of Lemma 5.13.

We denote

${\mathcal{R}}_{*,\Psi}\nu(x)=\sup_{{\varepsilon}>\Psi(x)}\big{|}{\mathcal{R}}_{\varepsilon}\nu(x)\big{|}.$

######  Lemma 5.9.

Suppose that (5.23) holds. Then,

(5.24)
$\nu\big{(}\big{\\{}x\in{\mathcal{S}}_{\eta}:\,{\mathcal{R}}_{*,\Psi}\nu(x)>M\,\Theta({\mathsf{H}})\big{\\}}\big{)}\leq
C_{2}\,\frac{\Lambda^{{\varepsilon}_{n}}}{M}\,\nu(H).$

###### Proof.

Recall that we denote by $V_{1}$ the $A_{0}^{-3}\ell(R)$-neighborhood of
${\mathcal{S}}_{\eta}$ and that

$G(x)\lesssim\Theta(R)\quad\mbox{ for all $x\in V_{1}$.}$

Then the assumption in the lemma implies that

(5.25)
$\int_{V_{1}}\big{|}({\mathcal{R}}\nu(x)-2c_{0}\,\Theta({\mathsf{H}}))_{+}|^{p}\,d\nu(x)\lesssim\sigma_{p}({\mathsf{H}}).$

By Remark 5.3, for any $x\in{\mathcal{S}}_{\eta}$ and ${\varepsilon}>\Psi(x)$,

$|{\mathcal{R}}_{\varepsilon}\nu(x)|\lesssim\;\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.57222pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.70335pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.59167pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.10834pt}}\\!\int_{B(x,2{\varepsilon}^{\prime})}|{\mathcal{R}}\nu|\,d\nu+\sup_{r\geq{\varepsilon}}\frac{\nu(B(x,r))}{r^{n}}+\frac{W_{\nu}(B(x,8{\varepsilon}^{\prime}))}{\nu(B(x,8{\varepsilon}^{\prime}))},$

where ${\varepsilon}^{\prime}=2^{7k}{\varepsilon}$, with $k\geq 0$ minimal
such that the ball $B(x,{\varepsilon}^{\prime})$ is
$(128,128^{n+2})$-doubling. In the case
${\varepsilon}^{\prime}\geq\frac{1}{2}A_{0}^{-3}\,\ell(R)$, by standard
arguments,

$|{\mathcal{R}}_{\varepsilon}\nu(x)|\leq|{\mathcal{R}}_{{\varepsilon}^{\prime}}\nu(x)|+C\,\frac{\nu(B(x,{\varepsilon}^{\prime}))}{({\varepsilon}^{\prime})^{n}}\leq
C\,\Theta(R)<M\,\Theta({\mathsf{H}}),$

for $\Lambda$ (or $M$) big enough.

In the case ${\varepsilon}^{\prime}<\frac{1}{2}A_{0}^{-3}\ell(R)$, we have
$B(x,2{\varepsilon}^{\prime})\subset V_{1}$ and thus

$|{\mathcal{R}}_{\varepsilon}\nu(x)|\lesssim{\mathcal{M}}_{\nu}^{(\Psi(x),16,128^{n+2})}(\chi_{V_{1}}{\mathcal{R}}\nu)(x)+\sup_{r\geq\Psi(x)}\frac{\nu(B(x,r))}{r^{n}}+\sup_{r\in
D(x):r\geq\Psi(x)}\\\ \frac{W_{\nu}(B(x,r))}{\nu(B(x,r))},$

where $D(x)$ denotes the set of radii $r>0$ such that $B(x,r)$ is
$(16,128^{n+2})$-doubling. Also, as shown in (5.22),

$\sup_{r\geq\Psi(x)}\frac{\nu(B(x,r))}{r^{n}}\lesssim\Theta({\mathsf{H}}).$

We deduce that in any case (i.e., for any ${\varepsilon}^{\prime}$), assuming
$M$ larger than some absolute constant,

$\displaystyle\nu\big{(}\big{\\{}x\in{\mathcal{S}}_{\eta}:\,{\mathcal{R}}_{*,\Psi}\nu(x)$
$\displaystyle>M\,\Theta({\mathsf{H}})\big{\\}}\big{)}$
$\displaystyle\leq\nu\Big{(}\Big{\\{}x\in{\mathcal{S}}_{\eta}:\,{\mathcal{M}}_{\nu}^{(\Psi(x),16,128^{n+2})}(\chi_{V_{1}}{\mathcal{R}}\nu)(x)>\frac{M\,\Theta({\mathsf{H}})}{3}\Big{\\}}\Big{)}$
$\displaystyle\\!\\!+\nu\Big{(}\Big{\\{}x\in{\mathcal{S}}_{\eta}:\\!\\!\\!\sup_{r\in
D(x):r\geq\Psi(x)}\\!\frac{W_{\nu}(B(x,r))}{\nu(B(x,r))}>\frac{M\,\Theta({\mathsf{H}})}{3}\Big{\\}}\Big{)}$
$\displaystyle=:T_{1}+T_{2}.$

To deal with the term $T_{1}$, notice that if
${\mathcal{M}}_{\nu}^{(\Psi(x),16,128^{n+2})}(\chi_{V_{1}}{\mathcal{R}}\nu)(x)>\frac{M}{3}\Theta({\mathsf{H}})$,
then

${\mathcal{M}}_{\nu}^{(\Psi(x),16,128^{n+2})}((|\chi_{V_{1}}{\mathcal{R}}\nu|-\frac{M}{6}\Theta({\mathsf{H}}))_{+})(x)>\frac{M}{6}\Theta({\mathsf{H}}),$

and thus using the weak $L^{p}(\nu)$ boundedness of
${\mathcal{M}}_{\nu}^{(\Psi(x),16,128^{n+2})}$ and (5.25) we obtain, assuming
$M$ big enough,

$T_{1}\leq\nu\Big{(}\Big{\\{}x\in{\mathcal{S}}_{\eta}:\,{\mathcal{M}}_{\nu}^{(\Psi(x),16,128^{n+2})}((|\chi_{V_{1}}{\mathcal{R}}\nu|-\frac{M}{6}\Theta({\mathsf{H}}))_{+})(x)>\frac{M}{6}\Theta({\mathsf{H}})\Big{\\}}\Big{)}\\\
\lesssim\frac{1}{(M\Theta({\mathsf{H}}))^{p}}\int_{V_{1}}\big{|}({\mathcal{R}}\nu(x)-\frac{M}{6}\,\Theta({\mathsf{H}}))_{+}|^{p}\,d\nu\lesssim\frac{1}{M^{p}}\,\nu(H).$

Concerning $T_{2}$, by Lemma 5.7,

$T_{2}\lesssim\frac{\Lambda^{{\varepsilon}_{n}}}{M}\,\nu(H).$

So we have

$\nu\big{(}\big{\\{}x\in{\mathcal{S}}_{\eta}:\,{\mathcal{R}}_{*,\Psi}\nu(x)>M\,\Theta({\mathsf{H}})\big{\\}}\big{)}\lesssim\frac{1}{M^{p}}\,\nu(H)+\frac{\Lambda^{{\varepsilon}_{n}}}{M}\,\nu(H)\lesssim\frac{\Lambda^{{\varepsilon}_{n}}}{M}\,\nu(H).$

∎

For $\gamma>1$, we let

$Z(\gamma)=\big{\\{}x\in{\mathcal{S}}_{\eta}^{\prime}:{\mathcal{R}}_{*,\Psi}\nu(x)>\gamma\Lambda^{{\varepsilon}_{n}}\,\Theta({\mathsf{H}})\big{\\}}.$

Notice that, under the assumption (5.23), by Lemma 5.9 (with
$M=\gamma\Lambda^{{\varepsilon}_{n}}$), we have

(5.26) $\nu(Z(\gamma)\cap{\mathcal{S}}_{\eta})\leq C_{2}\gamma^{-1}\,\nu(H).$

For $x\in Z(\gamma)$, let

(5.27)
$e(x,\gamma)=\sup\\{{\varepsilon}:{\varepsilon}>\Psi(x),\,|{\mathcal{R}}_{\varepsilon}\nu(x)|>\gamma\Lambda^{{\varepsilon}_{n}}\,\Theta({\mathsf{H}})\\}$

We define the exceptional set $Z^{\prime}(\gamma)$ as

$Z^{\prime}(\gamma)=\bigcup_{x\in Z(\gamma)}B(x,e(x,\gamma)).$

The next lemma shows that $\nu(Z^{\prime}(\gamma)\cap{\mathcal{S}}_{\eta})$ is
small if $\gamma$ is taken big enough.

######  Lemma 5.10.

If $y\in Z^{\prime}(\gamma)$, then
${\mathcal{R}}_{*,\Psi}\nu(y)>(\gamma\Lambda^{{\varepsilon}_{n}}-c_{2})\Theta({\mathsf{H}})$,
for some $c_{2}>0$. Thus, under the assumption (5.23), if $\gamma\geq 2c_{2}$,
then

(5.28)
$\nu(Z^{\prime}(\gamma)\cap{\mathcal{S}}_{\eta})\leq\frac{2C_{2}}{\gamma}\,\nu(H).$

###### Proof.

The arguments are similar to the ones in Lemma 5.2 from [To5]. However, for
the reader’s convenience we will explain here the basic details.

By standard arguments (which just use the fact that the Riesz kernel is a
Calderón-Zygmund kernel), for all $y\in B(x,e(x,\gamma))$, with $x\in
Z(\gamma)$, we have

$|{\mathcal{R}}_{e(x,\gamma)}\nu(x)-{\mathcal{R}}_{2e(x,\gamma)}\nu(y)|\lesssim\sup_{r\geq
e(x,\gamma)}\frac{\nu(B(x,r))}{r^{n}}\lesssim\Theta({\mathsf{H}}),$

taking into account that $e(x,\gamma)\geq\Psi(x)$ and recalling (5.22) for the
last estimate. So we have

$|{\mathcal{R}}_{2e(x,\gamma)}\nu(y)|\geq|{\mathcal{R}}_{e(x,\gamma)}\nu(x)|-c_{2}\,\Theta({\mathsf{H}})\geq\gamma\,\Lambda^{{\varepsilon}_{n}}\,\Theta({\mathsf{H}})-c_{2}\,\Theta({\mathsf{H}}).$

Observe now that

$\Psi(y)\leq\Psi(x)+|x-y|<2e(x,\gamma).$

So

${\mathcal{R}}_{*,\Psi}\nu(y)\geq|{\mathcal{R}}_{2e(x,\gamma)}\nu(y)|>(\gamma\Lambda^{{\varepsilon}_{n}}-c_{2})\Theta({\mathsf{H}}),$

which proves the first statement of the lemma.

In particular, taking $\gamma\geq 2c_{2}$, from the last estimate we derive

${\mathcal{R}}_{*,\Psi}\nu(y)\geq\frac{\gamma}{2}\,\Lambda^{{\varepsilon}_{n}}\,\Theta({\mathsf{H}}),$

and so $y\in Z(\gamma/2)$, which shows that $Z^{\prime}(\gamma)\subset
Z(\gamma/2)$. Thus, by (5.26),

$\nu(Z^{\prime}(\gamma)\cap{\mathcal{S}}_{\eta})\leq\nu(Z(\gamma/2)\cap{\mathcal{S}}_{\eta})\leq
2C_{2}\gamma^{-1}\,\nu(H).$

∎

We choose $\gamma=\max(1,2c_{2},4C_{2})$, so that, under the assumption
(5.23),

(5.29)
$\nu(Z^{\prime}(\gamma)\cap{\mathcal{S}}_{\eta})\leq\frac{1}{2}\,\nu(H).$

Also we define

$\Phi(x)=\max(\Psi(x),{\rm dist}(x,{\mathbb{R}}^{n+1}\setminus
Z^{\prime}(\gamma)).$

Notice that $\Phi$ is a $1$-Lipschitz function which coincides with $\Psi(x)$
away from $Z^{\prime}(\gamma)$ and that

(5.30) $\Phi(x)\geq e(x,\gamma)\quad\mbox{ for all $x\in Z(\gamma)$.}$

######  Lemma 5.11.

The suppressed operator ${\mathcal{R}}_{\Phi}$ is bounded in $L^{p}(\nu)$,
with $\|{\mathcal{R}}_{\Phi}\|_{L^{p}(\nu)\to
L^{p}(\nu)}\lesssim\Lambda^{{\varepsilon}_{n}}\Theta({\mathsf{H}}).$

###### Proof.

This is an immediate consequence of Theorem 5.1 and the construction of $\Phi$
above. Indeed, by (5.22),

$\nu(B(x,r))\lesssim\Theta({\mathsf{H}})\,r^{n}\quad\mbox{ for all
$r\geq\Phi(x)$.}$

Also, by (5.30),

$\sup_{{\varepsilon}>\Phi(x)}|{\mathcal{R}}_{\varepsilon}\nu(x)|\leq\sup_{{\varepsilon}>e(x,\gamma)}|{\mathcal{R}}_{\varepsilon}\nu(x)|\leq\gamma\Lambda^{{\varepsilon}_{n}}\,\Theta({\mathsf{H}})\quad\mbox{
for all $x\in Z(\gamma)$.}$

On the other hand, by the definition of $Z(\gamma)$ we also have

$\sup_{{\varepsilon}>\Phi(x)}|{\mathcal{R}}_{\varepsilon}\nu(x)|\leq\sup_{{\varepsilon}>\Psi(x)}|{\mathcal{R}}_{\varepsilon}\nu(x)|\leq\gamma\Lambda^{{\varepsilon}_{n}}\,\Theta({\mathsf{H}})\quad\mbox{
for $x\in{\mathcal{S}}_{\eta}^{\prime}\setminus Z(\gamma)$.}$

So we can apply Theorem 5.1, taking
$\omega=(C\Lambda^{{\varepsilon}_{n}}\Theta({\mathsf{H}}))^{-1}\,\nu$ with an
appropriate absolute constant $C$, and then the lemma follows. ∎

######  Lemma 5.12.

Under the assumption (5.23), there exists a subset $H_{0}\subset H$ such that:

  * (a)

$\eta(H_{0})\geq\frac{1}{2}\,\eta(H)$,

  * (b)

${\mathcal{R}}_{\Psi,\eta}$ is bounded from $L^{p}(\eta{\lfloor}_{H_{0}})$ to
$L^{p}(\eta{\lfloor}_{{\mathcal{S}}^{\prime}_{\eta}})$, with
$\|{\mathcal{R}}_{\Psi}\|_{L^{p}(\eta{\lfloor}_{H_{0}})\to
L^{p}(\eta{\lfloor}_{{\mathcal{S}}^{\prime}_{\eta}})}\lesssim\Lambda^{{\varepsilon}_{n}}\Theta({\mathsf{H}}).$

###### Proof.

We let

$H_{0}=H\setminus Z^{\prime}(\gamma),$

so that, by (5.29),

$\eta(H_{0})=\nu(H_{0})\geq\nu(H)-\nu(Z^{\prime}(\gamma)\cap{\mathcal{S}}_{\eta})\geq\frac{1}{2}\,\nu(H)=\frac{1}{2}\,\eta(H).$

To prove (b), take $f\in L^{p}(\eta{\lfloor}_{H_{0}})$ and observe that for
$x\in{\mathcal{S}}^{\prime}_{\eta}$, by (5.4),

$|{\mathcal{R}}_{\Psi}(f\eta)(x)|=|{\mathcal{R}}_{\Psi}(f\nu)(x)|\leq|{\mathcal{R}}_{\Psi(x)}(f\nu)(x)|+{\mathcal{M}}_{\Psi,n}(f\nu)(x),$

where ${\mathcal{R}}_{\Psi(x)}$ is the $\Psi(x)$-truncated Riesz transform and

(5.31)
${\mathcal{M}}_{\Psi,n}\alpha(x)=\sup_{r>\Psi(x)}\frac{|\alpha|(B(x,r))}{r^{n}}$

for any signed Radon measure $\alpha$. Taking into account that
$\Phi(x)\geq\Psi(x)$, we can split

${\mathcal{R}}_{\Psi(x)}(f\nu)(x)={\mathcal{R}}_{\Phi(x)}(f\nu)(x)+\int_{y\in
H_{0}:\Psi(x)<|x-y|\leq\Phi(x)}\frac{x-y}{|x-y|^{n+1}}\,f(y)\,d\nu(y).$

To estimate the last integral, notice that for $y\in H_{0}$,
$\Psi(y)=\Phi(y)$, and then the condition $\Psi(x)<|x-y|$ implies that

$\Phi(x)\leq\Phi(y)+|x-y|=\Psi(y)+|x-y|\leq\Psi(x)+2|x-y|<3|x-y|.$

So the last integral above is bounded by

$\int_{\Phi(x)/3<|x-y|\leq\Phi(x)}\frac{1}{|x-y|^{n}}\,|f(y)|\,d\nu(y)\lesssim{\mathcal{M}}_{\Psi,n}(f\nu)(x).$

From the preceding estimate and (5.4) we derive that

$|{\mathcal{R}}_{\Psi}(f\eta)(x)|\leq|{\mathcal{R}}_{\Phi(x)}(f\nu)(x)|+C\,{\mathcal{M}}_{\Psi,n}(f\nu)(x)\leq|{\mathcal{R}}_{\Phi}(f\nu)(x)|+C\,{\mathcal{M}}_{\Psi,n}(f\nu)(x).$

From the last inequality and Lemma 5.11, taking into account that $\eta$
coincides with $\nu$ on $V_{1}$, we deduce that

$\|{\mathcal{R}}_{\Psi}\|_{L^{p}(\eta{\lfloor}_{H_{0}})\to
L^{p}(\eta{\lfloor}_{V_{1}})}\lesssim\Lambda^{{\varepsilon}_{n}}\Theta({\mathsf{H}})+\|{\mathcal{M}}_{\Psi,n}\|_{L^{p}(\eta)\to
L^{p}(\eta)}.$

From the growth condition (5.22) and standard covering lemmas, it follows
easily that

$\|{\mathcal{M}}_{\Psi,n}\|_{L^{p}(\eta)\to
L^{p}(\eta)}\lesssim\Theta({\mathsf{H}}).$

On the other hand, using the fact that ${\rm
dist}(H_{0},{\mathbb{R}}^{n+1}\setminus V_{1})\gtrsim\ell(R)$, it is immediate
(by Schur’s criterion, for example) to check that also

$\|{\mathcal{R}}_{\Psi}\|_{L^{p}(\eta{\lfloor}_{H_{0}})\to
L^{p}(\eta{\lfloor}_{{\mathcal{S}}_{\eta}^{\prime}\setminus
V_{1}})}\lesssim\Theta(R)\leq\Theta({\mathsf{H}}).$

∎

###  5.5. The variational argument

######  Lemma 5.13.

There is a constant $c_{0}>0$, depending at most on $n$ and $A_{0}$, such that
for any $p\in(1,2]$, if $\Lambda$ is big enough, we have

$\int\big{|}(|{\mathcal{R}}\nu(x)|-G(x)-c_{0}\,\Theta({\mathsf{H}}))_{+}\big{|}^{p}\,d\nu(x)\gtrsim\Lambda^{-p^{\prime}{\varepsilon}_{n}}\sigma_{p}({\mathsf{H}}),$

where $p^{\prime}=p/(p-1)$ and the implicit constant depends on $p$.

###### Proof.

Suppose that, for a very small $0<\lambda<1$ to be fixed below,

(5.32)
$\int\bigl{|}\bigl{(}|{\mathcal{R}}\nu|-G-c_{0}\,\Theta({\mathsf{H}})\bigr{)}_{+}\bigr{|}^{p}\,d\nu\leq\lambda\,\sigma_{p}({\mathsf{H}}).$

Let $H_{0}\subset H$ be the set found in Lemma 5.12. Consider the measures of
the form $\tau=a\,\nu$, with $a\in L^{\infty}(\nu)$, $a\geq 0$, and let $F$ be
the functional

$\displaystyle F(\tau)$
$\displaystyle=\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-G-c_{0}\,\Theta({\mathsf{H}})\bigr{)}_{+}\bigr{|}^{p}\,d\tau+\lambda\,\|a\|_{L^{\infty}(\nu)}\,\sigma_{p}({\mathsf{H}})+\lambda\,\frac{\nu(H_{0})}{\tau(H_{0})}\,\sigma_{p}({\mathsf{H}}).$

Let

$m=\inf F(\tau),$

where the infimum is taken over all the measures $\tau=a\,\nu$, with $a\in
L^{\infty}(\nu)$. We call such measures admissible. Since $\nu$ is admissible
we infer that

(5.33) $m\leq F(\nu)\leq 3\lambda\,\sigma_{p}({\mathsf{H}}).$

So the infimum $m$ is attained over the collection of measures $\tau=a\nu$
such that $\|a\|_{L^{\infty}(\nu)}\leq 3$ and
$\tau(H_{0})\geq\frac{1}{3}\,\nu(H_{0})$. In particular, by taking a weak-$*$
limit in $L^{\infty}(\nu)$, this guaranties the existence of a minimizer.

Let $\tau$ be an admissible measure such that

(5.34) $m=F(\tau)\leq 3\lambda\,\sigma_{p}({\mathsf{H}}).$

To simplify notation, we denote

$f(x)=G+c_{0}\,\Theta({\mathsf{H}}).$

We claim that

(5.35)
$\bigl{|}\bigl{(}|{\mathcal{R}}\tau(x)|-f(x)\bigr{)}_{+}|^{p}+p\,{\mathcal{R}}_{\tau}^{*}\Bigl{[}\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p-1}|{\mathcal{R}}\tau|^{-1}{\mathcal{R}}\tau\Bigr{]}(x)\lesssim\lambda\,\Theta({\mathsf{H}})^{p}\quad\text{\\!\\!in
$\operatorname{supp}\tau$,}$

where for a vector field $U$, we wrote

${\mathcal{R}}_{\tau}^{*}U={\mathcal{R}}^{*}(U\tau)=-\sum_{i=1}^{n+1}{\mathcal{R}}_{i}(U_{i}\,\tau).$

Assume (5.35) for the moment. Since the function on the left hand side is
subharmonic in ${\mathbb{R}}^{n+1}\setminus\operatorname{supp}\tau$ and
continuous in ${\mathbb{R}}^{n+1}$ (recall that $\tau$ and $\eta$ are
absolutely continuous with respect to Lebesgue measure, with bounded
densities), we deduce that the same estimate holds in the whole
${\mathbb{R}}^{n+1}$.

Next we need to construct an auxiliary function ${\varphi}$. First we claim
that there exists a subfamily of cubes ${\mathsf{H}}^{0}\subset{\mathsf{H}}$
such that

  * (i)

the balls $3B_{Q}\equiv 3B_{Q^{(\mu)}}$, with $Q^{(\mu)}\in{\mathsf{H}}^{0}$,
are disjoint,

  * (ii)

$\frac{1}{12}\nu(Q)\leq\tau(Q\cap H_{0})\leq 3\nu(Q)$ for all
$Q^{(\mu)}\in{\mathsf{H}}^{0}$, and

  * (iii)

$\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\tau(Q\cap H_{0})\approx\nu(H)$.

