at any point of the boundary. We still can prove it in our context because the
property relies on the Hölder continuity of solutions at the boundary, that
holds in our context because we have (Harnack tubes and) Lemma 8.16.

######  Lemma 10.9.

The Green function satisfies

(10.110) $g(x,y)\leq C\delta(x)^{\alpha}|x-y|^{1-d-\alpha}\ \ \text{ for
$x,y\in\Omega$ such that }|x-y|\geq 4\delta(x),$

where $C>0$ and $\alpha>0$ depend only on $n$, $d$, $C_{0}$ and $C_{1}$.

Proof. Let $y\in\Omega$ be given. For any $x\in\Omega$, we write
$\mathfrak{g}(x)$ for $g(x,y)$. We want to prove that

(10.111) $\mathfrak{g}(x)\leq C\delta(x)^{\alpha}|x-y|^{1-d-\alpha}\ \ \text{
for $x\in\Omega$ such that }|x-y|\geq 4\delta(x),$

with constants $C>0$ and $\alpha>0$ that depend only on $n$, $d$, $C_{0}$ and
$C_{1}$. By Lemma 10.1-(v),

(10.112) $\mathfrak{g}(z)\leq C|z-y|^{1-d}\ \ \text{ for }z\in\Omega\setminus
B(y,\delta(y)/4).$

Let $x$ be such that $|x-y|\geq 4\delta(x)$, choose $x_{0}\in\Gamma$ such that
$|x-x_{0}|=\delta(x)$, and set $r=|x-y|$ and $B=B(x_{0},|x-y|/3)$; thus $x\in
B$. We shall need to know that

(10.113) $\delta(y)\leq\delta(x)+|x-y|\leq\frac{r}{4}+r=\frac{5r}{4}.$

Then let $z$ be any point of $\Omega\cap B$. Obviously
$|z-x|\leq|z-x_{0}|+|x_{0}-x|\leq\frac{r}{3}+\delta(x)\leq\frac{r}{3}+\frac{r}{4}=\frac{7r}{12}$,
which implies that

(10.114) $|y-z|\geq|y-x|-|z-x|\geq
r-\frac{7r}{12}=\frac{5r}{12}\geq\frac{\delta(y)}{3}.$

Hence by (10.112), $\mathfrak{g}(z)\leq C|z-y|^{1-d}$. Notice also that
$|y-z|\leq|y-x|+|z-x|\leq r+\frac{7r}{12}=\frac{19r}{12}$, so, with (10.114),
$\frac{5r}{12}\leq|y-z|\leq\frac{19r}{12}$ and

(10.115) $\mathfrak{g}(z)\leq Cr^{1-d}=C|x-y|^{1-d}\ \text{ for
}z\in\Omega\cap B,$

even if $d<1$, and with a constant $C>0$ that does not dependent on $x$, $y$,
or $x_{0}$.

We now use the fact that $\mathfrak{g}$ is a solution of $L\mathfrak{g}=0$ on
$\Omega\cap B$. Notice that its oscillation on $B$ is the same as its
supremum, because it is nonnegative and, by (i) of Lemma 10.5, its trace on
$\Gamma\cap B$ vanishes. Lemma 8.16 (the Hölder continuity of solutions at the
boundary) says that for some $\alpha>0$, that depends only on $n$, $d$,
$C_{0}$ and $C_{1}$,

(10.116)
$\begin{split}g(x,y)=\mathfrak{g}(x)&\leq\sup_{\overline{B}(x_{0},\delta(x))}\mathfrak{g}=\operatorname*{osc}_{\overline{B}(x_{0},\delta(x))}\mathfrak{g}\leq
C\left(\frac{3\delta(x)}{|x-y|}\right)^{\alpha}\operatorname*{osc}_{B(x_{0},|x-y|/3)}\mathfrak{g}\\\
&=C\left(\frac{3\delta(x)}{|x-y|}\right)^{\alpha}\sup_{B(x_{0},|x-y|/3)}\mathfrak{g}\leq
C\left(\frac{\delta(x)}{|x-y|}\right)^{\alpha}|x-y|^{1-d}.\end{split}$

because $B=B(x_{0},|x-y|/3)$. The lemma follows. $\Box$

##  Chapter 11 The Comparison Principle

In this section, we prove two versions of the comparison principle: one for
the harmonic measure (Lemma 11.16) and one for locally defined solutions
(Lemma 11.17). A big technical difference is that the former is a globally
defined solution, while the latter is local.

At the moment we write this manuscript, the proofs of the comparison principle
in codimension 1 that we are aware of cannot be straightforwardly adapted to
the case of higher codimension. To be more precise, we can indeed prove the
comparison principle (in higher codimension) for harmonic measures on $\Gamma$
by only slightly modifying the arguments of [CFMS, Ken]. However, the proof of
the comparison principle for solutions (of $Lu=0$) defined on a subset $D$ of
$\Omega$ in the case of codimension 1 relies on the use of the harmonic
measure on the boundary $\partial D$ (see for instance [CFMS, Ken]). In our
setting, in the case where the considered functions are non-negative and
solutions to $Lu=0$ only on a subset $D\subsetneq\Omega$, we are lacking a
definition for harmonic measures with mixed boundaries (some parts in
codimension 1 and some parts in higher codimension). The reader can imagine a
ball $B$ centered at a point of ${\partial\Omega}=\Gamma$. The boundary of the
$B\cap\Omega$ consists of $\Gamma\cap B$ and $\partial B$, the sets of
different co-dimension. For those reasons, our proof of the comparison
principle (in higher codimension) for locally defined functions nontrivially
differs from the one in [CFMS, Ken]. Therefore, in a first subsection, we
illustrate our arguments in the case of codimension 1 to build reader’s
intuition.

###  11.1. Discussion of the comparison theorem in codimension 1

We present here two proofs of the comparison principle in the codimension 1
case. The first proof of the one we can find in [CFMS, Ken] and the second one
is our alternative proof. We consider in this subsection that the reader knows
or is able to see the results in the three first sections of [Ken], that
contain the analogue in codimension 1 of the results proved in the previous
sections.

For simplicity, the domain $\Omega\subset\mathbb{R}^{n}$ that we study is a
special Lipschitz domain, that is

$\Omega=\\{(y,t)\in\mathbb{R}^{n-1}\times\mathbb{R},\,\varphi(y)<t\\}$

where $\varphi:\,\mathbb{R}^{n-1}\to\mathbb{R}$ is a Lipschitz function. The
elliptic operator that we consider is $L=-\operatorname{div}A\nabla$, where
$A$ is a matrix with bounded measurable coefficients satisfying the classical
elliptic condition (see for instance (1.1.1) in [Ken]). Yet, the change of
variable
$\rho:\mathbb{R}^{n-1}\times\mathbb{R}\to\mathbb{R}^{n-1}\times\mathbb{R}$
defined by

$\rho(y,t)=(y,t-\varphi(y))$

maps $\Omega$ into
$\widetilde{\Omega}=\mathbb{R}^{n}_{+}:=\\{(y,t)\in\mathbb{R}^{n-1}\times\mathbb{R},\,t>0\\}$
and changes the elliptic operator $L$ into
$\widetilde{L}=-\operatorname{div}\widetilde{A}\nabla$, where $\widetilde{A}$
is also a matrix with bounded measurable coefficients satisfying the elliptic
condition (1.1.1) in [Ken]. Therefore, in the sequel, we reduce our choices of
$\Omega$ and $\Gamma=\partial\Omega$ to $\mathbb{R}^{n}_{+}$ and
$\mathbb{R}^{n-1}=\\{(y,0)\in\mathbb{R}^{n},\,y\in\mathbb{R}^{n-1}\\}$
respectively.

Let us recall some facts that also hold in the present context. If $u\in
W^{1,2}(D)$ and $D$ is a Lipschitz set, then $u$ has a trace on the boundary
of $D$ and hence we can give a sense to the expression $u=h$ on $\partial D$.
If in addition, the function $u$ is a solution of $Lu=0$ in $D$ (the notion of
solution is taken in the weak sense, see for instance [Ken, Definition 1.1.4])
and $h$ is continuous on $\partial D$, then $u$ is continuous on
$\overline{D}$.

The Green function (associated to the domain $\Omega=\mathbb{R}^{n}_{+}$ and
the elliptic operator $L$) is denoted by $g(X,Y)$ \- with $X,Y\in\Omega$ \-
and the harmonic measure (associated to $\Omega$ and $L$) is written
$\omega^{X}(E)$ \- with $X\in\Omega$ and $E\subset\Gamma$. The notation
$\omega^{X}_{D}(E)$ \- where $X\in D$ and $E\subset\partial D$ \- denotes the
harmonic measure associated to the domain $D$ (and the operator $L$).

When $x_{0}=(y_{0},0)\in\Gamma$ and $r>0$, we use the notation $A_{r}(x_{0})$
for $(y_{0},r)$.

In this context, the comparison principle given in [Ken, Lemma 1.3.7] is

######  Lemma 11.1 (Comparison principle, codimension 1).

Let $x_{0}\in\Gamma$ and $r>0$. Let $u,v\in W^{1,2}(\Omega\cap B(x_{0},2r))$
be two non-negative solutions of $Lu=Lv=0$ in $\Omega\cap B(x_{0},2r)$
satisfying $u=v=0$ on $\Gamma\cap B(x_{0},2r)$. Then for any $X\in\Omega\cap
B(x_{0},r)$, we have

(11.1) $C^{-1}\frac{u(A_{r}(x_{0}))}{v(A_{r}(x_{0}))}\leq\frac{u(X)}{v(X)}\leq
C\frac{u(A_{r}(x_{0}))}{v(A_{r}(x_{0}))},$

where $C>0$ depends only on the dimension $n$ and the ellipticity constants of
the matrix $A$.

Proof. We recall quickly the ideas of the proof of the comparison principle
found in [Ken].

Let $x_{0}\in\Gamma$ and $r>0$ be given. We denote $A_{r}(x_{0})$ by $X_{0}$
and, for $\alpha>0$, $B(x_{0},\alpha r)$ by $B_{\alpha}$.

The proof of (11.1) is reduced to the proof of the upper bound

(11.2) $\frac{u(X)}{v(X)}\leq C\frac{u(X_{0})}{v(X_{0})}\ \ \text{ for }X\in
B_{1}\cap\Omega$

because of the symmetry of the role of $u$ and $v$.

Step 1: Upper bound on $u$.

By definition of the harmonic measure,

(11.3) $u(X)=\int_{\partial(\Omega\cap B_{3/2})}u(y)d\omega^{X}_{\Omega\cap
B_{3/2}}(y)\ \ \text{ for }X\in B_{3/2}\cap\Omega.$

Note that $\partial(\Omega\cap B_{3/2})=(\partial
B_{3/2}\cap\Omega)\cup(\Gamma\cap B_{3/2})$. Hence, for any $X\in
B_{3/2}\cap\Omega$,

(11.4) $u(X)=\int_{\partial B_{3/2}\cap\Omega}u(y)d\omega_{\Omega\cap
B_{3/2}}^{X}(y)+\int_{\Gamma\cap B_{3/2}}u(y)d\omega_{\Omega\cap
B_{3/2}}^{X}(y)=\int_{\partial B_{3/2}\cap\Omega}u(y)d\omega^{X}_{\Omega\cap
B_{3/2}}(y)$

because, by assumption, $u=0$ on $\Gamma\cap B_{2}$. Lemma 1.3.5 in [Ken]
gives now, for any $Y\in B_{7/4}$, the bound $u(Y)\leq Cu(X_{0})$ with a
constant $C>0$ which is independent of $Y$. So by the positivity of the
harmonic measure, we have for any $X\in B_{3/2}\cap\Omega$

(11.5) $\begin{split}u(X)&\leq Cu(X_{0})\int_{\partial
B_{3/2}\cap\Omega}d\omega^{X}_{\Omega\cap B_{3/2}}(y)\leq
Cu(X_{0})\omega^{X}_{\Omega\cap B_{3/2}}(\partial
B_{3/2}\cap\Omega).\end{split}$

Step 2: Lower bound on $v$.

First, again by definition of the harmonic measure, we have that for $X\in
B_{3/2}\cap\Omega$,

(11.6) $v(X)=\int_{\partial(\Omega\cap B_{3/2})}v(y)d\omega^{X}_{\Omega\cap
B_{3/2}}(y).$

Set $E=\\{y\in\partial
B_{3/2}\cap\Omega\,;\,\,\mathrm{dist}(y,\Gamma)\geq\frac{1}{2}r\\}$. By
assumption, $v\geq 0$ on $\partial(\Omega\cap B_{3/2})$. In addition, thanks
to the Harnack inequality, $v(y)\geq C^{-1}v(X_{0})$ for every $y\in E$, with
a constant $C>0$ that is independent of $y$. So the positivity of the harmonic
measure yields, for $X\in B_{3/2}\cap\Omega$,

(11.7) $\begin{split}v(X)&\geq C^{-1}v(X_{0})\int_{E}d\omega^{X}_{\Omega\cap
B_{3/2}}(y)\geq C^{-1}v(X_{0})\omega^{X}_{\Omega\cap B_{3/2}}(E).\end{split}$

Step 3: Conclusion.

From steps 1 and 2, we deduce that

(11.8) $\frac{u(X)}{v(X)}\leq
C\frac{u(X_{0})}{v(X_{0})}\frac{\omega^{X}_{\Omega\cap B_{3/2}}(\partial
B_{3/2}\cap\Omega)}{\omega^{X}_{\Omega\cap B_{3/2}}(E)}\ \ \ \text{ for
}X\in\Omega\cap B_{3/2}.$

The inequality (11.2) is now a consequence of the doubling property of the
harmonic measure (see for instance (1.3.7) in [Ken]), that gives

(11.9) $\omega^{X}_{\Omega\cap B_{3/2}}(\partial B_{3/2}\cap\Omega)\leq
C\omega^{X}_{\Omega\cap B_{3/2}}(E)\ \ \ \text{ for }X\in\Omega\cap B_{1}.$

The lemma follows. $\Box$

The proof above relies on the use of the harmonic measure for the domain
$\Omega\cap B_{3/2}$. We want to avoid this, and use only the Green functions
and harmonic measures related to the domain $\Omega$ itself.

First, we need a way to compare two functions in a domain, that is a suitable
maximum principle. In the previous proof of Lemma 11.1, the maximum principle
was replaced/hidden by the positivity of the harmonic measure, whose proof
makes a crucial use of the maximum principle for solutions. See [Ken,
Definition 1.2.6] for the construction of the harmonic measure, and [Ken,
Corollary 1.1.18] for the maximum principle. The maximum principle that we
will use is the following.

######  Lemma 11.2.

Let $F\subset E\subset\mathbb{R}^{n}$ be two sets such that $F$ is closed, $E$
is open, and $\,\mathrm{dist}(F,\mathbb{R}^{n}\setminus E)>0$. Let $u$ be a
solution in $E\cap\Omega$ such that

  1. (i)

$\displaystyle\int_{E}|\nabla u|^{2}\,dx<+\infty$,

  2. (ii)

$u\geq 0$ on $\Gamma\cap E$,

  3. (iii)

$u\geq 0$ in $(E\setminus F)\cap\Omega$.

Then $u\geq 0$ in $E\cap\Omega$.

In a more ‘classical’ maximum principle, assumption (iii) would be replaced by

(iii’) $u\geq 0$ in $\partial E\cap\Omega$.

Since this subsection aims to illustrate what we will do in the next
subsection, we state here a maximum principle which is as close as possible to
the one we will actually prove in higher codimension. Let us mention that
using (iii) instead of (iii’) will not make computations harder or easier.
However, (iii) is much easier to define and use in the higher codimension case
(to the point that we did not even try to give a precise meaning to (iii’)).

We do not prove Lemma 11.2 here, because the proof is the same as for Lemma
11.3 below, which is its higher codimension version.

