# Elliptic Theory for Sets with Higher Co-dimensional Boundaries

Guy David  Univ Paris-Sud, Laboratoire de Mathématiques, UMR8628, Orsay,
91405, France<EMAIL_ADDRESS>,  Joseph Feneuil  Department of
Mathematics, Temple University, 1805 N. Broad Street, Philadelphia, PA 19122,
USA<EMAIL_ADDRESS>and  Svitlana Mayboroda  School of Mathematics,
University of Minnesota, Minneapolis, MN 55455, USA<EMAIL_ADDRESS>
(Date: February 25, 2017)

###### Abstract.

Many geometric and analytic properties of sets hinge on the properties of
elliptic measure, notoriously missing for sets of higher co-dimension. The aim
of this manuscript is to develop a version of elliptic theory, associated to a
linear PDE, which ultimately yields a notion analogous to that of the harmonic
measure, for sets of codimension higher than 1.

To this end, we turn to degenerate elliptic equations. Let
$\Gamma\subset\mathbb{R}^{n}$ be an Ahlfors regular set of dimension $d<n-1$
(not necessarily integer) and $\Omega=\mathbb{R}^{n}\setminus\Gamma$. Let
$L=-\operatorname{div}A\nabla$ be a degenerate elliptic operator with
measurable coefficients such that the ellipticity constants of the matrix $A$
are bounded from above and below by a multiple of
$\,\mathrm{dist}(\cdot,\Gamma)^{d+1-n}$. We define weak solutions; prove trace
and extension theorems in suitable weighted Sobolev spaces; establish the
maximum principle, De Giorgi-Nash-Moser estimates, the Harnack inequality, the
Hölder continuity of solutions (inside and at the boundary). We define the
Green function and provide the basic set of pointwise and/or $L^{p}$ estimates
for the Green function and for its gradient. With this at hand, we define
harmonic measure associated to $L$, establish its doubling property, non-
degeneracy, change-of-the-pole formulas, and, finally, the comparison
principle for local solutions.

In another article to appear, we will prove that when $\Gamma$ is the graph of
a Lipschitz function with small Lipschitz constant, we can find an elliptic
operator $L$ for which the harmonic measure given here is absolutely
continuous with respect to the $d$-Hausdorff measure on $\Gamma$ and vice
versa. It thus extends Dahlberg’s theorem to some sets of codimension higher
than 1.

###### Key words and phrases:

harmonic measure, boundary of co-dimension higher than 1, trace theorem,
extension theorem, degenerate elliptic operators, maximum principle, Hölder
continuity of solutions, De Giorgi-Nash-Moser estimates, Green functions,
comparison principle, homogeneous weighted Sobolev spaces

###### 2010 Mathematics Subject Classification:

Primary 31B05, 35J25, Secondary 28A75, 28A78 , 35J70, 42B20, 42B25, 42B37

This research was supported in part by Fondation Jacques Hadamard and by CNRS.
The first author was supported in part by the ANR, programme blanc GEOMETRYA
ANR-12-BS01-0014, the European Community Marie Curie grant MANET 607643 and
H2020 grant GHAIA 777822, and the Simons Collaborations in MSP grant 601941,
GD. The third author was supported by the Alfred P. Sloan Fellowship, the NSF
INSPIRE Award DMS 1344235, NSF CAREER Award DMS 1220089, the NSF RAISE-TAQ
grant DMS 1839077, and the Simons Foundation grant 563916, SM.

We would like to thank the Department of Mathematics at Université Paris-Sud,
the Ecole des Mines, and the Mathematical Sciences Research Institute (NSF
grant DMS 1440140) for warm hospitality.

Finally, we would like to thank Alano Ancona for stimulating discussions at
the early stages of the project and for sharing with us the results of his
work.

###### Contents

  1. 1 Introduction
  2. 2 The Harnack Chain Condition and the Doubling Property
  3. 3 Traces
  4. 4 Poincaré Inequalities
  5. 5 Completeness and Density of Smooth Functions
  6. 6 The Chain Rule and Applications
  7. 7 The Extension Operator
  8. 8 Definition of Solutions
  9. 9 Harmonic Measure
  10. 10 Green Functions
  11. 11 The Comparison Principle
    1. 11.1 Discussion of the comparison theorem in codimension 1
    2. 11.2 The case of codimension higher than 1

##  Chapter 1 Introduction

The past few years have witnessed remarkable progress in the study of
relations between regularity properties of the harmonic measure $\omega$ on
the boundary of a domain of $\mathbb{R}^{n}$ (for instance, its absolute
continuity with respect to the Hausdorff measure $\mathcal{H}^{n-1}$) and the
regularity of the domain (for instance, rectifiability properties of the
boundary). In short, the emerging philosophy is that the rectifiability of the
boundary is necessary for the absolute continuity of $\omega$ with respect to
$\mathcal{H}^{n-1}$, and that rectifiability along with suitable connectedness
assumptions is sufficient. Omitting for now precise definitions, let us recall
the main results in this regard. The celebrated 1916 theorem of F. & M. Riesz
has established the absolute continuity of the harmonic measure for a simply
connected domain in the complex plane, with a rectifiable boundary [RR]. The
quantifiable analogue of this result (the $A^{\infty}$ property of harmonic
measure) was obtained by Lavrent’ev in 1936 [Lv] and the local version,
pertaining to subsets of a rectifiable curve which is a boundary of a simply
connected planar domain, was proved by Bishop and Jones in 1990 [BJ]. In the
latter work the authors also showed that some connectedness is necessary for
the absolute continuity of $\omega$ with respect to $\mathcal{H}^{n-1}$, for
there exists a planar set with rectifiable boundary for which the harmonic
measure is singular with respect to $\mathcal{H}^{n-1}$.

The situation in higher dimensions, $n\geq 3$, is even more complicated. The
absolute continuity of $\omega$ with respect to $\mathcal{H}^{n-1}$ was first
established by Dahlberg on Lipschitz graphs [Da] and was then extended to non-
tangentially accessible (NTA) domains with Ahlfors regular boundary in [DJ],
[Se], and to more general NTA domains in [Ba]. Roughly speaking, the non-
tangential accessibility is an assumption of quantifiable connectedness, which
requires the presence of interior and exterior corkscrew points, as well as
Harnack chains. Ahlfors regularity simply postulates that the measure of
intersection with the boundary of every ball of radius $r$ centered at the
boundary is proportional to $r^{n-1}$, i.e., that the boundary is in a certain
sense $n-1$ dimensional (we will provide a careful definition below).
Similarly to the lower-dimensional case, counterexamples show that some
topological restrictions are needed for the absolute continuity of $\omega$
with respect to $\mathcal{H}^{n-1}$ [Wu], [Z]. Much more recently, in [HM1],
[HMU], [AHMNT], the authors proved that, in fact, for sets with Ahlfors
regular boundaries, under a (weaker) 1-sided NTA assumption, the uniform
rectifiability of the boundary is equivalent to the complete set of NTA
conditions and hence, is equivalent to the absolute continuity of harmonic
measure with respect to the Lebesgue measure. Finally, in 2015 the full
converse, “free boundary” result was obtained and established that
rectifiability is necessary for the absolute continuity of harmonic measure
with respect to $\mathcal{H}^{n-1}$ in any dimension $n\geq 2$ (without any
additional topological assumptions) [AHM3TV]. It was proved simultaneously
that for a complement of an $(n-1)$-Ahlfors regular set the $A^{\infty}$
property of harmonic measure yields uniform rectifiability of the boundary
[HLMN]. Shortly after, it was established that in an analogous setting
$\varepsilon$-approximability and Carleson measure estimates for bounded
harmonic functions are equivalent to uniform rectifiability [HMM1], [GMT], and
that analogous results hold for more general elliptic operators [HMM2],
[AGMT].

The purpose of this work is to start the investigation of similar properties
for domains with a lower-dimensional boundary $\Gamma$. To the best of our
knowledge, the only known approach to elliptic problems on domains with higher
co-dimensional boundaries is by means of the $p$-Laplacian operator and its
generalizations [HKM], [LN]. In [LN] the authors worked with an associated
Wiener capacity, defined $p$-harmonic measure, and established boundary
Harnack inequalities for Reifenberg flat sets of co-dimension higher than one.
Our goals here are different.

We shall systematically assume that $\Gamma$ is Ahlfors-regular of some
dimension $d<n-1$, which does not need to be an integer. This means that there
is a constant $C_{0}\geq 1$ such that

(1.1) $C_{0}^{-1}r^{d}\leq\mathcal{H}^{d}(\Gamma\cap B(x,r))\leq C_{0}r^{d}\
\text{ for }x\in\Gamma\text{ and }r>0.$

We want to define an analogue of the harmonic measure, that will be defined on
$\Gamma$ and associated to a divergence form operator on
$\Omega=\mathbb{R}^{n}\setminus\Gamma$. We still write the operator as
$L=-\rm{div}A\nabla$, with $A:\Omega\to\mathbb{M}_{n}(\mathbb{R})$, and we
write the ellipticity condition with a different homogeneity, i.e., we require
that for some $C_{1}\geq 1$,

(1.2) $\displaystyle\,\mathrm{dist}(x,\Gamma)^{n-d-1}A(x)\xi\cdot\zeta\leq
C_{1}|\xi|\,|\zeta|\ \text{ for }x\in\Omega\text{ and
}\xi,\zeta\in\mathbb{R}^{n},$ (1.3)
$\displaystyle\,\mathrm{dist}(x,\Gamma)^{n-d-1}A(x)\xi\cdot\xi\geq
C_{1}^{-1}|\xi|^{2}\ \text{ for }x\in\Omega\text{ and }\xi\in\mathbb{R}^{n}.$

The effect of this normalization should be to incite the analogue of the
Brownian motion here to get closer to the boundary with the right probability;
for instance if $\Gamma=\mathbb{R}^{d}\subset\mathbb{R}^{n}$ and
$A(x)=\,\mathrm{dist}(x,\Gamma)^{-n+d+1}I$, it turns out that the effect of
$L$ on functions $f(x,t)$ that are radial in the second variable
$t\in\mathbb{R}^{n-d}$ is the same as for the Laplacian on
$\mathbb{R}^{d+1}_{+}$. In some sense, we create Brownian travelers which
treat $\Gamma$ as a “black hole”: they detect more mass and they are more
attracted to $\Gamma$ than a standard Brownian traveler governed by the
Laplacian would be.

The purpose of the present manuscript is to develop, with merely these
assumptions, a comprehensive elliptic theory. We solve the Dirichlet problem
for $Lu=0$, prove the maximum principle, the De Giorgi-Nash-Moser estimates
and the Harnack inequality for solutions, use this to define a harmonic
measure associated to $L$, show that it is doubling, and prove the comparison
principle for positive $L$-harmonic functions that vanish at the boundary. Let
us discuss the details.

We first introduce some notation. Set $\delta(x)=\,\mathrm{dist}(x,\Gamma)$
and $w(x)=\delta(x)^{-n+d+1}$ for $x\in\Omega=\mathbb{R}^{n}\setminus\Gamma$,
and denote by $\sigma$ the restriction to $\Gamma$ of $\mathcal{H}^{d}$.
Denote by $W=\dot{W}^{1,2}_{w}(\Omega)$ the weighted Sobolev space of
functions $u\in L^{1}_{loc}(\Omega)$ whose distribution gradient in $\Omega$
lies in $L^{2}(\Omega,w)$:

(1.4) $W=\dot{W}^{1,2}_{w}(\Omega):=\\{u\in L^{1}_{loc}(\Omega):\,\nabla u\in
L^{2}(\Omega,w)\\},$

and set $\|u\|_{W}=\big{\\{}\int_{\Omega}|\nabla
u(x)|^{2}w(x)dx\big{\\}}^{1/2}$ for $f\in W$. Finally denote by
${\mathcal{M}}(\Gamma)$ the set of measurable functions on $\Gamma$ and then
set

(1.5)
$H=\dot{H}^{1/2}(\Gamma):=\left\\{g\in{\mathcal{M}}(\Gamma):\,\int_{\Gamma}\int_{\Gamma}{|g(x)-g(y)|^{2}\over|x-y|^{d+1}}d\sigma(x)d\sigma(y)<\infty\right\\}.$

Before we solve Dirichlet problems we construct two bounded linear operators
$T:W\to H$ (a trace operator) and $E:H\to W$ (an extension operator), such
that $T\circ E=I_{H}$. The trace of $u\in W$ is such that for $\sigma$-almost
every $x\in\Gamma$,

(1.6) $Tu(x)=\lim_{r\to 0}\fint_{B(x,r)}u(y)dy:=\lim_{r\to
0}{1\over|B(x,r)|}\int u(y)dy,$

and even, analogously to the Lebesgue density property,

(1.7) $\lim_{r\to 0}\fint_{B(x,r)}|u(y)-Tu(x)|dy=0.$

Similarly, we check that if $g\in H$, then

(1.8) $\lim_{r\to 0}\fint_{\Gamma\cap B(x,r)}|g(y)-g(x)|d\sigma(y)=0$

for $\sigma$-almost every $x\in\Gamma$. We typically use the fact that
$|u(x)-u(y)|\leq\int_{[x,y]}|\nabla u|d\mathcal{L}^{1}$ for almost all choices
of $x$ and $y\in\Omega$, for which we can use the absolute continuity of $u\in
W$ on (almost all) line segments, plus the important fact that, by (1.1),
$\Gamma\cap\ell=\emptyset$ for almost every line $\ell$.

Note that the latter geometric fact is enabled specifically by the higher co-
dimension ($d<n-1$), even though our boundary can be quite wild. In fact, a
stronger property holds in the present setting and gives, in particular,
Harnack chains. There exists a constant $C>0$, that depends only on $C_{0}$,
$n$, and $d<n-1$, such that for $\Lambda\geq 1$ and $x_{1},x_{2}\in\Omega$
such that $\,\mathrm{dist}(x_{i},\Gamma)\geq r$ and $|x_{1}-x_{2}|\leq\Lambda
r$, we can find two points $y_{i}\in B(x_{i},r/2)$ such that
$\,\mathrm{dist}([y_{1},y_{2}],\Gamma)\geq C^{-1}\Lambda^{-d/(n-d-1)}r$. That
is, there is a thick tube in $\Omega$ that connects the two $B(x_{i},r/2)$.

Once we have trace and extension operators, we deduce from the Lax-Milgram
theorem that for $g\in H$, there is a unique weak solution $u\in W$ of $Lu=0$
such that $Tu=g$. For us a weak solution is a function $u\in W$ such that

(1.9) $\int_{\Omega}A(x)\nabla u(x)\cdot\nabla\varphi(x)dx=0$

for all $\varphi\in C_{0}^{\infty}(\Omega)$, the space of infinitely
differentiable functions which are compactly supported in $\Omega$.

Then we follow the Moser iteration scheme to study the weak solutions of
$Lu=0$, as we would do in the standard elliptic case in codimension $1$. This
leads to the quantitative boundedness (a.k.a. Moser bounds) and the
quantitative Hölder continuity (a.k.a. De Giorgi-Nash estimates), in an
interior or boundary ball $B$, of any weak solution of $Lu=0$ in $2B$ such
that $Tu=0$ on $\Gamma\cap 2B$ when the intersection is non-empty. Precise
estimates will be given later in the introduction. The boundary estimates are
trickier, because we do not have the conventional “fatness” of the complement
of the domain, and it is useful to know beforehand that suitable versions of
Poincaré and Sobolev inequalities hold. For instance,

(1.10) $\fint_{B(x,r)}|u(y)|dy\leq Cr^{-d}\int_{B(x,r)}|\nabla u(y)|w(y)dy$

for $u\in W$, $x\in\Gamma$, and $r>0$ such that $Tu=0$ on $\Gamma\cap B(x,r)$
and, if $m(B(x,r))$ denotes $\int_{B(x,r)}w(y)dy$,

(1.11)
$\Big{\\{}\frac{1}{m(B(x,r))}\int_{B(x,r)}\Big{|}u(y)-\fint_{B(x,r)}u\Big{|}^{p}w(y)dy\Big{\\}}^{1/p}\leq
Cr\Big{\\{}\frac{1}{m(B(x,r))}\int_{B(x,r)}|\nabla
u(y)|^{2}w(y)dy\Big{\\}}^{1/2}$

for $u\in W$, $x\in\overline{\Omega}=\mathbb{R}^{n}$, $r>0$, and
$p\in\left[1,\frac{2n}{n-2}\right]$ (if $n\geq 3$) or $p\in[1,+\infty)$ (if
$n=2$).

A substantial portion of the proofs lies in the analysis of the newly defined
Sobolev spaces. It is important to note, in particular, that we prove the
density of smooth functions on $\mathbb{R}^{n}$ (and not just $\Omega$) in our
weighted Sobolev space $W$. That is, for any function $f$ in $W$, there exists
a sequence $(f_{k})_{k\geq 1}$ of functions in
$C^{\infty}({\mathbb{R}^{n}})\cap W$ such that $\|f-f_{k}\|_{W}$ tends to 0
and $f_{k}$ converges to $f$ in $L^{1}_{loc}({\mathbb{R}^{n}})$. In
codimension $1$, this sort of property, just like (1.10) or (1.11), typically
requires a fairly nice boundary, e.g., Lipschitz, and it is quite remarkable
that here they all hold in the complement of any Ahlfors-regular set. This is,
of course, a fortunate outcome of working with lower dimensional boundary: we
can guarantee ample access to the boundary (cf., e.g., the Harnack “tubes”
discussed above), which turns out to be sufficient despite the absence of
traditionally required “massive complement”. Or rather one could say that the
boundary itself is sufficiently “massive” from the PDE point of view, due to
our carefully chosen equation and corresponding function spaces.

With all these ingredients, we can follow the standard proofs for elliptic
divergence form operators. When $u$ is a solution to $Lu=0$ in a ball
$2B\subset\Omega$, the De Giorgi-Nash-Moser estimates and the Harnack
inequality in the ball $B$ don’t depend on the properties of the boundary
$\Gamma$ and thus can be proven as in the case of codimension 1. When
$B\subset\mathbb{R}^{n}$ is a ball centered on $\Gamma$ and $u$ is a weak
solution to $Lu=0$ in $2B$ whose trace satisfies $Tu=0$ on $\Gamma\cap 2B$,
the quantitative boundedness and the quantitative Hölder continuity of the
solution $u$ are expressed with the help of the weight $w$. There holds, if
$m(2B)=\int_{2B}w(y)dy$,

(1.12) $\sup_{B}u\leq
C\left(\frac{1}{m(2B)}\int_{2B}|u(y)|^{2}w(y)dy\right)^{1/2}$

and, for any $\theta\in(0,1]$,

(1.13) $\sup_{\theta B}u\leq C\theta^{\alpha}\sup_{B}u\leq
C\theta^{\alpha}\left(\frac{1}{m(2B)}\int_{2B}|u(y)|^{2}w(y)dy\right)^{1/2},$

where $\theta B$ denotes the ball with same center as $B$ but whose radius is
multiplied by $\theta$, and $C,\alpha>0$ are constants that depend only on the
dimensions $d$ and $n$, the Ahlfors constant $C_{0}$ and the ellipticity
constant $C_{1}$.

We establish then the existence and uniqueness of a Green function $g$, which
is roughly speaking a positive function on $\Omega\times\Omega$ such that, for
all $y\in\Omega$, the function $g(.,y)$ solves $Lg(.,y)=\delta(y)$ and
$Tg(.,y)=0$. In particular, the following pointwise estimates are shown:

(1.14) $0\leq g(x,y)\leq\left\\{\begin{array}[]{ll}C|x-y|^{1-d}&\text{ if
}4|x-y|\geq\delta(y)\\\ \frac{C|x-y|^{2-n}}{w(y)}&\text{ if
}2|x-y|\leq\delta(y),\,n\geq 3\\\
\frac{C_{\epsilon}}{w(y)}\left(\frac{\delta(y)}{|x-y|}\right)^{\epsilon}&\text{
if }2|x-y|\leq\delta(y),\,n=2,\end{array}\right.$

where $C>0$ depends on $d$, $n$, $C_{0}$, $C_{1}$ and $C_{\epsilon}>0$ depends
on $d$, $C_{0}$, $C_{1}$, $\epsilon$. When $n\geq 3$, the pointwise estimates
can be gathered to a single one, and may look more natural for the reader: if
$m(B)=\int_{B}w(y)dy$,

(1.15) $0\leq g(x,y)\leq C\frac{|x-y|^{2}}{m(B(x,|x-y|))}$

whenever $x,y\in\Omega$. The bound in the case where $n=2$ and
$2|x-y|\leq\delta(y)$ can surely be improved into a logarithm bound, but the
bound given here is sufficient for our purposes. Also, our results hold for
any $d$ and any $n$ such that $d<n-1$, (i.e., even in the cases where $n=2$ or
$d\leq 1$), which proves that Ahlfors regular domains are ‘Greenian sets’ in
our adapted elliptic theory. Note that contrary to the codimension 1 case, the
notion of the fundamental solution in ${\mathbb{R}}^{n}$ is not accessible,
since the distance to the boundary of $\Omega$ is an integral part of the
definition of $L$.

We use the Harnack inequality, the De Giorgi-Nash-Moser estimates, as well as
a suitable version of the maximum principle, to solve the Dirichlet problem
for continuous functions with compact support on $\Gamma$, and then to define
harmonic measures $\omega^{x}$ for $x\in\Omega$ (so that
$\int_{\Gamma}gd\omega^{x}$ is the value at $x$ of the solution of the
Dirichlet problem for $g$). Note that we do not need an analogue of the Wiener
criterion (which normally guarantees that solutions with continuous data are
continuous up to the boundary and allows one to define the harmonic measure),
as we have already proved a stronger property, that solutions are Hölder
continuous up to the boundary. Then, following the ideas of [Ken, Section
1.3], we prove the following properties on the harmonic measure $\omega^{x}$.
First, the non-degeneracy of the harmonic measure states that if $B$ is a ball
centered on $\Gamma$,

(1.16) $\omega^{x}(B\cap\Gamma)\geq C^{-1}$

whenever $x\in\Omega\cap\frac{1}{2}B$ and

(1.17) $\omega^{x}(\Gamma\setminus B)\geq C^{-1}$

whenever $x\in\Omega\setminus 2B$, the constant $C>0$ depending as previously
on $d$, $n$, $C_{0}$ and $C_{1}$. Next, let us recall that any boundary ball
has a corkscrew point, that is for any ball
$B=B(x_{0},r)\subset\mathbb{R}^{n}$ centered on $\Gamma$, there exists
$\Delta_{B}\in B$ such that $\delta(\Delta_{B})$ is bigger than $\epsilon r$,
where $\epsilon>0$ depends only on $d$, $n$ and $C_{0}$. With this definition
in mind, we compare the harmonic measure with the Green function: for any ball
$B$ of radius $r$ centered on $\Gamma$,

(1.18) $C^{-1}r^{1-d}g(x,\Delta_{B})\leq\omega^{x}(B\cap\Gamma)\leq
Cr^{1-d}g(x,\Delta_{B})$

for any $x\in\Omega\setminus 2B$ and

(1.19) $C^{-1}r^{1-d}g(x,\Delta_{B})\leq\omega^{x}(\Gamma\setminus B)\leq
Cr^{1-d}g(x,\Delta_{B})$

for any $x\in\Omega\cap\frac{1}{2}B$ which is far enough from $\Delta_{B}$,
say $|x-\Delta_{B}|\geq\epsilon r/2$, where $\epsilon$ is the constant used to
define $\Delta_{B}$. The constant $C>0$ in (1.18) and (1.19) depends again
only on $d$, $n$, $C_{0}$ and $C_{1}$. The estimates (1.18) and (1.19) can be
seen as weak versions of the ‘comparison principle’, which deal only with the
Green functions and the harmonic measure and which can be proven by using the
specific properties of the latter objects. The inequalities (1.18) and (1.19)
are essential for the proofs of the next three results.

