(B.3)

A straightforward application of the triangle inequality and the previous
observations shows that then

$\eta_{j}:=\tilde{\eta}_{j}*\psi_{\epsilon_{j}}\rightarrow\eta\quad\textrm{in
}W^{1,q}(\Omega)\;\textrm{and }C(\overline{\Omega}).$ (B.4)

∎

######  Lemma B.2.

Let $\Omega\subset\mathbb{R}^{n}$ be a smooth domain. Then the closure of
$C^{2}(\overline{\Omega})\cap W_{0}^{1,2}(\Omega)$ with respect to the
$W^{2,2}$-norm is $W^{2,2}(\Omega)\cap W_{0}^{1,2}(\Omega)$.

###### Proof.

Let $v\in W^{2,2}(\Omega)\cap W_{0}^{1,2}(\Omega)$. Then $v$ lies in the
domain of the _Dirichlet Laplacian_ $\Delta_{0}$, a densely defined and
sectorial operator. Hence $\Delta_{0}$ generates an analytic semigroup
$(e^{t\Delta_{0}})_{t\geq 0}$ in $L^{2}(\Omega)$. Next we set
$v_{j}:=e^{\frac{1}{j}\Delta_{0}}v$ for all $j\in\mathbb{N}$. We claim that
$v_{j}\in C^{2}(\overline{\Omega})\cap W_{0}^{1,2}(\Omega)$ and
$v_{j}\rightarrow v$ in $W^{2,2}(\Omega)$. The first assertion follows easily
from the fact that by [Lunardi, Proposition 2.1.1] $v_{j}$ lies in the domain
of $\Delta_{0}^{k}$ for all $k\in\mathbb{N}$ and hence actually in
$C^{\infty}(\overline{\Omega})\cap W_{0}^{1,2}(\Omega)$. Moreover since $v$
lies in the domain of $\Delta_{0}$, [Lunardi, Proposition 2.1.4] yields

$||\Delta_{0}v-\Delta_{0}v_{j}||_{L^{2}}=||\Delta_{0}(e^{\frac{1}{j}\Delta_{0}}v-v)||_{L^{2}}\rightarrow
0\quad(j\rightarrow\infty).$ (B.5)

Since $\Delta_{0}:W^{2,2}(\Omega)\cap W_{0}^{1,2}(\Omega)\rightarrow
L^{2}(\Omega)$ is an isomorphism we infer that also
$||v-v_{j}||_{W^{2,2}}\rightarrow 0$ as $j\rightarrow\infty$. ∎

######  Lemma B.3.

Let $f:\mathbb{R}^{n}\setminus\\{0\\}\rightarrow\mathbb{R}$ be locally
Lipschitz continuous. Then the map $I:(0,\infty)\rightarrow\mathbb{R}$

$s\mapsto\int_{\partial B_{s}(0)}f(y)\;\mathrm{d}\mathcal{H}^{n-1}(y)$ (B.6)

is locally Lipschitz continuous on $(0,\infty)$.

###### Proof.

$\displaystyle I(r)-I(s)=\int_{\partial
B_{r}(0)}f(y)\;\mathrm{d}\mathcal{H}^{n-1}(y)-\int_{\partial
B_{s}(0)}f(y)\;\mathrm{d}\mathcal{H}^{n-1}(y)$ (B.7)
$\displaystyle=\int_{\partial
B_{r}(0)}f(y)\;\mathrm{d}\mathcal{H}^{n-1}(y)-\left(\frac{s}{r}\right)^{n-1}\int_{\partial
B_{r}(0)}f(\tfrac{s}{r}y)\;\mathrm{d}\mathcal{H}^{n-1}(y)$ (B.8)
$\displaystyle=\left(1-\left(\frac{s}{r}\right)^{n-1}\right)\int_{\partial
B_{r}(0)}f(y)\;\mathrm{d}\mathcal{H}^{n-1}(y)+\left(\frac{s}{r}\right)^{n-1}\int_{\partial
B_{r}(0)}(f(y)-f(\tfrac{s}{r}y))\;\mathrm{d}\mathcal{H}^{n-1}(y).$ (B.9)

Now for each $y\in\partial B_{r}(0)$ one has

$|f(y)-f(\tfrac{s}{r}y)|\leq||f||_{W^{1,\infty}(B_{s}(0)\setminus\overline{B_{r}(0)})}|y-\tfrac{s}{r}y|\leq||f||_{W^{1,\infty}(B_{s}(0)\setminus\overline{B_{r}(0)})}|r-s|$
(B.10)

and

$\displaystyle\left|\left(1-\left(\frac{s}{r}\right)^{n-1}\right)\int_{\partial
B_{r}(0)}f(y)\;\mathrm{d}\mathcal{H}^{n-1}(y)\right|$
$\displaystyle\leq||f||_{L^{\infty}(\partial
B_{r}(y))}\omega_{n}|r^{n-1}-s^{n-1}|$ (B.11)
$\displaystyle=||f||_{L^{\infty}(\partial
B_{r}(y))}\omega_{n}\left(\sum_{k=0}^{n-2}r^{k}s^{n-2-k}\right)|r-s|.$ (B.12)

With the previous two equations we infer

$|I(r)-I(s)|\leq L(r,s)|r-s|,$ (B.13)

where

$L(r,s):=||f||_{W^{1,\infty}(B_{s}(0)\setminus\overline{B_{r}(0)})}+||f||_{L^{\infty}(\partial
B_{r}(y))}\omega_{n}\left(\sum_{k=0}^{n-2}r^{k}s^{n-2-k}\right)$ (B.14)

is clearly locally bounded in $(0,\infty)^{2}$. The local Lipschitz property
follows. ∎

######  Lemma B.4.

One has $BV(\Omega)\cap W^{1,1}_{loc}(\Omega)=W^{1,1}(\Omega)$.

###### Proof.

If $w\in BV(\Omega)\cap W^{1,1}_{loc}(\Omega)$ then for all open sets
$U\subset\subset\Omega$ and $i=1,...,n$ we have

$\displaystyle\int_{U}|\partial_{i}w|\;\mathrm{d}x$
$\displaystyle=\sup_{\phi\in C_{0}^{1}(U),||\phi||_{\infty}\leq
1}\int_{U}\partial_{i}w\;\phi\;\mathrm{d}x=\sup_{\phi\in
C_{0}^{1}(U),||\phi||_{\infty}\leq
1}-\int_{U}w\;\partial_{i}\phi\;\mathrm{d}x$ (B.15)
$\displaystyle=\sup_{\phi\in C_{0}^{1}(U),||\phi||_{\infty}\leq
1}\int_{U}w\;\mathrm{div}(-\phi
e_{i})\;\mathrm{d}x\leq|Dw|(U)\leq|Dw|(\Omega).$ (B.16)

As $\Omega$ can be monotonically exhausted by countably many sets
$(U_{j})_{j\in\mathbb{N}},U_{j}\subset U_{j+1}\subset\subset\Omega$ we infer
by the monotonce convergence theorem that $|\partial_{i}w|\in L^{1}(\Omega)$.
The claim follows. ∎

######  Lemma B.5.

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded domain,
$(v_{k})_{k\in\mathbb{N}}\subset W^{1,\infty}(\Omega)$ be a sequence such that
$(||v_{k}||_{W^{1,\infty}(\Omega)})_{k\in\mathbb{N}}$ is uniformly bounded in
$k$. Then there exists a subsequence
$(v_{l_{k}})_{k\in\mathbb{N}}\in\mathbb{N}$ and some $v\in
W^{1,\infty}(\Omega)$ such that $(v_{l_{k}})_{k\in\mathbb{N}}$ converges to
$v$ uniformly on $\overline{\Omega}$, weakly in $W^{1,p}(\Omega)$ for all
$p\in(1,\infty)$. Further, one has

$||v||_{W^{1,\infty}(\Omega)}\leq\liminf_{k\rightarrow\infty}||v_{l_{k}}||_{W^{1,\infty}(\Omega)}.$
(B.17)

###### Proof.

One readily checks that $(v_{k})_{k\in\mathbb{N}}$ satisfies all the
prerequisites of the Arzela-Ascoli theorem, which explains the existence of a
uniformly convergent subsequence and a limit $v\in C^{0}(\overline{\Omega})$.
Choosing a further subsequence and using a standard diagonal sequence argument
one can find a subsequence that converges weakly in $W^{1,r}(\Omega)$ for all
$r\in\mathbb{N}$. One readily checks that the limit of this subsequence must
also be $v$. Let now $(v_{l_{k}})_{k\in\mathbb{N}}$ be this subsequence. If
now $p\in(1,\infty)$ is arbitrary we can choose some number $r\in\mathbb{N}$
such that $r>p$. Observing that $W^{1,r}(\Omega)\subset W^{1,p}(\Omega)$ (and
hence $W^{1,p}(\Omega)^{*}\subset W^{1,r}(\Omega)^{*}$) we obtain that
$v_{l_{k}}\rightharpoonup v$ in $W^{1,p}(\Omega)$. For the norm estimate
(B.17) observe that by uniform convergence we have

$||v||_{L^{\infty}(\Omega)}=\lim_{k\rightarrow\infty}||v_{l_{k}}||_{L^{\infty}(\Omega)}.$
(B.18)

Now for the gradients we observe

$\displaystyle||\nabla v||_{L^{\infty}(\Omega)}$
$\displaystyle=\lim_{p\rightarrow\infty}||\nabla
v||_{L^{p}(\Omega)}\leq\limsup_{p\rightarrow\infty}\liminf_{k\rightarrow\infty}||\nabla
v_{l_{k}}||_{L^{p}(\Omega)}$ (B.19)
$\displaystyle\leq\limsup_{p\rightarrow\infty}\liminf_{k\rightarrow\infty}||\nabla
v_{l_{k}}||_{L^{\infty}(\Omega)}|\Omega|^{\frac{1}{p}}=\liminf_{k\rightarrow\infty}||\nabla
v_{l_{k}}||_{L^{\infty}(\Omega)}.$ (B.20)

This and (B.18) yields

$||v||_{W^{1,\infty}(\Omega)}\leq\liminf_{k\rightarrow\infty}||v_{l_{k}}||_{W^{1,\infty}(\Omega)}.$
(B.21)

∎

######  Lemma B.6.

Let $y\in C^{1}([a,b],\mathbb{R})$ be such that $y^{\prime}(t)=f(t,y(t))$ for
all $t\in[a,b]$ for some $f\in C^{0,\alpha}(\mathbb{R}^{2})$. Then

$||y||_{C^{1,\alpha}([a,b])}\leq||y||_{\infty}+[2+(b-a)^{1-\alpha}](1+||f||_{C^{0,\alpha}(\mathbb{R}^{2})})^{2}$
(B.22)

###### Proof.

First derive the estimate

$|y(t)-y(s)|=\int_{t}^{s}|f(u,y(u))|\;\mathrm{d}u\leq||f||_{\infty}|t-s|.$
(B.23)

In particular

$|y(t)-y(s)|\leq||f||_{\infty}(b-a)^{1-\alpha}|t-s|^{\alpha}.$ (B.24)

Moreover we infer using (B.23)

$\displaystyle|y^{\prime}(t)-y^{\prime}(s)|$
$\displaystyle=|f(t,y(t))-f(s,y(s))|\leq||f||_{C^{0,\alpha}}(|t-s|^{2}+|y(t)-y(s)|^{2})^{\frac{\alpha}{2}}$
(B.25)
$\displaystyle=||f||_{C^{0,\alpha}}(1+||f||_{\infty}^{2})^{\frac{\alpha}{2}}|t-s|^{\alpha}.$
(B.26)

Now note that
$(1+||f||_{\infty}^{2})^{\frac{\alpha}{2}}\leq(1+||f||_{\infty})^{\alpha}\leq
1+||f||_{\infty}.$ The previous computations yield

$\displaystyle||y||_{C^{1,\alpha}}$
$\displaystyle\leq||y||_{\infty}+||y^{\prime}||_{\infty}+[y]_{C^{0,\alpha}}+[y^{\prime}]_{C^{0,\alpha}}$
(B.27)
$\displaystyle\leq||y||_{\infty}+||f||_{\infty}+(b-a)^{1-\alpha}||f||_{\infty}+||f||_{C^{0,\alpha}}(1+||f||_{\infty}).$
(B.28)

The claim follows estimating all $||f||_{\infty}$ by $||f||_{C^{0,\alpha}}$
and some elementary estimates. ∎

######  Lemma B.7.

Let $B\subset\mathbb{R}^{m}$ be an open ball and $h\in
C^{0,\alpha}(\overline{B};\mathbb{R})$. For $z\in B$ and
$v\in\mathbb{R}^{m},|v|=1$ we define $t_{+}(z,v)$ to be the supremum of all
$t\in\mathbb{R}$ such that $z+tv\in B$ and similarly $t_{-}(z,v)$ to be the
infimum. Define

$y_{z,v}(t):=h(z+tv)\quad t\in[t_{-}(z,v),t_{+}(z,v)].$ (B.29)

Then

$||h||_{C^{0,\alpha}(\overline{B})}\leq 2\sup_{z\in
B,|v|=1}||y_{z,v}||_{C^{0,\alpha}([t_{-}(z,l),t_{+}(z,l)])}.$ (B.30)

###### Proof.

It is straightforward to show that

$||h||_{\infty}=\sup_{z\in
B,|v|=1}||y_{z,v}||_{L^{\infty}([t_{-}(z,v),t_{+}(z,v)])}.$ (B.31)

We next show the corresponding estimate for
$[h]_{C^{0,\alpha}(\overline{B})}$. Let $x_{1},x_{2}\in B$, $x_{1}\neq x_{2}$.
Define $z:=x_{1}$ and $v:=\frac{x_{2}-x_{1}}{|x_{2}-x_{1}|}$ now observe that
by convexity of $B$

$z+tv=x_{1}+\tfrac{t}{|x_{2}-x_{1}|}(x_{2}-x_{1})\in B\quad\forall
t\in[0,|x_{2}-x_{1}|].$ (B.32)

this implies that $t_{-}(z,v)\leq 0\leq|x_{2}-x_{1}|\leq t_{+}(z,v)$. Now

$\displaystyle|h(x_{2})-h(x_{1})|$
$\displaystyle=|y_{z,v}(|x_{1}-x_{2}|)-y_{z,v}(0)|\leq[y_{z,v}]_{C^{0,\alpha}([t_{-}(z,v),t_{+}(z,v)]}|\;|x_{1}-x_{2}|-0|^{\alpha}$
(B.33) $\displaystyle\leq\sup_{z\in
B,|v|=1}[y_{z,v}]_{C^{0,\alpha}([t_{-}(z,v),t_{+}(z,v)]}|x_{1}-x_{2}|^{\alpha}.$
(B.34)

We infer that $[h]_{C^{0,\alpha}(B)}\leq\sup_{z\in
B,|v|=1}[y_{z,v}]_{C^{0,\alpha}([t_{-}(z,v),t_{+}(z,v)]}$ and the claim
follows. ∎

######  Lemma B.8.

Let for $A\subset\mathbb{R}$ closed and $\alpha>0$ $f:A\rightarrow\mathbb{R}$
be $\alpha-$Hölder continuous in the sense that

$|f(x)-f(y)|\leq H|x-y|^{\alpha}\quad\forall x,y\in A$ (B.35)

for some $H>0$. Then there exists some
$\bar{f}:\mathbb{R}^{n}\rightarrow\mathbb{R}$ such that $\bar{f}|_{A}=f$ and

$|\bar{f}(x)-\bar{f}(y)|\leq H|x-y|^{\alpha}\quad\forall
x,y\in\mathbb{R}^{n}.$ (B.36)

In particular $[\bar{f}]_{C^{0,\alpha}}=[f]_{C^{0,\alpha}}$.