Notice that Lemma 11.2 is really a maximum principle where we use the values
of $u$ on a boundary $(\Gamma\cap E)\cup(\Omega\cap F\setminus E)$ that
surrounds $E$ to control the values of $u$ in $\Omega\cap E$, but here the
boundary also has a thick part, $\Omega\cap F\setminus E$. This makes it
easier to define Dirichlet conditions on that thick set, which is the main
point of (iii).

The first assumption (i) is a technical hypothesis, it can be seen as a way to
control $u$ at infinity, which is needed because we actually do not require
$E$ or even $F$ to be bounded.

Lemma 11.2 will be used in different situations. For instance, we will use it
when $E=2B$ and $F=B$, where $B$ is a ball centered on $\Gamma$.

Step 1 (modified): We want to find an upper estimate for $u$ that avoids using
the measure $\omega^{X}_{\Omega\cap B_{3/2}}$. Lemma 1.3.5 in [Ken] gives, as
before, that $u(X)\leq Cu(X_{0})$ for any $X\in B_{7/4}\cap\Omega$. The
following result states the non-degeneracy of the harmonic measure.

(11.10) $\omega^{X}(\Gamma\setminus B_{5/4})\geq C^{-1}\ \ \ \text{ for
}X\in\Omega\setminus B_{3/2},$

where $C>0$ is independent of $x_{0}$, $r$ or $X$. Indeed, when
$X\in\Omega\setminus B_{3/2}$ is close to the boundary, the lower bound
(11.10) can be seen as a consequence of the Hölder continuity of solutions.
The proof for all $X\in\Omega\setminus B_{3/2}$ is then obtained with the
Harnack inequality. See [Ken, Lemma 1.3.2] or Lemma 11.10 below for the proof.

From there, we deduce that

(11.11) $u(X)\leq Cu(X_{0})\omega^{X}(\Gamma\setminus B_{5/4})\ \ \ \text{ for
}X\in\Omega\cap[B_{7/4}\setminus B_{3/2}].$

We want to use the maximum principle given above (Lemma 11.2), with
$E=B_{7/4}$ and $F=B_{3/2}$. However, the function
$X\to\omega^{X}(\Gamma\setminus B_{5/4})$ doesn’t satisfy the assumption (i)
of Lemma 11.2. So we take $h\in C^{\infty}(\mathbb{R}^{n})$ such that $0\leq
h\leq 1$, $h\equiv 1$ on $\mathbb{R}^{n}\setminus B_{5/4}$, and $h\equiv 0$ on
$\mathbb{R}^{n}\setminus B_{9/8}$. Define $u_{h}$ as the only solution of
$Lu_{h}=0$ in $\Omega$ with the Dirichlet condition $u_{h}=h$ on $\Gamma$. We
have $u_{h}(X)\geq\omega^{X}(\Gamma\setminus B_{5/4})$ by the positivity of
the harmonic measure, and thus the bound (11.11) yields the existence of
$K_{0}>0$ (independent of $x_{0}$, $r$, $X$) such that

(11.12) $u_{1}(X):=K_{0}u(X_{0})u_{h}(X)-u(X)\geq 0\ \ \ \text{ for
}X\in\Omega\cap[B_{7/4}\setminus B_{3/2}].$

It would be easy to check that $u_{1}\geq 0$ on $\Gamma\cap B_{7/4}$ and
$\int_{E}|\nabla u_{1}|<+\infty$, but we leave the details because they will
be done in the larger codimension case. So Lemma 11.2 gives that $u_{1}\geq 0$
in $\Omega\cap B_{7/4}$, that is

(11.13) $u(X)\leq K_{0}u(X_{0})u_{h}(X)\leq
K_{0}u(X_{0})\omega^{X}(\Gamma\setminus B_{9/8})\ \ \ \text{ for
}X\in\Omega\cap B_{7/4},$

by definition of $h$ and positivity of the harmonic measure.

Step 2 (modified): In the same way, we want to adapt Step 2 of the proof of
Lemma 11.1. If we want to proceed as in Step 1, we would like to find and use
a function $f$ that keeps the main properties of the object
$\omega^{X}_{\Omega\cap B_{7/4}}(E)$, where $E=\big{\\{}y\in\partial
B_{7/4},\,;\,\,\mathrm{dist}(y,\Gamma)\geq r/2\big{\\}}$. For instance, $f$
such that

  1. (a)

$f$ is a solution of $Lf=0$ in $\Omega\cap B_{7/4}$,

  2. (b)

$f\leq 0$ in $\Gamma\cap B_{7/4}$,

  3. (c)

$f\leq 0$ in
$\\{X\in\Omega,\,\,\mathrm{dist}(X,\Gamma)<r/2\\}\cap[B_{7/4}\setminus
B_{3/2}]$,

  4. (d)

$f(X)\approx\omega^{X}(\Gamma\setminus B_{9/8})$ in $\Omega\cap B_{1}$, in
particular $f>0$ in $\Omega\cap B_{1}$.

The last point is important to be able to conclude (in Step 3). It is given by
the doubling property of the harmonic measure (11.9) in the previous proof of
Lemma 11.1.

We were not able to find such a function $f$. However, we can construct an $f$
that satisfies some conditions close to (a), (b), (c) and (d) above. Since $f$
fails to verify exactly (a), (b), (c) and (d), extra computations are needed.

First, note that it is enough to prove that there exists $M>0$ depending only
on $n$ and the ellipticity constants of $A$, such that for $y_{0}\in\Gamma$,
$s>0$, and any non-negative solution $v$ to $Lv=0$ in $B(y_{0},Ms)$

(11.14) $v(X)\geq C^{-1}v(A_{s}(y))\omega^{X}(\Gamma\setminus B(y_{0},2s))\ \
\ \text{ for }X\in\Omega\cap B(y_{0},s),$

where here the the corkscrew point $A_{s}(y)$ is just $A_{s}(y)=(y,s)$.
Indeed, if we have (11.14), then we can prove that, in the situation of Step
2,

(11.15) $v(X)\geq C^{-1}v(X_{0})\omega^{X}(\Gamma\setminus B_{2})\ \ \ \text{
for }X\in\Omega\cap B_{1}$

by using a proper covering of the domain $\Omega\cap B_{1}$ (if
$X\in\Omega\cap B_{1}$ lies within $\frac{1}{4M}$ of $\Gamma$, say, we use
(11.14) with $y_{0}\in\Gamma$ close to $X$ and $s=\frac{1}{2M}$, and then the
Harnack inequality; if instead $X\in\Omega\cap B_{1}$ is far from the boundary
$\Gamma$, (11.15) is only a consequence of the Harnack inequality).

The conclusion (11.2) comes then from (11.13), (11.15) and the doubling
property of the harmonic measure (see for instance (1.3.7) in [Ken]).

It remains to prove the claim (11.14). Let $y_{0}\in\Gamma$, $s>0$, and $v$ be
given. Write $Y_{0}$ for $A_{s}(y_{0})$ and, for $\alpha>0$, write
$B^{\prime}_{\alpha}$ for $B(y_{0},\alpha s)$. Let $K_{1}$ and $K_{2}$ be some
positive constants that are independent of $y_{0}$, $s$, and $X$, and will be
chosen later. Pick $h_{K_{2}}\in C^{\infty}(\mathbb{R}^{n})$ such that
$h_{K_{2}}\equiv 1$ on $\mathbb{R}^{n}\setminus B^{\prime}_{K_{2}}$, $0\leq
h_{K_{2}}\leq 1$ everywhere, and $h_{K_{2}}\equiv 0$ on $B_{K_{2}/2}$. Define
$u_{K_{2}}$ as the solution of $Lu_{K_{2}}=0$ in $\Omega$ with the Dirichlet
condition $h_{K_{2}}$ on $\Gamma$, that will serve as a smooth substitute for
$X\to\omega^{X}(\Gamma\setminus B^{\prime}_{K_{2}})$. Define a function
$f_{y_{0},s}$ on $\Omega\setminus\\{Y_{0}\\}$ by

(11.16) $f_{y_{0},s}(X)=s^{n-2}g(X,Y_{0})-K_{1}u_{K_{2}}.$

When $|X-Y_{0}|\geq s/8$, the term $s^{n-2}g(X,Y_{0})$ is uniformly bounded:
this fact can be found in [HoK] (for $n\geq 3$) and [DK] (for $n=2$). In
addition, due to the non-degeneracy of the harmonic measure (same argument as
for (11.10), similar to [Ken, Lemma 1.3.2]), there exists $C>0$ (independent
of $K_{2}>0$) such that $\omega^{X}(\Gamma\setminus B^{\prime}_{K_{2}})\geq
C^{-1}$ for $X\in\Omega\setminus B^{\prime}_{2K_{2}}$. Hence we can find
$K_{1}>0$ such that for any choice of $K_{2}>0$, we have

(11.17) $f_{y_{0},s}(X)\leq 0\ \ \ \text{ for }X\in\Omega\setminus
B^{\prime}_{2K_{2}}.$

For the sequel, we state an important result. There holds

(11.18) $C^{-1}s^{n-2}g(X,Y_{0})\leq\omega^{X}(\Gamma\setminus
B^{\prime}_{2})\leq Cs^{n-2}g(X,Y_{0})\ \ \ \text{ for
}X\in\Omega\cap[B^{\prime}_{1}\setminus B(Y_{0},s/8)],$

where $C>0$ depends only on $n$ and the ellipticity constant of the matrix
$A$. This result can be seen as an analogue of [Ken, Corollary 1.3.6]. It is
proven in the higher codimension case in Lemma 11.11 below. The equivalence
(11.18) can be seen as a weak version of the comparison principle, dealing
only with harmonic measures and Green functions. It can be proven, like [Ken,
Corollary 1.3.6], before the full comparison principle by using the specific
properties of the Green functions and harmonic measures.

We want to take $K_{2}>0$ so large that

(11.19) $f_{y_{0},s}(X)\geq\frac{1}{2}s^{n-2}g(X,Y_{0})\ \ \ \text{ for
}X\in\Omega\cap[B^{\prime}_{1}\setminus B(Y_{0},s/8)].$

We build a smooth substitute $u_{4}$ for $\omega^{X}(\Gamma\setminus
B^{\prime}_{2})$, namely the solution of $Lu_{4}=0$ in $\Omega$ with the
Dirichlet condition $u_{4}=h_{4}$ on $\Gamma$, where $h_{4}\in
C^{\infty}(\mathbb{R}^{n})$, $h_{4}\equiv 1$ on $\mathbb{R}^{n}\setminus
B^{\prime}_{4}$, $0\leq h_{4}\leq 1$ everywhere, and $h_{4}\equiv 0$ on
$B^{\prime}_{2}$. Thanks to the Hölder continuity of solutions and the non-
degeneracy of the harmonic measure, we have that for
$X\in\Omega\cap[B^{\prime}_{10}\setminus B^{\prime}_{5}]$ and any $K_{2}\geq
20$,

(11.20) $C^{-1}u_{K_{2}}(X)\leq(K_{2})^{-\alpha}\leq
C(K_{2})^{-\alpha}u_{4}(X),$

with constants $C,\alpha>0$ independent of $K_{2}$, $y_{0}$, $s$ or $X$. Since
the functions $u_{K_{2}}$ and $u_{4}$ are smooth enough, and
$C^{-1}u_{K_{2}}=0\leq C(K_{2})^{-\alpha}u_{4}(X)$ on $\Gamma\cap
B^{\prime}_{10}$, the maximum principle (Lemma 11.2) implies that

(11.21) $u_{K_{2}}(X)\leq C(K_{2})^{-\alpha}u_{4}(X)\ \ \ \text{ for
}X\in\Omega\cap B^{\prime}_{10}.$

We use (11.18) to get that for $K_{2}\geq 20$,

(11.22) $K_{1}u_{K_{2}}(X)\leq CK_{1}(K_{2})^{-\alpha}s^{n-2}g(X,Y_{0})\ \ \
\text{ for }X\in\Omega\cap[B^{\prime}_{1}\setminus B(Y_{0},s/8)].$

The inequality (11.19) can be now obtained by taking $K_{2}\geq 20$ so that
$CK_{1}(K_{2})^{-\alpha}\leq\frac{1}{2}$. From (11.19) and (11.18), we deduce
that

(11.23) $f_{y_{0},s}(X)\geq C^{-1}\omega^{X}(\Gamma\setminus B^{\prime}_{2})\
\ \ \text{ for }X\in B^{\prime}_{1}\setminus B(Y_{0},s/8),$

where $C>0$ depends only on $n$ and the ellipticity constants of the matrix
$A$.

Recall that our goal is to prove the claim (11.14), which will be established
with $M=4K_{2}$. Let $v$ be a non-negative solution of $Lv=0$ in $\Omega\cap
B^{\prime}_{4K_{2}}$. We can find $K_{3}>0$ (independent of $y_{0}$, $s$ and
$X$) such that

(11.24) $v(X)\geq K_{3}v(Y_{0})f_{y_{0},s}(X)\ \ \ \text{ for }X\in
B(Y_{0},\frac{1}{4}s)\setminus B(Y_{0},\frac{1}{8}s).$

Indeed $f_{y_{0},s}(X)\leq s^{n-2}g(X,Y_{0})\leq C$ when $|X-Y_{0}|\geq s/8$,
thanks to the pointwise bounds on the Green function (see [HoK], [DK]) and
$v(X)\geq C^{-1}v(Y_{0})$ when $|X-Y_{0}|\leq s/4$ because of the Harnack
inequality. Also, thanks to (11.17),

(11.25) $v(X)\geq 0\geq K_{3}v(Y_{0})f_{y_{0},s}(X)\ \ \ \text{ for
}X\in\Omega\cap[B^{\prime}_{4K_{2}}\setminus B^{\prime}_{2K_{2}}]$

and it is easy to check that

(11.26) $v(y)\geq 0\geq K_{3}v(Y_{0})f_{y_{0},s}(y)\ \ \ \text{ for
}y\in\Gamma\cap B^{\prime}_{4K_{2}}.$

We can apply our maximal principle, that is Lemma 11.2, with
$E=B^{\prime}_{4K_{2}}\setminus B(Y_{0},\frac{1}{8}s)$ and
$F=B^{\prime}_{2K_{2}}\setminus B(Y_{0},\frac{1}{4}s)$ and get that

(11.27) $v(X)\geq K_{3}v(Y_{0})f_{y_{0},s}(X)\ \ \ \text{ for
}X\in\Omega\cap[B^{\prime}_{4K_{2}}\setminus B(Y_{0},\frac{1}{8}s)].$

In particular, thanks to (11.23),

(11.28) $v(X)\geq C^{-1}v(Y_{0})\omega^{X}(\Gamma\setminus B^{\prime}_{2})\ \
\ \text{ for }X\in\Omega\cap[B^{\prime}_{1}\setminus B(Y_{0},\frac{1}{8}s)].$

Since both $v$ and $X\to\omega^{X}(\Gamma\setminus B^{\prime}_{2})$ are
solutions in $\Omega\cap B^{\prime}_{2}$, the Harnack inequality proves

(11.29) $v(X)\geq C^{-1}v(Y_{0})\omega^{X}(\Gamma\setminus B^{\prime}_{2})\ \
\ \text{ for }X\in\Omega\cap B^{\prime}_{1}.$

The claim (11.14) follows, which ends our alternative proof of Lemma 11.1.

###  11.2. The case of codimension higher than 1

We need first the following version of the maximum principle.

######  Lemma 11.3.

Let $F\subset\mathbb{R}^{n}$ be a closed set and $E\subset\mathbb{R}^{n}$ an
open set such that $F\subset E\subset\mathbb{R}^{n}$ and
$\,\mathrm{dist}(F,\mathbb{R}^{n}\setminus E)>0$. Let $u\in W_{r}(E)$ be a
supersolution for $L$ in $\Omega\cap E$ such that

  1. (i)

$\displaystyle\int_{E}|\nabla u|^{2}\,dm<+\infty$,

  2. (ii)

$Tu\geq 0$ a.e. on $\Gamma\cap E$,

  3. (iii)

$u\geq 0$ a.e. in $(E\setminus F)\cap\Omega$.

Then $u\geq 0$ a.e. in $E\cap\Omega$.

