|},\frac{1}{|\hat{Y}-X|}\right\}|\bar{Y}-\hat{Y}|\le C|\bar{Y}-\hat{Y}|.$$

$$\hat{m}-\bar{m}=\frac{\hat{Y}-X}{|\hat{Y}-X|}-\frac{\bar{Y}-X}{|\bar{Y}-X|}=\frac{|\bar{Y}-X|(\hat{Y}-\bar{Y})+(\bar{Y}-X)(|\bar{Y}-X|-|\hat{Y}-X|)}{|\hat{Y}-X||\bar{Y}-X|},$$

$$|\bar{m}-\hat{m}|\le \frac{2|\hat{Y}-\bar{Y}|}{|\hat{Y}-X|},|\bar{m}-\hat{m}|\le \frac{2|\hat{Y}-\bar{Y}|}{|\bar{Y}-X|}.$$

$$\frac{Y-X}{|Y-X|}\cdot \frac{X}{|X|}\ge \kappa ,\text{ and }\frac{|X|}{|Y-X|}\le {\hat{C}}_{\tau ,\kappa },$$

$$C_{\tau ,\kappa }=\frac{\sqrt{1-{\kappa }^2}}{(\kappa +\tau )\sqrt{1-{\kappa }^2}-\kappa \sqrt{1-(\kappa +\tau {)}^2}},{\hat{C}}_{\tau ,\kappa }=\frac{1}{C_{\tau ,\kappa }-1}.$$

$${\nabla }^{T}h(x,Y,X_0)={\nabla }_{x}h(x,Y,X_0)-⟨{\nabla }_{x}h(x,Y,X_0),x⟩x,$$

$${\nabla }^{T}h(x_0,Y,X_0)=\kappa |X_0|\frac{m-⟨m,x_0⟩x_0}{1-\kappa ⟨m,x_0⟩}:=v=T_{X_0}(m),x_0=X_0/|X_0|.$$

$$\frac{1-\kappa ⟨m,x_0⟩}{\kappa |X_0|}=\frac{1-{\kappa }^2}{\kappa }\frac{1}{|X_0|+\sqrt{{\kappa }^2|X_0{|}^2-(1-{\kappa }^2)|v{|}^2}}:=t(v).$$

$$m=m(v):=\frac{1}{\kappa }{x}_0+t(v)(v-X_0).$$

$$\bar{v}=\kappa |X_0|\frac{\bar{m}-⟨\bar{m},x_0⟩x_0}{1-\kappa ⟨\bar{m},x_0⟩},\text{ and }\hat{v}=\kappa |X_0|\frac{\hat{m}-⟨\hat{m},x_0⟩x_0}{1-\kappa ⟨\hat{m},x_0⟩}.$$

$$\tilde{m}(\gamma )=m(v_{\gamma })=\frac{1}{\kappa }{x}_0+t(v_{\gamma })(v_{\gamma }-X_0),0<\gamma <1,$$

$$\bar{\beta }=(1-\lambda )\beta (\lambda )=\frac{t(v_{\gamma })(1-\gamma )}{t(\bar{v})}\text{and}\hat{\beta }=\lambda \beta (\lambda )=\frac{t(v_{\gamma })\gamma }{t(\hat{v})};$$

$$m(\lambda )$$

$$=\frac{1}{\kappa }{x}_0+\frac{1}{\kappa }t(v_{\gamma })\frac{(1-\gamma )t(\hat{v})+\gamma t(\bar{v})}{t(\hat{v})t(\bar{v})}\left((1-\lambda )(\kappa \bar{m}-x_0)+\lambda (\kappa \hat{m}-x_0)\right).$$

$$\left(\frac{1}{s_{X_0}(\tilde{m}(\gamma ))}+\frac{\mu }{|X_0|}\right)\frac{1}{t(v_{\gamma })}\ge (1-\gamma )\left(\frac{1}{s_{X_0}(\tilde{m}(0))}+\frac{\mu }{|X_0|}\right)\frac{1}{t(v_0)}+\gamma \left(\frac{1}{s_{X_0}(\tilde{m}(1))}+\frac{\mu }{|X_0|}\right)\frac{1}{t(v_1)},$$

$$\Phi (v)=\left(\frac{1}{s_{X_0}(m(v))}+\frac{\mu }{|X_0|}\right)\frac{1}{t(v)}$$

$${\frac{d^2}{d{t}^2}\left(\Phi (v+t\xi )\right)|}_{t=0}\le 0$$

$$\Sigma =\left\{Y=(Y^{\prime },Y_n):|Y^{\prime }|\le R{Y}_n=M\right\}.$$

Full text

The first author was partially supported by NSF grant DMS–1600578.

1. 1 Introduction

2. 2 Preliminaries, structural assumptions, and examples

3. 2.1 Structural assumptions on the target $\Sigma$

4. 2.2 Examples

5. 3 Preliminary results for ovals and a maximum principle

6. 4 Estimates for derivatives of ovals

7. 5 $C^{1,\alpha}$ estimates

8. 5.1 Regularity of weak solutions

1. Introduction

The main purpose in this paper is to prove Hölder estimates for gradients of weak solutions to the near field refractor problem introduced in [GH14], where existence of weak solutions is proved as a consequence of a general abstract method applicable also in other situations. The set up for the problem is as follows. Suppose we have a domain $\Omega\subset S^{n-1}$ and a domain $\Sigma$ contained in an $n$ dimensional surface in ${\mathbb{R}}^{n}$; here, $\Omega$ denotes the set of incident directions, and $\Sigma$ denotes the target domain, receiver, or screen to be illuminated. Let $n_{1}$ and $n_{2}$ be the indices of refraction of two homogeneous and isotropic media I and II, respectively. From a point source at the origin, surrounded by medium I, radiation emanates in each direction $x$ with intensity $f(x)$ for $x\in\Omega$, and the target $\Sigma$ is surrounded by medium II. A near field refractor is an optical surface $\mathcal{R}$, interface between media I and II, such that all rays refracted by $\mathcal{R}$ into medium II, in accordance with the Snell law, are received at the surface $\Sigma$ with prescribed radiation intensity distribution given by a measure $\nu$. Assuming no loss of energy in this process, we have the conservation of energy equation $\int_{\Omega}f(x)\,dx=\nu(\Sigma)$. Under visibility assumptions on the target and conditions to avoid total reflection, existence of solutions to this problem is proved in [GH14].

The problem solved in the present paper is that weak solutions are $C^{1}$ and their gradients are locally Hölder continuous under no smoothness assumptions on the density $f$ and the measure $\nu$. In fact, we prove a more general result, Theorem 5.5, valid for more general near field refractors in the sense of Definition 5.1. Our assumptions are of structural nature, that is, they depend on the relative location of the target, its visibility from the cone of incident directions, and its convexity; see Section 2.1. A major difficulty with the near field refractor problem is that solutions have a complicated structure given by Descartes ovals that often require difficult analytical estimates, and it does not have an optimal mass transport structure.

To place our results in perspective we mention that regularity results for one source far field reflectors are in [CGH08], results for near field parallel refractors are in [GT15] and [AGT16], and results for generated Jacobian equations, including reflector problems, are in [GK17]. Numerical methods are developed in [LGM17] to solve the one source far field refractor problem, in [GM19] to solve the near field, and in [AG17] to solve generated Jacobian equations.

The organization of the paper is as follows. Section 2 contains structural conditions on the target as well as a discussion on them and an example. Analytical estimates for ovals and a maximum principle, Lemma 3.4, of the type developed in [Loe09] and [KM10] are contained in Section 3. More analytical estimates for derivatives of ovals are in Section 4. Section 5 contains the Hölder estimates, where the main result is Theorem 5.4 from which we deduce as consequences Theorems 5.5 and 5.6.

2. Preliminaries, structural assumptions, and examples

Recall that a Descartes oval is the set $\mathcal{O}(Y,b)=\{X\in{\mathbb{R}}^{n}:|X|+\kappa|X-Y|=b\}$, with $\kappa|Y|<b<|Y|$. Here $\kappa=n_{2}/n_{1}$, where $n_{1}$ is the refractive index of the material inside the oval and $n_{2}$ is the refractive index of the material outside. We assume throughout that $\kappa<1$, which is the most interesting from an optical point of view (when $\kappa>1$ the arguments are similar). From the Snell law, a ray emanating from the origin with unit direction $x$ is refracted at the point $X\in\mathcal{O}(Y,b)$ into the point $Y$ provided that

[TABLE]

an inequality that by the equation of the oval is equivalent to $x\cdot Y\geq b$. The polar equation of the oval is $\mathcal{O}(Y,b)=\{\rho(x,Y,b)x:x\in S^{n-1}\}$ where

[TABLE]

For a geometric analysis and estimates for Descartes ovals we refer to [GH14, Sec. 4]. If we specify a point $X_{0}$ on the oval $\mathcal{O}(Y,b)$, then $b=|X_{0}|+\kappa|X_{0}-Y|$ and it will be useful to introduce the function

[TABLE]

with the point $X_{0}$ so that $\dfrac{X_{0}}{|X_{0}|}\cdot\dfrac{Y-X_{0}}{|Y-X_{0}|}\geq\kappa$. For $\Omega\subseteq S^{n-1}$ open and constants $0<c_{1}<c_{2}$, we let

[TABLE]

2.1. Structural assumptions on the target $\Sigma$

We begin introducing the following notion of curve in $S^{n-1}$ that will be used to state our assumptions.

Let $x_{0},\hat{m},\bar{m}\in S^{n-1}$ with $\bar{m}\cdot x_{0}\geq\kappa$ and $\hat{m}\cdot x_{0}\geq\kappa$. By definition, $[\bar{m},\hat{m}]_{x_{0}}$ denotes the curve obtained intersecting the triangle with vertices $\bar{m}$, $\hat{m}$, and $x_{0}/\kappa$ with the sphere $S^{n-1}$. Notice that since $\kappa<1$, the point $x_{0}/\kappa$ is outside the unit ball. In this triangle, the side joining $\hat{m}$ and $\bar{m}$, is given by $m_{\lambda}=(1-\lambda)\bar{m}+\lambda\hat{m}$, with $0\leq\lambda\leq 1$. Each point $m\in[\bar{m},\hat{m}]_{x_{0}}$ can then be obtained intersecting the line $\dfrac{x_{0}}{\kappa}+\beta\,\xi$ with the sphere $S^{n-1}$, where $\xi=m_{\lambda}-\dfrac{1}{\kappa}x_{0}$, $\beta\in{\mathbb{R}}$. Solving for $\beta$ yields

[TABLE]

since the point $\dfrac{x_{0}}{\kappa}+\beta\,\xi$ is inside the triangle so $0<\beta<1$. Therefore, we obtain the parametrization

[TABLE]

In particular, for $m\in[\bar{m},\hat{m}]_{x_{0}}$ we can write

[TABLE]

with $\bar{\beta},\hat{\beta}\geq 0$ and $\bar{\beta}+\hat{\beta}\leq 1$; $\bar{\beta}=(1-\lambda)\beta(\lambda)$, $\hat{\beta}=\lambda\,\beta(\lambda)$. Notice that $m(\lambda)\cdot x_{0}\geq\kappa$ for $0\leq\lambda\leq 1$ since $\beta(\lambda)\leq 1$ and $\kappa<1$.

