Let $T=T_{\beta}>10$ be big enough so that $\log x\leq x^{\beta/2}$ for every
$x>T$. Hence,

(5.16)
$\displaystyle\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}}K\log^{+}K$
$\displaystyle=\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}:K(\xi)\leq
T\right\\}}K\log^{+}K+\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}:K(\xi)>T\right\\}}K\log^{+}K$
$\displaystyle\leq\log
T\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}}K+\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set<EMAIL_ADDRESS>
By (5.13), if we take $\zeta$ small enough, we bound the first term in the
right-hand side by

(5.17) $\log
T\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}}K\leq
C_{1+\beta}\log
T\sigma(\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega})^{1-1/(1+\beta)}\leq
0.5.$

By Hölder’s inequality, the reverse Hölder inequality $\mathrm{RH}_{1+\beta}$
and taking small enough $\zeta$ respectively, the second term in the right-
hand side in (5.16) is controlled by

(5.18)
$\displaystyle\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}}K^{1+\beta/2}$
$\displaystyle\leq\left(\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}}K^{1+\beta}\right)^{\frac{1+\beta/2}{1+\beta}}\sigma\left(\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}\right)^{1-\frac{1+\beta/2}{1+\beta}}$
$\displaystyle\overset{\mathrm{RH}_{1+\beta}}{}$
$\displaystyle{\leq}C_{1+\beta}{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\omega}$}}}}(\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}})^{1+\beta/2}\sigma\left(\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}\right)^{1-\frac{1+\beta/2}{1+\beta}}\leq
C_{1+\beta}\sigma\left(\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\setminus\partial\widetilde{\Omega}\right)^{1-\frac{1+\beta/2}{1+\beta}}\leq
0.5.$

By 5.14, 5.15, 5.16, 5.18 and 5.17 we can control the term with $K\log^{+}K$
as

$\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}}K\log^{+}K\leq\log
M\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)\geq
1\right\\}}K\,d\sigma+C+1.$

From this and (5.10),

$\displaystyle\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}}K\log^{-}K\,d\sigma$
$\displaystyle=\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}}K\log^{+}K\,d\sigma-\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}}K\log
K\,d\sigma\leq\log
M\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)\geq
1\right\\}}K\,d\sigma+C_{\Omega,A}$ $\displaystyle\leq\log
M\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)\geq
M^{-\tau}\right\\}}K\,d\sigma+C_{\Omega,A}.$

Using this in the last inequality of the following computations, we obtain

$\displaystyle\tau\log M$ $\displaystyle-\tau\log
M\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)\geq
M^{-\tau}\right\\}}K\,d\sigma=\log
M^{\tau}\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}}K\,d\sigma-\log
M^{\tau}\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)\geq
M^{-\tau}\right\\}}K\,d\sigma$ $\displaystyle=\log
M^{\tau}\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)<M^{-\tau}\right\\}}K\,d\sigma\leq\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)<M^{-\tau}\right\\}}K\log^{-}K\,d\sigma\leq\int_{\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}}K\log^{-}K\,d\sigma$
$\displaystyle\leq\log
M\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)\geq
M^{-\tau}\right\\}}K\,d\sigma+C_{\Omega,A},$

which gives

$\int_{\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}:K(\xi)\geq
M^{-\tau}\right\\}}K\,d\sigma\geq\frac{\tau\log M-C_{\Omega,A}}{(1+\tau)\log
M}=\frac{\tau}{1+\tau}-\frac{C_{\Omega,A}}{(1+\tau)\log M}\geq\frac{\tau}{2},$

as claimed, by letting $M>0$ be big enough depending on $\tau$ and
$C_{\Omega,A}$. ∎

We are now in position to find the final set $F$ with the claimed properties.
Let $\\{B_{Q_{n}}\\}_{n}\subset\\{B_{Q}\\}_{Q\in\mathcal{I}}$ be the subfamily
satisfying either $({\mathsf{HD}})$ or $({\mathsf{TS}})$:

  1. (${\mathsf{HD}}$)

$h(z_{Q_{n}})>\rho$, where $z_{Q_{n}}=c(B_{Q_{n}})$ is the center of the ball
$B_{Q_{n}}$.

  2. (${\mathsf{TS}}$)

$h(z_{Q_{n}})=\rho$ and $\partial
B_{Q_{n}}\cap\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\cap\partial\widetilde{\Omega}:K(\xi)\geq
M^{-\tau}\right\\}\not=\emptyset$.

Here ${\mathsf{HD}}$ stands for ‘high density’ and ${\mathsf{TS}}$ for
‘touching set’.

######  Notation.

We write $B_{Q}\in{\mathsf{HD}}$ if $B_{Q}$ satisfies (${\mathsf{HD}}$), and
$B_{Q}\in{\mathsf{TS}}$ if $B_{Q}$ satisfies (${\mathsf{TS}}$).

With this choice,

$\frac{\tau}{4}\overset{\text{{(\ref{points with very small density has small
elliptic
measure})}}}{\leq}{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\omega}$}}}}\left(\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\cap\partial\widetilde{\Omega}:K(\xi)\geq
M^{-\tau}\right\\}\right)\leq{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\omega}$}}}}\left(\overline{\bigcup_{n}B_{Q_{n}}}\right)\leq\sum_{n}{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\omega}$}}}}\left(\overline{B_{Q_{n}}}\right),$

and by (5.6),

$\frac{\tau}{4}\leq\sum_{n}{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\omega}$}}}}\left(\overline{B_{Q_{n}}}\right)\lesssim\sum_{n}\omega(10B_{Q_{n}}).$

If $\varepsilon$ is small enough, then $10B_{Q}=B\left(z_{Q},10\varepsilon
d(z_{Q})\right)\subset\frac{\varepsilon^{-2}}{4}Q$ for each $Q\in\mathcal{I}$.
Moreover, since $\\{\frac{\varepsilon^{-2}}{4}Q\\}_{Q\in\mathcal{I}}$ has
finite overlapping by Lemma 5.3(3), $\\{10B_{Q}\\}_{Q\in\mathcal{I}}$ has also
finite overlapping with constant depending on $\varepsilon$ only. From this we
obtain,

(5.19)
$\tau\lesssim\sum_{n}\omega(10B_{Q_{n}})\lesssim\omega\left(\bigcup_{n}10B_{Q_{n}}\right).$

At this point we have found a subset of $\partial\Omega$ (covered by balls)
with elliptic measure bounded uniformly from below. Moreover, the radii
$\varepsilon d(z_{Q_{n}})$ of these balls $B_{Q_{n}}$ are smaller than the
‘high density value’ $h(z_{Q_{n}})$, which will allow us to have a control on
the sum of the radii.

First we need to define the set $F\subset\partial\Omega$ and its covering. For
each $B_{Q_{n}}\in{\mathsf{HD}}$, we have
$10B_{Q_{n}}=B\left(z_{Q_{n}},10\varepsilon d(z_{Q_{n}})\right)$, and since
$d(\cdot)\leq h(\cdot)$ on $\partial\Omega$, $10B_{Q_{n}}\subset
B\left(z_{Q_{n}},10\varepsilon h(z_{Q_{n}})\right)\subset
B\left(z_{Q_{n}},10h(z_{Q_{n}})\right)$. Since the family
$\left\\{B\left(z_{Q_{n}},10h(z_{Q_{n}})\right)\right\\}_{B_{Q_{n}}\in{\mathsf{HD}}}$
is finite, by means of the $3R$-covering theorem consider a disjoint subfamily

$\left\\{B_{m}\right\\}_{m}\subset\left\\{B\left(z_{Q_{n}},10h(z_{Q_{n}})\right)\right\\}_{B_{Q_{n}}\in{\mathsf{HD}}}$

such that

$\bigcup_{B_{Q_{n}}\in{\mathsf{HD}}}B\left(z_{Q_{n}},10h(z_{Q_{n}})\right)\subset\bigcup_{m}3B_{m}.$

Let us define

$F\coloneqq\left(\bigcup_{m}3B_{m}\cup\bigcup_{B_{Q_{n}}\in{\mathsf{TS}}}10B_{Q_{n}}\right)\cap\partial\Omega.$

Note that
$\bigcup_{n}10B_{Q_{n}}\subset\bigcup_{m}3B_{m}\cup\bigcup_{B_{Q_{n}}\in{\mathsf{TS}}}10B_{Q_{n}}$
implies $\omega(F)\gtrsim\tau$ by (5.19). Next we show that the covering

$\left\\{3B_{m}\right\\}_{m}\cup\left\\{10B_{Q_{n}}\right\\}_{B_{Q_{n}}\in{\mathsf{TS}}}$

satisfies the properties (1enumi) and (2enumi) in Lemma 3.2.

We can control the radius of the balls with high density

$\sum_{m}r(3B_{m})=30\sum_{m}h\left(c(B_{m})\right)\overset{\text{{(\ref{high
density
function})}}}{\leq}\frac{30}{M}\sum_{m}\omega\left(B_{m}\right)=\frac{30}{M}\omega\left(\bigcup_{m}B_{m}\right)\lesssim\frac{1}{M},$

by the definition of $h(\cdot)$ in (5.1), and the fact that the balls
$\\{B_{m}\\}_{m}$ are pairwise disjoint. Also,
$r\left(3B_{m}\right)=30h\left(c(B_{m})\right)>30\rho$. We have shown the
second property of the covering.

Recall that the balls in ${\mathsf{TS}}$ intersect the set
$\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\cap\partial\widetilde{\Omega}:K(\xi)\geq
M^{-\tau}\right\\}$. For each $B_{Q_{n}}\in{\mathsf{TS}}$ consider a point
$x_{Q_{n}}\in\partial
B_{Q_{n}}\cap\left\\{\xi\in\partial{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\Omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\Omega}$}}}}\cap\partial\widetilde{\Omega}:K(\xi)\geq
M^{-\tau}\right\\}$. Then, by (5.7) and (5.4) we have

$M^{-\tau}\leq\frac{d{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\omega}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\omega}$}}}}}{d\sigma}(x_{Q_{n}})\overset{\text{{(\ref{newnew
not excessive
density})}}}{\leq}\frac{d\widetilde{\omega}}{d\sigma}(x_{Q_{n}})\overset{\text{{(\ref{density
elliptic
measure})}}}{\lesssim}\frac{\omega\left(10B_{Q_{n}}\right)}{r(B_{Q_{n}})},$

which implies

$\sum_{B_{Q_{n}}\in{\mathsf{TS}}}r\left(10B_{Q_{n}}\right)\lesssim
M^{\tau}\sum_{B_{Q_{n}}\in{\mathsf{TS}}}\omega\left(10B_{Q_{n}}\right)\lesssim
M^{\tau}\omega\left(\bigcup_{{Q_{n}}\in{\mathsf{TS}}}10B_{Q_{n}}\right)\leq
M^{\tau},$

obtaining the first property of the covering. Note that for
$B_{Q_{n}}\in{\mathsf{TS}}$ we have $h(z_{Q_{n}})=\rho$, and in particular
$d(z_{Q_{n}})=\rho$, which implies $r(10B_{Q_{n}})=10\varepsilon
d(z_{Q})=10\varepsilon\rho$. ∎

##  6\. Proof of Theorem 1.1

In this section we will prove Theorem 1.1. First we make the reduction to the
case of bounded domains (Claim 6.4) and then we prove the theorem using 3.1.

###  6.1. Reduction to bounded domains

First, we state some lemmas.

Let $\Omega\subset{\mathbb{R}}^{n+1}$ with $n\geq 1$, and let $B$ be a ball
centered at $\partial\Omega$. By the maximum principle [GT01, p. 46] we have

(6.1) $\omega^{z}_{\Omega\cap
B,A}(E)\leq\omega^{z}_{\Omega,A}(E)\quad\text{for any
}E\subset\partial\Omega\cap B\text{ and }z\in\Omega\cap B.$

The converse inequality may fail. However, the following weaker relation
holds.

######  Lemma 6.1.

Let $\Omega\subset{\mathbb{R}}^{2}$ be a (possibly unbounded) Wiener regular
domain, $A$ be a real uniformly elliptic matrix and $B$ be a ball centered at
$\partial\Omega$ with $\operatorname{Cap}(B\cap\partial\Omega,4B)\not=0$. Let
$E\subset\partial\Omega\cap B$ be a Borel set and $z_{E}\in\partial 1.5B$ such
that $\omega^{z_{E}}_{\Omega,A}(E)=\max_{z\in\partial
1.5B}\omega^{z}_{\Omega,A}(E)$. Then

$\omega^{z_{E}}_{\Omega,A}(E)\lesssim\frac{\operatorname{Cap}(2B,4B)}{\operatorname{Cap}(B\cap\partial\Omega,4B)}\omega^{z_{E}}_{\Omega\cap
4B,A}(E),$

where the constant involved depends on the ellipticity constant of the matrix
$A$ and the dimension. The same also holds for bounded Wiener regular domains
$\Omega\subset{\mathbb{R}}^{n+1}$ when $n\geq 2$.

###### Proof.

During this proof we write $\omega_{\cdot}$ instead of $\omega_{\cdot,A}$.

By Lemma 4.6 we have

(6.2) $\omega^{z_{E}}_{\Omega}(E)-\omega^{z_{E}}_{\Omega\cap
4B}(E)=\int_{\partial
4B\cap\Omega}\omega^{\xi}_{\Omega}(E)\,d\omega^{z_{E}}_{\Omega\cap 4B}(\xi).$

By the maximum principle222In the planar case the maximum principle on
$\Omega\setminus 1.5\overline{B}$ holds even if $\Omega$ is not bounded, since
$\omega^{\cdot}_{\Omega}(E)\in[0,1]$. in $\Omega\setminus 1.5\overline{B}$,
$\omega^{\xi}_{\Omega}(E)\leq\omega^{z_{E}}_{\Omega}(E)$ for $\xi\in 4\partial
B\cap\Omega$. From this and (6.2), we get

(6.3) $\omega^{z_{E}}_{\Omega}(E)\leq\omega^{z_{E}}_{\Omega\cap 4B}(4\partial
B\cap\Omega)\cdot\omega^{z_{E}}_{\Omega}(E)+\omega^{z_{E}}_{\Omega\cap
4B}(E).$

It remains to bound $\omega^{z_{E}}_{\Omega\cap 4B}(4\partial B\cap\Omega)$.
By the maximum principle we have

$\omega^{z_{E}}_{\Omega\cap 4B}(4\partial
B\cap\Omega)\leq\omega^{z_{E}}_{4B\setminus(B\cap\partial\Omega)}(4\partial
B).$

By [HKM06, Lemma 6.21], since $z_{E}\in 1.5\partial B$, we have

(6.4) $1-\omega^{z_{E}}_{4B\setminus(B\cap\partial\Omega)}(4\partial
B)\geq\widetilde{c}\cdot\operatorname{Cap}(B\cap\partial\Omega,4B)/\operatorname{Cap}(2B,4B)\eqqcolon
c\in(0,1).$

In particular

(6.5) $\omega^{z_{E}}_{\Omega\cap 4B}(4\partial
B\cap\Omega)\leq\omega^{z_{E}}_{4B\setminus(B\cap\partial\Omega)}(4\partial
B)\leq 1-c.$

From 6.3 and 6.5 we get

$\omega^{z_{E}}_{\Omega}(E)\leq\omega^{z_{E}}_{\Omega\cap 4B}(4\partial
B\cap\Omega)\cdot\omega^{z_{E}}_{\Omega}(E)+\omega^{z_{E}}_{\Omega\cap
4B}(E)\leq(1-c)\cdot\omega^{z_{E}}_{\Omega}(E)+\omega^{z_{E}}_{\Omega\cap
4B}(E),$

obtaining

$c\cdot\omega^{z_{E}}_{\Omega}(E)\leq\omega^{z_{E}}_{\Omega\cap 4B}(E),$

as claimed, with $c$ as in step (6.4). ∎

We remark that $\operatorname{Cap}(B\cap\partial\Omega,4B)=0$ would imply
$\omega_{D}(B\cap\partial\Omega)=0$ for any domain $D$ with
$B\cap\partial\Omega\subset\partial D$, see [HKM06, Theorems 10.1 and 11.14].
For $(\delta,r_{0})$-Reifenberg flat domains,
$\operatorname{Cap}(B\cap\partial\Omega,4B)/\operatorname{Cap}(2B,4B)\approx
1$ whenever $r_{B}\leq r_{0}$. In fact it is only needed the exterior
Corkscrew condition, see Remark 2.5.

######  Lemma 6.2.

Let $\Omega\subset{\mathbb{R}}^{2}$ be a (possibly unbounded) Wiener regular
domain and $A$ be a real uniformly elliptic matrix. Let $\\{B_{i}\\}_{i}$ be a
pairwise disjoint collection of balls centered at $\partial\Omega$ with
$\omega_{\Omega,A}\left(\partial\Omega\setminus\bigcup_{i}B_{i}\right)=0$, and
$F_{i}\subset\partial(\Omega\cap 4B_{i})$ with $\omega_{\Omega\cap
4B_{i},A}(F_{i})=1$. Then $F\coloneqq\partial\Omega\cap\bigcup_{i}F_{i}$
satisfies $\omega_{\Omega,A}(F)=1$. The same also holds for bounded Wiener
regular domains $\Omega\subset{\mathbb{R}}^{n+1}$ when $n\geq 2$.

###### Proof.

In this proof we denote $\omega\coloneqq\omega_{\Omega,A}$ and
$\omega_{\Omega\cap 4B_{i}}\coloneqq\omega_{\Omega\cap 4B_{i},A}$.

Let $p\in\Omega\setminus\\{\infty\\}$. Abusing notation, we write
$F^{c}=\partial\Omega\setminus F$. Since
$\omega^{p}(F^{c}\setminus\bigcup_{i}B_{i})\leq\omega^{p}(\partial\Omega\setminus\bigcup_{i}B_{i})=0$
and the balls are pairwise disjoint, we can conclude

(6.6) $\omega^{p}(F^{c})=\omega^{p}\left(\bigcup_{i}F^{c}\cap
B_{i}\right)=\sum_{i}\omega^{p}\left(F^{c}\cap B_{i}\right).$

We claim that each term in the right-hand side is zero. Indeed, for each $i$
fix a pole $p_{i}\in\Omega\cap 4B_{i}$, by the Borel regularity of
$\omega_{\Omega\cap 4B_{i}}^{p_{i}}$ let $E_{i}\supset F^{c}\cap B_{i}$ be a
Borel set with $\omega_{\Omega\cap 4B_{i}}^{p_{i}}(E_{i})=\omega_{\Omega\cap
4B_{i}}^{p_{i}}(F^{c}\cap B_{i})$, and finally let $z_{i}\in 1.5\partial
B_{i}$ such that $\omega^{z_{i}}(E_{i})=\max_{z\in 1.5\partial
B_{i}}\omega^{z}(E_{i})$. With this choice of $z_{i}$, by Lemma 6.1 we have

(6.7) $\omega^{z_{i}}(E_{i})\lesssim\omega^{z_{i}}_{\Omega\cap
4B_{i}}(E_{i}).$

Since the balls are pairwise disjoint we have that $F^{c}\cap
B_{i}\subset(F_{i}^{c}\cap B_{i})\cap\partial\Omega\subset F_{i}^{c}$, so

$\omega_{\Omega\cap 4B_{i}}^{p_{i}}(E_{i})=\omega^{p_{i}}_{\Omega\cap
4B_{i}}(F^{c}\cap B_{i})=0.$

Hence by the Harnack inequality (denoting its use with $H$), (6.7) and
$\omega_{\Omega\cap 4B_{i}}^{p_{i}}(E_{i})=0$, for every index $i$ there holds

(6.8) $\omega^{p}(F^{c}\cap
B_{i})\leq\omega^{p}(E_{i})\overset{(H)}{\lesssim}\omega^{z_{i}}(E_{i})\overset{\text{{(\ref{finding
the set for each ball})}}}{\lesssim}\omega_{\Omega\cap
4B_{i}}^{z_{i}}(E_{i})\overset{(H)}{\lesssim}\omega_{\Omega\cap
4B_{i}}^{p_{i}}(E_{i})=0.$

Notice that for each $i$ the constants involved in the use of Harnack
inequality and Lemma 6.1 in 6.7 and (6.8) depend on $i$, but the right-hand
side in (6.7) is zero.