Proof. The present proof is a slight variation of the proof of Lemma 9.2.

Set $v:=\min\\{u,0\\}$ in $E\cap\Omega$ and $v:=0$ in $\Omega\setminus E$.
Note that $v\leq 0$. We want to use $v$ as a test function. We claim that

(11.30) $v$ lies in $W_{0}$ and is supported in $F$.

Pick $\eta\in C^{\infty}_{0}(E)$ such that $\eta=1$ in $F$ and $\eta\geq 0$
everywhere, it is indeed possible because
$\,\mathrm{dist}(F,\mathbb{R}^{n}\setminus E)>0$. Since $u\in W_{r}(E)$, we
have $\eta u\in W$, from which we deduce $\min\\{0,\eta u\\}Â \in W$ by Lemma
6.1. By (iii), $v=\min\\{0,\eta u\\}$ almost everywhere and hence $v\in W$.

Notice that $T(\eta u)\geq 0$ because of Assumption (ii) (and Lemma 8.1).
Hence $v=\min\\{\eta u,0\\}\in W_{0}$. And since (iii) also proves that $v$ is
supported in $F$, the claim (11.30) follows.

Since $v$ is in $W_{0}$, Lemma 5.5 proves that $v$ can be approached in $W$ by
a sequence of functions $(v_{k})_{k\geq 1}$ in $C^{\infty}_{0}(\Omega)$ (i.e.,
that are compactly supported in $\Omega$; see (5.24)). Note also that the
construction used in Lemma 5.5 allows us, since $v\leq 0$ is supported in $F$,
to take $v_{k}\leq 0$ and compactly supported in $E$. Definition 8.13 gives

(11.31) $\int_{E}\mathcal{A}\nabla u\cdot\nabla
v_{k}\,dm=\int_{\Omega}\mathcal{A}\nabla u\cdot\nabla v_{k}\,dm\leq 0$

and since the map

(11.32) $\varphi\in W\to\int_{E}\mathcal{A}\nabla u\cdot\nabla\varphi\,dm$

is bounded on $W$ thanks to assumption (i) and (8.8), we deduce that

(11.33) $\int_{E}\mathcal{A}\nabla u\cdot\nabla v\,dm\leq 0.$

Now Lemma 6.1 gives

(11.34) $\nabla v=\left\\{\begin{array}[]{ll}\nabla u&\text{ if }u<0\\\
0&\text{ if }u\geq 0\end{array}\right.$

and so (11.33) becomes

(11.35) $\int_{\Omega}\mathcal{A}\nabla v\cdot\nabla
v\,dm=\int_{E}\mathcal{A}\nabla u\cdot\nabla v\,dm\leq 0.$

Together with the ellipticity condition (8.9), we obtain $\|v\|_{W}\leq 0$.
Recall that $\|.\|_{W}$ is a norm on $W_{0}\ni v$, hence $v=0$ a.e. in
$\Omega$. We conclude from the definition of $v$ that $u\geq 0$ a.e. in
$E\cap\Omega$. $\Box$

Let us use the maximum principle above to prove the following result on the
Green function.

######  Lemma 11.4.

We have

(11.36) $g(x,y)\leq C\min\\{\delta(y),\delta(x)\\}^{1-d}\ \ \ \text{ for
}x,y\in\Omega\text{ such that }|x-y|\geq\delta(y)/4,$

where the constant $C>0$ depends only on $d$, $n$, $C_{0}$ and $C_{1}$.

######  Remark 11.5.

Lemma 11.4 is an improvement on the pointwise bounds (10.6) only when $d<1$.

Proof. Let $y\in\Omega$. Lemma 10.1 (v) gives

(11.37) $g(x,y)\leq K_{1}\delta(y)^{1-d}\ \ \ \text{ for }x\in
B(y,\delta(y)/4)\setminus B(y,\delta(y)/8)$

for some $K_{1}>0$ that is independent of $x$ and $y$. Define $u$ on
$\Omega\setminus\\{y\\}$ by $u(x)=K_{1}\delta(y)^{1-d}-g(x,y)$. Notice that
$u$ is a solution in $\Omega\setminus\overline{B(y,\delta(y)/4)}$, by (10.4).
Also, thanks to Lemma 10.1 (i), the integral
$\displaystyle\int_{\Omega\setminus B(y,\delta(y)/8)}|\nabla
u|^{2}\,dm=\int_{\Omega\setminus B(y,\delta(y)/8)}|\nabla g(.,y)|^{2}\,dm$ is
finite, and $Tu=K_{1}\delta(y)^{1-d}$ is non-negative a.e. on $\Gamma$. In
addition, due to (11.37), we have $u\geq 0$ on $B(y,\delta(y)/4)\setminus
B(y,\delta(y)/4)$. Thus $u$ satisfies all the assumption of Lemma 11.3 (the
maximum principle), where we choose
$E=\mathbb{R}^{n}\setminus\overline{B(y,\delta(y)/8)}$ and
$F=\mathbb{R}^{n}\setminus B(y,\delta(y)/8)$, and which yields

(11.38) $g(x,y)\leq C\delta(y)^{1-d}\ \ \ \text{ for }x\in\Omega\setminus
B(y,\delta(y)/8).$

It remains to prove that

(11.39) $g(x,y)\leq C\delta(x)^{1-d}\ \ \ \text{ for }x,y\in\Omega\text{ such
that }|x-y|\geq\delta(y)/4.$

But Lemma 10.6 says that $g(x,y)=g_{T}(y,x)$, where $g_{T}$ is the Green
function associated to the operator $L_{T}=-\operatorname{div}A^{T}\nabla$.
The above argument proves that

(11.40) $g(x,y)=g_{T}(y,x)\leq C\delta(x)^{1-d}\ \ \ \text{ for
}x,y\in\Omega\text{ such that }|x-y|\geq\delta(x)/8,$

which is (11.39) once we remark that $|x-y|\geq\delta(y)/4$ implies that
$|x-y|\geq\delta(x)/8$. $\Box$

Let us prove the existence of “corkscrew points” in $\Omega$.

######  Lemma 11.6.

There exists $\epsilon>0$, that depends only upon the dimensions $d$ and $n$
and the constant $C_{0}$, such that for $x_{0}\in\Gamma$ and $r>0$, there
exists a point $A_{r}(x_{0})\in\Omega$ such that

  1. (i)

$|A_{r}(x_{0})-x_{0}|\leq r$,

  2. (ii)

$\delta(A_{r}(x_{0}))\geq\epsilon r$.

In particular, $\delta(A_{r}(x_{0}))\approx|A_{r}(x_{0})-x_{0}|\approx r$.

In the sequel, for any $s>0$ and $y\in\Gamma$, $A_{s}(y)$ will denote any
point in $\Omega$ satisfying the conditions (i) and (ii) of Lemma 11.6.

Proof. Let $x_{0}\in\Gamma$ and $r>0$ be given. Let $\epsilon\in(0,1/8)$ be
small, to be chosen soon. Let $z_{1},\cdots z_{N}$ be a maximal collection of
points of $B(x_{0},(1-2\epsilon)r)$ that lie at mutual distances at least
$4\epsilon r$. Set $B_{i}=B(z_{i},\epsilon r)$; notice that the
$2B_{i}=B(z_{i},2\epsilon r)$ are disjoint and contained in $B(x_{0},r)$, and
the $5B_{i}$ cover $B(x_{0},(1-2\epsilon)r)$ (by maximality), so
$\sum_{i}|5B_{i}|\geq|B(x_{0},(1-2\epsilon)r)|\geq C^{-1}r^{n}$ and hence
$N\geq C^{-1}\epsilon^{-n}$.

Suppose for a moment that every $B_{i}$ meets $\Gamma$. Pick
$y_{i}\in\Gamma\cap B_{i}$, notice that $B(y_{i},\epsilon r)\subset 2B_{i}$,
and then use the Ahlfors-regularity property (1.1) to prove that

(11.41) $C_{0}^{-1}(\epsilon
r)^{d}N\leq\sum_{i=1}^{N}\mathcal{H}^{d}(\Gamma\cap B(y_{i},\epsilon
r))\leq\sum_{i=1}^{N}\mathcal{H}^{d}(\Gamma\cap
2B_{i})\leq\mathcal{H}^{d}(\Gamma\cap B(x_{0},r))\leq C_{0}r^{d}$

because the $2B_{i}$ are disjoint and contained in $B(x_{0},r)$. Thus $N\leq
C_{0}^{2}\epsilon^{-d}$, which makes our initial estimate on $N$ impossible if
we choose $\epsilon$ such that $\epsilon^{n-d}<C^{-1}C_{0}^{-2}$.

We pick $\epsilon$ like this, and by contraposition get that at least one
$B_{i}$ does not meet $\Gamma$. We choose $A_{r}(x_{0})=z_{i}$, and notice
that $\delta(x_{i})\geq\epsilon r$ because $B_{i}\cap\Gamma=\emptyset$, and
$|z_{i}-x_{0}|\leq r$ by construction. The lemma follows. $\Box$

We also need the following slight improvement of Lemma 11.6.

######  Lemma 11.7.

Let $M_{1}\geq 1$ be given. There exists $M_{2}>M_{1}$ (depending on $d$, $n$,
$C_{0}$ and $M_{1}$) such that for any ball $B$ of radius $r$ and centered on
$\Gamma$ and any $x\in B\setminus\Gamma$ such that
$\delta(x)\leq\frac{r}{M_{2}}$, we can find $y\in B\setminus\Gamma$ such that

  1. (i)

$\delta(y)\geq M_{1}\delta(x)$,

  2. (ii)

$|x-y|\leq M_{2}\delta(x)$.

Proof. The proof is almost the same. Let $M_{1}\geq 1$ be given, and let
$M_{2}\geq 10M_{1}$ be large, to be chosen soon. Then let $B=B(x_{0},r)$ and
$x\in B$ be as in the statement. Set
$B^{\prime}=B(x_{0},r-2M_{1}\delta(x))\cap B(x,(M_{2}-2M_{1})\delta(x))$;
notice that the two radii are larger than $M_{2}\delta(x)/2$, because $r\geq
M_{2}\delta(x)\geq 10M_{1}\delta(x)$, so $|B^{\prime}|\geq
C^{-1}(M_{2}\delta(x))^{n}$ for a constant $C>0$ independent of $x$.

Pick a maximal family $(z_{i})$, $1\leq i\leq N$, of points of $B^{\prime}$
that lie at mutual distances at least $M_{1}\delta(x)$ from each other, and
set $B_{i}=B(z_{i},M_{1}\delta(x))$ for $1\leq i\leq N$. The $5B_{i}$ cover
$B^{\prime}$ by maximality, so $N\geq
C^{-1}(M_{1}\delta(x))^{-n}|B^{\prime}|\geq C^{-1}(M_{2}/M_{1})^{n}$.

Suppose for a moment that every $B_{i}$ meets $\Gamma$. Then pick $y_{i}\in
B_{i}\cap\Gamma$ and use the Ahlfors regularity property (1.1) and the fact
that the $2B_{i}$ contain the $B(y_{i},M_{1}\delta(x))$ and are disjoint to
prove that

(11.42) $\displaystyle C_{0}^{-1}(M_{1}\delta(x))^{d}N$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{N}\mathcal{H}^{d}(\Gamma\cap
B(y_{i},M_{1}\delta(x)))$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{N}\mathcal{H}^{d}(\Gamma\cap
2B_{i})\leq\mathcal{H}^{d}(\Gamma\cap B(x,M_{2}\delta(x)))\leq
C_{0}(M_{2}\delta(x))^{d}.$

That is, $M_{1}^{d}N\leq C_{0}^{2}M_{2}^{d}$, and this contradicts our other
bound for $N$ if $M_{2}/M_{1}$ is large enough. We choose $M_{2}$ like this;
then some $B_{i}$ doesn’t meet $\Gamma$, and we can take $y=z_{i}$. $\Box$

Before we prove the comparison theorem, we need a substitute for [Ken, Lemma
1.3.4].

######  Lemma 11.8.

Let $x_{0}\in\Gamma$ and $r>0$ be given, and let $X_{0}:=A_{r}(x_{0})$ be as
in Lemma 11.6. Let $u\in W_{r}(B(x_{0},2r))$ be a non-negative, non
identically zero, solution of $Lu=0$ in $B(x_{0},2r)\cap\Omega$, such that
$Tu\equiv 0$ on $B(x_{0},2r)\cap\Gamma$. Then

(11.43) $u(X)\leq Cu(X_{0})\ \ \ \ \text{ for }X\in B(x_{0},r),$

where $C>0$ depends only on $d$, $n$, $C_{0}$ and $C_{1}$.

Proof. We follow the proof of [KJ, Lemma 4.4].

Let $x\in\Gamma$ and $s>0$ such that $Tu\equiv 0$ on $B(x,s)\cap\Gamma$. Then
the Hölder continuity of solutions given by Lemma 8.33 proves the existence of
$\epsilon>0$ (that depends only on $d$, $n$, $C_{0}$, $C_{1}$) such that

(11.44) $\sup_{B(x,\epsilon s)}u\leq\frac{1}{2}\sup_{B(x,s)}u.$

Without loss of generality, we can choose $\epsilon<\frac{1}{2}$.

A rough idea of the proof of (11.43) is that $u(x)$ should not be near the
maximum of $u$ when $x$ lies close to $B(x_{0},r)\cap\Gamma$, because of
(11.44). Then we are left with points $x$ that lie far from the boundary, and
we can use the Harnack inequality to control $u(x)$. The difficulty is that
when $x\in B(x_{0},r)$ lies close to $\Gamma$, $u(x)$ can be bounded by values
of $u$ inside the domain, and not by values of $u$ near $\Gamma$ but from the
exterior of $B(x_{0},r)$. We will prove this latter fact by contradiction: we
show that if $\sup_{B(x_{0},r)}u$ exceeds a certain bound, then we can
construct a sequence of points $X_{k}\in B(x_{0},\frac{3}{2}r)$ such that
$\delta(X_{k})\to 0$ and $u(X_{k})\to+\infty$, and hence we contradict the
Hölder continuity of solutions at the boundary.

Since $u(X)>0$ somewhere, the Harnack inequality (Lemma 8.9), maybe applied a
few times, yields $u(X_{0})>0$. We can rescale $u$ and assume that
$u(X_{0})=1$. We claim that there exists $M>0$ such that for any integer
$N\geq 1$ and $Y\in B(x_{0},\frac{3}{2}r)$,

(11.45) $\delta(Y)\geq\epsilon^{N}r\Longrightarrow u(Y)\leq M^{N},$

where $\epsilon$ comes from (11.44) and the constant $M$ depends only upon
$d$, $n$, $C_{0}$, $C_{1}$. We will prove this by induction. The base case
(and in fact we will manage to start directly from some large integer $N_{0}$)
is given by the following. Let $M_{2}>0$ be the value given by Lemma 11.7 when
$M_{1}:=\frac{1}{\epsilon}$. Let $N_{0}\geq 1$ be the smallest integer such
that $M_{2}\leq\epsilon^{-N_{0}}$. We want to show the existence of $M_{3}\geq
1$ such that

(11.46) $u(Y)\leq M_{3}\ \ \ \ \ \text{ for every }Y\in
B(x_{0},\frac{3}{2}r)\text{ such that }\delta(Y)\geq\epsilon^{N_{0}}r.$

Indeed, if $Y\in B(x_{0},\frac{3}{2}r)$ satisfies
$\delta(Y)\geq\epsilon^{N_{0}}r$, Lemma 2.2 and the fact that
$|x_{0}-X_{0}|\approx r$ (by Lemma 11.6) imply the existence of a Harnack
chain linking $Y$ to $X_{0}$. More precisely, we can find balls
$B_{1},\dots,B_{h}$ with a same radius, such that $Y\in B_{1}$, $X_{0}\in
B_{h}$, $3B_{i}\subset B(x_{0},2r)\setminus\Gamma$ for $i\in\\{1,\dots,h\\}$,
and $B_{i}\cap B_{i+1}\neq\emptyset$ for $i\in\\{1,\dots,h-1\\}$, and in
addition $h$ is bounded independently of $x_{0}$, $r$ and $Y$. Together with
the Harnack inequality (Lemma 8.9), we obtain (11.46). This proves (11.45) for
$N=N_{0}$, but also directly for $1\leq N\leq N_{0}$, if we choose $M\geq
M_{3}$.

