# On the semi-additivity of the $1/2$-symmetric caloric capacity in the plane

Joan Hernández, Joan Mateu, and Laura Prat

###### Abstract.

In this paper we study properties of a variant of the $1/2$-caloric capacity,
called $1/2$-symmetric caloric capacity. The latter is associated
simultaneously with the $1/2$-fractional heat equation and its conjugate. We
establish its semi-additivity in $\mathbb{R}^{2}$ and, moreover, we compute
explicitly the $1/2$-symmetric caloric capacity of rectangles, which
illustrates its anisotropic behaviour.

AMS 2020 Mathematics Subject Classification: 42B20 (primary); 28A12
(secondary).

Keywords: Fractional heat equation, singular integrals, caloric capacity.

J. H. has been supported by PID2020-114167GB-I00 (Mineco, Spain). J. M. has
been partially supported by 2021SGR-00071 (Catalonia) and PID2020-112881GB-I00
(Mineco, Spain). L.P. has been partially supported by 2021SGR-00071
(Catalonia) and PID2020-114167GB-I00 (Mineco, Spain)

##  1\. Introduction

The study of removable subsets for caloric equations done by Mateu, Prat &
Tolsa in [MPrTo] and by Mateu & Prat in [MPr] introduces, naturally, caloric
capacities associated to such PDE’s. In the first reference
$(1,1/2)$-Lipschitz caloric capacity was introduced, as well as the notion of
equivalence between the nullity of this quantity and the removability of
compact subsets for the heat equation, i.e. the one associated with the
differential operator $\Theta:=(-\Delta_{x})+\partial_{t}$, where
$(x,t)\in\mathbb{R}^{n}\times\mathbb{R}$. In [MPr], Mateu & Prat studied the
corresponding caloric capacities associated with the fractional heat equation.
That is, the equation associated with the pseudo-differential operator
$\Theta^{s}:=(-\Delta_{x})^{s}+\partial_{t}$, for $0<s<1$. The authors
distinguished the cases $s=1/2$, $1/2<s<1$ and $0<s<1/2$, focusing mainly on
the first. The study for $s=1/2$ was quite successful, obtaining a
removability result for the $\Theta^{1/2}$-equation, as it was done for the
heat equation. For instance, if $f$ is a solution of the $1/2$-heat equation
in $\mathbb{R}^{n+1}\setminus{E}$ satisfying

$\|f\|_{L^{\infty}(\mathbb{R}^{n+1})}<\infty,$

then $E$ will be removable if and only if the $1/2$-caloric capacity of $E$ is
null. One of the main features that characterized the case $s=1/2$ is that the
authors could work with an explicit fundamental solution for the operator
$\Theta^{1/2}$, namely

$P(x,t):=\frac{t}{\big{(}|x|^{2}+t^{2}\big{)}^{(n+1)/2}}\chi_{\\{t>0\\}}(x,t),$

a continuous function in $\mathbb{R}^{n+1}\setminus{\\{0\\}}$, harmonic in
$\mathbb{R}^{n+1}\setminus{\\{t=0\\}}$. This, together with the fact that the
Fourier symbols of $(-\Delta_{x})^{1/2}$ and $\partial_{t}$ share the same
homogeneity, Mateu & Prat were able to characterize the removability of
compact sets in a simpler way than the cases $s\neq 1/2$. More precisely, they
established that the removability condition for $E$ can be studied by
controlling the supremum of expressions of the form $|\langle T,1\rangle|$,
where $T$ is a distribution supported on $E$ satisfying the normalization
condition $\|P\ast T\|_{L^{\infty}(\mathbb{R}^{n+1})}\leq 1$. The latter
supremum is known as the $1/2$-caloric capacity of $E$, written
$\gamma_{\Theta^{1/2}}(E)$. On the other hand, for $s\neq 1/2$, although there
is a fundamental solution $P_{s}(x,t)$, its expression is no longer explicit.
However, there are precise estimates computed by the probabilists Blumenthal &
Getoor [BlGe] who described the power-like tail behaviour of $P_{s}$ and
allowed the authors to carry out a similar study. Nevertheless, there was
still a technical obstacle: the fact that the Fourier symbols of
$(-\Delta_{x})^{s}$ and $\partial_{t}$ do not share homogeneity. This feature,
essentially, ended up by limiting the amount of results obtained.
In the present paper we aim at obtaining a more precise description of a
smaller variant of the $1/2$-caloric capacity, introduced in [MPr, §4] only in
the context of positive measures. Denoted by
$\widetilde{\gamma}_{\Theta^{1/2}}$, we call it $1/2$-symmetric caloric
capacity and it is the supremum of expressions of the form $|\langle
T,1\rangle|$, where $T$ is a distribution supported on a compact set now
satisfying both conditions

$\|P\ast T\|_{L^{\infty}(\mathbb{R}^{n+1})}\leq 1\quad\text{ and
}\quad\|P^{\ast}\ast T\|_{L^{\infty}(\mathbb{R}^{n+1})}\leq 1,$

where $P^{\ast}(x,t):=P(-x,-t)$ is the conjugate kernel of $P$. If the
supremum is only taken among positive Borel measures (with the same
normalization conditions), we will write
$\widetilde{\gamma}_{\Theta^{1/2},+}$. The main result of this text is found
in Section 6, where we prove:

######  Theorem.

For $E\subset\mathbb{R}^{2}$ compact,

$\widetilde{\gamma}_{\Theta^{1/2}}(E)\approx\gamma_{\Theta^{1/2},+}(E).$

Such result implies, bearing in mind the results of [He], the semi-additivity
of $\widetilde{\gamma}_{\Theta^{1/2}}$ in $\mathbb{R}^{2}$ (Theorem 6.8). We
also give a result in the general $\mathbb{R}^{n+1}$ setting, also proved in
Section 6, where the same type of estimate is established for the typical
multidimensional corner-like Cantor set. Sections 2, 3, 4 and 5 are devoted to
present the notation and preliminary results necessary to prove the above
estimates. In Section 2 we introduce basic terminology and properties, in
Section 3 we prove Theorem 3.5, which encapsulates all the different ways to
define, equivalently, $\gamma_{\Theta^{1/2},+}$. In Section 4 we use a
particular characterization of the latter via a variational approach. With it,
we construct a Whitney decomposition of a certain family of compact sets, so
that we gain control of their $\widetilde{\gamma}_{\Theta^{1/2}}$ capacity.
Finally, in Section 5, we prove Theorem 5.19, a general comparability result
between $\widetilde{\gamma}_{\Theta^{1/2}}$ and $\gamma_{\Theta^{1/2}}$ if an
additional assumption $\text{{{A}}}_{\boldsymbol{3}}$ is satisfied. In
general, the arguments carried out are influenced by those of Tolsa [To3,
Ch.5] and Volberg [Vo], where the same type of comparability results are
studied for analytic and Lipschitz harmonic capacity respectively.
In our setting, taking into account that $P$ is not an anti-symmetric kernel,
such arguments can be applied with some modifications inspired by the
arguments presented to prove general $Tb$-theorems such as those found in
[NTVo2] and [HyMar]. The fact that $P$ is harmonic outside the hyperplane
$\\{t=0\\}$, an $\pazocal{L}^{n+1}$-null set, and that satisfies being an
$n$-dimensional Calderón-Zygmund kernel [MPr, Lemma 2.1] are essentially the
features that allow to carry out similar arguments. Finally, Section 7 is
devoted to the computation of the $\widetilde{\gamma}_{\Theta^{1/2}}$ capacity
of a rectangle $R\subset\mathbb{R}^{2}$, obtaining:

$\widetilde{\gamma}_{\Theta^{1/2}}(R)\approx\ell_{t}\,\Bigg{[}\frac{1}{2}\ln\bigg{(}1+\frac{\ell_{t}^{2}}{\ell_{x}^{2}}\bigg{)}+\frac{\ell_{t}}{\ell_{x}}\arctan\bigg{(}\frac{\ell_{x}}{\ell_{t}}\bigg{)}\Bigg{]}^{-1},$

where $\ell_{x}$ and $\ell_{t}$ are the horizontal and vertical side lengths
of $R$. The above behavior differs substantially, for example, with that
obtained for $\gamma(R)$, the analytic capacity of a rectangle. For the latter
one has, as a consequence of the $1/4$-Koebe theorem,
$\gamma(R)\approx\text{diam}(R)$ (see, for example, [To3, Proposition 1.5]),
which is a significantly simpler expression.
Let us also stress one last feature of the above capacities associated to the
$1/2$-fractional heat equations: the kernels $P$ and $P^{\ast}$ are both
nonnegative. This suggests that, maybe, using classical arguments of potential
theory (see [La, Ch.I & II], [Ra, Ch.3] or [Ki]), the comparability between
capacities defined through positive measures and distributions should follow
by a simpler argument similar to that of [Ve, p.10], provided that one is able
to prove the existence of an equilibrium measure. The author has not been able
to deduce its existence for $\gamma_{\Theta^{1/2},+}$, the main obstacle being
the incapability of obtaining an equivalent formulation of the latter capacity
in terms of an (inverted) infimum over energies. The following simple example
already illustrates some problems that arise by proceeding this way: pick
$E\subset\mathbb{R}^{2}$ the horizontal line segment $[0,1]\times\\{0\\}$ and
$\mu:=\pazocal{H}^{1}|_{E}$. In [MPr, §6] the authors bound explicitly the
potential $P\ast\mu$ by $\pi$, obtaining that
$\gamma_{\Theta,+}(E)\geq\pi^{-1}$. However, by the definition of $P$ and the
choice of $\mu$, it is also clear that $P\ast\mu(x,0)=0$, for any $x\in[0,1]$.
Therefore, if we were to compute the energy of $\mu$ we would obtain

$I[\mu]:=\int
P\ast\mu\mathop{}\\!\mathrm{d}\mu=\int_{0}^{1}P\ast\mu(x,0)\mathop{}\\!\mathrm{d}\pazocal{H}^{1}(x)=0,$

which would imply that the $\gamma_{\Theta^{1/2},+}$ capacity of $E$ is
infinite. The latter example also shows that the potentials $P\ast\mu$ do not
obey the so called maximum principle, i.e. they do not attain their maximum
values at the support of $\mu$. To avoid this problem, the author has tried to
work with the auxiliary potentials
$U_{\mu}(x,t):=\limsup_{(y,s)\to(x,t)}P\ast\mu(y,s)$. For these the maximum
principle holds (this can be deduced, essentially, from the fact that $P$ is
subharmonic in $\mathbb{R}^{n+1}\setminus{\\{0\\}}$). However, the potentials
$U_{\mu}$ still lack the continuity principle, i.e.: if $U_{\mu}$ restricted
to $\text{supp}(\mu)$ is continuous, then $U_{\mu}$ is continuous everywhere
else; which is an essential tool in order to carry out the construction of
possible equilibrium measures. In any case, the above classical methods were
finally discarded and we chose to follow arguments similar to those of [To3,
Ch.5] and [Vo].

About the notation used in the sequel: absolute positive constants will be
those that can depend on the dimension of the ambient space and possibly on
the Calderón-Zygmund (C-Z) constants of $P$, and whose value may change at
different occurrences. The notation $A\lesssim B$ means that there exists such
a constant $C$, so that $A\leq CB$. Moreover, $A\approx B$ is equivalent to
$A\lesssim B\lesssim A$. Also, $A\simeq B$ will mean $A=CB$. We will write
$B_{r}(x,t)$ to denote the usual Euclidean ball in $\mathbb{R}^{n+1}$ of
radius $r>0$ and center $(x,t)$. We also emphasize that the gradient symbol
$\nabla$ will refer to $(\nabla_{x},\partial_{t})$, with $x\in\mathbb{R}^{n}$
and $t\in\mathbb{R}$.

##  2\. Notation and basic definitions

Our ambient space will be $\mathbb{R}^{n+1}$, and a generic point will be
denoted as $\overline{x}=(x,t)\in\mathbb{R}^{n+1}$, where $x\in\mathbb{R}^{n}$
will be usually referred to as the spatial variable, and $t\in\mathbb{R}$ the
time variable. Most of the results we will present in this text will be valid
in a general multidimensional context, i.e. $n$ is any integer bigger or equal
than $1$. However, some of them will only hold in the planar setting $(n=1)$,
and the author will make clear when this is the case. Let $\Theta^{1/2}$ be
the $1/2$-heat operator,

$\Theta^{1/2}:=(-\Delta)^{1/2}+\partial_{t},$

where $(-\Delta)^{1/2}=(-\Delta_{x})^{1/2}$ is a pseudo-differential operator
known as the $1/2$-Laplacian with respect to the spatial variable. It may be
defined through its Fourier transform,

$\widehat{(-\Delta)^{1/2}}f(\xi,t)=|\xi|\widehat{f}(\xi,t),$

or by its integral representation

$\displaystyle(-\Delta)^{1/2}f(x,t)$
$\displaystyle\simeq\text{p.v.}\int_{\mathbb{R}^{n}}\frac{f(x,t)-f(y,t)}{|x-y|^{n+1}}\text{d}\pazocal{L}^{n}(y)$
$\displaystyle\simeq\int_{\mathbb{R}^{n}}\frac{f(x+y,t)-2f(x,t)+f(x-y,t)}{|y|^{n+1}}\text{d}\pazocal{L}^{n}(y).$

The reader may find more details about the properties of such operator in
[DPV, §3] or [St]. Borrowing the notation of [MPr], let $P$ be the fundamental
solution of the $1/2$-heat equation in $\mathbb{R}^{n+1}$, which is given by
[Va, Eq. 2.2]

$P(\overline{x})=\frac{t}{\big{(}t^{2}+|x|^{2}\big{)}^{(n+1)/2}}\,\chi_{\\{t>0\\}}(\overline{x})=\frac{t}{|\overline{x}|^{n+1}}\,\chi_{\\{t>0\\}}(\overline{x}).$

where $\chi$ is the usual indicator function. Notice that the previous kernel
is continuous in $\mathbb{R}^{n+1}\setminus{\\{0\\}}$ and not differentiable
at any point of the hyperplane $\\{t=0\\}$. Another fundamental function that
will frequently appear is $P^{\ast}$,

$P^{\ast}(\overline{x}):=P(-\overline{x})=\frac{-t}{|\overline{x}|^{n+1}}\,\chi_{\\{t<0\\}}(\overline{x}).$

More generally, for any function $f:\mathbb{R}^{n+1}\to\mathbb{R}$, the
notation $f^{\ast}$ will always denote
$f^{\ast}(\overline{x}):=f(-\overline{x})$. Hence $f$ being an even function
is equivalent to $f=f^{\ast}$. We will also extend this notion for
distributions as follows: given $T$ any distribution in $\mathbb{R}^{n+1}$, we
define $T^{\ast}$ to be the distribution acting on test functions as

$\langle T^{\ast},\varphi\rangle:=\langle
T,\varphi^{\ast}\rangle,\quad\forall\varphi\in\pazocal{C}^{\infty}_{c}(\mathbb{R}^{n+1}).$

We will say that a distribution is even if $T=T^{\ast}$. It is not hard to
check (approximating via test functions and using the associativity of [Ca,
Theorem 8.15]) that for any $T$ distribution and $S$ distribution with compact
support,

(2.1) $\langle T\ast S,\varphi\rangle=\langle
S,T^{\ast}\ast\varphi\rangle,\quad\forall\varphi\in\pazocal{C}^{\infty}_{c}(\mathbb{R}^{n+1}).$

With this notions we observe that on the one hand, for any
$\overline{x}=(x,t)\in\mathbb{R}^{n+1}$ we have

$\displaystyle(-\Delta)^{1/2}$ $\displaystyle
P^{\ast}(\overline{x})=C^{\prime}\int_{\mathbb{R}^{n}}\frac{P^{\ast}(x+y,t)-2P^{\ast}(x,t)+P^{\ast}(x-y,t)}{|y|^{n+1}}\text{d}\pazocal{L}^{n}(y)$
$\displaystyle=C^{\prime}\int_{\mathbb{R}^{n}}\frac{P(-x-y,-t)-2P(-x,-t)+P(-x+y,-t)}{|y|^{n+1}}\text{d}\pazocal{L}^{n}(y)=(-\Delta)^{1/2}P(-\overline{x}).$

That is, we have

$(-\Delta)^{1/2}P^{\ast}=\big{[}(-\Delta)^{1/2}P\big{]}^{\ast}.$

On the other hand, understanding $\partial_{t}P$ distributionally (that makes
sense, since $P$ is a locally integrable function in $\mathbb{R}^{n+1}$ and
thus a distribution) we have, for any $\varphi$ test function in
$\mathbb{R}^{n+1}$,

$\langle(\partial_{t}P)^{\ast},\varphi\rangle:=\langle\partial_{t}P,\varphi^{\ast}\rangle:=-\langle
P,\partial_{t}\varphi^{\ast}\rangle=\langle
P,(\partial_{t}\varphi)^{\ast}\rangle=\langle
P^{\ast},\partial_{t}\varphi\rangle=:-\langle\partial_{t}P^{\ast},\varphi\rangle.$

Therefore, the following distributional identity holds

$(\partial_{t}P)^{\ast}=-\partial_{t}P^{\ast}.$

Hence, if we define the operator

$\overline{\Theta}^{1/2}:=(-\Delta)^{1/2}-\partial_{t},$

we have that

$\overline{\Theta}^{1/2}P^{\ast}=\big{[}\Theta^{1/2}P\big{]}^{\ast}=\delta_{0}^{\ast}=\delta_{0},$

implying that $P^{\ast}$ is the fundamental solution of
$\overline{\Theta}^{1/2}$. We will refer to $P^{\ast}$ and
$\overline{\Theta}^{1/2}$ as the conjugates of $P$ and $\Theta^{1/2}$
respectively.
We recall the definition of the $1/2$-caloric capacity and some of its
fundamental variants introduced in [MPr]. For $E\subset\mathbb{R}^{n+1}$
compact, define its $1/2$-caloric capacity as

$\gamma_{\Theta^{1/2}}(E)=\sup|\langle T,1\rangle|,$

where the supremum is taken over all distributions $T$ with
$\text{supp}(T)\subset E$ satisfying

$\|P\ast T\|_{\infty}:=\|P\ast T\|_{L^{\infty}(\mathbb{R}^{n+1})}\leq 1.$

Such distributions will be called admissible for $\gamma_{\Theta^{1/2}}(E)$.
In [MPr, Lemma 3.4] it is proved that for a distribution $T$ in
$\mathbb{R}^{n+1}$ with $\|P\ast T\|_{\infty}\leq 1$, if $\varphi$ is a
$\pazocal{C}^{1}$ function supported on $Q\subset\mathbb{R}^{n+1}$ a cube,
with $\|\nabla\varphi\|_{\infty}\leq\ell(Q)^{-1}$, then

$|\langle T,\varphi\rangle|\leq C\ell(Q)^{n},$

for some absolute constant $C>0$, that we may assume $C\geq 1$ without loss of
generality. If the previous property holds for a distribution $T$, we say that
$T$ has $n$-growth (with constant $C$). It is easy to see that the previous
definition of growth for a distribution generalizes the usual notion of growth
if $T$ coincides with a positive measure. Recall that a positive measure $\mu$
is said to have $n$-growth with constant $C$ if

$\mu\big{(}B(\overline{x},r)\big{)}\leq Cr^{n},\hskip 14.22636pt\text{for all
}\,\overline{x}\in\mathbb{R}^{n+1},\,r>0.$

It is clear that the above properties are invariant if formulated using cubes
instead of balls.
The fact that $C$ depends only on the dimension of the ambient space and the
C-Z constants of $P$, motivates the following redefinition of
$\gamma_{\Theta^{1/2}}$, that will be the one used throughout the whole text:

######  Definition 2.1 ($1/2$-caloric capacity).

For a compact subset $E\subset\mathbb{R}^{n+1}$, define its $1/2$-caloric
capacity as

$\gamma(E)=\gamma_{\Theta^{1/2}}(E):=\sup|\langle T,1\rangle|,$

with the supremum taken over distributions satisfying $\|P\ast
T\|_{\infty}\leq 1$ and belonging to $\pazocal{T}(E)$, the set of
distributions in $\mathbb{R}^{n+1}$ with $\text{supp}(T)\subset E$ and
$n$-growth with constant 1.

We also define the ($1/2,+$)-caloric capacity, denoted by
$\gamma_{\Theta^{1/2},+}$, in the same way as $\gamma_{\Theta^{1/2}}$, but
with the supremum only taken with respect to positive measures. More
precisely,

$\gamma_{+}(E)=\gamma_{\Theta^{1/2},+}(E):=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,\|P\ast\mu\|_{\infty}\leq
1\big{\\}},$

where $\Sigma(E)$ is the collection of positive Borel measures supported on
$E$, with $n$-growth with constant 1. We also define the following capacity:

$\gamma^{\prime}_{+}(E)=\gamma_{\Theta^{1/2},+}^{\prime}(E):=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,\|P\ast\mu\|_{L^{\infty}(\mu)}\leq
1\big{\\}}.$

Analogous definitions are associated with the operator
$\overline{\Theta}^{1/2}$, giving rise to the objects

$\overline{\gamma}=\gamma_{\overline{\Theta}^{1/2}},\quad\overline{\gamma}_{+}=\gamma_{\overline{\Theta}^{1/2},+}\quad\text{and}\quad\overline{\gamma}^{\prime}_{+}=\gamma_{\overline{\Theta}^{1/2},+}^{\prime}.$

In some of the results presented in [MPr, §3], one encounters expressions of
the form $\langle T,\varphi\rangle$, where $\varphi$ is a compactly supported
$\pazocal{C}^{1}$ function, and $T$ is a compactly supported distribution
satisfying $\|P\ast T\|_{\infty}\leq 1$. This last estimate and the fact that
$(-\Delta_{x})^{1/2}\varphi\in L^{1}(\mathbb{R}^{n+1})$, proved in [MPr, §3],
allows to give meaning to $\langle T,\varphi\rangle$ (that a priori may not
have sense since $\varphi$ is not a test function) as follows:

$\langle T,\varphi\rangle:=\big{\langle}P\ast
T,\overline{\Theta}^{1/2}\varphi\big{\rangle}=\big{\langle}P\ast
T,(-\Delta_{x})^{1/2}\varphi+\partial_{t}\varphi\big{\rangle}.$

We need to extend such definition to a slightly wider class of functions. We
do not claim that such class will be the largest where $\langle
T,\varphi\rangle$ can be defined if $T$ is a compactly supported distribution
with $\|P\ast T\|_{\infty}\leq 1$, but it will suffice for our purposes.

