Local $C^{1,\beta }$-regularity at the boundary of two dimensional sliding almost minimal sets in $\mathbb{R}^{3}$

Abstract

In this paper, we will give a $C^{1,\beta }$-regularity result on the boundary for two dimensional sliding almost minimal sets in $\mathbb{R}^3$. This effect may apply to the regularity of the soap films at the boundary, and may also lead to the existence of a solution to the Plateau problem with sliding boundary conditions proposed by Guy David in the case that the boundary is a 2-dimensional smooth submanifold.

1. Introduction

Jean Taylor, in 13, proved a celebrated regularity result of Almgren almost minimal sets, that gives a complete classification of the local structure of 2-dimensional (almost) minimal sets, that is, every $2$-dimensional almost minimal set $E$, in an open set $U\subseteq \mathbb{R}^3$ with gauge function $h(t)\leq Ct^{\alpha }$, is local $C^{1,\beta }$ equivalent to a $2$-dimensional minimal cone. This result may apply to many actual surfaces, soap films are considered as typical examples. In 5, Guy David gave a new proof of this result and generalized it to any codimension. Even with this very nice regularity property, we still do not know the behavior of almost minimal sets $E$ at the boundary $\overline{E}\cap \partial U$, since it could be more and more complicated when points tend to the boundary, that is, the behaves of soap films at the boundary is not clear.

In 7, Guy David proposed to consider the Plateau Problem with sliding boundary conditions, since it is very natural to the soap films, here we mean that the soap films can be consider as sliding almost minimal sets. We see that, away from the boundary, sliding almost minimal sets are almost minimal, Jean Taylor’s regularity also applies, so that we already know the behavior of sliding almost minimal sets except at the boundary. Indeed, the feature that allow surfaces moving along the boundary could make the local structure more simple. Motivated by these, the regularity at the boundary would be well worth our considering. In fact, we are looking for a result similar to Jean Talyor’s, for which together with Jean Taylor’s theorem will imply the local Lipschitz retract property of sliding (almost) minimal sets, and the existence of minimizers for the sliding Plateau Problem will easily follows. Certainly we will get the whole story about the regularity of the soap films.

In 13, Jean Taylor gave a full list of a two dimensional minimal cones in $\mathbb{R}^3$, that is, planes, cones of type $\mathbb{Y}$, and cones of type $\mathbb{T}$. One of the advantages for the sliding boundary conditions is that we perceived the chance to determine the possibility of minimal cones in the upper half space $\Omega _0$ of $\mathbb{R}^{3}$, where minimal cone is a cone which is minimal under the sliding deformations. Indeed, there are seven kinds of cones which are minimal, they are $\partial \Omega _0$, cones of type $\mathbb{V}$, cones of type $\mathbb{P}_+$, cones of type $\mathbb{Y}_+$, cones of type $\mathbb{T}_+$ and cones $\partial \Omega _0 \cup Z$ where $Z$ are cones of type $\mathbb{P}_+$ or $\mathbb{Y}_+$, see Section 3 in 9 for the precise definition of cones of type $\mathbb{P}_+$, $\mathbb{Y}_+$, $\mathbb{T}_+$ and $\mathbb{V}$. Let us refer to Remark 3.11 in 9 for the claim there are at most seven, Theorem 3.10 in 9 proved some cones are minimal, and the rest is proved by Cavallotto 2. We ascertain that there are only three kinds of cones which are minimal and contains the boundary $\partial \Omega _0$, they are $\partial \Omega _0$ and $\partial \Omega _0 \cup Z$ where $Z$ is cone of type $\mathbb{P}_+$ or $\mathbb{Y}_+$, see Theorem 3.10 in 9 for the statement.

Another advantages of the sliding almost minimal sets is that they are not far from usual almost minimal sets, away from the boundary, they are almost minimal, we have also the monotonicity of density property, and at the boundary we can establish a similar monotonicity of density property without too much effort, see Theorem 2.3 for precise statement. But in fact, the monotonicity of density property is not enough, we have estimated the decay of the almost density, and that is also possible with sliding on the boundary, see Corollary 3.16.

In 9, we proved a Hölder regularity of two dimensional sliding almost minimal set at the boundary. That is, suppose that $\Omega \subseteq \mathbb{R}^{3}$ is a closed domain with boundary $\partial \Omega$ a $C^1$ manifold of dimension 2, $E\subseteq \Omega$ is a 2 dimensional sliding almost minimal set with sliding boundary $\partial \Omega$, and that $\partial \Omega \subseteq E$. Then $E$, at the boundary, is locally biHölder equivalent to a sliding minimal cone in the upper half space $\Omega _0$. In this paper, we will generalized the biHölder equivalence to a $C^{1,\beta }$ equivalence when the gauge function $h$ satisfies that $h(t)\leq Ct^{\alpha _1}$ and $\partial \Omega$ is a 2 dimensional $C^{1,\alpha }$ manifold. Let us refer to Theorem 1.2 for details. Where the sliding minimal cones always contain the boundary $\partial \Omega _0$, namely only there kinds of cones can appear: $\partial \Omega _0$ and $\partial \Omega _0\cup Z$, where $Z$ are cones of type $\mathbb{P}_+$ or $\mathbb{Y}_+$.

Let us introduce some notation and definitions before state our main theorem. A gauge function is a nondecreasing function $h:[0,\infty )\to [0,\infty ]$ with $\lim _{t\to 0} h(t)=0$. Let $\Omega$ be a closed domain of $\mathbb{R}^{3}$, $L$ be a closed subset in $\mathbb{R}^{3}$, $E\subseteq \Omega$ be a given set. Let $U\subseteq \mathbb{R}^3$ be an open set. A family of mappings $\{ \varphi _t \}_{0\leq t\leq 1}$, from $E$ into $\Omega$, is called a sliding deformation of $E$ in $U$, while $\varphi _1(E)$ is called a competitor of $E$ in $U$, if following properties hold:

•

$\varphi _t(x)=x$ for $x\in E\setminus U$, $\varphi _t(x)\subseteq U$ for $x\in E\cap U$, $0\leq t\leq 1$,

•

$\varphi _t(x)\in L$ for $x\in E\cap L$, $0\leq t \leq 1$,

•

the mapping $[0,1]\times E\to \Omega , (t,x)\mapsto \varphi _{t}(x)$ is continuous,

•

$\varphi _1$ is Lipschitz and $\varphi _0=\operatorname {id}_E$.

Definition 1.1.

Let $L, \Omega \subseteq \mathbb{R}^3$ be two closed sets, $L\subseteq \Omega$. We say that an nonempty set $E\subseteq \Omega$ is locally sliding almost minimal at $x\in E$ with sliding boundary $L$ and with gauge function $h$, called $(\Omega ,L,h)$ locally sliding almost at $x\in E$ for short, if $\mathcal{H}^2\mathbin{\rightangle }E$ is locally finite, and for any sliding deformation $\{ \varphi _t \}_{0\leq t\leq 1}$ of $E$ in $B(x,r)$, we have that

$$\begin{equation*} \mathcal{H}^2(E\cap B(x,r))\leq \mathcal{H}^2(\varphi _1(E)\cap B(x,r)) +h(r)r^2. \end{equation*}$$

We say that $E$ is sliding almost minimal with sliding boundary $L$ and gauge function $h$, denote by $SAM(\Omega ,L,h)$ the collection of all such sets, if $E$ is locally sliding almost minimal at all points $x\in E$.

For any $x\in \mathbb{R}^3$, we let $\boldsymbol{\tau }_{x}: \mathbb{R}^3\to \mathbb{R}^3$ be the translation defined by $\boldsymbol{\tau }_{x}(y)=y+x$, and let $\boldsymbol{\mu }_{r}:\mathbb{R}^{3}\to \mathbb{R}^{3}$ be the mapping defined by $\boldsymbol{\mu }_{r}(y)=ry$ for any $r>0$. For any $S\subseteq \mathbb{R}^3$ and $x\in S$, a blow-up limit of $S$ at $x$ is any closed set in $\mathbb{R}^3$ that can be obtained as the Hausdorff limit of a sequence $\boldsymbol{\mu }_{1/r_k}\circ \boldsymbol{\tau }_{-x}(S)$ with $\lim _{k\to \infty }r_k=0$. A set $X$ in $\mathbb{R}^3$ is called a cone centered at the origin $0$ if for any $\boldsymbol{\mu }_{t}(X)=X$ for any $t\geq 0$; in general, we call a cone $X$ centered at $x$ if $\boldsymbol{\tau }_{-x}(X)$ is a cone centered at $0$. We denote by $\operatorname {Tan}(S,x)$ the tangent cone of $S$ at $x$, see Section 2.1 in 1. We see that if there is unique blow-up limit of $S$ at $x$, then it coincide with the tangent cone $\operatorname {Tan}(S,x)$. Our main theorem is the following.

Theorem 1.2.

Let $\Omega \subseteq \mathbb{R}^{3}$ be a closed set such that the boundary $\partial \Omega$ is a $2$-dimensional manifold of class $C^{1, \alpha }$ for some $\alpha >0$ and $\operatorname {Tan}(\Omega ,z)$ is a half space for any $z\in \partial \Omega$. Let $E\subseteq \Omega$ be a closed set such that $E\supseteq \partial \Omega$ and $E$ is a sliding almost minimal set with sliding boundary $\partial \Omega$ and with gauge function $h$ satisfying that

$$\begin{equation*} h(t)\leq C_h t^{\alpha _1},\ 0<t\leq t_0, \text{ for some }C_h>0, \alpha _1>0 \text{ and }t_0>0. \end{equation*}$$

Then for any $x_0\in \partial \Omega$, there is unique blow-up limit of $E$ at $x_0$; moreover, there exist a radius $r>0$, a sliding minimal cone $Z$ in $\Omega _0$ with sliding boundary $\partial \Omega _0$, and a mapping $\Phi :\Omega _0\cap B(0,2r)\to \Omega$ of class $C^{1,\beta }$, which is a diffeomorphism between its domain and image, such that $\Phi (0)=x_0$, $\Phi (\partial \Omega _0\cap B(0,2r))\subseteq \partial \Omega$, $|\Phi (x)-x_0-x|\leq 10^{-2}r$ for $x\in B(0,2r)$, and

$$\begin{equation*} E\cap B(x_0,r)=\Phi (Z)\cap B(x_0,r). \end{equation*}$$

The theorem above, together with the Jean Taylor’s theorem, will imply that any sliding almost minimal set $E$ as in the theorem is local Lipschitz neighborhood retract. This effect may gives the existence of a solution to the Plateau problem with sliding boundary conditions in a special case, see Theorem 8.1.

2. Lower bound of the decay for the density

In this section, we will consider a simple case that $\Omega$ is a half space and $L$ is its boundary; without loss of generality, we assume that $\Omega$ is the upper half space, and change the notation to be $\Omega _0$ for convenience, i.e.

$$\begin{equation*} \Omega _0=\{(x_1,x_2,x_3)\in \mathbb{R}^{3}\mid x_3\geq 0\}, L_0=\partial \Omega _0. \end{equation*}$$

It is well known that for any 2-rectifiable set $E$, there exists an approximate tangent plane $\operatorname {Tan}(E,y)$ of $E$ at $y$ for $\mathcal{H}^2$-a.e. $y\in E$. We will denote by $\theta (y)\in [0,\pi /2]$ the angle between the segment $[0,y]$ and the plane $\operatorname {Tan}(E,y)$, by $\theta _x(y)\in [0,\pi /2]$ the angle between the segment $[x,y]$ and the plane $\operatorname {Tan}(E,y)$, for $x\in \mathbb{R}^{3}$.

For any gauge function $h$ in this paper, we always assume that there is a number $r_h>0$ such that

$$\begin{equation} \int _{0}^{r_h}\frac{h(2t)}{t}dt<\infty , {\tag{2.1}} \end{equation}$$

and put

$$\begin{equation*} h_1(t)=\int _{0}^{t}\frac{h(2s)}{s}ds ,\text{ for }0\leq t\leq r_h. \end{equation*}$$

For any mapping $f:E\subseteq \mathbb{R}^m\to \mathbb{R}^n$, we denote by $\operatorname {ap}J_kf(x)=\|\wedge _k \operatorname {ap}Df(x)\|$ the $k$ dimensional approximate Jacobian of $f$ at $x$, if $f$ is approximate differentiable at $x$, see Section 3.2.1 in 10

In this section, we will compare a set $E$ to the cone over $E\cap \partial B(0,r)$, then establish a monotonicity of density formula for any $2$-rectifiable set $E$ which is locally sliding almost minimal at $0$, see Theorem 2.3.

Lemma 2.1.

Let $E\subseteq \Omega _0$ be any $2$-rectifiable set. Then, by putting $u(r)= \mathcal{H}^2(E\cap B(x,r))$, we have that $u$ is differentiable almost every $r>0$, and for such $r$,

$$\begin{equation*} \mathcal{H}^1(E\cap \partial B(x,r))\leq u'(r). \end{equation*}$$

Proof.

Considering the function $\psi :\mathbb{R}^{3}\to \mathbb{R}$ defined by $\psi (y)=|y-x|$, we have that, for any $y\neq x$ and $v\in \mathbb{R}^{3}$,

$$\begin{equation*} D\psi (y)v=\left\langle \frac{y-x}{|y-x|},v \right\rangle , \end{equation*}$$

thus

$$\begin{equation} \operatorname {ap}J_1(\psi \vert _{E})(y)=\sup \{ |D\psi (y)v|: v\in \operatorname {Tan}(E,x),|v|=1 \}=\cos \theta _x(y). {\tag{2.2}} \end{equation}$$

Employing Theorem 3.2.22 in 10, we have that, for any $0<r<R<\infty$,

$$\begin{equation*} \int _{r}^{R}\mathcal{H}^1(E\cap \partial B(x,t))dt=\int _{E\cap B(x,R)\setminus B(x,r)}\cos _x(y)d\mathcal{H}^2(y)\leq u(R)-u(r), \end{equation*}$$

we get so that, for almost every $r\in (0,\infty )$,

$$\begin{equation*} \mathcal{H}^1(E\cap \partial B(x,t))\leq u'(r). \end{equation*}$$ ■

Lemma 2.2.

Let $E$ be a $2$-rectifiable $(\Omega _0,L_0,h)$ locally sliding almost minimal at $x\in E$.

•

If $x\in E\cap L_0$, then for $\mathcal{H}^1$-a.e. $r\in (0,\infty )$,$$\begin{equation} \mathcal{H}^2(E\cap B(x,r))\leq \frac{r}{2}\mathcal{H}^1(E\cap \partial B(x,r))+h(2r)(2r)^2. {\tag{2.3}} \end{equation}$$

•

If $x\in E\setminus L_0$, then inequality 2.3 holds for $\mathcal{H}^1$-a.e. $r\in (0,\operatorname {dist}(x,L_0))$.

Proof.

If $\mathcal{H}^2(E\cap \partial B(x,r))>0$, then $\mathcal{H}^1(E\cap \partial B(x,r))=\infty$, and nothing need to do. We assume so that $\mathcal{H}^2(E\cap \partial B(x,r))=0$.

Let $f:[0,\infty )\to [0,\infty )$ be any Lipschitz function, we let $\phi :\Omega _0\to \Omega _0$ be defined by

$$\begin{equation*} \phi (y)=f(|y-x|)\frac{y-x}{|y-x|}. \end{equation*}$$

Then, for any $y\neq x$ and any $v\in \mathbb{R}^{3}$, by putting $\tilde{y}=y-x$, we have that

$$\begin{equation*} D\phi (y)v=\frac{f(|\tilde{y}|)}{|\tilde{y}|}v+\frac{|\tilde{y}|f'(|\tilde{y}|)-f(|\tilde{y}|)}{|\tilde{y}|^{2}} \left\langle \frac{\tilde{y}}{|\tilde{y}|},v \right\rangle \tilde{y} \end{equation*}$$

If the tangent plane $\operatorname {Tan}^{2}(E,y)$ of $E$ at $y$ exists, we take $v_1,v_2\in \operatorname {Tan}^{2}(E,y)$ such that $|v_1|=|v_2|=1$, $v_1$ is perpendicular to $y=x$, and that $v_2$ is perpendicular to $v_1$, let $v_3$ be a vector in $\mathbb{R}^{3}$ which is perpendicular to $\operatorname {Tan}^{2}(E,y)$ and $|v_3|=1$, then

$$\begin{equation*} \tilde{y}=\langle \tilde{y},v_2 \rangle v_2+\langle \tilde{y},v_3 \rangle v_3=|\tilde{y}|\cos \theta _x(y) v_2+|\tilde{y}|\sin \theta _x(y) v_3, \end{equation*}$$

and

$$\begin{equation*} D\phi (y)v_1\wedge D\phi (y)v_2= \frac{f(|\tilde{y}|)^2}{|\tilde{y}|^2}v_1\wedge v_2+\frac{|\tilde{y}|f'(|\tilde{y}|)f(|\tilde{y}|) -f(|\tilde{y}|)^2}{|\tilde{y}|^3}\cos \theta _x(y) v_1\wedge \tilde{y}, \end{equation*}$$

thus

$$\begin{equation*} \begin{aligned} \operatorname {ap}J_{2}(\phi \vert _{E})(y) &=\|D\phi (y)v_1\wedge D\phi (y)v_2\|\\ &=\frac{f(|\tilde{y}|)}{|\tilde{y}|}\left(f'(|\tilde{y}|)^2 \cos ^2 \theta _x(y)+\frac{f(|\tilde{y}|)^2}{|\tilde{y}|^2}\sin ^2\theta _x(y) \right)^{1/2}. \end{aligned} \end{equation*}$$

We consider the function $\psi :\mathbb{R}^{3}\to \mathbb{R}$ defined by $\psi (y)=|y-x|$. Then, by 2.2, we have that

$$\begin{equation*} \operatorname {ap}J_{1}(\psi \vert _{E})(y)=\cos \theta _x(y). \end{equation*}$$

For any $\xi \in (0,r/2)$, we consider the function $f$ defined by

$$\begin{equation*} f(t)=\begin{cases} 0,& 0\leq t\leq r-\xi \\ \frac{r}{\xi }(t-r+\xi ),& r-\xi < t\leq r\\ t,&t>r. \end{cases} \end{equation*}$$

Then we have that

$$\begin{equation*} \operatorname {ap}J_{2}(\phi \vert _{E})(y)\leq \frac{f(|\tilde{y}|)f'(|\tilde{y}|)}{|\tilde{y}|}\cos \theta _x (y)+\frac{f(|\tilde{y}|)^2}{|\tilde{y}|^2}\sin \theta _x(y). \end{equation*}$$

Applying Theorem 3.2.22 in 10, by putting $A_{\xi }=E\cap B(0,r)\setminus B(0,r-\xi )$, we get that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(\phi (E\cap B(0,r)))&\leq \int _{A_{\xi }}\frac{r^2}{\xi ^2}\cdot \frac{|\tilde{y}|-r+\xi }{|\tilde{y}|}\cos \theta _x(y)d\mathcal{H}^2(y) +\frac{r^2}{(r-\xi )^2}\mathcal{H}^2(A_{\xi })\\ &=\int _{r-\xi }^{r}\frac{r^2(t-r+\xi )}{\xi ^2 t}\mathcal{H}^1(E\cap \partial B(x,t))dt+4\mathcal{H}^2(A_{\xi }), \end{aligned} \end{equation*}$$

thus

$$\begin{equation*} \mathcal{H}^2(E\cap B(0,r))\leq (2r)^2h(2r)+\lim _{\xi \to 0+}r^2 \int _{r-\xi }^r\frac{t-r+\xi }{t\xi ^2}\mathcal{H}^1(E\cap \partial B(x,t))dt. \end{equation*}$$

Since the function $g(t)=\mathcal{H}^1(E\cap B(x,t))/t$ is a measurable function, we have that, for almost every $r$,

$$\begin{equation*} \lim _{\xi \to 0+}\int _{0}^{\xi }\frac{tg(t-r+\xi )}{\xi ^2}dt=\frac{1}{2}g(r), \end{equation*}$$

thus for such $r$,

$$\begin{equation*} \mathcal{H}^2(E\cap B(x,r))\leq (2r)^2h(2r)+\frac{r}{2}\mathcal{H}^1(E\cap \partial B(x,r)). \end{equation*}$$ ■

For any set $E\subseteq \mathbb{R}^3$, we set

$$\begin{equation*} \Theta _E(x,r)=r^{-2}\mathcal{H}^2(E\cap B(x,r)),\ \text{for any } r>0, \end{equation*}$$

and denote by $\Theta _E(x)=\lim _{r\to 0+}\Theta _E(x,r)$ if the limit exist, we may drop the script $E$ if there is no danger of confusion.

Theorem 2.3.

Let $E$ be a $2$-rectifiable $(\Omega _0,L_0,h)$ locally sliding almost minimal at $x\in E$.

•

If $x\in L_0$, then $\Theta (x,r)+8 h_1(r)$ is nondecreasing as $r\in (0,r_h)$.

•

If $x\not \in L_0$, then $\Theta (x,r)+8 h_1(r)$ is nondecreasing as $r\in (0,\min \{r_h,\operatorname {dist}(x,L)\})$.

Proof.

From Lemma 2.2 and Lemma 2.1, by putting $u(r)=\mathcal{H}^2(E\cap B(x,r))$, we get that, if $x\in L$,

$$\begin{equation} u(r)\leq \frac{r}{2}u'(r)+h(2r)(2r)^2, {\tag{2.4}} \end{equation}$$

for almost every $r\in (0,\infty )$; if $x\not \in L$, then 2.4 holds for almost every $r\in (0,\min \{r_h,\operatorname {dist}(x,L)\})$.

We put $v(r)=r^{-2}u(r)$, then $v'(r)\geq -8r^{-1}h(2r)$, we get that $\Theta (x,r)+8 h_1(r)$ is nondecreasing.

■

Remark 2.4.

Let $E$ be a 2-rectifiable $(\Omega _0,L_0,h)$ locally sliding almost minimal at some point $x\in E$. Then by Theorem 2.3, we get that $\Theta _E(x)$ exists.

3. Estimation of upper bound

In the previous section, we get a monotonicity of density formula, that is $\Omega _E(x,r)-\Theta _E(x)+8h_1(r)$ is nondecreasing, thus we get the estimation $\Theta _E(x,r)-\Theta _E(x)\geq -8h_1(r)$ when $r$ small. But in fact we need a good estimation for $|\Omega _E(x,r)-\Theta _E(x)|$, so we have to get some estimation for upper bound. The main purpose of this section is get the control of $\mathcal{H}^2(E\cap B(0,r))$ by a convex combination of $\mathcal{H}^2(Z\cap B(0,r))$ and $\Theta _E(0)r^2$, where $Z$ is the cone over $E\cap \partial B(0,r)$, see Theorem 3.15 and Corollary 3.16.

Let $\mathcal{Z}$ be a collection of cones. We say that a set $E\subseteq \mathbb{R}^{3}$ is locally $C^{k,\alpha }$-equivalent (resp. $C^{k}$-equivalent) to a cone in $\mathcal{Z}$ at $x\in E$ for some nonnegative integer $k$ and some number $\alpha \in (0,1]$, if there exist $\varrho _0>0$ and $\tau _0>0$ such that for any $\tau \in (0,\tau _0)$ there is $\varrho \in (0,\varrho _0)$, a cone $Z\in \mathcal{Z}$ and a mapping $\Phi :B(0,2\varrho )\to \mathbb{R}^{3}$, which is a homeomorphism of class $C^{k,\alpha }$ (resp. $C^k$) between $B(0,2\varrho )$ and its image $\Phi (B(0,2\varrho ))$ with $\Phi (0)=x$, satisfying that

$$\begin{equation} \| \Phi -\operatorname {id}-\Phi (0) \|_{\infty }\leq \varrho \tau {\tag{3.1}} \end{equation}$$

and

$$\begin{equation} E\cap B(x,\varrho )\subseteq \Phi \left(Z\cap B\left( 0,2\varrho \right)\right)\subseteq E\cap B(x,3\varrho ). {\tag{3.2}} \end{equation}$$

Similarly, if $\Omega \subseteq \mathbb{R}^{3}$ is a closed set with the boundary $\partial \Omega$ is a 2-dimensional manifold, a set $E\subseteq \Omega$ is called locally $C^{k,\alpha }$-equivalent to a sliding minimal cone $Z$ in $\Omega _0$ at $x\in E\cap \partial \Omega$, if there exist $\varrho _0>0$ and $\tau _0>0$ such that for any $\tau \in (0,\tau _0)$ there is $\varrho \in (0,\varrho _0)$ and a mapping $\Phi : B(0,2\varrho )\cap \Omega _0\to \Omega$, which is a diffeomorphism of class $C^{k,\alpha }$ between its domain and image with $\Phi (0)=x$ satisfying that $\Phi (L_0\cap B(0,2\varrho ))\subseteq \partial \Omega$ and 3.1 and 3.2.

Suppose that $\Omega \subseteq \mathbb{R}^3$ is closed set with the boundary $\partial \Omega$ is a $2$-dimensional $C^1$ manifold. Suppose that $E\subseteq \Omega$ is sliding almost minimal with sliding boundary $\partial \Omega$ and gauge function $h$. Then, by putting $U=\Omega \setminus \partial \Omega$, we see that $E\cap U$ is almost minimal in $U$, applying Jean Taylor’s theorem, $E$ is locally $C^{1,\beta }$-equivalent to a minimal cone at each point $x\in E\cap U$ for some $\beta >0$ in case $h(r)\leq cr^{\alpha }$ for some $c>0$, $\alpha >0$, $r_0>0$ and $0<r<r_0$. We see from 9, Theorem 6.1 that, at $x\in E\cap \partial \Omega$, $E$ is locally $C^{0,\beta }$-equivalent to a sliding minimal cone in $\Omega _0$ in case the gauge function $h$ satisfying 2.1.

3.1. Approximation of $E\cap \partial B(0,r)$ by rectifiable curves

For any sets $X,Y\subseteq \mathbb{R}^{3}$, any $z\in \mathbb{R}^{3}$ and any $r>0$, we denote by $d_{z,r}$ the normalized local Hausdorff distance defined by

$$\begin{align*} d_{z,r}(X,Y)=&\frac{1}{r}\sup \left\{\operatorname {dist}(x,Y):x\in X\cap \overline{B(z,r)}\right\}\\ &+ \frac{1}{r}\sup \left\{\operatorname {dist}(y,X):y\in Y\cap \overline{B(z,r)}\right\}. \end{align*}$$

It is quite easy to see that for $r>0$,

•

$d_{z,r}(X,Y)\leq d_{z,r}(X,Z)+d_{z,r}(Z,Y)$ if $Z$ is a cone centered at $z$;

•

$d_{z,r}(X,Y)=d_{z,1}(X,Y)$, if $X$ and $Y$ are cones centered at $z$;

•

$d_{z,r}(X,Y)\leq d_{z,r}(X,E)+d_{z,r}(E,Y)$, if $X$ and $Y$ are cones centered at $z$, $E\cap \overline{B(0,r)}\neq \emptyset$, $d_{z,r}(X,E)\ll 1$ and $d_{z,r}(X,E)\ll 1$.

A cone in $\mathbb{R}^{3}$ is called of type $\mathbb{Y}$ if it is the union of three half planes with common boundary line and that make $120^{\circ }$ angles along the boundary line. A cone $Z\subseteq \Omega _0$ is called of type $\mathbb{P}_+$ is if it is a half plane perpendicular to $L_0$; a cone $Z\subseteq \Omega _0$ is called of type $\mathbb{Y}_+$ is if $Z=\Omega _0\cap Y$, where $Y$ is a cone of type $\mathbb{Y}$ perpendicular to $L_0$; for convenient, we will also use the notation $\mathbb{P}_+$, to denote the collection of all of cones of type $\mathbb{P}_+$, and $\mathbb{Y}_{+}$ to denote the collection of all of cones of type $\mathbb{Y}_{+}$.

For any set $E\subseteq \Omega _0$ with $0\in E$, and any $r>0$, we set

$$\begin{equation*} \begin{gathered} \varepsilon _{P}(r)=\inf \{ d_{0,r}(E,Z): Z \in \mathbb{P}_{+}\},\\ \varepsilon _{Y}(r)=\inf \{ d_{0,r}(E,Z): Z \in \mathbb{Y}_{+}\}. \end{gathered} \end{equation*}$$

If $E$ is 2-rectifiable and $\mathcal{H}^2(E)<\infty$, then $E\cap \partial B(0,r)$ is 1-rectifiable and $\mathcal{H}^1(E\cap \partial B(0,r))<\infty$ for $\mathcal{H}^1$-a.e. $r\in (0,\infty )$, we denote by $\mathscr{R}_0$ the collection of such $r$; we now consider the function $u:(0,\infty )\to \mathbb{R}$ which is defined by $u(r)=\mathcal{H}^2(E\cap B(0,r))$, it is quite easy to see that $u$ is nondecreasing, thus $u$ is differentiable for $\mathcal{H}^1$-a.e.; we will denote by $\mathscr{R}$ the set $r\in (0,\infty )$ such that $\mathcal{H}^1(E\cap \partial B(0,r))<\infty$, $u$ is differentiable at $r$, and for any continuous nonnegative function $f$

$$\begin{equation} \lim _{\xi \to 0+}\frac{1}{\xi }\int _{t\in (r-\xi ,r)}\int _{E\cap \partial B(0,t)}f(z)d\mathcal{H}^1(z)dt=\int _{E\cap \partial B(0,r)}f(z)d\mathcal{H}^1(z), {\tag{3.3}} \end{equation}$$

and

$$\begin{equation} \sup _{\xi >0}\frac{1}{\xi }\int _{t\in (r-\xi ,r)}\mathcal{H}^1(E\cap \partial B(0,t))dt<+\infty . {\tag{3.4}} \end{equation}$$

It is not hard to see that $\mathcal{H}^1((0,\infty )\setminus \mathscr{R})=0$, see for example Lemma 4.12 in 5.

Lemma 3.1.

Let $E\subseteq \mathbb{R}^{3}$ be a connected set. If $\mathcal{H}^1(E)<\infty$, then $E$ is path connected.

For a proof, see for example Lemma 3.12 in 8, so we omit it here.

Lemma 3.2.

Let $\mathbb{X}$ be a locally connected and simply connected compact metric space. Let $A$ and $B$ be two connected subsets of $\mathbb{X}$. If $F$ is a closed subset of $\mathbb{X}$ such that $A$ and $B$ are contained in two different connected components of $\mathbb{X}\setminus F$, then there exists a connected closed set $F_0\subseteq F$ such that $A$ and $B$ still lie in two different connected components of $\mathbb{X}\setminus F_0$.

Proof.

See for example 52.III.1 on page 335 in 12, so we omit the proof here.

■

For any $r>0$, we put $\mathfrak{z}_r=(0,0,r)\in \mathbb{R}^{3}$.

Lemma 3.3.

Let $E\subseteq \Omega _0$ be a $2$-rectifiable set with $\mathcal{H}^2(E)<\infty$. Suppose that $0\in E$, and that $E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{P}_+$ at $0$. Then for any $\tau \in (0,\tau _0)$ there exist $\mathfrak{r}=\mathfrak{r}(\tau )>0$ such that, for any $r\in (0,\mathfrak{r})\cap \mathscr{R}_0$ and $\varepsilon >\varepsilon _{P}(r)$, we can find $y_r\in E\cap \partial B(0,r)\setminus L_0$ , $x_{r,1}, x_{r,2}\in E\cap L_0\cap \partial B(0,r)$ and two simple curves $\gamma _{r,1},\gamma _{r,2}\subseteq E\cap \partial B(0,r)$ satisfying that

(1)

$|y_r-\mathfrak{z}_r|\leq \varepsilon r$ and $|x_{r,1}-x_{r,2}|\geq (2-2\varepsilon )r$;

(2)

$\gamma _{r,i}$ joins $y_r$ and $x_{r,i}$, $i=1,2$;

(3)

$\gamma _{r,1}$ and $\gamma _{r,2}$ are disjoint except for point $y_r$.

Proof.

Since $E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{P}_+$ at $0$, for any $\tau \in (0,\tau _0)$, there exist $\varrho >0$, sliding minimal cone $Z$ of type $\mathbb{P}_+$, and a mapping $\Phi :\Omega _0\cap B(0,2\varrho )\to \Omega _0$ which is a homeomorphism between $\Omega _0\cap B(0,2\varrho )$ and $\Phi (\Omega _0\cap B(0,2\varrho ))$ with $\Phi (0)=0$ and $\Phi (\partial \Omega _0\cap B(0,2\varrho ))\subseteq \partial \Omega _0$ such that 3.1 and 3.2 hold. We new take $\mathfrak{r}=\varrho$. Then for any $r\in (0,\mathfrak{r})$,

$$\begin{equation*} \Phi ^{-1}\left[ E\cap \partial B(0,r) \right]\subseteq Z \cap B(0,3\varrho ). \end{equation*}$$

Without loss of generality, we assume that $Z=\{(x_1,0,x_3)\mid x_1\in \mathbb{R}, x_3\geq 0 \}$. Applying Lemma 3.2 with $\mathbb{X}=Z\cap \overline{B(0,3\varrho )}$, $F=\Phi ^{-1}\left[ E\cap \partial B(0,r)\right]$, $A=\{ 0 \}$ and $B=Z\cap \partial B(0,3\varrho )$, we get that there is a connected closed set $F_0\subseteq F$ such that $A$ and $B$ lie in two different connected components of $\mathbb{X}\setminus F_0$, thus $\Phi (F_0) \subseteq E\cap \partial B(0,r)$ is connected. We put $a_1=\{ (x_1,0,0)\mid x_1< 0 \}$ and $a_2=\{ (x_1,0,0)\mid x_1> 0 \}$. Then $F_0\cap a_i\neq \emptyset$, $i=1,2$; otherwise $A$ and $B$ are contained in a same connected component of $\mathbb{X}\setminus F_0$. We take $z_{r,i}\in F_0\cap a_i$, and let $x_{r,i}=\Phi (z_{r,i})\in E\cap \partial B(0,r)$. Then $|x_{r,1}-x_{r,2}|\geq (2-2\varepsilon )r$.

Since $\Phi (F_0)$ is connected and $\mathcal{H}^1(\Phi (F_0))\leq \mathcal{H}^1(E\cap \partial B(0,r))<\infty$, by Lemma 3.1, $\Phi (F_0)$ is path connected. But $\Phi$ is a homeomorphism, we get that $F_0=\Phi ^{-1}(\Phi (F_0))$ is path connected. Let $\gamma$ be a simple curve which joins $z_{r,1}$ and $z_{r,2}$. We see that ${B(\mathfrak{z}_r,\varepsilon r)}\cap \gamma \neq \emptyset$, because $\varepsilon _P(r)<\varepsilon$ and $\mathfrak{z}_r\in Z$ for sliding minimal cone $Z$ of type $\mathbb{P}_{+}$. We take $y_r\in {B(\mathfrak{z}_r,\varepsilon r)}\cap \gamma$.

■

Lemma 3.4.

Let $E\subseteq \Omega _0$ be a $2$-rectifiable set with $\mathcal{H}^2(E)<\infty$. Suppose that $0\in E$, and that $E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{Y}_+$ at $0$. Then for any $\tau \in (0,\tau _0)$ there exist $\mathfrak{r}=\mathfrak{r}(\tau )>0$ such that, for any $r\in (0,\mathfrak{r})\cap \mathscr{R}_0$ and $\varepsilon >\varepsilon _{Y}(r)$, we can find $y_r\in E\cap \partial B(0,r)\setminus L_0$ , $x_{r,1},x_{r,2},x_{r,3}\in E\cap L_0\cap \partial B(0,r)$ and three simple curves $\gamma _{r,1},\gamma _{r,2}, \gamma _{r,3}\subseteq E\cap \partial B(0,r)$ satisfying that

(1)

$|\mathfrak{z}_r-y_r|\leq \varepsilon r$, and there exists $Z\in \mathbb{Y}_{+}$ through 0 such that $\operatorname {dist}(x,Z)\leq \varepsilon r$ for $x\in \gamma _{r,i}$;

(2)

$\gamma _{r,i}$ join $y_r$ and $x_{r,i}$;

(3)

$\gamma _{r,i}$ and $\gamma _{r,j}$ are disjoint except for point $y_r$.

Proof.

