$\mathfrak{g}^{\rho}-\sup_{B^{j+2}\setminus B^{j+1}}\mathfrak{g}^{\rho}$. We
also write $\Omega_{s,j}$ for $\\{x\in\Omega,\,\mathfrak{g}^{\rho}_{j}>s\\}$.
Notice that $B^{j+2}\subset\Omega$, and $\sup_{B^{j+2}\setminus
B^{j+1}}=\sup_{\Omega\setminus B^{j+1}}$ by the maximum principle. Hence,
$\Omega_{s,j}\subset B^{j+1}$ when $s>0$. Let $t>0$ be given; we use again
(14.3), but with the test function

$\varphi_{j}(z):=\left(\frac{2}{t}-\frac{1}{(\mathfrak{g}^{\rho}_{j}(z))^{+}}\right)^{+},$

to get, as we did for (14.9),

$a(\mathfrak{g}^{\rho},\varphi)=\int_{\Omega_{t/2,j}}\frac{\mathcal{A}\nabla\mathfrak{g}^{\rho}_{j}\cdot\nabla\mathfrak{g}^{\rho}_{j}}{(\mathfrak{g}^{\rho}_{j})^{2}}\,dm\leq\fint_{B_{\rho}}\varphi_{j}\,dm\leq\frac{2}{t},$

and by (11.3)

(14.18)
$\int_{\Omega_{t/2,j}}\frac{|\nabla\mathfrak{g}^{\rho}_{j}|^{2}}{(\mathfrak{g}^{\rho}_{j})^{2}}\,dm\leq\frac{C}{t}.$

Set $v_{j}(z):=(\ln(\mathfrak{g}^{\rho}_{j}(z))-\ln t+\ln 2)^{+}$; as before
$v_{j}\in W$, it is supported in $\Omega_{t/2,j}\subset B^{j+1}$, and $|\nabla
v_{j}|=|\nabla\mathfrak{g}^{\rho}_{j}|/\mathfrak{g}^{\rho}_{j}$ on
$\Omega_{t/2}$. Since $v_{j}=0$ on $B^{j+2}\setminus B^{j+1}\subset\Omega$, we
can apply Theorem 5.24 and Remark 5.27, and we get that

(14.19)
$\begin{split}\left(\int_{\Omega_{t/2,j}}|v_{j}|^{p}\,dm\right)^{\frac{1}{p}}&\leq
C2^{j+1}|x-y|\,m(B^{j+1})^{\frac{1}{p}-\frac{1}{2}}\left(\int_{\Omega_{t/2,j}}|\nabla
v_{j}|^{2}\,dm\right)^{\frac{1}{2}}\\\ &\leq
C2^{j}|x-y|\,m(B^{j})^{\frac{1}{p}-\frac{1}{2}}t^{-\frac{1}{2}}\end{split}$

by (14.18) and (H4), where $p\in[k,2k]$ plays the role of $kp$ in Theorem 5.24
and Remark 5.27, and we are mostly interested in $p=2$ there which yields
$p=2k$ here. Of course $C$ is independent of $j$. Since $|v_{j}|>\ln 2$ on
$\Omega_{t,j}$ and $\Omega_{t,j}\subset\Omega_{t/2,j}$, (14.19) implies that

(14.20) $\frac{m(\Omega_{t,j})}{m(B^{j+2})}\leq
C\left(\frac{t\,m(B^{j})}{(2^{j}|x-y|)^{2}}\right)^{-\frac{p}{2}}.$

The rest of the proof of (14.17) is similar to what we did for Lemma 14.6.
Since $\mathfrak{g}^{\rho}_{j}$ is a solution in $B^{j+2}\setminus B^{j-1}$,
we can use the Moser inequality inside $B^{j+2}\setminus B^{j-1}$ to get that
for $z\in B^{j+1}\setminus B^{j}$, $\mathfrak{g}^{\rho}_{j}(z)$ is smaller -
up to a constant - than its average on $B(z,2^{j-10}|x-y|)$. The measure of
this last ball is equivalent, by (H4), to the measure of $B_{j+2}$ and thus
for any $z\in B^{j+1}\setminus B^{j}$

(14.21)
$\mathfrak{g}^{\rho}_{j}(z)\lesssim\fint_{B^{j+2}}\mathfrak{g}^{\rho}_{j}\,dm=\int_{0}^{+\infty}\frac{m(\Omega_{t,j})}{m(B^{j+2})}dt$

(compare with (14.14)). Again we split the last integral into two pieces, and
for the second one we use (14.20); we obtain that for all $z\in
B^{j+1}\setminus B^{j}$

(14.22)
$\begin{split}\mathfrak{g}^{\rho}_{j}(z)&\lesssim\int_{0}^{s}1\,dt+\left(\frac{m(B^{j}))}{(2^{j}|x-y|)^{2}}\right)^{-\frac{p}{2}}\int_{s}^{+\infty}t^{-\frac{p}{2}}dt\\\
&\lesssim
s\left[1+\left(\frac{s\,m(B^{j})}{(2^{j}|x-y|)^{2}}\right)^{-\frac{p}{2}}\right].\end{split}$

where we can choose $p$, which comes from Theorem 5.24 (for instance applied
with the exponent $2$), strictly bigger than 2. We optimize in $s$ and take
the supremum in $z$ to get the desired estimate (14.17).

We can use (H4) to rewrite (14.17) as

(14.23) $\sup_{B^{j+1}\setminus
B^{j}}\mathfrak{g}^{\rho}-\sup_{B^{j+2}\setminus
B^{j+1}}\mathfrak{g}^{\rho}\leq
C\int_{2^{j}|x-y|}^{2^{j+1}|x-y|}\frac{r^{2}}{m(B(y,r))}\frac{dr}{r}.$

This was for $j<j_{0}$, but for $j=j_{0}$ we will be able to apply Lemma 14.6.
Recall that $2^{j_{0}+2}|x-y|>\delta(y)$ by definition of $j_{0}$; hence for
$z\in B^{j_{0}+1}\setminus B^{j_{0}}$, $|z-y|\geq 2^{j_{0}}|x-y|>\delta(y)/10$
and by Lemma 14.6 and the same trick as for (14.23),

$\sup_{B^{j_{0}+1}\setminus B^{j_{0}}}\mathfrak{g}^{\rho}\leq
C\int_{2^{j_{0}}|x-y|}^{2^{j_{0}+1}|x-y|}\frac{r^{2}}{m(B(y,r))}\frac{dr}{r}\leq
C\int_{2^{j_{0}}|x-y|}^{\delta(y)}\frac{r^{2}}{m(B(y,r))}\frac{dr}{r}.$

Now, since $\mathfrak{g}_{\rho}$ is continuous around $x$ (by the interior
Hölder estimates and (14.4)),

$\begin{split}\mathfrak{g}^{\rho}(x)&\leq\sup_{B^{1}\setminus
B^{0}}\mathfrak{g}^{\rho}\leq\sup_{B^{j_{0}+1}\setminus
B^{j_{0}}}\mathfrak{g}+\sum_{j=0}^{j_{0}-1}\left(\sup_{B^{j+1}\setminus
B^{j}}\mathfrak{g}^{\rho}-\sup_{B^{j+2}\setminus
B^{j+1}}\mathfrak{g}^{\rho}\right)\\\
&\lesssim\int_{2^{j_{0}}|x-y|}^{\delta(y)}\frac{r^{2}}{m(B(y,r))}\frac{dr}{r}+\sum_{j=0}^{j_{0}-1}\int_{2^{j}|x-y|}^{2^{j+1}|x-y|}\frac{r^{2}}{m(B(y,r))}\frac{dr}{r}\\\
&\lesssim\int_{|x-y|}^{\delta(y)}\frac{r^{2}}{m(B(y,r))}\frac{dr}{r}\,;\end{split}$

Lemma 14.16 follows. $\Box$

Before we continue to prove estimate about Green functions, we take a little
time to talk about cut-off functions. Pick $\phi\in
C^{\infty}(\mathbb{R}_{+})$ such that $0\leq\phi\leq 1$, $\phi\equiv 0$ on
$(2,+\infty)$, $\phi\equiv 1$ on $[0,1)$, and $|\phi^{\prime}|\leq 2$. If we
want a cut-off function adapted to the ball $B(x,r)$, the first choice will be

(14.24) $\alpha^{1}(y):=\phi\left(\frac{|x-y|}{r}\right).$

If the above cut-off function fails to work, we might try to use a cut-off
function that involves logarithms, in the spirit of the one used by Sobolev.
For instance, if we work on the Green function for the unit disk (with the
classical Lebesgue measure) on $\mathbb{R}^{2}$, the good cut-off may be

(14.25)
$\alpha^{2}(y):=\phi\left(\frac{\ln(\delta(y)/r)}{\ln(\delta(y)/|x-y|)}\right).$

In the classical theory, where the domains are equipped with the usual
Lebesgue measure, we would use $\alpha^{1}$ when $n\geq 3$ and $\alpha^{2}$
when $n=2$. In our article, $\alpha^{1}$ or $\alpha^{2}$ may be needed, or
something different. The cut-off functions that we shall need depend on the
measure $m$ and the purpose of the next lines is to define them.

We define the function $\gamma=\gamma_{y}$ on $(0,\delta(y))$ by

(14.26)
$\gamma(s):=\int_{s}^{\delta(y)}\frac{t^{2}}{m(B(y,t))}\,\frac{dt}{t}.$

The function $\gamma$ is well defined (because $m(B(y,t))>0$ since $m$ is
doubling on $\Omega$), and decreasing. In addition, $t\mapsto m(B(y,t))$ is
continuous, because $m$ is absolutely continuous with respect to the Lebesgue
measure (and not because $m$ is doubling), and so $\gamma$ is of class
$C^{1}$, with a derivative equal to

(14.27) $\gamma^{\prime}(s):=-\frac{s}{m(B(y,s))}.$

Next we use our function $\gamma$ and set

(14.28) $\alpha_{r}(s):=\phi\left(\frac{\gamma(r)}{\gamma(s)}\right)\hbox{ for
}0<r,s<\gamma(y).$

By construction,

(14.29) $\alpha_{r}\equiv 1\text{ on }[0,r),$ (14.30) $\alpha_{r}(s)=0$ when
$\gamma(s)<\frac{1}{2}\gamma(r)$,

and $\alpha_{r}$ is of class $C^{1}$ on $(0,\delta(y))$, with a derivative
equal to
$-\frac{\gamma^{\prime}(s)\gamma(r)}{\gamma(s)^{2}}\,\phi^{\prime}\left(\frac{\gamma(r)}{\gamma(s)}\right)$.
Thus

(14.31) $\alpha^{\prime}_{r}\text{ is supported on the interval where }\
\frac{1}{2}\gamma(r)\leq\gamma(s)\leq\gamma(r)$

(recall that $\gamma$ is decreasing), and

(14.32) $|\alpha^{\prime}_{r}(s)|\leq
8\,\frac{\gamma^{\prime}(s)}{\gamma(r)}.$

We shall also need a variation of the maximum principle (Lemma 12.6).

######  Lemma 14.33.

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 $u\in
W_{r}(E\cap\overline{\Omega})$ be a supersolution for $L$ in $\Omega\cap E$
such that

  1. (i)

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

  2. (ii)

$\operatorname{Tr}u\geq 0$ 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 proof of this result is the same as the one of [DFM2, Lemma 11.3].
$\Box$

We are now ready to establish a lower bound on $\mathfrak{g}^{\rho}(x)$ when
$x$ and $y$ are close. Those lower bounds are not necessary in our article to
prove the existence of the Green function, and a reader who is only interested
in existence can skip the next lemma.

######  Lemma 14.34.

If $x,y\in\Omega$ are such that $40\rho\leq 2|x-y|\leq\delta(y)$, then

$\mathfrak{g}^{\rho}(x)\geq
C^{-1}\int_{|x-y|}^{\delta(y)}\frac{t^{2}}{m(B(y,t))}\,\frac{dt}{t},$

where $C>0$ depends on $C_{4}$, $C_{6}$, $C_{A}$, and $n$.

Proof. The first point that we need to verify is that $\mathfrak{g}^{\rho}(x)$
is increasing when $x\to y$, at least in a weak sense, and when $|x-y|$ is way
bigger than $\rho$. Pick $r>10\rho$. The function
$v_{r}:=(\sup_{B_{r}\setminus
B_{r/2}}\mathfrak{g}^{\rho})-\mathfrak{g}^{\rho}$ \- when $B_{s}$ denotes as
usual $B(y,s)$ \- is a solution to $Lu=0$ in
$\Omega\setminus\overline{B_{r/2}}$. Moreover we can easily observe that
$\operatorname{Tr}v_{r}\geq 0$ and $v_{r}\geq 0$ on $B_{r}\setminus B_{r/2}$.
We can apply Lemma 14.33 with $E=\mathbb{R}^{n}\setminus\overline{B_{r/2}}$
and $F=\mathbb{R}^{n}\setminus B_{r}$, and in particular (i) holds because
$\mathfrak{g}^{\rho}\in W_{0}$; we deduce that $v_{r}\geq 0$ in
$\Omega\setminus\overline{B_{r/2}}$, that is,

(14.35) $\sup_{\Omega\setminus
B_{r/2}}\mathfrak{g}^{\rho}\leq\sup_{B_{r}\setminus
B_{r/2}}\mathfrak{g}^{\rho}.$

For the rest of the proof, we write $r$ for $|x-y|$, and for $i\geq 0$, we set
$r_{i}$ as the only value such that

(14.36) $\gamma(r_{i})=2^{-i}\gamma(r).$

Such a point exists, because $\gamma$ is a (strictly) decreasing continuous
function with $\gamma(\delta(y))=0$. Notice that $r_{0}=r$, and $\\{r_{i}\\}$
is an increasing sequence whoes limit is $\delta(y)$ (but we won’t go that
far). First, we use the test function on $\eta_{1}(x):=\alpha_{r_{1}}(|x-y|)$
in (14.3). Thanks to (14.31), $\nabla\eta_{1}$ is supported in
$B_{r_{2}}\setminus B_{r_{1}}$ and one obtains

$\displaystyle 1$ $\displaystyle=\int_{B_{r_{2}}\setminus
B_{r_{1}}}\mathcal{A}\nabla\mathfrak{g}^{\rho}\cdot\nabla\eta_{1}\,dm\leq\frac{C}{\gamma(r_{1})}\int_{B_{r_{2}}\setminus
B_{r_{1}}}|\nabla\mathfrak{g}^{\rho}||\gamma^{\prime}(|x-y|)|\,dm$ (14.37)
$\displaystyle\leq\frac{C}{\gamma(r_{1})}\left(\int_{B_{r_{2}}\setminus
B_{r_{1}}}|\nabla\mathfrak{g}^{\rho}|^{2}\,dm\right)^{\frac{1}{2}}\left(\int_{B_{r_{2}}\setminus
B_{r_{1}}}\frac{|x-y|^{2}}{m(B(y,|x-y|))^{2}}\,dm\right)^{\frac{1}{2}}$

We now want to prove that

(14.38) $\int_{B_{r_{2}}\setminus
B_{r_{1}}}\frac{|x-y|^{2}}{m(B(y,|x-y|))^{2}}\,dm\leq
C\int_{r_{1}}^{r_{2}}\frac{s^{2}}{m(B_{s})}\frac{ds}{s}\leq C\gamma(r_{1}).$

The second part is just the definition of $\gamma(r_{1})$ (we integrate
further); the first inequality will be a little longer to prove, because we
want to avoid the unpleasant situation where $B_{r_{2}}\setminus B_{r_{1}}$ is
a very thin annulus.

Let $C_{4}$ denote the doubling constant for $m$, as in (2.3), then set
$C^{\prime}_{4}=2C_{4}+4$, and let $\tau>0$ be small, to be chosen soon,
depending on $C_{4}$.