###### Proof.

This is readily checked choosing

$\bar{f}(z):=\inf_{x\in
A}\left\\{f(x)+H|z-x|^{\alpha}\right\\}\quad(z\in\mathbb{R}^{n}).$ (B.37)

We remark that the subadditivity of $z\mapsto z^{\alpha}$ plays an important
role in this proof. ∎

######  Lemma B.9.

Let $K\subset\Omega$ be compact and $Q\in C^{0,\alpha}(K)$. Then there exists
some $(Q_{j})_{j\in\mathbb{N}}\subset C^{\infty}(\Omega)$ such that
$Q_{j}\rightarrow Q$ uniformly on $K$ and
$\lim_{j\rightarrow\infty}||Q_{j}||_{C^{0,\alpha}(K)}=||Q||_{C^{0,\alpha}(K)}$.

###### Proof.

By Lemma B.8 we may actually assume that $Q\in C^{0,\alpha}(\mathbb{R}^{n})$
(without making the Hölder seminorm larger). Next let
$Q_{\epsilon}:=Q*\phi_{\epsilon}$ where $(\phi_{\epsilon})_{\epsilon>0}$ is
the standard mollifier. A standard result yields that $Q_{\epsilon}\rightarrow
Q$ uniformly on $K$. The lower semicontinuity of $[\cdot]_{C^{0,\alpha}}$ with
respect to uniform convergence yields

$[Q]_{C^{0,\alpha}(K)}\leq\liminf_{\epsilon\rightarrow
0}[Q_{\epsilon}]_{C^{0,\alpha}(K)}.$ (B.38)

Now observe also that for all $x_{1},x_{2}\in K$ there holds

$\displaystyle|Q_{\epsilon}(x_{1})-Q_{\epsilon}(x_{2})|$
$\displaystyle=\left|\int Q(y)\phi_{\epsilon}(x_{1}-y)\;\mathrm{d}y-\int
Q(y)\phi_{\epsilon}(x_{2}-y)\;\mathrm{d}y\right|$ (B.39)
$\displaystyle\leq\left|\int(Q(x_{1}-z)-Q(x_{2}-z))\phi_{\epsilon}(z)\;\mathrm{d}z\right|$
(B.40)
$\displaystyle\leq\int|Q(x_{1}-z)-Q(x_{2}-z)|\phi_{\epsilon}(z)\;\mathrm{d}z$
(B.41)
$\displaystyle\leq\int[Q]_{C^{0,\alpha}(\mathbb{R}^{n})}|x_{1}-z-(x_{2}-z)|^{\alpha}\phi_{\epsilon}(z)\;\mathrm{d}z\leq[Q]_{C^{0,\alpha}(K)}|x_{1}-x_{2}|^{\alpha}.$
(B.42)

This yields $[Q_{\epsilon}]_{C^{0,\alpha}(K)}\leq[Q]_{C^{0,\alpha}(K)}$ and
together with (B.38) we conclude that
$[Q]_{C^{0,\alpha}(K)}=\lim_{\epsilon\rightarrow
0}[Q_{\epsilon}]_{C^{0,\alpha}(K)}$. All in all we infer that

$||Q||_{C^{0,\alpha}(K)}=||Q||_{\infty}+[Q]_{C^{0,\alpha}(K)}=\lim_{\epsilon\rightarrow
0}||Q_{\epsilon}||_{\infty}+[Q_{\epsilon}]_{C^{0,\alpha}(K)}=\lim_{\epsilon\rightarrow
0}||Q_{\epsilon}||_{C^{0,\alpha}(K)}.$ (B.43)

∎

##  Appendix C $C^{1,\alpha}$-hypersurfaces

In this section we collect some facts about $C^{1,\alpha}$-boundaries that we
use. In the following we will always look at $\Gamma=\partial\Omega^{\prime}$
for a compact $C^{1,\alpha}$-domain
$\Omega^{\prime}\subset\subset\mathbb{R}^{n}$.

An important tool that we use is the _regularized signed distance function_ ,
$\rho:\mathbb{R}^{n}\rightarrow\mathbb{R}$ introduced in [Lieberman]. This is
a function $\rho\in C^{\infty}(\mathbb{R}^{n}\setminus\Gamma)\cap
C^{0,1}(\mathbb{R}^{n})$ such that $\rho<0$ on $\Omega^{\prime}$, $\rho>0$ on
$\mathbb{R}^{n}\setminus\Omega^{\prime}$ and

$\frac{1}{2}\leq\frac{|\rho(x)|}{\mathrm{dist}(x,\Gamma)}\leq
2,\quad\textrm{and}\quad\frac{1}{2}\leq|\nabla\rho(x)|\leq 2\quad\forall
x\in\mathbb{R}^{n}\setminus\Gamma.$ (C.1)

We see from [Lieberman, Theorem 2.1] (used with $\zeta(z)=z^{\alpha}$), that
for $\Gamma=\partial\Omega^{\prime}\in C^{1,\alpha}$ there exists a
regularized signed distance function $\rho$, which fulfills in addition

$|\nabla\rho(x)-\nabla\rho(y)|\leq
10[\Gamma]_{1,\alpha}|x-y|^{\alpha}\quad\forall x,y\in\mathbb{R}^{n}.$ (C.2)

(We remark that for obtaining the previous equation we need to demand that the
graph representations $f_{i}$ of $\Gamma$ (cf. (2.2)) must lie in
$C^{1,\alpha}(4\overline{U}_{i})$ instead of just
$C^{1,\alpha}(\overline{U}_{i})$. This is why the counterintuitive constant
$4$ appears in our definition of a representation). In particular $\rho$
extends to a $C^{1,\alpha}$-function on $\mathbb{R}^{n}$. As $\rho\equiv 0$ on
$\Gamma$ the derivative $\nabla\rho$ must point in normal direction, which
means that the outward unit normal $\nu:\Gamma\rightarrow\mathbb{R}^{n}$
satisfies

$\nu(x)=\frac{\nabla\rho(x)}{|\nabla\rho(x)|}\quad\forall x\in\Gamma.$ (C.3)

This, (C.2) and the fact that $\frac{1}{2}\leq|\nabla\rho|\leq 2$ on
$\mathbb{R}^{n}$ implies

$|\nu(x)-\nu(y)|\leq C([\Gamma]_{1,\alpha})|x-y|^{\alpha}\quad\forall
x,y\in\Gamma.$ (C.4)

As a consequence one finds that if $\Pi_{x}$ denotes the orthogonal projection
on $T_{x}\Gamma$ then (with $|\cdot|$ denoting the operator norm) there holds

$|\Pi_{x}-\Pi_{y}|\leq C([\Gamma]_{1,\alpha})|x-y|^{\alpha}\quad\forall
x,y\in\Gamma.$ (C.5)

In the following we prove some further geometric estimates for
$C^{1,\alpha}$-hypersurfaces and check that the constants appearing in them
only depend on $[\Gamma]_{1,\alpha}$. In the end we will show that one can
approximate each $C^{1,\alpha}$-hypersurface $\Gamma$ with
$C^{\infty}$-hypersurfaces $(\Gamma_{j})_{j\in\mathbb{N}}$ such that
$[\Gamma_{j}]_{1,\alpha}$ is uniformly bounded in terms of
$[\Gamma]_{1,\alpha}$.

######  Lemma C.1.

Let $\Gamma$ be a $C^{1,\alpha}$-hypersurface. Then there exists
$C([\Gamma]_{1,\alpha})>0$ such that

$(\nu(x),x-y)\leq C([\Gamma]_{1,\alpha})|x-y|^{1+\alpha}\quad\forall
x,y\in\Gamma.$ (C.6)

###### Proof.

Let $x,y\in\Gamma$ be arbitrary. The mean value theorem on $\mathbb{R}^{n}$
yields that there exists some $\xi\in\mathbb{R}^{n}$ on the line segement
connecting $x$ and $y$ which satisfies

$(\nabla\rho(\xi),x-y)=\rho(x)-\rho(y)=0.$ (C.7)

We infer

$(\nabla\rho(x),x-y)=(\nabla\rho(x)-\nabla\rho(\xi),x-y)\leq|\nabla\rho(x)-\nabla\rho(\xi)|\;|x-y|.$
(C.8)

Using (C.2) we can bound the first factor by
$10[\Gamma]_{1,\alpha}|x-\xi|^{\alpha}$. Since $\xi$ lies on the line segment
connecting $x$ and $y$ we also have $|x-\xi|^{\alpha}\leq|x-y|^{\alpha}$.
Hence we obtain

$(\nabla\rho(x),x-y)\leq 10[\Gamma]_{1,\alpha}|x-y|^{1+\alpha}.$ (C.9)

The claim follows then from $\nu=\frac{\nabla\rho}{|\nabla\rho|}$ and
$|\nabla\rho|\geq\frac{1}{2}$. ∎

Next we show that the intrinsic distance and the Euclidean norm are comparable
on a $C^{1,\alpha}$ hypersurface and the constant in the comparison estimate
only depends on $[\Gamma]_{1,\alpha}$.

######  Lemma C.2.

Let $\Gamma\subset\subset\mathbb{R}^{n}$ be an $n-1$ dimensional
$C^{1,\alpha}$-hypersurface. Then there exists $C=C([\Gamma]_{1,\alpha})>1$
such that for all $y,z\in\Gamma$ one has
$|y-z|\leq\mathrm{dist}_{\Gamma}(y,z)\leq C([\Gamma]_{1,\alpha})|y-z|$, where

$\mathrm{dist}_{\Gamma}(y,z):=\inf\left\\{\int_{0}^{1}|\gamma^{\prime}(s)|\;\mathrm{d}s:\gamma\in
C^{1,\alpha}([0,1];\Gamma):\gamma(0)=y,\gamma(1)=z\right\\}.$ (C.10)

###### Proof.

We use [Blatt, Theorem 1.1]. For a compact $C^{1}$-submanifold $S$ of
$\mathbb{R}^{n}$ one can define

$\gamma_{0}(S):=\max\\{\gamma_{1}(S),\gamma_{2}(S)\\}$ (C.11)

where

$\gamma_{1}(S):=\sup_{x\in S,R>0}\inf_{v\in\mathbb{R}^{n},|v|=1}I_{x,R}(v),$
(C.12) $I_{x,R}(v):=\frac{1}{\mathcal{H}^{n-1}(S\cap B_{R}(x))}\int_{S\cap
B_{R}(x)}|\nu(y)-v|\;\mathrm{d}\mathcal{H}^{n-1}(y)\quad(v\in\mathbb{R}^{n},|v|=1)$
(C.13)

and

$\gamma_{2}(S):=\inf_{x\in S,R>0}\sup_{y\in S\cap
B_{R}(x),v\in\mathrm{arginf}I_{x,R}}\frac{(v,x-y)}{R},$ (C.14)

where $\mathrm{arginf}$ is meant among all $v\in\mathbb{R}^{n}$ s.t. $|v|=1$.
Now [Blatt, Theorem 1.1] states that there exists $\epsilon=\epsilon(n)>0$ and
$C=C(n)>0$ such that $\gamma_{0}(S)\leq\epsilon$ implies that

$\left|\frac{\mathrm{dist}_{\Gamma}(x,y)}{|x-y|}-1\right|\leq
C(n)\gamma_{0}(S)\log\tfrac{1}{\gamma_{0}(S)}\quad\forall x,y\in\Gamma.$
(C.15)

Next we look at the compact submanifold
$S:=\Gamma\cap\overline{B_{s}(z)}\subset\Gamma$ for some arbitrary
$z\in\mathbb{R}^{n}$, $s>0$. Fix $x\in S$. One has (with
$C=C([\Gamma]_{1,\alpha})$ defined as in (C.4))

$\displaystyle\inf_{|v|=1}I_{x,R}(v)\leq I_{x,R}(\nu(x))$
$\displaystyle=\fint_{S\cap
B_{R}(x)}|\nu(y)-\nu(x)|\;\mathrm{d}\mathcal{H}^{n-1}(y)$ (C.16)
$\displaystyle\leq C\fint_{S\cap B_{R}(x)}|y-x|^{\alpha}\leq C(2s)^{\alpha}.$
(C.17)

and thus $\gamma_{1}(\Gamma\cap\overline{B_{s}(z)})\leq
D([\Gamma]_{1,\alpha})s^{\alpha}$. Moreover for each
$v\in\mathrm{arginf}I_{x,R}$ we have