The first one is the doubling property of the harmonic measure, which
guarantees that, if $B$ is a ball centered on $\Gamma$,
$\omega^{x}(2B\cap\Gamma)\leq C\omega^{x}(B\cap\Gamma)$ whenever
$x\in\Omega\setminus 4B$. It has an interesting counterpart:
$\omega^{x}(\Gamma\setminus B)\leq C\omega^{x}(\Gamma\setminus 2B)$ whenever
$x\in\Omega\cap\frac{1}{2}B$.

The second one is the change-of-the-pole estimates, which can be stated as

(1.20)
$C^{-1}\omega^{\Delta_{B}}(E)\leq\frac{\omega^{x}(E)}{\omega^{x}(\Gamma\cap
B)}\leq C\omega^{\Delta_{B}}(E)$

when $B$ is a ball centered on $\Gamma$, $E\subset B\cap\Gamma$ is a Borel
set, and $x\in\Omega\setminus 2B$.

The last result is the comparison principle, that says that if $u$ and $v$ are
positive weak solutions of $Lu=Lv=0$ such that $Tu=Tv=0$ on $2B\cap\Gamma$,
where $B$ is a ball centered on $\Gamma$, then $u$ and $v$ are comparable in
$B$, i.e.,

(1.21) $\sup_{z\in B\setminus\Gamma}\frac{u(z)}{v(z)}\leq C\inf_{z\in
B\setminus\Gamma}\frac{u(z)}{v(z)}.$

In each case, i.e., for the doubling property of the harmonic measure, the
change of pole, or the comparison principle, the constant $C>0$ depends only
on $d$, $n$, $C_{0}$ and $C_{1}$.

It is difficult to survey a history of the subject that is so classical (in
the co-dimension one case). In that setting, that is, in co-dimension one and
reasonably nice geometry, e.g., of Lipschitz domains, the results have largely
become folklore and we often follow the exposition in standard texts [GT],
[HL], [Maz], [MZ], [Sta2], [GW], [CFMS]. The general order of development is
inspired by [Ken]. Furthermore, let us point out that while the invention of a
harmonic measure which serves the higher co-dimensional boundaries, which is
associated to a linear PDE, and which is absolutely continuous with respect to
the Lebesgue measure on reasonably nice sets, is the main focal point of our
work, various versions of degenerate elliptic operators and weighted Sobolev
spaces have of course appeared in the literature over the years. Some versions
of some of the results listed above or similar ones can be found, e.g., in
[A], [FKS], [Haj], [HaK2], [HKM], [Kil], [JW]. However, the presentation here
is fully self-contained, and since we did not rely on previous work, we hope
to be forgiven for not providing a detailed review of the corresponding
literature. Also, the context of the present paper often makes it possible to
have much simpler proofs than a more general setting of not necessarily
Ahlfors regular sets. It is perhaps worth pointing out that we work with
homogeneous Sobolev spaces. Unfortunately, those are much less popular in the
literature that their non-homogeneous counterparts, while they are more
suitable for PDEs on unbounded domains.

As outlined in [DFM], , we intend in subsequent publications to take stronger
assumptions, both on the geometry of $\Gamma$ and the choice of $L$, and prove
that the harmonic measure defined here is absolutely continuous with respect
to $\mathcal{H}^{d}_{|\Gamma}$. For instance, we will assume that $d$ is an
integer and $\Gamma$ is the graph of a Lipschitz function
$F:\mathbb{R}^{d}\to\mathbb{R}^{n-d}$, with a small enough Lipschitz constant.
As for $A$, we will assume that $A(x)=D(x)^{-n+d+1}I$ for $x\in\Omega$, with

(1.22)
$D(x)=\Big{\\{}\int_{\Gamma}|x-y|^{-d-\alpha}d\mathcal{H}^{d}(y)\Big{\\}}^{-1/\alpha}$

for some constant $\alpha>0$. Notice that because of (1.1), $D(x)$ is
equivalent to $\delta(x)$; when $d=1$ we can also take
$A(x)=\delta(x)^{-n+d+1}I$, but when $d\geq 2$ we do not know whether
$\delta(x)$ is smooth enough to work. In (1.22), we could also replace
$\mathcal{H}^{d}$ with another Ahlfors-regular measure on $\Gamma$.

With these additional assumptions we will prove that the harmonic measure
described above is absolutely continuous with respect to
$\mathcal{H}^{d}_{|\Gamma}$, with a density which is a Muckenhoupt
$A_{\infty}$ weight. In other words, we shall establish an analogue of
Dahlberg’s result [Da] for domains with a higher co-dimensional boundary given
by a Lipschitz graph with a small Lipschitz constant. It is not so clear what
is the right condition for this in terms of $A$, but the authors still hope
that a good condition on $\Gamma$ is its uniform rectifiability. Notice that
in remarkable contrast with the case of codimension 1, we do not state an
additional quantitative connectedness condition on $\Omega$, such as the
Harnack chain condition in codimension $1$; this is because such conditions
are automatically satisfied when $\Gamma$ is Ahlfors-regular with a large
codimension.

The present paper is aimed at giving a fairly pleasant general framework for
studying a version of the harmonic measure in the context of Ahlfors-regular
sets $\Gamma$ of codimension larger than $1$, but it will probably be
interesting and hard to understand well the relations between the geometry of
$\Gamma$, the regularity properties of $A$ (which has to be linked to $\Gamma$
through the distance function), and the regularity properties of the
associated harmonic measure.

##  Chapter 2 The Harnack Chain Condition and the Doubling Property

We keep the same notation as in Section 1, concerning
$\Gamma\subset\mathbb{R}^{n}$, a closed set that satisfies (1.1) for some
$d<n-1$, $\Omega=\mathbb{R}^{n}\setminus\Gamma$, then
$\sigma=\mathcal{H}^{d}_{|\Gamma}$, $\delta(z)=\,\mathrm{dist}(z,\Gamma)$, and
the weight $w(z)=\delta(z)^{d+1-n}$.

Let us add the notion of measure. The measure $m$ is defined on
(Lebesgue-)measurable subset of $\mathbb{R}^{n}$ by $m(E)=\int_{E}w(z)dz$. We
may write $dm(z)$ for $w(z)dz$. Since $0<w<+\infty$ a.e. in $\mathbb{R}^{n}$,
$m$ and the Lebesgue measure are mutually absolutely continuous, that is they
have the same zero sets. Thus there is no need to specify the measure when
using the expressions almost everywhere and almost every, both abbreviated
a.e.; conversely the expression $a.e.$ without any measure refers to Lebesgue
almost everywhere or equivalently $m$-almost everywhere.

In the sequel of the article, $C$ will denote a real number (usually big) that
can vary from one line to another. The parameters which the constant $C$
depends on are either obvious from context or recalled. Besides, the notation
$A\approx B$ will be used to replace $C^{-1}A\leq B\leq CA$.

This section is devoted to the proof of the very first geometric properties on
the space $\Omega$ and the weight $w$. We will prove in particular that $m$ is
a doubling measure and $\Omega$ satisfies the Harnack chain condition.

First, let us prove the Harnack chain condition we stated in Section 1.

######  Lemma 2.1.

Let $\Gamma$ be a $d$-Ahlfors regular set in ${\mathbb{R}}^{n}$, $d<n-1$, that
is, assume that (1.1) is satisfied. Then there exists a constant $c>0$, that
depends only on $C_{0}$, $n$, and $d<n-1$, such that for $\Lambda\geq 1$ and
$x_{1},x_{2}\in\Omega$ such that $\,\mathrm{dist}(x_{i},\Gamma)\geq r$ and
$|x_{1}-x_{2}|\leq\Lambda r$, we can find two points $y_{i}\in B(x_{i},r/3)$
such that $\,\mathrm{dist}([y_{1},y_{2}],\Gamma)\geq c\Lambda^{-d/(n-d-1)}r$.
That is, there is a thick tube in $\Omega$ that connects the two
$B(x_{i},r/3)$.

Proof. Indeed, suppose $x_{2}\neq x_{1}$, set $\ell=[x_{1},x_{2}]$, and denote
by $P$ the vector hyperplane with a direction orthogonal to $x_{2}-x_{1}$. Let
$\varepsilon\in(0,1)$ be small, to be chosen soon. We can find $N\geq
C^{-1}\varepsilon^{1-n}$ points $z_{j}\in P\cap B(0,r/3)$, such that
$|z_{j}-z_{k}|\geq 4\varepsilon r$ for $j\neq k$. Set $\ell_{j}=z_{j}+\ell$,
and suppose that $\,\mathrm{dist}(\ell_{j},\Gamma)\leq\varepsilon r$ for all
$j$. Then we can find points $w_{j}\in\Gamma$ such that
$\,\mathrm{dist}(w_{j},\ell_{j})\leq\varepsilon r$. Notice that the balls
$B_{j}=B(w_{j},\varepsilon r)$ are disjoint because
$\,\mathrm{dist}(\ell_{j},\ell_{k})\geq 4\varepsilon r$, and by (1.1)

$NC_{0}^{-1}(\varepsilon
r)^{d}\leq\sum_{j}\sigma(B_{j})=\sigma\big{(}\bigcup_{j}B_{j}\big{)}\leq\sigma(B(w,2r+|x_{2}-x_{1}|))\leq
C_{0}(2+\Lambda)^{d}r^{d}$

where $w$ is any of the $w_{j}$. Thus $\varepsilon^{1-n}\varepsilon^{d}\leq
CC_{0}^{2}\Lambda^{d}$ (recall that $\Lambda\geq 1$), a contradiction if we
take $\varepsilon\leq c\Lambda^{-d/(n-d-1)}$, where $c>0$ depends on $C_{0}$
too. Thus we can find $j$ such that
$\,\mathrm{dist}(\ell_{j},\Gamma)\geq\varepsilon r$, and the desired
conclusion holds with $y_{i}=x_{i}+z_{j}$. $\Box$

######  Lemma 2.2.

Let $\Gamma\subset\mathbb{R}^{n}$ be an Ahlfors regular set of dimension $d$,
$d<n-1$. Then for any $\Lambda\geq 1$, we can find a large integer
$N_{\Lambda}$ and a small real $\epsilon_{\Lambda}$ such that for any couple
of points $x_{1},x_{2}\in\Omega$ satisfying $\,\mathrm{dist}(x_{i},\Gamma)\geq
r$ and $|x_{1}-x_{2}|\leq\Lambda r$, there exists a sequence of balls
$(B_{i})_{0\leq i\leq N_{\Lambda}}$ \- called Harnack chain - verifying:

  1. (i)

$B_{0}$ is centered on $x_{1}$ and $B_{N_{\Lambda}}$ is centered on $x_{2}$,

  2. (ii)

for $0\leq i\leq N_{\Lambda}$, $3B_{i}\subset\Omega$ and $B_{i}$ has radius at
least $\epsilon_{\Lambda}r$,

  3. (iii)

for $1\leq i\leq N_{\Lambda}$, $B_{i-1}\cap B_{i}\neq\emptyset$.

Proof. This lemma is a corollary of Lemma 2.1. First, in the proof, $c$ shall
be the constant given by Lemma 2.1.

Let $\Lambda\geq 1$ and $x_{1},x_{2}$ as in the lemma. Let $y_{1},y_{2}$ be
the associated points given by Lemma 2.1, which thus verify $y_{1}\in
B(x_{1},r/3)$, $y_{2}\in B(x_{2},r/3)$, and
$\,\mathrm{dist}([y_{1},y_{2}],\Gamma)\geq c\Lambda^{-d/(n-d-1)}r$.

From there, we define $\epsilon_{\Lambda}$ as $c\Lambda^{-d/(n-d-1)}/3$ and
$N_{\Lambda}$ as the smallest integer bigger than
$(\Lambda+2)r/\epsilon_{\Lambda}+2$. The sequence $(B_{i})_{0\leq i\leq
N_{\Lambda}}$ is then constructed as follows. First, $B_{0}$ is
$B(x_{1},r/3)$. Then $(B_{i})_{1\leq i<N_{\Lambda}-1}$ are balls centered on
the segment $[y_{1},y_{2}]$ and of radius $\epsilon_{\Lambda}r$ \- precisely
$B_{i}$ is centered on $[(N_{\Lambda}-i-1)y_{1}+(i-1)y_{2}]/[N_{\Lambda}-2]$
\- and finally $B_{N_{\Lambda}}$ is $B(x_{2},r/3)$. $\Box$

Then, we give estimates on the weight $w$.

######  Lemma 2.3.

There exists $C>0$ such that

  1. (i)

for any $x\in\mathbb{R}^{n}$ and any $r>0$ satisfying $\delta(x)\geq 2r$,

(2.1) $C^{-1}r^{n}w(x)\leq m(B(x,r))=\int_{B(x,r)}w(z)dz\leq Cr^{n}w(x),$

  2. (ii)

for any $x\in\mathbb{R}^{n}$ and any $r>0$ satisfying $\delta(x)\leq 2r$,

(2.2) $C^{-1}r^{d+1}\leq m(B(x,r))=\int_{B(x,r)}w(z)dz\leq Cr^{d+1}.$

######  Remark 2.4.

In the above lemma, the estimates are different if $\delta(x)$ is bigger or
smaller than $2r$. Yet the critical ratio $\frac{\delta(x)}{r}=2$ is not
relevant: for any $\alpha>0$, we can show as well that (2.1) holds whenever
$\delta(x)\geq\alpha r$ and (2.2) holds if $\delta(x)\leq\alpha r$, with a
constant $C$ that depends on $\alpha$.

Indeed, we can replace $2$ by $\alpha$ if we can prove that for any $K>1$
there exists $C>0$ such that for any $x\in\mathbb{R}^{n}$ and $r>0$ satisfying

(2.3) $K^{-1}r\leq\delta(x)\leq Kr$

we have

(2.4) $C^{-1}r^{d+1}\leq r^{n}w(x)\leq Cr^{d+1}.$

However, since $w(x)=\delta(x)^{d+1-n}$, (2.3) implies $w(x)\approx r^{d+1-n}$
which in turn gives (2.4).

Proof. First suppose that $\delta(x)\geq 2r$. Then for any $z\in B(x,r)$,
$\frac{1}{2}\delta(x)\leq\delta(z)\leq\frac{3}{2}\delta(x)$ and hence
$C^{-1}w(x)\leq w(z)\leq Cw(x)$; (2.1) follows.

The lower bound in (2.2) is also fairly easy, just note that when
$\delta(x)\leq 2r$, $\delta(z)\leq 3r$ for any $z\in B(x,r)$ and hence

(2.5) $m(B(x,r))\geq\int_{B(x,r)}(3r)^{1+d-n}dz\geq C^{-1}r^{d+1}.$

Finally we check the upper bound in (2.2). We claim that for any $y\in\Gamma$
and any $r>0$,

(2.6) $m(B(y,r))=\int_{B(y,r)}\delta(\xi)^{d+1-n}d\xi\leq Cr^{d+1}.$

From the claim, let us prove the upper bound in (2.2). Let
$x\in\mathbb{R}^{n}$ and $r>0$ be such that $\delta(x)\leq 2r$. Thus there
exists $y\in\Gamma$ such that $B(x,r)\subset B(y,3r)$ and thanks to (2.6)

(2.7) $m(B(x,r))\leq\int_{B(y,3r)}w(z)dz\leq C(3r)^{d+1}\leq Cr^{d+1},$

which gives the upper bound in (2.2).

Let us now prove the claim. By translation invariance, we can choose
$y=0\in\Gamma$. Note that $\delta(\xi)\leq r$ in the domain of integration.
Let us evaluate the measure of the set $Z_{k}=\big{\\{}\xi\in
B(0,r)\,;\,2^{-k-1}r<\delta(\xi)\leq 2^{-k}r\big{\\}}$. We use (1.1) to cover
$\Gamma\cap B(0,2r)$ with less than $C2^{kd}$ balls $B_{j}$ of radius
$2^{-k}r$ centered on $\Gamma$; then $Z_{k}$ is contained in the union of the
$3B_{j}$, so $|Z_{k}|\leq C2^{kd}(2^{-k}r)^{n}$ and $\int_{\xi\in
Z_{k}}\delta(\xi)^{1+d-n}d\xi\leq
C2^{kd}(2^{-k}r)^{n}(2^{-k}r)^{d+1-n}=C2^{-k}r^{d+1}$. We sum over $k\geq 0$
and get (2.6). $\Box$

A consequence of Lemma 2.3 is that $m$ is a doubling measure, that is for any
$x\in\mathbb{R}^{n}$ and $r>0$, one has $m(B(x,2r))\leq Cm(B(x,r))$ for a
constant $C>0$ independent of $x$ and $r$. Actually, we can prove the
following stronger fact: for any $x\in\mathbb{R}^{n}$ and any $r>s>0$, there
holds

(2.8) $C^{-1}\left(\frac{r}{s}\right)^{d+1}\leq\frac{m(B(x,r))}{m(B(x,s))}\leq
C\left(\frac{r}{s}\right)^{n}.$

Three cases may happen. First, $\delta(x)\geq 2r\geq 2s$ and then with (2.1),

(2.9)
$\frac{m(B(x,r))}{m(B(x,s))}\approx\frac{r^{n}w(x)}{s^{n}w(x)}=\left(\frac{r}{s}\right)^{n}.$

Second, $\delta(x)\leq 2s\leq 2r$. In this case, note that (2.2) implies

(2.10)
$\frac{m(B(x,r))}{m(B(x,s))}\approx\frac{r^{d+1}}{s^{d+1}}=\left(\frac{r}{s}\right)^{d+1}.$

At last, $2s\leq\delta(x)\leq 2r$. Note that (2.1) and (2.2) yield

(2.11) $\frac{m(B(x,r))}{m(B(x,s))}\approx\frac{r^{d+1}}{s^{n}w(x)}.$

Yet, $2s\leq\delta(x)\leq 2r$ implies $C^{-1}r^{d+1-n}\leq w(x)\leq
Cs^{d+1-n}$ and thus

(2.12)
$C^{-1}\left(\frac{r}{s}\right)^{d+1}\leq\frac{m(B(x,r))}{m(B(x,s))}\leq
C\left(\frac{r}{s}\right)^{n}.$

which finishes the proof of (2.8).

One can see that the coefficients $d+1$ and $n$ are optimal in (2.8). The fact
that the volume of a ball with radius $r$ is not equivalent to $r^{\alpha}$
for some $\alpha>0$ will cause some difficulties. For instance, regardless of
the choice of $p$, we cannot have a Sobolev embedding $W\hookrightarrow L^{p}$
and we have to settle for the Sobolev-Poincaré inequality (1.11).

Another consequence of Lemma 2.3 is that for any $x\subset\mathbb{R}^{n}$, any
$r>0$, and any nonnegative function $g\in L^{1}_{loc}(\mathbb{R}^{n})$,

(2.13) $\frac{1}{|B(x,r)|}\int_{B(x,r)}g(z)dz\leq
C\frac{1}{m(B(x,r))}\int_{B}g(z)w(z)dz.$

Indeed, the inequality (2.13) holds if we can prove that

(2.14) $\frac{m(B(x,r))}{|B(x,r)|}\leq Cw(z)\qquad\forall z\in B(x,r).$

This latter fact can be proven as follows:

(2.15) $\frac{m(B(x,r))}{|B(x,r)|}\leq\frac{m(B(z,2r))}{|B(x,r)|}\leq
Cr^{-n}m(B(z,2r))$

If $\delta(z)\geq 4r$, then Lemma 2.3 gives $r^{-n}m(B(z,2r))\leq Cw(z)$. If
$\delta(z)\leq 4r$, then $w(z)\geq C^{-1}r^{d+1-n}$ and Lemma 2.3 entails
$r^{-n}m(B(z,2r))\leq Cr^{d+1-n}\leq Cw(z)$. In both cases, we obtain (2.14)
and thus (2.13).

We end the section with a corollary of Lemma 2.3.

######  Lemma 2.5.

The weight $w$ is in the $\mathcal{A}_{2}$-Muckenhoupt class, i.e. there
exists $C>0$ such that for any ball $B\subset\mathbb{R}^{n}$,

(2.16) $\fint_{B}w(z)dz\fint_{B}w^{-1}(z)dz\leq C.$

Proof. Let $B:=B(x,r)$. If $\delta(x)\geq 2r$, then for any $z\in B(x,r)$,
$C^{-1}w(x)\leq w(z)\leq Cw(x)$ and thus
$\fint_{B}w(z)dz\cdot\fint_{B}w^{-1}(z)dz\leq Cw(x)w^{-1}(x)=C$. If
$\delta(x)\leq 2r$, then (2.2) implies that $\fint_{B}w(z)dz\leq
Cr^{-n}r^{d+1}=Cr^{d+1-n}$. Besides, for any $z\in B(x,r)$, $\delta(z)\leq 3r$
and hence $w^{-1}(z)\leq Cr^{n-d-1}$. It follows that if $\delta(x)\leq 2r$,
$\fint_{B}w(z)dz\cdot\fint_{B}w^{-1}(z)dz\leq C$. The assertion (2.16)
follows. $\Box$

##  Chapter 3 Traces

The weighted Sobolev space $W=\dot{W}_{w}^{1,2}(\Omega)$ and
$H=\dot{H}^{1/2}(\Gamma)$ are defined as in Section 1 (see (1.4), (1.5)). Let
us give a precison. Any $u\in W$ has a distributional derivative in $\Omega$
that belongs to $L^{2}(\Omega,w)$, that is there exists a vector valued
function $v\in L^{2}(\Omega,w)$ such that for any $\varphi\in
C^{\infty}_{0}(\Omega,\mathbb{R}^{n})$

(3.1)
$\int_{\Omega}v\cdot\varphi\,dx=-\int_{\Omega}u\operatorname{div}\varphi\,dx.$

This definition make sense since $v\in L^{2}(\Omega,w)\subset
L^{1}_{loc}(\Omega)$. For the proof of the latter inclusion, use for instance
Cauchy-Schwarz inequality and (2.13).