First assume that $(1+\tau)r_{1}\leq r_{2}\leq 2r_{1}$. Then the integrand on
the left is comparable to $r_{1}^{2}m(B(y,r_{2})^{-2}$ and
$m(B_{r_{2}}\setminus B_{r_{1}})$ is comparable to $m(B(y,r_{2})$; the desired
inequality follows, with a constant that depends on $\tau$, because
$\int_{r_{1}}^{r_{2}}\frac{ds}{s}\geq C\tau^{-1}$. When $r_{2}>2r_{1}$, this
is also easy: cut $B_{r_{2}}\setminus B_{r_{1}}$ into annuli of modulus
comparable to $1$, and prove the inequality separately on each one as we just
did.

We may now assume that $r_{2}\leq(1+\tau)r_{1}$; our defense will be that this
does not happen in the present circumstances. We claim that if $\tau$ is
chosen so small that $C^{\prime}_{4}\tau<1$, $r_{2}\leq(1+\tau)r_{1}$ implies
that $(1+C^{\prime}_{4}\tau)r_{1}\geq\delta(y)$.

Indeed, suppose instead that $(1+C^{\prime}_{4}\tau)r_{1}<\delta(y)$. By
definition, $\gamma(r_{1})=2\gamma(r_{1})$, hence by (14.26)

(14.39)
$\int_{r_{1}}^{r_{2}}\frac{t^{2}}{m(B(y,t))}\,\frac{dt}{t}=\int_{r_{2}}^{\delta(y)}\frac{t^{2}}{m(B(y,t))}\,\frac{dt}{t}\geq\int_{(1+\tau)r_{1}}^{(1+C^{\prime}_{4}\tau)r_{1}}\frac{t^{2}}{m(B(y,t))}\,\frac{dt}{t}.$

The left-hand side is at most $(r_{2}-r_{1})r_{2}m(B_{r_{1}})^{-1}\leq 2\tau
r_{1}^{2}m(B_{r_{1}})^{-1}$ because $\tau<1$, and the left-hand side is at
least $[(C^{\prime}_{4}-1)\tau
r_{1}]\,[(1+C^{\prime}_{4}\tau)r_{1}]\,m(B_{2r_{1}})^{-1}$ because
$C^{\prime}_{4}\tau<1$; since $m(B_{2r_{1}})\leq C_{4}m(B_{r_{1}})$ because
$m$ is doubling, this is at least $C_{4}^{-1}(C^{\prime}_{4}-1)\tau
r_{1}^{2}m(B_{r_{1}})^{-1}$. We chose $C^{\prime}_{4}=2C+4$, and the ensuing
contradiction proves the claim.

For the purposes of this lemma, we can take $\tau=(100C^{\prime}_{4})^{-1}$,
and then we just proved that (14.38) holds as soon as
$(1+10^{-2})r_{1}\leq\delta(y)$. Similarly,

(14.40) $\int_{B_{r_{i+1}}\setminus
B_{r_{i}}}\frac{|x-y|^{2}}{m(B(y,|x-y|))^{2}}\,dm\leq
C\int_{r_{i}}^{r_{i+1}}\frac{s^{2}}{m(B_{s})}\frac{ds}{s}\leq C\gamma(r_{i})$

as long as $(1+10^{-2})r_{i}\leq\delta(y)$. We want to show that this does not
happen for $i\leq 2$, so we need a control on the variations of
$\delta(y)-r_{i}$ along our sequence. Let us check that

(14.41) $\delta(y)-r_{i}\leq 3(\delta(y)-r_{i+1})\,\text{ for $i\geq 0$ such
that }r_{i+1}\geq\frac{5}{6}\delta(y).$

Suppose not, set $R=\delta(y)-3(\delta(y)-r_{i+1})>r_{i}$, and observe that
(as in (14.39))

(14.42)
$\int_{R}^{r_{i+1}}\frac{t^{2}}{m(B(y,t))}\,\frac{dt}{t}<\int_{r_{i}}^{r_{i+1}}\frac{t^{2}}{m(B(y,t))}\,\frac{dt}{t}=\int_{r_{i+1}}^{\delta(y)}\frac{t^{2}}{m(B(y,t))}\,\frac{dt}{t}.$

When we replace $m(B(y,t))$ by the larger number $m(B(y,r_{i+1}))$ on the
left-hand side, we get a smaller integral; similarly, when we replace
$m(B(y,t))$ by the smaller number $m(B(y,r_{i+1}))$ on the right-hand side, we
get a larger integral. Hence
$\int_{R}^{r_{i+1}}tdt<\int_{r_{i+1}}^{\delta(y)}tdt$. Notice that
$R\geq\delta(y)/2$ because $r_{i+1}\geq\frac{5}{6}\delta(y)$ and the interval
on the left is twice as long as on the right; this gives a contradiction, and
(14.41) follows.

We may now prove that in the present circumstances, (14.38) holds, and even
(14.40) for $0\leq i\leq 2$. Indeed, we start from
$r_{0}=r=|x-y|\leq\delta(y)/2$, so $\delta(y)-r_{0}\geq\delta(y)/2$, and it
follows from a short iteration of (14.41) that
$\delta(y)-r_{2}\geq\delta(y)/24$, and so $(1+10^{-2})r_{2}\leq\delta(y)$.

We may now return to (14). Since $\gamma(r_{1})\approx\gamma(r_{0})=\gamma(r)$
by (14.36), (14) and (14.38) imply that

(14.43) $1\leq\frac{C}{\gamma(r)}\int_{B_{r_{2}}\setminus
B_{r_{1}}}|\nabla\mathfrak{g}^{\rho}|^{2}\,dm.$

Next, since $\mathfrak{g}^{\rho}$ is a solution to $Lu=0$ in
$B_{r_{3}}\setminus B_{r}$, we can use the Cacciopoli inequality (Lemma 11.12)
with the test function
$\eta_{2}(x):=\alpha_{r_{2}}(|x-y|)[1-\alpha_{r_{0}}(|x-y|)]$, which lies in
$C^{\infty}_{0}(B_{r_{3}}\setminus B_{r})$ and satisfies $\eta_{2}\equiv 1$ on
$B_{r_{2}}\setminus B_{r_{1}}$. This yields

$\int_{B_{r_{2}}\setminus B_{r_{1}}}|\nabla\mathfrak{g}|^{2}\,dm\leq
C\int_{B_{r_{3}}\setminus
B_{r_{0}}}|\mathfrak{g}^{\rho}|^{2}|\nabla\eta_{2}|^{2}\,dm\leq
C\sup_{B_{r_{3}}\setminus
B_{r_{0}}}(\mathfrak{g}^{\rho})^{2}\int_{B_{r_{3}}\setminus
B_{r_{0}}}|\nabla\eta_{2}|^{2}\,dm.$

We bound the gradient of $\eta_{2}$ with the help of (14.32), (14.27), and
(14.36). We find that
$|\nabla\eta_{2}(x)|\leq\frac{C}{\gamma(r)}\frac{|x-y|}{m(B_{|x-y|})}$. So by
(14.40)

$\int_{B_{r_{3}}\setminus
B_{r_{0}}}|\nabla\eta_{2}|^{2}\,dm\leq\frac{C}{\gamma(r)^{2}}\int_{B_{r_{3}}\setminus
B_{r_{0}}}\frac{|x-y|^{2}}{m(B(y,|x-y|))^{2}}\,dm\leq\frac{C}{\gamma(r)}.$

As a consequence,

$\int_{B_{r_{2}}\setminus
B_{r_{1}}}|\nabla\mathfrak{g}^{\rho}|^{2}\,dm\leq\frac{C}{\gamma(r)}\sup_{B_{r_{3}}\setminus
B_{r_{0}}}(\mathfrak{g}^{\rho})^{2}$

and, together with (14.43),

(14.44) $1\leq\frac{C}{\gamma(r)^{2}}\sup_{B_{r_{3}}\setminus
B_{r_{0}}}(\mathfrak{g}^{\rho})^{2}.$

To conclude, we invoke (14.35) and then the interior Harnack inequality (Lemma
11.35), to get (since $r_{0}=r$)

$\sup_{B_{r_{3}}\setminus
B_{r_{0}}}\mathfrak{g}^{\rho}\leq\sup_{B_{r}\setminus
B_{r/2}}\mathfrak{g}^{\rho}\leq C\mathfrak{g}^{\rho}(x).$

We use the above estimate in (14.44) and notice $\gamma(r)$ is exactly the
bound required for Lemma 14.34 (see (14.26)); the lemma follows. $\Box$

######  Lemma 14.45.

Suppose that $\rho\leq 10^{-2}\delta(y)$ and $r<\delta(y)/2$, and set
$B_{r}=B(y,r)$ as usual. Then

$\int_{\Omega\setminus B_{r}}|\nabla\mathfrak{g}^{\rho}|^{2}\,dm\leq
C\int_{r}^{\delta(y)}\frac{s^{2}}{m(B(y,s))}\,\frac{ds}{s},$

where $C>0$ depends on $C_{1}$ to $C_{6}$, $C_{A}$, and $n$.

Proof. By (14.3) for the test function $v:=\mathfrak{g}^{\rho}$ (and the
elliptic condition (11.3)),

(14.46)
$\int_{\Omega}|\nabla\mathfrak{g}^{\rho}|^{2}dm\lesssim\int_{\Omega}\mathcal{A}\nabla\mathfrak{g}^{\rho}\cdot\nabla\mathfrak{g}^{\rho}\,dm=\fint_{B_{\rho}}\mathfrak{g}^{\rho}\,dm.$

Then cut the last integral in two; this yields

$\begin{split}\int_{\Omega}|\nabla\mathfrak{g}^{\rho}|^{2}dm&\lesssim\fint_{B_{\rho}}\left(\mathfrak{g}^{\rho}-\fint_{B_{50\rho}\setminus
B_{25\rho}}\mathfrak{g}^{\rho}\,dm\right)\,dm+\fint_{B_{50\rho}\setminus
B_{25\rho}}\mathfrak{g}^{\rho}\,dm\\\
&\lesssim\fint_{B_{50\rho}}\left|\mathfrak{g}^{\rho}-\fint_{B_{50\rho}\setminus
B_{25\rho}}\mathfrak{g}^{\rho}\,dm\right|\,dm+\fint_{B_{50\rho}\setminus
B_{25\rho}}\mathfrak{g}^{\rho}\,dm.\end{split}$

We use the Poincaré inequality (Theorem 5.24) to bound the first integral, and
Lemma 14.16 to bound the last one. Notice in particular that $40\rho\leq
2|x-y|<\delta(y)$ for $x\in B_{50\rho}\setminus B_{25\rho}$, so the lemma
applies. This gives

$\begin{split}\int_{\Omega}|\nabla\mathfrak{g}^{\rho}|^{2}dm&\lesssim\rho\left(\fint_{B_{50\rho}}\left|\nabla\mathfrak{g}^{\rho}\right|^{2}\,dm\right)^{\frac{1}{2}}+\int_{25\rho}^{\delta(y)}\frac{s^{2}}{m(B_{s})}\frac{ds}{s}\\\
&\lesssim\frac{25\rho}{m(B_{25\rho})^{1/2}}\left(\int_{\Omega}\left|\nabla\mathfrak{g}^{\rho}\right|^{2}\,dm\right)^{\frac{1}{2}}+\int_{25\rho}^{\delta(y)}\frac{s^{2}}{m(B_{s})}\frac{ds}{s}.\end{split}$

Since $m$ is doubling, $m(B_{t})\leq C_{4}m(B_{25\rho})$ for $25\rho\leq t\leq
50\rho$, so

$\frac{(25\rho)^{2}}{m(B_{25\rho})}\leq
C_{4}\int_{25\rho}^{50\rho}\frac{t^{2}}{m(B_{t})}\frac{dt}{t}\leq
C_{4}\gamma(25\rho)$

(because $50\rho<\delta(y)$ and by the definition (14.26)). So the above
estimate can be written

$\int_{\Omega}|\nabla\mathfrak{g}^{\rho}|^{2}dm\lesssim\gamma(25\rho)^{\frac{1}{2}}\left(\int_{\Omega}\left|\nabla\mathfrak{g}^{\rho}\right|^{2}\,dm\right)^{\frac{1}{2}}+\gamma(25\rho),$

or

(14.47)
$\int_{\Omega}|\nabla\mathfrak{g}^{\rho}|^{2}dm\lesssim\gamma(25\rho):=\int_{25\rho}^{\delta(y)}\frac{s^{2}}{m(B_{s})}\frac{ds}{s}.$

The estimate (14.47) is already good, and it proves Lemma 14.45 for any $r>0$
such that $\gamma(r)\geq\gamma(25\rho)/2$. Assume now that
$\gamma(r)\leq\gamma(25\rho)/2$. Since $\gamma$ is decreasing, there is a
unique $R\in[25\rho,r)$ such that $\gamma(R)=2\gamma(r)$. Then
$\mathfrak{g}^{\rho}$ is a solution to $Lu=0$ in $\Omega\setminus B_{R}$. By
Lemma 14.16 and the proof of (14.35), $\mathfrak{g}^{\rho}(x)\leq
C\gamma(R)=2C\gamma(|x-y|)$ for any $x\in\Omega\setminus B_{R}$. We claim that
if $\eta_{3}$ is a nonnegative smooth function that satisfies $\eta_{3}\equiv
0$ on $B_{R}$, we have the following Cacciopoli-type inequality:

(14.48) $\int_{\Omega}|\nabla\mathfrak{g}^{\rho}|^{2}|\eta_{3}|^{2}\,dm\leq
C\int_{\Omega}(\mathfrak{g}^{\rho})^{2}|\nabla\eta_{3}|^{2}dm.$

The above bound is not an application of the Cacciopoli inequalities stated in
Section 11, because the test function $\eta_{3}$ is not contained in a ball
$2B$ such that $\mathfrak{g}^{\rho}$ is a solution in $2B$. However, the proof
of (14.48) is very similar to the proof of the usual Cacciopoli inequality,
and we leave it to the reader. It relies on the fact that
$\mathfrak{g}^{\rho}\eta_{3}\in W_{0}$ even though $\eta_{3}$ is not compactly
supported.

We use (14.48) with $\eta_{3}(x):=1-\alpha_{R}(|x|)$, where $\alpha_{R}$ is
the function in (14.28). Notice that $\eta_{3}(x)=0$ on $B_{R}$ and
$\eta_{3}(x)=1$ on $\mathbb{R}^{n}\setminus B_{r}$ (because
$\gamma(|x|)<\gamma(r)=\gamma(R)/2$). So we obtain

(14.49) $\int_{\Omega\setminus
B_{r}}|\nabla\mathfrak{g}^{\rho}|^{2}\,dm\leq\frac{C}{\gamma(r)^{2}}\int_{B_{r}\setminus
B_{R}}|\mathfrak{g}^{\rho}(x)|^{2}\frac{|x-y|^{2}}{m(B(y,|x-y|))^{2}}dm(x).$

by (14.32) and (14.27). But for $x\in B_{r}\setminus B_{R}$, Lemma 14.16 says
that $\mathfrak{g}^{\rho}(x)\leq C\gamma(R)=2C\gamma(r)$; hence by the proof
of (14.38),

(14.50) $\int_{\Omega\setminus
B_{r}}|\nabla\mathfrak{g}^{\rho}|^{2}\,dm\lesssim\int_{B_{r}\setminus
B_{R}}\frac{|x-y|^{2}}{m(B(y,|x-y|))^{2}}dm(x)\lesssim\gamma(R)\lesssim\gamma(r).$

Lemma 14.45 follows. $\Box$

######  Lemma 14.51.

There exists $q>1$, that depends only on $C_{4}$, such that for any
$y\in\Omega$ and any $\rho\leq\delta(y)/100$,

$\int_{B(y_{0},2\delta(y))}|\nabla\mathfrak{g}^{\rho}|^{q}\,dm\leq
Cm(B_{\delta(y)})\left(\frac{\delta(y)}{m(B_{\delta(y)})}\right)^{q},$

where $y_{0}\in\Gamma$ is such that $|y-y_{0}|=\delta(y)$, and where $C>0$
depends only on $C_{1}$ to $C_{6}$, $C_{A}$, and $n$.