We next introduce our structural assumptions.

1. H.A

For each $X\in\Gamma_{c_{1}c_{2}}$, let $C_{X}=\left\{Y:\dfrac{X}{|X|}\cdot\dfrac{Y-X}{|Y-X|}\geq\kappa\right\}$ be the cone with vertex $X$, axis $X/|X|$, and opening $\arccos\kappa$. Set

[TABLE]

We assume the following:

1. (a)

$\Sigma\subset\mathcal{C}_{\Omega}$, so (2.1) holds for all $Y\in\Sigma$ and $X\in\Gamma_{c_{1}c_{2}}$; 2. (b)

For each $X\in\Gamma_{c_{1}c_{2}}$ there exists a set $E(X)\subset\{m\in S^{n-1}:m\cdot x\geq\kappa,x=X/|X|\}$ and a continuous function $s_{X}:E(X)\to{\mathbb{R}}^{+}$ such that

[TABLE]

with the set $E(X)$ satisfying $[\bar{m},\hat{m}]_{x}\subset E(X)$ for all $\bar{m},\hat{m}\in E(X)$, with $x=X/|X|$; 3. (c)

The family of functions $\{s_{X}\}_{X\in\Gamma_{c_{1}c_{2}}}$ is uniformly Lipschitz continuous, i.e., there exists a constant $C>0$ such that $|s_{X}(m_{1})-s_{X}(m_{2})|\leq C\,|m_{1}-m_{2}|$ for all $m_{1},m_{2}\in E(X)$ and $X\in\Gamma_{c_{1}c_{2}}$. 2. H.B

Let $C(\kappa)=\kappa\left(\sqrt{1+(1+\kappa)^{-2}}-1\right)$. We assume $\dfrac{|X|}{|Y-X|}\leq C(\kappa)$ for all $X\in\Gamma_{c_{1}c_{2}}$ and $Y\in\Sigma$. Notice that this holds if $\text{dist}(\Gamma,\Sigma)\geq c_{2}/C(\kappa).$ 3. H.C

There exists a constant $0\leq\mu<\kappa$ such that for all $X_{0}\in\Gamma_{c_{1}c_{2}}$ and $\bar{m},\hat{m}\in E(X_{0})$, the function $s_{X_{0}}$ satisfies the following concavity condition

[TABLE]

for $0\leq\lambda\leq 1$, with $\bar{\beta}(\lambda)=(1-\lambda)\beta(\lambda)$ and $\hat{\beta}(\lambda)=\lambda\beta(\lambda)$, $\beta(\lambda)$ defined in (2.4) (depending on $x_{0}$), and $m(\lambda)$ from (2.5). 4. H.D

Given $X_{0}\in\Gamma_{c_{1}c_{2}}$, $\bar{Y},\hat{Y}\in\Sigma$, let $\bar{m}=\dfrac{\bar{Y}-X_{0}}{|\bar{Y}-X_{0}|}$ and $\hat{m}=\dfrac{\hat{Y}-X_{0}}{|\hat{Y}-X_{0}|}$; $x_{0}=X_{0}/|X_{0}|$. Let $[\bar{Y},\hat{Y}]_{X_{0}}$ be the curve defined by

[TABLE]

where $m(\lambda)$ is the parametrization of $[\bar{m},\hat{m}]_{x_{0}}$ defined in (2.5). We assume that there exist positive constants $\mu_{0}$ and $C$ such that for all $X_{0}\in\Gamma_{c_{1}c_{2}}$, $\bar{Y},\hat{Y}\in\Sigma$, we have

[TABLE]

for each $\mu\leq\mu_{0}$, where $H^{n-1}$ denotes the $n-1$ dimensional Hausdorff measure in ${\mathbb{R}}^{n}$ and $N_{\mu}$ denotes the $\mu$-neighborhood in ${\mathbb{R}}^{n}$.

Throughout the paper, a structural constant refers to a constant depending only on some or all of the constants in the structural conditions above.

Remark 2.1**.**

We begin noticing that from H.A and H.B we get that $s_{X}$ is bounded below:

[TABLE]

for all $X\in\Gamma_{c_{1}c_{2}}$ and $m\in E(X)$.

Also from H.A and H.B we get for $\hat{Y}=X+s_{X}(\hat{m})\,\hat{m}$ and $\bar{Y}=X+s_{X}(\bar{m})\,\bar{m}$ that

[TABLE]

Indeed, from H.A

[TABLE]

so

[TABLE]

Therefore from H.B the desired inequality follows since $X\in\Gamma_{c_{1}c_{2}}$.

Concerning each of our assumptions we mention the following. Assumption H.A guarantees that each ray from [math] striking $X\in\Gamma_{c_{1}c_{2}}$ can be refracted into $\Sigma$ and the refracted ray intersects $\Sigma$ at only one point. Assumption H.B says that $\Gamma_{c_{1}c_{2}}$ is sufficiently far from the target $\Sigma$***The value of the constant $C(\kappa)$ in H.B is only needed in Lemma 3.1.and it will be applied to show that the ovals used in the definition of refractor have controlled derivatives. Assumption H.C is crucial to obtain regularity of refractors and is akin to the condition (AW) first introduced in [MTW05] and later considered in [Loe09] and [KM10]. Assumption H.D is a form of convexity of $\Sigma$ with respect to points $X\in\Gamma_{c_{1}c_{2}}$.

Remark 2.2**.**

We relate now the structural assumptions introduced with the following assumptions needed to prove existence of refractors [GH14, Sect. 5]:

1. H.1

there exists $\tau$, with $0<\tau<1-\kappa$, such that $x\cdot Y\geq(\kappa+\tau)|Y|$ for all $x\in\Omega$ and $Y\in\Sigma$; 2. H.2

if $0<r_{0}<\dfrac{\tau}{1+\kappa}\,\text{dist}(0,\Sigma)$ and $Q_{r_{0}}=\{t\,x:x\in\Omega,0<t<r_{0}\}$, then given $X\in Q_{r_{0}}$ each ray emanating from $X$ intersects $\Sigma$ in at most one point.

We show that if $\tau$ is sufficiently small, then H.1 and H.2 imply H.A (a) and H.B. We first claim that there are positive constants $C_{\tau,\kappa}$ and $\hat{C}_{\tau,\kappa}$ such that if $\dfrac{X}{|X|}\cdot\dfrac{Y}{|Y|}\geq\kappa+\tau,$ and $|Y|\geq C_{\tau,\kappa}|X|,$ for all $Y\in\Sigma$ and $X\in\Gamma_{c_{1}c_{2}}$, then

[TABLE]

with

[TABLE]

Then the desired relation between the assumptions follows noticing that $C_{\tau,\kappa}\rightarrow\infty$ and $\hat{C}_{\tau,\kappa}\rightarrow 0$ as $\tau\rightarrow 0$. To prove the claim, fix $X$ and calculate the intersection between the cones $\mathcal{C}_{1}=\left\{Y:\dfrac{X}{|X|}\cdot\dfrac{Y}{|Y|}=\kappa+\tau\right\}$, and $\mathcal{C}_{2}=\left\{Y:\dfrac{Y-X}{|Y-X|}\cdot\dfrac{X}{|X|}=\kappa\right\}$. From the sine law, it is easy to see that if $Y$ is in the intersection of these cones, then $|Y|=C_{\tau,\kappa}|X|$. So $|Y|\geq C_{\tau,\kappa}|X|$ and $Y$ is in the interior of $\mathcal{C}_{1}$, then $Y$ is in the interior of $\mathcal{C}_{2}$, and $|Y-X|\geq|Y|-|X|\geq\left(C_{\tau,\kappa}-1\right)|X|$.

Remark 2.3**.**

When the target $\Sigma$ is $C^{2}$ one can give a differential condition that is equivalent to H.C. To do this, we first need to have another parametrization of the curve $[\bar{m},\hat{m}]_{x_{0}}$. For $Y\in\Sigma$, recall that from (5.5)

[TABLE]

and since $Y=X_{0}+s\,m$, for some $m\in E(X_{0})$, we have from (4.4) that

[TABLE]

Notice that $v\perp x_{0}$ and $|v|^{2}\leq\dfrac{\kappa^{2}|X_{0}|^{2}}{1-\kappa^{2}}$. We will write $m$ in terms of $v$, with $\langle m,x_{0}\rangle\geq\kappa$ and $|m|=1$. First note that $\langle m,x_{0}\rangle=\dfrac{|v|^{2}+|X_{0}|\sqrt{\kappa^{2}|X_{0}|^{2}-(1-\kappa^{2})|v|^{2}}}{\kappa\left(|v|^{2}+|X_{0}|^{2}\right)}$, and thus

[TABLE]

We can then write $m=\langle m,x_{0}\rangle x_{0}+t(v)\,v$ and so

[TABLE]

Given $\bar{m},\hat{m}\in S^{n-1}$ with $\langle\bar{m},x_{0}\rangle\geq\kappa$ and $\langle\hat{m},x_{0}\rangle\geq\kappa$, let

[TABLE]

Letting $v_{\gamma}=(1-\gamma)\bar{v}+\gamma\hat{v}$, we show that the curve $[\bar{m},\hat{m}]_{x_{0}}$ in (2.5) can be parametrized as follows:

[TABLE]

that is, $\tilde{m}(\gamma)=m(\lambda)$ with the change of parameter $\lambda=\dfrac{\gamma t(\bar{v})}{(1-\gamma)t(\hat{v})+\gamma t(\bar{v})}$ (we are abusing the notation $m(\lambda)$ and $m(v)$). In fact, from the definition of $\beta(\lambda)$

[TABLE]

see the end of the proof of Lemma 3.4 for similar calculations with $\beta(\lambda)$. Also $\bar{m}=\frac{1}{\kappa}x_{0}+t(\bar{v})(\bar{v}-X_{0})$ and $\hat{m}=\frac{1}{\kappa}x_{0}+t(\hat{v})(\hat{v}-X_{0})$. Then,

[TABLE]

Since $\kappa\bar{m}-x_{0}=\kappa t(\bar{v})(\bar{v}-X_{0})$ and $\kappa\hat{m}-x_{0}=\kappa t(\hat{v})(\hat{v}-X_{0})$, substituting and simplifying yields $m(\lambda)=\tilde{m}(\gamma)$ as desired.