For any point $Y\in B(x_{0},\frac{3}{2}r)$ such that
$\delta(Y)\leq\epsilon^{N_{0}}r\leq\frac{r}{M_{2}}$, Lemma 11.7 (and our
choice of $M_{2}$) gives the existence of $Z\in B(x_{0},\frac{3}{2}r)\cap
B(Y,M_{2}\delta(Y))$ such that $\delta(Z)\geq M_{1}\delta(Y)$. Since $Z\in
B(Y,M_{2}\delta(Y))$ and $\delta(Z)>\delta(Y)>0$, Lemma 2.2 implies the
existence of a Harnack chain whose length is bounded by a constant depending
on $d$, $n$, $C_{0}$ (and $M_{2}$ \- but $M_{2}$ depends only on the three
first parameters) and together with the Harnack inequality (Lemma 8.9), we
obtain the existence of $M_{4}\geq 1$ (that depends only on $d$, $n$, $C_{0}$
and $C_{1}$) such that $u(Y)\leq M_{4}u(Z)$. So we just proved that

(11.47) $\begin{array}[]{l}\text{for any $Y\in B(x_{0},\frac{3}{2}r)$ such
that $\delta(Y)\leq\epsilon^{N_{0}}r$, }\\\ \qquad\text{there exists $Z\in
B(x_{0},\frac{3}{2}r)$ such that $\delta(Z)\geq M_{1}\delta(Y)$ and $u(Y)\leq
M_{4}u(Z)$.}\end{array}$

We turn to the main induction step. Set $M=\max\\{M_{3},M_{4}\\}\geq 1$ and
let $N\geq N_{0}$ be given. Assume, by induction hypothesis, that for any
$Z\in B(x_{0},\frac{3}{2}r)$ satisfying $\delta(Z)\geq\epsilon^{N}r$, we have
$u(Z)\leq M^{N}$. Let $Y\in B(x_{0},\frac{3}{2}r)$ be such that
$\delta(Y)\geq\epsilon^{N+1}r$ . The assertion (11.47) yields the existence of
$Z\in B(x_{0},\frac{3}{2}r)$ such that $\delta(Z)\geq
M_{1}\delta(Y)=\epsilon^{-1}\delta(Y)\geq\epsilon^{N}r$ and $u(Y)\leq
M_{4}u(Z)\leq Mu(Z)$. By the induction hypothesis, $u(Y)\leq M^{N+1}$. This
completes our induction step, and the proof of (11.45) for every $N\geq 1$.

Choose an integer $i$ such that $2^{i}\geq M$, where $M$ is the constant of
(11.45) that we just found, and then set $M^{\prime}=M^{i+3}$. We want to
prove by contradiction that

(11.48) $u(X)\leq M^{\prime}u(X_{0})=M^{\prime}\ \ \ \text{ for every }X\in
B(x_{0},r).$

So we assume that

(11.49) there exists $X_{1}\in B(x_{0},r)$ such that $u(X_{1})>M^{\prime}$

and we want to prove by induction that for every integer $k\geq 1$,

(11.50) there exists $X_{k}\in B(x_{0},\frac{3}{2}r)$ such that
$u(X_{k})>M^{i+2+k}$ and $|X_{k}-x_{0}|\leq\frac{3}{2}r-2^{-k}r$.

The base step of the induction is given by (11.49) and we want to do the
induction step. Let $k\geq 1$ be given and assume that (11.50) holds. From the
contraposition of (11.45), we deduce that $\delta(X_{k})<\epsilon^{i+2+k}r$.
Choose $x_{k}\in\Gamma$ such that
$|X_{k}-x_{k}|=\delta(X_{k})<\epsilon^{i+2+k}r$. By the induction hypothesis,

(11.51)
$|x_{k}-x_{0}|\leq|x_{k}-X_{k}|+|X_{k}-x_{0}|\leq\frac{3r}{2}-2^{-k}r+\epsilon^{i+2+k}r$

and, since $\epsilon\leq\frac{1}{2}$,

(11.52) $|x_{k}-x_{0}|\leq\frac{3r}{2}-2^{-k}r+2^{-2-k}r.$

Now, due to (11.44), we can find $X_{k+1}\in B(x_{k},\epsilon^{2+k}r)$ such
that

(11.53) $u(X_{k+1})\geq 2^{i}\sup_{X\in B(x_{k},\epsilon^{i+2+k}r)}u(X)\geq
2^{i}u(X_{k})\geq M^{i+2+(k+1)}.$

The induction step will be complete if we can prove that
$|X_{k+1}-x_{0}|\leq\frac{3}{2}r-2^{-(k+1)}r$. Indeed,

(11.54)
$\begin{split}|X_{k+1}-x_{0}|&\leq|X_{k+1}-x_{k}|+|x_{k}-x_{0}|\leq\epsilon^{2+k}r+\frac{3r}{2}-2^{-k}r+2^{-2-k}r\\\
&\leq\frac{3r}{2}-2^{-k}r+2^{-1-k}r=\frac{3}{2}r-2^{-k-1}r\end{split}$

by (11.52) and because $\epsilon\leq\frac{1}{2}$.

Let us sum up. We assumed the existence of $X_{1}\in B(x_{0},r)$ such that
$u(X_{1})>M^{\prime}$ and we end up with (11.50), that is a sequence $X_{k}$
of values in $B(x_{0},\frac{3}{2}r)$ such that $u(X_{k})$ increases to
$+\infty$. Up to a subsequence, we can thus find a point in
$\overline{B(x_{0},\frac{3}{2}r)}$ where $u$ is not continuous, which
contradicts Lemma 8.16. Hence $u(X)\leq M^{\prime}=M^{\prime}u(X_{0})$ for
$X\in B(x_{0},r)$. Lemma 11.8 follows. $\Box$

######  Lemma 11.9.

Let $x_{0}\in\Gamma$ and $r>0$ be given, and set $X_{0}:=A_{r}(x_{0})$ as in
Lemma 11.6. Then for all $X\in\Omega\setminus B(X_{0},\delta(X_{0})/4)$,

(11.55) $r^{d-1}g(X,X_{0})\leq C\omega^{X}(B(x_{0},r)\cap\Gamma)$

and

(11.56) $r^{d-1}g(X,X_{0})\leq C\omega^{X}(\Gamma\setminus B(x_{0},2r)),$

where $C>0$ depends only on $d$, $n$, $C_{0}$ and $C_{1}$.

Proof. We prove (11.55) first. Let $h\in C^{\infty}_{0}(B(x_{0},r))$ satisfy
$h\equiv 1$ on $B(x_{0},r/2)$ and $0\leq h\leq 1$. Define then $u\in W$ as the
solution of $Lu=0$ with data $Th$ given by Lemma 9.1. Set $v(X)=1-u(X)\in W$
and observe that $0\leq v\leq 1$ and $Tv=0$ on $B(x_{0},r/2)\cap\Gamma$.

By Lemma 8.16, we can find $\epsilon>0$ (that depends only on $d$, $n$,
$C_{0}$, $C_{1}$) such that $v(A_{\epsilon r}(x_{0}))\leq\frac{1}{2}$, i.e.
$u(A_{\epsilon r}(x_{0}))\geq\frac{1}{2}$. The existence of Harnack chains
(Lemma 2.2) and the Harnack inequality (Lemma 8.9) give

(11.57) $C^{-1}\leq u(X)\ \ \ \text{ for }X\in B(X_{0},\delta(X_{0})/2).$

By Lemma 10.1 (v), $g(X,X_{0})\leq C|X-X_{0}|^{1-d}$ for $X\in\Omega\setminus
B(X_{0},\delta(X_{0})/4)$. Since $\delta(X_{0})\approx r$ by construction of
$X_{0}$,

(11.58) $r^{d-1}g(X,X_{0})\leq C\ \ \ \text{ for }X\in
B(X_{0},\delta(X_{0})/2)\setminus B(X_{0},\delta(X_{0})/4).$

The combination of (11.57) and (11.58) yields the existence of $K_{1}>0$
(depending only on $n$, $d$, $C_{0}$ and $C_{1}$) such that

(11.59) $r^{d-1}g(X,X_{0})\leq K_{1}u(X)\ \ \ \text{ for }X\in
B(X_{0},\delta(X_{0})/2)\setminus B(X_{0},\delta(X_{0})/4).$

We claim that $K_{1}u(X)-r^{d-1}g(X,X_{0})$ satisfies the assumptions of Lemma
11.3, with $E=\mathbb{R}^{n}\setminus\overline{B(X_{0},\delta(X_{0})/4)}$ and
$F=\mathbb{R}^{n}\setminus B(X_{0},\delta(X_{0})/2)$. Indeed Assumption (i) of
Lemma 11.3 is satisfied because $u\in W$ and by Lemma 10.1 (i). Assumption
(ii) of Lemma 11.3 holds because $Tu=h\geq 0$ by construction and also
$Tg(.,X_{0})=0$ thanks to Lemma 10.1 (i). Assumption (iii) of Lemma 11.3 is
given by (11.59). The lemma yields

(11.60) $r^{d-1}g(X,X_{0})\leq K_{1}u(X)\ \ \ \text{ for }X\in\Omega\setminus
B(X_{0},\delta(X_{0})/4).$

By the positivity of the harmonic measure,
$u(X)\leq\omega^{X}(B(x_{0},r)\cap\Gamma)$ for $X\in\Omega$; (11.55) follows.

Let us turn to the proof of (11.56). We want to find two points
$x_{1},x_{2}\in\Gamma\cap[B(x_{0},Kr)\setminus B(x_{0},4r)]$, where the
constant $K\geq 10$ depends only on $C_{0}$ and $d$, such that
$X_{1}:=A_{r}(x_{1})$ and $X_{2}:=A_{r}(x_{2})$ satisfy

(11.61) $B(X_{1},\delta(X_{1})/4)\cap B(X_{2},\delta(X_{2})/4)=\emptyset.$

To get such points, we use the fact that $\Gamma$ is Ahlfors regular to find
$M\geq 3$ (that depends only on $C_{0}$ and $d$) such that
$\Gamma_{1}:=\Gamma\cap[B(x_{0},2Mr)\setminus B(x_{0},6r)]\neq\emptyset$ and
$\Gamma_{2}:=\Gamma\cap[B(x_{0},2M^{2}r)\setminus B(x_{0},6Mr)]\neq\emptyset$.
Any choice of points $x_{1}\in\Gamma_{1}$ and $x_{2}\in\Gamma_{2}$ verifies
(11.61).

Let $X\in\Omega\setminus B(X_{0},\delta(X_{0})/4)$. Thanks to (11.61), there
exists $i\in\\{1,2\\}$ such that $X\notin B(X_{i},\delta(X_{i})/4)$. The
existence of Harnack chains (Lemma 2.2), the Harnack inequality (Lemma 8.9),
and the fact that $Y\to g(X,Y)$ is a solution of
$L_{T}u:=-\operatorname{div}A^{T}\nabla u=0$ in $\Omega\setminus\\{X\\}$
(Lemma 10.1 and Lemma 10.6) yield

(11.62) $r^{d-1}g(X,X_{0})\leq Cr^{d-1}g(X,X_{i}).$

By (11.55) and the positivity of the harmonic measure,

(11.63) $r^{d-1}g(X,X_{0})\leq Cr^{d-1}g(X,X_{i})\leq
Cw^{X}(B(x_{i},r)\cap\Gamma)\leq Cw^{X}(\Gamma\setminus B(x_{0},r)).$

The lemma follows. $\Box$

We turn now to the non-degeneracy of the harmonic measure.

######  Lemma 11.10.

Let $\alpha>1$, $x_{0}\in\Gamma$, and $r>0$ be given, and let
$X_{0}:=A_{r}(x_{0})\in\Omega$ be as in Lemma 11.6. Then

(11.64) $\omega^{X}(B(x_{0},r)\cap\Gamma)\geq C_{\alpha}^{-1}\ \ \ \text{ for
}X\in B(x_{0},r/\alpha),$ (11.65) $\omega^{X}(B(x_{0},r)\cap\Gamma)\geq
C^{-1}_{\alpha}\ \ \ \text{ for }X\in B(X_{0},\delta(X_{0})/\alpha),$ (11.66)
$\omega^{X}(\Gamma\setminus B(x_{0},r))\geq C^{-1}_{\alpha}\ \ \ \text{ for
}X\in\Omega\setminus B(x_{0},\alpha r),$

and

(11.67) $\omega^{X}(\Gamma\setminus B(x_{0},r))\geq C^{-1}_{\alpha}\ \ \
\text{ for }X\in B(X_{0},\delta(X_{0})/\alpha),$

where $C_{\alpha}>0$ depends only upon $d$, $n$, $C_{0}$, $C_{1}$ and
$\alpha$.

Proof. Let us first prove (11.64). Set
$u(X)=1-\omega^{X}(B(x_{0},r)\cap\Gamma)$. By Lemma 9.7, $u$ lies in
$W_{r}(B(x_{0},r))$, is a solution of $Lu=0$ in $\Omega\cap B(x_{0},r)$, and
has a vanishing trace on $\Gamma\cap B(x_{0},r)$. So the Hölder continuity of
solutions at the boundary (Lemma 8.16) gives the existence of an $\epsilon>0$,
that depends only on $d$, $n$, $C_{0}$, $C_{1}$ and $\alpha$, such that
$u(X)\leq\frac{1}{2}$ for every $X\in
B(x_{0},\frac{1}{2}[1+\frac{1}{\alpha}]r)$ such that $\delta(X)\leq\epsilon
r$. Thus $v(X):=\omega^{X}(B(x_{0},r)\geq\frac{1}{2}$ for $X\in
B(x_{0},\frac{1}{2}[1+\frac{1}{\alpha}]r)$ such that $\delta(X)\leq\epsilon
r$. We now deduce (11.64) from the existence of Harnack chains (Lemma 2.2) and
the Harnack inequality (Lemma 8.9).

The assertion (11.65) follows from (11.64). Indeed, (11.64) implies that
$\omega^{A_{r/2}(x_{0})}(B(x_{0},r)\cap\Gamma)\geq C^{-1}$. The existence of
Harnack chains (Lemma 2.2) and the Harnack inequality (Lemma 8.9) allow us to
conclude. Finally (11.66) and (11.67) can be proved as above, and we leave the
details to the reader. $\Box$

######  Lemma 11.11.

Let $x_{0}\in\Gamma$ and $r>0$ be given, and set $X_{0}=A_{r}(x_{0})$. Then

(11.68) $C^{-1}r^{d-1}g(X,X_{0})\leq\omega^{X}(B(x_{0},r)\cap\Gamma)\leq
Cr^{d-1}g(X,X_{0})\ \ \ \text{ for }X\in\Omega\setminus B(x_{0},2r),$

and

(11.69) $\begin{split}C^{-1}r^{d-1}g(X,X_{0})\leq\omega^{X}(\Gamma\setminus
B(x_{0},2r))&\leq Cr^{d-1}g(X,X_{0})\\\ &\text{ for }X\in B(x_{0},r)\setminus
B(X_{0},\delta(X_{0})/4),\end{split}$

where $C>0$ depends only upon $d$, $n$, $C_{0}$ and $C_{1}$.

Proof. The lower bounds are a consequence of Lemma 11.9; the one in (11.68)
also requires to notice that $\delta(X_{0})\leq r$ and thus
$B(X_{0},\delta(X_{0})/4)\subset B(x_{0},2r)$.

