# Regularity estimates for fractional orthotropic $p$-Laplacians of mixed
order

Jamil Chaker  Fakultät für Mathematik, Universität Bielefeld, 33615 Bielefeld,
Germany<EMAIL_ADDRESS>and  Minhyun Kim  Fakultät für
Mathematik, Universität Bielefeld, 33615 Bielefeld, Germany  minhyun.kim@uni-
bielefeld.de

###### Abstract.

We study robust regularity estimates for a class of nonlinear integro-
differential operators with anisotropic and singular kernels. In this paper,
we prove a Sobolev-type inequality, a weak Harnack inequality, and a local
Hölder estimate.

###### Key words and phrases:

nonlocal operators, divergence form, regularity theory, anisotropic measures,
weak Harnack inequality

###### 2020 Mathematics Subject Classification:

35B65, 47G20, 31B05, 42B25

Jamil Chaker gratefully acknowledges support by the Deutsche
Forschungsgemeinschaft (SFB 1283/2 2021 – 317210226). Minhyun Kim gratefully
acknowledges funding by the Deutsche Forschungsgemeinschaft (GRK 2235/2 2021 -
282638148).

##  1\. Introduction

In this paper we investigate regularity estimates for weak solutions to
nonlocal equations

(1.1) $Lu=f\quad\text{in }Q=(-1,1)^{n},$

where $L$ is a nonlinear integro-differential operator of the form

(1.2)
$Lu(x)=\mathrm{PV}\int_{\mathbb{R}^{n}}|u(y)-u(x)|^{p-2}(u(y)-u(x))\mu(x,\mathrm{d}y)$

for $p>1$ and $f\in L^{q}(Q)$ for some sufficiently large $q$. The operator
$L$ is clearly determined by the family of measures
$(\mu(x,\textnormal{d}y))_{x\in\mathbb{R}^{n}}$. In the special case, when $L$
is the generator of a Lévy process, $\mu(x,A)$ measures the number of expected
jumps from $x$ into the set $A$ within the unit time interval. However, the
class of operators that we consider in this paper is more involved and for
that reason we first take a look at an important example. Let
$n\in\mathbb{N}$. For $s_{1},\cdots,s_{n}\in(0,1)$, we define

$\mu_{\mathrm{axes}}(x,\mathrm{d}y)=\sum_{k=1}^{n}s_{k}(1-s_{k})|x_{k}-y_{k}|^{-1-s_{k}p}\mathrm{d}y_{k}\prod_{i\neq
k}\delta_{x_{i}}(\mathrm{d}y_{i}).$

This family plays a central role in our paper, since admissible operators
resp. families of measures will be defined on the basis of
$\mu_{\mathrm{axes}}$. Given $x\in\mathbb{R}^{n}$, the measure
$\mu_{\mathrm{axes}}(x,\cdot)$ only charges differences that occur along the
axes

$\\{x+te_{k}\,|\,t\in\mathbb{R}\\}\quad\text{for }k\in\\{1,\dots,n\\}.$

Hence, we can think of the operator $Lu$ for
$\mu(x,\cdot)=\mu_{\mathrm{axes}}(x,\cdot)$ as the sum of one-dimensional
fractional $p$-Laplacian in $\mathbb{R}^{n}$ with orders of differentiability
$s_{1},\dots,s_{n}\in(0,1)$ depending on the respective direction. In
particular $\mu_{\mathrm{axes}}(x,\cdot)$ does not possess a density with
respect to the Lebesgue measure. An interesting phenomenon for the case $p=2$
and $s=s_{1}=\dots=s_{n}$ is that on one hand the corresponding energies for
the fractional Laplacian and $L$ are comparable. On the other hand (for
sufficiently good functions) both operators converge to the Laplace operator
as $s\nearrow 1$. It is known that the fractional $p$-Laplacian converges to
the $p$-Laplacian (see [11, Theorem 2.8] or [20, Lemma 5.1] for details), that
is defined by

$\Delta_{p}u(x)=\textnormal{div}\left(|\nabla u(x)|^{p-2}\nabla u(x)\right).$

However, the operator $L$ for $\mu(x,\cdot)=\mu_{\mathrm{axes}}(x,\cdot)$
converges for any $p>1$ and $s=s_{1}=\dots=s_{n}$ to the following local
operator (up to a constant depending on $p$ only)