Since $E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{Y}_+$ at $0$, for any $\tau \in (0,\tau _0)$, there exist $\tau >0$, $\varrho >0$, sliding minimal cone $Z$ of type $\mathbb{Y}_+$, and a mapping $\Phi :\Omega _0\cap B(0,2\varrho )\to \Omega _0$ which is a homeomorphism between $\Omega _0\cap B(0,2\varrho )$ and $\Phi (\Omega _0\cap B(0,2\varrho ))$ with $\Phi (0)=0$ and $\Phi (\partial \Omega _0\cap B(0,2\varrho ))\subseteq \partial \Omega _0$ such that 3.1 and 3.2 hold. We now take $\mathfrak{r}=\varrho$. Then for any $r\in (0,\mathfrak{r})$,

$$\begin{equation*} \Phi ^{-1}\left[ E\cap \partial B(0,r) \right]\subseteq Z \cap B(0,3\varrho ). \end{equation*}$$

Applying Lemma 3.2 with $\mathbb{X}=Z\cap \overline{B(0,3\varrho )}$, $F=\Phi ^{-1}\left[ E\cap \partial B(0,r)\right]$, $A=\{ 0 \}$ and $B=Z\cap \partial B(0,3\varrho )$, we get that there is a connected closed set $F_0\subseteq F$ such that $A$ and $B$ lie in two different connected components of $\mathbb{X}\setminus F_0$, thus $\Phi (F_0) \subseteq E\cap \partial B(0,r)$ is connected. We let $a_i$, $i=1,2,3$, be the three component of $Z\cap L_0\setminus A$. Then $F_0\cap a_i\neq \emptyset$, $i=1,2,3$; otherwise $A$ and $B$ are contained in a same connected component of $\mathbb{X}\setminus F_0$. We take $z_{r,i}\in F_0\cap a_i$, and let $x_{r,i}=\Phi (z_{r,i})\in E\cap \partial B(0,r)$. Then $|x_{r,1}-x_{r,2}|\geq (\sqrt {3}-2\varepsilon )r$.

Using the same arguments as in the proof of Lemma 3.3, we get that $F_0$ is path connected. We see that $Z$ is of type $\mathbb{Y}_+$, denote by $\ell (Z)$ the spine of $Z$, that is, the half line through 0 and perpendicular to $\partial \Omega _0$. We find a point $y_r\in F_0\cap \ell (Z)$ and curves $\gamma _{r,i}$ satisfying the conditions.

■

3.2. Approximation of rectifiable curves in $\mathbb{S}^2$ by Lipschitz graph

We denote by $\mathbb{S}^2$ the unit sphere in $\mathbb{R}^{3}$. We say that a simple rectifiable curve $\gamma \subseteq \mathbb{S}^2$ is a Lipschitz graph with constant at most $\eta$, if it can be parametrized, after a rotation, by

$$\begin{equation*} z(t)=\left(\sqrt {1-v(t)^2}\cos \theta (t),\sqrt {1-v(t)^2}\sin \theta (t),v(t)\right), \end{equation*}$$

where $v$ is Lipschitz with $\operatorname {Lip}(v)\leq \eta$.

Lemma 3.5.

Let $T\in [\pi /3,2\pi /3]$ be a number, and $\gamma :[0,T]\to \mathbb{S}^2$ a simple rectifiable curve given by

$$\begin{equation*} \gamma (t)=\left(\sqrt {1-v(t)^2}\cos \theta (t),\sqrt {1-v(t)^2}\sin \theta (t), v(t)\right), \end{equation*}$$

where $v$ is a Lipschitz function with $v(0)=v(T)=0$, $\theta$ is a continuous function with $\theta (0)=0$ and $\theta (T)=T$. Then there is a small number $\tau _0\in (0,1)$ such that whenever $|v(t)|\leq \tau _0$, we have that

$$\begin{equation} |v(t)|\leq 10\sqrt {\mathcal{H}^1(\gamma )-T}. {\tag{3.5}} \end{equation}$$

Moreover, there is an $\varepsilon _0>0$ such that 3.5 holds whenever $\mathcal{H}^1(\gamma )-T\leq \varepsilon _0$.

Proof.

We let $A=\gamma (0)=(1,0,0)$, $B=\gamma (T)=(\cos T,\sin T,0)$, and let $C=\gamma (t_0)$ be a point in $\gamma$ such that

$$\begin{equation*} |v(t_0)|=\max \{ |v(t)|: t\in [0,T] \}. \end{equation*}$$

We let $\gamma _i$, $i=1,2$, be two curves such that $\gamma _1(0)=A$, $\gamma _1(1)=C$, $\gamma _2(0)=B$, $\gamma _2(1)=C$ and $\gamma _i\subseteq \gamma$, let $s=\inf \{s\in [0,1]:\gamma _1(s)\in \gamma _2\}$, and put $D=\gamma _1(s)$. By setting $\mathfrak{C}_1$, $\mathfrak{C}_2$ and $\mathfrak{C}_3$ the arcs $AD$, $BD$ and $CD$ respectively, i.e., $AD$ is the arc of the great circle on the unity sphere which joint the points $A$ and $D$. Then we have that

$$\begin{equation*} \mathcal{H}^1(\gamma )\geq \mathcal{H}^1(\gamma _1\cup \gamma _2)\geq \mathcal{H}^1(\mathfrak{C}_1)+\mathcal{H}^1(\mathfrak{C}_2)+\mathcal{H}^1(\mathfrak{C}_3). \end{equation*}$$

We see that $\mathfrak{C}_1\cup \mathfrak{C}_2$ is a simple Lipschitz curve joining $A$ and $B$, and let $\gamma _3:[0,\ell ]\to \mathbb{S}^2$ giving by

$$\begin{equation*} \gamma _3(t)=\left( \sqrt {1-w(t)^2}\cos \theta (t), \sqrt {1-w(t)^2}\sin \theta (t), w(t)\right) \end{equation*}$$

be its parametrization by length. We assume that $\gamma _3(t_1)=D$, then $w'(t)>0$ on $(0,t_1)$, or $w'(t)<0$ on $(0,t_1)$, thus $|w(t)|=\int _0^{t_1}|w'(t)|d t$.

We let the number $\tau _0\in (0,1)$ to be the small number $\tau _1$ in Lemma 7.8 in 5. If $\mathcal{H}^1(\gamma )-T\leq \tau _0$, then we have that

$$\begin{equation*} \int _0^{\ell } |w'(t)|^2dt\leq 14 (\ell -T), \end{equation*}$$

thus

$$\begin{equation*} |w(t_1)|=\int _0^{t_1}|w'(t)|d t\leq \left( t_1 \int _0^{t_1} |w'(t)|^2dt\right)^{1/2}\leq \sqrt {14\ell (\ell -T)}. \end{equation*}$$

We get so that

$$\begin{equation*} \begin{aligned} |v(t_0)|&\leq \mathcal{H}^1(\mathfrak{C}_3)+|w(t_1)|\leq (\mathcal{H}^1(\gamma )-\ell )+\sqrt {14\ell (\ell -T)}\\ &\leq \sqrt {14\mathcal{H}^1(\gamma )(\mathcal{H}^1(\gamma )-T)}\leq 10 \sqrt {\mathcal{H}^1(\gamma )-T}. \end{aligned} \end{equation*}$$

If $\mathcal{H}^1(\gamma )-T> \tau _0$, then $v(t)\leq \tau _0\leq 10\sqrt {\tau _0}\leq 10 \sqrt {\mathcal{H}^1(\gamma )-T}$.

■

Lemma 3.6.

Let $a$ and $b$ be two points in $\Omega _0\cap \partial B(0,1)$ satisfying

$$\begin{equation*} \frac{\pi }{3}\leq \operatorname {dist}_{\mathbb{S}^2}(a,b)\leq \frac{2\pi }{3}. \end{equation*}$$

Let $\gamma$ be a simple rectifiable curve in $\Omega _0\cap \partial B(0,1)$ which joins $a$ and $b$, and satisfies

$$\begin{equation*} \operatorname {length}(\gamma )\leq \operatorname {dist}_{\mathbb{S}^2}(a,b)+\tau _0, \end{equation*}$$

where $\tau _0>0$ is as in Lemma 3.5. Then there is a constant $C>0$ such that, for any $\eta >0$, we can find a simple curve $\gamma _{\ast }$ in $\Omega _0\cap \partial B(0,1)$ which is a Lipschitz graph with constant at most $\eta$ joining $a$ and $b$, and satisfies that

$$\begin{equation*} \mathcal{H}^1(\gamma _{\ast }\setminus \gamma )\leq \mathcal{H}^1(\gamma \setminus \gamma _{\ast })\leq C\eta ^{-2}(\operatorname {length}(\gamma )-\operatorname {dist}_{\mathbb{S}^2}(a,b)). \end{equation*}$$

Moreover, if we denote by $\Gamma _{a,b}$ the geodesic joining $a$ and $b$, then we can assume that

$$\begin{equation} \operatorname {dist}(x,\Gamma _{a,b})\leq \eta \operatorname {dist}(x,\{a,b\}), \forall x\in \gamma _{\ast }. {\tag{3.6}} \end{equation}$$

The proof will be the same as in 5, p.875-p.878, so we omit it.

3.3. Comparison surfaces

Let $\Gamma$ be a Lipschitz curve in $\mathbb{S}^2$. We assume for simplicity that its extremities $a$ and $b$ lie in the horizontal plane. Let us assume that $a=(1,0,0)$ and $b=(\cos T, \sin T, 0)$ for some $T\in [\pi /3,2\pi /3]$. We also assume that $\Gamma$ is a Lipschitz graph with constant at most $\eta$, i.e. there is a Lipschitz function $s:[0,T]\to \mathbb{R}$ with $s(0)=s(T)=0$ and $\operatorname {Lip}(s)\leq \eta$, such that $\Gamma$ is parametrized by

$$\begin{equation*} z(t)=(w(t)\cos t,w(t)\sin t, s(t)) \text{ for }t\in [0,T], \end{equation*}$$

where $w(t)=(1-|s(t)|^2)^{1/2}$.

We set

$$\begin{equation*} D_T=\{ (r\cos t, r\sin t)|\mid 0< r <1 ,0 <t < T\}, \end{equation*}$$

and consider the function $v:\overline{D}_T\to \mathbb{R}$ defined by

$$\begin{equation*} v(r\cos t,r\sin t)=\frac{rs(t)}{w(t)} \text{ for } 0\leq r\leq 1 \text{ and } 0\leq t\leq T. \end{equation*}$$

For any function $f:\overline{D}_T\to \mathbb{R}$, we denote by $\Sigma _f$ the graphs of $f$ over $\overline{D}_T$.

Lemma 3.7.

There is a universal constant $\kappa >0$ such that we can find a Lipschitz function $u$ on $\overline{D}_T$ satisfying that

$$\begin{equation} \begin{gathered} \operatorname {Lip}(u)\leq C\eta ,\\ u(r,0)=u(r\cos T,r\sin T)=0, \text{ for }0\leq r\leq 1,\\ u(r\cos t,r\sin t)=v(r\cos t,r\sin t)\text{ for }1-2\kappa \leq r\leq 1, 0\leq t\leq T,\\ u(r\cos t,r\sin t)=0,\text{ for }0\leq r\leq 2\kappa , 0\leq t\leq T \end{gathered} {\tag{3.7}} \end{equation}$$

and

$$\begin{equation} \mathcal{H}^2(\Sigma _v)-\mathcal{H}^2(\Sigma _u)\geq 10^{-4}(\mathcal{H}^1(\Gamma )-T). {\tag{3.8}} \end{equation}$$

The proof is the same as Lemma 8.8 in 5, we omit it here.

3.4. Retractions

In this subsection, we assume that $E\subseteq \Omega _0$ is a $2$-rectifiable set satisfying that

(a)

$\mathcal{H}^2(E)<\infty$, $0\in E$,

(b)

$E$ is locally $(\Omega _0,L_0,h)$ sliding almost minimal at $0$,

(c)

$E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{P}_+$ or $\mathbb{Y}_+$.

For any $r>0$, we let $\varepsilon (r)=\varepsilon _P(r)$ if $E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{P}_+$, and let $\varepsilon (r)=\varepsilon _Y(r)$, if $E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{Y}_+$. Recall that $\mathscr{R}_0$ is denoted by the collection of radii $r\in (0,\infty )$ such that $\mathcal{H}^1(E\cap \partial B(0,r))<\infty$. For any $r\in (0,\mathfrak{r})\cap \mathscr{R}_0$, we will discuss two situations: first, if $Z$ is a sliding minimal cone of type $\mathbb{P}_+$, we put $\mathcal{X}_r=\{ x_{r,1},x_{r,2} \}$, where $x_{r,1}$ and $x_{r,2}$ are considered as in Lemma 3.3. Second, if $Z$ is a sliding minimal cone of type $\mathbb{Y}_+$, we put $\mathcal{X}_r=\{ x_{r,1}, x_{r,2},x_{r,3} \}$, where $x_{r,1}$, $x_{r,2}$ and $x_{r,3}$ are consider as in Lemma 3.4. We see that $\mathcal{X}_r\subseteq E\cap \partial B(0,r)\cap L_0$.

We take $y_r$ as in Lemma 3.3 or Lemma 3.4. For any $x\in \mathcal{X}_r$, we let $\gamma _x$ be the curve joining $x$ and $y_r$ which is considered as in Lemma 3.3 or Lemma 3.4, put $\Gamma _x=\boldsymbol{\mu }_{1/r}(\gamma _x)$. Then by Lemma 3.6, there is a curve $\Gamma _{x,\ast }$ on $\Omega _0\cap \partial B(0,1)$ joining $\boldsymbol{\mu }_{1/r}(x)$ and $\boldsymbol{\mu }_{1/r}(y_r)$ which is a Lipschitz graph with constant at most $\eta \leq 10^{-6}$. Let $\mathfrak{C}_x$ be the arc on $\partial B(0,1)$ joining $\boldsymbol{\mu }_{1/r}(x)$ and $\boldsymbol{\mu }_{1/r}(y_r)$, let $D_{x}$ and $\mathcal{M}_x$ be the cone over $\mathfrak{C}_x$ and $\Gamma _{x,\ast }$ respectively. By Lemma 3.7, we can find Lipschitz graph $\Sigma _x$ corresponding to $\mathcal{M}_x$ such that 3.7 and 3.8 hold, that is,

$$\begin{equation} \begin{gathered} \Sigma _x\cap B(0,2\kappa )=D_x\cap B(0,2\kappa ),\\ \Sigma _x\cap \overline{B(0,1)}\setminus B(0,1-2\kappa )= \mathcal{M}_x\cap \overline{B(0,1)}\setminus B(0,1-2\kappa ),\\ \mathcal{H}^2(\mathcal{M}_x\cap B(0,1))-\mathcal{H}^2(\Sigma _x)\geq 10^{-4}(\mathcal{H}^1(\Gamma _{x,\ast })-\mathcal{H}^1(\mathfrak{C}_x)). \end{gathered} {\tag{3.9}} \end{equation}$$

We put

$$\begin{equation} X=\bigcup _{x\in \mathcal{X}_r} D_{x},\ \Gamma =\bigcup _{x\in \mathcal{X}_r} \Gamma _{x},\ \Gamma _{\ast }=\bigcup _{x\in \mathcal{X}_r} \Gamma _{x,\ast },\ \mathfrak{C}=\bigcup _{x\in \mathcal{X}_r} \mathfrak{C}_x,\ \mathcal{M}=\bigcup _{x\in \mathcal{X}_r} \mathcal{M}_x,\ \Sigma =\bigcup _{x\in \mathcal{X}_r} \Sigma _x. {\tag{3.10}} \end{equation}$$

From 3.9, we see that

$$\begin{equation} \mathcal{H}^2(\mathcal{M}\cap B(0,1))-\mathcal{H}^2(\Sigma )\geq 10^{-4}\left( \mathcal{H}^1(\Gamma _{\ast })-\mathcal{H}^1(\mathfrak{C})\right). {\tag{3.11}} \end{equation}$$

By Lemma 3.5 and Lemma 3.6, we have that

$$\begin{equation*} d_H(\mathfrak{C}_x,\Gamma _{x,\ast })\leq 10 \left(\mathcal{H}^1(\Gamma _{x,\ast })-\mathcal{H}^1(\mathfrak{C}_x)\right)^{1/2}\leq 10 \left(\mathcal{H}^1(\Gamma _{\ast })-\mathcal{H}^1(\mathfrak{C})\right)^{1/2} \end{equation*}$$

and

$$\begin{equation} d_{0,1}(X,\mathcal{M})\leq d_H(\mathfrak{C},\Gamma _{\ast })\leq \max _{x\in \mathcal{X}_r} d_H(\mathfrak{C}_x,\Gamma _{x,\ast })\leq 10 \left(\mathcal{H}^1(\Gamma _{\ast })-\mathcal{H}^1(\mathfrak{C})\right)^{1/2}. {\tag{3.12}} \end{equation}$$

For any $r \in (0,\mathfrak{r})\cap \mathscr{R}_0$, we put $j(r)=r^{-1}\mathcal{H}^1 (E\cap \partial B(0,r))-\mathcal{H}^1(\mathfrak{C})$, and denote by $\mathscr{R}_1$ the set $\{ r\in (0,\mathfrak{r})\cap \mathscr{R}:j(r) \leq \tau _0\}$, where $\tau _0$ is the small number considered as in Lemma 3.5, $\mathscr{R}\subseteq \mathscr{R}_0$ is defined in 3.3 and 3.4. Then 3.12 implies that

$$\begin{equation} d_{0,r}(X,M)\leq 10 j(r)^{1/2}. {\tag{3.13}} \end{equation}$$

Lemma 3.8.

If $\varepsilon (r)<1/2$, then for any $\varepsilon \in (\varepsilon (r),1/2)$, there is a sliding minimal cone $Z=Z_r$ such that

$$\begin{equation*} d_{0,1}(X,Z)\leq 4 \varepsilon . \end{equation*}$$

Moreover, we have that

$$\begin{equation*} d_{0,r}(E,X)\leq 5 \varepsilon (r). \end{equation*}$$

Proof.

There exists sliding minimal cone $Z$ such that $d_{0,r}(E,Z)\leq \varepsilon$, thus for any $x\in \mathcal{X}_r$, there is $x_z\in Z\cap (L_0\cap \partial B_r)$ satisfying that $|x-x_z|\leq 2 \varepsilon r$. We get so that

$$\begin{equation*} d_H([x,y_r],[x_z,\mathfrak{z}_r])\leq 2 \varepsilon r. \end{equation*}$$

Since $\operatorname {dist}(0,[x,y_r])>r/2$ for any $x\in \mathcal{X}_r$, we have that

$$\begin{equation*} d_H(X\cap B(0,r/2) ,Z\cap B(0,r/2))\leq 2 \varepsilon r. \end{equation*}$$

Thus

$$\begin{equation*} d_{0,1}(X,Z)=d_{0,r/2}(X,Z)\leq 4 \varepsilon , \end{equation*}$$

and

$$\begin{equation*} d_{0,r}(E,X)\leq d_{0,r}(E,Z)+d_{0,r}(Z,X)\leq 5 \varepsilon . \end{equation*}$$ ■

Lemma 3.9.

Let $0<\delta ,\varepsilon < 1/2$ be positive numbers. Let $v_1,v_2,v_3\in \mathbb{R}^{3}$ be three unit vectors.

•

If $|\langle v_2,v_i \rangle |\leq \delta$ for $i=1,3$, then for any $v\in \mathbb{R}^{3}$ with $\langle v,v_2 \rangle =0$ and $\operatorname {dist}(v,{\mathrm{span}}\{v_1,v_2\})\leq \varepsilon |v|$, we have that$$\begin{equation*} |\langle v,v_3 \rangle -\langle v_1,v_3 \rangle \langle v,v_1 \rangle |\leq (\varepsilon +\delta )|v|, \text{ and }|\langle v,v_1 \rangle |\geq (1-\varepsilon -\delta )|v|. \end{equation*}$$

•

If $\langle v_1,v_3 \rangle <1$ and $0<64\delta <1-\langle v_1,v_3 \rangle$, then for any $w_1,w_3\in \mathbb{R}^{3}$ with $\langle v_i,w_i \rangle \geq (1-\delta ) |w_i|$, $i=1,3$, we have that$$\begin{equation} |w_1|+|w_3|\leq 2\cdot \left(1-\langle v_1,v_3 \rangle \right)^{-1/2}|w_1-w_3|. {\tag{3.14}} \end{equation}$$

Proof.

We write $v=v^{\perp }+\lambda _1 v_1+\lambda _2 v_2$, $\lambda _i\in \mathbb{R}$, $\langle v^{\perp },v_i \rangle =0$. Since $\langle v,v_2 \rangle =0$, we have that $\lambda _2=-\lambda _1\langle v_1,v_2 \rangle$, thus

$$\begin{equation*} \lambda _1=\frac{\langle v,v_1 \rangle }{1-\langle v_1,v_2 \rangle ^2},\ \lambda _2=-\frac{\langle v,v_1 \rangle \langle v_1,v_2 \rangle }{1-\langle v_1,v_2 \rangle ^2}, \end{equation*}$$

we get so that

$$\begin{equation} v=v^{\perp }+\frac{\langle v,v_1 \rangle v_1- \langle v,v_1 \rangle \langle v_1,v_2 \rangle v_2}{1-\langle v_1,v_2 \rangle ^2}, {\tag{3.15}} \end{equation}$$

and then

$$\begin{equation*} \langle v,v_3 \rangle =\langle v^{\perp },v_3 \rangle +\frac{\langle v_1,v_3 \rangle -\langle v_2,v_3 \rangle \langle v_1,v_2 \rangle }{1-\langle v_1.v_2 \rangle ^2}\langle v,v_1 \rangle , \end{equation*}$$

thus

$$\begin{equation*} |\langle v,v_3 \rangle -\langle v_1,v_3 \rangle \langle v,v_1 \rangle |\leq \varepsilon |v|+\frac{\delta ^2+\delta }{1-\delta ^2}|v|\leq (\varepsilon +2\delta )|v|. \end{equation*}$$

We get also, from 3.15, that

$$\begin{equation*} |v|\leq |v^{\perp }|+\frac{1+|\langle v_1.v_2 \rangle |}{1-\langle v_1,v_2 \rangle ^2}|\langle v,v_1 \rangle |\leq \varepsilon |v|+\frac{1}{1-\delta }|\langle v,v_1 \rangle |, \end{equation*}$$

thus

$$\begin{equation*} |\langle v,v_1 \rangle |\geq (1-\varepsilon )(1-\delta ) |v|\geq (1-\varepsilon -\delta )|v|. \end{equation*}$$

We can certainly assume $w_i\neq 0$, otherwise the inequality 3.14 will be trivial true. Since $\langle v_i,w_i \rangle \geq (1-\delta )|w_i|$, we have that $\langle v_i,w_i/|w_i| \rangle \geq 1-\delta$, and

$$\begin{equation*} \Big |v_i-w_i/|w_i|\Big |^2=2-2 \langle v_i,w_i/|w_i| \rangle \leq 2 \delta , \end{equation*}$$

we have so that

$$\begin{equation*} \begin{aligned} \left| \frac{w_1}{|w_1|}-\frac{w_3}{|w_3|}\right|^2&=\left| \left(\frac{w_1}{|w_1|}-v_1\right)-\left(\frac{w_3}{|w_3|}-v_3\right)+ (v_1-v_3)\right|^2\\ &\geq |v_1-v_3|^2-2|v_1-v_3|\left(\left|\frac{w_1}{|w_1|}-v_1\right|+ \left|\frac{w_3}{|w_3|}-v_3\right|\right)\\ &\geq | v_1-v_3|^2-4\sqrt {2 \delta }|v_1-v_3|, \end{aligned} \end{equation*}$$

and

$$\begin{equation*} \langle w_1,w_3 \rangle =|w_1| |w_3|\left\langle \frac{w_1}{|w_1|},\frac{w_3}{|w_3|} \right\rangle \leq |w_1| |w_3| \left(\langle v_1,v_3\rangle +2\sqrt { 2\delta }|v_1-v_3|\right). \end{equation*}$$

Thus

$$\begin{equation*} |w_1-w_3|^2\geq |w_1|^2+|w_3|^2-2|w_1| |w_3|\left(\langle v_1,v_3\rangle +2\sqrt {2 \delta }|v_1-v_3|\right)\geq (1-s)(|w_1|+|w_3|)^2, \end{equation*}$$

where $s=\frac{1}{2}(1+\langle v_1,v_3 \rangle + 2\sqrt {2 \delta }|v_1-v_3|)\leq \frac{1}{2}(1+\langle v_1,v_3 \rangle )+\frac{1}{4}(1-\langle v_1,v_3 \rangle )$. Hence

$$\begin{equation*} |w_1|+|w_3|\leq (1-s)^{-1/2}|w_1-w_3|\leq 2(1-\langle v_1,v_3 \rangle )^{-1/2}|w_1-w_3|. \end{equation*}$$ ■

Lemma 3.10.

For any $r\in (0,\mathfrak{r})\cap \mathscr{R}_1$, we let $\Sigma$ be as in 3.10. Then there is a universal constant $C>0$ and a Lipschitz mapping $p:\Omega _0\to \Sigma$ with $\operatorname {Lip}(p)\leq C$, such that $p(z)\in L$ for $z\in L$, and that $p(z)=z$ for $z\in \Sigma$.

Proof.

We see from 3.9 that

$$\begin{equation*} \Sigma \cap \overline{B(0,1)}\setminus B(0,1-2\kappa )=\mathcal{M}\cap \overline{B(0,1)}\setminus B(0,1-2\kappa )), \end{equation*}$$

and

$$\begin{equation*} \Sigma \cap B(0,2\kappa )=X\cap B(0,2\kappa ). \end{equation*}$$

For any $z\in \Omega _0\setminus \{0\}$, we denote by $\ell (z)$ the line which goes through $0$ and $z$, and denote $\partial D_{x}=\ell (x)\cup \ell (y_r)$. Let $\sigma \in (0,10^{-3})$ be fixed. We put

$$\begin{equation*} \begin{gathered} R^x=\{ z\in \Omega _0\mid \operatorname {dist}(z,D_{x})\leq \sigma \operatorname {dist}(z,\partial D_{x}) \}, \\ R_1^x=\{ z\in \Omega _0\mid \operatorname {dist}(z,D_{x})\leq \sigma \operatorname {dist}(z,\ell (y_r)) \}, \end{gathered} \end{equation*}$$

and

$$\begin{equation*} R=\bigcup _{x\in \mathcal{X}_r}R^x, R_1=\bigcup _{x\in \mathcal{X}_r}R_1^x. \end{equation*}$$

Then we see that $R^x\subseteq R_1^x$, and that both of them are cones,

$$\begin{equation*} R^{x_i}\cap R^{x_j}=R_1^{x_i}\cap R_1^{x_j}=\ell (y_r) \text{ for } x_i,x_j\in \mathcal{X}_r, x_i\neq x_j. \end{equation*}$$

Since $\Gamma _{x,\ast }$ is a Lipschitz graph with constant at most $\eta$ such that 3.6 hold, we have that

$$\begin{equation*} \mathcal{M}_x\subseteq R^x \text{ and }\Sigma _{x}\subseteq R^x, \end{equation*}$$

when $\eta$ small enough.

We will construct a Lipschitz retraction $p_0:\Omega _0\to R_1$ such that $p_0(z)=z$ for $z\in R_1$, $p_0(z)\in L_0$ for $z\in L_0$, and $\operatorname {Lip}(p_0)\leq 25$. We now distinguish two cases, depending on cardinality of $\mathcal{X}_r$.

Case 1.

$\operatorname {card}(\mathcal{X}_r)=2$. We assume that $\mathcal{X}_r=\{ x_1,x_2\}$. Then $|y_r|=|x_1|=|x_2|=r$, and

$$\begin{equation*} 0\leq \langle x_1,x_2 \rangle +r^2\leq 2\varepsilon ^2r^2. \end{equation*}$$

Since $|y_r-\mathfrak{z}_r|\leq \varepsilon r$, we have that $|\langle y_r,x \rangle |\leq \varepsilon r^2$ for any $x\in L\cap \partial B(0,r)$.

We now let $e_1$ and $e_2$ be two unit vectors in $L_0$ such that $\langle x_1,e_1 \rangle =\langle x_2,e_1 \rangle \geq 0$ and $e_2=-e_1$. Then

$$\begin{equation*} 0\leq \langle x_i,e_1 \rangle \leq \varepsilon r,\ i\in \{1,2\}. \end{equation*}$$

We let $\Omega _1'$ and $\Omega _2'$ be the two connected components of $\Omega _0\setminus (\cup _{i} D_{x_i})$ such that $e_i\in \Omega _i'$. We put $\Omega _i=\Omega _i'\setminus R_1$. We claim that

$$\begin{equation} |\langle z_1-z_2, e_i\rangle |\leq 10(\sigma +\varepsilon )|z_1-z_2| {\tag{3.16}} \end{equation}$$

whenever $z_1,z_2\in \partial \Omega _i$, $z_1\neq z_2$, $i\in \{1,2\}$.

Without loss of generality, we assume $z_1,z_2\in \partial \Omega _1$, because for another case we will use the same treatment. We see that

$$\begin{equation*} \operatorname {dist}(z_i,D_{x_j})=\sigma \operatorname {dist}(z_i,\ell (y_r)). \end{equation*}$$

Figure 1.

The angle between $z_1-z_2$ and $D_{x}$ is small.

(1)

In case $z_1,z_2\in \partial R_1^{x_i}\cap \overline{\Omega }_1$, without loss of generality, we assume that $z_1,z_2\in \partial R_1^{x_1}\cap \overline{\Omega }_1$. We let $\widetilde{z}_i\in D_{x_1}$ and $z_i'\in \ell (y_r)$ be such that$$\begin{equation*} |z_i-\widetilde{z}_i|=\operatorname {dist}(z_i,D_{x_1}),\ |z_i-z_i'|=\operatorname {dist}(z_i,\ell (y_r)),\ i\in \{1,2\}. \end{equation*}$$

We put$$\begin{equation*} w_1=z_1-\widetilde{z}_1+\widetilde{z}_2,\ w_2=z_1-z_1'+z_2', \end{equation*}$$

then we get that $z_1-z_2=(z_1-w_2)+(w_2-z_2)$. Moreover, we have that $z_1-w_2$ is perpendicular to $w_2-z_2$ and parallel to $y_r$. Thus $|w_2-z_2|\leq |z_1-z_2|$, $|z_1-w_2|\leq |z_1-z_2|$ and$$\begin{equation*} \operatorname {dist}(w_2-z_2,{\mathrm{span}}\{x_1,y_r\})=\sigma |w_2-z_2|. \end{equation*}$$

Applying Lemma 3.9, we get that$$\begin{equation*} |\langle z_1-w_2,e_1\rangle | \leq \varepsilon |z_1-w_2 |\text{ and } |\langle w_2-z_2,e_1 \rangle |\leq \left( \sigma +3\varepsilon \right)|w_2-z_2|, \end{equation*}$$

thus$$\begin{equation*} |\langle z_1-z_2,e_1 \rangle |\leq |\langle z_1-w_2,e_1 \rangle |+|\langle w_2-z_2,e_1 \rangle |\leq \left( \sigma +4\varepsilon \right)|z_1-z_2|. \end{equation*}$$

(2)

In case $z_1\in \partial R^{x_1}\cap \overline{\Omega }_1$, $z_2\in \partial R^{x_2}\cap \overline{\Omega }_1$. We let $\widetilde{z}_i\in D_{x_i}$ and $z_{i}'\in \ell (y_r)$ be such that$$\begin{equation*} |z_i-\widetilde{z}_i|=\operatorname {dist}(z_i,D_{x_i}),\ |z_i-z_i'|=\operatorname {dist}(z_i,\ell (y_r)),\ i=1,2. \end{equation*}$$

Then by Lemma 3.9, we have that$$\begin{equation*} \left\langle z_i-z_i',\frac{x_i}{|x_i|} \right\rangle \geq (1-\sigma -\varepsilon )|z_i-z_i'|,\ i=1,2. \end{equation*}$$

Since $z_1-z_2=(z_1-z_1')+(z_2'-z_2)+(z_1'-z_2')$, we have that$$\begin{equation*} |\langle z_1'-z_2',e_1 \rangle |\leq \varepsilon |z_1'-z_2'|\leq \varepsilon |z_1-z_2| \end{equation*}$$

and$$\begin{equation*} |\langle z_i-z_i',e_1 \rangle |\leq \left( \sigma +\varepsilon \right)|z_i-z_i'|, \end{equation*}$$

we get so that$$\begin{equation} \begin{aligned} |\langle z_1-z_2,e_1 \rangle |&\leq |\langle z_1-z_1',e_1 \rangle |+|\langle z_2'-z_2,e_1 \rangle |+|\langle z_1'-z_2',e \rangle |\\ &\leq 2\cdot \left( \sigma +\varepsilon \right) (|z_1-z_1'|+|z_2-z_2'|)+\varepsilon |z_1-z_2|. \end{aligned} {\tag{3.17}} \end{equation}$$

Since $z_1'-z_2'$ is perpendicular to $z_1-z_1'$ and $z_2-z_2'$, $\left\langle x_1/|x_1|,x_2/|x_2| \right\rangle \leq -1+2\varepsilon ^2$ and$$\begin{equation*} \left\langle z_i-z_i',\frac{x_i}{|x_i|} \right\rangle \geq (1-\sigma -\varepsilon )|z_i-z_i'|,\ i=1,2, \end{equation*}$$

by 3.14 in Lemma 3.9, we get that$$\begin{equation*} |z_1-z_1'|+|z_2-z_2'|\leq 2\cdot \left( 2-2\varepsilon ^2 \right)^{-1/2}|(z_1-z_1')-(z_2-z_2')|\leq 4 |z_1-z_2|. \end{equation*}$$

Thus inequality 3.17 implies that$$\begin{equation*} |\langle z_1-z_2,e_1 \rangle |\leq (8\sigma +9\varepsilon ) |z_1-z_2|\leq 10(\sigma +\varepsilon )|z_1-z_2|, \end{equation*}$$

and we finished the proof of the claim 3.16.

We now define $p_0:\Omega _0\to R_1$ as follows: for any $z\in \Omega _i$, we let $p_0(z)$ be the unique point in $\partial \Omega _i$ such that $p_0(z)-z$ parallels $e$; and for any $z\in R_1$, we let $p_0(z)=z$. Since $p_0(z)-z$ parallels $e$, we see that $p_0(L_0)\subseteq L_0$. We will check that

$$\begin{equation*} p_0\text{ is Lipschitz with }\operatorname {Lip}(p_0)\leq \frac{2}{1-10(\sigma +\varepsilon )}. \end{equation*}$$

Indeed, for any $z_1,z_2\in \Omega _0$, we put

$$\begin{equation*} p_0(z_i)=z_i+t_ie,\ t_i\in \mathbb{R}, \end{equation*}$$

then

$$\begin{equation*} \begin{aligned} |t_1-t_2|&=|\langle (t_1-t_2)e,e \rangle | \leq |\langle p_0(z_1)-p_0(z_2),e \rangle |+|\langle z_1-z_2,e \rangle |\\ &\leq 10(\sigma +\varepsilon )|p_0(z_1)-p_0(z_2)|+|z_1-z_2|, \end{aligned} \end{equation*}$$

and

$$\begin{equation*} |p_0(z_1)-p_0(z_2)|\leq |z_1-z_2|+|t_1-t_2|\leq 10(\sigma +\varepsilon )|p_0(z_1)-p_0(z_2)|+2|z_1-z_2|, \end{equation*}$$

thus

$$\begin{equation*} |p_0(z_1)-p_0(z_2)|\leq \frac{2}{1-10(\sigma +\varepsilon )}|z_1-z_2|. \end{equation*}$$

Case 2.

$\operatorname {card}(\mathcal{X}_r)=3$. We assume that $\mathcal{X}_r=\{ x_1,x_2,x_3\}$, then

$$\begin{equation*} |\langle x_i,y_r \rangle |\leq \varepsilon r^2, \left(-\frac{1}{2}- \sqrt {3}\varepsilon \right) r^2\leq \langle x_i,x_j \rangle \leq \left(-\frac{1}{2}+ 2\varepsilon \right) r^2. \end{equation*}$$

We put

$$\begin{equation*} e_1=\frac{x_{2}+x_{3}}{|x_{2}+x_{3}|}, e_2=\frac{x_{1}+x_{3}}{|x_{1}+x_{3}|}, e_3=\frac{x_{2}+x_{1}}{|x_{2}+x_{1}|}, \end{equation*}$$

and let $\Omega _1'$, $\Omega _2'$ and $\Omega _3'$ be the three connected components of $\Omega _0\setminus (\cup _{i} D_{x_i})$ such that $e_i\in \Omega _i'$. By putting $\Omega _i=\Omega _i'\setminus R_1$, we claim that

$$\begin{equation} |\langle z_1-z_2,e_i\rangle |\leq \left( \frac{9}{10}+5\sigma +5\varepsilon \right)|z_1-z_2| {\tag{3.18}} \end{equation}$$

whenever $z_1,z_2\in \partial \Omega _i$, $z_1\neq z_2$, $i\in \{1,2,3\}$.