######  Definition 2.2.

Let $Q\subset\mathbb{R}^{n+1}$ be a cube, $\pazocal{N}\subset\mathbb{R}^{n+1}$
a $\pazocal{L}^{n+1}$-null set, and $\varphi:\mathbb{R}^{n+1}\to\mathbb{R}$ a
function. For $m\in\mathbb{N}\cup\\{\infty\\}$ we will write

$\varphi\in\pazocal{C}_{c}(Q)\cap\Big{(}\pazocal{C}_{x}^{m}(Q),\pazocal{C}_{t}^{m}(Q\setminus{\pazocal{N}})\Big{)},$

or simply

$\varphi\in\pazocal{C}^{m}_{c,\pazocal{N}}(Q),$

if $\varphi$ is continuous and compactly supported on $Q$, $m$ times
continuously differentiable with respect to the spatial variables, and $m$
times continuously differentiable with respect to the temporal variable $t$
except for a null set $\pazocal{N}$ (where it may not even be differentiable).
We also write

$\varphi\in\pazocal{C}(Q)\cap\Big{(}\pazocal{C}_{x}^{m}(Q),\pazocal{C}_{t}^{m}(Q\setminus{\pazocal{N}})\Big{)}=\pazocal{C}^{m}_{\pazocal{N}}(Q),$

if $\varphi$ satisfies the above regularity properties on $Q$, but we do not
require it to be compactly supported there (in fact, it may even have a larger
domain of definition). In the sequel we will only be interested in the case
$m=\infty$.

For $\varphi\in\pazocal{C}^{m}_{c,\pazocal{N}}(Q)$ one can define $\langle
T,\varphi\rangle$ as above, assuming $\|P\ast T\|_{\infty}\leq 1$ and $T$
compactly supported. Indeed, observe that we only have to give meaning to
$\langle P\ast T,(-\Delta_{x})^{1/2}\varphi\rangle$ and $\langle P\ast
T,\partial_{t}\varphi\rangle$. The former term can be defined as in [MPr, §3],
where the integrability of $(-\Delta_{x})^{1/2}\varphi$ is proved using only
regularity assumptions over the spatial variables. Regarding the term $\langle
P\ast T,\partial_{t}\varphi\rangle$, since $P\ast T\in
L^{\infty}(\mathbb{R}^{n+1})$, it can be simply defined as

$\langle P\ast
T,\partial_{t}\varphi\rangle:=\int_{\text{supp}(\varphi)\setminus{\pazocal{N}}}P\ast
T(\overline{x})\partial_{t}\varphi(\overline{x})\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}(\overline{x}),$

since the set $\pazocal{N}$ has null $\pazocal{L}^{n+1}$-measure.

######  Remark 2.1.

We notice that if for $\varphi\in\pazocal{C}_{c,\pazocal{N}}^{1}(Q)$ it also
happens that $\|\nabla\varphi\|_{\infty}\leq\ell(Q)^{-1}$, then [MPr,
Corollary 3.3] and [MPr, Lemma 3.5] also hold with exactly the same proofs for
such $\varphi$; and therefore also do [MPr, Theorem 3.1] (localization of
potentials) and [MPr, Lemma 3.4] ($n$-growth of admissible distributions).
Here, condition $\|\nabla\varphi\|_{\infty}\leq\ell(Q)^{-1}$ must be
understood as: for every test function $\psi$,

$|\langle\nabla\varphi,\psi\rangle|=\bigg{\rvert}\int_{Q}\varphi(\nabla\psi)\,\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}\bigg{\rvert}=\bigg{\rvert}\int_{Q\setminus{\pazocal{{N}}}}(\nabla\varphi)\psi\,\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}\bigg{\rvert}\leq\ell(Q)^{-1}\|\psi\|_{L^{1}}.$

To end this introductory section, we define more capacities that are already
presented in [MPr] and some other auxiliary ones. First the so called
1/2-symmetric caloric capacities, where the normalization conditions for the
potentials against $P$ and $P^{\ast}$ are both required:

$\displaystyle\widetilde{\gamma}(E)=\widetilde{\gamma}_{\Theta^{1/2}}(E)$
$\displaystyle:=\sup\big{\\{}|\langle
T,1\rangle|\,:\,T\in\pazocal{T}(E),\,\|P\ast T\|_{\infty}\leq 1,\|P^{\ast}\ast
T\|_{\infty}\leq 1\big{\\}},$
$\displaystyle\widetilde{\gamma}_{+}(E)=\widetilde{\gamma}_{\Theta^{1/2},+}(E)$
$\displaystyle:=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,\|P\ast\mu\|_{\infty}\leq
1,\|P^{\ast}\ast\mu\|_{\infty}\leq 1\big{\\}},$
$\displaystyle\widetilde{\gamma}^{\prime}_{+}(E)=\widetilde{\gamma}_{\Theta^{1/2},+}^{\prime}(E)$
$\displaystyle:=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,\|P\ast\mu\|_{L^{\infty}(\mu)}\leq
1,\|P^{\ast}\ast\mu\|_{L^{\infty}(\mu)}\leq 1\big{\\}}.$

We also have the following auxiliary capacities, that we only consider in the
context of positive measures:

$\displaystyle\gamma_{\text{{sy}},+}(E)$
$\displaystyle:=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,\|P_{\text{{sy}}}\ast\mu\|_{\infty}\leq
1\big{\\}},$ $\displaystyle\gamma_{\text{{sy}},+}^{\prime}(E)$
$\displaystyle:=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,\|P_{\text{{sy}}}\ast\mu\|_{L^{\infty}(\mu)}\leq
1\big{\\}},$

where $P_{\text{{sy}}}$ is the symmetric part of $P$, that is,

$P_{\text{{sy}}}(\overline{x}):=\frac{P(\overline{x})+P(-\overline{x})}{2}=\frac{|t|}{2|\overline{x}|^{n+1}}.$

Finally, we also consider the following capacities defined via a normalization
condition that involves an $L^{2}(\mu)$ operator bound:

$\displaystyle\gamma_{\text{op}}(E)=\gamma_{\Theta^{1/2},\text{op}}(E)$
$\displaystyle:=\sup\big{\\{}\mu(E)\;:\;\mu\in\Sigma(E),\;\|\pazocal{P}_{\mu}\|_{L^{2}(\mu)\to
L^{2}(\mu)}\leq 1\big{\\}},$ $\displaystyle\gamma_{\text{sy},\text{op}}(E)$
$\displaystyle:=\sup\big{\\{}\mu(E)\;:\;\mu\in\Sigma(E),\;\|\pazocal{P}_{\text{sy},\mu}\|_{L^{2}(\mu)\to
L^{2}(\mu)}\leq 1\big{\\}},$

where $\pazocal{P}_{\mu}$ and $\pazocal{P}_{\text{sy},\mu}$ are the
convolution operators associated to $P$ and $P_{\text{{sy}}}$ with respect to
the measure $\mu$. For any $\varepsilon>0$, we will also denote by
$\pazocal{P}_{\mu,\varepsilon}$ and $\pazocal{P}_{\text{sy},\mu,\varepsilon}$
the convolution operators associated to the truncated kernels
$P(\overline{x})\chi_{\\{|\overline{x}|>\varepsilon\\}}$ and
$P_{\text{{sy}}}(\overline{x})\chi_{\\{|\overline{x}|>\varepsilon\\}}$
respectively. With this definition, we will understand

$\|\pazocal{P}_{\mu}\|_{L^{2}(\mu)\to
L^{2}(\mu)}:=\sup_{\varepsilon>0}\|\pazocal{P}_{\mu,\varepsilon}\|_{L^{2}(\mu)\to
L^{2}(\mu)},$

and analogously for $P_{\text{{sy}}}$. We will frequently use the notation:

$\pazocal{P}(f\mu):=\pazocal{P}_{\mu}f:=P\ast(f\mu),$

and normally $f$ will happen to be the constant function $1$, so the reader
has to understand $\pazocal{P}\mu$ as $\pazocal{P}_{\mu}1$. The same
considerations will be done for $\pazocal{P}_{\text{sy}}$ and for
$\pazocal{P}^{\ast}$, the latter being the convolution operator associated to
the conjugate kernel $P^{\ast}$.
From the results of [MPr] and [He] one can deduce an important property of the
above different capacities: they are all comparable.

######  Theorem 2.1.

The previous capacities defined through positive measures are all comparable.
That is, for any compact set $E\subset\mathbb{R}^{n+1}$,

$\displaystyle\gamma_{+}(E)\approx\overline{\gamma}_{+}(E)\approx\widetilde{\gamma}_{+}(E)$
$\displaystyle\approx\gamma_{\text{{sy}},+}(E)$
$\displaystyle\approx\gamma^{\prime}_{+}(E)\approx\overline{\gamma}^{\prime}_{+}(E)\approx\widetilde{\gamma}^{\prime}_{+}(E)\approx\gamma_{\text{{sy}},+}^{\prime}(E)$
$\displaystyle\hskip
134.5814pt\approx\gamma_{\text{{op}}}(E)\approx\gamma_{\text{{sy,op}}}(E).$

######  Remark 2.2.

To be precise, the capacities
$\overline{\gamma}^{\prime}_{+},\gamma_{\text{{sy}},+}^{\prime}$ and
$\gamma_{\text{{sy,op}}}$ are not introduced in the aforementioned references.
To justify the comparability of the capacity $\gamma_{\text{{sy}},+}^{\prime}$
to all the rest, just observe that by the nonnegativity of $P_{\text{{sy}}}$
it is clear that

$\widetilde{\gamma}^{\prime}_{+}(E)\leq\gamma_{\text{sy},+}^{\prime}(E)\leq
2\widetilde{\gamma}^{\prime}_{+}(E).$

To do the same with $\overline{\gamma}^{\prime}_{+}$, we use that $P^{\ast}$
is the fundamental solution of the conjugate operator
$\overline{\Theta}^{1/2}$. By an analogous proof to [MPr, Lemma 3.4] one
deduces that admissible measures for $\overline{\gamma}_{+}$ present
$n$-growth with respect to a dimensional constant. Therefore [He, Lemma 3.10 &
Theorem 3.11] admit an analogous version for $P^{\ast}$, implying

$\overline{\gamma}_{+}(E)\lesssim\overline{\gamma}^{\prime}_{+}(E)\lesssim\gamma_{\text{{op}}}(E)\approx\widetilde{\gamma}_{+}\leq\overline{\gamma}_{+}(E).$

Hence, $\overline{\gamma}^{\prime}_{+}$ is also comparable to all the
capacities of Theorem 2.1. Finally, to deduce the comparability of
$\gamma_{\text{{sy,op}}}$, it suffices to check
$\gamma_{\text{sy,op}}\lesssim\gamma_{\text{op}}$, since the reverse
inequality can be verified to hold easily. Fix $\mu$ any admissible measure
for $\gamma_{\text{{sy,op}}}(E)$ with $\gamma_{\text{{sy,op}}}(E)\leq
2\mu(E)$. Notice that since $\pazocal{P}_{\mu}$ and $\pazocal{P}^{\ast}_{\mu}$
are $n$-dimensional C-Z operators, then $\pazocal{P}_{\text{sy},\mu}$ also
satisfies such property. Therefore, following an analogous proof to [To3,
Theorem 2.16] one deduces, from the $L^{2}(\mu)$ boundedness of
$\pazocal{P}_{\text{sy},\mu}$, its boundedness as an operator from the space
of Borel finite signed measures $M(\mathbb{R}^{n+1})$ to $L^{1,\infty}(\mu)$.
That is, there exists a constant $C>0$ such that for any $\nu\in
M(\mathbb{R}^{n+1})$,

$\mu\big{(}\big{\\{}|\pazocal{P}_{\text{sy},\varepsilon}\nu(\overline{x})|>\lambda\big{\\}}\big{)}\leq
C\frac{\|\nu\|}{\lambda},\qquad\forall\varepsilon>0,\,\forall\lambda>0.$

With this, and using that
$\pazocal{P}_{\text{sy}}=\pazocal{P}_{\text{sy}}^{\ast}$ (the operator is its
own adjoint), applying for example [Ch, Ch.VII, Theorem 23], we can find a
function $h:E\to[0,1]$ such that $\int_{E}h\text{d}\mu\geq\mu(E)/2$ and
$\|\pazocal{P}_{\text{sy}}(h\mu)\|_{\infty}<4C$. Then,

$\gamma_{\text{sy},+}(E)\geq\frac{1}{4C}\int_{E}h\text{d}\mu\geq\frac{\mu(E)}{8C}\geq\frac{1}{16C}\,\gamma_{\text{sy,op}}(E),$

that implies
$\gamma_{\text{sy,op}}(E)\lesssim\gamma_{\text{sy},+}(E)\approx\gamma_{\text{op}}(E)$,
and we are done.

##  3\. Even more capacities

The main goal of this section is to add three additional ways of understanding
$1/2$-caloric capacities: one via an $L^{2}(\mu)$ normalization condition,
another using a uniform bound at all points of the compact set, and a last one
formulated in a variational way. All the results proved in this section are
summarized in Theorem 3.5.

###  3.1. $L^{2}(\mu)$ normalization. Suppressed kernels.

Given a convolution kernel one may define its suppressed version in terms of a
general nonnegative 1-Lipschitz function (with constant 1). However, for our
purposes we do not need to work with such generality, and we will focus on a
particular type of functions. Given a closed set $F\subset\mathbb{R}^{n+1}$,
we will typically have

$\Lambda(\overline{x}):=\text{dist}(\overline{x},F),\quad\forall\overline{x}\in\mathbb{R}^{n+1}.$

The above $\Lambda$ is indeed a 1-Lipschitz function: take any
$\overline{x},\overline{y}\in\mathbb{R}^{n+1}$ and observe that for any
$\overline{z}\in F$ we have
$\Lambda(\overline{x})\leq|\overline{x}-\overline{z}|\leq|\overline{x}-\overline{y}|+|\overline{y}-\overline{z}|$.
Since $\overline{z}$ is arbitrary,
$\Lambda(\overline{x})\leq|\overline{x}-\overline{y}|+\Lambda(\overline{y})\,\leftrightarrow\,\Lambda(\overline{x})-\Lambda(\overline{y})\leq|\overline{x}-\overline{y}|$.
Repeating the same argument changing the roles of $\overline{x}$ and
$\overline{y}$ we deduce the desired Lipschitz property. We define the
suppressed kernel associated to $P$, that depends on $\Lambda$ and a
previously fixed closed set $F$, as:

(3.1)
$P_{\Lambda}(\overline{x},\overline{y}):=\frac{P(\overline{x}-\overline{y})}{1+P(\overline{x}-\overline{y})^{2}\Lambda(\overline{x})^{n}\Lambda(\overline{y})^{n}},$

for all the pair of points where such expression makes sense. Notice that it
is also a nonnegative kernel and differentiable, with respect to
$\overline{x}$ or $\overline{y}$, out of a set of null Lebesgue measure. We
may also define $P_{\Lambda}^{\ast}$ and $P_{\text{sy},\Lambda}$ in an
analogous way. In fact, for the sake of notation, we shall prove three basic
properties only for $P_{\Lambda}$, but they can be checked to hold also for
$P_{\Lambda}^{\ast}$ and $P_{\text{sy},\Lambda}$ with almost exactly the same
proofs. Essentially the convolution kernels $P_{\Lambda},P_{\Lambda}^{\ast}$
and $P_{\text{sy},\Lambda}$ in $\mathbb{R}^{n+1}$ define $n$-dimensional C-Z
operators. In fact, such suppressed kernels satisfy 3 as an additional
property.

######  Lemma 3.1.

Let $\Lambda:\mathbb{R}^{n+1}\to\mathbb{[}0,\infty)$ be a 1-Lipschitz
function. Then, kernels $P_{\Lambda},P_{\Lambda}^{\ast}$ and
$P_{\text{{sy}},\Lambda}$ define $n$-dimensional C-Z operators which satisfy
being “well suppressed” at the points where $\Lambda>0$. More precisely, if we
fix any $\overline{x}\neq\overline{y}$ points in $\mathbb{R}^{n+1}$, the
precise estimates that $P_{\Lambda}$ satisfies $($as well as
$P_{\Lambda}^{\ast}$ and $P_{\text{{sy}},\Lambda})$ are

  1. 1.

$|P_{\Lambda}(\overline{x},\overline{y})|\leq|\overline{x}-\overline{y}|^{-n}$.

  2. 2.

For any $\overline{x}^{\prime}\in\mathbb{R}^{n+1}$ satisfying
$|\overline{x}-\overline{x}^{\prime}|\leq|\overline{x}-\overline{y}|/2$,

$\big{\rvert}P_{\Lambda}(\overline{x},\overline{y})-P_{\Lambda}(\overline{x}^{\prime},\overline{y})\big{\rvert}\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|^{n+1}}\quad\text{and}\quad\big{\rvert}P_{\Lambda}(\overline{y},\overline{x})-P_{\Lambda}(\overline{y},\overline{x}^{\prime})\big{\rvert}\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|^{n+1}}.$

In particular, where the differentiation is well-defined,

$\big{\rvert}\nabla_{\overline{x}}P_{\Lambda}(\overline{x},\overline{y})\big{\rvert}+\big{\rvert}\nabla_{\overline{y}}P_{\Lambda}(\overline{x},\overline{y})\big{\rvert}\lesssim\frac{1}{|\overline{x}-\overline{y}|^{n+1}}.$

  3. 3.

$|P_{\Lambda}(\overline{x},\overline{y})|\lesssim\min\big{\\{}\Lambda(\overline{x})^{-n},\Lambda(\overline{y})^{-n}\big{\\}}$.