(1.3) $A_{\text{loc}}^{p}u(x)=\sum_{i=1}^{n}\frac{\partial}{\partial
x_{i}}\left(\left|\frac{\partial u(x)}{\partial
x_{i}}\right|^{p-2}\frac{\partial u(x)}{\partial
x_{i}}\right)=\textnormal{div}\left(a\left(\nabla u(x)\right)\right)$

as $s\nearrow 1$, where $a:\mathbb{R}^{n}\to\mathbb{R}^{n}$ with
$a(z)=(|z_{i}|^{p-2}z_{i})_{i\in\\{1,\dots,n\\}}$. This convergence is a
direct consequence of the convergence for the one-dimensional fractional
$p$-Laplacian and the summation structure of the operator for
$\mu_{\mathrm{axes}}$. For details, we refer the reader to Proposition D.1.
The operator $A_{\text{loc}}^{p}$ is known as orthotropic $p$-Laplacian and is
a well-known operator in analysis (see for instance [41, Chapter 1, Section
8]). This operator is sometimes also called pseudo $p$-Laplacian. Minimizers
for the corresponding energies have been studied in [6], where the authors
prove for instance Hölder continuity of minimizers. In [9], local Lipschitz
regularity for weak solutions to orthotropic $p$-Laplace equations for $p\geq
2$ and every dimension is proved. The case, when $p$ is allowed to be
different in each direction, is also studied in several papers. For instance
in [3], the authors introduce anisotropic De Giorgi classes and study related
problems. Another interesting paper studying such operators with nonstandard
growth condition is [8], where the authors show that bounded local minimizers
are locally Lipschitz continuous. For further results, we refer the reader to
the references given in the previously mentioned papers.

The two local operators $\Delta_{p}$ and $A_{\text{loc}}^{p}$ are
substantially different, as for instance $\Delta_{p}$ is invariant under
orthogonal transformation, while $A_{\text{loc}}^{p}$ is not. One strength of
our results is that they are robust and we can recover results for the
orthotropic $p$-Laplacian by taking the limit.

One way to deal with the anisotropy of $\mu_{\mathrm{axes}}$, is to consider
for given $s_{1},\dots,s_{n}\in(0,1)$ a class of suitable rectangles instead
of cubes or balls. For this purpose we define
$s_{\max}=\max\\{s_{1},\cdots,s_{n}\\}$.

######  Definition 1.1.

For $r>0$ and $x\in\mathbb{R}^{n}$ we define

$\displaystyle
M_{r}(x)=\mathop{\mathchoice{\raise-2.20001pt\hbox{\huge$\times$}}{\raise-0.50003pt\hbox{\Large$\times$}}{\hbox{\large$\times$}}{\times}}_{k=1}^{n}\left(x_{k}-r^{\frac{s_{\max}}{s_{k}}},x_{k}+r^{\frac{s_{\max}}{s_{k}}}\right)\quad\text{
and }M_{r}=M_{r}(0)\,.$

The advantage of taking these cubes is that they take the anisotropy of the
measures resp. operators into account and the underlying metric measure space
is a doubling space. The choice of $s_{\max}$ in the definition of $M_{r}(x)$
is not important. It can be replaced by any positive number $\varsigma\geq
s_{\max}$. We only need to ensure that $M_{r}(x)$ are balls in a metric
measure space with radius $r>0$ and center $x\in\mathbb{R}^{n}$. This allows
us to use known results on doubling spaces like the John–Nirenberg inequality
or results on the Hardy–Littlewood maximal function.

In the spirit of [15], for each $k\in\\{1,\dots,n\\}$, we define
$E_{r}^{k}(x)=\\{y\in\mathbb{R}^{n}:|x_{k}-y_{k}|<r^{s_{\max}/{s_{k}}}\\}$.
Note, that

(1.4) $\displaystyle M_{r}(x)=\bigcap_{k=1}^{n}E_{r}^{k}(x).$

We consider families of measures $\mu(x,\textnormal{d}y)$ which are given
through certain properties regarding the reference family
$\mu_{\mathrm{axes}}(x,\textnormal{d}y)$. Let us introduce and briefly discuss
our assumptions on the families $(\mu(x,\cdot))_{x\in\mathbb{R}^{n}}$.