By (6.8) we have that the sum in the right-hand side of (6.6) is zero as
claimed. Therefore the set $F\coloneqq\partial\Omega\cap\bigcup_{i}F_{i}$
satisfies $\omega^{p}(F)=1$. ∎

######  Lemma 6.3.

Let $r_{0}\in(0,\infty]$ and let $\varepsilon>0$ be small enough. There exists
$\delta_{0}=\delta_{0}(\varepsilon)>0$ such that if
$\Omega\in{\mathbb{R}}^{2}$ is $(\delta,r_{0})$-Reifenberg flat with
$\delta\in(0,\delta_{0})$ and $B$ is a ball centered at $\partial\Omega$ with
radius $r_{B}\leq r_{0}/100$, then there exists a bounded
$(\varepsilon,r)$-Reifenberg flat domain $D\subset\\{z:{\rm
dist}(z,\partial\Omega)<r_{0}/2\\}$ (for some $r\in(0,r_{0}/2)$) with
$\Omega\cap 4B=D\cap 4B$.

Note that we are not interested in the precise dependence of $r$ with respect
to $r_{0}$, because we seek for a qualitative result in Theorem 1.1. It is
quite likely that with some care the previous result could be made
quantitative.

###### Proof.

The proof uses the construction in [AMT16, Definition 2.1 and Lemma 2.2]. Set
$0<\varepsilon<1/100$ and $E\coloneqq\partial\Omega\cap 5\overline{B}$. Let
${\mathcal{W}}_{\varepsilon^{-2}}(E^{c})$ be the set of maximal dyadic cubes
$Q\subset E^{c}$ such that ${\rm diam}(\varepsilon^{-2}Q)\leq r_{0}$ and
$\varepsilon^{-2}Q\cap E=\emptyset$.

Denote $\mathcal{I}$ the family of cubes
$Q\in{\mathcal{W}}_{\varepsilon^{-2}}(E^{c})$ such that
$Q\cap\partial\Omega\not=\emptyset$. For each cube $Q\in\mathcal{I}$ fix a
point $z_{Q}\in Q\cap\partial\Omega$, and set
$r_{Q}\coloneqq\varepsilon\min\\{r_{0},{\rm dist}(z_{Q},E)\\}$ and
$B_{Q}\coloneqq B(z_{Q},r_{Q})$. Consider the enlarged domain

$\Omega_{\varepsilon}^{+}\coloneqq\Omega\cup\bigcup_{Q\in\mathcal{I}}B_{Q}\supset\Omega.$

By [AMT16, Lemma 2.2], this new domain is
$(c\varepsilon^{1/2},r_{0}/2)$-Reifenberg flat, where the constant $c$ depends
only on the dimension, provided the initial domain is
$(\delta,r_{0})$-Reifenberg flat with $\delta\in(0,\delta_{0})$ and
$\delta_{0}$ is small enough depending on $\varepsilon$.

Consider the domain $D_{0}\coloneqq\Omega_{\varepsilon}^{+}\cap 10B$. Clearly
$D_{0}\subset\\{z:{\rm dist}(z,\partial\Omega)<r_{0}/2\\}$ as $10r_{B}\leq
r_{0}/10$. Let us smooth the corners of $D_{0}$ out, where the
$(c\varepsilon^{1/2},s)$-Reifenberg flat condition fails for all $s>0$. Note
that this may only happen in a finite number of points $\\{\xi_{j}\\}_{j\in
J}\in\partial D_{0}\cap\partial 10B$ because $\Omega_{\varepsilon}^{+}$ is
contructed as a countable union of balls. Fix a small parameter $\tau$. For
each $\xi_{j}$ of these points in $\partial D_{0}\cap\partial 10B$, let
$B_{j}\in\\{B_{Q}:Q\in\mathcal{I}\\}$ such that $\xi_{j}\in\partial
10B\cap\partial B_{j}$, and let $c_{j}$ and $r_{j}$ denote its centers and
radii respectively. Consider now the unique point $p_{j}\in D_{0}\cap(\partial
B(c_{B},10r_{B}-\tau)\cap\partial B(c_{j},r_{j}-\tau))$. In particular, the
ball $B(p_{j},\tau)$ is tangent to $\partial 10B$ and $\partial B_{j}$. Let
$T_{j}=10B\cap B_{j}\cap B(p_{j},\tau)^{c}$ the bounded open region enclosed
between the previous balls. Taking $\tau$ to be small enough, the final domain

$D\coloneqq D_{0}\setminus\bigcup_{j\in J}T_{j}\subset D_{0}$

satisfies $D\cap 4B=\Omega_{\varepsilon}^{+}\cap 4B=\Omega\cap 4B$ and is
$(c\varepsilon^{1/2},r)$-Reifenberg flat for some $r>0$ depending on $\tau$. ∎

######  Claim 6.4.

If Theorem 1.1 holds for bounded $(\delta_{0},r_{0})$-Reifenberg flat domains,
then there exists $\delta_{1}=\delta_{1}(\delta_{0})>0$ such that Theorem 1.1
holds for unbounded $(\delta_{1},r_{0})$-Reifenberg flat domains.

###### Proof.

First we want to remark that if Theorem 1.1 holds for
$(\delta,r_{0})$-Reifenberg flat domains for a fixed $r_{0}>0$, then by means
of a dilation it holds for $(\delta,r)$-Reifenberg flat domains for any $r>0$.

Let $\varepsilon>0$ given by Lemma 6.3 and let
$\varepsilon^{\prime}\coloneqq\min\\{\varepsilon,\delta_{0}/2\\}$ be small
enough. Let $\delta_{1}=\delta_{1}(\varepsilon^{\prime})$ given by Lemma 6.3.
Let $\Omega\subset{\mathbb{R}}^{2}$ be an unbounded
$(\delta_{1},r_{0})$-Reifenberg flat domain. Let
$\\{B_{i}\\}_{i}\subset\\{B(\xi,r)\\}_{\xi\in\partial\Omega,0<r<r_{0}/100}$ be
a disjoint family with
$\omega_{\Omega,A}(\partial\Omega\setminus\bigcup_{i}B_{i})=0$, by Vitali’s
covering theorem. For each ball $B_{i}$ let $D_{i}$ be the bounded
$(\varepsilon,r_{i})$-Reifenberg flat domain from Lemma 6.3, for some
$r_{i}\in(0,r_{0}/2)$. As $\varepsilon^{\prime}<\delta_{0}$, in particular
each $D_{i}$ is a bounded $(\delta_{0},r_{i})$-Reifenberg flat domain.

As we are assuming that Theorem 1.1 holds for bounded
$(\delta_{0},r)$-Reifenberg flat domains for any $r>0$, for each $i$ take
$F_{i}\subset\partial D_{i}$ with $\omega_{D_{i},A}(F_{i})=1$ and
$\sigma$-finite one-dimensional Hausdorff measure. From $\Omega\cap
4B_{i}=D_{i}\cap 4B_{i}$, the maximum principle and $\omega_{D_{i}}(F_{i})=1$
we get

$\omega_{\Omega\cap 4B_{i},A}((\partial\Omega\setminus F_{i})\cap
4B_{i})=\omega_{D_{i}\cap 4B_{i},A}((\partial D_{i}\setminus F_{i})\cap
4B_{i})\leq\omega_{D_{i},A}((\partial D_{i}\setminus F_{i})\cap 4B_{i})=0.$

In particular $\omega_{\Omega\cap 4B_{i},A}((F_{i}\cap 4B_{i})\cup\partial
4B_{i})=1$. As $F_{i}$ has $\sigma$-finite length, so does
$\widetilde{F}_{i}\coloneqq(F_{i}\cap 4B_{i})\cup\partial 4B_{i}$. By Lemma
6.2, the set $F=\partial\Omega\cap\bigcup_{i}\widetilde{F}_{i}$ satisfies
$\omega_{\Omega}(F)=1$, and clearly has $\sigma$-finite length. ∎

###  6.2. Proof for bounded domains

Theorem 1.1 follows from 3.1 as it is done in [Wol93, Proof of Theorem 1],
with some small modifications. For the sake of completeness we give the
detailed proof.

###### Proof of Theorem 1.1.

By Claim 6.4 we can assume without loss of generality that $\Omega$ is
bounded. We denote $\omega\coloneqq\omega_{\Omega,A}$.

Let $\phi:[0,\infty)\to[0,\infty)$ be any increasing function with $\lim_{t\to
0}\phi(t)/t=0$, and consider the $\phi$-Hausdorff content

$h_{\phi}(E)=\inf\left\\{\sum_{i}\phi(r_{i}):E\subset\bigcup_{i}B(z_{i},r_{i})\right\\}.$

Now we claim that there exists $F_{\phi}\subset\partial\Omega$ with
$\omega(F_{\phi})=1$ and $h_{\phi}(F_{\phi})=0$. Indeed, suppose that the pole
$z\in\Omega$ is such that ${\rm dist}(z,\partial\Omega)>r_{0}$. Set $\tau=1/2$
and fix $0<r\leq 1$ to be small enough as in 3.1. Let $\varepsilon>0$ and fix
$M>\varepsilon^{-1}$ satisfying the hypothesis in 3.1. Take $\rho$ small
enough such that $0<\rho/r<1/M$ and $\phi(\gamma)<\varepsilon M^{-1/2}\gamma$
for all $\gamma\leq\rho$. Then, by 3.1 with these parameters, we obtain a set
$F_{\varepsilon}\subset\partial\Omega$ such that
$\omega^{z}(F_{\varepsilon})\geq C^{-1}$ and with a covering
$F_{\varepsilon}\subset\bigcup_{i}B(z_{i},r_{i})$ with $\sum_{i}r_{i}\leq
CM^{1/2}$ and $\sum_{\\{i\,:\,r_{i}>\rho\\}}r_{i}\leq CM^{-1}$. This covering
satisfies

$h_{\phi}(F_{\varepsilon})\leq\sum_{r_{i}\leq\rho}\phi(r_{i})+\sum_{r_{i}>\rho}\phi(r_{i})\leq\varepsilon
M^{-1/2}\sum_{r_{i}\leq\rho}r_{i}+\sum_{r_{i}>\rho}r_{i}\leq\varepsilon
M^{-1/2}CM^{1/2}+CM^{-1}\leq C\varepsilon.$

Define now $F_{\infty}\subset\partial\Omega$ as

$F_{\infty}\coloneqq\limsup_{j\to\infty}F_{1/j^{2}}=\bigcap_{j\geq
1}\bigcup_{k\geq j}F_{1/k^{2}}.$

With this choice we have

$\omega^{z}(F_{\infty})=\lim_{j\to\infty}\omega^{z}\left(\bigcup_{k\geq
j}F_{1/k^{2}}\right)\geq\limsup_{j\to\infty}\omega^{z}(F_{1/j^{2}})\geq
C^{-1},$

and as $F_{\infty}\subseteq\bigcup_{k\geq j}F_{1/k^{2}}$ for any $j\geq 1$,
then

$h_{\phi}(F_{\infty})\leq h_{\phi}\left(\bigcup_{k\geq
j}F_{1/k^{2}}\right)\leq\sum_{k\geq j}h_{\phi}(F_{1/k^{2}})\leq C\sum_{k\geq
j}\frac{1}{k^{2}},$

which gives $h_{\phi}(F_{\infty})=0$ letting $j\to\infty$.

Let $\\{z_{k}\\}_{k=1}^{\infty}$ be a countable dense subset of $\Omega$. Fix
$z_{k}$ and set $d_{k}={\rm dist}(z_{k},\partial\Omega)$. As $\Omega$ is
$(\delta,r_{0})$-Reifenberg flat and the matrix $A$ is Lipschitz in $\\{x:{\rm
dist}(x,\partial\Omega)<r_{0}\\}$, then $\Omega$ is
$(\delta,r_{k})$-Reifenberg flat and $A$ is Lipschitz in $\\{x:{\rm
dist}(x,\partial\Omega)<r_{k}\\}$ with $r_{k}=\min~\\{r_{0},d_{k}/2\\}$. By
the choice of $r_{k}$ we are in the situation ${\rm
dist}(z_{k},\partial\Omega)>r_{k}$. By the same argument done in the previous
paragraphs we get a set $F_{k}\subset\partial\Omega$ (relative to $z_{k}$)
such that

$\omega^{z_{k}}(F_{k})\geq C^{-1}\quad\text{and}\quad h_{\phi}(F_{k})=0.$

Define $F_{\phi}\coloneqq\bigcup_{k=1}^{\infty}F_{k}$. The condition
$h_{\phi}(F_{k})=0$ for every $k\geq 1$ means that for every $\epsilon>0$
there exists a covering
$F_{k}\subset\bigcup_{i}B(z_{i}^{k,\epsilon},r_{i}^{k,\epsilon})$ such that
$\sum_{i}\phi(r_{i}^{k,\epsilon})\leq\epsilon/2^{k}$. Hence,
$F_{\phi}=\bigcup_{k=1}^{\infty}F_{k}\subset\bigcup_{k=1}^{\infty}\bigcup_{i}B(z_{i}^{k,\epsilon},r_{i}^{k,\epsilon})$
which gives $h_{\phi}(F_{\phi})=0$ because

$h_{\phi}(F_{\phi})\leq\sum_{k=1}^{\infty}\sum_{i}\phi(r_{i}^{k,\epsilon})\leq\sum_{k=1}^{\infty}\frac{\epsilon}{2^{k}}=\epsilon.$

Moreover $\omega^{z_{k}}(F_{\phi})\geq\omega^{z_{k}}(F_{k})\geq C^{-1}$ for
any $z_{k}$. As $\\{z_{k}\\}_{k=1}^{\infty}\subset\Omega$ is dense and
$\omega^{z}(F_{\phi})$ is $L_{A}$-harmonic with respect to $z$ (in particular
continuous), then $\omega^{p}(F_{\phi})\geq C^{-1}$ for any $p\in\Omega$. By
[HKM06, Lemma 11.16] we conclude $\omega^{p}(F_{\phi})=1$ for any
$p\in\Omega$.

Finally let

$F=\left\\{\xi\in\partial\Omega:\limsup_{r\to
0}\frac{\omega^{p}(B(\xi,r))}{r}>0\right\\}.$

We claim that this set has $\sigma$-finite length and $\omega(F)=1$. We start
by proving $\omega(F)=1$, and later we will show that it has $\sigma$-finite
length.

Suppose to get a contradiction that $\omega^{p}(F)\not=1$, that is,

$\omega^{p}(F^{c})=\omega^{p}\left(\left\\{\xi\in\partial\Omega:\lim_{r\to
0}\frac{\omega^{p}(B(\xi,r))}{r}=0\right\\}\right)>0.$

Egorov’s theorem ensures that for every $s>0$ there exists a measurable set
$V=V_{s}\subset F^{c}$ such that $\omega^{p}(V)<s$ and
$\omega^{p}(B(\xi,r))/r\to 0$ as $r\to 0$ uniformly on $F^{c}\setminus V$.
Since $0<\omega^{p}(F^{c})=\omega^{p}(V)+\omega^{p}(F^{c}\setminus
V)<s+\omega^{p}(F^{c}\setminus V)$, if we take $s>0$ small enough, say
$s=\omega^{p}(F^{c})/2$, then $\omega^{p}(F^{c}\setminus V)>0$. The set
$Y\coloneqq F^{c}\setminus V$ has non zero elliptic measure and the limit

$\lim_{r\to
0}\left\|\frac{\omega^{p}(B(\cdot,r))}{r}\right\|_{L^{\infty}(Y)}=0.$

Note that the function $\phi(r)\coloneqq\sup_{\xi\in Y}\omega^{p}(B(\xi,r))$
satisfies the conditions of the rate function in the beginning of this proof,
and by definition it satisfies $\omega^{p}(B(\xi,r))\leq\phi(r)$ for all
$\xi\in Y$ and all $r>0$. All in all,

  1. (1)

$\omega^{p}(B(\xi,r))\leq\phi(r)$ for all $\xi\in Y$ and all $r>0$,

  2. (2)

$\phi$ is increasing, and

  3. (3)

$\phi(r)/r\to 0$ as $r\to 0$.

For this particular function $\phi$, let $F_{\phi}$ be the set constructed in
the beginning of this proof. That is, a set $F_{\phi}$ with
$h_{\phi}(F_{\phi})=0$ and $\omega^{p}(F_{\phi})=1$ for every $p\in\Omega$.
Hence $\omega^{p}(Y\cap F_{\phi})=\omega^{p}(Y)>0$, and moreover
$h_{\phi}(Y\cap F_{\phi})\leq h_{\phi}(F_{\phi})=0$. Consequently, we can
cover $Y\cap F_{\phi}$ with balls $B(\xi_{i},r_{i})$ centered at $Y\cap
F_{\phi}$ such that $\sum_{i}\phi(r_{i})<\omega^{p}(Y\cap F_{\phi})/2$. With
this we get

$\omega^{p}(Y\cap
F_{\phi})\leq\sum_{i}\omega^{p}(B(\xi_{i},r_{i}))\overset{\text{{(\ref{phi
prop 1})}}}{\leq}\sum_{i}\phi(r_{i})<\frac{\omega^{p}(Y\cap F_{\phi})}{2}.$

This is a contradiction, and hence $\omega^{p}(F)=1$.

It remains to prove that $F$ has $\sigma$-finite one-dimensional Hausdorff
measure. The set $F$ can be written as

$F=\bigcup_{j\geq 1}F^{1/j}\quad\text{with
}F^{1/j}\coloneqq\left\\{\xi\in\partial\Omega:\limsup_{r\to
0}\frac{\omega^{p}(B(\xi,r))}{r}>1/j\right\\}.$

Therefore, it suffices to see that every $F^{1/j}$ has finite one-dimensional
Hausdorff measure. Each point $\xi\in F^{1/j}$ has arbitrarily small
neighborhoods $B(\xi,r)$ such that $\omega^{p}(B(\xi,r))\geq r/j$. Given
$\varepsilon>0$ small, consider the family of these balls centered at
$F^{1/j}$ with radius at most $\varepsilon$, i.e.,

$\mathcal{B}_{\varepsilon}\coloneqq\\{B(\xi,r):\xi\in
F^{1/j},r<\varepsilon,\text{ and }\omega^{p}(B(\xi,r))\geq r/j\\}.$

By the Besicovitch covering theorem there is a countable subfamily
$\\{B(\xi_{i},r_{i})\\}_{i}\subset\mathcal{B}_{\varepsilon}$ such that no
point belongs to more than a fixed finite number $C$ (it depends on the
dimension only) of these balls. Hence,

$\sum_{i}r_{i}\leq j\sum_{i}\omega^{p}(B(\xi_{i},r_{i}))\leq Cj,$

and letting $\varepsilon\to 0$ we obtain ${\mathcal{H}}^{1}(F^{1/j})\leq Cj$,
as claimed. ∎

Hence Theorem 1.1 is proved under the assumption that (1.2) holds.