The aim of the section is to state and prove a trace theorem. But for the
moment, let us keep discussing about the space $W$. We say that $u$ is
absolutely continuous on lines in $\Omega$ if there exists $\bar{u}$ which
coincides with $u$ a.e. such that for almost every line $\ell$ (for the usual
invariant measure on the Grassman manifold, but we can also say, given any
choice of direction $v$ and and a vector hyperplane plane $P$ transverse to
$v$, for the line $x+\mathbb{R}v$ for almost every $x\in P$), we have the
following properties. First, the restriction of $\bar{u}$ to $\ell\cap\Omega$
(which makes sense, for a.e. line $\ell$, and is measurable, by Fubini) is
absolutely continuous, which means that it is differentiable almost everywhere
on $\ell\cap\Omega$ and is the indefinite integral of its derivative on each
component of $\ell\cap\Omega$. By the natural identification, the derivative
in question is obtained from the distributional gradient of $u$.

######  Lemma 3.1.

Every $u\in W$ is absolutely continuous on lines in $\Omega$.

Proof. This lemma can be seen as a consequence of [Maz, Theorem 1.1.3/1] since
the absolute continuity on lines is a local property and, thanks to (2.13),
$W\subset\\{u\in L^{1}_{loc}(\Omega),\,\nabla u\in L^{2}_{loc}(\Omega)\\}$.
Yet, the proof of Lemma 3.1 is classical: since the property is local, it is
enough to check the property on lines parallel to a fixed vector $e$, and when
$\Omega$ is the product of $n$ intervals, one of which is parallel to $e$.
This last amounts to using the definition of the distributional gradient,
testing on product functions, and applying Fubini. In addition, the derivative
of $u$ on almost every line $\ell$ of direction $e$ coincides with $\nabla
u\cdot e$ almost everywhere on $\ell$. $\Box$

######  Lemma 3.2.

We have the following equality of spaces

(3.2) $W=\\{u\in L^{1}_{loc}({\mathbb{R}^{n}}),\,\nabla u\in
L^{2}(\mathbb{R}^{n},w)\\},$

where the derivative of $u$ is taken in the sense of distribution in
$\mathbb{R}^{n}$, that is for any $\varphi\in
C^{\infty}_{0}(\mathbb{R}^{n},\mathbb{R}^{n})$,

$\int\nabla u\cdot\varphi\,dx=-\int u\operatorname{div}\varphi\,dx.$

Proof. Here and in the sequel, we will constantly use the fact that with
$\Omega=\mathbb{R}^{n}\setminus\Gamma$ and because (1.1) holds with $d<n-1$,

(3.3) almost every line $\ell$ is contained in $\Omega$.

Let us recall that it means that given any choice of direction $v$ and a
vector hyperplane $P$ transverse to $v$, the line $x+\mathbb{R}v\subset\Omega$
for almost every $x\in P$. In particular, for almost every
$(x,y)\in(\mathbb{R}^{n})^{2}$, there is a unique line going through $x$ and
$y$ and this line is included in $\Gamma$.

Lemma 3.1 and (3.3) implies that $u\in W$ is actually absolutely continuous on
lines in $\mathbb{R}^{n}$, i.e. any $u\in W$ (possibly modified on a set of
zero measure) is absolutely continuous on almost every line
$\ell\subset\mathbb{R}^{n}$. As we said before, $\nabla
u=(\partial_{1}u,\dots,\partial_{n}u)$, the distributional gradient of $u$ in
$\Omega$, equals the ‘classical’ gradient of $u$ defined in the following way.
If $e_{1}=(1,0,\dots,0)$ is the first coordinate vector, then
$\partial_{1}u(y,z)$ is the derivative at the point $y$ of the function
$u_{|(0,z)+\mathbb{R}e_{1}}$, the latter quantity being defined for almost
every $(y,z)\in\mathbb{R}\times\mathbb{R}^{n-1}$ because $u$ is absolutely
continuous on lines in $\mathbb{R}^{n}$. If $i>1$, $\partial_{i}u(x)$ is
defined in a similar way.

As a consequence, for almost any $(y,z)\in\mathbb{R}^{n}\times\mathbb{R}^{n}$,
$u(z)-u(y)=\int_{0}^{1}(z-y)\cdot\nabla u(y+t(z-y))dt$ and hence,

(3.4) $|u(y)-u(z)|\leq\int_{0}^{1}|z-y||\nabla u(y+t(z-y))|dt.$

Let us integrate this for $y$ in a ball $B$. We get that for almost every
$z\in\mathbb{R}^{n}$,

(3.5) $\fint_{y\in B}|u(y)-u(z)|dy\leq\fint_{y\in B}\int_{0}^{1}|z-y||\nabla
u(y+t(z-y))|dt.$

Let us further restrict to the case $z\in B=B(x,r)$; the change of variable
$\xi=z+t(y-z)$ shows that

(3.6) $\begin{split}\fint_{y\in B}|u(y)-u(z)|dy&=\int_{0}^{1}\fint_{y\in
B}|y-z||\nabla u(z+t(y-z))|dy\,dt\\\ &=\int_{0}^{1}\frac{1}{|B|}\int_{\xi\in
B(z+t(x-z),tr)}\frac{|z-\xi|}{t}|\nabla u(\xi)|\frac{d\xi}{t^{n}}dt\\\
&=\int_{\xi\in B}|\nabla
u(\xi)|\frac{|z-\xi|}{|B(z,r)|}d\xi\int_{|z-\xi|/2r}^{1}\frac{dt}{t^{n+1}}\\\
&\leq 2^{n}|B(0,1)|^{-1}\int_{\xi\in B}|\nabla u(\xi)||z-\xi|^{1-n}d\xi,\\\
\end{split}$

where the last but one line is due to the fact that $\xi\in B(z+t(x-z),tr)$ is
equivalent to $|\xi-z-t(x-z)|\leq tr$, which forces $|\xi-z|\leq tr+t|x-z|\leq
2rt$. Therefore, for almost any $z\in B$,

(3.7) $\fint_{y\in B}|u(y)-u(z)|dy\leq C\int_{\xi\in B}|\nabla
u(\xi)||z-\xi|^{1-n}d\xi,$

where $C$ depends on $n$, but not on $r$, $u$, or $z$. With a second
integration on $z\in B=B(x,r)$, we obtain

(3.8) $\fint_{z\in B}\fint_{y\in B}|u(y)-u(z)|dy\,dz\leq C\int_{\xi\in
B}|\nabla u(\xi)|\fint_{z\in B}|z-\xi|^{1-n}dz\,d\xi\leq Cr\fint_{\xi\in
B}|\nabla u(\xi)|d\xi.$

By Hölder’s inequality and (2.13), the right-hand side is bounded (up to a
constant depending on $r$) by $\|u\|_{W}$. As a consequence,

(3.9) $\fint_{z\in B}\fint_{y\in B}|u(y)-u(z)|\leq C_{r}\|u\|_{W}<+\infty.$

and thus, by Fubini’s lemma, $\fint_{y\in B}|u(y)-u(z)|<+\infty$ for a.e.
$z\in B$. In particular, the quantity $\fint_{y\in B}|u(y)|$ is finite for any
ball $B\subset\mathbb{R}^{n}$, that is $u\in L^{1}_{loc}(\mathbb{R}^{n})$.

Since $L^{1}_{loc}(\mathbb{R}^{n})\subset L^{1}_{loc}(\Omega)$, we just proved
that $W=\\{u\in L^{1}_{loc}(\mathbb{R}^{n}),\,\nabla u\in L^{2}(\Omega,w)\\}$,
where $\nabla u=(\partial_{1}u,\dots,\partial_{n}u)$ is distributional
gradient on $\Omega$. Let $u\in W$. Since $\Gamma$ has zero measure, $\nabla
u\in L^{2}(\mathbb{R}^{n},w)$ and thus it suffices to check that $u$ has
actually a distributional derivative in $\mathbb{R}^{n}$ and that this
derivative equals $\nabla u$. However, the latter fact is a simple consequence
of [Maz, Theorem 1.1.3/2], because $u$ is absolutely continuous on lines in
$\mathbb{R}^{n}$. The proof of Maz’ya’s result is basically the following: for
any $i\in\\{1,\dots,n\\}$ and any $\phi\in C^{\infty}_{0}(\mathbb{R}^{n})$, an
integration by part gives $\int u\partial_{i}\phi=-\int(\partial_{i}u)\phi$.
The two integrals in the latter equality make sense since both $u$ and
$\partial_{i}u$ are in $L^{1}_{loc}(\mathbb{R}^{n})$; the integration by part
is possible because $u$ is absolutely continuous on almost every line. $\Box$

######  Remark 3.3.

An important by-product of the proof is that Lemma 3.1 can be improved into:
for any $u\in W$ (possibly modified on a set of zero measure) and almost every
line $\ell\subset\mathbb{R}^{n}$, $u_{|\ell}$ is absolutely continuous. This
property will be referred to as (ACL).

######  Theorem 3.4.

There exists a bounded linear operator $T:W\to H$ (a trace operator) with the
following properties. The trace of $u\in W$ is such that, for $\sigma$-almost
every $x\in\Gamma$,

(3.10) $Tu(x)=\lim_{r\to 0}\fint_{B(x,r)}u(y)dy:=\lim_{r\to
0}{1\over|B(x,r)|}\int u(y)dy$

and, analogously to the Lebesgue density property,

(3.11) $\lim_{r\to 0}\fint_{B(x,r)}|u(y)-Tu(x)|dy=0.$

Proof. First, we want bounds on $\nabla u$ near $x\in\Gamma$, so we set

(3.12) $M_{r}(x)=\fint_{B(x,r)}|\nabla u|^{2}\,dz$

and estimate $\int_{\Gamma}M_{r}(x)d\sigma(x)$. We cover $\Gamma$ by balls
$B_{j}=B(x_{j},r)$ centered on $\Gamma$ such that the $2B_{j}=B(x_{j},2r)$
have bounded overlap (we could even make the $B(x_{j},r/5)$ disjoint), and
notice that for $x\in B_{j}$,

(3.13) $M_{r}(x)\leq Cr^{-n}\int_{2B_{j}}|\nabla u|^{2}\,dz.$

We sum and get that

(3.14) $\displaystyle\int_{\Gamma}M_{r}(x)d\sigma(x)$ $\displaystyle\leq$
$\displaystyle\sum_{j}\int_{B_{j}}M_{r}(x)d\sigma(x)\leq
C\sum_{j}\sigma(B_{j})\sup_{x\in B_{j}}M_{r}(x)$ $\displaystyle\leq$
$\displaystyle C\sum_{j}\sigma(B_{j})r^{-n}\int_{2B_{j}}|\nabla u|^{2}\,dz\leq
Cr^{d-n}\sum_{j}\int_{2B_{j}}|\nabla u|^{2}\,dz$ $\displaystyle\leq$
$\displaystyle Cr^{d-n}\int_{\Gamma(2r)}|\nabla u|^{2}\,dz$

because the $2B_{j}$ have bounded overlap and where $\Gamma(2r)$ denotes a
$2r$-neighborhood of $\Gamma$. Next set

(3.15) $N(x)=\sum_{k\geq 0}2^{-k}M_{2^{-k}}(x);$

then

(3.16) $\displaystyle\int_{\Gamma}N(x)d\sigma(x)$ $\displaystyle=$
$\displaystyle\sum_{k\geq 0}2^{-k}\int_{\Gamma}M_{2^{-k}}(x)d\sigma(x)\leq
C\sum_{k\geq 0}2^{k(n-d-1)}\int_{\Gamma(2^{-k+1})}|\nabla u(z)|^{2}dz$
$\displaystyle\leq$ $\displaystyle C\int_{\Gamma(2)}|\nabla u(z)|^{2}a(z)dz,$

where $a(z)=\sum_{k\geq 0}2^{k(n-d-1)}{\mathds{1}}_{z\in\Gamma(2^{-k+1})}$.
For a given $z\in\Omega$, $z\in\Gamma(2^{-k+1})$ only for $k$ so small that
$\delta(z)\leq 2^{-k+1}$. The largest values of $2^{k(n-d-1)}$ are for $k$ as
large as possible, when $2^{-k}\approx\delta(z)$; thus $a(z)\leq
C\delta(z)^{-n+d+1}=Cw(z)$, and

(3.17) $\int_{\Gamma}N(x)d\sigma(x)\leq C\int_{\Gamma(2)}|\nabla
u(z)|^{2}w(z)dz.$

Our trace function $g=Tu$ will be defined as the limit of the functions
$g_{r}$, where

(3.18) $g_{r}(x)=\fint_{z\in B(x,r)}u(z)dz.$

Our aim is to use the estimates established in the proof of Lemma 3.2. Notice
that for $x\in\Gamma$ and $r>0$,

(3.19) $\begin{split}\fint_{z\in B(x,r)}|u(z)-g_{r}(x)|dz&=\fint_{z\in
B(x,r)}\Big{|}u(z)-\fint_{\xi\in B(x,r)}u(y)dy\Big{|}dz\\\ &\leq\fint_{z\in
B(x,r)}\fint_{y\in B(x,r)}\Big{|}u(z)-u(y)\Big{|}dy\,dz.\\\ \end{split}$

By (3.8),

(3.20) $\displaystyle\fint_{z\in B(x,r)}|u(z)-g_{r}(x)|dz$ $\displaystyle\leq$
$\displaystyle\fint_{z\in B(x,r)}\fint_{y\in
B(x,r)}\Big{|}u(z)-u(y)\Big{|}dy\,dz$ $\displaystyle\leq$ $\displaystyle
Cr^{-n+1}\int_{\xi\in B(x,r)}|\nabla u(\xi)|d\xi.$

Thus for $r/10\leq s\leq r$,

(3.21) $\displaystyle|g_{s}(x)-g_{r}(x)|$ $\displaystyle=$
$\displaystyle\Big{|}\fint_{z\in B(x,s)}u(z)dz-g_{r}(x)\Big{|}\leq\fint_{z\in
B(x,s)}|u(z)-g_{r}(x)|dz$ $\displaystyle\leq$ $\displaystyle Cr\fint_{\xi\in
B(x,r)}|\nabla u(\xi)|d\xi\leq CrM_{r}(x)^{1/2}.$

Set $\Delta_{r}(x)=\sup_{r/10\leq s\leq r}|g_{s}(x)-g_{r}(x)|$; we just proved
that $\Delta_{r}(x)\leq CrM_{r}(x)^{1/2}$. Let $\alpha\in(0,1/2)$ be given. If
$N(x)<+\infty$, we get that

(3.22) $\displaystyle\sum_{k\geq 0}2^{\alpha k}\Delta_{2^{-k}}(x)$
$\displaystyle\leq$ $\displaystyle C\sum_{k\geq 0}2^{\alpha
k}2^{-k}M_{2^{-k}}(x)^{1/2}$ $\displaystyle\leq$ $\displaystyle
C\Big{\\{}\sum_{k\geq 0}2^{-k}M_{2^{-k}}(x)\Big{\\}}^{1/2}\Big{\\{}\sum_{k\geq
0}2^{2\alpha k}2^{-k}\Big{\\}}^{1/2}\leq CN(x)^{1/2}<+\infty.$

Therefore, $\sum_{k\geq 0}\Delta_{2^{-k-2}}(x)$ converges (rather fast), and
since (3.17) implies that $N(x)<+\infty$ for $\sigma$-almost every
$x\in\Gamma$, it follows that there exists

(3.23) $g(x)=\lim_{r\to 0}g_{r}(x)\ \text{ for $\sigma$-almost every
}x\in\Gamma.$

In addition, we may integrate (the proof of) (3.22) and get that for
$2^{-j-1}<r\leq 2^{-j}$,

(3.24) $\displaystyle\|g-g_{r}\|_{L^{2}(\sigma)}^{2}$ $\displaystyle=$
$\displaystyle\int_{\Gamma}|g(x)-g_{r}(x)|^{2}d\sigma(x)\leq\int_{\Gamma}\Big{\\{}\sum_{k\geq
j}\Delta_{2^{-k}}(x)\Big{\\}}^{2}d\sigma(x)$ $\displaystyle\leq$
$\displaystyle C2^{-2\alpha j}\int_{\Gamma}\Big{\\{}\sum_{k\geq j}2^{\alpha
j}\Delta_{2^{-k}}(x)\Big{\\}}^{2}d\sigma(x)$ $\displaystyle\leq$
$\displaystyle Cr^{2\alpha}\int_{\Gamma}N(x)d\sigma(x)\leq
Cr^{2\alpha}\|u\|_{W}^{2}$

by (3.23) and the definition of $\Delta_{r}(x)$, then (3.22) and (3.17). Thus
$g_{r}$ converges also (rather fast) to $g$ in $L^{2}$. Let us make an
additional remark. Fix $r>0$ and $\alpha\in(0,1/2)$. For any ball $B$ centered
on $\Gamma$,

(3.25) $\|g\|_{L^{1}(B,\sigma)}\leq
C_{B}\|g-g_{r}\|_{L^{2}(\sigma)}+\|g_{r}\|_{L^{1}(B,\sigma)}$

by Hölder’s inequality. The first term is bounded with (3.24). Use (1.1) and
Fubini’s theorem to bound the second one by
$C_{r}\|g\|_{L^{1}(\widetilde{B})}$, where $\widetilde{B}$ is a large ball
(that depends on $r$ and contains $B$). As a consequence,

(3.26) for any $u\in W$, $g=Tu\in L^{1}_{loc}(\Gamma,\sigma)$.

This completes the definition of the trace $g=T(u)$. We announced (as a
Lebesgue property) that

(3.27) $\lim_{r\to 0}\fint_{B(x,r)}|u(y)-Tu(x)|dy$

for $\sigma$-almost every $x\in\Gamma$, and indeed

(3.28) $\displaystyle\fint_{B(x,r)}|u(y)-Tu(x)|dy$ $\displaystyle=$
$\displaystyle\fint_{B(x,r)}|u(y)-g(x)|dy\leq|g(x)-g_{r}(x)|+\fint_{B(x,r)}|u(y)-g_{r}(x)|\,dy$
$\displaystyle\leq$ $\displaystyle|g(x)-g_{r}(x)|+Cr\fint_{B(x,3r)}|\nabla
u|\,dy\leq|g(x)-g_{r}(x)|+CrM_{4r}(x)^{1/2}$

by (3.20) and the second part of (3.21). The first part tends to $0$ for
$\sigma$-almost every $x\in\Gamma$, by (3.23), and the second part tends to
$0$ as well, because $N(x)<+\infty$ almost everywhere and by the definition
(3.15).

Next we show that $g=Tu$ lies in the Sobolev space $H=H^{1/2}(\Gamma)$, i.e.,
that

(3.29)
$\|g\|_{H}^{2}=\int_{\Gamma}\int_{\Gamma}{|g(x)-g(y)|^{2}\over|x-y|^{d+1}}d\sigma(x)d\sigma(y)<+\infty.$

The simplest will be to prove uniform estimates on the $g_{r}$, and then go to
the limit. Let us fix $r>0$ and consider the integral

(3.30) $I(r)=\int_{x\in\Gamma}\int_{y\in\Gamma;|y-x|\geq
r}{|g_{r}(x)-g_{r}(y)|^{2}\over|x-y|^{d+1}}d\sigma(x)d\sigma(y).$

Set
$Z_{k}(r)=\big{\\{}(x,y)\in\Gamma\times\Gamma\,;\,2^{k}r\leq|y-x|<2^{k+1}r\big{\\}}$
and
$I_{k}(r)=\int\int_{Z_{k}(r)}{|g_{r}(x)-g_{r}(y)|^{2}\over|x-y|^{d+1}}d\sigma(x)d\sigma(y)$.
Thus $I(r)=\sum_{k\geq 0}I_{k}(r)$ and

(3.31)
$I_{k}(r)\leq(2^{k}r)^{-d-1}\int\int_{Z_{k}(r)}|g_{r}(x)-g_{r}(y)|^{2}d\sigma(x)d\sigma(y).$

Fix $k\geq 0$, set $\rho=2^{k+1}r$, and observe that for $(x,y)\in Z_{k}(r)$,

(3.32) $\displaystyle|g_{r}(x)-g_{\rho}(y)|$ $\displaystyle=$
$\displaystyle\Big{|}\fint_{z\in B(x,r)}\fint_{\xi\in
B(y,\rho)}[u(z)-u(\xi)]dzd\xi\Big{|}\leq\fint_{z\in B(x,r)}\fint_{\xi\in
B(y,\rho)}|u(z)-u(\xi)|d\xi dz$ $\displaystyle\leq$ $\displaystyle
3^{n}\fint_{z\in B(x,r)}\Big{\\{}\fint_{\xi\in
B(z,3\rho)}|u(z)-u(\xi)|d\xi\Big{\\}}dz$ $\displaystyle\leq$ $\displaystyle
C\rho^{n}\fint_{z\in B(x,r)}\fint_{\zeta\in B(z,3\rho)}|\nabla
u(\zeta)||z-\zeta|^{1-n}d\zeta dz$

because $B(y,\rho)\subset B(z,3\rho)$ and by (3.7). We apply Cauchy-Schwarz,
with an extra bit $|z-\zeta|^{-\alpha}$, where $\alpha>0$ will be taken small,
and which will be useful for convergence later

(3.33) $\displaystyle|g_{r}(x)-g_{\rho}(y)|^{2}$ $\displaystyle\leq$
$\displaystyle C\rho^{2n}\Big{\\{}\fint_{z\in B(x,r)}\fint_{\zeta\in
B(z,3\rho)}|\nabla u(\zeta)|^{2}|z-\zeta|^{1-n+\alpha}\Big{\\}}$
$\displaystyle\hskip 170.71652pt\Big{\\{}\fint_{z\in B(x,r)}\fint_{\zeta\in
B(z,3\rho)}|z-\zeta|^{1-n-\alpha}\Big{\\}}$ $\displaystyle\leq$ $\displaystyle
C\rho^{n+1-\alpha}\fint_{z\in B(x,r)}\fint_{\zeta\in B(z,3\rho)}|\nabla
u(\zeta)|^{2}|z-\zeta|^{1-n+\alpha}d\zeta dz.$

The same computation, with $g_{r}(y)$, yields

(3.34) $|g_{r}(y)-g_{\rho}(y)|^{2}\leq C\rho^{n+1-\alpha}\fint_{z\in
B(y,r)}\fint_{\zeta\in B(z,3\rho)}|\nabla
u(\zeta)|^{2}|z-\zeta|^{1-n+\alpha}d\zeta dz.$

We add the two and get an estimate for $|g_{r}(x)-g_{r}(y)|^{2}$, which we can
integrate to get that