######  Assumption 1.

We assume

$\sup_{x\in\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}(|x-y|^{p}\land
1)\mu(x,\mathrm{d}y)<\infty$

and for all sets $A,B\in\mathcal{B}(\mathbb{R}^{n})$:

$\displaystyle\int_{A}\int_{B}\mu(x,\textnormal{d}y)\textnormal{d}x=\int_{B}\int_{A}\mu(x,\textnormal{d}y)\textnormal{d}x.$

1 provides integrability and symmetry of the family of measures. Furthermore,
we assume the following tail behavior of
$(\mu(x,\cdot))_{x\in\mathbb{R}^{n}}$.

######  Assumption 2.

There is $\Lambda\geq 1$ such that for every $x_{0}\in\mathbb{R}^{n}$,
$k\in\\{1,\dots,n\\}$ and all $r>0$

$\displaystyle\mu(x_{0},\mathbb{R}^{n}\setminus
E_{r}^{k}(x_{0}))\leq\Lambda(1-s_{k})r^{-ps_{\max}}.$

Note that 2 is a stronger assumption than an assumption on the volume on the
complement of every $M_{r}(x_{0})$. It gives an appropriate tail behavior for
the family of measures in each direction separately and allows us to control
the appearing constants in our tail estimate in all directions. This is
necessary to prove robust estimates for the corresponding operators.
Note that by 2 and (1.4), we have

(1.5) $\displaystyle\mu(x_{0},\mathbb{R}^{n}\setminus
M_{\rho}(x_{0}))\leq\sum_{k=1}^{n}\mu(x_{0},\mathbb{R}^{n}\setminus
E_{\rho}^{k}(x_{0}))\leq\Lambda\sum_{k=1}^{n}(1-s_{k})\rho^{-ps_{\max}}\leq\Lambda
n\rho^{-ps_{\max}}.$

Hence, (1.5) shows that 2 implies $\mu(x_{0},\mathbb{R}^{n}\setminus
M_{\rho}(x_{0}))\leq c\mu_{\mathrm{axes}}(x_{0},\mathbb{R}^{n}\setminus
M_{\rho}(x_{0}))$ for all $x_{0}\in\mathbb{R}^{n}$.
Finally, we assume local comparability of corresponding functionals. For this
purpose, we define for any open and bounded $\Omega\subset\mathbb{R}^{n}$

$\mathcal{E}_{\Omega}^{\mu}(u,v)=\int_{\Omega}\int_{\Omega}|u(y)-u(x)|^{p-2}(u(y)-u(x))(v(y)-v(x))\mu(x,\mathrm{d}y)\mathrm{d}x$

and $\mathcal{E}^{\mu}(u,v)=\mathcal{E}_{\mathbb{R}^{n}}^{\mu}(u,v)$ whenever
these quantities are finite.

######  Assumption 3.

There is $\Lambda\geq 1$ such that for every $x_{0}\in\mathbb{R}^{n}$,
$\rho\in(0,3)$ and every
$u\in L^{p}(M_{\rho}(x_{0}))$:

(1.6)
$\displaystyle\Lambda^{-1}\mathcal{E}^{\mu}_{M_{\rho}(x_{0})}(u,u)\leq\mathcal{E}^{\mu_{\mathrm{axes}}}_{M_{\rho}(x_{0})}(u,u)\leq\Lambda\mathcal{E}^{\mu}_{M_{\rho}(x_{0})}(u,u).$

Local comparability of the functionals is an essential assumption on the
family of measures. It tells us that our family of measures can vary from our
reference family in the given sense of local functionals without losing
crucial information on $(\mu(x,\cdot))_{x\in\mathbb{R}^{n}}$ like functional
inequalities, which we deduce for the explicitly known family
$(\mu_{\mathrm{axes}}(x,\cdot))_{x\in\mathbb{R}^{n}}$. This assumption allows
us for instance to study operators of the form (1.2) for $\mu_{\mathrm{axes}}$
in the general framework of bounded and measurable coefficients. We emphasize
that further examples of families of measures satisfying (1.6) can be
constructed similarly to the case $p=2$ (see [15, Section 9]).
In this paper, we study nonlocal operators of the form (1.2) for families of
measures that satisfy the previously given assumptions.

######  Definition 1.2.

Let $p>1$, $\Lambda\geq 1$, and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for
some $s_{0}\in(0,1)$. We call a family of measures
$(\mu(x,\cdot))_{x\in\mathbb{R}^{n}}$ admissible with regard to
$(\mu_{\mathrm{axes}}(x,\cdot))_{x\in\mathbb{R}^{n}}$, if it satisfies 1, 2,
and 3. We denote the class of such measures by $\mathcal{K}(p,s_{0},\Lambda)$.