The existence of the family ${\mathsf{H}}^{0}$ follows easily from the fact
that $\tau(H_{0})\geq\frac{1}{3}\nu(H_{0})\geq\frac{1}{6}\,\nu(H)$ and
$\tau=a\,\nu$ with $\|a\|_{L^{\infty}(\nu)}\leq 3$. Indeed, notice that the
family $I$ of cubes $Q^{(\mu)}\in{\mathsf{H}}$ such that $\tau(Q\cap
H_{0})\leq\frac{1}{12}\nu(Q)$ satisfies

$\sum_{Q^{(\mu)}\in I}\tau(Q\cap H_{0})\leq\frac{1}{12}\sum_{Q^{(\mu)}\in
I}\nu(Q)\leq\frac{1}{12}\,\nu(H)\leq\frac{1}{2}\,\tau(H_{0}).$

So

$\sum_{Q^{(\mu)}\in{\mathsf{H}}\setminus
I}\tau(Q)\geq\sum_{Q^{(\mu)}\in{\mathsf{H}}\setminus I}\tau(Q\cap
H_{0})\geq\frac{1}{2}\,\tau(H_{0}).$

By Vitali’s 5r-covering lemma, there exists a subfamily
${\mathsf{H}}^{0}\subset{\mathsf{H}}\setminus I$ such that the balls
$\\{3B_{Q}\\}_{Q^{(\mu)}\in{\mathsf{H}}^{0}}$ are disjoint and the balls
$\\{15B_{Q}\\}_{Q^{(\mu)}\in{\mathsf{H}}^{0}}$ cover
$\bigcup_{Q^{(\mu)}\in{\mathsf{H}}\setminus I}Q$. From the fact that the cubes
from ${\mathsf{H}}^{0}$ are ${\mathcal{P}}$-doubling and the properties of the
family ${\mathsf{Reg}}(e^{\prime}(R))$ in Lemma 5.4, we have
$\nu(Q)\approx\nu(15B_{Q})$ if $Q^{(\mu)}\in{\mathsf{H}}^{0}$, and thus

$\displaystyle\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\tau(Q\cap H_{0})$
$\displaystyle\geq\frac{1}{12}\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\nu(Q)\approx\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\nu(15B_{Q})$
$\displaystyle\geq\sum_{Q^{(\mu)}\in{\mathsf{H}}\setminus
I}\nu(Q)\geq\frac{1}{3}\,\sum_{Q^{(\mu)}\in{\mathsf{H}}\setminus
I}\tau(Q)\geq\frac{1}{6}\,\tau(H_{0})\geq\frac{1}{36}\,\nu(H).$

The converse estimate $\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\tau(Q\cap
H_{0})\lesssim\nu(H)$ follows trivially from $\|a\|_{L^{\infty}(\nu)}\leq 3$.

We consider the function

(5.36)
${\varphi}=\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\Theta(Q)\,{\varphi}_{Q},$

where $\chi_{B_{Q}}\leq{\varphi}_{Q}\leq\chi_{1.1B_{Q}}$,
$\|\nabla{\varphi}_{Q}\|_{\infty}\lesssim\ell(Q)^{-1}$. Remark that

${\mathcal{R}}^{*}(\nabla{\varphi}\,{\mathcal{L}}^{n+1})=\tilde{c}\,{\varphi},$

where $\tilde{c}$ is some non-zero absolute constant. Notice also that

(5.37)
$\|\nabla{\varphi}\|_{1}\lesssim\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\Theta(Q)\,\ell(Q)^{n}\approx\tau(H_{0}).$

Multiplying the left hand side of (5.35) by $|\nabla{\varphi}|$, integrating
with respect to the Lebesgue measure ${\mathcal{L}}^{n+1}$, taking into
account that the estimate in (5.35) holds on the whole ${\mathbb{R}}^{n+1}$,
and using (5.37), we get

(5.38)
$\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p}\,|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1}+p\\!\int{\mathcal{R}}_{\tau}^{*}\Bigl{[}\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p-1}|{\mathcal{R}}\tau|^{-1}{\mathcal{R}}\tau\Bigr{]}\,|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1}\\\
\lesssim\lambda\,\sigma_{p}({\mathsf{H}}).$

Next we estimate the second integral on the left hand side of the preceding
inequality, which we denote by $I$. For that purpose we will also need the
estimate

(5.39)
$\int|{\mathcal{R}}(|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1})|^{p}\,d\tau\lesssim\Lambda^{p{\varepsilon}_{n}}\sigma_{p}({\mathsf{H}}),$

which we defer to Lemma 5.14. Using this inequality we get

$|I|=\left|\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p-1}|{\mathcal{R}}\tau|^{-1}{\mathcal{R}}\tau\cdot{\mathcal{R}}(|\nabla{\varphi}|\,{\mathcal{L}}^{n+1})\,d\tau\right|\\\
\leq\left(\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p}\,d\tau\right)^{1/p^{\prime}}\left(\int|{\mathcal{R}}(|\nabla{\varphi}|\,{\mathcal{L}}^{n+1})|^{p}\,d\tau\right)^{1/p}\\\
\leq(F(\tau))^{1/p^{\prime}}\left(\int|{\mathcal{R}}(|\nabla{\varphi}|\,{\mathcal{L}}^{n+1})|^{p}\,d\tau\right)^{1/p}\\\
\overset{{(\ref{e.admis2})},\eqref{eqacpsi}}{\lesssim}\bigl{(}\lambda\,\sigma_{p}({\mathsf{H}})\bigr{)}^{1/p^{\prime}}\bigl{(}\Lambda^{p{\varepsilon}_{n}}\sigma_{p}({\mathsf{H}})\bigr{)}^{1/p}=\lambda^{1/p^{\prime}}\,\Lambda^{{\varepsilon}_{n}}\sigma_{p}({\mathsf{H}}).$

From (5.38) and the preceding estimate we deduce that

$\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p}\,|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1}\lesssim\lambda^{1/p^{\prime}}\,\Lambda^{{\varepsilon}_{n}}\sigma_{p}({\mathsf{H}}).$

Notice now that, for all $x\in\operatorname{supp}{\varphi}$,

$G(x)=2A_{0}^{3}\int_{{\mathcal{S}}_{\eta}^{\prime}\setminus
V_{2}}\frac{1}{\ell(R)\,|x-y|^{n-1}}\,d\eta(y)\lesssim\frac{\eta({\mathcal{S}}_{\eta}^{\prime})}{\ell(R)\,{\rm
dist}({\mathcal{S}}_{\eta}^{\prime}\setminus
V_{2},{\mathcal{S}}_{\eta})^{n-1}}\lesssim\Theta(R),$

and so

$|f(x)|\leq C\,\Theta(R)+c_{0}\,\Theta({\mathsf{H}})\leq
2c_{0}\,\Theta({\mathsf{H}})\qquad\mbox{for all
$x\in\operatorname{supp}{\varphi}$,}$

taking into account that $\Lambda\gg 1$. So we have

(5.40)
$\displaystyle\int|{\mathcal{R}}\tau|^{p}\,|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1}$
$\displaystyle\lesssim\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p}\,|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1}+\int|f|^{p}\,|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1}$
$\displaystyle\lesssim\bigl{(}\lambda^{1/p^{\prime}}\Lambda^{{\varepsilon}_{n}}+c_{0}^{p}\bigr{)}\,\sigma_{p}({\mathsf{H}}).$

To get a contradiction, note that by the construction of ${\varphi}$ and by
the properties of $\tau$, we have

(5.41)
$\left|\int{\mathcal{R}}\tau\cdot\nabla{\varphi}\,d{\mathcal{L}}^{n+1}\right|=\left|\int{\mathcal{R}}^{*}(\nabla{\varphi}\,{\mathcal{L}}^{n+1})\,d\tau\right|=|\tilde{c}|\int{\varphi}\,d\tau\\\
\geq|\tilde{c}|\,\Theta({\mathsf{H}})\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\tau(B_{Q})\gtrsim|\tilde{c}|\,\Theta({\mathsf{H}})\tau(H_{0})\geq\frac{|\tilde{c}|}{6}\,\Theta({\mathsf{H}})\nu(H).$

However, by Hölder’s inequality, (5.37), and (5.40), we have

$\displaystyle\left|\int{\mathcal{R}}\tau\cdot\nabla{\varphi}\,d{\mathcal{L}}^{n+1}\right|$
$\displaystyle\leq\left(\int|{\mathcal{R}}\tau|^{p}\,|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1}\right)^{1/p}\left(\int|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1}\right)^{1/p^{\prime}}$
$\displaystyle\lesssim\bigl{(}\lambda^{1/(p\,p^{\prime})}\Lambda^{{\varepsilon}_{n}/p}+c_{0}\bigr{)}\,\Theta({\mathsf{H}})\nu(H),$

which contradicts (5.41) if $c_{0}$ is chosen small enough and

$\lambda\leq c\,\Lambda^{-p^{\prime}{\varepsilon}_{n}},$

with $c$ small enough. ∎

###### Proof of (5.35).

Recall that $\tau=a\,\nu$ is a minimizer for $F(\cdot)$ that in particular
satisfies $F(\tau)\leq 3\lambda\,\sigma_{p}({\mathsf{H}}),$ by (5.34). We have
to show that, for all $x\in\operatorname{supp}\tau$,

$\bigl{|}\bigl{(}|{\mathcal{R}}\tau(x)|-f(x)\bigr{)}_{+}|^{p}+p\,{\mathcal{R}}_{\tau}^{*}\Bigl{[}\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p-1}|{\mathcal{R}}\tau|^{-1}{\mathcal{R}}\tau\Bigr{]}(x)\lesssim\lambda\,\Theta({\mathsf{H}})^{p}.$

Let $x_{0}\in\operatorname{supp}\tau$ and let $B=B(x_{0},\rho)$, with $\rho>0$
small. For $0<t<1$ we consider the competing measure $\tau_{t}=a_{t}\,\tau$,
where $a_{t}$ is defined as follows:

$a_{t}=(1-t\,\chi_{B})a.$

Notice that, for each $0<t<1$, $a_{t}$ is a non-negative function such that

$\|a_{t}\|_{L^{\infty}(\nu)}\leq\|a\|_{L^{\infty}(\nu)}.$

Taking the above into account we deduce that

$\displaystyle F(\tau_{t})$
$\displaystyle=\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau_{t}|-f\bigr{)}_{+}\bigr{|}^{p}\,d\tau_{t}+\lambda\,\|a_{t}\|_{L^{\infty}(\nu)}\,\sigma_{p}({\mathsf{H}})+\lambda\,\frac{\nu(H_{0})}{\tau_{t}(H_{0})}\,\sigma_{p}({\mathsf{H}})$
$\displaystyle\leq\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau_{t}|-f\bigr{)}_{+}\bigr{|}^{p}\,d\tau_{t}+\lambda\,\|a\|_{L^{\infty}(\nu)}\,\sigma_{p}({\mathsf{H}})+\lambda\,\frac{\nu(H_{0})}{\tau_{t}(H_{0})}\,\sigma_{p}({\mathsf{H}})$
$\displaystyle=:{\widetilde{F}}(\tau_{t}).$

Since ${\widetilde{F}}(\tau)=F(\tau)\leq
F(\tau_{t})\leq{\widetilde{F}}(\tau_{t})$ for $t\geq 0$, we infer that

(5.42)
$\frac{1}{\tau(B)}\,\frac{d}{dt}\,{\widetilde{F}}(\tau_{t})\biggr{|}_{t=0+}\geq
0.$

Using that

$\frac{d}{dt}|{\mathcal{R}}\tau_{t}(x)|\biggr{|}_{t=0}=\frac{1}{|{\mathcal{R}}\tau(x)|}{\mathcal{R}}\tau(x)\cdot{\mathcal{R}}(-\chi_{B}\tau)(x),$

an easy computation gives

$\displaystyle\frac{d}{dt}\,{\widetilde{F}}(\tau_{t})\biggr{|}_{t=0}$
$\displaystyle=-\int_{B}\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p}\,d\tau$
$\displaystyle\quad+p\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p-1}\,\frac{1}{|{\mathcal{R}}\tau|}{\mathcal{R}}\tau\cdot{\mathcal{R}}(-\chi_{B}\tau)\,d\tau$
$\displaystyle\quad+\lambda\,\frac{\nu(H_{0})}{\tau(H_{0})^{2}}\,\sigma_{p}({\mathsf{H}})\,\tau(B).$

Recalling that
$\tau(H_{0})\geq\frac{1}{3}\,\nu(H_{0})\geq\frac{1}{6}\,\nu(H)$, from (5.42)
and the preceding calculation we derive

(5.43)
$\frac{1}{\tau(B)}\int_{B}\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p}\,d\tau+\frac{p}{\tau(B)}\int\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p-1}\,\frac{1}{|{\mathcal{R}}\tau|}{\mathcal{R}}\tau\cdot{\mathcal{R}}(\chi_{B}\tau)\,d\tau\\\
\leq
3\lambda\,\frac{\sigma_{p}({\mathsf{H}})}{\tau(H_{0})}=3\lambda\,\Theta({\mathsf{H}})^{p}\,\frac{\nu(H)}{\tau(H_{0})}\leq
18\lambda\,\Theta({\mathsf{H}})^{p}.$

We rewrite the left hand side of (5.43) as

$\frac{1}{\tau(B)}\int_{B}\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p}\,d\tau+\frac{p}{\tau(B)}\int_{B}{\mathcal{R}}_{\tau}^{*}\Bigl{[}\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p-1}\,|{\mathcal{R}}\tau|^{-1}{\mathcal{R}}\tau\Bigr{]}\,d\tau.$

Taking into account that the functions in the integrands are continuous on
$\operatorname{supp}(\tau)$, letting the radius $\rho$ of $B$ tend to $0$, it
turns out that the above expression converges to

$\bigl{|}\bigl{(}|{\mathcal{R}}\tau(x_{0})|-f(x_{0})\bigr{)}_{+}|^{p}+p\,{\mathcal{R}}_{\tau}^{*}\Bigl{[}\bigl{|}\bigl{(}|{\mathcal{R}}\tau|-f\bigr{)}_{+}|^{p-1}\,|{\mathcal{R}}\tau|^{-1}{\mathcal{R}}\tau\Bigr{]}(x_{0}).$

The desired estimate (5.35) follows from the above and (5.43). ∎

In order to complete the proof of Lemma 5.13 we need the following technical
result.

######  Lemma 5.14.

Suppose that (5.32) holds with $\lambda\leq 1$ and let ${\varphi}$ be as in
(5.36). Then,

$\int|{\mathcal{R}}(|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1})|^{p}\,d\nu\lesssim\Lambda^{p{\varepsilon}_{n}}\sigma_{p}({\mathsf{H}}),$

###### Proof.

Recall that

${\varphi}=\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\Theta(Q)\,{\varphi}_{Q},$

with $\chi_{B_{Q}}\leq{\varphi}_{Q}\leq\chi_{1.1B_{Q}}$,
$\|\nabla{\varphi}_{Q}\|_{\infty}\lesssim\ell(Q)^{-1}$. We consider the
function

$g=\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}g_{Q},$

where $g_{Q}=c_{Q}\,\chi_{Q\cap H_{0}}$, with
$c_{Q}=\Theta(Q)\int|\nabla{\varphi}_{Q}|\,d{\mathcal{L}}^{n+1}\,\eta(Q\cap
H_{0})^{-1}$. Observe that

(5.44) $\int
g_{Q}\,d\eta=\Theta(Q)\int|\nabla{\varphi}_{Q}|\,d{\mathcal{L}}^{n+1}$

and that

$0\leq c_{Q}\lesssim\frac{\Theta(Q)\,\ell(Q)^{n}}{\eta(Q\cap
H_{0})}\approx\frac{\mu(Q^{(\mu)})}{\eta(Q)}=1.$

The first step of our arguments consists in comparing
${\mathcal{R}}(|\nabla{\varphi}|\,d{\mathcal{L}}^{n+1})(x)$ to
${\mathcal{R}}_{\Psi(x)}(g\,\eta)(x)$, with $\Psi$ given by (5.21). We will
prove that, for each $Q^{(\mu)}\in{\mathsf{H}}^{0}$,

(5.45) $\displaystyle|{\mathcal{R}}(\Theta(Q)|\nabla{\varphi}_{Q}|$
$\displaystyle\,{\mathcal{L}}^{n+1})(x)-{\mathcal{R}}_{\Psi(x)}(g_{Q}\,\eta)(x)|$
$\displaystyle\lesssim\frac{\Theta(Q)\,\ell(Q)^{n+1}}{{\rm
dist}(x,Q)^{n+1}+\ell(Q)^{n+1}}+\chi_{2B_{Q}}(x)\,|{\mathcal{R}}_{\Psi(x)}(g_{Q}\,\eta)(x)|$
$\displaystyle\quad+\int_{c\Psi(x)\leq|x-y|\leq\Psi(x)}\frac{g_{Q}(y)}{|x-y|^{n}}\,d\eta(y),$

for some fixed $c>0$. The arguments to prove this estimate are quite standard.

Suppose first that $x\in 2B_{Q}$. In this case, we have

$\displaystyle|{\mathcal{R}}(\Theta(Q)|\nabla{\varphi}_{Q}|\,{\mathcal{L}}^{n+1})(x)|$
$\displaystyle\lesssim\Theta(Q)\int_{1.1B_{Q}}\frac{1}{\ell(Q)\,|x-y|^{n}}\,d{\mathcal{L}}^{n+1}(y)\lesssim\Theta(Q),$

which yields

$|{\mathcal{R}}(\Theta(Q)|\nabla{\varphi}_{Q}|\,{\mathcal{L}}^{n+1})(x)-{\mathcal{R}}_{\Psi(x)}(g_{Q}\,\eta)(x)|\lesssim\Theta(Q)+|{\mathcal{R}}_{\Psi(x)}(g_{Q}\,\eta)(x)|,$

and shows that (5.45) holds in this situation.

In the case $x\not\in 2B_{Q}$ we write

$\displaystyle{\mathcal{R}}(\Theta(Q)|$
$\displaystyle\nabla{\varphi}_{Q}|\,{\mathcal{L}}^{n+1})(x)-{\mathcal{R}}_{\Psi(x)}(g_{Q}\,\eta)(x)$
$\displaystyle={\mathcal{R}}\big{(}\Theta(Q)|\nabla{\varphi}_{Q}|\,{\mathcal{L}}^{n+1}-g_{Q}\,\eta\big{)}(x)+\int_{|x-y|\leq\Psi(x)}\frac{x-y}{|x-y|^{n+1}}g_{Q}(y)\,d\eta(y)$
$\displaystyle\overset{\eqref{eq:gQintegral}}{=}\int\left(\frac{x-y}{|x-y|^{n+1}}-\frac{x-x_{Q}}{|x-x_{Q}|^{n+1}}\right)\bigl{[}\Theta(Q)|\nabla{\varphi}_{Q}(y)|\,d{\mathcal{L}}^{n+1}(y)-g_{Q}(y)\,d\eta(y)\bigr{]}$
$\displaystyle\quad\quad\quad+\int_{|x-y|\leq\Psi(x)}\frac{x-y}{|x-y|^{n+1}}g_{Q}(y)\,d\eta(y)$
$\displaystyle=:I_{1}(x)+I_{2}(x).$

Concerning the term $I_{1}(x)$, recalling that
$\operatorname{supp}{\varphi}_{Q}\cup\operatorname{supp}g_{Q}\subset
1.1\overline{B_{Q}}$, we obtain

$\displaystyle|I_{1}(x)|$
$\displaystyle\lesssim\int\frac{\ell(Q)}{|x-x_{Q}|^{n+1}}\bigl{[}\Theta(Q)|\nabla{\varphi}_{Q}(y)|\,d{\mathcal{L}}^{n+1}(y)+g_{Q}(y)\,d\eta(y)\bigr{]}\lesssim\frac{\ell(Q)}{|x-x_{Q}|^{n+1}}\,\eta(Q).$

Regarding $I_{2}(x)$, we write

$I_{2}(x)\leq\int_{y\in
Q:|x-y|\leq\Psi(x)}\frac{1}{|x-y|^{n}}\,g_{Q}(y)\,d\eta(y).$

Notice that for $y\in Q$, since $x\notin 2B_{Q}$, we have

$C|x-y|\geq\ell(Q)\overset{\eqref{eqdefPsi1}}{\geq}\Psi(y)\geq\Psi(x)-|x-y|.$

Thus, $|x-y|\geq c\,\Psi(x)$, and so

$I_{2}(x)\leq\int_{c\Psi(x)\leq|x-y|\leq\Psi(x)}\frac{1}{|x-y|^{n}}\,g_{Q}(y)\,d\eta(y).$

Gathering the estimates for $I_{1}(x)$ and $I_{2}(x)$ we see that (5.45) also
holds in this case.