Indeed, we only need to check the case $z_1,z_2\in \partial \Omega _1$, and the other two cases will be the same. Since $(-1/2-\sqrt {3}\varepsilon )r^2 \leq \langle x_i,x_j \rangle \leq (-1/2+ 2\varepsilon )r^2$, we have that $(1/2-\varepsilon )r\leq \langle x_i,e_1 \rangle \leq (1/2+\varepsilon )r$ for $i=2,3$. In case $z_1,z_2\in \partial R_1^{x_2}\cap \overline{\Omega }_1$ or $z_1,z_2\in \partial R_1^{x_3}\cap \overline{\Omega }_1$. Let us assume that $z_1,z_2\in \partial R_1^{x_2}\cap \overline{\Omega }_1$. Let $\widetilde{z}_i\in D_{x_2}$ and $z_i'\in \ell (y_r)$ be such that

$$\begin{equation*} |z_i-\widetilde{z}_i|=\operatorname {dist}(z_i,D_{x_2}),\ |z_i-z_i'|=\operatorname {dist}(z_i,\ell (y_r)),\ i=1,2. \end{equation*}$$

We put

$$\begin{equation*} w_1=z_1-\widetilde{z}_1+\widetilde{z}_2,\ w_2=z_1-z_1'+z_2', \end{equation*}$$

then we get that $z_1-w_2$ is perpendicular to $w_2-z_2$ and parallel to $y_r$. Since $z_1-z_2=(z_1-w_2)+(w_2-z_2)$, we have that $|w_2-z_2|\leq |z_1-z_2|$, $|z_1-w_2|\leq |z_1-z_2|$ and

$$\begin{equation*} \operatorname {dist}(w_2-z_2,{\mathrm{span}}\{x_1,y_r\})=\sigma |w_2-z_2|. \end{equation*}$$

We apply Lemma 3.9 to get that $|\langle z_1-w_2,e_1\rangle |\leq \varepsilon |z_1-w_2 |$ and

$$\begin{equation*} |\langle w_2-z_2,e_1 \rangle |\leq \left( \frac{1}{2}+\varepsilon +\sigma +\varepsilon \right)|w_2-z_2|, \end{equation*}$$

thus

$$\begin{equation*} |\langle z_1-z_2,e_1 \rangle |\leq |\langle z_1-w_2,e_1 \rangle |+|\langle w_2-z_2,e_1 \rangle |\leq \left( \frac{1}{2}+ \sigma +3\varepsilon \right)|z_1-z_2|. \end{equation*}$$

If $z_1\in \partial R^{x_2}\cap \Omega _1$, $z_2\in \partial R^{x_3}\cap \Omega _1$, we let $\widetilde{z}_i\in D_{x_i}$ and $z_{i}'\in \ell (y_r)$ be such that

$$\begin{equation*} |z_1-\widetilde{z}_1|=\operatorname {dist}(z_1,D_{x_2}),\ |z_2-\widetilde{z}_2|=\operatorname {dist}(z_2,D_{x_3}) \end{equation*}$$

and

$$\begin{equation*} |z_i-z_i'|=\operatorname {dist}(z_i,\ell (y_r)),\ i=1,2, \end{equation*}$$

then $z_1'-z_2'$ is perpendicular to $z_1-z_1'$ and $z_2-z_2'$, and we get that $|(z_1-z_1')-(z_2-z_2')|\leq |z_1-z_2|$, since $z_1-z_2=(z_1-z_1')-(z_2-z_2')+(z_1'-z_2')$. We see that $|\langle z_1'-z_2',e_1 \rangle |\leq \varepsilon |z_1'-z_2'|\leq \varepsilon |z_1-z_2|$ and

$$\begin{equation*} |\langle z_i-z_i',e_1 \rangle |\leq \left(\frac{1}{2}+\varepsilon + \sigma +\varepsilon \right)|z_i-z_i'|, \end{equation*}$$

thus

$$\begin{equation} \begin{aligned} |\langle z_1-z_2,e_1 \rangle |&\leq |\langle z_1-z_1',e_1 \rangle |+|\langle z_2-z_2',e_1 \rangle |+|\langle z_1'-z_2',e \rangle |\\ &\leq \left( \frac{1}{2}+ \sigma +2\varepsilon \right) (|z_1-z_1'|+|z_2-z_2'|)+\varepsilon |z_1-z_2|.\\ \end{aligned} {\tag{3.19}} \end{equation}$$

By Lemma 3.9, we get that

$$\begin{equation*} \left\langle z_1-z_1',\frac{x_2}{|x_2|}\right\rangle \geq (1-\sigma -\varepsilon )|z_1-z_1'| \text{ and } \left\langle z_2-z_2',\frac{x_3}{|x_3|}\right\rangle \geq (1-\sigma -\varepsilon )|z_2-z_2'|, \end{equation*}$$

and applying Lemma 3.9 again with $\langle x_2/|x_2|,x_3/|x_3| \rangle \leq -1/2+2\varepsilon$, we have that

$$\begin{equation*} |z_1-z_1'|+|z_2-z_2'|\leq 2\left(3/2-2 \varepsilon \right)^{-1/2}|(z_1-z_1')-(z_2-z_2')| \leq \frac{9}{5} |z_1-z_2| \end{equation*}$$

We get, from 3.19, that

$$\begin{equation*} |\langle z_1-z_2,e_1 \rangle |\leq \left(\frac{9}{10}+2 \sigma + 5 \varepsilon \right)|z_1-z_2|, \end{equation*}$$

and we proved our claim 3.18.

For any $z\in \Omega _i$, we now let $p_0(z)$ be the unique point in $\partial \Omega _i$ such that $p_0(z)-z$ parallels $e_i$; and for $z\in R_1$, we let $p_0(z)=z$. Then $p_0(L_0)\subseteq L_0$. We will check that

$$\begin{equation*} p_0\text{ is Lipschitz with }\operatorname {Lip}(p_0)\leq 25. \end{equation*}$$

Indeed, for any $z_1,z_2\in \Omega _i$, we put

$$\begin{equation*} p_0(z_j)=z_j+t_je_i,\ t_i\in \mathbb{R},\ j=1,2, \end{equation*}$$

then

$$\begin{equation*} \begin{aligned} |t_1-t_2|&=|\langle (t_1-t_2)e_i,e_i \rangle | \leq |\langle p_0(z_1)-p_0(z_2),e_i \rangle |+|\langle z_1-z_2,e_i \rangle |\\ &\leq \left(\frac{9}{10}+5 \sigma + 5 \varepsilon \right)|p_0(z_1)-p_0(z_2)|+|z_1-z_2|, \end{aligned} \end{equation*}$$

and

$$\begin{equation*} |p_0(z_1)-p_0(z_2)|\leq |z_1-z_2|+|t_1-t_2|\leq \left(\frac{9}{10}+5 \sigma + 5 \varepsilon \right)|p_0(z_1)-p_0(z_2)|+2|z_1-z_2|, \end{equation*}$$

thus

$$\begin{equation*} |p_0(z_1)-p_0(z_2)|\leq \frac{2}{1/10-5(\sigma +\varepsilon )}|z_1-z_2|. \end{equation*}$$

By the definition of $R^x$ and $R_1^x$, we have that

$$\begin{equation*} R^x=\{ z\in R_1^x \mid \operatorname {dist}(z,D_{x})\leq \sigma \operatorname {dist}(z,\ell (x))\}. \end{equation*}$$

Similar as above, we will get that, for any $z_1,z_2\in R_1^x \cap \partial R^{x}$ with $[z_1,z_2]\cap D_{x,y_r}=\emptyset$, if $\operatorname {card}(\mathcal{X}_r)=2$ then

$$\begin{equation*} |\langle z_1-z_2,e_i \rangle |\leq 10(\sigma +\varepsilon )|z_1-z_2|; \end{equation*}$$

if $\operatorname {card}(\mathcal{X}_r)=3$ then

$$\begin{equation*} |\langle z_1-z_2,e_i \rangle |\leq \left( \frac{9}{10}+5\sigma +5\varepsilon \right)|z_1-z_2|, \end{equation*}$$

where $e_i$ is the vector in 2 with $z_1,z_2\in \Omega _i$.

We now consider the mapping $p_1:R_1\to R$ defined by

$$\begin{equation*} p_1(z)= \begin{cases} z,& \text{ for }z\in R,\\ z-te_i\in \partial R \cap \Omega _i,&\text{ for }z\in \Omega _i. \end{cases} \end{equation*}$$

By the same reason as above, we get that

$$\begin{equation*} \operatorname {Lip}(p_1)\leq \frac{2}{1/10-5\sigma -5\varepsilon }\leq 25. \end{equation*}$$

We define a mapping $p_2:R\cap \overline{B(0,1)}\to \Sigma$ as follows: we see that $\Sigma _x$ is the graph over $D_{x}$, thus for any $z\in R^x$, there is only one point in the intersection of $\Sigma _x$ and the line which is perpendicular to $D_{x}$ and through $z$, we let $p_2(z)$ to be the unique intersection point. That is, $p_2(z)$ is the unique point in $\Sigma _x$ such that $p_2(z)-z$ is perpendicular to $D_{x}$. We will show that $p_2$ is Lipschitz and $\operatorname {Lip}(p_2)\leq 1+10^{4}C\eta$. Indeed, we assume that $\Sigma _x$ is the graph of founction $u$ on $D_x$, then by Lemma 3.7 we have that $\operatorname {Lip}(u)\leq C\eta$. For any points $z_1,z_2\in R^{x}$, we let $\widetilde{z}_i$, $i=1,2$, be the points in $D_{x}$ such that $z_i-\widetilde{z}_i$ is perpendicular to $D_{x}$, then

$$\begin{equation*} |(p_2(z_1)-z_1)-(p_2(z_2)-z_2)|=\left|u\left( \widetilde{z}_1\right)-u\left( \widetilde{z}_2 \right)\right|\leq \operatorname {Lip}(u)|\widetilde{z}_1-\widetilde{z}_2|\leq \operatorname {Lip}(u)|z_1-z_2|, \end{equation*}$$

thus

$$\begin{equation*} |p_2(z_1)-p_2(z_2)|\leq (1+\operatorname {Lip}(u))|z_1-z_2|\leq (1+10^{4}C\eta )|z_1-z_2|. \end{equation*}$$

Let $p_3:\mathbb{R}^{3}\to \mathbb{R}^{3}$ be the mapping defined by

$$\begin{equation*} p_3(x)=\begin{cases} x,& |x|\leq 1\\ \frac{x}{|x|},&|x|>1. \end{cases} \end{equation*}$$

Then $p=p_3\circ p_2\circ p_3\circ p_1\circ p_0$ is our desire mapping.■

Lemma 3.11.

For any $r\in (0,\mathfrak{r})\cap \mathscr{R}_1$, we let $\Sigma$ be as in 3.10, and let $\Sigma _r$ be given by $\boldsymbol{\mu }_{r}(\Sigma )$. Then we have that

$$\begin{equation*} \mathcal{H}^2(E\cap B(0,r))\leq \mathcal{H}^2(\Sigma _r)+C\int _{E\cap \partial B(0,r)}\operatorname {dist}(z,\Sigma _r)d\mathcal{H}^1(z) + (2r)^2h(2r), \end{equation*}$$

where $C>0$ is a universal constant.

Proof.

For any $\xi >0$, we consider the function $\psi _{\xi }:[0,\infty )\to \mathbb{R}$ defined by

$$\begin{equation*} \psi _{\xi }(t)=\begin{cases} 1,& 0\leq t\leq 1-\xi \\ -\frac{t-1}{\xi },&1-\xi <t\leq 1\\ 0,&t> 1, \end{cases} \end{equation*}$$

and the mapping $\phi _{\xi }:\Omega _0\to \Omega _0$ defined by

$$\begin{equation*} \phi _{\xi }(z)=\psi _{\xi }(|z|)p(z)+(1-\psi _{\xi }(|z|))z, \end{equation*}$$

where $p$ is the Lipschitz mapping considered in Lemma 3.10. We see that $\phi _{\xi }(L)\subseteq L$. For any $t\in [0,1]$, we put

$$\begin{equation*} \varphi _t(z)=tr\phi _{\xi }\left( z/r \right)+(1-t)z,\text{ for } z\in \Omega _0. \end{equation*}$$

Then $\{ \varphi _t \}_{0\leq t\leq 1}$ is a sliding deformation, and we get that

$$\begin{equation*} \mathcal{H}^2(E\cap \overline{B(0,r)})\leq \mathcal{H}^2(\varphi _1(E)\cap \overline{B(0,r)})+(2r)^2h(2r). \end{equation*}$$

Since $\psi _{\xi }(t)=1$ for $t\in [0,1-\xi ]$, we get that

$$\begin{equation*} \varphi _1(E\cap B(0,(1-\xi )r))=p(E\cap B(0,(1-\xi )r))\subseteq \Sigma _r. \end{equation*}$$

We set $A_{\xi }=B(0,r)\setminus B(0,(1-\xi )r)$. By Theorem 3.2.22 in 10, we get that

$$\begin{equation} \mathcal{H}^2(\varphi _1(E\cap A_{\xi }))\leq \int _{E\cap A_{\xi }}\operatorname {ap}J_{2}(\varphi _1\vert _{E})(z)d\mathcal{H}^2(z). {\tag{3.20}} \end{equation}$$

For any $z\in A_{\xi }$ and $v\in \mathbb{R}^{3}$, by setting $z'=z/r$, we have that

$$\begin{equation*} D\varphi _1(z)v=\psi _{\xi }(|z'|)Dp(z')v+(1-\psi _{\xi }(|z'|))v+ \psi _{\xi }'(|z'|)\langle z/|z|,v\rangle (rp(z')-z). \end{equation*}$$

For any $z\in A_{\xi }\cap E$, we let $v_1,v_2\in \operatorname {Tan}(E,x)$ be such that

$$\begin{equation*} |v_1|=|v_2|=1,\ v_1\perp z\text{ and } v_2\perp v_1, \end{equation*}$$

then we have that $\langle z/|z|,v\rangle =\cos \theta (z)$, and that

$$\begin{equation*} |\psi _{\xi }(|z'|)Dp(z')v_i+(1-\psi _{\xi }(|z'|))v_i|\leq |Dp(z')v_i|\leq \operatorname {Lip}(p), \end{equation*}$$

thus

$$\begin{equation} \begin{aligned} \operatorname {ap}J_{2}(\varphi _1\vert _{E})(z)&=|D\varphi _1(z)v_1\wedge D\varphi _1(z)v_2|\\ &\leq \operatorname {Lip}(p)^2+\frac{1}{\xi }\operatorname {Lip}(p)\cos \theta (z) |rp(z')-z|. \end{aligned} {\tag{3.21}} \end{equation}$$

Since $p(\widetilde{z})=\widetilde{z}$ for any $\widetilde{z}\in \Sigma$, we have that

$$\begin{equation*} |p(z')-z'|=|p(z')-p(\widetilde{z})+\widetilde{z}-z'|\leq (\operatorname {Lip}(p)+1)|\widetilde{z}-z'|, \end{equation*}$$

then we get that

$$\begin{equation*} |p(z')-z'|\leq (\operatorname {Lip}(p)+1)\operatorname {dist}(z,\Sigma ). \end{equation*}$$

We now get, from 3.21, that

$$\begin{equation*} \operatorname {ap}J_{2}(\varphi _1\vert _{E})(z)\leq \operatorname {Lip}(p)^2+\frac{1}{\xi }\operatorname {Lip}(p)(\operatorname {Lip}(p)+1) \operatorname {dist}(z,\Sigma _r)\cos \theta (z), \end{equation*}$$

plug that into 3.20 to get that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(\varphi _1(E\cap A_{\xi }))&\leq C\mathcal{H}^2(E\cap A_{\xi })+\frac{C}{\xi }\int _{E\cap A_{\xi }}\operatorname {dist}(z,\Sigma _r)\cos \theta (z)d\mathcal{H}^2(z)\\ &\leq C\mathcal{H}^2(E\cap A_{\xi })+\frac{C}{\xi }\int _{(1-\xi )r}^r\int _{E\cap \partial B(0,t)}\operatorname {dist}(z,\Sigma _r)d\mathcal{H}^1(z)dt, \end{aligned} \end{equation*}$$

we let $\xi \to 0+$, then we get that, for such $r$,

$$\begin{equation*} \lim _{\xi \to 0+}\mathcal{H}^2(\varphi _1(E\cap A_{\xi }))\leq Cr\int _{E\cap \partial B(0,r)}\operatorname {dist}(z,\Sigma _r)d\mathcal{H}^1(z), \end{equation*}$$

thus

$$\begin{equation*} \mathcal{H}^2(E\cap B(0,r))\leq \mathcal{H}^2(\Sigma _r)+Cr\int _{E\cap \partial B(0,r)}\operatorname {dist}(z,\Sigma _r)d\mathcal{H}^1(z) + (2r)^2h(2r). \end{equation*}$$ ■

3.5. The comparison statement

For any $x,y\in \Omega _0\cap \partial B(0,1)$, if $|x-y|<2$, we denote by $g_{x,y}$ the unique geodesic on $\Omega _0\cap \partial B(0,1)$ which join $x$ and $y$. We will denote by $B_t$ the open ball $B(0,t)$ sometimes for short.

Lemma 3.12.

Let $\tau \in (0,10^{-4})$ be a given. Then there is a constant $\vartheta >0$ such that the following hold. Let $a\in \partial B(0,1)$ and $b,c\in L_0\cap \partial B(0,1)$ be such that $\operatorname {dist}(a,(0,0,1))\leq \tau$, $\operatorname {dist}(b,(1,0,0))\leq \tau$ and $\operatorname {dist}(c,(-1,0,0))\leq \tau$. Let $X$ be the cone over $g_{a,b}\cup g_{a,c}$. Then there is a Lipschitz mapping $\varphi :\Omega _0\to \Omega _0$ with $\varphi ( L_0)\subseteq L_0$, $|\varphi (z)|\leq 1$ when $|z|\leq 1$, and $\varphi (z)=z$ when $|z|>1$, such that

$$\begin{equation*} \mathcal{H}^2(\varphi (X)\cap \overline{B(0,1)})\leq (1-\vartheta )\mathcal{H}^2(X \cap B(0,1))+\frac{\vartheta \pi }{2}. \end{equation*}$$

Proof.

We let $b_0$ a unit vector in $L_0$ which is perpendicular to $b$, and let $c_0$ be a unit vector in $L_0$ which is perpendicular to $c$, such that $b_0+c_0$ is parallel to $b+c$, and take

$$\begin{equation*} u_i=\frac{a-\langle a,i \rangle i}{|a-\langle a,i \rangle i|},\ e_i=\frac{i-\langle i,a \rangle a}{|i-\langle i,a \rangle a|},\ \text{ for }i\in \{b,c\}, \end{equation*}$$

$v_a=\lambda _a (e_b+e_c)$, $v_b=\lambda _b b_0$ and $v_c=\lambda _c c_0$, where $\lambda _j\in \mathbb{R}$, $j\in \{a,b,c\}$, will be chosen later. We let $\psi _1:\mathbb{R}\to \mathbb{R}$ be a function of class $C^1$ such that $0\leq \psi _1\leq 1$, $\psi _1(x)=0$ for $x\in (-\infty ,1/4)\cup (3/4,+\infty )$, $\psi _1(x)=1$ for $x\in [2/5,3/5]$, and $|\psi _1'|\leq 10$. We let $\psi _2:\mathbb{R}\to \mathbb{R}$ be a non increasing function of class $C^1$ such that $0\leq \psi _2\leq 1$, $\psi _2(x)=1$ for $x\in (-\infty ,0]$, $\psi _2(x)=0$ for $x\in [1/5,+\infty )$, and $|\psi _2'|\leq 10$. We let $\psi : \mathbb{R}^3 \times \mathbb{R}^3\to \mathbb{R}$ be a function defined by

$$\begin{equation} \psi (z,v)=\psi _1(\langle z,v \rangle ) \psi _2(|z-\langle z,v \rangle v|). {\tag{3.22}} \end{equation}$$

We now consider the mapping $\varphi :\mathbb{R}^3\to \mathbb{R}^3$ defined by

$$\begin{equation*} \varphi (z)=z+\psi (z,a)v_a+\psi (z,b)v_b+\psi (z,c)v_c. \end{equation*}$$

We see that $\operatorname {supp}(\psi (\cdot ,a))$, $\operatorname {supp}(\psi (\cdot ,b))$ and $\operatorname {supp}(\psi (\cdot ,c))$ are mutually disjoint, and that

$$\begin{equation*} \overline{\{z\in \mathbb{R}^3:\varphi (z)\neq z\}}\subseteq B(0,1),\ \varphi (\Omega _0)\subseteq \Omega _0,\ \varphi (L_0)\subseteq L_0 . \end{equation*}$$

We have that

$$\begin{equation*} D \varphi (z)w=w+\langle D\psi (\cdot ,a),w \rangle v_a+\langle D\psi (\cdot ,b),w \rangle v_b+\langle D\psi (\cdot ,c),w \rangle v_c. \end{equation*}$$

By setting $z_v^{\perp }=z-\langle z,v \rangle v$ for convenient, if $w\neq 0$ and $z_v^{\perp }\neq 0$, we have that

$$\begin{equation*} D\psi (\cdot ,v)w=\psi _1'(\langle z,v \rangle )\psi _2(|z_v^{\perp }|)\langle w/|w|,v \rangle +\psi _1(\langle z,v \rangle )\psi _2'(|z_v^{\perp }|)\langle w_v^{\perp },z_v^{\perp } /|z_v^{\perp }|\rangle . \end{equation*}$$

If $w$ is perpendicular to $v$, then $w_v^{\perp }=w$; if $w$ is parallel to $v$ and $|v|=1$ , then $w_v^{\perp }=0$. We denote by $W_j=\operatorname {supp}(\psi (\cdot ,j))$ for $j\in \{a,b,c\}$. Then

$$\begin{equation*} D\psi (\cdot ,v)w=\begin{cases} w,&z\notin W_a\cup W_b\cup W_c,\\ w+\langle D\psi (\cdot ,v),w \rangle v_j, & z\in W_a\cup W_b\cup W_c. \end{cases} \end{equation*}$$

But

$$\begin{equation*} \langle D\psi (\cdot ,j),j \rangle = \psi _1'(\langle z,j \rangle ) \psi _2(|z_j^{\perp }|), \ j\in \{a,b,c\}, \end{equation*}$$

$$\begin{equation*} \langle D\psi (\cdot ,i),u_i \rangle = \psi _1(\langle z,i \rangle ) \psi _2'(|z_i^{\perp }|) \langle u_i, z_i^{\perp } /|z_i^{\perp }|\rangle , \ i\in \{b,c\}, \end{equation*}$$ and

$$\begin{equation*} \langle D\psi (\cdot ,a),e_i \rangle = \psi _1(\langle z,a \rangle ) \psi _2'(|z_a^{\perp }|) \langle e_i, z_a^{\perp } /|z_a^{\perp }|\rangle , \ i\in \{b,c\}, \end{equation*}$$

by putting

$$\begin{equation*} g_j(z)=\psi _1'(\langle z,j \rangle ) \psi _2(|z_j^{\perp }|),\ j\in \{a,b,c\}, \end{equation*}$$

$$\begin{equation*} g_{a,i}(z)=\psi _1(\langle z,a \rangle ) \psi _2'(|z_a^{\perp }|) \langle e_i, z_a^{\perp } /|z_a^{\perp }|\rangle , \ i\in \{b,c\} \end{equation*}$$ and

$$\begin{equation*} g_{i,i}(z)=\psi _1(\langle z,i \rangle ) \psi _2'(|z_i^{\perp }|) \langle v_i, z_i^{\perp } /|z_i^{\perp }|\rangle , \ i\in \{b,c\}, \end{equation*}$$

and denote by $X_i$ the cone over $g_{a,i}$, $i\in \{b,c\}$, we have that

$$\begin{equation*} D \varphi (z)a \wedge D \varphi (z)e_i = a \wedge e_i + g_a(z) v_a \wedge e_i + g_{a,i}(z) a \wedge v_a, \ z\in X_i\cap W_a \end{equation*}$$

and

$$\begin{equation*} D \varphi (z)i \wedge D \varphi (z)u_i =i \wedge u_i + g_i(z) v_i \wedge u_i + g_{i,i}(z) i \wedge v_i,\ z\in X_i\cap W_i. \end{equation*}$$

If $z\in X_i\cap W_a$, $i\in \{b,c\}$, we have that

$$\begin{equation*} \begin{aligned} J_2 \varphi \vert _{X}(z)&=\|D \varphi (z)a \wedge D \varphi (z)e_i\|\\ &\leq 1+\left\langle a \wedge e_i,g_a(z) v_a \wedge e_i + g_{a,i}(z) a \wedge v_a\right\rangle + \frac{1}{2}\|g_a(z) v_a \wedge e_i + g_{a,i}(z) a \wedge v_a\|^2\\ &= 1+g_a(z)\langle a,v_a \rangle + g_{a,i}(z) \langle e_i,v_a \rangle + \frac{1}{2}\left( g_a(z)^2\|v_a\wedge e_i\|^2+g_{a,i}(z)^2|v_a|^2\right)\\ &\leq 1+ g_{a,i}(z) \langle e_i,v_a \rangle + 100|v_a|^2. \end{aligned} \end{equation*}$$

Similarly, we have that, for $z\in X_i\cap W_i$,

$$\begin{equation*} J_2 \varphi \vert _{X}(z)=\|D \varphi (z)i \wedge D \varphi (z)u_i\|\leq 1+ g_{i,i}(z) \langle u_i,v_i \rangle + 100|v_i|^2. \end{equation*}$$

We see that $z_a^{\perp }/|z_a^{\perp }|=e_i$ when $z\in X_i\setminus {\mathrm{span}}\{a\}$, and $z_i^{\perp }/|z_i^{\perp }|=u_i$ in case $z\in X_i\setminus {\mathrm{span}}\{i\}$, thus

$$\begin{equation*} g_{a,i}(z)=\psi _1(\langle z,a \rangle ) \psi _2'(|z_a^{\perp }|) \text{ and } g_{i,i}(z)=\psi _1(\langle z,i \rangle ) \psi _2'(|z_i^{\perp }|). \end{equation*}$$

Hence, for $j=a$ or $i$, we have that

$$\begin{equation*} \begin{aligned} \int _{z\in X_i \cap W_j} g_{j,i}(z)d\mathcal{H}^2(z)&= \int _{z\in X_i\cap W_j} \psi _1(\langle z,j \rangle )\psi _2'(|z_j^{\perp }|)d\mathcal{H}^2(z)\\ &= \int _{0}^{+\infty }\int _{0}^{+\infty }\psi _1(t) \psi _2'(s) dt ds = -\int _{0}^{+\infty }\psi _1(t) dt <-\frac{1}{5}, \end{aligned} \end{equation*}$$

Thus

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(\varphi (X\cap B_1))&= \int _{z\in X\cap B(0,1)}J_2 \varphi \vert _{X}(z) d\mathcal{H}^2(z)\\ &\leq (1+100\sum _j|v_j|^2)\mathcal{H}^2(X\cap B_1)-\frac{1}{5}(\langle v_a,e_b+e_c \rangle +\sum _{i}\langle u_i,v_i \rangle ) \end{aligned} \end{equation*}$$

If we take $\lambda _a=10^{-3}\mathcal{H}^2(X\cap B_1)^{-1}$ and $\lambda _i=10^{-3}\mathcal{H}^2(X\cap B_1)^{-1}\langle u_i,i_0 \rangle$, $i\in \{b,c\}$, then

$$\begin{equation*} \mathcal{H}^2(\varphi (X\cap B_1)) \leq \mathcal{H}^2(X\cap B_1)- 10^{-4}(|e_b+ e_c|^2+ \langle u_b,b_0 \rangle ^2 + \langle u_c,c_0 \rangle ^2). \end{equation*}$$

Since $|\langle a,w \rangle |\leq \tau |w|$ for $w\in L_0$, and $-1\leq \langle b,c \rangle \leq -1+2\tau ^2$, we get that

$$\begin{equation*} \begin{aligned} |e_b+e_c|^{2}&=2(1+\langle e_b,e_c \rangle )= \frac{2}{1-\langle e_b,e_c \rangle }(1-\langle e_b,e_c \rangle ^2)\\ &\geq 1-\frac{(\langle b,c \rangle -\langle a,b \rangle \langle a,c \rangle )^2}{(1-\langle a,b \rangle ^2)(1-\langle a,c \rangle ^2)}\\ &\geq 1-\langle a,b \rangle ^2-\langle a,c \rangle ^2-\langle b,c \rangle ^2+2\langle a,b \rangle \langle b,c \rangle \langle c,a \rangle \\ &=(1-\langle b,c \rangle +2 \langle a,b \rangle \langle a,c \rangle )(1+\langle b,c \rangle ) -\langle a,b+c \rangle ^2\\ &\geq (1-3\tau ^2)|b+c|^2. \end{aligned} \end{equation*}$$

Since $\arcsin x= x+\sum _{n\geq 1}C_nx^{2n+1}$ for $|x|\leq 1$, where $C_n=\frac{(2n)!}{4^n(n!)^2(2n+1)}$, we have that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(X\cap B_1)-\frac{\pi }{2}&=\frac{1}{2}(\arccos \langle a,b \rangle + \arccos \langle a,c \rangle )-\frac{\pi }{2}\\ &=-\frac{1}{2} (\arcsin \langle a,b \rangle + \arcsin \langle a,c \rangle ) \leq \frac{1}{2}(1+\tau )|\langle a,b+c \rangle |. \end{aligned} \end{equation*}$$

If $b+c\neq 0$, then $|b_0+c_0|\geq 1$, and we have that

$$\begin{equation*} \left\langle a,\frac{b+c}{|b+c|} \right\rangle ^2=\left\langle a,\frac{b_0+c_0}{|b_0+c_0|} \right\rangle ^2 \leq 2 \left(\langle a,b_0 \rangle ^2 + \langle a,c_0 \rangle ^2\right). \end{equation*}$$

We get so that in any case

$$\begin{equation*} |\langle a,b+c \rangle | \leq \frac{1}{2} \left(|b+c|^2+ 2 \langle a,b_0 \rangle ^2 + 2\langle a,c_0 \rangle ^2\right). \end{equation*}$$

Since

$$\begin{equation*} \langle u_b,b_0 \rangle ^2+\langle u_c,c_0 \rangle ^2= \frac{ \langle a,b_0 \rangle ^2}{1-\langle a,b \rangle ^2}+ \frac{ \langle a,c_0 \rangle ^2}{1-\langle a,c \rangle ^2} \geq \langle a,b_0 \rangle ^2 + \langle a,c_0 \rangle ^2, \end{equation*}$$

we get that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(\varphi (X\cap B_1)) &\leq \mathcal{H}^2(X\cap B_1)- 10^{-4}\left(\frac{1}{2}|b+c|^2+ \langle a,b_0 \rangle ^2 + \langle a,c_0 \rangle ^2 \right)\\ &\leq \mathcal{H}^2(X\cap B_1)-10^{-4} \left(\mathcal{H}^2(X\cap B_1)-\frac{\pi }{2}\right) . \end{aligned} \end{equation*}$$ ■

Lemma 3.13.

Let $\tau \in (0,10^{-4})$ be a given. Then there is a constant $\vartheta >0$ such that the following hold. Let $a\in \partial B(0,1)$ and $b,c,d\in L_0\cap \partial B(0,1)$ be such that $\operatorname {dist}(a,(0,0,1))\leq \tau$, $\operatorname {dist}(b,(-1/2,\sqrt {3}/2,0))\leq \tau$, $\operatorname {dist}(c,(-1/2,-\sqrt {3}/2,0))\leq \tau$ and $\operatorname {dist}(d,(1,0,0))\leq \tau$. Let $X$ be the cone over $g_{a,b}\cup g_{a,c}\cup g_{a,d}$. Then there is a Lipschitz mapping $\varphi :\Omega _0\to \Omega _0$ with $\varphi (E\cap L)\subseteq L$, $|\varphi (z)|\leq 1$ when $|z|\leq 1$, and $\varphi (z)=z$ when $|z|>1$, such that

$$\begin{equation*} \mathcal{H}^2(\varphi (X)\cap \overline{B(0,1)})\leq (1-\vartheta )\mathcal{H}^2(X\cap B(0,1))+\vartheta \frac{3\pi }{4}. \end{equation*}$$

Proof.