It is not hard to see that the family
$(\mu_{\mathrm{axes}}(x,\cdot))_{x\in\mathbb{R}^{n}}$ is admissible in the
above sense. Note that 1 and 3 are clearly satisfied. Furthermore, for every
$x_{0}\in\mathbb{R}^{n}$, $k\in\\{1,\dots,n\\}$ and all $r>0$

$\mu_{\mathrm{axes}}(x_{0},\mathbb{R}^{n}\setminus
E_{r}^{k}(x_{0}))=2s_{k}(1-s_{k})\int_{r^{s_{\max}/s_{k}}}^{\infty}h^{-1-s_{k}p}=\frac{2(1-s_{k})}{p}r^{-s_{\max}p},$

which shows 2 for $\Lambda=\frac{2}{p}.$

The purpose of this paper is to study weak solutions to nonlocal equations
governed by the class of operators $L$ as in (1.2). In order to study weak
solutions, we need appropriate Sobolev-type function spaces which guarantee
regularity and integrability with respect to $\mu$.

######  Definition 1.3.

Let $\Omega\subset\mathbb{R}^{n}$ open and $p>1$. We define the function
spaces

$\displaystyle V^{p,\mu}(\Omega|\mathbb{R}^{n})$
$\displaystyle=\Big{\\{}u:\,\mathbb{R}^{n}\to\mathbb{R}\text{
meas.}\,|\,u\bigr{|}_{\Omega}\in
L^{p}(\Omega),(u,u)_{V^{p,\mu}(\Omega|\mathbb{R}^{n})}<\infty\Big{\\}}\,,$
$\displaystyle H^{p,\mu}_{\Omega}(\mathbb{R}^{n})$
$\displaystyle=\Big{\\{}u:\,\mathbb{R}^{n}\to\mathbb{R}\text{
meas.}\,|\,u\equiv 0\text{ on
}\mathbb{R}^{n}\setminus\Omega,\|u\|_{H^{p,\mu}_{\Omega}(\mathbb{R}^{n})}<\infty\Big{\\}},$

where

$\displaystyle(u,v)_{V^{p,\mu}(\Omega|\mathbb{R}^{n})}$
$\displaystyle=\int_{\Omega}\int_{\mathbb{R}^{n}}|u(y)-u(x)|^{p-2}(u(y)-u(x))(v(y)-v(x))\,\mu(x,\textnormal{d}y)\,\textnormal{d}x\,,$
$\displaystyle\|u\|_{H^{p,\mu}_{\Omega}(\mathbb{R}^{n})}^{p}$
$\displaystyle=\|u\|_{L^{p}(\Omega)}^{p}+\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}|u(y)-u(x)|^{p}\mu(x,\textnormal{d}y)\,\textnormal{d}x\,.$

The space $V^{p,\mu}(\Omega|\mathbb{R}^{n})$ can be seen as a nonlocal analog
of the space $H^{1,p}(\Omega)$. It provides fractional regularity (measured in
terms of $\mu$) inside of $\Omega$ and integrability on
$\mathbb{R}^{n}\setminus\Omega$. The space $V^{p,\mu}(\Omega|\mathbb{R}^{n})$
will serve as solution space. On the other hand, the space
$H^{p,\mu}_{\Omega}(\mathbb{R}^{n})$ can be seen as a nonlocal analog of
$H^{1,p}_{0}(\Omega)$. See [28] and [24] for further studies of these spaces
in the case $p=2$.

We are interested in finding robust regularity estimates for weak solutions to
a class of nonlocal equations. This means that the constants in the regularity
estimates do not depend on the orders of differentiability of the integro-
differential operator itself but only on a lower bound of the orders. Let us
formulate the main results of this paper. For this purpose we define $\bar{s}$
to be the harmonic mean of the orders $s_{1},\dots,s_{n}$, that is

$\bar{s}=\left(\frac{1}{n}\sum_{k=1}^{n}\frac{1}{s_{k}}\right)^{-1}.$

It is well known that the Harnack inequality fails for weak solutions to
singular equations of the type (1.1). Even in the case $p=2$ and
$s_{1}=\dots=s_{n}$, the Harnack inequality does not hold (See for instance
[7, 4]). Our first main result is a weak Harnack inequality for weak
supersolutions to equations of the type (1.1). Throughout the paper, we denote
by $p_{\star}=np/(n-p\bar{s})$ the Sobolev exponent, which will appear in
Theorem 2.7.