##  7\. $L\log L(d\sigma)$ type estimate for small densities: Proof of (1.2)

The purpose of this section is to prove (1.2), under the hypothesis of Lemma
3.2. More specifically, we prove the following result.

######  Lemma 7.1.

Let $\Omega\subset{\mathbb{R}}^{2}$ be a bounded $(\delta,r_{0})$-Reifenberg
flat domain, $p\in\Omega$ with ${\rm dist}(p,\partial\Omega)>r_{0}$, and $A$
be a real uniformly elliptic (not necessarily symmetric) matrix with
ellipticity constant $\lambda$. Suppose also that $A$ is $\kappa$-Lipschitz in
$U_{r_{0}}(\partial\Omega)\coloneqq\\{x\in{\mathbb{R}}^{2}:{\rm
dist}(x,\partial\Omega)<r_{0}\\}$ and that its symmetric part
$A_{0}=\frac{A+A^{T}}{2}$ is of the form $A_{0}=R^{T}BR$, with $R\in
C^{0,1}(U_{r_{0}}(\partial\Omega))$ a rotation and $B\in
C^{0,1}(U_{r_{0}}(\partial\Omega))$ diagonal.

Then there exists
$\delta_{0}=\delta_{0}(\lambda,\kappa\|A\|_{L^{\infty}({\mathbb{R}}^{2})})>0$
and $C=C(\lambda,\kappa,{\rm diam}\,\partial\Omega)\in(0,\infty)$ such that if
$\delta\leq\delta_{0}$, then for any $(\delta,r_{0})$-Reifenberg flat domain
$\widetilde{\Omega}\subset\Omega$ with smooth boundary
$\partial\widetilde{\Omega}$ and small enough ${\rm
dist}_{\mathcal{H}}(\partial\Omega,\partial\widetilde{\Omega})$, we have

(7.1) $\left|\int_{\partial\widetilde{\Omega}}\log|S\nabla
g_{p}^{T}(\xi)|^{2}\,d\widetilde{\omega}^{p}(\xi)\right|\leq C<+\infty,$

where $\widetilde{\omega}^{p}=\omega^{p}_{\widetilde{\Omega},A}$ is the
elliptic measure in $\widetilde{\Omega}$ with respect to the matrix $A$,
$g_{p}^{T}$ is the Green function in $\widetilde{\Omega}$ with respect to the
matrix $A^{T}$, and $S$ is the square root matrix of the symmetric part
$A_{0}=(A+A^{T})/2$, i.e., $S^{T}S=A_{0}$.

######  Remark 7.2.

As argued from (5.8) to (5.10), this implies that the Radon-Nykodym derivative
$\frac{d\widetilde{\omega}^{p}}{d\sigma}$ satisfies the following $L\log
L(d\sigma)$ type estimate

$-\infty<C^{\prime}(\lambda,\kappa,{\rm
diam}\,\partial\Omega)\leq\int_{\partial\widetilde{\Omega}}\frac{d\widetilde{\omega}^{p}}{d\sigma}(\xi)\log\frac{d\widetilde{\omega}^{p}}{d\sigma}(\xi)\,d\sigma(\xi).$

By symmetry, estimate (7.1) is equivalent to the existence of a constant
$C=C(\lambda,\kappa,{\rm diam}\partial\Omega)$ depending only on the
ellipticity constant, the Lipschitz seminorm of the matrix $A$, and the
diameter ${\rm diam}\,\partial\Omega$ (but not on $\widetilde{\Omega}$) such
that

(7.2) $\left|\int_{\partial\widetilde{\Omega}}\log|S\nabla
g(\xi)|^{2}\,d\widetilde{\omega}_{T}^{p}(\xi)\right|\leq C<\infty,$

where $S=A_{0}^{1/2}$, i.e., $S^{T}S=A_{0}$, $g=g_{p}$ is the Green function
in $\widetilde{\Omega}$ with respect to the matrix $A$ with pole $p$, and
$\widetilde{\omega}_{T}=\omega_{\widetilde{\Omega},A^{T}}^{p}$ denotes the
elliptic measure in $\widetilde{\Omega}$ with respect to the matrix $A^{T}$
and a pole $p\in\widetilde{\Omega}$ such that ${\rm
dist}(p,\partial\Omega)>r_{0}$. The existence of the matrix $S$ is granted by
the fact that the symmetric matrix $A_{0}$ is uniformly elliptic with the same
ellipticity constant as $A$, and hence positive definite. In fact,
$S=\sqrt{B}R$, where $\sqrt{B}=(\delta_{i=j}\sqrt{b_{ij}})_{1\leq i,j\leq 2}$.

Throughout all this section, when dealing with terms within solid integrals we
will use the following notation:

  * •

We write $h(x)={\mathcal{O}}(f(x))$ to denote “$|h(x)|\leq Cf(x)$ almost
everywhere with respect to the Lebesgue measure”.

  * •

Given a Lipschitz function $h$, we write $|\nabla h(x)|\leq C$ instead of
“$|\nabla h(x)|\leq C$ almost everywhere with respect to the Lebesgue
measure”. Recall that Lipschitz functions are differentiable almost everywhere
by Rademacher’s theorem, see [Mat95, Theorem 7.3].

###  7.1. Directional derivatives and the dual space

We introduce some extra notation regarding the directional derivatives.

######  Notation.

Definition of directional derivatives.

  * •

The (vertical) vector
$e_{i}\coloneqq(0,\ldots,0,\overset{\mathop{\stackMath\stackinset{c}{0pt}{c}{-0.6ex}{{\scriptscriptstyle\smile}}{{\scriptscriptstyle
i}}}}{1},0,\ldots,0)^{T}$ has $1$ in position $i$ and $0$’s otherwise. In
${\mathbb{R}}^{2}$ there are only two such vectors, namely $e_{1}=(1,0)^{T}$
and $e_{2}=(0,1)^{T}$.

  * •

$R_{i}\coloneqq e_{i}^{T}\cdot R$ corresponds to the $i$-th row of $R$.

  * •

The $\partial^{i}$-directional derivative
$\partial^{i}\coloneqq\partial_{R^{T}_{i}}$ (superscript) is defined as

$\partial^{i}f\coloneqq\partial_{R^{T}_{i}}f=\langle\nabla
f,R^{T}_{i}\rangle=R_{i}\cdot\nabla f.$

Here $\langle\cdot,\cdot\rangle$ denotes the standard scalar product in
${\mathbb{R}}^{n+1}$. The derivative $\partial_{i}=\partial_{e_{i}}$
(subscript) is the usual one in the direction $e_{i}$.

  * •

The $R$-directional gradient $\nabla_{R}$ is defined as

$\nabla_{R}f\coloneqq R\cdot\nabla f=\left(\partial^{i}f\right)_{i=1}^{n+1}.$

######  Remark 7.3.

The directional derivative $\partial^{i}$ preserves the usual properties in
sums, products, and logarithms, i.e.,
$\partial^{i}(f+g)=\partial^{i}f+\partial^{i}g$,
$\partial^{i}(fg)=g\partial^{i}f+f\partial^{i}g$, and $\partial^{i}\log
f=\frac{\partial^{i}f}{f}$.

Note that

(7.3) $\left|\partial^{ij}f\right|=\left|\partial^{i}(R_{j}\cdot\nabla
f)\right|=\left|\partial^{i}R_{j}\cdot\nabla f+R_{j}\partial^{i}\nabla
f\right|\lesssim|\nabla f|+\left|\nabla^{2}f\right|,$

where we denote
$|\nabla^{2}f|^{2}\coloneqq\sum_{i,j=1}^{n+1}(\partial_{i,j}f)^{2}$.

Since the directions $R_{i}$ change depending on the point, the usual
integration by parts formula does not apply. Instead we have the following
formula.

######  Claim 7.4 (Integration by parts formula).

Let $U\subset{\mathbb{R}}^{n+1}$ be an open set. For $f,h\in W^{1,2}(U)$, if
$fh\in W_{0}^{1,2}(U)$ then

(7.4)
$\int_{U}\partial^{i}f(x)h(x)\,dx=-\int_{U}f(x)h(x)\operatorname{div}R^{T}_{i}(x)\,dx-\int_{U}f(x)\partial^{i}h(x)\,dx.$

###### Proof.

The product rule gives
$\partial^{i}f(x)h(x)=\partial^{i}(f(x)h(x))-f(x)\partial^{i}h(x)$. By
definition of the directional derivative $\partial^{i}$ we can write

$\displaystyle\partial^{i}(f(x)h(x))$
$\displaystyle=R_{i}(x)\cdot\nabla(f(x)h(x))=\sum_{j=1}^{n+1}\partial_{j}(f(x)h(x))R_{i,j}(x)$
$\displaystyle=\sum_{j=1}^{n+1}\partial_{j}\left(f(x)h(x)R_{i,j}(x)\right)-f(x)h(x)\sum_{j=1}^{n+1}\partial_{j}R_{i,j}(x)$
$\displaystyle=\operatorname{div}(fhR^{T}_{i})(x)-f(x)h(x)\operatorname{div}R^{T}_{i}(x),$

which imply

$\int\partial^{i}f(x)h(x)\,dx=\int\operatorname{div}(fhR^{T}_{i})(x)\,dx-\int
f(x)h(x)\operatorname{div}R^{T}_{i}(x)\,dx-\int f(x)\partial^{i}h(x)\,dx.$

Using that $C^{\infty}_{c}(U)$ is dense in $W_{0}^{1,2}(U)$ and the divergence
theorem, the first element in the right-hand side is
$\int\operatorname{div}(fhR^{T}_{i})(x)\,dx=0$. ∎

Again, since the directions $R$ depend on the point, the directional
derivatives do not commute. Instead, the following formula relating
$\partial^{\alpha,\beta}$ and $\partial^{\beta,\alpha}$ is available, where
$\partial^{\alpha,\beta}\coloneqq\partial^{\alpha}\partial^{\beta}$. We refer
to this as the ‘almost’-commutative property.

######  Claim 7.5.

Let $U\subset{\mathbb{R}}^{n+1}$ be an open set and $f\in W^{2,2}(U)$. For
$\alpha,\beta\in\\{1,\ldots,n+1\\}$,

$\partial^{\alpha,\beta}f-\partial^{\beta,\alpha}f=(\partial^{\alpha}R_{\beta}-\partial^{\beta}R_{\alpha})\cdot\nabla
f.$

###### Proof.

By definition we have $\partial^{\beta}f=R_{\beta}\cdot\nabla
f=\sum_{i=1}^{n+1}\partial_{i}fR_{\beta,i}$. Applying $\partial^{\alpha}$
gives

$\partial^{\alpha,\beta}f=R_{\alpha}\cdot\nabla\left(\sum_{i=1}^{n+1}\partial_{i}fR_{\beta,i}\right)=\sum_{j=1}^{n+1}\partial_{j}\left(\sum_{i=1}^{n+1}\partial_{i}fR_{\beta,i}\right)R_{\alpha,j}.$

Expanding the derivative of this last expression and arranging we obtain

$\partial^{\alpha,\beta}f=\sum_{i,j=1}^{n+1}\partial_{j,i}fR_{\beta,i}R_{\alpha,j}+\sum_{i=1}^{n+1}\partial_{i}f\sum_{j=1}^{n+1}\partial_{j}R_{\beta,i}R_{\alpha,j}.$

The sum inside the second term in the right-hand side is precisely
$\sum_{j=1}^{n+1}\partial_{j}R_{\beta,i}R_{\alpha,j}=R_{\alpha}\cdot\nabla
R_{\beta,i}=\partial^{\alpha}R_{\beta,i}$. Hence

$\partial^{\alpha,\beta}f=\sum_{i,j=1}^{n+1}\partial_{j,i}fR_{\beta,i}R_{\alpha,j}+\sum_{i=1}^{n+1}\partial_{i}f\partial^{\alpha}R_{\beta,i}=\sum_{i,j=1}^{n+1}\partial_{j,i}fR_{\beta,i}R_{\alpha,j}+\partial^{\alpha}R_{\beta}\cdot\nabla
f.$

By symmetry we obtain
$\partial^{\beta,\alpha}f=\sum_{i,j=1}^{n+1}\partial_{j,i}fR_{\alpha,i}R_{\beta,j}+\partial^{\beta}R_{\alpha}\cdot\nabla
f$, and subtracting $\partial^{\alpha,\beta}f-\partial^{\beta,\alpha}f$ we get
Claim 7.5. ∎

In particular, setting $\alpha=1$ and $\beta=2$ in the planar case we have

$\partial^{1,2}f-\partial^{2,1}f=(\partial^{1}R_{2}-\partial^{2}R_{1})\cdot\nabla
f,$

and applying it to the Green function $g$, we get

(7.5)
$\partial^{1,2}g-\partial^{2,1}g=(\partial^{1}R_{2}-\partial^{2}R_{1})\cdot\nabla
g={\mathcal{O}}(|\nabla g|).$

####  7.1.1. Definition of the dual space

Let $U\subset\widetilde{\Omega}$ and $u\in L^{1}_{\operatorname{loc}}(U)$.
Define the linear functional

$T_{u}(\psi)\coloneqq\int_{U}u\psi,$

whenever it makes sense. In particular, $T_{u}\in L^{\infty}_{c}(U)^{\prime}$.
To simplify the notation, we will also write $u(\psi)$ to denote
$T_{u}(\psi)$.

The $\partial_{i}$-derivative functional is defined as

$T_{\partial_{i}u}(\psi)\coloneqq-\int_{U}u\partial_{i}\psi,$

whenever it makes sense. Note that $T_{\partial_{i}u}\in
W^{1,\infty}_{c}(U)^{\prime}$ for $u\in L^{1}_{\operatorname{loc}}(U)$.
Moreover, we also have $T_{\partial_{i}u}\in W^{1,2}_{c}(U)^{\prime}$ whenever
$u\in L^{2}_{\operatorname{loc}}(U)$. As before, we will also write $Du(\psi)$
to denote $T_{Du}(\psi)$ for any differential operator $D$. In general, the
functional

$T_{fD(u)}(\psi)\coloneqq T_{D(u)}(f\psi),$

is in $W^{1,\infty}_{c}(U)^{\prime}$ as long as either $f\in
W^{1,2}_{\operatorname{loc}}(U)$ and $u\in L^{2}_{\operatorname{loc}}(U)$, or
$f\in W^{1,\infty}_{\operatorname{loc}}(U)$ and $u\in
L^{1}_{\operatorname{loc}}(U)$.

We claim that whenever $u\in L^{2}_{\operatorname{loc}}(U)$ and $f\in
W^{1,2}_{\operatorname{loc}}(U)$, in the dual space
$W^{1,\infty}_{c}(U)^{\prime}$, the $\partial_{i}$-derivative functional
satisfies the product rule

(7.6) $T_{\partial_{i}(fu)}=T_{\partial_{i}fu}+T_{f\partial_{i}u}.$

Indeed, for any $\psi\in W_{c}^{1,\infty}(U)$, the previous equality reads as

$-\int_{U}fu\partial_{i}\psi=\int_{U}\partial_{i}fu\psi-\int_{U}u\partial_{i}(f\psi),$

which holds by the Leibniz’s rule a.e. for $W^{1,2}$-functions.

Using the definitions of $T_{\partial_{i}}$ and the directional derivative
$\partial^{i}(\cdot)=R_{i}\nabla(\cdot)$, see the beginning of Section 7.1,
the directional $\partial^{i}$-derivative functional
$T_{\partial^{i}u}:W^{1,\infty}_{c}(U)\to{\mathbb{R}}$, for $u\in
L^{1}_{\operatorname{loc}}(U)$, is defined as

(7.7) $T_{\partial^{i}u}\coloneqq T_{R_{i}\nabla
u}=\sum_{j}T_{R_{i,j}\partial_{j}u}.$

Equivalently, for $\psi\in W^{1,\infty}_{c}(U)$,

(7.8) $\displaystyle T_{\partial^{i}u}(\psi)$
$\displaystyle=\sum_{j}T_{\partial_{j}u}(R_{i,j}\psi)=-\sum_{j}\int_{U}u\partial_{j}(R_{i,j}\psi)=-\sum_{j}\int_{U}u\psi\partial_{j}R_{i,j}-\sum_{j}\int_{U}uR_{i,j}\partial_{j}\psi$
$\displaystyle=-\int_{U}u\psi\operatorname{div}R_{i}^{T}-\int_{U}u\partial^{i}\psi.$

This agrees with the integration by parts formula in (7.4) when $u\in
W^{1,2}(U)$.

Next we claim that, in the dual space $W_{c}^{1,\infty}(U)^{\prime}$, for
$u\in L^{2}_{\operatorname{loc}}(U)$ and $f\in
W^{1,2}_{\operatorname{loc}}(U)$, the directional $\partial^{i}$-derivative
functional satisfies the product rule

(7.9) $T_{\partial^{i}(fu)}=T_{\partial^{i}fu}+T_{f\partial^{i}u}.$

Indeed, $R_{i,j}\psi\in W_{c}^{1,\infty}(U)$ when $\psi\in
W^{1,\infty}_{c}(U)$, and hence, for $f\in W^{1,2}_{\operatorname{loc}}(U)$,

$T_{\partial^{i}(fu)}(\psi)\overset{\text{{(\ref{def:distributional
directional
derivative})}}}{=}\sum_{j}T_{\partial_{j}(fu)}(R_{i,j}\psi)\overset{\text{{(\ref{product
rule distributional
final})}}}{=}\sum_{j}T_{\partial_{j}fu}(R_{i,j}\psi)+T_{f\partial_{j}u}(R_{i,j}\psi)\overset{\text{{(\ref{def:distributional
directional
derivative})}}}{=}T_{\partial^{i}fu}(\psi)+T_{f\partial^{i}u}(\psi),$

as claimed.