We let $b_0$, $c_0$ and $d_0$ be unit vectors in $L_0$ such that

$$\begin{equation*} b_0 \perp b, c_0\perp c, d_0\perp d. \end{equation*}$$

For $i\in \{b,c,d\}$, we put

$$\begin{equation*} u_i=\frac{a-\langle a,i \rangle i}{|a-\langle a,i \rangle i|},\ e_i=\frac{i-\langle i,a \rangle a}{|i-\langle i,a \rangle a|}. \end{equation*}$$

We take $v_a=\lambda _a (e_b+e_c+e_d)$ and $v_i=\lambda _i i_0$, where $\lambda _i>0$, $i\in \{b,c,d\}$, will be chosen later. We let $\psi$ be the same as in 3.22, and consider the mapping $\varphi :\mathbb{R}^3\to \mathbb{R}^3$ defined by

$$\begin{equation*} \varphi (z)=z+\psi (z,a)v_a+\psi (z,b)v_b+\psi (z,c)v_c+\psi (z,d)v_d. \end{equation*}$$

We see that $\operatorname {supp}(\psi (\cdot ,a))$, $\operatorname {supp}(\psi (\cdot ,b))$, $\operatorname {supp}(\psi (\cdot ,c))$ and $\operatorname {supp}(\psi (\cdot ,d))$ are mutually disjoint, and that

$$\begin{equation*} \overline{\{z\in \mathbb{R}^3:\varphi (z)\neq z\}}\subseteq B(0,1),\ \varphi (\Omega _0)\subseteq \Omega _0,\ \varphi (L_0)\subseteq L_0 . \end{equation*}$$

By putting $W_j=\operatorname {supp}(\psi (\cdot ,j))$ for $j\in \{a,b,c,d\}$, we have that

$$\begin{equation*} D\psi (\cdot ,v)w=\begin{cases} w,&z\notin W_a\cup W_b\cup W_c \cup W_d,\\ w+\langle D\psi (\cdot ,v),w \rangle v_j, & z\in W_a\cup W_b\cup W_c\cup W_d, \end{cases} \end{equation*}$$

and

$$\begin{equation*} \begin{gathered} \langle D\psi (\cdot ,j),j \rangle = \psi _1'(\langle z,j \rangle ) \psi _2(|z_j^{\perp }|), \ j\in \{a,b,c,d\},\\ \langle D\psi (\cdot ,i),u_i \rangle = \psi _1(\langle z,i \rangle ) \psi _2'(|z_i^{\perp }|) \langle u_i, z_i^{\perp } /|z_i^{\perp }|\rangle ,\\ \langle D\psi (\cdot ,a),e_i \rangle = \psi _1(\langle z,a \rangle ) \psi _2'(|z_a^{\perp }|) \langle e_i, z_a^{\perp } /|z_a^{\perp }|\rangle , \ i\in \{b,c,d\}, \end{gathered} \end{equation*}$$

where $z_w=z-\langle z,w \rangle w$. By putting

$$\begin{equation*} \begin{gathered} g_j(z)=\psi _1'(\langle z,j \rangle ) \psi _2(|z_j^{\perp }|),\ j\in \{a,b,c,d\},\\ g_{a,i}(z)=\psi _1(\langle z,a \rangle ) \psi _2'(|z_a^{\perp }|) \langle e_i, z_a^{\perp } /|z_a^{\perp }|\rangle ,\\ g_{i,i}(z)=\psi _1(\langle z,i \rangle ) \psi _2'(|z_i^{\perp }|) \langle v_i, z_i^{\perp } /|z_i^{\perp }|\rangle , \ i\in \{b,c,d\}, \end{gathered} \end{equation*}$$

and denote by $X_i$ the cone over $g_{a,i}$, $i\in \{b,c,d\}$, we have that

$$\begin{equation*} \begin{gathered} D \varphi (z)a \wedge D \varphi (z)e_i = a \wedge e_i + g_a(z) v_a \wedge e_i + g_{a,i}(z) a \wedge v_a, \ z\in X_i\cap W_a,\\ D \varphi (z)i \wedge D \varphi (z)u_i =i \wedge u_i + g_i(z) v_i \wedge u_i + g_{i,i}(z) i \wedge v_i,\ z\in X_i\cap W_i. \end{gathered} \end{equation*}$$

We have that, for $i\in \{b,c,d\}$,

$$\begin{equation*} \begin{gathered} J_2 \varphi \vert _{X}(z)=\|D \varphi (z)a \wedge D \varphi (z)e_i\|\leq 1+ g_{a,i}(z) \langle e_i,v_a \rangle + 100|v_a|^2, z\in X_i\cap W_a,\\ J_2 \varphi \vert _{X}(z)=\|D \varphi (z)i \wedge D \varphi (z)u_i\|\leq 1+ g_{i,i}(z) \langle u_i,v_i \rangle + 100|v_i|^2,z\in X_i\cap W_i. \end{gathered} \end{equation*}$$

Since $z_a^{\perp }/|z_a^{\perp }|=e_i$ when $z\in X_i\setminus {\mathrm{span}}\{a\}$, and $z_i^{\perp }/|z_i^{\perp }|=u_i$ in case $z\in X_i\setminus {\mathrm{span}}\{i\}$, we have that

$$\begin{equation*} g_{a,i}(z)=\psi _1(\langle z,a \rangle ) \psi _2'(|z_a^{\perp }|) \text{ and } g_{i,i}(z)=\psi _1(\langle z,i \rangle ) \psi _2'(|z_i^{\perp }|). \end{equation*}$$

Thus, for $j=a$ or $i$,

$$\begin{equation*} \int _{z\in X_i \cap W_j} g_{j,i}(z)d\mathcal{H}^2(z) = -\int _{0}^{+\infty }\psi _1(t) dt <-\frac{1}{5}. \end{equation*}$$

Hence

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(\varphi (X\cap B_1))&= \int _{z\in X\cap B_1}J_2 \varphi \vert _{X}(z) d\mathcal{H}^2(z)\\ &\leq \left(1+100(|v_a|^2+|v_b|^2+|v_c|^2+|v_d|^2)\right)\mathcal{H}^2(X\cap B_1)\\ &\quad -\frac{1}{5}\left(\langle v_a,e_b+e_c+e_d \rangle +\langle u_b,v_b \rangle +\langle u_c,v_c \rangle +\langle u_d,v_d \rangle \right). \end{aligned} \end{equation*}$$

If we take $\lambda _a=10^{-3}\mathcal{H}^2(X\cap B_1)^{-1}$ and $\lambda _i=10^{-3}\mathcal{H}^2(X\cap B_1)^{-1}\langle u_i,i_0 \rangle$, $i\in \{b,c,d\}$, then

$$\begin{equation*} \mathcal{H}^2(\varphi (X\cap B_1)) \leq \mathcal{H}^2(X\cap B_1)- 10^{-4}\left(|e_b+ e_c+e_d|^2+ \sum _{i}\langle u_i,i_0 \rangle ^2 \right). \end{equation*}$$

Since $|\langle a,w \rangle |\leq \tau |w|$, for $w\in L_0$, and $-1/2-\sqrt {3}\tau \leq \langle i_1,i_2 \rangle \leq -1/2+\sqrt {3}\tau$, $i_1,i_2\in \{b,c,d\}$, $i_1\neq i_2$, we get that $\langle i,j \rangle - \langle a,i \rangle \langle a,j \rangle < 0$. By putting $e=(0,0,1)$, it is evident that

$$\begin{equation*} \langle a,w \rangle ^2\leq 1-\langle a,e \rangle ^2, \text{ for any }w\in L_0 \text{ with }|w|=1. \end{equation*}$$

We put $N=\langle a,b \rangle ^2+\langle a,c \rangle ^2+\langle a,d \rangle ^2$, and we claim that

$$\begin{equation} N\leq \left(3/2+25\tau \right) \left(1- \langle a,e \rangle ^2\right). {\tag{3.23}} \end{equation}$$

Indeed, for any $w=\lambda b + \mu c$ with $\lambda ,\mu \geq 0$, we have that

$$\begin{equation*} |w|^2=\lambda ^2+\mu ^2+2 \lambda \mu \langle b,c \rangle \geq \lambda ^2+\mu ^2-(1+4\tau ) \lambda \mu , \end{equation*}$$

$$\begin{equation*} \langle w,d \rangle ^2\leq (1/2+\sqrt {3}\tau )^2(\lambda +\mu )^2\leq (1/4+2\tau )(\lambda +\mu )^2 \end{equation*}$$ and

$$\begin{equation*} \begin{aligned} \langle w,b \rangle ^2+\langle w,b \rangle ^2+\langle w,b \rangle ^2&=(\lambda ^2+\mu ^2)(1+\langle b,c \rangle ^2)+4 \lambda \mu \langle b,c \rangle +\langle w,d \rangle ^2\\ &\leq \left(3/2+4\tau \right)(\lambda ^2+\mu ^2)-(3/2-10\tau )\lambda \mu \\ &\leq (3/2+25\tau )|w|^2. \end{aligned} \end{equation*}$$

Hence, for any $w\in L_0$, we have that

$$\begin{equation*} \langle w,b \rangle ^2+\langle w,b \rangle ^2+\langle w,b \rangle ^2 \leq (3/2+25\tau )|w|^2, \end{equation*}$$

we now take $w=a-\langle a,e \rangle e$, then

$$\begin{equation*} N\leq (3/2+25\tau )|a-\langle a,e \rangle e|^2=(3/2+25\tau )(1-\langle a,e \rangle ^2), \end{equation*}$$

the claim 3.23 follows.

Since $(1-x)^{1/2}\leq 1-x/2-x^2/8$ for any $x\in (0,1)$, and

$$\begin{equation*} (1-\langle a,b \rangle ^2)(1-\langle a,c \rangle ^2)(1-\langle a,d \rangle ^2) \geq 1-N, \end{equation*}$$

we have that, for $\{i,j,k\}=\{b,c,d\}$,

$$\begin{align*} \langle e_i,e_j \rangle =&\frac{\langle i,j \rangle - \langle a,i \rangle \langle a,j \rangle }{(1-\langle a,i \rangle ^2)^{1/2}(1-\langle a,j \rangle ^2)^{1/2}}\\ &\geq \frac{(\langle i,j \rangle - \langle a,i \rangle \langle a,j \rangle )(1-\langle a,k \rangle ^2/2-\langle a,k \rangle ^4/8)}{(1-N)^{1/2}}. \end{align*}$$

Note that

$$\begin{equation*} \langle a,b \rangle ^4+\langle a,c \rangle ^4+\langle a,d \rangle ^4\geq N^2/3, \end{equation*}$$

and

$$\begin{equation*} |\langle a,b+c+d \rangle |\leq \frac{1}{2}\left(|b+c+d|^2+1-\langle a,e \rangle ^2\right), \end{equation*}$$

we get so that

$$\begin{equation*} \begin{aligned} |e_b+e_c+e_d|^2&\geq 3+(1-N)^{-1/2}\Big (-3+(3/2-\sqrt {3}\tau )N+ \frac{1}{12}(1/2-\sqrt {3}\tau )N^2\\ &\quad + |b+c+d|^2-\langle a,b+c+d \rangle ^2+\langle a,b \rangle \langle a,c \rangle \langle a,d \rangle \langle a,b+c+d\rangle \\ &\quad +\frac{1}{4}\langle a,b \rangle \langle a,c \rangle \langle a,d \rangle \big ( \langle a,b \rangle ^3+\langle a,c \rangle ^3+\langle a,d \rangle ^3\big ) \Big )\\ &\geq (1-N)^{-1/2} \left((1-\tau ^2)|b+c+d|^2- 2\tau N-2\tau ^3| \langle a,b+c+d \rangle |\right)\\ &\geq (1-\tau )|b+c+d|^2-6\tau (1-\langle a,e \rangle ^2). \end{aligned} \end{equation*}$$

Since $1/(1-x)= 1+x+x^2/(1-x)$ for $x\in [0,1)$, and $\langle a,i \rangle ^2\leq 1-\langle a,e \rangle ^2$ for $i\in \{b,c,d\}$, we have that

$$\begin{equation*} \frac{\langle a,e \rangle ^2}{1-\langle a,i \rangle ^2}=\langle a,e \rangle ^2+\frac{\langle a,e \rangle ^2 \langle a,i \rangle ^2}{1-\langle a,i \rangle ^2}\leq \langle a,e \rangle ^2 +\langle a,i \rangle ^2 \end{equation*}$$

and

$$\begin{equation*} \begin{aligned} \langle u_b,b_0 \rangle ^2+\langle u_c,c_0 \rangle ^2+ \langle u_d,d_0 \rangle ^2 &= \sum _{i\in \{b,c,d\}} \frac{1-\langle a,e \rangle ^2-\langle a,i \rangle ^2}{1-\langle a,i \rangle ^2}\\ &=3(1- \langle a,e \rangle ^2)- N \geq (1-\tau )(1-\langle a,e \rangle ^2). \end{aligned} \end{equation*}$$

We get so that

$$\begin{equation*} \mathcal{H}^2(\varphi (X\cap B_1)) \leq \mathcal{H}^2(X\cap B_1)- 10^{-4}(1-10\tau )\left( |b+c+d|^2+1-\langle a,e \rangle ^2 \right) \end{equation*}$$

Since $\arcsin x= x+\sum _{n\geq 1}C_nx^{2n+1}$ for $|x|\leq 1$, where $C_n=\frac{(2n)!}{4^n(n!)^2(2n+1)}$, we have that $\arcsin \langle a,i \rangle \geq \langle a,i \rangle -\tau \langle a,i \rangle ^2$, thus

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(X\cap B_1)-\frac{3\pi }{4}&=-\frac{1}{2}\left(\arcsin \langle a,b \rangle +\arcsin \langle a,c \rangle + \arcsin \langle a,c \rangle \right)\\ &\leq -\frac{1}{2}\langle a,b+c+d \rangle +\frac{\tau }{2}N\\ &\leq \frac{1}{2} \left(|b+c+d|^2+ 1-\langle a,e\rangle ^2 \right)+\tau \left(1- \langle a,e \rangle ^2\right). \end{aligned} \end{equation*}$$

Thus

$$\begin{equation*} \mathcal{H}^2(\varphi (X\cap B_1)) \leq (1- 10^{-4}) \mathcal{H}^2(X\cap B_1)-10^{-4}\cdot \frac{3\pi }{4}. \end{equation*}$$ ■

Let $E\subseteq \Omega _0$ be a $2$-rectifiable set satisfying (a), (b) and (c). We will denote by $\mathscr{R}_2$ the set $\left\{ r\in \mathscr{R}_1 : 10 C(1+C\eta ^{-2})\left(\varepsilon (r)+j(r)^{1/2}\right)\leq 1/2 - 10^{-4} \right\}$, where we take constant $C$ to be the maximum value of the constants in Lemma 3.6 and Lemma 3.11.

Lemma 3.14.

For any $r\in (0,\mathfrak{r})\cap \mathscr{R}_2$, we have that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E\cap B_r)&\leq (1-2\cdot 10^{-4})\frac{r}{2}\mathcal{H}^1(E\cap \partial B_r)+(2\cdot 10^{-4}-\vartheta \kappa ^2)\frac{r^2}{2}\mathcal{H}^1(X\cap \partial B_1)\\ &\quad +\vartheta \kappa ^2r^2\Theta (0)+(2r)^2h(2r). \end{aligned} \end{equation*}$$

Proof.

Let $\Sigma$, $\Sigma _r$, $\xi$, $\psi _{\xi }$, $\phi _{\xi }$ and $\{ \varphi _t \}_{0\leq t\leq 1}$ be the same as in the proof of Lemma 3.11. We see that

$$\begin{equation*} \varphi _1(E\cap B(0,(1-\xi )r))=p(E\cap B(0,(1-\xi )r))\subseteq \Sigma _r, \end{equation*}$$

and that $\Sigma \cap B(0,2\kappa )=X\cap B(0,2\kappa )$, where $X$ is a cone defined in 3.10. We see that if $\Theta (0)=\pi /2$, then $X$ satisfies the conditions in Lemma 3.12; if $\Theta (0)=3\pi /4$, then $X$ satisfies the conditions in Lemma 3.13. Thus we can find a Lipschitz mapping $\Omega _0\to \Omega _0$ with $\varphi (E\cap L)\subseteq L$, $|\varphi (z)|\leq 1$ when $|z|\leq 1$, and $\varphi (z)=z$ when $|z|>1$, such that

$$\begin{equation*} \mathcal{H}^2\left(\varphi (X)\cap \overline{B(0,1)}\right)\leq (1-\vartheta )\mathcal{H}^2(X\cap B(0,1))+\vartheta \Theta (x). \end{equation*}$$

Let $\widetilde{\varphi }:\Omega _0\to \Omega _0$ be the mapping defined by $\widetilde{\varphi }(x)=r\varphi (x/r)$, then

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E\cap B(0,r))&\leq \mathcal{H}^2(\widetilde{\varphi }\circ \varphi _1(E)\cap \overline{B(0,r)})+(2r)^2h(2r)\\ &\leq \mathcal{H}^2(\widetilde{\varphi }\circ \varphi _1(E\cap B(0,(1-\xi )r)))+ \mathcal{H}^2(\varphi _1(E\cap A_{\xi }))\\ &\leq \mathcal{H}^2(\Sigma _r\setminus \overline{B(0,\kappa r)})+(1-\vartheta )(\kappa r)^2\mathcal{H}^2(X\cap B(0,1))\\ &\quad +\vartheta \cdot (\kappa r)^2\Theta (0)+\mathcal{H}^2(\varphi _1(E\cap A_{\xi })). \end{aligned} \end{equation*}$$

But we see that $\Sigma _r=\{ rx:x\in \Sigma \}$, $\Sigma \cap B(0,2\kappa )=X\cap B(0,2\kappa )$, and

$$\begin{equation*} \lim _{\xi \to 0+}\mathcal{H}^2(\varphi _1(E\cap A_{\xi }))\leq C\int _{E\cap \partial B(0,r)}\operatorname {dist}(z,\Sigma _r)d\mathcal{H}^1(z), \end{equation*}$$

we get so that

$$\begin{equation*} \mathcal{H}^2(\Sigma _r\setminus \overline{B(0,\kappa r)})=r^2\left(\mathcal{H}^2(\Sigma )-\mathcal{H}^2(X\cap B(0,\kappa ))\right), \end{equation*}$$

and

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E\cap B(0,r))&\leq r^2\mathcal{H}^2(\Sigma )-(\kappa r)^2\mathcal{H}^2(X\cap B(0,1))\\ &\quad +(1-\vartheta )(\kappa r)^2\mathcal{H}^2(X\cap B(0,1))+(\kappa r)^2\vartheta \cdot \Theta (0)\\ &\quad +C\int _{E\cap \partial B(0,r)}\operatorname {dist}(z,\Sigma _r)d\mathcal{H}^1(z) +(2r)^2h(2r). \end{aligned} \end{equation*}$$

Since $\mathcal{M}$ is the cone over $\Gamma _{\ast }$, and $X$ is the cone over $\mathfrak{C}$, by 3.11, we get that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(\Sigma )&\leq \mathcal{H}^2(\mathcal{M}\cap B(0,1))-10^{-4}\Big (\mathcal{H}^1(\Gamma _{\ast })-\mathcal{H}^1(\mathfrak{C})\Big )\\ &= (1/2-10^{-4})\mathcal{H}^1(\Gamma _{\ast })+ 10^{-4}\mathcal{H}^1(\mathfrak{C}), \end{aligned} \end{equation*}$$

and then

$$\begin{equation} \begin{aligned} \mathcal{H}^2(E\cap B_r)&\leq (1/2-10^{-4})r^2\mathcal{H}^1(\Gamma _{\ast })+ (10^{-4}-\vartheta \kappa ^2/2)r^2\mathcal{H}^1(\mathfrak{C})\\ &\quad +\vartheta \kappa ^2 r^2\Theta (0) +C\int _{E\cap \partial B_r}\operatorname {dist}(z,\Sigma _r)d\mathcal{H}^1(z)+(2r)^2h(2r). \end{aligned} {\tag{3.24}} \end{equation}$$

By 3.13 and Lemma 3.8, we have that

$$\begin{equation*} d_{0,r}(E,\mathcal{M})\leq d_{0,r}(E,X)+d_{0,r}(X,\mathcal{M})\leq 5 \varepsilon (r)+ 10 j(r)^{1/2}, \end{equation*}$$

thus for any $z\in E\cap \partial B(0,r)$,

$$\begin{equation*} \operatorname {dist}(\boldsymbol{\mu }_{1/r}(z),\mathcal{M})=r^{-1}\operatorname {dist}(z,\mathcal{M})\leq 5\varepsilon (r)+10j(r)^{1/2}. \end{equation*}$$

Since $\Sigma \setminus B(0,1-2\kappa )=\mathcal{M}\cap \overline{B(0,1)}\setminus B(0,1-2\kappa )$, we have that

$$\begin{equation*} \operatorname {dist}(z,\Sigma _r)=r\operatorname {dist}(\boldsymbol{\mu }_{1/r}(z),\Sigma )= r\operatorname {dist}(\boldsymbol{\mu }_{1/r}(z),\mathcal{M}) \leq 5r\varepsilon (r)+10rj(r)^{1/2}, \end{equation*}$$

and we get so that

$$\begin{equation} \int _{E\cap \partial B(0,r)}\operatorname {dist}(z,\Sigma _r)d\mathcal{H}^1(z) \leq r\big (5\varepsilon (r)+10j(r)^{1/2}\big ) \mathcal{H}^1\big ((E\cap \partial B_r)\setminus (\Sigma _r\cap \partial B_r)\big ). {\tag{3.25}} \end{equation}$$

By Lemma 3.6, we have that

$$\begin{equation*} \mathcal{H}^1(\Gamma _{\ast }\setminus \Gamma )\leq \mathcal{H}^1(\Gamma \setminus \Gamma _{\ast })\leq C\eta ^{-2} (\mathcal{H}^1(\Gamma )-\mathcal{H}^1(\mathfrak{C})), \end{equation*}$$

thus

$$\begin{equation*} \mathcal{H}^1(\mathfrak{C})\leq \mathcal{H}^1(\Gamma _{\ast })\leq \mathcal{H}^1(\Gamma )\leq \mathcal{H}^1(\boldsymbol{\mu }_{1/r}(E\cap \partial B_r)). \end{equation*}$$

Since $\boldsymbol{\mu }_{1/r}(\Gamma ) \subseteq E\cap \partial B_r$ and $\Sigma _r\cap \partial B_r=\boldsymbol{\mu }_{1/r}(\Sigma \cap \partial B_1)=\boldsymbol{\mu }_{1/r}(\Gamma _{\ast })$, by setting $\Gamma _r=\boldsymbol{\mu }_{1/r}(\Gamma )$ and $\Gamma _{r,\ast }=\boldsymbol{\mu }_{1/r}(\Gamma _{\ast })$, we have that

$$\begin{equation} \begin{aligned} \mathcal{H}^1\big ((E\cap \partial B_r)\setminus (\Sigma _r\cap \partial B_r)\big )&\leq \mathcal{H}^1\big ((E\cap \partial B_r)\setminus \Gamma _r\big )+\mathcal{H}^1\big (\Gamma _r\setminus \Gamma _{r,\ast }\big )\\ &\leq \mathcal{H}^1(E\cap \partial B_r)-\mathcal{H}^1( \Gamma _r)+C\eta ^{-2}r(\mathcal{H}^1(\Gamma )-\mathcal{H}^1(\mathfrak{C}))\\ &\leq (1+C \eta ^{-2})(\mathcal{H}^1(E\cap \partial B_r)-r\mathcal{H}^1(\mathfrak{C})). \end{aligned} {\tag{3.26}} \end{equation}$$

We obtain, from 3.24, 3.25 and 3.26, that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E\cap B_r)&\leq (1/2-10^{-4})r^2\mathcal{H}^1(\Gamma _{\ast })+ (10^{-4}-\vartheta \kappa ^2/2)r^2\mathcal{H}^1(\mathfrak{C})\\ &\quad +10C(1+C\eta ^{-2})(\varepsilon (r)+j(r)^{1/2})r ( \mathcal{H}^1(E\cap \partial B_r)-r\mathcal{H}^1(\Gamma _{\ast }))\\ &\quad +\vartheta \kappa ^2 r^2\Theta (0) +(2r)^2h(2r). \end{aligned} \end{equation*}$$

Since $r\in (0,\mathfrak{r})\cap \mathscr{R}_2$, we have that $10C(1+C\eta ^{-2})\left(\varepsilon (r)+j(r)^{1/2}\right)\leq 1/2- 10^{-4}$, thus

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E\cap B_r)&\leq (1-2\cdot 10^{-4})\frac{r}{2}\mathcal{H}^1(E\cap \partial B_r)+(2\cdot 10^{-4}-\vartheta \kappa ^2)\frac{r^2}{2}\mathcal{H}^1(\mathfrak{C})\\ &\quad +\vartheta \kappa ^2r^2\Theta (0)+(2r)^2h(2r). \end{aligned} \end{equation*}$$ ■

Theorem 3.15.

There exist $\lambda ,\mu \in (0,10^{-3})$ and $\mathfrak{r}_1>0$ such that, for any $0<r<\mathfrak{r}_1$,

$$\begin{equation*} \mathcal{H}^2(E\cap B_r)\leq (1-\mu -\lambda )\frac{r}{2}\mathcal{H}^1(E\cap \partial B_r)+\mu \frac{r^2}{2}\mathcal{H}^1(X\cap \partial B_1)+\lambda \Theta (0)r^2+4r^2h(2r). \end{equation*}$$

Proof.

Recall that $\mathscr{R}_1=\{r\in (0,\mathfrak{r})\cap \mathscr{R}:j(r)\leq \tau _0\}$ and

$$\begin{equation*} \mathscr{R}_2=\left\{ r\in \mathscr{R}_1 : 10 C(1+C\eta ^{-2})\left(\varepsilon (r)+j(r)^{1/2}\right)\leq 1/2 - 10^{-4} \right\}. \end{equation*}$$

We put $\tau _1=\min \{ \tau _0,(100C(1+C\eta ^{-2}))^{-2}\}$, and take $\delta$ such that

$$\begin{equation} \kappa <\delta <\kappa +(10\Theta (0)\vartheta )^{-1}(1-2\cdot 10^{-4})\tau _1. {\tag{3.27}} \end{equation}$$

We see that $\varepsilon (r)\to 0$ as $r\to 0+$, there exist $\mathfrak{r}_2\in (0,\mathfrak{r})$ such that, for any $r\in (0,\mathfrak{r}_2)$,

$$\begin{equation} \varepsilon (r)\leq 10^{-2}\min \{\tau _1,\vartheta (\delta ^2-\kappa ^2)\}. {\tag{3.28}} \end{equation}$$

If $r\in (0,\mathfrak{r}_2)$ and $j(r)\leq \tau _1$, then $r\in \mathscr{R}_2$, then by Lemma 3.14, we have that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E\cap B_r)&\leq (1-2\cdot 10^{-4})\frac{r}{2}\mathcal{H}^1(E\cap \partial B_r)+(2\cdot 10^{-4}-\vartheta \kappa ^2)\frac{r^2}{2}\mathcal{H}^1(X\cap \partial B_1)\\ &\quad +\vartheta \kappa ^2r^2\Theta (0)+(2r)^2h(2r). \end{aligned} \end{equation*}$$

We only need to consider the case $r\in (0,\mathfrak{r}_2)$, $j(r)>\tau _1$ and $\mathcal{H}^1(E\cap \partial B_r)<+\infty$, thus

$$\begin{equation} \mathcal{H}^1(X\cap \partial B_1)+\tau _1\leq \frac{1}{r}\mathcal{H}^1(E\cap \partial B_r). {\tag{3.29}} \end{equation}$$

By the construction of $X$, we see that $X\cap B(0,1)$ is local Lipschitz neighborhood retract, let $U$ be a neighborhood of $X\cap B(0,1)$ and $\varphi _0:U\to X\cap B(0,1)$ be a retraction such that $|\varphi _0(x)-x|\leq r/2$. We put $U_1=\boldsymbol{\mu }_{8r/9}(U)$, $\varphi _1=\boldsymbol{\mu }_{8r/9}\circ \varphi _0\circ \boldsymbol{\mu }_{9/(8r)}$, and let $s:[0,\infty )\to [0,1]$ be a function given by

$$\begin{equation*} s(t)=\begin{cases} 1,&0\leq t\leq 3r/4,\\ -(8/r)(t-7r/8),& 3r/4<t \leq 7r/8,\\ 0,&t>7r/8. \end{cases} \end{equation*}$$

We see, from Lemma 3.8, that there exist sliding minimal cone $Z$ such that $d_{0,r}(E,X)\leq 5\varepsilon (r)$, then for any $x\in E\cap B(0,r)\setminus B(0,3r/4)$,

$$\begin{equation*} \operatorname {dist}(x,X)\leq 5\varepsilon (r) r\leq \frac{20\varepsilon (r)}{3}|x|\leq 7 \varepsilon (r)|x|. \end{equation*}$$

We consider the mapping $\psi :\Omega _0\to \Omega _0$ defined by

$$\begin{equation*} \psi (x)=s(|x|)\varphi _1(x)+(1-s(|x|))x, \end{equation*}$$

then $\psi (L)=L$ and $\psi (x)=x$ for $|x|\geq 8r/9$.

Since $\varepsilon (r)\to 0$ and $U$ is a neighborhood of $X\cap B(0,1)$, we can find $\mathfrak{r}_1\in (0,\mathfrak{r}_2)$ such that, for any $r\in (0,\mathfrak{r}_1)$, $\{ x\in \Omega _0\cap B(0,1):\operatorname {dist}(x,X)\leq 7\varepsilon (r) \}\subseteq U$. Then we get that $\psi (x)\in X$ for any $x\in E\cap B(0,3r/4)$;

$$\begin{equation*} \operatorname {dist}(\psi (x),X)\leq 7\varepsilon (r)|x|\text{ for any }x\in E\cap B(0,r)\setminus B(0,3r/4); \end{equation*}$$

and $\psi (E\cap B_r)\cap B(0,r/4)=X\cap B(0,r/4)$. We now consider the mapping $\Pi _1:\Omega _0\to \Omega _0$ defined by

$$\begin{equation*} \Pi _1(x)=s(4|x|)x+(1-s(4|x|))\frac{x}{|x|}, \end{equation*}$$

and the mapping $\psi _1:\Omega _0\to \Omega _0$ defined by

$$\begin{equation*} \psi _1(x)=\begin{cases} \Pi _1\circ \psi (x),& |x|\leq r,\\ x,&|x|\geq r. \end{cases} \end{equation*}$$

We have that $\psi _1$ is Lipschitz, $\psi _1(L_0)=L_0$ and $\psi _1(B(0,r))\subseteq \overline{B(0,r)}$,

$$\begin{equation*} \psi _1(E\cap B(0,r))\subseteq (X\cap B(0,r))\cup \{ x\in \partial B_r: \operatorname {dist}(x,X) \leq 7r\varepsilon (r)\}. \end{equation*}$$

Let $\varphi$ be the same as in Lemma 3.12 and Lemma 3.13, and let $\psi _2=\boldsymbol{\mu }_{\delta }\circ \varphi \circ \boldsymbol{\mu }_{1/\delta }\circ \psi _1$. Then we have that

$$\begin{equation} \begin{aligned} \mathcal{H}^2(E\cap \overline{B(0,r)})&\leq \mathcal{H}^2\left(\psi _2\left(E\cap \overline{B(0,r)}\right)\right)+(2r)^2h(2r)\\ &\leq (1-\vartheta \delta ^2)\mathcal{H}^2(X\cap B(0,r))+\vartheta \delta ^2 \Theta (0)r^2\\ &\quad +\mathcal{H}^2(\{ x\in \partial B_r: \operatorname {dist}(x,X) \leq 7r\varepsilon (r)\})+4r^2h(2r)\\ &\leq (1-\vartheta \delta ^2)\mathcal{H}^2(X\cap B(0,r))+\vartheta \delta ^2 \Theta (0)r^2 \\ &\quad + 8r\varepsilon (r)\mathcal{H}^1(X\cap \partial B_r)+4r^2h(2r)\\ &\leq (1-\vartheta \delta ^2+16\varepsilon (r))\frac{r^2}{2}\mathcal{H}^1(X\cap \partial B_1)+\vartheta \delta ^2 \Theta (0)r^2 +4r^2h(2r)\\ \end{aligned} {\tag{3.30}} \end{equation}$$

We take $\mu =2\cdot 10^{-4}-\vartheta \kappa ^2$ and $\lambda =\vartheta \kappa ^2$, then by 3.27 and 3.28, we have that

$$\begin{equation*} 16\varepsilon (r)<\vartheta (\delta ^2-\kappa ^2) \text{ and } \vartheta (\delta ^2-\kappa ^2)\Theta (0)\leq (1-2\cdot 10^{-4})\frac{\tau _1}{2}. \end{equation*}$$

We obtain from 3.29 and 3.30 that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E\cap \overline{B_r})&\leq (1-2\cdot 10^{-4}) \frac{r^2}{2}(\mathcal{H}^1(X\cap \partial B_1)+\tau _1)-(1-2\cdot 10^{-4})\frac{\tau _1r^2}{2}\\ &\quad +\mu \frac{r^2}{2}\mathcal{H}^1(X\cap \partial B_1)+ \vartheta \kappa ^2\Theta (0)r^2 +4r^2h(2r)\\ &\quad +(16\varepsilon (r)-\vartheta \delta ^2+\vartheta \kappa ^2)\frac{r^2}{2}\mathcal{H}^1(X\cap \partial B_1)+(\vartheta \delta ^2-\vartheta \kappa ^2)\Theta (0)r^2\\ &\leq (1-\lambda -\mu )\frac{r}{2}\mathcal{H}^1(E\cap \partial B_r)+\mu \frac{r^2}{2}\mathcal{H}^1(X\cap \partial B_1)+ \lambda \Theta (0)r^2 +4r^2h(2r). \end{aligned} \end{equation*}$$ ■

For convenient, we put $\lambda _0=\lambda /(1-\lambda )$, $f(r)=\Theta (0,r)-\Theta (0)$ and $u(r)=\mathcal{H}^1(E\cap B(0,r))$ for $r>0$. Since $f(r)=r^{-2}u(r)-\Theta (0)$ and $u$ is a nondecreasing function, we have that, for any $\lambda _1\in \mathbb{R}$ and $0<r\leq R<+\infty$,

$$\begin{equation*} R^{\lambda _1}f(R)-r^{\lambda _1}f(r)\geq \int _{r}^{R}\left( t^{\lambda _1}f(t) \right)' dt, \end{equation*}$$

thus

$$\begin{equation} f(r)\leq r^{-\lambda _1}R^{\lambda _1}f(R)+r^{-\lambda _1}\int _{r}^{R}\left( t^{\lambda _1}f(t) \right)' dt. {\tag{3.31}} \end{equation}$$

Corollary 3.16.

If the gauge function $h$ satisfy

$$\begin{equation*} h(t)\leq C_ht^{\alpha },\ 0<t\leq \mathfrak{r}_1\text{ for some }C_h>0,\ \alpha >0, \end{equation*}$$

then for any $0<\beta <\min \{ \alpha , 2\lambda _0\}$, there is a constant $C=C(\lambda _0,\alpha ,\beta ,\mathfrak{r}_1,C_h)>0$ such that

$$\begin{equation} |\Theta (0,\rho )-\Theta (0)|\leq C\rho ^{\beta },\text{ for any }0<\rho \leq \mathfrak{r}_1. {\tag{3.32}} \end{equation}$$

Proof.

For any $r>0$, we put $u(r)=\mathcal{H}^2(E\cap B(0,r))$. Then $u$ is differentiable for $\mathcal{H}^1$-a.e. $r\in (0,\infty )$. By Theorem 3.15 and Lemma 2.1, we have that for any $r\in (0,\mathfrak{r}_1)\cap \mathscr{R}$,

$$\begin{equation*} \begin{aligned} u(r)&\leq (1-\lambda )\frac{r}{2}\mathcal{H}^1(E\cap \partial B(0,r))+\lambda \Theta (0)r^2+4r^2h(2r)\\ &\leq (1-\lambda )\frac{r}{2}u'(r)+\lambda \Theta (0)r^2+4r^2h(2r), \end{aligned} \end{equation*}$$

thus

$$\begin{equation*} rf'(r)\geq \frac{2\lambda }{1-\lambda }f(r)-\frac{8}{1-\lambda }h(2r)=2\lambda _0 f(r)-8(1+\lambda _0)h(2r), \end{equation*}$$

and

$$\begin{equation*} \left( r^{-2\lambda _0}f(r) \right)'= r^{-1-2\lambda _0}\left( rf'(r)-2\lambda _0 \right)\geq -8(1+\lambda _0)r^{-1-2\lambda _0}h(2r). \end{equation*}$$

Recall that $\mathcal{H}^1((0,\infty )\setminus \mathscr{R})=0$. We get so that, from 3.31, for any $0<r<R\leq \mathfrak{r}_1$,

$$\begin{equation} f(r)\leq r^{2\lambda _0}R^{-2\lambda _0}f(R) +8(1+\lambda _0)r^{2\lambda _0} \int _r^{R}t^{-1-2\lambda _0}h(2t)dt. {\tag{3.33}} \end{equation}$$

Since $h(t)\leq C_ht^{\alpha }$, we have that

$$\begin{equation*} f(r)\leq (r/R)^{-2\lambda _0}f(R) +2^{3+\alpha }(1+\lambda _0)C_hr^{2\lambda _0} \int _r^{R}t^{\alpha -2\lambda _0-1}dt. \end{equation*}$$

If $\alpha >2\lambda _0$, then

$$\begin{equation} f(r)\leq \left( f(R) +2^{3+\alpha }(1+\lambda _0)(1+\lambda _0) (\alpha -2\lambda _0)^{-1}C_h R^{\alpha }\right)(r/R)^{2\lambda _0}; {\tag{3.34}} \end{equation}$$

if $\alpha =2\lambda _0$, then

$$\begin{equation*} f(r)\leq f(R)(r/R)^{\alpha } +2^{\alpha +3}(1+\lambda _0)C_h r^{\alpha } \ln (R/r), \end{equation*}$$

thus, for any $\beta \in (0,\alpha )$,

$$\begin{equation} \begin{aligned} f(r)&\leq f(R)r^{\alpha } +2^{\alpha +3}(1+\lambda _0)C_h r^{\beta }R^{\alpha -\beta } \frac{\ln (R/r)}{(R/r)^{\alpha -\beta }}\\ &\leq \left(f(R) + 2^{\alpha +3}(1+\lambda _0)C_h (\alpha -\beta )^{-1}e^{-1} R^{\alpha } \right)(r/R)^{\beta }; \end{aligned} {\tag{3.35}} \end{equation}$$

if $\alpha <2\lambda _0$, then

$$\begin{equation} \begin{aligned} f(r)&\leq f(R)(r/R)^{2\lambda _0} +2^{\alpha +3}(1-\lambda _0)C_h r^{2\lambda _0} \cdot (2\lambda _0-\alpha )^{-1}\left( r^{\alpha -2\lambda _0} - R^{\alpha -2\lambda _0}\right)\\ &\leq \left((r/R)^{2\lambda _0-\alpha }f(R) + 2^{\alpha +3}(1-\lambda _0)C_h (2\lambda _0-\alpha )^{-1}R^{\alpha } \right)(r/R)^{\alpha }. \end{aligned} {\tag{3.36}} \end{equation}$$

Hence 3.32 follows from 3.34, 3.35, 3.36 and Theorem 2.3. Indeed, there is a constant $C_1(\alpha ,\beta ,\lambda _0)>0$ such that

$$\begin{equation} r^{2\lambda _0}\int _{r}^{R}t^{\alpha -2\lambda _0-1}dt\leq C_1(\alpha ,\beta ,\lambda _0) R^{\alpha }\cdot (r/R)^{\beta }, {\tag{3.37}} \end{equation}$$

and there is a constant $C_2(\alpha ,\beta ,\lambda _0)>0$ such that

$$\begin{equation*} f(r)\leq \left( f(R)+C_2(\alpha ,\beta ,\lambda _0)C_h \cdot R^{\alpha } \right)(r/R)^{\beta }. \end{equation*}$$ ■

Remark 3.17.