######  Theorem 1.4 (Weak Harnack inequality).

Let $\Lambda\geq 1$ and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some
$s_{0}\in(0,1)$. Let $1<p<n/\bar{s}$ and $f\in L^{q/(p\bar{s})}(M_{1})$ for
some $q>n$. There are $p_{0}=p_{0}(n,p,p_{\star},s_{0},q,\Lambda)\in(0,1)$ and
$C=C(n,p,p_{\star},s_{0},q,\Lambda)>0$ such that for each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $u\in
V^{p,\mu}(M_{1}|\mathbb{R}^{n})$ satisfying $u\geq 0$ in $M_{1}$ and

$\mathcal{E}^{\mu}(u,\varphi)\geq(f,\varphi)\quad\text{for every non-negative
}\varphi\in H^{p,\mu}_{M_{1}}(\mathbb{R}^{n}),$

the following holds:

(1.7) $\displaystyle\inf_{M_{1/4}}u\geq
C\left(\fint_{M_{1/2}}u^{p_{0}}(x)\,\mathrm{d}x\right)^{1/p_{0}}$
$\displaystyle-\sup_{x\in M_{15/16}}2\left(\int_{\mathbb{R}^{n}\setminus
M_{1}}u^{-}(z)^{p-1}\mu(x,\mathrm{d}z)\right)^{1/(p-1)}$
$\displaystyle-\|f\|_{L^{q/(p\bar{s})}(M_{15/16})}.$

Although the weak Harnack inequality provides an estimate on the infimum only,
it is sufficient to prove a decay of oscillation for bounded weak solutions
and therefore a local Hölder estimate.

######  Theorem 1.5 (Local Hölder estimate).

Let $\Lambda\geq 1$ and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some
$s_{0}\in(0,1)$. Let $1<p<n/\bar{s}$ and $f\in L^{q/(p\bar{s})}(M_{1})$ for
some $q>n$. There are $\alpha=\alpha(n,p,p_{\star},s_{0},q,\Lambda)\in(0,1)$
and $C=C(n,p,p_{\star},s_{0},q,\Lambda)>0$ such that for each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $u\in
V^{p,\mu}(M_{1}|\mathbb{R}^{n})\cap L^{\infty}(\mathbb{R}^{n})$ satisfying

$\mathcal{E}^{\mu}(u,\varphi)=(f,\varphi)\quad\text{for every }\varphi\in
H^{p,\mu}_{M_{1}}(\mathbb{R}^{n}),$

the following holds: $u\in C^{\alpha}(\overline{M}_{1/2})$ and

$\|u\|_{C^{\alpha}(\overline{M}_{1/2})}\leq
C\left(\|u\|_{L^{\infty}(\mathbb{R}^{n})}+\|f\|_{L^{q/(p\bar{s})}(M_{15/16})}\right).$

Note that the result needs global boundedness of weak solutions. The same
assumption is also needed in the previous works [15, 25, 14]. Global and local
boundedness of weak solutions to anisotropic nonlocal equations are nontrivial
open questions.

Furthermore, note that the general case, replacing
$M_{\frac{1}{4}},M_{\frac{1}{2}},M_{\frac{15}{16}},M_{1}$ by
$M_{\frac{r}{4}}(x_{0})$, $M_{\frac{r}{2}}(x_{0})$,
$M_{\frac{15r}{16}}(x_{0}),$ $M_{r}(x_{0})$, for $x_{0}\in\mathbb{R}^{n}$ and
$r\in(0,1]$ follows by a translation and anisotropic scaling argument
introduced in Section 4. See also [14].

Let us comment on related results in the literature. The underlying ideas in
developing regularity results for uniformly elliptic operators in divergence
form with bounded and measurable coefficients go back to the influential
contributions by De Giorgi, Nash and Moser (See [19, 46, 45]). These works led
to many further results in various directions. Similar results for nonlocal
operators in divergence form have been obtained by several authors including
the works [1, 2, 5, 12, 16, 17, 18, 22, 24, 26, 27, 33, 35, 38, 42, 43, 44,
54, 55]. See also the references therein. For further regularity results
concerning nonlocal equations governed by fractional $p$-Laplacians, we refer
the reader to [38, 39, 48, 49, 51, 50, 10],
In [22], the authors extend the De Giorgi–Nash–Moser theory to a class of
fractional $p$-Laplace equations. They provide the existence of a unique
minimizer to homogenous equations and prove local regularity estimates for
weak solutions. Moreover, in [21], the same authors prove a general Harnack
inequality for weak solutions.