From (5.45) we infer that

$\displaystyle|{\mathcal{R}}(|\nabla{\varphi}|\,{\mathcal{L}}^{n+1})$
$\displaystyle(x)-{\mathcal{R}}_{\Psi(x)}(g\,\eta)(x)|$
$\displaystyle\lesssim\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\frac{\Theta(Q)\,\ell(Q)^{n+1}}{{\rm
dist}(x,Q)^{n+1}+\ell(Q)^{n+1}}+\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\chi_{2B_{Q}}(x)\,|{\mathcal{R}}_{\Psi(x)}(g_{Q}\,\eta)(x)|$
$\displaystyle\quad+\int_{c\Psi(x)\leq|x-y|\leq\Psi(x)}\frac{1}{|x-y|^{n}}\,g(y)\,d\eta(y)$
$\displaystyle=S_{1}(x)+S_{2}(x)+S_{3}(x).$

We split

(5.46)
$\int|{\mathcal{R}}(|\nabla{\varphi}|\,{\mathcal{L}}^{n+1})(x)-{\mathcal{R}}_{\Psi(x)}(g\,\eta)(x)|^{p}\,d\eta(x)\lesssim\sum_{i=1}^{3}\int|S_{i}(x)|^{p}\,d\eta(x).$

We estimate the first summand by duality. Consider a function $h\in
L^{p^{\prime}}(\eta)$. Then,

(5.47) $\int
S_{1}(x)\,h(x)\,d\eta(x)=\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\eta(Q)\int\frac{\ell(Q)}{{\rm
dist}(x,Q)^{n+1}+\ell(Q)^{n+1}}\,h(x)\,d\eta(x).$

For each $Q^{(\mu)}\in{\mathsf{H}}^{0}$, using the fact that $\eta(\lambda
Q)\leq C\,\Theta({\mathsf{H}})\,\ell(\lambda Q)^{n}$ for every $\lambda\geq
1$, we obtain

$\int\frac{\ell(Q)}{{\rm
dist}(x,Q)^{n+1}+\ell(Q)^{n+1}}\,h(x)\,d\eta(x)\lesssim
C\,\Theta({\mathsf{H}})\,\inf_{y\in Q}{\mathcal{M}}_{\eta}h(y).$

Therefore, the right side of (5.47) does not exceed

$\displaystyle
C\Theta({\mathsf{H}})\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\eta(Q)\,\inf_{y\in
Q}{\mathcal{M}}_{\eta}h(y)$
$\displaystyle\lesssim\Theta({\mathsf{H}})\int_{H}{\mathcal{M}}_{\eta}h(y)\,d\eta(y)$
$\displaystyle\leq\Theta({\mathsf{H}})\,\eta(H)^{1/p}\,\|{\mathcal{M}}_{\eta}h\|_{L^{p^{\prime}}(\eta)}\lesssim\Theta({\mathsf{H}})\,\eta(H)^{1/p}\,\|h\|_{L^{p^{\prime}}(\eta)}.$

So we deduce that

$\int|S_{1}(x)|^{p}\,d\eta(x)\lesssim\sigma_{p}({\mathsf{H}}).$

Regarding the summand in (5.46) involving $S_{2}$, since the balls $2B_{Q}$
are disjoint, we have

$\int|S_{2}(x)|^{p}\,d\eta(x)=\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\int_{2B_{Q}}|{\mathcal{R}}_{\Psi(x)}(g_{Q}\,\eta)(x)|^{p}\,d\eta(x)\leq\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\|{\mathcal{R}}_{\Psi(\cdot)}(g_{Q}\,\eta)\|_{L^{p}(\eta)}^{p},$

where
${\mathcal{R}}_{\Psi(\cdot)}(g_{Q}\,\eta)(x)={\mathcal{R}}_{\Psi(x)}(g_{Q}\,\eta)(x)$.
Finally, to estimate the last summand in (5.46), we take into account that
$|S_{3}(x)|\lesssim{\mathcal{M}}_{\Psi,n}(g\,\eta)(x),$ where
${\mathcal{M}}_{\Psi,n}(g\,\eta)$ is the maximal operator from (5.31). Hence,

$\int|S_{3}(x)|^{p}\,d\eta(x)\lesssim\int|{\mathcal{M}}_{\Psi,n}(g\,\eta)|^{p}\,d\eta\lesssim\Theta({\mathsf{H}})^{p}\,\|g\|_{L^{p}(\eta)}^{p}\lesssim\sigma_{p}({\mathsf{H}}).$

Gathering the estimates obtained for $S_{1}$, $S_{2}$, $S_{3}$, by (5.46) we
get

$\|{\mathcal{R}}(|\nabla{\varphi}|\,{\mathcal{L}}^{n+1})\|_{L^{p}(\eta)}^{p}\lesssim\sigma_{p}({\mathsf{H}})+\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\|{\mathcal{R}}_{\Psi(\cdot)}(g_{Q}\,\eta)\|_{L^{p}(\eta)}^{p}+\|{\mathcal{R}}_{\Psi(\cdot)}(g\eta)\|_{L^{p}(\eta)}^{p}.$

From (5.4), we deduce that

$\displaystyle\|{\mathcal{R}}(|\nabla{\varphi}|\,{\mathcal{L}}^{n+1})\|_{L^{p}(\eta)}^{p}$
$\displaystyle\lesssim\sigma_{p}({\mathsf{H}})+\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\|{\mathcal{R}}_{\Psi}(g_{Q}\,\eta)\|_{L^{p}(\eta)}^{p}+\|{\mathcal{R}}_{\Psi}(g\eta)\|_{L^{p}(\eta)}^{p}$
$\displaystyle\quad+\sum_{Q^{(\mu)}\in{\mathsf{H}}^{0}}\|{\mathcal{M}}_{\Psi,n}(g_{Q}\,\eta)\|_{L^{p}(\eta)}^{p}+\|{\mathcal{M}}_{\Psi,n}(g\,\eta)\|_{L^{p}(\eta)}^{p}.$

Using now that, by Lemma 5.12, ${\mathcal{R}}_{\Psi,\eta}$ is bounded from
$L^{p}(\eta{\lfloor}_{H_{0}})$ to
$L^{p}(\eta{\lfloor}_{{\mathcal{S}}^{\prime}_{\eta}})$ with

$\|{\mathcal{R}}_{\Psi}\|_{L^{p}(\eta{\lfloor}_{H_{0}})\to
L^{p}(\eta{\lfloor}_{{\mathcal{S}}^{\prime}_{\eta}})}\lesssim\Lambda^{{\varepsilon}_{n}}\Theta({\mathsf{H}}),$

and that the same happens for ${\mathcal{M}}_{\Psi,n}$, we get

$\|{\mathcal{R}}(|\nabla{\varphi}|\,{\mathcal{L}}^{n+1})\|_{L^{p}(\eta)}^{p}\lesssim\sigma_{p}({\mathsf{H}})+\Lambda^{p{\varepsilon}_{n}}\Theta({\mathsf{H}})^{p}\,\|g\|_{L^{p}(\eta)}^{p}\lesssim\Lambda^{p{\varepsilon}_{n}}\sigma_{p}({\mathsf{H}}).$

∎

###  5.6. Lower estimates for ${\mathcal{R}}\eta$

######  Lemma 5.15.

Let $R\in{\mathsf{MDW}}$ be such that $R\in{\mathsf{Trc}}$ and let $V_{4}$ and
$\eta$ be as in Section 5.4. Assume $\Lambda>0$ big enough and let $c_{0}$ be
as in Lemma 5.13. Then we have

$\int_{V_{4}}\big{|}(|{\mathcal{R}}\eta(x)|-\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta(x)\gtrsim\Lambda^{-(p^{\prime}+1){\varepsilon}_{n}}\sigma_{p}({\mathsf{HD}}_{1}(e(R))),$

for any $p\in(1,\infty)$, with the implicit constant depending on $p$.

###### Proof.

Recall that $\nu={\varphi}_{R}\,\eta$. For all $x\in\operatorname{supp}\eta$,
using that $\|\nabla{\varphi}_{R}\|_{\infty}\leq 2A_{0}^{-3}\ell(R)^{-1}$, we
obtain

$\displaystyle\big{|}{\varphi}_{R}(x)\,{\mathcal{R}}\eta(x)-{\mathcal{R}}\nu(x)\big{|}$
$\displaystyle=\big{|}{\varphi}_{R}(x)\,{\mathcal{R}}\eta(x)-{\mathcal{R}}({\varphi}_{R}\eta)(x)\big{|}$
$\displaystyle=\left|\int\frac{({\varphi}_{R}(x)-{\varphi}_{R}(y))\,(x-y)}{|x-y|^{n+1}}\,d\eta(y)\right|$
$\displaystyle\leq 2A_{0}^{3}\int_{y\in{\mathcal{S}}_{\eta}\setminus
V_{2}}\frac{1}{\ell(R)\,|x-y|^{n-1}}\,d\eta(y)$
$\displaystyle\quad+2A_{0}^{3}\int_{y\in
V_{2}:{\varphi}_{R}(y)\neq{\varphi}_{R}(x)}\frac{1}{\ell(R)\,|x-y|^{n-1}}\,d\eta(y).$

Recall that ${\varphi}_{R}$ equals $1$ on $V_{3}$ and vanishes out of $V_{4}$.
Then it is clear that the last integral on the right hand side above vanishes
if $x\in V_{3}$ and it does not exceed $C\,\Theta(R)$ if $x\in V_{3}^{c}$. So
in any case

$\big{|}{\varphi}_{R}(x)\,{\mathcal{R}}\eta(x)-{\mathcal{R}}\nu(x)\big{|}\leq
G(x)+C_{3}\,\Theta(R).$

From the preceding estimate we infer that

$\displaystyle|{\varphi}_{R}(x)\,{\mathcal{R}}\eta(x)|-\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1})$
$\displaystyle\geq|{\mathcal{R}}\nu(x)|-G(x)-C_{3}\,\Theta(R)-\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1})$
$\displaystyle\geq|{\mathcal{R}}\nu(x)|-G(x)-c_{0}\,\Theta({\mathsf{HD}}_{1}).$

Therefore, since $|(\;\cdot\;)_{+}|^{p}$ is non-decreasing,

$\displaystyle\int_{V_{4}}\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta$
$\displaystyle\geq\int\big{|}(|{\varphi}_{R}\,{\mathcal{R}}\eta|-\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta$
$\displaystyle\geq\int\big{|}(|{\mathcal{R}}\nu|-G-c_{0}\,\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta$
$\displaystyle\geq\int\big{|}(|{\mathcal{R}}\nu|-G-c_{0}\,\Theta({\mathsf{H}}))_{+}\big{|}^{p}\,d\nu.$

By Lemmas 5.13 and 5.6,

$\int\big{|}(|{\mathcal{R}}\nu|-G-c_{0}\,\Theta({\mathsf{H}}))_{+}\big{|}^{p}\,d\nu\geq
c\Lambda^{-p^{\prime}{\varepsilon}_{n}}\sigma_{p}({\mathsf{H}})\geq
c\Lambda^{-(p^{\prime}+1){\varepsilon}_{n}}\sigma_{p}({\mathsf{HD}}_{1}(e(R))),$

and so the lemma follows. ∎

##  6\. Lower estimates for the Haar coefficients of ${\mathcal{R}}\mu$ for
cubes near a typical tractable tree

For $R\in{\mathsf{MDW}}$ and $P\in{\mathcal{D}}_{\mu}$, we write
$P\sim{\mathcal{T}}(e^{\prime}(R))$ if there exists some
$P^{\prime}\in{\mathcal{T}}(e^{\prime}(R))$ such that

(6.1) $A_{0}^{-2}\ell(P)\leq\ell(P^{\prime})\leq A_{0}^{2}\,\ell(P)\quad\text{
and }\quad 20P^{\prime}\cap 20P\neq\varnothing.$

We say that ${\mathcal{T}}(e^{\prime}(R))$ is a typical tractable tree, and we
write $R\in{\mathsf{Ty}}$ if

(6.2)
$\sum_{P\in{\mathsf{DB}}:P\sim{\mathcal{T}}(e^{\prime}(R))}{\mathcal{E}}_{\infty}(9P)\leq\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}(e(R))).$

The reason for the terminology “typical tree” is due to the fact that such
trees are prevalent, in a sense, because of Lemma 12.4 below.

In all this section we assume that $R\in{\mathsf{MDW}}$ is such that
$R\in{\mathsf{Trc}}\cap{\mathsf{Ty}}$, i.e., ${\mathcal{T}}(e^{\prime}(R))$ is
tractable and typical, and we consider the measure $\eta$ constructed in
Section 5.2. Roughly speaking, our objective is to transfer the lower estimate
we obtained for ${\mathcal{R}}\eta$ in Lemma 5.15 to the Haar coefficients of
${\mathcal{R}}\mu$ for cubes close to ${\mathcal{T}}(e^{\prime}(R))$. Recall
that ${\mathcal{R}}\mu(x)$ exists $\mu$-a.e. as a principal value under the
assumptions of Proposition 3.6.

###  6.1. The operators ${\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}$,
${\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}$, and
$\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}$

To simplify notation, in this section we will write

${\mathsf{End}}={\mathsf{End}}(e^{\prime}(R)),\quad\;{\mathsf{Reg}}={\mathsf{Reg}}(e^{\prime}(R)),\;\quad\;{\mathcal{T}}={\mathcal{T}}(e^{\prime}(R)),\quad\,\text{and}\quad\;{\mathcal{T}}_{\mathsf{Reg}}={\mathcal{T}}_{\mathsf{Reg}}(e^{\prime}(R)).$

We need to consider an enlarged version of the generalized tree
${\mathcal{T}}$, due to some technical difficulties that arise because the
cubes from ${\mathsf{Neg}}:={\mathsf{Neg}}(e^{\prime}(R))\cap{\mathsf{End}}$
are not ${\mathcal{P}}$-doubling. To this end, denote by
${\mathsf{Reg}}_{\mathsf{Neg}}$ the family of the cubes from ${\mathsf{Reg}}$
which are contained in some cube from ${\mathsf{Neg}}$. Let
${\mathsf{D}}_{\mathsf{Neg}}$ be the subfamily of the cubes
$P\in{\mathsf{Reg}}_{\mathsf{Neg}}$ for which there exists some
${\mathcal{P}}$-doubling cube $S\in{\mathcal{T}}_{\mathsf{Reg}}$ that contains
$P$. By the definition of ${\mathsf{Neg}}$, such cube $S$ should be contained
in the cube from $Q\in{\mathsf{Neg}}$ such that $P\subset Q$. We also denote
by ${\mathsf{M}}_{\mathsf{Neg}}$ the family of maximal
${\mathcal{P}}$-doubling cubes which belong to ${\mathcal{T}}_{\mathsf{Reg}}$
and are contained in some cube from ${\mathsf{Neg}}$, so that, in particular,
any cube from ${\mathsf{D}}_{\mathsf{Neg}}$ is contained in another from
${\mathsf{M}}_{\mathsf{Neg}}$.

We define

${\widetilde{{\mathsf{End}}}}=({\mathsf{End}}\setminus{\mathsf{Neg}})\cup{\mathsf{M}}_{\mathsf{Neg}},$

and we let
${\widetilde{{\mathcal{T}}}}={\widetilde{{\mathcal{T}}}}(e^{\prime}(R))$ be
the family of cubes that belong to ${\mathcal{T}}_{\mathsf{Reg}}$ but are not
strictly contained in any cube from ${\widetilde{{\mathsf{End}}}}$. Further,
we write

(6.3)
$Z=Z(e^{\prime}(R))=e^{\prime}(R)\setminus\bigcup_{Q\in{\mathsf{End}}}Q\quad\text{
and
}\quad{\widetilde{Z}}={\widetilde{Z}}(e^{\prime}(R))=e^{\prime}(R)\setminus\bigcup_{Q\in{\widetilde{{\mathsf{End}}}}}Q$

Then we denote

${\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)=\sum_{Q\in{\mathsf{Reg}}}\chi_{Q}(x)\,{\mathcal{R}}(\chi_{2R\setminus
2Q}\mu)(x),$
${\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)=\sum_{Q\in{\widetilde{{\mathsf{End}}}}}\chi_{Q}(x)\,{\mathcal{R}}(\chi_{2R\setminus
2Q}\mu)(x),$

and

$\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu(x)=\sum_{Q\in{\widetilde{{\mathsf{End}}}}}\chi_{Q}(x)\,\big{(}m_{\mu,Q}({\mathcal{R}}\mu)-m_{\mu,2R}({\mathcal{R}}\mu)\big{)}+\chi_{Z}(x)\big{(}{\mathcal{R}}\mu(x)-m_{\mu,2R}({\mathcal{R}}\mu)\big{)}.$

Remark that the cubes from ${\mathsf{Reg}}$ have the advantage over the cubes
from ${\widetilde{{\mathsf{End}}}}$ that their sizes change smoothly so that,
for example, neighboring cubes have comparable side lengths. However, they
need not be ${\mathcal{P}}$-doubling or doubling, unlike the cubes from
${\widetilde{{\mathsf{End}}}}$. In particular, it is not clear how to estimate
their energy ${\mathcal{E}}_{\infty}$ in terms of the condition
${\mathsf{DB}}$ (recall that the cubes from ${\mathsf{DB}}$ are asked to be
${\mathcal{P}}$-doubling).

For $Q\in{\mathcal{D}}_{\mu}$, we define

${\mathcal{Q}}_{\mathsf{Reg}}(Q)=\sum_{P\in{\mathsf{Reg}}}\frac{\ell(P)}{D(P,Q)^{n+1}}\,\mu(P),$

where

$D(P,Q)=\ell(P)+{\rm dist}(P,Q)+\ell(Q).$

The coefficient ${\mathcal{Q}}_{\mathsf{Reg}}(Q)$ will be used to bound some
“error terms” in our transference arguments. We will see later how they can be
estimated in terms of the coefficients ${\mathcal{P}}(Q)$. Notice that, unlike
${\mathcal{P}}(Q)$, the coefficients ${\mathcal{Q}}_{\mathsf{Reg}}(Q)$ depend
on the family ${\mathsf{Reg}}$.

######  Lemma 6.1.

For any $Q\in{\mathsf{Reg}}$ such that $(Q\cup\frac{1}{2}B(Q))\cap
V_{4}\neq\varnothing$ and $x\in Q$, $y\in\frac{1}{2}B(Q)$,

$\big{|}{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)-{\mathcal{R}}\eta(y)\big{|}\lesssim\Theta(R)+{\mathcal{P}}(Q)+{\mathcal{Q}}_{\mathsf{Reg}}(Q).$

###### Proof.

By the triangle inequality, for $x$, $y$ and $Q$ as in the lemma, we have

$\displaystyle\big{|}{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)-{\mathcal{R}}\eta(y)\big{|}$
$\displaystyle\leq\big{|}{\mathcal{R}}_{\mu}\chi_{2R\setminus
e^{\prime}(R)}(x)\big{|}+\big{|}{\mathcal{R}}_{\mu}\chi_{e^{\prime}(R)\setminus
2Q}(x)-{\mathcal{R}}_{\mu}\chi_{e^{\prime}(R)\setminus 2Q}(x_{Q})\big{|}$
$\displaystyle\quad+\big{|}{\mathcal{R}}_{\mu}\chi_{2Q\setminus
Q}(x_{Q})\big{|}+\big{|}{\mathcal{R}}_{\mu}\chi_{e^{\prime}(R)\setminus
Q}(x_{Q})-{\mathcal{R}}_{\eta}\chi_{(\frac{1}{2}B(Q))^{c}}(x_{Q})\big{|}$
$\displaystyle\quad+\big{|}{\mathcal{R}}_{\eta}\chi_{(\frac{1}{2}B(Q))^{c}}(x_{Q})-{\mathcal{R}}_{\eta}\chi_{(\frac{1}{2}B(Q))^{c}}(y)\big{|}+\big{|}{\mathcal{R}}_{\eta}\chi_{\frac{1}{2}B(Q)}(y)\big{|}$
$\displaystyle=I_{1}+\ldots+I_{6}.$

To estimate $I_{1}$, notice that, by (5.15), ${\rm
dist}(Q,\operatorname{supp}\mu\setminus e^{\prime}(R))\gtrsim\ell(R)$. Then it
follows that

$I_{1}\lesssim\frac{\mu(2R)}{\ell(R)^{n}}\lesssim\Theta(R).$

Arguing similarly one also gets

$I_{3}\lesssim\frac{\mu(2Q)}{\ell(Q)^{n}}\lesssim{\mathcal{P}}(Q).$

By standard arguments involving continuity of the Riesz kernel, we also deduce
that

$I_{2}\lesssim{\mathcal{P}}(Q),\quad I_{5}\lesssim{\mathcal{P}}(Q).$

Concerning the term $I_{6}$, using that $\eta$ is a constant multiple of
${\mathcal{L}}^{n+1}$ on $\frac{1}{2}B(Q)$, we get

$I_{6}\leq\int_{\frac{1}{2}B(Q)}\frac{1}{|x-y|^{n}}\,d\eta(y)\lesssim\frac{\eta(\tfrac{1}{2}B(Q))}{\ell(Q)^{n}}\lesssim{\mathcal{P}}(Q).$

Finally we deal with the term $I_{4}$. To this end, recall that
$\mu(P)=\eta(\frac{1}{2}B(P))$ for all $P\in{\mathsf{Reg}}$, and so

$\displaystyle I_{4}$ $\displaystyle\leq\sum_{P\in{\mathsf{Reg}}:P\neq
Q}\left|\int
K(x_{Q}-z)\,d\big{(}\eta{\lfloor}_{\frac{1}{2}B(P)}-\mu{\lfloor}_{P}\big{)}(z)\right|$
$\displaystyle\leq\sum_{P\in{\mathsf{Reg}}:P\neq
Q}\int|K(x_{Q}-z)-K(x_{Q}-x_{P})|\,d\big{(}\eta{\lfloor}_{\frac{1}{2}B(P)}+\mu{\lfloor}_{P}\big{)}(z)$
$\displaystyle\lesssim\sum_{P\in{\mathsf{Reg}}:P\neq
Q}\frac{\ell(P)}{|x_{Q}-x_{P}|^{n+1}}\,\mu(P)\approx\sum_{P\in{\mathsf{Reg}}:P\neq
Q}\frac{\ell(P)}{D(Q,P)^{n+1}}\,\mu(P)\leq{\mathcal{Q}}_{\mathsf{Reg}}(Q).$

For the estimates in the last line we took into account the properties of the
family ${\mathsf{Reg}}$ described in Lemma 5.4. Gathering the estimates above,
the lemma follows. ∎

######  Lemma 6.2.

For any $Q\in{\widetilde{{\mathsf{End}}}}$ and $x,y\in Q$,

$\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)-\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu(y)\big{|}\lesssim\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}+\left(\frac{{\mathcal{E}}(2Q)}{\mu(Q)}\right)^{1/2}.$

###### Proof.

For $Q\in{\widetilde{{\mathsf{End}}}}$ and $x,y\in Q$, we have

(6.4)
$\displaystyle{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)-\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu(y)$
$\displaystyle={\mathcal{R}}(\chi_{2R\setminus
2Q}\mu)(x)-\big{(}m_{\mu,Q}({\mathcal{R}}\mu)-m_{\mu,2R}({\mathcal{R}}\mu)\big{)}$
$\displaystyle=\big{(}{\mathcal{R}}(\chi_{2R\setminus
2Q}\mu)(x)-m_{\mu,Q}({\mathcal{R}}\chi_{2R\setminus
2Q}\mu)\big{)}-m_{\mu,Q}({\mathcal{R}}(\chi_{2Q}\mu))$
$\displaystyle\quad-\big{(}m_{\mu,Q}({\mathcal{R}}(\chi_{2R^{c}}\mu))-m_{\mu,2R}({\mathcal{R}}\mu)\big{)}.$

To estimate the first term on the right hand side, notice that for all $x,z\in
Q$, by standard arguments we have

$\big{|}{\mathcal{R}}(\chi_{2R\setminus
2Q}\mu)(x)-{\mathcal{R}}(\chi_{2R\setminus
2Q}\mu)(z)\big{|}\lesssim{\mathcal{P}}(Q).$

Averaging over $z\in Q$, we get

$\big{|}{\mathcal{R}}(\chi_{2R\setminus
2Q}\mu)(x)-m_{\mu,Q}({\mathcal{R}}\chi_{2R\setminus
2Q}\mu)\big{|}\lesssim{\mathcal{P}}(Q).$