If the gauge function $h$ satisfy that

$$\begin{equation*} h(t)\leq C\left( \ln \left( \frac{A}{t} \right) \right)^{-b} \end{equation*}$$

for some $A,b,C>0$, then 3.33 implies that there exist $R>0$ and constant $C(R,\lambda ,b)$ such that

$$\begin{equation*} f(r)\leq C(R,\lambda ,b)\left( \ln \left( \frac{A}{r} \right) \right)^{-b}\text{ for }0<r\leq R. \end{equation*}$$

4. Approximation of $E$ by cones at the boundary

In the previous section, we get a power decay of the almost density, and in this section we will use that to get the uniqueness of blow-up limit of $E$ at $0$, and also the estimation $d_{0,r}(E,Z)\leq Cr^{\beta }$ for $r$ small, where $Z$ is the unique blow-up limit, see Theorem 4.14.

We also assume that $E\subseteq \Omega _0$ is a $2$-rectifiable set satisfying (a), (b) and (c). We let $\varepsilon (r)=\varepsilon _P(r)$ if $E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{P}_+$; and let $\varepsilon (r)=\varepsilon _Y(r)$ if $E$ is locally $C^0$-equivalent to a sliding minimal cone of type $\mathbb{Y}_+$.

For any $r>0$, we put

$$\begin{equation*} f(r)=\Theta (0,r)-\Theta (0),\ F(r)=f(r)+8h_1(r),\ F_1(r)=F(r)+8h_1(r), \end{equation*}$$

and for $r\in \mathscr{R}$, we put

$$\begin{equation*} \Xi (r)=rf'(r)+2f(r)+16h(2r)+32h_1(r). \end{equation*}$$

We see from Theorem 2.3 that $F$ is nondecreasing, and $\lim _{r\to 0+}F(r)=0$, thus $\Xi (r)\geq 0$.

We denote by $X(r)$ and $\Gamma (r)$, respectively, the cone $X$ and the set $\Gamma$ which are defined in 3.10, and by $\gamma (r)$ the set $\boldsymbol{\mu }_{r}(\Gamma (r))$. Let $\Pi :\mathbb{R}^3\setminus \{0\}\to \partial B(0,1)$ be the mapping defined by $\Pi (x)=x/|x|$. For any $r_2>r_1>0$, we put $A(r_1,r_2)=\{ x\in \mathbb{R}^{3}: r_1\leq |x|\leq r_2 \}$. Let $\lambda$, $\mu$ and $\mathfrak{r}_1$ be the constants in Theorem 3.15.

Lemma 4.1.

For any $0<r<R<\infty$ with $\mathcal{H}^2(E\cap \partial B_r)=\mathcal{H}^2(E\cap \partial B_R)=0$, we have that

$$\begin{equation} \int _{E\cap A(r,R)} \frac{1-\cos \theta (x)}{|x|^2}d\mathcal{H}^2(x)\leq F(R)-F(r), {\tag{4.1}} \end{equation}$$

and

$$\begin{equation} \mathcal{H}^2\left( \Pi (E\cap A(r,R)) \right)\leq \int _{E\cap A(r,R)}\frac{\sin \theta (x)}{|x|^2}d\mathcal{H}^2(x). {\tag{4.2}} \end{equation}$$

Proof.

We see that for $\mathcal{H}^2$-a.e. $x\in E$, the tangent plane $\operatorname {Tan}(E,x)$ exists, we will denote by $\theta (x)$, the angle between the line $[0,x]$ and the plane $\operatorname {Tan}(E,x)$. For any $t>0$, we put $u(t)=\mathcal{H}^2(E\cap B(0,t))$, then $u:(0,\infty )\to [0,\infty ]$ is a nondecreasing function. By Lemma 2.2, we have that

$$\begin{equation*} u(t)\leq \frac{t}{2}\mathcal{H}^1(E\cap \partial B(0,t))+4t^2h(2t), \end{equation*}$$

for $\mathcal{H}^1$-a.e. $t\in (0,\infty )$. Considering the mapping $\phi :\mathbb{R}^{3}\to [0,\infty )$ given by $\phi (x)=|x|$, we have, by 2.2, that for $\mathcal{H}^2$-a.e. $x\in E$,

$$\begin{equation*} \operatorname {ap}J_1(\phi |_{E})(x)=\cos \theta (x). \end{equation*}$$

Apply Theorem 3.2.22 in 10, we get that

$$\begin{equation*} \begin{aligned} &\int _{E\cap A(r,R)}\frac{1}{|x|^2}\cos \theta (x)d\mathcal{H}^2(x) =\int _{r}^{R}\frac{1}{t^2}\mathcal{H}^1(E\cap \partial B(0,t)dt\\ &\geq 2\int _{r}^{R}\frac{u(t)}{t^3} dt-8\int _{r}^{R}\frac{h(2t)}{t}dt =2\int _{r}^{R}\frac{1}{t^3}\left(\int _{E\cap B(0,t)}d\mathcal{H}^2(x)\right) dt-8(h_1(R)-h_1(r))\\ &=2\int _{E\cap B(0,R)}\left(\int _{\max \{ r,|x| \}}^{R}\frac{1}{t^3} dt\right) d\mathcal{H}^2(x) -8(h_1(R)-h_1(r))\\ &=\int _{E\cap A(r,R)}\frac{1}{|x|^2}d\mathcal{H}^2(x)+r^{-2}u(r)-R^{-2}u(R)- 8(h_1(R)-h_1(r)), \end{aligned} \end{equation*}$$

thus 4.1 holds.

By a simple computation, we get that

$$\begin{equation*} \operatorname {ap}J_2\Pi (x)=\frac{\sin \theta (x)}{|x|^2}, \end{equation*}$$

then applying Theorem 3.2.22 in 10, we will get that 4.2 hold.

■

For any $0<r<R$, if $\mathcal{H}^2(E\cap \partial B_r)=\mathcal{H}^2(E\cap \partial B_R)=0$, by Cauchy-Schwarz inequality, we get from above Lemma that

$$\begin{equation*} \mathcal{H}^2(\Pi (E\cap A(r,R)))\leq \frac{R}{r}\left(2\Theta (0,R) \right)^{1/2}\left( F(R)-F(r) \right)^{1/2}\leq \frac{R}{r}\left(2\Theta (0,R) \right)^{1/2}F(R)^{1/2}. \end{equation*}$$

Lemma 4.2.

For any $r\in (0,\mathfrak{r}_1)\cap \mathscr{R}$, if $\Xi (r)\leq \mu \tau _0$, then

$$\begin{equation*} d_H(\Gamma (r), X(r)\cap \partial B(0,1))\leq 10\mu ^{-1/2}\Xi (r)^{1/2}. \end{equation*}$$

Proof.

By lemma 2.1, we get that

$$\begin{equation*} \frac{1}{r}\mathcal{H}^1(E\cap \partial B(0,r))\leq 2\Theta (0)+rf'(r)+2f(r), \end{equation*}$$

By Theorem 3.15, we get that

$$\begin{align*} &r^2\Theta (0,r)\\ &\qquad \leq (1-\lambda -\mu )\frac{r}{2}\mathcal{H}^1(E\cap \partial B_r)+\mu \frac{r^2}{2}\mathcal{H}^1(X\cap \partial B_1)+\lambda \Theta (0)r^2+4r^2h(2r)\\ &\qquad \leq \frac{(1-\lambda -\mu )r^2}{2}(2\Theta (0)+rf'(r)+2f(r))+ \frac{\mu r^2}{2}\mathcal{H}^1(X\cap \partial B_1) +\lambda \Theta (0)r^2+4r^2h(2r), \end{align*}$$

thus

$$\begin{equation*} \mathcal{H}^1(X\cap \partial B_1)\geq 2\Theta (0)+\frac{2(\lambda +\mu )}{\mu }f(r)-\frac{1-\lambda -\mu }{\mu }rf'(r)- \frac{\mu }{8}h(2r). \end{equation*}$$

By Theorem 2.3, we see that $f(r)+8h_1(r)$ is nondecreasing, thus $f(r)+8h_1(r)\geq 0$ and $rf'(r)+8h(2r)\geq 0$. Hence

$$\begin{equation*} \begin{aligned} j(r)&=\frac{1}{r}\mathcal{H}^1(E\cap B_r)-\mathcal{H}^1(X\cap \partial B_1) \leq \frac{1-\lambda }{\mu }rf'(r)-\frac{2\lambda }{\mu }f(r)+ \frac{8}{\mu }h(2r)\\ &\leq \frac{1}{\mu }(rf'(r)+8h_1(r)+16h(2r))\leq \frac{1}{\mu }\Xi (r). \end{aligned} \end{equation*}$$

Since

$$\begin{equation*} \mathcal{H}^1(X\cap \partial B_1)\leq \mathcal{H}^1(\Gamma _{\ast }(r))\leq \mathcal{H}^1(\Gamma (r))\leq \mathcal{H}^1(\boldsymbol{\mu }_{1/r}(E\cap \partial B_r)), \end{equation*}$$

we have that

$$\begin{equation*} 0\leq \mathcal{H}^1(\Gamma (r))-\mathcal{H}^1(X\cap B_1)\leq j(r)\leq \frac{1}{\mu }\Xi (r), \end{equation*}$$

by Lemma 3.5, we get that for any $z\in \Gamma (r)$,

$$\begin{equation*} \operatorname {dist}\left( z, X\cap \partial B(0,1)\right)\leq 10 \left(\frac{\Xi (r)}{\mu }\right)^{1/2}. \end{equation*}$$ ■

Lemma 4.3.

For any $0<r_1<r_2< (1-\tau )\mathfrak{r}_1$, if $P$ is a plane such that $\mathcal{H}^1(E\cap P\cap B_{\mathfrak{r}_1})<\infty$ and $P\cap \mathcal{X}_r=\emptyset$ for any $r\in [r_1,r_2]$, then there is a compact path connected set

$$\begin{equation*} \mathcal{C}_{P,r_1,r_2}\subseteq E\cap P\cap A(r_2,r_1) \end{equation*}$$

such that

$$\begin{equation*} \mathcal{C}_{P,r_1,r_2}\cap \gamma (t)\neq \emptyset \text{ for }r_1\leq t\leq r_2. \end{equation*}$$

Proof.

We let $\varrho$ be the same as in 3.1. Since $\|\Phi -\operatorname {id}\|_{\infty }\leq \tau \varrho$, we get that

$$\begin{equation*} \Phi ^{-1}\left( E\cap \overline{B(0,r_2)} \right)\subseteq Z_{0,\varrho }\cap \overline{B(0,r_2+\tau \varrho )}. \end{equation*}$$

We put

$$\begin{equation*} \mathbb{X}=Z_{0,\varrho }\cap \overline{B(0,r_2+\tau \varrho )},\ F=\mathbb{X}\cap \Phi ^{-1}(E\cap P_z). \end{equation*}$$

We take $x_1,x_2\in \mathcal{X}_r$, $x_2\neq x_1$, such that $\Phi ^{-1}(x_1)$ and $\Phi ^{-1}(x_2)$ are contained in two different connected components of $\mathbb{X}\setminus F$. By Lemma 3.2, there is a connected closed subset $F_0$ of $F$ such that $\Phi ^{-1}(x)$ and $\Phi ^{-1}(x_2)$ are still contained in two different connected components of $\mathbb{X}\setminus F_0$. Then $F_0\cap \phi ^{-1}(\gamma (t))\neq \emptyset$ for $0<t\leq r_2$; otherwise, if $F_0\cap \phi ^{-1}(\gamma (t_0))=\emptyset$, then $x_1$ and $x_2$ are in the same connected component of $\Phi (\mathbb{X})\setminus \Phi (F_0)$, thus $\Phi ^{-1}(x_1)$ and $\Phi ^{-1}(x_2)$ are in the same connected component of $\mathbb{X}\setminus F_0$, absurd!

Since $\mathcal{H}^1(\Phi (F_0))\leq \mathcal{H}^1(E\cap P_z\cap B_{\varrho })<\infty$, we get that $\Phi (F_0)$ is path connected. We take $z_1\in \Phi (F_0)\cap \gamma (r_1)$ and $z_2\in \Phi (F_0)\cap \gamma (r_2)$, and let $g:[0,1]\to \Phi (F_0)$ be a path such that $g(0)=z_1$ and $g(1)=z_2$. We take $t_1=\sup \{ t\in [0,1]: |g(t)|\leq r_1 \}$ and $t_2=\inf \{ t\in [t_1,1]:|g(t)|\geq r_2 \}$. Then $\mathcal{C}_{z,r_1,r_2}=g([t_1,t_2])$ is our desire set.

■

Lemma 4.4.

Let $T\in [\pi /4,3\pi /4]$ and $\varepsilon \in (0,1/2)$ be given. Suppose that $F$ a $2$-rectifiable set satisfying

$$\begin{equation*} F\subseteq \partial B(0,1)\cap \{ (t\cos \theta ,t\sin \theta ,x_3)\in \mathbb{R}^{3}\mid t\geq 0, |\theta | \leq T/2, |x_3|\leq \varepsilon \}. \end{equation*}$$

Then we have, by putting $\mathcal{P}_{\theta }=\{ (t\cos \theta ,t\sin \theta , x_3)\mid t\geq 0, x_3\in \mathbb{R} \}$, that

$$\begin{equation*} \int _{-T/2}^{T/2}\mathcal{H}^1(F\cap \mathcal{P}_{\theta })d\theta \leq (1+\varepsilon ) \mathcal{H}^2(F) \end{equation*}$$

Proof.

For any $x=(x_1,x_2,x_3)\in F$, we have that $x_1^2+x_2^2+x_3^2=1$ and $|x_3|\leq \varepsilon$, thus $x_1^2+x_2^2\geq 1-\varepsilon ^2$. Since $|\theta |\leq T/2\leq 3\pi /8$, we get that the mapping $\phi :F\to \mathbb{R}$ given by

$$\begin{equation*} \phi (x_1,x_2,x_3)=\arctan \frac{x_2}{x_1} \end{equation*}$$

is well defined and Lipschitz. Moreover, we have that

$$\begin{equation*} \operatorname {ap}J_1\phi (x)=(x_1^2+x_2^2)^{-1/2}\leq (1-\varepsilon ^2)^{-1/2}\leq 1+\varepsilon . \end{equation*}$$

Hence

$$\begin{equation*} \int _{-T/2}^{T/2}\mathcal{H}^1(F\cap \mathcal{P}_{\theta })d\theta =\int _{F}\operatorname {ap}J_1\phi (x)d\mathcal{H}^2(x)\leq (1+\varepsilon )\mathcal{H}^2(F). \end{equation*}$$ ■

For any $0< t_1\leq t_2$, we put $E_{t_1,t_2}=\Pi \left(\{ x\in E:t_1\leq |x|\leq t_2 \}\right)$. For any $t>0$, we put

$$\begin{equation*} \bar{\varepsilon }(t)=\sup \{ \varepsilon (r):r\leq t \}. \end{equation*}$$

Lemma 4.5.

If $r_2> r_1>0$ satisfy that $10(1+r_2/r_1)\bar{\varepsilon }(r_2)<1/2$, then we have that

$$\begin{equation*} \int _{X(t)\cap \partial B(0,1)}\mathcal{H}^1\left( P_z\cap E_{r_1,r_2}\right) d\mathcal{H}^1(z)\leq 2\mathcal{H}^2\left(E_{r_1,r_2}\right),\ \forall r_1\leq t\leq r_2. \end{equation*}$$

Proof.

By Lemma 3.8, we have that, for any $r>0$, if $\varepsilon (r)<1/2$, then

$$\begin{equation*} d_{0,r}(E,X(r))\leq 5\varepsilon (r). \end{equation*}$$

We get so that

$$\begin{align*} d_{0,1}(X(t),X(r_2))=&d_{0,t}(X(t),X(r_2))\leq d_{0,t}(E,X(t))+d_{0,t}(E,X(r_2))\\ &\leq 5\bar{\varepsilon }(r_2)+5\frac{r_2}{t}\bar{\varepsilon }(r_2). \end{align*}$$

Since

$$\begin{equation*} \operatorname {dist}(x,X(r_2))\leq 5r_2\varepsilon (r_2), \text{ for any }x\in E\cap B(0,r_2), \end{equation*}$$

we have that

$$\begin{equation*} \operatorname {dist}(\Pi (x),X(r_2))\leq \frac{5r_2\varepsilon (r_2)}{|x|}, \text{ for any }x\in E\cap A(r_1,r_2), \end{equation*}$$

we get so that

$$\begin{equation*} \operatorname {dist}(\Pi (x),X(t))\leq \frac{5r_2\varepsilon (r_2)}{|x|}+5\bar{\varepsilon }(r_2)+5\frac{r_2}{t}\bar{\varepsilon }(r_2)\leq 10(r_2/r_1+1)\bar{\varepsilon }(r_2)<\frac{1}{2}. \end{equation*}$$

Applying Lemma 4.4, we will get the result.

■

Lemma 4.6.

Let $\varepsilon \in (0,1/2)$ be given. Let $A\subseteq \partial B(0,1)$ be an arc of a great circle such that $0<\mathcal{H}^1(A)\leq \pi$ and

$$\begin{equation*} \operatorname {dist}(x,L_0)\leq \varepsilon , \forall x\in A. \end{equation*}$$

Then

$$\begin{equation*} \operatorname {dist}(x,L_0)\leq \frac{\pi ^2}{2\mathcal{H}^1(A)^2}\int _{A}\operatorname {dist}(x,L_0)d\mathcal{H}^1(x),\ \forall x\in A. \end{equation*}$$

Proof.

We let $P$ be the plane such that $A\subseteq P$, let $v_0\in P\cap L_0 \cap \partial B(0,1)$ and $v_2\in P\cap \partial B(0,1)$ be two vectors such that $v_0$ is perpendicular to $v_1$. Then $A$ can be parametrized as $\gamma :[\theta _1,\theta _2]\to A$ given by

$$\begin{equation*} \gamma (t)=v_0\cos t+ v_1\sin t, \end{equation*}$$

where $\theta _2-\theta _1=\mathcal{H}^1(A)$. We write $v_1=w+w^{\perp }$ with $w\in L_0$ and $w^{\perp }$ perpendicular to $L_0$. Since $\operatorname {ap}J_{1}\gamma (t)=1$ for any $t\in [\theta _1,\theta _2]$, by Theorem 3.2.22 in 10, we have that

$$\begin{equation*} \begin{aligned} \int _{A}\operatorname {dist}(x,L_0)\mathcal{H}^1(x)&=\int _{\theta _1}^{\theta _2}\operatorname {dist}(\gamma (t),L_0)dt= \int _{\theta _1}^{\theta _2}|w^{\perp }\sin t|dt\\ &\geq 2|w^{\perp }|\left(1-\cos \frac{\theta _2-\theta _1}{2}\right)\geq \frac{2(\theta _2-\theta _1)^2}{\pi ^2}|w^{\perp }|, \end{aligned} \end{equation*}$$

and that

$$\begin{equation*} \operatorname {dist}(x,L_0)\leq |w^{\perp }|\leq \frac{\pi ^2}{2\mathcal{H}^1(A)^2}\int _{A} \operatorname {dist}(x,L_0)d\mathcal{H}^1(x). \end{equation*}$$ ■

Lemma 4.7.

Let $r_1$ and $r_2$ be the same as in Lemma 4.3. If $\Xi (r_i)\leq \mu \tau _0$, $10(1+r_2/r_1)\bar{\varepsilon }(r_2)\leq 1$, then we have that

$$\begin{equation*} d_{0,1}(X(r_1),X(r_2))\leq \frac{30r_2}{r_1}\Theta (0,r_2)^{1/2}\cdot F(r_2)^{1/2}+ 20\pi \mu ^{-1/2}\cdot \left( \Xi (r_1)^{1/2}+\Xi (r_2)^{1/2} \right). \end{equation*}$$

Proof.

For $z\in X(r_2)\cap \partial B_1$, if $z\notin \{ y_r \}\cup \mathcal{X}_r$, we will denote by $P_z$ the plane which is through $0$ and $z$ and perpendicular to $\operatorname {Tan}(X(r_2)\cap \partial B_1, z)$. By Lemma 4.2, we have that

$$\begin{equation*} |z-a|\leq 10\mu ^{-1/2}\Xi (r_1)^{1/2},\forall a\in \Gamma (r_2)\cap P_z. \end{equation*}$$

Since $\mathcal{C}_{P_z,r_1,r_2}\cap \gamma (r_i)\neq \emptyset$, $i=1,2$, we take $b_i\in \mathcal{C}_{P_z,r_1,r_2}\cap \gamma (r_i)$, then

$$\begin{equation*} |\Pi (b_1)-\Pi (b_2)|\leq \mathcal{H}^1(\Pi (\mathcal{C}_{P_z,r_1,r_2}))\leq \mathcal{H}^1(P_z\cap E_{r_1,r_2}), \end{equation*}$$

thus

$$\begin{equation*} \begin{aligned} \operatorname {dist}(z,X(r_1)\cap \partial B_1)&\leq |z-\Pi (b_2)|+|\Pi (b_2)-\Pi (b_1)|+\operatorname {dist}(\Pi (b_1),X(r_1)\cap \partial B_1)\\ &\leq \mathcal{H}^1(P_z\cap E_{r_1,r_2})+10\mu ^{-1/2}\left( \Xi (r_1)^{1/2}+ \Xi (r_2)^{1/2} \right). \end{aligned} \end{equation*}$$

For any $x\in \mathcal{X}_r$, we let $A_x$ be the arc in $\partial B(0,1)$ which join $\Pi (x)$ and $\Pi (y_r)$, We see that $X(r_2)\cap \partial B(0,1)=\cup _{x\in \mathcal{X}_r}A_x$, and $\mathcal{H}^1(A_x)\geq (1/2-\bar{\varepsilon }(r_2))\pi \geq \pi /4$. Suppose $z\in A_x$, then

$$\begin{equation*} \begin{aligned} &\operatorname {dist}(z,X(r_1))\\ &\qquad \leq \frac{\pi ^2}{2\mathcal{H}^1(A_x)^2}\int _{A_x}\operatorname {dist}(z,X(r_1))d\mathcal{H}^1(x)\\ &\qquad \leq \frac{2\pi }{\mathcal{H}^1(A_x)} \int _{A_x}\mathcal{H}^1(P_z\cap E_{r_1,r_2})d\mathcal{H}^1(x)+ 20\pi \mu ^{-1/2}\left( \Xi (r_1)^{1/2}+ \Xi (r_2)^{1/2} \right)\\ &\qquad \leq 16 \mathcal{H}^2(E_{r_1,r_2})+20\pi \mu ^{-1/2}\left( \Xi (r_1)^{1/2}+ \Xi (r_2)^{1/2} \right)\\ &\qquad \leq \frac{16r_2}{r_1}\left(2\Theta (0,r_2) \right)^{1/2}F(r_2)^{1/2}+ 20\pi \mu ^{-1/2}\left( \Xi (r_1)^{1/2}+ \Xi (r_2)^{1/2} \right). \end{aligned} \end{equation*}$$ ■

Remark 4.8.

It is easy to see that, for any cones $X_1$ and $X_2$,

$$\begin{equation*} d_H(X_1\cap \partial B(0,1),X_2\cap \partial B(0,1))\leq 2d_{0,1}(X_1,X_2). \end{equation*}$$

Since $\Xi (r)=rf'(r)+2f(r)+16h(2r)+32h_1(r)$ and $F_1(r)=f(r)+16h_1(r)$, we see that $\Xi (r)=[rF_1(r)]'$ for any $r\in \mathscr{R}$, we get so that

$$\begin{equation*} \int _{r_1}^{r_2}\Xi (t)dt\leq r_2F_1(r_2)-r_1F_1(r_1). \end{equation*}$$

For any $\zeta >2$, if $r_1\leq r_2\leq r$, then by Chebyshev’s inequality, we get that,

$$\begin{equation*} \mathcal{H}^1\left( \left\{ t\in [r_1,r_2] \middle | \Xi (t)\leq \zeta F_1(r)^{2/3} \right\} \right)\geq r_2-r_1-\frac{1}{\zeta }rF_1(r)^{1/3}, \end{equation*}$$

thus $\left\{ t\in [r_1,r_2] \middle | \Xi (t)\leq \zeta F_1(r)^{2/3} \right\}\neq \emptyset$ when $r_2-r_1>(1/\zeta ) rF_1(r)^{1/3}$.

Lemma 4.9.

Let $R_0<(1-\tau )\mathfrak{r}_1$ be a positive number such that $F(R_0)\leq \mu \tau _0/4$ and $\bar{\varepsilon }(R_0)\leq 10^{-4}$. For any $r\in \mathscr{R}\cap (0,R_0)$, if $\Xi (r)\leq \mu \tau _0$, then there is a constant $C=C(\mu ,\Theta (0))$ such that

$$\begin{equation*} \operatorname {dist}(x, E)\leq C r\left(F_1(r)^{1/3}+\Xi (r)^{1/2}\right), \ x\in X(r)\cap B_r. \end{equation*}$$

Proof.

For any $k\geq 0$, we take $r_k=2^{-k}r$. Then there exists $t_k\in [r_k,r_{k-1}]$ such that

$$\begin{equation*} \Xi (t_k)\leq \frac{\int _{r_k}^{r_{k-1}}\Xi (t)dt}{r_{k-1}-r_k}\leq \frac{r_{k-1}F_1(r_{k-1})}{r_{k-1}/2}=2F_1(r_{k-1}). \end{equation*}$$

We let $X_k=X(t_k)$, then for any $j>i\geq 1$, we have that

$$\begin{equation} \begin{aligned} &d_{0,1}(X_i,X_j)\\ &\qquad \leq \sum _{k=i}^{j-1}d_{0,1}(X_k,X_{k+1})\\ &\qquad \leq 60 \left( \Theta (0)+\mu \tau _0/4 \right)^{1/2} \sum _{k=i}^{j-1}F_1(t_k)^{1/2}+ 20\pi \mu ^{-1/2}\sum _{k=i}^{j-1} \left( \Xi (t_k)^{1/2}+\Xi (t_{k+1})^{1/2} \right) \\ &\qquad \leq \left( 60 \left( \Theta (0)+\mu \tau _0/4 \right)^{1/2}+40\pi \mu ^{-1/2} \right)\sum _{k=i}^{j-1}2F_1(t_k)^{1/2}+F_1(t_{k-1})^{1/2}\\ &\qquad \leq C_1(\mu ,\Theta (0))(j-i)F_1(r_{i-1})^{1/2}=C_1(\mu ,\Theta (0)) F_1(r_{i-1})^{1/2}\log _2 (r_i/r_j), \end{aligned} {\tag{4.3}} \end{equation}$$

where $C_1(\mu ,\Theta (0))=3\left( 60 \left( \Theta (0)+\mu \tau _0/4 \right)^{1/2}+40\pi \mu ^{-1/2} \right)$.

For any $x\in X(r)\cap B_r$ with $\Xi (|x|)\leq \mu \tau _0$, we assume that $t_{k+1}\leq |x|< t_k$, then

$$\begin{equation*} \begin{aligned} &\operatorname {dist}(x,E)\\ &\qquad \leq d_H(X(r)\cap B_{|x|},X(|x|)\cap B_{|x|})+d_H(X(|x|)\cap B_{|x|},\gamma (|x|))\\ &\qquad \leq 2|x|d_{0,1}(X(r),X(|x|))+10\mu ^{-1/2}|x|\Xi (|x|)^{1/2} \\ &\qquad \leq 2|x|(d_{0,1}(X(|x|),X_k)+d_{0,1}(X_k,X_1)+d_{0,1}(X_1,X(r))) +10\mu ^{-1/2}|x|\Xi (|x|)^{1/2}\\ &\qquad \leq (40\pi +10)\mu ^{-1/2}|x| \left( \Xi (|x|)^{1/2}+\Xi (r)^{1/2}\right)+ C_2(\mu ,\Theta (0))|x|F_1(r)^{1/2}\log _{2}(r/|x|)\\ &\qquad \leq (40\pi +10)\mu ^{-1/2} |x|\Xi (|x|)^{1/2}+C_3(\mu ,\Theta (0))r \left( \Xi (r)^{1/2}+ F_1(r)^{1/2} \right) \end{aligned} \end{equation*}$$

For any $0\leq a\leq b\leq r$, we put

$$\begin{equation*} I(a,b)=\left\{ t\in [a,b] \middle | \Xi (t)\leq F_1(r)^{2/3} \right\}, \end{equation*}$$

then $I(a,b)\neq \emptyset$ when $b-a> rF_1(r)^{1/3}$. If $|x|\in I(0,r)$, then

$$\begin{equation*} \operatorname {dist}(x,E)\leq C_4(\mu ,\Theta (0))r\left(F_1(r)^{1/3} +\Xi (r)^{1/2}\right). \end{equation*}$$

We let $\{ s_i \}_{i=0}^{m+1}\subseteq [0,r]$ be a sequence such that

$$\begin{equation*} 0=s_0<s_1<\cdots <s_m<s_{m+1}=r,\ s_i\in I(0,r), \end{equation*}$$

and

$$\begin{equation*} s_{i+1}-s_i\leq 2rF_1(r)^{1/3}. \end{equation*}$$

For any $x\in X(r)\cap B_r$, if $s_i\leq |x|<s_{i+1}$ for some $0\leq i\leq m$, we have that

$$\begin{equation*} \begin{aligned} \operatorname {dist}(x,E)&\leq \left|x-\frac{s_i}{|x|}x\right| +\operatorname {dist}\left( \frac{s_i}{|x|}x,E\right)\\ &\leq (s_{i+1}-s_i)+ C_4(\mu ,\Theta (0))r\left(F_1(r)^{1/3} +\Xi (r)^{1/2}\right)\\ &\leq (C_4(\mu ,\Theta (0))+2) r \left(F_1(r)^{1/3} +\Xi (r)^{1/2}\right). \end{aligned} \end{equation*}$$ ■

Definition 4.10.

Let $U\subseteq \mathbb{R}^{3}$ be an open set, $E\subseteq \mathbb{R}^{3}$ be a set of Hausdorff dimension $2$. $E$ is called Ahlfors-regular in $U$ if there is a $\delta >0$ and $\xi _0\geq 1$ such that, for any $x\in E\cap U$, if $0<r<\delta$ and $B(x,r)\subseteq U$, we have that

$$\begin{equation*} \xi _0^{-1} r^2\leq \mathcal{H}^2(E\cap B(x,r))\leq \xi _0 r^2. \end{equation*}$$

Lemma 4.11.

Let $R_0$ be the same as in Lemma 4.9. If $E$ is Ahlfors-regular, and $r\in \mathscr{R}\cap (0,R_0)$ satisfies $\Xi (r)\leq \mu \tau _0$, then there is a constant $C=C(\mu ,\xi _0,\Theta (0))$ such that

$$\begin{equation*} \operatorname {dist}(x,X(r))\leq Cr\left(F_1(r)^{1/4}+\Xi (r)^{1/2}\right),\ x\in E\cap B(0,9r/10). \end{equation*}$$

Proof.

Let $\{X_k\}_{k\geq 1}$ be the same as in 4.3. For any $t\in \mathscr{R}$ with $t_{k+1}\leq t< t_k$, $\Xi (t)\leq \mu \tau _0$ and $x\in \gamma (t)$, we have that

$$\begin{equation*} \begin{aligned} \operatorname {dist}(x,X(r))&\leq d_H(\gamma (t),X(|x|)\cap B_{|x|})+d_H(X(|x|)\cap B_{|x|},X(r))\\ &\leq (40\pi +10)\mu ^{-1/2} |x|\Xi (|x|)^{1/2}+C_3(\mu ,\Theta (0))r \left( \Xi (r)^{1/2}+ F_1(r)^{1/2} \right) \end{aligned} \end{equation*}$$

We put

$$\begin{equation*} J(0,r)=\{ t\in [0,r]:\Xi (t)> F_1(r)^{1/2} \}. \end{equation*}$$

For any $x\in \gamma (t)$ with $t\in (0,r)\setminus J(0,r)$, we have that

$$\begin{equation*} \operatorname {dist}(x,X(r))\leq C_5(\mu ,\Theta (0))r\left( \Xi (r)^{1/2} + F_1(r)^{1/4}\right). \end{equation*}$$

We put

$$\begin{equation*} E_1=\bigcup _{t\in J(0,r)}(E\cap \partial B_t),\ E_2= \bigcup _{t\in (0,r)\setminus J(0,r)}(E\cap B_t\setminus \gamma (t)), \end{equation*}$$

and

$$\begin{equation*} E_3=E\cap B_r\setminus (E_1\cup E_2)=\bigcup _{t\in (0,r)\setminus J(0,r)}\gamma (t). \end{equation*}$$

Then

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E_1\cup E_2)&=\int _{E\cap B_r}d \mathcal{H}^2(x)-\int _{E_3}d\mathcal{H}^2(x) \leq \int _{E\cap B_r}d \mathcal{H}^2(x)-\int _{E_3}\cos \theta (x)d\mathcal{H}^2(x)\\ &=\int _{E\cap B_r}(1-\cos \theta (x))d \mathcal{H}^2(x)+\int _{ E_1\cup E_2}\cos \theta (x)d\mathcal{H}^2(x)\\ &\leq r^2F(r)+\int _{0}^{r}\mathcal{H}^1(E_1\cap \partial B_t)d t +\int _{0}^{r}\mathcal{H}^1(E_2\cap \partial B_t) dt\\ &\leq r^2F(r)+\int _{J(0,r)}(2\Theta (0)+tf'(t)+2f(t))td t + \mu ^{-1}\int _{0}^rt\Xi (t) d t\\ &\leq (2+\mu ^{-1})r^2F_1(r)+2\Theta (0)\int _{\{ t\in [0,r]:\Xi (t)> F_1(r)^{1/2} \}} t dt\\ &\leq (2+\mu ^{-1})r^2F_1(r)+\frac{2\Theta (0)}{ F_1(r)^{1/2}}\int _0^rt\Xi (t) d t \leq C_6(\mu ,\Theta (0))r^2F_1(r)^{1/2}, \end{aligned} \end{equation*}$$

where $C_6(\mu ,\Theta (0))=(2+\mu ^{-1})(\mu \tau _0/4)^{1/2}+2\Theta (0)$.

We see that, for any $x\in E_3$,

$$\begin{equation*} \operatorname {dist}(x,X(r))\leq C_5(\mu ,\Theta (0)) r\left(\Xi (r)^{1/2} + F_1(r)^{1/4} \right). \end{equation*}$$

If $x\in E\cap B(0,9r/10)$ with

$$\begin{equation*} \operatorname {dist}(x,X(r))>C_5(\mu ,\Theta (0)) r\left(\Xi (r)^{1/2} + F_1(r)^{1/4} \right)+s \end{equation*}$$

for some $s\in (0,r/10)$, then $E\cap B(x,s)\subseteq E_1\cup E_2$, thus

$$\begin{equation*} \mathcal{H}^2(E\cap B(x,s))\leq C_6(\mu ,\Theta (0))r^2F_1(r)^{1/2}. \end{equation*}$$

But on the other hand, by Ahlfors-regular property of $E$, we have that

$$\begin{equation*} \mathcal{H}^2(E\cap B(x,s))\geq \xi _0^{-1}s^2. \end{equation*}$$

We get so that

$$\begin{equation*} s\leq C_6(\mu ,\Theta (0))^{1/2}\cdot \xi _0^{1/2}\cdot r F_1(r)^{1/4} . \end{equation*}$$

Therefore, for $x\in E\cap B(0,9r/10)$,

$$\begin{equation*} \operatorname {dist}(x,X(r))\leq \left(C_6(\mu ,\Theta (0))^{1/2}\cdot \xi _0^{1/2}+C_5(\mu ,\Theta (0))\right) \left(\Xi (r)^{1/2} + F_1(r)^{1/4} \right) . \end{equation*}$$ ■

For any $k\geq 0$, we take $R_k=2^{-k}R_0$ and $s_k\in [R_{k+1},R_{k}]$ such that

$$\begin{equation*} \Xi (s_k)\leq \frac{\int _{R_{k+1}}^{R_{k}}\Xi (t) d t}{R_{k}-R_{k+1}}\leq 2F_1(R_{k}). \end{equation*}$$

We put $X_k=X(s_k)$. Then for any $j\geq i\geq 2$, we have that

$$\begin{equation*} \begin{aligned} &d_{0,1}(X_i,X_j)\\ &\qquad \leq \frac{C_1(\mu ,\Theta (0))}{3}\sum _{k=i}^{j-1} \left(2F_1(s_k)^{1/2}+F_1(s_{k-1})^{1/2}\right) \leq C_1(\mu ,\Theta (0))\sum _{k=i-1}^{j-1} F_1(R_{k})^{1/2}\\ &\qquad \leq \frac{C_1(\mu ,\Theta (0))}{\ln 2}\sum _{k=i-1}^{j-1}\int _{R_k}^{R_{k-1}}\frac{F_1(t)^{1/2}}{t}dt =\frac{C_1(\mu ,\Theta (0))}{\ln 2} \int _{R_{i-2}}^{R_{j-1}}\frac{F_1(t)^{1/2}}{t}dt. \end{aligned} \end{equation*}$$

If the gauge function $h$ satisfy that

$$\begin{equation} \int _0^{R_0}\frac{F_1(t)^{1/2}}{t}dt < +\infty , {\tag{4.4}} \end{equation}$$

then $X_k$ converges to a cone $X(0)$, and

$$\begin{equation*} d_{0,1}(X(0),X_k)\leq \frac{C_1(\mu ,\Theta (0))}{\ln 2}\int _0^{R_{k-2}}\frac{F_1(t)^{1/2}}{t}dt. \end{equation*}$$

Remark 4.12.