Nonlocal operators with anisotropic and singular kernels of the type
$\mu_{\mathrm{axes}}$ are studied in various mathematical areas such as
stochastic differential equations and potential theory. In [4], the authors
study regularity estimates for harmonic functions for systems of stochastic
differential equations $\textnormal{d}X_{t}=A(X_{t-})\textnormal{d}Z_{t}$
driven by Lévy processes $Z_{t}$ with Lévy measure
$\mu_{\mathrm{axes}}(0,\textnormal{d}y)$, where $2s_{1}=\dots=2s_{n}=\alpha$
and $p=2$. See also [56, 13, 52]. Sharp two sided bounds for the heat kernels
are established in [36, 34]. In [37], the authors prove the existence of
transition density of the process $X_{t}$ and establish semigroup properties
of solutions. The existence of densities for solutions to stochastic
differential equations with Hölder continuous coefficients driven by Lévy
processes with anisotropic jumps has been proved in [30]. Such type of
anisotropies also appear in the study of the anisotropic stable JCIR process,
see [29].

Our approach follows mainly the ideas of [25, 14] and [15]. In [25], the
authors develop a local regularity theory for a class of linear nonlocal
operators which covers the case $s=s_{1}=\dots=s_{n}\in(0,1)$ and $p=2$. Based
on the ideas of [25], the authors in [14] establish regularity estimates in
the case $p=2$ for weak solutions in a more general framework which allows the
orders of differentiability $s_{1},\dots,s_{n}$ to be different. In [15]
parabolic equations in the case $p=2$ and possible different orders of
differentiability are studied. That paper provides robust regularity
estimates, which means the constants in the weak Harnack inequality and Hölder
regularity estimate do not depend on the orders of differentiability but on
their lower one, only. This allows us to recover regularity results for local
operators from the theory of nonlocal operators by considering the limit.

The purpose of this paper is to provide local regularity estimates as in [14]
for operators which are allowed to be nonlinear. This nonlinearity leads to
several difficulties like the need for a different proof for the discrete
gradient estimate (See Lemma 3.4). Since we cannot use the helpful properties
of Hilbert spaces (like Plancherel’s theorem), we also need an approach
different from the one in [14] to prove a Sobolev-type inequality. One
strength of this paper is the robustness of all results. This allows us to
recover regularity estimates for the limit operators such as for the
orthotropic $p$-Laplacian.

Finally, we would like to point out that it is also interesting to study such
operators in non-divergence form. We refer the reader to [53] for regularity
results concerning the fractional Laplacian and to [40] for the fractional
$p$-Laplacian. See also [23] for the anisotropic case.
Even in the most simple case, that is $p=2$ and $s=s_{1}=s_{2}=\dots=s_{n}$,
regularity estimates for operators in non-divergence form of the type (1.2)
with $\mu=\mu_{\mathrm{axes}}$ lead to various open problems such as an
Alexandrov–Bakelmann–Pucci estimate.

The authors wish to express their thanks to Lorenzo Brasco for helpful
comments.

### Outline

This paper is organized as follows. In Section 2, we introduce appropriate
cut-off functions and prove auxiliary results concerning functionals for
admissible families of measures. One main result of that section is a Sobolev-
type inequality (See Theorem 2.7). In Section 3, we prove the weak Harnack
inequality and Section 4 contains the proof of the local Hölder estimate. In
Appendix A, we prove some auxiliary algebraic inequalities, and Appendix B, we
briefly sketch the construction of appropriate anisotropic “dyadic”
rectangles. In Appendix C, we use the anisotropic dyadic rectangles to sketch
the proof of a suitable sharp maximal function theorem.

##  2\. Auxiliary results

This section is devoted to providing some general properties for the class of
nonlocal operators that we study in the scope of this paper. The main
auxiliary result is a robust Sobolev-type inequality.

Let us first introduce a class of suitable cut-off functions that will be
useful for appropriate localization.