It remains to check the upper bounds. But we first prove an intermediate
result. We claim that for $\phi\in C^{\infty}(\mathbb{R}^{n})\cap W$ and
$X\notin\mathrm{supp}\,\phi$,

(11.70) $u_{\phi}(X)=-\int_{\Omega}A\nabla\phi(Y)\cdot\nabla_{y}g(X,Y)dY,$

where $u_{\phi}\in W$ is the solution of $Lu_{\phi}=0$, with the Dirichlet
condition $Tu_{\phi}=T\phi$ on $\Gamma$, given by Lemma 9.1. Indeed, recall
that by (8.8) and (8.9) the map

(11.71) $u,v\in W_{0}\to\int_{\Omega}A\nabla u\cdot\nabla
v\,dY=\int_{\Omega}\mathcal{A}\nabla u\cdot\nabla v\,dm$

is bounded and coercive on $W_{0}$ and the map

(11.72) $\varphi\in
W_{0}\to\int_{\Omega}A\nabla\phi\cdot\nabla\varphi\,dY=\int_{\Omega}\mathcal{A}\nabla\phi\cdot\nabla\varphi\,dm$

is bounded on $W_{0}$. So the Lax-Milgram theorem yields the existence of
$\mathfrak{v}\in W_{0}$ such that

(11.73)
$\int_{\Omega}A\nabla\phi\cdot\nabla\varphi\,dY=\int_{\Omega}A\nabla\mathfrak{v}\cdot\nabla\varphi\,dY\qquad\forall\varphi\in
W_{0}.$

Let $s>0$ such that $B(X,2s)\cap(\mathrm{supp}\,\phi\cup\Gamma)=\emptyset$.
For any $\rho>0$ we define, as we did in (10.16), the function
$\mathfrak{g}^{\rho}_{T}=g^{\rho}_{T}(.,X)$ on $\Omega$ as the only function
in $W_{0}$ such that

(11.74)
$\int_{\Omega}A\nabla\varphi\cdot\nabla\mathfrak{g}^{\rho}_{T}\,dY=\int_{\Omega}A^{T}\nabla\mathfrak{g}^{\rho}_{T}\cdot\nabla\varphi\,dY=\fint_{B(X,\rho)}\varphi\,dY\qquad\forall\varphi\in
W_{0}.$

We take $\varphi=\mathfrak{g}^{\rho}_{T}$ in (11.73) to get

(11.75)
$\int_{\Omega}A\nabla\phi\cdot\nabla\mathfrak{g}^{\rho}_{T}\,dY=\int_{\Omega}A\nabla\mathfrak{v}\cdot\nabla\mathfrak{g}^{\rho}_{T}\,dY=\fint_{B(X,\rho)}\mathfrak{v}\,dY.$

We aim to take the limit as $\rho\to 0$ in (11.75). Since $\mathfrak{v}$
satisfies

(11.76)
$\int_{\Omega}A\nabla\mathfrak{v}\cdot\nabla\varphi\,dY=\int_{\Omega}A\nabla\phi\cdot\nabla\varphi\,dY=0\qquad\forall\varphi\in
C^{\infty}_{0}(B(X,2s)),$

$\mathfrak{v}$ is a solution of $L\mathfrak{v}=0$ on $B(X,2s)$ and thus Lemma
8.8 proves that $\mathfrak{v}$ is continuous at $X$. As a consequence,

(11.77) $\lim_{\rho\to 0}\fint_{B(X,\rho)}\mathfrak{v}\,dY=\mathfrak{v}(X).$

Recall that the $\mathfrak{g}^{\rho}_{T}$, $\rho>0$, are the same functions as
in in the proof of Lemma 10.1, but for the transpose matrix $A^{T}$. Let
$\alpha\in C^{\infty}_{0}(B(x,2s))$ be such that $\alpha\equiv 1$ on $B(x,s)$.
By (10.78) and Lemma 10.6, there exists a sequence $(\rho_{\eta})$ tending to
0, such that $(1-\alpha)\mathfrak{g}^{\rho_{\eta}}_{T}$ converges weakly to
$(1-\alpha)g_{T}(.,X)=(1-\alpha)g(X,.)$ in $W_{0}$. As a consequence,

(11.78)
$\displaystyle\lim_{\eta\to+\infty}\int_{\Omega}A\nabla\phi\cdot\nabla\mathfrak{g}^{\rho_{\eta}}_{T}\,dY$
$\displaystyle=$
$\displaystyle\lim_{\eta\to+\infty}\int_{\Omega}A\nabla\phi\cdot\nabla[(1-\alpha)\mathfrak{g}^{\rho_{\eta}}_{T}]\,dY$
$\displaystyle=$
$\displaystyle\int_{\Omega}A\nabla\phi(Y)\cdot\nabla_{y}[(1-\alpha)g(X,Y)]dY$
$\displaystyle=$
$\displaystyle\int_{\Omega}A\nabla\phi(Y)\cdot\nabla_{y}g(X,Y)dY.$

The combination of (11.75), (11.77) and (11.78) yields

(11.79) $\int_{\Omega}A\nabla\phi(Y)\cdot\nabla_{y}g(X,Y)dY=\mathfrak{v}(X).$

Since $\mathfrak{v}\in W_{0}$ satisfies (11.73), the function
$u_{\phi}=\phi-\mathfrak{v}$ lies in $W$ and is a solution of $Lu_{\phi}=0$
with the Dirichlet condition $Tu_{\phi}=T\phi$. Hence

(11.80)
$\int_{\Omega}A\nabla\phi(Y)\cdot\nabla_{y}g(X,Y)dY=\mathfrak{v}(X)=\phi(X)-u_{\phi}(X)=-u_{\phi}(X),$

by (11.79) and because $X\notin\mathrm{supp}\,\phi$. The claim (11.70)
follows.

We turn to the proof of the upper bound in (11.68), that is,

(11.81) $\omega^{X}(B(x_{0},r)\cap\Gamma)\leq Cr^{d-1}g(X,X_{0})\ \ \ \text{
for }X\in\Omega\setminus B(x_{0},2r).$

Let $X\in\Omega\setminus B(x_{0},2r)$ be given, and choose $\phi\in
C^{\infty}_{0}(\mathbb{R}^{n})$ such that $0\leq\phi\leq 1$, $\phi\equiv 1$ on
$B(x_{0},r)$, $\phi\equiv 0$ on $\mathbb{R}^{n}\setminus
B(x_{0},\frac{5}{4}r)$, and $|\nabla\phi|\leq\frac{10}{r}$. We get that

(11.82)
$u_{\phi}(X)\leq\frac{C}{r}\int_{B(x_{0},\frac{5}{4}r)}|\nabla_{y}g(X,Y)|dm(Y)$

by (11.70) and (8.8), and since $\omega^{X}(B(x_{0},r)\cap\Gamma)\leq
u_{\phi}(X)$ by the positivity of the harmonic measure,

(11.83)
$\begin{split}\omega^{X}(B(x_{0},r)\cap\Gamma)&\leq\frac{C}{r}\int_{B(x_{0},\frac{5}{4}r)}|\nabla_{y}g(X,Y)|dm(Y).\\\
&\leq\frac{C}{r}r^{\frac{d+1}{2}}\left(\int_{B(x_{0},\frac{5}{4}r)}|\nabla_{y}g(X,Y)|^{2}dm(Y)\right)^{\frac{1}{2}}\end{split}$

by Cauchy-Schwarz’ inequality and Lemma 2.3. Since $X\in\Omega\setminus
B(x_{0},2r)$, Lemma 10.6 and Lemma 10.1 (iii) say that the function $Y\to
g(X,Y)$ is a solution of $L_{T}u:=-\operatorname{div}A^{T}\nabla u$ on
$B(x_{0},2r)$, with a vanishing trace on $\Gamma\cap B(x_{0},2r)$. So the
Caccioppoli inequality at the boundary (see Lemma 8.11) applies and yields

(11.84)
$\begin{split}\omega^{X}(B(x_{0},r)\cap\Gamma)&\leq\frac{C}{r^{2}}\,r^{\frac{d+1}{2}}\left(\int_{B(x_{0},\frac{3}{2}r)}|g(X,Y)|^{2}dm(Y)\right)^{\frac{1}{2}}.\end{split}$

Then by Lemma 11.8,

(11.85)
$\begin{split}\omega^{X}(B(x_{0},r)\cap\Gamma)&\leq\frac{C}{r^{2}}\,r^{d+1}g(X,X_{0})=Cr^{d-1}g(X,X_{0});\end{split}$

the bound (11.81) follows.

It remains to prove the upper bound in (11.69), i.e., that

(11.86) $\omega^{X}(\Gamma\setminus B(x_{0},2r))\leq Cr^{d-1}g(X,X_{0})\ \ \
\text{ for }X\in B(x_{0},r)\setminus B(X_{0},\delta(X_{0})/4).$

The proof will be similar to the upper bound in (11.68) once we choose an
appropriate function $\phi$ in (11.70). Let us do this rapidly. Let $X\in
B(x_{0},r)\setminus B(X_{0},\delta(X_{0})/4)$ be given and take $\phi\in
C^{\infty}(\mathbb{R}^{n})$ such that $0\leq\phi\leq 1$, $\phi\equiv 1$ on
$\mathbb{R}^{n}\setminus B(x_{0},\frac{8}{5}r)$, $\phi\equiv 0$ on
$B(x_{0},\frac{7}{5}r)$ and $|\nabla\phi|\leq\frac{10}{r}$. Notice that
$X\notin\mathrm{supp}(\phi)$, so (11.70) applies and yields

(11.87) $u_{\phi}(X)\leq\frac{C}{r}\int_{B(x_{0},\frac{8}{5}r)\setminus
B(x_{0},\frac{7}{5}r)}|\nabla_{y}g(X,Y)|dm(Y).$

By the positivity of the harmonic measure, $\omega^{X}(\Gamma\setminus
B(x_{0},2r))\leq u_{\phi}(X)$. We use the Cauchy-Schwarz and Caccioppoli
inequalities (see Lemma 8.11), as above, and get that

(11.88) $\displaystyle\omega^{X}(\Gamma\setminus B(x_{0},2r))$
$\displaystyle\leq$
$\displaystyle\frac{C}{r}m(B(x_{0},\frac{8}{5}r))\left(\frac{1}{m(B(x_{0},\frac{8}{5}r))}\int_{B(x_{0},\frac{8}{5}r)\setminus
B(x_{0},\frac{7}{5}r)}|\nabla_{y}g(X,Y)|^{2}dm(Y)\right)^{\frac{1}{2}}$
$\displaystyle\leq$
$\displaystyle\frac{C}{r^{2}}r^{d+1}\left(\frac{1}{m(B(x_{0},\frac{8}{5}r))}\int_{B(x_{0},\frac{9}{5}r)\setminus
B(x_{0},\frac{6}{5}r)}|g(X,Y)|^{2}dm(Y)\right)^{\frac{1}{2}}.$

We claim that

(11.89) $g(X,Y)\leq Cg(X,X_{0})\qquad\forall Y\in
B(x_{0},\frac{9}{5}r)\setminus B(x_{0},\frac{6}{5}r)$

where $C>0$ depends only on $d$, $n$, $C_{0}$ and $C_{1}$. Two cases may
happen. If $\delta(Y)\geq\frac{r}{20}$, (11.89) is only a consequence of the
existence of Harnack chains (Lemma 2.2) and the Harnack inequality (Lemma
8.9). Otherwise, if $\delta(Y)<\frac{r}{20}$ then Lemma 11.8 says that
$g(X,Y)\leq Cg(X,X_{Y})$ for some point $X_{Y}\in
B(x_{0},\frac{9}{5}r)\setminus B(x_{0},\frac{6}{5}r)$ that lies at distance at
least $\epsilon r$ from $\Gamma$. Here $\epsilon$ comes from Lemma 11.6 and
thus depends only on $d$, $n$ and $C_{0}$. Together with the existence of
Harnack chains (Lemma 2.2) and the Harnack inequality (Lemma 8.9), we find
that $g(X,X_{Y})$, or $g(X,Y)$, is bounded by $Cg(X,X_{0})$.

We use (11.89) in the right hand side of (11.88) to get that

(11.90) $\omega^{X}(\Gamma\setminus B(x_{0},2r))\leq
Cr^{d-1}\frac{m(B(x_{0},\frac{9}{5}r))}{m(B(x_{0},\frac{8}{5}r))}g(X,X_{0})\leq
Cr^{d-1}g(X,X_{0}),$

by the doubling property of the measure $m$. The second and last assertion of
the lemma follows. $\Box$

######  Lemma 11.12 (Doubling volume property for the harmonic measure).

For $x_{0}\in\Gamma$ and $r>0$, we have

(11.91) $\omega^{X}(B(x_{0},2r)\cap\Gamma)\leq
C\omega^{X}(B(x_{0},r)\cap\Gamma)\ \ \ \text{ for }X\in\Omega\setminus
B(x_{0},4r)$

and

(11.92) $\omega^{X}(\Gamma\setminus B(x_{0},r))\leq
C\omega^{X}(\Gamma\setminus B(x_{0},2r))\ \ \ \text{ for }X\in B(x_{0},r/2),$

where $C>0$ depends only on $n$, $d$, $C_{0}$ and $C_{1}$.

Proof. Let us prove (11.91) first. Lemma 11.11 says that for
$X\in\Omega\setminus B(x_{0},4r)$,

(11.93) $\omega^{X}(B(x_{0},2r)\cap\Gamma)\approx r^{d-1}g(X,A_{2r}(x_{0}))$

and

(11.94) $\omega^{X}(B(x_{0},r)\cap\Gamma)\approx r^{d-1}g(X,A_{r}(x_{0})),$

where $A_{2r}(x_{0})$ and $A_{r}(x_{0})$ are the points of $\Omega$ given by
Lemma 11.6. The bound (11.91) will be thus proven if we can show that

(11.95) $g(X,A_{2r}(x_{0}))\approx g(X,A_{r}(x_{0}))\ \ \ \text{ for
}X\in\Omega\setminus B(x_{0},4r).$

Yet, since $Y\to g(X,Y)$ belongs to $W_{r}(\Omega\setminus\\{X\\})$ and is a
solution of $L_{T}u:=-\operatorname{div}A^{T}\nabla u=0$ in
$\Omega\setminus\\{X\\}$ (see Lemma 10.1 and Lemma 10.6), the equivalence in
(11.95) is an easy consequence of the properties of $A_{r}(x_{0})$ (Lemma
11.6), the existence of Harnack chains (Lemma 2.2) and the Harnack inequality
(Lemma 8.9).

We turn to the proof of (11.92). Set $X_{1}:=A_{r}(x_{0})$ and
$X_{\frac{1}{2}}:=A_{r/2}(x_{0})$. Call $\Xi$ the set of points $X\in
B(x_{0},r/2)$ such that $|X-X_{1}|\geq\frac{1}{4}\delta(X_{1})$ and
$|X-X_{\frac{1}{2}}|\geq\frac{1}{4}\delta(X_{\frac{1}{2}})$, and first
consider $X\in\Xi$. By Lemma 11.11 again,

(11.96) $\omega^{X}(\Gamma\setminus B(x_{0},2r))\approx r^{d-1}g(X,X_{1})$

and

(11.97) $\omega^{X}(\Gamma\setminus B(x_{0},r))\approx
r^{d-1}g(X,X_{\frac{1}{2}}).$

Since $\delta(X_{1})\approx\delta(X_{\frac{1}{2}})\approx r$ and $Y\to g(X,Y)$
is a solution of $L_{T}u=-\operatorname{div}A^{T}\nabla u=0$, the existence of
Harnack chains (Lemma 2.2) and the Harnack inequality (Lemma 8.9) give
$g(X,X_{1})\approx g(X,X_{\frac{1}{2}})$ for $X\in\Xi$. Hence

(11.98) $\omega^{X}(\Gamma\setminus
B(x_{0},2r))\approx\omega^{X}(\Gamma\setminus B(x_{0},r)),$

with constants that do not depend on $X$, $x_{0}$, or $r$. The equivalence in
(11.98) also holds for all $X\in B(x_{0},r/2)$, and not only for $X\in\Xi$, by
Harnack’s inequality (Lemma 8.9). This proves (11.92). $\Box$

######  Remark 11.13.

The following results also hold for every $\alpha>1$. For $x_{0}\in\Gamma$ and
$r>0$,

(11.99) $\omega^{X}(B(x_{0},2r)\cap\Gamma)\leq
C_{\alpha}\omega^{X}(B(x_{0},r)\cap\Gamma)\ \ \ \text{ for
}X\in\Omega\setminus B(x_{0},2\alpha r),$

and

(11.100) $\omega^{X}(\Gamma\setminus B(x_{0},r))\leq
C_{\alpha}\omega^{X}(\Gamma\setminus B(x_{0},2r))\ \ \ \text{ for }X\in
B(x_{0},r/\alpha),$

where $C_{\alpha}>0$ depends only on $n$, $d$, $C_{0}$, $C_{1}$ and $\alpha$.