(3.35) $\displaystyle I_{k}(r)$ $\displaystyle\leq$ $\displaystyle
C\rho^{-d-1}\rho^{n+1-\alpha}\int\int_{(x,y)\in Z_{k}(r)}\fint_{z\in
B(x,r)}\fint_{\zeta\in B(z,3\rho)}|\nabla
u(\zeta)|^{2}|z-\zeta|^{1-n+\alpha}d\zeta dzd\sigma(x)d\sigma(y)$
$\displaystyle\leq$ $\displaystyle C\rho^{-d-\alpha}r^{-n}\int\int_{(x,y)\in
Z_{k}(r)}\int_{z\in B(x,r)}\int_{\zeta\in B(z,3\rho)}|\nabla
u(\zeta)|^{2}|z-\zeta|^{1-n+\alpha}d\zeta dzd\sigma(x)d\sigma(y)$

by (3.31), (3.33), and (3.34), and where we can drop the part that comes from
(3.34) by symmetry. We integrate in $y\in\Gamma$ such that
$2^{k}r\leq|x-y|\leq 2^{k+1}r$ and get that

(3.36) $\displaystyle I_{k}(r)$ $\displaystyle\leq$ $\displaystyle
C\rho^{-\alpha}r^{-n}\int_{x\in\Gamma}\int_{z\in B(x,r)}\int_{\zeta\in
B(z,3\rho)}|\nabla u(\zeta)|^{2}|z-\zeta|^{1-n+\alpha}d\zeta dzd\sigma(x)$
$\displaystyle\leq$ $\displaystyle C\int_{\zeta\in\Omega}|\nabla
u(\zeta)|^{2}h_{k}(\zeta)d\zeta,$

with

(3.37) $h_{k}(\zeta)=\rho^{-\alpha}r^{-n}\int_{x\in\Gamma}\int_{z\in
B(x,r)\cap B(\zeta,3\rho)}|z-\zeta|^{1-n+\alpha}dzd\sigma(x).$

We start with the contribution $h_{k}^{0}(\zeta)$ of the region where
$|x-\zeta|\geq 2r$, where the computation is simpler because
$|z-\zeta|\geq{1\over 2}|x-\zeta|$ there. We get that

(3.38) $\displaystyle h_{k}^{0}(\zeta)$ $\displaystyle\leq$ $\displaystyle
C\rho^{-\alpha}r^{-n}\int_{x\in\Gamma}\int_{z\in B(x,r)\cap
B(\zeta,3\rho)}|x-\zeta|^{1-n+\alpha}dzd\sigma(x)$ $\displaystyle\leq$
$\displaystyle C\rho^{-\alpha}\int_{x\in\Gamma\cap
B(\zeta,4\rho)}|x-\zeta|^{1-n+\alpha}d\sigma(x).$

With $\zeta$, $r$, and $\rho$ fixed, $h_{k}^{0}(\zeta)$ vanishes unless
$\delta(\zeta)=\,\mathrm{dist}(\zeta,\Gamma)<4\rho$. The region where
$|x-\zeta|$ is of the order of $2^{m}\delta(\zeta)$, $m\geq 0$, contributes
less than $C(2^{m}\delta(\zeta))^{d+1-n+\alpha}$ to the integral (because
$\sigma$ is Ahlfors-regular). If $\alpha$ is chosen small enough, the exponent
is still negative, the largest contribution comes from $m=0$, and
$h_{k}^{0}(\zeta)\leq C\rho^{-\alpha}\delta(\zeta)^{d+1-n+\alpha}$. Recall
that $\rho=2^{k+1}r$, and $k$ is such that $\delta(\zeta)<4\rho$; we sum over
$k$ and get that

(3.39) $\sum_{k}h_{k}^{0}(\zeta)\leq C\sum_{k\geq
0\,;\,\delta(\zeta)<4\rho}\rho^{-\alpha}\delta(\zeta)^{d+1-n+\alpha}\leq
C\delta(\zeta)^{d+1-n},$

because this time the smallest values of $\rho$ give the largest
contributions. We are left with

(3.40)
$h_{k}^{1}(\zeta)=h_{k}(\zeta)-h_{k}^{0}(\zeta)=\rho^{-\alpha}r^{-n}\int_{x\in\Gamma\cap
B(\zeta,2r)}\int_{z\in B(x,r)\cap
B(\zeta,3\rho)}|z-\zeta|^{1-n+\alpha}dzd\sigma(x).$

Notice that $|z-\zeta|\leq|z-x|+|x-\zeta|\leq 3r$; we use the local Ahlfors-
regularity to get rid of the integral on $\Gamma$, and get that

(3.41) $h_{k}^{1}(\zeta)\leq C\rho^{-\alpha}r^{-n}r^{d}\int_{z\in
B(\zeta,3r)}|z-\zeta|^{1-n+\alpha}dz\leq C\rho^{-\alpha}r^{d+1-n+\alpha}.$

We sum over $k$ and get that $\sum_{k}h_{k}^{1}(\zeta)\leq Cr^{d+1-n}\leq
C\delta(\zeta)^{d+1-n}$, because if $\delta(\zeta)\geq 2r$, we simply get that
$h_{k}^{1}(\zeta)=0$ for all $k$, because $\Gamma\cap B(\zeta,2r)=\emptyset$
and by (3.40). Altogether, $\sum_{k}h_{k}(\zeta)\leq C\delta(\zeta)^{d+1-n}$,
and

(3.42) $I(r)=\sum_{k}I_{k}(r)\leq C\sum_{k}\int_{\zeta\in\Omega}|\nabla
u(\zeta)|^{2}h_{k}(\zeta)d\zeta\leq C\int_{\zeta\in\Omega}|\nabla
u(\zeta)|^{2}\delta(\zeta)^{d+1-n}d\zeta=C\|u\|_{W}^{2}$

by definition of the $I_{k}(r)$, then (3.36) and the definition of $W$. We may
now look at the definition (3.30) of $I(r)$, let $r$ tend to $0$, and get that

(3.43) $\|g\|_{H}\leq C\|u\|_{W}^{2}$

by Fatou’s lemma, as needed for the trace theorem. $\Box$

##  Chapter 4 Poincaré Inequalities

######  Lemma 4.1.

Let $\Gamma$ be a $d$-ADR set in ${\mathbb{R}}^{n}$, $d<n-1$, that is, assume
that (1.1) is satisfied. Then

(4.1) $\fint_{B(x,r)}|u(y)|dy\leq Cr^{-d}\int_{B(x,r)}|\nabla u(y)|w(y)dy$

for $u\in W$, $x\in\Gamma$, and $r>0$ such that $Tu=0$ on $\Gamma\cap B(x,r)$.

Proof. To simplify the notation we assume that $x=0$.

We should of course observe that the right-hand side of (4.1) is finite.
Indeed, recall that Lemma 2.3 gives

(4.2) $\int_{\xi\in B(0,r)}w(\xi)d\xi\leq Cr^{1+d};$

then by Cauchy-Schwarz

(4.3) $\displaystyle r^{-d}\int_{\xi\in B(0,r)}|\nabla u(\xi)|w(\xi)d\xi$
$\displaystyle\leq$ $\displaystyle r^{-d}\Big{\\{}\int_{\xi\in B(0,r)}|\nabla
u(\xi)|^{2}w(\xi)\,d\xi\Big{\\}}^{1/2}\Big{\\{}\int_{\xi\in
B(0,r)}w(\xi)\,d\xi\Big{\\}}^{1/2}$ $\displaystyle\leq$ $\displaystyle
r^{1-d\over 2}\Big{\\{}\int_{\xi\in B(0,r)}|\nabla
u(\xi)|^{2}w(\xi)\,d\xi\Big{\\}}^{1/2}.$

The homogeneity still looks a little weird because of the weight (but things
become simpler if we think that $\delta(\xi)$ is of the order of $r$), but at
least the right-hand side is finite because $u\in W$.

Turning to the proof of (4.1), to avoid complications with the fact that (3.4)
and (3.5) do not necessarily hold $\sigma$-almost everywhere on $\Gamma$, let
us use the $g_{s}$ again. We first prove that for $s<r$ small,

(4.4) $\fint_{y\in B(0,r)}\fint_{x\in\Gamma\cap
B(0,r/2)}|u(y)-g_{s}(x)|d\sigma(x)\,dy\leq Cr^{-d}\fint_{B(0,r)}|\nabla
u(\xi)|\delta(\xi)^{1+d-n}d\xi.$

Denote by $I(s)$ the left-hand side. By (3.18),

(4.5) $I(s)\leq\fint_{y\in B(0,r)}\fint_{x\in\Gamma\cap B(0,r/2)}\fint_{z\in
B(x,s)}|u(y)-u(z)|dz\,d\sigma(x)\,dy.$

For $x$ fixed, we can still prove as in (3.7) that

(4.6) $\fint_{y\in B(0,r)}|u(y)-u(z)|dy\leq C\int_{B(0,r)}|\nabla
u(\xi)||z-\xi|^{1-n}d\xi$

(for $x\in\Gamma\cap B(0,r/2)$ and $z\in B(x,s)$, there is even a bilipschitz
change of variable that sends $z$ to $0$ and maps $B(0,r)$ to itself). We are
left with

(4.7) $I(s)\leq C\fint_{x\in\Gamma\cap B(0,r/2)}\fint_{z\in
B(x,s)}\int_{\xi\in B(0,r)}|\nabla u(\xi)||z-\xi|^{1-n}d\xi dzd\sigma(x).$

The main piece of the integral will again be called $I_{0}(s)$, where we
integrate in the region where $|\xi-x|\geq 2s$ and hence $|z-\xi|^{1-n}\leq
2^{n}|x-\xi|^{1-n}$. Thus

(4.8) $\displaystyle I_{0}(s)$ $\displaystyle\leq$ $\displaystyle
C\int_{\xi\in B(0,r)}\fint_{x\in\Gamma\cap B(0,r/2)}\fint_{z\in B(x,s)}|\nabla
u(\xi)||x-\xi|^{1-n}dzd\sigma(x)d\xi$ $\displaystyle\leq$ $\displaystyle
Cr^{-d}\int_{\xi\in B(0,r)}\int_{x\in\Gamma\cap B(0,r/2)}|\nabla
u(\xi)||x-\xi|^{1-n}d\sigma(x)d\xi$ $\displaystyle\leq$ $\displaystyle
Cr^{-d}\int_{\xi\in B(0,r)\setminus\Gamma}|\nabla u(\xi)|h(\xi)d\xi,$

where for $\xi\in B(0,r)\setminus\Gamma$ we set

(4.9) $h(\xi)=\int_{x\in\Gamma\cap B(0,r/2)}|x-\xi|^{1-n}d\sigma(x)\leq
C\delta(\xi)^{1-n+d}$

where for the last inequality we cut the domain of integration into pieces
where $|x-\xi|\approx 2^{m}\delta(\xi)$ and use (1.1). For the other piece of
(4.7) where $|\xi-x|<2s$, we get the integral

(4.10) $\displaystyle I_{1}(s)$ $\displaystyle\leq$ $\displaystyle
Cr^{-d}s^{-n}\int_{\xi\in B(0,r)}\int_{x\in\Gamma\cap B(0,r/2)\cap
B(\xi,2s)}\int_{z\in B(x,s)}|\nabla u(\xi)||z-\xi|^{1-n}d\xi dzd\sigma(x)$
$\displaystyle\leq$ $\displaystyle Cr^{-d}s^{d-n}\int_{\xi\in
B(0,r);\delta(\xi)\leq 2s}\int_{z\in B(\xi,3s)}|\nabla
u(\xi)||z-\xi|^{1-n}d\xi dz$ $\displaystyle\leq$ $\displaystyle
Cr^{-d}s^{1+d-n}\int_{\xi\in B(0,r);\delta(\xi)\leq 2s}|\nabla u(\xi)|d\xi\leq
Cr^{-d}\int_{\xi\in B(0,r)}|\nabla u(\xi)|\delta(\xi)^{1+d-n}d\xi.$

Altogether

(4.11) $I(s)\leq Cr^{-d}\int_{\xi\in B(0,r)}|\nabla
u(\xi)|\delta(\xi)^{1+d-n}d\xi,$

which is (4.4). When $s$ tends to $0$, $g_{s}(x)$ tends to $g(x)=Tu(x)=0$ for
$\sigma$-almost every $x\in\Gamma\cap B(0,r/2)$, and we get (4.1) by Fatou.
$\Box$

######  Lemma 4.2.

Let $\Gamma$ be a $d$-ADR set in ${\mathbb{R}}^{n}$, $d<n-1$, that is, assume
that (1.1) is satisfied. Let $p\in\left[1,\frac{2n}{n-2}\right]$ (or
$p\in[1,+\infty)$ if $n=2$). Then for any $u\in W$, $x\in\mathbb{R}^{n}$ and
$r>0$

(4.12)
$\Big{\\{}\frac{1}{m(B(x,r))}\int_{B(x,r)}\left|u(y)-u_{B(x,r)}\right|^{p}w(y)dy\Big{\\}}^{1/p}\leq
Cr\Big{\\{}\frac{1}{m(B(x,r))}\int_{B(x,r)}|\nabla
u(y)|^{2}w(y)dy\Big{\\}}^{1/2},$

where $u_{B}$ denotes either $\fint_{B}u\,dz$ or $m(B)^{-1}\int_{B}u\,w(z)dz$.
If $x\in\Gamma$ and, in addition, $Tu=0$ on $\Gamma\cap B(x,r)$ then

(4.13) $\Big{\\{}r^{-d-1}\int_{B(x,r)}|u(y)|^{p}w(y)dy\Big{\\}}^{1/p}\leq
Cr\Big{\\{}r^{-d-1}\int_{B(x,r)}|\nabla u(y)|^{2}w(y)dy\Big{\\}}^{1/2}.$

Proof. In the proof, we will use $dm(z)$ for $w(z)dz$ and hence, for instance
$\int_{B}u\,dm$ denotes $\int_{B}u(z)w(z)dz$. We start with the following
inequality. Let $p\in[1,+\infty)$. If $u\in
L^{p}_{loc}(\mathbb{R}^{n},w)\subset L^{1}_{loc}(\mathbb{R}^{n})$, then for
any ball $B$,

(4.14)
$\int_{B}\left|u-\fint_{B}u\,dy\right|^{p}dm\approx\int_{B}\left|u-\frac{1}{m(B)}\int_{B}u\,dm\right|^{p}dm.$

First we bound the left-hand side. We introduce $m(B)^{-1}\int_{B}u\,dm$
inside the absolute values and then use the triangle inequality:

(4.15) $\displaystyle\int_{B}\left|u-\fint_{B}u\,dy\right|^{p}dm$
$\displaystyle\leq$ $\displaystyle
C\int_{B}\left|u(z)-\frac{1}{m(B)}\int_{B}u\,dm\right|^{p}dm+Cm(B)\left|\fint_{B}u-\frac{1}{m(B)}\int_{B}u\,dm\right|^{p}$
$\displaystyle\leq$ $\displaystyle
C\int_{B}\left|u(z)-\frac{1}{m(B)}\int_{B}u\,dm\right|^{p}dm+C\frac{m(B)}{|B|}\int_{B}\left|u-\frac{1}{m(B)}\int_{B}u\,dm\right|^{p}$
$\displaystyle\leq$ $\displaystyle
C\int_{B}\left|u(z)-\frac{1}{m(B)}\int_{B}u\,dm\right|^{p}dm,$

where the last line is due to (2.13). The reverse estimate is quite immediate

(4.16)
$\begin{split}&\int_{B}\left|u-\frac{1}{m(B)}\int_{B}u\,dm\right|^{p}dm\\\
&\qquad\leq
C\int_{B}\left|u(z)-\fint_{B}u\,dy\right|^{p}dm+Cm(B)\left|\frac{1}{m(B)}\int_{B}u\,dm-\fint_{B}u\,dy\right|^{p}\\\
&\qquad\leq
C\int_{B}\left|u(z)-\fint_{B}u\,dy\right|^{p}dm+Cm(B)\left|\frac{1}{m(B)}\int_{B}\left(u-\fint_{B}u\,dy\right)\,dm\right|^{p}\\\
&\qquad\leq C\int_{B}\left|u(z)-\fint_{B}u\,dy\right|^{p}dm,\end{split}$

which finishes the proof of (4.14).

In the sequel of the proof, we write $u_{B}$ for $m(B)^{-1}\int_{B}u\,dm$.
Thanks to (4.14), it suffices to prove (4.12) only for this particular choice
of $u_{B}$. We now want to prove a (1,1) Poincaré inequality, that is

(4.17) $\int_{B}\left|u(z)-u_{B}\right|dm\leq Cr\int_{B}|\nabla u(z)|dm.$

for any $u\in W$ and any ball $B\subset\mathbb{R}^{n}$ of radius $r$. In
particular, $u\in L^{1}_{loc}(\mathbb{R}^{n},w)$.

Let $B\subset\mathbb{R}^{n}$ of radius $r$. Recall first that thanks to Lemma
3.2, $\fint_{B}u$ makes sense for every ball $B$. If we prove for $u\in W$ the
estimate

(4.18) $\int_{B}\left|u-\fint_{B}u\,dy\right|dm\leq Cr\int_{B}|\nabla u|dm,$

then (4.17) will follows. Indeed, assume (4.18) holds for any ball $B$. The
left-hand of (4.18) is then bounded, up to a constant, by $r\|u\|_{W}$ and is
thus finite. Therefore, for any ball $B$,
$\int_{B}|u|dm\leq\int_{B}|u-\fint_{B}u|dm+|\fint_{B}u\,dy|<+\infty$, i.e.
$u\in L^{1}_{loc}(\mathbb{R}^{n},w)$. But now, $u\in
L^{1}_{loc}(\mathbb{R}^{n},w)$, so we can use (4.14). Together with (4.18), it
implies (4.17).

We want to prove (4.18). The estimate (3.7) yields

(4.19) $\begin{split}\int_{B}\left|u-\fint_{B}u\,dy\right|dm&\leq
C\int_{B}\int_{B}|\nabla u(\xi)||z-\xi|^{1-n}w(z)d\xi\,dz\\\ &\leq
C\int_{B}|\nabla u(\xi)|\int_{B(\xi,2r)}|z-\xi|^{1-n}w(z)dz\,d\xi\end{split}$

and thus it remains to check that for $\xi\in\mathbb{R}^{n}$ and $r>0$,

(4.20) $I=\int_{B(\xi,2r)}|z-\xi|^{1-n}w(z)dz\leq Crw(\xi).$

First, note that if $\delta(\xi)\geq 4r$, then for all $z\in B(\xi,2r)$, one
has

$\frac{1}{2}\delta(\xi)\leq\delta(\xi)-2r\leq\delta(z)\leq\delta(\xi)+2r\leq
2\delta(\xi)$

and so $w(z)$ is equivalent to $w(\xi)$. Thus $I\leq
Cw(\xi)\int_{B(\xi,2r)}|z-\xi|^{1-n}dz\leq Crw(\xi)$. It remains to prove the
case $\delta(\xi)<4r$. We split $I$ into $I_{1}+I_{2}$ where, for $I_{1}$, the
domain of integration is restrained to $B(\xi,\delta(\xi)/2)$. For any $z\in
B(\xi,\delta(\xi)/2)$, we have $w(z)\leq Cw(\xi)$ and thus

(4.21) $I_{1}\leq Cw(\xi)\int_{B(\xi,\delta(\xi)/2)}|z-\xi|^{1-n}dz\leq
Cw(\xi)\delta(\xi)\leq Crw(\xi).$

It remains to bound $I_{2}$. In order to do it, we decompose the remaining
domain into annuli
$C_{j}(\xi):=\\{z\in\mathbb{R}^{n},\,2^{j-1}\delta(\xi)\leq|\xi-z|\leq
2^{j}\delta(\xi)\\}$. We write $\kappa$ for the smallest integer bigger than
$\log_{2}(2r/\delta(\xi))$, which is the highest value for which
$C_{\kappa}\cap B(\xi,2r)$ is non-empty. We have

(4.22) $\begin{split}I_{2}\leq
C\sum_{j=0}^{\kappa}2^{j(1-n)}\delta(\xi)^{1-n}\int_{C_{j}(\xi)}dm(z)\leq
C\sum_{j=0}^{\kappa}2^{j(1-n)}\delta(\xi)^{1-n}m(B(\xi,2^{j}\delta(\xi))).\end{split}$

The ball $B(\xi,2^{j}\delta(\xi))$ is close to $\Gamma$ and thus Lemma 2.3
gives that the quantity $m(B(\xi,2^{j}\delta(\xi)))$ is bounded, up to a
constant, by $2^{j(d+1)}\delta(x)^{d+1}$. We deduce, since $2+d-n\leq 1$, that

(4.23) $\begin{split}I_{2}&\leq
C\sum_{j=0}^{\kappa}2^{j(2+d-n)}\delta(\xi)^{2+d-n}\leq
C\delta(\xi)^{2+d-n}\sum_{j=0}^{\kappa}2^{j}\\\ &\leq
C\delta(\xi)^{2+d-n}\left(\frac{2r}{\delta(\xi)}\right)\leq
Cr\delta(\xi)^{1+d-n}=Crw(\xi),\end{split}$

which ends the proof of (4.20) and thus also the one of the Poincaré
inequality (4.17).

Now we want to establish (4.12). The quickest way to do it is to use some
results of Hajłasz and Koskela. We say that $(u,g)$ forms a Poincaré pair if
$u$ is in $L^{1}_{loc}(\mathbb{R}^{n},w)$, $g$ is positive and measurable and
for any ball $B\subset\mathbb{R}^{n}$ of radius $r$, we have

(4.24) $m(B)^{-1}\int_{B}|u(z)-u_{B}|dm(z)\leq Crm(B)^{-1}\int_{B}g\,dm(z).$

In this context, Theorem 5.1 (and Corollary 9.8) in [HaK2] states that the
Poincaré inequality (4.24) can be improved into a Sobolev-Poincaré inequality.
More precisely, if $s$ is such that, for any ball $B_{0}$ of radius $r_{0}$,
any $x\in B_{0}$ and any $r\leq r_{0}$,

(4.25) $\frac{m(B(x,r))}{m(B_{0})}\geq C^{-1}\left(\frac{r}{r_{0}}\right)^{s}$

then (4.24) implies for any $1<q<s$

(4.26)
$\left(m(B)^{-1}\int_{B}|u(z)-u_{B}|^{q^{*}}dm(z)\right)^{\frac{1}{q^{*}}}\leq
Cr\left(m(B)^{-1}\int_{B}g^{q}dm(z)\right)^{\frac{1}{q}}$

where $q^{*}=\frac{qs}{s-q}$ and $B$ is a ball of radius $r$. Combined with
Hölder’s inequality, we get

(4.27) $\left(m(B)^{-1}\int_{B}|u(z)-u_{B}|^{p}dm(z)\right)^{\frac{1}{p}}\leq
Cr\left(m(B)^{-1}\int_{B}g^{2}dm(z)\right)^{\frac{1}{2}}$

for any $p\in[1,2s/(s-2)]$ if $s>2$ or any $p<+\infty$ if $s\leq 2$.