Therefore, with this reparametrization of the curve $[\bar{m},\hat{m}]_{x_{0}}$ assumption H.C is then equivalent to

[TABLE]

for $0<\gamma<1$, with $\tilde{m}(0)=\bar{m}$, $\tilde{m}(1)=\hat{m}$, $v_{0}=\bar{v}$, and $v_{1}=\hat{v}$. In other words, the function

[TABLE]

is a concave function of $v$ for $|v|^{2}\leq\dfrac{\kappa^{2}|X_{0}|^{2}}{1-\kappa^{2}}$ and $v\perp x_{0}$, i.e., concave in a $n-1$-dimensional disk. Thus, when $\Sigma$ is $C^{2}$, we obtain that H.C is equivalent to

[TABLE]

for all $v\perp x_{0}$ with $|v|^{2}\leq\dfrac{\kappa^{2}|X_{0}|^{2}}{1-\kappa^{2}}$ and for all $\xi\perp x_{0}$. The domain of $s_{X_{0}}$ is $E(X_{0})$, and the domain of $\Phi(v)$ is $T_{X_{0}}(E(X_{0}))$. The fact that $E(X_{0})$ satisfies the convexity assumption that $[\bar{m},\hat{m}]_{x_{0}}\subset E(X_{0})$ for all $\bar{m},\hat{m}\in E(X_{0})$ is equivalent that $T_{X_{0}}(E(X_{0}))$ is a convex set in the classical sense on the hyperplane perpendicular to $x_{0}$.

2.2. Examples

We will construct $\Omega\subset S^{n-1}$ and a target $\Sigma$ so that the structural assumptions are satisfied. Notice that if $\Omega\subset\Omega^{\prime}$, then $\mathcal{C}_{\Omega^{\prime}}\subset\mathcal{C}_{\Omega}$, with $c_{1},c_{2}$ fixed. To do this construction, we will first choose $\Omega^{\prime}$ and calculate $\mathcal{C}_{\Omega^{\prime}}$. We will then choose a target $\Sigma\subset\mathcal{C}_{\Omega^{\prime}}$ and next pick $\Omega\subset\Omega^{\prime}$. It will then follow that $\Sigma\subset\mathcal{C}_{\Omega}$.

Let $\theta=\arccos\kappa$ and $\Omega^{\prime}=\left\{x\in S^{n-1}:x\cdot e_{n}\geq\cos(\theta/2)\right\}$ where $e_{n}$ is the unit vector in the vertical direction $x_{n}$.

Pick constants $c_{1}=1$ and $c_{2}>1$, and let $Y_{0}=2\,c_{2}\cos(\theta/2)\,e_{n}$. We claim that $\mathcal{C}_{\Omega^{\prime}}=\left\{Y:\dfrac{Y-Y_{0}}{|Y-Y_{0}|}\cdot e_{n}\geq\cos(\theta/2)\right\}:=E$, the cone with vertex at $Y_{0}$ direction $e_{n}$ and opening $\theta/2$. To prove this, let $Y\in E$ and we want to show that $Y\in C_{X}$ for all $X\in\Gamma_{c_{1}c_{2}}$ (defined with $\Omega^{\prime}$). That $Y\in E$ means $\angle\left(Y-Y_{0},e_{n}\right)\leq\theta/2$, where $\angle$ denotes the angle between the vectors. Obviously, $Y\in C_{X}$ if and only if $\angle(Y-X,X)\leq\theta$. From the choice of $Y_{0}$, it is easy to see that $\angle(Y_{0}-X,X)\leq\theta$ for all $X\in\Gamma_{c_{1}c_{2}}$, i.e., $Y_{0}\in C_{X}$.

We have $Y=Y_{0}+v$ with $\angle(v,e_{n})\leq\theta/2$. Let $\bar{Y}=X+v$. Since $\angle(\bar{Y}-X,X)=\angle(v,X)\leq\angle(v,e_{n})+\angle(e_{n},X)\leq\frac{\theta}{2}+\frac{\theta}{2}$, it follows that $\bar{Y}\in C_{X}$. Then from the convexity of $C_{X}$ we obtain $\dfrac{Y_{0}+\bar{Y}}{2}\in C_{X}$. Since $Y=\bar{Y}+Y_{0}-X$, it follows that $\angle(Y-X,X)=\angle(\bar{Y}+Y_{0}-2X,X)=\angle\left(\dfrac{\bar{Y}+Y_{0}}{2}-X,X\right)\leq\theta$. So $Y\in C_{X}$ and the claim is proved.

Now, we choose $\Sigma$ the planar disk centered at [math] with radius $R$ at height $M$, that is,

[TABLE]

If we pick $M=C+2\,c_{2}\,\cos(\theta/2)=C+2\,c_{2}\,\sqrt{\frac{1+\kappa}{2}}$ with $C$ any positive constant and pick $R\leq\sqrt{\dfrac{1-\kappa}{1+\kappa}}C$, then it is easy to verify that $\Sigma\subset E=\mathcal{C}_{\Omega^{\prime}}$.

Next, we will choose $\Omega\subset\Omega^{\prime}$ so that if $\bar{Y},\hat{Y}\in\Sigma$, then $[\bar{m},\hat{m}]_{x}\subseteq E(X)$, for all $X\in\Gamma_{c_{1}c_{2}}$, where $\bar{m}=\dfrac{\bar{Y}-X}{|\bar{Y}-X|}$, $\hat{m}=\dfrac{\hat{Y}-X}{|\hat{Y}-X|}$ and $E(X)$ is the set of visibility directions in $S^{n-1}$ of the target $\Sigma$ from the point $X$. First notice that since $\cos(\theta/2)=\sqrt{\dfrac{1+\kappa}{2}}$ and $\sin(\theta/2)=\sqrt{\dfrac{1-\kappa}{2}}$, we have $x\in\Omega^{\prime}$ if and only if $\dfrac{|x^{\prime}|}{x_{n}}\leq\sqrt{\dfrac{1-\kappa}{1+\kappa}}$, with $x_{n}>0$. Now define

[TABLE]

Since $\dfrac{R}{M}\leq\dfrac{\sqrt{\dfrac{1-\kappa}{1+\kappa}}C}{C+2c_{2}\sqrt{\dfrac{1-\kappa}{1+\kappa}}}$, we obtain that $\Omega\subset\Omega^{\prime}$.

If $X\in\Gamma_{c_{1}c_{2}}$ and $\bar{Y},\hat{Y}\in\Sigma$, then we show $[\bar{m},\hat{m}]_{x}\subseteq E(X)$, where $\bar{m}=\dfrac{\bar{Y}-X}{|\bar{Y}-X|}$ and $\hat{m}=\dfrac{\hat{Y}-x}{|\hat{Y}-X|}$ and $x=\dfrac{X}{|X|}$. Since $\bar{Y},\hat{Y}\in\mathcal{C}_{\Omega}$, we have $\bar{m}\cdot x\geq\kappa$ and $\hat{m}\cdot x\geq\kappa$. We have from (2.6) that $m=\frac{1}{\kappa}x+\bar{\beta}(\bar{m}-\frac{1}{\kappa}x)+\hat{\beta}(\hat{m}-\frac{1}{\kappa}x)$ for $m\in[\bar{m},\hat{m}]_{x}$, and we need to show that the ray $X+s\,m$ strikes $\Sigma$ for some $s$ (that is, $s=s_{X}(m)$). If $s=\dfrac{M-X_{n}}{m_{n}}$, then will show that $Y=X+\dfrac{M-X_{n}}{m_{n}}m\in\Sigma$. Indeed, write $Y=(Y^{\prime},Y_{n})$. Clearly $Y_{n}=M$. If $|\bar{Y}^{\prime}|,|\hat{Y}^{\prime}|\leq R$, will prove that $|Y^{\prime}|\leq R$. We have with $m=(m^{\prime},m_{n})$ that

[TABLE]

and

[TABLE]

Combining the last two terms and simplifying yields

[TABLE]

Therefore, $|Y^{\prime}|\leq R\dfrac{\bar{\beta}\bar{m}_{n}+\hat{\beta}\hat{m}_{n}}{m_{n}}+\dfrac{1}{\kappa}|x^{\prime}|\,\dfrac{1-\bar{\beta}-\hat{\beta}}{m_{n}}\,M\leq R$, where we have used that $|x^{\prime}|\leq\dfrac{x_{n}R}{M}$ since $x\in\Omega$. Thus, $[\bar{m},\hat{m}]_{x}\subseteq E(X)$.

In addition,

[TABLE]

and so the concavity assumption in H.C holds with $\mu=0$.

Therefore the example described satisfies the assumptions H.A, and H.C. In order to satisfy H.B, it is enough to keep $c_{2}$ fixed and pick $C$ large enough. It remains to verify that example satisfies H.D. For this we use the following lemma.

Lemma 2.4**.**

Let $\gamma:[a,b]\rightarrow R^{n}$ be a smooth curve such that $|\gamma^{\prime}(t)|=1$ and $|\gamma^{\prime\prime}(t)|\leq M_{1}$ for all $t\in[a,b]$. In addition, assume $M_{2}|t_{1}-t_{2}|\leq|\gamma(t_{1})-\gamma(t_{2})|$ for all $t_{1},t_{2}\in[a,b]$. Let $T_{t}$ denote the hyperplane passing through $\gamma(t)$ with normal $\gamma^{\prime}(t)$ and let $D_{\mu}(t)=B_{\mu}(\gamma(t))\cap T_{t}$, and $N_{\mu}=\bigcup_{t\in[a,b]}D_{\mu}(t)$. Then, there exists $\mu_{0}$ and $C$ depending only on $M_{1},M_{2}$ such that for $\mu\leq\mu_{0}$, we have $H^{n}(N_{\mu})\geq C\mu^{n-1}|\gamma(b)-\gamma(a)|$.