If $h(r)\leq C(\ln (A/r))^{-b}$, $0<r\leq R_0$, for some $A>R_0$, $C>0$ and $b>3$, then 4.4 holds.

Indeed,

$$\begin{equation*} h_1(r)=\int _{0}^{r}\frac{h(2t)}{t}dt\leq \frac{C}{b-1}\left( \ln \left( \frac{A}{r} \right)\right)^{-b+1}, \end{equation*}$$

and then Remark 3.17 implies that

$$\begin{equation*} F(r)\leq C_1\left( \ln \left( \frac{A}{r}\right)\right)^{-b}+ \frac{C}{b-1}\left( \ln \left( \frac{A}{r} \right)\right)^{-b+1}\leq C_2\left( \ln \left( \frac{A}{r} \right)\right)^{-b+1}, \end{equation*}$$

thus 4.4 holds.

Lemma 4.13.

If 4.4 holds, then $X(0)$ is a minimal cone.

Proof.

By Lemma 3.8, for any $r\in (0,\mathfrak{r}_1)\cap \mathscr{R}$, there exist sliding minimal cone $Z(r)$ such that $d_{0,1}(X(r),Z(r))\leq 4 \varepsilon (r)$. But $\varepsilon (r)\to 0$ as $r\to 0+$, we get that

$$\begin{equation*} d_{0,1}(Z(s_k),X(0))\to 0. \end{equation*}$$

Since $Z(s_k)$ is sliding minimal for any $k$, we get that $X(0)$ is also sliding minimal.

■

For any $r\in \mathscr{R}\cap (0,R_0)$ with $\Xi (r)\leq \mu \tau _0$, we assume $R_{k+1}\leq r<R_k$, by Lemma 4.7, we have that

$$\begin{equation} \begin{aligned} d_{0,1}(X(0),X(r))&\leq d_{0,1}(X(0),X_{k+3})+d_{0,1}(X_{k+3},X(r))\\ &\leq \frac{C_1(\mu ,\Theta (0))}{\ln 2}\int _0^{R_{k+1}}\frac{F_1(t)^{1/2}}{t}dt\\ &\quad +\frac{30r}{s_{k+3}}\Theta (0,r)^{1/2}F_1(r)^{1/2}+20\pi \mu ^{-1/2}\left( \Xi (s_{k+3})^{1/2}+\Xi (r)^{1/2} \right)\\ &\leq 10C_1(\mu ,\Theta (0))\left(\Xi (r)^{1/2}+F_1(r)^{1/2}+ \int _0^{r}\frac{F_1(t)^{1/2}}{t}dt\right). \end{aligned} {\tag{4.5}} \end{equation}$$

Theorem 4.14.

If 4.4 holds, and $E$ is Ahlfors-regular, then $E$ has unique blow-up limit $X(0)$ at $0$, and there is a constant $C=C_{10}(\mu ,\Theta , \xi _0)$ such that

$$\begin{equation} d_{0,9r/10}(E,X(0))\leq C \left(F_1(r)^{1/4}+ \int _0^{r}\frac{F(t)^{1/2}}{t}dt \right),\ 0<r< R_0, {\tag{4.6}} \end{equation}$$

where $0<R_0<(1-\tau )\mathfrak{r}_1$ satisfying that $F(R_0)\leq \mu \tau _0/4$ and $\bar{\varepsilon }(R_0)\leq 10^{-4}$. In particular,

•

if $h(r)\leq C_h(\ln (A/r))^{-b}$ for some $A,C_h>0$, $b>3$ and $0<r\leq R_0<A$, then$$\begin{equation*} d_{0,r}(E,X(0))\leq C'(\ln (A_1/r))^{-(b-3)/4},\ 0<r\leq 9R_0/10, \ A_1\leq 10A/9; \end{equation*}$$

•

if $h(r)\leq C_hr^{\alpha _1}$ for some $C_h,\alpha _1>0$, and $0<r\leq r_0$, $0<r_0\leq \min \{ 1,R_0 \}$, then$$\begin{equation*} d_{0,r}(E,X(0))\leq C (r/r_0)^{\beta },\ 0<r\leq 9r_0/10,\ 0<\beta <\alpha _1, \end{equation*}$$

where$$\begin{equation*} C\leq C_{11}(\mu ,\lambda _0,\alpha _1,\beta ,C_h,\xi _0,\Theta (0))\left( F(r_0)^{1/4}+r_0^{\alpha _1/4}\right). \end{equation*}$$

Proof.

From 4.5 and Lemma 4.9, we get that, for any $x\in X(0)\cap B_r$ where $r\in \mathscr{R}\cap (0,R_0)$ such that $\Xi (r)\leq \mu \tau _0$,

$$\begin{equation*} \operatorname {dist}(x,E)\leq C_7(\mu ,\xi _0,\Theta (0))r\left( \Xi (r)^{1/2}+F_1(r)^{1/4}+ \int _0^{r}\frac{F_1(t)^{1/2}}{t}dt \right). \end{equation*}$$

Similarly to the proof of Lemma 4.9, we still consider

$$\begin{equation*} I(a,b)=\left\{ t\in [a,b] \middle | \Xi (t)\leq F_1(r)^{2/3} \right\},\ 0\leq a\leq b\leq r, \end{equation*}$$

we have that $I(a,b)\neq \emptyset$ whenever $b-a>rF_1(r)^{1/3}$. We let $\{ s_i \}_{0}^{m+1}\subseteq [0,r]$ be a sequence such that

$$\begin{equation*} 0=s_0<s_1<\cdots <s_m<s_{m+1}=r,\ s_i\in I(0,r), \end{equation*}$$

and

$$\begin{equation*} s_{i+1}-s_i\leq 2rF_1(r)^{1/3}. \end{equation*}$$

For any $r\in (0,R_0)$, we assume that $s_i\leq r<s_{i+1}$, $x\in X(0)\cap \partial B_r$.

$$\begin{equation} \begin{aligned} \operatorname {dist}(x,E)&\leq \left| x-\frac{s_i}{|x|}x\right|+\operatorname {dist}\left( \frac{s_i}{|x|}x,E \right)\\ &\leq C_8(\mu ,\xi _0,\Theta (0)) r \left( F_1(r)^{1/4}+ \int _0^{r}\frac{F_1(t)^{1/2}}{t}dt\right) \end{aligned} {\tag{4.7}} \end{equation}$$

From 4.5 and Lemma 4.11, we have that, for any $x\in X(0)\cap B(0,9r/10)$ where $r\in \mathscr{R}\cap (0,R_0)$ such that $\Xi (r)\leq \mu \tau _0$,

$$\begin{equation*} \operatorname {dist}(x,X(0))\leq C_9(\mu ,\xi _0,\Theta (0))\left( \Xi (r)^{1/2}+F_1(r)^{1/4} + \int _0^{r}\frac{F_1(t)^{1/2}}{t}dt\right). \end{equation*}$$

Similarly to the proof of 4.7, we can get that

$$\begin{equation} \operatorname {dist}(x,X(0))\leq C_{10}(\mu ,\xi _0,\Theta (0)) \left( F_1(r)^{1/4} + \int _0^{r}\frac{F_1(t)^{1/2}}{t}dt \right). {\tag{4.8}} \end{equation}$$

We get, from 4.7 and 4.8, that 4.6 holds.

If $h(r)\leq C_h(\ln (A/r))^{-b}$ for some $A,C_h>0$ and $b>3$ and $0<r\leq R_0<A$, then

$$\begin{equation*} h_1(r)=\int _{0}^{r}\frac{h(2t)}{t}dt \leq \frac{C_h}{b-1}\left( \ln \left( \frac{A}{r} \right) \right)^{-b+1}, \end{equation*}$$

and by Remark 3.17 we have that

$$\begin{equation*} F(r)\leq C''\left( \ln \frac{A}{r} \right)^{-b+1} \end{equation*}$$

where

$$\begin{equation*} C''\leq C(R_0,\lambda ,b)\left(\ln \frac{A}{r}\right)^{-1}+ \frac{C_1}{b-1} \leq C(R_0,\lambda ,b)\left(\ln \frac{A}{R_0}\right)^{-1}+ \frac{C_1}{b-1} \end{equation*}$$

is bounded, thus

$$\begin{equation*} \int _{0}^{r}\frac{F_1(t)^{1/2}}{t}dt \leq C''' \left( \ln \frac{A}{r} \right)^{(-b+3)/2} \end{equation*}$$

Hence we get that

$$\begin{equation*} d_{0,9r/10}(E,X(0))\leq C_{10}(\mu ,\xi _0,\Theta (0)) \left( F_1(r)^{1/4}+\int _{0}^{r}\frac{F_1(t)^{1/2}}{t}dt \right) \leq C' \left( \ln \frac{A}{r} \right)^{-\frac{b-3}{4}}. \end{equation*}$$

If $h(r)\leq C_hr^{\alpha _1}$ for some $C_h,\alpha _1>0$ and $0<r\leq r_0$, then

$$\begin{equation*} h_1(r)=\int _{0}^{r}\frac{h(2t)}{t}dt\leq \frac{C_h}{\alpha _1}(2r)^{\alpha _1}. \end{equation*}$$

We see, from the proof of Corollary 3.16, that

$$\begin{equation*} f(r)\leq \left( f(r_0)+C_2(\alpha _1,\beta ,\lambda _0)C_hr_0^{\alpha _1} \right)(r/r_0)^{\beta },\ \forall 0<\beta <\alpha _1, \end{equation*}$$

thus

$$\begin{equation*} F_1(r)=f(r)+16h_1(r)\leq (f(r_0)+C_2'(\alpha _1,\beta ,\lambda _0) C_hr_0^{\alpha _1})(r/r_0)^{\beta }. \end{equation*}$$

Then

$$\begin{equation*} d_{0,9r/10}(E,X(0))\leq C_{10}(\mu ,\xi _0,\Theta (0)) \left(F_1(r)^{1/4}+\int _{0}^{r}\frac{F_1(t)^{1/2}}{t}dt\right) \leq C (r/r_0)^{\beta /4}, \end{equation*}$$

where

$$\begin{equation*} C\leq C_{10}'(\mu ,\xi _0,\Theta (0)) (F(r_0)^{1/4}+C_{2}''(\alpha _1,\beta ,\lambda _0,C_h)r_0^{1/4}). \end{equation*}$$ ■

5. Parameterization of well approximate sets

Recall that a cone in $\mathbb{R}^3$ is called of type $\mathbb{P}$ if it is a plane; a cone is called of type $\mathbb{Y}$ if it is the union of three half planes with common boundary line and that make $120^{\circ }$ angles along the boundary line; a cone of type $\mathbb{T}$ if it is the cone over the union of the edges of a regular tetrahedron.

Theorem 5.1.

Let $E\subseteq \Omega _0$ be a set with $0\in E$. Suppose that there exist $C>0$, $r_0>0$, $\beta >0$ and $0<\eta \leq 1$ such that, for any $x\in E\cap B(0,r_0)$ and $0<r\leq 2r_0$, we can find cone $Z_{x,r}$ through $x$ such that

$$\begin{equation*} d_{x,r}(E,Z_{x,r})\leq Cr^{\beta }, \end{equation*}$$

where $Z_{x,r}$ is a minimal cone in $\mathbb{R}^3$ of type $\mathbb{P}$ or $\mathbb{Y}$ when $x\notin \partial \Omega _0$ and $0<r<\eta \operatorname {dist}(x,\partial \Omega _0)$, and otherwise, $Z_{x,r}$ is a sliding minimal cone of type $\mathbb{P}_+$ or $\mathbb{Y}_+$ in $\Omega _0$ with sliding boundary $\partial \Omega _0$ centered at some point in $\partial \Omega _0$. Then there exist a radius $r_1\in (0,r_0/2)$, a sliding minimal cone $Z$ centered at $0$ and a mapping $\Phi :\Omega _0\cap B(0,r_1)\to \Omega _0$, which is a $C^{1,\beta }$-diffeomorphism between its domain and image, such that $\Phi (0)=0$, $\Phi (\partial \Omega _0\cap B(0,2r_1))\subseteq \partial \Omega _0$, $\|\Phi -\operatorname {id}\|_{\infty }\leq 10^{-2}r_1$ and

$$\begin{equation*} E\cap B(0,r_1)=\Phi (Z)\cap B(0,r_1). \end{equation*}$$

Proof.

Let $\sigma :\mathbb{R}^3\to \mathbb{R}^3$ be given by $\sigma (x_1,x_2,x_3)=(x_1,x_2,-x_3)$. By setting $E_1=E\cup \sigma (E)$, we have that, for any $x\in E_1\cap B(0,r_0)$ and $0<r\leq 2r_0$, there exist minimal cone $Z(x,r)$ in $\mathbb{R}^3$ centered at $x$ of type $\mathbb{P}$ or $\mathbb{Y}$ such that $Z(\sigma (x),r)=\sigma (Z(x,r))$ and

$$\begin{equation*} d_{x,r}(E,Z(x,r))\leq Cr^{\beta }. \end{equation*}$$

By Theorem 4.1 in 9, there exist $r_1\in (0,r_0)$, $\tau \in (0,1)$, a cone $Z$ centered at $0$ of type $\mathbb{P}$ or $\mathbb{Y}$, and a mapping $\Phi _1:B(0,3r_1/2)\to B(0,2r_1)$ such that

$$\begin{equation*} \begin{gathered} \sigma (Z)=Z,\ \sigma \circ \Phi _1=\Phi _1\circ \sigma ,\ \|\Phi _1-\operatorname {id}\|\leq r_0\tau ,\\ C_1|x-y|^{1+\tau }\leq |\Phi (x)-\Phi (y)|\leq C_1^{-1}|x-y|^{1/(1+\tau )},\\ E_1\cap B(0,r_1)\subseteq \Phi _1(Z\cap B(0,3r_1/2))\subseteq E_1\cap B(0,2r_1). \end{gathered} \end{equation*}$$

Using the same argument as in Section 10 in 3, we get that $\Phi _1$ is of class $C^{1,\beta }$.

■

6. Approximation of $E$ by cones away from the boundary

In this section, we let $\Omega \subseteq \mathbb{R}^3$ be a closed set. Let $E\in SAM(\Omega ,\partial \Omega ,h)$ be a sliding almost minimal set, $x_0\in E\setminus \partial \Omega$. Then $E\cap B(x,r)$ is almost minimal with gauge function $h$ for any $0<r<\operatorname {dist}(x_0,\partial \Omega )$. We put

$$\begin{equation*} F(x,r)=\Theta (x,r)-\Theta (x)+8h_1(r). \end{equation*}$$

We see from Theorem 2.3 that $F(x,r)\geq 0$ and $F(x,\cdot )$ is nondecreasing for $0<r<\operatorname {dist}(x_0,L)$.

Theorem 6.1.

If $\int _{0}^{R_0}r^{-1}F(x,r)^{1/3}dr<\infty$ for some $R_0>0$, then $E$ has unique blow-up limit $T$ at $x$. Moreover there is a constant $C>0$ and a radius $\rho _0=\rho _0(x)>0$ such that

$$\begin{equation*} d_{x,r}(E,T)\leq C \int _{0}^{200r}\frac{F(x,t)^{1/3}}{t}dt,\ 0<r\leq \rho _0. \end{equation*}$$

In particular, if the gauge function $h$ satisfies that

$$\begin{equation*} h(t)\leq C_ht^{\alpha _1} \text{ for some }\alpha _1>0 \text{ and }0<t\leq R_0, \end{equation*}$$

then there exists $\beta _0>0$ such that, for any $0<\beta <\beta _0$,

$$\begin{equation*} d_{x,r}(E,T)\leq C(\alpha _1,\beta )\left(F(x,\rho _0)+ C_h\rho _0^{\alpha _1}\right)^{1/3}(r/\rho _0)^{\beta /3}. \end{equation*}$$

Proof.

By Theorem 16.1 in 4, we get that $E$ is a locally $C^{0,\alpha _2}$-equivalent to a two dimensional minimal cone for some $0<\alpha _2<1$. Let $\varrho$ be the radius defines as in 3.2. We take $\rho _0=10^{-3}\min \{R_0,\operatorname {dist}(x_0,\partial \Omega ),\varrho \}$. By Theorem 11.4 in 5, there is a constant $C>0$ and cone $Z_r$ for each $0<r<\rho _0$ such that

$$\begin{equation*} d_{x,r}(E,Z_r)\leq CF(x,110r)^{1/3}. \end{equation*}$$

We put $\rho _k=2^{-k}\rho _0$, and $Z_k=Z_{\rho _k}$. Then

$$\begin{equation*} \begin{aligned} d_{x,1}(Z_k,Z_{k+1})&=d_{x,\rho _{k+1}}(Z_{k},Z_{k+1})\leq d_{x,\rho _{k+1}}(Z_k,E)+d_{x,\rho _{k+1}}(E,Z_{k+1})\\ &\leq CF(x,110\rho _{k+1})^{1/3}+2CF(x,110\rho _{k})^{1/3}. \end{aligned} \end{equation*}$$

For any $1\leq i<j$, we have that

$$\begin{equation*} \begin{aligned} d_{x,1}(Z_i,Z_j)&\leq 2C \sum _{k=i}^{j-1}F(x,110\rho _{k})^{1/3}+C \sum _{k=i+1}^{j}F(x,110\rho _{k})^{1/3}\\ &\leq 3C \sum _{k=i}^{j}F(x,110\rho _{k})^{1/3}\\ &\leq \frac{3C}{\ln 2} \int _{\rho _j}^{\rho _{i-1}}\frac{F(x,110t)^{1/3}}{t}dt. \end{aligned} \end{equation*}$$

Let $Z_0$ be the limit of $\{Z_k\}_{k=1}^{\infty }$. Then we have that

$$\begin{equation*} d_{x,1}(Z_0,Z_i)\leq \frac{3C}{\ln 2} \int _{0}^{\rho _{i-1}}\frac{F(x,110t)^{1/3}}{t}dt. \end{equation*}$$

For any $0<r<\rho _0$, we assume that $\rho _{k+1}\leq r< \rho _k$, then

$$\begin{equation*} \begin{aligned} d_{x,1}(Z_r,Z_0)&\leq d_{x,\rho _{k+1}}(Z_r,Z_{k+1})+d_{x,1}(Z_{k+1},Z_0)\\ &\leq d_{x,1}(Z_{k+1},Z_0)+d_{x,\rho _{k+1}}(Z_r,E)+ d_{x,\rho _{k+1}}(E,Z_{k+1})\\ &\leq d_{x,1}(Z_{k+1},Z_0)+\frac{r}{\rho _{k+1}}d_{x,r}(Z_r,E)+ d_{x,\rho _{k+1}}(E,Z_{k+1})\\ &\leq 3CF(x,110r)^{1/3}+\frac{3C}{\ln 2} \int _{0}^{\rho _k}\frac{F(x,110t)^{1/3}}{t}dt. \end{aligned} \end{equation*}$$

Hence

$$\begin{equation} d_{x,r}(E,Z_0)\leq d_{x,r}(E,Z_r)+ d_{x,r}(Z_r,Z_0)\leq \frac{10C}{\ln 2} \int _{0}^{200r}\frac{F(x,t)^{1/3}}{t}dt {\tag{6.1}} \end{equation}$$

and $T=\boldsymbol{\tau }_{x}(Z_0)$ is the only blow up limit of $E$ at $x$, which is a minimal cone.

By Theorem 4.5 in 5, we have that

$$\begin{equation*} \Theta _E(x,r)\leq \left(\frac{1}{2}-\alpha _0\right)\frac{\mathcal{H}^1(E\cap \partial B(x,r))}{r} +2 \alpha _0\Theta _E(x)+4h(r), \end{equation*}$$

where we take $\alpha _0$ the constant $\alpha$ in Theorem 4.5 in 5. For our convenient, we denote $u(r)=\mathcal{H}^2(E\cap B(x,r))$ and $f(r)=\Theta _E(x,r)- \Theta _E(x)$, then we have $\mathcal{H}^1(E\cap \partial B(x,r))\leq u'(r)$ and

$$\begin{equation*} \begin{aligned} f(r)+\Theta _E(x)&\leq \left(\frac{1}{2}-\alpha _0\right)\frac{u'(r)}{r}+ 2 \alpha _0 \Theta _E(x)+4h(r)\\ &= \left(\frac{1}{2}-\alpha _0\right)(2f(r)+rf'(r)+2 \Theta _E(x))+ 2 \alpha _0 \Theta _E(x)+4h(r), \end{aligned} \end{equation*}$$

thus

$$\begin{equation*} rf'(r)\geq \frac{4 \alpha _0}{1-2 \alpha _0}f(r)-\frac{8}{1-2 \alpha _0}h(r), \end{equation*}$$

and

$$\begin{equation*} \left( r^{-\frac{4 \alpha _0}{1-2 \alpha _0}}f(r)\right)'\geq -\frac{8}{1-2 \alpha _0}r^{-\frac{1+2 \alpha _0}{1-2 \alpha _0}}h(r). \end{equation*}$$

We take $\beta _0=\min \{4 \alpha _0/(1-2 \alpha _0),\alpha _1\}$. Then for any $0<\beta <\beta _0$, we have that

$$\begin{equation*} \begin{aligned} f(r)&\leq (r/\rho _0)^{\frac{4 \alpha _0}{1-2 \alpha _0}} f(\rho _0)+\frac{8}{1-2 \alpha _0}r^{\frac{4 \alpha _0}{1-2 \alpha _0}}\int _{r}^{\rho _0}t^{-\frac{1+2 \alpha _0}{1-2 \alpha _0}}h(t) dt\\ &\leq (r/\rho _0)^{\frac{4 \alpha _0}{1-2 \alpha _0}} f(\rho _0) + C_1'(\alpha _1,\beta ,\alpha _0) \rho _0^{\alpha _1} \cdot (r/\rho _0)^{\beta }. \end{aligned} \end{equation*}$$

We get so that

$$\begin{equation*} F(x,r)\leq C(\alpha _1,\beta ,\alpha _0)(F(x,\rho _0)+C_h\rho _0^{\alpha _1}) (r/\rho _0)^{\beta }, \end{equation*}$$

combine this with 6.1, we get the conclusion.

■

7. Parameterization of sliding almost minimal sets

Let $n$, $d\leq n$ and $k$ be nonnegative integers, $\alpha \in (0,1)$. By a $d$-dimensional submanifold of class $C^{k,\alpha }$ of $\mathbb{R}^{n}$ we mean a subset $M$ of $\mathbb{R}^{n}$ satisfying that for each $x\in M$ there exist s neighborhood $U$ of $x$ in $\mathbb{R}^{n}$, a mapping $\Phi :U\to \mathbb{R}^{n}$ which is a diffeomorphism of class $C^{k,\alpha }$ between its domain and image, and a $d$ dimensional vector subspace $Z$ of $\mathbb{R}^{n}$ such that

$$\begin{equation*} \Phi (M\cap U)= Z\cap \Phi (U). \end{equation*}$$

In Section 4, we get the estimation $d_{x,r}(E,Z)\leq Cr^{\beta }$ for $x\in E\cap \partial \Omega$ and $0<r<r(x)$, where $\Omega$ is a half space, $E$ is locally sliding almost minimal at $x$, and $r(x)>0$ depends on $x$. In Section 6, we get the estimation $d_{x,r}(E,Z)\leq Cr^{\beta }$ for $x\in E\setminus \partial \Omega$ and $0<r<r(x)$.

In this section, we assume that $\Omega \subseteq \mathbb{R}^{3}$ is a closed set whose boundary $\partial \Omega$ is a 2-dimensional submanifold of class $C^{1,\alpha }$ for some $\alpha \in (0,1)$, and suppose that $\Omega$ has tangent cone a half space at any point in $\partial \Omega$. We will show that $\Omega$ is locally $C^{1,\alpha }$ diffeomorphic to a half space at any point $x_0\in \partial \Omega$, see Lemma 7.1, and after the diffeomorphism $\Psi$, $\Psi (E)$ become a locally sliding almost set at $0$, see Lemma 7.2, so we can apply the results in Section 4 to see that the estimation $d_{x,r}(E,Z)\leq Cr^{\beta }$ for $x\in E\cap \partial \Omega$ and $0<r<r(x)$ is still valid, see Theorem 7.4. But the problem is that $r(x)$ depends on $x$. In fact, we need a uniform control of radius $r(x)$ to apply the Reifenberg’s parameterization theorem, Theorem 5.1, to get our main result Theorem 1.2, and that will be done in Lemma 7.9 and Lemma 7.10.

Let $E\subseteq \Omega$ be a closed set such that $E\in SAM(\Omega ,\partial \Omega ,h)$ and $\partial \Omega \subseteq E$, $x_0\in \partial \Omega$. We always assume that the gauge function $h$ satisfies that

$$\begin{equation} \int _{0}^{R_0}\frac{1}{r}\left(\int _{0}^{r}\frac{h(2t)}{t}dt\right)^{1/2}dr <+\infty {\tag{7.1}} \end{equation}$$

and

$$\begin{equation} \int _{0}^{R_0}r^{-1+\frac{\lambda }{1-\lambda }} \left(\int _{r}^{R_0} t^{-1-\frac{2\lambda }{1-\lambda }}h(2t)dt\right)^{1/2}dr<+\infty , {\tag{7.2}} \end{equation}$$

for some $R_0>0$, where $\lambda$ is the same constant as in Theorem 3.15. It is easy to see that if $h(t)\leq Ct^{\alpha _1}$ for some $\alpha _1>0$, $C>0$ and $0<t\leq R_0$, then 7.1 and 7.2 hold. For our convenient, we still put $\lambda _0=\lambda /(1-\lambda )$, and put

$$\begin{align*} h_2(\rho )&=\int _{0}^{\rho }\frac{1}{r}\left( \int _{0}^{r}\frac{h(2t)}{t} dt\right)^{1/2}dr,\\ h_3(\rho )&=\int _{0}^{\rho }r^{-1+\lambda _0}\left( \int _{r}^{R_0}t^{-1-2\lambda _0}h(2t) dt\right)^{1/2}dr. \end{align*}$$

We see, from Proposition 4.1 in 6, that $E$ is Ahlfors-regular in $B(x_0,R_0)$, i.e. there exist $\delta _1>0$ and $\xi _1\geq 1$ such that for any $x\in E\cap B(x_0,R_0)$, if $0<r<\delta _1$ and $B(x,r)\subseteq B(x_0,R_0)$, we have that

$$\begin{equation*} \xi _1^{-1}r^2\leq \mathcal{H}^2(E\cap B(x,r))\leq \xi _1 r^{2} . \end{equation*}$$

We see from Theorem 3.10 in 9 that there only there kinds of possibility for the blow-up limits of $E$ at $x_0$, they are the plane $\operatorname {Tan}(\partial \Omega ,x_0)$, cones of type $\mathbb{P}_+$ union $\operatorname {Tan}(\partial \Omega ,x_0)$, and cones of type $\mathbb{Y}_+$ union $\operatorname {Tan}(\partial \Omega ,x_0)$. By Proposition 29.53 in 6, we get so that

$$\begin{equation*} \Theta _E(x_0)=\pi ,\ \frac{3\pi }{2}, \text{ or } \frac{7\pi }{4}. \end{equation*}$$

If $\Theta _E(x_0)=\pi$, then there is a neighborhood $U_0$ of $x_0$ in $\mathbb{R}^{3}$ such that $E\cap U_0=\partial \Omega \cap U_0$, see Lemma 5.2 in 9. In the next content of this section, we put ourself in the case $\Theta _E(x_0)=3\pi /2$ or $7\pi /4$.

By Theorem 4.14 and Theorem 1.15 in 5, we see that, for any $x\in E$, there is unique blow-up limit of $E$ at $x$, which coincide with the tangent cone $\operatorname {Tan}(E,x)$.

Lemma 7.1.

For any $R_0>0$, there exist $r_0=r_0(x_0)>0$ and a mapping $\Psi =\Psi _{x_0}:B(0,r_0)\to \mathbb{R}^{3}$, which is a diffeomorphism of class $C^{1,\alpha }$ from $B(0,r_0)$ to $\Psi (B(0,r_0))$, such that

$$\begin{equation*} \Psi (0)=x_0, \Psi (\Omega _0\cap B_{r_0})\subseteq \Omega \cap B(x_0,R_0), \Psi (L_0\cap B_{r_0})\subseteq \partial \Omega \cap B(x_0,R_0), \end{equation*}$$

and that $D\Psi (0)$ is a rotation satisfying that

$$\begin{equation*} D\Psi (0)(\Omega _0)=\operatorname {Tan}(\Omega ,x_0)\text{ and }D\Psi (0)(L_0)=\operatorname {Tan}(\partial \Omega ,x_0). \end{equation*}$$

Proof.

By definition, there exist open sets $U,V \subseteq \mathbb{R}^{3}$ and a diffeomorphism $\Phi :U\to V$ of class $C^{1,\alpha }$ such that $x_0\in U$, $0=\Phi (x_0)\in V$ and

$$\begin{equation*} \Phi (U\cap \partial \Omega )=Z\cap V, \end{equation*}$$

where $Z$ is a plane through $0$. Indeed, we have that

$$\begin{equation*} Z=D\Phi (x_0)\operatorname {Tan}(\partial \Omega ,x_0) \end{equation*}$$

and

$$\begin{equation*} \Phi (U\cap \Omega )=V\cap D\Phi (x_0)\operatorname {Tan}(\Omega ,x_0). \end{equation*}$$

We will denote by $A$ the linear mapping given by $A(v)=D\Phi (x_0)^{-1}v$, and assume that $A(V)=B(0,r)$ is a ball. Let $\Phi _1$ be a rotation such that $\Phi _1(\operatorname {Tan}(\partial \Omega ,x_0))=L_0$ and $\Phi _1(\operatorname {Tan}(\Omega ,x_0))=\Omega _0$. Then we get that $\Phi _1\circ A\circ \Phi$ is also $C^{1,\alpha }$ mapping which is a diffeomorphism between $U$ and $B(0,r)$,

$$\begin{equation*} \begin{gathered} D(\Phi _1\circ A\circ \Phi )(x_0)\operatorname {Tan}( \Omega ,x_0)=\Phi _1(\operatorname {Tan}(\Omega ,x_0))=\Omega _0,\\ D(\Phi _1\circ A\circ \Phi )(x_0)\operatorname {Tan}(\partial \Omega ,x_0)=\Phi _1(\operatorname {Tan}(\partial \Omega ,x_0))=L_0, \end{gathered} \end{equation*}$$

and

$$\begin{equation*} \begin{gathered} \Phi _1\circ A\circ \Phi (U\cap \partial \Omega )=\Phi _1\circ A(Z\cap V)= L_0\cap B(0,r),\\ \Phi _1\circ A\circ \Phi (U\cap \partial \Omega )=\Phi _1\circ A(V\cap D\Phi (x_0)\operatorname {Tan}(\Omega ,x_0))=\Omega _0\cap B(0,r). \end{gathered} \end{equation*}$$

We now take $r_0=r$ and $\Psi =(\Phi _1\circ A\circ \Phi )^{-1}\vert _{B(0,r)}$ to get the result.

■

Let $U\subseteq \mathbb{R}^{n}$ be an open set. For any mapping $\Psi :U\to \mathbb{R}^{n}$ of class $C^{1,\alpha }$, we will denote by $C_{\Psi }$ the constant $C_{\Psi }=\sup \left\{\|D\Psi (x)-D\psi (y)\|/|x-y|^{\alpha }: x,y\in U,x\neq y\right\}$. Then we have that

$$\begin{equation*} \Psi (x)-\Psi (y)=\left\langle x-y,\int _{0}^{1}D\Psi (y+t(x-y))dt\right\rangle , \end{equation*}$$

and thus

$$\begin{equation} |\Psi (x)-\Psi (y)-D\Psi (y)(x-y)|\leq |x-y|\int _{0}^{1}C_{\Psi }(t|x-y|)^{\alpha }dt \leq \frac{C_\Psi }{\alpha +1}|x-y|^{1+\alpha }. {\tag{7.3}} \end{equation}$$

For any $0<\rho \leq r_0$, we set $U_{\rho }=\Psi (B_{\rho })$, $M_{\rho }=\Psi ^{-1}(E\cap U_{\rho })$ and

$$\begin{equation} \Lambda (\rho )=\max \left\{ \operatorname {Lip}\left(\Psi _{B_{\rho }}\right), \operatorname {Lip}\left(\Psi ^{-1}_{U_{\rho }}\right)\right\}. {\tag{7.4}} \end{equation}$$

Then

$$\begin{equation*} \|D\Psi (0)\|-\|D\Psi (x)-D\Psi (0)\|\leq \|D\Psi (x)\|\leq \|D\Psi (0)\|+\|D\Psi (x)-D\Psi (0)\|, \end{equation*}$$

thus $1-C_{\Psi }\rho ^{\alpha }\leq \|D\Psi (x)\|\leq 1+C_{\Psi }\rho ^{\alpha }$ for $x\in B_{\rho }$, and we have that

$$\begin{equation} \Lambda (\rho )\leq 1/(1-C_{\Psi }\rho ^{\alpha }) \text{ whenever } C_{\Psi }\rho ^{\alpha }<1. {\tag{7.5}} \end{equation}$$

Lemma 7.2.

For any $1<\rho \leq \min \{r_0, C_{\Psi }^{-1/\alpha }\}$, $M_{\rho }$ is local almost minimal in $B_{\rho }$ at $0$ with gauge function $H$ satisfying that

$$\begin{equation*} H(2r)\leq 4 \Lambda (r)^2h(2\Lambda (r)r)+ 4\xi _1 C_{\Psi } \Lambda (\rho )r^{\alpha } \text{ for }0<r<(1-C_{\Psi }\rho ^{\alpha })\delta _1. \end{equation*}$$

Proof.