To estimate $m_{\mu,Q}({\mathcal{R}}(\chi_{2Q}\mu))$ we use the fact that, by
the antisymmetry of the Riesz kernel,
$m_{\mu,Q}({\mathcal{R}}(\chi_{Q}\mu))=0$ and333Here we use the assumption
that ${\mathcal{R}}_{*}\mu\in L^{1}(\mu)$. Indeed, this implies that
${\mathcal{R}}_{*}(\chi_{Q}\mu)\in L^{1}(\mu)$ (using Lemma 2.6, for example).
Also, by antisymmetry we have
$\int_{Q}{\mathcal{R}}_{\varepsilon}(\chi_{Q}\mu)\,d\mu=0$ for any
${\varepsilon}>0$. Then the $\mu$-a.e. convergence of
${\mathcal{R}}_{\varepsilon}(\chi_{Q}\mu)(x)$ to the principal value
${\mathcal{R}}(\chi_{Q}\mu)(x)$ as ${\varepsilon}\to 0$ and the dominated
convergence theorem ensure that $\int_{Q}{\mathcal{R}}(\chi_{Q}\mu)\,d\mu=0$.
thus

$\big{|}m_{\mu,Q}({\mathcal{R}}(\chi_{2Q}\mu))\big{|}=\big{|}m_{\mu,Q}({\mathcal{R}}(\chi_{2Q\setminus
Q}\mu))\big{|}\leq\;\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.57222pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.70335pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.59167pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.10834pt}}\\!\int_{z\in Q}\int_{\xi\in 2Q\setminus
Q}\frac{1}{|z-\xi|^{n}}\,d\mu(\xi)d\mu(z).$

By Cauchy-Schwarz and Lemma 2.6, we have

(6.5) $\;\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.57222pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.70335pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.59167pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.10834pt}}\\!\int_{z\in Q}\int_{\xi\in 2Q\setminus
Q}\frac{1}{|z-\xi|^{n}}\,d\mu(\xi)d\mu(z)\\\
\leq\;\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.57222pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.70335pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.59167pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.10834pt}}\\!\int_{z\in Q}\int_{\xi\in 2Q\setminus
2B_{Q}}\frac{1}{|z-\xi|^{n}}\,d\mu(\xi)d\mu(z)+\;\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.57222pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.70335pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.59167pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.10834pt}}\\!\int_{z\in Q}\int_{\xi\in 2B_{Q}\setminus
Q}\frac{1}{|z-\xi|^{n}}\,d\mu(\xi)d\mu(z)\\\
\leq{\mathcal{P}}(Q)+\bigg{(}\;\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.57222pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.70335pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.59167pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.10834pt}}\\!\int_{z\in Q}\bigg{(}\int_{\xi\in 2B_{Q}\setminus
Q}\frac{1}{|z-\xi|^{n}}\,d\mu(\xi)\bigg{)}^{2}d\mu(z)\bigg{)}^{1/2}\lesssim{\mathcal{P}}(Q)+\left(\frac{{\mathcal{E}}(2Q)}{\mu(Q)}\right)^{1/2}.$

Finally we deal with the last term on the right hand side of (6.4). We split
it:

$\displaystyle\big{|}m_{\mu,Q}({\mathcal{R}}(\chi_{2R^{c}}\mu))-m_{\mu,2R}({\mathcal{R}}\mu)\big{|}$
$\displaystyle\leq\big{|}m_{\mu,Q}({\mathcal{R}}(\chi_{3R\setminus
2R}\mu))|+\big{|}m_{\mu,2R}({\mathcal{R}}(\chi_{3R}\mu))\big{|}$
$\displaystyle\quad+\big{|}m_{\mu,Q}({\mathcal{R}}(\chi_{3R^{c}}\mu))-m_{\mu,2R}({\mathcal{R}}(\chi_{3R^{c}}\mu))\big{|}$
$\displaystyle=J_{1}+J_{2}+J_{3}.$

Clearly,

$J_{1}\lesssim\frac{\mu(3R)}{\ell(R)^{n}}\lesssim{\mathcal{P}}(R).$

Also, by the antisymmetry of the Riesz kernel and arguing as in (6.5),

$\displaystyle J_{2}=$
$\displaystyle\big{|}m_{\mu,2R}({\mathcal{R}}(\chi_{3R}\mu))\big{|}=\big{|}m_{\mu,2R}({\mathcal{R}}(\chi_{3R\setminus
2R}\mu))\big{|}$ $\displaystyle\leq\;\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.57222pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.70335pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.59167pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.10834pt}}\\!\int_{z\in 2R}\int_{\xi\in 3R\setminus
2R}\frac{1}{|z-\xi|^{n}}\,d\mu(\xi)d\mu(z)\lesssim{\mathcal{P}}(R)+\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}.$

Regarding $J_{3}$, by standard methods, for all $z,z^{\prime}\in 2Q$,

$\big{|}{\mathcal{R}}(\chi_{3R^{c}}\mu)(z)-{\mathcal{R}}(\chi_{3R^{c}}\mu)(z^{\prime})\big{|}\lesssim{\mathcal{P}}(R).$

Hence averaging for $z\in Q$, $z^{\prime}\in 2R$, we obtain

$J_{3}\lesssim{\mathcal{P}}(R).$

We showed that

$\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)-\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu(y)\big{|}\lesssim\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}+\left(\frac{{\mathcal{E}}(2Q)}{\mu(Q)}\right)^{1/2}+{\mathcal{P}}(Q)+{\mathcal{P}}(R).$

To get rid of the last two terms, recall that the cubes in
${\widetilde{{\mathsf{End}}}}$ are ${\mathcal{P}}$-doubling, and so is $R$.
Hence, by the definition of the energy ${\mathcal{E}}(2Q)$, we have

${\mathcal{P}}(Q)\lesssim\left(\frac{{\mathcal{E}}(2Q)}{\mu(Q)}\right)^{1/2},$

and an analogous estimate for ${\mathcal{P}}(R)$. This finishes the proof. ∎

######  Lemma 6.3.

Let $1<p\leq 2$. For any $Q\in{\widetilde{{\mathsf{End}}}}$ such that
$\ell_{0}\leq\ell(Q)$,

$\int_{Q}\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{p}\,d\mu\lesssim{\mathcal{E}}(2Q)^{\frac{p}{2}}\,\mu(Q)^{1-\frac{p}{2}}.$

###### Proof.

The lemma follows from Hölder’s inequality and the estimate

(6.6)
$\int_{Q}\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{2}\,d\mu\lesssim{\mathcal{E}}(2Q),$

which we prove below.

Notice first that the cube $Q$ as above is covered by the cubes
$P\in{\mathcal{D}}_{\mu}(Q)\cap{\mathsf{Reg}}$. Indeed, if
$Q\in{\mathsf{M}}_{\mathsf{Neg}}$, then $Q\in{\mathcal{T}}_{{\mathsf{Reg}}}$
and this is trivially true. On the other hand, if
$Q\in{\widetilde{{\mathsf{End}}}}\setminus{\mathsf{M}}_{\mathsf{Neg}}\subset{\mathsf{End}}$,
then we have $Q\in{\mathcal{T}}$, and the condition $\ell_{0}\leq\ell(Q)$
implies that $d_{R,\ell_{0}}(x)\leq\ell(Q)$. Hence, $Q$ is covered by the
cubes $P\in{\mathcal{D}}_{\mu}(Q)\cap{\mathsf{Reg}}$.

Observe now that, for $Q\in{\widetilde{{\mathsf{End}}}}$, $P\in{\mathsf{Reg}}$
such that $P\subset Q$, and $x\in P$,

${\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)-{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)={\mathcal{R}}(\chi_{2R\setminus
2P}\mu)(x)-{\mathcal{R}}(\chi_{2R\setminus
2Q}\mu)(x)={\mathcal{R}}(\chi_{2Q\setminus 2P}\mu)(x).$

Thus,

$\big{|}{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)-{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)\big{|}\lesssim\sum_{S\in{\mathcal{D}}_{\mu}:P\subset
S\subset Q}\theta_{\mu}(2B_{S}).$

Notice now that, if ${\widehat{Q}}$ denotes the cube from ${\mathsf{End}}$
that contains $Q$ (which coincides with $Q$ when
$P\not\in{\mathsf{Reg}}_{\mathsf{Neg}}$), we have

$d_{R}(x)\gtrsim{\rm
dist}(P,\operatorname{supp}\mu\setminus{\widehat{Q}})\geq{\rm
dist}(P,\operatorname{supp}\mu\setminus Q)\quad\mbox{ for all $x\in P$.}$

So by Lemma 5.4 we also have

(6.7) $\ell(P)\gtrsim{\rm dist}(P,\operatorname{supp}\mu\setminus Q)$

Thus,

$\sum_{S\in{\mathcal{D}}_{\mu}:P\subset S\subset
Q}\theta_{\mu}(2B_{S})\lesssim\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q):S\supset
P}\theta_{\mu}(2B_{S}),$

with ${\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q)$ defined in (2.9).

So we have

$\displaystyle\int_{Q}\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{2}\,d\mu$
$\displaystyle=\sum_{P\in{\mathsf{Reg}}:P\subset
Q}\int_{P}\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{2}\,d\mu$
$\displaystyle\lesssim\sum_{P\in{\mathsf{Reg}}:P\subset
Q}\bigg{(}\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q):S\supset
P}\theta_{\mu}(2B_{S})\bigg{)}^{2}\,\mu(P).$

By Hölder’s inequality, for any $\alpha>0$,

$\displaystyle\bigg{(}\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q):S\supset
P}$ $\displaystyle\theta_{\mu}(2B_{S})\bigg{)}^{2}$
$\displaystyle\leq\bigg{(}\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q):S\supset
P}\left(\frac{\ell(Q)}{\ell(S)}\right)^{\alpha}\theta_{\mu}(2B_{S})^{2}\bigg{)}\bigg{(}\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q):S\supset
P}\left(\frac{\ell(S)}{\ell(Q)}\right)^{\alpha}\bigg{)}$
$\displaystyle\lesssim_{\alpha}\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q):S\supset
P}\left(\frac{\ell(Q)}{\ell(S)}\right)^{\alpha}\theta_{\mu}(2B_{S})^{2}.$

Therefore,

$\displaystyle\int_{Q}\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{2}\,d\mu$
$\displaystyle\lesssim_{\alpha}\sum_{P\in{\mathsf{Reg}}:P\subset
Q}\,\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q):S\supset
P}\left(\frac{\ell(Q)}{\ell(S)}\right)^{\alpha}\theta_{\mu}(2B_{S})^{2}\mu(P)$
$\displaystyle=\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q)}\left(\frac{\ell(Q)}{\ell(S)}\right)^{\alpha}\theta_{\mu}(2B_{S})^{2}\sum_{P\in{\mathsf{Reg}}:P\subset
S}\mu(P)$
$\displaystyle\leq\sum_{S\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q)}\left(\frac{\ell(Q)}{\ell(S)}\right)^{\alpha}\theta_{\mu}(2B_{S})^{2}\,\mu(S).$

By Lemma 2.9, for $\alpha$ small enough, the right hand side is bounded above
by $C{\mathcal{E}}(2Q)$, so that (6.6) holds. ∎

###  6.2. Estimates for the ${\mathcal{P}}$ and ${\mathcal{Q}}_{\mathsf{Reg}}$
coefficients of some cubes from ${\widetilde{{\mathsf{End}}}}$ and
${\mathsf{Reg}}$

We will transfer the lower estimate obtained for the $L^{p}(\eta)$ norm of
${\mathcal{R}}\eta$ in Lemma 5.15 to ${\mathcal{R}}_{{\mathcal{T}}}\mu$,
${\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu$, and
$\Delta_{{\mathcal{T}}}{\mathcal{R}}\mu$ by means of Lemmas 6.1, 6.2, and 6.3.
To this end, we will need careful estimates for the ${\mathcal{P}}$ and
${\mathcal{Q}}_{\mathsf{Reg}}$ coefficients of cubes from
${\widetilde{{\mathsf{End}}}}$ and ${\mathsf{Reg}}$. This is the task we will
perform in this section.

Given $R\in{\mathsf{MDW}}$, recall that
${\mathsf{HD}}_{1}(e^{\prime}(R))={\mathsf{HD}}(R)\cap{\mathsf{Stop}}(e^{\prime}(R))$.
To shorten notation, we will write
${\mathsf{HD}}_{1}={\mathsf{HD}}_{1}(e^{\prime}(R))$ in this section. Recall
that
${\widetilde{{\mathsf{End}}}}={\mathsf{End}}\setminus{\mathsf{Neg}}\cup{\mathsf{M}}_{{\mathsf{Neg}}}$.
By (4.8), we have

(6.8)
${\widetilde{{\mathsf{End}}}}={\mathsf{LD}}_{1}\cup{\mathsf{OP}}_{1}\cup{\mathsf{LD}}_{2}\cup{\mathsf{OP}}_{2}\cup{\mathsf{HD}}_{2}\cup{\mathsf{M}}_{\mathsf{Neg}},$

where we introduced the following notations:

  * •

${\mathsf{LD}}_{1}$ is the subfamily of ${\widetilde{{\mathsf{End}}}}$ of
those maximal ${\mathcal{P}}$-doubling cubes which are contained both in
$e^{\prime}(R)$ and in some cube from
${\mathsf{LD}}(R)\cap{\mathsf{Stop}}_{1}(e^{\prime}(R))\setminus{\mathsf{Neg}}$.

  * •

${\mathsf{OP}}_{1}$ is the subfamily of ${\widetilde{{\mathsf{End}}}}$ of
those maximal ${\mathcal{P}}$-doubling cubes which are contained both in
$e^{\prime}(R)$ and in some cube from
${\mathsf{OP}}(R)\cap{\mathsf{Stop}}_{1}(e^{\prime}(R))\setminus{\mathsf{Neg}}$.

  * •

${\mathsf{LD}}_{2}$ is the subfamily of ${\widetilde{{\mathsf{End}}}}$ of
those maximal ${\mathcal{P}}$-doubling cubes which are contained in some cube
$Q\in{\mathsf{LD}}(Q^{\prime})\cap{\mathsf{Stop}}(Q^{\prime})\setminus{\mathsf{Neg}}$
for some $Q^{\prime}\in{\mathsf{HD}}_{1}$.

  * •

${\mathsf{OP}}_{2}$ is the subfamily of ${\widetilde{{\mathsf{End}}}}$ of
those maximal ${\mathcal{P}}$-doubling cubes which are contained in some cube
$Q\in{\mathsf{OP}}(Q^{\prime})\cap{\mathsf{Stop}}(Q^{\prime})$ for some
$Q^{\prime}\in{\mathsf{HD}}_{1}$.

  * •

${\mathsf{HD}}_{2}=\bigcup_{Q^{\prime}\in{\mathsf{HD}}_{1}}({\mathsf{HD}}(Q^{\prime})\cap{\mathsf{Stop}}(Q^{\prime}))$.

Remark that the splitting in (6.8) is disjoint. Indeed, notice that, by the
definition of ${\mathsf{M}}_{\mathsf{Neg}}$, the cubes from
${\mathsf{OP}}_{2}\cup{\mathsf{HD}}_{2}$ do not belong to
${\mathsf{M}}_{\mathsf{Neg}}$, since they are strictly contained in some cube
from ${\mathsf{HD}}_{1}$, which is ${\mathcal{P}}$-doubling, in particular.

For $i=1,2$, we also denote by ${\mathsf{Reg}}_{{\mathsf{LD}}_{i}}$ the
subfamily of the cubes from ${\mathsf{Reg}}$ which are contained in some cube
from ${\mathsf{LD}}_{i}$, and we define ${\mathsf{Reg}}_{{\mathsf{OP}}_{i}}$,
${\mathsf{Reg}}_{{\mathsf{HD}}_{2}}$, ${\mathsf{Reg}}_{\mathsf{Neg}}$, and
${\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}}$ analogously.444Notice that
${\mathsf{D}}_{\mathsf{Neg}}={\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}}$. We
let ${\mathsf{Reg}}_{\mathsf{Ot}}$ be the “other” cubes from ${\mathsf{Reg}}$:
the ones which are not contained in any cube from ${\mathsf{End}}$ (which, in
particular, have side length comparable to $\ell_{0}$). Also, we let
${\mathsf{Reg}}_{{\mathsf{DB}}}$ be the subfamily of the cubes from
${\mathsf{Reg}}$ which are contained in some cube from
${\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}}$. Notice that we have the
splitting

(6.9)
${\mathsf{Reg}}={\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{OP}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{LD}}_{2}}\cup{\mathsf{Reg}}_{{\mathsf{OP}}_{2}}\cup{\mathsf{Reg}}_{{\mathsf{HD}}_{2}}\cup{\mathsf{Reg}}_{{\mathsf{Neg}}}\cup{\mathsf{Reg}}_{{\mathsf{Ot}}}.$

The families above may intersect the family ${\mathsf{Reg}}_{{\mathsf{DB}}}$.

Given a family ${\mathcal{I}}\subset{\mathcal{D}}_{\mu}$ and $1<p\leq 2$, we
denote

$\Sigma_{p}^{\mathcal{P}}({\mathcal{I}})=\sum_{Q\in{\mathcal{I}}}{\mathcal{P}}(Q)^{p}\,\mu(Q),\qquad\Sigma_{p}^{\mathcal{Q}}({\mathcal{I}})=\sum_{Q\in{\mathcal{I}}}{\mathcal{Q}}_{\mathsf{Reg}}(Q)^{p}\,\mu(Q).$

We also write
$\Sigma^{\mathcal{P}}({\mathcal{I}})=\Sigma_{2}^{\mathcal{P}}({\mathcal{I}})$,
$\Sigma^{\mathcal{Q}}({\mathcal{I}})=\Sigma_{2}^{\mathcal{Q}}({\mathcal{I}})$.

######  Lemma 6.4.

For any $Q\in{\widetilde{{\mathsf{End}}}}$,

$\Sigma^{\mathcal{P}}({\mathsf{Reg}}\cap{\mathcal{D}}_{\mu}(Q))\lesssim{\mathcal{E}}(2Q).$

###### Proof.

For all $S\in{\mathsf{Reg}}\cap{\mathcal{D}}_{\mu}(Q)$, by Hölder’s
inequality, we have

$\displaystyle{\mathcal{P}}(S)^{2}$
$\displaystyle\lesssim\bigg{(}\sum_{P:S\subset P\subset
Q}\frac{\ell(S)}{\ell(P)}\,\Theta(P)\bigg{)}^{2}+\bigg{(}\frac{\ell(S)}{\ell(Q)}\,{\mathcal{P}}(Q)\bigg{)}^{2}$
$\displaystyle\leq\bigg{(}\sum_{P:S\subset P\subset
Q}\frac{\ell(S)}{\ell(P)}\,\Theta(P)^{2}\bigg{)}\bigg{(}\sum_{P:S\subset
P\subset Q}\frac{\ell(S)}{\ell(P)}\bigg{)}+{\mathcal{P}}(Q)^{2}$
$\displaystyle\lesssim\sum_{P:S\subset P\subset
Q}\frac{\ell(S)}{\ell(P)}\,\Theta(P)^{2}+{\mathcal{P}}(Q)^{2}.$

We deduce that

$\displaystyle\Sigma^{\mathcal{P}}({\mathsf{Reg}}\cap{\mathcal{D}}_{\mu}(Q))$
$\displaystyle=\sum_{S\in{\mathsf{Reg}}\cap{\mathcal{D}}_{\mu}(Q)}{\mathcal{P}}(S)^{2}\,\mu(S)$
$\displaystyle\lesssim\sum_{S\in{\mathsf{Reg}}\cap{\mathcal{D}}_{\mu}(Q)}\sum_{P:S\subset
P\subset
Q}\frac{\ell(S)}{\ell(P)}\,\Theta(P)^{2}\,\mu(S)+\sum_{S\in{\mathsf{Reg}}\cap{\mathcal{D}}_{\mu}(Q)}{\mathcal{P}}(Q)^{2}\,\mu(S).$

Clearly, the last sum is bounded by ${\mathcal{P}}(Q)^{2}\,\mu(Q),$ which in
turn is bounded by ${\mathcal{E}}(2Q)$ since $Q$ is ${\mathcal{P}}$-doubling.
Concerning the first term, arguing as in (6.7), we deduce that the cubes $P$
in the sum belong to ${\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q)$. Thus, by
Fubini,

$\displaystyle\Sigma^{\mathcal{P}}({\mathsf{Reg}}\cap{\mathcal{D}}_{\mu}(Q))$
$\displaystyle\lesssim\sum_{P\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q)}\Theta(P)^{2}\sum_{S\in{\mathsf{Reg}}:S\subset
P}\frac{\ell(S)}{\ell(P)}\,\mu(S)+{\mathcal{E}}(2Q)$
$\displaystyle\leq\sum_{P\in{\widetilde{{\mathcal{D}}}}_{\mu}^{int}(Q)}\\!\Theta(P)^{2}\,\mu(P)+{\mathcal{E}}(2Q)\lesssim{\mathcal{E}}(2Q).$

∎

######  Lemma 6.5.

We have:

  * (i)

If $Q\in{\mathsf{LD}}_{1}$, then
${\mathcal{P}}(Q)\lesssim\delta_{0}\,\Theta(R)$.

  * (ii)

If $Q\in{\mathsf{LD}}_{2}$, then
${\mathcal{P}}(Q)\lesssim\delta_{0}\,\Lambda\,\Theta(R)$.

  * (iii)

If $Q\in{\mathsf{OP}}_{1}$, then ${\mathcal{P}}(Q)\lesssim\Lambda\,\Theta(R)$.

  * (iv)

If $Q\in{\mathsf{OP}}_{2}\cup{\mathsf{HD}}_{2}$, then
${\mathcal{P}}(Q)\lesssim\Lambda^{2}\,\Theta(R)$.

  * (v)

If $Q\in{\mathsf{Neg}}\cup{\mathsf{M}}_{\mathsf{Neg}}$, then
${\mathcal{P}}(Q)\lesssim\left(\frac{\ell(Q)}{\ell(R)}\right)^{1/3}\,\Theta(R)$.

###### Proof.

The statements (i) and (ii) follow directly from the definitions of
${\mathsf{LD}}_{1}$ and ${\mathsf{LD}}_{2}$ and Lemma 3.3. The statement (iii)
is due to the fact that, by the stopping conditions, the cubes
$Q^{\prime}\in{\mathsf{Stop}}_{1}(e^{\prime}(R))$ satisfy
${\mathcal{P}}(Q^{\prime})\lesssim\Lambda\,{\mathcal{P}}(R)\approx\Lambda\,\Theta(R)$.
Then it just remains to notice that if $Q$ is a maximal doubling cube
contained in $Q^{\prime}$, from Lemma 3.3 it follows that
${\mathcal{P}}(Q)\lesssim{\mathcal{P}}(Q^{\prime})$.