Another important property is the following ‘almost’-commutative property of
the directional derivatives (compare to Claim 7.5). For $u\in
W^{1,1}_{\operatorname{loc}}(U)$, in the dual space
$W^{1,\infty}_{c}(U)^{\prime}$, we claim that

(7.10)
$T_{\partial^{1}(\partial^{2}u)}-T_{\partial^{2}(\partial^{1}u)}=T_{(\partial^{1}R_{2}-\partial^{2}R_{1})\cdot\nabla
u}.$

Indeed, for $\psi\in W^{1,\infty}_{c}(U)$ and $\alpha,\beta\in\\{1,2\\}$,

$T_{\partial^{\alpha}(\partial^{\beta}u)}(\psi)\overset{\text{{(\ref{def2:distributional
directional
derivative})}}}{=}-\int_{U}\partial^{\beta}u\psi\operatorname{div}R_{\alpha}^{T}-\int_{U}\partial^{\beta}u\partial^{\alpha}\psi.$

Given $\\{u_{k}\\}_{k\geq 1}\subset C_{c}^{\infty}(U)$ with
$\lim_{k\to\infty}\|u_{k}-u\|_{W^{1,1}(\operatorname{supp}\psi)}=0$, this
equals

(7.11) $\displaystyle T_{\partial^{\alpha}(\partial^{\beta}u)}(\psi)=$
$\displaystyle-\int_{U}\partial^{\beta}u_{k}\psi\operatorname{div}R_{\alpha}^{T}-\int_{U}\partial^{\beta}u_{k}\partial^{\alpha}\psi$
$\displaystyle-\int_{U}\partial^{\beta}(u-u_{k})\psi\operatorname{div}R_{\alpha}^{T}-\int_{U}\partial^{\beta}(u-u_{k})\partial^{\alpha}\psi.$

By the integration by parts in (7.4) applied to $\partial^{\beta}u_{k}$, we
have that the first row in the right-hand side is precisely
$\int_{U}\partial^{\alpha}(\partial^{\beta}u_{k})\psi$. Applying now the
‘almost’-commutative property (Claim 7.5) here we obtain

$\displaystyle T_{\partial^{\alpha}(\partial^{\beta}u)}(\psi)=$
$\displaystyle\int_{U}\partial^{\beta}(\partial^{\alpha}u_{k})\psi+\int_{U}((\partial^{\alpha}R_{\beta}-\partial^{\beta}R_{\alpha})\cdot\nabla
u_{k})\psi$
$\displaystyle-\int_{U}\partial^{\beta}(u-u_{k})\psi\operatorname{div}R_{\alpha}^{T}-\int_{U}\partial^{\beta}(u-u_{k})\partial^{\alpha}\psi.$

Again, by the integration by parts in (7.4), we can replace the first term in
the right-hand side to obtain

$\displaystyle T_{\partial^{\alpha}(\partial^{\beta}u)}(\psi)=$
$\displaystyle-\int_{U}\partial^{\alpha}u_{k}\psi\operatorname{div}R_{\beta}^{T}-\int_{U}\partial^{\alpha}u_{k}\partial^{\beta}\psi$
$\displaystyle+\int_{U}((\partial^{\alpha}R_{\beta}-\partial^{\beta}R_{\alpha})\cdot\nabla
u_{k})\psi$
$\displaystyle-\int_{U}\partial^{\beta}(u-u_{k})\psi\operatorname{div}R_{\alpha}^{T}-\int_{U}\partial^{\beta}(u-u_{k})\partial^{\alpha}\psi.$

Now, adding and subtracting $u$ in the second row we get

$\displaystyle T_{\partial^{\alpha}(\partial^{\beta}u)}(\psi)=$
$\displaystyle-\int_{U}\partial^{\alpha}u_{k}\psi\operatorname{div}R_{\beta}^{T}-\int_{U}\partial^{\alpha}u_{k}\partial^{\beta}\psi$
$\displaystyle+T_{(\partial^{\alpha}R_{\beta}-\partial^{\beta}R_{\alpha})\nabla
u}(\psi)+\int_{U}((\partial^{\alpha}R_{\beta}-\partial^{\beta}R_{\alpha})\cdot\nabla(u_{k}-u))\psi,$
$\displaystyle-\int_{U}\partial^{\beta}(u-u_{k})\psi\operatorname{div}R_{\alpha}^{T}-\int_{U}\partial^{\beta}(u-u_{k})\partial^{\alpha}\psi.$

By symmetry in (7.11) and subtracting we obtain

$\displaystyle
T_{\partial^{\alpha}(\partial^{\beta}u)}(\psi)-T_{\partial^{\beta}(\partial^{\alpha}u)}(\psi)=$
$\displaystyle\
T_{(\partial^{\alpha}R_{\beta}-\partial^{\beta}R_{\alpha})\nabla
u}(\psi)+\int_{U}((\partial^{\alpha}R_{\beta}-\partial^{\beta}R_{\alpha})\cdot\nabla(u_{k}-u))\psi,$
$\displaystyle-\int_{U}\partial^{\beta}(u-u_{k})\psi\operatorname{div}R_{\alpha}^{T}-\int_{U}\partial^{\beta}(u-u_{k})\partial^{\alpha}\psi$
$\displaystyle+\int_{U}\partial^{\alpha}(u-u_{k})\psi\operatorname{div}R_{\beta}^{T}+\int_{U}\partial^{\alpha}(u-u_{k})\partial^{\beta}\psi.$

By the convergence
$\lim_{k\to\infty}\|u_{k}-u\|_{W^{1,1}(\operatorname{supp}\psi)}=0$ we get
(7.10).

####  7.1.2. Properties of directional derivatives and the dual space

The following claim allows us to move from the initial matrix $A$ to its
symmetric part $A_{0}$, which we will relate to the directional derivatives in
Claim 7.7.

######  Claim 7.6.

Let $U\subset\widetilde{\Omega}$ be an open set and $u\in
W_{\operatorname{loc}}^{1,2}(U)$. Every $\psi\in W^{1,\infty}_{c}(U)$
satisfies

$\left(\operatorname{div}A\nabla
u\right)(\psi)=\left(\operatorname{div}A_{0}\nabla
u\right)(\psi)+\left(\sum_{i,j=1}^{n+1}\partial_{i}\left(\frac{a_{i,j}-a_{j,i}}{2}\right)\partial_{j}u\right)(\psi).$

###### Proof.

Let $u\in C^{\infty}(U)$. By definition of the divergence and differentiating
we have

$\operatorname{div}A\nabla
u=\sum_{i,j=1}^{n+1}\partial_{i}a_{i,j}\partial_{j}u+\sum_{i,j=1}^{n+1}a_{i,j}\partial_{i,j}u=\sum_{i,j=1}^{n+1}\partial_{i}a_{i,j}\partial_{j}u+\sum_{i,j=1}^{n+1}\frac{a_{i,j}+a_{j,i}}{2}\partial_{i,j}u.$

Note that $a^{0}_{i,j}\coloneqq\frac{a_{i,j}+a_{j,i}}{2}$ are precisely the
coefficients of the symmetric matrix $A_{0}$. The product derivative rule
gives

$a^{0}_{i,j}\partial_{i,j}u=\partial_{i}\left(a^{0}_{i,j}\partial_{j}u\right)-\partial_{i}a^{0}_{i,j}\partial_{j}u,$

and applying this relation to each pair of indexes $i,j\in\\{1,\ldots,n+1\\}$
we get

$\sum_{i,j=1}^{n+1}a^{0}_{i,j}\partial_{i,j}u=\operatorname{div}A_{0}\nabla
u-\sum_{i,j=1}^{n+1}\partial_{i}a^{0}_{i,j}\partial_{j}u.$

Summing up,

$\displaystyle\operatorname{div}A\nabla u$
$\displaystyle=\operatorname{div}A_{0}\nabla
u+\sum_{i,j=1}^{n+1}\partial_{i}a_{i,j}\partial_{j}u-\sum_{i,j=1}^{n+1}\partial_{i}\left(\frac{a_{i,j}+a_{j,i}}{2}\right)\partial_{j}u$
$\displaystyle=\operatorname{div}A_{0}\nabla
u+\sum_{i,j=1}^{n+1}\partial_{i}\left(\frac{a_{i,j}-a_{j,i}}{2}\right)\partial_{j}u,$

as claimed, for functions in $C^{\infty}(U)$.

Let us check this in the dual sense for a function $u\in
W^{1,2}_{\operatorname{loc}}(U)$. Let $\psi\in W^{1,\infty}_{c}(U)$, and hence
$u\in W^{1,2}(\operatorname{supp}\psi)$. Take $\\{u_{k}\\}_{k\geq 1}\subset
C^{\infty}_{c}(U)$ such that
$\lim_{k\to\infty}\|u_{k}-u\|_{W^{1,2}(\operatorname{supp}\psi)}=0$. As we
have the claim for each function $u_{k}$, in particular

$\displaystyle\int_{U}A\nabla u\nabla\psi=$
$\displaystyle\int_{U}A\nabla(u-u_{k})\nabla\psi+\int_{U}A\nabla
u_{k}\nabla\psi$ $\displaystyle=$
$\displaystyle\int_{U}A\nabla(u-u_{k})\nabla\psi+\int_{U}A_{0}\nabla
u_{k}\nabla\psi-\int_{U}\sum_{i,j=1}^{n+1}\partial_{i}\left(\frac{a_{i,j}-a_{j,i}}{2}\right)\partial_{j}u_{k}\cdot\psi$
$\displaystyle=$ $\displaystyle\int_{U}A_{0}\nabla
u\nabla\psi-\int_{U}\sum_{i,j=1}^{n+1}\partial_{i}\left(\frac{a_{i,j}-a_{j,i}}{2}\right)\partial_{j}u\cdot\psi$
$\displaystyle+\int_{U}A\nabla(u-u_{k})\nabla\psi+\int_{U}A_{0}\nabla(u_{k}-u)\nabla\psi-\int_{U}\sum_{i,j=1}^{n+1}\partial_{i}\left(\frac{a_{i,j}-a_{j,i}}{2}\right)\partial_{j}(u_{k}-u)\psi.$

Now, since the matrix is Lipschitz and
$\lim_{k\to\infty}\|u_{k}-u\|_{W^{1,2}(\operatorname{supp}\psi)}=0$, we
conclude

$\int_{U}A\nabla u\nabla\psi=\int_{U}A_{0}\nabla
u\nabla\psi-\int_{U}\sum_{i,j=1}^{n+1}\partial_{i}\left(\frac{a_{i,j}-a_{j,i}}{2}\right)\partial_{j}u\cdot\psi,$

as claimed. ∎

We treat one of the main terms in (7.1) by means of a perturbation argument.
To do that, we note that we can write $\operatorname{div}(A_{0}\nabla u)$ as
in the diagonal case using the $R$-directional derivatives, plus an error
term. More precisely, we have the following claim.

######  Claim 7.7.

Let $A_{0}=R^{T}BR$ satisfy the conditions in Lemma 7.1, let
$U\subset\widetilde{\Omega}$ be an open set and $u\in
W^{1,2}_{\operatorname{loc}}(U)$. Every $\psi\in W^{1,\infty}_{c}(U)$
satisfies

$\left(\operatorname{div}(A_{0}\nabla
u)\right)(\psi)=\left(\sum_{i=1}^{n+1}\partial^{i}(b_{i}\partial^{i}u)\right)(\psi)+\left(\sum_{i=1}^{n+1}b_{i}\partial^{i}u\cdot\operatorname{div}R^{T}_{i}\right)(\psi).$

###### Proof.

Assume first $u\in C^{\infty}(U)$. Let us write $R\nabla u=(R_{j}\cdot\nabla
u)_{j=1}^{n+1}=(\partial^{j}u)_{j=1}^{n+1}$, and so $BR\nabla
u=(b_{j}\partial^{j}u)_{j=1}^{n+1}$. Hence,

$R^{T}BR\nabla
u=R^{T}(b_{j}\partial^{j}u)_{j=1}^{n+1}=\left(\sum_{j=1}^{n+1}R^{T}_{i,j}b_{j}\partial^{j}u\right)_{i=1}^{n+1}.$

Therefore,

$\operatorname{div}(A_{0}\nabla
u)=\sum_{i=1}^{n+1}\partial_{i}\left(\sum_{j=1}^{n+1}R^{T}_{i,j}b_{j}\partial^{j}u\right)=\sum_{j=1}^{n+1}b_{j}\partial^{j}u\sum_{i=1}^{n+1}\partial_{i}(R_{j,i})+\sum_{j=1}^{n+1}\sum_{i=1}^{n+1}R_{j,i}\partial_{i}(b_{j}\partial^{j}u).$

Note that the sum inside the first term in the right-hand side is precisely
$\sum_{i=1}^{n+1}\partial_{i}(R_{j,i})=\operatorname{div}R^{T}_{j}$, and the
sum inside the second term in the right-hand side is
$\sum_{i=1}^{n+1}R_{j,i}\partial_{i}(b_{j}\partial^{j}u)=R_{j}\cdot\nabla(b_{j}\partial^{j}u)=\partial^{j}(b_{j}\partial^{j}u)$.
Thus, the claim follows for $u\in C^{\infty}(U)$.

From the definition of the directional $\partial^{i}$-derivative in 7.7 and
7.8, the claim follows by a density argument as in Claim 7.6. ∎

###  7.2. Sketch of the proof

Throughout this section the pole $p\in\Omega$ so that ${\rm
dist}(p,\partial\Omega)>r_{0}$ is fixed unless it is otherwise stated (see
Lemma 7.10 below). In any case, it will be far from $\partial\Omega$ and so
from $\partial\widetilde{\Omega}$. Recall also that $S\coloneqq A_{0}^{1/2}$,
i.e., $S^{T}S=A_{0}$.

Fix $R\coloneqq\min\\{1,r_{0}/2\\}/2$, and so we have
$0<R<\min\\{1,r_{0}/2\\}$. By [LLN08, Lemma 3.35], see Lemma 2.13 and Remark
2.14 above, if $\widetilde{\Omega}$ is Reifenberg flat enough, depending on
the ellipticity constant $\lambda$ and the value
$\kappa\|A\|_{L^{\infty}({\mathbb{R}}^{n+1})}$, then there exists a constant
$c=c(\lambda,\kappa\|A\|_{L^{\infty}({\mathbb{R}}^{n+1})})\geq 1$ such that

$c^{-1}|\nabla g(y)|\leq\frac{g(y)}{{\rm
dist}\left(y,\partial\widetilde{\Omega}\right)}\leq c|\nabla
g(y)|\quad\text{for every }y\in B(\xi,R/c)\text{ with
}\xi\in\partial\widetilde{\Omega}.$

Here we need to work with $(\delta,r_{0})$-Reifenberg flat domains with
flatness parameter $\delta$ smaller than some constant depending on the
ellipticity constant $\lambda$ and the value
$\kappa\|A\|_{L^{\infty}({\mathbb{R}}^{2})}$.

Let us now fix the support function $\varphi$ with $\varphi(p)=0$:

######  Remark 7.8.

Let $\varphi\in C_{c}^{\infty}({\mathbb{R}}^{2})$ with

  * •

$\varphi=1$ in
$U_{\frac{R}{2000c}}(\partial\Omega)\coloneqq\\{x\in{\mathbb{R}}^{2}:{\rm
dist}(x,\partial\Omega)\leq\frac{R}{2000c}\\}$,

  * •

$\varphi=0$ in $U_{\frac{R}{1500c}}(\partial\Omega)^{c}$, and

  * •

$|\nabla\varphi|\lesssim 1$.

So $\operatorname{supp}\varphi\subseteq U_{\frac{R}{1500c}}(\partial\Omega)$.
Note that
$\partial\widetilde{\Omega}\subset\Omega\cap\operatorname{supp}\varphi$ if we
assume that ${\rm
dist}_{\mathcal{H}}(\partial\widetilde{\Omega},\partial\Omega)$ is small
enough. With this choice of $\varphi$ we have the comparability

(7.12) $|\nabla g(y)|\approx\frac{g(y)}{{\rm
dist}(y,\partial\widetilde{\Omega})}\quad\text{for }y\in\widetilde{\Omega}\cap
U_{\frac{R}{1500c}}(\partial\Omega)\supset\widetilde{\Omega}\cap\operatorname{supp}\varphi.$

We claim

(7.13) $\log|S\nabla g|^{2}\in
W^{1,2}_{\operatorname{loc}}(\widetilde{\Omega}\cap
U_{\frac{R}{1500c}}(\partial\Omega)).$

Indeed, this follows as $|\nabla g|\approx g/{\rm
dist}(\cdot,\partial\widetilde{\Omega})>0$ in $\widetilde{\Omega}\cap
U_{\frac{R}{1500c}}(\partial\Omega)$ and $g\in W^{2,2}(\widetilde{\Omega}\cap
U_{\frac{R}{1500c}}(\partial\Omega))$, see (7.20).

######  Notation.

From now on the variables and the region of integration will not be written
unless they are not clear from the context.

Estimate (7.2) will follow from the following partial results.

For $a\geq 10$, let $\log_{(a)}x\coloneqq\max\\{\log x,-a\\}$. Note that in
the weak sense $(\log_{(a)}(x))^{\prime}=\mathbf{1}_{\\{x\geq
e^{-a}\\}}(x)\log x$.

######  Lemma 7.9 (Step 1).

For $a\geq 10$ big enough,

$\int_{\partial\widetilde{\Omega}}\log|S\nabla
g|^{2}\,d\widetilde{\omega}_{T}^{p}=\int_{\partial\widetilde{\Omega}}\log_{(a)}|S\nabla
g|^{2}\,d\widetilde{\omega}_{T}^{p}+\sigma(\partial\widetilde{\Omega}){\mathcal{O}}(ae^{-a/2}).$

Since $\sigma(\partial\widetilde{\Omega})<\infty$, the second term in the
right-hand side tends to zero as $a\to\infty$.

We prove the preceding lemma in Section 7.5.1.

######  Lemma 7.10 (Step 2).

For any $a\geq 10$ so that
$e^{-a}<\min_{z\in\operatorname{supp}\nabla\varphi}|S\nabla g(z)|^{2}$, and
for a.e. $p\in\Omega$ with ${\rm dist}(p,\partial\Omega)>r_{0}$,

$\int_{\partial\widetilde{\Omega}}\log_{(a)}|S\nabla
g|^{2}\,d\widetilde{\omega}_{T}^{p}=-\int_{\widetilde{\Omega}}\langle
A^{T}\nabla\log_{(a)}|S\nabla g|^{2},\nabla(\varphi
g)\rangle+{\mathcal{O}}\left(1+\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla
g|}g\right).$

We prove the preceding lemma in Section 7.5.2. Note that the left-hand side
integral is supported on the boundary, while the one on the right-hand side is
a solid integral.

Next, fix $\widetilde{\Omega}_{\varphi}\subset\widetilde{\Omega}$ a smooth
domain satisfying

$\widetilde{\Omega}\cap\operatorname{supp}\varphi\subset\widetilde{\Omega}_{\varphi}\subset
U_{\frac{R}{1000c}}(\partial\Omega)\cap\widetilde{\Omega}.$

With this choice we have the comparability $|\nabla g|\approx g/{\rm
dist}(\cdot,\partial\widetilde{\Omega})$ in $\widetilde{\Omega}_{\varphi}$,
see (7.12). We prove the following two lemmas in Section 7.5.1.

######  Lemma 7.11 (Step 3).

For $a\geq 10$ big enough,

$\displaystyle-\int_{\widetilde{\Omega}}\langle A^{T}\nabla\log_{(a)}|S\nabla
g|^{2},\nabla(\varphi g)\rangle=$
$\displaystyle-\int_{\widetilde{\Omega}}\langle A^{T}\nabla\log|S\nabla
g|^{2},\nabla(\varphi g)\rangle$
$\displaystyle+{\mathcal{O}}\left({\mathcal{H}}^{2}(\Omega)e^{-a/2}+\int_{\widetilde{\Omega}\cap\operatorname{supp}\varphi\cap\\{|S\nabla
g|^{2}\leq e^{-a}\\}}|\nabla^{2}g|\right).$

Note that the last term in the right-hand side also tends to zero as
$a\to\infty$ because $g\in W^{2,2}(\widetilde{\Omega}_{\varphi})$ and
$\bigcap_{a\geq 10}\\{|S\nabla g|^{2}\leq
e^{-a}\\}\cap\widetilde{\Omega}_{\varphi}=\emptyset$.

For $\varepsilon>0$ as small as desired, we consider a given function
$\psi_{\varepsilon}\in C^{\infty}({\mathbb{R}}^{2})$ satisfying

  1. (1)

$0\leq\psi_{\varepsilon}\leq 1$ everywhere,

  2. (2)

$\psi_{\varepsilon}=0$ in $U_{\varepsilon}(\partial\widetilde{\Omega})$,

  3. (3)

$\psi_{\varepsilon}=1$ in $U_{3\varepsilon}(\partial\widetilde{\Omega})^{c}$,
and

  4. (4)

$|\nabla\psi_{\varepsilon}|\lesssim\frac{1}{\varepsilon}$.