$\displaystyle|v-\nu(x)|$ $\displaystyle=\fint_{S\cap
B_{r}(x)}|v-\nu(x)|\;\mathrm{d}\mathcal{H}^{n-1}(y)$ (C.18)
$\displaystyle\leq\fint_{S\cap
B_{r}(x)}|\nu(y)-v|\;\mathrm{d}\mathcal{H}^{n-1}(y)+\fint_{S\cap
B_{r}(x)}|\nu(y)-\nu(x)|\;\mathrm{d}\mathcal{H}^{n-1}(y)$ (C.19)
$\displaystyle\leq 2\fint_{S\cap B_{r}(x)}|\nu(y)-\nu(x)|\leq
2D([\Gamma]_{1,\alpha})s^{\alpha}.$ (C.20)

Therefore we have that for each $v\in\mathrm{arginf}I_{x,R}$ and $y\in S\cap
B_{R}(x)$ there holds

$\displaystyle\frac{(v,x-y)}{R}$
$\displaystyle=\frac{(v-\nu(x),x-y)}{R}+\frac{(\nu(x),x-y)}{R}$ (C.21)
$\displaystyle\leq
2D([\Gamma]_{1,\alpha})s^{\alpha}\frac{|x-y|}{R}+C([\Gamma]_{1,\alpha})\frac{|x-y|^{1+\alpha}}{R}$
(C.22) $\displaystyle\leq
2D([\Gamma]_{1,\alpha})s^{\alpha}\frac{|x-y|}{R}+C([\Gamma]_{1,\alpha})(2s)^{\alpha}\frac{|x-y|}{R}$
(C.23) $\displaystyle\leq
2D([\Gamma]_{1,\alpha})s^{\alpha}+C([\Gamma]_{1,\alpha})(2s)^{\alpha}.$ (C.24)

This implies that

$\gamma_{2}(\Gamma\cap\overline{B_{s}(z)})=E([\Gamma]_{1,\alpha})s^{\alpha}.$
(C.25)

From this and the discussion after (C.17) we infer that
$s<s_{0}([\Gamma]_{1,\alpha})$ implies that
$\gamma_{0}(\Gamma\cap\overline{B_{s}(z)})\leq\epsilon(n)$ and hence for all
$s<s_{0}([\Gamma]_{1,\alpha})$ one has

$\left|\frac{\mathrm{dist}_{\Gamma\cap\overline{B_{s}(z)}}(x,y)}{|x-y|}-1\right|\leq
C(n)\sup_{t\in(0,\epsilon(n))}t\log\tfrac{1}{t}<\infty.$ (C.26)

Now suppose that $x,y\in\Gamma$. If
$|x-y|\geq\frac{1}{2}s_{0}([\Gamma]_{1,\alpha})$ then one has

$\frac{\mathrm{dist}_{\Gamma}(x,y)}{|x-y|}\leq\frac{2\mathrm{diam}_{\Gamma}(\Gamma)}{s_{0}([\Gamma]_{1,\alpha})},$
(C.27)

where $\mathrm{diam}_{\Gamma}$ denotes the intrinsic diameter of $\Gamma$,
which we bound later in terms of $[\Gamma]_{1,\alpha}$. If
$|x-y|<\frac{1}{2}s_{0}([\Gamma]_{1,\alpha})$ then one has by (C.26)

$\frac{\mathrm{dist}_{\Gamma}(x,y)}{|x-y|}=\frac{\mathrm{dist}_{\Gamma\cap
B_{s}(x)}(x,y)}{|x-y|}\leq C(n)\sup_{t\in(0,\epsilon(n))}t\log\tfrac{1}{t}.$
(C.28)

Choosing

$C_{1}:=\frac{2\mathrm{diam}_{\Gamma}(\Gamma)}{s_{0}([\Gamma]_{1,\alpha})}+C(n)\sup_{t\in(0,\epsilon(n))}t\log\tfrac{1}{t}$
(C.29)

we infer from the previous computations that $\mathrm{dist}_{\Gamma}(x,y)\leq
C_{1}|x-y|$. To show that $C_{1}$ depends only on $[\Gamma]_{1,\alpha}$ it
remains to prove that $\mathrm{diam}_{\Gamma}(\Gamma)$ is bounded in terms of
$[\Gamma]_{1,\alpha}$. To this end pick a representation
$(O_{i},U_{i},V_{i},f_{i}$, $i=1,...,M)$ and let $\Gamma_{i}$ be as in (2.2),
i.e. $\Gamma_{i}\cap\Gamma_{i+1}\neq\emptyset$. Choose arbitrary
$x_{i}\in\Gamma_{i}\cap\Gamma_{i+1}$ for all $i=1,...,M$. Now let
$y,z\in\Gamma$ be arbitrary. Without loss of generality we may assume
$y\in\Gamma_{i}$ and $z\in\Gamma_{j}$ for some $i\leq j$. Now we estimate

$\mathrm{dist}_{\Gamma}(y,z)\leq\mathrm{dist}_{\Gamma}(y,x_{i})+\sum_{l=i}^{j-1}\mathrm{dist}_{\Gamma}(x_{l+1},x_{l})+\mathrm{dist}_{\Gamma}(z,x_{j}).$
(C.30)

Observe that each summand is the distance of two points lying in the same
subset $\Gamma_{i}$. For any two $y_{1},y_{2}\in\Gamma_{i}$, corresponding to
the values $(z_{1},f(z_{1})),(z_{2},f(z_{2}))$ for some $z_{1},z_{2}\in U_{i}$
one sees that

$\gamma(t):=O_{i}((z_{1}+t(z_{2}-z_{1}),f(z_{1}+t(z_{2}-z_{1})))$ (C.31)

lies in $C^{1,\alpha}([0,1];\Gamma)$ and connects $y_{1},y_{2}$. (Notice that
we have used here that $U_{i}$ is a ball). Therefore

$\displaystyle\mathrm{dist}_{\Gamma}(y_{1},y_{2})$
$\displaystyle\leq\int_{0}^{1}|\gamma^{\prime}(t)|\;\mathrm{d}t\leq|z_{1}-z_{2}|+|f(z_{1})-f(z_{2})|\leq(1+||Df_{i}||_{\infty})|z_{1}-z_{2}|$
(C.32)
$\displaystyle\leq(1+||Df_{i}||_{\infty})|y_{1}-y_{2}|\leq(1+||Df_{i}||_{\infty})\mathrm{diam}(\Gamma)\leq
C([\Gamma]_{1,\alpha}).$ (C.33)

We infer from (C.30) (and the fact that $M$ is bounded by
$[\Gamma]_{1,\alpha}$, cf. (2.2) f.) that

$\mathrm{dist}_{\Gamma}(y,z)\leq
MC([\Gamma]_{1,\alpha})\leq\tilde{C}([\Gamma]_{1,\alpha}).$ (C.34)

Since $y,z\in\Gamma$ were arbitrary the diameter bound follows. ∎

######  Lemma C.3.

Let $\Gamma\subset\mathbb{R}^{n}$ be an $(n-1)$-dimensional
$C^{1,\alpha}$-hypersurface. Then there exists some
$r_{1}=r_{1}([\Gamma]_{1,\alpha})>0$ such that for all $r<r_{1}$ and for all
$x_{0}\in\Gamma$ the hypersurface $\Gamma\cap B_{r}(x_{0})$ can be covered by
a single graph of a $C^{1,\alpha}$-function $u:W\rightarrow\mathbb{R}$ with
vanishing gradient at the preimage of $x_{0}$ and Lipschitz-norm bounded by 1.
The radius $r_{1}$ and $||u||_{C^{1,\alpha}}$ depend only on
$[\Gamma]_{1,\alpha}$.

###### Proof.

Let $C_{1}=C_{1}([\Gamma]_{1,\alpha})$ be the constant from Lemma C.2 and
$C_{2}=C_{2}([\Gamma]_{1,\alpha})$ be the constant from (C.5). Next we let
$r_{1}=r_{1}([\Gamma]_{1,\alpha})>0$ be chosen such that

$r_{1}^{\alpha}\min\\{2^{\alpha}(1+C_{1})^{\alpha}C_{1}C_{2},C_{2}\\}=\frac{1}{2}.$
(C.35)

Now fix $x\in\Gamma$ arbitrary. Let $\Pi_{x}^{\perp}=I-\Pi_{x}$ be the
orthogonal projection on $T_{x}\Gamma^{\perp}$. Let $r\leq r_{1}$ and
$y,z\in\Gamma\cap B_{r}(x)$ be arbitrary. Fix $\epsilon>0$ and choose some
$\gamma\in C^{1,\alpha}([0,1];\Gamma)$ such that $\gamma(0)=y$, $\gamma(1)=z$,
$L(\gamma):=\int_{0}^{1}|\gamma^{\prime}(s)|\;\mathrm{d}s\leq(1+\epsilon)\mathrm{dist}_{\Gamma}(y,z)$.
Observe that for all $t\in[0,1]$ one has $\gamma^{\prime}(t)\in
T_{\gamma(t)}\Gamma$ and $\mathrm{dist}_{\Gamma}(\gamma(t),y)\leq
L(\gamma|_{[0,t]})\leq L(\gamma)\leq(1+\epsilon)\mathrm{dist}(y,z)$. We
compute

$\displaystyle|\Pi_{x}^{\perp}y-\Pi_{x}^{\perp}z|=\left|\Pi_{x}^{\perp}\int_{0}^{1}\gamma^{\prime}(s)\;\mathrm{d}s\right|=\left|\int_{0}^{1}\Pi_{x}^{\perp}\gamma^{\prime}(s)\;\mathrm{d}s\right|=\left|\int_{0}^{1}\Pi_{x}^{\perp}\Pi_{\gamma(s)}\gamma^{\prime}(s)\;\mathrm{d}s\right|$
(C.36)
$\displaystyle=\left|\int_{0}^{1}\Pi_{x}^{\perp}(\Pi_{\gamma(s)}-\Pi_{x})\gamma^{\prime}(s)\;\mathrm{d}s\right|\leq\int_{0}^{1}|\Pi_{x}^{\perp}||\Pi_{\gamma(s)}-\Pi_{x}||\gamma^{\prime}(s)|\;\mathrm{d}s\underset{\eqref{eq:C2}}{\leq}C_{2}\int_{0}^{1}|\gamma(s)-x|^{\alpha}|\gamma^{\prime}(s)|\;\mathrm{d}s$
(C.37) $\displaystyle\leq
C_{2}\int_{0}^{1}(|\gamma(s)-y|+|y-x|)^{\alpha}|\gamma^{\prime}(s)|\;\mathrm{d}s\leq
C_{2}\int_{0}^{1}(\mathrm{dist}_{\Gamma}(\gamma(s),y)+r)^{\alpha}|\gamma^{\prime}(s)|\;\mathrm{d}s$
(C.38) $\displaystyle\leq
C_{2}((1+\epsilon)\mathrm{dist}_{\Gamma}(z,y)+r)^{\alpha}\int_{0}^{1}|\gamma^{\prime}(s)|\;\mathrm{d}s\leq
C_{2}((1+\epsilon)\mathrm{dist}_{\Gamma}(z,y)+r)^{\alpha}(1+\epsilon)\mathrm{dist}_{\Gamma}(y,z)$
(C.39) $\displaystyle\leq
C_{2}((1+\epsilon)(\mathrm{dist}_{\Gamma}(z,x)+\mathrm{dist}_{\Gamma}(x,y))+r)^{\alpha}(1+\epsilon)\mathrm{dist}_{\Gamma}(y,z)$
(C.40) $\displaystyle\leq
C_{2}C_{1}((1+\epsilon)C_{1}(|x-z|+|y-z|)+r)^{\alpha}(1+\epsilon)|y-z|$ (C.41)
$\displaystyle\leq(1+\epsilon)C_{1}C_{2}(2(1+\epsilon)C_{1}+1)^{\alpha}r^{\alpha}|y-z|.$
(C.42)

Letting $\epsilon\rightarrow 0$ and using $r\leq r_{1}$ we obtain

$|\Pi_{x}^{\perp}y-\Pi_{x}^{\perp}z|\leq\frac{1}{2}|y-z|\leq\frac{1}{2}|\Pi_{x}y-\Pi_{x}z|+\frac{1}{2}|\Pi_{x}^{\perp}y-\Pi_{x}^{\perp}z|.$
(C.43)

Rearranging we obtain

$|\Pi_{x}^{\perp}y-\Pi_{x}^{\perp}z|\leq|\Pi_{x}y-\Pi_{x}z|\quad\forall
y,z\in\Gamma\cap B_{r}(x),$ (C.44)

and one also concludes immediately

$|y-z|\leq\sqrt{2}|\Pi_{x}y-\Pi_{x}z|\quad\forall y,z\in\Gamma\cap B_{r}(x).$
(C.45)

We conclude from (C.44) that $\Pi_{x}$ is injective on $\Gamma\cap B_{r}(x)$,
call $W$ its image, and infer from (C.45) that the inverse of
$(\Pi_{x}|_{\Gamma\cap B_{r}(x)})^{-1}:W\rightarrow\Gamma\cap B_{r}(x)$ is
Lipschitz continuous. We infer that $\Pi_{x}:\Gamma\cap B_{r}(x)\rightarrow W$
is a homeomorphism. Hence $\Gamma\cap B_{r}(x)$ is a Lipschitz graph over
$T_{x}\Gamma$. Since $\Gamma\cap B_{r}(x)$ is a $C^{1,\alpha}$-submanifold,
one obtains that this graph representation must be a $C^{1,\alpha}$-graph.
This means