We will use the result of Hajłasz and Koskela with $g=|\nabla u|$. We need to
check the assumptions of their result. The bound (4.24) is exactly (4.17) and
we already proved it. The second and last thing we need to verify is that
(4.25) holds with $s=n$. This fact is an easy consequence of (2.8). Indeed, if
$B_{0}$ is a ball of radius $r_{0}$, $x\in B_{0}$ and $r\leq r_{0}$

(4.28) $\frac{m(B(x,r))}{m(B_{0})}\geq\frac{m(B(x,r))}{m(B(x,2r_{0}))}.$

Yet, (2.8) implies that $\frac{m(B(x,r))}{m(B(x,2r_{0}))}$ is bounded from
below by $C^{-1}(\frac{r}{2r_{0}})^{n}$, that is
$C^{-1}(\frac{r}{r_{0}})^{n}$. Then

(4.29) $\frac{m(B(x,r))}{m(B_{0})}\geq
C^{-1}\left(\frac{r}{r_{0}}\right)^{n},$

which is the desired conclusion. We deduce that (4.27) holds with $g=|\nabla
u|$ and for any $p\in[1,\frac{2n}{n-2}]$ ($1\leq p<+\infty$ if $n=2$), which
is exactly (4.12).

To finish to prove the lemma, it remains to establish (4.13). Let $B=B(x,r)$
be a ball centered on $\Gamma$. However, since $x\in\Gamma$, (2.2) entails
that $m(B)$ is equivalent to $r^{d+1}$. Thus, thanks to (4.12) and Lemma 4.1,

(4.30)
$\displaystyle\left(r^{-d-1}\int_{B}|u(z)|^{p}dm(z)\right)^{\frac{1}{p}}$
$\displaystyle\leq$ $\displaystyle
C\left(m(B)^{-1}\int_{B}\Big{|}u(z)-\fint_{B}u\,dy\Big{|}^{p}dm(z)\right)^{\frac{1}{p}}+C\fint_{B}|u(z)|dz$
$\displaystyle\leq$ $\displaystyle Cr\left(r^{-d-1}\int_{B}|\nabla
u|^{2}dm(z)\right)^{\frac{1}{2}},$

which proves Lemma 4.2. $\Box$

######  Remark 4.3.

If $B\subset\mathbb{R}^{n}$ and $u\in W$ is supported in $B$, then for any
$p\in[1,2n/(n-2)]$ (or $p\in[1,+\infty)$ if $n=2$), there holds

(4.31)
$\Big{\\{}\frac{1}{m(B)}\int_{B}\left|u(y)\right|^{p}w(y)dy\Big{\\}}^{1/p}\leq
Cr\Big{\\{}\frac{1}{m(B)}\int_{B}|\nabla u(y)|^{2}w(y)dy\Big{\\}}^{1/2}.$

That is, we can choose $u_{B}=0$ in (4.12).

To prove (4.31), the main idea is that we can replace in (4.12) the quantity
$u_{B}=m(B)^{-1}\int_{B}u$ by the average $u_{\bar{B}}$, where $\bar{B}$ is a
ball near $B$. We choose for $\bar{B}$ a ball with same radius as $B$ and
contained in $3B\setminus B$, because this way $u_{\bar{B}}=0$ since $u$ is
supported in $B$. Then

(4.32)
$\displaystyle\Big{\\{}\frac{1}{m(3B)}\int_{3B}|u(y)|^{p}w(y)dy\Big{\\}}^{1/p}$
$\displaystyle=$
$\displaystyle\Big{\\{}\frac{1}{m(3B)}\int_{3B}|u(y)-u_{\bar{B}}|^{p}w(y)dy\Big{\\}}^{1/p}$
$\displaystyle\leq$
$\displaystyle\Big{\\{}\frac{1}{m(3B)}\int_{3B}|u(y)-u_{3B}|^{p}w(y)dy\Big{\\}}^{1/p}+|u_{3B}-u_{\bar{B}}|$

Yet, using Jensen’s inequality and then Hölder’s inequality,
$|u_{3B}-u_{\bar{B}}|$ is bounded by
$\Big{\\{}\frac{1}{m(\bar{B})}\int_{3B}|u(y)-u_{3B}|^{p}w(y)dy\Big{\\}}^{1/p}$.
If we use in addition the doubling property given by (4.29), we get that
$|u_{3B}-u_{\bar{B}}|$ is bounded by
$\Big{\\{}\frac{1}{m(3B)}\int_{3B}|u(y)-u_{3B}|^{p}w(y)dy\Big{\\}}^{1/p}$,
that is,

(4.33)
$\begin{split}\Big{\\{}\frac{1}{m(3B)}\int_{3B}|u(y)|^{p}w(y)dy\Big{\\}}^{1/p}\leq\Big{\\{}\frac{1}{m(3B)}\int_{3B}|u(y)-u_{3B}|^{p}w(y)dy\Big{\\}}^{1/p}.\end{split}$

We conclude thanks to (4.12) and the doubling property (2.8).

##  Chapter 5 Completeness and Density of Smooth Functions

In later sections we shall work with various dense classes. We prepare the job
in this section, with a little bit of work on function spaces and
approximation arguments. Most results in this section are basically
unsurprising, except perhaps the fact that when $d\leq 1$, the test functions
are dense in $W$ (with no decay condition at infinity).

Let $\dot{W}$ be the factor space $W/{\mathbb{R}}$, equipped with the norm
$\|\cdot\|_{W}$. The elements of $\dot{W}$ are classes
$\dot{u}=\\{u+c\\}_{c\in{\mathbb{R}}}$, where $u\in W$.

######  Lemma 5.1.

The space $\dot{W}$ is complete. In particular, if a sequence of elements of
$W$, $\\{v_{k}\\}_{k=1}^{\infty}$, and $u\in W$ are such that
$\|v_{k}-u\|_{W}\to 0$ as $k\to\infty$, then there exist constants
$c_{k}\in{\mathbb{R}}$ such that $v_{k}-c_{k}\to u$ in
$L^{1}_{loc}({\mathbb{R}^{n}}).$

Proof. Let $(\dot{u}_{k})_{k\in\mathbb{N}}$ be a Cauchy sequence in $\dot{W}$.
We need to show that

  1. (i)

for every sequence $(v_{k})_{k\in\mathbb{N}}$ in $W$, with
$v_{k}\in\dot{u}_{k}$ for $k\in\mathbb{N}$, there exists $u\in W$ and
$(c_{k})_{k\in\mathbb{N}}$ such that $v_{k}-c_{k}\to u$ in
$L^{1}_{loc}(\mathbb{R}^{n})$ and

(5.1) $\lim_{k\to\infty}\|v_{k}-u\|_{W}=0;$

  2. (ii)

if $u$ and $u^{\prime}$ are such that there exist $(v_{k})_{k\in\mathbb{N}}$
and $(v^{\prime}_{k})_{k\in\mathbb{N}}$ such that
$v_{k},v^{\prime}_{k}\in\dot{u}_{k}$ for all $k\in\mathbb{N}$ and

(5.2)
$\lim_{k\to\infty}\|v_{k}-u\|_{W}=\lim_{k\to\infty}\|v^{\prime}_{k}-u^{\prime}\|_{W}=0,$

then $\dot{u}^{\prime}=\dot{u}$.

First assume that (i) is true and let us prove (ii). Let $u$, $u^{\prime}$,
$(v_{k})_{k\in\mathbb{N}}$ and $(v^{\prime}_{k})_{k\in\mathbb{N}}$ be such
that $v_{k},v^{\prime}_{k}\in\dot{u}_{k}$ for any $k\in\mathbb{N}$ and (5.2)
holds. Then the sequence $(\nabla v_{k}-\nabla
v^{\prime}_{k})_{k\in\mathbb{N}}$ converges in $L^{2}(\Omega,w)$ to
$\nabla(u-u^{\prime})$ on one hand and is constant equal to 0 on the other
hand. Thus $\nabla(u-u^{\prime})=0$ and $u$ and $u^{\prime}$ differ only by a
constant, hence $\dot{u}^{\prime}=\dot{u}$.

Now we prove (i). By translation invariance, we may assume that $0\in\Gamma$.
Let the $v_{k}\in\dot{u}_{k}$ be given, and choose
$c_{k}=\fint_{B(0,1)}v_{k}$. We want to show that $v_{k}-c_{k}$ converges in
$L^{1}_{loc}(\mathbb{R}^{n})$.

Set $B_{j}=B(0,2^{j})$ for $j\geq 0$; let us check that for $f\in W$ and
$j\geq 0$,

(5.3) $\fint_{B_{j}}\Big{|}f(x)-\fint_{B_{0}}f(y)\,dy\Big{|}\,dx\leq
C2^{(n+1)j}\|f\|_{W}.$

Set $m_{j}=\fint_{B_{j}}f\,dy$; observe that

$\displaystyle\fint_{B_{j}}|f-m_{j}|\,dx$ $\displaystyle\leq$
$\displaystyle\frac{1}{m(B_{j})}\int_{B_{j}}|f(x)-m_{j}|w(x)dx$
$\displaystyle\leq$ $\displaystyle
C2^{j}m(B_{j})^{-1/2}\Big{\\{}\int_{B_{j}}|\nabla
f(y)|^{2}w(y)dy\Big{\\}}^{1/2}$ $\displaystyle\leq$ $\displaystyle
C2^{j}m(B_{j})^{-1/2}||f||_{W}\leq C2^{j}||f||_{W}$

by (2.13), the Poincaré inequality (4.12) with $p=1$, and a brutal estimate
using (2.2), our assumption that $0\in\Gamma$, and the fact that $B_{j}\supset
B_{0}$. In addition,

(5.5) $\begin{split}|m_{0}-m_{j}|&=\Big{|}\fint_{B_{0}}f(x)\,dx-
m_{j}\Big{|}\leq\fint_{B_{0}}|f(x)-m_{j}|\,dx\\\ &\leq
2^{jn}\fint_{B_{j}}|f(x)-m_{j}|\,dx\leq C2^{(n+1)j}\|f\|_{W}\end{split}$

by (5). Finally

(5.6)
$\begin{split}\fint_{B_{j}}\Big{|}f(x)-\fint_{B_{0}}f(y)\,dy\Big{|}\,dx&=\fint_{B_{j}}\Big{|}f(x)-m_{0}\Big{|}\,dx\leq\fint_{B_{j}}|f(x)-m_{j}|\,dx+|m_{0}-m_{j}|\\\
&\leq C2^{(n+1)j}||f||_{W},\end{split}$

as needed for (5.3).

Return to the convergence of $v_{k}$. Recall that $c_{k}=\fint_{B_{1}}v_{k}$.
By (5.3) with $f=v_{k}-c_{k}-v_{l}+c_{l}$ (so that $m_{0}=0$), $v_{k}-c_{k}$
is a Cauchy sequence in $L^{1}_{loc}(B_{j})$ for each $j\geq 0$, hence there
exists $u^{j}\in L^{1}(B_{j})$ such that $v_{k}-c_{k}$ converges to $u^{j}$.
By uniqueness of the limit, we have that for $1\leq j\leq j_{0}$,

(5.7) $u^{j_{0}}=u^{j}\ \text{ a.e. in }B_{j}$

and thus we can define a function $u$ on $\mathbb{R}^{n}$ as $u(x)=u^{j}(x)$
if $x\in B_{j}$. By construction $u\in L^{1}_{loc}(\mathbb{R}^{n})$ and
$v_{k}-c_{k}\to u$ in $L^{1}_{loc}(\mathbb{R}^{n})$.

It remains to show that $u$ is actually in $W$ and $v_{k}\to u$ in $W$. First,
since $L^{2}(\Omega,w)$ is complete, there exists $V$ such that $\nabla v_{k}$
converges to $V$ in $L^{2}(\Omega,w)$. Then observe that for $\varphi\in
C^{\infty}_{0}(B_{j}\setminus\Gamma,\mathbb{R}^{n})$,

$\int_{B_{j}}V\cdot\varphi=\lim_{k\to\infty}\int_{B_{j}}\nabla
v_{k}\cdot\varphi=-\lim_{k\to\infty}\int_{B_{j}}(v_{k}-c_{k})\operatorname{div}\varphi=-\int_{B_{j}}u^{j}\operatorname{div}\varphi.$

Hence by definition of a weak derivative,

$\nabla u=\nabla u^{j}=V\ \text{ a.e. in }B_{j}.$

Since the result holds for any $j\geq 1$,

$\nabla u=V\ \text{ a.e. in }\mathbb{R}^{n},$

that is, by construction of $V$, $u\in W$ and $\|v_{k}-u\|_{W}$ converges to
0. $\Box$

######  Lemma 5.2.

The space

(5.8) $W_{0}=\big{\\{}u\in W\,;\,Tu=0\big{\\}},$

equipped with the scalar product $\left<u,v\right>_{W}:=\int_{\Omega}\nabla
u(z)\cdot\nabla v(z)\,dm(z)$ (and the norm $\|.\|_{W}$) is a Hilbert space.

Moreover, for any ball $B$ centered on $\Gamma$, the set

(5.9) $W_{0,B}=\big{\\{}u\in W\,;\,Tu=0\ \text{ $\mathcal{H}^{d}$-almost
everywhere on }\Gamma\cap B\big{\\}},$

equipped with the scalar product $\left<.,.\right>_{W}$, is also a Hilbert
space.

Proof. Observe that $W_{0}$ and $W_{0,B}$ are no longer spaces of functions
defined modulo an additive constant. That is, if $f\in W_{0}$ (or $W_{0,B}$)
is a constant $c$, then $c=0$ because (3.10) says that $Tu=c$ almost
everywhere on $\Gamma$. Thus $\|.\|_{W}$ is really a norm on $W_{0}$ and
$W_{0,B}$, and we only need to prove that these spaces are complete. We first
prove this for $W_{0,B}$; the case of $W_{0}$ will be easy deal with
afterwards.

Let $B$ be a ball centered on $\Gamma$, and consider $W_{0,B}$. By translation
and dilation invariance of the result, we can assume that $B=B(0,1)$.

Let $(v_{k})_{k\in\mathbb{N}}$ be a Cauchy sequence of functions in $W_{0,B}$.
We want first to show that $v_{k}$ has a limit in
$L^{1}_{loc}(\mathbb{R}^{n})$ and $W$. We use Lemma 5.1 and so there exists
$\bar{u}\in W$ and $c_{k}\in\mathbb{R}$ such that

(5.10) $\|v_{k}-\bar{u}\|_{W}\to 0$

and

(5.11) $v_{k}-c_{k}\to\bar{u}\ \ \text{ in }L^{1}_{loc}(\mathbb{R}^{n}).$

By looking at the proof of Lemma 5.1, we can take $c_{k}=\fint_{B}v_{k}$. Let
us prove that $(c_{k})$ is a Cauchy sequence in $\mathbb{R}$. We have for any
$k,l\geq 0$

(5.12) $|c_{k}-c_{l}|\leq\fint_{B}|v_{k}(z)-v_{l}(z)|\,dz\leq
Cm(B)^{-1}\int_{B}|v_{k}(z)-v_{l}(z)|dm(z)$

with (2.13). Since $T(v_{k}-v_{l})=0$ on $B$ and since we assumed that $B$ is
the unit ball, Lemma 4.2 entails

(5.13) $|c_{k}-c_{l}|\leq C\|v_{k}-v_{l}\|_{W}.$

Since $(v_{k})_{k\in\mathbb{N}}$ is a Cauchy sequence in W,
$(c_{k})_{k\in\mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$ and thus
converges to some value $c\in\mathbb{R}$. Set $u=\bar{u}-c$. We deduce from
(5.11) that

(5.14) $v_{k}\to u\ \ \text{ in }L^{1}_{loc}(\mathbb{R}^{n}),$

and since $u$ and $\bar{u}$ differ only from a constant, (5.10) can be
rewritten as

(5.15) $\|v_{k}-u\|_{W}\to 0.$

We still need to show that $u\in W_{0,B}$, i.e., that $Tu=0$ a.e. on $B$. We
will actually prove something a bit stronger. We claim that if $u,v_{k}\in W$,
then the convergence of $v_{k}$ to $u$ in both $W$ and
$L^{1}_{loc}(\mathbb{R}^{n})$ implies the convergence of the traces $Tv_{k}\to
Tu$ in $L^{1}_{loc}(\Gamma,\sigma)$. That is,

(5.16) $v_{k}\to u\text{ in $W$ and in $L^{1}_{loc}(\mathbb{R}^{n})$
}\Longrightarrow Tv_{k}\to Tu\text{ in $L^{1}_{loc}(\Gamma,\sigma)$}.$

Recall that by (3.26), $Tf\in L^{1}_{loc}(\Gamma,\sigma)$ whenever $f\in W$.
Our result, that is $Tu=0$ a.e. on $B$, follows easily from the claim: we
already established that $v_{k}\to u$ in $W$ and in
$L^{1}_{loc}(\mathbb{R}^{n})$ and thus (5.16) gives that
$\|Tu\|_{L^{1}(B,\sigma)}=\lim_{k\to\infty}\|Tv_{k}\|_{L^{1}(B,\sigma)}=0$,
i.e., that $Tu=0$ $\sigma$-a.e. in $B$.

We turn to the proof of (5.16). Since $T$ is linear, we may subtract $u$, and
assume that $v_{k}$ tends to $0$ and $u=0$. Let us use the notation of Theorem
3.4, and set $g^{k}=Tv_{k}$ and $g^{k}_{r}(x)=\fint_{B(x,r)}v_{k}(z)\,dz$.
Since $\|v_{k}\|_{W}$ tends to $0$, we may assume without loss of generality
that $\|v_{k}\|_{W}\leq 1$ for $k\in\mathbb{N}$. We want to prove that for
every ball $\widetilde{B}\subset\mathbb{R}^{n}$ centered on $\Gamma$ and every
$\epsilon>0$, we can find $k_{0}$ such that

(5.17) $\|g^{k}\|_{L^{1}(\widetilde{B},\sigma)}\leq\epsilon\ \ \text{ for
}k\geq k_{0}.$

We may also assume that the radius of $\widetilde{B}$ is larger than $1$ (as
it makes (5.17) harder to prove).

Fix $\widetilde{B}$ and $\varepsilon$ as above, and $\alpha\in(0,1/2)$, and
observe that for $r\in(0,1)$,

(5.18)
$\begin{split}\int_{\widetilde{B}}\left|g^{k}\right|d\sigma&\leq\int_{\widetilde{B}}|g^{k}-g^{k}_{r}|d\sigma+\int_{\widetilde{B}}|g^{k}_{r}|d\sigma\\\
&\leq
C(\widetilde{B})\|g^{k}-g^{k}_{r}\|_{L^{2}(\sigma)}+\int_{x\in\widetilde{B}}\fint_{y\in
B(x,r)}|v_{k}(y)|dyd\sigma(x)\\\ &\leq
C(\widetilde{B},\alpha)r^{2\alpha}\|v_{k}\|_{W}+Cr^{d-n}\int_{2\widetilde{B}}|v_{k}(y)|dy,\end{split}$

where for the last line we used (3.24), Fubini, and the condition (1.1) on
$\Gamma$. Recall that $\|v_{k}\|_{W}\leq 1$; we choose $r$ so small that
$C(\widetilde{B},\alpha)r^{2\alpha}\|u\|_{W}\leq\epsilon/2$, and since by
assumption $v_{k}$ tends to $0$ in $L^{1}_{loc}$, we can find $k_{0}$ such
that $Cr^{d-n}\int_{2\widetilde{B}}|v_{k}(y)|dy\leq\varepsilon/2$ for $k\geq
k_{0}$, as needed for (5.17).

This completes the proof of (5.16), and we have seen that the completeness of
$W_{0,B}$ follows. Since $W_{0}$ is merely an intersection of spaces
$W_{0,B}$, it is complete as well, and Lemma 5.2 follows. $\Box$

######  Lemma 5.3.

Choose a non-negative function $\rho\in C^{\infty}_{0}(\mathbb{R}^{n})$ such
that $\int\rho=1$ and $\rho$ is supported in $\overline{B(0,1)}$. Furthermore
let $\rho$ be radial and nonincreasing, i.e. $\rho(x)=\rho(y)\geq\rho(z)$ if
$|x|=|y|\leq|z|$. Define $\rho_{\epsilon}$, for $\epsilon>0$, by
$\rho_{\epsilon}(x)=\epsilon^{-n}\rho(\epsilon^{-1}x)$. For every $u\in W$, we
have:

  1. (i)

$\rho_{\epsilon}*u\in C^{\infty}(\mathbb{R}^{n})$ for every $\epsilon>0$;

  2. (ii)

If $x\in\mathbb{R}^{n}$ is a Lebesgue point of $u$, then
$\rho_{\epsilon}*u(x)\to u(x)$ as $\epsilon\to 0$; in particular,
$\rho_{\epsilon}*u\to u$ a.e. in $\mathbb{R}^{n}$;

  3. (iii)

$\nabla(\rho_{\epsilon}*u)=\rho_{\epsilon}*\nabla u$ for $\varepsilon>0$;

  4. (iv)

$\lim_{\epsilon\to 0}\|\rho_{\epsilon}*u-u\|_{W}=0$;

  5. (v)

$\rho_{\epsilon}*u\to u$ in $L^{1}_{loc}(\mathbb{R}^{n})$.

Proof. Recall that $W\subset L^{1}_{loc}(\mathbb{R}^{n})$ (see Lemma 3.2).
Thus conclusions (i) and (ii) are classical and can be found as Theorem 1.12
in [MZ].

Let $u\in W$ and write $u_{\epsilon}$ for $\rho_{\epsilon}*u$. We have seen
that $u_{\epsilon}\in C^{\infty}(\mathbb{R}^{n})$, so $\nabla u_{\epsilon}$ is
defined on $\mathbb{R}^{n}$. One would like to say that $\nabla
u_{\epsilon}=\rho_{\epsilon}*\nabla u$, i.e. point (iii). Here $\nabla
u_{\epsilon}$ is the classical gradient of $u_{\epsilon}$ on $\mathbb{R}^{n}$,
thus a fortiori  also the distributional gradient on $\mathbb{R}^{n}$ of
$u_{\epsilon}$. That is, for any $\varphi\in
C^{\infty}_{0}(\mathbb{R}^{n},\mathbb{R}^{n})$, there holds

(5.19) $\begin{split}\int_{\mathbb{R}^{n}}\nabla
u_{\epsilon}\cdot\varphi&=-\int_{\mathbb{R}^{n}}u_{\epsilon}(x)\operatorname{div}\varphi(x)dx=-\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\rho_{\epsilon}(y)u(x-y)\operatorname{div}\varphi(x)dy\,dx\\\
&=\int_{B(0,\epsilon)}\rho_{\epsilon}(y)\left(-\int_{\mathbb{R}^{n}}u(z)\operatorname{div}\varphi(z+y)dz\right)dy.\end{split}$

The function $\varphi$ lies in
$C^{\infty}_{0}(\mathbb{R}^{n},\mathbb{R}^{n})$, and so does, for any
$y\in\mathbb{R}^{n}$, the function $z\mapsto\varphi(z+y)$. Recall that $\nabla
u$ is the distributional derivative of $u$ on $\Omega$ but yet also the
distributional derivative of $u$ on $\mathbb{R}^{n}$ (see Lemma 3.2).
Therefore

(5.20) $\begin{split}\int_{\mathbb{R}^{n}}\nabla
u_{\epsilon}\cdot\varphi&=\int_{B(0,\epsilon)}\rho_{\epsilon}(y)\int_{\mathbb{R}^{n}}\nabla
u(z)\cdot\varphi(z+y)dz\,dy\\\
&=\int_{\mathbb{R}^{n}}\rho_{\epsilon}(y)\int_{\mathbb{R}^{n}}\nabla
u(x-y)\cdot\varphi(x)dx\,dy=\int_{\mathbb{R}^{n}}(\rho_{\epsilon}*\nabla
u)\cdot\varphi,\end{split}$

which gives (iii).