The property (iv) for the cubes in ${\mathsf{OP}}_{2}$ follows by arguments
analogous to the ones for (iii), taking into account that such cubes are
maximal doubling cubes contained in cubes from ${\mathsf{Stop}}(R^{\prime})$,
for some $R^{\prime}\in{\mathsf{HD}}_{1}$ with
${\mathcal{P}}(R^{\prime})\approx\Theta(R^{\prime})=\Lambda\,\Theta(R)$.555This
argument shows that, in fact, ${\mathcal{P}}(Q)\lesssim\Lambda^{2}\Theta(R)$
for all $Q\in{\mathcal{T}}(e^{\prime}(R))$. On the other hand the cubes
$Q\in{\mathsf{HD}}_{2}$ satisfy $\Theta(Q)=\Lambda^{2}\,\Theta(R)$ and are
${\mathcal{P}}$-doubling by construction.

Finally, we turn our attention to (v). By the definitions of ${\mathsf{Neg}}$
and ${\mathsf{M}}_{\mathsf{Neg}}$ and Lemma 3.3, for all
$S\in{\mathcal{D}}_{\mu}$ such that $Q\subset S\subset R$, we have

$\Theta(S)\lesssim\left(\frac{\ell(S)}{\ell(R)}\right)^{1/2}\,{\mathcal{P}}(R).$

Thus,

$\displaystyle{\mathcal{P}}(Q)$
$\displaystyle\approx\frac{\ell(Q)}{\ell(R)}\,{\mathcal{P}}(R)+\sum_{S:Q\subset
S\subset R}\frac{\ell(Q)}{\ell(S)}\,\Theta(S)$
$\displaystyle\lesssim\frac{\ell(Q)}{\ell(R)}\,\Theta(R)+\sum_{S:Q\subset
S\subset
R}\frac{\ell(Q)}{\ell(S)}\,\left(\frac{\ell(S)}{\ell(R)}\right)^{1/2}\,\Theta(R)$
$\displaystyle\lesssim\frac{\ell(Q)}{\ell(R)}\,\Theta(R)+\sum_{S:Q\subset
S\subset R}\,\left(\frac{\ell(Q)}{\ell(R)}\right)^{1/2}\,\Theta(R)$
$\displaystyle\lesssim\frac{\ell(Q)}{\ell(R)}\,\Theta(R)+\log\left(\frac{\ell(R)}{\ell(Q)}\right)\,\left(\frac{\ell(Q)}{\ell(R)}\right)^{1/2}\,\Theta(R)\lesssim\left(\frac{\ell(Q)}{\ell(R)}\right)^{1/3}\,\Theta(R).$

∎

######  Remark 6.6.

We will assume that

$\delta_{0}\leq\Lambda^{-2}.$

With this choice, it follows that, by the preceding lemma,

${\mathcal{P}}(Q)\lesssim\delta_{0}^{1/2}\,\Theta(R)\quad\mbox{ for all
$Q\in{\mathsf{LD}}_{1}\cup{\mathsf{LD}}_{2}$.}$

######  Lemma 6.7.

Suppose that $R\in{\mathsf{Trc}}\cap{\mathsf{Ty}}$. Then

(6.10)
$\Sigma^{\mathcal{P}}({\widetilde{{\mathsf{End}}}})\approx\sigma({\widetilde{{\mathsf{End}}}})\lesssim
B\,\sigma({\mathsf{HD}}_{1}),$

(6.11)
$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{DB}})\lesssim\sum_{Q\in{\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q)\leq\sum_{Q\in{\mathsf{DB}}:Q\sim{\mathcal{T}}}{\mathcal{E}}_{\infty}(9Q)\lesssim\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}),$

(6.12)
$\Sigma^{\mathcal{P}}({\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}})\approx\sigma({\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}})\leq
M^{-2}\,\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}),$

(6.13)
$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{1}})+\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})\lesssim\sum_{Q\in{\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2}}{\mathcal{E}}_{\infty}(9Q)\lesssim\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}),$

(6.14)
$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{LD}}_{2}})\lesssim\sum_{Q\in{\mathsf{LD}}_{1}\cup{\mathsf{LD}}_{2}}{\mathcal{E}}_{\infty}(9Q)\lesssim\big{(}B\,M^{2}\,\delta_{0}+\Lambda^{\frac{-1}{3n}}\big{)}\,\sigma({\mathsf{HD}}_{1}).$

Also, for $1<p\leq 2$,

(6.15)
$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{2}})\lesssim\Big{(}B\,\Lambda^{2}\,\delta_{0}^{\frac{p}{2}}+\Lambda^{\frac{-p}{6n}}\Big{)}\,\sigma_{p}({\mathsf{HD}}_{1}),$

(6.16)
$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{HD}}_{2}})\lesssim\Sigma_{p}^{\mathcal{P}}({\mathsf{HD}}_{2})\approx\sigma_{p}({\mathsf{HD}}_{2})\lesssim
B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}),$

(6.17)
$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})\lesssim\Lambda^{\frac{-p}{6n}}\,\sigma_{p}({\mathsf{HD}}_{1}).$

###### Proof.

To prove (6.10), notice first that
$\Sigma^{\mathcal{P}}({\widetilde{{\mathsf{End}}}})\approx\sigma({\widetilde{{\mathsf{End}}}})$
because the cubes from ${\widetilde{{\mathsf{End}}}}$ are
${\mathcal{P}}$-doubling. Also, by (6.8) we have

$\Sigma^{\mathcal{P}}({\widetilde{{\mathsf{End}}}})=\Sigma^{\mathcal{P}}({\mathsf{LD}}_{1})+\Sigma^{\mathcal{P}}({\mathsf{LD}}_{2})+\Sigma^{\mathcal{P}}({\mathsf{OP}}_{1})+\Sigma^{\mathcal{P}}({\mathsf{OP}}_{2})+\Sigma^{\mathcal{P}}({\mathsf{HD}}_{2})+\Sigma^{\mathcal{P}}({\mathsf{M}}_{\mathsf{Neg}}).$

By Lemma 6.5 and the subsequent remark, we have

(6.18)
$\Sigma^{\mathcal{P}}({\mathsf{LD}}_{1})+\Sigma^{\mathcal{P}}({\mathsf{LD}}_{2})\lesssim\delta_{0}\,\Theta(R)^{2}\,\mu(R)=\delta_{0}\,\sigma(R).$

Again by Lemma 6.5, we have ${\mathcal{P}}(Q)\lesssim\Theta(R)$ for all
$Q\in{\mathsf{M}}_{\mathsf{Neg}}$ and thus666We will obtain better estimates
for $\Sigma^{\mathcal{P}}({\mathsf{M}}_{\mathsf{Neg}})$ below.

$\Sigma^{\mathcal{P}}({\mathsf{M}}_{\mathsf{Neg}})\lesssim\sigma(R).$

Also, since ${\mathcal{T}}$ is a tractable tree, we have

$\sigma(R)\leq B\,\sigma({\mathsf{HD}}_{1})\quad\text{ and
}\quad\Sigma^{\mathcal{P}}({\mathsf{HD}}_{2})\approx\sigma({\mathsf{HD}}_{2})\lesssim
B\,\sigma({\mathsf{HD}}_{1}).$

Regarding the family ${\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2}$, denote

${\mathsf{Stop}}_{\mathsf{OP}}={\mathsf{Stop}}_{1}(e^{\prime}(R))\cap{\mathsf{OP}}(R)\cup\bigcup_{S\in{\mathsf{HD}}_{1}}\big{(}{\mathsf{Stop}}(S)\cap{\mathsf{OP}}(S)\big{)},$

so that the cubes from $P\in{\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2}$ are
maximal ${\mathcal{P}}$-doubling cubes contained in some cube from
${\mathsf{Stop}}_{\mathsf{OP}}$. By Lemma 3.3, if
$P\in{\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2}$ is contained in
$Q\in{\mathsf{Stop}}_{\mathsf{OP}}$, then
${\mathcal{P}}(P)\lesssim{\mathcal{P}}(Q)$. In this case,
${\mathsf{OP}}(\,\cdot\,)={\mathsf{NDB}}(\,\cdot\,)$, and so there exists
$Q^{\prime}=Q^{\prime}(Q)\in{\mathsf{DB}}$ such that
$\ell(Q^{\prime})=\ell(Q)$ and $Q^{\prime}\subset 20Q$. In particular, this
implies that $Q^{\prime}\sim{\mathcal{T}}(e^{\prime}(R))$ and that
${\mathcal{P}}(Q)\approx{\mathcal{P}}(Q^{\prime})\approx\Theta(Q^{\prime})$,
and by the definition of ${\mathsf{DB}}$,

(6.19) ${\mathcal{E}}_{\infty}(9Q^{\prime})\gtrsim M^{2}\,\sigma(Q^{\prime}).$

Thus,

(6.20)
$\displaystyle\Sigma^{\mathcal{P}}({\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2})$
$\displaystyle=\sum_{Q\in{\mathsf{Stop}}_{\mathsf{OP}}}\sum_{P\in{\widetilde{{\mathsf{End}}}}\cap{\mathcal{D}}_{\mu}(Q)}{\mathcal{P}}(P)^{2}\,\mu(P)$
$\displaystyle\lesssim\sum_{Q\in{\mathsf{Stop}}_{\mathsf{OP}}}{\mathcal{P}}(Q)^{2}\,\mu(Q)\approx\sum_{Q\in{\mathsf{Stop}}_{\mathsf{OP}}}\sigma(Q^{\prime}(Q))$
$\displaystyle\lesssim\frac{1}{M^{2}}\sum_{Q\in{\mathsf{Stop}}_{\mathsf{OP}}}{\mathcal{E}}_{\infty}(9Q^{\prime}(Q))\lesssim\frac{1}{M^{2}}\sum_{Q^{\prime}\in{\mathsf{DB}}:Q^{\prime}\sim{\mathcal{T}}}{\mathcal{E}}_{\infty}(9Q^{\prime}),$

where in the last estimate we took into account that for each
$Q^{\prime}\in{\mathsf{DB}}$ there is at most a bounded number of cubes
$Q\in{\mathsf{Stop}}_{\mathsf{OP}}$ such that $Q^{\prime}=Q^{\prime}(Q)$
(possibly depending on $n$ and $A_{0}$). Since ${\mathcal{T}}$ is a typical
tree, by the definition in (6.2), we have

(6.21)
$\sum_{Q^{\prime}\in{\mathsf{DB}}:Q^{\prime}\sim{\mathcal{T}}}{\mathcal{E}}_{\infty}(9Q^{\prime})\leq\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}).$

So we have

(6.22) $\Sigma^{\mathcal{P}}({\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2})\lesssim
M^{-2}\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}).$

Gathering the estimates above, and recalling that
$B=\Lambda^{\frac{1}{100n}}$, the estimate (6.10) follows.

To prove (6.11), we apply Lemma 6.4 and the fact that ${\mathcal{T}}$ is a
typical tree again:

$\displaystyle\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{DB}})$
$\displaystyle=\sum_{Q\in{\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}}}\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{DB}}\cap{\mathcal{D}}_{\mu}(Q))$
$\displaystyle\lesssim\sum_{Q\in{\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}}}{\mathcal{E}}(9Q)\lesssim\sum_{Q\in{\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q).$

Recall that
${\widetilde{{\mathsf{End}}}}={\mathsf{End}}\setminus{\mathsf{Neg}}\cup{\mathsf{M}}_{\mathsf{Neg}}$.
For $Q\in{\mathsf{End}}$ we have have $Q\in{\mathcal{T}}$, and so in
particular $Q\sim{\mathcal{T}}$ (see (6.1)). On the other hand, if
$Q\in{\mathsf{M}}_{\mathsf{Neg}}$, then by the construction of the family
${\mathsf{Reg}}$ there exists some cube $Q^{\prime}\in{\mathcal{T}}$ such that

(6.23) $\ell(Q)\leq\ell(Q^{\prime})\leq A_{0}^{2}\,\ell(Q)\quad\text{ and
}\quad 20Q^{\prime}\cap 20Q\neq\varnothing.$

Hence, $Q\sim{\mathcal{T}}$, and so

$\displaystyle\sum_{Q\in{\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q)$
$\displaystyle=\sum_{Q\in{\mathsf{End}}\cap{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q)+\sum_{Q\in{\mathsf{M}}_{\mathsf{Neg}}\cap{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q)$
$\displaystyle\lesssim\sum_{S\in{\mathsf{DB}}:S\sim{\mathcal{T}}}{\mathcal{E}}_{\infty}(9S)\leq\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}),$

taking into account that ${\mathcal{T}}$ is a typical tree. This completes the
proof of (6.11). Notice also that (6.12) follows from (6.11) and the fact that
the cubes from ${\mathsf{DB}}$ satisfy (6.19).

Next we turn our attention to (6.13). By Lemma 6.4 we have

$\displaystyle\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{OP}}_{2}})$
$\displaystyle\lesssim\sum_{Q\in{\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2}}{\mathcal{E}}_{\infty}(9Q)$
$\displaystyle\leq\sum_{Q\in{\widetilde{{\mathsf{End}}}}\cap{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q)+\sum_{Q\in({\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2})\setminus{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q).$

By (6.11), the first term on the right hand side above does not exceed
$C\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})$. Concerning the second
term, we use the fact for the cubes $Q\not\in{\mathsf{DB}}$, we have
${\mathcal{E}}_{\infty}(9Q)\lesssim M^{2}\,\sigma(Q)$ together with (6.22).
Then we get

$\sum_{Q\in({\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2})\setminus{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q)\lesssim
M^{2}\,\sigma({\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2})\lesssim\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}).$

Regarding the estimate (6.14), by (6.18) and Lemma 6.4, we obtain

$\displaystyle\Sigma^{\mathcal{P}}(({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{LD}}_{2}})\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})$
$\displaystyle\lesssim\sum_{Q\in({\mathsf{LD}}_{1}\cup{\mathsf{LD}}_{2})\setminus{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9Q)\lesssim
M^{2}\,\Sigma^{\mathcal{P}}({\mathsf{LD}}_{1}\cup{\mathsf{LD}}_{2})$
$\displaystyle\lesssim M^{2}\,\delta_{0}\,\sigma(R)\leq
B\,M^{2}\,\delta_{0}\,\sigma({\mathsf{HD}}_{1}).$

Together with (6.11), this yields (6.14).

To get (6.15), we apply Hölder’s inequality in the preceding estimate:

$\displaystyle\Sigma_{p}^{\mathcal{P}}(({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{LD}}_{2}})\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})$
$\displaystyle\leq(\Sigma^{\mathcal{P}}(({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{LD}}_{2}})\setminus{\mathsf{Reg}}_{{\mathsf{DB}}}))^{\frac{p}{2}}\,\mu(e^{\prime}(R))^{1-\frac{p}{2}}$
$\displaystyle\lesssim(B\,M^{2}\,\delta_{0}\,\sigma({\mathsf{HD}}_{1}))^{\frac{p}{2}}\,\mu(R)^{1-\frac{p}{2}}.$

Observe now that, writing $HD_{i}=\bigcup_{Q\in{\mathsf{HD}}_{i}}Q$,

(6.24)
$\mu(HD_{1})=\frac{1}{\Lambda^{2}\,\Theta(R)^{2}}\,\sigma({\mathsf{HD}}_{1})\geq\frac{1}{B\,\Lambda^{2}\,\Theta(R)^{2}}\,\sigma(R)=\frac{1}{B\,\Lambda^{2}}\,\mu(R).$

Thus, using also that $M\leq\Lambda$,

(6.25)
$\displaystyle\Sigma_{p}^{\mathcal{P}}(({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{LD}}_{2}})\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})$
$\displaystyle\lesssim(B\,M^{2}\,\delta_{0}^{\frac{2}{n+2}}\sigma({\mathsf{HD}}_{1}))^{\frac{p}{2}}\,(B\,\Lambda^{2}\,\mu(HD_{1}))^{1-\frac{p}{2}}$
$\displaystyle\leq
B\,\Lambda^{2}\,\delta_{0}^{\frac{p}{n+2}}\sigma_{p}({\mathsf{HD}}_{1}).$

On the other hand, if we let ${\mathsf{Reg}}_{{\mathsf{DB}}_{2}}$ be the
subfamily of the cubes from ${\mathsf{Reg}}_{{\mathsf{DB}}}$ which are
contained in some cube from ${\mathsf{HD}}_{1}$, by Hölder’s inequality, Lemma
6.4, and (6.21), we get

$\displaystyle\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{DB}}_{2}})$
$\displaystyle\lesssim\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{DB}}_{2}})^{\frac{p}{2}}\,\mu(HD_{1})^{1-\frac{p}{2}}$
$\displaystyle\lesssim\bigg{(}\sum_{Q^{\prime}\in{\mathsf{DB}}:Q^{\prime}\sim{\mathcal{T}}}{\mathcal{E}}_{\infty}(9Q^{\prime})\bigg{)}^{\frac{p}{2}}\,\mu(HD_{1})^{1-\frac{p}{2}}$
$\displaystyle\lesssim\big{(}\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})\big{)}^{\frac{p}{2}}\,\mu(HD_{1})^{1-\frac{p}{2}}=\Lambda^{\frac{-p}{6n}}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Gathering (6.25) and the last estimate, we get (6.15).

To prove (6.16), recall that $\Theta(Q)\leq\Lambda^{2}\,\Theta(R)$ for all
$Q\in{\mathcal{T}}$. This implies that also
${\mathcal{P}}(Q)\lesssim\Lambda^{2}\,\Theta(R)$ for all $Q\in{\mathsf{Reg}}$,
by Lemma 3.3. Consequently,

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{HD}}_{2}})\lesssim\Lambda^{2p}\,\Theta(R)^{p}\,\mu(HD_{2})=\sigma_{p}({\mathsf{HD}}_{2})\approx\Sigma_{p}^{\mathcal{P}}({\mathsf{HD}}_{2}).$

On the other hand, since $R\in{\mathsf{Trc}}$,

$\displaystyle\sigma_{p}({\mathsf{HD}}_{2})$
$\displaystyle=\Theta({\mathsf{HD}}_{2})^{p-2}\,\sigma({\mathsf{HD}}_{2})\leq
B\,\Theta({\mathsf{HD}}_{2})^{p-2}\,\sigma({\mathsf{HD}}_{1})$
$\displaystyle=B\,\frac{\Theta({\mathsf{HD}}_{2})^{p-2}}{\Theta({\mathsf{HD}}_{1})^{p-2}}\,\sigma_{p}({\mathsf{HD}}_{1})=B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}),$

which completes the proof of (6.16).

Finally, observe that (6.17) follows from (6.13) using Hölder’s inequality and
the fact that the cubes from ${\mathsf{Reg}}_{{\mathsf{OP}}_{2}}$ are
contained in cubes from ${\mathsf{HD}}_{1}$:

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})\leq\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})^{\frac{p}{2}}\mu(HD_{1})^{1-\frac{p}{2}}\lesssim\big{(}\Lambda^{\frac{-1}{3n}}\sigma({\mathsf{HD}}_{1})\big{)}^{\frac{p}{2}}\mu(HD_{1})^{1-\frac{p}{2}}=\Lambda^{\frac{-p}{6n}}\,\sigma_{p}({\mathsf{HD}}_{1}).$

∎

For the record, notice that (6.20) also shows that the family
${\mathsf{Stop}}_{\mathsf{OP}}$ defined just above (6.19) satisfies

$\Sigma^{\mathcal{P}}({\mathsf{Stop}}_{\mathsf{OP}})=\sum_{Q\in{\mathsf{Stop}}_{\mathsf{OP}}}{\mathcal{P}}(Q)^{2}\,\mu(Q)\lesssim\frac{1}{M^{2}}\sum_{Q^{\prime}\in{\mathsf{DB}}:Q^{\prime}\sim{\mathcal{T}}}{\mathcal{E}}_{\infty}(9Q^{\prime}),$

which together with (6.21) yields

(6.26) $\Sigma^{\mathcal{P}}({\mathsf{Stop}}_{\mathsf{OP}})\lesssim
M^{-2}\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}).$

######  Lemma 6.8.

Suppose that $R\in{\mathsf{Trc}}\cap{\mathsf{Ty}}$. Then

(6.27)
$\Sigma^{\mathcal{P}}({\mathsf{M}}_{\mathsf{Neg}})\lesssim\Sigma^{\mathcal{P}}({\mathsf{Neg}})\lesssim\big{(}\delta_{0}^{2}\,B+M^{-2}\Lambda^{\frac{-1}{3n}}\big{)}\,\sigma({\mathsf{HD}}_{1}),$

and

(6.28)
$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Neg}})\lesssim\Sigma^{\mathcal{P}}({\mathsf{Neg}})+\sum_{Q\in{\mathsf{M}}_{\mathsf{Neg}}}{\mathcal{E}}_{\infty}(9Q)\lesssim\big{(}\delta_{0}^{2}\,B\,\Lambda^{2}+\Lambda^{\frac{-1}{3n}}\big{)}\,\sigma({\mathsf{HD}}_{1}).$

###### Proof.

Recall that
${\mathsf{Neg}}={\mathsf{End}}(e^{\prime}(R))\cap{\mathsf{Neg}}(e^{\prime}(R))={\mathsf{Stop}}_{2}(e^{\prime}(R))\cap{\mathsf{Neg}}(e^{\prime}(R)),$
by the definition of ${\mathsf{End}}(e^{\prime}(R))$. By Lemma 4.3, the cubes
from ${\mathsf{Neg}}$ are not contained in any cube from ${\mathsf{HD}}_{1}$.
Consequently, each cube from ${\mathsf{Neg}}$ belongs either to
${\mathsf{LD}}(R)$ or to ${\mathsf{OP}}(R)$, and in the latter case
${\mathsf{OP}}(R)={\mathsf{NDB}}(R)$. Thus, if we denote

${\mathsf{Neg}}_{\mathsf{LD}}={\mathsf{Neg}}\cap{\mathsf{LD}}(R)\cap{\mathsf{Stop}}(e^{\prime}(R)),\quad\;{\mathsf{Neg}}_{\mathsf{OP}}={\mathsf{Neg}}\cap{\mathsf{OP}}(R)\cap{\mathsf{Stop}}(e^{\prime}(R)),$

it is clear that

${\mathsf{Neg}}={\mathsf{Neg}}_{\mathsf{LD}}\cup{\mathsf{Neg}}_{\mathsf{OP}}.$

We can split ${\mathsf{Reg}}_{\mathsf{Neg}}$ in an analogous way:

${\mathsf{Reg}}_{\mathsf{Neg}}={\mathsf{Reg}}_{{\mathsf{Neg}}_{\mathsf{LD}}}\cup{\mathsf{Reg}}_{{\mathsf{Neg}}_{\mathsf{OP}}}.$

Recall that the cubes from ${\mathsf{Neg}}$ are not ${\mathcal{P}}$-doubling
and that
${\mathcal{P}}(Q)\lesssim\left(\frac{\ell(Q)}{\ell(R)}\right)^{1/3}\,\Theta(R)$
for all $Q\in{\mathsf{Neg}}$.