This can be deduced from Lemma 11.12 \- that corresponds to the case
$\alpha=2$ \- by applying it to smaller balls.

Let us prove for instance (11.100). Let $X\in B(x_{0},r/\alpha)$ be given. We
only need to prove (11.100) when $\delta(X)<\frac{r}{4}(1-\frac{1}{\alpha})$,
because as soon as we do this, the other case when
$\delta(X)\geq\frac{r}{4}(1-\frac{1}{\alpha})$ follows, by Harnack’s
inequality (Lemma 8.9).

Let $x\in\Gamma$ such that $|x-X|=\delta(X)$; then set
$r_{k}=2^{k-1}r[1-\frac{1}{\alpha}]$ and $B_{k}=B(x,r_{k})$ for
$k\in{\mathbb{Z}}$. We wish to apply the doubling property (11.92) and get
that

(11.101) $\omega^{X}(\Gamma\setminus B_{k})\leq C\omega^{X}(\Gamma\setminus
B_{k+1}),$

and we can do this as long as $X\in B_{k-1}$. With our extra assumption that
$|x-X|=\delta(X)<\frac{r}{4}(1-\frac{1}{\alpha})$, this is possible for all
$k\geq 0$. Notice that

(11.102)
$|x-x_{0}|\leq\delta(X)+|X-x_{0}|\leq\frac{r}{4}(1-\frac{1}{\alpha})+\frac{r}{\alpha}\leq\frac{r}{2}(1-\frac{1}{\alpha})+\frac{r}{\alpha}=\frac{r}{2}[1+\frac{1}{\alpha}]$

and then
$|x-x_{0}|+r_{0}\leq\frac{r}{2}[1+\frac{1}{\alpha}]+\frac{r}{2}[1-\frac{1}{\alpha}]=r$,
so $B_{0}=B(x,r_{0})\subset B(x_{0},r)$ and, by the monotonicity of the
harmonic measure,

(11.103) $\omega^{X}(\Gamma\setminus B(x_{0},r))\leq\omega^{X}(\Gamma\setminus
B_{0}).$

Let $k$ be the smallest integer such that $2^{k-1}(1-\frac{1}{\alpha})\geq 3$;
obviously $k$ depends only on $\alpha$, and $r_{k}\geq 3r$. Then
$|x-x_{0}|+2r<3r\leq r_{k}$ by (11.102), hence $B(x_{0},2r)\subset B_{k}$ and
$\omega^{X}(\Gamma\setminus B_{k})\leq\omega^{X}(\Gamma\setminus B(x_{0},2r))$
because the harmonic measure is monotone. Together with (11.103) and (11.101),
this proves that $\omega^{X}(\Gamma\setminus B(x_{0},r))\leq
C^{k}\omega^{X}(\Gamma\setminus B(x_{0},2r))$, and (11.100) follows because
$k$ depends only on $\alpha$. The proof of (11.99) would be similar.

######  Lemma 11.14 (Comparison principle for global solutions).

Let $x_{0}\in\Gamma$ and $r>0$ be given, and let
$X_{0}:=A_{r}(x_{0})\in\Omega$ be the point given in Lemma 11.6. Let $u,v\in
W$ be two non-negative, non identically zero, solutions of $Lu=Lv=0$ in
$\Omega$ such that $Tu=Tv=0$ on $\Gamma\setminus B(x_{0},r)$. Then

(11.104) $C^{-1}\frac{u(X_{0})}{v(X_{0})}\leq\frac{u(X)}{v(X)}\leq
C\frac{u(X_{0})}{v(X_{0})}\ \ \ \text{ for }X\in\Omega\setminus B(x_{0},2r),$

where $C>0$ depends only on $n$, $d$, $C_{0}$ and $C_{1}$.

######  Remark 11.15.

We also have (11.104) for any $X\in\Omega\setminus B(x_{0},\alpha r)$, where
$\alpha>1$. In this case, the constant $C$ depends also on $\alpha$. We let
the reader check that the proof below can be easily adapted to prove this too.

Proof. By symmetry and as before, it is enough to prove that

(11.105) $\frac{u(X)}{v(X)}\leq C\frac{u(X_{0})}{v(X_{0})}\ \ \ \text{ for
}X\in\Omega\setminus B(x_{0},2r).$

Notice also that thanks to the Harnack inequality (Lemma 8.9), $v(X)>0$ on the
whole $\Omega\setminus\overline{B(x_{0},r)}$, so we we don’t need to be
careful when we divide by $v(X)$.

Set $\Gamma_{1}:=\Gamma\cap B(x_{0},r)$ and $\Gamma_{2}:=\Gamma\cap
B(x_{0},\frac{15}{8}r)$. Lemma 11.12 \- or more exactly (11.99) - gives the
following fact that will be of use later on:

(11.106) $\omega^{X}(\Gamma_{2})\leq C\omega^{X}(\Gamma_{1})\qquad\forall
X\in\Omega\setminus B(x_{0},2r).$

with a constant $C>0$ which depends only on $d$, $n$, $C_{0}$ and $C_{1}$.

We claim that

(11.107) $v(X)\geq C^{-1}\omega^{X}(\Gamma_{1})v(X_{0})\ \ \ \text{ for
}X\in\Omega\setminus B(x_{0},2r).$

Indeed, by Harnack’s inequality (Lemma 8.9),

(11.108) $v(X)\geq C^{-1}v(X_{0})\ \ \ \text{ for }X\in
B(X_{0},\delta(X_{0})/2).$

Together with Lemma 11.4, which states that $g(X,X_{0})\leq
C\delta(X_{0})^{1-d}\leq Cr^{1-d}$ for any $X\in\Omega\setminus
B(X_{0},\delta(X_{0})/4)$, we deduce the existence of $K_{1}>0$ (that depends
only on $d$, $n$, $C_{0}$ and $C_{1}$) such that

(11.109) $v(X)\geq K_{1}^{-1}r^{d-1}v(X_{0})g(X,X_{0})\ \ \ \text{ for }X\in
B(X_{0},\frac{1}{2}\delta(X_{0}))\setminus B(X_{0},\frac{1}{4}\delta(X_{0}))$

Let us apply the maximum principle (Lemma 11.3, with
$E=\mathbb{R}^{n}\setminus\overline{B(X_{0},\delta(X_{0})/4)}$ and
$F=\mathbb{R}^{n}\setminus B(X_{0},\delta(X_{0}/2))$), to the function $X\to
v(X)-K_{1}^{-1}r^{d-1}v(X_{0})g(X,X_{0})$. The assumptions are satisfied
because of (11.109), the properties of the Green function given in Lemma 10.1,
and the fact that $v\in W$ is a non-negative solution of $Lv=0$ on $\Omega$.
We get that

(11.110) $v(X)\geq K_{1}^{-1}r^{d-1}v(X_{0})g(X,X_{0})\ \ \ \text{ for
}X\in\Omega\setminus B(X_{0},\frac{1}{4}\delta(X_{0}))\supset\Omega\setminus
B(x_{0},2r).$

The claim (11.107) is now a straightforward consequence of (11.110) and Lemma
11.11.

We want to prove now that

(11.111) $u(X)\leq Cu(X_{0})\omega^{X}(\Gamma_{2})\ \ \ \text{ for
}X\in\Omega\setminus B(x_{0},2r).$

First, we need to prove that

(11.112) $u(X)\leq Cu(X_{0})\ \ \ \text{ for
}X\in\left[B(x_{0},\frac{13}{8}r)\setminus
B(x_{0},\frac{11}{8}r)\right]\cap\Omega.$

We split $\left[B(x_{0},\frac{13}{8}r)\setminus
B(x_{0},\frac{11}{8}r)\right]\cap\Omega$ into two sets:

(11.113) $\Omega_{1}:=\Omega\cap\\{X\in B(x_{0},\frac{13}{8}r)\setminus
B(x_{0},\frac{11}{8}r),\,\delta(X)<\frac{1}{8}r\\}$

and

(11.114) $\Omega_{2}:=\\{X\in B(x_{0},\frac{13}{8}r)\setminus
B(x_{0},\frac{11}{8}r),\,\delta(X)\geq\frac{1}{8}r\\}.$

The proof of (11.112) for $X\in\Omega_{2}$ is a consequence of the existence
of Harnack chain (Lemma 2.2) and the Harnack inequality (Lemma 8.9). So it
remains to prove (11.112) for $X\in\Omega_{1}$. Let thus $X\in\Omega_{1}$ be
given. We can find $x\in\Gamma$ such that $X\in B(x,\frac{1}{8}r)$. Notice
that $x\in B(x_{0},\frac{7}{4}r)$ because $X\in B(x_{0},\frac{13}{8}r)$. Yet,
since $u$ is a non-negative solution of $Lu=0$ in
$B(x,\frac{1}{4}r)\cap\Omega$ satisfying $Tu=0$ on
$B(x,\frac{1}{4}r)\cap\Gamma$, Lemma 11.8 gives that $u(Y)\leq Cu(A_{r/8}(x))$
for $Y\in B(x,\frac{1}{8}r)$ and thus in particular $u(X)\leq Cu(A_{r/8}(x))$.
By the existence of Harnack chains (Lemma 2.2) and the Harnack inequality
(Lemma 8.9) again, $u(A_{r/8}(x))\leq Cu(X_{0})$. The bound (11.112) for all
$X\in\Omega_{1}$ follows.

We proved (11.112) and now we want to get (11.111). Recall from Lemma 11.10
that $\omega^{X}(B(x_{0},\frac{7}{4}r)\cap\Gamma)\geq C^{-1}$ for $X\in
B(x_{0},\frac{13}{8}r)\setminus\Gamma$. Hence, by (11.112),

(11.115) $u(X)\leq Cu(X_{0})\omega^{X}(B(x_{0},\frac{7}{4}r)\cap\Gamma)\ \ \
\text{ for }X\in\left[B(x_{0},\frac{13}{8}r)\setminus
B(x_{0},\frac{11}{8}r)\right]\cap\Omega.$

Let $h\in C^{\infty}_{0}(B(x_{0},\frac{15}{8}r))$ be such that $0\leq h\leq 1$
and $h\equiv 1$ on $B(x_{0},\frac{7}{4}r)$. Then let $u_{h}\in W$ be the
solution of $Lu_{h}=0$ with the Dirichlet condition $Tu_{h}=Th$. By the
positivity of the harmonic measure,

(11.116) $u(X)\leq Cu(X_{0})u_{h}(X)\ \ \ \text{ for
}X\in\left[B(x_{0},\frac{13}{8}r)\setminus
B(x_{0},\frac{11}{8}r)\right]\cap\Omega.$

The maximum principle given by Lemma 11.3 \- where we take
$E=\mathbb{R}^{n}\setminus\overline{B(x_{0},\frac{11}{8}r)}$ and
$F=\mathbb{R}^{n}\setminus B(x_{0},\frac{13}{8}r)$ \- yields

(11.117) $u(X)\leq Cu(X_{0})u_{h}(X)\ \ \ \text{ for }X\in\Omega\setminus
B(x_{0},\frac{13}{8}r)$

and hence

(11.118) $u(X)\leq Cu(X_{0})\omega^{X}(\Gamma_{2})\ \ \ \text{ for
}X\in\Omega\setminus B(x_{0},\frac{13}{8}r),$

where we use again the positivity of the harmonic measure. The assertion
(11.111) is now proven.

We conclude the proof of the lemma by gathering the previous results. Because
of (11.107) and (11.111),

(11.119) $\frac{u(X)}{v(X)}\leq
C\frac{u(X_{0})}{v(X_{0})}\frac{\omega^{X}(\Gamma_{2})}{\omega^{X}(\Gamma_{1})}\
\ \ \text{ for }X\in\Omega\setminus B(x_{0},2r),$

and (11.105) follows from (11.106). Lemma 11.14 follows. $\Box$

Note that the functions $X\to\omega^{X}(E)$, where $E\subset\Gamma$ is a non-
trivial Borel set, do not lie in $W$ and thus cannot be used directly in Lemma
11.14. The following lemma solves this problem.

######  Lemma 11.16 (Comparison principle for harmonic measures / Change of
poles).

Let $x_{0}\in\Gamma$ and $r>0$ be given, and let
$X_{0}:=A_{r}(x_{0})\in\Omega$ be as in Lemma 11.6. Let $E,F\subset\Gamma\cap
B(x_{0},r)$ be two Borel subsets of $\Gamma$ such that $\omega^{X_{0}}(E)$ and
$\omega^{X_{0}}(F)$ are positive. Then

(11.120)
$C^{-1}\frac{\omega^{X_{0}}(E)}{\omega^{X_{0}}(F)}\leq\frac{\omega^{X}(E)}{\omega^{X}(F)}\leq
C\frac{\omega^{X_{0}}(E)}{\omega^{X_{0}}(F)}\ \ \ \text{ for
}X\in\Omega\setminus B(x_{0},2r),$

where $C>0$ depends only on $n$, $d$, $C_{0}$ and $C_{1}$. In particular, with
the choice $F=B(x_{0},r)\cap\Gamma$,

(11.121)
$C^{-1}\omega^{X_{0}}(E)\leq\frac{\omega^{X}(E)}{\omega^{X}(B(x_{0},r)\cap\Gamma)}\leq
C\omega^{X_{0}}(E)\ \ \ \text{ for }X\in\Omega\setminus B(x_{0},2r),$

where again $C>0$ depends only on $n$, $d$, $C_{0}$ and $C_{1}$.

Proof. The second part of the lemma, that is (11.121) an immediate consequence
of (11.120) and the non-degeneracy of the harmonic measure (Lemma 11.10). In
addition, it is enough to prove

(11.122)
$C^{-1}\frac{\omega^{X_{0}}(E)}{u(X_{0})}\leq\frac{\omega^{X}(E)}{u(X)}\leq
C\frac{\omega^{X_{0}}(E)}{u(X_{0})},$

where $u\in W$ is any non-negative non-zero solution of $Lu=0$ in $\Omega$
satisfying $Tu=0$ on $\Gamma\setminus B(x_{0},r)$, and $C>0$ depends only on
$n$, $d$, $C_{0}$ and $C_{1}$. Indeed, (11.120) follows by applying (11.122)
to both $E$ and $F$. Incidentally, it is very easy to find $u$ like this: just
apply Lemma 9.4 to a smooth bump function $g$ with a small compact support
near $x_{0}$.

Assume first that $E=K$ is a compact set. Let $X\in\Omega\setminus
B(x_{0},2r)$ be given. Thanks to Lemma 9.7 (i), the assumption
$\omega^{X_{0}}(K)>0$ implies that $\omega^{X}(K)>0$. By the the regularity of
the harmonic measure (see (9.27)), we can find an open set $U_{X}\supset K$
such that

(11.123) $\omega^{X_{0}}(U_{X})\leq 2\omega^{X_{0}}(K)$ and
$\omega^{X}(U_{X})\leq 2\omega^{X}(K)$.