$\Gamma\cap B_{r}(x)=O\\{(x^{\prime},f(x^{\prime})):x^{\prime}\in W\\}$ (C.46)

for some $C^{1,\alpha}$ function $f:W\rightarrow\mathbb{R}$ and some
orthogonal matrix $O$ such that
$O\cdot\mathrm{span}(e_{1},...,e_{n-1})=T_{x}\Gamma$. By the choice of $O$ we
also infer $\nabla f(x^{\prime})=0$ if $x^{\prime}$ is the preimage of $x$.
The Lipschitz continuity with constant $1$ of $f$ follows immediately from
(C.44). It only remains to show that the $C^{1,\alpha}$-norm of $f$ can be
bounded in terms of $[\Gamma]_{1,\alpha}$. Notice first that the Lipschitz
estimate implies $||\nabla f||_{L^{\infty}(W)}\leq 1$. Define
$h(z):=\frac{z}{\sqrt{1+|z|^{2}}}$ and observe that for all $z_{1},z_{2}\in
B_{1}(0)$ one has

$|z_{1}-z_{2}|=|h^{-1}(h(z_{1}))-h^{-1}(h(z_{2}))|\leq||\nabla
h^{-1}||_{\infty}|h(z_{1})-h(z_{2})|\leq\sqrt{2}|h(z_{1})-h(z_{2})|.$ (C.47)

Therefore we obtain for $x^{\prime},y^{\prime}\in W$

$\displaystyle|\nabla f(x^{\prime})-\nabla f(y^{\prime})|$
$\displaystyle\leq\sqrt{2}|h(\nabla f(x^{\prime}))-h(\nabla
f(y^{\prime}))|\leq\sqrt{2}|\nu((x^{\prime},f(x^{\prime})))-\nu((y^{\prime},f(y^{\prime}))|$
(C.48)
$\displaystyle\leq\sqrt{2}C([\Gamma]_{1,\alpha})|(x^{\prime},f(x^{\prime}))-(y^{\prime},f(y^{\prime}))|^{\alpha}\leq\tilde{C}([\Gamma]_{1,\alpha})\
|x^{\prime}-y^{\prime}|^{\alpha},$ (C.49)

where we used in the last step again that $f$ is Lipschitz continuous with
$[f]_{0,1}\leq 1$. ∎

######  Lemma C.4.

Suppose that $\Gamma\subset\mathbb{R}^{n}$ is an $(n-1)$-dimensional
$C^{1,\alpha}$-hypersurface. Let $r_{1}=r_{1}([\Gamma]_{1,\alpha})$ be as in
Lemma C.3. Then there exists $C=C([\Gamma]_{1,\alpha})$ such that for all
$x_{0}\in\Gamma$ one has

$\left|\int_{\Gamma\cap
B_{r}(x_{0})}(x-x_{0})\;\mathrm{d}\mathcal{H}^{n-1}(x)\right|\leq
Cr^{n+\alpha}\quad\forall r<r_{1}.$ (C.50)

###### Proof.

After performing a translation and rotation we may assume $x_{0}=0$ and that
$\Gamma\cap B_{r_{1}}(0)$ can be written as

$\Gamma\cap B_{r_{1}}(0)=\\{(x,u(x)):x\in W_{1}\\}.$ (C.51)

for some $u\in C^{1,\alpha}(W_{1})$ such that $u(0)=\nabla u(0)=0$, $u$ is
Lipschitz with Lipschitz constant 1 and $||u||_{C^{1,\alpha}(W)}\leq
C([\Gamma]_{1,\alpha})$. As a consequence, for all $r<r_{1}$ there holds

$\Gamma\cap B_{r}(0)=\\{(x,u(x)):x\in W\\}$ (C.52)

with $W:=(\mathrm{id}+ue_{n})^{-1}(B_{r}(0))$. We now claim that for all
$x^{\prime}\in W$ one has
$|u(x^{\prime})|\leq\tilde{C}([\Gamma]_{1,\alpha})|x^{\prime}|^{1+\alpha}.$
Indeed, for all $x^{\prime}\in W$ there holds with Lemma C.1

$\displaystyle|u(x^{\prime})-\nabla u(x^{\prime})\cdot x^{\prime}|$
$\displaystyle=|(\nu((x^{\prime},u(x^{\prime})),(x^{\prime},u(x^{\prime})))|\sqrt{1+|\nabla
u(x^{\prime})|^{2}}$ (C.53)
$\displaystyle=|(\nu((x^{\prime},u(x^{\prime})),(x^{\prime},u(x^{\prime}))-(0,u(0)))|\sqrt{1+|\nabla
u(x^{\prime})|^{2}}$ (C.54) $\displaystyle\leq
C([\Gamma]_{1,\alpha})|(x^{\prime},u(x^{\prime}))|^{1+\alpha}.$ (C.55)

Using $u(0)=0$ and that $u$ is Lipschitz with Lipschitz constant $1$ one finds
that the right hand side is bounded by
$\sqrt{2}^{1+\alpha}C([\Gamma]_{1,\alpha})|x^{\prime}|^{1+\alpha}=:\bar{C}([\Gamma]_{1,\alpha})|x^{\prime}|^{1+\alpha}$.
Using that by Lemma C.3 the $C^{1,\alpha}$-norm of $u$ can be bounded in terms
of $[\Gamma]_{1,\alpha}$ we find

$\displaystyle|u(x^{\prime})|$ $\displaystyle\leq|\nabla u(x^{\prime})\cdot
x^{\prime}|+\bar{C}([\Gamma]_{1,\alpha})|x^{\prime}|^{1+\alpha}\leq|\nabla
u(x^{\prime})|\;|x^{\prime}|+\bar{C}([\Gamma]_{1,\alpha})|x|^{1+\alpha}$
(C.56) $\displaystyle\leq|\nabla u(x^{\prime})-\nabla
u(0)|\;|x^{\prime}|+\bar{C}([\Gamma]_{1,\alpha})|x^{\prime}|^{1+\alpha}\leq\tilde{C}([\Gamma]_{1,\alpha})|x^{\prime}|^{1+\alpha}.$
(C.57)

Now define $f(s):=s\sqrt{1+\tilde{C}([\Gamma]_{1,\alpha})^{2}s^{2\alpha}}$. We
claim that then

$B_{f^{-1}(r)}(0)\subset W\subset B_{r}(0).$ (C.58)

Indeed, for the second inclusion observe that if $x\in W$ then $(x,u(x))\in
B_{r}(0)$ and hence $r^{2}>|x|^{2}+u(x)^{2}$. In particular $r^{2}>|x|^{2}$
and thus $|x|<r$. Now we turn to the first inclusion. If $f(|x|)<r$ then

$|x|^{2}+u(x)^{2}\leq|x|^{2}+\tilde{C}([\Gamma]_{1,\alpha})^{2}|x|^{2+2\alpha}=|x|^{2}(1+\tilde{C}([\Gamma]_{1,\alpha})^{2}|x|^{2\alpha})=f(|x|)^{2}<r^{2}$
(C.59)

and hence $(x,u(x))\in B_{r}(0)$, implying $x\in W$ by the definition of $W$.
Now we turn to the integral we want to estimate

$\int_{\Gamma\cap
B_{r}(0)}x\;\mathrm{d}\mathcal{H}^{n-1}(x)=\int_{W}\begin{pmatrix}x^{\prime}\\\
u(x^{\prime})\end{pmatrix}\sqrt{1+|\nabla
u(x^{\prime})|^{2}}\;\mathrm{d}x^{\prime}.$ (C.60)

For the last component observe that

$\displaystyle\left|\int_{W}u(x^{\prime})\sqrt{1+|\nabla
u(x^{\prime})|^{2}}\;\mathrm{d}x^{\prime}\right|$
$\displaystyle\leq\int_{W}|u(x^{\prime})|\sqrt{1+|\nabla
u(x^{\prime})|^{2}}\;\mathrm{d}x^{\prime}$ (C.61)
$\displaystyle\leq\sqrt{1+||\nabla
u||_{\infty}^{2}}\int_{W}|u(x^{\prime})|\;\mathrm{d}x^{\prime}$ (C.62)
$\displaystyle=\sqrt{2}\int_{B_{r}(0)}\tilde{C}([\Gamma]_{1,\alpha})|x^{\prime}|^{1+\alpha}\;\mathrm{d}x^{\prime}$
(C.63)
$\displaystyle\leq\sqrt{2}\tilde{C}([\Gamma]_{1,\alpha})c(n)r^{n-1}r^{1+\alpha}\leq\tilde{C}([\Gamma]_{1,\alpha})r^{n+\alpha}.$
(C.64)

For the first $n-1$ components observe

$\displaystyle\int_{W}x_{i}^{\prime}\sqrt{1+|\nabla
u(x^{\prime})|^{2}}\;\mathrm{d}x^{\prime}$ (C.65)
$\displaystyle=\sqrt{1+|\nabla
u(0)|^{2}}\int_{W}x_{i}^{\prime}\;\mathrm{d}x^{\prime}+\int_{W}x_{i}^{\prime}(\sqrt{1+|\nabla
u(x^{\prime})|^{2}}-\sqrt{1+|\nabla u(0)|^{2}})\;\mathrm{d}x^{\prime}.$ (C.66)

Now observe that

$(\sqrt{1+|\nabla u(x^{\prime})|^{2}}-\sqrt{1+|\nabla u(0)|^{2}})\leq
D([\Gamma]_{1,\alpha})|x^{\prime}|^{\alpha}$ (C.67)

and hence the absolute value of last summand is bounded by

$\displaystyle\left|\int_{W}x_{i}^{\prime}(\sqrt{1+|\nabla
u(x^{\prime})|^{2}}-\sqrt{1+|\nabla
u(0)|^{2}})\;\mathrm{d}x_{i}^{\prime}\right|$
$\displaystyle\leq\int_{B_{r}(0)}D|x^{\prime}||x^{\prime}|^{\alpha}\;\mathrm{d}x^{\prime}\leq
Dr^{n+\alpha}.$ (C.68)

For the first summand we conclude by symmetry of $B_{f^{-1}(r)}(0)$ with
respect to $x\mapsto-x$

$\displaystyle\left|\int_{W}x_{i}^{\prime}\;\mathrm{d}x^{\prime}\right|$
$\displaystyle=\left|\int_{W}x_{i}^{\prime}\;\mathrm{d}x^{\prime}-\int_{B_{f^{-1}(r)}(0)}x_{i}^{\prime}\;\mathrm{d}x^{\prime}\right|\leq\int_{W\setminus
B_{f^{-1}(r)}(0)}|x_{i}^{\prime}|\;\mathrm{d}x^{\prime}$ (C.69)
$\displaystyle\leq\int_{B_{r}(0)\setminus
B_{f^{-1}(r)}(0)}|x^{\prime}|\;\mathrm{d}x^{\prime}\leq\alpha_{n}r(r^{n-1}-f^{-1}(r)^{n-1}).$
(C.70)

Now we estimate

$\displaystyle
r-f^{-1}(r)=f^{-1}(f(r))-f^{-1}(r)=\int_{r}^{f(r)}(f^{-1})^{\prime}(s)\;\mathrm{d}s=\int_{r}^{f(r)}\frac{1}{f^{\prime}(f^{-1}(s))}\;\mathrm{d}s.$
(C.71)

Recalling that $f(s)=s\sqrt{1+\tilde{C}([\Gamma]_{1,\alpha})^{2}s^{2\alpha}}$
we find $f^{\prime}(s)\geq 1$ for all $s>0$. Therefore the previous two
equations yield

$\displaystyle r-f^{-1}(r)$ $\displaystyle\leq
f(r)-r=r\left(\sqrt{1+\tilde{C}([\Gamma]_{1,\alpha})^{2}r^{2\alpha}}-1\right)$
(C.72)
$\displaystyle=\frac{\tilde{C}([\Gamma]_{1,\alpha})^{2}r^{1+2\alpha}}{1+\sqrt{1+\tilde{C}([\Gamma]_{1,\alpha})^{2}r^{2\alpha}}}\leq\tilde{C}([\Gamma]_{1,\alpha})^{2}r^{1+2\alpha}.$
(C.73)

We conclude (using $f^{-1}(r)\leq r$) that

$|r^{n-1}-f^{-1}(r)^{n-1}|=(r-f^{-1}(r))\left|\sum_{i=0}^{n-2}r^{i}f^{-1}(r)^{n-2-i}\right|\leq(r-f^{-1}(r))nr^{n-2}\leq\tilde{C}([\Gamma]_{1,\alpha})^{2}r^{n-1+2\alpha}.$
(C.74)

Together with (C.70) the claim follows. ∎

######  Lemma C.5.

Let $\Gamma=\partial\Omega^{\prime}\subset\subset\Omega$ be a
$C^{1,\alpha}$-boundary. Then there exists a sequence
$(\Gamma_{j}=\partial\Omega_{j}^{\prime})_{j\in\mathbb{N}}$ of
$C^{\infty}$-boundaries such that $\Omega_{j}^{\prime}\supset\Omega^{\prime}$
and

  1. (1)

$\sup_{x\in\Gamma_{j}}\mathrm{dist}(x,\Gamma)\rightarrow 0$ as
$j\rightarrow\infty$.

  2. (2)

$\mathcal{H}^{n-1}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\Gamma_{j}\overset{*}{\rightharpoonup}\mathcal{H}^{n-1}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\Gamma$ in
$C(\overline{\Omega})^{*}$ in the sense that for each $f\in
C(\overline{\Omega})$ one has
$\int_{\Gamma_{j}}f\;\mathrm{d}\mathcal{H}^{n-1}\rightarrow\int_{\Gamma}f\;\mathrm{d}\mathcal{H}^{n-1}$.

  3. (3)

There exists some $C=C([\Gamma]_{1,\alpha})>0$ independent of $j$ such that
$[\Gamma_{j}]_{1,\alpha}\leq C$.