###### Proof.

The proof of 1 is trivial, since $|P_{\Lambda}(\overline{x},\overline{y})|\leq
P(\overline{x}-\overline{y})$. We move on by proving the first inequality in
2, and we will do it in a way that the arguments will also be valid for
$P^{\ast}$, implying the validity of the second estimate of 2. Let us first
assume $\Lambda(\overline{x}^{\prime})\leq\Lambda(\overline{x})$ and define

$g(\overline{x},\overline{y}):=1+P(\overline{x}-\overline{y})^{2}\Lambda(\overline{x})^{n}\Lambda(\overline{y})^{n}.$

Using [MPr, Lemma 2.1] (that is also valid for $P^{\ast}$ and hence also for
$P_{\text{{sy}}}$) we get

$\displaystyle|P_{\Lambda}(\overline{x},$
$\displaystyle\overline{y})-P_{\Lambda}(\overline{x}^{\prime},\overline{y})|$
$\displaystyle\leq\frac{|P(\overline{x}-\overline{y})-P(\overline{x}^{\prime}-\overline{y})|}{g(\overline{x},\overline{y})}+P(\overline{x}^{\prime}-\overline{y})\frac{|P(\overline{x}-\overline{y})^{2}\Lambda(\overline{x})^{n}-P(\overline{x}^{\prime}-\overline{y})^{2}\Lambda(\overline{x}^{\prime})^{n}|\Lambda(\overline{y})^{n}}{g(\overline{x},\overline{y})g(\overline{x}^{\prime},\overline{y})}$
$\displaystyle\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|^{n+1}}+P(\overline{x}^{\prime}-\overline{y})\frac{|P(\overline{x}-\overline{y})^{2}-P(\overline{x}^{\prime}-\overline{y})^{2}|\Lambda(\overline{x}^{\prime})^{n}\Lambda(\overline{y})^{n}}{g(\overline{x},\overline{y})g(\overline{x}^{\prime},\overline{y})}$
$\displaystyle\hskip
99.58464pt+P(\overline{x}^{\prime}-\overline{y})\frac{P(\overline{x}-\overline{y})^{2}|\Lambda(\overline{x})^{n}-\Lambda(\overline{x}^{\prime})^{n}|\Lambda(\overline{y})^{n}}{g(\overline{x},\overline{y})g(\overline{x}^{\prime},\overline{y})}=:\text{\@slowromancap
i@}+\text{\@slowromancap ii@}+\text{\@slowromancap <EMAIL_ADDRESS>
We only have to deal with ii@ and<EMAIL_ADDRESS>For the first, we rewrite it as

$\text{\@slowromancap
ii@}=|P(\overline{x}-\overline{y})-P(\overline{x}^{\prime}-\overline{y})|\cdot\text{\@slowromancap
iv@},$

where

$\text{\@slowromancap
iv@}:=\frac{P(\overline{x}^{\prime}-\overline{y})|P(\overline{x}-\overline{y})+P(\overline{x}^{\prime}-\overline{y})|\Lambda(\overline{x}^{\prime})^{n}\Lambda(\overline{y})^{n}}{g(\overline{x},\overline{y})g(\overline{x}^{\prime},\overline{y})}.$

It is clear that by the definition of $g$

$\text{\@slowromancap iv@}\leq
1+\frac{P(\overline{x}^{\prime}-\overline{y})P(\overline{x}-\overline{y})\Lambda(\overline{x}^{\prime})^{n}\Lambda(\overline{y})^{n}}{g(\overline{x},\overline{y})g(\overline{x}^{\prime},\overline{y})}.$

For the remaining fraction, if $P(\overline{x}-\overline{y})\leq
P(\overline{x}^{\prime}-\overline{y})$ we obtain the same bound by 1. If on
the other hand
$P(\overline{x}-\overline{y})>P(\overline{x}^{\prime}-\overline{y})$, using
that $\Lambda(\overline{x}^{\prime})\leq\Lambda(\overline{x})$ we also obtain
the same estimate and we are done with<EMAIL_ADDRESS>Regarding iii@, observe that there
exists a point $\overline{\xi}$ in the line segment joining $\overline{x}$ and
$\overline{x}^{\prime}$ such that

$\text{\@slowromancap
iii@}\leq|\overline{x}-\overline{x}^{\prime}|P(\overline{x}^{\prime}-\overline{y})\frac{n\Lambda(\overline{\xi})^{n-1}\Lambda(\overline{y})^{n}P(\overline{x}-\overline{y})^{2}}{g(\overline{x},\overline{y})g(\overline{x}^{\prime},\overline{y})}.$

We distinguish two cases: if
$\Lambda(\overline{\xi})\leq|\overline{x}-\overline{y}|$. By the Lipschitz
property of $\Lambda$, $\Lambda(\overline{y})\leq
2|\overline{x}-\overline{y}|$. Thus,

iii@
$\displaystyle\leq|\overline{x}-\overline{x}^{\prime}|\frac{n\Lambda(\overline{\xi})^{n-1}\Lambda(\overline{y})^{n}P(\overline{x}-\overline{y})^{2}}{g(\overline{x},\overline{y})g(\overline{x}^{\prime},\overline{y})}\Big{[}P(\overline{x}-\overline{y})+\big{\rvert}P(\overline{x}-\overline{y})-P(\overline{x}^{\prime}-\overline{y})\big{\rvert}\Big{]}$
$\displaystyle\lesssim|\overline{x}-\overline{x}^{\prime}|\frac{|\overline{x}-\overline{y}|^{2n-1}}{|\overline{x}-\overline{y}|^{2n}}\Bigg{[}\frac{1}{|\overline{x}-\overline{y}|^{n}}+\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|^{n+1}}\Bigg{]}=\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|^{n+1}}\Bigg{[}1+\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|}\Bigg{]}$
$\displaystyle\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|^{n+1}},$

that is the desired estimate. If on the other hand we had
$\Lambda(\overline{\xi})>|\overline{x}-\overline{y}|$, then notice that by the
Lipschitz property of $\Lambda$,

$\Lambda(\overline{\xi})\leq\Lambda(\overline{x})+|\overline{x}-\overline{\xi}|\leq\Lambda(\overline{x})+|\overline{x}-\overline{x}^{\prime}|\leq\Lambda(\overline{x})+\frac{|\overline{x}-\overline{y}|}{2}<\Lambda(\overline{x})+\frac{\Lambda(\overline{\xi})}{2},$

that is $\Lambda(\overline{x})>\Lambda(\overline{\xi})/2$, and in particular
$g(\overline{x},\overline{y})\gtrsim g(\overline{\xi},\overline{y})$.
Therefore,

iii@ $\displaystyle\lesssim
P(\overline{x}^{\prime}-\overline{y})\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|}\frac{P(\overline{x}-\overline{y})^{2}\Lambda(\overline{\xi})^{n}\Lambda(\overline{y})^{n}}{g(\overline{\xi},\overline{y})g(\overline{x}^{\prime},\overline{y})}\leq
P(\overline{x}^{\prime}-\overline{y})\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|}$
$\displaystyle\leq\frac{1}{2}\big{\rvert}P(\overline{x}^{\prime}-\overline{y})-P(\overline{x}-\overline{y})\big{\rvert}+\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|^{n+1}}\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}-\overline{y}|^{n+1}},$

that is what we wanted to prove. With this, we have proved 2 for the case
$\Lambda(\overline{x}^{\prime})\leq\Lambda(\overline{x})$. The case
$\Lambda(\overline{x})\leq\Lambda(\overline{x}^{\prime})$ can be treated
analogously, using symmetric considerations with respect to $\overline{x}$ and
$\overline{x}^{\prime}$.
To prove the gradient estimates, let us fix $j\in\\{1,\ldots,n+1\\}$ and
consider $\partial_{x_{j}}P_{\Lambda}(\overline{x},\overline{y})$ at the
points where this expression is well defined. Observe that since $\Lambda$ is
a Lipschitz function and $P$ can be differentiated except for null Lebesgue
sets, $\partial_{x_{j}}P_{\Lambda}$ makes sense out of sets of null Lebesgue
measure. In any case, choose $0<|h|\ll 1$ so that
$|h|\leq|\overline{x}-\overline{y}|/2$, where $|\overline{x}-\overline{y}|$ is
strictly positive since $\overline{x}\neq\overline{y}$ are two fixed different
points. Hence, applying the already proved estimates of 2 we get

$\frac{|P_{\Lambda}(\overline{x}+h\widehat{e}_{j},\overline{y})-P_{\Lambda}(\overline{x},\overline{y})|}{|h|}\lesssim\frac{1}{|h|}\frac{|h|}{|\overline{x}-\overline{y}|^{n+1}}=\frac{1}{|\overline{x}-\overline{y}|^{n+1}},$

so taking the limit as $h\to 0$ the result follows.
Finally we prove 3. Assume that
$\Lambda(\overline{x})=\max\big{\\{}\Lambda(\overline{x}),\Lambda(\overline{y})\big{\\}}$.
In this setting,

(3.2)
$\displaystyle\big{\rvert}P_{\Lambda}(\overline{x},\overline{y})\big{\rvert}=\frac{1}{P(\overline{x}-\overline{y})}\Bigg{[}\frac{1}{P(\overline{x}-\overline{y})^{-2}+\Lambda(\overline{x})^{n}\Lambda(\overline{y})^{n}}\Bigg{]}\leq\frac{1}{P(\overline{x}-\overline{y})}\frac{1}{\Lambda(\overline{x})^{n}\Lambda(\overline{y})^{n}}.$

Assume that

(3.3)
$\frac{1}{P(\overline{x}-\overline{y})}\leq\frac{\Lambda(\overline{x})^{n}}{2^{n}}.$

In this case, $|\overline{x}-\overline{y}|^{n}\leq
P(\overline{x}-\overline{y})^{-1}\leq\Lambda(\overline{x})^{n}/2^{n}$, that is
$|\overline{x}-\overline{y}|\leq\Lambda(\overline{x})/2$. By the Lipschitz
property we also have
$\Lambda(\overline{y})\geq\Lambda(\overline{x})-|\overline{x}-\overline{y}|$,
so $\Lambda(\overline{y})\geq\Lambda(\overline{x})/2$, and applying this
estimate to (3.2) the result would follow. If (3.3) did not hold, we would
simply have

$|P_{\Lambda}(\overline{x}-\overline{y})|\leq|P(\overline{x}-\overline{y})|\leq\frac{2^{n}}{\Lambda(\overline{x})^{n}},$

and we would be also done. Moreover, it is clear that if
$\Lambda(\overline{y})=\max\big{\\{}\Lambda(\overline{x}),\Lambda(\overline{y})\big{\\}}$,
the above proof can be carried out in an analogous way, obtaining the desired
result. ∎

For a fixed $E\subset\mathbb{R}^{n+1}$ compact set, we will use the above
suppressed kernels to study the capacity

$\gamma_{\text{sy},2}(E):=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,\|P_{\text{{sy}}}\ast\mu\|_{L^{2}(\mu)}\leq\mu(E)^{1/2}\big{\\}}.$

It is clear that
$\gamma_{\text{sy},+}^{\prime}(E)\leq\gamma_{\text{sy},2}(E)$. The fundamental
lemma that will imply the reverse inequality is the following:

######  Lemma 3.2.

For $E\subset\mathbb{R}^{n+1}$ compact, the following inequality holds:

$\gamma_{\text{{sy}},2}(E)\lesssim\gamma_{\text{{sy}},\text{{op}}}(E).$

###### Proof.

Let $\mu$ be an admissible measure for $\gamma_{\text{{sy}},2}(E)$. Then, by
definition,

$\int_{E}|\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{x})|^{2}\text{d}\mu(\overline{x})\leq\mu(E),\quad\forall\varepsilon>0.$

Consider the maximal operator

$\pazocal{P}_{\text{sy},\ast}\mu(\overline{x}):=\sup_{\varepsilon>0}|\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{x})|,$

that by the nonnegativity of $P_{\text{{sy}}}$ and $\mu$ is such that

$\pazocal{P}_{\text{sy},\ast}\mu(\overline{x})=\lim_{\varepsilon\to
0}\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{x}),$

which is well-defined for every point $\overline{x}\in\mathbb{R}^{n+1}$. In
fact, by the monotone convergence theorem, it is clear that the above
$L^{2}(\mu)$ uniform estimate with respect to $\varepsilon>0$, can be
rewritten as

$\int_{E}|\pazocal{P}_{\text{sy},\ast}\mu(\overline{x})|^{2}\text{d}\mu(\overline{x})\leq\mu(E).$

Now we define

$F:=\Big{\\{}\overline{x}\in
E\,:\,\pazocal{P}_{\text{sy},\ast}\mu(\overline{x})\leq\sqrt{2}\Big{\\}},$

and apply Chebyshev’s inequality to obtain

$\mu(E\setminus{F})\leq\frac{1}{2}\int_{E}|\pazocal{P}_{\text{sy},\ast}\mu(\overline{x})|^{2}\text{d}\mu(\overline{x})\leq\frac{\mu(E)}{2}\quad\text{and
then}\quad\mu(F)\geq\frac{\mu(E)}{2}.$

If we are able to prove

$\|\pazocal{P}_{\text{sy},\mu|_{F}}\|_{L^{2}(\mu|_{F})\to
L^{2}(\mu|_{F})}\lesssim 1$

we will be done, since the latter would mean
$\gamma_{\text{sy},\text{op}}(E)\gtrsim\mu(F)\geq\mu(E)/2$, and by the
arbitrariness of $\mu$ the result would follow.
To proceed, let us observe that $\pazocal{P}_{\text{sy},\ast}\mu$ can be
understood as a non-decreasing limit of continuous functions. Indeed, we may
rewrite, for each $\varepsilon>0$ and $\overline{x}\in\mathbb{R}^{n+1}$,

$\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{x})=\int_{\mathbb{R}^{n+1}}P_{\text{{sy}}}(\overline{x}-\overline{y})\chi_{\\{|\overline{x}-\overline{y}|>\varepsilon\\}}\mathop{}\\!\mathrm{d}\mu(\overline{y})=:P_{\text{sy},\chi_{\varepsilon}}\ast\mu(\overline{x}),$

where
$P_{\text{sy},\chi_{\varepsilon}}(\cdot):=P_{\text{{sy}}}(\cdot)\chi_{\\{|\cdot|>\varepsilon\\}}$.
Now, let us fix $\psi$ any test function such that $0\leq\psi\leq 1$,
$\psi\equiv 0$ in $B_{1}(0)$ and $\psi\equiv 1$ in
$\mathbb{R}^{n+1}\setminus{B_{2}(0)}$. Assume also that
$\|\nabla\psi\|_{\infty}\leq 1$. Write
$\psi_{\varepsilon}(\cdot):=\psi(\cdot/\varepsilon)$ and define, for each
$\overline{x}\in\mathbb{R}^{n+1}$,

$P_{\text{sy},\psi_{\varepsilon}}\ast\mu(\overline{x}):=\int_{\mathbb{R}^{n+1}}P_{\text{{sy}}}(\overline{x}-\overline{y})\psi_{\varepsilon}(\overline{x}-\overline{y})\text{d}\mu(\overline{y}).$

Observe that since $P_{\text{{sy}}}\geq 0$,

$P_{\text{sy},\chi_{2\varepsilon}}\ast\mu(\overline{x})\leq
P_{\text{sy},\psi_{\varepsilon}}\ast\mu(\overline{x})\leq
P_{\text{sy},\chi_{\varepsilon}}\ast\mu(\overline{x}),\hskip
7.11317pt\forall\overline{x}\in\mathbb{R}^{n+1},$

which implies, by definition,

$\lim_{\varepsilon\to
0}P_{\text{sy},\psi_{\varepsilon}}\ast\mu(\overline{x})=\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{x}),\hskip
7.11317pt\forall\overline{x}\in\mathbb{R}^{n+1}.$

Now, since $\mu$ is in particular a finite measure, it is not difficult to
prove using Fatou’s lemma and its reverse version, that for each
$\varepsilon>0$ fixed, $P_{\text{sy},\psi_{\varepsilon}}\ast\mu$ is a
continuous function. Moreover, since $\mu$ and $P_{\text{{sy}}}$ are
nonnegative, it is clear that for $\varepsilon_{1}<\varepsilon_{2}$, then
$P_{\text{sy},\psi_{\varepsilon_{2}}}\ast\mu\leq
P_{\text{sy},\psi_{\varepsilon_{1}}}\ast\mu$. Hence, indeed,
$\pazocal{P}_{\text{sy},\ast}\mu$ can be defined as a non-decreasing limit of
continuous functions. Therefore, $\pazocal{P}_{\text{sy},\ast}\mu$ is lower
semicontinuous [Ba, Ch.4] and thus $F$ is a closed set. Now consider $B$ a
large open ball containing $E$ inside and define $G:=B\setminus{F}$. For any
$\overline{z}\in G\cap E=E\setminus{F}$ we define

$\varepsilon(\overline{z}):=\text{dist}(\overline{z},F)=\Lambda(\overline{z}),$

recalling the notation used when defining the suppressed kernels. First we
estimate for any $\varepsilon\geq\varepsilon(\overline{z})$:

$\displaystyle\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{z})=\int_{|\overline{y}-\overline{z}|\geq
2\varepsilon}P_{\text{{sy}}}(\overline{z}-\overline{y})\mathop{}\\!\mathrm{d}\mu(\overline{y})+\int_{\varepsilon<|\overline{y}-\overline{z}|\leq
2\varepsilon}P_{\text{{sy}}}(\overline{z}-\overline{y})\mathop{}\\!\mathrm{d}\mu(\overline{y})=:\text{\@slowromancap
i@}+\text{\@slowromancap <EMAIL_ADDRESS>
Notice that by the $n$-growth of $\mu$ we have

$\displaystyle|\text{\@slowromancap
ii@}|\leq\frac{\mu(B_{2\varepsilon}(\overline{z}))}{\varepsilon^{n}}\lesssim
1.$

To deal with i@, let $\overline{x}_{0}$ be a closest point to $\overline{z}$
on $F$, so that $\varepsilon(\overline{z})=|\overline{x}_{0}-\overline{z}|$.
Then

$\displaystyle|\text{\@slowromancap
i@}|\leq\int_{|\overline{y}-\overline{z}|\geq
2\varepsilon}P_{\text{{sy}}}(\overline{x}_{0}-\overline{y})\mathop{}\\!\mathrm{d}\mu(\overline{y})+\int_{|\overline{y}-\overline{z}|\geq
2\varepsilon}|P_{\text{{sy}}}(\overline{x}_{0}-\overline{y})-P_{\text{{sy}}}(\overline{z}-\overline{y})|\mathop{}\\!\mathrm{d}\mu(\overline{y})=:\text{\@slowromancap
iii@}+\text{\@slowromancap <EMAIL_ADDRESS>
It is clear that $\text{\@slowromancap iii@}\leq\sqrt{2}$, since
$\overline{x}_{0}\in F$. On the other hand, since
$|(\overline{z}-\overline{y})-(\overline{x}_{0}-\overline{y})|\leq\varepsilon\leq|\overline{z}-\overline{y}|/2$,
integration over annuli yields

$\displaystyle|\text{\@slowromancap
iv@}|\lesssim\varepsilon\int_{|\overline{y}-\overline{z}|\geq\varepsilon}\frac{\mathop{}\\!\mathrm{d}\mu(\overline{y})}{|\overline{z}-\overline{y}|^{n+1}}\lesssim
1.$

So we have $|\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{z})|\lesssim 1$,
for any $\varepsilon\geq\varepsilon(\overline{z})$. Now we shall prove

$A:=\big{\rvert}\pazocal{P}_{\text{sy},\Lambda,\varepsilon}\mu(\overline{z})-\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{z})\big{\rvert}\lesssim
1,\quad\forall\varepsilon\geq\varepsilon(\overline{z}),$

where
$\pazocal{P}_{\text{sy},\Lambda,\varepsilon}\mu(\overline{z}):=\int_{|\overline{z}-\overline{y}|>\varepsilon}P_{\text{sy},\Lambda}(\overline{z},\overline{y})\mathop{}\\!\mathrm{d}\mu(\overline{y})$.
Since $\varepsilon(\overline{z})=\Lambda(\overline{z})$ by definition,

$\displaystyle
A\leq\int_{|\overline{y}-\overline{z}|\geq\Lambda(\overline{z})}\big{\rvert}P_{\text{sy},\Lambda}(\overline{z},\overline{y})-P_{\text{{sy}}}(\overline{z}-\overline{y})\big{\rvert}\mathop{}\\!\mathrm{d}\mu(\overline{y}).$

Using the definition of $P_{\text{sy},\Lambda}$ we know

(3.4)
$\big{\rvert}P_{\text{sy},\Lambda}(\overline{z},\overline{y})-P_{\text{{sy}}}(\overline{z}-\overline{y})\big{\rvert}\leq
P_{\text{{sy}}}(\overline{z}-\overline{y})\Bigg{[}\frac{P_{\text{{sy}}}(\overline{z}-\overline{y})^{2}\Lambda(\overline{z})^{n}\Lambda(\overline{y})^{n}}{1+P_{\text{{sy}}}(\overline{z}-\overline{y})^{2}\Lambda(\overline{z})^{n}\Lambda(\overline{y})^{n}}\Bigg{]}.$

In the region of integration
$\Lambda(\overline{z})\leq|\overline{y}-\overline{z}|$, then
$\Lambda(\overline{y})\leq 2|\overline{y}-\overline{z}|$, by the Lipschitz
property. Therefore, returning to (3.4) we get

$\displaystyle\big{\rvert}P_{\text{sy},\Lambda}(\overline{z},\overline{y})-P_{\text{{sy}}}(\overline{z}-\overline{y})\big{\rvert}\leq
P_{\text{{sy}}}(\overline{z}-\overline{y})$
$\displaystyle\Big{[}P_{\text{{sy}}}(\overline{z}-\overline{y})^{2}\Lambda(\overline{z})^{n}\Lambda(\overline{y})^{n}\Big{]}$
$\displaystyle\leq
2^{n}P_{\text{{sy}}}(\overline{z}-\overline{y})^{2}\Lambda(\overline{z})^{n}\lesssim\frac{\Lambda(\overline{z})^{n}}{|\overline{z}-\overline{y}|^{2n}},$

so by the $n$-growth of $\mu$ we obtain (again, integrating over annuli)

$A\lesssim\int_{|\overline{y}-\overline{z}|\geq\Lambda(\overline{z})}\frac{\Lambda(\overline{z})^{n}}{|\overline{z}-\overline{y}|^{2n}}\mathop{}\\!\mathrm{d}\mu(\overline{y})\lesssim
1,$

that is what we wanted to prove. Hence, combining this last estimate with
$|\pazocal{P}_{\text{sy},\varepsilon}\mu(\overline{z})|\lesssim 1$, for any
$\varepsilon\geq\varepsilon(\overline{z})$ we deduce

$|\pazocal{P}_{\text{sy},\Lambda,\varepsilon}\mu(\overline{z})|\lesssim
1,\quad\forall\varepsilon\geq\varepsilon(\overline{z}).$

Now, fixing $\eta\in(0,\varepsilon(\overline{z}))$, using property 3 in Lemma
3.1 together with $\eta<\Lambda(\overline{z})$, we get

$\bigg{\rvert}\int_{|\overline{y}-\overline{z}|\leq\eta}P_{\text{sy},\Lambda}(\overline{z},\overline{y})\mathop{}\\!\mathrm{d}\mu(\overline{y})\bigg{\rvert}\leq\frac{1}{\Lambda(\overline{z})^{n}}\mu(B_{\eta}(\overline{z}))\leq
1.$

All in all, we have proved

$\big{\rvert}\pazocal{P}_{\text{sy},\Lambda,\ast}\mu(\overline{z})\big{\rvert}\lesssim
1,\quad\forall\overline{z}\in E\setminus{F}.$

In fact, this last estimate also holds for $\overline{z}\in F$, since in this
case $\Lambda(\overline{z})=0$ and $P_{\text{sy},\Lambda}=P_{\text{sy}}$,
implying