Urysohn’s lemma (see Lemma 2.12 in [Rud]) gives a function $h\in
C^{0}_{0}(\Gamma)$ such that ${\mathds{1}}_{K}\leq h\leq{\mathds{1}}_{U_{X}}$.
Write $v^{h}=U(h)$ for the image of the function $h$ by the map given in Lemma
9.4. We have seen for the proof of Lemma 9.4 that $h$ can be approximated, in
the supremum norm, by smooth, compactly supported functions $h_{k}$, and that
the corresponding solutions $v_{k}=U(h_{k})$, and that can also obtained
through 9.1, lie in $W$ and converge to $v^{h}$ uniformly on $\Omega$. Hence
we can find $k>0$ such that

(11.124) $\frac{1}{2}\,v_{k}\leq v^{h}\leq 2v_{k}$

everywhere in $\Omega$. Write $v$ for $v_{k}$. Notice that $v$ depends on $X$,
but it has no importance. The estimates (11.123) and (11.124) give

(11.125) $\frac{1}{4}v(X_{0})\leq\omega^{X_{0}}(K)\leq 2v(X_{0})\quad$ and
$\quad\frac{1}{4}v(X)\leq\omega^{X}(K)\leq 2v(X)$.

We can even choose $U_{X}\supset K$ so small, and then $g_{k}$ with a barely
larger support, so that $Tv=g_{k}$ is supported in $B(x_{0},r)$. As a
consequence, the solution $v$ satisfies the assumption of Lemma 11.14. Hence,
the latter entails

(11.126) $C^{-1}\frac{v(X_{0})}{u(X_{0})}\leq\frac{v(X)}{u(X)}\leq
C\frac{v(X_{0})}{u(X_{0})}$

with a constant $C>0$ that depends only on $d$, $n$, $C_{0}$ and $C_{1}$.
Together with (11.125), we get that

(11.127)
$C^{-1}\frac{\omega^{X_{0}}(K)}{u(X_{0})}\leq\frac{\omega^{X}(K)}{u(X)}\leq
C\frac{\omega^{X_{0}}(K)}{u(X_{0})}$

with a constant $C>0$ that still depends only on $d$, $n$, $C_{0}$ and $C_{1}$
(and thus is independent of $X$). Thus the conclusion (11.120) holds whenever
$E=K$ is a compact set.

Now let $E$ be any Borel subset of $\Gamma\cap B(x_{0},r)$. Let
$X\in\Omega\setminus B(x_{0},2r)$. According to the regularity of the harmonic
measure (9.27), there exists $K_{X}\subset E$ (depending on $X$) such that

(11.128) $\omega^{X_{0}}(K_{X})\leq\omega^{X_{0}}(E)\leq
2\omega^{X_{0}}(K_{X})\quad$ and $\quad\omega^{X}(K_{X})\leq\omega^{X}(E)\leq
2\omega^{X}(K_{X})$.

The combination of (11.128) and (11.127) (applied to $K_{X}$) yields

(11.129)
$C^{-1}\frac{\omega^{X_{0}}(E)}{u(X_{0})}\leq\frac{\omega^{X}(E)}{u(X)}\leq
C\frac{\omega^{X_{0}}(E)}{u(X_{0})}$

where the constant $C>0$ depends only upon $d$, $n$, $C_{0}$ and $C_{1}$. The
lemma follows. $\Box$

Let us prove now a comparison principle for the solution that are not defined
in the whole domain $\Omega$.

######  Theorem 11.17 (Comparison principle for locally defined functions).

Let $x_{0}\in\Gamma$ and $r>0$ and let $X_{0}:=A_{r}(x_{0})\in\Omega$ be the
point given in Lemma 11.6. Let $u,v\in W_{r}(B(x_{0},2r))$ be two non-
negative, not identically zero, solutions of $Lu=Lv=0$ in $B(x_{0},2r)$, such
that $Tu=Tv=0$ on $\Gamma\cap B(x_{0},2r)$. Then

(11.130) $C^{-1}\frac{u(X_{0})}{v(X_{0})}\leq\frac{u(X)}{v(X)}\leq
C\frac{u(X_{0})}{v(X_{0})}\ \ \ \text{ for }X\in\Omega\cap B(x_{0},r),$

where $C>0$ depends only on $n$, $d$, $C_{0}$ and $C_{1}$.

Proof. The plan of the proof is as follows: first, for $y_{0}\in\Gamma$ and
$s>0$, we construct a function $f_{y_{0},s}$ on $\Omega$ such that (i)
$f_{y_{0},s}(X)$ is equivalent to $\omega^{X}(\Gamma\setminus B(y_{0},2s))$
when $X\in B(y_{0},s)$ is close to $\Gamma$ and (ii) $f_{y_{0},s}(X)$ is
negative when $X\in\Omega\setminus B(y_{0},Ms)$ \- with $M$ depending only on
$d$, $n$, $C_{0}$ and $C_{1}$. We use $f_{y_{0},s}$ to prove that $v(X)\geq
v(A_{s}(y_{0}))\omega^{X}(\Gamma\setminus B(y_{0},2s))$ whenever $X\in
B(y_{0},s)$ and $B(y_{0},Ms)\subset B(x_{0},2r)$ is a ball centered on
$\Gamma$. We use then an appropriate covering of $B(x_{0},r)$ by balls and the
Harnack inequality to get the lower bound $v(X)\geq
v(X_{0})\omega^{X}(\Gamma\setminus B(x_{0},4r))$, which is the counterpart of
(11.107) in our context. We conclude as in Lemma 11.14 by using Lemma 11.8 and
the doubling property for the harmonic measure (Lemma 11.12)

Let $y_{0}\in\Gamma$ and $s>0$. Write $Y_{0}$ for $A_{s}(y_{0})$. The main
idea is to take

(11.131) $f_{y_{0},s}(X):=s^{d-1}g(X,Y_{0})-K_{1}\omega^{X}(\Gamma\setminus
B(y_{0},K_{2}s))$

for some $K_{1},K_{2}>0$ that depend only on $n$, $d$, $C_{0}$ and $C_{1}$.
With good choices of $K_{1}$ and $K_{2}$, the function $f_{y_{0},s}$ is
positive in $B(y_{0},s)$ and negative outside of a big ball
$B(y_{0},2K_{2}s)$. However, with this definition involving the harmonic
measure, the function $f_{y_{0},s}$ doesn’t satisfy the appropriate estimates
required for the use of the maximum principle given as Lemma 11.3. So we shall
replace $\omega^{X}(\Gamma\setminus B(y_{0},K_{2}s))$ by some solution of
$Lu=0$, with smooth Dirichlet condition.

Let $h\in C^{\infty}(\mathbb{R}^{n})$ be such that $0\leq h\leq 1$, $h\equiv
0$ on $B(0,1/2)$ and $h\equiv 1$ on the complement of $B(0,1)$. For $\beta>0$
(which will be chosen large), we define $h_{\beta}$ by
$h_{\beta}(x)=h(\frac{x-y_{0}}{\beta s})$. Let $u_{\beta}$ be the solution,
given by Lemma 9.1, of $Lu_{\beta}=0$ with the Dirichlet condition
$Tu_{\beta}=Th_{\beta}$. Notice that $u_{\beta}\in W$ because $1-u_{\beta}$ is
the solution of $L$ with the smooth and compactly supported trace $1-h$.
Observe that for any $X\in\Omega$ and $\beta>0$,

(11.132) $\omega^{X}(\Gamma\setminus B(y_{0},\beta s))\leq
u_{\beta}(X)\leq\omega^{X}(\Gamma\setminus B(y_{0},\beta s/2)).$

The functions $u_{\beta}$ will play the role of harmonic measures but, unlike
these, the functions $u_{\beta}$ lie in $W$ and are thus suited for the use of
Lemma 11.3.

By Lemma 11.4, there exists $C>0$, that depends only on $d$, $n$, $C_{0}$ and
$C_{1}$, such that

(11.133) $g(X,Y_{0})\leq C\delta(Y_{0})^{1-d}\ \ \ \text{ for
}X\in\Omega\setminus B(Y_{0},\delta(Y_{0})/4).$

Moreover, since $Y_{0}$ comes from Lemma 11.6, we have $\epsilon
s\leq\delta(Y_{0})\leq s$ with an $\epsilon>0$ that does not depend on $s$ or
$y_{0}$, and hence

(11.134) $s^{d-1}g(X,Y_{0})\leq C\ \ \ \text{ for }X\in\Omega\setminus
B(y_{0},2s).$

From this and the non-degeneracy of the harmonic measure (Lemma 11.10), we
deduce that for $\beta\geq 1$,

(11.135) $s^{d-1}g(X,Y_{0})\leq K_{1}\omega^{X}(\Gamma\setminus B(y_{0},\beta
s))\leq K_{1}u_{\beta}(X)\ \ \ \text{ for }X\in\Omega\setminus B(y_{0},2\beta
s),$

where the constant $K_{1}>0$, depends only on $d$, $n$, $C_{0}$ and $C_{1}$.

Our aim now is to find $K_{2}\geq 20$ such that

(11.136) $K_{1}u_{K_{2}}(X)\leq\frac{1}{2}s^{d-1}g(X,Y_{0})\ \ \ \text{ for
}X\in\Omega\cap[B(y_{0},s)\setminus B(Y_{0},\delta(Y_{0})/4)].$

By construction, the function $u_{\beta}$ is such that $T(u_{\beta})=0$ on
$\Gamma\cap B(y_{0},\beta s/2)$, so according to the Hölder continuity at the
boundary (Lemma 8.16), we have

(11.137) $\sup_{B(y_{0},10s)}u_{\beta}\leq C\beta^{-\alpha}$

for any $\beta\geq 20$, where $C$ and $\alpha>0$ depend only on $d$, $n$,
$C_{0}$ and $C_{1}$. Moreover, due to (11.132) and the non-degeneracy of the
harmonic measure (Lemma 11.10),

(11.138) $u_{4}(X)\geq C^{-1}\ \ \ \text{ for }X\in\Omega\setminus
B(y_{0},8s)$

where $u_{4}$ is defined as $u_{\beta}$ (with $\beta=4$). As a consequence,
there exists $K_{3}>0$, that depends only on $d$, $n$, $C_{0}$, and $C_{1}$,
such that for $\beta\geq 20$,

(11.139) $u_{\beta}(X)\leq K_{3}\beta^{-\alpha}u_{4}(X)\ \ \ \text{ for
}X\in\Omega\cap[B(y_{0},10s)\setminus B(y_{0},8s)].$

We just proved that for $\beta\geq 20$, the function
$u^{\prime}=K_{3}\beta^{-\alpha}u_{4}-u_{\beta}$ satisfies the assumption
(iii) of Lemma 11.3, with $E=B(y_{0},10s)$ and $F=\overline{B(y_{0},8s)}$. The
other assumptions of Lemma 11.3 are satisfied as well, since $u^{\prime}\in W$
is smooth and $T(u^{\prime})=K_{3}\beta^{-\alpha}Tu_{4}\geq 0$ on $\Gamma\cap
E$. Therefore, Lemma 11.3 gives

(11.140) $u_{\beta}(X)\leq K_{3}\beta^{-\alpha}u_{4}(X)\ \ \ \text{ for
}X\in\Omega\cap B(y_{0},10s).$

Use now (11.132) and Lemma 11.11 to get for
$X\in\Omega\cap[B(y_{0},s)\setminus B(Y_{0},\delta(Y_{0})/4)]$,

(11.141) $u_{\beta}(X)\leq K_{3}\beta^{-\alpha}\omega^{X}(\Gamma\setminus
B(y_{0},2s))\leq C\beta^{-\alpha}s^{d-1}g(X,Y_{0}),$

where $C>0$ depends only on $d$, $n$, $C_{0}$ and $C_{1}$. The existence of
$K_{2}\geq 20$ satisfying (11.136) is now immediate.

Define the function $f_{y_{0},s}$ on $\Omega\setminus\\{Y_{0}\\}$ by

(11.142) $f_{y_{0},s}(X):=s^{d-1}g(X,Y_{0})-K_{1}u_{K_{2}}(X).$

The inequality (11.135) gives

(11.143) $f_{y_{0},s}(X)\leq 0\ \ \ \text{ for }X\in\Omega\setminus
B(y_{0},2K_{2}s),$

and the estimates (11.136) and (11.69) imply that

(11.144) $f_{y_{0},s}(X)\geq\frac{1}{2}s^{d-1}g(X,Y_{0})\geq
C^{-1}\omega^{X}(\Gamma\setminus B(y_{0},2s))\ \ \ \text{ for
}X\in\Omega\cap[B(y_{0},s)\setminus B(Y_{0},\delta(Y_{0})/4)].$

Let us turn to the proof of the comparison principle. By symmetry and as in
Lemma 11.14, it suffices to prove the upper bound in (11.130), that is

(11.145) $\frac{u(X)}{v(X)}\leq C\frac{u(X_{0})}{v(X_{0})}\ \ \ \text{ for
}X\in\Omega\cap B(x_{0},r).$

We claim that

(11.146) $v(X)\geq C^{-1}v(X_{0})\omega^{X}(\Gamma\setminus B(x_{0},2r))\ \ \
\text{ for }X\in\Omega\cap B(x_{0},r),$

where $C>0$ depends only on $n$, $d$, $C_{0}$ and $C_{1}$. So let
$X\in\Omega\cap B(x_{0},r)$ be given. Two cases may happen. If
$\delta(X)\geq\frac{r}{8K_{2}}$, where $K_{2}$ comes from (11.136) and is the
same as in the definition of $f_{y_{0},s}$, the existence of Harnack chains
(Lemma 2.2), the Harnack inequality (Lemma 8.9) and the non-degeneracy of the
harmonic measure (Lemma 11.10) give

(11.147) $v(X)\approx v(X_{0})\approx v(X_{0})\frac{\omega^{X}(\Gamma\setminus
B(x_{0},2r))}{\omega^{X_{0}}(\Gamma\setminus B(x_{0},2r))}\approx
v(X_{0})\omega^{X}(\Gamma\setminus B(x_{0},2r))$

by (11.67). The more interesting remaining case is when
$\delta(X)<\frac{r}{8K_{2}}$. Take $y_{0}\in\Gamma$ such that
$|X-y_{0}|=\delta(X)$. Set $s:=\frac{r}{8K_{2}}$ and $Y_{0}=A_{s}(y_{0})$. The
ball $B(y_{0},\frac{1}{2}r)=B(y_{0},4K_{2}s)$ is contained in
$B(x_{0},\frac{7}{4}r)$. The following points hold :

  * •

The quantity $\int_{B(y_{0},4K_{2}s)\setminus B(Y_{0},\delta(Y_{0})/4)}|\nabla
v|^{2}dm$ is finite because $v\in W_{r}(B(x_{0},2r))$. The fact that
$\int_{B(y_{0},4K_{2}s)\setminus B(Y_{0},\delta(Y_{0})/4)}|\nabla
f_{y_{0},s}|^{2}dm$ is finite as well follows from the property (10.2) of the
Green function.

  * •

There exists $K_{4}>0$ (depending only on $d$, $n$, $C_{0}$ and $C_{1}$) such
that

(11.148) $v(Y)-K_{4}v(Y_{0})f_{y_{0},s}(Y)\geq 0\ \ \ \text{ for }Y\in
B(Y_{0},\delta(Y_{0})/2)\setminus B(Y_{0},\delta(Y_{0})/4).$

This latter inequality is due to the following two bounds: the fact that

(11.149) $f_{y_{0},s}(Y)\leq s^{1-d}g(Y,Y_{0})\leq C\ \ \ \text{ for }Y\in
B(Y_{0},\delta(Y_{0})/2)\setminus B(Y_{0},\delta(Y_{0})/4),$

which is a consequence of the definition (11.142) and (10.6), and the bound

(11.150) $v(Y)\geq C^{-1}v(Y_{0})\ \ \ \text{ for }Y\in
B(Y_{0},\delta(Y_{0})/2),$

which comes from the Harnack inequality (Lemma 8.9).

  * •

The function $v-K_{4}v(Y_{0})f_{y_{0},s}$ is nonnegative on
$\Omega\cap[B(y_{0},4K_{2}s)\setminus B(y_{0},2K_{2}s)]$. Indeed, $v\geq 0$ on
$B(y_{0},4K_{2}s)$ and, thanks to (11.143), $f_{y_{0},s}\leq 0$ on
$\Omega\setminus B(y_{0},2K_{2}s)$.