######  Lemma 7.12 (Step 4).

For $\psi_{\varepsilon}$ as above we have

$\lim_{\varepsilon\to 0}\left|\int_{\widetilde{\Omega}}\langle
A^{T}\nabla\log|S\nabla g|^{2},\nabla((1-\psi_{\varepsilon})\varphi
g)\rangle\right|\lesssim\lim_{\varepsilon\to
0}\frac{1}{\varepsilon}\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}\frac{|\nabla^{2}g|}{|\nabla
g|}g.$

Thus,

$\int_{\widetilde{\Omega}}\langle A^{T}\nabla\log|S\nabla
g|^{2},\nabla(\varphi g)\rangle=\lim_{\varepsilon\to
0}\left\\{\int_{\widetilde{\Omega}}\langle A^{T}\nabla\log|S\nabla
g|^{2},\nabla(\psi_{\varepsilon}\varphi
g)\rangle+{\mathcal{O}}\left(\frac{1}{\varepsilon}\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}\frac{|\nabla^{2}g|}{|\nabla
g|}g\right)\right\\}.$

In the weak sense, we can write the first term in the right-hand side in Lemma
7.12 as $\operatorname{div}(A^{T}\nabla\log|S\nabla g|^{2})$ acting on
$\psi_{\varepsilon}\varphi g\in
W_{c}^{1,\infty}(\widetilde{\Omega}_{\varphi})$, which can be understood as
test functions. When studying the harmonic measure in the plane, i.e., $A=Id$,
this term is $0$ by the harmonicity of $\log|\nabla g|$ far from the critical
points, a property that does not hold in general in higher dimensions. This
was a key point to establish (1.2) for the Laplacian in [JW88, Lemma 3.1].

In the following remark we discuss this argument in the constant coefficient
case.

######  Remark 7.13 (Constant matrix and $L_{A}$-harmonic functions).

Suppose now that the matrix $A$ is constant. Given any function $u$, by Claim
7.6,

$\operatorname{div}(A^{T}\nabla u)=\operatorname{div}(A_{0}\nabla u),$

i.e., we can reduce the study to the symmetric part. Moreover, by means of a
linear change of variables, see Lemma 3.3, we have that
$\operatorname{div}(A^{T}\nabla u)=0$ if and only if $\widetilde{u}=u\circ D$
satisfies
$\operatorname{div}\left(\widetilde{A_{0}}\nabla\widetilde{u}\right)=0$ where
$\widetilde{A_{0}}=D^{-1}A_{0}(D^{-1})^{T}$, for any constant matrix $D$ with
$\det D\not=0$. Since $A$ is constant now, if we choose $D=S^{T}$ where
$S=A_{0}^{1/2}$, i.e., $S^{T}S=A_{0}$, then
$\widetilde{A_{0}}=(S^{T})^{-1}A_{0}S^{-1}=(S^{T})^{-1}S^{T}SS^{-1}=Id$, and
so $\Delta\widetilde{u}=0$. It is known that if $\Delta\widetilde{u}=0$, then
$\Delta\log|\nabla\widetilde{u}|^{2}=0$ (only) in the plane, whenever
$\nabla\widetilde{u}\not=0$. If we undo the previous change of variables then
we obtain

$\operatorname{div}\left(A_{0}\nabla\left(\log\left|S\nabla
u\right|^{2}\right)\right)=0,$

off the critical points.

Back to the general case, when the matrix $A$ is non-constant, instead of
moving from our matrix to the Laplacian by means of a change of variables as
it is done in the constant case (meaning that we would need a non-constant
change of variables), we will work with its symmetric part. Using the
particular form $A_{0}=R^{T}BR$, the rotation matrices will play the role of
the directional derivatives, so we will be left with the diagonal matrix $B$,
and this allows us to apply the previous strategy.

Now we study the first term appearing in the right-hand side in Lemma 7.12,

$-\int_{\widetilde{\Omega}}\langle A^{T}\nabla\log|S\nabla
g|^{2},\nabla(\psi_{\varepsilon}\varphi g)\rangle,$

which is the most delicate, and the key point in the different behavior
between the planar case and higher dimensions. Recall that, in the dual space,
this term is

$\left(\operatorname{div}(A^{T}\nabla\log|S\nabla
g|^{2})\right)(\psi_{\varepsilon}\varphi g)\quad\text{where
}\psi_{\varepsilon}\varphi g\in
W^{1,\infty}_{c}(\widetilde{\Omega}_{\varphi}).$

######  Notation.

From now on, a boxed term $\boxed{X}$ (representing the functional $T_{X}$)
can be understood as a pointwise function, while a functional written as
$T_{X}$ needs to be understood strictly in the dual sense.

In the following lemma we decompose this last term.

######  Lemma 7.14 (Step 5).

In the dual space $W_{c}^{1,\infty}(\widetilde{\Omega}_{\varphi})^{\prime}$,
we have

$\operatorname{div}(A^{T}\nabla\log|S\nabla
g|^{2})=\boxed{M1}+\boxed{M2}+T_{M3}+T_{E},$

where

$\displaystyle\boxed{M1}$
$\displaystyle\coloneqq-\sum_{i=1}^{2}\frac{4b_{i}}{|S\nabla
g|^{4}}\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle^{2},$
$\displaystyle\boxed{M2}$
$\displaystyle\coloneqq\sum_{i=1}^{2}\frac{2b_{i}}{|S\nabla
g|^{2}}\langle\partial^{i}\nabla_{R}g,B\partial^{i}\nabla_{R}g\rangle,$
$\displaystyle T_{M3}$ $\displaystyle\coloneqq\frac{2}{|S\nabla
g|^{2}}\left\langle\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}\nabla_{R}g\right),B\nabla_{R}g\right\rangle,$

and $T_{E}$ is an error term (involving derivatives of the matrix $A$) of the
form

(7.14) $T_{E}=\sum_{i=1}^{2}\frac{1}{|S\nabla
g|^{2}}\partial^{i}\left({\mathcal{O}}(|\nabla
g|^{2})\right)+{\mathcal{O}}\left(1+\frac{|\nabla^{2}g|}{|\nabla g|}\right).$

We prove the preceding lemma in Section 7.5.3.

Note that, using (7.3), we have that the two main terms satisfy

(7.15) $\left|\boxed{M1}\right|+\left|\boxed{M2}\right|\lesssim
1+\left(\frac{|\nabla^{2}g|}{|\nabla g|}\right)^{2},$

which is bad for our purposes. A similar behavior occurs with the term
$T_{M3}$. Instead, we need to exploit their cancellation, as illustrated in
the constant matrix case:

######  Remark 7.15 (Constant matrix and main terms).

If the matrix $A$ were constant, and hence $R$ and $B$ were constant as well,
then we would have $T_{E}=0$. That is,

$\operatorname{div}\left(A^{T}\nabla\log|S\nabla
g|^{2}\right)=\boxed{M1}+\boxed{M2}+T_{M3}.$

This suggests that these 3 terms are the main terms, and the others must be
bounded error terms. Moreover, we have the following points:

  * •

The key point in the plane is that
$\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i}g)=0$ would imply
$b_{1}\partial^{1,1}g=-b_{2}\partial^{2,2}g$ (off the critical points). One
can show that, as a consequence, the main terms $None$ and $None$ would cancel
each other in the sense $\boxed{M1}+\boxed{M2}=0$ off the critical points of
$g$.

  * •

Since the derivatives would commute, i.e., $\partial^{1,2}=\partial^{2,1}$, we
would have

(7.16)
$\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}\nabla_{R}g\right)=\nabla_{R}\left(\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}g\right)\right)=\nabla_{R}(0)=0,$

where we used that the Green function $g$ satisfies
$\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i}g)=0$. This is why we call
$T_{M3}$ the “zero” term. In contrast to the previous point, here it is not
needed that we are in the plane. However, we use that the Green function is
$L_{A}$-harmonic, that is, this term is zero even in higher dimensions.

All in all, in the constant matrix case we would have

$\operatorname{div}\left(A^{T}\nabla\log|S\nabla
g|^{2}\right)=\boxed{M1}+\boxed{M2}+T_{M3}=0$

off the critical points of $g$.

The strategy explained in the previous remark is, morally, what we will do to
prove the following two lemmas.

######  Lemma 7.16 (Step 6).

Pointwise $\left|\boxed{M1}+\boxed{M2}\right|\lesssim 1+|\nabla^{2}g|/|\nabla
g|$ in $\widetilde{\Omega}_{\varphi}$.

We prove the preceding lemma in Section 7.5.4.

######  Lemma 7.17 (Step 7).

In the dual space $W_{c}^{1,\infty}(\widetilde{\Omega}_{\varphi})^{\prime}$,
the functional $T_{M3}$ is of the form

$T_{M3}=\sum_{j=1}^{2}\frac{2b_{j}\partial^{j}g}{|S\nabla
g|^{2}}\left({\mathcal{O}}(|\nabla
g|)+{\mathcal{O}}(|\nabla^{2}g|)+\partial^{1}\left({\mathcal{O}}(|\nabla
g|)\right)+\partial^{2}\left({\mathcal{O}}(|\nabla g|)\right)\right).$

Recall that this equality must be understood at a functional level, see (7.36)
below.

We prove the preceding lemma in Section 7.5.5.

In the following lemma, we bound the error terms that have appeared in the
previous computations.

######  Lemma 7.18 (Step 8).

For small enough $\varepsilon>0$,

$|T_{M3}(\psi_{\varepsilon}\varphi g)|+|T_{E}(\psi_{\varepsilon}\varphi
g)|\lesssim 1+\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla
g|}g.$

We prove the preceding lemma in Section 7.5.6.

######  Lemma 7.19 (Step 9).

$\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla g|}g\leq
C<\infty,$ and for small enough $\varepsilon>0$,
$\int_{U_{\varepsilon}(\partial\widetilde{\Omega})}\frac{|\nabla^{2}g|}{|\nabla
g|}g\lesssim\varepsilon$.

We prove the preceding lemma at the end of Section 7.4.

Using the previous lemmas we obtain (7.2).

###### Proof of (7.2)..

Let $p\in\Omega$ with ${\rm dist}(p,\partial\Omega)>r_{0}$ so that Lemma 7.10
holds for all $a>10$ integer. By Lemmas 7.9, 7.10 and 7.11 we have

$\left|\int_{\partial\widetilde{\Omega}}\log|S\nabla
g|^{2}\,d\widetilde{\omega}_{T}^{p}\right|\leq\left|-\int\langle
A^{T}\nabla\log|S\nabla g|^{2},\nabla(\varphi
g)\rangle\right|+{\mathcal{O}}\left(1+\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla
g|}g\right).$

By Lemmas 7.12, 7.14, 7.16 and 7.18, we get

$\left|-\int\langle A^{T}\nabla\log|S\nabla g|^{2},\nabla(\varphi
g)\rangle\right|\lesssim
1+\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla
g|}g+{\mathcal{O}}\left(\lim_{\varepsilon\to
0}\frac{1}{\varepsilon}\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}\frac{|\nabla^{2}g|}{|\nabla
g|}g\right).$

In particular,

$\left|\int_{\partial\widetilde{\Omega}}\log|S\nabla
g|^{2}\,d\widetilde{\omega}_{T}^{p}\right|\lesssim
1+\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla
g|}g+{\mathcal{O}}\left(\lim_{\varepsilon\to
0}\frac{1}{\varepsilon}\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}\frac{|\nabla^{2}g|}{|\nabla
g|}g\right).$

Finally, (7.2) follows from this and Lemma 7.19. Note that by Harnack
inequality this is true for every $p\in\Omega$ with ${\rm
dist}(p,\partial\Omega)>r_{0}$. ∎

###  7.3. Second derivatives of the Green function

In this subsection, we collect a fundamental property of the Green function in
our situation, and some useful equalities and bounds that we will use later on
involving the second order derivatives of the Green function.

The following claim points out the main difference between the planar and the
higher dimensional case. This is a key point in the proof of Lemmas 7.16 and
7.17.

######  Claim 7.20.

The Green function $g$ satisfies

$\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i}g)=-\sum_{i=1}^{2}b_{i}\partial^{i}g\operatorname{div}R_{i}^{T}-\sum_{i,j=1}^{2}\partial_{i}\left(\frac{a_{ij}-a_{ji}}{2}\right)\partial_{j}g={\mathcal{O}}(|\nabla
g|).$

a.e. in $U_{r_{0}}(\partial\Omega)\cap\widetilde{\Omega}$. In particular,

(7.17) $b_{1}\partial^{1,1}g=-b_{2}\partial^{2,2}g+{\mathcal{O}}(|\nabla g|).$

###### Proof.

By Claims 7.7 and 7.6,

$\displaystyle\operatorname{div}A\nabla g$
$\displaystyle=\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i}g)+\sum_{i=1}^{2}b_{i}\partial^{i}g\operatorname{div}R_{i}^{T}+\sum_{i,j=1}^{2}\partial_{i}\left(\frac{a_{ij}-a_{ji}}{2}\right)\partial_{j}g$
$\displaystyle=\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i}g)+{\mathcal{O}}(|\nabla
g|),$

where we used that the matrices $A$, $B$ and $R$ are Lipschitz. The claim
follows from the fact that the Green function satisfies
$\operatorname{div}A\nabla g=0$ a.e. in
$U_{r_{0}}(\partial\Omega)\cap\widetilde{\Omega}$, see Remark 2.11.

As a consequence of
$\partial^{i}(b_{i}\partial^{i}g)=\partial^{i}b_{i}\partial^{i}g+b_{i}\partial^{i,i}g$
and $\partial^{i}b_{i}\partial^{i}g\in{\mathcal{O}}(|\nabla g|)$, we have
(7.17). ∎

A close look to the proof reveals that if the matrix $A$ were constant, then
we would have $b_{1}\partial^{1,1}g=-b_{2}\partial^{2,2}g$ in the planar case,
as it was explained in Remark 7.15. Instead, we get the analog with some error
terms in 7.17.

In the following claim we collect some basic equalities and estimates. These
equalities will be useful in the decomposition procedure in Lemma 7.14.

######  Claim 7.21.

Using the notation $\nabla_{R}f=R\cdot\nabla f$, see Section 7.1, we have

(7.18) $\partial^{i}|S\nabla
g|^{2}=2\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle+\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle.$

Consequently,

(7.19) $\partial^{i}\left(\frac{1}{|S\nabla
g|^{2}}\right)=\frac{-\partial^{i}|S\nabla g|^{2}}{|S\nabla
g|^{4}}\overset{\text{{(\ref{derivative product of dot
product})}}}{=}-\frac{2\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle+\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle}{|S\nabla
g|^{4}},$

and

(7.20) $|\nabla\log|S\nabla g|^{2}|=\frac{|\nabla|S\nabla g|^{2}|}{|S\nabla
g|^{2}}\lesssim 1+\frac{|\nabla^{2}g|}{|\nabla g|}\quad\text{in
}\widetilde{\Omega}_{\varphi}.$

###### Proof.

Since $S^{T}S=A_{0}$ and $A_{0}=R^{T}BR$, we can write

$|S\nabla g|^{2}=\langle S\nabla g,S\nabla g\rangle=\langle\nabla
g,A_{0}\nabla g\rangle=\langle R\nabla g,BR\nabla
g\rangle=\langle\nabla_{R}g,B\nabla_{R}g\rangle.$

Recall that $R$ and $B$ are Lipschitz regular, thus we can apply the usual
derivative rules a.e. to obtain

$\partial^{i}|S\nabla
g|^{2}=\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle+\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle+\langle\nabla_{R}g,B\partial^{i}\nabla_{R}g\rangle,$

and (7.18) follows from the symmetry of $B$.

The inequality in (7.20) follows from

$\displaystyle\left|\nabla|S\nabla g|^{2}\right|$
$\displaystyle{\approx}\sum_{i=1}^{2}\left|2\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle+\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle\right|$
$\displaystyle\leq\sum_{i=1}^{2}2|\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle|+\sum_{i=1}^{2}|\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle|$
$\displaystyle\lesssim|\nabla
g|^{2}+\sum_{i=1}^{2}|\partial^{i}\nabla_{R}g||\nabla
g|\overset{\text{{(\ref{two directional derivatives})}}}{\lesssim}|\nabla
g|^{2}+|\nabla g||\nabla^{2}g|,$

and this concludes the proof of the claim. ∎

######  Remark 7.22.

$1/|S\nabla g|^{2}\in
W^{1,2}_{\operatorname{loc}}(\widetilde{\Omega}_{\varphi})$ because of (7.19),
$|\nabla g|\approx g/{\rm dist}(\cdot,\partial\widetilde{\Omega})>0$ in
$\widetilde{\Omega}_{\varphi}$ and $g\in
W^{2,2}(\widetilde{\Omega}_{\varphi})$. By the same reason, $\log|S\nabla
g|^{2}\in W^{1,2}_{\operatorname{loc}}(\widetilde{\Omega}_{\varphi})$.

###  7.4. Whitney cubes and proof of Step 9

A classical way to compute an integral over a given set is to discretize it.
We will do so using Whitney cubes, that is, we divide the domain into regions
(cubes) which have diameter comparable to their distance to the boundary, so
that Harnack’s and Caccioppoli’s inequalities can be used locally.

Let us start by defining the Whitney covering of $\widetilde{\Omega}$, and
then we will move to study the properties of those cubes touching
$\operatorname{supp}\varphi$.

######  Definition 7.23.

There exists a collection $W_{\widetilde{\Omega}}$ of dyadic cubes satisfying
the following properties:

  1. (W1)

$\sqrt{2}\ell(Q)\leq{\rm dist}(Q,\partial\widetilde{\Omega})\leq
4\sqrt{2}\ell(Q)$, equivalently, ${\rm diam}Q\leq{\rm
dist}(Q,\partial\widetilde{\Omega})\leq 4{\rm diam}Q$.

  2. (W2)

$1.5Q\cap\partial\widetilde{\Omega}=\emptyset$, since ${\rm
dist}(1.5Q,\partial\widetilde{\Omega})\geq{\rm
dist}(Q,\partial\widetilde{\Omega})-\sqrt{2}\cdot 0.25\ell(Q)\geq
0.75\sqrt{2}\ell(Q)>0$,

which we call _Whitney cubes_ , see [Gra08, Appendix J]. We define
$W_{\varphi}\coloneqq\\{Q\in
W_{\widetilde{\Omega}}:Q\cap\operatorname{supp}\varphi\not=\emptyset\\}$.

######  Lemma 7.24.

Every $Q\in W_{\varphi}$ satisfies also:

  1. (W3)

${\rm dist}(x,\partial\widetilde{\Omega})\leq 7\frac{R}{2000c}\leq R/c$ for
$x\in 1.5Q$.

  2. (W4)

$\ell(Q)\lesssim 1$.

  3. (W5)

If $Q\cap\operatorname{supp}\nabla\varphi\not=\emptyset$, then $\ell(Q)\gtrsim
1$.

  4. (W6)

For $x\in 1.5Q$ we have

$g(x)\approx
g(A(\xi,10\ell(Q)))\lesssim\widetilde{\omega}_{T}(B(\xi,10\ell(Q)))\leq\widetilde{\omega}_{T}(50Q)\leq
1,$

where $\xi\in\partial\widetilde{\Omega}$ is such that ${\rm dist}(\xi,Q)={\rm
dist}(\partial\widetilde{\Omega},Q)$, and $A(\xi,10\ell(Q))$ is the Corkscrew
point at $\xi$ with radius $10\ell(Q)$.