$\displaystyle\big{\rvert}\pazocal{P}_{\text{sy},\Lambda,\ast}\mu(\overline{z})\big{\rvert}$
$\displaystyle=\sup_{\eta>0}\bigg{\rvert}\int_{|\overline{y}-\overline{z}|\geq\eta}P_{\text{sy},\Lambda}(\overline{z},\overline{y})\mathop{}\\!\mathrm{d}\mu(\overline{y})\bigg{\rvert}$
$\displaystyle=\sup_{\eta>0}\bigg{\rvert}\int_{|\overline{y}-\overline{z}|\geq\eta}P_{\text{{sy}}}(\overline{z}-\overline{y})\mathop{}\\!\mathrm{d}\mu(\overline{y})\bigg{\rvert}=\pazocal{P}_{\text{sy},\ast}\mu(\overline{z})\leq\sqrt{2},$

by definition of $F$. Hence, we get
$\|\pazocal{P}_{\text{sy},\Lambda}\mu\|_{L^{\infty}(\mu)}\lesssim 1$, where
this estimate has to be understood as
$\|\pazocal{P}_{\text{sy},\Lambda,\varepsilon}\mu\|_{L^{\infty}(\mu)}\lesssim
1$, uniformly on $\varepsilon>0$. The same estimate also holds for
$\pazocal{P}_{\text{sy},\Lambda}^{\ast}\mu$, since
$P_{\text{sy},\Lambda}(\overline{x},\overline{y})=P_{\text{sy},\Lambda}(\overline{y},\overline{x})$,
by the symmetry of $P_{\text{{sy}}}$. Moreover, it is also clear that for any
cube $Q\subset\mathbb{R}^{n+1}$ and any $\varepsilon>0$,

$\displaystyle\big{\rvert}\big{\langle}\pazocal{P}_{\text{sy},\Lambda,\mu,\varepsilon}\chi_{Q},\chi_{Q}\big{\rangle}\big{\rvert}=\int_{Q}\bigg{(}\int_{Q\cap\\{|\overline{y}-\overline{z}|>\varepsilon\\}}$
$\displaystyle
P_{\text{sy},\Lambda}(\overline{z},\overline{y})\mathop{}\\!\mathrm{d}\mu(\overline{y})\bigg{)}\mathop{}\\!\mathrm{d}\mu(\overline{z})$
$\displaystyle\leq\int_{Q}\pazocal{P}_{\text{sy},\Lambda,\varepsilon}\mu(\overline{z})\mathop{}\\!\mathrm{d}\mu(\overline{z})\lesssim\mu(Q),$

by the nonnegativity of $P_{\text{sy},\Lambda}$. So bearing in mind Remark
Lemma 3.1, we may apply a suitable $T1$-theorem, namely [To3, Theorem 1.3], to
deduce $\|\pazocal{P}_{\text{sy},\Lambda,\mu}\|_{L^{2}(\mu)\to
L^{2}(\mu)}\lesssim 1$. But for $f,g\in L^{2}(\mu)$ supported on $F$ one has
$\langle\pazocal{P}_{\text{sy},\Lambda}f,g\rangle=\langle\pazocal{P}_{\text{sy}}f,g\rangle$,
meaning that

$\|\pazocal{P}_{\text{sy},\mu|_{F}}\|_{L^{2}(\mu|_{F})\to
L^{2}(\mu|_{F})}\lesssim 1,$

and we are done. ∎

######  Remark 3.1.

Due to the above Lemma 3.2, we may add $\gamma_{\text{sy},2}$ to the statement
of Theorem 2.1. In fact, defining

$\widetilde{\gamma}_{2}(E):=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,\|P\ast\mu\|_{L^{2}(\mu)}\leq\mu(E)^{1/2},\|P^{\ast}\ast\mu\|_{L^{2}(\mu)}\leq\mu(E)^{1/2}\big{\\}},$

it is clear that $\widetilde{\gamma}_{2}(E)\leq\gamma_{\text{sy},2}(E)$ and
that $\widetilde{\gamma}^{\prime}_{+}(E)\leq\widetilde{\gamma}_{2}(E)$.
Meaning that $\widetilde{\gamma}_{2}$ can be also added to Theorem 2.1.

###  3.2. Normalization by a uniform bound at all points of the support

For a fixed $E\subset\mathbb{R}^{n+1}$ compact set, let us consider now the
capacity

$\gamma^{\star}_{+}(E):=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,|P\ast\mu(\overline{x})|\leq
1,\,\forall\overline{x}\in E\big{\\}},$

as well as all the variants
$\overline{\gamma}^{\star}_{+},\widetilde{\gamma}^{\star}_{+}$ and
$\gamma^{\star}_{\text{sy},+}$. It is clear that
$\gamma^{\star}_{+}(E)\leq\gamma_{+}(E)\approx\gamma_{\text{op}}(E)$. We claim
that the following holds:

$\gamma_{\text{op}}(E)\lesssim\gamma_{+}^{\star}(E).$

Indeed, given $\mu\in\Sigma(E)$ admissible for $\gamma_{\text{op}}(E)$ with
$\gamma_{\text{op}}(E)\leq 2\mu(E)$, proceeding as in Remark 2.2 we can find a
function $h:E\to[0,1]$ such that $\int_{E}h\text{d}\mu\geq\mu(E)/2$ and
$\|\pazocal{P}(h\mu)\|_{\infty}\leq C$, for some $C>0$ absolute constant. We
know that the latter estimate implies
$\|\pazocal{P}(h\mu)\|_{L^{\infty}(\mu)}\leq C^{\prime}$ for some other
absolute constant $C^{\prime}>0$ (by a Cotlar type inequality analogous to
that of [MaP, Lemma 5.4]). Applying now Cotlar’s inequality of [To3, Theorem
2.18], for example, we have

$\sup_{\varepsilon>0}\big{\rvert}\pazocal{P}_{\varepsilon}(h\mu)(\overline{x})\big{\rvert}\leq
C^{\prime\prime}\big{(}\widetilde{M}_{\mu}|\pazocal{P}(h\mu)|^{\delta}\big{)}^{1/\delta}(\overline{x})+C^{\prime\prime\prime}(\widetilde{M}_{\mu}h)(\overline{x}),\quad\forall\overline{x}\in\mathbb{R}^{n+1},$

where

$\widetilde{M}_{\mu}h(\overline{x}):=\sup_{r>0}\frac{1}{\mu(B_{3r}(\overline{x}))}\int_{B_{r}(\overline{x})}|h(\overline{y})|\mathop{}\\!\mathrm{d}\mu(\overline{y}),$

$\delta\in(0,1)$ is arbitrarily fixed and
$C^{\prime\prime},C^{\prime\prime\prime}$ are positive constants that depend
on the dimension, $\delta$ and the $L^{2}(\mu)$-norm of $\pazocal{P}_{\mu}$,
that equals 1 by hypothesis. Since $0\leq h\leq 1$, it is clear that
$\widetilde{M}_{\mu}h\leq 1$; and since
$\|\pazocal{P}(h\mu)\|_{L^{\infty}(\mu)}\leq C^{\prime}$, we deduce for
$\delta=1/2$

$\widetilde{M}_{\mu}|\pazocal{P}(h\mu)|^{1/2}(\overline{x})=\sup_{r>0}\frac{1}{\mu(B_{3r}(\overline{x}))}\int_{B_{r}(\overline{x})}\big{\rvert}\pazocal{P}(h\mu)(\overline{x})\big{\rvert}^{1/2}\mathop{}\\!\mathrm{d}\mu(\overline{y})\leq(C^{\prime})^{1/2}.$

Therefore, setting
$\widetilde{C}:=\max\\{1,C^{\prime}C^{\prime\prime}+C^{\prime\prime\prime}\\}$
we get

$\big{\rvert}\pazocal{P}(\widetilde{C}^{-1}h\mu)(\overline{x})\big{\rvert}=\sup_{\varepsilon>0}\big{\rvert}\pazocal{P}_{\varepsilon}(\widetilde{C}^{-1}h\mu)(\overline{x})\big{\rvert}\leq
1,\quad\forall\overline{x}\in\mathbb{R}^{n+1}.$

So in particular $\widetilde{C}^{-1}h\mu$ is an admissible measure (up to a
dimensional factor that makes it an $n$-growth measure with constant 1) for
$\gamma^{\star}_{+}(E)$. Thus,

$\gamma^{\star}_{+}(E)\gtrsim\frac{1}{\widetilde{C}}\int_{E}h\mathop{}\\!\mathrm{d}\mu\geq\frac{1}{2\widetilde{C}}\mu(E),$

and by the arbitrariness of $\mu$ we conclude that
$\gamma_{\text{op}}(E)\lesssim\gamma_{+}^{\star}(E)$.

######  Remark 3.2.

Therefore, we conclude that the capacity $\gamma_{+}^{\star}$ can be added to
the statement of Theorem 2.1. In fact, since the operator $L^{2}(\mu)$-norm of
$\pazocal{P}_{\mu}^{\ast}$ is the same as that of $\pazocal{P}_{\mu}$,
following an analogous argument we may also add
$\overline{\gamma}^{\star}_{+}$ as well as $\widetilde{\gamma}^{\star}_{+}$.
Further, the same arguments can also be followed to compare
$\gamma^{\star}_{\text{sy},+}$ with $\gamma_{\text{sy},\text{op}}$, allowing
us to also consider $\gamma^{\star}_{\text{sy},+}$ in Theorem 2.1.

###  3.3. An equivalent variational capacity

We shall consider a construction already presented in the proof of Lemma 3.2:
let $\psi$ be a radial test function with $0\leq\psi\leq 1$, $\psi\equiv 0$ in
$B_{1/2}(0)$, $\psi\equiv 1$ in $\mathbb{R}^{n+1}\setminus{B_{1}(0)}$ and such
that $\|\nabla\psi\|_{\infty}\leq 1$. Fix $E\subset\mathbb{R}^{n+1}$ compact
set and $\mu\in\Sigma(E)$. For each $\tau>0$, write
$\psi_{\tau}(\cdot):=\psi(\cdot/\tau)$ and define, for each
$\overline{x}\in\mathbb{R}^{n+1}$ and $f\in L^{1}_{\text{loc}}(\mu)$.

$\pazocal{P}_{\text{sy},\psi_{\tau}}(f\mu)(\overline{x}):=P_{\text{{sy}}}\psi_{\tau}\ast(f\mu)(\overline{x})=\int_{\mathbb{R}^{n+1}}P_{\text{{sy}}}(\overline{x}-\overline{y})\psi_{\tau}(\overline{x}-\overline{y})f(\overline{y})\text{d}\mu(\overline{y}).$

Let us prove, using [MPr, Lemma 2.1] (also valid for $P^{\ast}$ and
$P_{\text{{sy}}}$) , that the regularized (continuous) kernel
$P_{\text{{sy}}}\psi_{\tau}$ defines a C-Z convolution operator with constants
not depending on $\tau$. It is clear that
$|P_{\text{{sy}}}\psi_{\tau}(\overline{x})|\leq|\overline{x}|^{-n}$. Then, by
the symmetry of $P_{\text{{sy}}}\psi_{\tau}$, it suffices to check that for
any $\overline{x},\overline{x}^{\prime}\in\mathbb{R}^{n+1}$ with
$\overline{x}\neq 0$ and
$|\overline{x}-\overline{x}^{\prime}|\leq|\overline{x}|/2$,

(3.5)
$\big{\rvert}P_{\text{{sy}}}\psi_{\tau}(\overline{x})-P_{\text{{sy}}}\psi_{\tau}(\overline{x}^{\prime})\big{\rvert}\leq
C\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}|^{n+1}},$

where $C>0$ is an absolute constant independent of $\tau$. To prove this, we
distinguish two cases: if $\tau\geq|\overline{x}|/4$,

$\displaystyle\big{\rvert}P_{\text{{sy}}}\psi_{\tau}(\overline{x})-P_{\text{{sy}}}\psi_{\tau}(\overline{x}^{\prime})\big{\rvert}$
$\displaystyle\leq\big{\rvert}P_{\text{{sy}}}(\overline{x})-P_{\text{{sy}}}(\overline{x}^{\prime})\big{\rvert}\psi_{\tau}(\overline{x}^{\prime})+\big{\rvert}\psi_{\tau}(\overline{x})-\psi_{\tau}(\overline{x}^{\prime})\big{\rvert}P_{\text{{sy}}}(\overline{x})$
$\displaystyle\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}|^{n+1}}+\frac{|\overline{x}-\overline{x}^{\prime}|}{\tau}\frac{1}{|\overline{x}|^{n}}\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}|^{n+1}},$

where we have applied [MPr, Lemma 2.1] and $\|\nabla\psi\|_{\infty}\leq 1$. If
on the other hand $\tau<|\overline{x}|/4$, by definition of $\psi_{\tau}$ we
have $\psi_{\tau}(\overline{x})=1$. In addition, by the triangle inequality,

(3.6)
$|\overline{x}^{\prime}|\geq\big{\rvert}|\overline{x}-\overline{x}^{\prime}|-|\overline{x}|\big{\rvert}$

If
$||\overline{x}-\overline{x}^{\prime}|-|\overline{x}||=|\overline{x}-\overline{x}^{\prime}|-|\overline{x}|$,
then $|\overline{x}-\overline{x}^{\prime}|\geq|\overline{x}|$, so
$|\overline{x}-\overline{x}^{\prime}|\approx|\overline{x}|$. So in this case,

$\displaystyle\big{\rvert}P_{\text{{sy}}}\psi_{\tau}(\overline{x})-P_{\text{{sy}}}\psi_{\tau}(\overline{x}^{\prime})\big{\rvert}$
$\displaystyle\leq\big{\rvert}P_{\text{{sy}}}(\overline{x})-P_{\text{{sy}}}(\overline{x}^{\prime})\big{\rvert}\psi_{\tau}(\overline{x}^{\prime})+\big{\rvert}1-\psi_{\tau}(\overline{x}^{\prime})\big{\rvert}P_{\text{{sy}}}(\overline{x})$
$\displaystyle\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}|^{n+1}}+\frac{2}{|\overline{x}|^{n}}\leq\frac{3|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}|^{n+1}},$

and we are done. If on the other hand
$||\overline{x}-\overline{x}^{\prime}|-|\overline{x}||=|\overline{x}|-|\overline{x}-\overline{x}^{\prime}|$,
then
$|\overline{x}^{\prime}|\geq|\overline{x}|-|\overline{x}-\overline{x}^{\prime}|\geq|\overline{x}|/2$
which implies $\psi_{\tau}(\overline{x}^{\prime})=1$. Then, in this case,

$\displaystyle\big{\rvert}P_{\text{{sy}}}\psi_{\tau}(\overline{x})-P_{\text{{sy}}}\psi_{\tau}(\overline{x}^{\prime})\big{\rvert}=\big{\rvert}P_{\text{{sy}}}(\overline{x})-P_{\text{{sy}}}(\overline{x}^{\prime})\big{\rvert}\lesssim\frac{|\overline{x}-\overline{x}^{\prime}|}{|\overline{x}|^{n+1}},$

and the proof of (3.5) is complete.
Let us also observe that for any $f\in L^{1}_{\text{loc}}(\mu)$ any $\tau>0$
and any $\overline{x}\in E$,

$\displaystyle\big{\rvert}\pazocal{P}_{\text{sy},\tau}(f\mu)(\overline{x})-\pazocal{P}_{\text{sy},\psi_{\tau}}(f\mu)(\overline{x})\big{\rvert}$
$\displaystyle\leq\int_{\tau/2<|\overline{x}-\overline{y}|<\tau}P_{\text{{sy}}}(\overline{x}-\overline{y})\big{\rvert}1-\psi_{\tau}(\overline{x}-\overline{y})\big{\rvert}|f(\overline{y})|\mathop{}\\!\mathrm{d}\mu(\overline{y})$
$\displaystyle\lesssim\sup_{\tau>0}\frac{1}{\tau^{n}}\int_{|\overline{x}-\overline{y}|<\tau}|f(\overline{y})|\mathop{}\\!\mathrm{d}\mu(\overline{y}).$

In particular, choosing $f\equiv 1$, by the $n$-growth of $\mu$ we get

(3.7)
$\big{\rvert}\pazocal{P}_{\text{sy},\tau}\mu(\overline{x})-\pazocal{P}_{\text{sy},\psi_{\tau}}\mu(\overline{x})\big{\rvert}\lesssim
1,\quad\forall\overline{x}\in E,\,\tau>0.$

Let us define the auxiliary capacity, for each $\tau>0$,

$\gamma^{\star}_{\text{sy},\psi_{\tau},+}(E):=\sup\big{\\{}\mu(E)\,:\,\mu\in\Sigma(E),\,|P_{\text{{sy}}}\psi_{\tau}\ast\mu(\overline{x})|\leq
1,\,\forall\overline{x}\in E\big{\\}}.$

######  Lemma 3.3.

The following estimates hold:

  1. 1.

$\limsup\limits_{\tau\to
0}\gamma^{\star}_{\text{{sy}},\psi_{\tau},+}(E)\lesssim\gamma^{\star}_{\text{{sy}},+}(E)$.

  2. 2.

$\liminf\limits_{\tau\to
0}\gamma^{\star}_{\text{{sy}},\psi_{\tau},+}(E)\gtrsim\gamma^{\star}_{\text{{sy}},+}(E)$.

###### Proof.

We begin by proving 1. Let $(\tau_{k})_{k}$ be a monotonically decreasing
sequence to 0, and let $\mu_{k}\in\Sigma(E)$ be admissible for
$\gamma^{\star}_{\text{{sy}},\psi_{\tau_{k}},+}(E)$ and such that
$\gamma^{\star}_{\text{{sy}},\psi_{\tau_{k}},+}(E)\leq 2\mu_{k}(E)$. Passing
to a subsequence if necessary, assume $\mu_{k}\rightharpoonup\mu_{0}$ weakly
[Ma, Theorem 1.23], i.e.

$\lim_{k\to\infty}\int\varphi\mathop{}\\!\mathrm{d}\mu_{k}=\int\varphi\mathop{}\\!\mathrm{d}\mu_{0},\quad\forall\varphi\in\pazocal{C}_{c}(\mathbb{R}^{n+1}).$

It is not hard to prove that $\mu_{0}\in\Sigma(E)$. Now fix $m>0$ integer and
$\tau\in[\tau_{m+1},\tau_{m}]$, and also assume $k>m$. By (3.7) we get that
for $\overline{x}\in E$,

$\displaystyle\big{\rvert}\pazocal{P}_{\text{sy},\psi_{\tau}}\mu_{k}(\overline{x})\big{\rvert}\leq\big{\rvert}\pazocal{P}_{\text{sy},\psi_{\tau_{k}}}\mu_{k}(\overline{x})\big{\rvert}+\big{\rvert}\pazocal{P}_{\text{sy},\psi_{\tau}}\mu_{k}(\overline{x})-\pazocal{P}_{\text{sy},\psi_{\tau_{k}}}\mu_{k}(\overline{x})\big{\rvert}\leq
1+C,$

for some finite absolute constant $C>0$. Since $\mu_{k}\rightharpoonup\mu_{0}$
we obtain
$\big{\rvert}\pazocal{P}_{\text{sy},\psi_{\tau}}\mu_{0}\big{\rvert}\leq 1+C$
on $E$ and for each $\tau>0$. In addition, using (3.7) again we get

$\big{\rvert}\pazocal{P}_{\text{sy},\tau}\mu_{0}(\overline{x})\big{\rvert}\leq
1+2C,\quad\forall\overline{x}\in E,\,\tau>0,$

that is $|\pazocal{P}_{\text{sy}}\mu_{0}(\overline{x})\big{|}\leq 1+2C$ for
all $\overline{x}\in E$. Thus, by the semicontinuity properties of weak
convergence of [Ma, Theorem 1.24] we get

$\displaystyle\gamma^{\star}_{\text{sy},+}(E)\geq\frac{\mu_{0}(E)}{1+2C}\geq\frac{1}{1+2C}\limsup_{k\to\infty}\mu_{k}(E)\gtrsim\limsup_{\tau\to
0}\gamma^{\star}_{\text{{sy}},\psi_{\tau_{k}},+}(E),$

and 1 follows. To prove 2, take $\mu$ admissible for
$\gamma^{\star}_{\text{sy},+}(E)$. Using (3.7) we deduce

$\big{\rvert}\pazocal{P}_{\text{sy},\psi_{\tau}}\mu(\overline{x})\big{\rvert}\leq
1+C,\quad\forall\overline{x}\in E,\,\tau>0,$

implying that
$\gamma^{\star}_{\text{sy},\psi_{\tau},+}(E)\geq\frac{\mu(E)}{1+C}$, and by
the arbitrariness of $\mu$ we have

$\inf_{\tau>0}\gamma^{\star}_{\text{sy},\psi_{\tau},+}(E)\gtrsim\gamma^{\star}_{\text{sy},+}(E),$

which implies the desired result. ∎

######  Remark 3.3.

Let us observe that, for each $\tau>0$, using the regularized (continuous)
kernel $P_{\text{{sy}}}\psi_{\tau}$ we may also define the rest of
corresponding capacities, requiring different normalization conditions over
the potentials. In fact, the following chain of estimates holds for any
compact set $E\subset\mathbb{R}^{n+1}$:

$\gamma_{\text{sy},\psi_{\tau},+}^{\star}(E)\underset{(1)}{\leq}\gamma_{\text{sy},\psi_{\tau},+}^{\prime}(E)\underset{(2)}{\leq}\gamma_{\text{sy},\psi_{\tau},2}(E)\underset{(3)}{\lesssim}\gamma_{\text{sy},\psi_{\tau},\text{op}}(E)\underset{(4)}{\lesssim}\gamma_{\text{sy},\psi_{\tau},+}(E)\underset{(5)}{\leq}\gamma_{\text{sy},\psi_{\tau},+}^{\star}(E).$

Inequalities (1) and (2) are trivial. To verify (3), proceed, for example, as
in the proof of
$\gamma_{\text{sy},2}(E)\lesssim\gamma_{\text{sy},\text{op}}(E)$ in Lemma 3.2
(although the arguments could be possibly simplified using the continuity of
$P_{\text{{sy}}}\psi_{\tau}$). The proof of (4) goes as in Remark 2.2, and it
relies on [To3, Theorem 2.16] and [Ch, Ch.VII, Theorem 23]. it is important to
notice that a fundamental property that ensures the validity of (3) and (4) is
that the C-Z constants of $P_{\text{{sy}}}\psi_{\tau}$ are independent of
$\tau$. Finally, (5) holds by the continuity of
$\pazocal{P}_{\text{sy},\psi_{\tau}}\mu$ for each $\tau>0$, that has already
been argued in the proof of Lemma 3.2. Observe that the previous chain of
estimates also holds changing every $\gamma_{\text{sy}}$ for
$\widetilde{\gamma}$, where we ask the respective normalization conditions in
each case for both kernels $P\psi_{\tau}$ and $P^{\ast}\psi_{\tau}$.