######  Remark 14.52.

If $\gamma(r):=\int_{r}^{\delta}(y)\frac{s^{2}}{m(B_{s})}\frac{ds}{s}$ is
uniformly bounded, or in other words if
$\int_{0}^{\delta}(y)\frac{s^{2}}{m(B_{s})}$ is finite, then by (14.47) we can
take $q=2$ in Lemma 14.51.

Proof. We first start by proving general results, which are only consequences
of the doubling property (H4). There exists $\alpha\in(0,1)$ such that

(14.53) $m(B)\leq\alpha m(2B)$

for every ball $B\subset\mathbb{R}^{n}$ such that $2B\subset\Omega$. Indeed,
if $r$ denotes the radius of $B$, then we can find a ball $B_{0}$ of radius
$r/2$ in $2B\setminus B$. Then $B\subset 3B_{0}$, hence $m(B)\leq
C_{4}^{2}m(B_{0})$ by (H4), and now $m(2B)-m(B)\geq m(B_{0})\geq
C_{4}^{-2}m(B)$, and (14.53) holds with $\alpha=(1+C_{4}^{-2})^{-1}$.

Similarly to (2.5), the estimate (14.53) can be improved into

(14.54) $m(B(y,r))\leq
C\left(\frac{r}{s}\right)^{2\epsilon}m(B(y,s))\qquad\text{ for }r\leq
s\leq\delta(y),$

where $C,\epsilon>0$ depends only on $C_{4}$, and we use $2\epsilon$ instead
of $\epsilon$ to simplify the later computations. Indeed, let $k$ be the
integer such that $2^{-k-1}<r/s\leq 2^{-k}$. Then by (14.53)

$\begin{split}m(B(y,r))&\leq
m(B(y,2^{-k}s))\leq\alpha^{k}m(B(y,s))\leq(2^{-k})^{\ln_{2}(1/\alpha)}m(B(y,s))\\\
&\leq\frac{1}{\alpha}\left(\frac{r}{s}\right)^{\ln_{2}(1/\alpha)}m(B(y,s)).\end{split}.$

The claim (14.54) follows by taking $\epsilon=\frac{1}{2}\ln_{2}(1/\alpha)>0$.

Let us use again $B_{s}$ for $B(y,s)$. The inequality (14.54) implies in
particular that

$\frac{r^{\epsilon}}{m(B_{r})}\geq
C\left(\frac{r}{s}\right)^{-\epsilon}\frac{s^{\epsilon}}{m(B_{s})}\qquad\text{
for }r\leq s\leq\delta(y),$

which proves that the function
$r\to\delta(y)^{1-\epsilon}r^{\epsilon}/m(B_{r})$ reaches all the values
between $\delta(y)/m(B_{\delta(y)})$ and $+\infty$. Moreover if $t$ is in the
given range, the values of $t$ that
satisfy$t=\delta(y)^{1-\epsilon}r^{\epsilon}/m(B_{r})$ are all the same up to
a harmless constant.

For the next step we prove weak $L^{q}$ estimates on the gradient of
$\mathfrak{g}^{\rho}$. Set $\widehat{\Omega}_{t}:=\\{x\in\Omega,\,|\nabla
g^{\rho}(x)|>t\\}$. Thanks to Lemma 14.45, for all $r\in(0,\delta(y)/2)$, we
have

$m(\widehat{\Omega}_{t})\leq
m(B_{r})+\frac{C}{t^{2}}\int_{r}^{\delta(y)}\frac{s^{2}}{m(B_{s})}\frac{ds}{s}.$

Then by (14.54)

$m(\widehat{\Omega}_{t})\leq
m(B_{r})+\frac{C}{t^{2}}\frac{1}{m(B_{r})}\int_{r}^{\delta(y)}\left(\frac{r}{s}\right)^{2\epsilon}s\,ds.$

Or, since we can always chose $\epsilon<1$,

(14.55) $m(\widehat{\Omega}_{t})\leq
m(B_{r})+\frac{C}{t^{2}}\frac{\delta(y)^{2(1-\epsilon)}r^{2\epsilon}}{m(B_{r})}.$

We aim to optimize the above expression in $r$. But we shall only care about
big values of $t$, so let us only consider $t\geq\delta(y)/m(B_{\delta(y)})$
for the moment. First assume that
$2^{-\epsilon}\delta(y)/m(B_{\delta(y)/2})\geq
t\geq\delta(y)/m(B_{\delta(y)})$. Then we choose $r=\delta(y)/2$ in (14.55),
and it is easy to see, using (H4) and (2.5) in particular, that

$m(\widehat{\Omega}_{t})\leq
Cm(B_{\delta(y)})^{-\frac{\epsilon}{d-\epsilon}}\left(\frac{\delta(y)}{t}\right)^{\frac{d}{d-\epsilon}},$

where $d=d_{m}>0$ is the (possibly large) exponent of (2.5). Notice that we
may always replace $d_{m}$ with a larger exponent in (2.5), so we may assume
that $d\geq 2\epsilon$, and this way the exponent
$-\frac{\epsilon}{d-\epsilon}$ is rather small and negative. We strive for the
same bound when $t\geq 2^{-\epsilon}\delta(y)m(B_{\delta(y)/2})$. We then take
$r$ such that

(14.56) $t=\delta(y)^{1-\epsilon}r^{\epsilon}/m(B_{r}),$

and we have seen in the previous paragraph that, even if we may have different
choices for $r$, they are all the same up to a constant. Using $r$ as in
(14.56) in (14.55), we obtain that

(14.57) $m(\widehat{\Omega}_{t})\lesssim
m(B_{r})=\frac{\delta(y)^{1-\epsilon}r^{\epsilon}}{t}.$

Yet, by (14.56) and (2.5),

$\begin{split}m(B_{r})&=\frac{\delta(y)}{t}\left(\frac{r}{\delta(y)}\right)^{\epsilon}\leq
C\frac{\delta(y)}{t}\left(\frac{m(B_{r})}{m(B_{\delta(y)})}\right)^{\frac{\epsilon}{d}},\end{split}$

or equivalently

$m(B_{r})^{\frac{d-\epsilon}{d}}\leq
C\,\frac{\delta(y)}{t}m(B_{\delta(y)})^{-\frac{\epsilon}{d}}.$

Using this bound on $m(B_{r})$ in (14.57), we obtain that

(14.58) $m(\widehat{\Omega}_{t})\leq Cm(B_{r})\leq
C^{\prime}m(B_{\delta(y)})^{-\frac{\epsilon}{d-\epsilon}}\left(\frac{\delta(y)}{t}\right)^{\frac{d}{d-\epsilon}}$

as desired.

We are ready to conclude. We write $q_{0}$ for $\frac{d}{d-\epsilon}>1$. The
bound (14.58) becomes

(14.59) $m(\widehat{\Omega}_{t})\leq
Cm(B_{\delta(y)})^{1-q_{0}}\left(\frac{\delta(y)}{t}\right)^{q_{0}}.$

We take $q=(1+q_{0})/2>1$. Then

$\begin{split}\frac{1}{q}\int_{B(y_{0},2\delta(y))}|\nabla\mathfrak{g}^{\rho}|^{q}\,dm&=\int_{0}^{\infty}t^{q-1}m(\widehat{\Omega}_{t}\cap
B(y_{0},2\delta(y)))dt\\\ &\leq
m(B(y_{0},2\delta(y)))\int_{0}^{\delta(y)/m(B_{\delta(y)})}t^{q-1}dt+\int_{\delta(y)/m(B_{\delta(y)})}^{\infty}t^{q-1}m(\widehat{\Omega}_{t})dt.\end{split}$

Then by (H4) and (14.59),

$\begin{split}\int_{B(y_{0},2\delta(y))}|\nabla\mathfrak{g}^{\rho}|^{q}\,dm&\lesssim
m(B_{\delta(y)})\left(\frac{\delta(y)}{m(B_{\delta(y)})}\right)^{q}+m(B_{\delta(y)})\left(\frac{\delta(y)}{m(B_{\delta(y)})}\right)^{q_{0}}\int_{\delta(y)/m(B_{\delta(y)})}^{\infty}t^{q-q_{0}-1}dt\\\
&\lesssim
m(B_{\delta(y)})\left(\frac{\delta(y)}{m(B_{\delta(y)})}\right)^{q}\end{split}$

since $q<q_{0}$. Lemma 14.51 follows. $\Box$

We are now ready for the big theorem.

######  Theorem 14.60.

There exists a non-negative function $g:\
\Omega\times\Omega\to\mathbb{R}\cup\\{+\infty\\}$ with the following
properties.

  1. (i)

For any $y\in\Omega$ and any function $\alpha\in
C_{0}^{\infty}(\mathbb{R}^{n})$ such that $\alpha\equiv 1$ in a neighborhood
of $y$

(14.61) $(1-\alpha)g(.,y)\in W_{0}.$

In particular, $g(.,y)\in W_{r}(\overline{\Omega}\setminus\\{y\\})\subset
L^{1}_{loc}(\overline{\Omega}\setminus\\{y\\},dm)$ and
$\operatorname{Tr}[g(.,y)]=0$ on $\Gamma$.

  2. (ii)

There exists $q>1$ that depends only on $C_{4}$ such that for every choice of
$y\in\Omega$,

(14.62) $\nabla g(.,y)\in L^{q}(B(y,\delta(y)),dm).$

  3. (iii)

For $y\in\Omega$ and $\varphi\in C^{\infty}_{0}(\Omega)$,

(14.63) $\int_{\Omega}A\nabla_{x}g(x,y)\cdot\nabla\varphi(x)dx=\varphi(y).$

In particular, $g(.,y)$ is a solution of $Lu=0$ in $\Omega\setminus\\{y\\}$.

In addition, the following bounds hold.

  1. (iv)

For $x,y\in\Omega$ such that $|x-y|\geq\delta(y)/10$,

(14.64) $0\leq g(x,y)\leq\frac{|x-y|^{2}}{m(B(y,|x-y|)\cap\Omega)},$

where $C>0$ depends on $C_{1}$ to $C_{6}$, $C_{A}$, and $n$.

  2. (v)

For $x,y\in\Omega$ such that $|x-y|\leq\delta(y)/2$,

(14.65) $c\int_{|x-y|}^{\delta(y)}\frac{s^{2}}{m(B(y,s))}\frac{ds}{s}\leq
g(x,y)\leq C\int_{|x-y|}^{\delta(y)}\frac{s^{2}}{m(B(y,s))}\frac{ds}{s},$

where $C>0$ depends on $C_{1}$ to $C_{6}$, $C_{A}$, and $n$; and where $c>0$
depends on $C_{4}$, $C_{6}$, $C_{A}$, and $n$.

  3. (vi)

For $r\in(0,\delta(y)/2)$ and $y\in\Omega$,

(14.66) $\int_{\Omega\setminus B(y,r)}|\nabla_{x}g(x,y)|^{2}dm(x)\leq
C\int_{|x-y|}^{\delta(y)}\frac{s^{2}}{m(B(y,s))}\frac{ds}{s},$

where $C>0$ depends again on $C_{1}$ to $C_{6}$, $C_{A}$, and $n$.

  4. (vii)

If $q>1$ is the exponent in (14.62)

(14.67)
$\left(\fint_{B(y,\delta(y))}|\nabla_{x}g(x,y)|^{q}dm(x)\right)^{\frac{1}{q}}\leq
C\frac{\delta(y)}{m(B(y,\delta(y))},$

where $C>0$ depends as usual on $C_{1}$ to $C_{6}$, $C_{A}$, and $n$.

Proof. As we shall see, we already have all the desired estimates on the
$\mathfrak{g}^{\rho}:=g^{\rho}(x,y)$; the proof will mainly consist in
choosing a right limit to those $\mathfrak{g}^{\rho}$.

We start with a standard exercise on compactness. For every compact set $K$ in
$\overline{\Omega}\setminus\\{y\\}$, Lemmas 14.5, 14.6, and 14.16 prove that
the set $F_{K}:=\big{\\{}\mathfrak{g}^{\rho}(x),\,;\,x\in K\text{ and
}0<\rho<\,\mathrm{dist}(y,K)/100\big{\\}}$ is bounded; then by Lemma 11.32 the
functions $\mathfrak{g}^{\rho}$, $\rho<\,\mathrm{dist}(y,K)/100$ are Hölder
continuous on $K$ (on a slightly smaller compact set), with uniform bounds. In
particular, for every compact set $K\subset\overline{\Omega}\setminus\\{y\\}$
the set
$A_{K}:=\big{\\{}\mathfrak{g}^{\rho},\,;\,0<\rho\leq\,\mathrm{dist}(y,K)/200\big{\\}}$
\- seen as a subset of the continuous functions on $K$ \- is equicontinuous.
Ascoli’s theorem entails that $A_{K}$ is relatively compact in $C^{0}(K)$,
that is we can find a sequence of radii $\rho$, that tends to $0$, such that
the corresponding $\mathfrak{g}^{\rho}$ converge, uniformly on $K$, to a
(continuous) function written $\mathfrak{g}_{K}$ . We take a sequence of
compacts sets $K_{i}$ such that $K_{i}\subset K_{i+1}$ and
$\bigcup_{i}K_{i}=\overline{\Omega}\setminus\\{y\\}$, and by a diagonal
process, we can find a sequence $(\rho_{\eta})_{\eta\in\mathbb{N}}$ and a
continuous function $\mathfrak{g}$ on $\overline{\Omega}\setminus\\{y\\}$ such
that $\rho_{\eta}\to 0$ and

(14.68) $\mathfrak{g}^{\rho_{\eta}}$ converges to $\mathfrak{g}$ uniformly on
every compact set of $\overline{\Omega}\setminus\\{y\\}$.