  * •

The trace of $v-K_{4}v(Y_{0})f_{y_{0},s}$ is non-negative on
$B(y_{0},4K_{2}s)\cap\Gamma$ again because $Tv=0$ on
$B(y_{0},4K_{2}s)\cap\Gamma$ and $T[f_{y_{0},s}]\leq 0$ on
$B(y_{0},4K_{2}s)\cap\Gamma$ by construction.

The previous points prove that $v-K_{4}v(Y_{0})f_{y_{0},s}$ satisfies the
assumptions of Lemma 11.3 with
$E=B(y_{0},4K_{2}s)\setminus\overline{B(Y_{0},\delta(Y_{0})/4)}$ and
$F=\overline{B(y_{0},2K_{2}s)}\setminus B(Y_{0},\delta(Y_{0})/2)$. As a
consequence, for any $Y\in B(y_{0},4K_{2}s)\setminus B(Y_{0},\delta(Y_{0})/4)$

(11.151) $v(Y)-K_{4}v(Y_{0})f_{y_{0},s}(Y)\geq 0,$

and hence, for any $Y\in B(y_{0},s)\setminus B(Y_{0},\delta(Y_{0})/4)$

(11.152) $v(Y)\geq K_{4}v(Y_{0})f_{y_{0},s}(Y)\geq
C^{-1}v(Y_{0})\omega^{Y}(\Gamma\setminus B(y_{0},2s))$

by (11.144). Since both $v$ and $Y\to\omega^{Y}(\Gamma\setminus B(y_{0},2s))$
are solutions on $B(y_{0},2s)$, we can use the Harnack inequality (Lemma 8.9)
to deduce, first, that (11.152) holds for any $Y\in B(y_{0},s)$ and second,
that we can replace $v(Y_{0})$ by $v(X_{0})$ (recall that at this point,
$\frac{s}{r}=\frac{1}{8K_{2}}$ is controlled by the usual constants).
Therefore,

(11.153) $v(Y)\geq C^{-1}v(X_{0})\omega^{Y}(\Gamma\setminus B(y_{0},2s))\ \ \
\text{ for }Y\in B(y_{0},s).$

In particular, with our choice of $y_{0}$ and $s$, the inequality is true when
$X=Y$, that is,

(11.154) $v(X)\geq C^{-1}v(X_{0})\omega^{X}(\Gamma\setminus B(y_{0},2s))\geq
C^{-1}v(X_{0})\omega^{X}(\Gamma\setminus B(x_{0},2r))$

where $C>0$ depends only on $d$, $n$, $C_{0}$ and $C_{1}$. The claim (11.146)
follows.

Now we want to prove that

(11.155) $u(X)\leq Cu(X_{0})\omega^{X}(\Gamma\setminus B(x_{0},\frac{5}{4}r))\
\ \ \text{ for }X\in\Omega\cap B(x_{0},r).$

By Lemma 11.8,

(11.156) $u(X)\leq Cu(X_{0})\ \ \ \text{ for }X\in\Omega\cap
B(x_{0},\frac{7}{4}r).$

Pick $h^{\prime}\in C^{\infty}(\mathbb{R}^{n})$ such that $0\leq
h^{\prime}\leq 1$, $h^{\prime}\equiv 1$ outside of $B(x_{0},\frac{3}{2}r)$,
and $h^{\prime}\equiv 0$ on $B(x_{0},\frac{5}{4}r)$. Let
$u_{h^{\prime}}=U(h^{\prime})$ be the solution of $Lu_{h^{\prime}}=0$ with the
data $Tu_{h^{\prime}}=Th^{\prime}$ (given by Lemma 9.1). As before,
$u_{h^{\prime}}\in W$ because $1-u_{h^{\prime}}=U(1-h)$ and $1-h$ is a test
function. Also, $u_{h^{\prime}}(X)\geq\omega^{X}(\Gamma\setminus
B(x_{0},\frac{3}{2}r))$ by monotonicity. So (11.66), which states the non-
degeneracy of the harmonic measure, gives

(11.157) $u_{h^{\prime}}(X)\geq C^{-1}\ \ \ \text{ for }X\in\Omega\setminus
B(x_{0},\frac{13}{8}r).$

The combination of (11.156) and (11.157) yields the existence of $K_{5}>0$
(that depends only on $d$, $n$, $C_{0}$ and $C_{1}$) such that
$K_{5}u(X_{0})u_{h^{\prime}}-u\geq 0$ on
$\Omega\cap[B(x_{0},\frac{7}{4}r)\setminus B(x_{0},\frac{13}{8}r)]$. It is
easy to check that $K_{5}u(X_{0})u_{h^{\prime}}-u$ satisfies all the
assumptions of Lemma 11.3, with $E=B(x_{0},\frac{7}{4}r)$ and
$F=\overline{B(x_{0},\frac{13}{8}r)}$. This is because $u\in
W_{r}(B(x_{0},2r))$, $u_{h^{\prime}}\in W$, $Tu_{h^{\prime}}\geq 0$, and
$Tu=0$ on $\Gamma\cap B(x_{0},2r)$. Then by Lemma 11.3

(11.158) $u\leq K_{5}u(X_{0})u_{h^{\prime}}\ \ \ \text{ for }X\in\Omega\cap
B(x_{0},\frac{7}{4}r),$

and since $u_{h^{\prime}}(X)\leq\omega^{X}(\Gamma\setminus
B(x_{0},\frac{5}{4}r))$ for all $X\in\Omega$,

(11.159) $u(X)\leq Cu(X_{0})\omega^{X}(\Gamma\setminus B(x_{0},\frac{5}{4}r))\
\ \ \text{ for }X\in\Omega\cap B(x_{0},\frac{7}{4}r).$

The claim (11.155) follows.

The bounds (11.146) and (11.155) imply that

(11.160) $\frac{u(X)}{v(X)}\leq
C\frac{u(X_{0})}{v(X_{0})}\frac{\omega^{X}(\Gamma\setminus
B(x_{0},\frac{5}{4}r))}{\omega^{X}(\Gamma\setminus B(x_{0},2r))}\ \ \ \text{
for }X\in\Omega\cap B(x_{0},r).$

The bound (11.145) is now a consequence of the above inequality and the
doubling property of the harmonic measure (Lemma 11.12, or more exactly
(11.100)). $\Box$

## References

  * [A] A. Ancona. Théorie du potentiel sur les graphes et les variétés, pp 1–112, École d’Été de Probabilités de Saint-Flour XVIII - 1988. Ed. Hennequin, Paul-Louis. Lecture Notes in Mathematics 1427, Springer Berlin Heidelberg, 1989.
  * [AGMT] J. Azzam, J. Garnett, M. Mourgoglou, X. Tolsa. Uniform rectifiability, elliptic measure, square functions, and $\varepsilon$-approximability via an ACF monotonicity formula, preprint.
  * [AHMNT] J. Azzam, S. Hofmann, J.M. Martell, K. Nyström, T. Toro. A new characterization of chord-arc domains, JEMS, to appear.
  * [AHM3TV] J. Azzam, S. Hoffman, M. Mourgoglou, J. M. Martell, S. Mayboroda, X. Tolsa, A. Volberg. Rectifiability of harmonic measure, Geom. Funct. Anal. 26 (2016), no. 3, 703–728.
  * [Ba] M. Badger. Null sets of harmonic measure on NTA domains: Lipschitz approximation revisited. Math. Z. 270 (2012), no. 1-2, 241–262.
  * [BJ] C. Bishop, P. Jones. Harmonic measure and arclength, Ann. of Math. (2), 132 (1990), 511–547.
  * [CFK] L. Caffarelli, E. Fabes, C. Kenig. Completely singular elliptic-harmonic measures. Indiana Univ. Math. J., 30 (1981), no. 6, 917–924.
  * [CFMS] L. Caffarelli, E. Fabes, S. Mortola, S. Salsa. Boundary behavior of nonnegative solutions of elliptic operators in divergence form. Indiana Univ. Math. J., 30 (1981), no. 4, 621–640.
  * [Da] B. E. Dahlberg. On the absolute continuity of elliptic measures. Amer. J. Math., 108 (1986), no. 5, 1119–1138.
  * [DJ] G. David, D. Jerison. Lipschitz approximation to hypersurfaces, harmonic measure, and singular integrals. Indiana Univ. Math. J., 39 (1990), no. 3, 831–845.
  * [DFM] G. David, J. Feneuil, S. Mayboroda. Harmonic measure on sets of codimension larger than one. Comptes Rendus Mathematique, 355(4), 406–410.
  * [DK] H. Dong, S. Kim. Green’s matrices of second order elliptic systems with measurable coefficients in two dimensional domains. Trans. Amer. Math. Soc., 361 (2009), no. 6, 3303–3323.
  * [Duo] J. Duoandikoetxea. Fourier analysis. Graduate Studies in Mathematics 29. American Mathematical Society, Providence, RI, 2001.
  * [Eva] L. C. Evans. Partial differential equations. Graduate Studies in Mathematics 19. Second edition. American Mathematical Society, Providence, RI, 2010.
  * [FKS] E. Fabes, C. Kenig, R. Serapioni. The local regularity of solutions of degenerate elliptic equations. Comm. Partial Differential Equations, 7 (1982), no. 1, 77–116.
  * [FJK] E. Fabes, D. Jerison, C. Kenig. The Wiener test for degenerate elliptic equations. Ann. Inst. Fourier (Grenoble), 32 (1982), no. 3, 151–182.
  * [Fed] H. Federer. Geometric measure theory. Die Grundlehren der mathematischen Wissenschaften, Band 153. Springer-Verlag New York Inc., New York, 1996.
  * [GMT] J. Garnett, M. Mourgoglou, X. Tolsa, Uniform rectifiability in terms of Carleson measure estimates and $\varepsilon$-approximability of bounded harmonic functions, Duke Math. J. Vol. 167 (2018), no. 8, 1473–1524.
  * [GT] D. Gilbarg, N. S. Trudinger. Elliptic partial differential equations of second order. Classics in Mathematics. Springer-Verlag, Berlin (2001). Reprint of the 1998 edition.
  * [GW] M. Grüter and K.-O. Widman. The Green function for uniformly elliptic equations. Manuscripta Math., 37 (1982), no. 3, 303–342.
  * [Haj] P. Hajłasz. Geometric approach to Sobolev spaces and badly degenerated elliptic equations. Nonlinear analysis and applications (Warsaw, 1994), 141–168, GAKUTO Internat. Ser. Math. Sci. Appl., 7, Gakkotosho, Tokyo, 1996.
  * [HaK] P. Hajłasz and P. Koskela. Sobolev meets Poincaré. C. R. Acad. Sci. Paris Sér. I Math., 320 (1995), no. 10, 1211–1215.
  * [HaK2] by same authorSobolev met Poincaré. Mem. Amer. Math. Soc., 145 (2000), no. 688.
  * [HL] Q. Han and F. Lin. Elliptic partial differential equations. Courant Lecture Notes in Mathematics. Courant Institute of Mathematical Sciences, New York; American Mathematical Society, Providence, RI, 2011. Second edition.
  * [HKM] J. Heinonen, T. KilpeläÂinen, O. Martio. Nonlinear potential theory of degenerate elliptic equations. Oxford Mathematical Monographs. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1993.
  * [HoK] S. Hofmann, S. Kim. The Green function estimates for strongly elliptic systems of second order Manuscripta Math., 124 (2007), no. 2, 139–172.
  * [HM1] S. Hofmann, J.M. Martell. Uniform rectifiability and harmonic measure I: uniform rectifiability implies Poisson kernels in $L^{p}$. Ann. Sci. Éc. Norm. Supér. (4), 47 (2014), no. 3, 577–654.
  * [HLMN] S. Hofmann, P. Le, J. M. Martell, K. Nyström. The weak-$A_{\infty}$ property of harmonic and $p$-harmonic measures implies uniform rectifiability, Anal. PDE 10 (2017), no. 3, 513–558.
  * [HMM1] S. Hofmann, J.-M. Martell, S. Mayboroda. Uniform rectifiability, Carleson measure estimates, and approximation of harmonic functions. Duke Math. J. 165 (2016), no. 12, 2331–2389.
  * [HMM2] by same authorTransference of scale-invariant estimates from Lipschitz to Non-tangentially accessible to Uniformly rectifiable domains, preprint.
  * [HMU] S. Hofmann, J.M. Martell, I. Uriarte-Tuero. Uniform rectifiability and harmonic measure, II: Poisson kernels in $L^{p}$ imply uniform rectifiability. Duke Math. J., 163 (2014), no. 8, 1601–1654.
  * [JW] A. Jonsson, H. Wallin, Function spaces on subsets of ${\mathbb{R}}^{n}$. Math. Rep. 2 (1984), no. 1, xiv+221 pp.
  * [Ken] C. E. Kenig. Harmonic analysis techniques for second order elliptic boundary value problems. CBMS Regional Conference Series in Mathematics 83 (AMS, Providence, RI, 1994).
  * [KKPT] C. Kenig, H. Koch, J. Pipher, T. Toro. A new approach to absolute continuity of elliptic measure, with applications to non-symmetric equations. Adv. Math., 153 (2000), no. 2, 231–298.
  * [KKiPT] C. Kenig, B. Kirchheim, J. Pipher, T. Toro. Square Functions and the $A^{\infty}$ Property of Elliptic Measures. J. Geom. Anal., 26 (2016), no. 3, 2383–2410.
  * [KJ] D. Jerison, C. Kenig. Boundary behavior of harmonic functions in nontangentially accessible domains. Adv. in Math., 46 (1982), no. 1, 80–147.
  * [KP] C. Kenig, J. Pipher. The Dirichlet problem for elliptic equations with drift terms. Publ. Mat., 45 (2001), no. 1, 199–217.
  * [Kil] T. Kilpeläinen. Weighted Sobolev spaces and capacity. Ann. Acad. Sci. Fenn. Ser. A I Math., 19 (1994), no. 1, 95–113.
  * [Lv] M. Lavrent’ev. Boundary problems in the theory of univalent functions. Amer. Math. Soc. Transl. (2), 32 (1963), 1–35.
  * [LN] J. Lewis, K. Nyström. Quasi-linear PDEs and low-dimensional sets, JEMS, to appear.
  * [MZ] J. Malý, W. P. Ziemer. Fine regularity of solutions of elliptic partial differential equations. Mathematical Surveys and Monographs. American Mathematical Society, Providence, RI, 1997.
  * [Maz] V. Maz’ya. Sobolev spaces with applications to elliptic partial differential equations. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], Springer, Heidelberg. no. 342 (2011).
  * [RR] F. & M. Riesz. Über die randwerte einer analtischen funktion. Compte Rendues du Quatrième Congrès des Mathématiciens Scandinaves, Stockholm 1916, Almqvists and Wilksels, Upsala, 1920.
  * [Rud] W. Rudin. Real and complex analysis. McGraw-Hill Book Co., New York-Toronto, Ont.-London, 1966.
  * [Se] S. Semmes. Analysis vs. geometry on a class of rectifiable hypersurfaces in $\mathbb{R}^{n}$. Indiana Univ. Math. J., 39 (1990), no. 4, 1005–1035.
  * [Sta] G. Stampacchia. Le problème de Dirichlet pour les équations elliptiques du second ordre à coefficients discontinus. Ann. Inst. Fourier, 15 (1965), 189–258.
  * [Sta2] by same authorFormes bilinéaires coercitives sur les ensembles convexes. C. R. Acad. Sci. Paris, 258 (1964), 4413–4416.
  * [Ste] E. M. Stein. Singular integrals and differentiability properties of functions. Princeton Mathematical Series, No. 30. Princeton University Press, Princeton, N.J., 1970.
  * [Wu] J.-M. Wu. On singularity of harmonic measure in space. Pacific J. Math., 121 (1986), no. 2, 485–496.
  * [Z] W.P. Ziemer. Some remarks on harmonic measure in space. Pacific J. Math., 55 (1974), 629–637.

## Index