From there, our point (iv), that is the convergence of $\rho_{\epsilon}*u$ to
$u$ in $W$, can be deduced with, for instance, [Kil, Lemma 1.5]. The latter
states that, under our assumptions on $\rho$, the convergence
$\rho_{\epsilon}*g\to g$ holds in $L^{2}(\mathbb{R}^{n},w)$ whenever $g\in
L^{2}(\mathbb{R}^{n},w)$ and $w$ is in the Muckenhoupt class $\mathcal{A}_{2}$
(we already proved this fact, see Lemma 2.5). Note that Kilpelai’s result is
basically a consequence of a result from Muckenhoupt about the boundedness of
the (unweighted) Hardy-Littlewood maximal function in weighted $L^{p}$.

Finally we need to prove (v). Just notice that $u\in
L^{1}_{loc}(\mathbb{R}^{n})$, and apply the standard proof of the fact that
$\rho_{\epsilon}*u\to u$ in $L^{1}$ for $f\in L^{1}$. The lemma follows $\Box$

######  Lemma 5.4.

Let $u\in W$ and $\varphi\in C^{\infty}_{0}(\mathbb{R}^{n})$. Then
$u\varphi\in W$ and for any point $x\in\Gamma$ satisfying (3.11)

(5.21) $T(u\varphi)(x)=\varphi(x)Tu(x).$

Proof. The function $u$ lies in $L^{1}_{loc}(\mathbb{R}^{n})$ and thus defines
a distribution on $\mathbb{R}^{n}$ (see Lemma 3.2). Multiplication by smooth
functions and (distributional) derivatives are always defined for
distributions and, in the sense of distribution,
$\nabla(u\varphi)=\varphi\nabla u+u\nabla\varphi$. Let
$B\subset\mathbb{R}^{n}$ be a big ball such that
$\mathrm{supp}\,\varphi\subset B$. Then

(5.22) $\begin{split}\|u\varphi\|_{W}&\leq\|\varphi\|_{\infty}\|\nabla
u\|_{L^{2}(\Omega,w)}+\|\nabla\varphi\|_{\infty}\Big{\|}u-\fint_{B}u\Big{\|}_{L^{2}(B,w)}+\|\nabla\varphi\|_{\infty}\Big{\|}\fint_{B}u\Big{\|}_{L^{2}(B,w)}\\\
&\leq\|\varphi\|_{\infty}\|\nabla
u\|_{L^{2}(\Omega,w)}+C_{B}\|\nabla\varphi\|_{\infty}\|u\|_{W}+C_{B}\|\nabla\varphi\|_{\infty}\|u\|_{L^{1}(B)}<+\infty\end{split}$

by the Poincaré inequality (4.12). We deduce $u\varphi\in W$.

Let take a Lebesgue point $x$ satisfying (3.11). We have

$\begin{split}\fint_{B(x,r)}|u(z)\varphi(z)&-\varphi(x)Tu(x)|\,dz\\\
&\leq\fint_{B(x,r)}|u(z)-Tu(x)||\varphi(z)|\,dz+|Tu(x)|\fint_{B(x,r)}|\varphi(z)-\varphi(x)|\,dz\\\
&\leq\|\varphi\|_{\infty}\fint_{B(x,r)}|u(z)-Tu(x)|\,dz+|Tu(x)|\fint_{B(x,r)}|\varphi(z)-\varphi(x)|\,dz.\end{split}$

The first term of the right-hand side converges to 0 because $x$ is a Lebesgue
point. The second term in the right-hand side converges to 0 because $\varphi$
is continuous. The equality (5.21) follows. $\Box$

Let $F$ be a closed set in $\mathbb{R}^{n}$ and $E=\mathbb{R}^{n}\setminus F$.
In the sequel, we let

(5.23) $C^{\infty}_{c}(E)=\big{\\{}f\in
C^{\infty}(E),\,\exists\epsilon>0\text{ such that $f(x)=0$ whenever
}\,\mathrm{dist}(x,F)\leq\varepsilon\big{\\}}$

denote the set of functions in $C^{\infty}(E)$ that equal 0 in a neighborhood
of $F$. Furthermore, we use the notation $C^{\infty}_{0}(E)$ for the set of
functions that are compactly supported in $E$, that is

(5.24) $C^{\infty}_{0}(E)=\\{f\in C^{\infty}_{c}(E),\,\exists
R>0:\,\mathrm{supp}f\subset B(0,R)\\}.$

######  Lemma 5.5.

The completion of $C_{0}^{\infty}(\Omega)$ for the norm $\|.\|_{W}$ is the set

(5.25) $W_{0}=\big{\\{}u\in W\,;\,Tu=0\big{\\}}$

of (5.8). Moreover, if $u\in W_{0}$ is supported in a compact subset of the
open ball $B\subset\mathbb{R}^{n}$, then $u$ can be approximated in the
$W$-norm by functions of $C^{\infty}_{0}(B\setminus\Gamma)$.

Proof. The proof of this result will use two main steps, where

  1. (i)

we use cut-off functions $\varphi_{r}$ to approach any function $u\in W_{0}$
by functions in $W$ that equal 0 on a neighborhood of $\Gamma$;

  2. (ii)

we use cut-off functions $\phi_{R}$ to approach any function $u\in W_{0}$ by
functions in $W$ that are compactly supported in $\mathbb{R}^{n}$.

Part (i): For $r>0$ small, we choose a smooth function $\varphi_{r}$ such that
$\varphi(x)=0$ when $\delta(x)\leq r$, $\varphi(x)=1$ when $\delta(x)\geq 2r$,
$0\leq\varphi\leq 1$ everywhere, and $|\nabla\varphi(x)|\leq 10r^{-1}$
everywhere.

Let $u\in W_{0}$ be given. We want to show that for $r$ small, $\varphi_{r}u$
lies in $W$ and

(5.26) $\lim_{r\to 0}\|u-\varphi_{r}u\|_{W}^{2}=0.$

Notice that $\varphi_{r}u\in L^{1}_{loc}(\Omega)$, just like $u$, and its
distribution gradient on $\Omega$ is locally in $L^{2}$ and given by

(5.27) $\nabla(\varphi_{r}u)(x)=\varphi_{r}(x)\nabla
u(x)+u(x)\nabla\varphi_{r}(x).$

So we just need to show that

(5.28) $\lim_{\begin{subarray}{c}r\to
0\end{subarray}}\int|\nabla(\varphi_{r}u)(x)-\nabla u(x)|^{2}w(x)dx=\lim_{r\to
0}\int|u(x)\nabla\varphi_{r}(x)+(1-\varphi_{r}(x))\nabla u(x)|^{2}w(x)dx=0.$

Now $\int|\nabla u(x)|^{2}w(x)dx=\|u\|_{W}^{2}<+\infty$, so
$\int|(1-\varphi_{r})\nabla u(x)|^{2}w(x)dx$ tends to $0$, by the dominated
convergence theorem, and it is enough to show that

(5.29) $\lim_{r\to 0}\int|u(x)\nabla\varphi_{r}(x)|^{2}w(x)dx=0.$

Cover $\Gamma$ with balls $B_{j}$, $j\in J$, of radius $r$, centered on
$\Gamma$, and such that the $3B_{j}$ have bounded overlap, and notice that the
region where $\nabla\varphi_{r}\neq 0$ is contained in $\cup_{j\in J}3B_{j}$.
In addition, if $x\in 3B_{j}$ is such that $\nabla\varphi_{r}\neq 0$, then
$|\nabla\varphi_{r}(x)|\leq 10r^{-1}$, so that

(5.30) $\int_{3B_{j}}|u(x)\nabla\varphi_{r}(x)|^{2}w(x)dx\leq
100r^{-2}\int_{3B_{j}}|u(x)|^{2}w(x)dx\leq C\int_{3B_{j}}|\nabla
u(x)|^{2}w(x)dx,$

where the last part comes from (4.13), applied with $p=2$ and justified by the
fact that $Tu=0$ on the whole $\Gamma$. We may now sum over $j$. Denote by
$A_{r}$ the union of the $3B_{j}$; then

(5.31) $\displaystyle\int_{\Omega}|u(x)\nabla\varphi_{r}(x)|^{2}w(x)dx$
$\displaystyle\leq$ $\displaystyle\sum_{j\in
J}\int_{3B_{j}}|u(x)\nabla\varphi_{r}(x)|^{2}w(x)dx\leq C\sum_{j\in
J}\int_{3B_{j}}|\nabla u(x)|^{2}w(x)dx$ $\displaystyle\leq$ $\displaystyle
C\int_{A_{r}}|\nabla u(x)|^{2}w(x)dx$

because the $3B_{j}$ have bounded overlap. The right-hand side of (5.31) tends
to $0$, because $\int_{\Omega}|\nabla u(x)|^{2}w(x)dx=\|u\|_{W}^{2}<+\infty$
and by the dominated convergence theorem. The claim (5.29) follows, and so
does (5.26). This completes Part (i).

Part (ii). By translation invariance, we may assume that $0\in\Gamma$. Let $R$
be a big radius; we want to define a cut-off function $\phi_{R}$.

If we used the classical cut-off function built as
$\bar{\phi}_{R}=\bar{\phi}\left(\frac{x}{R}\right)$ with $\bar{\phi}$
supported in $B(0,1)$, the convergence would work with the help of Poincaré’s
inequality on annuli. But since we did not prove this inequality, we will
proceed differently and use the ‘better’ cut-off functions defined as follows.

Set $\phi_{R}(x)=\phi\left(\frac{\ln|x|}{\ln R}\right)$, where $\phi$ is a
smooth function defined on $[0,+\infty)$, supported in $[0,1]$ and such that
$\phi\equiv 1$ on $[0,1/2]$. In particular, one can see that
$\nabla\phi_{R}(x)\leq\frac{C}{\ln R}\frac{1}{|x|}$ and that $\nabla\phi_{R}$
is supported in $\\{x\in\mathbb{R}^{n},\,\sqrt{R}\leq|x|\leq R\\}$. We take
$\widehat{u}:=\phi_{R}u$ and we want to show that $\widehat{u}\in W$ and
$\|u-\widehat{u}\|_{W}$ is small. Notice that $\widehat{u}\in W_{0}$, by Lemma
5.4, and in addition $\widehat{u}$ is supported in $B(0,R)$. We want to show
that

(5.32) $\lim_{R\to+\infty}\|\widehat{u}-u\|_{W}^{2}=0.$

But $\widehat{u}\in L^{1}_{loc}(\Omega)$, just like $u$, and its distribution
gradient on $\Omega$ is locally in $L^{1}$ and given by

(5.33) $\nabla\widehat{u}(x)=\phi_{R}(x)\nabla u(x)+u(x)\nabla\phi_{R}(x).$

Hence

(5.34) $\displaystyle\|\widehat{u}-u\|_{W}^{2}$ $\displaystyle=$
$\displaystyle\int|\nabla\widehat{u}(x)-\nabla u(x)|^{2}w(x)dx$
$\displaystyle=$
$\displaystyle\int|u(x)\nabla\phi_{R}(x)+(1-\phi_{R}(x))\nabla
u(x)|^{2}w(x)dx.$

Now $\int|\nabla u(x)|^{2}w(x)dx=\|u\|_{W}^{2}<+\infty$, so
$\int|(1-\phi_{R})\nabla u(x)|^{2}w(x)dx$ tends to $0$, by the dominated
convergence theorem, and it is enough to show that

(5.35) $\lim_{R\to+\infty}\int|u(x)\nabla\phi_{R}(x)|^{2}w(x)dx=0.$

Let $C_{j}$ be the annulus $\\{x\in\mathbb{R}^{n},\,2^{j}<|x|\leq 2^{j+1}\\}$.
The bounds on $\nabla\phi_{R}$ yield

(5.36) $\int|u(x)\nabla\phi_{R}(x)|^{2}w(x)dx\leq\frac{C}{(\ln
R)^{2}}\sum_{j=0}^{1+\log_{2}R}2^{-2j}\int_{C_{j}}|u(x)|^{2}w(x)dx.$

The integral on the annulus $C_{j}$ is smaller than the integral in the ball
$B(0,2^{j+1})$. Since $u\in W_{0}$ and $0\in\Gamma$, (4.13) yields

(5.37) $\begin{split}\int|u(x)\nabla\phi_{R}(x)|^{2}w(x)dx&\leq\frac{C}{(\ln
R)^{2}}\sum_{j=0}^{1+\log_{2}R}\int_{B(0,2^{j+1})}|\nabla u(x)|^{2}w(x)dx\\\
&\leq\frac{C}{(\ln
R)^{2}}\|u\|_{W}^{2}\sum_{j=0}^{1+\log_{2}R}1\leq\frac{C}{|\ln
R|}\|u\|_{W}.\\\ \end{split}$

Thus $\int|u(x)\nabla\phi_{R}(x)|^{2}w(x)dx$ converges to 0 as $R$ goes to
$+\infty$, which proves (5.35) and ends Part (ii).

We are now ready to prove the lemma. If $u\in W_{0}$ and $\varepsilon>0$ is
given, we can find $R$ such that $\|\phi_{R}u-u\|_{W}^{2}\leq\varepsilon$ (by
(5.32)). Notice that $\phi_{R}u\in W_{0}$, by Lemma 5.4, and now we can find
$r$ such that $\|\varphi_{r}\phi_{R}u-\phi_{R}u\|_{W}^{2}\leq\varepsilon$ (by
(5.26)). In turn $\varphi_{r}\phi_{R}u$ is compactly supported away from
$\Gamma$, and we may now use Lemma 5.3 to approximate it with smooth functions
with compact support in $\Omega$. It follows that $W_{0}$ is included in the
completion of $C^{\infty}_{0}(\Omega)$. Since $W_{0}$ is complete (see Lemma
5.2), the reverse inclusion is immediate.

For the second part of the lemma, we are given $u\in W_{0}$ with a compact
support inside $B$, we can use Part (i) to approximate it by some
$\varphi_{r}u$ with a compact support inside $B$. A convolution as in Lemma
5.3 then makes it smooth without destroying the support property; Lemma 5.5
follows. $\Box$

######  Remark 5.6.

We don’t know how to prove exactly the same result for the spaces $W_{0,B}$ of
(5.9). However, we have the following weaker result. Let
$B\subset\mathbb{R}^{n}$ be a ball and $B_{\frac{1}{2}}$ denotes the ball with
same center as $B$ but half its radius. For any function $u\in W_{0,B}$, there
exists a sequence $(u_{k})_{k\in\mathbb{N}}$ of functions in
$C^{\infty}_{c}(\mathbb{R}^{n}\setminus\overline{B_{\frac{1}{2}}\cap\Gamma})$
such that $\|u_{k}-u\|_{W}$ converges to 0.

Indeed, take $\eta\in C^{\infty}_{0}(B)$ such that $\eta=1$ on
$B_{\frac{3}{4}}$. Write $u=\eta u+(1-\eta)u$; it is enough to prove that both
$\eta u$ and $(1-\eta)u$ can be approximated by functions in
$C^{\infty}_{c}(\mathbb{R}^{n}\setminus\overline{B_{\frac{1}{2}}\cap\Gamma})$.
Notice first that $\eta u\in W_{0}$ and thus can be approximated by functions
in $C^{\infty}_{0}(\Omega)\subset
C^{\infty}_{c}(\mathbb{R}^{n}\setminus\overline{B_{\frac{1}{2}}\cap\Gamma})$,
according to Lemma 5.5. Besides, $(1-\eta)u$ is supported outside of
$B_{\frac{3}{4}}$ and thus, if $\epsilon$ is smaller than a quarter of the
radius of $B$, then the functions $\rho_{\epsilon}*[(1-\eta)u]$ are in
$C^{\infty}_{c}(\mathbb{R}^{n}\setminus\overline{B_{\frac{1}{2}}})\subset
C^{\infty}_{c}(\mathbb{R}^{n}\setminus\overline{B_{\frac{1}{2}}\cap\Gamma})$.
Lemma 5.3 gives then that the family $\rho_{\epsilon}*[(1-\eta)u]$ approaches
$(1-\eta)u$ as $\epsilon$ goes to 0.

Next we worry about the completion of $C^{\infty}_{0}(\mathbb{R}^{n})$ for the
norm $\|.\|_{W}$. We start with the case when $d>1$; when $0<d\leq 1$, things
are a little different and they will be discussed in Lemma 5.11.

######  Lemma 5.7.

Let $d>1$. Choose $x_{0}\in\Gamma$ and write $B_{j}$ for $B(x_{0},2^{j})$.
Then for any $u\in W$

(5.38) $u^{0}:=\lim_{j\to+\infty}\fint_{B_{j}}u(z)\,dz\text{ exists and is
finite.}$

The completion of $C^{\infty}_{0}(\mathbb{R}^{n})$ for the norm $\|.\|_{W}$
can be identified to a subspace of $L^{1}_{loc}(\mathbb{R}^{n})$, which is

(5.39) $W^{0}=\\{u\in W,\,u^{0}=0\\}.$

######  Remark 5.8.

Since $C_{0}^{\infty}(\Omega)\subset C_{0}^{\infty}(\mathbb{R}^{n})$, Lemmata
5.5 and 5.7 imply that $W_{0}\subset W^{0}$. In particular, we get that

(5.40) $\lim_{j\to+\infty}\fint_{B_{j}}u(z)\,dz=0\ \ \text{ for }u\in W_{0}.$

######  Remark 5.9.

Since the completion of $C^{\infty}_{0}(\mathbb{R}^{n})$ doesn’t depend on our
choice of $x_{0}$, the value $u^{0}$ doesn’t depend on $x_{0}$ either.
Similarly, with a small modification in the proof, we could replace $(2^{j})$
with any other sequence that tends to $+\infty$.

######  Remark 5.10.

The lemma immediately implies the following result: for any $u\in W$,
$u-u^{0}\in W^{0}$ and thus can be approximated in
$L^{1}_{loc}(\mathbb{R}^{n})$ and in the $W$-norm by function in
$C^{\infty}_{0}(\mathbb{R}^{n})$.

Proof. Let $d>1$ and choose $u\in W$. Let us first prove that $u^{0}$ is well
defined. By translation invariance, we can choose $x_{0}=0$, that is
$B_{j}=B(0,2^{j})$. For $j\in\mathbb{N}$, set $u_{j}=\fint_{B_{j}}f(z)\,dz$
and $V_{j}=\int_{B_{j}}dm$. The bounds (2.2) give that $V_{j}$ is equivalent
to $2^{j(1+d)}$ and (2.14) gives that for any $z\in B_{j}$,
$\frac{V_{j}}{|B_{j}|}\leq Cw(z)$. Then by Lemma 4.2

(5.41) $\begin{split}|u_{j+1}-u_{j}|&\leq
C\fint_{B_{j+1}}|u(z)-u_{j+1}|\,dz\leq
CV_{j+1}^{-1}\int_{B_{j+1}}|u(z)-u_{j+1}|dm(z)\\\ &\leq
C2^{j(1-\frac{d+1}{2})}\left(\int_{B_{j+1}}|\nabla
u(z)|^{2}dm(z)\right)^{\frac{1}{2}}\leq
C2^{j\frac{1-d}{2}}\|u\|_{W}.\end{split}$

Since $d>1$, $(u_{j})_{j\in\mathbb{N}}$ is a Cauchy sequence and converges to
some value

(5.42) $u^{0}=\lim_{j\to+\infty}u_{j}.$

Moreover (5.41) also entails

(5.43) $|u_{j}-u^{0}|\leq C2^{j\frac{1-d}{2}}\|u\|_{W}.$

Let us prove additional properties on $u^{0}$. Set $v=|u|$. Notice that

(5.44) $|u_{j}|\leq
v_{j}:=\fint_{B_{j}}|u(z)|\,dz\leq|u_{j}|+\fint_{B_{j}}|u(z)-u_{j}|\,dz\leq|u_{j}|+C2^{j\frac{1-d}{2}}\|u\|_{W},$

where the last inequality follows from (5.41) (with $j-1$). As a consequence,
for any $j\geq 1$, $|v_{j}-|u_{j}||\leq C2^{j\frac{1-d}{2}}\|u\|_{W}$ and by
taking the limit as $j\to+\infty$,

(5.45) $|u^{0}|=\lim_{j\to+\infty}\fint_{B_{j}}|u(z)|\,dz.$

In addition,

(5.46)
$\begin{split}\left|\fint_{B_{j}}|u(z)|\,dz-|u^{0}|\right|&\leq|v_{j}-|u_{j}||+||u_{j}|-|u^{0}||\leq|v_{j}-|u_{j}||+|u_{j}-u^{0}|\leq
C2^{j\frac{1-d}{2}}\|u\|_{W}.\end{split}$

Let us show that $\|.\|_{W}$ is a norm for $W^{0}$. Let $u\in W^{0}$ be such
that $\|u\|_{W}=0$, then since $W^{0}\subset W$, $u\equiv c$ is a constant
function. Yet, observe that in this case, $u^{0}=c$. The assumption $u\in
W^{0}$ forces $u\equiv c\equiv 0$, that is $\|.\|_{W}$ is a norm on $W^{0}$.