For any open set $U\subseteq \mathbb{R}^3$, $M\geq 1$, $\delta >0$ and $\epsilon >0$, we let $GSAQ(U,M,\delta ,\epsilon )$ be the collection of generalized sliding Almgren quasiminimal sets which is defined in Definition 2.3 in 6. We see that

$$\begin{equation*} \operatorname {diam}(U_{\rho })\leq 2\rho \operatorname {Lip}\left( \Psi \vert _{B_{\rho }} \right)\leq 2\rho \Lambda (\rho ) \end{equation*}$$

and

$$\begin{equation*} E\cap U_{\rho }\in GSAQ(U_{\rho },1,\operatorname {diam}(U_{\rho }),h(2\operatorname {diam}(U_{\rho }))), \end{equation*}$$

By Proposition 2.8 in 6, we have that

$$\begin{equation*} M_{\rho }\in GSAQ\left(B_{\rho },\Lambda (\rho )^4,2\rho ,\Lambda (\rho )^4 h\left(2\rho \Lambda (\rho )\right)\right) \end{equation*}$$

By Proposition 4.1 in 6, we get that $M_\rho$ is Ahlfors-regular in $B_{\rho }$. Indeed, we can get a little more, that is, for any $x\in M_{\rho }$ with $0<r\Lambda (\rho )<\delta _1$ and $B(x,r)\subseteq B(0,\rho )$, we have that

$$\begin{equation} \left( \xi _1 \Lambda (\rho ) \right)^{-1}r^2\leq \mathcal{H}^2(M_{\rho }\cap B(x,r))\leq \left( \xi _1 \Lambda (\rho ) \right)r^2. {\tag{7.6}} \end{equation}$$

Let $\{ \varphi _t \}_{0\leq t\leq 1}$ be any sliding deformation of $M_{\rho }$ in $B_{r}$. Then

$$\begin{equation*} \left\{ \Psi \circ \varphi _t\circ \Psi ^{-1} \right\}_{0\leq t\leq 1} \end{equation*}$$

is a sliding deformation of $E$ in $U_{r}$. Hence we get that

$$\begin{equation} \mathcal{H}^2(E\cap U_{r})\leq \mathcal{H}^2(\Psi \circ \varphi _1\circ \Psi ^{-1}(E\cap U_{r}))+ h(2\operatorname {diam}(U_{r}))^2\operatorname {diam}(U_{r})^2 {\tag{7.7}} \end{equation}$$

For any $2$-rectifiable set $A\subseteq B_{\rho }$, by Theorem 3.2.22 in 10, we have that

$$\begin{equation*} \operatorname {ap}J_2(\Psi \vert _{A})(x)= \left\|\wedge _2\left( D\Psi (x)\vert _{\operatorname {Tan}(A,x)} \right)\right\| \end{equation*}$$

and

$$\begin{equation*} \mathcal{H}^2(\Psi (A\cap B_{r}))=\int _{A\cap B_{r}}\operatorname {ap}J_2(\Psi \vert _{A})(x) d\mathcal{H}^2(x) \end{equation*}$$

By 7.5, we get that

$$\begin{equation*} \int _{A\cap B_{r}}(1-C_{\Psi }|x|^{\alpha })^2 d\mathcal{H}^2\leq \mathcal{H}^2(\Psi (A\cap B_{r}))\leq \int _{A\cap B_{r}}(1+C_{\Psi }|x|^{\alpha })^2 d\mathcal{H}^2. \end{equation*}$$

Thus, by taking $A=M_{\rho }$, we have that $M_r=M_{\rho }\cap B_r$, $\Psi (M_r)=E\cap U_r$ and

$$\begin{equation*} \mathcal{H}^2(\Psi (M_{r}))\geq (1-C_{\Psi }\rho ^{\alpha })^2 \mathcal{H}^2(M_{r}) ; \end{equation*}$$

by taking $A=\varphi _1(M_{\rho })$, we have that

$$\begin{equation*} \mathcal{H}^2(\Psi (\varphi _1(M_{\rho })\cap B_r))\leq (1+C_{\Psi }r^{\alpha })^2\mathcal{H}^2(\varphi _1(M_{\rho })\cap B_r) . \end{equation*}$$

Combine these two equations with 7.7 and 7.6, we get that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(\varphi _1(M_{\rho })\cap B_r)&\geq (1+C_{\Psi }r^{\alpha })^{-2}\mathcal{H}^2(\Psi (\varphi _1(M_{\rho })\cap B_r)) \\ &\geq (1+C_{\Psi }r^{\alpha })^{-2} \left(\mathcal{H}^2(E\cap U_{r}) - h(4r \Lambda (r))(2 r\Lambda (r))^2\right) \\ &\geq \left(\frac{1-C_{\Psi }\rho ^{\alpha }}{1+C_{\Psi }r^{\alpha }}\right)^2\mathcal{H}^2(M_r) -\left(\frac{2 r\Lambda (r)}{1+C_{\Psi }r^{\alpha }}\right)^2h(4 r\Lambda (r))\\ &\geq \mathcal{H}^2(M_r)-H(2r)r^2. \end{aligned} \end{equation*}$$ ■

Lemma 7.3.

Let $E_1\subseteq \Omega _0$ be a $2$-rectifiable set, $x\in E_1$, $X$ a cone centered at $0$, $\Phi :\mathbb{R}^{3}\to \mathbb{R}^{3}$ a diffeomorphism of class $C^{1,\alpha }$. Then there exist $C>0$ such that, for any $r>0$ and $\rho >0$ with $B(\Phi (x),\rho )\subseteq \Phi (B(x,r))$,

$$\begin{equation*} d_{\Phi (x),\rho }\left(\Phi (E_1),\Phi (x)+D\Phi (x)X\right)\leq \left(Cr^{\alpha }+\|D\Phi (x)\| d_{x,r}(E_1,x+X)\right)\frac{r}{\rho }. \end{equation*}$$

Proof.

Since $\Phi$ is of class $C^{1,\alpha }$, by 7.3, we have that

$$\begin{equation*} |\Phi (y)-\Phi (x)-D\Phi (x)(y-x)|\leq \frac{C_{\Phi }}{\alpha +1}|x-y|^{1+\alpha }, \end{equation*}$$

by putting $C_1=C_{\Phi }/(\alpha +1)$, we get that

$$\begin{equation*} \operatorname {dist}(\Phi (y),\Phi (x)+D\Phi (x)X)\leq C_1 |y-x|^{1+\alpha } \text{ for } y\in x+X. \end{equation*}$$

For any $z\in E_1\cap B_r$ and $y\in x+X$, we have that

$$\begin{equation*} \begin{aligned} |\Phi (z)-\Phi (y)|&\leq |\Phi (z)-\Phi (y)-D\Phi (x)(z-y)|+\|D\Phi (x)\|\cdot |z-y|\\ &\leq \|D\Phi (x)\|\cdot |z-y| + C_1|z-x|^{1+\alpha }+C_{1}|y-x|^{1+\alpha }, \end{aligned} \end{equation*}$$

thus

$$\begin{equation*} \operatorname {dist}(\Phi (z),\Phi (x+X))\leq \|D\Phi (x)\|rd_{x,r}(E_1,x+X)+2C_1r^{1+\alpha }, \end{equation*}$$

hence

$$\begin{equation} \operatorname {dist}(\Phi (z),\Phi (x)+D\Phi (x)X)\leq \|D\Phi (x)\|rd_{x,r}(E_1,x+X)+3C_1r^{1+\alpha }. {\tag{7.8}} \end{equation}$$

For any $z\in X\cap B_r$, $\Phi (x)+D\Phi (x)z\in \Phi (x)+D\Phi (x)X$, and

$$\begin{equation} \begin{aligned} \operatorname {dist}(\Phi (x)+D\Phi (x)z, \Phi (E_1)) &=\inf \{ |\Phi (y)-\Phi (x)-D\Phi (x)z|:y\in E_1 \}\\ &\leq \inf \{C_1r^{1+\alpha }+\|D\Phi (x)\|\cdot |y-x-z|:y\in E_1\}\\ &\leq \|D\Phi (x)\|rd_{x,r}(x+X,E_1)+C_1r^{1+\alpha }. \end{aligned} {\tag{7.9}} \end{equation}$$

We get from 7.8 and 7.9 that

$$\begin{equation*} d_{\Phi (x),\rho }(\Phi (E_1),\Phi (x)+D\Phi (x)X)\leq \frac{r}{\rho }\left( 3C_1r^{\alpha }+\|D\Phi (x)\|\cdot d_{x,r}(E_1,x+X) \right) \end{equation*}$$ ■

Theorem 7.4.

Let $\Omega$, $E\subseteq \Omega$, $x_0\in \partial \Omega$ and $h$ be the same as in the beginning of this section. Then there is a unique blow-up limit $X$ of $E$ at $x_0$; moreover, if the gauge function $h$ satisfy that

$$\begin{equation} h(t)\leq C_h t^{\alpha _1} \text{ for some }C_h>0, \alpha _1>0 \text{ and }0<t<t_0, {\tag{7.10}} \end{equation}$$

then there exists $\rho _0>0$ such that, for any $0<\beta <\min \{ \alpha ,\alpha _1,2\lambda _0 \}$,

$$\begin{equation*} d_{x_0,\rho }(E,x_0+X)\leq C(\rho /\rho _0)^{\beta /4},\ 0<\rho \leq 9\rho _0/20, \end{equation*}$$

where $C$ is a constant satisfying that

$$\begin{equation*} C\leq C_{20}(\mu ,\lambda _0,\alpha ,\alpha _1,\beta ,\xi _1) (F_E(x_0,2\rho _0)+ C_{\Psi }\rho _0^{\alpha }+C_h\rho _0^{\alpha _1})^{1/4}, \end{equation*}$$

and $F_E(x_0,r)=r^{-2}\mathcal{H}^2(E\cap B(x_0,r))-\Theta _E(x_0)+16h_1(r)$.

Proof.

We take $R_0>0$ such that $R_0<(1-\tau )\mathfrak{r}_1$ and $\bar{\varepsilon }(R_0)\leq 10^{-4}$, let $\Psi$, $r_0$ be the same as in Lemma 7.1. Let $r\in (0,r_0)$ be such that $C_{\Psi }r^{\alpha }\leq 1/2$ and $2r\leq R_0$. Then $\Lambda (r)\leq 2$, see 7.4 and 7.5. By Lemma 7.2, we have that $M_r$ is local almost minimal at 0 with gauge function $H$ satisfying that

$$\begin{equation} H(t)\leq 16 h(2t)+C_rt^{\alpha } ,\ 0<t<r, {\tag{7.11}} \end{equation}$$

where $C_r\in (0, 2^{3-\alpha }\xi _1C_{\Psi })$ is a constant.

We put $f_{M_r}(\rho )=\Theta _{M_r}(0,\rho )-\Theta _{M_r}(0)$, $0<\rho \leq r$. From 3.33 and 3.37, we get that

$$\begin{equation*} \begin{aligned} f_{M_r}(\rho )&\leq \left( r^{-2\lambda _0}f_{M_r}(r) \right) \rho ^{2\lambda _0}+8(1+\lambda _0)\rho ^{2\lambda _0}\int _{\rho }^{r} t^{-1-2\lambda _0}H(2t)dt\\ &\leq \left(r^{-2\lambda _0}f_{M_r}(r)\right) \rho ^{2\lambda _0}+ 2^{7+2\lambda _0}(1+\lambda _0) \rho ^{2\lambda _0} \int _{2\rho }^{2 r} \frac{h(2t)}{t^{1+2\lambda _0}}dt \\ &\quad + 2^{\alpha +3}(1+\lambda _0)C_r\cdot C_1(\alpha ,\beta ,\lambda _0)r^{\alpha }\cdot (\rho /r)^{\beta }, \end{aligned} \end{equation*}$$

where $C_1(\alpha ,\beta ,\lambda _0)$ is the constant in 3.37.

We get from 7.11 that

$$\begin{equation*} H_1(\rho )=\int _{0}^{\rho }\frac{H(2s)}{s}ds\leq 16h_1(2\rho )+\frac{C_r}{\alpha }(2\rho )^{\alpha }, \end{equation*}$$

by setting $F_1(\rho )=f_{M_r}(\rho )+16H_1(\rho )$, we have that

$$\begin{equation*} \begin{aligned} F_1(\rho )&\leq C_{12}(\lambda _0,\alpha ,\beta ,r)(\rho /r)^{\beta } + 2^8h_1(2\rho )+2^{4+\alpha }C_r\alpha ^{-1}\rho ^{\alpha }\\ &\quad + 2^{7+2\lambda _0}(1+\lambda _0) \rho ^{2\lambda _0} \int _{2\rho }^{2 r} \frac{h(2t)}{t^{1+2\lambda _0}}dt, \end{aligned} \end{equation*}$$

where

$$\begin{equation*} C_{12}(\lambda _0,\alpha ,\beta ,r)\leq f_{M_r}(r)+ 2^{\alpha +3}(1+\lambda _0)C_rC_1(\alpha ,\beta ,\lambda _0) r^{\alpha }. \end{equation*}$$

Hence

$$\begin{equation*} \begin{aligned} \int _{0}^{t}\frac{F_1(\rho )^{1/2}}{\rho }d\rho &\leq C_{12}(\lambda _0,\alpha ,\beta ,r)^{1/2}(\beta /2) (t/r)^{\beta } + 16 h_2(2t) +C_{13}(\alpha ,r)t^{\alpha /2}\\ & + 2^{4+\lambda _0}(1+\lambda _0)^{1/2}\int _{0}^{t} \rho ^{-1+\lambda _0}\left(\int _{2\rho }^{2r}\frac{h(2s)}{s^{1+2\lambda _0}}ds \right)^{1/2}d\rho , \end{aligned} \end{equation*}$$

where $C_{13}(\alpha ,r)\leq 2^{3+\alpha /2}\alpha ^{-3/2}C_r^{1/2}$, thus

$$\begin{equation*} \int _{0}^{t}\frac{F_1(\rho )^{1/2}}{\rho }d\rho <+\infty ,\text{ for }0<t\leq r. \end{equation*}$$

We now apply Theorem 4.14, there is a unique blow-up limit $T$ of $M_r$ at $0$, thus there is a unique blow-up limit $X$ of $E$ at $x_0$.

For any $R\in (0,R_0)$, we put

$$\begin{equation*} f_E(x_0,R)=R^{-2}\mathcal{H}^2(E\cap B(x_0,R))-\Theta _E(x_0) \end{equation*}$$

and

$$\begin{equation*} F_E(x_0,R)=f_E(x_0,R)+16h_1(R), \end{equation*}$$

where $h_1(r)=\int _0^r t^{-1}h(2t)dt$. From 7.7 and $B(x_0,\rho /\Lambda (\rho ))\subseteq U_{\rho }\subseteq B(x_0,\rho \Lambda (\rho ))$, we see that

$$\begin{equation*} (1-C_{\Psi }\rho ^{\alpha })^2(f_{M_r}(\rho )+\Theta _E(x_0))\leq \rho ^{-2}\mathcal{H}^2(E\cap U_{\rho })\leq (1+C_{\Psi }\rho ^{\alpha })^2(f_{M_r}(\rho )+\Theta _E(x_0)), \end{equation*}$$

since $\Lambda (\rho )\leq 1/(1-C_{\Psi }\rho ^{\alpha })$, we get so that

$$\begin{equation*} f_{M_r}(\rho )\leq (1-C_{\Psi }\rho ^{\alpha })^{-4}f_E(x_0,\rho \Lambda (\rho ))+ 4\Theta _E(x_0)C_{\Psi }\rho ^{\alpha }, \end{equation*}$$

and

$$\begin{equation*} f_{M_r}(\rho )\geq (1-C_{\Psi }^2\rho ^{2\alpha })^2f_E(x_0,\rho /\Lambda (\rho ))+ 2\Theta _E(x_0)C_{\Psi }^2\rho ^{2\alpha }. \end{equation*}$$

Since $\rho <r$, $C_{\Psi }r^{\alpha }\leq 1/2$, $h_1\geq 0$, $\Theta _E(x_0)\leq 7\pi /4$ and $\Lambda (r)\leq 2$, we get that

$$\begin{equation*} f_{M_r}(\rho )\leq (1-C_{\Psi }\rho ^{\alpha })^{-4}F_E(x,\rho \Lambda (\rho ))+4\Theta _E(x_0)C_{\Psi }\rho ^{\alpha }\leq 16 F_E(x, 2\rho )+(7\pi /2) (\rho /r)^{\alpha }, \end{equation*}$$

and

$$\begin{equation*} C_{12}(\lambda _0,\alpha ,\beta ,r)\leq 16 F_E(x_0,2r)+ 9\xi _1\cdot 2^{\alpha +2} (1+\lambda _0) C_1(\alpha ,\beta ,\lambda _0)+2\Theta _E(x_0). \end{equation*}$$

If $h$ satisfy 7.10, we take $0<\rho _0\leq \min \left\{r,t_0,r_0(x_0),R_0/2,(2C_{\Psi })^{-1/\alpha },1 \right\}$ such that

$$\begin{equation} F_E(x_0,2\rho _0)\leq 10^{-2}\mu \tau _0,\ h_1(2\rho )\leq 10^{-2}\mu \tau _0 \text{ and }(\rho _0/r)^{\alpha }\leq 10^{-2}\mu \tau _0, {\tag{7.12}} \end{equation}$$

then

$$\begin{equation*} h_1(\rho )\leq \frac{C_h}{\alpha _1}(2\rho )^{\alpha _1},\ H_1(\rho )\leq \frac{2^{4+2\alpha _1}C_h}{\alpha _1}\rho ^{\alpha _1}+ \frac{2^{\alpha }C_r}{\alpha }\rho ^{\alpha },\ 0<\rho \leq \rho _0, \end{equation*}$$

and

$$\begin{equation} F_1(\rho )\leq C_{13}(\lambda _0,\alpha ,\beta ,\rho _0,C_h)(\rho /\rho _0)^{\beta }+ 2^{8+\alpha _1}\alpha _1^{-1}C_h\rho ^{\alpha _1}+ C_{14}(\alpha ,\xi _1,C_{\Psi })\rho ^{\alpha }, {\tag{7.13}} \end{equation}$$

where $C_{13}(\lambda _0,\alpha _1,\beta ,\rho _0,C_h)$ and $C_{14}(\alpha ,\xi _1,C_{\Psi })$ are constant satisfying that

$$\begin{equation*} C_{13}(\lambda _0,\alpha _1,\beta ,\rho _0,C_h)\leq C_{12}(\lambda _0,\alpha ,\rho _0)+ 2^{7+4\alpha _1}(1+\lambda _0) C_1(\alpha _1,\beta ,\lambda _0)C_h\rho _0^{\alpha _1} \end{equation*}$$

and

$$\begin{equation*} C_{14}(\alpha ,\xi _1,C_{\Psi })\leq 2^{8+\alpha }\alpha ^{-1}\xi _1C_{\Psi }. \end{equation*}$$

We get so that 7.13 can be rewrite as

$$\begin{equation*} F_{1}(\rho )\leq C_{15}(\lambda _0,\alpha ,\alpha _1,\beta ,\xi _1) (F_E(x_0,2\rho _0)+C_{\Psi }\rho _0^{\alpha }+C_h\rho _0^{\alpha _1}) (\rho /\rho _0)^{\beta /4}. \end{equation*}$$

By Theorem 4.14, we have that

$$\begin{equation*} \begin{aligned} d_{0,9\rho /10}(M_r,T)&\leq C_{16}(\mu ,\xi _0) \left(F_{1}(\rho )^{1/4}+ \int _{0}^{\rho }\frac{F_{1}(t)^{1/2}}{t}dt\right)\\ &\leq C_{17}(\mu ,\lambda _0,\alpha ,\alpha _1,\beta ,\xi _1) G_E(x_0,\rho _0) (\rho /\rho _0)^{\beta /4}, \end{aligned} \end{equation*}$$

where

$$\begin{equation*} G_E(x_0,\rho _0)= (F_E(x_0,2\rho _0)+C_{\Psi }\rho _0^{\alpha }+C_h\rho _0^{\alpha _1})^{1/4}. \end{equation*}$$

Applying Lemma 7.3 with $\Phi =\Psi$, by setting $X=D\Psi (0)T$, we get that for any $\rho \in (0,9\rho _0/10)$,

$$\begin{equation*} \begin{aligned} d_{x_0,\rho /2}(E,x_0+X)&\leq d_{x_0,\rho /\Lambda (\rho )}(E,x_0+D\Psi (0)T)\\ &\leq 6C_{\Psi }\rho ^{\alpha }+2d_{x,\rho }(M_r,T)\\ &\leq 6C_{\Psi }\rho ^{\alpha }+ C_{18}(\mu ,\lambda _0,\alpha ,\alpha _1,\beta ,\xi _1)G_E(x_0,\rho _0) (\rho /\rho _0)^{\beta /4}\\ &\leq C_{19}(\mu ,\lambda _0,\alpha ,\alpha _1,\beta ,\xi _1) G_E(x_0,\rho _0) (\rho /\rho _0)^{\beta /4}. \end{aligned} \end{equation*}$$ ■

Lemma 7.5.

For any $\tau >0$ small enough, there exists $\varepsilon _2=\varepsilon _2(\tau )>0$ such that the following hold: $E$ is an sliding almost minimal set in $\Omega$ with sliding boundary $\partial \Omega$ and gauge function $h$, $x_0\in E\cap \partial \Omega$, $\Psi$ is a mapping as in Lemma 7.1 and $C_{\Psi }$ is the constant as in 7.5, if $r_1>0$ satisfy that $C_{\Psi }r_1^{\alpha }\leq \varepsilon _2$, $h(2r_1)\leq \varepsilon _2$ and $F_E(x_0,r_1)\leq \varepsilon _2$, then for any $r\in (0,9r_1/10)$, we can find sliding minimal cone $Z_{x_0,r}$ in $\operatorname {Tan}(\Omega ,x_0)$ with sliding boundary $\operatorname {Tan}(\partial \Omega ,x_0)$ such that

$$\begin{equation*} \begin{gathered} \operatorname {dist}(x,Z_{x_0,r})\leq \tau r,\ x\in E\cap B(x_0,(1-\tau )r)\\ \operatorname {dist}(x,E)\leq \tau r,\ x\in Z_{x_0,r}\cap B(x_0,(1-\tau )r), \end{gathered} \end{equation*}$$

and for any ball $B(x,t)\subseteq B(x_0,(1-\tau )r)$,

$$\begin{equation*} |\mathcal{H}^2(Z_{x_0,r}\cap B(x,t))-\mathcal{H}^2(E\cap B(x,t))|\leq \tau r^2. \end{equation*}$$

Moreover, if $E\supseteq \partial \Omega$, then $Z_{x_0,r}\supseteq \operatorname {Tan}(\partial \Omega ,x_0)$.

Proof.

It is a consequence of Proposition 30.19 in 6.

■

Corollary 7.6.

Let $\Omega$, $E\subseteq \Omega$, $x_0\in \partial \Omega$, $h$ and $F_E$ be the same as in Theorem 7.4. Suppose that the gauge function $h$ satisfying

$$\begin{equation*} h(t)\leq C_h t^{\alpha _1} \text{ for some }C_h>0, \alpha _1>0 \text{ and }0<t<t_0. \end{equation*}$$

Then there exists $\delta >0$ and constant $C=C_{20}(\mu ,\lambda _0,\alpha ,\alpha _1, \beta ,\xi _1)>0$ for $0<\beta <\min \{ \alpha ,\alpha _1,2\lambda _0 \}$ such that, whenever

$$\begin{equation*} 0<\rho _0\leq \min \left\{r,t_0,r_0(x_0),R_0/2,(2C_{\Psi })^{-1/\alpha },1 ,(1-\tau )\mathfrak{r}_1\right\} \end{equation*}$$

satisfying

$$\begin{equation*} F_E(x_0,2\rho _0)+ C_{\Psi }\rho _0^{\alpha }+C_h\rho _0^{\alpha _1}\leq \delta , \end{equation*}$$

we have that, for $0<\rho \leq 9\rho _0/20$,

$$\begin{equation*} d_{x_0,\rho }(E,x_0+\operatorname {Tan}(E,x_0))\leq C(F_E(x_0,2\rho _0)+ C_{\Psi } \rho _0^{\alpha }+C_h\rho _0^{\alpha _1})^{1/4}(\rho /\rho _0)^{\beta /4}. \end{equation*}$$

Proof.

By Theorem 7.4, there exist $\rho _0>0$ such that

$$\begin{equation*} d_{x_0,\rho }(E,x_0+\operatorname {Tan}(E,x_0))\leq C(\rho /\rho _0)^{\beta /4}, \ 0<\rho \leq 9\rho _0/20, \end{equation*}$$

where $\rho _0>0$ is chosen to be as in Theorem 7.4.

By Lemma 7.5, there exists $\delta >0$ such that if $F_E(x_0,2\rho _0)+C_{\Psi }\rho _0^{\alpha }+C_h\rho _0^{\alpha _1}\leq \delta$, then 7.12 holds, and we get the result.

■

Lemma 7.7.

Let $\Omega ,E$ and $h$ be the same as in Theorem 7.4. We have that

$$\begin{equation*} \overline{E\setminus \partial \Omega }\in SAM(\Omega , \partial \Omega ,h). \end{equation*}$$

Proof.

We will put $E_1=\overline{E\setminus \partial \Omega }$ for convenient. We first show that $\mathcal{H}^2(E_1\cap \partial \Omega )=0$. Indeed, for any $x\in E_1\cap \partial \Omega$, $\Theta _{E}(x)\geq 3\pi /2$. It follows from the fact that for $\mathcal{H}^2$-a.e. $x\in E$, $\Theta _E(x)=\pi$ that $\mathcal{H}^2(E_1\cap \partial \Omega )=0$.

Let $\{\varphi _t\}_{0\leq t\leq 1}$ be any sliding deformation in some ball $B=B(y,r)$. Since $E\supseteq \partial \Omega$ and $E\in SAM(\Omega , \partial \Omega ,h)$, we have that

$$\begin{equation*} \begin{aligned} \mathcal{H}^2(E_1)&=\mathcal{H}^2(E\setminus \partial \Omega )\leq \mathcal{H}^2(\varphi _1(E)\setminus \partial \Omega )+4h(2r)r^2\\ &= \mathcal{H}^2(\varphi _1(E_1)\setminus \partial \Omega )+4h(2r)r^2\\ &\leq \mathcal{H}^2(\varphi _1(E_1))+4h(2r)r^2. \end{aligned} \end{equation*}$$

Thus $E_1\in SAM(\Omega , \partial \Omega ,h)$.

■

Lemma 7.8.

Let $\Omega ,E$, $x_0$ and $h$ be the same as in Theorem 7.4. For any $\varepsilon >0$ small enough, there exists a $\rho _0>0$ such that for any $0<\rho <\rho _0$ and $x\in E\cap B(x_0,\rho )$, there exists $x_1\in B(x_0,5\rho )\cap \partial \Omega$ with $x_1\in \overline{E\setminus \partial \Omega }$ such that

$$\begin{equation*} |x-x_1|\leq (1+\varepsilon )\operatorname {dist}(x,\partial \Omega ). \end{equation*}$$

Proof.

If $\Theta _E(x_0)=\pi$, then there is an open ball $B=B(x_0,r)$ such that $E\cap B=\partial \Omega \cap B$, and we have nothing to prove.

We assume that $\Theta _E(x_0)=3\pi /2$ or $7\pi /4$. We put $E_1=\overline{E\setminus \partial \Omega }$. Then $x_0\in E_1$ and $\Theta _E(x_0)=\pi /2$ or $3\pi /4$, and by Lemma 7.7, we have that $E_1\in SAM(\Omega ,\partial \Omega ,h)$. By Lemma 7.5, for any $\varepsilon \in (0,10^{-3})$, there exists $\rho _0\in (0,r_0)$ such that, for any $0<\rho <\rho _0$, we can find sliding minimal cone $Z_{\rho }$ centered at $x_0$ of type $\mathbb{P}_+$ or $\mathbb{Y}_+$ satisfying that

$$\begin{equation*} d_{x_0,\rho }(E_1,Z_{\rho })\leq \varepsilon . \end{equation*}$$

Let $\Psi :B(0,r_0)\to \mathbb{R}^3$ be the mapping defined in Lemma 7.1, and let $\Lambda$ be the same as in 7.4. We put $U_{\rho }=\Psi (B_{\rho })$, $A_1=\Psi ^{-1}(E_1\cap U_{\rho _0})$. By Lemma 7.3, for any $0<r\leq \rho /\Lambda (\rho )$, there exist sliding minimal cone $X_{r}$ in $\Omega _0$ such that

$$\begin{equation*} d_{0,r}(A_1,X_{r})\leq (C\rho ^{\alpha }+\varepsilon )\frac{\rho }{r}. \end{equation*}$$

Thus there exists $\rho _1>0$ such that for any $0<r\leq \rho _1$, we can find sliding minimal cone $X_r$ of type $\mathbb{P}_+$ or $\mathbb{Y}_+$ such that

$$\begin{equation*} d_{0,r}(A_1,X_r)\leq 2 \varepsilon . \end{equation*}$$

Using the same argument as in the proof Lemma 5.4 in 9, we get that there exists $\rho _2>0$ such that for any $x\in A_1\cap B(0,\rho )$ with $0<\rho \leq \rho _2$, we can find $a\in A_1\cap L_0\cap B(0,3\rho )$ such that

$$\begin{equation*} |P_{L_0}(x)-a|\leq 8 \varepsilon |x-a|, \end{equation*}$$

where we denote by $P_{L_0}$ the orthogonal projection from $\mathbb{R}^3$ to $L_0$. Thus

$$\begin{equation*} |x-a|\leq |x-P_{L_0}(x)|+|P_{L_0}(x)-a|\leq \operatorname {dist}(x,L_0)+8 \varepsilon |x-a|, \end{equation*}$$

and we get that

$$\begin{equation*} \operatorname {dist}(x,A_1\cap L_0\cap B(0,3\rho ))\leq \frac{1}{1-8 \varepsilon } \operatorname {dist}(x,L_0\cap B(0,3\rho )). \end{equation*}$$

We take $\rho _3=\operatorname {dist}(x_0, \mathbb{R}^3\setminus U_{\rho _2})/10$. Then, for any $0<\rho \leq \rho _3$ and $z\in E_1\cap B(x_0,\rho )$,

$$\begin{equation*} \begin{aligned} \operatorname {dist}(z,E_1\cap \partial \Omega \cap B(x_0,5\rho ))&\leq \operatorname {Lip}\left(\Psi \vert _{B(0,3\rho _2)}\right)\operatorname {dist}(\Psi ^{-1}(z),A_1\cap L_0 \cap B(0,3\rho ))\\ &\leq (1-8 \varepsilon )^{-1}\Lambda (3\rho ) \operatorname {dist}(\Psi ^{-1}(z),A_1\cap L_0\cap B(0,3\rho )) \\ &\leq (1-8 \varepsilon )^{-1}\Lambda (3\rho )^2\operatorname {dist}(z,\partial \Omega \cap B(x_0,5\rho )). \end{aligned} \end{equation*}$$

We assume $\rho _2$ to be small enough such that $(1-8 \varepsilon )^{-1}\Lambda (3\rho _2)^2<1+10 \varepsilon$, then

$$\begin{equation*} \operatorname {dist}(z,E_1\cap \partial \Omega \cap B(x_0,5\rho ) )\leq (1+10 \varepsilon ) \operatorname {dist}(z,\partial \Omega \cap B(x_0,5\rho )). \end{equation*}$$ ■

Lemma 7.9.

Let $\Omega ,E$, $x_0$ and $h$ be the same as in Theorem 7.4. Suppose that $\Theta _E(x_0)=3\pi /2$. Then, by putting $E_1= \overline{E\setminus \partial \Omega }$, there exist a radius $r>0$, a number $\beta >0$ and a constant $C>0$ such that, for any $x\in B(x_0,r)\cap E_1$ and $0<\rho <2r$, we can find cone $Z_{x,\rho }$ such that

$$\begin{equation*} d_{x,\rho }(E_1,Z_{x,\rho })\leq C\rho ^{\beta }, \end{equation*}$$

where $Z_{x,\rho }=y+\operatorname {Tan}(E_1,y)$, $y\in E_1\cap B(x,C\rho )$, and $y\in E_1\cap \partial \Omega \cap B(x,C\rho )$ in case $\rho \geq \operatorname {dist}(x,\partial \Omega )/10$.

Proof.

We see that $E=E_1\cup \partial \Omega$, and $F_E(x_0,\rho )=F_{E_1}(x,\rho )+F_{\partial \Omega }(x_0,\rho )$. By Corollary 7.6, there exist $\delta >0$ and $C>0$ such that whenever $0<\rho _0\leq \min \{ 1,t_0,r_0(x_0) \}$ satisfying

$$\begin{equation*} F_{E_1}(x_0,2\rho _0)+ C_{\Psi _{x_0}}\rho _0^{\alpha }+C_h\rho _0^{\alpha _1}\leq \delta . \end{equation*}$$

we have that, for $0<\rho \leq 9\rho _0/20$,

$$\begin{equation*} d_{x_0,\rho }(E_{1},x_0+\operatorname {Tan}(E_{1},x_0))\leq C\delta ^{1/4}(\rho /\rho _0)^{\beta }, \end{equation*}$$

where $0<\beta <\min \{ \alpha ,\alpha _1,2\lambda _0 ,\beta _0\}/4$. We take $\rho _1\in (0,\rho _0)$ such that

$$\begin{equation*} F_{E_1}(x_0,2\rho )+ C_{\Psi _{x_0}}\rho ^{\alpha }+C_h\rho ^{\alpha _1}\leq \min \{\delta /2,\varepsilon _2(\tau )\}, \forall 0<\rho \leq \rho _1. \end{equation*}$$

If $x\in \partial \Omega \cap B(x_0,\rho _1/10)$, we take $t=\rho _1/2$, then apply Lemma 7.5 with $r=|x-x_0|+t$ to get that

$$\begin{equation*} \mathcal{H}^2(E_{1}\cap B(x,t))\leq \mathcal{H}^2(Z_{x,r}\cap B(x,t))+\tau r^2, \end{equation*}$$

thus

$$\begin{equation*} \Theta _{E_1}(x,t)\leq \frac{1}{t^2}\mathcal{H}^2(Z_{x,r}\cap B(x,t))+4\tau \leq \frac{\pi }{2}+C_{\Psi _{x_0}}r^{\alpha }+4\tau , \end{equation*}$$

and

$$\begin{equation*} F_{E_1}(x,t)\leq C_{\Psi _{x_0}}r^{\alpha }+4\tau +16 h_1(t). \end{equation*}$$

We get that $F_{E_1}(x,2\rho )+ C_{\Psi _{x}}\rho ^{\alpha }+C_h\rho ^{\alpha _1}\leq \delta$ for $0<\rho \leq t/2$. Thus

$$\begin{equation} d_{x,r}(E_{1},x+\operatorname {Tan}(E_{1},x))\leq C\delta ^{1/4} (r/t)^{\beta },\ 0<r<9t/20. {\tag{7.14}} \end{equation}$$

By Lemma 7.8, we assume that for any $x\in E_{1}\cap B(x_0,\rho _1/10)$, there exists $x_1\in E_1\cap B(x_0,\rho _1/2)\cap \partial \Omega$ such that

$$\begin{equation*} |x-x_1|\leq 2\operatorname {dist}(x,\partial \Omega ). \end{equation*}$$

If $x\in E_{1}\cap B(x_0,\rho _1/10)\setminus \partial \Omega$, we take $t=t(x)=10^{-3}\operatorname {dist}(x,\partial \Omega )$, then apply Lemma 7.5 with $r=|x-x_1|+t$ to get that

$$\begin{equation*} \mathcal{H}^2(E_{1}\cap B(x,t))\leq \mathcal{H}^2(Z_{x_1,r}\cap B(x,t))+\tau r^2, \end{equation*}$$

thus

$$\begin{equation*} \Theta _{E_1}(x,t)\leq \frac{1}{t^2}\mathcal{H}^2(Z_{x_1,r}\cap B(x,t))+ (1+2\cdot 10^3)^2\tau \leq \pi /2+(1+2\cdot 10^3)^2\tau , \end{equation*}$$

and

$$\begin{equation*} F(x,t)\leq (1+2\cdot 10^3)^2\tau +8h_1(t). \end{equation*}$$

By Theorem 6.1, there is a constent $C_1>0$ such that

$$\begin{equation} d_{x,r}(E_{1},x+\operatorname {Tan}(E_{1},x))\leq C_1 (r/t)^{\beta },\ 0<r<t. {\tag{7.15}} \end{equation}$$

Hence we get from 7.14 and 7.15 that

$$\begin{equation} d_{x,r}(E_{1},x+\operatorname {Tan}(E_{1},x))\leq C_2 (r/t_0)^{\beta },\forall x\in E_{1}\cap B(x_0,\rho _1/10), 0<r<t_0, {\tag{7.16}} \end{equation}$$

where

$$\begin{equation*} t_0=\begin{cases} \rho _1/10,& x\in \partial \Omega ,\\ 10^{-3}\operatorname {dist}(x,\partial \Omega ),&x\notin \partial \Omega . \end{cases} \end{equation*}$$

We take $0<a<\beta /(1+\beta )$. For any $x\in B(x_0,\rho _1/10)\setminus \partial \Omega$, if $r\leq t_0^{1/(1-a)}$, then we get from 7.16 that

$$\begin{equation} d_{x,r}(E_{1},x+\operatorname {Tan}(E_{1},x))\leq C_2r^{a\beta }; {\tag{7.17}} \end{equation}$$

if $t_0^{1/(1-a)}<r<\rho _1/5$, then by 7.16, we have that

$$\begin{equation} \begin{aligned} d_{x,r}(E_{1},x_1+\operatorname {Tan}(E_{1},x_1))&\leq \frac{|x-x_1|+r}{r} d_{x_1,|x-x_1|+r}(E_{1},x_1+\operatorname {Tan}(E_{1},x_1))\\ &\leq C_2\left( 1+\frac{2\cdot 10^3t_0}{r} \right)\left( \frac{r+2\cdot 10^3t_0}{\rho _1/10} \right)^{\beta }\\ &\leq C_5(1+C_6r^{-a})^{\beta +1}r^{\beta }\leq C_7 r^{\beta -a\beta -a}. \end{aligned} {\tag{7.18}} \end{equation}$$

From 7.17 and 7.18, we get so that, for any $0<\beta _1<\min \{a\beta , \beta -a\beta -a \}$ there is a constant $C_8>0$ such that for any $x\in E_{1}\cap B(x_0,\rho _1/10)$ and $0<\rho <\rho _1/5$, we can find cone $Z_{x,\rho }$ such that

$$\begin{equation*} d_{x,\rho }(E_{1},Z_{x,\rho })\leq C_8\rho ^{\beta _1}, \end{equation*}$$

where $Z_{x,\rho }=y+\operatorname {Tan}(E_{1},y)$, $y\in E_{1}\cap B(x,C_8\rho )$, and $y\in E_{1}\cap \partial \Omega \cap B(x,C_8\rho )$ in case $\rho \geq t_0^{1/(1-a)}$.