Regarding $\Sigma^{\mathcal{P}}({\mathsf{Neg}}_{\mathsf{LD}})$, from the
definition of ${\mathsf{LD}}(R)$ we deduce that

$\Sigma^{\mathcal{P}}({\mathsf{Neg}}_{\mathsf{LD}})\leq\Sigma^{\mathcal{P}}({\mathsf{LD}}(R)\cap{\mathsf{Stop}}(e^{\prime}(R))\leq\delta_{0}^{2}\,\Theta(R)^{2}\,\mu(e^{\prime}(R))\approx\delta_{0}^{2}\,\sigma(R)\leq\delta_{0}^{2}B\,\sigma({\mathsf{HD}}_{1}),$

where in the last estimate we used the ${\mathsf{MDW}}$ condition.

To estimate $\Sigma^{\mathcal{P}}({\mathsf{Neg}}_{\mathsf{OP}})$ we just take
into account that
${\mathsf{Neg}}_{\mathsf{OP}}\subset{\mathsf{Stop}}_{\mathsf{OP}}$, and then
by (6.26) we have

$\Sigma^{\mathcal{P}}({\mathsf{Neg}}_{\mathsf{OP}})\leq\Sigma^{\mathcal{P}}({\mathsf{Stop}}_{\mathsf{OP}})\lesssim
M^{-2}\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}).$

Gathering the estimates obtained for
$\Sigma^{\mathcal{P}}({\mathsf{Neg}}_{\mathsf{LD}})$ and
$\Sigma^{\mathcal{P}}({\mathsf{Neg}}_{\mathsf{OP}})$ we get

$\Sigma^{\mathcal{P}}({\mathsf{Neg}})\lesssim\big{(}\delta_{0}^{2}\,B+M^{-2}\Lambda^{\frac{-1}{3n}}\big{)}\,\sigma({\mathsf{HD}}_{1}),$

as wished. The fact that
$\Sigma^{\mathcal{P}}({\mathsf{M}}_{\mathsf{Neg}})\lesssim\Sigma^{\mathcal{P}}({\mathsf{Neg}})$
follows from the fact that if $P\in{\mathsf{M}}_{\mathsf{Neg}}$ satisfies
$P\subset Q\in{\mathsf{Neg}}$, then ${\mathcal{P}}(P)\lesssim{\mathcal{P}}(Q)$
because $P$ is a maximal ${\mathcal{P}}$-doubling cube contained in $Q$.

Next we deal with (6.28). We split

$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Neg}})=\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}})+\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Neg}}\setminus{\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}}).$

Notice that, for all
$P\in{\mathsf{Reg}}_{\mathsf{Neg}}\setminus{\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}}$
such that $P\subset Q\in{\mathsf{Neg}}$, by Lemma 3.3,
${\mathcal{P}}(P)\lesssim{\mathcal{P}}(Q)$. Then it follows that

$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Neg}}\setminus{\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}})\lesssim\Sigma^{\mathcal{P}}({\mathsf{Neg}}).$

To estimate
$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}})$, we apply
Lemma 6.4:

(6.29)
$\displaystyle\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}})$
$\displaystyle=\sum_{S\in{\mathsf{M}}_{\mathsf{Neg}}}\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}}\cap{\mathcal{D}}_{\mu}(S))\lesssim\sum_{S\in{\mathsf{M}}_{\mathsf{Neg}}}{\mathcal{E}}_{\infty}(9S)$
$\displaystyle\leq\sum_{S\in{\mathsf{M}}_{\mathsf{Neg}}\setminus{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9S)+\sum_{S\in{\mathsf{DB}}\cap{\widetilde{{\mathsf{End}}}}}{\mathcal{E}}_{\infty}(9S).$

By the definition of ${\mathsf{DB}}$ and the fact that for
$S\in{\mathsf{M}}_{{\mathsf{Neg}}}$ with $S\subset Q\in{\mathsf{Neg}}$ we have
$\Theta(S)\lesssim{\mathcal{P}}(Q)$, we derive

$\sum_{S\in{\mathsf{M}}_{\mathsf{Neg}}\setminus{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9S)\lesssim
M^{2}\sum_{S\in{\mathsf{M}}_{\mathsf{Neg}}\setminus{\mathsf{DB}}}\sigma(S)\lesssim
M^{2}\,\Sigma^{\mathcal{P}}({\mathsf{Neg}}).$

On the other hand, by (6.11),

$\sum_{S\in{\mathsf{DB}}\cap{\widetilde{{\mathsf{End}}}}}{\mathcal{E}}_{\infty}(9S)\leq\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}).$

Therefore,

$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}})\lesssim
M^{2}\,\Sigma^{\mathcal{P}}({\mathsf{Neg}})+\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}).$

Gathering the estimates above, and taking into account that $M<\Lambda$ (by
(4.1)), we deduce

$\displaystyle\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Neg}})$
$\displaystyle\lesssim
M^{2}\,\Sigma^{\mathcal{P}}({\mathsf{Neg}})+\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})$
$\displaystyle\lesssim\big{(}\delta_{0}^{2}\,B\,M^{2}+\Lambda^{\frac{-1}{3n}}\big{)}\,\sigma({\mathsf{HD}}_{1})+\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})\leq\big{(}\delta_{0}^{2}\,B\,\Lambda^{2}+\Lambda^{\frac{-1}{3n}}\big{)}\,\sigma({\mathsf{HD}}_{1}).$

∎

Next lemma deals with
$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}})$. Below we write
${\mathsf{Reg}}_{\mathsf{Ot}}(\ell_{0})$ to recall the dependence of the
family ${\mathsf{Reg}}_{\mathsf{Ot}}$ on the parameter $\ell_{0}$ in (5.8).
Recall that the set $Z$ was defined in (6.3).

######  Lemma 6.9.

For $1\leq p\leq 2$, we have

(6.30) $\limsup_{\ell_{0}\to
0}\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}}(\ell_{0}))\lesssim\Lambda^{2p}\,\Theta(R)^{p}\,\mu(Z).$

Consequently, if $\mu(Z)\leq{\varepsilon}_{Z}\,\mu(R)$, then

(6.31) $\limsup_{\ell_{0}\to
0}\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}}(\ell_{0}))\lesssim
B\,\Lambda^{4}\,{\varepsilon}_{Z}\,\sigma_{p}({\mathsf{HD}}_{1}).$

###### Proof.

If $x\in Q\in{\mathsf{End}}$ with $\ell(Q)\geq\ell_{0}$, then

$d_{R,\ell_{0}}(x)\leq\max(\ell_{0},\ell(Q))=\ell(Q)$

(recall that $d_{R,\ell_{0}}$ is defined in (5.8)), and thus $x$ is contained
in some cube $Q^{\prime}\in{\mathsf{Reg}}$ with $\ell(Q^{\prime})\leq\ell(Q)$,
by the definition of the family ${\mathsf{Reg}}$. So $Q^{\prime}\subset Q$ and
then $Q^{\prime}\in{\mathsf{Reg}}\setminus{\mathsf{Reg}}_{\mathsf{Ot}}$.
Therefore,

(6.32) $\bigcup_{P\in{\mathsf{Reg}}_{\mathsf{Ot}}}P\subset
e^{\prime}(R)\setminus\bigcup_{Q\in{\mathsf{End}}:\ell(Q)>\ell_{0}}Q,$

and so

(6.33) $\limsup_{\ell_{0}\to
0}\mu\bigg{(}\bigcup_{P\in{\mathsf{Reg}}_{\mathsf{Ot}}(\ell_{0})}P\bigg{)}\leq\mu\bigg{(}e^{\prime}(R)\setminus\bigcup_{Q\in{\mathsf{End}}}Q\bigg{)}=\mu(Z).$

To complete the proof of (6.30) it just remains to notice that, using the fact
that ${\mathcal{P}}(P)\lesssim\Lambda^{2}\,\Theta(R)$ for all
$P\in{\mathsf{Reg}}$, we have

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}}(\ell_{0}))\lesssim\Lambda^{2p}\,\Theta(R)^{p}\,\mu\bigg{(}\bigcup_{P\in{\mathsf{Reg}}_{\mathsf{Ot}}(\ell_{0})}P\bigg{)}.$

Regarding the second statement of the lemma recall that, as in (6.24),
$\mu(HD_{1})\geq\frac{1}{B\,\Lambda^{2}}\,\mu(R)$, which implies that

$\mu(Z)\leq B\,\Lambda^{2}\,{\varepsilon}_{Z}\,\mu(HD_{1}).$

Plugging this estimate into (6.30), we get

$\limsup_{\ell_{0}\to
0}\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}}(\ell_{0}))\lesssim
B\,\Lambda^{2+p}\,{\varepsilon}_{Z}\,\Theta({\mathsf{HD}}_{1})^{p}\,\mu(HD_{1})\leq
B\,\Lambda^{4}\,{\varepsilon}_{Z}\,\sigma_{p}({\mathsf{HD}}_{1}).$

∎

######  Remark 6.10.

By Lemma 6.7, for $p\in(1,2)$, if $\Lambda\gg B\gg 1$ and
$\delta_{0}\ll\Lambda^{-1}$, we have

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{HD}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})\ll\sigma_{p}({\mathsf{HD}}_{1}).$

Analogously, by Lemma 6.9, if ${\varepsilon}_{Z}\ll(B\Lambda^{4})^{-1}$ and
$\ell_{0}$ is small enough,

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{Ot}}})\ll\sigma_{p}({\mathsf{HD}}_{1}).$

To transfer the lower estimates for the $L^{p}(\eta)$ norm of
${\mathcal{R}}\eta$ in Lemma 5.15 to ${\mathcal{R}}_{{\mathcal{T}}}\mu$,
${\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu$, and
$\Delta_{{\mathcal{T}}}{\mathcal{R}}\mu$, it would be useful to have estimates
for $\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{1}})$,
$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{DB}}})$, and
$\Sigma_{p}^{\mathcal{P}}({\mathsf{Neg}})$ analogous to the ones above.
However, we have not been able to get them. This fact originates some
technical difficulties, which will be solved in the next lemmas.

######  Remark 6.11.

By Lemma 6.7, for $1<p\leq 2$, and $\ell_{0}$ small enough we have

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})\lesssim\Big{(}B\,\Lambda^{2}\,\delta_{0}^{\frac{1}{2}}+\Lambda^{\frac{-1}{6n}}\Big{)}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Also, by Lemma 6.9, for $\ell_{0}$ small enough, if
$\mu(Z)\leq{\varepsilon}_{Z}\,\mu(R)$,

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}}(\ell_{0}))\lesssim
B\,\Lambda^{4}\,{\varepsilon}_{Z}\,\sigma_{p}({\mathsf{HD}}_{1}).$

From now on we will assume that $\ell_{0}$ small enough so that this holds.
Remember also that we chose

$B=\Lambda^{\frac{1}{100n}}.$

We assume also $\delta_{0}$ and ${\varepsilon}_{Z}$ small enough so that

(6.34)
$B\,\Lambda^{2}\,\delta_{0}^{\frac{1}{2}}+B\,\Lambda^{4}\,{\varepsilon}_{Z}+B\,\Lambda^{2}\,\delta_{0}\leq\Lambda^{\frac{-1}{3n}}.$

In this way, we have

(6.35)
$\displaystyle\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}})\lesssim\Lambda^{\frac{-1}{6n}}\,\sigma_{p}({\mathsf{HD}}_{1}),$

under the assumption that $\mu(Z)\leq{\varepsilon}_{Z}\,\mu(R)$. Also, from
(6.28), it follows that

(6.36)
$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Neg}})\lesssim\big{(}\delta_{0}\,B\,\Lambda^{2}+\Lambda^{\frac{-1}{3n}}\big{)}\,\sigma({\mathsf{HD}}_{1})\lesssim\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}).$

######  Lemma 6.12.

Suppose that $R\in{\mathsf{Trc}}\cap{\mathsf{Ty}}$ and that
$\mu(Z)\leq{\varepsilon}_{Z}\,\mu(R)$. We have

$\Sigma^{\mathcal{P}}({\mathsf{Reg}})\lesssim B\,\sigma({\mathsf{HD}}_{1}).$

For $p_{0}=2-\frac{1}{18n}$, we have

$\Sigma_{p_{0}}^{\mathcal{P}}({\mathsf{Reg}})\lesssim\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

###### Proof.

By (6.35) and (6.16), for $1<p\leq 2$,

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{HD}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}})\\\
\lesssim\Big{(}\Lambda^{\frac{-1}{6n}}+B\,\Lambda^{p-2}\Big{)}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Also, by (6.11), (6.13), and (6.36),

$\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{DB}})+\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{1}})+\Sigma^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{Neg}}})\lesssim\big{(}\delta_{0}\,B\,\Lambda^{2}+\Lambda^{\frac{-1}{3n}}\big{)}\,\sigma({\mathsf{HD}}_{1})\lesssim\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}),$

by the assumptions on $\delta_{0}$, $\Lambda$, and $B$ in Remarks 6.6 and
6.11. Choosing $p=2$ above and adding the preceding estimates, we get the
first statement in the lemma.

To get the second inequality in the lemma, we apply Hölder’s inequality and
(6.24) in the last estimate, and we get

(6.37)
$\displaystyle\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{DB}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{1}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{Neg}}})$
$\displaystyle\lesssim\big{(}\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})\big{)}^{\frac{p}{2}}\,\mu(R)^{1-\frac{p}{2}}$
$\displaystyle\lesssim\big{(}\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})\big{)}^{\frac{p}{2}}\,\big{(}B\,\Lambda^{2}\mu(HD_{1})\big{)}^{1-\frac{p}{2}}$
$\displaystyle\leq\Lambda^{\frac{-1}{6n}}\,\big{(}\Lambda^{3}\big{)}^{1-\frac{p}{2}}\,\sigma_{p}({\mathsf{HD}}_{1}).$

For $p=2-\frac{1}{18n}$, we have

$\big{(}\Lambda^{3}\big{)}^{1-\frac{p}{2}}=\Lambda^{\frac{1}{12n}},$

and thus

(6.38)
$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{DB}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{1}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{Neg}}})\lesssim\Lambda^{\frac{-1}{12n}}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Also,

$B\,\Lambda^{p-2}=\Lambda^{\frac{1}{100n}}\,\Lambda^{\frac{-1}{18n}}=\Lambda^{\frac{-41}{900n}}<\Lambda^{\frac{-1}{25n}},$

and thus

$\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\setminus{\mathsf{Reg}}_{{\mathsf{DB}}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{LD}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{OP}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{HD}}_{2}})+\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{\mathsf{Ot}})\\\
\lesssim\Lambda^{\frac{-1}{25n}}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Gathering the estimates above, the lemma follows. ∎

Remark that in order to transfer the lower estimates for ${\mathcal{R}}\eta$
to ${\mathcal{R}}\mu$ we will need to take $p$ very close to $1$ below, in
particular with $p<p_{0}$.

Next lemma shows how one can estimate the ${\mathcal{Q}}_{\mathsf{Reg}}$
coefficients in terms of the ${\mathcal{P}}$ coefficients.

######  Lemma 6.13.

For all $p\in(1,\infty)$,

$\Sigma_{p}^{\mathcal{Q}}({\mathsf{Reg}})\lesssim\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}).$

###### Proof.

By duality,

(6.39)
$\Sigma_{p}^{\mathcal{Q}}({\mathsf{Reg}})^{1/p}=\bigg{(}\sum_{Q\in{\mathsf{Reg}}}{\mathcal{Q}}_{\mathsf{Reg}}(Q)^{p}\,\mu(Q)\bigg{)}^{1/p}=\sup\sum_{Q\in{\mathsf{Reg}}}{\mathcal{Q}}_{\mathsf{Reg}}(Q)\,g_{Q}\,\mu(Q),$

where the supremum is taken over all sequences
$g=\\{g_{Q}\\}_{Q\in{\mathsf{Reg}}}$ such that

(6.40) $\sum_{Q\in{\mathsf{Reg}}}|g_{Q}|^{p^{\prime}}\,\mu(Q)\leq 1.$

We will identify the sequence $g$ with the function

${\widetilde{g}}=\sum_{Q\in{\mathsf{Reg}}}g_{Q}\,\chi_{Q},$

so that the sum in (6.40) equals
$\|{\widetilde{g}}\|_{L^{p^{\prime}}(\mu)}^{p^{\prime}}$. By the definition of
${\mathcal{Q}}_{\mathsf{Reg}}$ and Fubini we have

(6.41)
$\displaystyle\sum_{Q\in{\mathsf{Reg}}}{\mathcal{Q}}_{\mathsf{Reg}}(Q)\,g_{Q}\,\mu(Q)$
$\displaystyle=\sum_{Q\in{\mathsf{Reg}}}\sum_{P\in{\mathsf{Reg}}}\frac{\ell(P)}{D(P,Q)^{n+1}}\,\mu(P)\,g_{Q}\,\mu(Q)$
$\displaystyle=\sum_{P\in{\mathsf{Reg}}}\bigg{(}\sum_{Q\in{\mathsf{Reg}}}\frac{\ell(P)}{D(P,Q)^{n+1}}\,g_{Q}\,\mu(Q)\bigg{)}\,\mu(P).$

For each $P\in{\mathsf{Reg}}$, we have

(6.42)
$\displaystyle\sum_{Q\in{\mathsf{Reg}}}\frac{\ell(P)}{D(P,Q)^{n+1}}\,|g_{Q}|\,\mu(Q)$
$\displaystyle=\sum_{j\geq 0}\;\sum_{Q\in{\mathsf{Reg}}:2^{j}\ell(P)\leq
D(P,Q)\leq 2^{j+1}\ell(P)}\frac{\ell(P)}{D(P,Q)^{n+1}}\,|g_{Q}|\,\mu(Q)$
$\displaystyle\leq\sum_{j\geq 0}\;\sum_{Q\in{\mathsf{Reg}}:D(P,Q)\leq
2^{j+1}\ell(P)}\frac{2^{-j}}{(2^{j}\ell(P))^{n}}\,|g_{Q}|\,\mu(Q).$

Observe now that the condition $\ell(Q)+{\rm dist}(P,Q)\leq D(P,Q)\leq
2^{j+1}\ell(P),$ implies that $Q\subset B(x_{Q},\ell(Q))\subset
B(x_{P},2^{j+3}\ell(P)).$ From (6.42) and this fact, we infer that

$\displaystyle\sum_{Q\in{\mathsf{Reg}}}\frac{\ell(P)}{D(P,Q)^{n+1}}\,|g_{Q}|\,\mu(Q)$
$\displaystyle\leq\sum_{j\geq 0}\;\sum_{Q\in{\mathsf{Reg}}:Q\subset
B(x_{P},C2^{j}\ell(P))}\frac{2^{-j}}{(2^{j}\ell(P))^{n}}\,|g_{Q}|\,\mu(Q)$
$\displaystyle\leq\sum_{j\geq
0}\frac{2^{-j}}{(2^{j}\ell(P))^{n}}\,\int_{B(x_{P},C2^{j}\ell(P))}|{\widetilde{g}}|\,d\mu$

Notice now that, for all $x\in P$,

$\displaystyle\int_{B(x_{P},C2^{j}\ell(P))}|{\widetilde{g}}|\,d\mu$
$\displaystyle\leq\int_{B(x,C^{\prime}2^{j}\ell(P))}|{\widetilde{g}}|\,d\mu$
$\displaystyle\leq\mu(B(x,C^{\prime}2^{j}\ell(P)))\,{\mathcal{M}}_{\mu}{\widetilde{g}}(x)\leq\mu(B(x_{P},C^{\prime\prime}2^{j}\ell(P)))\,{\mathcal{M}}_{\mu}{\widetilde{g}}(x),$

where ${\mathcal{M}}_{\mu}$ is the centered Hardy-Littlewood maximal operator.
Thus,

$\displaystyle\sum_{Q\in{\mathsf{Reg}}}\frac{\ell(P)}{D(P,Q)^{n+1}}\,|g_{Q}|\,\mu(Q)$
$\displaystyle\leq\sum_{j\geq
0}\frac{2^{-j}\mu(B(x_{P},C^{\prime\prime}2^{j}\ell(P)))}{(2^{j}\ell(P))^{n}}\,{\mathcal{M}}_{\mu}{\widetilde{g}}(x)$
$\displaystyle\approx\sum_{k\geq
0}2^{-k}\theta_{\mu}(2^{k}B_{P})\,{\mathcal{M}}_{\mu}{\widetilde{g}}(x)\approx{\mathcal{P}}(P)\,\,{\mathcal{M}}_{\mu}{\widetilde{g}}(x).$

Plugging this estimate into (6.41), and taking the infimum for $x\in P$, we
get

$\displaystyle\sum_{Q\in{\mathsf{Reg}}}{\mathcal{Q}}_{\mathsf{Reg}}(Q)\,|g_{Q}|\,\mu(Q)$
$\displaystyle\lesssim\sum_{P\in{\mathsf{Reg}}}{\mathcal{P}}(P)\,\inf_{x\in
P}{\mathcal{M}}_{\mu}{\widetilde{g}}(x)\,\mu(P)$
$\displaystyle\leq\bigg{(}\sum_{P\in{\mathsf{Reg}}}{\mathcal{P}}(P)^{p}\,\mu(P)\bigg{)}^{1/p}\bigg{(}\sum_{P\in{\mathsf{Reg}}}\int_{P}|{\mathcal{M}}_{\mu}{\widetilde{g}}|^{p^{\prime}}\,d\mu\bigg{)}^{1/p^{\prime}}$
$\displaystyle=\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}})^{1/p}\,\|{\mathcal{M}}_{\mu}{\widetilde{g}}\|_{L^{p^{\prime}}(\mu{\lfloor}_{e^{\prime}(R)})}$
$\displaystyle\lesssim\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}})^{1/p}\,\|{\widetilde{g}}\|_{L^{p^{\prime}}(\mu)}\leq\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}})^{1/p},$

which concludes the proof of the lemma, by (6.39). ∎

###  6.3. Transference of the lower estimates for ${\mathcal{R}}\eta$ to
$\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu$

Denote

$F=e^{\prime}(R)\setminus\bigcup_{Q\in{\mathsf{Reg}}_{{\mathsf{HD}}_{2}}}Q.$

By the splitting (6.9), it is clear that $F$ coincides with the union of the
cubes in the family

${\mathsf{Reg}}_{F}:={\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{LD}}_{2}}\cup{\mathsf{Reg}}_{{\mathsf{OP}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{OP}}_{2}}\cup{\mathsf{Reg}}_{{\mathsf{Neg}}}\cup{\mathsf{Reg}}_{{\mathsf{Ot}}}.$

We also write

$F_{\eta}=\bigcup_{Q\in{\mathsf{Reg}}_{F}}\frac{1}{2}B(Q),$

so that the measure $\mu|_{F}$ is well approximated by $\eta|_{F_{\eta}}$, in
a sense.