We shall use again the cut-off functions $\alpha_{r}$ defined in (14.28) and
their properties. Set $\widetilde{\alpha}_{r}(x)=\alpha_{r}(|x-y|)$; we want
to prove that the
$\\{\mathfrak{g}^{\rho_{\eta}}(1-\widetilde{\alpha}_{r})\\}_{\eta\in\mathbb{N}}$
form a Cauchy sequence in $W_{0}$. For any $r<\delta(y)/2$, define
$r_{1}\in(r,\delta(y))$ as the only value such that
$\gamma(r_{1})=\gamma(r)/2$; then for $\eta,\nu\in\mathbb{N}$,

$\displaystyle\int_{\Omega}|\nabla[(\mathfrak{g}^{\rho_{\eta}}-\mathfrak{g}^{\rho_{\nu}})$
$\displaystyle(1-\widetilde{\alpha}_{r})]|^{2}\,dm$ $\displaystyle\leq
2\int_{\Omega}|\nabla[\mathfrak{g}^{\rho_{\eta}}-\mathfrak{g}^{\rho_{\nu}}]|^{2}|1-\widetilde{\alpha}_{r}|^{2}\,dm+2\int_{\Omega}|\mathfrak{g}^{\rho_{\eta}}-\mathfrak{g}^{\rho_{\nu}}|^{2}|\nabla\widetilde{\alpha}_{r}|^{2}\,dm$
(14.69) $\displaystyle\leq
C\int_{\Omega}|\mathfrak{g}^{\rho_{\eta}}-\mathfrak{g}^{\rho_{\nu}}|^{2}|\nabla\widetilde{\alpha}_{r}|^{2}\,dm$

where, for the last line, we take $\eta,\nu$ big enough so that
$\rho_{\eta},\rho_{\nu}$ are way smaller than $r$ and we use the Cacciopoli-
type inequality (14.48). Since $\nabla\widetilde{\alpha}_{r}$ is supported in
$B_{r_{1}}\setminus B_{r}$, and the later is a compact set in
$\Omega\setminus\\{y\\}$, the convergence (14.68) forces the right-hand side
of (14) to tend to $0$. In addition, all the $\mathfrak{g}^{\rho}_{\eta}$ have
a vanishing trace, and so do the
$\mathfrak{g}^{\rho_{\eta}}(1-\widetilde{\alpha}_{r})$ (see Lemma 6.20). We
deduce that
$\\{\mathfrak{g}^{\rho_{\eta}}(1-\widetilde{\alpha}_{r})\\}_{\eta\in\mathbb{N}}$
is indeed a Cauchy sequence in $W_{0}$, so it converges strongly in $W_{0}$ to
a function $g^{(r)}$. By uniqueness of the limit,
$\mathfrak{g}^{(r)}=\mathfrak{g}(1-\widetilde{\alpha}_{r})$. In short, we
proved that for $0<r<\delta(y)/2$

(14.70) $\mathfrak{g}^{\rho_{\eta}}(1-\widetilde{\alpha}_{r})$ converges
strongly to $\mathfrak{g}(1-\widetilde{\alpha}_{r})$ in $W_{0}$.

Notice that $\mathfrak{g}$ has a gradient in
$L^{2}_{loc}(\Omega\setminus\\{y\\},dm)$ defined as

(14.71) $\nabla\mathfrak{g}(x)=\nabla[\mathfrak{g}(1-\widetilde{\alpha}_{r})]$
if $\widetilde{\alpha}_{r}(x)=0$.

We still need a last convergence, one that goes across the pole $\\{y\\}$.
Lemma 14.51 provides us with the uniform bound

$\left(\fint_{B_{\delta(y)}}|\nabla
g^{\rho_{\eta}}|^{q}\,dm\right)^{\frac{1}{q}}\leq
C\frac{\delta(y)}{m(B_{\delta(y)})}.$

So, up to a subsequence, the quantities $\nabla\mathfrak{g}^{\rho_{\eta}}$
converges weakly to a function $h\in L^{q}(B_{\delta(y)},dm)$. But since
$\nabla\mathfrak{g}^{\rho_{\eta}}$ already converges to $\nabla\mathfrak{g}$
in $L^{2}_{loc}(B_{\delta(y)}\setminus\\{y\\},dm)$, it forces
$\nabla\mathfrak{g}=h$ except maybe at the point $y$, but it has no importance
because $m(\\{y\\})=0$. To sum up,

(14.72) $\nabla\mathfrak{g}^{\rho_{\eta}}$ converges weakly to
$\nabla\mathfrak{g}$ in $L^{q}(B_{\delta(y)},dm)$.

Now let us show $(i)$–$(vii)$ of the theorem. For the first statement $(i)$,
let us start with the more likely situation where $\lim_{r\to
0}\gamma(r)=+\infty$. Since $\alpha=1$ near $y$, we can we can find $s>0$ such
that $|y-x|>s$ when $\alpha(x)\neq 1$. Choose $r$ so small that
$\gamma(r)>2\gamma(s)$; then for $x$ such that $\alpha(x)\neq 1$,
$\gamma(|x-y|)\leq\gamma(s)<\frac{1}{2}\gamma(r)$, so
$\widetilde{\alpha}_{r}(x)=\alpha_{r}(|x-y|)=0$ by (14.30). Because of this,
$(1-\alpha)\mathfrak{g}=(1-\alpha)(1-\widetilde{\alpha}_{r})\mathfrak{g}$.
This function lies in $W_{0}$, as needed, by (14.70) and Lemma 6.20.

In the other case when $\lim_{r\to 0}\gamma(r)<+\infty$, we are in the happy
situation where (14.47) says that
$\int_{\Omega}|\nabla\mathfrak{g}^{\rho}|^{2}dm\leq C$, with a constant that
depends on $y$, but not on $\rho$; then the almost everywhere pointwise limit
$\mathfrak{g}^{\rho}$ satisfies $\int_{\Omega}|\nabla\mathfrak{g}|^{2}dm\leq
C$ too, and its trace is still $0$ on $\Gamma$. Finally
$(1-\alpha)\mathfrak{g}$ does the same; see for instance the proof of Lemma
9.15 for the limit, and Lemma 10.6 for the product. This takes care of $(i)$.

The statement $(ii)$ is part of (14.72).

For the identity $(iii)$, we take $r_{0}$ so that
$\gamma(r_{0})=\frac{1}{2}\gamma(\delta(y)/2)$. We then write
$\varphi=\varphi_{1}+\varphi_{2}$ where
$\varphi_{1}=\varphi\widetilde{\alpha}_{r_{0}}$ and
$\varphi_{2}=\varphi(1-\widetilde{\alpha}_{r_{0}})$. The function
$\varphi_{1}$ is continuous and smooth enough for $\nabla\varphi_{1}$ to lie
in $L^{q^{\prime}}(B_{\delta(y)},dm)$, and so by (14.72) and then definition
(14.3),

(14.73)
$\int_{\Omega}A\nabla\mathfrak{g}\cdot\nabla\varphi_{1}\,dx=\lim_{\eta\to\infty}\int_{\Omega}\mathcal{A}\nabla\mathfrak{g}^{\rho_{\eta}}\cdot\nabla\varphi_{1}\,dm=\lim_{\eta\to\infty}\fint_{B_{\rho_{\eta}}}\varphi_{1}\,dm=\varphi_{1}(x)=\varphi(x).$

When $x\in B_{r_{0}}$, $\widetilde{\alpha}_{r_{0}}(x)=\alpha_{r_{0}}(|x-y|)=1$
by (14.29), and hence $\varphi_{2}(x)=0$. But otherwise
$\widetilde{\alpha}_{\delta(y)/2}(x)=\alpha_{\delta(y)/2}(|x-y|)=0$ because
$\gamma(|x-y|)\leq\gamma(r_{0})=\frac{1}{2}\gamma(\delta(y)/2)$, and by
(14.30). Hence $\mathfrak{g}(1-\alpha_{\delta(y)/2})=\mathfrak{g}$ on the
support of $\nabla\varphi_{2}$, and so

(14.74)
$\begin{split}\int_{\Omega}A\nabla\mathfrak{g}\cdot\nabla\varphi_{2}\,dx&=\int_{\Omega}A\nabla[\mathfrak{g}(1-\widetilde{\alpha}_{\delta(y)/2})]\cdot\nabla\varphi_{2}\,dx=\lim_{\eta\to\infty}\int_{\Omega}\mathcal{A}\nabla[\mathfrak{g}^{\rho_{\eta}}(1-\widetilde{\alpha}_{\delta(y)/2})]\cdot\nabla\varphi_{2}\,dm\\\
&=\lim_{\eta\to\infty}\int_{\Omega}\mathcal{A}\nabla\mathfrak{g}^{\rho_{\eta}}\cdot\nabla\varphi_{2}\,dm=\lim_{\eta\to\infty}\fint_{B_{\rho_{\eta}}}\varphi_{2}\,dm=\varphi_{2}(x)=0\end{split}$

where we used (14.70) for the second equality and then returned by the same
path. The combination of (14.73) and (14.74) infers $(iii)$.

The estimates given in $(iv)$ and $(v)$ are direct consequences of Lemmas
14.5, 14.6, 14.16, 14.34, and the convergence (14.68). The bound found in
$(vi)$ is due to Lemma 14.45 and (14.70), while $(vii)$ comes from Lemma 14.51
and (14.72). Theorem 14.60 follows. $\Box$

######  Remark 14.75.

Before stating the next result, let us comment a bit on Theorem 14.60. One can
easily see that $g(.,y)$ lies in
$L^{1}_{loc}(\overline{\Omega}\setminus\\{y\\},dm)$, since the latter is
bigger than the space of continuous functions on
$\overline{\Omega}\setminus\\{y\\}$ (and $g(.,y)$ is continuous on
$\overline{\Omega}\setminus\\{y\\}$ thanks to (14.68)). However, we said
nothing about the integration of $g(.,y)$ on a neighborhood of $\\{y\\}$. The
fact that $g(.,y)$ can be integrated over a bounded region that covers
$\\{y\\}$ is a simple consequence of (14.65). Indeed, if $y\in\Omega$ and
$r\leq\delta(y)/2$, first observe that

$\int_{B(y,r)}g(x,y)\,dm(x)\leq
C\int_{B(y,r)}\int_{|x-y|}^{\delta(y)}\frac{s^{2}}{m(B(y,s))}\frac{ds}{s}\,dm(x)\leq
C\int_{t=0}^{r}\int_{s=t}^{\delta(y)}\frac{m(B(y,t))}{m(B(y,s))}\,s\,ds\frac{dt}{t}.$

Let $\epsilon\in(0,1)$ be the constant in (14.54); then

$\int_{B(y,r)}g(x,y)\,dm(x)\leq
C\int_{0}^{r}\int_{t}^{\delta(y)}\,s^{1-2\epsilon}ds\,t^{2\epsilon-1}dt\leq
C\delta(y)^{2-2\epsilon}r^{2\epsilon}.$

In addition, by (14.64)

$\int_{B(y,\delta(y))\setminus B(y,\delta(y)/2)}g(x,y)\,dm\leq
Cm(B(y,\delta(y))\sup_{x\in B(y,\delta(y))\setminus
B(y,\delta(y)/2)}g(x,y)\leq C\delta(y)^{2}.$

The combination of the last two estimates implies that

(14.76) $\int_{B(y,\delta(y))}g(x,y)\,dm\leq C\delta(y)^{2}.$

Due to (14.68), the functions $g^{\rho_{\eta}}(.,y)$ converges pointwise a.e.
to $g(.,y)$ on $B(y,\delta(y))$. So by the Lebesgue domination theorem (and
the fact that bounds above are also valid for the $g^{\rho_{\eta}}$), we even
have

$\lim_{\eta\to\infty}\int_{B(y,\delta(y))}|g^{\rho_{\eta}}(x,y)-g(x,y)|\,dm(x)=0.$

Together with (14.68), we proved that

(14.77) the functions $g^{\rho_{\eta}}(.,y)$ converge to $g(.,y)$ in
$L^{1}_{loc}(\overline{\Omega})$.

For the next lemma, we need some additional notation. We write $A^{T}$ for the
transpose matrix of $A$, i.e. $(A^{T})_{ij}(x)=A_{ji}(x)$ for all $1\leq
i,j\leq n$ and $x\in\Omega$. Obviously, $A^{T}$ satisfies the ellipticity and
boundedness conditions (11.1)–(11.2) with the same constant as $A$. The
elliptic operator $L_{T}:=-\mathop{\operatorname{div}}A^{T}\nabla$ enjoys the
very same properties as $L$, in particular, Theorem 14.60 yields the existence
of $g^{T}:\,\Omega\cap\Omega\to\mathbb{R}\cup\\{+\infty\\}$ with the same
properties as $g$ (except for (14.63), where $A$ is replaced by $A^{T}$).

######  Lemma 14.78.

With the notation above,

(14.79) $g(x,y)=g_{T}(x,y)\quad\text{ for }x,y\in\Omega.$

In particular, the functions $x\to g(y,x)$ satisfy the estimates in Theorem
14.60.

Proof. The result is the same as the one of [GW, Theorem 1.3] (or Lemma 10.6
in [DFM2]). Yet, the limits we took in the proof of Theorem 14.60 is a bit
different to the one in [GW] and [DFM2]. So our result deserves a proof.

Actually, the convergence property (14.72) will make the proof very easy for
us. Let $x,y\in\Omega$ be such that $x\neq y$. By our construction of the
Green functions, there exist two sequences $(\rho_{\eta})_{\eta\in\mathbb{N}}$
and $(\sigma_{\nu})_{\nu\in\mathbb{N}}$ such that $\rho_{\eta},\sigma_{\nu}\to
0$ and

(14.80) $\mathfrak{g}^{\rho_{\eta}}$ converges to $\mathfrak{g}$ uniformly on
any compact set of $\overline{\Omega}\setminus\\{y\\}$

and

(14.81) $\mathfrak{g}_{T}^{\sigma_{\nu}}$ converges to $\mathfrak{g}_{T}$
uniformly on any compact set of $\overline{\Omega}\setminus\\{x\\}$.

Using (14.2) and (14.4) for both $g(.,y)$ and $g_{T}(.,x)$, we see that for
any $\eta,\nu\in\mathbb{N}$

(14.82) $\int_{\Omega}\mathcal{A}\nabla g^{\rho_{\eta}}(z,y)\cdot\nabla
g_{T}^{\sigma_{\nu}}(z,x)\,dm(z)=\fint_{B(y,\rho_{\eta})}g_{T}^{\sigma_{\nu}}(z,x)\,dm(z)=\fint_{B(x,\sigma_{\nu})}g^{\rho_{\eta}}(z,y)\,dm(z)$

We use the uniform convergence of $\mathfrak{g}_{T}^{\sigma_{\nu}}$ on
$\overline{B(y,|x-y|/2)}\subset\Omega\setminus\\{x\\}$ and the uniform
convergence of $g(.,y)^{\rho_{\eta}}$ on
$\overline{B(x,|x-y|/2)}\subset\Omega\setminus\\{y\\}$ given by
(14.80)–(14.81) in the last equality of (14.82). We get that
$g_{T}(y,x)=g(x,y)$ as desired. $\Box$

######  Lemma 14.83.

The Green function satisfies

(14.84) $g(x,y)\leq C\delta(x)^{\alpha}\frac{|x-y|^{2-\alpha}}{m(B(x,|x-y|))}\
\ \text{ for $x,y\in\Omega$ such that }|x-y|\geq 4\delta(x),$

where $C>0$ and $\alpha>0$ depend only on $C_{1}$ to $C_{6}$, $C_{A}$, and
$n$.

Proof. See the proof of [DFM2, Lemma 10.9]. The arguments are based on the
pointwise bounds (14.64) and the Hölder regularity at the boundary (Lemma
11.32). Actually, the coefficient $\alpha$ is the one of Lemma 11.32. $\Box$

The next result that we wanted is the representation of solutions by Green
functions. More precisely, we wanted to take a smooth function $f\in
C^{\infty}_{0}(\Omega)$ and construct $u(x)$ for $x\in\Omega$ as

(14.85) $u(x)=\int_{\Omega}g(x,y)f(y)dm(y).$

We have seen in Remark 14.75 that $g(.,y)$ lies in
$L^{1}_{loc}(\overline{\Omega},dm)$. Moreover, due to Lemma 14.78, we also
have that $g(x,.)$ is in $L^{1}_{loc}(\overline{\Omega},dm)$. Yet, in the case
of a general weights $w$, we do not know if $g(x,.)$ lies in the unweighted
space $L^{1}_{loc}(\overline{\Omega})$. That is why, in the definition
(14.85), the function $u$ has to be defined as an integral over the measure
$m$.

In doing so, the formal identity satisfied by $u$ is not $Lu=f$ but $Lu=fw$,
where $w$ is the weight used to define the measure $m$. Another way to see it,
maybe more relevant, is to say that $w^{-1}Lu=f$. That is, we are solving the
Dirichlet problem $\mathcal{L}u=f$ for an elliptic operator
$\mathcal{L}=-w^{-1}\mathop{\operatorname{div}}[\mathcal{A}w\nabla]$ where
$\mathcal{A}$ satisfies the classical elliptic conditions (11.3)–(11.4).