###### Proof.

Let $(O_{i},U_{i},V_{i},f_{i}$, $i=1,...,M)$ represent $\Gamma$ in the sense
of (2.2) and be such that

$\sum_{i=1}^{M}\sup_{x\in
V_{i}}|x|+1+|U_{i}|+||f_{i}||_{C^{1,\alpha}(4U_{i})}\leq[\Gamma]_{1,\alpha}+1.$
(C.75)

Fix also a regularized signed distance function $\rho$. We will show that
$\Omega_{j}^{\prime}:=\\{\rho<\frac{1}{j}\\}$ and
$\Gamma_{j}=\partial\Omega_{j}^{\prime}$ satisfies the desired properties.
Property (1) is clear since
$|\rho|\geq\frac{1}{2}\mathrm{dist}(\cdot,\Gamma)$. Property (2) is
straightforward to show with the definition of $\rho$ and the fact that the
level sets of $\rho$ are regular. Next we proceed to property (3). To this end
we first show the following
Intermediate claim. $(\nabla\rho(x),O_{i}e_{n})\neq 0$ for all $x\in
O_{i}(4U_{i}\times V_{i})\cap\Gamma$ and $i=1,...,M$. To show the intermediate
claim notice that for each $x^{\prime}\in 4U_{i}$ one has
$\rho(O_{i}(x^{\prime},f_{i}(x^{\prime})))=0$. We infer by taking the
derivative with respect to $x_{l}^{\prime}$ $(l=1,...,n-1)$

$\partial_{n}(\rho\circ
O_{i})\partial_{x_{l}^{\prime}}f_{i}(x^{\prime})=-\partial_{l}(\rho\circ
O_{i}).$ (C.76)

Taking squares and summing over all $l$ we find

$[\partial_{n}(\rho\circ O_{i})]^{2}|\nabla f_{i}|^{2}=|\nabla(\rho\circ
O_{i})|^{2}-[\partial_{n}(\rho\circ O_{i})]^{2}.$ (C.77)

This and $|\nabla\rho|^{2}\geq\frac{1}{4}$ yields

$[\partial_{n}(\rho\circ O_{i})]^{2}\geq\frac{1}{4(1+||\nabla
f_{i}||_{\infty}^{2})}\geq\frac{1}{4[\Gamma]_{1,\alpha}^{2}}.$ (C.78)

since $\partial_{n}(\rho\circ O_{i})=(\nabla\rho\circ O_{i},O_{i}e_{n})$, the
intermediate claim is shown.

For $s\in(0,\infty)$ define now

$\zeta(s):=\frac{1}{2\sqrt{1+[\Gamma]_{1,\alpha}^{2}}}s-10[\Gamma]_{1,\alpha}s^{1+\alpha}.$
(C.79)

Note that there exists some $\delta_{0}=\delta_{0}([\Gamma]_{[1,\alpha]})$
such that for each $\delta<\delta_{0}$ the equation $\zeta(s)=\delta$ has two
positive solutions $s_{1},s_{2}\in(0,\infty)$. Let $s_{\delta}>0$ be the
unique smallest positive solution of $\zeta(s)=\delta$. One readily checks
that $\lim_{\delta\rightarrow 0}s_{\delta}=0$.

Now choose some $\delta\in(0,\delta_{0})$ small enough such that

  * •

On $B_{3\delta}(\Gamma)\cap O_{i}(4U_{i}\times V_{i})$ there holds
$|(\nabla\rho,O_{i}e_{n})|\geq\frac{1}{8[\Gamma]_{1,\alpha}^{2}}$ for all
$i\in\\{1,...,M\\}$,

  * •

$B_{3\delta}(\Gamma)\subset\bigcup_{i=1}^{M}O_{i}(U_{i}\times V_{i}),$

  * •

$2s_{\delta}<\mathrm{diam}(V_{i}\cap\\{y:y>f_{i}(x^{\prime})\\})$ for all
$i\in\\{1,....,M\\}$.

Without loss of generality we assume in the following
$(\nabla\rho,O_{i}e_{n})>0$ on $B_{3\delta}(\Gamma)$. We next derive a uniform
bound for $[\Gamma_{j}]_{1,\alpha}$ for all $j\in\mathbb{N}$ such that
$\frac{1}{j}<\delta$. We do so by giving an explicit representation for
$\Gamma_{j}$. Let $i\in\\{1,...,M\\}$. For $x^{\prime}\in 4U_{i}$ we define

$h(x^{\prime}):=\inf\\{y\in
V_{i},y>f_{i}(x^{\prime}):\rho(O_{i}(x^{\prime},y))=\delta\\}.$ (C.80)

We claim that the number $h(x^{\prime})$ is well-defined and lies in $V_{i}$.
Indeed, if we assume $\rho(O_{i}(x^{\prime},y))<\delta$ for all $y\in
V_{i},y>f_{i}(x^{\prime})$ then one has (with $\xi$ being a certain point on
the line segment between $O_{i}(x^{\prime},y)$ and
$O_{i}(x^{\prime},f_{i}(x^{\prime}))$)

$\displaystyle\delta$
$\displaystyle>\rho(O_{i}(x^{\prime},y))=\rho(O_{i}(x^{\prime},y))-\rho(O_{i}(x^{\prime},f_{i}(x^{\prime})))=\nabla\rho(\xi)\cdot
O_{i}\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$ (C.81)
$\displaystyle=\nabla\rho(O_{i}(x^{\prime},f_{i}(x^{\prime})))\cdot
O_{i}\begin{pmatrix}0\\\
y-f_{i}(x^{\prime})\end{pmatrix}-(\nabla\rho(\xi)-\nabla\rho(O_{i}(x^{\prime},f_{i}(x^{\prime}))))\cdot
O_{i}\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$ (C.82)
$\displaystyle=|\nabla\rho|(O_{i}(x^{\prime},f_{i}(x^{\prime})))\nu(O_{i}(x^{\prime},f_{i}(x^{\prime}))\cdot
O_{i}\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$ (C.83)
$\displaystyle\quad-(\nabla\rho(\xi)-\nabla\rho(O_{i}(x^{\prime},f_{i}(x^{\prime}))))\cdot
O_{i}\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$ (C.84)
$\displaystyle=|\nabla\rho|(O_{i}(x^{\prime},f_{i}(x^{\prime})))\frac{1}{\sqrt{1+|\nabla
f_{i}(x^{\prime})|^{2}}}O_{i}\begin{pmatrix}-\nabla f_{i}(x^{\prime})\\\
1\end{pmatrix}\cdot O_{i}\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$
(C.85)
$\displaystyle\quad-(\nabla\rho(\xi)-\nabla\rho(O_{i}(x^{\prime},f_{i}(x^{\prime}))))\cdot
O_{i}\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$ (C.86)
$\displaystyle=|\nabla\rho|(O_{i}(x^{\prime},f_{i}(x^{\prime})))\frac{1}{\sqrt{1+|\nabla
f_{i}(x^{\prime})|^{2}}}\begin{pmatrix}-\nabla f_{i}(x^{\prime})\\\
1\end{pmatrix}\cdot\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$
(C.87)
$\displaystyle\quad-(\nabla\rho(\xi)-\nabla\rho(O_{i}(x^{\prime},f_{i}(x^{\prime}))))\cdot
O_{i}\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$ (C.88)
$\displaystyle=|\nabla\rho|(O_{i}(x^{\prime},f_{i}(x^{\prime})))\frac{1}{\sqrt{1+|\nabla
f_{i}(x^{\prime})|^{2}}}(y-f_{i}(x^{\prime}))$ (C.89)
$\displaystyle\quad-(\nabla\rho(\xi)-\nabla\rho(O_{i}(x^{\prime},f_{i}(x^{\prime}))))\cdot
O_{i}\begin{pmatrix}0\\\ y-f_{i}(x^{\prime})\end{pmatrix}$ (C.90)
$\displaystyle\geq|\nabla\rho|(O_{i}(x^{\prime},f_{i}(x^{\prime})))\frac{1}{\sqrt{1+|\nabla
f_{i}(x^{\prime})|^{2}}}(y-f_{i}(x^{\prime}))$ (C.91)
$\displaystyle\quad-||\nabla\rho||_{C^{0,\alpha}}|y-f_{i}(x^{\prime})|^{1+\alpha}$
(C.92)
$\displaystyle\geq\frac{1}{2\sqrt{1+[\Gamma]_{1,\alpha}^{2}}}|y-f_{i}(x^{\prime})|-10[\Gamma]_{1,\alpha}|y-f_{i}(x^{\prime})|^{1+\alpha}=\zeta(|y-f_{i}(x^{\prime})|).$
(C.93)

We have shown that for all $y\in V_{i},y>f_{i}(x^{\prime})$ one has
$|y-f_{i}(x^{\prime})|\in\zeta^{-1}((-\infty,\delta))\cap(0,\infty)$. Notice
that
$\zeta^{-1}((-\infty,\delta))\cap(0,\infty)=(0,s_{\delta})\cup(s_{\delta}^{(2)},\infty)$
for some $s_{\delta}^{(2)}>s_{\delta}$ (the other solution of
$\zeta(s)=\delta$). Notice that $V_{i}\ni y\mapsto|y-f_{i}(x^{\prime})|$ is
continuous and takes arbitrarily small values for $y\in V_{i}$ close to
$f_{i}(x^{\prime})$. Since $V_{i}\cap\\{y:y>f_{i}(x^{\prime})\\}$ is an
interval (i.e. connected) and $(0,s_{\delta})\cup(s_{\delta}^{(2)},\infty)$ is
disconnected we conclude

$|y-f_{i}(x^{\prime})|<s_{\delta}\qquad\forall y\in
V_{i},\;y>f_{i}(x^{\prime}).$ (C.94)

In particular, for all $y_{1},y_{2}\in V_{i}\cap\\{y:y>f_{i}(x^{\prime})\\}$
one has

$|y_{1}-y_{2}|\leq|y_{1}-f_{i}(x^{\prime})|+|y_{2}-f_{i}(x^{\prime})|<2s_{\delta},$
(C.95)

contradicting
$\mathrm{diam}(V_{i}\cap\\{y:y<f_{i}(x^{\prime})\\})>2s_{\delta}$. Hence
$h(x^{\prime})\in V_{i}$ is well-defined.

We also note that $h(x^{\prime})$ is the only value $y\in V_{i}$ such that
$\rho(O_{i}(x^{\prime},y))=\delta$. Indeed,
$\partial_{y}\rho(O_{i}(x^{\prime},y))=(\nabla\rho(O_{i}(x^{\prime},y)),O_{i}e_{n})>0$
if $O_{i}(x^{\prime},y)\in B_{3\delta}(\Gamma)$ and if $O_{i}(x^{\prime},y)\in
B_{3\delta}(\Gamma)^{C}$ one has
$\rho(O_{i}(x^{\prime},y))\geq\frac{1}{2}\mathrm{dist}(O_{i}(x^{\prime},y),\Gamma)\geq\frac{3}{2}\delta$.
Hence $y\mapsto\rho(O_{i}(x^{\prime},y))$ must strictly increase until it
reaches the level $\frac{3}{2}\delta$ and can never drop below
$\frac{3}{2}\delta$ from there. This shows that the value $\delta$ is attained
exactly once. We therefore obtain

$\\{\rho=\delta\\}\cap O_{i}(4U_{i}\times
V_{i})=O_{i}\\{(x^{\prime},h(x^{\prime})):x^{\prime}\in 4U_{i}\\}\quad\forall
i=1,...,M.$ (C.96)

In particular, $\rho(O_{i}(x^{\prime},h(x^{\prime})))=\delta$ for all
$x^{\prime}\in 4\overline{U_{i}}$. This also implies that $h\in
C^{1,\alpha}(4\overline{U_{i}})$. Moreover for all
$v\in\mathbb{R}^{n-1}:|v|=1$ one has (if $\partial_{v}$ denotes the
directional derivative with respect to $v$)

$\partial_{v}h(x^{\prime})=-\frac{\partial_{v}(\rho\circ
O_{i})(x^{\prime},h(x^{\prime}))}{\partial_{n}(\rho\circ
O_{i})(x^{\prime},h(x^{\prime}))}\quad(x^{\prime}\in 4\overline{U_{i}}).$
(C.97)

We now intend to bound $||h||_{C^{1,\alpha}(4\overline{U_{i}})}$ independently
of $\delta$ with a constant in terms of $[\Gamma]_{1,\alpha}$. For a fixed
$z\in 4U_{i}$ and $v\in\mathbb{R}^{n-1}:|v|=1$ look at the function
$y_{z,v}(t):=h(z+tv)$, $t\in[-a^{-}_{z,v},a^{+}_{z,v}]$ where (if $z_{i}$ is
the center and $r_{i}$ is the radius of $4U_{i}$) one defines

$a^{\pm}_{z,v}=-\frac{(z-z_{i},v)}{2}\pm\sqrt{\frac{(z-z_{i},v)^{2}}{4}+r_{i}^{2}-|z-z_{i}|^{2}}$
(C.98)

which are exactly the two unique numbers $t_{\pm}\in\mathbb{R}$ such that
$z+t_{\pm}v$ lie in $\partial(4U_{i})$. Then (C.97) yields that

$y_{z,v}^{\prime}(t)=-\frac{\partial_{v}(\rho\circ
O_{i})(z+tv,y_{z,v}(t))}{\partial_{n}(\rho\circ
O_{i})(z+tv,y_{z,v}(t))}\quad(t\in[a_{z,v}^{-},a_{z,v}^{+}])$ (C.99)

and therefore by Lemma B.6 one has

$||y_{z,v}^{\prime}||_{C^{1,\alpha}}\leq||y_{z,v}||_{\infty}+(2+(a_{z,l}^{+}-a_{z,l}^{-})^{1-\alpha})(1+||\tfrac{\partial_{v}(\rho\circ
O_{i})}{\partial_{n}(\rho\circ
O_{i})}||_{C^{0,\alpha}(O_{i}^{T}\\{\rho=\delta\\})})^{2},$ (C.100)

where we have used that $(x^{\prime},h(x^{\prime}))\in
O_{i}^{T}\\{\rho=\delta\\}$ for all $x^{\prime}\in 4\overline{U_{i}}$ by
(C.96). Noticing now that on $O_{i}^{T}\\{\rho=\delta\\}\subset
O_{i}^{T}B_{3\delta}(\Gamma)$ one has