Therefore, in particular, we shall restate Lemma 3.3 as

  1. 1.

$\limsup\limits_{\tau\to
0}\gamma_{\text{{sy}},\psi_{\tau},2}(E)\lesssim\gamma_{\text{{sy}},2}(E)$,

  2. 2.

$\liminf\limits_{\tau\to
0}\gamma_{\text{{sy}},\psi_{\tau},2}(E)\gtrsim\gamma_{\text{{sy}},2}(E)$.

Now, bearing in mind that we have found

  * $\star$

$\limsup\limits_{\tau\to 0}\gamma_{\text{{sy}},\psi_{\tau},2}(E)\leq
C_{1}\,\gamma_{\text{{sy}},2}(E)$,

  * $\star$

$\gamma_{\text{{sy}},2}(E)\leq C_{2}\,\gamma_{\text{sy},\text{op}}(E)$,

for some positive absolute constants $C_{1},C_{2}$, we define the variational
capacity:

######  Definition 3.1 (Operator $\pazocal{S}$ and variational capacity).

Let $E\subset\mathbb{R}^{n+1}$ be a compact set and choose $\tau_{0}$ small
enough so that

$\gamma_{\text{{sy}},\psi_{\tau_{0}},2}(E)\leq
2C_{1}C_{2}\,\gamma_{\text{sy},\text{op}}(E).$

Let $\pazocal{S}$ be the convolution operator associated to the kernel
$P_{\text{{sy}}}\psi_{\tau_{0}}$ (that depends on $\tau_{0}$ and thus on $E$).
We define the variational capacity of $E$ as

$\gamma_{\text{var}}(E):=\sup\Bigg{\\{}\frac{\mu(E)^{2}}{\mu(E)+\int_{E}|\pazocal{S}\mu|^{2}\mathop{}\\!\mathrm{d}\mu}\,:\,\mu\in\Sigma(E)\Bigg{\\}}.$

We convey that expressions of the form $\frac{0}{0}$ equal 0.

######  Lemma 3.4.

The supremum in $\gamma_{\text{{var}}}(E)$ is attained. Moreover,
$\gamma_{\text{{var}}}(E)\approx\gamma_{\text{{sy}},\text{{op}}}(E)$, where we
remark that the implicit constants do not depend on $E$.

###### Proof.

For each $\mu\in\Sigma(E)$ define

$F(\mu):=\frac{\mu(E)^{2}}{\mu(E)+\int_{E}|\pazocal{S}\mu|^{2}\mathop{}\\!\mathrm{d}\mu}.$

As it has already been mentioned in the proof of Lemma 3.3, it is not
difficult to prove that $\Sigma(E)$ is closed and sequentially compact with
respect to weak convergence of measures. In fact, $\Sigma(E)$ is compact
(since it is contained in the space of finite signed Radon measures on $E$,
which is metrizable, thought as the dual of the separable space
$\pazocal{C}_{c}(\mathbb{R}^{n+1})$). Moreover, it is clear that if
$\mu_{k}\rightharpoonup\mu_{0}$ on $\Sigma(E)$, then
$\mu_{k}(E)\to\mu_{0}(E)$. In this setting, we also claim that

(3.8)
$\lim_{k\to\infty}\int_{E}|\pazocal{S}\mu_{k}|^{2}\mathop{}\\!\mathrm{d}\mu_{k}=\int_{E}|\pazocal{S}\mu_{0}|^{2}\mathop{}\\!\mathrm{d}\mu_{0}.$

To prove the claim (3.8) we argue as follows: given continuous functions
$\varphi_{i},\psi_{i}$ for $i=1,\ldots N$ in
$\pazocal{C}_{c}(\mathbb{R}^{n+1})$, by definition of weak convergence of
measures we get

(3.9) $\displaystyle\lim_{k\to\infty}$
$\displaystyle\int_{E}\bigg{(}\int_{E}\sum_{i=1}^{N}\varphi_{i}(\overline{x})\psi_{i}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})$
$\displaystyle=\lim_{k\to\infty}\bigg{\\{}\sum_{i=1}^{N}\int_{E}\bigg{(}\int_{E}\varphi_{i}(\overline{x})\psi_{i}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})$
$\displaystyle\hskip 49.79231pt+2\sum_{1\leq i<j\leq
N}\int_{E}\bigg{(}\int_{E}\varphi_{i}(\overline{x})\psi_{i}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}\bigg{(}\int_{E}\varphi_{j}(\overline{x})\psi_{j}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})\bigg{\\}}$
$\displaystyle=\lim_{k\to\infty}\bigg{\\{}\sum_{i=1}^{N}\bigg{(}\int_{E}\varphi_{i}(\overline{x})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}^{2}\bigg{(}\int_{E}\psi_{i}^{2}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})\bigg{)}$
$\displaystyle\hskip 31.2982pt+2\sum_{1\leq i<j\leq
N}\bigg{(}\int_{E}\varphi_{i}(\overline{x})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}\bigg{(}\int_{E}\varphi_{j}(\overline{x})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}\bigg{(}\int_{E}\psi_{i}(\overline{y})\psi_{j}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})\bigg{)}\bigg{\\}}$
$\displaystyle=\int_{E}\bigg{(}\int_{E}\sum_{i=1}^{N}\varphi_{i}(\overline{x})\psi_{i}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y}).$

Observe that the collection of continuous functions on the compact set
$E\times E$ of the form
$\sum_{i=1}^{N}\varphi_{i}(\overline{x})\psi_{i}(\overline{y})$ is an algebra
that contains the constant functions (on $E$) and separates points. Therefore,
by the Stone-Weierstrass theorem, every continuous function $f:E\times
E\to\mathbb{R}$ can be uniformly approximated by such sums. Namely, consider

$\big{(}g_{j}(\overline{x},\overline{y})\big{)}_{j}:=\bigg{(}\sum_{i=1}^{N_{j}}\varphi_{i,j}(\overline{x})\psi_{i,j}(\overline{y})\bigg{)}_{j},$

so that $\|f-g_{j}\|_{\infty}:=\|f-g_{j}\|_{L^{\infty}(E\times E)}\to 0$ as
$j\to\infty$. Observe:

$\displaystyle\bigg{\rvert}$
$\displaystyle\int_{E}\bigg{(}\int_{E}f(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})-\int_{E}\bigg{(}\int_{E}f(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})\bigg{\rvert}$
$\displaystyle\leq\bigg{\rvert}\int_{E}\bigg{(}\int_{E}f(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})-\int_{E}\bigg{(}\int_{E}g_{j}(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})\bigg{\rvert}$
$\displaystyle\hskip
28.45274pt+\bigg{\rvert}\int_{E}\bigg{(}\int_{E}g_{j}(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})-\int_{E}\bigg{(}\int_{E}g_{j}(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})\bigg{\rvert}$
$\displaystyle\hskip
28.45274pt+\bigg{\rvert}\int_{E}\bigg{(}\int_{E}g_{j}(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})-\int_{E}\bigg{(}\int_{E}f(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})\bigg{\rvert}$
$\displaystyle\hskip 28.45274pt=:\text{\@slowromancap i@}+\text{\@slowromancap
ii@}+\text{\@slowromancap <EMAIL_ADDRESS>
Choose $j$ large enough so that $\|g_{j}\|_{\infty}\leq 2\|f\|_{\infty}$. We
estimate i@ as follows

i@
$\displaystyle=\Bigg{\rvert}\int_{E}\bigg{(}\int_{E}(f+g_{j})(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}\bigg{(}\int_{E}(f-g_{j})(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}\mathop{}\\!\mathrm{d}\mu({\overline{y}})\Bigg{\rvert}$
$\displaystyle\leq
3\|f\|_{\infty}\cdot\text{diam}(E)^{3n}\|f-g_{j}\|_{\infty}\xrightarrow[]{j\to\infty}0.$

Term iii@ can be treated the same way. For ii@ use that
$\mu_{k}\rightharpoonup\mu_{0}$ as $k\to\infty$ and apply (3.9) to deduce that
ii@ also tends to 0 as $j\to\infty$. The latter can be done since the supports
of the measures $\mu_{k}$ are all contained in $E$, so we may understand each
factor $\varphi_{i,j}$ and $\psi_{i,j}$ of the summands that define $g_{j}$
once extended continuously onto $\pazocal{C}_{c}(\mathbb{R}^{n+1})$. Hence, by
the arbitrariness of $\varepsilon$ we deduce

$\lim_{k\to\infty}\int_{E}\bigg{(}\int_{E}f(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{k}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{k}(\overline{y})=\int_{E}\bigg{(}\int_{E}f(\overline{x},\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{x})\bigg{)}^{2}\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y}),$

for any $f$ continuous on $E\times E$. Therefore, applying this result to
$f:=P_{\text{{sy}}}\psi_{\tau_{0}}|_{E}\geq 0$ we get (3.8). All in all, we
have proved that $F$ defines a continuous functional on a compact space,
meaning that it attains its maximum and thus the supremum that defines
$\gamma_{\text{var}}$ is indeed attained.
In order to prove
$\gamma_{\text{{var}}}(E)\approx\gamma_{\text{{sy}},\text{{op}}}(E)$, we claim
that it suffices to prove

$\gamma_{\text{{var}}}(E)\approx\gamma_{\text{{sy}},\psi_{\tau_{0}},2}(E).$

Indeed, since $0\leq P_{\text{{sy}}}\psi_{\tau_{0}}\leq P_{\text{{sy}}}$, we
trivially have
$\gamma_{\text{{sy}},\text{{op}}}(E)\leq\gamma_{\text{{sy}},\psi_{\tau_{0}},2}(E)$;
and by the choice of $\tau_{0}$, we also have
$\gamma_{\text{{sy}},\psi_{\tau_{0}},2}(E)\leq
2C_{1}C_{2}\,\gamma_{\text{{sy}},\text{{op}}}(E)$, so the claim follows.
Further, observe that we can restrict ourselves to the case that there exists
$\mu\in\Sigma(E)$ such that $F(\mu)>0$. If this was not the case, since
$\int_{E}|\pazocal{S}\mu|\mathop{}\\!\mathrm{d}\mu\lesssim\text{diam}(E)^{2n}/\tau_{0}^{n}<\infty$,
we would have, necessarily, $\mu(E)=0$, implying, by definition,
$\gamma_{\text{var}}(E)=0=\gamma_{\text{{sy}},\psi_{\tau_{0}},2}(E)$.
So let us assume that there exists $\mu\in\Sigma(E)$ such that $F(\mu)>0$,
meaning $\gamma_{\text{var}}(E)>0$ . Let us begin by proving that any extremal
$\mu_{0}$ for $F$ in this setting satisfies

(3.10)
$\int_{E}|\pazocal{S}\mu_{0}|^{2}\mathop{}\\!\mathrm{d}\mu_{0}\leq\mu_{0}(E).$

If this was not the case, then
$\int_{E}|\pazocal{S}\mu_{0}|^{2}\mathop{}\\!\mathrm{d}\mu_{0}=M\mu_{0}(E)$
for some $M>1$. Define $\mu_{1}:=M^{-1/2}\mu_{0}\in\Sigma(E)$ and notice that

$\displaystyle F(\mu_{1})$
$\displaystyle=\frac{M^{-1}\mu_{0}(E)^{2}}{M^{-1/2}\mu_{0}(E)+\int_{E}|\pazocal{S}\mu_{1}|^{2}\mathop{}\\!\mathrm{d}\mu_{1}}$
$\displaystyle=\frac{M^{-1}\mu_{0}(E)^{2}}{M^{-1/2}\mu_{0}(E)+M^{-1/2}\mu_{0}(E)}=\frac{\mu_{0}(E)}{2M^{1/2}}>\frac{\mu_{0}(E)}{1+M}=F(\mu_{0}),$

that is a contradiction. Then, (3.10) holds, and it implies, by definition,
$\mu_{0}(E)\leq\gamma_{\text{{sy}},\psi_{\tau_{0}},2}(E)$. So we have

$\gamma_{\text{var}}(E)=F(\mu_{0})<\mu_{0}(E)\leq\gamma_{\text{{sy}},\psi_{\tau_{0}},2}(E).$

On the other hand, for any $\mu$ admissible for
$\gamma_{\text{{sy}},\psi_{\tau_{0}},2}(E)$,

$\frac{\mu(E)}{2}=\frac{\mu(E)^{2}}{\mu(E)+\mu(E)}\leq\frac{\mu(E)^{2}}{\mu(E)+\int_{E}|\pazocal{S}\mu|^{2}\mathop{}\\!\mathrm{d}\mu}\leq\gamma_{\text{var}}(E),$

and the proof is complete. ∎

######  Remark 3.4.

Let us notice that (3.10) holds in general for any extremal measure $\mu_{0}$
for $F$. Indeed, in the case that for any $\mu\in\Sigma(E)$ one had
$F(\mu)=0$, then $\mu(E)=0$. So any measure would be extremal and (3.10) would
hold trivially.

Finally we give a new version of Theorem 2.1 that relates the capacities
presented in this section:

######  Theorem 3.5.

All of the above capacities defined via positive Borel measures are
comparable. More precisely, for $E\subset\mathbb{R}^{n+1}$ compact,

$\gamma_{+}(E)\approx\gamma^{\prime}_{+}(E)\approx\gamma^{\star}_{+}(E),$

where the respective bounds of the potentials can be taken indifferently and
independently for $P,P^{\ast},P_{\text{{sy}}}$ or both $P$ and $P^{\ast}$
simultaneously. Moreover, they are also comparable to

$\widetilde{\gamma}_{2}(E),\;\gamma_{\text{{sy}},2}(E),\;\gamma_{\text{{op}}}(E),\;\gamma_{\text{{sy,op}}}(E)\quad\text{and}\quad\gamma_{\text{{var}}}(E).$

In addition, for each $\tau>0$ the following holds

$\displaystyle\gamma_{\text{{sy}},\psi_{\tau},+}(E)$
$\displaystyle\approx\gamma_{\text{{sy}},\psi_{\tau},+}^{\prime}(E)\approx\gamma_{\text{{sy}},\psi_{\tau},+}^{\star}(E)\approx\gamma_{\text{{sy}},\psi_{\tau},2}(E)\approx\gamma_{\text{{sy}},\psi_{\tau},\text{{op}}}(E)$

as well as

$\limsup_{\tau\to
0}\gamma_{\text{{sy}},\psi_{\tau},+}(E)\lesssim\gamma_{\text{{sy}},+}(E)\lesssim\liminf_{\tau\to
0}\gamma_{\text{{sy}},\psi_{\tau},+}(E).$

The latter relations for the capacities depending on $\psi_{\tau}$ also hold
changing each $\gamma_{\text{{sy}}}$ for $\widetilde{\gamma}$.

##  4\. The construction of the cubes

The goal of this section is to use Theorem 3.5 to carry out the construction
done in [Vo, §5.2] for Riesz kernels. The aim of this section is to prove the
following result:

######  Theorem 4.1.

Let $E\subset B_{1}(0)\subset\mathbb{R}^{n+1}$ be a compact set consisting of
a finite union of closed cubes, with sides parallel to the axes. Then, there
exists a finite collection of dyadic cubes
$\\{\pazocal{Q}_{1},\ldots,\pazocal{Q}_{N}\\}$ that cover $E$ and such that
$\frac{1}{2}\pazocal{Q}_{1},\ldots,\frac{1}{2}\pazocal{Q}_{N}$ have disjoint
interiors. Moreover, if $\pazocal{F}:=\cup_{i=1}^{N}\pazocal{Q}_{i}$,

  1. $\text{{{P}}}_{\boldsymbol{1}}$.

$\frac{5}{8}\pazocal{Q}_{i}\cap E\neq\varnothing$, for each $i=1,\ldots,N$.

  2. $\text{{{P}}}_{\boldsymbol{2}}$.

$\widetilde{\gamma}_{+}(\pazocal{F})\leq C_{0}\widetilde{\gamma}_{+}(E)$.

  3. $\text{{{P}}}_{\boldsymbol{3}}$.

$\sum_{i=1}^{N}\widetilde{\gamma}_{+}(2\pazocal{Q}_{i}\cap E)\leq
C_{1}\widetilde{\gamma}_{+}(E)$.

  4. $\text{{{P}}}_{\boldsymbol{4}}$.

If $\widetilde{\gamma}_{+}(E)\leq C_{+}\text{{diam}}(E)^{n}$, then
$\text{{diam}}(\pazocal{Q}_{i})\leq\frac{1}{10}\text{{diam}}(E)$, for each
$i=1,\ldots,N$.

  5. $\text{{{P}}}_{\boldsymbol{5}}$.

The family $\\{5\pazocal{Q}_{1},\ldots,5\pazocal{Q}_{N}\\}$ has bounded
overlap with constant $C_{2}$.

Constants $C_{0},C_{1},C_{2},C_{+}$ are absolute.

Before proceeding, let us clarify the from this point on
$E\subset\mathbb{R}^{n+1}$ will be a fixed compact set and $\mu_{0}$ will
always denote a maximizer of $\gamma_{\text{var}}(E)$. Let us also write
explicitly the following expression, that will appear repeatedly, for the sake
of clarity: for $\mu,\nu$ say positive finite Borel measures supported on $E$,
we have, in light of Definition 3.1,

$\pazocal{S}_{\mu}(\pazocal{S}\nu)(\overline{x})=\int_{E}\int_{E}P_{\text{{sy}}}(\overline{x}-\overline{y})P_{\text{{sy}}}(\overline{y}-\overline{z})\psi_{\tau_{0}}(\overline{x}-\overline{y})\psi_{\tau_{0}}(\overline{y}-\overline{z})\mathop{}\\!\mathrm{d}\nu(\overline{z})\mathop{}\\!\mathrm{d}\mu(\overline{y}).$

######  Lemma 4.2.

Let $H$ be a positive Borel measure supported on $E$ such that for any
$\lambda_{0}>0$ and any $\lambda\in[0,\lambda_{0}]$,
$\mu_{\lambda}:=\mu_{0}+\lambda H\in\Sigma(E)$. Then,

$H(E)\mu_{0}(E)\bigg{(}\mu_{0}(E)+2\int_{E}|\pazocal{S}\mu_{0}|^{2}\mathop{}\\!\mathrm{d}\mu_{0}\bigg{)}\leq\mu_{0}(E)^{2}\int_{E}\Big{(}|\pazocal{S}\mu_{0}|^{2}+2\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})\Big{)}\mathop{}\\!\mathrm{d}H.$

###### Proof.

Let us begin by noticing that

$F(\mu_{\lambda})=\frac{\big{[}\mu_{0}(E)+\lambda
H(E)\big{]}^{2}}{\mu_{0}(E)+\lambda
H(E)+\int_{E}|\pazocal{S}\mu_{\lambda}|^{2}\mathop{}\\!\mathrm{d}\mu_{\lambda}},$

where the integral in the denominator can be expanded as

$\displaystyle\int_{E}|\pazocal{S}\mu_{\lambda}$
$\displaystyle|^{2}\mathop{}\\!\mathrm{d}\mu_{\lambda}=\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}\mu_{0}+\lambda\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}H+2\lambda\int_{E}(\pazocal{S}\mu_{0})(\pazocal{S}H)\mathop{}\\!\mathrm{d}\mu_{0}$
$\displaystyle\hskip
71.13188pt+2\lambda^{2}\int_{E}(\pazocal{S}\mu_{0})(\pazocal{S}H)\mathop{}\\!\mathrm{d}H+\lambda^{2}\int_{E}(\pazocal{S}H)^{2}\mathop{}\\!\mathrm{d}\mu_{0}+\lambda^{3}\int_{E}(\pazocal{S}H)^{2}\mathop{}\\!\mathrm{d}H.$

Observe that for $\nu\in\\{\mu_{0},H\\}$, since $P_{\text{{sy}}}$ and
$\psi_{\tau_{0}}$ are symmetric functions,

$\displaystyle\int_{E}\pazocal{S}\mu_{0}$ $\displaystyle
1(\overline{x})\pazocal{S}H(\overline{x})\mathop{}\\!\mathrm{d}\nu(\overline{x})$
$\displaystyle=\int_{E}\bigg{(}\int_{E}P_{\text{{sy}}}(\overline{x}-\overline{y})\psi_{\tau_{0}}(\overline{x}-\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})\bigg{)}\bigg{(}\int_{E}P_{\text{{sy}}}(\overline{x}-\overline{z})\psi_{\tau_{0}}(\overline{x}-\overline{z})\mathop{}\\!\mathrm{d}H(\overline{z})\bigg{)}\mathop{}\\!\mathrm{d}\nu(\overline{x})$
$\displaystyle=\int_{E}\bigg{(}\int_{E}\bigg{(}\int_{E}P_{\text{{sy}}}(\overline{x}-\overline{y})\psi_{\tau_{0}}(\overline{x}-\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})\bigg{)}P_{\text{{sy}}}(\overline{z}-\overline{x})\psi_{\tau_{0}}(\overline{z}-\overline{x})\mathop{}\\!\mathrm{d}\nu(\overline{x})\bigg{)}\mathop{}\\!\mathrm{d}H(\overline{z})$
$\displaystyle=\int_{E}\pazocal{S}_{\nu}(\pazocal{S}\mu_{0})(\overline{z})\mathop{}\\!\mathrm{d}H(\overline{z}).$