At last, by using $L$ instead of $\mathcal{L}:=w^{-1}L$, we are somehow
linking the measure $m$ to the plain Lebesgue measure on $\mathbb{R}^{n}$. So
some readers may want to use $\mathcal{L}$ all the time. The theory is
identical to what we have done until now, since we only worked with solutions
to $Lu=0$ before the Green functions, and $Lu=0$ is obviously equivalent to
$\mathcal{L}u=0$.

We expect from the Green representation of solutions that the function
$u=u_{f}$ constructed in (14.85) lies in $W_{0}$ and is a weak solution to
$Lu=fw$ in the sense that

(14.86) $\int_{\Omega}A\nabla u\cdot\nabla\varphi=\int_{\Omega}f\varphi
w=\int_{\Omega}f\varphi\,dm\ \ \text{ for every }\varphi\in
C^{\infty}_{0}(\mathbb{R}^{n}).$

Our assumptions (H1)–(H6) are enough to have (14.86) and the fact that
$\operatorname{Tr}u\equiv 0$ (for the former, we still need to be careful
about our weird definition of the gradient, and for the later, just use Lemma
14.83). However, we did not succeed to prove that $u\in W$. That is why our
next results will be restricted to the case where the weight $w$ is nice
enough, that is when (H6’) is satisfied instead of (H6)

######  Lemma 14.87.

Assume that $(\Omega,m,\mu)$ satisfies (H1)–(H5) and (H6’). Let
$g:\Omega\times\Omega\to\mathbb{R}\cup\\{+\infty\\}$ be the non-negative Green
function constructed in Lemma 14.60. Take $f\in C^{\infty}_{0}(\Omega)$ and
construct $u(x)$ for $x\in\Omega$ as

(14.88) $u(x)=\int_{\Omega}g(x,y)f(y)dm(y).$

Then $u$ belongs to $W_{0}$ and is the solution to $Lu=fw$ (given by Lemma
12.2) in the sense that

(14.89) $\int_{\Omega}A\nabla u\cdot\nabla\varphi=\int_{\Omega}f\varphi
w=\int_{\Omega}f\varphi\,dm\ \ \text{ for every }\varphi\in W_{0}.$

######  Remark 14.90.

In (14.88) and (14.89), we can replace $dm$ by the classical $n$-dimensional
Lebesgue measure $dx$.

Proof. The proof is the same as the one of Lemma 10.7 in [DFM2]. It relies on
the fact that the solutions to $Lu=fw$ are continuous inside $\Omega$, because
as long as we consider inside estimates, (H6’) implies that the classical
unweighted elliptic theory can be applied. See Theorem 8.22 in [GT] for the
theorem in the classical case.

We also need the fact that the approximations $g^{\rho}(.,y)$ converges in
$L^{1}_{loc}(\Omega)$ to $g(.,y)$. Under (H6’), this result is a consequence
of the weak $L^{q}$ convergence of the gradients and the $L^{1}$-Poincaré
inequality for inside balls, the latter is true because inside estimates works
here exactly like the classical unweighted case. With only (H6), we can only
use $L^{2}$ \- or $L^{2-\epsilon}$ \- Poincaré inequalities. $\Box$

######  Lemma 14.91.

Assume that $(\Omega,m,\mu)$ satisfies (H1)–(H5) and (H6’). There exists a
unique function $g:\Omega\times\Omega\mapsto\mathbb{R}\cup\\{+\infty\\}$ such
that $g(x,.)$ is continuous on $\Omega\setminus\\{x\\}$ and locally integrable
in $\Omega$ for every $x\in\Omega$, and such that for every $f\in
C^{\infty}_{0}(\Omega)$ the function $u$ given by

(14.92) $u(x):=\int_{\Omega}g(x,y)f(y)dm(y)$

belongs to $W_{0}$ and is a solution of $Lu=f$ in the sense that

(14.93) $\int_{\Omega}A\nabla
u\cdot\nabla\varphi=\int_{\Omega}\mathcal{A}\nabla
u\cdot\nabla\varphi\,dm=\int_{\Omega}f\varphi dm\qquad\text{for every
}\varphi\in W_{0}.$

Proof. See the proof of Lemma 10.8 in [DFM2]. In short, the existence is due
to Theorem 14.60, and Lemmas 14.78 and 14.87, while the uniqueness of $g$
comes from the uniqueness of $u\in W_{0}$ satisfying $Lu=fw$, and the latter
is due to Lemma 12.2. $\Box$

##  15\. Comparison principle

First, let us state the non-degeneracy of the harmonic measure.

######  Lemma 15.1.

Let $\alpha>1$, $B:=B(x_{0},r)$ be a ball. Take $X_{0}\in\Omega$ be any
corkscrew point associated to $x_{0}$ and $r$ given by the assumption (H1).
Then

(15.2) $\omega^{X}(B\cap\Gamma)\geq C_{\alpha}^{-1}\ \ \ \text{ for
}X\in\frac{1}{\alpha}B$

and

(15.3) $\omega^{X}(B\cap\Gamma)\geq C^{-1}_{\alpha}\ \ \ \text{ for }X\in
B(X_{0},\delta(X_{0})/\alpha),$ (15.4) $\omega^{X}(\Gamma\setminus
B)+\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\geq C_{\alpha}^{-1}\ \ \ \text{ for
}X\in\Omega\setminus\alpha B,$

and

(15.5) $\omega^{X}(\Gamma\setminus
B)+\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\geq C^{-1}_{\alpha}\ \ \ \text{ for
}X\in B(X_{0},\delta(X_{0})/\alpha),$

where in the four estimates, $C_{\alpha}$ depends on $C_{1}$ to $C_{6}$,
$C_{A}$, $n$, and $\alpha$.

######  Remark 15.6.

The estimates (15.2)–(15.3) are classical results about the non-degeneracy of
the harmonic measure. However, the reader can be at first surprised by the
appearance of the Green functions in (15.4)–(15.5). The terms that involves
the Green functions are yet necessary. Indeed, none of our assumptions stops
the boundary $\Gamma$ to be a bounded and $\Omega$ to be still infinite.
Simply take for instance $\Omega=\mathbb{R}^{n}\setminus\\{0\\}$ and
$\Gamma=\\{0\\}$ with appropriate measure $\mu$ and $m$. Under those
conditions, we can actually have $\Gamma\setminus B=\emptyset$, which leads to
$\omega^{X}(\Gamma\setminus B)=0$ for all $X\in\Omega$.

We claim - without proof but we pretend that there are real difficulties to it
- that the estimates

(15.7) $\omega^{X}(\Gamma\setminus B)\geq C_{\alpha}^{-1}\ \ \ \text{ for
}X\in\Omega\setminus\alpha B$

holds whenever $\Omega$ is bounded (since the $\Omega\setminus\alpha B$ would
be empty when $\Gamma\setminus B=\emptyset$) or when we can find a point in
$\Gamma$ close to $\alpha B$ yet outside of $\alpha B$, i.e. whenever
$[100B\setminus\alpha B]\cap\Gamma\neq\emptyset$.

At last, the estimate (15.5) is given to make it similar to (15.4). The
harmonic measure is actually unnecessary in (15.5).

Proof. The proof of (15.2)–(15.3) is the same as the one of [DFM2, Lemma
11.10], and relies on the Hölder continuity at the boundary (Lemma 11.32), the
existence of Harnack chains (Proposition 2.18), and the Harnack inequality
(Lemma 11.35).

Rapidly, the Hölder inequality at the boundary will imply that
$\omega^{X}(B\cap\Gamma)$ is bigger than $1/2$ for any points “close” to
$\Gamma\setminus\frac{1}{\alpha}B$. Then we use Harnack chains of balls to
link any point in $\frac{1}{\alpha}B$ to one of the previous points, and the
Harnack inequality repeatedly on the balls of the Harnack chain.

Let us make the proof of (15.4)–(15.5). First, let us prove (15.5). Thanks to
(14.65), we have

(15.8) $g(X,X_{0})\geq
C^{-1}\int_{\delta(X_{0})/2}^{\delta(X_{0})}\frac{s^{2}}{m(B(X_{0},s))}\frac{ds}{s}\qquad\text{
for all }X\in B(X_{0},\delta(X_{0})/2).$

Using the doubling property, since $\delta(X_{0})\approx r$, we have
$m(B(X_{0},s))/s^{2}\approx m(B\cap\Omega)/r^{2}$ for all
$s\in(\delta(X_{0})/2,\delta(X_{0}))$. The estimate (15.8) becomes

(15.9) $\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\geq C^{-1}\qquad\text{ for all
}X\in B(X_{0},\delta(X_{0})/2).$

We let the reader check that $\Omega\setminus\\{X_{0}\\}$, obtained from
$\Omega$ by removing a single point, will still satisfy (H1)–(H2), maybe with
some constant $C^{\prime}_{1},C^{\prime}_{2}$ smaller than $C_{1},C_{2}$.
Indeed, if $X_{0}$ is close to a Corkscrew point for $\Omega$ associated to
$(x,r)$, then the Corkscrew point for $\Omega$ associated to $(x,C_{1}^{-1}r)$
will be far from $X_{0}$ and so Corkscrew point for
$\Omega\setminus\\{X_{0}\\}$ with a constant $C^{\prime}_{1}=(C_{1})^{2}$. The
Harnack chains in $\Omega\setminus\\{X_{0}\\}$ are the same as in $\Omega$,
except if they got close to $X_{0}$. In this case, we consider smaller balls,
and we avoid $X_{0}$ by taking balls in $B(X_{0},\delta(X_{0})/2)\setminus
B(X_{0},\delta(X_{0})/4)\subset\Omega$. As a consequence, we can link any
point from

$[\\{X\in\Omega,\,\,\mathrm{dist}(X,\Gamma)>\eta r\\}\cap 4B]\setminus
B(X_{0},\delta(X_{0})/4)$

to a point in $B(X_{0},\delta(X_{0})/2)\setminus B(X_{0},\delta(X_{0})/4)$ by
a Harnack chain of ball with uniformly finite length (the length of the chain
is bounded by a constant that depends only on $\eta$ and $n$). For the sequel,
we write $\Omega_{\eta}$ for $\\{X\in\Omega,\,\,\mathrm{dist}(X,\Gamma)>\eta
r\\}$. We use the fact that $g(.,X_{0})$ is a solution to $Lu=0$ on
$\Omega\setminus\\{X_{0}\\}$ and the Harnack inequality (Lemma 11.35) on each
balls of those Harnack chain to improve (15.9) into

(15.10) $\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\geq C_{\eta}^{-1}\qquad\text{
for all }X\in\Omega_{\eta}\cap 4B.$

In particular, if $\eta=(\alpha-1)/C_{1}$, we get (15.5) without the harmonic
measure, so we get (15.5) since $\omega^{X}$ is non-negative.

The proof of (15.4) needs additional computations. We write $h$ for the smooth
function in $C^{\infty}_{0}((\alpha+1)B/2))$ satisfying $0\leq h\leq 1$ and
$h\equiv 1$ on $B$. We set $u_{h}\in W$ for the solution to $Lu_{h}=0$ with
$\operatorname{Tr}u_{h}=1-\operatorname{Tr}h$. By positivity of the harmonic
measure,

(15.11) $\omega^{X}(\Gamma\setminus B)\geq
u_{h}(X)\geq\omega^{X}(\Gamma\setminus(\alpha+1)B/2)\geq 0\qquad\text{ for
}X\in\Omega.$

We prefer $u_{h}$ to $\omega^{X}(\Gamma\setminus B)$ because $u_{h}$ is in
$W$, which makes him suitable for the use of Lemma 14.33 (our maximum
principle). The first estimate that we state comes from (15.10) without
difficulty:

(15.12) $u_{h}(X)+\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\geq
C_{\eta}^{-1}\qquad\text{ for all }X\in\Omega_{\eta}\cap[4B\setminus\alpha
B].$

We want the estimate on the larger set $\Omega\cap[4B\setminus\alpha B]$, so
we need to prove that (15.12) is also true when $X$ is close to $\Gamma$. Let
$\eta>0$ be small and to be fixed. Let
$X\in[\Omega\setminus\Omega_{\eta}]\cap[4B\setminus\alpha B]$. Take
$x\in\Gamma$ so that $|X-x|\leq\eta r$, which is possible since
$X\in\Omega\setminus\Omega_{\eta}$. Due to the fact that $X$ is also in
$2B\setminus\alpha B$, it forces $x$ to be in $\Gamma\setminus(\alpha-\eta)B$.
We chose then $\eta=(\alpha-1)/8$, which makes
$x\in\Gamma\setminus\frac{7\alpha+1}{8}B$. Consequently, for
$X\in[\Omega\setminus\Omega_{\eta}]\cap[4B\setminus\alpha B]$,

$u_{h}(X)\geq\omega^{X}(\Gamma\setminus(\alpha+1)B/2)\geq\omega^{X}(B(x,3\eta
r)\geq C^{-1}$

by (15.11), the construction of $\eta$ and $x$, and (15.2). The combination of
the last estimate with (15.12) entails

(15.13) $u_{h}(X)+\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\geq
C_{\alpha}^{-1}\qquad\text{ for all }X\in\Omega\cap[4B\setminus\alpha B]$

since Green functions are non-negative. We finish the proof with the maximum
principle given by Lemma 14.33, which will become our favorite tool for the
section. Indeed, we define $v$ as

$v:=u_{h}(X)+\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})-C_{\alpha}^{-1}$

where $C_{\alpha}^{-1}$ is the constant in the right-hand side of (15.13), and
we aim to apply Lemma (14.33) for the solution $v$ with the sets
$E=\mathbb{R}^{n}\setminus\alpha B$ and $F=4B\setminus\alpha B$. Recall that
the term $u_{h}(X)$ lies in $W$. Together with (14.67), we deduce that
assumption (i) of Lemma 14.33 is true. The other assumptions required by the
lemma are given by (15.13) and the fact that
$\operatorname{Tr}v=1-C_{\alpha}^{-1}>0$ on $\Gamma\setminus\alpha B$. We
deduce that $v\geq 0$ on $E$, which is exactly the desired estimate (15.4).
The lemma follows. $\Box$

If $B$ is a ball centered on the boundary $\Gamma$, we bound the values in
$B\cap\Omega$ of a solution $u$ (to $Lu=0$ in $KB\cap\Omega$) by the value of
$u$ at a Corkscrew point associated to the ball $B$.

######  Lemma 15.14.

There exists $K:=K(C_{1},C_{2},n)$ such that the following holds.

Let $B=B(x_{0},r)$ be a ball centered on $\Gamma$ and let $X_{0}$ be a
Corkscrew point associated to $x_{0}$ and $r$ given by (H1). Let $u\in
W_{r}(KB\cap\overline{\Omega})$ be a non-negative, non identically zero,
solution of $Lu=0$ in $KB\cap\Omega$, such that $\operatorname{Tr}u\equiv 0$
on $KB\cap\Gamma$. Then

(15.15) $u(X)\leq Cu(X_{0})\ \ \ \ \text{ for }X\in B\cap\Omega,$

where $C>0$ depends on $C_{1}$ to $C_{6}$, $C_{A}$, and $n$.

Proof. We get inspiration from the proof of [KJ, Lemma 4.4] (see also [DFM2,
Lemma 11.8]). Lemma 11.8 in [DFM2] deals with balls centered at the boundary,
and $K=2$. However, in [DFM2], the connectedness is not a issue, while we need
to be careful here. Indeed, taking the universal constant $K=2$ in Lemma 15.14
is not possible, since nothing garantees that we can link 2 points in
$B\cap\Omega$ by a path that stays in $2B$.

We solve this problem by taking the tent sets constructed in Section 5, which
can be seen as connected substitute of the sets $B\cap\Omega$. Then we use the
property (5.15) to conclude.