$\partial_{n}(\rho\circ O_{i})=(\nabla\rho\circ
O_{i},O_{i}e_{n})\geq\frac{1}{8[\Gamma]_{1,\alpha}^{2}}$ (C.101)

and $\nabla\rho\in C^{0,\alpha}(\mathbb{R}^{n})$ one readily checks that
$||\tfrac{\partial_{l}(\rho\circ O_{i})}{\partial_{n}(\rho\circ
O_{i})}||_{C^{0,\alpha}(O_{i}^{T}\\{\rho=\delta\\})}\leq
C([\Gamma]_{1,\alpha})$. Going back to (C.100) and using that
$y_{z,v}=h(z+tv)\in V_{i}$ for all $t\in[a_{z,v}^{-},a_{z,v}^{+}]$ as well as
by (C.98) $|a_{z,l}^{+}-a_{z_{l}}^{-}|\leq
2r_{i}=2(\frac{1}{\omega_{n}}|4U_{i}|)^{\frac{1}{n}}=8(\frac{1}{\omega_{n}}|U_{i}|)^{\frac{1}{n}}$
one has

$||y_{z,v}^{\prime}||_{C^{1,\alpha}}\leq\sup_{x\in
V_{i}}|x|+(2+8\tfrac{1}{\omega_{n}^{1/n}}(|U_{i}|)^{\frac{1-\alpha}{n}})C([\Gamma]_{1,\alpha})\leq
C([\Gamma]_{1,\alpha}).$ (C.102)

This and Lemma B.7 yield that also
$||h||_{C^{1,\alpha}(4\overline{U_{i}})}\leq C([\Gamma]_{1,\alpha})$. From now
on we will explicitly write the dependence of $h$ on $i$ and $\delta$, i.e.
$h=h_{i}^{\delta}:4U_{i}\rightarrow\mathbb{R}$ for any $i=1,...,M$ and
$\delta$ as above. Observe by the fact that $\\{\rho=\delta\\}\subset
B_{3\delta}(\Gamma)\subset\bigcup_{i=1}^{M}O_{i}(U_{i}\times V_{i})$ and
(C.96) that

$\displaystyle\\{\rho=\delta\\}$
$\displaystyle=\bigcup_{i=1}^{M}\\{\rho=\delta\\}\cap O_{i}(U_{i}\times
V_{i})$ (C.103) $\displaystyle=\bigcup_{i=1}^{M}O_{i}[(U_{i}\times
V_{i})\cap\\{(x^{\prime},h_{i}^{\delta}(x^{\prime})):x^{\prime}\in U_{i}\\}].$
(C.104)

Recalling that $h_{i}^{\delta}\in C^{1,\alpha}$ and
$||h_{i}^{\delta}||_{C^{1,\alpha}(4\overline{U}_{i})}\leq
C([\Gamma]_{1,\alpha})$ we find that $O_{i},U_{i},V_{i},h_{i}^{\delta}$,
$i=1,...,M$ is a representation of $\\{\rho=\delta\\}$. Therefore

$[\\{\rho=\delta\\}]_{1,\alpha}\leq\sum_{i=1}^{M}\sup_{x\in
V_{i}}|x|+1+|U_{i}|+||h_{i}^{\delta}||_{C^{1,\alpha}(U_{i})}\leq
C([\Gamma]_{1,\alpha}),$ (C.105)

where we used (C.75) and the fact that $M$ is bounded by $[\Gamma]_{1,\alpha}$
in the last step. Observing that $\delta=\frac{1}{j_{0}}$ for $j_{0}$ large
enough is a possible choice, Property (3) follows. ∎

######  Lemma C.6.

Let $\Gamma=\partial\Omega^{\prime}\in C^{1,\alpha}$. Then
$\mathcal{H}^{n-1}(\Gamma)\leq(1+[\Gamma]_{1,\alpha})^{2}$.

###### Proof.

Let $\epsilon>0$ and $O_{i},U_{i},V_{i},f_{i}$, $i=1,...,M$ be a
representation of $\Gamma$ such that

$\sum_{i=1}^{M}\sup_{x\in
V_{i}}|x|+|U_{i}|+1+||f_{i}||_{C^{1,\alpha}(4U_{i})}\leq[\Gamma]_{1,\alpha}+1.$
(C.106)

Then one infers from (2.2)

$\displaystyle\mathcal{H}^{n-1}(\Gamma)$
$\displaystyle\leq\sum_{i=1}^{M}\int_{U_{i}}\sqrt{1+|\nabla
f_{i}|^{2}}\;\mathrm{d}x^{\prime}\leq\sum_{i=1}^{M}\int_{U_{i}}(1+|\nabla
f_{i}|)\;\mathrm{d}x^{\prime}$ (C.107)
$\displaystyle\leq\sum_{i=1}^{M}|U_{i}|+|U_{i}|\;||f_{i}||_{C^{1,\alpha}}\leq\sum_{i=1}^{M}|U_{i}|+(1+[\Gamma]_{1,\alpha})||f_{i}||_{C^{1,\alpha}(U_{i})}$
(C.108)
$\displaystyle\leq(1+[\Gamma]_{1,\alpha})\sum_{i=1}^{M}(|U_{i}|+||f_{i}||_{C^{1,\alpha}(U_{i})})\leq(1+[\Gamma]_{1,\alpha})([\Gamma]_{1,\alpha}+1).$
(C.109)

The claim follows. ∎

##  Appendix D The space $L^{\infty}(\mathbb{R}^{n})\cap BV(\mathbb{R}^{n})$

In this section we prove some useful integral formulas that hold for functions
in the set $L^{\infty}(\mathbb{R}^{n})\cap BV(\mathbb{R}^{n})$. Here we derive
some important tools to deal with functions in this class.

We first recall some facts about $BV$-functions. For $v\in BV(\mathbb{R}^{n})$
and $U\subset\mathbb{R}^{n}$ open one can define

$|Dv|(U):=\sup_{\phi\in C^{1}_{c}(U;\mathbb{R}^{n}),||\phi||\leq
1}\int_{\mathbb{R}^{n}}v\;\mathrm{div}(\phi)\;\mathrm{d}x.$ (D.1)

By [EvGar, Chapter 5] $|Dv|$ extends uniquely to a finite Radon measure on
$\mathbb{R}^{n}$, called the _total variation measure_. This measure has the
property that there exists a Borel measurable function
$\sigma_{v}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$, $|\sigma_{v}|=1$ a.e.
and

$\int_{\mathbb{R}^{n}}v\;\mathrm{div}(\phi)\;\mathrm{d}x=\int_{\mathbb{R}^{n}}(\phi,\sigma_{v})\;\mathrm{d}|Dv|\quad\forall\phi\in
C_{c}^{1}(\mathbb{R}^{n}).$ (D.2)

We further define for $v\in BV(\mathbb{R}^{n})$ the _precise representative_

$v^{*}(x):=\lim_{r\rightarrow 0}\fint_{B_{r}(x)}v(y)\;\mathrm{d}y,$ (D.3)

which exists for $\mathcal{H}^{n-1}$ a.e. $x\in\mathbb{R}^{n}$ by [EvGar,
Theorem 5.20] (and not just Lebesgue almost-everywhere). If
$(\phi_{\epsilon})_{\epsilon>0}$ is the standard mollifier (defined as in
[EvGar, Section 4.2]) one can define $v_{\epsilon}:=v*\phi_{\epsilon}$ which
then lies in $C^{\infty}(\mathbb{R}^{n})\cap BV(\mathbb{R}^{n})$. For $v\in
BV(\mathbb{R}^{n})$ it is shown in [EvGar, Theorem 5.20] that
$v_{\epsilon}\rightarrow v^{*}$ as $\mathcal{H}^{n-1}$ a.e. as
$\epsilon\rightarrow 0$ (and not just Lebesgue almost everywhere). We will
need also the following

######  Lemma D.1.

Let $v\in BV(\mathbb{R}^{n})$ and $U\subset\mathbb{R}^{n}$ be open. Then for
any $\epsilon>0$ there holds

$\int_{U}|\nabla v_{\epsilon}(x)|\;\mathrm{d}x\leq|Dv|(B_{\epsilon}(U))$ (D.4)

###### Proof.

Let $U,v$ be as in the statement. Then

$\displaystyle\int_{U}|\partial_{i}v_{\epsilon}(x)|\;\mathrm{d}x$
$\displaystyle=\int_{U}\left|\int_{\mathbb{R}^{n}}v(y)\partial_{x_{i}}\phi_{\epsilon}(x-y)\;\mathrm{d}y\right|\;\mathrm{d}x=\int_{U}\left|\int_{\mathbb{R}^{n}}v(y)\partial_{y_{i}}\phi_{\epsilon}(x-y)\;\mathrm{d}y\right|\;\mathrm{d}x$
(D.5)
$\displaystyle=\int_{U}\left|\int_{\mathbb{R}^{n}}v(y)\mathrm{div}(\phi_{\epsilon}(x-\cdot)e_{i})(y)\;\mathrm{d}y\right|\;\mathrm{d}x$
(D.6)
$\displaystyle=\int_{U}\left|\int_{\mathbb{R}^{n}}\phi_{\epsilon}(x-y)(\sigma_{v}(y),e_{i})\;\mathrm{d}|Dv|(y)\right|\;\mathrm{d}x\leq\int_{U}\int_{\mathbb{R}^{n}}\phi_{\epsilon}(x-y)\mathrm{d}|Dv|(y)\;\mathrm{d}x$
(D.7)
$\displaystyle=\int_{\mathbb{R}^{n}}\left(\int_{U}\phi_{\epsilon}(x-y)\;\mathrm{d}x\right)\;\mathrm{d}|Dv|(y)\leq\int_{\mathbb{R}^{n}}\chi_{\overline{B_{\epsilon}(U)}}(y)\;\mathrm{d}|Dv|(y)$
(D.8) $\displaystyle\leq|Dv|(\overline{B_{\epsilon}(U)}),$ (D.9)

where the last step is due to the fact that

$0\leq\int_{U}\phi_{\epsilon}(x-y)\;\mathrm{d}x\leq\int_{\mathbb{R}^{n}}\phi_{\epsilon}(x-y)\;\mathrm{d}x=1$
(D.10)

and if $y\not\in\overline{B_{\epsilon}(U)}$ then the fact that
$\mathrm{spt}(\phi_{\epsilon})\subset\overline{B_{\epsilon}(0)}$ implies

$\int_{U}\phi_{\epsilon}(x-y)\;\mathrm{d}x=0.$ (D.11)

∎

######  Lemma D.2 (Radial integration for $BV$-functions).

Let $v\in BV(\mathbb{R}^{n})\cap L^{\infty}(\mathbb{R}^{n})$ have compact
support in $\mathbb{R}^{n}$. Then

$\int_{\mathbb{R}^{n}}v(x)\;\mathrm{d}x=\int_{0}^{\infty}\int_{\partial
B_{s}(0)}v^{*}(x)\;\mathrm{d}\mathcal{H}^{n-1}(x)\;\mathrm{d}s$ (D.12)

###### Proof.

Notice that $\fint_{B_{r}(x)}v(y)\;\mathrm{d}y\rightarrow v^{*}(x)$ as
$r\rightarrow 0$ for Lebesgue a.e $x\in\mathbb{R}^{n}$ and
$|v^{*}(x)|\leq||v||_{L^{\infty}}\chi_{B_{1}(\mathrm{spt}(v))}(x)$ for all
$x\in\mathbb{R}^{n}$ and $r\in(0,1)$. Thus the dominated convergence theorem
yields

$\int_{\mathbb{R}^{n}}v(x)\;\mathrm{d}x=\int_{\mathbb{R}^{n}}v^{*}(x)\;\mathrm{d}x=\lim_{r\rightarrow
0}\int_{\mathbb{R}^{n}}\left(\fint_{B_{r}(x)}v(y)\;\mathrm{d}y\right)\;\mathrm{d}x.$
(D.13)

Now observe that for fixed $r\in(0,1)$ the map
$f:\mathbb{R}^{n}\rightarrow\mathbb{R},x\mapsto\fint_{B_{r}(x)}v(y)\;\mathrm{d}y$
is continuous. Indeed this follows immediately from the estimate

$|f(x_{1})-f(x_{2})|\leq\frac{|B_{r}(x_{1})\Delta
B_{r}(x_{2})|}{|B_{r}(0)|}||v||_{L^{\infty}}.$ (D.14)

Since radial integration is available for continuous functions one has

$\int_{\mathbb{R}^{n}}v(x)\;\mathrm{d}x=\lim_{r\rightarrow
0}\int_{0}^{\infty}\int_{\partial
B_{s}(0)}\left(\fint_{B_{r}(x)}v(y)\;\mathrm{d}y\right)\;\mathrm{d}\mathcal{H}^{n-1}(x)\;\mathrm{d}s.$
(D.15)

We can now use the dominated convergence theorem again to switch the limit and
the first two integrals (with the same dominating function as above) and
obtain

$\int_{\mathbb{R}^{n}}v(x)\;\mathrm{d}x=\int_{0}^{\infty}\int_{\partial
B_{s}(0)}v^{*}(x)\;\mathrm{d}\mathcal{H}^{n-1}(x)\;\mathrm{d}s.$ (D.16)

∎

######  Lemma D.3.

Let $v\in BV(\mathbb{R}^{n})\cap L^{\infty}(\mathbb{R}^{n})$ and $f\in
C^{0}(\mathbb{R}^{n}\times\mathbb{R})$. Let

$A:=\\{s\in(0,\infty):t\mapsto|Dv|(B_{t}(0))\textrm{is continuous at $s$}\\}.$

Then $(0,\infty)\setminus A$ is countable and the map
$I:A\rightarrow\mathbb{R}$

$I(r):=\int_{\partial B_{r}(0)}f(x,v^{*}(x))\;\mathrm{d}\mathcal{H}^{n-1}(x)$
(D.17)

is continuous on $A$.