Proof.

First observe that there exists $\mu_{0}$ such that if $\mu\leq\mu_{0}$, then $D_{\mu}(t_{1})\cap D_{\mu}(t_{2})=\emptyset$, for $t_{1}\neq t_{2}$.

Consider the cylinder in ${\mathbb{R}}^{n}$ given by $D\times[a,b]=\left\{(x^{\prime},t):|x^{\prime}|\leq\mu;\;t\in[a,b]\right\}$, where $D=\{(x^{\prime},0):|x^{\prime}|\leq\mu\}$, and define $F:D\times[a,b]\rightarrow N_{\mu}$ by

[TABLE]

where $A(t)$ is the $n\times n$ matrix whose column vectors are $\{\eta_{1}(t),...,\eta_{n-1}(t),\gamma^{\prime}(t)\}$ where $\eta_{i}(t)$ are chosen so that they are smooth with $A(t)A^{T}(t)=I$; here $(x^{\prime},0)$ is a column vector. Notice that $F$ is one to one and each disk $D\times\{t\}$ is mapped to $D_{\mu}(t)$. By the formula of change of variables

[TABLE]

provided that $|\det DF(x^{\prime},t)|\geq C$ for some $C>0$. Indeed, note that the matrix $DF(x^{\prime},t)$ has column vectors given by $\eta_{1}(t),...,\eta_{n-1}(t)$ and its last column vector is $\gamma^{\prime}(t)+x_{1}\eta_{1}^{\prime}(t)+...+x_{n-1}\eta_{n-1}^{\prime}(t)$, $x^{\prime}=(x_{1},\cdots,x_{n-1})$. Therefore, we can expand $\det DF(x^{\prime},t)=\det A(t)+\sum_{k=1}^{n-1}x_{k}\det\Lambda_{k}(t)$, where $\Lambda_{k}(t)$ is the matrix whose column vectors are $\eta_{1}(t),...,\eta_{n-1}(t),\eta_{k}^{\prime}(t)$. It follows that $|\det DF(x^{\prime},t)|\geq|\det A(t)|-\sum_{k=1}^{n-1}|x_{k}||\det\Lambda_{k}(t)|\geq 1-\mu_{0}\sum_{k=1}^{n-1}|\det\Lambda_{k}(t)|\geq 1-\mu_{0}\,C$, with $C$ depending only on $M_{1},M_{2}$ and $n$. Then choosing $\mu_{0}$ sufficiently small the lemma follows.

∎

Finally, to verify that our example satisfies H.D, we notice that the curves $[\bar{Y},\hat{Y}]_{X_{0}}$ in the example satisfy the assumptions of the last lemma (the curves can be reparametrized to have $|\gamma^{\prime}|=1$) in ${\mathbb{R}}^{n-1}$ so it is applicable to our case, obtaining constants that depend only on the structure. Also, varying the parameters $c_{2}$, $C$ and $R$ in the construction, we obtain a family of examples.

3. Preliminary results for ovals and a maximum principle

We analyze the function $h(x,Y,X_{0})$ for $Y\in\Sigma$ and $X_{0}\in\Gamma_{c_{1}c_{2}}$; $x\in S^{n-1}$, corresponding to the oval $\mathcal{O}(Y,b)$. From H.A, we can write $Y=X_{0}+s\,m$ with $m\in S^{n-1}$, $s=s_{X_{0}}(m)>0$, $x_{0}\cdot m\geq\kappa$, and recall that $b=|X_{0}|+\kappa|Y-X_{0}|=|X_{0}|+\kappa\,s$; $x_{0}=X_{0}/|X_{0}|$. Hence

[TABLE]

Settting

[TABLE]

we then can write

[TABLE]

In order to get to our crucial Lemma 3.4, first we need to prove three auxiliary lemmas.

Lemma 3.1**.**

Assume H.A and H.B. Let $\bar{Y},\hat{Y}\in\Sigma$, and consider the ovals $\mathcal{O}(\bar{Y},\bar{b}),\mathcal{O}(\hat{Y},\hat{b})$. If $X_{0}\in\Gamma_{c_{1}c_{2}}$ is a common point to both ovals, then with the notation above we have

[TABLE]

Proof.

Since $h(x,\bar{Y},X_{0})>0$, it follows that $\bar{B}-\sqrt{\bar{B}^{2}-\bar{C}}=\dfrac{\bar{C}}{\bar{B}+\sqrt{\bar{B}^{2}-\bar{C}}}\leq\dfrac{\bar{C}}{\bar{B}}$. So, it is enough to show $\bar{C}\leq\bar{B}\,\hat{B}$, which is equivalent to show that

[TABLE]

where we have used the notation $\bar{Y}=X_{0}+\bar{s}\,\bar{m}$ and $\hat{Y}=X_{0}+\hat{s}\,\hat{m}$, $\bar{s}=s_{X_{0}}(\bar{m}),\hat{s}=s_{X_{0}}(\hat{m})>0$, $x_{0}=X_{0}/|X_{0}|$ with $x_{0}\cdot\bar{m}\geq\kappa$, $x_{0}\cdot\hat{m}\geq\kappa$. The last inequality is equivalent to

[TABLE]

The left hand side of the last inequality is $\leq|X_{0}|^{2}+2\kappa|X_{0}|\bar{s}$ and the right hand side is $\geq\dfrac{\kappa^{2}}{(1-\kappa^{2})^{2}}\,\bar{s}\,\hat{s}\,(1-\kappa)^{2}=\dfrac{\kappa^{2}}{(1+\kappa)^{2}}\,\bar{s}\,\hat{s}$. Therefore, if $|X_{0}|^{2}+2\kappa|X_{0}|\bar{s}\leq\dfrac{\kappa^{2}}{(1+\kappa)^{2}}\,\bar{s}\,\hat{s}$, then the desired inequality follows. This is equivalent to

[TABLE]

which follows from H.B.

∎

A second auxiliary calculus lemma is as follows.

Lemma 3.2**.**

Consider the two variable function $f(B,C)=B-\sqrt{B^{2}-C}$ on the set $0\leq C\leq B^{2}$ and $B\geq 0$. Fix $(\bar{B},\bar{C})$ in that set, and suppose that $f(\bar{B},\bar{C})\leq B$. Then $f(B,C)\leq f(\bar{B},\bar{C})$ if and only if $C-\bar{C}\leq 2(B-\bar{B})f(\bar{B},\bar{C})$. In addition, if $C-\bar{C}\leq 2(B-\bar{B})f(\bar{B},\bar{C})-E$ for some $E\geq 0$, then $f(B,C)\leq f(\bar{B},\bar{C})-\dfrac{E}{B+\sqrt{B^{2}-C}-f(\bar{B},\bar{C})}$.

Proof.

Assume that $C-\bar{C}\leq 2(B-\bar{B})f(\bar{B},\bar{C})-E$, for some $E\geq 0$. Then

[TABLE]

Therefore,

[TABLE]

which implies

[TABLE]

Conversely, assume $f(B,C)\leq f(\bar{B},\bar{C})$, that is, $B-\sqrt{B^{2}-C}\leq f(\bar{B},\bar{C})$ which implies

[TABLE]

where the first inequality is from the assumption. Hence,

[TABLE]

∎

The third auxiliary lemma says that the oval passing thru $X_{0}$ is enclosed by the ellipsoid with axis $m$ and eccentricity $\kappa$ passing thru $X_{0}$ when $x_{0}\cdot m\geq\kappa$.

Lemma 3.3**.**

Suppose $x_{0}\cdot m\geq\kappa$ and let $Y=X_{0}+s\,m$ with $s>0$; $x_{0}=X_{0}/|X_{0}|$. Then

[TABLE]

In particular,

[TABLE]

for all $x\in S^{n-1}$.

Proof.

Let $X$ with $|X|+\kappa|X-Y|\leq|X_{0}|+\kappa|X_{0}-Y|$. Then

[TABLE]

∎

We are now ready to prove a crucial lemma akin to [Loe09, Prop. 5.1] and [KM10, Thm. 4.10 (DMASM)] in optimal mass transport.

Lemma 3.4**.**

Assume H.A, H.B, and H.C. There exists a structural constant $C_{0}>0$ such that if $\bar{Y},\hat{Y}\in\Sigma$, $X_{0}\in\Gamma_{c_{1}c_{2}}$ with $\bar{Y}=X_{0}+\bar{s}\,\bar{m}$, $\bar{s}=s_{X_{0}}(\bar{m})$, $\hat{Y}=X_{0}+\hat{s}\,\hat{m}$, $\hat{s}=s_{X_{0}}(\hat{m})$, then

[TABLE]

for all $x\in S^{n-1}$, $Y=X_{0}+s_{X_{0}}(m(\lambda))$, $m(\lambda)\in[\bar{m},\hat{m}]_{x_{0}}$ and $0<\lambda<1$.

Proof.

Fix $x\in S^{n-1}$ and assume without lost of generality that $h(x,\bar{Y},X_{0})\geq h(x,\hat{Y},X_{0})$, that is, $f(\hat{B},\hat{C})\leq f(\bar{B},\bar{C})$. We will show that

[TABLE]

By Lemma 3.1, we have $\hat{B}\geq\bar{B}-\sqrt{\bar{B}^{2}-\bar{C}}=f(\bar{B},\bar{C})$ so we can apply Lemma 3.2 to obtain

[TABLE]

This means

[TABLE]

which is equivalently to

[TABLE]

We will show that

[TABLE]

with some $E$ to be chosen at the end, where $B$ and $C$ are given in (3.3) corresponding to $Y=X_{0}+s_{X_{0}}(m(\lambda))$. To show (3.5), is equivalent to show that

[TABLE]

for $m=m(\lambda)\in[\bar{m},\hat{m}]_{x_{0}}$, $s=s_{X_{0}}(m(\lambda))$, and $0<\lambda<1$. Equivalently, we will show

[TABLE]

Indeed, first recall that $m$ can be written as in (2.6) with $\bar{\beta}(\lambda)=(1-\lambda)\beta(\lambda)$ and $\hat{\beta}(\lambda)=\lambda\beta(\lambda)$ with $\beta(\lambda)$ defined in (2.4). From (2.6)

[TABLE]

Again from (2.6)

[TABLE]

From (3.4)

[TABLE]

If we now let

[TABLE]

$\bar{\beta}(\lambda)+\hat{\beta}(\lambda)=\beta(\lambda)$, then with simplified notation H.C reads

[TABLE]

We also notice that since $f(\bar{B},\bar{C})=h(x,\bar{Y},X_{0})$ and $x_{0}\cdot\bar{m}\geq\kappa$, by Lemma 3.3

[TABLE]

Therefore

[TABLE]

To estimate the middle term in the last inequality we shall prove that for some $\delta>0$

[TABLE]

where $h(x,\bar{Y},X_{0})=f(\bar{B},\bar{C})$. From the definition of $K$ this inequality is equivalent to

[TABLE]

Let

[TABLE]

Writing $X=h(x,\bar{Y},X_{0})x$ with $\bar{Y}=X_{0}+\bar{s}\,\bar{m}$, we have $|X|+\kappa|X-\bar{Y}|=|X_{0}|+\kappa|X_{0}-\bar{Y}|$, which after simplification implies that

[TABLE]

By calculation, the right hand side of the last identity is equal to

[TABLE]

implying (3.9) with $\delta=\mu/(2\kappa)$. Therefore inserting (3.9) in (3) yields

[TABLE]

Therefore we have proved (3.6) with

[TABLE]

and consequently (3.5).