First, let us recall the following fact. Let $x\in\Gamma$ and $s>0$ such that
$\operatorname{Tr}u\equiv 0$ on $B(x,s)\cap\Gamma$. Then the Hölder continuity
of solutions given by Lemma 11.31 proves the existence of $\epsilon>0$ (that
depends on $C_{1}$ to $C_{6}$, $C_{A}$, and $n$) such that

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

Without loss of generality, we can choose $\epsilon<1/1000C_{1}<\frac{1}{2}$.

A rough idea of the proof of (15.15) is that $u(X)$ should not be near the
maximum of $u$ when $X$ lies close to $B\cap\Gamma$, because of (15.16). 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 $X\in
B\cap\Gamma$ 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 $B$. We will prove this latter fact by contradiction: we show that
if $\sup_{B}u$ exceeds a certain bound, then we can construct a sequence of
points $X_{k}\in\frac{K}{2}B$, where $K$ is large enough, 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.

As said in the beginning of the proof, the quantities $\lambda B\cap\Omega$
lack connectedness, and it will be more convenient to work with a tent set
$T_{2Q^{*}}$, which has all the desired connectedness by Lemma 5.23. The cube
$Q^{*}$ and the constant $K$ are defined as follow. Let $k\in\mathbb{Z}$ be
such that $2^{-k-2}\leq r\leq 2^{-k-1}$, and we write $Q$ for the unique cube
in $\mathbb{D}_{k}$ containing $x$. Notice that $2Q\supset B\cap\Gamma$, but
$T_{2Q}$ is not necessarily bigger than $B$, and so we take $Q^{*}$ the first
ancestor of $Q$ such that $\,\mathrm{dist}(T_{2Q},T_{2Q^{*}}^{c})>\ell(Q)$.
Check that the difference of generations between $Q$ and $Q^{*}$ is uniformly
finite, and so combined with (5.15), we obtain that we can find $K$ that
depends only on $n$, $c_{1}$, and $C_{2}$ such that

(15.17) $2B\cap\Omega\subset T_{2Q^{*}}\subset\frac{1}{2}B^{*}:=\frac{K}{2}B.$

We can link any couple of points in $T_{2Q^{*}}$ by a chain of balls $B_{i}$
that satisfies $2B_{i}\subset 2B^{*}\cap\Omega$. The proof of this fact is
similar to the proof of fact that $T_{2Q^{*}}$ satisfies the chain condition
$C(\kappa,M)$ for some $\kappa,M$ (see Lemma 5.23) and thus will be omitted.
Therefore, the fact that $u(X)>0$ somewhere, that $T_{2Q^{*}}$ is connected,
and the Harnack inequality (Lemma 11.35), maybe applied a few times, yield
$u(X_{0})>0$. 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 $Y\in
T_{2Q^{*}}$,

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

where $\epsilon$ comes from (15.16) and the constant $M$ depends only upon
$n$, $C_{1}$ to $C_{6}$, and $C_{A}$. We will prove the claim by induction.
The base case is given by the following. We want to show the existence of
$M_{1}\geq 1$ such that

(15.19) $u(Y)\leq M_{1}\ \ \ \ \ \text{ for every }Y\in T_{2Q^{*}}\text{ such
that }\delta(Y)\geq\epsilon^{2}r.$

Indeed, if $Y\in T_{2Q^{*}}$ satisfies $\delta(Y)\geq\epsilon^{2}r$,
Proposition 2.18 implies that we can link $Y$ to $X_{0}$ by a chain of balls
that stay away from the boundary and with length uniformly bounded by
$C(\epsilon)$. We can construct the chain such that it stays also far from the
boundary of $B^{*}$111This fact is true because by construction of
$T_{2Q^{*}}$, one can see that the center of the balls constituting the chain
can be taken in $T_{2Q^{*}}$, and (15.17) let us a bit of freedom, but if may
be easier for the reader to think that this statement would be also true by
taking a larger $K$. Together with the Harnack inequality (Lemma 11.35), we
obtain (15.19), and hence (15.18) for $N=1,2$ as long as $M$ is chosen bigger
than $M_{1}$.

Now, let $Y\in T_{2Q^{*}}$ such that $\delta(Y)\leq\epsilon^{2}r$. By
construction of $T_{2Q^{*}}$, $Y$ belongs to some $\gamma^{*}_{Q^{*}}(z)$ for
some $z\in 2Q^{*}$. We take $Z$ a Corkscrew point associated to $z$ and
$C_{1}\delta(Y)/\epsilon$. Since $\epsilon<C_{1}^{-1}$, $Z\in B(z,r)$, and so
$Z$ stays in $T_{2Q^{*}}$. In addition, by construction,
$\delta(Z)\geq\delta(Y)/\epsilon$ and $|Z-Y|\leq\delta(Y)/\epsilon^{2}$; these
two estimates can be combined to Proposition 2.18 (existence of Harnack
chains, as before the chain can stay far from $\partial B^{*}$) and Lemma
11.35 (Harnack inequality) to get we the existence of $M_{2}\geq 1$ such that
$u(Y)\leq M_{2}u(Z)$. So we just proved that

(15.20) $\begin{array}[]{l}\text{for any $Y\in T_{Q^{*}}$ such that
$\delta(Y)\leq\epsilon^{-2}r$, }\\\ \qquad\text{there exists $Z\in T_{2Q^{*}}$
such that $\delta(Z)\geq\delta(Y)/\epsilon$ and $u(Y)\leq
M_{2}u(Z)$.}\end{array}$

We turn to the main induction step. Set $M=\max\\{M_{1},M_{2}\\}\geq 1$ and
let $N\geq 2$ be given. Assume, by induction hypothesis, that for any $Z\in
T_{2Q^{*}}$ satisfying $\delta(Z)\geq\epsilon^{N}\ell(Q)$, we have $u(Z)\leq
M^{N}$. Let $Y\in T_{2Q^{*}}$ be such that
$\delta(Y)\geq\epsilon^{N+1}\ell(Q)$ . The assertion (15.20) yields the
existence of $Z\in T_{2Q^{*}}$ such that
$\delta(Z)\geq\delta(Y)/\epsilon\geq\epsilon^{N}\ell(Q)$ and $u(Y)\leq
M_{2}u(Z)\leq Mu(Z)$. By the induction hypothesis, $u(Y)\leq M^{N+1}$. This
completes our induction step, and the proof of (15.18) for every $N\geq 1$.

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

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

So we assume that

(15.22) 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 $k\geq 1$,

(15.23) there exists $X_{k}\in T_{2Q^{*}}$ such that $u(X_{k})>M^{i+2+k}$ and
$\,\mathrm{dist}(X_{k},B)\leq(1-2^{1-k})r$.

The base step of the induction is given by (15.22) and we want to do the
induction step. Let $k\geq 1$ be given and assume that (15.23) holds. From the
contraposition of (15.18), 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,

(15.24)
$\,\mathrm{dist}(x_{k},B)\leq|x_{k}-X_{k}|+\,\mathrm{dist}(X_{k},B)\leq(1-2^{1-k})r+\epsilon^{i+2+k}r$

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

(15.25) $|x_{k}-x_{0}|\leq(1-2^{1-k}-2^{-2-k})r.$

Now, due to (15.16), we can find $X_{k+1}\in B(x_{k},\epsilon^{2+k}r)$ such
that

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

The induction step will be complete if we can prove that
$\,\mathrm{dist}(X_{k+1},T_{2Q})\leq(1-2^{-k})r$. Indeed,

(15.27)
$\begin{split}\,\mathrm{dist}(X_{k+1},B)&\leq|X_{k+1}-x_{k}|+\,\mathrm{dist}(x_{k},B)\leq\epsilon^{2+k}r+(1-2^{1-k}-2^{-2-k})r\\\
&\leq(1-2^{-k})r\end{split}$

by (15.25) and because $\epsilon\leq\frac{1}{2}$.

Let us sum up. We assumed the existence of $X_{1}\in B$ such that
$u(X_{1})>M^{\prime}$ and we end up with (15.23), that is a sequence $X_{k}$
of values in $2B$ such that $u(X_{k})$ increases to $+\infty$. Up to a
subsequence, we can thus find a point in $\overline{2B}\subset B^{*}$ where
$u$ is not continuous, which contradicts Lemma 11.32. Hence $u(X)\leq
M^{\prime}=M^{\prime}u(X_{0})$ for $X\in B$. Lemma 15.14 follows. $\Box$

We can now compare the harmonic measure with the Green function, that can be
seen as a weak version of the comparison principle.

######  Lemma 15.28.

Let $B:=B(x_{0},r)$ be a ball centered on $\Gamma$. Let $X_{0}$ is a corkscrew
point associated to $x_{0}$ and $r$. Then one has

(15.29)
$C^{-1}\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\leq\omega^{X}(B\cap\Gamma)\leq
C\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\ \ \ \text{ for }X\in\Omega\setminus
2B,$

and

(15.30) $\omega^{X}(\Gamma\setminus\frac{5}{4}B)\leq
C\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\ \ \ \text{ for
}X\in[B\cap\Omega]\setminus B(X_{0},\delta(X_{0})/4),$

where $C>0$ depends only upon $n$, $C_{1}$ to $C_{6}$, and $C_{A}$.

######  Remark 15.31.

The bound (15.30) may look a bit artificial. There is nothing deep about the
constant $\frac{5}{4}$ in the left-hand side. We could have used $2$ instead,
and obtain a statement which looks a little weaker but is actually equivalent
(we leave the proof of this to the reader); we simply reproduced $\frac{5}{4}$
in the form given by our general comparison principle (Theorem 15.64).

Observe also that we do not necessarily have the lower bound in (15.30). See
Remark 15.6.

Proof. This lemma is the analogue of [DFM2, Lemmas 11.9 and 11.11].

First, we quickly prove the first inequality in (15.29). The upper bound
(14.64) for the Green function, together with (H4), implies that

(15.32) $0\leq\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\leq C\qquad\text{ for
}X\in B(X_{0},\delta(X_{0})/2)\setminus B(X_{0},\delta(X_{0})/4).$

As in the proof of Lemma 15.1, we take $h\in C^{\infty}_{0}(B)$ such that
$0\leq h\leq 1$ and $h\equiv 1$ on $\frac{1}{2}B$. We set $u_{h}\in W$ for the
solution to $Lu_{h}=0$ with $\operatorname{Tr}u_{h}=\operatorname{Tr}h$. By
the positivity of the harmonic measure,

(15.33) $\omega^{X}(\Gamma\cap\frac{1}{2}B)\leq
u_{h}(X)\leq\omega^{X}(\Gamma\cap B)\qquad\text{ for }X\in\Omega.$

We combine (15.33) with the non-degeneracy of the harmonic measure (15.3) to
get that

(15.34) $u_{h}\geq C^{-1}\qquad\text{ for }X\in B(X_{0},\delta(X_{0})/2).$

The estimates (15.32) and (15.34) easily infer the existence of a constant
$\kappa$ such that

$v(X):=\kappa u_{h}(X)-\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\geq
0\qquad\text{ for }X\in B(X_{0},\delta(X_{0})/2)\setminus
B(X_{0},\delta(X_{0})/4).$

The assumptions of Lemma 14.33 for the function $v$ and the sets
$E=\mathbb{R}^{n}\setminus B(X_{0},\delta(X_{0})/4)$ and
$F=B(X_{0},\delta(X_{0})/2)\setminus B(X_{0},\delta(X_{0})/4)$ are satisfied,
which implies that $v\geq 0$ on $E$, i.e.,

(15.35) $\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\leq\kappa
u_{h}\leq\omega^{X}(\Gamma\cap B)\qquad\text{ for }X\in\Omega\setminus
B(X_{0},\delta(X_{0})/4).$

This is stronger than the first inequality in (15.29).

We shall also use the following result on Green functions: for $\phi\in
C^{\infty}(\mathbb{R}^{n})\cap W$ and $X\notin\operatorname{supp}\,\phi$,

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

where $u_{\phi}\in W$ is the solution ro $Lu=0$, with the Dirichlet condition
$\operatorname{Tr}u_{\phi}=\operatorname{Tr}\phi$ on $\Gamma$, given by Lemma
12.2. The identity (15.36) is the same as (11.70) in [DFM2], and its proof -
which only relies on the properties on the Green functions given in Section 14
\- is the same as in [DFM2].

We turn to the proof of the upper bound in (15.29), that is,

(15.37) $\omega^{X}(B(x_{0},r)\cap\Gamma)\leq
C\,\frac{m(B\cap\Omega)}{r^{2}}\,g(X,X_{0})\ \ \ \text{ for
}X\in\Omega\setminus 2B.$

For the rest of the proof, $K$ is the constant in Lemma 15.14. Let
$X\in\Omega\setminus 2B$ be given, and choose $\phi\in
C^{\infty}_{0}(\mathbb{R}^{n})$ such that $0\leq\phi\leq 1$, $\phi\equiv 1$ on
$B\cap\Gamma$, $\operatorname{supp}\,\phi\subset
E_{B}:=\\{Y\in\Omega,\,\,\mathrm{dist}(Y,B\cap\Gamma)\geq(100K)^{-1}r\\}$, and
$|\nabla\phi|\leq 200K/r$. We get that

(15.38) $u_{\phi}(X)\leq\frac{C}{r}\int_{E_{B}}|\nabla_{y}g(X,Y)|dm(Y)$

by (15.36) and (11.2), and since $\omega^{X}(B\cap\Gamma)\leq u_{\phi}(X)$ by
the positivity of the harmonic measure,

(15.39)
$\begin{split}\omega^{X}(B\cap\Gamma)&\leq\frac{C}{r}\int_{E_{B}}|\nabla_{y}g(X,Y)|dm(Y).\end{split}$

We cover $E_{B}$ by a finitely overlapping collection of balls
$(B_{i})_{i\in\mathcal{I}}$ centered on $B\cap\Gamma$ and of radius
$(10K)^{-1}r$. Then

(15.40)
$\begin{split}\omega^{X}(B\cap\Gamma)&\leq\frac{C}{r}\sum_{i\in\mathcal{I}}\int_{B_{i}\cap\Omega}|\nabla_{y}g(X,Y)|dm(Y)\\\
&\lesssim\sum_{i\in\mathcal{I}}\frac{m(B_{i}\cap\Omega)^{1/2}}{r}\left(\int_{B_{i}\cap\Omega}|\nabla_{y}g(X,Y)|^{2}dm(Y)\right)^{\frac{1}{2}}\\\
&\lesssim\sum_{i\in\mathcal{I}}\frac{m(B_{i}\cap\Omega)^{1/2}}{r^{2}}\left(\int_{2B_{i}\cap\Omega}|g(X,Y)|^{2}dm(Y)\right)^{\frac{1}{2}},\end{split}$

where we use successively the Cauchy-Schwarz inequality and Cacciopoli’s
inequality at the boundary (Lemma 11.15); the use of Cacciopoli’s inequality
is indeed allowed because $Y\to g(X,Y)$ is a solution of
$L_{T}u:=-\mathop{\operatorname{div}}A^{T}\nabla u$ in $2B_{i}\cap\Omega$ by
Lemmas 14.78 and 14.60 (iii). Observe that $Y\to g(X,Y)$ is more generally a
solution of $L_{T}u:=-\mathop{\operatorname{div}}A^{T}\nabla u$ in each set
$2KB_{i}\cap\Omega$, because the radius of $2KB_{i}$ is less than $r/2$ and
hence $2KB_{i}\subset 2B\not\ni X$. As a consequence, Lemma 15.14 yields that

(15.41)
$\omega^{X}(B\cap\Gamma)\lesssim\sum_{i\in\mathcal{I}}\frac{m(B_{i})}{r^{2}}\,g(X,X_{i}),$

where $X_{i}$ is a corkscrew point associated to the ball $B_{i}$. Hence

(15.42) $\omega^{X}(B\cap\Gamma)\lesssim\frac{m(B)}{r^{2}}g(X,X_{0})$

because of the finite overlapping of the $(B_{i})_{i}$, the Harnack
inequality, and the fact that we can easily find Harnack chains of balls that
link $X_{i}$ to $X_{0}$ and that avoids $X$. The bounds (15.37) and then
(15.29) follow.