■

Lemma 7.10.

Let $\Omega ,E$, $x_0$ and $h$ be the same as in Theorem 7.4. Suppose that $\Theta _E(x_0)=7\pi /4$. Then, by putting $E_1= \overline{E\setminus \partial \Omega }$, there exist a radius $r>0$, a number $\beta >0$ and a constant $C>0$ such that, for any $x\in B(x_0,r)\cap E_1$ and $0<\rho <2r$, we can find a cone $Z_{x,\rho }$ such that

$$\begin{equation*} d_{x,\rho }(E_1,Z_{x,\rho })\leq C\rho ^{\beta }, \end{equation*}$$

where $Z_{x,\rho } = y+\operatorname {Tan}(E_1,y)$, $y\in E_1\cap B(x_0,C\rho )$, and $y\in E_1\cap \partial \Omega \cap B(x_0,C\rho )$ in case $\rho \geq \operatorname {dist}(x,\partial \Omega )/10$.

Proof.

By Corollary 7.6, there exist $\delta >0$ and $C>0$ such that whenever $0<\rho _0\leq \min \{ 1,t_0,r_0(x_0) \}$ satisfying

$$\begin{equation*} F_{E_1}(x_0,2\rho _0)+ C_{\Psi _{x_0}}\rho _0^{\alpha }+C_h\rho _0^{\alpha _1}\leq \delta , \end{equation*}$$

we have that, for $0<\rho \leq 9\rho _0/20$,

$$\begin{equation*} d_{x_0,\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))\leq C\delta ^{1/4}(\rho /\rho _0)^{\beta }, \end{equation*}$$

where $0<\beta <\min \{ \alpha ,\alpha _1,2\lambda _0 \}/4$. We take $\rho _1\in (0,\rho _0)$ such that

$$\begin{equation*} F_{E_1}(x_0,2\rho )+ C_{\Psi _{x_0}}\rho ^{\alpha }+C_h\rho ^{\alpha _1}\leq \min \{\delta /2,\varepsilon _2(\tau )\}, \forall 0<\rho \leq \rho _1. \end{equation*}$$

If $x\in \partial \Omega \cap B(x_0,\rho _1/10)$, we take $t=|x-x_0|/2$, then apply Lemma 7.5 with $r=|x-x_0|+t$ to get that

$$\begin{equation*} \mathcal{H}^2(E_1\cap B(x,t))\leq \mathcal{H}^2(Z_{x,r}\cap B(x,t))+\tau r^2, \end{equation*}$$

thus

$$\begin{equation*} \Theta _{E_1}(x,t)\leq \frac{1}{t^2}\mathcal{H}^2(Z_{x,r}\cap B(x,t))+9\tau \leq \frac{\pi }{2}+C_{\Psi _{x_0}}r^{\alpha }+9\tau , \end{equation*}$$

and

$$\begin{equation*} F_{E_1}(x,t)\leq C_{\Psi _{x_0}}r^{\alpha }+9\tau +16 h_1(t). \end{equation*}$$

We get that $F_{E_1}(x,2\rho )+ C_{\Psi _{x}}\rho ^{\alpha }+C_h\rho ^{\alpha _1}\leq \delta$ for $0<\rho \leq t/2$. Thus

$$\begin{equation} d_{x,r}(E_1,x+\operatorname {Tan}(E_1,x))\leq C\delta ^{1/4} (r/t)^{\beta },\ 0<r<9t/20. {\tag{7.19}} \end{equation}$$

By Lemma 7.8, we assume that for any $x\in E_1\cap B(x_0,\rho _1/10)$, there exists $x_1\in E_1\cap B(x_0,\rho _1/5)\cap \partial \Omega$ such that

$$\begin{equation} |x-x_1|\leq 2\operatorname {dist}(x,\partial \Omega ). {\tag{7.20}} \end{equation}$$

If $x\in E_1\cap B(x_0,\rho _1/10)\setminus \partial \Omega$, then $\Theta _{E_1}(x)=\pi$ or $3\pi /2$. We put $t(x)=\operatorname {dist}(x,\partial \Omega )$. If $\Theta _{E_1}(x)=3\pi /2$, we take $t=10^{-3}t(x)$, then apply Lemma 7.5 with $r=|x-x_1|+t$ to get that

$$\begin{equation*} \mathcal{H}^2(E_1\cap B(x,t))\leq \mathcal{H}^2(Z_{x_1,r}\cap B(x,t))+\tau r^2, \end{equation*}$$

thus

$$\begin{equation*} \Theta _{E_1}(x,t)\leq \frac{1}{t^2}\mathcal{H}^2(Z_{x_1,r}\cap B(x,t))+(1+2\cdot 10^3)^2\tau \leq \frac{3\pi }{2}+(1+2\cdot 10^3)^2\tau . \end{equation*}$$

and

$$\begin{equation*} F_{E_1}(x,t)\leq (1+2\cdot 10^3)^2\tau +8h_1(t). \end{equation*}$$

By Theorem 6.1, we have that

$$\begin{equation} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_1 (\rho /t)^{\beta },\ 0<\rho <t. {\tag{7.21}} \end{equation}$$

We put $E_Y=\{x_0\}\cup \{ x\in E\setminus \partial \Omega :\Theta _{E_1}(x)=\pi \}$. If $\Theta _{E_1}(x)=\pi$ and $\operatorname {dist}(x,E_Y)\leq 10^{-2}\operatorname {dist}(x,\partial \Omega )$, we take $x_2\in E_Y$ such that $|x-x_2|\leq 2\operatorname {dist}(x,E_Y)$ and $t=10^{-1}\operatorname {dist}(x,E_Y)$, then apply Lemma 7.24 in 4 with $r=|x-x_2|+t$ to get that

$$\begin{equation*} \mathcal{H}^2(E_1\cap B(x,t))\leq \mathcal{H}^2(Z_{x_2,r}\cap B(x,t))+\tau r^2, \end{equation*}$$

thus

$$\begin{equation*} \Theta _{E_1}(x,t)\leq \frac{1}{t^2}\mathcal{H}^2(Z_{x_2,r}\cap B(x,t))+400\tau \leq \pi +400\tau , \end{equation*}$$

and

$$\begin{equation*} F_{E_1}(x,t)\leq 4\tau +8h_1(t). \end{equation*}$$

By Theorem 6.1, we have that

$$\begin{equation} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_2 (\rho /t)^{\beta },\ 0<\rho <t. {\tag{7.22}} \end{equation}$$

If $\Theta _{E_1}(x)=\pi$ and $\operatorname {dist}(x,E_Y)> 10^{-2}\operatorname {dist}(x,\partial \Omega )$, we take $t=10^{-3}\operatorname {dist}(x,\partial \Omega )$, then apply Lemma 7.5 with $r=|x-x_1|+t$ to get that

$$\begin{equation*} \mathcal{H}^2(E_1\cap B(x,t))\leq \mathcal{H}^2(Z_{x_1,r}\cap B(x,t))+\tau r^2, \end{equation*}$$

thus

$$\begin{equation*} \Theta _{E_1}(x,t)\leq \frac{1}{t^2}\mathcal{H}^2(Z_{x_1,r}\cap B(x,t))+(1+2\cdot 10^3)^2\tau \leq \pi +(1+2\cdot 10^3)^2\tau . \end{equation*}$$

and

$$\begin{equation*} F_{E_1}(x,t)\leq (1+2\cdot 10^3)^2\tau +8h_1(t). \end{equation*}$$

By Theorem 6.1, we have that

$$\begin{equation} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_3 (\rho /t)^{\beta },\ 0<\rho <t. {\tag{7.23}} \end{equation}$$

We get, from 7.19, 7.21, 7.22 and 7.23, so that

$$\begin{equation} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_4 (\rho /t_0)^{\beta },\ x\in E_1\cap B(x_0,\rho _1/10),\ 0<\rho <t_0, {\tag{7.24}} \end{equation}$$

where

$$\begin{equation*} t_0=\begin{cases} \rho _1/2,& x=x_0,\\ |x-x_0|/10,& x\in \partial \Omega \setminus \{x_0\},\\ 10^{-3}\operatorname {dist}(x,\partial \Omega ),&x\notin \partial \Omega , \Theta _{E_1}(x)=3\pi /2\\ 10^{-1}\min \{10^{-2}\operatorname {dist}(x,\partial \Omega ),\operatorname {dist}(x,E_Y)\},&x\notin \partial \Omega , \Theta _{E_1}(x)=\pi .\\ \end{cases} \end{equation*}$$

Claim.

$E_Y\cap B(x_0,\rho _1/2)$ is a $C^{1}$ curve which is perpendicular to $\operatorname {Tan}(\Omega ,x_0)$. Indeed, by biHölder regularity at the boundary, we see that $E_Y\cap B(x_0,\rho _1/2)$ is a curve, and by J. Taylor’s regularity theorem 13, we get that $E_Y\cap B(x_0,\rho _1/2)$ is of class $C^{1}$.

By the claim, we can assume that, there is a constant $\eta _3>0$ such that

$$\begin{equation} \operatorname {dist}(x,\partial \Omega )\geq \eta _3|x-x_0|, \ \forall x\in E_Y\cap B(x_0,\rho _1/10). {\tag{7.25}} \end{equation}$$

We fix $0<\beta _1< \beta _2<\beta /(1+\beta )$ such that $\beta _1\leq \beta _2\beta /(1+\beta )$. By 7.24, we have that, for any $x\in \partial \Omega \cap B(x_0,\rho _1/10)\setminus \{x_0\}$, and any $0<\rho <|x-x_0|/10$,

$$\begin{equation*} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_4 (10\rho /|x-x_0|)^{\beta }. \end{equation*}$$

If $0<\rho \leq (|x-x_0|/10)^{1/(1-\beta _1)}$, then

$$\begin{equation*} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_4 (10\rho /|x-x_0|)^{\beta }=C_6\rho ^{\beta _1\beta }; \end{equation*}$$

if $(|x-x_0|/10)^{1/(1-\beta _1)}<\rho \leq \rho _1/5$, then

$$\begin{equation*} \begin{aligned} d_{x,\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))&\leq \frac{|x-x_0|+\rho }{\rho }d_{x_0,|x-x_0|+\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))\\ &\leq \left(1+10\rho ^{-\beta _1}\right)C_4 \left(\frac{10\rho ^{1-\beta _1}+\rho }{\rho _1/2}\right)^{\beta } \leq C_7\rho ^{\beta -\beta _1-\beta \beta _1}. \end{aligned} \end{equation*}$$

Thus we get that, for any $0<\beta _3\leq \min \{\beta \beta _1,\beta -\beta _1-\beta \beta _1)\}$, there is a constant $C_8$ such that for any $x\in \partial \Omega \cap B(x_0,\rho _1/10)$ and $0<\rho \leq \rho _1/5$ we can find cone $Z_{x,\rho }$ satisfying that

$$\begin{equation} d_{x,\rho }(E_1,Z_{x,\rho })\leq C_8\rho ^{\beta _3}. {\tag{7.26}} \end{equation}$$

If $x\in E_1\cap B(x_0,\rho _1/10)\setminus \partial \Omega$ and $\Theta _{E_1}(x)=3\pi /2$, then for any $0<\rho \leq (10^{-3}\eta _3|x-x_0|)^{ 1/(1-\beta _1)}$, from 7.24 and 7.25, we get that

$$\begin{equation*} 0<\rho < 10^{-3}\eta _3 |x-x_0|\leq t_0 \end{equation*}$$

and

$$\begin{equation*} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_4 (10^3\rho /\operatorname {dist}(x,\partial \Omega ))^{\beta }=C_9\rho ^{\beta _1\beta }; \end{equation*}$$

for any $(10^{-3}\eta _3|x-x_0|)^{1/(1-\beta _1)}<\rho \leq \rho _1/5$, we have that

$$\begin{equation*} \begin{aligned} d_{x,\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))&\leq \frac{|x-x_0|+\rho }{\rho }d_{x_0,|x-x_0|+\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))\\ &\leq \left(1+10^3\eta _3^{-1}\rho ^{-\beta _1}\right)C_4 \left(\frac{10^3\eta _3^{-1}\rho ^{1-\beta _1}+\rho }{\rho _1/2}\right)^{\beta }\\ &\leq C_{10}\rho ^{\beta -\beta _1-\beta \beta _1}. \end{aligned} \end{equation*}$$

Thus we get that, for any $0<\beta _4\leq \min \{\beta \beta _1,\beta -\beta _1-\beta \beta _1)\}$, there is a constant $C_{11}$ such that for any $x\in E_1 \cap B(x_0,\rho _1/10)\setminus \partial \Omega$ with $\Theta _{E_1}(x)=3\pi /2$, and $0<\rho \leq \rho _1/5$ we can find cone $Z_{x,\rho }$ satisfying that

$$\begin{equation} d_{x,\rho }(E_1,Z_{x,\rho })\leq C_{11}\rho ^{\beta _4}. {\tag{7.27}} \end{equation}$$

If $x\in E_1\cap B(x_0,\rho _1/10)\setminus \Omega$, $\Theta _{E_1}(x)=\pi$ and $\operatorname {dist}(x,\partial \Omega )<100\operatorname {dist}(x,E_Y)$, then

$$\begin{equation*} t_0=10^{-3}\operatorname {dist}(x,\partial \Omega ). \end{equation*}$$

From 7.24, we get that, for any $0<\rho < 10^{-3}\operatorname {dist}(x,\partial \Omega )^{1/(1-\beta _1)}$,

$$\begin{equation} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_4(10^3\rho /\operatorname {dist}(x,\partial \Omega ))^{\beta }=10^{3\beta _1\beta }C_4\cdot \rho ^{\beta _1\beta }. {\tag{7.28}} \end{equation}$$

For any $10^{-3}\operatorname {dist}(x,\partial \Omega )^{1/(1-\beta _1)}\leq \rho \leq \rho _1/5$, we take $C_{12}=10^{-3}\cdot 10^{2/(1-\beta _1)}$, then we see that $10^{-3}\operatorname {dist}(x,\partial \Omega )^{1/(1-\beta _1)}\leq C_{12}|x-x_0|^{1/(1-\beta _1)}$. If $\rho \leq C_{12}|x-x_0|^{1/(1-\beta _2)}$, we let $x_1$ be the point chose in 7.20, then we have that $|x-x_1|\leq 2\operatorname {dist}(x,\partial \Omega )\leq 2 (10^3\rho )^{1-\beta _1}$ and

$$\begin{equation*} |x-x_0|\geq \operatorname {dist}(x,E_Y)\geq 100\operatorname {dist}(x,\partial \Omega )\geq 50|x-x_1|, \end{equation*}$$

thus

$$\begin{equation*} |x_1-x_0|\geq |x-x_0|-|x-x_1|\geq (1-1/50) |x-x_0|\geq \frac{49}{50}(C_{12}^{-1}\rho )^{1-\beta _2}, \end{equation*}$$

From 7.24, we get that

$$\begin{equation} \begin{aligned} d_{x,\rho }(E_1,x_1+\operatorname {Tan}(E_1,x_1))&\leq \frac{|x-x_1|+\rho }{\rho }d_{x_1,|x-x_1|+\rho }(E_1,x_1+\operatorname {Tan}(E_1,x_1))\\ &\leq \left(1+2\cdot 10^{3-3\beta _1}\rho ^{-\beta _1}\right)C_4 \left(\frac{2\cdot 10^{3-3\beta _1}\rho ^{1-\beta _1}+\rho }{|x_0-x_1|/10}\right)^{\beta }\\ &\leq 101\cdot C_4\cdot C_{12}^{1-\beta _2}\cdot (2\cdot 10^{3-3\beta _1}+\rho ^{\beta _1})^{1+\beta } \rho ^{\beta \beta _2-\beta \beta _1-\beta _1}\\ &\leq C_{13}\rho ^{\beta \beta _2-\beta \beta _1-\beta _1}. \end{aligned} {\tag{7.29}} \end{equation}$$

If $\rho > C_{12}|x-x_0|^{1/(1-\beta _2)}$, we have that

$$\begin{equation} \begin{aligned} d_{x,\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))&\leq \frac{|x-x_0|+\rho }{\rho }d_{x_0,|x-x_0|+\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))\\ &\leq \left(1+C_{12}^{-1+\beta _2}\rho ^{-\beta _2}\right)C_4 \left(\frac{C_{12}^{-1+\beta _2}\rho ^{1-\beta _2}+\rho }{\rho _1/2}\right)^{\beta }\\ &= 2C_4\rho _1^{-1}(C_{12}^{-1+\beta _2}+\rho )^{1+\beta } \rho ^{\beta -\beta _2-\beta \beta _2}\leq C_{14}\rho ^{\beta -\beta _2-\beta \beta _2}. \end{aligned} {\tag{7.30}} \end{equation}$$

If $x\in E_1\cap B(x_0,\rho _1/10)\setminus \partial \Omega$, $\Theta _{E_1}(x)=\pi$ and $\operatorname {dist}(x,\partial \Omega )\geq 100\operatorname {dist}(x,E_Y)$, then from 7.24, we have that for any $0<\rho <10^{-1}\operatorname {dist}(x,E_Y)^{1/(1-\beta _1)}$,

$$\begin{equation} d_{x,\rho }(E_1,x+\operatorname {Tan}(E_1,x))\leq C_4(10\rho /\operatorname {dist}(x,E_Y))^{\beta }\leq 10^{\beta \beta _1}\cdot C_4\cdot \rho ^{\beta _1\beta }. {\tag{7.31}} \end{equation}$$

For any $10^{-1}\operatorname {dist}(x,E_Y)^{1/(1-\beta _1)}\leq \rho \leq \rho _1/5$, we take $y\in E_Y$ such that $|x-y|\leq 2\operatorname {dist}(x,E_Y)$. We put $C_{16}=10^{-1-2/(1-\beta _1)}$. We see that $10^{-1}\operatorname {dist}(x,E_Y)^{1/(1-\beta _1)}\leq C_{16}\operatorname {dist}(x,\partial \Omega )^{1/(1-\beta _1)}$. If $\rho \leq C_{16}\operatorname {dist}(y,\partial \Omega )^{ 1/(1-\beta _2)}$, then $|x-y|\leq 2\operatorname {dist}(x,E_Y)\leq 2(10\rho )^{1-\beta _1}$. From 7.24, we get that

$$\begin{equation} \begin{aligned} d_{x,\rho }(E_1,y+\operatorname {Tan}(E_1,y))&\leq \frac{|x-y|+\rho }{\rho }d_{y,|x-y|+\rho }(E_1,y+\operatorname {Tan}(E_1,y))\\ &\leq \left(1+2\cdot 10^{1-\beta _1}\rho ^{-\beta _1}\right)C_4 \left(\frac{2\cdot 10^{1-\beta _1}\rho ^{1-\beta _1}+\rho }{10^{-3} \operatorname {dist}(y,\partial \Omega )}\right)^{\beta }\\ &=10^{3\beta }C_4C_{16}^{\beta (1-\beta _1)} \left(2\cdot 10^{1-\beta _1}+\rho ^{\beta _1}\right)^{1+\beta }\rho ^{\beta (\beta _2-\beta _1)- \beta _1}\\ &\leq C_{17}\rho ^{\beta \beta _2-\beta _1-\beta \beta _1}. \end{aligned} {\tag{7.32}} \end{equation}$$

If $\rho > C_{16}\operatorname {dist}(y,\partial \Omega )^{1/(1-\beta _2)}$, we have that $|x-x_0|\geq \operatorname {dist}(x,\partial \Omega )\geq 100\operatorname {dist}(x,E_Y) \geq 50 |x-y|$. Since $y\in E_y$, we see from 7.25 that

$$\begin{equation*} \operatorname {dist}(y,\partial \Omega )\geq \eta _3|y-x_0|\geq \eta _3(|x-x_0|-|x-y|)\geq \eta _3\cdot \frac{49}{50}|x-x_0|, \end{equation*}$$

and $|x-x_0|\leq \frac{50}{49}\eta _3^{-1}\operatorname {dist}(y,\partial \Omega )\leq 2\eta _3^{-1} (C_{16}^{-1}\rho )^{1-\beta _2}$. We get from 7.24 that

$$\begin{equation} \begin{aligned} &d_{x,\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))\\ &\qquad \leq \frac{|x-x_0|+\rho }{\rho }d_{x_0,|x-x_0|+\rho }(E_1,x_0+\operatorname {Tan}(E_1,x_0))\\ &\qquad \leq \left(1+2\eta _3^{-1}C_{16}^{-1+\beta _2}\rho ^{-\beta _2}\right)C_4 \left(\frac{2\eta _3^{-1}C_{16}^{-1+\beta _2}\rho ^{1-\beta _2}+\rho }{\rho _1/2}\right)^{\beta }\\ &\qquad =2C_4\rho _1^{-1}(2\eta _3^{-1}C_{16}^{-1+\beta _2}+\rho ^{\beta _2})^{1+\beta } \rho ^{\beta (1-\beta _2)-\beta _2}\\ &\qquad \leq C_{18}\rho ^{\beta -\beta _2-\beta \beta _2}. \end{aligned} {\tag{7.33}} \end{equation}$$

We get, from 7.28, 7.29, 7.30, 7.31,7.32 and 7.33, that for any $0<\beta _5\leq \min \{\beta \beta _1,\beta \beta _2-\beta _1-\beta \beta _1, \beta -\beta _2-\beta \beta _2\}$, there is a constant $C_{19}>0$ such that for any $x\in E_1\cap B(x_0,\rho _1/10)\setminus \partial \Omega$ with $\Theta _{E_1}(x)=\pi$ and $0<\rho \leq \rho _1/5$, we can find cone $Z_{x,\rho }$ such that

$$\begin{equation} d_{x,\rho }(E_1,Z_{x,\rho })\leq C_{19}\rho ^{\beta _5}. {\tag{7.34}} \end{equation}$$

Hence we get, from 7.26, 7.27 and 7.34, there is a constant $C_{20}>0$ and $C_{21}>0$ such that for any $x\in E_1\cap B(x_0,\rho _1/10)$ and $0<\rho \leq \rho _1/5$, we can find cone $Z_{x,\rho }$ such that

$$\begin{equation*} d_{x,\rho }(E_1,Z_{x,\rho })\leq C_{20}\rho ^{\beta _6}, \end{equation*}$$

where $Z_{x,\rho }=z+\operatorname {Tan}(E_1,z)$ for some $z\in E_1\cap B(x,C_{21}\rho )$, and $z\in E_1\cap \partial \Omega \cap B(x,C_{21}\rho )$ in case

$$\begin{equation*} \rho \geq \max \{ (10^{-3}\eta _3|x-x_0|)^{1/(1-\beta _1)},10^{-3}\operatorname {dist}(x,\partial \Omega )^{1/(1-\beta _1)},C_{16}\operatorname {dist}(y,\partial \Omega )^{1/(1-\beta _2)}\}. \end{equation*}$$ ■

Corollary 7.11.

Let $\Omega$, $E$ and $h$ be the same as in Theorem 7.4. Let $E_1=\overline{E\setminus \partial \Omega }$ and $x_0\in E_1\cap \partial \Omega$. Then there exist a radius $r>0$, a number $\beta >0$ and a constant $C>0$ such that, for any $x\in E_1\cap B(x_0,r)$ and $0<\rho <2r$, we can find cone $Z_{x,\rho }$ such that

$$\begin{equation*} d_{x,\rho }(E_1,Z_{x,\rho })\leq C\rho ^{\beta }, \end{equation*}$$

where $Z_{x,\rho } = y+\operatorname {Tan}(E_1,y)$, $y\in E_1\cap B(x,C\rho )$, and $y\in E_1\cap \partial \Omega \cap B(x,C\rho )$ in case $\rho \geq \operatorname {dist}(x,\partial \Omega )/10$.

Proof.

It is follow from Lemma 7.9 and Lemma 7.10.

■

Lemma 7.12.

Let $\Omega ,E$, $x_0$ and $h$ be the same as in Corollary 7.11. Let $\Psi :B(0,r_0)\to \mathbb{R}^3$ be the mapping defined in Lemma 7.1. Let $R>0$ be such that $\Psi (B(0,R))\subseteq B(x_0,r)$, where $B(x_0,r)$ is the ball considered as in Corollary 7.11. By putting $U=\Psi (B(0,R))$, $M_1=\Psi ^{-1}(E_1\cap U)$, we have that there exist $\rho _3>0$, $\beta >0$, and constant $C>0$ such that for any $z\in M_1\cap B(0,\rho _3)$ and $0<t<2\rho _3$, we can find cone $Z(z,t)$ through $z$ such that

$$\begin{equation*} d_{z,t}(M_1,Z(z,t))\leq Ct^{\beta }, \end{equation*}$$

where $Z(z,t)$ is a minimal cone of type $\mathbb{P}$ or $\mathbb{Y}$ in case $z\in M_1\setminus L_0$ and $0<t<\operatorname {dist}(z,L_0)/2$; and in case $t\geq \operatorname {dist}(z,L_0)/2$ or $z\in L_0$, $Z(z,t)$ is a sliding minimal cone in $\Omega _0$ with sliding boundary $L_0$, if $Z(z,t)\setminus L_0\neq \emptyset$, we can be written as $Z(z,t)=L_0\cup Z'(z,t)$, $Z'(z,t)$ is a sliding minimal cone of type $\mathbb{P}_+$ or $\mathbb{Y}_+$.

Proof.

For any $x\in B(x_0,r)\cap E_1$ and $0<\rho <2r$, we let $Z_{x,\rho }$ be the same cone considered as in Corollary 7.11. We put $\Phi =\Psi ^{-1}\vert _{B(x_0,r)}$ for convenient. For any $y\in E_1\cap B(x_0,r)$, and $z\in E_1\cap B(x,\rho )$, we put $X=\operatorname {Tan}(E_1,y)$, then

$$\begin{equation} \operatorname {dist}(\Phi (z),\Phi (y+X))\leq \operatorname {Lip}(\Phi )\operatorname {dist}(z,y+X)\leq C \operatorname {Lip}(\Phi )\rho ^{1+\beta }. {\tag{7.35}} \end{equation}$$

Since

$$\begin{equation*} |\Phi (z_1)-\Phi (z_2)-D\Phi (z_2)(z_1-z_2)|\leq C_1|z_1-z_2|^{1+\alpha }, \end{equation*}$$

we have that, for any $z_1\in y+X$,

$$\begin{equation} \operatorname {dist}(\Phi (z_1),\Phi (y)+D\Phi (y)X)\leq C_1 |z_1-y|^{1+\alpha }. {\tag{7.36}} \end{equation}$$

Hence, from 7.35 and 7.36, we have that

$$\begin{equation} \operatorname {dist}(\Phi (z),\Phi (y)+D\Phi (y)X)\leq C\operatorname {Lip}(\Phi )\rho ^{1+\beta } + C_1(\rho +C\rho +C\rho ^{1+\beta })^{1+\alpha }\leq C_2 \rho ^{1+\beta }. {\tag{7.37}} \end{equation}$$

For any $v\in X$, we see that $\Phi (y)+D\Phi (y)v\in \Phi (y)+D\Phi (y)X$, and we have that

$$\begin{equation*} \begin{aligned} \operatorname {dist}(\Phi (y)+D\Phi (y)v, M_1)&\leq \operatorname {dist}(\Phi (y)+D\Phi (y)v, \Phi (E_1\cap B(x,\rho )))\\ &=\inf \{|\Phi (z)-\Phi (y)-D\Phi (y)v|:z\in E_1\cap B(x,\rho )\}\\ &\leq \inf \{C_1|z-y|^{1+\alpha }+\operatorname {Lip}(\Phi )|z-y-v|:z\in E_1\cap B(x,\rho )\}\\ &\leq C_1(\rho +C\rho )^{1+\alpha }+\operatorname {Lip}(\Phi )\operatorname {dist}(y+v,E_1). \end{aligned} \end{equation*}$$

Thus there exist $C_3>0$ such that, for any $v\in X$ with $|y+v-x|\leq \rho$,

$$\begin{equation} \operatorname {dist}(\Phi (y)+D\Phi (y)v, M_1)\leq C_3\rho ^{1+\beta }. {\tag{7.38}} \end{equation}$$

We take $0<C_5<C_4<1$ small enough, for example $C_4<(10\operatorname {Lip}(\Phi ))^{-1}$, then for any $C_5\rho \leq t\leq C_4\rho \leq \rho /\operatorname {Lip}(\Phi )-C_1(C\rho )^{1+\alpha }$, we have that $M_1\cap B(\Phi (x),t)\subseteq \Phi (E_1\cap B(x,\rho ))$ and

$$\begin{equation*} [\Phi (y)+D\Phi (y)X]\cap B(\Phi (x),t)\subseteq \{\Phi (y)+D\Phi (y)v:v\in X,y+v\in B(x,\rho )\}. \end{equation*}$$

From 7.37 and 7.38, we get so that

$$\begin{equation*} d_{\Phi (x),t}(M_1,\Phi (y)+D\Phi (y)X)\leq C_6\rho ^{\beta }\leq C_7t^{\beta }, \end{equation*}$$

and

$$\begin{equation*} |\Phi (x)-\Phi (y)|\leq \operatorname {Lip}(\Phi )|x-y|\leq (\operatorname {Lip}(\Phi )CC_5^{-1})t. \end{equation*}$$

Hence

$$\begin{equation*} d_{\Phi (x),t}(M_1,\Phi (y)+D\Phi (y)X)\leq C_7t^{\beta },\ \text{ for any }0<t<C_4\rho _1, \end{equation*}$$

where $\rho _1\in (0,2r)$ satisfy that $C_1C^{1+\alpha }\rho _1\leq \operatorname {Lip}(\Phi )^{-1}-C_4$.

We take $\rho _2>0$ such that, for any $x\in E_1\cap \Phi (B(x_0,\rho _2))$ and $0<\rho <2\rho _2$, $Z_{x,\rho }$ can be expressed as $Z_{x,\rho }=y+\operatorname {Tan}(E_1,y)$ with $y\in E_1\cap U$. Since $D\Phi (y)X=D\Phi (y) \operatorname {Tan}(E_1,y)=\operatorname {Tan}(M_1,\Phi (y))$ in case $y\in E_1\cap U$, by putting $\rho _3=\min \{\rho _2,C_4\rho _1/2,R\}$, we have that, for any $z\in M_1\cap B(0,\rho _3)$ and $0<t<2\rho _3$, there exist cone $Z'(z,t)$ in $\Omega _0$ with sliding boundary $L_0=\partial \Omega _0$, such that

$$\begin{equation*} d_{x,t}(M_1,Z'(z,t))\leq C_7t^{\beta }. \end{equation*}$$

For such cone $Z'(z,t)$, we have that $Z'(z,t)=w+\operatorname {Tan}(M_1,w)$, $w\in M_1$, $|w-z|\leq C_8 t$, and $w\in L_0\cap B(z,C_8t)$ in case $t\geq \operatorname {dist}(z,L_0)/2$. $Z'(z,t)$ may not pass through $z$, but the cone $Z(z,t)=Z'(z,t)-w+z$ pass through $z$, and

$$\begin{equation*} d_{x,t}(M_1,Z(z,t))\leq C_7t^{\beta } + C_8t\leq C_9 t^{\beta }. \end{equation*}$$ ■

Proof of Theorem 1.2.

Let $M_1$ be the same as in Lemma 7.12, and let $M=\Psi ^{-1}(E\cap U)$. Then by Lemma 7.12, we have that for any $x\in M_1\cap B(0,\rho _3)$ and $0<r<2\rho _3$, there exist cone $Z(x,r)$ such that

$$\begin{equation*} d_{x,r}(M_1,Z(x,r))\leq Cr^{\beta }, \end{equation*}$$

where $Z(x,r)$ is a minimal cone in $\mathbb{R}^3$ of type $\mathbb{P}$ or $\mathbb{Y}$ in case $x\not \in L_0$ and $t\leq \operatorname {dist}(x,L_0)/2$; and $Z(x,r)$ is a sliding minimal cone in $\Omega _0$ with sliding boundary $L_0$ of type $\mathbb{P}_+$ or $\mathbb{Y}_+$ in other case. We apply Theorem 5.1 to get that there exist $\rho _4>0$, a sliding minimal cone $Z'$ centered at 0, and a mapping $\Phi _1:\Omega _0\cap B(0,\rho _4)\to \Omega _0$, which is a $C^{1,\beta }$-diffeomorphism such that $\Phi _1(0)=0$, $\Phi _1(\partial \Omega _0\cap B(0,\rho _4))\subseteq L_0$, $\|\Phi -\operatorname {id}\|\leq 10^{-1}\rho _4$ and

$$\begin{equation*} M_1\cap B(0,\rho _4)=\Phi (Z')\cap B(0,\rho _4). \end{equation*}$$

We take $Z=Z'\cup L_0$, then we get that

$$\begin{equation*} M\cap B(0,\rho _4)=\Phi (Z)\cap B(0,\rho _4). \end{equation*}$$ ■

8. Existence of the plateau problem with sliding boundary conditions

The Plateau Problem with sliding boundary conditions arise in 7, proposed by Guy David. That is, given an initial set $E_0$, and boundary $\Gamma$, to find the minimizers among all competitors. The author of the paper 7 also gives the sketch to the existence in Section 6, and later on in 6, he pave the way. We will give an existence result in case the boundary is nice enough.

Let $\Omega \subseteq \mathbb{R}^{3}$ be a closed domain such that the boundary $\partial \Omega$ is a $2$-dimensional manifold of class $C^{1,\alpha }$ for some $\alpha >0$. Let $E_0\subseteq \Omega$ be a closed set with $E_0\supseteq \partial \Omega$. We denote by $\mathscr{C}(E_0)$ the collection of all competitors of $E_0$.

Theorem 8.1.

If there is a bounded minimizing sequence of competitors, then there exists $E\in \mathscr{C}(E_0)$ such that

$$\begin{equation*} \mathcal{H}^{2}(E\setminus \partial \Omega )=\inf \{\mathcal{H}^2(S\setminus \partial \Omega ): S\in \mathscr{C}(E_0)\} \end{equation*}$$

Proof.

We put

$$\begin{equation*} m_0=\inf \{\mathcal{H}^2(S\setminus \partial \Omega ): S\in \mathscr{C}(E_0)\}. \end{equation*}$$

If $m_0=+\infty$, we have nothing to do. We now assume that $0\leq m_0<+\infty$.

Let $\{S_i\}\subseteq \mathscr{C}_0$ be a sequence of competitors bounded by $B(0,R)$ such that

$$\begin{equation*} \lim _{i\to \infty }\mathcal{H}^2(S_i\setminus \partial \Omega ) = m_0. \end{equation*}$$

Apply Lemme 5.2.6 in 11, we can fined a sequence of open sets $\{ U_i \}$ and a sequence of competitors $\{ E_i \} \subseteq \mathscr{C}(E_0)$ of $E_0$ bounded by $B(0,R+1)$ such that

•

$U_i\subseteq U_{i+1}$, $\cup _{i\geq 1} U_i=B(0,R+2)\setminus \partial \Omega$;

•

$E_i\cap U_i\in QM(U_i,M,\operatorname {diam}(U_i))$ for constant $M>0$;

•

$\mathcal{H}^2(E_i)\leq \mathcal{H}^2(S_i)+2^{-i}$.

We assume that $E_i$ converge locally to $E_{\infty }$ in $B(0,R+2)$, pass to subsequence if necessary, then by Corollary 21.15 in 6, we get that $E_{\infty }$ is sliding minimal.

Since $E_i\cap U_i\in QM(U_i,M,\operatorname {diam}(U_i))$, by Lemma 3.3 in 4, we have that

$$\begin{equation*} \mathcal{H}^2(E_{\infty }\cap U_i)\leq \liminf _{k\to \infty }\mathcal{H}^2(E_k\cap U_i)\leq m_0, \end{equation*}$$

thus

$$\begin{equation*} \mathcal{H}^2(E_{\infty }\setminus \partial \Omega )\leq m_0. \end{equation*}$$

By Theorem 1.2 and Theorem 1.15 in 5, we get that $E_{\infty }$ is local Lipschitz neighborhood retract. We denote by $\varphi$ a Lipschitz neighborhood retraction of $E_{\infty }$, since $E_i$ converges to $E_{\infty }$, we get that $\varphi (E_k)\subseteq E_{\infty }$ for $k$ large enough. Thus $\varphi (E_k)$ are minimizers.

■