Since $X_{0}$ is on both ovals $\mathcal{O}(Y,b),\mathcal{O}(\bar{Y},\bar{b})$, then by Lemma 3.1, $B\geq f(\bar{B},\bar{C})$ . So from (3.5) we can apply the last part of Lemma 3.2 to get

[TABLE]

that is,

[TABLE]

Finally, to complete the proof of the lemma, we estimate $\dfrac{E}{B+\sqrt{B^{2}-C}-f(\bar{B},\bar{C})}$ from below. We shall first prove that $1-(\bar{\beta}+\hat{\beta})\geq C_{\kappa}\,\lambda\,(1-\lambda)\,|\bar{m}-\hat{m}|^{2}$. In fact,

[TABLE]

We have $(\kappa|\xi|^{2}+\langle x_{0},\xi\rangle)^{2}-(\langle x_{0},\xi\rangle^{2}-(1-\kappa^{2})|\xi|^{2})=|\xi|^{2}(\kappa^{2}|\xi|^{2}+2\kappa\langle x_{0},\xi\rangle+1-\kappa^{2})=|\xi|^{2}(|\kappa\xi+x_{0}|^{2}-\kappa^{2})=|\xi|^{2}\kappa^{2}(|m_{\lambda}|^{2}-1)$. Therefore

[TABLE]

Since $1-\beta(\lambda)>0$ and $|m_{\lambda}|<1$, for $0<\lambda<1$, it follows that $\Delta:=-\kappa|\xi|^{2}-\langle x_{0},\xi\rangle+\sqrt{\langle x_{0},\xi\rangle^{2}-(1-\kappa^{2})|\xi|^{2}})>0$ and since $|\xi|\leq 1+(1/\kappa)$, $\Delta$ is bounded above by a constant depending only on $\kappa$. Since $1-|m_{\lambda}|^{2}=\lambda\,(1-\lambda)\,|\bar{m}-\hat{m}|^{2}$, the desired lower bound for $1-(\bar{\beta}+\hat{\beta})$ follows.

Next, we show that $f(\bar{B},\bar{C})$ is bounded below by a structural constant. In fact, from [GH14, first identity in (4.7)], $f(\bar{B},\bar{C})=h(x,\bar{Y},X_{0})\geq\dfrac{\bar{b}-\kappa\,|\bar{Y}|}{1+\kappa}$ for all $x\in S^{n-1}$ where $\bar{b}=|X_{0}|+\kappa\,|\bar{Y}-X_{0}|$. So $\dfrac{\bar{b}-\kappa\,|\bar{Y}|}{1+\kappa}\geq\dfrac{1-\kappa}{1+\kappa}|X_{0}|\geq c_{1}\dfrac{1-\kappa}{1+\kappa}$, since $X_{0}\in\Gamma_{c_{1}c_{2}}$. Thus,

[TABLE]

with $C>0$ a structural constant.

It remains to estimate $B+\sqrt{B^{2}-C}-f(\bar{B},\bar{C})$ from above. We have from (3.1) that $B+\sqrt{B^{2}-C}-f(\bar{B},\bar{C})\leq B+\sqrt{B^{2}-C}\leq 2B\leq C\,\left(|X_{0}|+\kappa\,s\right)$. Since $s=s_{X_{0}}(m)=|Y-X_{0}|$ we obtain from H.B that $|X_{0}|+\kappa\,s\leq C\,s$ with a structural constant $C>0$. Therefore

[TABLE]

for $0<\lambda<1$ with $C>0$ a structural constant (since $\mu<\kappa$). The proof of the lemma is then complete. ∎

4. Estimates for derivatives of ovals

We analyze now the derivatives of the function $h(x,Y,X_{0})$ for $Y\in\Sigma$ and $X_{0}\in\Gamma_{c_{1}c_{2}}$. To differentiate the function $h$ with respect to the variables $x$ and $Y$ we will extend $h(x,Y,X_{0})$ for $x$ in a neighborhood of the unit ball and $Y$ in a neighborhood of $\Sigma$. In order to do this, we first need to bound from below the quantity inside the square root in (2.2).

Lemma 4.1**.**

Let $X_{0}\in\Gamma_{c_{1}c_{2}}$. There exist $\epsilon>0$ sufficiently small depending only on $\kappa$ and constants $C_{0},C_{1}$ depending only on $\kappa$ and $c_{2}$ such that if $b=|X_{0}|+\kappa|Y-X_{0}|<|Y|$ and $|Y|\geq C_{1}$ then

[TABLE]

Then by continuity there is a small neighborhood $V$ of $Y$ such that (4.1) holds for all $Y\in V$ with a smaller positive constant $C$. This implies that under this configuration, the formula defining $h$ in (2.3) can be extended for $|x|\leq 1+\epsilon$ and $Y\in V$.

Proof.

By calculation

[TABLE]

From the estimate for the ovals [GH14, first identity in (4.7)], $|X_{0}|=h(x_{0},Y,X_{0})\geq\dfrac{b-\kappa\,|Y|}{1+\kappa}$, so $b\leq\kappa\,|Y|+(1+\kappa)|X_{0}|$. Therefore $|Y|-b\geq(1-\kappa)\,|Y|-(1+\kappa)\,|X_{0}|$. Clearly, the last quantity is non negative if $|Y|\geq\dfrac{1+\kappa}{1-\kappa}|X_{0}|$.

We have $\displaystyle\min_{|x|\leq 1+\epsilon}\Delta(x\cdot Y)=\min_{-(1+\epsilon)|Y|\leq t\leq(1+\epsilon)|Y|}\Delta(t)$. The function $\Delta(t)$ is decreasing in the interval $\left(-\infty,b/\kappa^{2}\right)$. Let $\epsilon>0$ be such that $\kappa\,(1+\epsilon)<1$. Since $b>\kappa\,|Y|$ we then have $\left[-(1+\epsilon)|Y|,(1+\epsilon)|Y|\right]\subset\left(-\infty,b/\kappa^{2}\right)$. Therefore

[TABLE]

Let us estimate $\Delta\left((1+\epsilon)|Y|\right)$ from below:

[TABLE]

From the choice of $\epsilon$, $\alpha_{2}\leq(1-\kappa^{2})\dfrac{1+\kappa}{\kappa}:=\beta_{2}$ and taking $\epsilon$ small we have $\alpha_{1}\geq(1-\kappa)^{2}/2:=\beta_{1}$. Hence

[TABLE]

We now choose $\delta>0$ sufficiently small depending only on $\kappa$ such that

[TABLE]

for some $C_{i}$ positive constants. Thus

[TABLE]

and the desired inequality follows.

∎

With Lemma 4.1 in hand we proceed to prove estimates for $h$ and its derivatives.

Lemma 4.2**.**

There exists a structural constant $C>0$ such that if $Y\in\Sigma$, $t>0$ and $(1+t)X_{0}\in\Gamma_{c_{1}c_{2}}$, then $0\leq h(x,Y,(1+t)X_{0})-h(x,Y,X_{0})\leq C\,t\,|X_{0}|.$

Proof.

If $b(t)=(1+t)|X_{0}|+\kappa|Y-(1+t)X_{0}|$, then $0\leq b(t)-b(0)\leq(1+\kappa)t|X_{0}|$. †††We can see $b(0)\leq b(t)$ for $t>0$ because this is equivalent to $(1+t)|X_{0}|+\kappa|Y-(1+t)X_{0}|\geq|X_{0}|+\kappa|Y-X_{0}|$ which is equivalent to show $t|X_{0}|+\kappa|Y-(1+t)X_{0}|\geq\kappa|Y-X_{0}|$. But by triangle inequality $\kappa|Y-(1+t)X_{0}|\geq\kappa|Y-X_{0}|-\kappa t|X_{0}|$ which implies $t|X_{0}|+\kappa|Y-(1+t)X_{0}|\geq t|X_{0}|+\kappa|Y-X_{0}|-\kappa t|X_{0}|=(1-\kappa)t|X_{0}|+\kappa|Y-X_{0}|>\kappa|Y-X_{0}|$ since $\kappa<1$. Let $Q(t)=(b(t)-\kappa^{2}\,x\cdot Y)^{2}-(1-\kappa^{2})(b(t)^{2}-\kappa^{2}\,|Y|^{2})$. We have

[TABLE]

From the definition of $h$

[TABLE]

Since $\rho(x,Y,b)$ is increasing in $b$ and $b(0)<b(t)$, it follows from (2.3) that $h(x,Y,X_{0})\leq h(x,Y,(1+t)X_{0})$. From Lemma 4.1 the denominator in the last string of expressions is bounded away from zero and we obtain

[TABLE]

∎

Lemma 4.3**.**

Suppose H.A and H.B hold. There exist a structural constant $C>0$ such that if $\bar{Y},Y\in\Sigma$ and $X_{0}\in\Gamma_{c_{1}c_{2}}$, then $|\nabla_{x}h(x_{0},Y,X_{0})-\nabla_{x}h(x_{0},\bar{Y},X_{0})|\leq C\,|Y-\bar{Y}|$.

Proof.