###### Proof.

Note that for $x\in 1.5Q$, Harnack’s inequality gives $g(x)\approx
g(A(\xi,10\ell(Q)))$. Using the relation between the elliptic measure and
Green’s function in NTA domains (see [Ken94, Lemma 1.3.3]), we get

$g(A(\xi,10\ell(Q)))\lesssim\widetilde{\omega}_{T}(B(\xi,10\ell(Q))).$

Here we need $p\not\in B(\xi,3\cdot 10\ell(Q))$, which is granted as long as
$\partial\widetilde{\Omega}$ and $\partial\Omega$ are close enough.

The remaining estimates in the lemma follow from the definition by standard
arguments. The details are left to the reader. ∎

In order to bound the error terms arising in the proofs of Section 7.2, we
will use (without mention) the following remark.

######  Remark 7.25.

If $f\in{\mathcal{O}}(|\nabla g|^{k})$ then $\left|\int\frac{f}{|\nabla
g|^{k}}\varphi g\right|\leq C<\infty$.

###### Proof.

Since $g\lesssim 1$ in $\operatorname{supp}\varphi$ (see Lemma 7.24(W6)), we
obtain

$\left|\int\frac{f}{|\nabla g|^{k}}\varphi g\right|\leq\int\frac{|f|}{|\nabla
g|^{k}}|\varphi g|\lesssim\int\left|\varphi
g\right|\lesssim{\mathcal{H}}^{2}(\Omega)\leq({\rm
diam}\partial\Omega)^{2}<\infty,$

where we used that $\Omega$ is bounded. ∎

Next we continue by controlling the Green function over Whitney cubes:

######  Lemma 7.26.

We have:

  1. (1)

$\sum_{Q\in W_{\varphi}}\ell(Q)\cdot\widetilde{\omega}_{T}(50Q)\lesssim 1$.

  2. (2)

For $\varepsilon>0$, $\sum_{\left\\{Q\in W_{\widetilde{\Omega}}:Q\cap
U_{\varepsilon}(\partial\widetilde{\Omega})\not=\emptyset\right\\}}\ell(Q)\cdot\widetilde{\omega}_{T}(50Q)\lesssim\varepsilon$.

  3. (3)

For $\varepsilon>0$,
$\int_{U_{\varepsilon}(\partial\widetilde{\Omega})}g\lesssim\varepsilon^{2}$.

  4. (4)

For each $Q\in W_{\varphi}$,
$\left(\int_{1.5Q}g^{2}\right)^{1/2}\lesssim\ell(Q)\cdot\widetilde{\omega}_{T}(50Q)$.

  5. (5)

For each $Q\in W_{\varphi}$, $\int_{Q}|\nabla
g|\lesssim\ell(Q)\cdot\widetilde{\omega}_{T}(50Q)$.

  6. (6)

For each $Q\in W_{\varphi}$,
$\int_{Q}|\nabla^{2}g|\lesssim\widetilde{\omega}_{T}(50Q)$.

###### Proof.

Item 1: As $\ell(Q)\lesssim 1$ for every $Q\in W_{\varphi}$ (see Lemma
7.24(W4)), there is $k_{0}$ such that every $Q\in W_{\varphi}$ satisfies
$\ell(Q)=2^{-k}$ with $k\geq k_{0}$. Note also that, for each scale $k\geq
k_{0}$, the family $\\{50Q\\}_{\\{Q\in W_{\varphi}\,:\,\ell(Q)=2^{-k}\\}}$ has
finite overlapping. Therefore,

$\displaystyle\sum_{Q\in W_{\varphi}}\ell(Q)\cdot\widetilde{\omega}_{T}(50Q)$
$\displaystyle=\sum_{k\geq k_{0}}\sum_{\begin{subarray}{c}Q\in W_{\varphi}\\\
\ell(Q)=2^{-k}\end{subarray}}\ell(Q)\cdot\widetilde{\omega}_{T}(50{Q})=\sum_{k\geq
k_{0}}2^{-k}\sum_{\begin{subarray}{c}Q\in W_{\varphi}\\\
\ell(Q)=2^{-k}\end{subarray}}\widetilde{\omega}_{T}(50{Q})$
$\displaystyle\lesssim\sum_{k\geq
k_{0}}2^{-k}\widetilde{\omega}_{T}\Biggl{(}\bigcup_{\begin{subarray}{c}Q\in
W_{\varphi}\\\ \ell(Q)=2^{-k}\end{subarray}}50{Q}\Biggr{)}\leq\sum_{k\geq
k_{0}}2^{-k}<\infty.$

Item 2: Recall that the cubes $Q\in W_{\widetilde{\Omega}}$ with $Q\cap
U_{\varepsilon}(\partial\widetilde{\Omega})\not=\emptyset$ satisfy
$\ell(Q)\lesssim\varepsilon$, and $\ell(Q)\approx\varepsilon$ if
$Q\cap(\partial
U_{\varepsilon}(\partial\widetilde{\Omega})\setminus\partial\widetilde{\Omega})\not=\emptyset$,
see (W1). Hence, as in the proof of item 1, taking
$k_{0}(\varepsilon)\approx-\log_{2}\varepsilon$ we have

$\sum_{\begin{subarray}{c}Q\in W_{\widetilde{\Omega}}\\\ Q\cap
U_{\varepsilon}(\partial\widetilde{\Omega})\not=\emptyset\end{subarray}}\ell(Q)\widetilde{\omega}_{T}(50Q)\\\
\leq\sum_{k\geq k_{0}(\varepsilon)}2^{-k}\sum_{\begin{subarray}{c}Q\in
W_{\widetilde{\Omega}}\\\
\ell(Q)=2^{-k}\end{subarray}}\widetilde{\omega}_{T}(50Q)\lesssim\sum_{k\geq
k_{0}(\varepsilon)}2^{-k}\approx 2^{\log_{2}\varepsilon}=\varepsilon.$

Item 3: Using that $g\lesssim\widetilde{\omega}_{T}(50Q)$ in $Q$, see (W6),
that the cubes $Q\in W_{\widetilde{\Omega}}$ with $Q\cap
U_{\varepsilon}(\partial\widetilde{\Omega})\not=\emptyset$ satisfy
$\ell(Q)\lesssim\varepsilon$, see (W1), and item 2, we have

$\int_{U_{\varepsilon}(\partial\widetilde{\Omega})}g\leq\sum_{\begin{subarray}{c}Q\in
W_{\widetilde{\Omega}}\\\ Q\cap
U_{\varepsilon}(\partial\widetilde{\Omega})\not=\emptyset\end{subarray}}\int_{Q}g\overset{\text{{(\ref{green
bounded by measure and 1 in the supp of varphi})},{(\ref{whitney
p1})}}}{\lesssim}\varepsilon\sum_{\begin{subarray}{c}Q\in
W_{\widetilde{\Omega}}\\\ Q\cap
U_{\varepsilon}(\partial\widetilde{\Omega})\not=\emptyset\end{subarray}}\ell(Q)\widetilde{\omega}_{T}(50Q)\overset{\text{{(\ref{sum
over whitney region p1 epsilon})}}}{\lesssim}\varepsilon^{2}.$

Item 4 follows since $g(x)\lesssim\widetilde{\omega}_{T}(50Q)$ for $x\in 1.5Q$
(see Lemma 7.24(W6)). Item 5 follows from Cauchy-Schwarz and Caccioppoli’s
inequalities and item 4.

Item 6: Since the Green function $g$ of $\widetilde{\Omega}$ is solution of
$\operatorname{div}(A\nabla\cdot)=0$ in $1.1Q$, the function
$\widehat{g}(\cdot)=g(\ell(Q)\cdot)$ is solution of
$\operatorname{div}(\widehat{A}\nabla\cdot)$ in $1.1\widehat{Q}$, where
$\widehat{Q}\coloneqq\\{x/\ell(Q):x\in Q\\}$ has side-length one, and
$\widehat{A}(\cdot)=A(\ell(Q)\cdot)$. Moreover, since
$\partial_{i}\partial_{j}\widehat{g}(\cdot)=\ell(Q)^{2}\partial_{i}\partial_{j}g(\ell(Q)\cdot)$,
we get

$\int_{Q}|\nabla^{2}g(x)|^{2}\,dx=\ell(Q)^{2}\int_{\widehat{Q}}|\nabla^{2}g(\ell(Q)x)|^{2}\,dx=\frac{1}{\ell(Q)^{2}}\int_{\widehat{Q}}|\nabla^{2}\widehat{g}(x)|^{2}\,dx.$

The matrix $\widehat{A}$ has the same ellipticity constant as $A$, but as
$\ell(Q)\lesssim 1$ for every $Q\in W_{\varphi}$, see Lemma 7.24(W4), the
Lipschitz norm cannot grow too much:

$\|\widehat{A}\|_{C^{0,1}\left(1.1\widehat{Q}\right)}\coloneqq\|\widehat{A}\|_{L^{\infty}\left(1.1\widehat{Q}\right)}+[\widehat{A}]_{C^{0,1}\left(1.1\widehat{Q}\right)}=\|A\|_{L^{\infty}(1.1Q)}+\ell(Q)[A]_{C^{0,1}(1.1Q)}\leq
C\|A\|_{C^{0,1}\left(1.1Q\right)}.$

Moreover, ${\rm dist}\left(\widehat{Q},(1.1\widehat{Q})^{c}\right)\approx 1$.
Therefore we can apply (2.2) in Theorem 2.9 to obtain

$\int_{Q}|\nabla^{2}g|^{2}\,dx=\frac{1}{\ell(Q)^{2}}\int_{\widehat{Q}}|\nabla^{2}\widehat{g}|^{2}\,dx\overset{\text{{(\ref{cacciopoli
type ineq for weak twice diff
solutions})}}}{\lesssim}\frac{1}{\ell(Q)^{2}}\left[\left(\int_{1.1\widehat{Q}}|\nabla\widehat{g}|^{2}\right)^{1/2}+\left(\int_{1.1\widehat{Q}}\widehat{g}^{2}\right)^{1/2}\right]^{2}.$

Applying Caccioppolli’s inequality and item 4, this implies

$\int_{Q}|\nabla^{2}g|^{2}\,dx\lesssim\frac{1}{\ell(Q)^{2}}\int_{1.5\widehat{Q}}\widehat{g}^{2}=\frac{1}{\ell(Q)^{4}}\int_{1.5Q}g^{2}\overset{\text{{(\ref{sum
over whitney region
p2})}}}{\lesssim}\frac{\widetilde{\omega}_{T}(50Q)^{2}}{\ell(Q)^{2}}.$

Using the Cauchy-Schwarz inequality and this, we get

$\int_{Q}|\nabla^{2}g|\leq\ell(Q)\left(\int_{Q}|\nabla^{2}g|^{2}\right)^{1/2}\lesssim\widetilde{\omega}_{T}(50{Q}).$

∎

Using this lemma we obtain Lemma 7.19.

###### Proof of Lemma 7.19.

Recall that, by (7.12) we have

$|\nabla g(x)|\gtrsim\frac{g(x)}{{\rm
dist}\left(x,\partial\widetilde{\Omega}\right)}\quad\text{for
}x\in\operatorname{supp}\varphi,$

provided the domain $\widetilde{\Omega}$ is $(\delta,r_{0}/2)$-Reifenberg flat
with $\delta$ small enough depending only on $\lambda$ and
$\kappa\|A\|_{L^{\infty}({\mathbb{R}}^{2})}$. With this it suffices to show

$\int_{\operatorname{supp}\varphi}|\nabla^{2}g(x)|{\rm
dist}\left(x,\partial\widetilde{\Omega}\right)\,dx\leq C.$

Summing in $W_{\varphi}$, by (6) and (1) in Lemma 7.26 we obtain

$\int_{\operatorname{supp}\varphi}|\nabla^{2}g(x)|{\rm
dist}(x,\partial\Omega)\,dx\lesssim\sum_{Q\in
W_{\varphi}}\ell(Q)\int_{Q}|\nabla^{2}g|\overset{\text{{(\ref{sum over whitney
region p4})}}}{\lesssim}\sum_{Q\in
W_{\varphi}}\ell(Q)\cdot\widetilde{\omega}_{T}(50{Q})\overset{\text{{(\ref{sum
over whitney region p1})}}}{\lesssim}1,$

as claimed.

Let us see the localized version. As above, using (7.12), and (6) and (2) in
Lemma 7.26,

$\displaystyle\int_{U_{\varepsilon}(\partial\widetilde{\Omega})}\frac{|\nabla^{2}g|}{|\nabla
g|}g$
$\displaystyle{\lesssim}\int_{U_{\varepsilon}(\partial\widetilde{\Omega})}|\nabla^{2}g|{\rm
dist}(x,\partial\widetilde{\Omega})\leq\sum_{\begin{subarray}{c}Q\in
W_{\widetilde{\Omega}}\\\ Q\cap
U_{\varepsilon}(\partial\widetilde{\Omega})\not=\emptyset\end{subarray}}\ell(Q)\int_{Q}|\nabla^{2}g|$
$\displaystyle{\lesssim}\sum_{\begin{subarray}{c}Q\in
W_{\widetilde{\Omega}}\\\ Q\cap
U_{\varepsilon}(\partial\widetilde{\Omega})\not=\emptyset\end{subarray}}\ell(Q)\widetilde{\omega}_{T}(50Q)\overset{\text{{(\ref{sum
over whitney region p1 epsilon})}}}{\lesssim}\varepsilon.$

∎

###  7.5. Proof of Steps 1 to 8

In this subsection we prove remaining Steps 1 to 8 in Section 7.2.

####  7.5.1. Proof of Lemmas 7.9, 7.11 and 7.12

###### Proof of Lemma 7.9.

We have

$\left|\int_{\partial\widetilde{\Omega}}\log|S(\xi)\nabla
g(\xi)|^{2}\,d\widetilde{\omega}_{T}^{p}(\xi)-\int_{\partial\widetilde{\Omega}}\log_{(a)}|S(\xi)\nabla
g(\xi)|^{2}\,d\widetilde{\omega}_{T}^{p}(\xi)\right|\\\
=\left|\int_{\\{\xi\in\partial\widetilde{\Omega}:|S(\xi)\nabla g(\xi)|^{2}\leq
e^{-a}\\}}(\log|S(\xi)\nabla
g(\xi)|^{2}+a)\,d\widetilde{\omega}_{T}^{p}(\xi)\right|$

Since the domain is smooth, we can use (2.8) to write and bound the right-hand
side term as

$\displaystyle\Biggl{|}\int_{\\{\xi\in\partial\widetilde{\Omega}:|S(\xi)\nabla
g(\xi)|^{2}\leq e^{-a}\\}}$ $\displaystyle-\langle A(\xi)\nabla
g(\xi),\nu\rangle(\log|S(\xi)\nabla g(\xi)|^{2}+a)\,d\sigma(\xi)\Biggr{|}$
$\displaystyle\lesssim\int_{\\{\xi\in\partial\widetilde{\Omega}:|S(\xi)\nabla
g(\xi)|^{2}\leq e^{-a}\\}}|S(\xi)\nabla g(\xi)|(\left|\log|S(\xi)\nabla
g(\xi)|\right|)\,d\sigma(\xi)$
$\displaystyle\leq\sigma(\partial\widetilde{\Omega})\left(e^{-a/2}|{\log
e^{-a/2}}|\right).$

∎

###### Proof of Lemma 7.11.

We have

$-\int\langle A^{T}\nabla\log_{(a)}|S\nabla g|^{2},\nabla(\varphi
g)\rangle+\int\langle A^{T}\nabla\log|S\nabla g|^{2},\nabla(\varphi
g)\rangle\\\ =\int\langle A^{T}\nabla\min\\{0,a+\log|S\nabla
g|^{2}\\},\nabla(\varphi g)\rangle.$

On the other hand, by (7.12) we have that if $a\geq 10$ is big enough then
$\varphi=1$ in $\widetilde{\Omega}_{\varphi}\cap\\{|S\nabla g|^{2}\leq
e^{-a}\\}$, and hence

$\int\langle A^{T}\nabla\min\\{0,a+\log|S\nabla g|^{2}\\},\nabla(\varphi
g)\rangle=\int\langle A^{T}\nabla\log|S\nabla g|^{2},\nabla
g\rangle\mathbf{1}_{\\{|S\nabla g|^{2}\leq e^{-a}\\}}.$

By (7.20), this is controlled in absolute value by

$\int_{\\{|S\nabla g|^{2}\leq e^{-a}\\}}\left(1+\frac{|\nabla^{2}g|}{|\nabla
g|}\right)|\nabla g|\leq\int_{\\{|S\nabla g|^{2}\leq e^{-a}\\}}|\nabla
g|+\int_{\\{|S\nabla g|^{2}\leq e^{-a}\\}}|\nabla^{2}g|.$

The lemma follows because the first term in the right-hand side is bounded by
a constant times ${\mathcal{H}}^{2}(\Omega)e^{-a/2}$. ∎

###### Proof of Lemma 7.12.

For $\varepsilon>0$ small we have
$(1-\psi_{\varepsilon})\varphi=(1-\psi_{\varepsilon})$. Hence

$-\int\langle A^{T}\nabla\log|S\nabla
g|^{2},\nabla((1-\psi_{\varepsilon})\varphi g)\rangle=-\int\langle
A^{T}\nabla\log|S\nabla g|^{2},\nabla((1-\psi_{\varepsilon})g)\rangle\\\
=\int\langle A^{T}\nabla\log|S\nabla g|^{2},\nabla\psi_{\varepsilon}\rangle
g-\int\langle A^{T}\nabla\log|S\nabla g|^{2},\nabla
g\rangle(1-\psi_{\varepsilon}).$

Using (7.20), the second term in the right-hand side is controlled in absolute
value by

$\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}\left(1+\frac{|\nabla^{2}g|}{|\nabla
g|}\right)|\nabla
g|=\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}|\nabla
g|+\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}|\nabla^{2}g|\overset{\varepsilon\to
0}{\longrightarrow}0,$

and the first term is controlled in absolute value by

$\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}\left(1+\frac{|\nabla^{2}g|}{|\nabla
g|}\right)g|\nabla\psi_{\varepsilon}|=\frac{1}{\varepsilon}\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}g+\frac{1}{\varepsilon}\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}\frac{|\nabla^{2}g|}{|\nabla
g|}g.$

The lemma follows by applying (3) in Lemma 7.26 in the last line. ∎

####  7.5.2. Proof of Lemma 7.10

Recall that the Green function $g$ in $\widetilde{\Omega}$ for the operator
$L_{A}u=\operatorname{div}A\nabla u$ satisfies, for every $\phi\in
C^{0}(\overline{\widetilde{\Omega}})\cap W^{1,2}(\widetilde{\Omega})$,

(7.21)
$\int_{\partial\widetilde{\Omega}}\phi(\xi)\,d\widetilde{\omega}_{T}^{x}(\xi)-\phi(x)=-\int_{\widetilde{\Omega}}\langle
A^{T}(y)\nabla\phi(y),\nabla g_{x}(y)\rangle\,dy,\quad\text{ for a.e.\
}x\in\widetilde{\Omega},$

see (2.4).