The proof of (15.30) can be treated in a similar manner, and we refer to
[DFM2, Lemma 11.11] for additional ideas on the proof. $\Box$

######  Lemma 15.43 (Doubling volume property for the harmonic measure).

Let $\alpha>1$, and take a ball $B:=B(x_{0},r)$ in $\mathbb{R}^{n}$ centered
on $\Gamma$. One has

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

where $C_{\alpha}>0$ depends only upon $n$, $C_{1}$ to $C_{6}$, $C_{A}$, and
$\alpha$.

Proof. The proof is the same as the one of [DFM2, Lemma 11.102]. Here are some
ideas.

When $\alpha=2$, we use Lemma 15.28, the doubling property (H4), the Harnack
inequality, and the existence of Harnack chains of balls to write

(15.45)
$\omega^{X}(2B\cap\Gamma)\lesssim\frac{m(2B)}{(2r)^{2}}g(X,X_{2})\lesssim\frac{m(B)}{r^{2}}g(X,X_{1})\lesssim\omega^{X}(B\cap\Gamma),$

where $X_{1}$ and $X_{2}$ are corkscrew points associated to respectively
$(x_{0},r)$ and $(x_{0},2r)$.

When $1<\alpha<2$, we cover $2B\cap\Gamma$ by a collection of finitely
overlapping balls $2B_{i}$ of radius $2r_{\alpha}:=(\alpha-1)r$ and centered
on $B(x_{0},2r-\frac{3}{2}r_{\alpha})\cap\Gamma$. In this case, for any
$X\in\Omega\setminus 2\alpha B$

(15.46)
$\omega^{X}(2B\cap\Gamma)\leq\sum_{i}\omega^{X}(2B_{i}\cap\Gamma)\lesssim\sum_{i}\omega^{X}(B_{i}\cap\Gamma)\lesssim\omega^{X}(B(x_{0},2r-\frac{1}{2}r_{\alpha})\cap\Gamma),$

where the second estimate is due to (15.45). We repeat the argument a finite
number of time (depending in $\alpha-1>0$) to get (15.44). The lemma follows.
$\Box$

######  Lemma 15.47 (Comparison principle for global solutions).

Let $B:=B(x_{0},r)$ be a ball centered on $\Gamma$, and let $X_{0}\in\Omega$
be a corkscrew point associated to $(x_{0},r)$. Let $u,v\in W$ be two non-
negative, non identically zero, solutions to $Lu=Lv=0$ in $\Omega$ such that
$\operatorname{Tr}u=\operatorname{Tr}v=0$ on $\Gamma\setminus B(x_{0},r)$.
Then

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

where $C>0$ depends only on $n$, $C_{1}$ to $C_{6}$, and $C_{A}$.

######  Remark 15.49.

We also have (15.48) 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. The proof is very similar to the one of [DFM2, Lemma 11.14]. Let us
recall the main steps and show the differences.

By symmetry of the roles of $u$ and $v$, we only need to show the upper bound

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

Notice also that thanks to the Harnack inequality (Lemma 11.35), $v(X)>0$ on
the whole $\Omega$, so we we don’t need to be careful when we divide by
$v(X)$.

We introduce some notation for two boundary balls: set $\Gamma_{1}:=\Gamma\cap
B$ and $\Gamma_{2}:=\Gamma\cap\frac{15}{8}B$. The proof of the lemma is
composed of three steps :

  1. (a)

we prove the lower bound

(15.51) $v(X)\geq C^{-1}\omega^{X}(\Gamma_{1})v(X_{0})\ \ \ \text{ for
}X\in\Omega\setminus 2B;$

  2. (b)

we prove the upper bound

(15.52) $u(X)\leq Cu(X_{0})\omega^{X}(\Gamma_{2})\ \ \ \text{ for
}X\in\Omega\setminus 2B;$

  3. (c)

we conclude by using the fact that the harmonic measure is doubling (Lemma
15.43).

The proof of (15.51) is can be done exactly as in [DFM2, Lemma 11.14]. Let us
quickly sketch it. By the Harnack inequality (Lemma 11.35), for all $X\in
B(X_{0},\delta(X_{0})/2)$, we have $v(X)\gtrsim v(X_{0})$. Together with the
upper bound (14.64), we get the existence of a constant $K_{1}$ such that the
function

$v_{1}(X):=K_{1}v(X)-\frac{m(B\cap\Omega)}{r^{2}}v(X_{0})g(X,X_{0})$

satisfies all the assumption of the maximum principle (Lemma 14.33) with
$E=\mathbb{R}^{n}\setminus B(X_{0},\delta(X_{0})/4)$ and
$F=B(X_{0},\delta(X_{0})/2)\setminus B(X_{0},\delta(X_{0})/4)$. Indeed, since
$v$ is non-negative everywhere, it forces
$\operatorname{Tr}v_{1}=\operatorname{Tr}v\geq 0$. We deduce that $v_{1}\geq
0$ on $E$, i.e.

$\frac{m(B\cap\Omega)}{r^{2}}v(X_{0})g(X,X_{0})\leq K_{1}v(X)\qquad\text{ for
}X\in\Omega\setminus B(X_{0},\delta(X_{0})/4).$

The claim (15.51) is now an immediate consequence of Lemma 15.28.

We turn to the proof of (15.52). We first check that

(15.53) $u(X)\leq Cu(X_{0})\ \ \ \text{ for
}X\in\frac{13}{8}B\setminus\frac{11}{8}B.$

Let $K$ as in Lemma 15.14. We want to establish (15.53) in the two sets:

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

and

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

The proof of (15.53) on $\Omega_{2}$ is easy, it is only a consequence of the
Harnack inequality (Lemma 11.35) and the existence of Harnack chains.

Then, we prove (15.53) for $X\in\Omega_{1}$. Let thus $X\in\Omega_{1}$ be
given. Take $x\in\Gamma$ such that $|X-x|=\delta(X)$; in particular,
particular, $|X-x|\leq\frac{r}{8}$, and hence $x\in\frac{7}{4}B$. Now let
$X_{1}$ be a Corkscrew point for $(x,\frac{r}{4K})$. Since $u$ is a non-
negative solution of $Lu=0$ in $B(x,\frac{r}{4})\cap\Omega$ satisfying
$\operatorname{Tr}u=0$ on $B(x,\frac{r}{4})\cap\Gamma$, Lemma 15.14 gives that
$u(Y)\leq Cu(X_{1})$ for $Y\in B(x,\frac{r}{4K})$ and thus $u(X)\leq
Cu(X_{1})$. By the existence of Harnack chains (Proposition 2.18) and the
Harnack inequality (Lemma 11.35), $u(X_{1})\leq Cu(X_{0})$. Hence $u(X)\leq
u(X_{1})$, which completes the proof of (15.53) on $\Omega_{1}$

The end of the proof is as in in [DFM2, Lemma 11.117], but let us recall it.
We proved (15.53) and now we want to get (15.52). Recall from Lemma 15.1 that
$\omega^{X}(\frac{7}{4}B\cap\Gamma)\geq C^{-1}$ for
$X\in\frac{13}{8}B\cap\Omega$. Hence, by (15.53),

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

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
$\operatorname{Tr}u_{h}=\operatorname{Tr}h$. By the positivity of the harmonic
measure,

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

The maximum principle given by Lemma 14.33 \- where we take
$E=\mathbb{R}^{n}\setminus\overline{\frac{11}{8}B}$ and
$F=\mathbb{R}^{n}\setminus\frac{13}{8}B$ \- yields

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

and hence

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

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

We conclude as follows. Because of (15.51) and (15.52),

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

The bound (15.50) - and thus the lemma - follows then from the fact that
$\omega^{X}(\Gamma_{2})\lesssim\omega^{X}(\Gamma_{1})$, which is given by
Lemma 15.43. $\Box$

######  Lemma 15.61 (Comparison principle for harmonic measures / Change of
poles).

Let $B:=B(x_{0},r)$ be a ball centered on $\Gamma$ and let $X_{0}$ be a
corkscrew point associated to $(x_{0},r)$. Let $E,F\subset\Gamma\cap B$ be two
Borel subsets of $\Gamma$ such that $\omega^{X_{0}}(E)$ and
$\omega^{X_{0}}(F)$ are positive. Then

(15.62)
$C^{-1}\frac{\omega^{X_{0}}(E)}{\omega^{X_{0}}(F)}\leq\frac{\omega^{X}(E)}{\omega^{X}(F)}\leq
C\frac{\omega^{X_{0}}(E)}{\omega^{X_{0}}(F)}\ \ \ \text{ for
}X\in\Omega\setminus 2B,$

where $C>0$ depends only on $n$, $C_{1}$ to $C_{6}$, and $C_{A}$. In
particular, with the choice $F=B\cap\Gamma$,

(15.63)
$C^{-1}\omega^{X_{0}}(E)\leq\frac{\omega^{X}(E)}{\omega^{X}(B\cap\Gamma)}\leq
C\omega^{X_{0}}(E)\ \ \ \text{ for }X\in\Omega\setminus 2B,$

where $C>0$ depends on the same quantity as for (15.62).

Proof. This result can be deduced from Lemma 15.47 with the same proof we
obtained [DFM2, Lemma 11.135] from [DFM2, Lemma 11.117]. It relies on
approximations of $X\mapsto\omega^{X}(E)$ and $X\mapsto\omega^{X}(F)$ by
solutions in $W$ to $Lu=0$ in $\Omega$. $\Box$

######  Theorem 15.64 (Comparison principle for locally defined functions).

There exists $K\geq 2$ depending only on $n$, $C_{1}$, and $C_{2}$ such that
the following holds.

Let $B:=B(x_{0},r)$ be a ball centered on $\Gamma$, and let $X_{0}\in\Omega$
be a corkscrew point associated to $(x_{0},r)$. Take $u,v\in
W_{r}(KB\cap\overline{\Omega})$ to be two non-negative, not identically zero,
solutions to $Lu=Lv=0$ in $KB\cap\Omega$, such that
$\operatorname{Tr}u=\operatorname{Tr}v=0$ on $KB\cap\Gamma$. Then

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

where $C>0$ depends only on depends only on $n$, $C_{1}$ to $C_{6}$, and
$C_{A}$.

Proof. Two strategies can be used to prove this theorem:

  1. (i)

If we mimic the classical proof from the codimension 1 case, we need to find a
(good enough) domain $D$ such that $B\cap\Omega\subset D\subset KB\cap\Omega$,
and we work with the harmonic measure on $\partial D$ of the operator $L$
restricted to $D$. One might think that for instance $D=2B\cap\Omega$ will
work out, but this choice of $D$ will not be “good enough” if it is not
connected (and it can easily happen).

The difficulty is first to construct such a $D$ that satisfies the corkscrew
point condition and the Harnack chain condition, but this part could be
possibly done by considering the tents sets constructed in Section 5. Yet,
even with such good $D$, we still need to build a measure $\mu_{D}$ on
$\partial D$ which is suitable, in particular satisfies (H5) for this
particular domain. Well, at the present moment, we don’t even know if building
such $\mu_{D}$ is possible with our assumptions.

  2. (ii)

The second strategy, that we shall apply, is to follow the ideas used in
[DFM2, Theorem 11.146]. In this strategy, we are not allowed to consider a
harmonic measure different from the one we defined on $\Gamma$. The main
pitfall in the present theory which did not exist in [DFM2] is the fact that
$\omega^{X}(\Gamma\setminus B)$ can be null because $\Gamma\setminus B$ is
empty, and so we do not necessary have the non-degeneracy of the harmonic
measure, and we shall use the estimate (15.4) involving the Green function
instead.

We may have chosen to restrict our attention to the balls $B$ that do not
cover entirely $\Gamma$. Here we decided to allow more balls, but then when we
take $r$ large in our theorem - when $\Omega$ is unbounded and $\Gamma$ is
bounded - we need to impose stronger conditions (and we get a stronger
conclusion).

Step 1: Construction of a function $f_{y_{0},s}$.

Let $y_{0}\in\Gamma$ and $s>0$. We write $Y_{0}$ for a corkscrew point
associated to $y_{0}$ and $s$. Roughly speaking, we would like to use the
function $f_{y_{0},s}(X)$ defined by

(15.66)
$f_{y_{0},s}(X):=\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})-K_{2}\left[\omega^{X}(\Gamma\setminus
B(y_{0},K_{1}s))+\frac{m(B(y_{0},K_{1}s)\cap\Omega)}{(K_{1}s)^{2}}g(X,Y_{K_{1}})\right]$

where $Y_{K_{1}}$ is a Corkscrew point associated to $y_{0}$ and $K_{1}s$, and
where $K_{1},K_{2}>0$ are some large constants that depend only on $n$,
$C_{1}$ to $C_{6}$, and$C_{A}$. We could show that with large enough choices
of $K_{1}$ and $K_{2}$, $f_{y_{0},s}$ is positive in $B(y_{0},s)$ and negative
outside of a big ball $B(y_{0},2K_{1}s)$. However, we want to use the maximum
principle to extend such inequalities to larger regions, and with this
definition involving the harmonic measure, our $f_{y_{0},s}$ is not smooth
enough to be used in Lemma 14.33. So we shall first replace
$\omega^{X}(\Gamma\setminus B(y_{0},K_{1}s))$ in (15.66) 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 $\beta>1$
(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 12.2, of $Lu_{\beta}=0$ with the Dirichlet condition
$\operatorname{Tr}u_{\beta}=\operatorname{Tr}h_{\beta}$. Notice that
$u_{\beta}\in W$ because $1-u_{\beta}$ is the solution of $L$ with the smooth
and compactly supported trace of $1-h_{\beta}$. By the positivity of the
harmonic measure, it holds that for any $X\in\Omega$ and $\gamma>0$,

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

and we can see here that $u_{\beta}$ will be used as a smooth substitute of
the harmonic measure.

Similarly to (15.32), using the Green function upper bounds and (H4), we have

(15.68) $\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})\leq C\qquad\ \text{
for }X\in B(Y_{0},\delta(Y_{0})/2)\setminus B(Y_{0},\delta(Y_{0})/4).$

Then by Lemma 14.33 with $E=\mathbb{R}^{n}\setminus B(Y_{0},\delta(Y_{0})/4)$
and $F=\mathbb{R}^{n}\setminus\overline{B(Y_{0},\delta(Y_{0})/2)}$,

(15.69) $\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})\leq C\ \ \ \text{ for
}X\in\Omega\setminus B(Y_{0},\delta(Y_{0})/4).$

From this and the non-degeneracy of the harmonic measure (Lemma 15.1), we
deduce that for $\beta\in(1,\infty)$ and $X\in\Omega\setminus B(y_{0},2\beta
s)$

(15.70) $\begin{split}\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})&\leq
K_{2}\left[\omega^{X}(\Gamma\setminus B(y_{0},\beta s))+\frac{m(B(y_{0},\beta
s)\cap\Omega)}{(\beta s)^{2}}g(X,Y_{\beta}))\right]\\\ &\leq
K_{2}\left[u_{\beta}(X)+\frac{m(B(y_{0},\beta s)\cap\Omega)}{(\beta
s)^{2}}g(X,Y_{\beta}))\right],\end{split}$

where $Y_{\beta}$ is a corkscrew point associated to $(y_{0},\beta s)$, and
where the constant $K_{2}>0$ depends only on $n$, $C_{1}$ to $C_{6}$, and
$C_{A}$; in particular, $K_{2}$ does not depend on $\beta$.