Let $Y=X_{0}+sm$ and $\bar{Y}=X_{0}+\bar{s}\bar{m}$ and $b=|X_{0}|+\kappa s$. From Lemma 4.1 we can take derivatives of $h$ with respect to $x$ for $x$ in a neighborhood of the unit ball, and by calculation

[TABLE]

So at $x=x_{0}$

[TABLE]

since $\sqrt{(b-\kappa^{2}\,x_{0}\cdot Y)^{2}-(1-\kappa^{2})(b^{2}-\kappa^{2}\,|Y|^{2})}=\kappa s(1-\kappa\,x_{0}\cdot m)$ from (3.1) and (3.2).

Therefore

[TABLE]

From (2.7), $|A|\leq C\,|Y-\bar{Y}|$. In addition

[TABLE]

since $|m-\bar{m}|\leq C|Y-\bar{Y}|$ from (2.8). ∎

Lemma 4.4**.**

There exists a structural constant $M$ such that if $X_{0}\in\Gamma_{c_{1}c_{2}}$, $Y\in\Sigma$ and $x_{0}=X_{0}/|X_{0}|$, then

[TABLE]

for all $x\in S^{n-1}$.

Proof.

We first calculate $\dfrac{\partial^{2}}{\partial x_{j}\partial x_{i}}$. From (4.2) and (4.3)

[TABLE]

From Lemma 4.1 we obtain that $\left|\dfrac{\partial^{2}h}{\partial x_{j}\partial x_{i}}(x,Y,X_{0})\right|\leq C_{1}$ for all $x$ in a neighborhood of the unit ball $|x|\leq 1$. From Taylor’s formula

[TABLE]

with $\xi$ between $x_{0}$ and $x$. The lemma then follows. ∎

Lemma 4.5**.**

Suppose H.A and H.B hold. There exists a structural constant $C>0$ such that if $X_{0}\in\Gamma_{c_{1}c_{2}}$ and $\bar{Y},Y\in\Sigma$, then

[TABLE]

for $x\in S^{n-1}$.

Proof.

Using Lemma 4.1 we shall first estimate the derivatives of $h$ with respect to $Y_{k}$; $Y=(Y_{1},\cdots,Y_{n})$. Recall $h(x,Y,X_{0})=\dfrac{1}{1-\kappa^{2}}\left(b-\kappa^{2}\,x\cdot Y-\sqrt{\Delta(x\cdot Y,b,|Y|)}\right)$ where $\Delta(t,b,|Y|)=\Delta(t)$ is given by (4.2) and $b=|X_{0}|+\kappa\,|X_{0}-Y|$. By Lemma 4.1, $h$ can be differentiated with respect to $Y_{k}$ since is defined in an open neighborhood of the target $\Sigma$. Then

[TABLE]

Now

[TABLE]

and $\dfrac{\partial b}{\partial Y_{k}}=-\kappa\,\dfrac{X_{0}^{k}-Y_{k}}{|X_{0}-Y|}$. We next differentiate (4.3) with respect to $Y_{k}$. Recall that from Lemma 4.1, the right hand side of (4.3) is well defined for $x$ in a neighborhood of the unit ball and for $Y$ in a neighborhood of the target $\Sigma$. We then have

[TABLE]

From Lemma 4.1, $\Delta\geq C$ so $\dfrac{\partial h}{\partial Y_{k}}$ is bounded, and therefore $\dfrac{\partial^{2}h}{\partial Y_{k}\partial x_{i}}(x,Y,X_{0})$ is also bounded.

Therefore we can write for some $\tilde{Y}\in\overline{\bar{Y}Y}$, the straight segment, and for some $\tilde{x}\in\overline{x_{0}x}$

[TABLE]

where we have used that $h(x_{0},Y,X_{0})=|X_{0}|$, for all $Y$ so $\dfrac{\partial h}{\partial Y_{k}}(x_{0},\tilde{Y},X_{0})=0$. It remains to show that $\Delta(x\cdot\tilde{Y},\tilde{b},|\tilde{Y}|)\geq C$ and $\Delta(\tilde{x}\cdot\tilde{Y},\tilde{b},|\tilde{Y}|)\geq C$ so the application of the mean value theorem above is justified and we can apply the bounds for the derivatives. We have $\tilde{Y}=(1-\lambda)\bar{Y}+\lambda Y$ for some $\lambda\in[0,1]$. From H.A, we can write $Y=X_{0}+s\,m$ and $\bar{Y}=X_{0}+\bar{s}\,\bar{m}$ with $x_{0}\cdot\bar{m}\geq\kappa$, $x_{0}\cdot m\geq\kappa$, $x_{0}=X_{0}/|X_{0}|$. So $\tilde{Y}=X_{0}+(1-\lambda)\bar{s}\,\bar{m}+\lambda s\,m:=X_{0}+w$, and

[TABLE]

and

[TABLE]

Thus

[TABLE]

Since $\bar{m}\cdot x_{0}\geq\kappa$ and $m\cdot x_{0}\geq\kappa$ with $\kappa<1$, it follows that $\bar{m}\cdot m\geq-\delta$ for some $0<\delta=\delta(\kappa)<1$. Then

[TABLE]

where the last expression attains its minimum when $\lambda=\dfrac{\bar{s}^{2}+\delta\,\bar{s}\,s}{\bar{s}^{2}+2\,\delta\,\bar{s}\,s+s^{2}}$. Since $\bar{s},s$ are bounded, at this minimum the expression is larger than or equal to $C\,(1-\delta^{2})\,\min\left\{\bar{s}^{2},s^{2}\right\}$, with $C>0$ structural. From (2.7) we then obtain

[TABLE]

Using the argument the proof of Lemma 4.1 with $\epsilon=0$, it follows that $\Delta\left(x\cdot\tilde{Y},b,|\tilde{Y}|\right)$ and $\Delta\left(\tilde{x}\cdot\tilde{Y},b,|\tilde{Y}|\right)$ are both greater than or equal to $\kappa^{2}\left(|\tilde{Y}|-b\right)^{2}$ obtaining the desired estimate.

∎

5. $C^{1,\alpha}$ estimates

We now turn to the definition of refractor and prove our main theorem.

Definition 5.1**.**

We say $u:\Omega\rightarrow[c_{1},c_{2}]$ is a refractor from $\Omega$ to $\Sigma$ if for each $x_{0}\in\Omega$, there exists $Y\in\Sigma$ such that

[TABLE]

for all $x\in\Omega$ with $X_{0}=u(x_{0})x_{0}$. If this holds, then we say $Y\in\partial u(x_{0})$. Notice that $X=u(x)x\in\Gamma_{c_{1}c_{2}}$ for all $x\in\Omega$.

We will show $u\in C^{1,\alpha}(\Omega)$, which will follow from the following two lemmas.

Lemma 5.2**.**

Assume H.A, H.B, and H.C, and let $u$ be a refractor from $\Omega$ to $\Sigma$. There exist structural constants $K_{1},K_{2}$ such that if $B_{2\delta}\cap S^{n-1}\subseteq\Omega$, $\bar{x},\hat{x}\in B_{\delta}\cap S^{n-1}$, $\bar{Y}\in\partial u(\bar{x})$ and $\hat{Y}\in\partial u(\hat{x})$, with $|\bar{Y}-\hat{Y}|\geq|\bar{x}-\hat{x}|$, then, there exists $x_{0}\in B_{\delta}\cap S^{n-1}$ such that, letting $X_{0}=u(x_{0})x_{0}$, if $Y(\lambda)\in[\bar{Y},\hat{Y}]_{X_{0}}$ we have

[TABLE]

for all $x\in\Omega$, $0<\lambda<1$.

Proof.

Let $\bar{X}=u(\bar{x})\bar{x}$ and $\hat{X}=u(\hat{x})\hat{x}$. We have $u(x)\geq h(x,\bar{Y},\bar{X})$ and $u(x)\geq h(x,\hat{Y},\hat{X})$, for all $x\in\Omega$. Let $\varphi(x)=h(x,\bar{Y},\bar{X})-h(x,\hat{Y},\hat{X})$. Since $\varphi(\bar{x})\geq 0$ and $\varphi(\hat{x})\leq 0$, by continuity there exists $x_{0}\in[\bar{x},\hat{x}]$, the geodesic segment in the unit sphere, such that $h(x_{0},\bar{Y},\bar{X})=h(x_{0},\hat{Y},\hat{X}):=\rho_{0}$. Set $\tilde{X}_{0}=\rho_{0}x_{0}$ and $X_{0}=u(x_{0})x_{0}$ and notice $\rho_{0}\leq u(x_{0})$ and by definition of refractor $C_{1}\leq u(x_{0})\leq C_{2}$, i.e., $X_{0}\in\Gamma_{C_{1}C_{2}}$. Also, the oval with focus $\bar{Y}$ that passes through $\bar{X}$ then also passes through $\tilde{X}_{0}$, i.e., $h(x,\bar{Y},\bar{X})=h(x,\bar{Y},\tilde{X}_{0})$ for all $x\in S^{n-1}$; and similarly $h(x,\hat{Y},\hat{X})=h(x,\hat{Y},\tilde{X}_{0})$. Hence $h\left(x_{0},\bar{Y},\bar{X}\right)=h\left(x_{0},\bar{Y},\tilde{X}_{0}\right)=|\tilde{X}_{0}|$. From [GH14, first identity in (4.7)], $h\left(x_{0},\bar{Y},\bar{X}\right)\geq\dfrac{b-\kappa\,|\bar{Y}|}{1+\kappa}$ where $b=|\bar{X}|+\kappa\,|\bar{X}-\bar{Y}|$. Therefore, $h\left(x_{0},\bar{Y},\bar{X}\right)\geq\dfrac{1-\kappa}{1+\kappa}|\bar{X}|=\dfrac{1-\kappa}{1+\kappa}\,u(\bar{x})\geq\dfrac{1-\kappa}{1+\kappa}\,C_{1}$. Thus, $|\tilde{X}_{0}|\geq\dfrac{1-\kappa}{1+\kappa}\,C_{1}$.

We claim

[TABLE]

for some structural constant $C$. Suppose for a moment the claim holds true. We can write $X_{0}=(1+t)\tilde{X}_{0}\in\Gamma_{C_{1}C_{2}}$ with $t=\dfrac{u(x_{0})-\rho_{0}}{\rho_{0}}$. So applying Lemma 4.2 yields

[TABLE]

and

[TABLE]

for all $x\in\Omega$. Thus

[TABLE]

where in the last inequality we have used Lemma 3.4 and renamed the resulting constants.