Recall that in Lemma 5.15 we showed that

$\int_{V_{4}}\big{|}(|{\mathcal{R}}\eta(x)|-\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta(x)\gtrsim\Lambda^{-(p^{\prime}+1){\varepsilon}_{n}}\sigma_{p}({\mathsf{HD}}_{1}(e(R))),$

for any $p\in(1,2]$, with $c_{0}$ as in Lemma 5.13. We will show below that a
similar lower estimate holds if we restrict the integral on the left side to
$HD_{2}$. The first step is the next lemma.

We let ${\widetilde{V}}_{4}$ be the union of the balls $\frac{1}{2}B(Q)$, with
$Q\in{\mathsf{Reg}}$, that intersect $V_{4}$.

######  Lemma 6.14.

Suppose that $R\in{\mathsf{Trc}}\cap{\mathsf{Ty}}$ and also that
$\mu(Z)\leq{\varepsilon}_{Z}\,\mu(R)$ and

$\|\Delta_{\widetilde{{\mathcal{T}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{2}\leq\Lambda^{-1}\,\sigma({\mathsf{HD}}_{1}).$

Then, for $p_{0}=2-\frac{1}{18n}$, assuming
${\varepsilon}_{Z}\leq\Lambda^{-300n}$ and $\ell_{0}$ small enough,

$\int_{F_{\eta}\cap{\widetilde{V}}_{4}}\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\eta\lesssim\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

Remark that the proof of this lemma takes advantage of the good estimates we
have obtained for $\Sigma^{\mathcal{P}}_{p_{0}}({\mathsf{Reg}})$ in Lemma
6.12. Later on we will show that an analogous estimate holds for all
$p\in(1,p_{0})$ (see Lemma 6.15 below).

###### Proof.

Denote

$F_{a}=\bigcup_{Q\in{\widetilde{{\mathsf{End}}}}\setminus{\mathsf{HD}}_{2}}Q,\qquad
F_{b}=\bigcup_{Q\in{\mathsf{Reg}}_{\mathsf{Neg}}\setminus{\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}}}Q,\qquad
F_{\mathsf{Ot}}=\bigcup_{Q\in{\mathsf{Reg}}_{\mathsf{Ot}}}Q,$

so that $F=F_{a}\cup F_{b}\cup F_{\mathsf{Ot}}$. Write also

${\mathsf{End}}_{a}={\mathsf{LD}}_{1}\cup{\mathsf{LD}}_{2}\cup{\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2}\cup{\mathsf{M}}_{\mathsf{Neg}}$

and

${\mathsf{Reg}}_{a}={\mathsf{Reg}}_{{\mathsf{LD}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{LD}}_{2}}\cup{\mathsf{Reg}}_{{\mathsf{OP}}_{1}}\cup{\mathsf{Reg}}_{{\mathsf{OP}}_{2}}\cup{\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}},$

so that

$F_{a}=\bigcup_{Q\in{\mathsf{End}}_{a}}Q=\bigcup_{Q\in{\mathsf{Reg}}_{a}}Q.$

We also consider

$F_{a,\eta}=\bigcup_{Q\in{\mathsf{Reg}}_{a}}\frac{1}{2}B(Q),\qquad
F_{b,\eta}=\bigcup_{Q\in{\mathsf{Reg}}_{\mathsf{Neg}}\setminus{\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}}}\frac{1}{2}B(Q),\qquad
F_{{\mathsf{Ot}},\eta}=\bigcup_{Q\in{\mathsf{Reg}}_{\mathsf{Ot}}}\frac{1}{2}B(Q),$

so that these sets approximate $F_{a},F_{b},F_{\mathsf{Ot}}$ at the level of
the family ${\mathsf{Reg}}$, in a sense. Moreover, we have
$F_{\eta}=F_{a,\eta}\cup F_{b,\eta}\cup F_{{\mathsf{Ot}},\eta}$.

We split

$\int_{F_{\eta}\cap{\widetilde{V}}_{4}}\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\eta=\int_{F_{a,\eta}\cap{\widetilde{V}}_{4}}\ldots+\int_{F_{b,\eta}\cap{\widetilde{V}}_{4}}\ldots+\int_{F_{{\mathsf{Ot}},\eta}\cap{\widetilde{V}}_{4}}\ldots=:I_{a}+I_{b}+I_{\mathsf{Ot}},$

where “$\ldots$” stands for
$\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\eta$.

Estimates for $I_{b}$. We claim that $I_{b}=0$. Indeed, given
$Q\in{\mathsf{Reg}}_{\mathsf{Neg}}\setminus{\mathsf{Reg}}_{{\mathsf{M}}_{\mathsf{Neg}}}$,
notice that all the cubes $S$ such that $Q\subset S\subset R$ satisfy

$\Theta(S)\lesssim\left(\frac{\ell(S)}{\ell(R)}\right)^{1/2}\,\Theta(R),$

and so, for any $x\in Q$,

$\displaystyle|{\mathcal{R}}\eta(x)|$ $\displaystyle\lesssim\sum_{S:Q\subset
S\subset R}\theta_{\eta}(4B_{S})\lesssim\sum_{S:Q\subset S\subset
R}\theta_{\mu}(CB_{S})$ $\displaystyle\lesssim\sum_{S:Q\subset S\subset
R}\left(\frac{\ell(S)}{\ell(R)}\right)^{1/2}\,\Theta(R)+\Theta(R)\lesssim\Theta(R).$

Hence, for $\Lambda$ big enough,
$(|{\mathcal{R}}\eta(x)|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}=0$,
which proves our claim.

Estimates for $I_{a}$. By Lemma 6.1, for all $Q\in{\mathsf{Reg}}_{a}$ such
that $\frac{1}{2}B(Q)\subset{\widetilde{V}}_{4}$, all $x\in\frac{1}{2}B(Q)$,
and all $y\in Q$,

$|{\mathcal{R}}\eta(x)|\leq|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|+C\Theta(R)+C{\mathcal{P}}(Q)+C{\mathcal{Q}}_{\mathsf{Reg}}(Q).$

Thus, for $\Lambda$ big enough, since
$C\Theta(R)=C\Lambda^{-1}\Theta({\mathsf{HD}}_{1})<\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1})$,

$(|{\mathcal{R}}\eta(x)|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\leq(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}+C{\mathcal{P}}(Q)+C{\mathcal{Q}}_{\mathsf{Reg}}(Q).$

Consequently,

(6.43) $\displaystyle I_{a}$
$\displaystyle\lesssim\sum_{Q\in{\mathsf{Reg}}_{a}}\int_{Q}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu(y)+\\!\sum_{Q\in{\mathsf{Reg}}_{a}}({\mathcal{P}}(Q)^{p_{0}}+{\mathcal{Q}}_{\mathsf{Reg}}(Q)^{p_{0}})\,\mu(Q)$
$\displaystyle\lesssim\sum_{S\in{\mathsf{End}}_{a}}\int_{S}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu+\Sigma_{p_{0}}^{\mathcal{P}}({\mathsf{Reg}}),$

where, in the last inequality, we used the fact that all the cubes from
${\mathsf{Reg}}_{a}$ are contained in some cube $S\in{\mathsf{End}}_{a}$ and
we applied Lemma 6.13.

For each $S\in{\mathsf{End}}_{a}$, by the triangle inequality and the fact
that $(\;\cdot\;)_{+}$ is a $1$-Lipschitz function, we get

$\displaystyle\int_{S}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu$
$\displaystyle\lesssim\int_{S}\big{|}(|{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu$
$\displaystyle\quad+\int_{S}\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{p_{0}}\,d\mu$

By Lemma 6.3, the last integral does not exceed
$C{\mathcal{E}}(2S)^{\frac{p_{0}}{2}}\,\mu(S)^{1-\frac{p_{0}}{2}}$, and thus
we deduce that

(6.44)
$I_{a}\lesssim\sum_{S\in{\mathsf{End}}_{a}}\int_{S}\big{|}(|{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu+\sum_{S\in{\mathsf{End}}_{a}}{\mathcal{E}}(2S)^{\frac{p_{0}}{2}}\,\mu(S)^{1-\frac{p_{0}}{2}}+\Sigma_{p_{0}}^{\mathcal{P}}({\mathsf{Reg}}).$

Next we apply Lemma 6.2, which ensures that for any
$S\in{\mathsf{End}}_{a}\subset{\widetilde{{\mathsf{End}}}}$ and all $x\in S$,

(6.45)
$|{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)|\leq|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu(x)|+C\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}+C\left(\frac{{\mathcal{E}}(2S)}{\mu(S)}\right)^{1/2}.$

In case that $R\not\in{\mathsf{DB}}$, recalling that
$\Lambda\geq\max(M^{\frac{8n-1}{8n-2}},CM)\gg M$ by (4.1), we obtain

(6.46) $C\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}\leq
C\,M\,\Theta(R)\leq\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}).$

In case that $R\in{\mathsf{DB}}$, since $R\in{\mathsf{Ty}}$, we have

${\mathcal{E}}(4R)\lesssim\sum_{Q\in{\mathsf{DB}}:Q\sim{\mathcal{T}}}{\mathcal{E}}_{\infty}(9Q)\leq\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}),$

and so we also get

(6.47) $C\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}\leq
C\left(\frac{\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})}{\mu(R)}\right)^{1/2}\leq
C\,\Lambda^{\frac{-1}{6n}}\,\Theta({\mathsf{HD}}_{1})\leq\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}).$

Thus, in any case,

$(|{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\leq|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu(x)|+C\left(\frac{{\mathcal{E}}(2S)}{\mu(S)}\right)^{1/2}.$

Plugging this estimate into (6.44), we get

(6.48)
$I_{a}\lesssim\sum_{S\in{\mathsf{End}}_{a}}\int_{S}|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu|^{p_{0}}\,d\mu+\sum_{S\in{\mathsf{End}}_{a}}{\mathcal{E}}(2S)^{\frac{p_{0}}{2}}\,\mu(S)^{1-\frac{p_{0}}{2}}+\Sigma_{p_{0}}^{\mathcal{P}}({\mathsf{Reg}}).$

We deal with each term on the right hand side of the preceding inequality
separately. First, by Hölder’s inequality and the assumptions in the lemma, we
have

$\int|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu|^{p_{0}}\,d\mu\lesssim\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{p_{0}}\,\mu(R)^{1-\frac{p_{0}}{2}}\leq\Lambda^{-\frac{p_{0}}{2}}\,\sigma({\mathsf{HD}}_{1})^{\frac{p_{0}}{2}}\,\mu(R)^{1-\frac{p_{0}}{2}}.$

Regarding the second term in (6.48), by Hölder’s inequality again,

$\displaystyle\sum_{S\in{\mathsf{End}}_{a}}{\mathcal{E}}(2S)^{\frac{p_{0}}{2}}\,\mu(S)^{1-\frac{p_{0}}{2}}$
$\displaystyle\leq\bigg{(}\sum_{S\in{\mathsf{End}}_{a}}{\mathcal{E}}(2S)\bigg{)}^{\frac{p_{0}}{2}}\,\bigg{(}\sum_{S\in{\mathsf{End}}_{a}}\mu(S)\bigg{)}^{1-\frac{p_{0}}{2}}$
$\displaystyle\lesssim\bigg{(}\sum_{S\in{\mathsf{End}}_{a}}{\mathcal{E}}(2S)\bigg{)}^{\frac{p_{0}}{2}}\,\mu(R)^{1-\frac{p_{0}}{2}}.$

We estimate the first factor on the right hand side using (6.13), (6.14), and
(6.28):

$\displaystyle\bigg{(}\sum_{S\in{\mathsf{End}}_{a}}{\mathcal{E}}(2S)\bigg{)}^{\frac{p_{0}}{2}}$
$\displaystyle\leq\bigg{(}\sum_{S\in{\mathsf{OP}}_{1}\cup{\mathsf{OP}}_{2}}{\mathcal{E}}_{\infty}(9S)+\sum_{S\in{\mathsf{LD}}_{1}\cup{\mathsf{LD}}_{2}}{\mathcal{E}}_{\infty}(9S)+\sum_{S\in{\mathsf{M}}_{\mathsf{Neg}}}{\mathcal{E}}_{\infty}(9S)\bigg{)}^{\frac{p_{0}}{2}}$
$\displaystyle\lesssim\big{(}B\Lambda^{-1}\\!+\Lambda^{\frac{-1}{3n}}+B\,M^{2}\,\delta_{0}^{\frac{2}{n+2}}+B\,\Lambda^{7}\delta_{0}\big{)}^{\frac{p_{0}}{2}}\,\sigma({\mathsf{HD}}_{1})^{\frac{p_{0}}{2}}\leq\Lambda^{\frac{-1}{6n}}\,\sigma({\mathsf{HD}}_{1})^{\frac{p_{0}}{2}},$

by the assumption (6.34) on $\delta_{0}$. In connection with the last summand
on the right hand side of (6.48), by Lemma 6.12 we have

$\Sigma_{p_{0}}^{\mathcal{P}}({\mathsf{Reg}})\lesssim\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

Therefore,

$I_{a}\lesssim\Lambda^{\frac{-1}{6n}}\,\sigma({\mathsf{HD}}_{1})^{\frac{p_{0}}{2}}\,\mu(R)^{1-\frac{p_{0}}{2}}+\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

From (6.24) we derive that $\mu(HD_{1})\geq\Lambda^{-3}\,\mu(R)$, and then

(6.49)
$\Lambda^{\frac{-1}{6n}}\,\sigma({\mathsf{HD}}_{1})^{\frac{p_{0}}{2}}\,\mu(R)^{1-\frac{p_{0}}{2}}\leq\Lambda^{\frac{-1}{6n}}\,\big{(}\Lambda^{3}\big{)}^{1-\frac{p_{0}}{2}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1})=\Lambda^{\frac{-1}{12n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1})\leq\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

So we get

(6.50)
$I_{a}\lesssim\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

Estimate of $I_{\mathsf{Ot}}$. By the same arguments as in (6.43), just
replacing ${\mathsf{Reg}}_{a}$ by ${\mathsf{Reg}}_{\mathsf{Ot}}$, we obtain

$I_{\mathsf{Ot}}\lesssim\sum_{Q\in{\mathsf{Reg}}_{\mathsf{Ot}}}\int_{Q}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu(y)+\\!\sum_{Q\in{\mathsf{Reg}}_{\mathsf{Ot}}}({\mathcal{P}}(Q)^{p_{0}}+{\mathcal{Q}}_{\mathsf{Reg}}(Q)^{p_{0}})\,\mu(Q).$

Thus, using again Lemmas 6.13 and 6.12, we get

(6.51) $\displaystyle I_{\mathsf{Ot}}$
$\displaystyle\lesssim\sum_{Q\in{\mathsf{Reg}}_{\mathsf{Ot}}}\int_{Q}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu(y)+\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1})$
$\displaystyle=:{\widetilde{I}}_{\mathsf{Ot}}+\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

To estimate the integral ${\widetilde{I}}_{\mathsf{Ot}}$ on the right hand
side, we split

$\displaystyle{\widetilde{I}}_{\mathsf{Ot}}$
$\displaystyle=\sum_{Q\in{\mathsf{Reg}}_{{\mathsf{Ot}}}\setminus{\mathsf{Neg}}(e^{\prime}(R))}\int_{Q}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu(x)$
$\displaystyle\quad+\sum_{Q\in{\mathsf{Reg}}_{{\mathsf{Ot}}}\cap{\mathsf{Neg}}(e^{\prime}(R))}\int_{Q}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\mu(x)$
$\displaystyle={\widetilde{I}}_{{\mathsf{Ot}},1}+{\widetilde{I}}_{{\mathsf{Ot}},2}.$

Notice that, by definition we have
${\mathsf{Reg}}_{\mathsf{Ot}}\subset{\mathcal{T}}$. Recall that
${\mathsf{Neg}}={\mathsf{Neg}}(e^{\prime}(R))\cap{\mathsf{End}}$ and thus we
may have
${\mathsf{Reg}}_{{\mathsf{Ot}}}\cap{\mathsf{Neg}}(e^{\prime}(R))\neq\varnothing$.
In this case, we have
${\mathsf{Reg}}_{{\mathsf{Ot}}}\cap{\mathsf{Neg}}(e^{\prime}(R))\subset{\mathcal{T}}_{\mathsf{Stop}}(e^{\prime}(R))$
(because
${\mathsf{Neg}}(e^{\prime}(R))\subset{\mathcal{T}}_{\mathsf{Stop}}(e^{\prime}(R))$
by construction).

The same argument used to show that $I_{b}=0$ shows that

${\widetilde{I}}_{{\mathsf{Ot}},2}=0.$

To estimate ${\widetilde{I}}_{{\mathsf{Ot}},1}$, denote by
${\mathsf{M}}_{\mathsf{Ot}}$ the family of maximal ${\mathcal{P}}$-doubling
cubes which are contained in some cube from
${\mathsf{Reg}}_{{\mathsf{Ot}}}\setminus{\mathsf{Neg}}(e^{\prime}(R))$ and let

$N_{\mathsf{Ot}}=\bigcup_{Q\in{\mathsf{Reg}}_{{\mathsf{Ot}}}\setminus{\mathsf{Neg}}(e^{\prime}(R))}Q\setminus\bigcup_{P\in{\mathsf{M}}_{{\mathsf{Ot}}}}P.$

We claim that

(6.52) ${\mathsf{M}}_{\mathsf{Ot}}\subset{\mathcal{T}}\quad\text{ and }\quad
N_{\mathsf{Ot}}\subset Z.$

To check this, for a given $P\in{\mathsf{M}}_{\mathsf{Ot}}$ with $P\subset
Q\in{\mathsf{Reg}}_{{\mathsf{Ot}}}\setminus{\mathsf{Neg}}(e^{\prime}(R))$,
suppose there exists $S\in{\mathsf{End}}$ such that $S\supset P$. As $P$ is a
contained in some
$Q\in{\mathsf{Reg}}_{{\mathsf{Ot}}}\setminus{\mathsf{Neg}}(e^{\prime}(R))$, we
have $S\not\in{\mathsf{Neg}}$. Further, $S\subsetneq Q$ because
$Q\in{\mathsf{Reg}}_{{\mathsf{Ot}}}$ implies that $Q\not\subset S$. Since $S$
is ${\mathcal{P}}$-doubling, we deduce that $P=S$, by the maximality of $P$ as
${\mathcal{P}}$-doubling cube contained in $Q$. An analogous argument shows
that $N_{\mathsf{Ot}}\subset Z$.

By Hölder’s inequality and (6.33), for $\ell_{0}$ small enough we have

$\displaystyle{\widetilde{I}}_{\mathsf{Ot}}$
$\displaystyle\leq\bigg{(}\sum_{Q\in{\mathsf{Reg}}_{{\mathsf{Ot}}}}\int_{Q}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)|-\frac{c_{0}}{8}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{2}\,d\mu(x)\big{)}^{\frac{p_{0}}{2}}\bigg{(}\sum_{Q\in{\mathsf{Reg}}_{{\mathsf{Ot}}}}\mu(Q)\bigg{)}^{1-\frac{p_{0}}{2}}$
$\displaystyle\leq\bigg{(}\sum_{P\in{\mathsf{M}}_{{\mathsf{Ot}}}}\int_{P}\big{|}{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{2}\,d\mu+\int_{N_{\mathsf{Ot}}}\big{|}{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{2}\,d\mu\bigg{)}^{\frac{p_{0}}{2}}\bigg{(}\mu(Z)+o(\ell_{0})\bigg{)}^{1-\frac{p_{0}}{2}},$

with $o(\ell_{0})\to 0$ as $\ell_{0}\to 0$.