Therefore, we may rewrite
$\int_{E}|\pazocal{S}\mu_{\lambda}|^{2}\mathop{}\\!\mathrm{d}\mu_{\lambda}$ as

$\displaystyle\int_{E}|\pazocal{S}\mu_{\lambda}$
$\displaystyle|^{2}\mathop{}\\!\mathrm{d}\mu_{\lambda}=\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}\mu_{0}+\lambda\bigg{[}\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}H+2\int_{E}\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})\mathop{}\\!\mathrm{d}H\bigg{]}$
$\displaystyle\hskip
71.13188pt+\lambda^{2}\bigg{[}\int_{E}(\pazocal{S}H)^{2}\mathop{}\\!\mathrm{d}\mu_{0}+2\int_{E}\pazocal{S}_{H}(\pazocal{S}\mu_{0})\mathop{}\\!\mathrm{d}H\bigg{]}+\lambda^{3}\int_{E}(\pazocal{S}H)^{2}\mathop{}\\!\mathrm{d}H.$

So condition $F(\mu_{0})\geq F(\mu_{\lambda})$ can be written as

$\displaystyle\frac{\mu_{0}(E)^{2}}{\mu_{0}(E)+\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}\mu_{0}}$
$\displaystyle\geq\frac{\mu_{0}(E)^{2}+2\lambda\mu_{0}(E)H(E)+\lambda^{2}H(E)^{2}}{\mu_{0}(E)+\lambda
H(E)+\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}\mu_{0}+\lambda\big{[}\cdots\big{]}+\lambda^{2}\big{[}\cdots\big{]}+\lambda^{3}\big{[}\cdots\big{]}},$

or equivalently

$\displaystyle\lambda\Bigg{\\{}\mu_{0}(E)^{2}\bigg{(}-H(E)$
$\displaystyle+\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}H+2\int_{E}\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})\mathop{}\\!\mathrm{d}H\bigg{)}$
$\displaystyle-2\mu_{0}(E)H(E)\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}\mu_{0}\Bigg{\\}}+\lambda^{2}\big{[}\cdots\big{]}+\lambda^{3}\int_{E}(\pazocal{S}H)^{2}\mathop{}\\!\mathrm{d}H\geq
0.$

Assume that $\lambda>0$ and divide both sides of the previous inequality by
this factor and make $\lambda\to 0$ to obtain the desired estimate. Let us
notice that the kernel associated to $\pazocal{S}$ is continuous, meaning that
the terms in $[\cdots]$ and
$\int_{E}(\pazocal{S}H)^{2}\mathop{}\\!\mathrm{d}H$ are finite, since
$H(E)<\infty$ due to the fact that $\mu_{\lambda}\in\Sigma(E)$ for any
$\lambda\in[0,\lambda_{0}]$ with $\lambda_{0}>0$. ∎

We remark that in the particular case we will apply this lemma, $E$ will be a
finite union of cubes, so $\mu_{0}(E)>0$. Moreover, we will choose a specific
measure $H$ so that $H(E)>0$. Therefore, in this setting, the inequality in
Lemma 4.3 can be rewritten as

(4.1)
$\frac{\mu_{0}(E)+2\int_{E}|\pazocal{S}\mu_{0}|^{2}\mathop{}\\!\mathrm{d}\mu_{0}}{\mu_{0}(E)}\leq\frac{1}{H(E)}\int_{E}\Big{(}|\pazocal{S}\mu_{0}|^{2}+2\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})\Big{)}\mathop{}\\!\mathrm{d}H,$

which is a similar estimate that resembles that of [Vo, Lemma 5.6].
We shall apply (4.1) to prove an auxiliary lemma, but first let us introduce
the following notation for a certain maximal function applied to a measure:

$M\nu(\overline{x}):=\sup_{r>0}\frac{\nu(B_{r}(\overline{x}))}{r^{n}}$

######  Lemma 4.3.

Let $E$ be a finite union of cubes. Then, for the potential

$\pazocal{U}^{\mu_{0}}:=M\mu_{0}+\pazocal{S}\mu_{0}+\pazocal{S}_{\mu_{0}}\pazocal{S}\mu_{0},$

there is an absolute constant $\alpha_{0}>0$ such that

$\pazocal{U}^{\mu_{0}}(\overline{x})\geq\alpha_{0},\quad\forall\overline{x}\in
E.$

###### Proof.

Choose $\alpha_{1}\in(0,80^{-n})$. If $\overline{x}_{0}\in E$ is such that
$M\mu_{0}(\overline{x}_{0})\geq\alpha_{1}$ then we are done at
$\overline{x}_{0}$. So assume $M\mu_{0}(\overline{x}_{0})<\alpha_{1}$. Pick
$\varepsilon_{1}\leq\tau_{0},R_{0}:=\varepsilon_{1}/10$ and
$B_{0}:=B_{R_{0}}(\overline{x}_{0})$. Set $\mu_{00}:=\mu_{0}|_{2B_{0}}$ and
define

$G:=\big{\\{}\overline{y}\in B_{0}\cap E\,:\,M\mu_{0}(\overline{y})\leq
4^{-n}\big{\\}}.$

We shall prove that the complement of $G$ in $B_{0}\cap E$ is small. If
$\overline{y}\in(B_{0}\cap E)\setminus{G}$, then there exists
$r=r(\overline{y})>0$ so that

$\frac{\mu_{0}(B_{r}(\overline{y}))}{r^{n}}>4^{-n}.$

If we had $r>\varepsilon_{1}/20=R_{0}/2$, then

$\frac{1}{80^{n}}\leq\frac{\mu_{0}(B_{r}(\overline{y}))}{20^{n}r^{n}}\leq\frac{\mu_{0}(B_{20r}(\overline{x}_{0}))}{(20r)^{n}}<\alpha_{1},\quad\text{since}\;M\mu_{0}(\overline{x}_{0})<\alpha_{1},$

and this cannot be. Then $r\leq\varepsilon_{1}/20$, implying
$B_{r}(\overline{y})\cap(2B_{0})=B_{r}(\overline{y})$, which in turn implies
$\mu_{00}(B_{r}(\overline{y}))=\mu_{0}(B_{r}(\overline{y}))$. So we have found

(4.2) $\forall\overline{y}\in(B_{0}\cap E)\setminus{G},\,\exists
r(\overline{y})>0\;\text{ such that
}\;\frac{\mu_{00}(B_{r}(\overline{y}))}{r^{n}}>4^{-n}.$

We continue by choosing $\varepsilon_{1}$ (that was already smaller than
$\tau_{0}$) small enough so that

$\pazocal{L}^{n+1}(B_{0}\cap E)\geq a(n)R_{0}^{n+1}=:aR_{0}^{n+1},$

for some dimensional constant $a>0$. This can be done since $E$ is a finite
union of cubes. Notice also that the dependence on $E$ of the previous
estimate is in $R_{0}$, since $\varepsilon_{1}$ depends on $\tau_{0}$, that
depends on $E$. Using (4.2) and [Ma, Theorem 2.1] we obtain a countable
covering $\\{B_{5r_{j}}(\overline{y}_{j})\\}$ of $(B_{0}\cap E)\setminus{G}$
with $\\{B_{r_{j}}(\overline{y}_{j})\\}$ disjoint and also satisfying (4.2).
Now,

$\displaystyle\pazocal{L}^{n+1}\big{(}(B_{0}\cap E)\setminus{G}\big{)}$
$\displaystyle\leq 5^{n+1}\sum_{j}r_{j}^{n+1}\leq 2\cdot
5^{n+1}R_{0}\sum_{j}r_{j}^{n}$ $\displaystyle<10\cdot 4^{n}\cdot
5^{n+1}R_{0}\sum_{j}\mu_{00}\big{(}B_{r_{j}}(\overline{y}_{j})\big{)}=10\cdot
20^{n}R_{0}\,\mu_{00}\bigg{(}\bigcup_{j}B_{r_{j}}(\overline{y}_{j})\bigg{)}$
$\displaystyle\leq 10\cdot 20^{n}R_{0}\,\mu_{0}(2B_{0})\leq 10\cdot
40^{n}R_{0}^{n+1}\alpha_{1}=:A(n)R_{0}^{n+1}\alpha_{1}$
$\displaystyle\leq\frac{A}{a}\alpha_{1}\pazocal{L}^{n+1}(B_{0}\cap E),$

where we have applied the hypothesis $M\mu_{0}(\overline{x}_{0})<\alpha_{1}$.
Therefore, by definition of $G$,

$\pazocal{L}^{n+1}(G)\geq\bigg{(}\frac{a}{A\alpha_{1}}-1\bigg{)}\pazocal{L}^{n+1}\big{(}(B_{0}\cap
E)\setminus{G}\big{)}.$

We pick $\alpha_{1}\in(0,80^{-n})$ depending only on $n$ so that

$\pazocal{L}^{n+1}(G):=\pazocal{L}^{n+1}\big{(}\big{\\{}\overline{y}\in
B_{0}\cap E\,:\,M\mu_{0}(\overline{y})\leq 4^{-n}\big{\\}}\big{)}>0.$

We will now prove that for all sufficiently small $\lambda>0$,
$\mu_{0}+\lambda\pazocal{L}^{n+1}|_{G}$ belongs to $\Sigma(E)$, i.e.

$M\big{(}\mu_{0}+\lambda\pazocal{L}^{n+1}|_{G}\big{)}\leq 1,\quad\text{on
}\,E\,\text{ if }\,\lambda\in(0,\lambda_{0}).$

Let us fix $\overline{z}\in E,r>0$ and write
$\mu_{\lambda}:=\mu_{0}+\lambda\pazocal{L}^{n+1}|_{G}$. Distinguish three
cases: first, if $B_{r}(\overline{z})\cap G=\varnothing$ then it is clear that

$\frac{\mu_{\lambda}(B_{r}(\overline{z}))}{r^{n}}=\frac{\mu_{0}(B_{r}(\overline{z}))}{r^{n}}\leq
1,$

that is the desired estimate. If $B_{r}(\overline{z})\cap G\neq\varnothing$
and $\overline{z}\in G$, then

$\displaystyle\frac{\mu_{\lambda}(B_{r}(\overline{z}))}{r^{n}}=\frac{\mu_{0}(B_{r}(\overline{z}))}{r^{n}}+\lambda\frac{\pazocal{L}^{n+1}(B_{r}(\overline{z})\cap
G)}{r^{n}}\leq\frac{1}{4^{n}}+\lambda\frac{\min\big{\\{}r^{n+1},R_{0}^{n+1}\big{\\}}}{r^{n}}.$

If $r\leq R_{0}$, then

$\frac{\mu_{\lambda}(B_{r}(\overline{z}))}{r^{n}}\leq\frac{1}{4^{n}}+\lambda
r\leq\frac{1}{4^{n}}+\lambda R_{0}<1,\quad\text{for $\lambda$ small enough}.$

If on the other hand $r>R_{0}$, we also have

$\frac{\mu_{\lambda}(B_{r}(\overline{z}))}{r^{n}}\leq\frac{1}{4^{n}}+\lambda\frac{R_{0}^{n+1}}{r^{n}}\leq\frac{1}{4^{n}}+\lambda
R_{0}<1,\quad\text{for $\lambda$ small enough}.$

Finally, for the third case, that is, if $B_{r}(\overline{z})\cap
G\neq\varnothing$ and $\overline{z}\not\in G$, take $\overline{y}\in
B_{r}(\overline{z})\cap G$ so that

$\displaystyle\frac{\mu_{\lambda}(B_{r}(\overline{z}))}{r^{n}}$
$\displaystyle=\frac{\mu_{0}(B_{r}(\overline{z}))}{r^{n}}+\lambda\frac{\min\big{\\{}r^{n+1},R_{0}^{n+1}\big{\\}}}{r^{n}}\leq
2^{n}\frac{\mu_{0}(B_{2r}(\overline{y}))}{(2r)^{n}}+\lambda R_{0}$
$\displaystyle\leq\frac{1}{2^{n}}+(2^{n}+1)\lambda R_{0},$

where for the last inequality we have applied the reasoning of the second
case. Hence, in general, we deduce that for $\lambda>0$ small enough,
$\mu_{\lambda}\in\Sigma(E)$.
Now we argue as follows: choose $\varepsilon_{1}^{(k)}\leq\varepsilon_{1}$ a
sequence converging to 0. For each $k$ set
$R_{0}^{(k)}:=\varepsilon_{1}^{(k)}/10$ and build a set $G_{k}$ in
$B_{R_{0}^{(k)}}(\overline{x}_{0})$ as above. Let
$H_{k}:=\pazocal{L}^{n+1}|_{G_{k}}$ and apply Lemma 4.2, or in this particular
case (4.1), to get

$1\leq\frac{\mu_{0}(E)+2\int_{E}|\pazocal{S}\mu_{0}|^{2}\mathop{}\\!\mathrm{d}\mu_{0}}{\mu_{0}(E)}\leq\frac{1}{H_{k}(E)}\int_{E}\Big{(}|\pazocal{S}\mu_{0}|^{2}+2\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})\Big{)}\mathop{}\\!\mathrm{d}H_{k}.$

Since the measures $H_{k}/H_{k}(E)$ converge weakly to a point mass
probability measure at $\overline{x}_{0}$, by the continuity of the kernel
associated to $\pazocal{S}$ we have

$1\leq|\pazocal{S}\mu_{0}|^{2}(\overline{x}_{0})+2\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{x}_{0})\quad\rightarrow\quad\frac{1}{2}\leq|\pazocal{S}\mu_{0}|^{2}(\overline{x}_{0})+\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{x}_{0}).$

Therefore, either $|\pazocal{S}\mu_{0}|^{2}(\overline{x}_{0})\geq 1/4$ or
$\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{x}_{0})\geq 1/4$, so in
any case we get

$\pazocal{S}\mu_{0}(\overline{x}_{0})+\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{x}_{0})\geq\frac{1}{4},$

by the nonnegativity of each term. Since this last estimate follows if
$M\mu_{0}(\overline{x}_{0})$ is smaller than $\alpha_{1}$, chosen to depend
only on $n$, setting $\alpha_{0}:=\min\\{\alpha_{1},1/4\\}$ the result
follows. ∎

###  4.1. The construction

Both of the above lemmas are enough to carry out the construction of the
desired cubes in a similar way to [Vo, pp. 38-42]. Fix $E\subset
B_{1}(0)\subset 2Q_{0}$ compact consisting of a finite union of cubes, and
define the auxiliary potential

$\widetilde{\pazocal{U}}^{\mu_{0}}(\overline{x}):=\pazocal{U}^{\mu_{0}}(\overline{x})+M(\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0})(\overline{x}).$

Apply Lemma 4.3 to $E$ and obtain its corresponding constant $\alpha_{0}$, and
pick $0<\beta\ll 1$ depending only on $n$, to be fixed later on, and set

$\pazocal{G}:=\big{\\{}\overline{y}\in\mathbb{R}^{n+1}\,:\,\widetilde{\pazocal{U}}^{\mu_{0}}(\overline{y})>\beta\alpha_{0}\big{\\}}\supset
E.$

Consider $\\{Q_{j}\\}_{j}$ a Whitney decomposition of $\pazocal{G}$ (see [St,
pp. 167-169], for example), that is: a countable family of closed dyadic cubes
with disjoint interiors that cover $\pazocal{G}$ and such that for some
absolute (possibly dimension dependent) constant $A>0$,

  * •

$20Q_{j}\subset\pazocal{G}$ for each $j$,

  * •

$(AQ_{j})\cap\pazocal{G}^{c}\neq\varnothing$ for each $j$,

  * •

The family $\\{10Q_{j}\\}_{j}$ has bounded overlap. Moreover, if $10Q_{j}\cap
10Q_{i}\neq\varnothing$, then $\ell(Q_{j})\approx\ell(Q_{i})$.

From those $Q_{j}$ satisfying $\frac{5}{4}Q_{j}\cap E\neq\varnothing$, choose
a finite subcovering and enumerate them as $Q_{j_{1}},\ldots,Q_{j_{N}}$. We
shall check that

$\pazocal{Q}_{i}:=2Q_{j_{i}},\quad i=1,\ldots,N$

is the desired family of cubes that satisfies the properties in Theorem 4.1.
It is clear that properties $\text{{{P}}}_{\boldsymbol{1}}$ and
$\text{{{P}}}_{\boldsymbol{5}}$ are satisfied by construction, so we are only
left to verify $\text{{{P}}}_{\boldsymbol{2}}$,
$\text{{{P}}}_{\boldsymbol{3}}$ and $\text{{{P}}}_{\boldsymbol{4}}$. The first
two will be checked for $\gamma_{\text{sy},+}$ instead of
$\widetilde{\gamma}_{+}$, which is enough due to the comparability of both
capacities.

###### Proof of $\text{{{P}}}_{\boldsymbol{2}}$ in Theorem 4.1..

We shall prove that property
$\widetilde{\pazocal{U}}^{\mu_{0}}(\overline{x})\geq\beta\alpha_{0}$ on
$\pazocal{F}$, implies

$\gamma_{\text{sy},\text{op}}(\pazocal{F})\leq\frac{C}{\beta\alpha_{0}}\mu_{0}(E),$

where $\mu_{0}$ is an extremal measure for $\gamma_{\text{var}}(E)$ and $C>0$
is absolute. If this is the case, we are done, since we have

$\gamma_{\text{sy},+}(\pazocal{F})\approx\gamma_{\text{sy},\text{op}}(\pazocal{F})\leq\frac{C}{\beta\alpha_{0}}\gamma_{\text{var}}(E)\leq\frac{C}{\beta\alpha_{0}}\gamma_{\text{sy},\psi_{\tau_{0}},2}(E)\leq\frac{C}{\beta\alpha_{0}}\,2C_{1}C_{2}\gamma_{\text{sy},\text{op}}(E)\approx\gamma_{\text{sy},+}(E),$

where for the third estimate recall Remark 3.4 and the fact that relation
(3.10) holds in general, and in the fourth inequality we have used the
definition of $\tau_{0}$.
So let us choose $\mu$ admissible for
$\gamma_{\text{sy},\text{op}}(\pazocal{F})$ with
$\gamma_{\text{sy},\text{op}}(\pazocal{F})\leq 2\mu(\pazocal{F})$ and observe
that

$\|\pazocal{S}_{\mu}\|_{L^{2}(\mu)\to
L^{2}(\mu)}\leq\|\pazocal{P}_{\text{sy},\mu}\|_{L^{2}(\mu)\to L^{2}(\mu)}\leq
1,$

since $P_{\text{{sy}}}$ and the measure $\mu$ are nonnegative. Therefore,
$\pazocal{S}_{\mu}$ becomes a C-Z convolution operator (with continuous
kernel), so that it satisfies the weak estimate

$\mu\big{(}\\{\pazocal{S}\mu_{0}\geq\lambda\\}\big{)}\leq
C^{\prime}\frac{\mu_{0}(E)}{\lambda},$

where $C^{\prime}>0$ is absolute. By definition, for each
$\overline{x}\in\pazocal{F}$,

$\displaystyle\widetilde{\pazocal{U}}^{\mu_{0}}(\overline{x})$
$\displaystyle:=M\mu_{0}(\overline{x})+\pazocal{S}\mu_{0}(\overline{x})+\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{x})+M(\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0})(\overline{x})$
$\displaystyle=:\text{\@slowromancap i@}(\overline{x})+\text{\@slowromancap
ii@}(\overline{x})+\text{\@slowromancap
iii@}(\overline{x})+\text{\@slowromancap iv@}(\overline{x}),$

so at least one of the four terms is larger than
$\frac{\beta\alpha_{0}}{4}=:a$. In fact, if we name

$\displaystyle\pazocal{F}_{1}:=\big{\\{}\overline{x}\in
F\,:\,\text{\@slowromancap i@}(\overline{x})\geq a\big{\\}},$
$\displaystyle\qquad\pazocal{F}_{2}:=\big{\\{}\overline{x}\in
F\,:\,\text{\@slowromancap ii@}(\overline{x})\geq a\big{\\}},$
$\displaystyle\pazocal{F}_{3}:=\big{\\{}\overline{x}\in
F\,:\,\text{\@slowromancap iii@}(\overline{x})\geq a\big{\\}},$
$\displaystyle\qquad\pazocal{F}_{4}:=\big{\\{}\overline{x}\in
F\,:\,\text{\@slowromancap iv@}(\overline{x})\geq a\big{\\}},$

we have $\pazocal{F}=\cup_{i=1}^{4}\pazocal{F}_{i}$. So in particular
$\mu(\pazocal{F})\leq\sum_{i=1}^{4}\mu(\pazocal{F}_{i})$ and then,
necessarily, there exists $i\in\\{1,2,3,4\\}$ such that
$\mu(\pazocal{F}_{i})\geq\mu(\pazocal{F})/4$. Let us study separately each
possibility:

  1. a)

If $i=1$. In this case $M\mu_{0}\geq a$ on $\pazocal{F}_{1}$ with
$\mu(\pazocal{F}_{1})\geq\mu(\pazocal{F})/4$. For each
$\overline{x}\in\pazocal{F}_{1}$ choose a ball
$B_{\overline{x}}:=B_{r(\overline{x})}(\overline{x})$ so that
$\mu_{0}(B_{\overline{x}})\geq ar(\overline{x})^{n}$. Apply Besicovitch’s
covering theorem [Ma, Theorem 2.7] to extract a countable collection of balls
$\\{B_{\overline{x}_{j}}\\}_{j}$ with bounded multiplicity of overlapping
(with constant depending only on the dimension) that covers $\pazocal{F}_{1}$.
Then,

$\displaystyle\hskip
42.67912pt\frac{\mu(\pazocal{F})}{4}\leq\mu(\pazocal{F}_{1})\leq\sum_{j}\mu(B_{\overline{x}_{j}})\leq\sum_{j}r^{n}(\overline{x}_{j})\leq\frac{1}{a}\sum_{j}\mu_{0}(B_{\overline{x}_{j}})\lesssim\frac{\mu_{0}(\pazocal{F}_{1})}{a}\leq\frac{\mu_{0}(E)}{a},$

since $\mu_{0}$ is supported on $E$.