It then remains to prove the claim. Since $x_{0}\in[\bar{x},\hat{x}]$, we can write $x_{0}=\dfrac{(1-t)\bar{x}+t\hat{x}}{|(1-t)\bar{x}+t\hat{x}|}:=\dfrac{x_{t}}{|x_{t}|}$, for some $t\in[0,1]$. If $Y_{0}\in\partial u(x_{0})$, then

[TABLE]

from Lemma 4.4. Therefore

[TABLE]

By calculation

[TABLE]

From (4.3) $|\nabla_{x}h(x_{0},Y_{0},X_{0})|\leq C$ and since $|x_{t}-x_{0}|\leq 2\,|\bar{x}-\hat{x}|^{2}$ it then follows from (5.1) that

[TABLE]

Next, since as proved above, $|\tilde{X}_{0}|\geq\dfrac{1-\kappa}{1+\kappa}\,C_{1}$, we can apply Lemma 4.4 with $X_{0}$ replaced by $\tilde{X}_{0}$ to obtain

[TABLE]

and similarly

[TABLE]

Therefore

[TABLE]

The last term is bounded above by $2\,M\,|\bar{x}-\hat{x}|^{2}$. To estimate the middle term we write

[TABLE]

so

[TABLE]

Since $|\tilde{X}_{0}|\geq\dfrac{1-\kappa}{1+\kappa}\,C_{1}$, from (5.2) and (4.3) the absolute value of the last term is $\leq C|\bar{x}-\hat{x}|^{2}$; and we can apply Lemma 4.3 to obtain that the absolute value of the first term is bounded by $C|\bar{Y}-\hat{Y}||\bar{x}-\hat{x}|$. Since $|\bar{x}-\hat{x}|\leq|\bar{Y}-\hat{Y}|$, the claim is proved, and the lemma follows.

∎

Now using Lemmas 5.2 and 4.5, we obtain the following.

Lemma 5.3**.**

Under the hypotheses of Lemma 5.2, there exist structural constants $K_{1},K_{2},K_{3}$ and $x_{0}\in B_{\sigma}\cap S^{n-1}$ such that for all $Y(\lambda)\in[\bar{Y},\hat{Y}]_{X_{0}}$, $Y\in\Sigma$ and $x\in\Omega$,

[TABLE]

where $X_{0}=u(x_{0})x_{0}$, $0<\lambda<1$.

Our main theorem is then the following.

Theorem 5.4**.**

Suppose that H.A, H.B, H.C, and H.D hold. Let $u$ be a refractor from $\Omega$ to $\Sigma$ and assume that there is a constant $C$ such that for all balls $B_{\sigma}$ such that $B_{\sigma}\cap S^{n-1}\subseteq\Omega$, we have

[TABLE]

where $H^{n-1}$ is the $(n-1)$-dimensional Hausdorff measure in ${\mathbb{R}}^{n}$.

Assume $B_{2\delta}\cap S^{n-1}\subseteq\Omega$. There exist constants $\tilde{C}_{1},\tilde{C}_{2}$ depending on $\delta$ and structure, such that if $\bar{x},\hat{x}\in B_{\delta}\cap S^{n-1}$, $\bar{Y}\in\partial u(\bar{x}),\hat{Y}\in\partial u(\hat{x})$ with $|\bar{Y}-\hat{Y}|\geq\tilde{C}_{1}|\bar{x}-\hat{x}|$, then $|\bar{Y}-\hat{Y}|\leq\tilde{C}_{2}\,|\bar{x}-\hat{x}|^{\alpha}$ where $\alpha=\dfrac{1}{4n-5}$, $n>1$.

Proof.

By Lemma 5.3, there exists $x_{0}\in[\bar{x},\hat{x}]\subseteq B_{\delta}$, such that for all $Y(\lambda)\in[\bar{Y},\hat{Y}]_{X_{0}}$ with $\frac{1}{4}\leq\lambda\leq\frac{3}{4}$, for all $Y\in\Sigma$ and for all $x\in\Omega$, we have

[TABLE]

where $X_{0}=u(x_{0})x_{0}$ and $K_{i},i=1,2,3$ are structural constants. Let

[TABLE]

If $|x-x_{0}|\geq t_{0}$, then $K_{1}|\bar{Y}-\hat{Y}|^{2}|x-x_{0}|^{2}-K_{2}|Y-Y(\lambda)||x-x_{0}|-K_{3}|\bar{Y}-\hat{Y}||\bar{x}-\hat{x}|\geq 0$. Let $\mu=\sqrt{|\bar{Y}-\hat{Y}|^{3}|\bar{x}-\hat{x}|}$ and suppose $|Y-Y(\lambda)|\leq\mu$, then

[TABLE]

Let $C\geq 1$ be large enough constant depending on $\delta$ and the structural constants such that $\dfrac{K}{\sqrt{C}}\leq\dfrac{\delta}{2}$ and $\dfrac{(\text{diam}(\Sigma))^{2}}{\sqrt{C}}\leq\mu_{0}$, with $\mu_{0}$ the constant in H.D. Set $\tilde{C}_{1}:=C$. If $|\bar{Y}-\hat{Y}|\geq\tilde{C}_{1}\,|\bar{x}-\hat{x}|$, then

[TABLE]

and

[TABLE]

Let $Y\in\Sigma$ and $|Y-Y(\lambda)|\leq\mu$ for some $\frac{1}{4}\leq\lambda\leq\frac{3}{4}$. We will show that

[TABLE]

Notice that $B(x_{0},\sigma)\cap S^{n-1}\subseteq B_{2\delta}\cap S^{n-1}\subseteq\Omega$, and if $|x-x_{0}|\geq\sigma$ and $x\in\Omega$, then $u(x)\geq h(x,Y,X_{0})$. If $X=u(x)x$, then this implies that $X$ is outside the region enclosed by the oval $\mathcal{O}\left(Y,|X_{0}|+\kappa|X_{0}-Y|\right)$ thorough $X_{0}$ and focus $Y$ which implies that the oval through $X$ with focus $Y$ encloses $\mathcal{O}\left(Y,|X_{0}|+\kappa|X_{0}-Y|\right)$. Therefore $|X|+\kappa|X-Y|\geq|X_{0}|+\kappa|X_{0}-Y|$ for $|x-x_{0}|\geq\sigma$ and $x\in\Omega$, and by continuity

[TABLE]

for some $\tilde{X}=u(\tilde{x})\tilde{x}$ with $\tilde{x}\in\bar{B}(x_{0},\sigma)\cap S^{n-1}$. So each $X=u(x)x$, with $x\in\Omega$, is outside the interior of the region enclosed by oval $\mathcal{O}\left(Y,|\tilde{X}|+\kappa|\tilde{X}-Y|\right)$ which implies that $u(x)\geq h(x,Y,\tilde{X})$, for all $x\in\Omega$. Since $u(\tilde{x})=|\tilde{X}|$ we obtain that $Y\in\partial u(\tilde{x})$ and (5.4) is proved.

Therefore

[TABLE]

Taking $H^{n-1}$-measures on both sides, using H.D on the left hand side and (5.3) on the right hand side yields

[TABLE]

which from the definitions of $\mu$ and $\sigma$ implies $|\bar{Y}-\hat{Y}|\leq\tilde{C}_{2}\,|\bar{x}-\hat{x}|^{\alpha},$ with $\tilde{C}_{2}$ an structural constant. ∎

We can now deduce Hölder estimates for the gradients of refractors.

Theorem 5.5**.**

If H.A, H.B, H.C, and H.D hold, and $u$ is a refractor from $\Omega$ to $\Sigma$ in the sense of Definition 5.1 satisfying (5.3), then $u\in C_{\text{\rm loc}}^{1,\alpha}(\Omega)$.

Proof.

Let $x_{0}\in\Omega$. We first show that $\partial u(x_{0})$ is singleton. Fix $\delta>0$ such that $B(x_{0},2\delta)\cap S^{n-1}\subseteq\Omega$ and suppose $Y_{0},Y_{1}\in\partial u(x_{0})$, with $Y_{1}\neq Y_{0}$. Let $\bar{x}\in B(x_{0},\delta)\cap S^{n-1}$ and $\bar{Y}\in\partial u(\bar{x})$. By Theorem 5.4, $|\bar{Y}-Y_{0}|\leq C|\bar{x}-x_{0}|^{\alpha}$ and $|\bar{Y}-Y_{1}|\leq C|\bar{x}-x_{0}|^{\alpha}$ where the constant $C$ depends on $\delta$. Hence, $|Y_{1}-Y_{0}|\leq 2C|\bar{x}-x_{0}|^{\alpha}$, so if we take $\bar{x}$ close enough to $x_{0}$ we get a contradiction.

Let $Y\in\partial u(x_{0})$. We first claim that for any $\eta\perp x_{0}$, $|\eta|=1$, we have $D_{\eta}u(x_{0})=\langle\nabla h(x_{0},Y,X_{0}),\eta\rangle$, where $X_{0}=u(x_{0})x_{0}$. To see this, let $c$ be any curve such that $c(0)=x_{0}$ and $c^{\prime}(0)=\eta$ and $c(t)\in B(x_{0},\delta)\cap S^{n-1}$ for all $t$ near [math]. Since $u$ is a refractor

[TABLE]

for all $t$ near [math]. Let $Y(t)\in\partial u(c(t))$ and $X(t)=u(c(t))c(t)$. Since $u(x)\geq h(x,Y(t),X(t))$ for all $x\in\Omega$, we get

[TABLE]

for all $t$ near zero. Therefore, we have for all $t>0$ small

[TABLE]

Note that for each $t$

[TABLE]

for some $\tilde{x}\in[x_{0},c(t)]$. From Theorem 5.4, $Y(t)\rightarrow Y$ as $t\rightarrow 0$, and $X(t)\rightarrow X_{0}$ by continuity of $u$. Letting $t\rightarrow 0$ the claim follows.

Define $\tilde{u}(X)=u\left(X/|X|\right)$ for $X$ with $X/|X|\in\Omega$. We will show that for each $x_{0}\in\Omega$

[TABLE]

Indeed, let $c(t)=\dfrac{x_{0}+te_{i}}{|x_{0}+te_{i}|}$ and notice that $c(0)=x_{0}$ and $c^{\prime}(0)=e_{i}-\langle x_{0},e_{i}\rangle x_{0}$. Since $\dfrac{\tilde{u}(x_{0}+te_{i})-\tilde{u}(x_{0})}{t}=\dfrac{u(c(t))-u(x_{0})}{t}$, letting $t\rightarrow 0$ and using the first part we get

[TABLE]

and the desired formula follows.