We claim that

$\varphi\log_{(a)}|S\nabla g|^{2}\in C^{0}(\overline{\widetilde{\Omega}})\cap
W^{1,2}(\widetilde{\Omega}).$

Indeed, $\varphi\log_{(a)}|S\nabla g|^{2}\in
C^{0}(\overline{\widetilde{\Omega}})$ because the function $\varphi$ avoids
the pole and $\nabla g$ is continuous up to the boundary, see Theorem 2.12.
Let us see that $\varphi\log_{(a)}|S\nabla g|^{2}\in
W^{1,2}(\widetilde{\Omega})$. Using Jensen’s inequality as the function
$x\mapsto\left|\log x\right|^{2}$ is concave for $x>e$, the $L^{2}$-norm is
controlled (depending on $a$) by

(7.22)
$\displaystyle\int_{\widetilde{\Omega}\cap\operatorname{supp}\varphi}\left|\log_{(a)}|S\nabla
g|^{2}\right|^{2}$ $\displaystyle\leq\int_{\widetilde{\Omega}\cap\\{|S\nabla
g|^{2}\leq e\\}}\left|\log_{(a)}|S\nabla
g|^{2}\right|^{2}+\int_{\widetilde{\Omega}\cap\operatorname{supp}\varphi\cap\\{|S\nabla
g|^{2}>e\\}}\left|\log|S\nabla g|^{2}\right|^{2}$
$\displaystyle\lesssim{\mathcal{H}}^{2}(\widetilde{\Omega})\left(a^{2}+\left|\log\left(\;\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.57222pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.70335pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.59167pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.10834pt}}\\!\int_{\widetilde{\Omega}\cap\operatorname{supp}\varphi\cap\\{|S\nabla
g|^{2}>e\\}}|S\nabla g|^{2}\right)\right|^{2}\right)<\infty,$

which is bounded since $g\in
W^{1,2}(\widetilde{\Omega}\cap\operatorname{supp}\varphi)$. The $L^{2}$ norm
of $\nabla(\varphi\log_{(a)}|S\nabla g|^{2})$ is

$\int_{\widetilde{\Omega}}|\nabla(\varphi\log_{(a)}|S\nabla
g|^{2})|^{2}\leq\int_{\widetilde{\Omega}}|\nabla\varphi|^{2}\left|\log_{(a)}|S\nabla
g|^{2}\right|^{2}+\int_{\widetilde{\Omega}}\varphi^{2}|\nabla\log_{(a)}|S\nabla
g|^{2}|^{2}.$

The first term is finite by (7.22) because $\varphi\in
C^{\infty}_{c}({\mathbb{R}}^{2})$. From the definition of
$\log_{(a)}x\coloneqq\max\\{\log x,-a\\}$ and by (7.20), we get

$|\nabla\log_{(a)}|S\nabla g|^{2}|\lesssim\mathbf{1}_{\\{x\geq
e^{-a}\\}}(|S\nabla g|^{2})\left(1+\frac{|\nabla^{2}g|}{|\nabla g|}\right),$

and hence the second term is controlled by

$\int_{\widetilde{\Omega}\cap\operatorname{supp}\varphi}|\nabla\log_{(a)}|S\nabla
g|^{2}|^{2}\lesssim\int_{\widetilde{\Omega}\cap\operatorname{supp}\varphi\cap\\{|S\nabla
g|^{2}\geq e^{-a}\\}}1+\frac{|\nabla^{2}g|^{2}}{|\nabla
g|^{2}}\lesssim{\mathcal{H}}^{2}(\widetilde{\Omega})+e^{a}\int_{\widetilde{\Omega}\cap\operatorname{supp}\varphi}|\nabla^{2}g|^{2}<\infty,$

which is bounded as $g\in
W^{2,2}(\widetilde{\Omega}\cap\operatorname{supp}\varphi)$.

By the choice of $\varphi$ with $\varphi(p)=0$ and
$\varphi_{|\partial\widetilde{\Omega}}=1$, and plugging
$\phi=\varphi\log_{(a)}|S\nabla g|^{2}\in
C^{0}(\overline{\widetilde{\Omega}})\cap W^{1,2}(\widetilde{\Omega})$ in
(7.21), for a.e. $p\in\Omega$ with ${\rm dist}(p,\partial\Omega)>r_{0}$ we
have

(7.23) $\displaystyle\int_{\partial\widetilde{\Omega}}\log_{(a)}|S\nabla
g|^{2}\,d\widetilde{\omega}_{T}^{p}$
$\displaystyle=-\int_{\widetilde{\Omega}\cap\operatorname{supp}\varphi}\langle
A^{T}\nabla\left(\varphi\log_{(a)}|S\nabla g|^{2}\right),\nabla g\rangle$
$\displaystyle=-\int\langle A^{T}\nabla\varphi,\nabla
g\rangle\cdot\log_{(a)}|S\nabla g|^{2}-\int\langle
A^{T}\nabla\log_{(a)}|S\nabla g|^{2},\varphi\nabla g\rangle$
$\displaystyle=\boxed{1}+\boxed{2}+\boxed{3},$

with

$\displaystyle\boxed{1}$ $\displaystyle\coloneqq-\int\langle
A^{T}\nabla\varphi,\nabla g\rangle\cdot\log_{(a)}|S\nabla g|^{2},$
$\displaystyle\boxed{2}$ $\displaystyle\coloneqq\int\langle
A^{T}~\nabla\log_{(a)}|S\nabla g|^{2},\nabla\varphi\rangle\cdot g,\text{ and}$
$\displaystyle\boxed{3}$ $\displaystyle\coloneqq-\int\langle
A^{T}\nabla\log_{(a)}|S\nabla g|^{2},\nabla(\varphi g)\rangle.$

We claim that $\left|\boxed{1}\right|\lesssim 1$ and
$\left|\boxed{2}\right|\lesssim
1+\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla g|}g$, and
hence Lemma 7.10 follows.

We start by controlling the term $\boxed{1}$. Recall that we have fixed
$\varphi$ so that $|\nabla\varphi|\lesssim 1$, and assumed that $a$ is big
enough so that $e^{-a}<\min_{z\in\operatorname{supp}\nabla\varphi}|S\nabla
g(z)|^{2}$, that is, $\log_{(a)}|S\nabla g(z)|^{2}=\log|S\nabla g(z)|^{2}$ for
$z\in\operatorname{supp}\nabla\varphi$. Thus,

$\left|\boxed{1}\right|\lesssim\int_{\operatorname{supp}\nabla\varphi}|\nabla
g|\left|\log|S\nabla
g|\right|\approx\int_{\operatorname{supp}\nabla\varphi}|S\nabla
g|\left|\log|S\nabla g|\right|.$

Since $\operatorname{supp}\nabla\varphi$ is far from
$\partial\widetilde{\Omega}$, in particular ${\rm
dist}(x,\partial\widetilde{\Omega})\approx 1$ for
$x\in\operatorname{supp}\nabla\varphi$. Consequently, from this and (7.12) we
have $|\nabla g|\approx g/{\rm dist}(\cdot,\partial\widetilde{\Omega})\approx
g$ in the support of $\nabla\varphi$, and hence

$|S\nabla g(x)|\approx|\nabla g(x)|\approx g(x)\overset{\text{{(\ref{green
bounded by measure and 1 in the supp of varphi})}}}{\lesssim}1\quad\text{for
}x\in\operatorname{supp}\nabla\varphi.$

Note that for $0<t<t_{0}$ we have $t\left|\log
t\right|\leq\max\\{1,t_{0}\left|\log t_{0}\right|\\}$. Therefore,

$\boxed{1}\lesssim\int_{\operatorname{supp}\nabla\varphi}|S\nabla
g|\left|\log|S\nabla
g|\right|\lesssim{\mathcal{H}}^{2}(\operatorname{supp}\nabla\varphi)\leq
C<\infty.$

Now we study the term $\boxed{2}$. Since $\varphi$ is fixed, we have
$|\nabla\varphi|\lesssim 1$ and so

$\left|\boxed{2}\right|\lesssim\int_{\operatorname{supp}\nabla\varphi}g\cdot|\nabla\log_{(a)}|S\nabla
g|^{2}|\leq\int_{\operatorname{supp}\varphi}g\cdot|\nabla\log_{(a)}|S\nabla
g|^{2}|.$

Plugging (7.20) in here we get

$\left|\boxed{2}\right|\lesssim\int_{\operatorname{supp}\varphi}g+\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla
g|}g\overset{\text{{(\ref{green bounded by measure and 1 in the supp of
varphi})}}}{\lesssim}1+\int_{\operatorname{supp}\varphi}\frac{|\nabla^{2}g|}{|\nabla
g|}g.$

∎

####  7.5.3. Proof of Lemma 7.14

In this subsection we study, via a perturbation argument, the functional term

$\operatorname{div}\left(A^{T}\nabla\log|S\nabla g|^{2}\right)\in
W^{1,\infty}_{c}(\widetilde{\Omega}_{\varphi})^{\prime}.$

Note that the action of this functional on $\varphi g$ gives rise to a
modified version of the term $\boxed{3}$ in (7.23).

First, we move from the matrix $A$ to its symmetric part $A_{0}$ by Claim 7.6,
and we write its divergence in terms of the directional derivatives using
Claim 7.7 (see Remark 7.22):

$\displaystyle\operatorname{div}\left(A^{T}\nabla\log|S\nabla g|^{2}\right)=$
$\displaystyle\operatorname{div}\left(A_{0}\nabla\log|S\nabla
g|^{2}\right)+\sum_{i,j=1}^{2}\partial_{i}\left(\frac{a_{ij}^{T}-a_{ji}^{T}}{2}\right)\partial_{j}\log|S\nabla
g|^{2}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i}(\log|S\nabla
g|^{2}))+\sum_{i=1}^{2}b_{i}\partial^{i}(\log|S\nabla
g|^{2})\operatorname{div}R^{T}_{i}$
$\displaystyle+\sum_{i,j=1}^{2}\partial_{i}\left(\frac{a_{ji}-a_{ij}}{2}\right)\partial_{j}(\log|S\nabla
g|^{2})\eqqcolon T_{3.A}+\boxed{3.B}+\boxed{3.C}.$

Note that the terms $None$ and $None$ contain derivatives of the matrix $A$.
In particular, if the matrix were constant then these two terms would be zero,
which suggests that the terms $None$ and $None$ must be bounded error terms.

Next we deal with $T_{3.A}$. As $1/|S\nabla g|^{2}\in
W^{1,2}_{\operatorname{loc}}(\widetilde{\Omega}_{\varphi})$ (see Remark 7.22),
by (7.9) we have

(7.24) $\displaystyle T_{3.A}=$
$\displaystyle\sum_{i=1}^{2}\partial^{i}\left(b_{i}\frac{\partial^{i}|S\nabla
g|^{2}}{|S\nabla g|^{2}}\right)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{2}\partial^{i}\left(\frac{1}{|S\nabla
g|^{2}}\right)b_{i}\partial^{i}|S\nabla g|^{2}+\sum_{i=1}^{2}\frac{1}{|S\nabla
g|^{2}}\partial^{i}(b_{i}\partial^{i}|S\nabla g|^{2}).$

We infer that each element in the sum in the first term in the right-hand side
in (7.24) is

(7.25) $\partial^{i}\left(\frac{1}{|S\nabla
g|^{2}}\right)b_{i}\partial^{i}|S\nabla g|^{2}\overset{\text{{(\ref{derivative
quotient})}}}{=}\frac{-b_{i}\left(\partial^{i}|S\nabla
g|^{2}\right)^{2}}{|S\nabla g|^{4}}.$

Therefore, using identities 7.18 and 7.25 and expanding, the first term in the
right-hand side of (7.24) can be written as

$\sum_{i=1}^{2}\partial^{i}\left(\frac{1}{|S\nabla
g|^{2}}\right)b_{i}\partial^{i}|S\nabla
g|^{2}=\boxed{M1}+\boxed{E1}+\boxed{E2},$

where

$\displaystyle\boxed{M1}$
$\displaystyle\coloneqq-\sum_{i=1}^{2}\frac{4b_{i}}{|S\nabla
g|^{4}}\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle^{2},$
$\displaystyle\boxed{E1}$
$\displaystyle\coloneqq-\sum_{i=1}^{2}\frac{4b_{i}}{|S\nabla
g|^{4}}\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle,\text{
and}$ $\displaystyle\boxed{E2}$
$\displaystyle\coloneqq-\sum_{i=1}^{2}\frac{b_{i}}{|S\nabla
g|^{4}}\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle^{2}.$

Note that $None$ and $None$ can be considered “error terms” from a
perturbation point of view, since in case $B$ were constant we would get
$\boxed{E1}=\boxed{E2}=0$, while $None$ can be considered a “main term”.

Next we perform a similar decomposition for
$\partial^{i}(b_{i}\partial^{i}|S\nabla g|^{2})$, in the second term in the
right-hand side in (7.24). In this case, we need to pay attention to the first
term in the right-hand side of (7.18):

(7.26)
$\partial^{i}\left(b_{i}\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle\right)=\left\langle\partial^{i}\left(b_{i}\partial^{i}\nabla_{R}g\right),B\nabla_{R}g\right\rangle+b_{i}\langle\partial^{i}\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle+b_{i}\langle\partial^{i}\nabla_{R}g,B\partial^{i}\nabla_{R}g\rangle.$

Then using 7.18 and 7.26 we get that the second term in the right-hand side of
(7.24) can be written as

$\sum_{i=1}^{2}\frac{1}{|S\nabla g|^{2}}\partial^{i}(b_{i}\partial^{i}|S\nabla
g|^{2})=\boxed{M2}+T_{M3}+\boxed{E3}+T_{E4},$

where

$\displaystyle\boxed{M2}$
$\displaystyle\coloneqq\sum_{i=1}^{2}\frac{2b_{i}}{|S\nabla
g|^{2}}\langle\partial^{i}\nabla_{R}g,B\partial^{i}\nabla_{R}g\rangle,$
$\displaystyle T_{M3}$ $\displaystyle\coloneqq\frac{2}{|S\nabla
g|^{2}}\left\langle\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}\nabla_{R}g\right),B\nabla_{R}g\right\rangle,$
$\displaystyle\boxed{E3}$
$\displaystyle\coloneqq\sum_{i=1}^{2}\frac{2b_{i}}{|S\nabla
g|^{2}}\langle\partial^{i}\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle,\text{
and}$ $\displaystyle T_{E4}$
$\displaystyle\coloneqq\sum_{i=1}^{2}\frac{1}{|S\nabla
g|^{2}}\partial^{i}\left(b_{i}\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle\right).$

All together we have

$\displaystyle\operatorname{div}\left(A^{T}\nabla\log|S\nabla g|^{2}\right)=$
$\displaystyle\ T_{3.A}+\boxed{3.B}+\boxed{3.C}$ $\displaystyle=$
$\displaystyle\ \boxed{M1}+\boxed{M2}+T_{M3}$ (Main terms)
$\displaystyle+\boxed{3.B}+\boxed{3.C}+\boxed{E1}+\boxed{E2}+\boxed{E3}+T_{E4},$
(Error terms)

and the first part of the lemma follows taking

$T_{E}\coloneqq\boxed{3.B}+\boxed{3.C}+\boxed{E1}+\boxed{E2}+\boxed{E3}+T_{E4}.$

Let us see now the second part of the lemma. Indeed, $\boxed{E2}$ is
${\mathcal{O}}(1)$ because
$|\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle|\lesssim|\nabla g|^{2}$.
By (7.20) we get that $\boxed{3.B}$ and $\boxed{3.C}$ are in
${\mathcal{O}}\left(1+\frac{|\nabla^{2}g|}{|\nabla g|}\right)$, and (7.3)
implies that $\boxed{E1}$ and $\boxed{E3}$ are of the form
${\mathcal{O}}\left(1+\frac{|\nabla^{2}g|}{|\nabla g|}\right)$. All in all,

$T_{E}=T_{E4}+{\mathcal{O}}\left(1+\frac{|\nabla^{2}g|}{|\nabla g|}\right),$

as claimed.∎

####  7.5.4. Proof of Lemma 7.16

In this subsection we prove that the two main terms cancel out in the sense

$\left|\boxed{M1}+\boxed{M2}\right|\lesssim 1+\frac{|\nabla^{2}g|}{|\nabla
g|}.$

This cancellation is related to what happens in the constant case, see Remark
7.15.

We will see that they both have a common term in opposite sign, and exploiting
this cancellation we will obtain a sum of error terms. The key identity is
(7.17), which only applies in the plane.

The key idea to prove the lemma is that using (7.5), which relates
$\partial^{1,2}g$ and $\partial^{2,1}g$, and (7.17), which relates
$b_{1}\partial^{1,1}g$ and $b_{2}\partial^{2,2}g$, we will be able to write
$\partial^{1,1}$, $\partial^{2,2}$, $\partial^{1,2}$ and $\partial^{2,1}$ in
terms of $\partial^{1,1}$ and $\partial^{2,1}$ only.

We start by studying the term
$\boxed{M1}\coloneqq-\sum_{i}\frac{4b_{i}}{|S\nabla
g|^{4}}\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle^{2}$ in
$\widetilde{\Omega}_{\varphi}$. Expanding the numerator,

$\displaystyle\sum_{i=1}^{2}b_{i}\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle^{2}$
$\displaystyle=b_{1}\langle\partial^{1}\nabla_{R}g,B\nabla_{R}g\rangle^{2}+b_{2}\langle\partial^{2}\nabla_{R}g,B\nabla_{R}g\rangle^{2}$
$\displaystyle=b_{1}(b_{1}\partial^{1}g\partial^{1,1}g+b_{2}\partial^{2}g~\partial^{1,2}g)^{2}+b_{2}(b_{1}\partial^{1}g\partial^{2,1}g+b_{2}\partial^{2}g~\partial^{2,2}g)^{2}.$

By (7.5) and (7.17), which read as
$\partial^{1,2}g=\partial^{2,1}g+{\mathcal{O}}(|\nabla g|)$ and
$b_{2}\partial^{2,2}g=-b_{1}\partial^{1,1}g+{\mathcal{O}}(|\nabla g|)$
respectively, we have

$\displaystyle\sum_{i=1}^{2}b_{i}\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle^{2}=$
$\displaystyle\
b_{1}(b_{1}\partial^{1}g\partial^{1,1}g+b_{2}\partial^{2}g\partial^{2,1}g+{\mathcal{O}}(|\nabla
g|^{2}))^{2}$
$\displaystyle+b_{2}(b_{1}\partial^{1}g\partial^{2,1}g-b_{1}\partial^{2}g\partial^{1,1}g+{\mathcal{O}}(|\nabla
g|^{2}))^{2}.$

Expanding, and using
$b_{1}(\partial^{1}g)^{2}+b_{2}(\partial^{2}g)^{2}=\langle\nabla_{R}g,B\nabla_{R}g\rangle=|S\nabla
g|^{2}$ and the cancellation of cross terms, we get

$\sum_{i=1}^{2}b_{i}\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle^{2}=|S\nabla
g|^{2}\left(b_{1}^{2}(\partial^{1,1}g)^{2}+b_{1}b_{2}(\partial^{2,1}g)^{2}\right)+(\partial^{1,1}g+\partial^{2,1}g){\mathcal{O}}(|\nabla
g|^{3})+{\mathcal{O}}(|\nabla g|^{4}).$

In conclusion, $None$ is of the form

(7.27)
$\boxed{M1}=-4\frac{b_{1}^{2}(\partial^{1,1}g)^{2}+b_{1}b_{2}(\partial^{2,1}g)^{2}}{|S\nabla
g|^{2}}+\frac{{\mathcal{O}}(\partial^{1,1}g+\partial^{2,1}g)}{|S\nabla
g|}+{\mathcal{O}}(1).$