  2. b)

If $i=2$. Now $\pazocal{S}\mu_{0}\geq a$ on $\pazocal{F}_{2}$ with
$\mu(\pazocal{F}_{2})\geq\mu(\pazocal{F})/4$. Using the weak estimate
presented above we get the desired estimate

$\frac{\mu(\pazocal{F})}{4}\leq\mu(\pazocal{F}_{2})=\mu\big{(}\\{\pazocal{S}\mu_{0}\geq
a\\}\big{)}\leq\frac{C^{\prime}}{a}\mu_{0}(E).$

  3. c)

If $i=3$. Now $\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})\geq a$ on
$\pazocal{F}_{3}$ with $\mu(\pazocal{F}_{3})\geq\mu(\pazocal{F})/4$. By (3.10)
we obtain

$\displaystyle\frac{\mu(\pazocal{F})}{4}\leq\mu(\pazocal{F}_{3})$
$\displaystyle=\mu\big{(}\\{\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})\geq
a\\}\big{)}\leq\frac{C^{\prime}}{a}\int_{E}\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0}$
$\displaystyle\leq\frac{C^{\prime}}{a}\mu_{0}(E)^{1/2}\bigg{(}\int_{E}(\pazocal{S}\mu_{0})^{2}\mathop{}\\!\mathrm{d}\mu_{0}\bigg{)}^{2}\leq\frac{C^{\prime}}{a}\mu_{0}(E).$

  4. d)

If $i=4$. Now $M(\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0})\geq a$ on
$\pazocal{F}_{4}$ with $\mu(\pazocal{F}_{4})\geq\mu(\pazocal{F})/4$.
Proceeding analogously as in the first case, we deduce

$\frac{\mu(\pazocal{F})}{4}\lesssim\frac{1}{a}\int_{E}\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0},$

and arguing as in the third case we obtain the desired result.

∎

To prove $\text{{{P}}}_{\boldsymbol{3}}$ we will need the following lemma:

######  Lemma 4.4.

For $i=1,\ldots,N$ define $\mu_{i}:=\mu_{0}|_{5Q_{j_{i}}}$ and consider the
auxiliary potential

$\pazocal{W}^{\mu_{i}}:=M\mu_{i}+\pazocal{S}_{\mu_{i}}1+\pazocal{S}_{\mu_{i}}(\pazocal{S}\mu_{0})+M(\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{i}).$

Then, there exists an absolute constant $\alpha_{0}^{\prime}>0$ such that for
each $i=1,\ldots,N$,

$\pazocal{W}^{\mu_{i}}(\overline{x})\geq\alpha_{0}^{\prime}\alpha_{0},\qquad\forall\overline{x}\in
4Q_{j_{i}}\cap E.$

###### Proof.

For the sake of notation, in this proof we rename $Q_{i}:=Q_{j_{i}}$. Fix
$i=1,\ldots,N$, $\overline{x}\in 4Q_{i}\cap E$ and
$\overline{z}\in(AQ_{i})\cap\pazocal{G}^{c}$, which is non-empty by
construction. So in particular

(4.3)
$M\mu_{0}(\overline{z})\leq\beta\alpha_{0},\quad\pazocal{S}\mu_{0}(\overline{z})\leq\beta\alpha_{0}.$

Now choosing, for example, $R:=2(4+A)\text{diam}(Q_{i})$ it is clear that we
have

$|\overline{x}-\overline{z}|\leq(4+A)\text{diam}(Q_{i})\leq\frac{R}{2}.$

Therefore, using the C-Z estimates for $P_{\text{{sy}}}\psi_{\tau_{0}}$, the
nonnegativity of the latter kernel and relations (4.3) we have

$\displaystyle\pazocal{S}_{R}\mu_{0}(\overline{x})$
$\displaystyle:=\int_{|\overline{x}-\overline{y}|>R}P_{\text{{sy}}}(\overline{x}-\overline{y})\psi_{\tau_{0}}(\overline{x}-\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})$
$\displaystyle\leq\beta\alpha_{0}+\int_{|\overline{x}-\overline{y}|>R}\Big{\rvert}P_{\text{{sy}}}(\overline{x}-\overline{y})\psi_{\tau_{0}}(\overline{x}-\overline{y})-P_{\text{{sy}}}(\overline{z}-\overline{y})\psi_{\tau_{0}}(\overline{z}-\overline{y})\Big{\rvert}\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})$
$\displaystyle\lesssim\beta\alpha_{0}+|\overline{x}-\overline{z}|\int_{|\overline{x}-\overline{y}|>R}\frac{\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})}{|\overline{x}-\overline{y}|^{n+1}}\lesssim\beta\alpha_{0}+|\overline{x}-\overline{z}|\int_{|\overline{z}-\overline{y}|>R/4}\frac{\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})}{|\overline{z}-\overline{y}|^{n+1}}$
(4.4)
$\displaystyle\leq\beta\alpha_{0}+\frac{R}{2}\sum_{j=0}^{\infty}\frac{\mu_{0}(B_{2^{j+1}R}(\overline{z}))}{(2^{j}R)^{n+1}}\leq\beta\alpha_{0}+\frac{R}{2}\sum_{j=0}^{\infty}\frac{\beta\alpha_{0}(2^{j+1}R)^{n}}{(2^{j}R)^{n+1}}\lesssim\beta\alpha_{0}.$

We also notice that there exists $\widetilde{A}=\widetilde{A}(n)>0$ such that

(4.5)
$M\mu_{0}(\overline{x})\leq\max\big{\\{}\widetilde{A}\,\beta\alpha_{0},M\mu_{i}(\overline{x})\big{\\}}.$

Indeed, if $r\geq\ell(Q_{i})/20$ then

$\frac{\mu_{0}(B_{r}(\overline{x}))}{r^{n}}\leq\frac{\mu_{0}(B_{40\sqrt{n}Ar}(\overline{z}))}{r^{n}}\leq\widetilde{A}(n)\beta\alpha_{0}.$

If $r<\ell(Q_{i})/20$ use that $\overline{x}\in 4Q_{i}$ and
$\mu_{i}:=\mu_{0}|_{5Q_{i}}$ to simply have

$\frac{\mu_{0}(B_{r}(\overline{x}))}{r^{n}}=\frac{\mu_{j}(B_{r}(\overline{x}))}{r^{n}},$

so indeed
$M\mu_{0}(\overline{x})\leq\max\big{\\{}\widetilde{A}\,\beta\alpha_{0},M\mu_{i}(\overline{x})\big{\\}}$.
Having made this observations, we recall that by Lemma 4.3 we always have

$M\mu_{0}+\pazocal{S}\mu_{0}+\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})\geq\alpha_{0},\quad\text{on
}\,E.$

Now we distinguish three cases:

  1. a)

If $M\mu_{0}(\overline{x})\geq\alpha_{0}/3$. Then (4.5) implies

$\frac{\alpha_{0}}{3}\leq\max\big{\\{}\widetilde{A}\,\beta\alpha_{0},M\mu_{i}(\overline{x})\big{\\}},$

so choosing $\beta\leq(6\widetilde{A})^{-1}$ we deduce
$M\mu_{i}(\overline{x})\geq\alpha_{0}/6$ and we would be done.

  2. b)

If $\pazocal{S}\mu_{0}(\overline{x})\geq\alpha_{0}/3$, by (4.4) there exists
$C>0$ absolute such that

$\pazocal{S}\big{(}\mu|_{B_{R}(\overline{x})}\big{)}(\overline{x})\geq\frac{\alpha_{0}}{3}-C\beta\alpha_{0}\geq\frac{\alpha_{0}}{4},$

for $\beta$ small enough. In addition, by definition of $R$,

$\pazocal{S}\big{(}\mu|_{B_{R}(\overline{x})\setminus{5Q_{i}}}\big{)}(\overline{x})\leq
C^{\prime}\frac{\mu_{0}(B_{2R}(\overline{z}))}{\ell(Q_{i})^{n}}\leq
2^{n}C^{\prime}\beta\alpha_{0}\frac{R^{n}}{\ell(Q_{i})^{n}}=C^{\prime\prime}\beta\alpha_{0}.$

Hence

$\pazocal{S}\mu_{i}(\overline{x})=\pazocal{S}\big{(}\mu_{0}|_{5Q_{i}}\big{)}(\overline{x})\geq\frac{\alpha_{0}}{4}-C^{\prime\prime}\beta\alpha_{0}\geq\frac{\alpha_{0}}{5},$

for $\beta$ small enough, and we are done.

  3. c)

If $\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{x})\geq\alpha_{0}/3$.
Let us observe that for $\overline{z}\in(AQ_{i})\cap\pazocal{G}^{c}$, by
definition of $\pazocal{G}$ we also have

$M(\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0})(\overline{z})\leq\beta\alpha_{0},\qquad\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{z})\leq\beta\alpha_{0}.$

So choosing again $R:=2(4+A)\text{diam}(Q_{i})$ we would be able to prove
analogously as in (4.4) that

$\pazocal{S}_{\mu_{0},R}(\pazocal{S}\mu_{0})(\overline{x})\leq
C\beta\alpha_{0},$

for some $C>0$ absolute. Moreover,

$\displaystyle\pazocal{S}_{\mu_{0}|_{B_{R}(\overline{x})\setminus{5Q_{i}}}}(\pazocal{S}\mu_{0})(\overline{x})\leq
C^{\prime}\frac{\int_{B_{2R}(\overline{z})}\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0}}{\ell(Q_{i})^{n}}\leq
C^{\prime\prime}\beta\alpha_{0},$

where we have used, again, the definition of $R$ and that
$M(\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0})(\overline{z})\leq\beta\alpha_{0}$.
Therefore,

$\displaystyle\pazocal{S}_{\mu_{i}}(\pazocal{S}\mu_{0})(\overline{x})$
$\displaystyle=\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{x})-\pazocal{S}_{\mu_{0}|_{B_{R}(\overline{x})\setminus{5Q_{i}}}}(\pazocal{S}\mu_{0})(\overline{x})-\pazocal{S}_{\mu_{0},R}(\pazocal{S}\mu_{0})(\overline{x})$
$\displaystyle\geq\frac{\alpha_{0}}{3}-C^{\prime\prime}\beta\alpha_{0}-C\beta\alpha_{0}\geq\frac{\alpha_{0}}{5},$

for $\beta$ small enough, and the proof is completed.

∎

###### Proof of $\text{{{P}}}_{\boldsymbol{3}}$ in Theorem 4.1..

Repeating the same arguments presented for the proof of
$\text{{{P}}}_{\boldsymbol{2}}$, one deduces that property
$\pazocal{W}^{\mu_{i}}(\overline{x})\geq\alpha_{0}^{\prime}\alpha$, for every
$\overline{x}\in 4Q_{j_{i}}\cap E$ and every $i=1,\ldots,N$, implies

$\gamma_{\text{sy},\text{op}}(4Q_{j_{i}}\cap
E)\leq\frac{C}{\alpha_{0}^{\prime}\alpha_{0}}\bigg{(}\mu_{0}(5Q_{j_{i}})+\int_{5Q_{j_{i}}}\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0}\bigg{)}.$

But notice that the bounded multiplicity of overlapping of the family
$\\{10Q_{j}\\}_{j}$ and the fact that $\mu_{0}$ is supported on $E$ yield

$\sum_{i=1}^{N}\bigg{(}\mu_{0}(Q_{j_{i}})+\int_{5Q_{j_{i}}}\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0}\bigg{)}\lesssim\mu_{0}(E)+\int_{E}\pazocal{S}\mu_{0}\mathop{}\\!\mathrm{d}\mu_{0}\leq
2\mu_{0}(E),$

where for the last inequality we have used (3.10). Therefore, by the latter
relation and using the definition of $\tau_{0}$ we get

$\displaystyle\sum_{i=1}^{N}\gamma_{\text{sy},+}(4Q_{j_{i}}\cap E)$
$\displaystyle\lesssim\sum_{i=1}^{N}\gamma_{\text{sy},\text{op}}(4Q_{j_{i}}\cap
E)\lesssim\mu_{0}(E)$
$\displaystyle=\gamma_{\text{var}}(E)\leq\gamma_{\text{sy},\psi_{\tau_{0}},2}(E)\leq
2C_{1}C_{2}\gamma_{\text{sy},\text{op}}(E)\lesssim\gamma_{\text{sy},+}(E),$

that is the desired estimate. ∎

Finally, we prove $\text{{{P}}}_{\boldsymbol{4}}$. Let us first observe that
the assumption

$\widetilde{\gamma}_{+}(E)\leq C_{+}\text{diam}(E)^{n},\quad\text{ for
$C_{+}>0$ absolute to be fixed below},$

is not at all restrictive. Indeed, if it failed, [MPr, Lemma 4.1] would imply

$\widetilde{\gamma}(E)\leq\gamma(E)\lesssim\pazocal{H}_{\infty}^{n}(E)\leq\text{diam}(E)^{n}<\frac{1}{C_{+}}\widetilde{\gamma}_{+}(E),$

and we would be done.

###### Proof of $\text{{{P}}}_{\boldsymbol{4}}$ in Theorem 4.1..

Fix $\overline{x}\not\in E$ and observe that the following estimates hold

$\displaystyle\pazocal{S}\mu_{0}(\overline{x})$
$\displaystyle\leq\frac{\mu_{0}(E)}{\text{dist}(\overline{x},E)^{n}},$
$\displaystyle\pazocal{S}_{\mu_{0}}(\pazocal{S}\mu_{0})(\overline{x})$
$\displaystyle\leq\frac{1}{\text{dist}(\overline{x},E)^{n}}\int_{E}\int_{E}P_{\text{{sy}}}(\overline{y}-\overline{z})\psi_{\tau_{0}}(\overline{y}-\overline{z})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{z})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})$
$\displaystyle=\frac{1}{\text{dist}(\overline{x},E)^{n}}\int_{E}\pazocal{S}\mu_{0}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})\leq\frac{\mu_{0}(E)}{\text{dist}(\overline{x},E)^{n}},$

where for the last inequality we have applied (3.10). Since
$\text{supp}(\mu_{0})\subset E$, we have

$\sup_{r>0}\frac{\mu_{0}(B_{r}(\overline{x}))}{r^{n}}=\sup_{r>\text{dist}(\overline{x},E)}\frac{\mu_{0}(B_{r}(\overline{x}))}{r^{n}}\leq\frac{\mu_{0}(E)}{\text{dist}(\overline{x},E)^{n}},$

and moreover, due to (3.10) again,

$\displaystyle\sup_{r>0}\frac{1}{r^{n}}\int_{B_{r}(\overline{x})}\pazocal{S}\mu_{0}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})$
$\displaystyle=\sup_{r>\text{dist}(\overline{x},E)}\frac{1}{r^{n}}\int_{B_{r}(\overline{x})}\pazocal{S}\mu_{0}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})$
$\displaystyle\leq\frac{1}{\text{dist}(\overline{x},E)^{n}}\int_{E}\pazocal{S}\mu_{0}(\overline{y})\mathop{}\\!\mathrm{d}\mu_{0}(\overline{y})\leq\frac{\mu_{0}(E)}{\text{dist}(\overline{x},E)^{n}}.$

All in all, we get

$\widetilde{\pazocal{U}}^{\mu_{0}}(\overline{x})\leq\frac{4\mu_{0}(E)}{\text{dist}(\overline{x},E)}.$

Let us pick $\overline{x}\in\pazocal{G}\cap E^{c}$ (which is non empty, since
$E\subset\pazocal{G}$ with $\pazocal{G}$ open, by the lower semicontinuity of
the potential $\widetilde{\pazocal{U}}^{\mu_{0}}$) so that
$\widetilde{\pazocal{U}}^{\mu_{0}}(\overline{x})>\beta\alpha_{0}$ and observe
that using the definition of $\mu_{0}$ and $\tau_{0}$ as in the previous
proofs, we get

$\displaystyle\text{dist}(\overline{x},E)^{n}$
$\displaystyle\leq\frac{8C_{1}C_{2}}{\alpha_{0}\beta}\gamma_{\text{sy},\text{op}}(E)\leq\frac{8CC_{1}C_{2}}{\alpha_{0}\beta}\gamma_{\text{sy},+}(E)$
$\displaystyle\leq\frac{16CC_{1}C_{2}}{\alpha_{0}\beta}\widetilde{\gamma}_{+}(E)\leq\frac{16CC_{1}C_{2}}{\alpha_{0}\beta}C_{+}\text{diam}(E)^{n}.$

Choosing $C_{+}$ appropriately to neglect the effect of all the above
constants that have already been fixed, the result follows. Indeed, this is
because the previous estimate is valid for all $\overline{x}\in\pazocal{G}\cap
E^{c}$, implying that one can make $\partial\pazocal{G}$ say
$\frac{\text{diam}(E)}{1000}$ close to $E$. Since we also know that
$20Q_{j}\subset\pazocal{G}$ we have

$\text{diam}(2Q_{j})\leq\frac{1}{10}\text{dist}(\partial\pazocal{G},Q_{j}),$

where $\text{dist}(\partial\pazocal{G},Q_{j})$ is comparable to
$\text{dist}(\partial\pazocal{G},E)$ since the cubes $Q_{j}$ have been also
chosen so that $\frac{5}{4}Q_{j}\cap E\neq\varnothing$. ∎

##  5\. A comparability result under an additional assumption

Let $E\subset\mathbb{R}^{n+1}$ be a compact set. As it is stated in
$\text{{{P}}}_{\boldsymbol{4}}$ in Theorem 4.1, we shall work under the
following additional assumption:

  1. $\text{{{A}}}_{\boldsymbol{1}}$:

$\widetilde{\gamma}_{+}(E)\leq C_{+}\text{diam}(E)^{n}$, with $C_{+}>0$
absolute.

By the translation invariance and $\widetilde{\gamma}(\lambda
E)=\lambda^{n}\widetilde{\gamma}(E)$, $\widetilde{\gamma}_{+}(\lambda
E)=\lambda^{n}\widetilde{\gamma}_{+}(E)$ (see [He, Proposition 3.1]), it is
clear that we may assume $E\subset B_{1}(0)$ without loss of generality. In
fact, in order to apply Theorem 4.1 let us check that we can assume:

  * $\text{{{A}}}_{\boldsymbol{2}}$:

$E$ is contained in the unit ball and consists of a finite union of dyadic
cubes belonging to a dyadic grid in $\mathbb{R}^{n+1}$ $($with sides parallel
to the coordinate axes$\,)$, all with the same size and with disjoint
interiors.

Let us verify that if we deduce the comparability between $\widetilde{\gamma}$
and $\widetilde{\gamma}_{+}$ for $E$ satisfying
$\text{{{A}}}_{\boldsymbol{2}}$, we obtain the same result for a general $E$.
So fix $E\subset\mathbb{R}^{n+1}$ any compact set and let
$\overline{\pazocal{U}_{\delta}(E)}$ be the closed $\delta$-neighbourhood of
$E$. Consider a grid of dyadic cubes in $\mathbb{R}^{n+1}$ with sides parallel
to the axes and diameter smaller than $\delta/2$. Name $E_{0}$ the collection
of dyadic cubes that intersect $E$ and notice that $E\subset
E_{0}\subset\overline{\pazocal{U}_{\delta}(E)}$. Now we would have

$\widetilde{\gamma}(E)\leq\widetilde{\gamma}(E_{0})\lesssim\widetilde{\gamma}_{+}(E_{0})\leq\widetilde{\gamma}_{+}(\overline{\pazocal{U}_{\delta}(E)})\leq\gamma_{+}(\overline{\pazocal{U}_{\delta}(E)}).$

Letting $\delta\to 0$ and using the outer regularity of $\gamma_{+}$ we
deduce, by Theorem 3.5,

$\widetilde{\gamma}(E)\lesssim\gamma_{+}(E)\lesssim\widetilde{\gamma}_{+}(E),$

and the result follows.
Finally, we present the final additional assumption that motivates the title
of this section:

  1. $\text{{{A}}}_{\boldsymbol{3}}$.

Let $\\{\pazocal{Q}_{1},\ldots,\pazocal{Q}_{N}\\}$ be the family of cubes
provided by Theorem 4.1 and $C_{1}$ the constant appearing in point
$\text{{{P}}}_{\boldsymbol{3}}$ of the same theorem. We will assume that

$\widetilde{\gamma}(E)\geq
C_{1}^{-1}\sum_{i=1}^{N}\widetilde{\gamma}(2\pazocal{Q}_{i}\cap E).$

The aim of this section is to prove the comparability between
$\widetilde{\gamma}$ and $\widetilde{\gamma}_{+}$ under assumption
$\text{{{A}}}_{\boldsymbol{3}}$. Later on, we will get rid of these hypotheses
in some particular cases, for which we will obtain the desired comparability
in full generality.