Next, let $\bar{x},\hat{x}\in B(x_{0},\delta)\cap S^{n-1}\subset\Omega$, and let $\bar{Y}\in\partial u(\bar{x})$ and $\hat{Y}\in\partial u(\hat{x})$. We shall prove that

[TABLE]

First notice that

[TABLE]

since $|\nabla h(\bar{x},\bar{Y},\bar{X})|$ is bounded. Next write

[TABLE]

First, by Lemma 4.3 $|\nabla h(\bar{x},\bar{Y},\bar{X})-\nabla h(\bar{x},\hat{Y},\bar{X})|\leq C\,|\bar{Y}-\hat{Y}|$. Second, that $|\nabla h(\bar{x},\hat{Y},\bar{X})-\nabla h(\hat{x},\hat{Y},\bar{X})|\leq C|\bar{x}-\hat{x}|$ follows using the mean value theorem in $x$ from the estimates in the proof of Lemma 4.4, i.e., from (4.1), (4.2) and (4.3). For the third term, from (4.3) we can write

[TABLE]

where $\bar{b}=|\bar{X}|+\kappa|\hat{Y}-\bar{X}|$ and $\hat{b}=|\hat{X}|+\kappa|\hat{Y}-\hat{X}|$. Since $\hat{Y}\in\Sigma$ and $\bar{X},\hat{X}\in\Gamma_{C_{1}C_{2}}$, and noticing that $|\bar{b}-\hat{b}|\leq C_{\kappa}\,|\bar{X}-\hat{X}|$, it follows from Definitions (2.2), (2.3), and Lemma 4.1 that

[TABLE]

Therefore

[TABLE]

We also have $|\bar{X}-\hat{X}|=|u(\bar{x})\bar{x}-u(\hat{x})\hat{x}|\leq C_{1}\,|\bar{x}-\hat{x}|+|u(\bar{x})-u(\hat{x})|\leq C\,|\bar{x}-\hat{x}|$, since $u$ is Lipschitz. From Theorem 5.4 we then obtain (5.6) and the proof is complete.

∎

5.1. Regularity of weak solutions

We now apply Theorem 5.5 to show that weak solutions to the near field refractor problem defined with the tracing map are $C_{loc}^{1,\alpha}$. Existence of weak solutions is proved in [GH14].

Recall that the tracing mapping $\mathcal{T}_{u}$ is defined as follows: given $Y\in\Sigma$, $\mathcal{T}_{u}(Y)=\{x\in\Omega:Y\in\partial u(x)\}$. A weak solution $u$ to the refractor problem from $\Omega$ to $\Sigma$ satisfies

[TABLE]

Here $\mu=f(x)\,dx$ with $f\in L^{1}(\Omega)$, $f>0$ a.e., and $\nu$ is a measure on the target $\Sigma$ so that the energy conservation condition $\int_{\Omega}f(x)\,dx=\nu(\Sigma)$ holds.

Theorem 5.6**.**

Assume that H.A, H.B, H.C, and H.D hold and the target $\Sigma$ is differentiable. If $f\in L^{\infty}(\Omega)$, $\nu\ll H^{n-1}$, and $H^{n-1}=g\,d\nu$ with $0\leq g(x)\leq\alpha$ for a.e. $x\in\Sigma$, then each weak solution $u$ to (5.7) satisfies (5.3), and therefore from Theorem 5.5 $u\in C^{1,\alpha}_{loc}$.

Proof.

Since $\Sigma$ is differentiable, then the visibility condition implies that the tangent plane to $\Sigma$ at each point cannot intersect the interior of $\Gamma_{c_{1},c_{2}}$. Indeed, suppose the tangent plane $T_{Y}$ to $\Sigma$ at $Y$ intersects $\Gamma_{c_{1},c_{2}}$ at $X_{0}$ and with a ball $B(X_{0},\epsilon)\subset\Gamma_{c_{1},c_{2}}$. The segment from $X_{0}$ to $Y$ is on $T_{Y}$ and by visibility for each $X\in B(X_{0},\epsilon)$, the segment from $X$ to $Y$ intersects $\Sigma$ only at $Y$. This implies that $\Sigma$ cannot be differentiable at $Y$, because if $\Sigma$ were differentiable at $Y$, then $T_{Y}\cap C=\{Y\}$ with $C$ the cone with vertex $Y$ and base $B(X_{0},\epsilon)$, but $\overline{X_{0}Y}\subset T_{Y}\cap C$.

Now let

[TABLE]

We shall prove that $H^{n-1}(S^{\star})=0$.

Define $u^{\star}:{\mathbb{R}}^{n}\rightarrow{\mathbb{R}}$ by

[TABLE]

It is easy to see that $u^{\star}$ is Lipschitz in ${\mathbb{R}}^{n}$.

If $\bar{Y}\in\partial u(\bar{x})$, $X=u(x)x$, $\bar{X}=u(\bar{x})\bar{x}$, with $u$ a refractor, then $X$ is outside the interior of the region enclosed by the oval $\mathcal{O}(\bar{Y},b)$ with $b=|\bar{X}|+\kappa\,|\bar{X}-\bar{Y}|$. This means that the region enclosed by an oval passing through $X$ with focus $\bar{Y}$ contains $\mathcal{O}(\bar{Y},b)$, that is,

[TABLE]

Hence

[TABLE]

and so

[TABLE]

for all $Y\in{\mathbb{R}}^{n}$. In particular, if $Y_{0}\in S^{\star}$ and say $Y_{0}\in\partial u(\bar{x})\cap\partial u(\hat{x})$, we then have

[TABLE]

and

[TABLE]

$\hat{X}=u(\hat{x})\hat{x}$, for all $Y\in{\mathbb{R}}^{n}$.

Let $O\subseteq{\mathbb{R}}^{n-1}$ be open and let $\psi:{\mathbb{R}}^{n-1}\rightarrow{\mathbb{R}}^{n}$ be Lipschitz such that $\Sigma=\psi(\bar{O})$ and $\psi$ is one to one in $\bar{O}$. Set $\tilde{S}=\psi^{-1}\left(S^{\star}\right)$. We show that $H^{n-1}(\tilde{S})=0$.

Define $h(Y^{\prime})=u^{\star}(\psi(Y^{\prime}))$. Since $u^{\star}$ is Lipschitz, $h$ is Lipschitz in ${\mathbb{R}}^{n-1}$. We claim that $h$ is not differentiable in $\tilde{S}$. Let $Y_{0}^{\prime}\in\tilde{S}$, so $Y_{0}=\psi(Y_{0}^{\prime})\in S^{\star}$, that is, there are $\bar{x}\neq\hat{x}$ in $\Omega$ with $Y_{0}\in\partial u(\bar{x})\cap\partial u(\hat{x})$. Then $h(Y^{\prime})\leq h(Y_{0}^{\prime})+|\bar{X}-\psi(Y^{\prime})|-|\bar{X}-\psi(Y_{0}^{\prime})|$ and $h(Y^{\prime})\leq h(Y_{0}^{\prime})+|\hat{X}-\psi(Y^{\prime})|-|\hat{X}-\psi(Y_{0}^{\prime})|$ for all $Y^{\prime}\in{\mathbb{R}}^{n}$ with $\bar{X}=u(\bar{x})\bar{x}$ and $\hat{X}=u(\hat{x})\hat{x}$. If $h$ were differentiable at $Y_{0}^{\prime}$, then we would have

[TABLE]

at $Y^{\prime}=Y_{0}^{\prime}$. Thus $D\psi(Y_{0}^{\prime})^{T}\dfrac{Y_{0}-\bar{X}}{|Y_{0}-\bar{X}|}=D\psi(Y_{0}^{\prime})^{T}\dfrac{Y_{0}-\hat{X}}{|Y_{0}-\hat{X}|}.$ Letting $w=\dfrac{Y_{0}-\bar{X}}{|Y_{0}-\bar{X}|}-\dfrac{Y_{0}-\hat{X}}{|Y_{0}-\hat{X}|}$, yields $D\psi(Y_{0}^{\prime})^{T}w=0$. If $v_{k}$ denote the columns of $D\psi(Y_{0}^{\prime})$, this means that $\langle v_{k},w\rangle=0$, for $1\leq k\leq n-1$. Since the $v_{k}$’s span the tangent plane to $\Sigma$ at $Y_{0}$, we get that $w$ is normal to the tangent plane to $\Sigma$ at $Y_{0}$. In particular, the line $Y_{0}+t\,\left(\dfrac{Y_{0}-\bar{X}}{|Y_{0}-\bar{X}|}+\dfrac{Y_{0}-\hat{X}}{|Y_{0}-\hat{X}|}\right)$ is contained in the tangent plane to $\Sigma$ at $Y_{0}$. But it is easy to see that this line intersects the straight segment $[\bar{X},\hat{X}]$, which implies that either both $\bar{X}$ and $\hat{X}$ are on the tangent plane or they are on opposite sides of the tangent plane. In either case, since $\bar{X}$ and $\hat{X}$ are on the graph of $u$, the tangent plane intersects the graph of $u$, which contradicts our initial assumption.

Since $h$ is Lipschitz we obtain that $H^{n-1}(\tilde{S})=0$. This implies, since $\psi$ is Lipschitz, that $H^{n-1}(S^{\star})=0$ as we wanted to show.

From the assumption, $\nu\ll H^{n-1}$, we will show first that $\nu(\partial u(B))\leq\mu(B)$ for each $B\subset\Omega$ Borel set. If

[TABLE]

let us see that $\mu(S)=0$. Indeed, since $\mathcal{T}_{u}(S^{\star})=S$, from the definition of weak solution $\nu(S^{\star})=\mu(\mathcal{T}_{u}(S^{\star}))=\mu(S)$. Since $H^{n-1}(S^{\star})=0$, we then get $\mu(S)=0$. On the other hand, $\mathcal{T}_{u}(\partial u(B))\subset B\cup S$ so $\nu(\partial u(B))=\mu(\mathcal{T}_{u}(\partial u(B)))\leq\mu(B\cup S)\leq\mu(B)$ and we are done.

Therefore, to conclude the proof of the theorem, we prove that $u$ verifies (5.3). Indeed, for each ball $B_{\sigma}$ with $B_{\sigma}\cap S^{n-1}\subset\Omega$ we have

[TABLE]

∎