We now prove that $(W^{0},\|.\|_{W})$ is complete. Let
$(v_{k})_{k\in\mathbb{N}}$ be a Cauchy sequence in $W^{0}$. Since
$(v_{k}-v_{l})^{0}=0$, we deduce from (5.46) that for $j\geq 1$ and
$k,l\in\mathbb{N}$,

(5.47) $\fint_{B_{j}}|v_{k}(z)-v_{l}(z)|\,dz\leq
C2^{j\frac{1-d}{2}}\|v_{k}-v_{l}\|_{W}.$

Consequently, $(v_{k})_{k\in\mathbb{N}}$ is a Cauchy sequence in $L^{1}_{loc}$
and thus there exists $u\in L^{1}_{loc}(\mathbb{R}^{n})$ such that $v_{k}\to
u$ in $L^{1}_{loc}(\mathbb{R}^{n})$. Since $(\nabla v_{k})_{k\in\mathbb{N}}$
is also a Cauchy sequence in $L^{2}(\Omega,w)$, there exists $V\in
L^{2}(\Omega,w)$ such that $\nabla v_{k}\to V$ in $\in L^{2}(\Omega,w)$. It
follows that $v_{k}$ and $\nabla v_{k}$ converge in the sense of distribution
to respectively $u$ and $V$, thus $u$ has a distributional derivative in
$\Omega$ and $\nabla u$ equals $V\in L^{2}(\Omega,w)$. In particular $u\in W$.
It remains to check that $u^{0}=0$. Yet, notice that

(5.48)
$|u^{0}|\leq\left|u^{0}-\int_{B_{j}}u(z)\,dz\right|+\left|\int_{B_{j}}(u-v_{k})(z)\,dz\right|+\left|\int_{B_{j}}v_{k}(z)\,dz\right|.$

The first term and the third term in the right-hand side are bounded by
$C2^{j\frac{1-d}{2}}\|u\|_{W}$ and $C2^{j\frac{1-d}{2}}\|u_{k}\|_{W}$
respectively (thanks to (5.43)), the second by
$C2^{j\frac{1-d}{2}}\|u-u_{k}\|_{W}$ (because of (5.47)). By taking $k$ and
$j$ big enough, we can make the right-hand side of (5.48) as small as we want.
It follows that $u^{0}=|u^{0}|=0$ and $u\in W^{0}$. The completeness of
$W^{0}$ follows.

It remains to check that the completion of $C^{\infty}_{0}(\mathbb{R}^{n})$ is
$W^{0}$. However, it is easy to see that any function $u$ in
$C^{\infty}_{0}(\mathbb{R}^{n})$ satisfies $u^{0}=0$ and thus lies in $W^{0}$.
Together with the fact that $W^{0}$ is complete, we deduce that the completion
of $C^{\infty}_{0}(\mathbb{R}^{n})$ with the norm $\|.\|_{W}$ is included in
$W^{0}$. The converse inclusion will hold once we establish that any function
in $W^{0}$ can be approached in the $W$-norm by functions in
$C^{\infty}_{0}(\mathbb{R}^{n})$. Besides, thanks to Lemma 5.3, it is enough
to prove that $u\in W^{0}$ can be approximated by functions in $W$ that are
compactly supported in $\mathbb{R}^{n}$.

Fix $\phi\in C^{\infty}((-\infty,+\infty))$ such that $\phi\equiv 1$ on
$(-\infty,1/2]$, $\phi\equiv 0$ on $[1,+\infty)$. For $R>0$ define $\phi_{R}$
by $\phi_{R}(x)=\phi(\ln|x|/\ln R)$. Observe that that $\phi_{R}(x)\equiv 1$
if $|x|\leq\sqrt{R}$, $\phi_{R}(x)\equiv 0$ if $|x|\geq R$ and, for any
$x\in\mathbb{R}^{n}$,

(5.49) $|\nabla\phi_{R}(x)|\leq\frac{C}{\ln R}\frac{1}{|x|}.$

The approximating functions will be the $\phi_{R}u$, which are compactly
supported in $\mathbb{R}^{n}$. Now

(5.50) $\begin{split}\|u\phi_{R}-u\|_{W}^{2}&=\|u(1-\phi_{R})\|_{W}^{2}\\\
&\leq\int_{\Omega}(1-\phi_{R}(z))^{2}|\nabla
u(z)|^{2}dm(z)+\int_{\Omega}|u(z)|^{2}|\nabla\phi_{R}(z)|^{2}dm(z)\\\
&\leq\int_{|z|\geq\sqrt{R}}|\nabla
u(z)|^{2}dm(z)+\int_{\Omega}|u(z)|^{2}|\nabla\phi_{R}(z)|^{2}dm(z).\\\
\end{split}$

By the dominated convergence theorem, the first term of the right-hand side
above converges to $0$ as $R$ goes to $+\infty$. It remains to check that the
second term also tends to $0$. Set $C_{j}=B_{j}\setminus B_{j-1}$. We have if
$R>1$,

(5.51)
$\begin{split}\int_{\Omega}|u(z)|^{2}|\nabla\phi_{R}(z)|^{2}dm(z)&\leq\frac{C}{|\ln
R|^{2}}\int_{\sqrt{R}<|z|<R}\frac{|u(z)|^{2}}{|z|^{2}}dm(z)\\\
&\leq\frac{C}{|\ln
R|^{2}}\sum_{j=0}^{\log_{2}R+1}2^{-2j}\int_{C_{j}}|u(z)|^{2}dm(z)\\\
&\leq\frac{C}{|\ln
R|^{2}}\sum_{j=0}^{\log_{2}R+1}2^{-2j}\int_{B_{j}}|u(z)|^{2}dm(z)\\\
&\leq\frac{C}{|\ln
R|^{2}}\sum_{j=0}^{\log_{2}R+1}2^{-2j}\left(\int_{B_{j}}|u(z)-u_{j}|^{2}dm(z)+V_{j}|u_{j}|^{2}\right).\\\
\end{split}$

Lemma 4.2 gives that $\int_{B_{j}}|u(z)-u_{j}|^{2}dm(z)$ is bounded, up to a
harmless constant, by $2^{2j}\int_{B_{j}}|\nabla u(z)|^{2}dm(z)\leq
2^{2j}\|u\|_{W}^{2}$. In addition, $V_{j}=m(B_{j})$ is bounded by
$C2^{j(1+d)}$ because of (2.2) and we get that $|u_{j}|^{2}\leq
2^{j(1-d)}\|u\|_{W}$, by (5.43). Hence

(5.52)
$\begin{split}\int_{\Omega}|u(z)|^{2}|\nabla\phi_{R}(z)|^{2}dm(z)&\leq\frac{C}{|\ln
R|^{2}}\|u\|_{W}^{2}\sum_{j=0}^{\log_{2}R+1}2^{-2j}\left(2^{2j}+2^{j(d+1)}2^{j(1-d)}\right)\\\
&\leq\frac{C}{|\ln
R|^{2}}\|u\|_{W}^{2}\sum_{j=0}^{\log_{2}R+1}1\leq\frac{C}{|\ln
R|}\|u\|_{W}^{2},\\\ \end{split}$

which converges to 0 as $R$ goes to $+\infty$. This concludes the proof of
Lemma 5.7. $\Box$

As we shall see now, the situation in low dimensions is different, essentially
because when $d\leq 1$, the constant function $1$ can be approximated by
functions of $C^{\infty}_{0}(\mathbb{R}^{n})$.

######  Lemma 5.11.

Let $d\leq 1$. For any function $u$ in $W$, we can find a sequence of
functions $(u_{k})_{k\in\mathbb{N}}$ in $C^{\infty}_{0}(\mathbb{R}^{n})$ such
that $u_{k}$ converges, in $L^{1}_{loc}(\mathbb{R}^{n})$ and and for the semi-
norm $\|.\|_{W}$, to $u$.

######  Remark 5.12.

The fact that the function $1$ can be approached with the norm $\|.\|_{W}$ by
functions in $C_{0}^{\infty}$ means that the completion of $C_{0}^{\infty}$
with the norm $\|.\|_{W}$ is not a space of distributions.

We can legitimately say that the completion of $C_{0}^{\infty}$ is embedded
into the space of distributions $D^{\prime}=(C_{0}^{\infty})^{\prime}\supset
L^{1}_{loc}$ if the convergence $u_{k}\in C_{0}^{\infty}\subset L^{1}_{loc}$
to $u\in W\subset L^{1}_{loc}$ in the norm $\|.\|_{W}$ implies, for
$\varphi\in C_{0}^{\infty}$, that $\int u_{k}\varphi$ tends to $\int
u\varphi$. Take $u_{k}\in C^{\infty}_{0}(\mathbb{R}^{n})$ such that $u_{k}$
tends to 1 in $L^{1}_{loc}(\mathbb{R}^{n})$ and $W$. Then since $\|.\|_{W}$
doesn’t see the constants, $u_{k}$ tends to 0 in $W$; but the convergence of
$u_{k}$ to 1 in $L^{1}_{loc}(\mathbb{R}^{n})$ implies that $\int u_{k}\varphi$
tends to $\int\varphi\neq 0$ for some function $\varphi\in
C^{\infty}_{0}(\mathbb{R}^{n})$.

Proof. As before, we may assume that $0\in\Gamma$. Let us first prove that for
$d\leq 1$, the constant function $1$ (and thus any constant function) is the
limit in $W$ and $L^{1}_{loc}(\mathbb{R}^{n})$ of test functions.

Choose $\phi\in C^{\infty}([0,+\infty))$ such that $\phi\equiv 1$ on $[0,1/2]$
and $\phi\equiv 0$ on $[1,+\infty)$. For $R>1$, define $\psi_{R}$ as
$\psi_{R}(x)=\phi(\ln\ln|x|/\ln\ln R)$ if $|x|>1$ and $\psi_{R}(x)=1$ if
$|x|\leq 1$. This cut-off function is famous for being used by Sobolev, and is
useful to handle the critical case (that is, for us, $d=1$). It can be avoided
if $d<1$ but we didn’t want to separate the cases $d<1$ and $d=1$. Let us
return to the proof of the lemma. We have: $\psi_{R}(x)\equiv 1$ if
$|x|\leq\exp(\sqrt{\ln R})$, $\psi_{R}(x)\equiv 0$ if $|x|\geq R$ and for any
$x\in\mathbb{R}^{n}$ satisfying $|x|>1$,

(5.53) $|\nabla\psi_{R}(x)|\leq\frac{C}{\ln\ln R}\frac{1}{|x|\ln|x|}.$

It is easy to see that $\psi_{R}$ converges to 1 in
$L^{1}_{loc}(\mathbb{R}^{n})$ as $R$ goes to $+\infty$. We claim that

(5.54) $\|\psi_{R}\|_{W}\text{ converges to 0 as $R$ goes to $+\infty$}.$

Let us prove (5.54). As in Lemma 5.7, we write $B_{j}$ for $B(0,2^{j})$ and
$C_{j}$ for $B_{j}\setminus B_{j-1}$. Then for $R$ large,

(5.55) $\begin{split}\|\psi_{R}\|_{W}^{2}&\leq\frac{C}{|\ln\ln
R|^{2}}\int_{2<|z|\leq R}\frac{1}{|z|^{2}|\ln|z||^{2}}dm(z)\\\
&\leq\frac{C}{|\ln\ln R|^{2}}\sum_{j=1}^{+\infty}2^{-2j}|\ln
2^{j}|^{-2}\int_{C_{j}}dm(z)\\\ &\leq\frac{C}{|\ln\ln
R|^{2}}\sum_{j=1}^{+\infty}\frac{1}{j^{2}}2^{-2j}2^{j(d+1)}\leq\frac{C}{|\ln\ln
R|^{2}}.\\\ \end{split}$

Our claim follows, and it implies that $\|1-\psi_{R}\|_{W}$ tends to $0$.

We will prove now that any function in $W$ can be approached by functions in
$C^{\infty}_{0}(\mathbb{R}^{n})$. Let $u\in W$ be given. Let
$u^{0}=\fint_{B_{0}}u$ denote the average of $u$ on the unit ball. We have
just seen how to approximate $u_{0}$ by test functions, so it will be enough
to show that $u-u^{0}$ can be approached by test functions.

For this we shall proceed as in Lemma 5.7. We shall use the product
$\psi_{R}(u-u^{0})$, where $\psi_{R}$ is the same cut-off function as above,
and prove that $\psi_{R}(u-u^{0})$ lies in $W$ and

(5.56) $\lim_{R\to+\infty}\|(u-u_{0})\psi_{R}\|_{W}=0.$

Notice that $\psi_{R}(u-u^{0})$ is compactly supported, and converges
(pointwise and in $L^{1}_{loc}$) to $u-u_{0}$. Thus, as soon as we prove
(5.56), Lemma 5.3 will allow us to approximate $(u-u_{0})\psi_{R}$ by smooth,
compactly supported functions, and the desired approximation result will
follow.

As for the proof of (5.56), of course we shall use Poincaré’s inequality, and
the the key point will be to get proper bounds on differences of averages of
$u$. These will not be as good as before, because now $d\leq 1$, and instead
of working directly on the balls $B_{j}$ we shall use strings of balls $D_{j}$
that do not contain the origin, so that their overlap is smaller.

Fix any unit vector $\xi\in\partial B(0,1)$, and consider the balls

(5.57) $D=D^{\xi}=B(\xi,9/10)\ \text{ and, for $j\in{\mathbb{N}}$, }\
D_{j}=D_{j}^{\xi}=B(2^{j}\xi,\frac{9}{10}2^{j}).$

We will later use the $D_{j}^{\xi}$ to cover our usual annuli $C_{j}$, but in
the mean time we fix $\xi$ and want estimates on the numbers
$m_{j}=\fint_{D_{j}}u_{j}$.

The Poincaré inequality (4.12), applied with with $p=1$, yields

(5.58) $m(D_{j})^{-1}\int_{D_{j}}|u-m_{j}|dm\leq
C2^{j}\left(m(D_{j})^{-1}\int_{D_{j}}|\nabla
u(z)|^{2}dm\right)^{\frac{1}{2}}.$

Of course we have a similar estimate on $D_{j+1}$; observe also that
$D_{j}\cap D_{j+1}$ contains a ball $D^{\prime}_{j}$ of radius $2^{j-2}$ (we
may even take it centered at $2^{j}\xi$); then

$\displaystyle|m_{j}-m_{j+1}|$ $\displaystyle=$ $\displaystyle
m(D^{\prime}_{j})^{-1}\int_{D^{\prime}_{j}}|m_{j}-m_{j+1}|dm$
$\displaystyle\leq$ $\displaystyle
m(D^{\prime}_{j})^{-1}\int_{D^{\prime}_{j}}(|u-m_{j}|+|u-m_{j+1}|)dm$
$\displaystyle\leq$ $\displaystyle
Cm(D_{j})^{-1}\int_{D_{j}}|u-m_{j}|dm+Cm(D_{j+1})^{-1}\int_{D_{j+1}}|u-m_{j}|dm$
$\displaystyle\leq$ $\displaystyle C2^{j}\left(m(D_{j})^{-1}\int_{D_{j}\cup
D_{j+1}}|\nabla u(z)|^{2}dm\right)^{\frac{1}{2}}$

because $m(D_{j})\leq Cm(D^{\prime}_{j})$ (and similarly for $m(D_{j+1})$),
since the measure $m$ is doubling by (2.8). By (2.2), $m(D_{j})\geq
C^{-1}2^{j(d+1)}$, so (5) yields

(5.60) $|m_{j}-m_{j+1}|\leq C2^{-j(d-1)/2}\left(\int_{D_{j}\cup
D_{j+1}}|\nabla u(z)|^{2}dm(z)\right)^{\frac{1}{2}}$

The same estimate, run with $B_{0}=B(0,1)$ and $D_{0}$ whose intersection also
contains a large ball, yields

(5.61) $|u^{0}-m_{0}|\leq C\left(\int_{B_{0}\cup D_{0}}|\nabla
u(z)|^{2}dm(z)\right)^{\frac{1}{2}}\leq C\|u\|_{W}.$

With $\xi$ fixed, the various $D_{j}\cup D_{j+1}$ have bounded overlap; thus
by (5.60) and Cauchy-Schwarz,

(5.62) $\begin{split}\left|m_{j+1}-m_{0}\right|^{2}&\leq
C\left(\sum_{i=0}^{j}2^{-i(d-1)/2}\|\nabla u\|_{L^{2}(D_{j}\cup
D_{j+1},w)}\right)^{2}\\\ &\leq C(j+1)\sum_{i=0}^{j}2^{i(1-d)}\|\nabla
u\|_{L^{2}(D_{j}\cup D_{j+1},w)}^{2}\leq
C(j+1)2^{j(1-d)}\|u\|_{W}^{2}.\end{split}$

Here we used our assumption that $d\leq 1$, and we are happy about our trick
with the bounded overlap because a more brutal estimate would lead to a factor
$(j+1)^{2}$ that would hurt us soon. Anyway, we add (5.61) and get that for
$j\geq 0$,

(5.63) $\left|m_{j}-u^{0}\right|^{2}\leq C(j+1)2^{j(1-d)}\|u\|_{W}^{2}.$

We are now ready to prove (5.56). Since the first part of the proof gives that
$\|u^{0}\psi_{R}\|_{W}$ tends to 0, we shall assume that $u^{0}=0$ to simplify
the estimates. By Lemma 5.4, $(u-u_{0})\psi_{R}=u\psi_{R}$ lies in $W$ and its
gradient is $u\nabla\psi_{R}+\psi_{R}\nabla u$. So we just need to show that
when $R$ tends to $+\infty$,

(5.64) $\|u\psi_{R}-u\|_{W}\leq\|(1-\psi_{R})\nabla
u\|_{L^{2}(\Omega,w)}+\|u\nabla\psi_{R}\|_{L^{2}(\Omega,w)}$

tends to $0$. The first term of the right-hand side converges to 0 as $R$ goes
to $+\infty$, thanks to the dominated convergence theorem, and for the second
term we use (5.53) and the fact that $\nabla\psi_{R}$ is supported in the
region $Z_{R}$ where $\exp(\sqrt{\ln R})\leq|x|\leq R$. Thus

(5.65)
$\|u\nabla\psi_{R}\|_{L^{2}(\Omega,w)}^{2}=\int_{\mathbb{R}^{n}}|u(z)|^{2}|\nabla\psi_{R}(z)|^{2}dm(z)\leq\frac{C}{|\ln\ln
R|^{2}}\int_{Z_{R}}\frac{|u(z)|^{2}}{|z|^{2}(\ln|z|)^{2}}dm(z)$

As usual, we cut $Z_{R}$ into annular subregions $C_{j}$, and then further
into balls like the $D_{j}$. We start with the $C_{j}=B_{j}\setminus B_{j-1}$.
For $R$ large, if $C_{j}$ meets $Z_{R}$, then $10\leq j\leq 1+\log_{2}R$ and

(5.66) $\int_{C_{j}}\frac{|u(z)|^{2}}{|z|^{2}(\ln|z|)^{2}}dm(z)\leq
j^{-2}2^{-2j}\int_{C_{j}}|u(z)|^{2}dm(z).$

We further cut $C_{j}$ into balls, because we want to apply Poincaré’s
inequality. Let the $D_{j}^{\xi}$ be as in the definition (5.57). We can find
a finite set $\Xi\subset\partial B(0,1)$ such that the balls $D^{\xi}$,
$\xi\in\Xi$, cover $B(0,1)\setminus B(0,1/2)$. Then for $j\geq 1$ the
$D^{\xi}$, $\xi\in\Xi$, cover $C_{j}$ and, by (5.65) and (5.66),

(5.67) $\|u\nabla\psi_{R}\|_{L^{2}(\Omega,w)}^{2}\leq\frac{C}{|\ln\ln
R|^{2}}\sum_{j=10}^{1+\log_{2}R}j^{-2}2^{-2j}\sum_{\xi\in\Xi}\int_{D_{j}^{\xi}}|u(z)|^{2}dm(z).$

Then by the Poincaré inequality (4.12) (with $p=2$),

(5.68) $\int_{D_{j}^{\xi}}|u(z)-m_{j}^{\xi}|^{2}dm(z)\leq
C2^{2j}\int_{D_{j}^{\xi}}|\nabla u(z)|^{2}dm(z),$

where $m_{j}^{\xi}=\fint_{D_{j}^{\xi}}$ as in the estimates above. Thus

(5.69) $\int_{D_{j}^{\xi}}|u(z)|^{2}dm(z)\leq C2^{2j}\int_{D_{j}^{\xi}}|\nabla
u(z)|^{2}dm(z)+Cm(D_{j}^{\xi})(j+1)2^{j(1-d)}\|u\|_{W}^{2}$

by (5.63) and because $u^{0}=0$. But $m(D_{j}^{\xi})\leq C2^{(d+1)j}$ by
(2.2), so

(5.70) $\int_{D_{j}^{\xi}}|u(z)|^{2}dm(z)\leq C(j+1)2^{2j}\|u\|_{W}^{2}.$

We return to (5.67) and get that

(5.71)
$\begin{split}\|u\nabla\psi_{R}\|_{L^{2}(\Omega,w)}^{2}&\leq\frac{C}{|\ln\ln
R|^{2}}\sum_{j=10}^{1+\log_{2}R}j^{-2}2^{-2j}\sum_{\xi\in\Xi}(j+1)2^{2j}\|u\|_{W}^{2}\\\
&\leq\frac{C}{|\ln\ln
R|^{2}}\sum_{j=10}^{1+\log_{2}R}j^{-1}\|u\|_{W}^{2}\leq\frac{C}{|\ln\ln
R|}\|u\|_{W}^{2}\end{split}$

because $\Xi$ is finite, and where we see that $j^{-1}$ is really useful.

We already took care of the other part of (5.64); thus $\|u\psi_{R}-u\|_{W}$
tends to $0$. This proves (5.56) (recall that $u^{0}=0$), and completes our
proof of Lemma 5.11. $\Box$

##  Chapter 6 The Chain Rule and Applications

We record here some basic (and not shocking) properties concerning the
derivative of $f\circ u$ when $u\in W$, and the fact that $uv\in W\cap
L^{\infty}$ when $u,v\in W\cap L^{\infty}$.

######  Lemma 6.1.

The following properties hold:

  1. (a)

Let $f\in C^{1}(\mathbb{R})$ be such that $f^{\prime}$ is bounded and let
$u\in W$. Then $f\circ u\in W$ and

(6.1) $\nabla(f\circ u)=f^{\prime}(u)\nabla u.$

Moreover, $T(f\circ u)=f\circ(Tu)$ $\sigma$-a.e. in $\Gamma$.

  2. (b)

Let $u,v\in W$. Then $\max\\{u,v\\}$ and $\min\\{u,v\\}$ belong to $W$ and,
for almost every $x\in\mathbb{R}^{n}$,

(6.2) $\nabla\max\\{u,v\\}(x)=\left\\{\begin{array}[]{ll}\nabla u(x)&\text{ if
}u(x)\geq v(x)\\\ \nabla v(x)&\text{ if }v(x)\geq u(x)\end{array}\right.$

and

(6.3) $\nabla\min\\{u,v\\}(x)=\left\\{\begin{array}[]{ll}\nabla u(x)&\text{ if
}u(x)\leq v(x)\\\ \nabla v(x)&\text{ if }v(x)\leq u(x).\end{array}\right.$

In particular, for any $\lambda\in\mathbb{R}$, $\nabla u=0$ (Lebesgue) a.e. on
$\\{x\in\mathbb{R}^{n},\,u(x)=\lambda\\}$.