###  5.1. Basic definitions and properties

Let $E\subset\mathbb{R}^{n+1}$ be a compact set verifying
$\text{{{A}}}_{\boldsymbol{1}},\text{{{A}}}_{\boldsymbol{2}}$ and
$\text{{{A}}}_{\boldsymbol{3}}$. We begin our argument by fixing a
distribution $T_{0}$ admissible for $\widetilde{\gamma}(E)$ so that $|\langle
T_{0},1\rangle|=\widetilde{\gamma}(E)/2$. We know that $T_{0}$ has $n$-growth
with constant 1 and

$\|P\ast T_{0}\|_{\infty}\leq 1,\qquad\|P^{\ast}\ast T_{0}\|_{\infty}\leq 1.$

Let $\pazocal{F}:=\cup_{i=1}^{N}\pazocal{Q}_{i}$ be the covering by cubes
provided by Theorem 4.1. Recall that $\pazocal{Q}_{i}=2Q_{i}$, where
$\\{Q_{i}\\}_{i=1}^{N}$ is the finite subset of dyadic cubes with disjoint
interiors of a Whitney decomposition of $\pazocal{G}\supset E$, with the
property $\frac{5}{4}Q_{i}\cap E\neq\varnothing$. We write

$F:=\bigcup_{i=1}^{N}Q_{i}\subset\pazocal{F}.$

It is clear that $E\subset F$ and moreover, by monotonicity and the properties
in Theorem 4.1 we also have

  1. $\text{{{P}}}_{\boldsymbol{2}}^{\prime}$.

$\widetilde{\gamma}_{+}(F)\leq C_{0}\widetilde{\gamma}_{+}(E)$,

  2. $\text{{{P}}}_{\boldsymbol{3}}^{\prime}$.

$\sum_{i=1}^{N}\widetilde{\gamma}_{+}(2Q_{i}\cap E)\leq
C_{1}\widetilde{\gamma}_{+}(E)$ and by $\text{{{A}}}_{\boldsymbol{3}}$ we also
have $\sum_{i=1}^{N}\widetilde{\gamma}(2Q_{i}\cap E)\leq
C_{1}\widetilde{\gamma}(E)$.

  3. $\text{{{P}}}_{\boldsymbol{4}}^{\prime}$.

$\text{{diam}}(Q_{i})\leq\frac{1}{20}\text{{diam}}(E)$, for each
$i=1,\ldots,N$,

  4. $\text{{{P}}}_{\boldsymbol{5}}^{\prime}$.

The family $\\{10Q_{1},\ldots,10Q_{N}\\}$ has bounded overlap. Moreover, if
$10Q_{j}\cap 10Q_{i}\neq\varnothing$, then $\ell(Q_{j})\approx\ell(Q_{i})$.

In order to simplify the arguments, in this section we will work with the
cubes $Q_{1},\ldots,Q_{N}$ instead of
$\pazocal{Q}_{1},\ldots,\pazocal{Q}_{N}$, and with $F$ instead of
$\pazocal{F}$. Let us choose for each $Q_{i}$ a ball
$B_{i}\subset\frac{1}{2}Q_{i}$ concentric with $Q_{i}$ with radius $r_{i}$
comparable to $\widetilde{\gamma}(2Q_{i}\cap E)^{1/n}$. Notice that

$\widetilde{\gamma}(2Q_{i}\cap
E)^{1/n}\leq\gamma(2Q_{i})^{1/n}\approx\ell(Q_{i}).$

Therefore, we may choose $r_{i}\approx\widetilde{\gamma}(2Q_{i}\cap E)^{1/n}$
with constants depending at most on the dimension and still have
$B_{i}\subset\frac{1}{2}Q_{i}$. Notice that
$\text{dist}(B_{i},B_{j})\geq\frac{1}{2}\min\\{\ell(Q_{i}),\ell(Q_{j})\\}$,
for $i\neq j$. We define the positive measure

(5.1)
$\mu:=\sum_{i=1}^{N}\mu_{i}:=\sum_{i=1}^{N}\frac{r_{i}^{n}}{\pazocal{L}^{n+1}(B_{i})}\pazocal{L}^{n+1}|_{B_{i}}.$

Apply [HPo, Lemma 3.1] to pick test functions
$\varphi_{i}\in\pazocal{C}^{\infty}_{c}(2Q_{i}),0\leq\varphi_{i}\leq 1$,
satisfying $\|\nabla\varphi_{i}\|_{\infty}\leq\ell(2Q_{i})^{-1}$ and with
$\sum_{i=1}^{N}\varphi_{i}\equiv 1$ in $F$. We also define the signed measure

(5.2) $\nu:=\sum_{i=1}^{N}\frac{\langle
T_{0},\varphi_{i}\rangle}{\pazocal{L}^{n+1}(B_{i})}\pazocal{L}^{n+1}|_{B_{i}},\qquad\text{that
is}\qquad\nu=\sum_{i=1}^{N}\frac{\langle
T_{0},\varphi_{i}\rangle}{r_{i}^{n}}\mu_{i}.$

Since $\text{supp}(\nu)\subset\text{supp}(\mu)\subset\pazocal{F}$ and
$\nu\ll\mu$, have $\nu=b\mu$, where $b$ is the Radon-Nikodym derivative.

######  Lemma 5.1.

The following hold for some positive absolute constants $c_{1},c_{3}$ and
$c_{4}$:

  1. 1.

$\|b\|_{\infty}\leq c_{1}$.

  2. 2.

$|\nu(F)|=\widetilde{\gamma}(E)/2$.

  3. 3.

$\widetilde{\gamma}(E)\leq c_{3}\mu(F)$.

  4. 4.

$|\nu(Q)|\leq c_{4}\ell(Q)^{n}$ for any cube $Q\subset\mathbb{R}^{n+1}$.

###### Proof.

Let us prove 1. By the localization result [MPr, Theorem 3.1] we get
$\|P\ast(\varphi_{i}T_{0})\|_{\infty}\leq A_{1}$, for every $i$, where
$A_{1}=A_{1}(n)>0$. Analogously, the same localization result holds for
$P^{\ast}$, so there also exists $A_{2}=A_{2}(n)>0$ with
$\|P^{\ast}\ast(\varphi_{i}T_{0})\|_{\infty}\leq A_{2}$, $\forall i$.
Therefore, since $\text{supp}(\varphi_{i}T_{0})\subset 2Q_{i}\cap E$ we have

(5.3) $|\langle
T_{0},\varphi_{i}\rangle|=|\langle\varphi_{i}T_{0},1\rangle|\lesssim\widetilde{\gamma}(2Q_{i}\cap
E)\approx r_{i}^{n},\quad\forall i=1,\ldots,n,$

which yields the desired result.
To prove 2 we simply use the definition of $\nu$, $B_{i}\cap F=B_{i}$ and
$\text{supp}(T_{0})\subset E\subset F$:

$\displaystyle|\nu(F)|=\Bigg{\rvert}\sum_{i=1}^{N}\frac{\langle
T_{0},\varphi_{i}\rangle}{\pazocal{L}^{n+1}(B_{i})}\int_{B_{i}}\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}\Bigg{\rvert}=\Bigg{\rvert}\Bigg{\langle}T_{0},\sum_{i=1}^{N}\varphi_{i}\Bigg{\rangle}\Bigg{\rvert}=|\langle
T_{0},1\rangle|=\frac{1}{2}\widetilde{\gamma}(E).$

Finally, 3 follows by the choice of radii $r_{i}$ and the admissibility of
$\varphi_{i}T_{0}$ for $\widetilde{\gamma}(2Q_{i}\cap E)$,

$\displaystyle\widetilde{\gamma}(E)=2|\langle T_{0},1\rangle|\leq
2\sum_{i=1}^{N}|\langle
T_{0},\varphi_{i}\rangle|\lesssim\sum_{i=1}^{N}\widetilde{\gamma}(2Q_{i}\cap
E)\approx\sum_{i=1}^{N}r_{i}^{n}=\mu(F).$

The proof of inequality 4 can be followed analogously to that given for
property $(g)$ of [To3, Lemma 6.8]. We refer the reader to this last reference
for the detailed arguments. ∎

######  Remark 5.1.

In the proof of inequality 4 it is used that for a Borel signed measure $\nu$
satisfying $\|P\ast\nu\|_{\infty}\leq 1$ one has $|\nu(Q)|\leq C\ell(Q)^{n}$,
for some absolute constant $C>0$ and any cube $Q\subset\mathbb{R}^{n+1}$. In
order to justify that this holds in our context we will prove the following:
given $\varphi\in\pazocal{C}^{\infty}_{c}(\mathbb{R}^{n+1})$ such that
$\|P\ast\varphi\|_{\infty}\leq 1$, then
$|\int_{Q}\varphi\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}|\leq C\ell(Q)^{n}$,
for some $C>0$ absolute. If this is the case, we will be done by mollifying
$\nu$ to be $\nu_{\varepsilon}:=\nu\ast\psi_{\varepsilon}$ with
$(\psi_{\varepsilon})_{\varepsilon}$ a proper approximation of the identity.
Let us estimate $|\int_{Q}\varphi\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}|$ as
follows:

$\displaystyle\bigg{\rvert}\int_{Q}\varphi(\overline{x})\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}(\overline{x})\bigg{\rvert}\leq\bigg{\rvert}\int_{Q}\partial_{t}(P\ast\varphi)(\overline{x})\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}(\overline{x})\bigg{\rvert}+\bigg{\rvert}\int_{Q}(\Delta_{x})^{1/2}(P\ast\varphi)(\overline{x})\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}(\overline{x})\bigg{\rvert}$

For the first term of the right-hand side we set $Q=Q_{1}\times
I_{Q}\subset\mathbb{R}^{n}\times\mathbb{R}$ where $I_{Q}$ is the temporal
interval, that we write as $I_{Q}=[a,b]$ for some $a,b\in\mathbb{R}$ with
$a<b$. This way, by Fubini’s theorem,

$\displaystyle\bigg{\rvert}\int_{Q}\partial_{t}(P\ast\varphi)(\overline{x})\mathop{}\\!\mathrm{d}\pazocal{L}^{n+1}(\overline{x})\bigg{\rvert}$
$\displaystyle=\bigg{\rvert}\int_{Q_{1}}\bigg{(}\int_{a}^{b}\partial_{t}(P\ast\varphi)(x,t)\mathop{}\\!\mathrm{d}t\bigg{)}\mathop{}\\!\mathrm{d}\pazocal{L}^{n}(x)\bigg{\rvert}$
$\displaystyle=\bigg{\rvert}\int_{Q_{1}}\Big{(}P\ast\varphi(x,b)-P\ast\varphi(x,a)\Big{)}\mathop{}\\!\mathrm{d}\pazocal{L}^{n}(x)\bigg{\rvert}$
$\displaystyle\lesssim\|P\ast\varphi\|_{\infty}\pazocal{L}^{n}(Q_{1})\leq\ell(Q)^{n}.$

For the second term we recall that we have the following representation

$(\Delta_{x})^{1/2}\approx\sum_{j=1}^{n}\partial_{j}R_{j}$

where $R_{j},\,1\leq j\leq n$ are the Riesz transforms with Fourier multiplier
$\xi_{j}/|\xi_{j}|$. Notice that if we prove that $P\ast(R_{j}\varphi)\in
L^{\infty}(\mathbb{R}^{n+1})$ for each $j=1,\ldots,n$, we are done, since we
may argue as above decomposing the cube $Q$ as a cartesian product of an
interval and an $n$-dimensional cube. To study $P\ast(R_{j}\varphi)$ we
consider its Fourier transform
$\frac{\xi_{j}}{|\xi|}\widehat{P}\widehat{\varphi}$. Notice that
$P\in(L^{1}+L^{2})(\mathbb{R}^{n+1})$ (just decompose it as
$P\chi_{|\overline{x}|\leq 1}+P\chi_{|\overline{x}|>1}$), and therefore
$\widehat{P}\in(L^{\infty}+L^{2})(\mathbb{R}^{n+1})$, and we write its
decomposition as $\widehat{P}_{\infty}+\widehat{P}_{2}$. Now it is immediate
to check that $\frac{\xi_{j}}{|\xi|}\widehat{P}_{\infty}\widehat{\varphi}\in
L^{1}(\mathbb{R}^{n+1})$ and we also have
$\frac{\xi_{j}}{|\xi|}\widehat{P}_{2}\widehat{\varphi}\in
L^{1}(\mathbb{R}^{n+1})$ using Cauchy-Schwarz inequality. Therefore the
Fourier transform of $P\ast(R_{j}\varphi)$ is integrable, implying
$P\ast(R_{j}\varphi)\in L^{\infty}(\mathbb{R}^{n+1})$ that is what we wanted
to see.

###  5.2. The exceptional sets $\boldsymbol{H_{\pazocal{D}}}$ and
$\boldsymbol{T_{\pazocal{D}}}$

In light of the third property in Lemma 5.1, assume we find a compact set
$G\subset F$ such that

  * i.

$\mu(F)\lesssim\mu(G)$,

  * ii.

$\mu|_{G}$ has $n$-growth,

  * iii.

$\pazocal{P}_{\mu|_{G}}$ is a bounded operator in $L^{2}(\mu|_{G})$.

Then, $\mu|_{G}$ is admissible for $\gamma_{\text{op}}(G)$, and we are done,
since we have

(5.4)
$\widetilde{\gamma}(E)\lesssim\mu(F)\lesssim\mu(G)\leq\gamma_{\text{op}}(G)\leq\gamma_{\text{op}}(F)\approx\widetilde{\gamma}_{+}(F)\lesssim\widetilde{\gamma}_{+}(E),$

where the first inequality is due to Lemma 5.1 and the last is by property
$\text{{{P}}}_{\boldsymbol{2}}^{\prime}$ above.
The construction of $G$ is inspired by the $Tb$-theorems found in [To3,
Theorem 5.1] and [Vo, Theorem 7.1], which are presented for the particular
case of Cauchy and Riesz potentials respectively. To apply this kind of
results we need to introduce the notions of Ahlfors ball, Ahlfors radius and
Ahlfors point in our context. Due to [MPr, Lemma 4.1] there are positive
dimensional constants $a_{1},a_{2}$ such that

$\widetilde{\gamma}(E)\leq a_{1}\,\text{diam}(E)^{n},$

and also

$a_{2}^{-1}r_{i}^{n}\leq\widetilde{\gamma}(2Q_{i}\cap E)\leq a_{2}r_{i}^{n},$

for each $i=1,\ldots,N$, where $r_{i}$ is the radius of the ball $B_{i}\subset
Q_{i}$ constructed at the beginning of Subsection 5.1. Let

$L:=100^{n}a_{1}a_{2}C_{1},$

where $C_{1}$ and $C_{2}$ are the constants appearing in Theorem 4.1. The
factor 100 is arbitrarily chosen; we may pick, for convenience, any other
large constant. Assumption $\text{{{A}}}_{\boldsymbol{3}}$ implies

$\displaystyle\mu(F)=\sum_{i=1}^{N}r_{i}^{n}\leq
a_{2}\sum_{i=1}^{N}\widetilde{\gamma}(2Q_{i}\cap E)\leq
a_{2}C_{1}\widetilde{\gamma}(E)\leq a_{1}a_{2}C_{1}\,\text{diam}(E)^{n}.$

Therefore:

$\forall\overline{x}\in F,\,\forall
R>\frac{1}{100}\text{diam}(F)\,:\quad\mu(B_{R}(\overline{x}))\leq LR^{n}.$

A ball $B_{R}(\overline{x})$, $\overline{x}\in F$, is an Ahlfors ball if
precisely the estimate $\mu(B_{R}(\overline{x}))\leq LR^{n}$ holds. Notice
that if $R>\frac{1}{100}\text{diam}(F)$, then the ball is an Ahlfors ball. We
also define the Ahlfors radius,

(5.5) $\pazocal{R}(\overline{x}):=\sup\big{\\{}r>0:B_{r}(\overline{x})\text{
is non-Ahlfors}\,\big{\\}}.$

Ahlfors points are those for which $\pazocal{R}(\overline{x})=0$. For every
$\overline{x}\in F$, we have
$\pazocal{R}(\overline{x})\leq\frac{1}{100}\text{diam}(F)$. Set

$H^{\prime}:=\bigcup_{\overline{x}\in
F}B_{\pazocal{R}(\overline{x})}(\overline{x}),$

where we convey $B_{0}(\overline{x}):=\varnothing$, and apply a $5r$-covering
theorem [Ma, Theorem 2.1] to choose a countable family of disjoint balls
$\\{B_{\pazocal{R}_{k}}\\}_{k}:=\\{B_{\pazocal{R}(\overline{x}_{k})}(\overline{x}_{k})\\}_{k}$
such that

(5.6) $H^{\prime}\subset H:=\bigcup_{k}B_{5\pazocal{R}_{k}},$

so that all non-Ahlfors balls are contained in $H$. Observe that since
$\mu(B_{\pazocal{R}_{k}})\geq L\pazocal{R}_{k}^{n}$ and the balls
$B_{\pazocal{R}_{k}}$ are disjoint, we have

$\sum_{k}\pazocal{R}_{k}^{n}\leq\frac{1}{L}\sum_{k}\mu(B_{\pazocal{R}_{k}})\leq\frac{1}{L}\mu(F).$

Let $\pazocal{D}^{0}$ be the usual dyadic lattice in $\mathbb{R}^{n+1}$ and
for $\overline{x}\in\mathbb{R}^{n+1}$ we write

$\pazocal{D}(\overline{x}):=\overline{x}+\pazocal{D}^{0},$

the translation of $\pazocal{D}^{0}$ by the vector $\overline{x}$. Let
$\pazocal{D}=\pazocal{D}(\overline{x})$ be any fixed dyadic lattice from the
family of lattices
$\\{\pazocal{D}(\overline{y})\\}_{\overline{y}\in\mathbb{R}^{n+1}}$. Consider
the subcollection of dyadic cubes $\pazocal{D}_{H}\subset\pazocal{D}$ defined
as follows: $Q\in\pazocal{D}_{H}$ if there is a ball
$B_{5\pazocal{R}_{k}}(\overline{x}_{k})$ satisfying

(5.7) $B_{5\pazocal{R}_{k}}(\overline{x}_{k})\cap
Q\neq\varnothing\qquad\text{and}\qquad 10\pazocal{R}_{k}<\ell(Q)\leq
20\pazocal{R}_{k}.$

Then, it is clear that

$\bigcup_{k}B_{5\pazocal{R}_{k}}(\overline{x}_{k})\subset\bigcup_{Q\in\pazocal{D}_{H}}Q.$

Pick a subfamily of disjoint maximal cubes $\\{Q_{k}\\}_{k}$ from
$\pazocal{D}_{H}$ so that we may define the exceptional set $H{D}$ as follows

$\bigcup_{Q\in\pazocal{D}_{H}}Q=\bigcup_{k}Q_{k}=:H_{\pazocal{D}}=H_{\pazocal{D}(\overline{x})}.$

By construction one has

$H\subset
H_{\pazocal{D}(\overline{x})},\quad\forall\overline{x}\in\mathbb{R}^{n+1},$

and since for each ball $B_{5\pazocal{R}_{k}}(\overline{x}_{k})$ conditions
(5.7) can only occur for a number of cubes in $\pazocal{D}_{H}$ which is
bounded by a dimensional constant, there is $c_{H}=c_{H}(n)>0$ so that

(5.8) $\sum_{k}\ell(Q_{k})^{n}\leq
c_{H}\sum_{k}\pazocal{R}_{k}^{n}\leq\frac{c_{H}}{L}\mu(F).$

We define the exceptional set $T{D}$ in a similar manner but imposing a proper
accretivity condition on the Radon-Nikodym derivative $b$. Consider the family
$\pazocal{D}_{T}\subset\pazocal{D}$ of cubes satisfying

(5.9) $\mu(Q)\geq c_{T}|\nu(Q)|,$

with $c_{T}=c_{T}(n)>0$ big constant to be fixed below. Let $\\{Q_{k}\\}_{k}$
be the subcollection of maximal (and thus disjoint) dyadic cubes from
$\pazocal{D}_{T}$. We define the exceptional set
$T{D}=T_{\pazocal{D}(\overline{x})}$ as

$T{D}:=\bigcup_{k}Q_{k}.$

The next result shows that $\mu(F\setminus{(H_{\pazocal{D}}\cup
T_{\pazocal{D}})})$ is comparable to $\mu(F)$. The reader can find its proof
in [To3, Lemma 6.12].

######  Lemma 5.2.

If $L$ and $c_{T}$ are chosen big enough, then

$|\nu(H_{\pazocal{D}}\cup
T_{\pazocal{D}})|\leq\frac{1}{2}|\nu(F)|\qquad\text{and}\qquad\mu(H_{\pazocal{D}}\cup
T_{\pazocal{D}})\leq\delta_{0}\mu(F),$

with $\delta_{0}<1$ absolute.

###  5.3. Verifying the hypotheses of a $Tb\,$-theorem

The goal of this subsection is to prove the lemma below, analogous to [To3,
Lemma 6.8 (Main Lemma)]. We remark that in order to apply the $Tb$-theorem
found in [NTVo1] we will still need to check an additional weak boundedness
property for a particular suppressed kernel, due to the lack of anti-symmetry
of our kernel $P$. Following the proof of the aforementioned reference or that
of [To3, Theorem 5.1], one notices that such condition needs not to be
verified, since the Cauchy kernel satisfies being anti-symmetric.

######  Lemma 5.3.

Let $E\subset\mathbb{R}^{n+1}$ be a compact set verifying