###### Proof.

Since $t\mapsto|Dv|(B_{t}(0))$ is monotone it can only have a countable set of
discontinuities. Hence $(0,\infty)\setminus A$ is countable. We notice that at
each point $s\in A$ there holds

$\displaystyle|Dv|(\partial B_{s}(0))$
$\displaystyle=|Dv|(\overline{B_{s}(0)}\setminus
B_{s}(0))=\lim_{n\rightarrow\infty}|Dv|(B_{s+1/n}(0)\setminus B_{s}(0))$
(D.18)
$\displaystyle=\lim_{n\rightarrow\infty}|Dv|(B_{s+1/n}(0))-|Dv|(B_{s}(0))=0.$
(D.19)

We first assume additionally that $f=f(x,p)\in
C^{1}(\mathbb{R}^{n}\times\mathbb{R})$. Let $(\phi_{\epsilon})_{\epsilon>0}$
be the standard mollifier. Since by [EvGar, Theorem 5.20]
$v_{\epsilon}:=v*\phi_{\epsilon}$ converges to $v^{*}$ $\mathcal{H}^{n-1}$
a.e. we have (by the dominated convergence theorem)

$\int_{\partial
B_{r}(0)}f(x,v^{*}(x))\;\mathrm{d}\mathcal{H}^{n-1}(x)=\lim_{\epsilon\downarrow
0}\int_{\partial
B_{r}(0)}f(x,v_{\epsilon}(x))\;\mathrm{d}\mathcal{H}^{n-1}(x).$ (D.20)

To find an integrable dominating function we have used that by Young’s
convolution inequality
$||v_{\epsilon}||_{L^{\infty}}=||v*\phi_{\epsilon}||_{L^{\infty}}\leq||v||_{L^{\infty}}||\phi_{\epsilon}||_{L^{1}}\leq||v||_{L^{\infty}}$
and $f$ is uniformly bounded on $\partial
B_{r}(0)\times[-||v||_{L^{\infty}},||v||_{L^{\infty}}]$. Now for
$0<\underline{r}\leq\overline{r}<\infty$ we estimate for $r,s\in
A\cap[\underline{r},\overline{r}],r\leq s$

$\displaystyle I(r)-I(s)=\left|\int_{\partial
B_{r}(0)}f(x,v^{*}(x))\;\mathrm{d}\mathcal{H}^{n-1}(x)-\int_{\partial
B_{s}(0)}f(x,v^{*}(x))\;\mathrm{d}\mathcal{H}^{n-1}(x)\right|$ (D.21)
$\displaystyle=\lim_{\epsilon\downarrow 0}\left|\int_{\partial
B_{r}(0)}f(x,v_{\epsilon}(x))\;\mathrm{d}\mathcal{H}^{n-1}(x)-\int_{\partial
B_{s}(0)}f(x,v_{\epsilon}(x))\;\mathrm{d}\mathcal{H}^{n-1}(x)\right|$ (D.22)
$\displaystyle=\lim_{\epsilon\downarrow 0}\left|\frac{1}{r}\int_{\partial
B_{r}(0)}(f(x,v_{\epsilon}(x))x,\tfrac{x}{r})\;\mathrm{d}\mathcal{H}^{n-1}(x)-\frac{1}{s}\int_{\partial
B_{s}(0)}(f(x,v_{\epsilon}(x))x,\tfrac{x}{s})\;\mathrm{d}\mathcal{H}^{n-1}(x)\right|$
(D.23) $\displaystyle=\lim_{\epsilon\downarrow
0}\left|\frac{1}{r}\int_{B_{r}(0)}\mathrm{div}(f(x,v_{\epsilon}(x))x)\;\mathrm{d}x-\frac{1}{s}\int_{B_{s}(0)}\mathrm{div}(f(x,v_{\epsilon}(x))x)\;\mathrm{d}x\right|$
(D.24) $\displaystyle\leq\lim_{\epsilon\downarrow
0}\left|\frac{n}{r}\int_{B_{r}(0)}f(x,v_{\epsilon}(x))\;\mathrm{d}x-\frac{n}{s}\int_{B_{s}(0)}f(x,v_{\epsilon}(x))\;\mathrm{d}x\right|$
(D.25) $\displaystyle\quad+\lim_{\epsilon\downarrow
0}\left|\frac{1}{r}\int_{B_{r}(0)}(D_{x}f(x,v_{\epsilon}(x)),x)\;\mathrm{d}x-\frac{1}{s}\int_{B_{s}(0)}(D_{x}f(x,v_{\epsilon}(x)),x)\;\mathrm{d}x\right|$
(D.26) $\displaystyle\quad+\lim_{\epsilon\downarrow
0}\left|\frac{1}{r}\int_{B_{r}(0)}\partial_{p}f(x,v_{\epsilon}(x))(\nabla
v_{\epsilon}(x),x)\;\mathrm{d}x-\frac{1}{s}\int_{B_{s}(0)}\partial_{p}f(x,v_{\epsilon}(x))(\nabla
v_{\epsilon}(x),x)\;\mathrm{d}x\right|$ (D.27) $\displaystyle=\textrm{(I) +
(II) + (III)}.$ (D.28)

In the first two terms one can (thanks to the fact that $f$ and
$(x,p)\mapsto(D_{x}f(x,p),x),$ is uniformly bounded on
$\overline{B_{\overline{r}}(0)}\times[-||v||_{L^{\infty}},||v||_{L^{\infty}}]$)
let $\epsilon\rightarrow 0$ and obtain

(I)+ (II)
$\displaystyle=\left|\frac{n}{r}\int_{B_{r}(0)}f(x,v^{*}(x))\;\mathrm{d}x-\frac{n}{s}\int_{B_{s}(0)}f(x,v^{*}(x))\;\mathrm{d}x\right|$
(D.29)
$\displaystyle+\left|\frac{1}{r}\int_{B_{r}(0)}(D_{x}f(x,v^{*}(x)),x)\;\mathrm{d}x-\frac{1}{s}\int_{B_{s}(0)}(D_{x}f(x,v^{*}(x)),x)\;\mathrm{d}x\right|.$
(D.30)

This expression is obviously continuous in $r,s$, again by boundedness of $f$
and $(x,p)\mapsto(D_{x}f(x,p),x)$ on
$\overline{B_{\overline{r}}(0)}\times[-||v||_{L^{\infty}},||v||_{L^{\infty}}]$.
For $(III)$ we estimate

$\displaystyle\textrm{(III)}=$
$\displaystyle\left|\frac{1}{r}\int_{B_{r}(0)}\partial_{p}f(x,v_{\epsilon}(x))(\nabla
v_{\epsilon}(x),x)\;\mathrm{d}x-\frac{1}{s}\int_{B_{s}(0)}\partial_{p}f(x,v_{\epsilon}(x))(\nabla
v_{\epsilon}(x),x)\;\mathrm{d}x\right|$ (D.31)
$\displaystyle\leq\left|\frac{1}{r}-\frac{1}{s}\right|\;\int_{B_{r}(0)}|\partial_{p}f(x,v_{\epsilon}(x))(\nabla
v_{\epsilon}(x),x)|\;\mathrm{d}x$ (D.32)
$\displaystyle\quad+\frac{1}{s}\int_{B_{s}(0)\setminus
B_{r}(0)}|\partial_{p}f(x,v_{\epsilon}(x))(\nabla
v_{\epsilon}(x),x)|\;\mathrm{d}x.$ (D.33)

Now we let
$M:=\sup_{(x,p)\in\overline{B_{\overline{r}}(0)}\times[-||v||_{L^{\infty}},||v||_{L^{\infty}}]}|\partial_{p}f(x,p)|$
and infer

(III) $\displaystyle\leq M\left|1-\frac{r}{s}\right|\int_{B_{r}(0)}|\nabla
v_{\epsilon}(x)|\;\mathrm{d}x+M\int_{B_{s}(0)\setminus B_{r}(0)}|\nabla
v_{\epsilon}(x)|\;\mathrm{d}x.$ (D.34)

We now use Lemma D.4 to estimate

(III) $\displaystyle\leq
M\left|1-\frac{r}{s}\right||Dv|(\mathbb{R}^{n})+M|Dv|(\overline{B_{\epsilon}(B_{s}(0)\setminus
B_{r}(0)})$ (D.35) $\displaystyle\leq
M\left|1-\frac{r}{s}\right||Dv|(\mathbb{R}^{n})+M|Dv|(\overline{B_{s+\epsilon}(0)}\setminus
B_{r-\epsilon}(0)).$ (D.36)

Letting $\epsilon\downarrow 0$ continuity properties of Radon measures and the
fact that $s\in A$ yield that

(III) $\displaystyle\leq
M\left|1-\frac{r}{s}\right||Dv|(\mathbb{R}^{n})+M|Dv|(\overline{B_{s}(0)}\setminus
B_{r}(0))$ (D.37)
$\displaystyle=M\left|1-\frac{r}{s}\right||Dv|(\mathbb{R}^{n})+M(|Dv|(B_{s}(0))-|Dv|(B_{r}(0)))+M|Dv|(\partial
B_{s}(0))$ (D.38)
$\displaystyle=M\left|1-\frac{r}{s}\right||Dv|(\mathbb{R}^{n})+M(|Dv|(B_{s}(0))-|Dv|(B_{r}(0))).$
(D.39)

Since $t\mapsto|Dv|(B_{t}(0))$ is continuous on $A$ we have thus found a
continuous function $g:A\rightarrow\mathbb{R}$ such that

$|I(r)-I(s)|\leq|g(r)-g(s)|\quad\forall r,s\in A.$ (D.40)

This implies continuity of $I$ on $A$. Finally, to obtain the result for any
continuius $f\in C^{0}(\mathbb{R}^{n}\times\mathbb{R})$, we approximate $f$
uniformly by a sequence of $C^{1}$ function and argue by the Definiton of $I$
that continuity of $I$ on $A$ is preserved by this uniform approximation. ∎

##  Appendix E The signed distance function

In this section let $\Gamma=\partial\Omega^{\prime}\in C^{k}$ for some $k\geq
2$. In this appendix we recall some properties of the _signed distance
function_ of $\Gamma$ and derive some useful formulas. The signed distance
function $d_{\Gamma}:\Omega\rightarrow\mathbb{R}$ is defined as

$d_{\Gamma}(x):=\begin{cases}\mathrm{dist}(x,\Gamma)&x\in\Omega^{\prime\prime}:=\Omega\setminus\overline{\Omega^{\prime}},\\\
0&x\in\Gamma,\\\ -\mathrm{dist}(x,\Gamma)&x\in\Omega^{\prime}.\end{cases}$
(E.1)

One readily checks that $d_{\Gamma}$ and
$|d_{\Gamma}|=\mathrm{dist}(\cdot,\Gamma)$ are Lipschitz continuous functions.
In a small tubular neighborhood
$B_{\epsilon_{0}}(\Gamma):=\\{x\in\mathbb{R}^{n}:\mathrm{dist}(x,\Gamma)<\epsilon_{0}\\}$
one can however still obtain more regularity for $d_{\Gamma}$. By [GilTru,
Lemma 14.16] one has that $d_{\Gamma}\in C^{k}(B_{\epsilon_{0}}(\Gamma))$ and
$\nabla d_{\Gamma}=\nu\circ\pi_{\Gamma}$, where $\nu$ denotes the outward
pointing unit normal of $\Omega^{\prime}$ on $\Gamma$ and
$\pi_{\Gamma}:B_{\epsilon_{0}}(\Gamma)\rightarrow\Gamma$ denotes the _nearest
point projection_ on $\Gamma$ which lies in
$C^{k-1}(B_{\epsilon_{0}}(\Gamma))$, cf. [GilTru, Eq. (14.96)]. Next we derive
some helpful formulas for test functions involving $d_{\Gamma}$. We remark
that $\epsilon_{0}=\epsilon_{0}(\Gamma)$ can be chosen independent of $k$.

######  Lemma E.1.

Let $Q_{0}\in W^{2,s}(\Omega)$ for some $s>n$ and let
$\Gamma=\partial\Omega^{\prime}\in C^{2}$ with outer unit normal $\nu$.
Further let $d_{\Gamma}:B_{\epsilon_{0}}(\Gamma)\rightarrow\mathbb{R}$ be the
signed distance function of $\Gamma$ and $\epsilon<\epsilon_{0}$. Then
(distributionally in $C_{0}^{\infty}(B_{\epsilon}(\Gamma))^{\prime}$) there
holds

$\partial^{2}_{ij}\left(\frac{Q_{0}}{2}|d_{\Gamma}|\right)=Q_{0}\nu_{i}\nu_{j}\mathcal{H}^{n-1}\mathbin{\vrule
height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=5.59721pt}\Gamma+\partial^{2}_{ij}(\frac{Q_{0}}{2}d_{\Gamma})(\chi_{\Omega^{\prime\prime}}-\chi_{\Omega^{\prime}}).$
(E.2)

###### Proof.

Let $\phi\in C_{0}^{\infty}(B_{\epsilon}(\Gamma))$ be arbitrary but fixed.
Denote by $\nu=\nu^{\Omega^{\prime}}$ and $\nu^{\Omega^{\prime\prime}}$ the
outer unit normals of $\Omega^{\prime},\Omega^{\prime\prime}$ respectively.
Then using integration by parts and $d_{\Gamma}|_{\Gamma}=0$