Our aim now is to find $K_{1}\geq 20C_{1}$ such that, for
$X\in\Omega\cap[B(y_{0},s)\setminus B(Y_{0},\delta(Y_{0})/4)]$,

(15.71)
$K_{2}\left[u_{K_{1}}(X)+\frac{m(B(y_{0},K_{1}s)\cap\Omega)}{(K_{1}s)^{2}}g(X,Y_{K_{1}})\right]\leq\frac{1}{2}\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0}).$

According to the Hölder continuity at the boundary (Lemma 11.32), we have

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

Moreover, Lemma 14.83 applied to $g(.,Y_{\beta})$ implies, that

(15.73) $\frac{m(B(y_{0},\beta s)\cap\Omega)}{(\beta s)^{2}}g(X,Y_{\beta})\leq
C\beta^{-\alpha}\qquad\text{ for }X\in B(y_{0},10s).$

If both cases, we need $\beta$ to be big enough, for instance $\beta\geq
20C_{1}$, and the constants $C$ and $\alpha>0$ depend only on $n$, $C_{1}$ to
$C_{6}$, and $C_{A}$. Due to the non-degeneracy of the harmonic measure given
by (15.4) - and (H4) - we have

$u_{4}+\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})\geq C^{-1}\quad\text{
for }X\in\Omega\setminus B(y_{0},5s).$

By the last estimate in (15.72)–(15.73), there exists $K_{3}>0$ such that

(15.74) $u_{\beta}(X)+\frac{m(B(y_{0},\beta s)\cap\Omega)}{(\beta
s)^{2}}g(X,Y_{\beta})\leq
K_{3}\beta^{-\alpha}\left[u_{4}(X)+\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})\right]$

for $X\in\Omega\cap[B(y_{0},10s)\setminus B(y_{0},5s)]$. In addition, by
increasing $K_{3}$ if needed, the estimates (15.72)–(15.73) and the lower
bound in (14.65) for the Green function imply that

(15.75) $u_{\beta}(X)+\frac{m(B(y_{0},\beta s)\cap\Omega)}{(\beta
s)^{2}}g(X,Y_{\beta})\leq
K_{3}\beta^{-\alpha}\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})$

when $X\in B(Y_{0},\delta(Y_{0})/2)\setminus B(Y_{0},\delta(Y_{0})/4)$. We
invoke then Lemma 14.33, used on the function

$X\to
K_{3}\beta^{-\alpha}\left[u_{4}(X)+\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})\right]-\left[u_{\beta}(X)+\frac{m(B(y_{0},\beta
s)\cap\Omega)}{(\beta s)^{2}}g(X,Y_{\beta})\right]$

and the sets $E=B(y_{0},10s)\setminus\overline{B}(Y_{0},\delta(Y_{0})/4)$ and
$F=\overline{B(y_{0},5s)}\setminus B(Y_{0},\delta(Y_{0})/2)$, to deduce that
for $X\in\Omega\cap B(y_{0},10s)\setminus B(Y_{0},\delta(Y_{0})/4)$,

$\begin{split}u_{\beta}(X)+\frac{m(B(y_{0},\beta s)\cap\Omega)}{(\beta
s)^{2}}g(X,Y_{\beta})&\leq
K_{3}\beta^{-\alpha}\left[u_{4}(X)+\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})\right]\\\
&\leq
K_{3}\beta^{-\alpha}\left[\omega^{X}(B(y_{0},4s))+\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})\right]\\\
&\leq
C\beta^{-\alpha}\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})\end{split}$

by Lemma 15.28. Therefore, (15.71) can be indeed achieved for some large
$K_{1}$, that depends only on $n$, $C_{1}$ to $C_{6}$, and $C_{A}$ (recall
that $K_{2}$ is already fixed, and depends on the same parameters).

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

(15.76)
$f(X):=\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})-K_{2}\left[u_{K_{1}}(X)+\frac{m(B(y_{0},K_{1}s)\cap\Omega)}{(K_{1}s)^{2}}g(X,Y_{K_{1}}))\right].$

The inequality (15.70) gives

(15.77) $f_{y_{0},s}(X)\leq 0\ \ \ \text{ for }X\in\Omega\setminus
B(y_{0},2K_{1}s),$

and the estimate (15.71) proves that

(15.78)
$f_{y_{0},s}(X)\geq\frac{1}{2}\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(X,Y_{0})$

for $X\in\Omega\cap[B(y_{0},s)\setminus B(Y_{0},\delta(Y_{0})/4)]$.

Step 2: End of the proof

Let us turn to the proof of the comparison principle. By symmetry and as in
Lemma 15.47, it suffices to prove the upper bound in (15.65), that is

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

We claim that

(15.80) $v(X)\geq C^{-1}\frac{m(B)}{r^{2}}v(X_{0})g(X,X_{0})\qquad\text{ for
}X\in[\Omega\cap B]\setminus B(X_{0},\delta(X_{0})/4),$

where $C>0$ depends only on $n$, $C_{1}$ to $C_{6}$, and $C_{A}$. So let
$X\in\Omega\cap B$ be given. Two cases may happen. If
$\delta(X)\geq\frac{r}{8K_{1}}$, where $K_{1}$ comes from (15.71) and is the
same as in the definition of $f_{y_{0},s}$, the existence of Harnack chains
(Proposition 2.18) and the Harnack inequality (Lemma 11.35) give that

$v(X)\approx v(X_{0})$

For the above inequality to hold, we need the Harnack chains to stay in the
area where $v$ is a solution; we take $K$ big enough to make sure that it
happens, and by Proposition 2.18, $K$ need to depend only on $C_{1}$ and
$C_{2}$. Similarly, Proposition 2.18 and Lemma 11.35, together with the bound
(14.65) on the Green function, give that

$\frac{m(B)}{r^{2}}g(X,X_{0})\approx 1\qquad\text{ on }[\Omega\cap
B(x_{0},r)]\setminus B(X_{0},\delta(X_{0})/4).$

We conclude that for all $X\in[\Omega\cap B]\setminus
B(X_{0},\delta(X_{0})/4)$ satisfying $\delta(X)\geq\frac{r}{8K_{1}}$.

(15.81) $v(X)\approx v(X_{0})\frac{m(B)}{r^{2}}g(X,X_{0}).$

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

  * •

The quantity $\int_{B(y_{0},4K_{1}s)\setminus B(Y_{0},\delta(Y_{0})/4)}|\nabla
v|^{2}dm$ is finite because $v\in W_{r}(B(x_{0},2r))$. The fact that
$\int_{B(y_{0},4K_{1}s)\setminus B(Y_{0},\delta(Y_{0})/4)}|\nabla
f_{y_{0},s}|^{2}dm$ is finite as well follows from the property (14.61) of the
Green function.

  * •

There exists $K_{4}>0$ such that

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

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

(15.83) $f_{y_{0},s}(Y)\leq\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}g(Y,Y_{0})\leq
C\ \ \ \text{ for }Y\in B(Y_{0},\delta(Y_{0})/2)\setminus
B(Y_{0},\delta(Y_{0})/4),$

which is a consequence of the definition (15.76) and (14.64), and the bound

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

which comes from the Harnack inequality (Lemma 11.35).

  * •

The function $K_{4}v-v(Y_{0})f_{y_{0},s}$ is nonnegative on
$\Omega\cap[B(y_{0},4K_{1}s)\setminus B(y_{0},2K_{1}s)]$. Indeed, $v\geq 0$ on
$\Omega\cap B(y_{0},4K_{1}s)$ and, thanks to (15.77), $f_{y_{0},s}\leq 0$ on
$\Omega\setminus B(y_{0},2K_{1}s)$.

  * •

The trace of $K_{4}v-v(Y_{0})f_{y_{0},s}$ is non-negative on
$B(y_{0},4K_{1}s)\cap\Gamma$ again because $\operatorname{Tr}v=0$ on
$B(y_{0},4K_{1}s)\cap\Gamma$ and $\operatorname{Tr}[f_{y_{0},s}]\leq 0$ on
$B(y_{0},4K_{1}s)\cap\Gamma$ by construction.

The previous points prove that $K_{4}v-v(Y_{0})f_{y_{0},s}$ satisfies the
assumptions of Lemma 14.33 with
$E=B(y_{0},4K_{1}s)\setminus\overline{B(Y_{0},\delta(Y_{0})/4)}$ and
$F=\overline{B(y_{0},2K_{1}s)}\setminus B(Y_{0},\delta(Y_{0})/2)$. As a
consequence, for any $Y\in B(y_{0},4K_{1}s)\setminus B(Y_{0},\delta(Y_{0})/4)$

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

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

(15.86) $v(Y)\geq(K_{4})^{-1}v(Y_{0})f_{y_{0},s}(Y)\geq
C^{-1}\frac{m(B(y_{0},s)\cap\Omega)}{s^{2}}v(Y_{0})g(Y,Y_{0})$

by (15.78). The points $X_{0}$ and $Y_{0}$ are both corkscrew points, and they
can be linked by a Harnack chain of balls whose length depends only on
$r/s=8K_{1}$, that is with uniformly bounded length. So using the Harnack
inequality on every ball of the chain, we deduce that $v(Y_{0})\approx
v(X_{0})$ and $g(Y,Y_{0})\approx g(Y,X_{0})$ whenever $Y\in B(y_{0},s)$ is far
from $Y_{0}$ and $X_{0}$ (but it cannot be close to $X_{0}$ since $K_{1}\geq
20C_{1}$). Moreover, $s^{-2}m(B(y_{0},s)\cap\Omega)\approx
r^{-2}m(B\cap\Omega)$ by (H4). Therefore (15.85) becomes

$v(Y)\geq C^{-1}\frac{m(B\cap\Omega)}{r^{2}}v(X_{0})g(Y,X_{0})\qquad\text{ for
}Y\in[\Omega\cap B(y_{0},s)]\setminus B(Y_{0},\delta(Y_{0})/4).$

Since the two functions of $Y$ in the inequality above are solutions in
$\Omega\cap B(y_{0},2s)$, the Harnack inequality yields the following
improvement:

$v(Y)\geq C^{-1}\frac{m(B\cap\Omega)}{r^{2}}v(X_{0})g(Y,X_{0})\qquad\text{ for
}Y\in\Omega\cap B(y_{0},s).$

Recall that $X\in B(y_{0},s)$ by construction of $y_{0}$ and $s$. We conclude,
at last, that even when $X\in\Omega\cap B$ is such that
$\delta(X)<\frac{r}{8K_{1}}$, we still have

$v(X)\geq C^{-1}\frac{m(B)}{r^{2}}v(X_{0})g(X,X_{0}).$

The claim (15.80) follows.

Now we want to prove that, for all $X\in[\Omega\cap B]\setminus
B(X_{0},\delta(X_{0})/4)$,

(15.87) $u(X)\leq
Cu(X_{0})\left[\omega^{X}(\Gamma\setminus\frac{5}{4}B)+\frac{m(B\cap\Omega)}{r^{2}}g(.,X_{0})\right],$

where the constant $C>0$ depends only on $n$, $C_{1}$ to $C_{6}$, and $C_{A}$.
By Lemma 15.14,

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

as long as $K$ is large enough so that Lemma 15.14 can be applied. But again,
$K$ does not need to depend on anything else than $n$, $C_{1}$ and $C_{2}$.
Pick $h^{\prime}\in C^{\infty}(\mathbb{R}^{n})$ such that $0\leq
h^{\prime}\leq 1$, $h^{\prime}\equiv 1$ outside of $\frac{3}{2}B$, and
$h^{\prime}\equiv 0$ on $\frac{5}{4}B$. Let $u_{h^{\prime}}=U(h^{\prime})$ be
the solution of $Lu_{h^{\prime}}=0$ with the data
$\operatorname{Tr}u_{h^{\prime}}=\operatorname{Tr}h^{\prime}$ (given by Lemma
12.2). As before, $u_{h^{\prime}}\in W$ because $1-u_{h^{\prime}}=U(1-h)$ and
$1-h$ is a test function. Also,
$u_{h^{\prime}}(X)\geq\omega^{X}(\Gamma\setminus\frac{3}{2}B)$ by
monotonicity. So (15.4), which states the non-degeneracy of the harmonic
measure, gives

(15.89)
$u_{h^{\prime}}(X)+\frac{m(\frac{3}{2}B\cap\Omega)}{(\frac{3}{2}r)^{2}}g(X,X_{3/2})\geq
C^{-1}\ \ \ \text{ for }X\in\Omega\setminus\frac{13}{8}B,$

where $X_{3/2}$ is a corkscrew point associated to $(x_{0},\frac{3}{2}r)$. The
doubling property (H4) and Harnack inequality for the function
$g(X,.)=g_{T}(.,X)$ entail now that

(15.90) $u_{h^{\prime}}(X)+\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\geq C^{-1}\
\ \ \text{ for }X\in\Omega\setminus\frac{13}{8}B.$

The combination of (15.88) and (15.90) yields the existence of $K_{5}>0$ such
that

$\tilde{u}:=K_{5}u(X_{0})\left[u_{h^{\prime}}+\frac{m(B\cap\Omega)}{r^{2}}g(.,X_{0})\right]-u\geq
0\quad\text{ on }\Omega\cap[\frac{7}{4}B\setminus\frac{13}{8}B].$

Moreover, using the Harnack inequality and the Green function lower bounds, by
increasing slightly $K_{5}$ if needed, we also have $\tilde{u}\geq 0$ in
$B(X_{0},\delta(X_{0})/2)\setminus B(X_{0},\delta(X_{0})/4)$. Now, it is not
very hard to see that $\tilde{u}$ satisfies all the assumptions of Lemma
14.33, with the sets
$E=\frac{7}{4}B\setminus\overline{B(X_{0},\delta(X_{0})/4)}$ and
$F=\overline{\frac{13}{8}B}\setminus B(X_{0},\delta(X_{0})/2)$. Observe in
particular that $u_{h^{\prime}}\in W$, $Tu_{h^{\prime}}\geq 0$, and as long as
we choose $K\geq 2$, $u\in W_{r}(2B)$ and $Tu=0$ on $\Gamma\cap 2B$. Then by
Lemma 14.33,

(15.91) $u(X)\leq
K_{5}u(X_{0})\left[u_{h^{\prime}}(X)+\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\right]\
\ \ \text{ for }X\in[\Omega\cap\frac{7}{4}B]\setminus
B(X_{0},\delta(X_{0})/4),$

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

$u(X)\leq
Cu(X_{0})\left[\omega^{X}(\Gamma\setminus\frac{5}{4}B)+\frac{m(B\cap\Omega)}{r^{2}}g(X,X_{0})\right]\
\ \ \text{ for }X\in[\Omega\cap\frac{7}{4}B]\setminus
B(X_{0},\delta(X_{0})/4).$

The claim (15.87) follows.

The bounds (15.80) and (15.87) imply that

(15.92) $\frac{u(X)}{v(X)}\leq
C\frac{u(X_{0})}{v(X_{0})}\left[\frac{r^{2}\omega^{X}(\Gamma\setminus\frac{5}{4}B)}{m(B\cap\Omega)g(X,X_{0})}+1\right]\
\ \ \text{ for }X\in[\Omega\cap B]\setminus B(X_{0},\delta(X_{0})/4).$

The bound (15.79) in the set $[\Omega\cap B]\setminus
B(X_{0},\delta(X_{0})/4)$ is now a consequence of the above inequality and
(15.30). The bound (15.79) in the full domain $\Omega\cap B$ is then an easy
consequence of the Harnack inequality (Lemma 11.35). $\Box$

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