With the same strategy we study the term
$\boxed{M2}\coloneqq\sum_{i}\frac{2b_{i}}{|S\nabla
g|^{2}}\langle\partial^{i}\nabla_{R}g,B\partial^{i}\nabla_{R}g\rangle$ in
$\widetilde{\Omega}_{\varphi}$. Using 7.5 and 7.17 as before and expanding,
the numerator can be written as

$\displaystyle\sum_{i=1}^{2}$ $\displaystyle
b_{i}\langle\partial^{i}\nabla_{R}g,B\partial^{i}\nabla_{R}g\rangle=b_{1}^{2}(\partial^{1,1}g)^{2}+b_{1}b_{2}(\partial^{1,2}g)^{2}+b_{1}b_{2}(\partial^{2,1}g)^{2}+b_{2}^{2}(\partial^{2,2}g)^{2}$
$\displaystyle=b_{1}^{2}(\partial^{1,1}g)^{2}+b_{1}b_{2}\left(\partial^{2,1}g+{\mathcal{O}}(|\nabla
g|)\right)^{2}+b_{1}b_{2}(\partial^{2,1}g)^{2}+\left(-b_{1}\partial^{1,1}g+{\mathcal{O}}(|\nabla
g|)\right)^{2}$
$\displaystyle=2\left(b_{1}^{2}(\partial^{1,1}g)^{2}+b_{1}b_{2}(\partial^{2,1}g)^{2}\right)+\left(\partial^{1,1}g+\partial^{2,1}g\right){\mathcal{O}}(|\nabla
g|)+{\mathcal{O}}(|\nabla g|^{2}).$

Hence, $None$ is of the form

(7.28)
$\boxed{M2}=4\frac{b_{1}^{2}(\partial^{1,1}g)^{2}+b_{1}b_{2}(\partial^{2,1}g)^{2}}{|S\nabla
g|^{2}}+\frac{{\mathcal{O}}(\partial^{1,1}g+\partial^{2,1}g)}{|S\nabla
g|}+{\mathcal{O}}(1).$

Note that both $None$ in (7.27) and $None$ in (7.28) have the term

$4\frac{b_{1}^{2}(\partial^{1,1}g)^{2}+b_{1}b_{2}(\partial^{2,1}g)^{2}}{|S\nabla
g|^{2}}$

in common with opposite sign, which allows us to remove the square in (7.15).
That is, adding up both terms,

$\boxed{M1}+\boxed{M2}=\frac{{\mathcal{O}}(\partial^{1,1}g+\partial^{2,1}g)}{|S\nabla
g|}+{\mathcal{O}}(1),$

whence we obtain

$\left|\boxed{M1}+\boxed{M2}\right|\overset{\text{{(\ref{two directional
derivatives})}}}{\lesssim}1+\frac{|\nabla^{2}g|}{|\nabla g|},$

as claimed. ∎

####  7.5.5. Proof of Lemma 7.17

We study, for $\psi\in W^{1,\infty}_{c}(\widetilde{\Omega}_{\varphi})$, the
term

$T_{M3}(\psi)\coloneqq\left(\frac{2}{|S\nabla
g|^{2}}\left\langle\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}\nabla_{R}g\right),B\nabla_{R}g\right\rangle\right)(\psi).$

In this subsection we don’t use the fact that we are in the plane. Instead,
the key ingredient is the $L_{A}$-harmonicity of the Green function far from
the pole, and so the computations in this subsection could be done also in
higher dimensions.

The term $T_{M3}$ must be read as

(7.29) $\displaystyle T_{M3}$ $\displaystyle=\frac{2}{|S\nabla
g|^{2}}\left\langle\sum_{i=1}^{2}\begin{pmatrix}\partial^{i}(b_{i}\partial^{i,1}g)\\\
\partial^{i}(b_{i}\partial^{i,2}g)\end{pmatrix},\begin{pmatrix}b_{1}\partial^{1}g\\\
b_{2}\partial^{2}g\end{pmatrix}\right\rangle$ $\displaystyle=\frac{2}{|S\nabla
g|^{2}}\sum_{i=1}^{2}\left\\{\partial^{i}(b_{i}\partial^{i,1}g)b_{1}\partial^{1}g+\partial^{i}(b_{i}\partial^{i,2}g)b_{2}\partial^{2}g\right\\}$
$\displaystyle=\frac{2}{|S\nabla
g|^{2}}\sum_{i=1}^{2}\sum_{j=1}^{2}\partial^{i}(b_{i}\partial^{i,j}g)b_{j}\partial^{j}g\eqqcolon\sum_{i=1}^{2}\sum_{j=1}^{2}T_{M3}^{(i,j)}.$

Here, for each $i,j\in\\{1,2\\}$,

(7.30)
$T_{M3}^{(i,j)}(\psi)=T_{\partial^{i}(b_{i}\partial^{i,j}g)}\left(\frac{2}{|S\nabla
g|^{2}}b_{j}\partial^{j}g\psi\right).$

We want to study the functionals $T_{M3}^{(i,j)}$ for $i,j\in\\{1,2\\}$. First
note that

$\frac{2}{|S\nabla g|^{2}}b_{j}\partial^{j}g\psi\in
W_{c}^{1,\infty}(\widetilde{\Omega}_{\varphi}),$

while $b_{i}\partial^{i,j}g\in L^{2}(\widetilde{\Omega}_{\varphi})$. Hence,
(7.30) makes sense and it suffices to study the functionals
$T_{\partial^{i}(b_{i}\partial^{i,j}g)}\in
W_{c}^{1,\infty}(\widetilde{\Omega})^{\prime}$. In fact, fixed
$j\in\\{1,2\\}$, we will exploit the cancellation of the functional
$\sum_{i=1}^{2}T_{\partial^{i}(b_{i}\partial^{i,j}g)}$. For simplicity we will
write $\partial^{i}(b_{i}\partial^{i,j}g)$ instead of
$T_{\partial^{i}(b_{i}\partial^{i,j}g)}$.

Compared to the strategy seen in Remark 7.15, now the matrix is not constant
and hence the directional derivatives do not commute. For this reason, some
error terms will appear in the procedure of extracting the gradient
$\nabla_{R}$ outside from

(7.31)
$\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}\nabla_{R}g\right)=\begin{pmatrix}\partial^{1}(b_{1}\partial^{1,1}g)+\partial^{2}(b_{2}\partial^{2,1}g)\\\
\partial^{1}(b_{1}\partial^{1,2}g)+\partial^{2}(b_{2}\partial^{2,2}g)\end{pmatrix},$

as it is done in the constant matrix case in (7.16). The idea is the same as
in the proof of Lemma 7.16, see Section 7.5.4. That is, using the
‘almost’-commutative property (7.5) (relating $\partial^{1,2}g$ and
$\partial^{2,1}g$) of the directional derivatives, and its functional version
in (7.10), we manage to extract the gradient $\nabla_{R}$ outside.

Let us fix $j\in\\{1,2\\}$. If $i=j$ (clearly $\partial^{i,j}=\partial^{j,i}$
in this case), then

(7.32)
$\partial^{i}(b_{i}\partial^{i,j}g)=\partial^{j}\left(\partial^{i}(b_{i}\partial^{i}g)\right)-\partial^{i}(\partial^{j}b_{i}\partial^{i}g),$

and when $i\not=j$, using (7.5) and (7.10) we get

(7.33)
$\displaystyle\partial^{i}(b_{i}\partial^{i,j}g)\overset{\text{{(\ref{derivative_commutator})}}}{}$
$\displaystyle{=}\partial^{i}\left(b_{i}\partial^{j,i}g\right)+\partial^{i}\left(b_{i}(\partial^{i}R_{j}-\partial^{j}R_{i})\nabla
g\right)$
$\displaystyle=\partial^{i}\left(\partial^{j}(b_{i}\partial^{i}g)\right)-\partial^{i}\left(\partial^{j}b_{i}\partial^{i}g\right)+\partial^{i}\left(b_{i}(\partial^{i}R_{j}-\partial^{j}R_{i})\nabla
g\right)$ $\displaystyle\overset{\text{{(\ref{directional derivatives does not
commute distribution})}}}{}$
$\displaystyle{=}\partial^{j}\left(\partial^{i}(b_{i}\partial^{i}g)\right)-(\partial^{j}R_{i}-\partial^{i}R_{j})\nabla(b_{i}\partial^{i}g)-\partial^{i}\left(\partial^{j}b_{i}\partial^{i}g\right)+\partial^{i}\left(b_{i}(\partial^{i}R_{j}-\partial^{j}R_{i})\nabla
g\right).$

Plugging 7.32 and 7.33 in (7.31), we get, for $j=1$,

(7.34) $\displaystyle\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i,1}g)=$
$\displaystyle\
\partial^{1}\left(\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}g\right)\right)-\sum_{i=1}^{2}\partial^{i}(\partial^{1}b_{i}\partial^{i}g)-(\partial^{1}R_{2}-\partial^{2}R_{1})\cdot\nabla(b_{2}\partial^{2}g)$
$\displaystyle+\partial^{2}\left(b_{2}(\partial^{2}R_{1}-\partial^{1}R_{2})\cdot\nabla
g\right),$

and for $j=2$,

(7.35) $\displaystyle\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i,2}g)=$
$\displaystyle\
\partial^{2}\left(\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}g\right)\right)-\sum_{i=1}^{2}\partial^{i}(\partial^{2}b_{i}\partial^{i}g)-(\partial^{2}R_{1}-\partial^{1}R_{2})\cdot\nabla(b_{1}\partial^{1}g)$
$\displaystyle+\partial^{1}\left(b_{1}(\partial^{1}R_{2}-\partial^{2}R_{1})\cdot\nabla
g\right).$

Note that in both cases there is the term
$\sum_{i=1}^{2}\partial^{i}\left(b_{i}\partial^{i}g\right)$, which is studied
in Claim 7.20. By (7.29) and using Claim 7.20 in both 7.34 and 7.35, in the
dual space we get

(7.36) $\displaystyle T_{M3}=$ $\displaystyle\frac{2}{|S\nabla
g|^{2}}\sum_{j=1}^{2}b_{j}\partial^{j}g\left(\sum_{i=1}^{2}\partial^{i}(b_{i}\partial^{i,j}g)\right)$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{2}\frac{2b_{j}\partial^{j}g}{|S\nabla
g|^{2}}\left\\{\partial^{j}\left(-\sum_{i=1}^{2}b_{i}\partial^{i}g\operatorname{div}R_{i}^{T}-\sum_{i,k=1}^{2}\partial_{i}\left(\frac{a_{ik}-a_{ki}}{2}\right)\partial_{k}g\right)-\sum_{i=1}^{2}\partial^{i}(\partial^{j}b_{i}\partial^{i}g)\right\\}$
$\displaystyle+\frac{2b_{1}\partial^{1}g}{|S\nabla
g|^{2}}\left\\{-(\partial^{1}R_{2}-\partial^{2}R_{1})\cdot\nabla(b_{2}\partial^{2}g)+\partial^{2}\left(b_{2}(\partial^{2}R_{1}-\partial^{1}R_{2})\cdot\nabla
g\right)\right\\}$ $\displaystyle+\frac{2b_{2}\partial^{2}g}{|S\nabla
g|^{2}}\left\\{-(\partial^{2}R_{1}-\partial^{1}R_{2})\cdot\nabla(b_{1}\partial^{1}g)+\partial^{1}\left(b_{1}(\partial^{1}R_{2}-\partial^{2}R_{1})\cdot\nabla
g\right)\right\\}.$

As a consequence of this, the right-hand equality in Claim 7.20 and
$|\nabla\partial^{i}g|={\mathcal{O}}(|\nabla
g|)+{\mathcal{O}}(|\nabla^{2}g|)$, see (7.3), we conclude that the functional
$T_{M3}$ is of the form

$T_{M3}=\sum_{j=1}^{2}\frac{2b_{j}\partial^{j}g}{|S\nabla
g|^{2}}\left({\mathcal{O}}(|\nabla
g|)+{\mathcal{O}}(|\nabla^{2}g|)+\partial^{1}\left({\mathcal{O}}(|\nabla
g|)\right)+\partial^{2}\left({\mathcal{O}}(|\nabla g|)\right)\right),$

as claimed. ∎

####  7.5.6. Proof of Lemma 7.18

First, we need the following claim, which we will prove later.

######  Claim 7.27.

Let $\psi\in W^{1,\infty}_{c}(\widetilde{\Omega}_{\varphi})$. Then, for
$i,j\in\\{1,2\\}$, both

$T_{\frac{1}{|S\nabla g|^{2}}\partial^{i}\left({\mathcal{O}}(|\nabla
g|^{2})\right)}(\psi)\quad\text{and}\quad
T_{\frac{b_{j}\partial^{j}g}{|S\nabla
g|^{2}}\partial^{i}\left({\mathcal{O}}(|\nabla g|)\right)}(\psi)$

are of the form

$\int\frac{{\mathcal{O}}(|\nabla^{2}g|)}{|\nabla
g|}\psi+\int{\mathcal{O}}(1)\psi+\int{\mathcal{O}}(1)\partial^{i}\psi.$

Granted this, by Lemma 7.17 and (7.14) we have

$|T_{M3}(\psi_{\varepsilon}\varphi g)|+|T_{E}(\psi_{\varepsilon}\varphi
g)|\lesssim\int\frac{|\nabla^{2}g|}{|\nabla g|}\psi_{\varepsilon}\varphi
g+\int\psi_{\varepsilon}\varphi g+\int|\partial^{i}(\psi_{\varepsilon}\varphi
g)|.$

The second term in the right-hand side is bounded by a constant times
${\mathcal{H}}^{2}(\Omega)<\infty$ as $\Omega$ is bounded and $g\lesssim 1$ in
$\operatorname{supp}\varphi$, see (W6). Hence, it suffices to prove

$\int\left|\partial^{i}(\psi_{\varepsilon}\varphi g)\right|\lesssim
1,\quad\text{for }i\in\\{1,2\\}.$

By the product derivative rule and the notation
$\partial^{i}f=R_{i}\cdot\nabla f$, we can use

$\int\left|\partial^{i}\left(\psi_{\varepsilon}\varphi
g\right)\right|\lesssim\int
g|\nabla\varphi|+\int_{\operatorname{supp}\varphi}|\nabla
g|+\frac{1}{\varepsilon}\int_{\operatorname{supp}\nabla\psi_{\varepsilon}}g.$

Since $|\nabla\varphi|\lesssim 1$ and $g(x)\lesssim 1$ in
$\operatorname{supp}\varphi$ (see (W6)), we get $\int g|\nabla\varphi|\lesssim
1$. For the second integral on the right-hand side we sum over Whitney cubes
in $W_{\varphi}$ and apply items 5 and 1 in Lemma 7.26

$\int_{\operatorname{supp}\varphi}|\nabla g|\leq\sum_{Q\in
W_{\varphi}}\int_{Q}|\nabla g|\overset{\text{{(\ref{sum over whitney region
p3})}}}{\lesssim}\sum_{Q\in
W_{\varphi}}\ell(Q)\cdot\widetilde{\omega}_{T}(50Q)\overset{\text{{(\ref{sum
over whitney region p1})}}}{\lesssim}1.$

On the other hand, using $\operatorname{supp}\nabla\psi_{\varepsilon}\subset
U_{3\varepsilon}(\partial\widetilde{\Omega})$ and Lemma 7.26(3) respectively,
we have

$\frac{1}{\varepsilon}\int_{\operatorname{supp}\nabla\psi_{\varepsilon}}g\leq\frac{1}{\varepsilon}\int_{U_{3\varepsilon}(\partial\widetilde{\Omega})}g\lesssim\varepsilon\leq
1,$

and Lemma 7.18 follows.∎

We now turn to the proof of Claim 7.27.

###### Proof of Claim 7.27.

First we study $T_{\frac{1}{|S\nabla
g|^{2}}\partial^{i}\left({\mathcal{O}}(|\nabla g|^{2})\right)}(\psi)$. We have

$\displaystyle T_{\frac{1}{|S\nabla g|^{2}}\partial^{i}({\mathcal{O}}(|\nabla
g|^{2}))}(\psi)$ $\displaystyle=T_{\partial^{i}({\mathcal{O}}(|\nabla
g|^{2}))}\left(\frac{1}{|S\nabla g|^{2}}\psi\right)$
$\displaystyle{=}-\int{\mathcal{O}}(|\nabla g|^{2})\frac{1}{|S\nabla
g|^{2}}\psi\operatorname{div}R_{i}^{T}-\int{\mathcal{O}}(|\nabla
g|^{2})\partial^{i}\left(\frac{1}{|S\nabla g|^{2}}\psi\right).$

The first term in the right-hand side is of the form
$\int{\mathcal{O}}(1)\psi$. Let us study the second term in the right-hand
side. By (7.19),

$\partial^{i}\left(\frac{1}{|S\nabla
g|^{2}}\psi\right)=-\frac{2\langle\partial^{i}\nabla_{R}g,B\nabla_{R}g\rangle}{|S\nabla
g|^{4}}\psi-\frac{\langle\nabla_{R}g,\partial^{i}B\nabla_{R}g\rangle}{|S\nabla
g|^{4}}\psi+\frac{\partial^{i}\psi}{|S\nabla g|^{2}},$

and hence we get

$\displaystyle-\int{\mathcal{O}}(|\nabla
g|^{2})\partial^{i}\left(\frac{1}{|S\nabla g|^{2}}\psi\right)$
$\displaystyle=\int\frac{{\mathcal{O}}(\partial^{i}\nabla_{R}g)}{|S\nabla
g|}\psi+\int{\mathcal{O}}(1)\psi+\int{\mathcal{O}}(1)\partial^{i}\psi$
$\displaystyle{=}\int\frac{{\mathcal{O}}(|\nabla^{2}g|)}{|\nabla
g|}\psi+\int{\mathcal{O}}(1)\psi+\int{\mathcal{O}}(1)\partial^{i}\psi.$

Next we study $T_{\frac{b_{j}\partial^{j}g}{|S\nabla
g|^{2}}\partial^{i}\left({\mathcal{O}}(|\nabla g|)\right)}(\psi)$, similarly
as we did with the previous term. Now we have

$\displaystyle T_{\frac{b_{j}\partial^{j}g}{|S\nabla
g|^{2}}\partial^{i}({\mathcal{O}}(|\nabla g|))}(\psi)$
$\displaystyle=T_{\partial^{i}({\mathcal{O}}(|\nabla
g|))}\left(\frac{b_{j}\partial^{j}g}{|S\nabla g|^{2}}\psi\right)$
$\displaystyle{=}-\int\frac{{\mathcal{O}}(|\nabla
g|)b_{j}\partial^{j}g}{|S\nabla
g|^{2}}\psi\operatorname{div}R_{i}-\int{\mathcal{O}}(|\nabla
g|)\partial^{i}\left(\frac{b_{j}\partial^{j}g}{|S\nabla g|^{2}}\psi\right).$

As before, the first term in the right-hand side is of the form
$\int{\mathcal{O}}(1)\psi$ and the second term is

$-\int{\mathcal{O}}(|\nabla
g|)\partial^{i}\left(\frac{b_{j}\partial^{j}g}{|S\nabla
g|^{2}}\psi\right)=\int\frac{{\mathcal{O}}(1)\psi}{|S\nabla
g|}\partial^{i}\left(b_{j}\partial^{j}g\right)+\int{\mathcal{O}}(|\nabla
g|^{2})\partial^{i}\left(\frac{1}{|S\nabla
g|^{2}}\right)\psi+\int{\mathcal{O}}(1)\partial^{i}\psi.$

The claim follows by using (7.3) and (7.19) in the previous line. ∎

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