$\displaystyle\int\frac{Q_{0}}{2}|d_{\Gamma}|\partial^{2}_{ij}\phi\;\mathrm{d}x$
$\displaystyle=-\int_{\Omega^{\prime}}\frac{Q_{0}}{2}d_{\Gamma}\partial^{2}_{ij}\phi\;\mathrm{d}x+\int_{\Omega^{\prime\prime}}\frac{Q_{0}}{2}d_{\Gamma}\partial^{2}_{ij}\phi\;\mathrm{d}x$
(E.3)
$\displaystyle=-\int_{\partial\Omega^{\prime}}\frac{Q_{0}}{2}d_{\Gamma}\nu_{i}^{\Omega^{\prime}}\partial_{j}\phi\;\mathrm{d}\mathcal{H}^{n-1}+\int_{\partial\Omega^{\prime\prime}}\frac{Q_{0}}{2}d_{\Gamma}\nu_{i}^{\Omega^{\prime\prime}}\partial_{j}\phi\;\mathrm{d}\mathcal{H}^{n-1}$
(E.4)
$\displaystyle\quad+\int_{\Omega^{\prime}}\partial_{i}\left(\frac{Q_{0}}{2}d_{\Gamma}\right)\partial_{j}\phi\;\mathrm{d}x-\int_{\Omega^{\prime\prime}}\partial_{i}\left(\frac{Q_{0}}{2}d_{\Gamma}\right)\partial_{j}\phi\;\mathrm{d}x$
(E.5)
$\displaystyle=\int_{\Omega^{\prime}}\partial_{i}\left(\frac{Q_{0}}{2}d_{\Gamma}\right)\partial_{j}\phi\;\mathrm{d}x-\int_{\Omega^{\prime\prime}}\partial_{i}\left(\frac{Q_{0}}{2}d_{\Gamma}\right)\partial_{j}\phi\;\mathrm{d}x$
(E.6)
$\displaystyle=\int_{\partial\Omega^{\prime}}\partial_{i}\left(\frac{Q_{0}}{2}d_{\Gamma}\right)\nu^{\Omega^{\prime}}_{j}\phi\;\mathrm{d}\mathcal{H}^{n-1}-\int_{\partial\Omega^{\prime\prime}}\partial_{i}\left(\frac{Q_{0}}{2}d_{\Gamma}\right)\nu^{\Omega^{\prime\prime}}_{j}\phi\;\mathrm{d}\mathcal{H}^{n-1}$
(E.7)
$\displaystyle\quad-\int_{\Omega^{\prime}}\partial^{2}_{ij}\left(\frac{Q_{0}}{2}d_{\Gamma}\right)\phi\;\mathrm{d}x+\int_{\Omega^{\prime\prime}}\partial^{2}_{ij}\left(\frac{Q_{0}}{2}d_{\Gamma}\right)\phi\;\mathrm{d}x.$
(E.8)

Noticing that
$\partial\Omega^{\prime}\cap\mathrm{supp}(\phi)=\partial\Omega^{\prime\prime}\cap\mathrm{supp}(\phi)=\Gamma$
and $\nu_{j}^{\Omega^{\prime\prime}}=-\nu_{j}^{\Omega^{\prime}}$ on $\Gamma$
we infer

$\int_{\Omega}\frac{Q_{0}}{2}|d_{\Gamma}|\partial^{2}_{ij}\phi\;\mathrm{d}x=\int_{\Gamma}\partial_{i}(Q_{0}d_{\Gamma})\nu_{j}\;\mathrm{d}\mathcal{H}^{n-1}+\int
g_{ij}\phi\;\mathrm{d}x,$ (E.9)

where
$g_{ij}:=\partial^{2}_{ij}(\frac{Q_{0}}{2}d_{\Gamma})(\chi_{\Omega^{\prime\prime}}-\chi_{\Omega^{\prime}})\in
L^{s}(B_{\epsilon}(\Gamma))$. Now notice that on $\Gamma$ one has

$\partial_{i}(Q_{0}d_{\Gamma})=\partial_{i}(Q_{0})d_{\Gamma}+Q_{0}\partial_{i}d_{\Gamma}=0+Q_{0}\nu_{i}=Q_{0}\nu_{i}.$
(E.10)

Thus we infer

$\int_{\Omega}\frac{Q_{0}}{2}|d_{\Gamma}|\partial^{2}_{ij}\phi\;\mathrm{d}x=\int_{\Gamma}Q_{0}\nu_{i}\nu_{j}\phi\;\mathrm{d}\mathcal{H}^{n-1}+\int_{\Omega}g_{ij}\phi\;\mathrm{d}x,$
(E.11)

which was asserted. ∎

######  Corollary E.2.

Let $\Gamma=\partial\Omega^{\prime}\in C^{2}$ and $Q_{0}\in W^{2,s}(\Omega)$
for some $s>n$. Then $\nabla\left(\frac{Q_{0}}{2}|d_{\Gamma}|\right)\in
BV(B_{\epsilon}(\Gamma))$.

###### Proof.

Let $\phi=(\phi_{1},...,\phi_{n})\in
C_{0}^{\infty}(B_{\epsilon}(\Gamma);\mathbb{R}^{n})$. Then we compute for
$j=1,...,n$

$\displaystyle\int_{\Omega}\partial_{j}\left(\tfrac{Q_{0}}{2}|d_{\Gamma}|\right)\mathrm{div}(\phi)\;\mathrm{d}x=\sum_{k=1}^{n}\int_{\Omega}\partial_{j}\left(\tfrac{Q_{0}}{2}|d_{\Gamma}|\right)\partial_{k}\phi_{k}\;\mathrm{d}x=-\sum_{k=1}^{n}\int_{\Omega}\left(\tfrac{Q_{0}}{2}|d_{\Gamma}|\right)\partial^{2}_{jk}\phi_{k}\;\mathrm{d}x.$
(E.12)

Now we use (E.2) to find (with the notation
$g_{ij}:=\partial^{2}_{ij}(\frac{Q_{0}}{2}d_{\Gamma})(\chi_{\Omega^{\prime\prime}}-\chi_{\Omega^{\prime}})\in
L^{s}(B_{\epsilon}(\Gamma))$ that

$\int_{\Omega}\partial_{j}\left(\tfrac{Q_{0}}{2}|d_{\Gamma}|\right)\mathrm{div}(\phi)\;\mathrm{d}x=-\sum_{k=1}^{n}\left(\int_{\Gamma}Q_{0}\nu_{j}\nu_{k}\phi_{k}\;\mathrm{d}\mathcal{H}^{n-1}+\int_{\Omega}g_{ik}\phi_{k}\right).$
(E.13)

One readily checks that the term on the right hand side is bounded by

$\left(\sum_{k=1}^{n}||Q_{0}||_{L^{\infty}}\mathcal{H}^{n-1}(\Gamma)+||g_{jk}||_{L^{1}(B_{\epsilon}(\Gamma))}\right)||\phi||_{L^{\infty}}=:C_{j}||\phi||_{L^{\infty}}.$
(E.14)

Notice that $C_{j}<\infty$ since $L^{s}(B_{\epsilon}(\Gamma))\subset
L^{1}(B_{\epsilon}(\Gamma))$. Using the fact that each
$C^{1}_{c}(B_{\epsilon}(\Gamma))$-function can be uniformly approximated by
$C^{\infty}_{c}(B_{\epsilon}(\Gamma))$-functions we find

$\int_{\Omega}\partial_{j}\left(\tfrac{Q_{0}}{2}|d_{\Gamma}|\right)\mathrm{div}(\phi)\;\mathrm{d}x\leq
C_{j}||\phi||_{\infty}\quad\forall\phi\in
C^{1}_{c}(B_{\epsilon}(\Gamma);\mathbb{R}^{n}),$ (E.15)

implying that $\partial_{j}\left(\tfrac{Q_{0}}{2}|d_{\Gamma}|\right)\in
BV(B_{\epsilon}(\Gamma))$. Since $j\in\\{1,...,n\\}$ was arbitrary the claim
is shown. ∎

######  Lemma E.3.

Let $\Gamma=\partial\Omega^{\prime}\in C^{k}$, $k\geq 2$ and $Q_{0}\in
W^{k,s}(\Omega)$, $s>n$. Then $\left(\frac{Q_{0}}{2}|d_{\Gamma}|\right)\in
W^{k,s}(\Omega^{\prime}\cap B_{\epsilon}(\Gamma))\cap
W^{k,s}(\Omega^{\prime\prime}\cap B_{\epsilon}(\Gamma))$.

###### Proof.

This follows immediately from the fact that on $\Omega^{\prime}\cap
B_{\epsilon}(\Gamma)$ there holds

$\left(\frac{Q_{0}}{2}|d_{\Gamma}|\right)=-\frac{Q_{0}}{2}d_{\Gamma},$ (E.16)

and $d_{\Gamma}$ extends to a function in
$C^{k}(\overline{\Omega}^{\prime}\cap
B_{\epsilon}(\Gamma))=C^{k}((\Omega^{\prime}\cap
B_{\epsilon}(\Gamma))\cup\Gamma)$. Hence all derivatives up to order $k$ are
bounded. The product rule in $W^{k,s}$ yields then the desired
$W^{k,s}(\Omega^{\prime}\cap B_{\epsilon}(\Gamma))$-regularity. On
$\overline{\Omega^{\prime\prime}}\cap B_{\epsilon}(\Gamma)$ the same
regularity can be obtained analogously. ∎

##  Appendix F A maximum principle for $BV$-solutions

In this appendix we introduce a nonclassical maximum principle for so-called
_$BV$ -solutions_ of

$\begin{cases}-\mathrm{div}(A(x)\nabla u)=0&\textrm{in }D,\\\ u=h&\textrm{on
}\partial D.\end{cases}$ (F.1)

This concept of solution is defined as follows.

######  Definition F.1.

Let $D\subset\mathbb{R}^{n}$ be a $C^{\infty}$-smooth domain, $A\in
W^{2,s}(D;\mathbb{R}^{n\times n})$ $(s>n)$ and $h\in L^{1}(\partial D)$. We
say $u\in BV(D)$ is a _$BV$ -solution_ of (F.1) if $\mathrm{tr}_{\partial
D}(u)=h$ and

$\int_{D}u(x)\;\mathrm{div}(A(x)\nabla\eta(x))\;\mathrm{d}x=0\quad\forall\eta\in
C_{0}^{\infty}(D).$ (F.2)

If a $BV$-solution $u$ lies additionally in $W^{1,p}(D)$ for some
$p\in(1,\infty)$ then it is easy to see that it is a usual weak solution of
the problem and hence established methods like regularity theory, maximum
principles etc. can be used. However it is not clear whether a $BV$-solution
lies in some higher Sobolev space. It was a longstanding conjecture by Serrin
(cf. [Serrin]) under which conditions on the coefficient matrix $A(x)$ further
interior regularity can be obtained. In [Ancona, Theorem A.1.2], Brezis
presents a presumably optimal condition: For Dini-continuous coefficients
$A(x)$ further interior regularity can be obtained. We will use this further
regularity to obtain a maximum principle.

######  Lemma F.2.

Let $D\subset\mathbb{R}^{n}$ be a $C^{\infty}$-smooth domain, $A\in
W^{2,s}(D;\mathbb{R}^{n\times n})$, $(s>n)$ and suppose that $\psi\in BV(D)$
is a BV-solution of (F.1) for some $h\in W^{1,s}(\mathbb{R}^{n})$. Then

$||\psi||_{L^{\infty}(D)}\leq||h||_{L^{\infty}(\partial D)}$ (F.3)

###### Proof.

Observe first that by [Ancona, Theorem A.1.2] $\psi\in BV(D)\cap
W^{1,2}_{loc}(D)$. This in particular implies (cf. Lemma B.4) that $\psi\in
W^{1,1}(D)$ and

$\int_{\Omega}(A(x)\nabla\psi(x),\nabla\eta(x))\;\mathrm{d}x=0\quad\forall\eta\in
C_{0}^{\infty}(\Omega).$ (F.4)

We infer that $\psi$ is a _weak $W^{1,1}$-solution_ of

$\begin{cases}\mathrm{div}(A(x)\nabla\psi(x))=0&\textrm{in }D,\\\
\psi=h&\textrm{on }\partial D.\end{cases}$ (F.5)

in the sense that $\mathrm{tr}_{\partial D}(\psi)=h$ and (F.4) holds. We next
observe that the concept of a $W^{1,1}$-solution just introduced is a notion
that characterises $\psi$ uniquely. Indeed, if $\psi_{1},\psi_{2}$ are two
$W^{1,1}$-solutions of (F.5) (i.e. satisfying (F.4) and having the same
boundary trace) then $\psi_{1}-\psi_{2}$ lies in $W_{0}^{1,1}(D)$ and also
satisfies (F.4). By [Ancona, Theorem A5.1, second sentence] we infer
$\psi_{1}-\psi_{2}=0$. We claim next that for each $h\in
W^{1,s}(\mathbb{R}^{n})$ there exists a weak $W^{1,1}$-solution of (F.5) which
additionally lies in $W^{1,s}(D)$. To this end notice that $A\nabla h\in
L^{s}(D)$ and hence by [Auscher, Theorem 1.1] there exists a unique
$\tilde{\psi}\in W_{0}^{1,s}(D)$ weak solution of

$\begin{cases}-\mathrm{div}(A\nabla\tilde{\psi})=\mathrm{div}(A\nabla
h)&\textrm{in }D,\\\ \tilde{\psi}=0&\textrm{on }\partial D.\end{cases}$ (F.6)

We now define $\bar{\psi}:=\tilde{\psi}+h$ and assert that $\bar{\psi}$ is a
weak $W^{1,1}$-solution. Indeed, $\mathrm{tr}_{\partial
D}(\bar{\psi})=\mathrm{tr}_{\partial D}(\tilde{\psi}+h)=0+h|_{\partial D}=h$
on $\partial D$. Moreover for each $\eta\in C_{0}^{\infty}(\Omega)$ there
holds by (F.6)

$\displaystyle\int_{D}(A(x)\nabla\bar{\psi}(x),\nabla\eta(x))\;\mathrm{d}x$
$\displaystyle=\int_{D}(A(x)\nabla\tilde{\psi}(x),\nabla\eta(x))\;\mathrm{d}x+\int_{D}(A\nabla
h(x),\nabla\eta(x))\;\mathrm{d}x$ (F.7) $\displaystyle=-\int_{D}(A\nabla
h(x),\nabla\eta(x))\;\mathrm{d}x+\int_{D}(A\nabla
h(x),\nabla\eta(x))\;\mathrm{d}x=0.$ (F.8)

By the previous uniqueness discussion for $W^{1,1}$-solutions of (F.5) we have
$\psi=\bar{\psi}$ and thus $\psi\in W^{1,s}(D)$. In particular, $\psi\in
C(\overline{D})$ by Sobolev embedding. Now, the classical maximum principle
implies the claim. ∎

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