Denote

${\mathcal{R}}_{{\mathsf{M}}_{\mathsf{Ot}}}\mu(x)=\sum_{P\in{\mathsf{M}}_{\mathsf{Ot}}}\chi_{P}(x)\,{\mathcal{R}}(\chi_{2R\setminus
2P}\mu)(x)$

and

$\Delta_{{\mathsf{M}}_{\mathsf{Ot}}}{\mathcal{R}}\mu(x)=\sum_{P\in{\mathsf{M}}_{\mathsf{Ot}}}\chi_{P}(x)\,\big{(}m_{\mu,P}({\mathcal{R}}\mu)-m_{\mu,2R}({\mathcal{R}}\mu)\big{)}+\chi_{Z}(x)\big{(}{\mathcal{R}}\mu(x)-m_{\mu,2R}({\mathcal{R}}\mu)\big{)}.$

Notice that, for $x\in P\in{\mathsf{M}}_{\mathsf{Ot}}$ and
$Q\in{\mathsf{Reg}}_{\mathsf{Ot}}\setminus{\mathsf{Neg}}(e^{\prime}(R))$ such
that $Q\supset P$, since there are no ${\mathcal{P}}$-doubling cubes
$P^{\prime}$ such that $P\subsetneq P^{\prime}\subset Q$,

$\big{|}{\mathcal{R}}_{{\mathsf{M}}_{\mathsf{Ot}}}\mu(x)-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)\big{|}=|{\mathcal{R}}(\chi_{2Q\setminus
2P}\mu)(x)|\lesssim\sum_{P:P\subset P^{\prime}\subset
Q}\Theta(P^{\prime})\lesssim{\mathcal{P}}(Q)\lesssim\Lambda\,\Theta({\mathsf{HD}}_{1}).$

Almost the same argument shows also that, for $x\in N_{\mathsf{Ot}}$,

$\big{|}{\mathcal{R}}(\chi_{2R}\mu)(x)-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(x)\big{|}\lesssim\Lambda\,\Theta({\mathsf{HD}}_{1}).$

Remark also that

$\int_{N_{\mathsf{Ot}}}\big{|}{\mathcal{R}}(\chi_{2R}\mu)\big{|}^{2}\,d\mu\leq\int_{Z}\big{|}{\mathcal{R}}(\chi_{2R}\mu)\big{|}^{2}\,d\mu.$

So we deduce that, for $\ell_{0}$ small enough,

${\widetilde{I}}_{\mathsf{Ot}}\lesssim\bigg{(}\sum_{P\in{\mathsf{M}}_{{\mathsf{Ot}}}}\int_{P}\big{|}{\mathcal{R}}_{{\mathsf{M}}_{\mathsf{Ot}}}\mu\big{|}^{2}\,d\mu+\int_{Z}\big{|}{\mathcal{R}}(\chi_{2R}\mu)\big{|}^{2}\,d\mu+\Lambda^{2}\,\Theta({\mathsf{HD}}_{1})^{2}\,\mu(R)\bigg{)}^{\frac{p_{0}}{2}}\big{(}{\varepsilon}_{Z}\,\mu(R)\big{)}^{1-\frac{p_{0}}{2}}.$

Almost the same arguments as in Lemma 6.2 show that for $x\in
P\in{\mathsf{M}}_{\mathsf{Ot}}$,

$\big{|}{\mathcal{R}}_{{\mathsf{M}}_{\mathsf{Ot}}}\mu(x)-\Delta_{{\mathsf{M}}_{\mathsf{Ot}}}{\mathcal{R}}\mu(x)\big{|}\lesssim{\mathcal{P}}(R)+\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}+{\mathcal{P}}(P)+\left(\frac{{\mathcal{E}}(2P)}{\mu(P)}\right)^{1/2}$

and that, for $x\in Z$,

$\big{|}{\mathcal{R}}(\chi_{2R}\mu)(x)-\Delta_{{\mathsf{M}}_{\mathsf{Ot}}}{\mathcal{R}}\mu(x)\big{|}\lesssim{\mathcal{P}}(R)+\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}.$

Therefore, by (6.46) and (6.47) and taking into account that
${\mathcal{P}}(P)\lesssim\Lambda\,\Theta({\mathsf{HD}}_{1})$ for
$P\in{\mathsf{M}}_{\mathsf{Ot}}$, we deduce

${\widetilde{I}}_{\mathsf{Ot}}\lesssim\bigg{(}\int|\Delta_{{\mathsf{M}}_{\mathsf{Ot}}}{\mathcal{R}}\mu|^{2}\,d\mu+\sum_{P\in{\mathsf{M}}_{{\mathsf{Ot}}}}{\mathcal{E}}(2P)+\Lambda^{2}\,\Theta({\mathsf{HD}}_{1})^{2}\,\mu(R)\bigg{)}^{\frac{p_{0}}{2}}\big{(}{\varepsilon}_{Z}\,\mu(R)\big{)}^{1-\frac{p_{0}}{2}}.$

By the orthogonality of the functions $\Delta_{Q}{\mathcal{R}}\mu$,
$Q\in{\mathcal{D}}_{\mu}$, the assumptions in the lemma, and (6.52), it is
clear that

$\int|\Delta_{{\mathsf{M}}_{\mathsf{Ot}}}{\mathcal{R}}\mu|^{2}\,d\mu\leq\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{2}\leq\Lambda^{-1}\,\sigma({\mathsf{HD}}_{1}).$

On the other hand, since the tree ${\mathcal{T}}$ is typical,

$\displaystyle\sum_{P\in{\mathsf{M}}_{{\mathsf{Ot}}}}{\mathcal{E}}(2P)$
$\displaystyle\leq\sum_{P\in{\mathsf{M}}_{{\mathsf{Ot}}}\setminus{\mathsf{DB}}}{\mathcal{E}}(2P)+\sum_{P\in{\mathcal{T}}\cap{\mathsf{DB}}}{\mathcal{E}}(2P)\leq
M^{2}\,\sigma({\mathsf{M}}_{{\mathsf{Ot}}})+\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})$
$\displaystyle\leq
M^{2}\Lambda^{2}\,\Theta({\mathsf{HD}}_{1})^{2}\,\mu(R)+\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})\lesssim\Lambda^{4}\,\Theta({\mathsf{HD}}_{1})^{2}\,\mu(R).$

Thus, using also (6.24),

$\displaystyle{\widetilde{I}}_{\mathsf{Ot}}$
$\displaystyle\lesssim\Lambda^{\frac{-p_{0}}{2}}\,\sigma({\mathsf{HD}}_{1})^{\frac{p_{0}}{2}}\,\mu(R)^{1-\frac{p_{0}}{2}}+{\varepsilon}_{Z}^{1-\frac{p_{0}}{2}}\,\Lambda^{2p_{0}}\,\Theta({\mathsf{HD}}_{1})^{p_{0}}\,\mu(R)$
$\displaystyle\lesssim(B\Lambda^{2})^{{}^{1-\frac{p_{0}}{2}}}\Lambda^{\frac{-p_{0}}{2}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1})+{\varepsilon}_{Z}^{1-\frac{p_{0}}{2}}\,B\,\Lambda^{2+{2p_{0}}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

By the choice of $p_{0}$, $\Lambda$, and $B$ and the assumption
${\varepsilon}_{Z}\leq\Lambda^{-300n},$ we have

${\widetilde{I}}_{\mathsf{Ot}}\lesssim\Lambda^{\frac{-1}{2}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1})+\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1})\lesssim\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

Together with (6.51), this yields

$I_{\mathsf{Ot}}\lesssim\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1}).$

Gathering the estimates obtained for $I_{a}$ and $I_{\mathsf{Ot}}$, the lemma
follows. ∎

######  Lemma 6.15.

Suppose that $R\in{\mathsf{Trc}}\cap{\mathsf{Ty}}$ and also that
$\mu(Z)\leq{\varepsilon}_{Z}\,\mu(R)$ and

$\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{2}\leq\Lambda^{-1}\,\sigma({\mathsf{HD}}_{1}).$

Assume also that ${\varepsilon}_{Z}\leq\Lambda^{-300n}$ and $\ell_{0}$ is
small enough. Then

$\eta\big{(}\big{\\{}x\in{\widetilde{V}}_{4}:|{\mathcal{R}}\eta(x)|>\frac{c_{0}}{2}\Theta({\mathsf{HD}}_{1})\big{\\}}\big{)}\lesssim\Lambda^{\frac{-1}{25n}}\,\mu(HD_{1}).$

Also, for all $p\in(1,p_{0})$,

$\int_{F_{\eta}\cap{\widetilde{V}}_{4}}\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{2}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta\lesssim\Lambda^{\frac{-1}{25n}}\,\sigma_{p}({\mathsf{HD}}_{1}).$

###### Proof.

Denote

$A=\big{\\{}x\in{\widetilde{V}}_{4}:|{\mathcal{R}}\eta(x)|>\frac{c_{0}}{2}\Theta({\mathsf{HD}}_{1})\big{\\}}.$

By the definition of $F_{\eta}$ we can split

$\eta(A)\leq\eta(A\cap
F_{\eta})+\eta\bigg{(}\bigcup_{Q\in{\mathsf{HD}}_{2}}\frac{1}{2}B(Q)\bigg{)}.$

Notice first that, since $R\in{\mathsf{Trc}}$,

(6.53)
$\eta\bigg{(}\bigcup_{Q\in{\mathsf{HD}}_{2}}\frac{1}{2}B(Q)\bigg{)}=\frac{1}{\Theta({\mathsf{HD}}_{2})^{2}}\,\sigma({\mathsf{HD}}_{2})\leq\frac{B}{\Lambda^{2}\,\Theta({\mathsf{HD}}_{1})^{2}}\,\sigma({\mathsf{HD}}_{1})\leq\Lambda^{-1}\,\mu(HD_{1}).$

On the other hand, for $x\in A$ we have

$(|{\mathcal{R}}\eta(x)|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\geq\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}).$

So, by Chebyshev and Lemma 6.14,

$\displaystyle\eta(A\cap F_{\eta})$
$\displaystyle\lesssim\frac{1}{\Theta({\mathsf{HD}}_{1})^{p_{0}}}\int_{F_{\eta}\cap{\widetilde{V}}_{4}}\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}\,d\eta$
$\displaystyle\lesssim\frac{\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1})}{\Theta({\mathsf{HD}}_{1})^{p_{0}}}=\Lambda^{\frac{-1}{25n}}\,\mu(HD_{1}),$

which, together with (6.53), proves the first assertion of the lemma.

For the second statement in the lemma we use Hölder and Lemma 6.14 again:

$\displaystyle\int_{F_{\eta}\cap{\widetilde{V}}_{4}}\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{2}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta$
$\displaystyle=\int_{A\cap
F_{\eta}}\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{2}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta$
$\displaystyle\leq\left(\int_{A\cap
F_{\eta}}\\!\big{|}(|{\mathcal{R}}\eta|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p_{0}}d\eta\right)^{\frac{p}{p_{0}}}\\!\eta(A\cap
F_{\eta})^{1-\frac{p}{p_{0}}}$
$\displaystyle\lesssim\big{(}\Lambda^{\frac{-1}{25n}}\,\sigma_{p_{0}}({\mathsf{HD}}_{1})\big{)}^{\frac{p}{p_{0}}}\,\big{(}\Lambda^{\frac{-1}{25n}}\,\mu(HD_{1})\big{)}^{1-\frac{p}{p_{0}}}$
$\displaystyle=\Lambda^{\frac{-1}{25n}}\,\sigma_{p}({\mathsf{HD}}_{1}).$

∎

Observe that, given $R\in{\mathsf{MDW}}\cap{\mathsf{Trc}}\cap{\mathsf{Ty}}$,
from Lemmas 5.15 and 6.15, under the assumptions in those lemmas, we derive
that

(6.54) $\displaystyle\int_{V_{4}\cap HD_{2}}\big{|}(|{\mathcal{R}}\eta(x)|-$
$\displaystyle\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta(x)$
$\displaystyle\geq
c\Lambda^{-(p^{\prime}+1){\varepsilon}_{n}}\,\sigma_{p}({\mathsf{HD}}_{1})-C\Lambda^{\frac{-1}{25n}}\,\sigma_{p}({\mathsf{HD}}_{1})\approx\Lambda^{-(p^{\prime}+1){\varepsilon}_{n}}\,\sigma_{p}({\mathsf{HD}}_{1}),$

for $p\in(1,p_{0}]$, assuming that ${\varepsilon}_{n}$ and $p$ are chosen so
that $(p^{\prime}+1){\varepsilon}_{n}\ll\frac{1}{25n}.$ This is the main
ingredient for the proof of the next lemma, which is the main result of this
section.

######  Lemma 6.16.

Let $R\in{\mathsf{MDW}}\cap{\mathsf{Trc}}\cap{\mathsf{Ty}}$. Let $\Lambda>0$
be big enough and suppose that ${\varepsilon}_{Z}\leq\Lambda^{-300n}$. Then
one of the following alternatives holds:

  * (a)

$\mu(Z)>{\varepsilon}_{Z}\,\mu(R)$, or

  * (b)

$\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{2}>\Lambda^{-1}\,\sigma({\mathsf{HD}}_{1}).$

###### Proof.

Suppose that none of the alternatives holds. Then, by Lemmas 5.15 and 6.15, as
in (6.54), we have

(6.55) $I_{{\mathsf{HD}}_{2}}:=\int_{V_{4}\cap
HD_{2}}\big{|}(|{\mathcal{R}}\eta(x)|-\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\eta(x)\gtrsim\Lambda^{-(p^{\prime}+1){\varepsilon}_{n}}\,\sigma_{p}({\mathsf{HD}}_{1}),$

for all $p\in(1,p_{0}]$, assuming that $\ell_{0}$ is chosen small enough and
that

(6.56) ${(p^{\prime}+1){\varepsilon}_{n}}\leq\frac{1}{50n}.$

The appropriate values of ${\varepsilon}_{n}$ and $p$ will be chosen at the
end of the proof.

By arguments analogous to the ones we used to estimate the integral $I_{a}$ in
the proof of Lemma 6.14 we will “transfer” the estimate for
${\mathcal{R}}\eta$ in (6.55) to
$\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu$, so that we will obtain
a lower estimate for
$\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}$ which
will contradict the assumption that (b) does not hold. Although some of the
estimates below are very similar to the ones to obtain the inequality (6.48)
in the proof of Lemma 6.14, we will include the full details here for the
reader’s convenience. On the other hand, an important difference between the
proof of that lemma and the current proof is that in Lemma 6.14 we took
advantage of the fact that $p_{0}$ is close to $2$, and the estimates there
would not work for the family ${\mathsf{Reg}}_{{\mathsf{HD}}_{2}}$, while in
the arguments below it is essential that we take $p$ close to $1$, and the
estimates work fine for the family ${\mathsf{Reg}}_{{\mathsf{HD}}_{2}}$, while
they would fail for the family ${\mathsf{Reg}}_{a}$.

By Lemma 6.1, for all $Q\in{\mathsf{Reg}}_{{\mathsf{HD}}_{2}}$ such that
$\frac{1}{2}B(Q)\subset{\widetilde{V}}_{4}$, all $x\in\frac{1}{2}B(Q)$, and
all $y\in Q$,

$|{\mathcal{R}}\eta(x)|\leq|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|+C\Theta(R)+C{\mathcal{P}}(Q)+C{\mathcal{Q}}_{\mathsf{Reg}}(Q).$

Thus, for $\Lambda$ big enough, since
$\Theta(R)=\Lambda^{-1}\Theta({\mathsf{HD}}_{1})<\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1})$,

$(|{\mathcal{R}}\eta(x)|-\frac{c_{0}}{2}\,\Theta({\mathsf{HD}}_{1}))_{+}\leq(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|-\frac{c_{0}}{4}\,\Theta({\mathsf{HD}}_{1}))_{+}+C{\mathcal{P}}(Q)+C{\mathcal{Q}}_{\mathsf{Reg}}(Q).$

Therefore,

$I_{{\mathsf{HD}}_{2}}\lesssim\sum_{Q\in{\mathsf{Reg}}_{{\mathsf{HD}}_{2}}}\\!\int_{Q}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\mu(y)+\\!\\!\sum_{Q\in{\mathsf{Reg}}_{{\mathsf{HD}}_{2}}}\\!\\!\\!\\!({\mathcal{P}}(Q)^{p}+{\mathcal{Q}}_{\mathsf{Reg}}(Q)^{p})\,\mu(Q).$

By (6.16) and Lemma 6.13, we have

$\displaystyle\sum_{Q\in{\mathsf{Reg}}_{{\mathsf{HD}}_{2}}}({\mathcal{P}}(Q)^{p}+{\mathcal{Q}}_{\mathsf{Reg}}(Q)^{p})\,\mu(Q)$
$\displaystyle=\Sigma_{p}^{\mathcal{P}}({\mathsf{Reg}}_{{\mathsf{HD}}_{2}})+\Sigma_{p}^{\mathcal{Q}}({\mathsf{Reg}}_{{\mathsf{HD}}_{2}})$
$\displaystyle\lesssim\sigma_{p}({\mathsf{HD}}_{2})+\Sigma^{\mathcal{Q}}({\mathsf{Reg}})^{\frac{p}{2}}\,\mu(HD_{2})^{1-\frac{p}{2}}$
$\displaystyle\lesssim
B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1})+\Sigma^{\mathcal{P}}({\mathsf{Reg}})^{\frac{p}{2}}\,\mu(HD_{2})^{1-\frac{p}{2}}.$

Also, recalling that $\sigma({\mathsf{HD}}_{2})\leq
B\,\sigma({\mathsf{HD}}_{1})$, we get $\mu(HD_{2})\leq
B\,\Lambda^{-2}\,\mu(HD_{1})$. Then, by Lemma 6.12, we obtain

$\Sigma^{\mathcal{P}}({\mathsf{Reg}})^{\frac{p}{2}}\,\mu(HD_{2})^{1-\frac{p}{2}}\leq\big{(}B\,\sigma({\mathsf{HD}}_{1})\big{)}^{\frac{p}{2}}\,\big{(}B\,\Lambda^{-2}\,\mu(HD_{1})\big{)}^{1-\frac{p}{2}}=B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}).$

So, since any cube from ${\mathsf{Reg}}_{{\mathsf{HD}}_{2}}$ is contained in
some cube $S\in{\mathsf{HD}}_{2}$:

$I_{{\mathsf{HD}}_{2}}\lesssim\sum_{S\in{\mathsf{HD}}_{2}}\\!\int_{S}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu(y)|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\mu(y)+B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}).$

For each $S\in{\mathsf{HD}}_{2}$, by the triangle inequality and the fact that
$(\;\cdot\;)_{+}$ is a $1$-Lipschitz function, we obtain

$\int_{S}\big{|}(|{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\mu\lesssim\int_{S}\big{|}(|{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\mu\\\
+\int_{S}\big{|}{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu-{\mathcal{R}}_{{\mathcal{T}}_{\mathsf{Reg}}}\mu\big{|}^{p}\,d\mu$

By Lemma 6.3, the last integral does not exceed
$C{\mathcal{E}}(2S)^{\frac{p}{2}}\,\mu(S)^{1-\frac{p}{2}}$, and thus

(6.57)
$I_{{\mathsf{HD}}_{2}}\lesssim\\!\sum_{S\in{\mathsf{HD}}_{2}}\int_{S}\big{|}(|{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\big{|}^{p}\,d\mu+\\!\\!\sum_{S\in{\mathsf{HD}}_{2}}\\!\\!{\mathcal{E}}(2S)^{\frac{p}{2}}\mu(S)^{1-\frac{p}{2}}+B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Next we apply Lemma 6.2, which implies that for any $S\in{\mathsf{HD}}_{2}$
and all $x\in S$,

(6.58)
$|{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)|\leq|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu(x)|+C\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}+C\left(\frac{{\mathcal{E}}(2S)}{\mu(S)}\right)^{1/2}.$

In case that $R\not\in{\mathsf{DB}}$, recalling that $\Lambda\gg M$ by (4.1),
for $\Lambda$ big enough we obtain

$C\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}\leq
C\,M\,\Theta(R)\leq\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}),$

If $R\in{\mathsf{DB}}$, then we use the fact that $R\in{\mathsf{Ty}}$, which
ensures that

${\mathcal{E}}(4R)\lesssim\sum_{P\sim{\mathcal{T}}:P\in{\mathsf{DB}}}{\mathcal{E}}_{\infty}(9P)\leq\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1}),$

and so we also get

$C\left(\frac{{\mathcal{E}}(4R)}{\mu(R)}\right)^{1/2}\leq
C\left(\frac{\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})}{\mu(R)}\right)^{1/2}\leq
C\,\Lambda^{\frac{-1}{6n}}\,\Theta({\mathsf{HD}}_{1})\leq\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}).$

Hence, in any case we have

$(|{\mathcal{R}}_{{\widetilde{{\mathcal{T}}}}}\mu(x)|-\frac{c_{0}}{4}\Theta({\mathsf{HD}}_{1}))_{+}\leq|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu(x)|+C\left(\frac{{\mathcal{E}}(2S)}{\mu(S)}\right)^{1/2}.$

Plugging this estimate into (6.57), we get

(6.59)
$I_{{\mathsf{HD}}_{2}}\lesssim\int_{HD_{2}}|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu|^{p}\,d\mu+\sum_{S\in{\mathsf{HD}}_{2}}{\mathcal{E}}(2S)^{\frac{p}{2}}\,\mu(S)^{1-\frac{p}{2}}+B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Next we will estimate each term on the right hand side. First, by Hölder’s
inequality, we have

$\displaystyle\int_{HD_{2}}|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu|^{p}\,d\mu$
$\displaystyle\lesssim\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{p}\,\mu(HD_{2})^{1-\frac{p}{2}}$
$\displaystyle\leq\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{p}\,(B\Lambda^{-2}\mu(HD_{1}))^{1-\frac{p}{2}}\leq\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{p}(\Lambda^{-1}\mu(HD_{1}))^{1-\frac{p}{2}}.$

Regarding the second term in (6.59), by Hölder’s inequality again,

(6.60)
$\displaystyle\sum_{S\in{\mathsf{HD}}_{2}}{\mathcal{E}}(2S)^{\frac{p}{2}}\,\mu(S)^{1-\frac{p}{2}}$
$\displaystyle\leq\bigg{(}\sum_{S\in{\mathsf{HD}}_{2}}{\mathcal{E}}(2S)\bigg{)}^{\frac{p}{2}}\,\mu(HD_{2})^{1-\frac{p}{2}}.$

We estimate now the first factor on the right hand side:

$\displaystyle\sum_{S\in{\mathsf{HD}}_{2}}{\mathcal{E}}(2S)$
$\displaystyle\leq\sum_{S\in{\mathsf{HD}}_{2}\setminus{\mathsf{DB}}}{\mathcal{E}}(2S)+\sum_{S\sim{\mathcal{T}}:S\in{\mathsf{DB}}}{\mathcal{E}}(2S)$
$\displaystyle\lesssim
M^{2}\,\sum_{S\in{\mathsf{HD}}_{2}\setminus{\mathsf{DB}}}\sigma(S)+\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})$
$\displaystyle\leq
M^{2}\,\sigma({\mathsf{HD}}_{2})+\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})$
$\displaystyle\leq
B\,M^{2}\,\sigma({\mathsf{HD}}_{1})+\Lambda^{\frac{-1}{3n}}\,\sigma({\mathsf{HD}}_{1})\lesssim
B\,M^{2}\,\sigma({\mathsf{HD}}_{1}).$

Hence, plugging this estimate into (6.60) and using that $\mu(HD_{2})\leq
B\,\Lambda^{-2}\mu(HD_{1})$,

$\sum_{S\in{\mathsf{HD}}_{2}}{\mathcal{E}}(2S)^{\frac{p}{2}}\,\mu(S)^{1-\frac{p}{2}}\lesssim\big{(}B\,M^{2}\,\sigma({\mathsf{HD}}_{1})\big{)}^{\frac{p}{2}}\,\big{(}B\,\Lambda^{-2}\mu(HD_{1})\big{)}^{1-\frac{p}{2}}=M^{p}\,B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Altogether, we deduce that

$I_{{\mathsf{HD}}_{2}}\lesssim\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{p}\,(\Lambda^{-1}\mu(HD_{1}))^{1-\frac{p}{2}}+M^{p}\,B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Recalling the lower estimate for $I_{{\mathsf{HD}}_{2}}$ in (6.55), using that
$p^{\prime}\geq 2$ we obtain

(6.61)
$\|\Delta_{{\widetilde{{\mathcal{T}}}}}{\mathcal{R}}\mu\|_{L^{2}(\mu)}^{p}\,(\Lambda^{-1}\mu(HD_{1}))^{1-\frac{p}{2}}\geq
c\,\Lambda^{-2p^{\prime}{\varepsilon}_{n}}\,\sigma_{p}({\mathsf{HD}}_{1})-C\,M^{p}\,B\,\Lambda^{p-2}\,\sigma_{p}({\mathsf{HD}}_{1}).$

Recall now that by (4.1)

$M\leq\Lambda^{1-\frac{1}{8n-1}}\ll\Lambda.$

Notice that for $p$ close enough to $1$, we have $M^{p}\,B\,\Lambda^{p-2}\ll
1$, so that the last term on the right hand side of (6.61) is much smaller
than the first one, assuming ${\varepsilon}_{n}$ close enough to $0$. To be
more precise, let us take

$p=1+\frac{1}{4(8n-1)}.$

A straightforward calculation gives
$M^{p}\,\Lambda^{p-2}\leq\Lambda^{-\frac{1}{2(8n-1)}-\frac{1}{4(8n-1)^{2}}}$,
so that

$B\,M^{p}\,\Lambda^{p-2}\leq\Lambda^{\frac{1}{100n}}\,\Lambda^{-\frac{1}{2(8n-1)}}\leq\Lambda^{-\frac{1}{4(8n-1)}}.$