In addition, $T\max\\{u,v\\}=\max\\{Tu,Tv\\}$ and
$T\min\\{u,v\\}=\min\\{Tu,Tv\\}$ $\sigma$-a.e. on $\Gamma$. Thus
$\max\\{u,v\\}$ and $\min\\{u,v\\}$ lie in $W_{0}$ as soon as $u,v\in W_{0}$.

######  Remark 6.2.

A consequence of Lemma 6.1 (b) is that, for example, $|u|\in W$ (resp. $|u|\in
W_{0}$) whenever $u\in W$ (resp. $u\in W_{0}$).

Proof. A big part of this proof follows the results from 1.18 to 1.23 in
[HKM].

Let us start with (a). More precisely, we aim for (6.1). Let $f\in
C^{1}(\mathbb{R})\cap Lip(\mathbb{R})$ and let $u\in W$. The idea of the proof
is the following: we approximate $u$ by smooth functions $\varphi_{k}$, for
which the result is immediate. Then we observe that both $\nabla(f\circ u)$
and $f^{\prime}(u)\nabla u$ are the limit (in the sense of distributions) of
the gradient of $f\circ\varphi_{k}$.

According to Lemma 5.3, there exists a sequence
$(\varphi_{k})_{k\in\mathbb{N}}$ of functions in
$C^{\infty}(\mathbb{R}^{n})\cap W$ such that $\varphi_{k}\to u$ in $W$ and in
$L^{1}_{loc}(\mathbb{R}^{n})$. The classical (thus distributional) derivative
of $f\circ\varphi_{k}$ is

(6.4) $\nabla[f\circ\varphi_{k}]=f^{\prime}(\varphi_{k})\nabla\varphi_{k}.$

In particular, since $\varphi_{k}\in W$ and $f^{\prime}$ is bounded,
$f\circ\varphi_{k}\in W$ and
$\|f\circ\varphi_{k}\|_{W}\leq\|\varphi_{k}\|_{W}\sup|f^{\prime}|$.

Notice that $|f(s)-f(t)|\leq|s-t|\sup|f^{\prime}|$. Therefore, since
$\varphi_{k}\to u$ in $L^{1}_{loc}(\mathbb{R}^{n})$, for any ball
$B\subset\mathbb{R}^{n}$

(6.5) $\int_{B}|f\circ\varphi_{k}-f\circ
u|\,dx\leq\sup|f^{\prime}|\int_{B}|\varphi_{k}-u|\,dx\longrightarrow 0.$

That is $f\circ\varphi_{k}\to f\circ u$ in $L^{1}_{loc}(\mathbb{R}^{n})$,
hence also in the sense of distributions. Besides,

(6.6)
$\displaystyle\left(\int_{\Omega}|f^{\prime}(\varphi_{k})\nabla\varphi_{k}-f^{\prime}(u)\nabla
u|^{2}dm\right)^{\frac{1}{2}}$
$\displaystyle\qquad\qquad\qquad\leq\left(\int_{\Omega}|f^{\prime}(\varphi_{k})[\nabla\varphi_{k}-\nabla
u]|^{2}dm\right)^{\frac{1}{2}}+\left(\int_{\Omega}|\nabla
u[f^{\prime}(\varphi_{k})-f^{\prime}(u)]|^{2}dm\right)^{\frac{1}{2}}$
$\displaystyle\qquad\qquad\qquad\leq\sup|f^{\prime}|\left(\int_{\Omega}|\nabla\varphi_{k}-\nabla
u|^{2}dm\right)^{\frac{1}{2}}+\left(\int_{\Omega}|\nabla
u|^{2}|f^{\prime}(\varphi_{k})-f^{\prime}(u)|^{2}dm\right)^{\frac{1}{2}}.$

The first term in the right-hand side is converges to 0 since $\varphi_{k}\to
u$ in $W$. Besides, $\varphi_{k}\to u$ a.e. in $\Omega$ and $f^{\prime}$ is
continuous, so $f^{\prime}(\varphi_{k})\to f^{\prime}(u)$ a.e. in $\Omega$.
Therefore, the second term also converges to 0 thanks to the dominated
convergence theorem. It follows that $\nabla[f\circ\varphi_{k}]\to
f^{\prime}(u)\nabla u$ in $L^{2}(\Omega,w)$, and hence also in the sense of
distributions. We proved that $f\circ\varphi_{k}\to f\circ u$ and
$\nabla[f\circ\varphi_{k}]\to f^{\prime}(u)\nabla u\in L^{2}(\Omega,w)$ in the
sense of distributions, and so the distributional derivative of $f\circ u$
lies in $L^{2}(\Omega,w)$ and is equal to $f^{\prime}(u)\nabla u$. In
particular, $f\circ u\in W$. Note that we also proved that
$f\circ\varphi_{k}\to f\circ u$ in $W$.

In order to finish the proof of (a), we need to prove that $T(f\circ u)=f(Tu)$
$\sigma$-a.e. in $\Gamma$. If $v\in W$ is also a continuous function on
$\mathbb{R}^{n}$, then it is easy to check from the definition of the trace
that $Tv(x)=v(x)$ for every $x\in\Gamma$. Since $f\circ\varphi_{k}$ and
$\varphi_{k}$ are both continuous functions, we get that

(6.7) $f\circ\varphi_{k}(x)=T(f\circ\varphi_{k})(x)=f(T\varphi_{k}(x))\ \text{
for $x\in\Gamma$ and }k\in\mathbb{N}.$

Hence for every ball $B$ centered on $\Gamma$ and every $k\geq 0$,

(6.8) $\begin{split}\int_{B}|T(f\circ u)-f(Tu)|d\sigma&\leq\int_{B}|T(f\circ
u)-T(f\circ\varphi_{k})|d\sigma+\int_{B}|f(T\varphi_{k})-f(Tu)|d\sigma\\\
&\leq\int_{B}|T(f\circ
u)-T(f\circ\varphi_{k})|d\sigma+\sup|f^{\prime}|\int_{B}|T\varphi_{k}-Tu|d\sigma.\end{split}$

Recall that each convergence $\varphi_{k}\to u$ and $f\circ\varphi_{k}\to
f\circ u$ holds in both $W$ and $L^{1}_{loc}(\mathbb{R}^{n})$. The assertion
(5.16) then gives that both convergences $T\varphi_{k}\to Tu$ and
$T(f\circ\varphi_{k})\to T(f\circ u)$ hold in $L^{1}_{loc}(\Gamma,\sigma)$.
Thus the right-hand side of (6.8) converges to 0 as $k$ goes to $+\infty$. We
obtain that for every ball $B$ centered on $\Gamma$,

(6.9) $\int_{B}|T(f\circ u)-f(Tu)|d\sigma=0;$

in particular, $T(f\circ u)=f(Tu)$ $\sigma$-a.e. in $\Gamma$.

Let us turn to the proof of (b). Set $u^{+}=\max\\{u,0\\}$. Then
$\max\\{u,v\\}Â =(u-v)^{+}+v$ and $\min\\{u,v\\}=u-(u-v)^{+}$. Thus is it
enough to show that for any $u\in W$, $u^{+}$ lies in $W$ and satisfies

(6.10) $\nabla u^{+}(x)=\left\\{\begin{array}[]{ll}\nabla u(x)&\text{ if
}u(x)>0\\\ 0&\text{ if }u(x)\leq 0\end{array}\right.\ \text{ for almost every
}x\in\mathbb{R}^{n}$

and

(6.11) $T(u^{+})=(Tu)^{+}\ \text{ $\sigma$-almost everywhere on }\Gamma.$

Note that in particular (6.10) implies that $\nabla u=0$ a.e. in
$\\{u=\lambda\\}$. Indeed, since $u=\lambda+(u-\lambda)_{+}-(\lambda-u)_{+}$,
(6.10) implies that for almost every $x\in\Omega$,

(6.12) $\nabla u(x)=\left\\{\begin{array}[]{ll}\nabla u(x)&\text{ if
}u(x)\neq\lambda\\\ 0&\text{ if }u(x)=\lambda.\end{array}\right.$

Let us prove the claim (6.10). Define $f$ and $g={\mathds{1}}_{(0,+\infty)}$
by $f(t)=\max\\{0,t\\}$ and $g(t)=0$ when $t\leq 0$ and $g(t)=1$ when $t>0$.
Our aim is to approximate $f$ by an increasing sequence of $C^{1}$-functions
and then to conclude by using (a) and the monotone convergence theorem. Define
for any integer $j\geq 1$ the function $f_{j}$ by

(6.13) $f_{j}(t)=\left\\{\begin{array}[]{ll}0&\text{ if }t\leq 0\\\
\frac{j}{j+1}t^{\frac{j+1}{j}}&\text{ if }0\leq t\leq 1\\\
t-\frac{1}{j+1}&\text{ if }t\geq 1.\end{array}\right.$

Notice that $f_{j}$ is non-negative and $(f_{j})$ is a nondecreasing sequence
that converges pointwise to $f$. Consequently, $f_{j}\circ u\geq 0$ and
$(f_{j}\circ u)$ is a nondecreasing sequence that converges pointwise to
$f\circ u=u^{+}\in L^{1}_{loc}(\mathbb{R}^{n})$. The monotone convergence
theorem implies that $f_{j}\circ u\to u^{+}$ in $L^{1}_{loc}(\mathbb{R}^{n})$.

Moreover, $f_{j}$ lies in $C^{1}(\mathbb{R})$ and its derivative is

(6.14) $f^{\prime}_{j}(t)=\left\\{\begin{array}[]{ll}0&\text{ if }t\leq 0\\\
t^{\frac{1}{j}}&\text{ if }0\leq t\leq 1\\\ 1&\text{ if }t\geq
1.\end{array}\right.$

Thus $f^{\prime}_{j}$ is bounded and part (a) of the lemma implies $f_{j}\circ
u\in W$ and $\nabla(f_{j}\circ u)=f^{\prime}_{j}(u)\nabla u$ almost everywhere
on $\mathbb{R}^{n}$. In addition, $f^{\prime}_{j}$ converges to $g$ pointwise
everywhere, so

(6.15) $\nabla(f_{j}\circ u)=f^{\prime}_{j}(u)\nabla u\to v:=(g\circ u)\nabla
u=\left\\{\begin{array}[]{ll}\nabla u&\text{ if }u>0\\\ 0&\text{ if }u\leq
0\end{array}\right.$

almost everywhere (i.e., wherever $\nabla(f_{j}\circ
u)=f^{\prime}_{j}(u)\nabla u$). The convergence also occurs in
$L^{2}(\Omega,w)$ and in $L^{1}_{loc}(\mathbb{R}^{n})$, because
$|\nabla(f_{j}\circ u)|\leq|\nabla u|$ and by the dominated convergence
theorem, and therefore also in the sense of distribution. Since $f_{j}\circ u$
converges to $u^{+}$ pointwise almost everywhere and hence (by the dominated
convergence theorem again) in $L^{1}_{loc}$ and in the sense of distributions,
we get that $v=(g\circ u)\nabla u$ is the distribution derivative of $u^{+}$.
This completes the proof of (6.10).

Finally, let us establish (6.11). The plan is to prove that we can find smooth
functions $\varphi_{k}$ such that $\varphi_{k}^{+}$ converges (in
$L^{1}_{loc}(\Gamma,\sigma)$) to both $Tu^{+}$ and $(Tu)^{+}$. We claim that
for $u\in W$ and any sequence $(u_{k})$ in $W$, the following implication
holds true:

(6.16) $u_{k}\to u\ \text{ pointwise a.e. and in }W\ \Longrightarrow\
u_{k}^{+}\to u^{+}\ \text{ pointwise a.e. and in }W.$

First we assume the claim and prove (6.11). With the help of Lemma 5.3, take
$(\varphi_{k})_{k\in\mathbb{N}}$ be a sequence of functions in
$C^{\infty}(\mathbb{R}^{n})$ such that $\varphi_{k}\to u$ in $W$, and in
$L^{1}_{loc}(\mathbb{R}^{n})$. We may also replace $(\varphi_{k})$ by a
subsequence, and get that $\varphi_{k}\to u$ pointwise a.e. The claim (6.16)
implies that $\varphi_{k}^{+}\to u^{+}$ in $W$. In addition,
$\varphi_{k}^{+}\to u^{+}$ in $L^{1}_{loc}(\mathbb{R}^{n})$, for instance
because $\varphi_{k}$ tends to $u$ in $L^{1}_{loc}(\mathbb{R}^{n})$ and by the
estimate $|\varphi_{k}^{+}-u^{+}|\leq|\varphi_{k}-u|$.

Thus we may apply (5.16), and we get that $T\varphi_{k}^{+}$ tends to $Tu^{+}$
in $L^{1}_{loc}(\Gamma)$. Since $\varphi_{k}^{+}$ is continuous,
$T\varphi_{k}^{+}=\varphi_{k}^{+}$ and

(6.17) $\varphi_{k}^{+}$ tends to $Tu^{+}$ in $L^{1}_{loc}(\Gamma)$.

We also need to check that $\varphi_{k}^{+}$ converges to $(Tu)^{+}$. Notice
that (5.16) also implies that $\varphi_{k}\to Tu$ in
$L^{1}_{loc}(\Gamma,\sigma)$. Together with the easy fact that
$|a^{+}-b^{+}|\leq|a-b|$ for $a,b\in\mathbb{R}$, this proves that
$\varphi_{k}^{+}\to(Tu)^{+}$ in $L^{1}_{loc}(\Gamma,\sigma)$.

We just proved that $\varphi_{k}^{+}$ converges to both $T(u^{+})$ and
$(Tu)^{+}$ in $L^{1}_{loc}(\Gamma,\sigma)$. By uniqueness of the limit,
$T(u^{+})=(Tu)^{+}$ $\sigma$-a.e. in $\Gamma$, as needed for (6.11). Thus the
proof of the lemma will be complete as soon as we establish the claim (6.16).

First notice that $|u_{j}^{+}-u^{+}|\leq|u_{j}-u|$ and thus the a.e. pointwise
convergence of $u_{j}\to u$ yields the a.e. pointwise convergence
$u_{j}^{+}\to u^{+}$. Let $g$ denote the characteristic function of
$(0,+\infty)$; then by (6.10)

(6.18) $\begin{split}&\left(\int_{\Omega}|\nabla u_{j}^{+}-\nabla
u^{+}|^{2}dm\right)^{\frac{1}{2}}=\left(\int_{\Omega}|g(u_{j})\nabla
u_{j}-g(u)\nabla u|^{2}dm\right)^{\frac{1}{2}}\\\
&\qquad\qquad\qquad\leq\left(\int_{\Omega}|g(u_{j})[\nabla u_{j}-\nabla
u]|^{2}dm\right)^{\frac{1}{2}}+\left(\int_{\Omega}|\nabla
u[g(u_{j})-g(u)]|^{2}dm\right)^{\frac{1}{2}}\\\
&\qquad\qquad\qquad\leq\left(\int_{\Omega}|\nabla u_{j}-\nabla
u|^{2}dm\right)^{\frac{1}{2}}+\left(\int_{\Omega}|\nabla
u|^{2}|g(u_{j})-g(u)|^{2}dm\right)^{\frac{1}{2}}.\\\ \end{split}$

The first term in the right-hand side converges to 0 since $u_{j}\to u$ in
$W$. Call $I$ the second term; $I$ is finite, since $u\in W$ and
$|g(u_{j})-g(u)|\leq 1$. Moreover, thanks to (6.12), $\nabla u=0$ a.e. on
$\\{u=0\\}$. So the square of $I$ can be written

(6.19) $\begin{split}I^{2}&=\int_{\big{\\{}|u|>0\big{\\}}}|\nabla
u|^{2}|g(u_{j})-g(u)|^{2}dm.\end{split}$

Let $x\in\big{\\{}|u|>0\big{\\}}$ be such that $u_{j}(x)$ converges to
$u(x)\neq 0$; then there exists $j_{0}\geq 0$ such that for $j\geq j_{0}$ the
sign of $u_{j}(x)$ is the same as the sign of $u(x)$. That is, $g(u_{j})(x)$
converges to $g(u)(x)$. Since $u_{j}\to u$ a.e. in $\Omega$, the previous
argument implies that $g(u_{j})\to g(u)$ a.e. in $\\{|u|>0\\}$. Then $I^{2}$
converges to $0$, by the dominated convergence theorem. Going back to (6.18),
we obtain that $u_{j}^{+}\to u^{+}$ in $W$, which concludes our proof of
(6.16); Lemma 6.1 follows. $\Box$

######  Lemma 6.3.

Let $u,v\in W\cap L^{\infty}(\Omega)$. Then $uv\in W\cap L^{\infty}(\Omega)$,
$\nabla(uv)=v\nabla u+u\nabla v$, and $T(uv)=Tu\cdot Tv$.

Proof. Without loss of generality, we can assume that
$\|u\|_{\infty},\|v\|_{\infty}\leq 1$. The fact that $uv\in
L^{\infty}(\Omega)$ is immediate. Let us now prove that $uv\in W$. According
to Lemma 5.3, there exists two sequences $(u_{j})_{j\in\mathbb{N}}$ and
$(v_{j})_{j\in\mathbb{N}}$ of functions in $C^{\infty}(\mathbb{R}^{n})\cap W$
such that $u_{j}\to u$ and $v_{j}\to v$ in $W$, in
$L^{1}_{loc}(\mathbb{R}^{n})$, and pointwise. Besides, the construction of
$u_{j},v_{j}$ given by Lemma 5.3 allows us to assume that
$\|u_{j}\|_{\infty}\leq\|u\|_{\infty}\leq 1$ and
$\|v_{j}\|_{\infty}\leq\|v\|_{\infty}\leq 1$. The distributional derivative of
$u_{j}v_{j}$ equals the classical derivative, which is

(6.20) $\nabla(u_{j}v_{j})=v_{j}\nabla u_{j}+u_{j}\nabla v_{j}.$

Since $u_{j}$ and $v_{j}$ lie in $W$, (6.20) says that $u_{j}v_{j}\in W$. The
bound

(6.21)
$\begin{split}\int_{B}|u_{j}v_{j}-uv|\leq\int_{B}|u_{j}||v_{j}-v|+\int_{B}|v||u_{j}-u|\leq\|v_{j}-v\|_{L^{1}(B)}+\|u_{j}-u\|_{L^{1}(B)},\end{split}$

which holds for any ball $B\subset\mathbb{R}^{n}$, shows that $u_{j}v_{j}\to
uv$ in $L^{1}_{loc}(\mathbb{R}^{n})$. Moreover,

(6.22) $\begin{split}&\left(\int_{B}|(u_{j}\nabla v_{j}+v_{j}\nabla
u_{j})-(u\nabla v+v\nabla u)|^{2}dm\right)^{\frac{1}{2}}\\\
&\qquad\qquad\leq\left(\int_{B}|u_{j}\nabla v_{j}-u\nabla
v|^{2}dm\right)^{\frac{1}{2}}+\left(\int_{B}|v_{j}\nabla u_{j}-v\nabla
u|^{2}dm\right)^{\frac{1}{2}}\\\
&\qquad\qquad\leq\left(\int_{B}|u_{j}|^{2}|\nabla v_{j}-\nabla
v|^{2}dm\right)^{\frac{1}{2}}+\left(\int_{B}|u_{j}-u|^{2}|\nabla
v|^{2}dm\right)^{\frac{1}{2}}\\\
&\qquad\qquad\qquad\qquad+\left(\int_{B}|v_{j}|^{2}|\nabla u_{j}-\nabla
u|^{2}dm\right)^{\frac{1}{2}}+\left(\int_{B}|v_{j}-v|^{2}|\nabla
u|^{2}dm\right)^{\frac{1}{2}}.\\\ \end{split}$

The first and third terms in the right-hand side converge to 0 as $j$ goes to
$+\infty$, because $|u_{j}|,|v_{j}|\leq 1$ and since $u_{j}\to u$ and
$v_{j}\to v$ in $W$. The second and forth terms also converge to 0 thanks to
the dominated convergence theorem (and the fact that $u_{j}\to u$ and
$v_{j}\to v$ pointwise a.e.). We deduce that $\nabla(u_{j}v_{j})=u_{j}\nabla
v_{j}+v_{j}\nabla u_{j}\to u\nabla v+v\nabla u$ in $L^{2}(\Omega,w)$. By the
uniqueness of the distributional derivative, $\nabla(uv)=u\nabla v+v\nabla
u\in L^{2}(\Omega,w)$. In particular, $uv\in W$. Note that we also proved that
$u_{j}v_{j}\to uv$ in $W$.

It remains to prove that $T(uv)=Tu\cdot Tv$. Since $u_{j}v_{j}$ is continuous
and $u_{j}v_{j}\to uv$ in $W$ and $L^{1}_{loc}(\mathbb{R}^{n})$, then by
(5.16), $u_{j}v_{j}=T(u_{j}v_{j})\to T(uv)$ in $L^{1}_{loc}(\Gamma,\sigma)$.
Moreover, for any ball $B$ centered on $\Gamma$,

(6.23) $\begin{split}\int_{B}|u_{j}v_{j}-Tu\cdot
Tv|d\sigma&\leq\int_{B}|u_{j}||v_{j}-Tv|d\sigma+\int_{B}|u_{j}-Tu||Tv|d\sigma\\\
&\leq\int_{B}|v_{j}-Tv|d\sigma+\int_{B}|u_{j}-Tu|d\sigma\\\ \end{split}$

where the last line holds because $|u_{j}|\leq 1$ and $|Tv|\leq\sup|v|\leq 1$,
where the later bound either follows from Lemma 6.1 or is easily deduced from
the definition of the trace. By construction, $u_{j}\to u$ and $v_{j}\to v$ in
$W$ and $L^{1}_{loc}(\mathbb{R}^{n})$. Then by (5.16) the right-hand side
terms in (6.23) converge to 0.

We proved that $u_{j}v_{j}$ converges in $L^{1}_{loc}(\Gamma,\sigma)$ to both
$T(uv)$ and $Tu\cdot Tv$. By uniqueness of the limit, $T(uv)=Tu\cdot Tv$
$\sigma$-a.e. in $\Gamma$. Lemma 6.3 follows. $\Box$

##  Chapter 7 The Extension Operator

The main point of this section is the construction of our extension operator
$E:H\to W$, which will be done naturally, with the Whitney extension scheme
that uses dyadic cubes.

Our main object will be a function $g$ on $\Gamma$, that typically lies in $H$
or in $L^{1}_{loc}(\Gamma,\sigma)$. We start with the Lebesgue density result
for $g\in L^{1}_{loc}(\Gamma,\sigma)$ that was announced in the introduction.

######  Lemma 7.1.

For any $g\in L^{1}_{loc}(\Gamma,\sigma)$ and $\sigma$-almost all
$x\in\Gamma$,

(7.1) $\lim_{r\to 0}\fint_{\Gamma\cap B(x,r)}|g(y)-g(x)|d\sigma(y)=0.$

Proof. Since $(\Gamma,\sigma)$ satisfies (1.1), the space $(\Gamma,\sigma)$
equipped with the metric induced by $\mathbb{R}^{n}$ is a doubling space.