######  Definition 2.1.

We say that $(\tau_{x_{0},r,\lambda})_{x_{0},r,\lambda}\subset
C^{0,1}(\mathbb{R}^{n})$ is an admissible family of cut-off functions if there
is $c\geq 1$ such that for all $x_{0}\in\mathbb{R}^{n}$, $r\in(0,1]$ and
$\lambda\in(1,2]$, it holds that

$\begin{cases}\operatorname{supp}(\tau)\subset M_{\lambda r}(x_{0}),\\\
\|\tau\|_{\infty}\leq 1,\\\ \tau\equiv 1\text{ on }M_{r}(x_{0}),\\\
\|\partial_{k}\tau\|_{\infty}\leq
c\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-1}r^{-s_{\max}/s_{k}}\text{ for
every }k\in\\{1\dots n\\}.\end{cases}$

For brevity, we simply write $\tau$ for any such function from
$(\tau_{x_{0},r,\lambda})_{x_{0},r,\lambda}$, if the respective choice of
$x_{0},r$ and $\lambda$ is clear. The existence of such functions is standard.
Recall the definition of admissible families of measures
$\mathcal{K}(p,s_{0},\Lambda)$ from Definition 1.2

######  Lemma 2.2.

Let $p>1$, $\Lambda\geq 1$, and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for
some $s_{0}\in(0,1)$. There is $C=C(n,p,\Lambda)>0$ such that for each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$, every $x_{0}\in\mathbb{R}^{n}$,
$r\in(0,1]$, $\lambda\in(1,2]$ and every admissible cut-off function $\tau$,
the following is true:

$\sup_{x\in\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}|\tau(y)-\tau(x)|^{p}\mu(x,\mathrm{d}y)\leq
C\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-ps_{k}}\right)r^{-ps_{\max}}.$

###### Proof.

We skip the proof. One can follow the lines of the proof from [15, Lemma 3.1]
and will get the same result with the factor $n^{p-1}$ instead of $n$. ∎

For future purposes, we deduce the following observation. It is an immediate
consequence of the foregoing lemma.

######  Corollary 2.3.

Let $p>1$, $\Lambda\geq 1$, and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for
some $s_{0}\in(0,1)$. There is a constant $C=C(n,p,\Lambda)>0$ such that for
each $\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $x_{0}\in\mathbb{R}^{n}$,
$r\in(0,1]$, $\lambda\in(1,2]$ and every admissible cut-off function $\tau$
and every $u\in L^{p}(M_{\lambda r}(x_{0}))$, it holds true that

$\displaystyle\int_{M_{\lambda r}(x_{0})}\int_{\mathbb{R}^{n}\setminus
M_{\lambda r}(x_{0})}$
$\displaystyle|u(x)|^{p}|\tau(x)|^{p}\mu(x,\textnormal{d}y)\textnormal{d}x$
$\displaystyle\leq
C\left(\sum_{k=1}^{n}(\lambda^{s_{\max}/s_{k}}-1)^{-ps_{k}}\right)r^{-ps_{\max}}\|u\|_{L^{p}(M_{\lambda
r}(x_{0}))}^{p}.$

Note that the constants in Lemma 2.2 and Corollary 2.3 do not depend on
$s_{0}$. Therefore, the lower bound $s_{0}\leq s_{k}$ for all
$k\in\\{1,\cdots,n\\}$ can be dropped here.

###  2.1. Functional inequalities

This subsection is devoted to the proofs of a Sobolev and a Poincaré-type
inequality. We start our analysis by first proving a technical lemma, see also
[15, Lemma 4.1].

######  Lemma 2.4.

Let $p>1$, $a\in(0,1]$, $b\geq 1$, $N\in\mathbb{N}$, $k\in\\{1,\cdots,n\\}$,
and $s_{k}\in(0,1)$. For any $u\in L^{p}(\mathbb{R}^{n})$

$\begin{split}&\int_{\mathbb{R}^{n}}\sup_{\rho>0}\frac{1}{\rho^{(1+ps_{k})b}}\int_{\mathbb{R}}|u(x)-u(x+he_{k})|^{p}{\bf
1}_{[a\rho^{b},2a\rho^{b})}(|h|)\,\mathrm{d}h\,\mathrm{d}x\\\
&\leq(2a)^{1+ps_{k}}N^{p(1-s_{k})}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(x)-u(x+he_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{[\frac{a}{N}\rho^{b},\frac{2a}{N}\rho^{b})}(|h|)\,\mathrm{d}h\,\mathrm{d}x.\end{split}$

###### Proof.

Let $I_{a}=[a\rho^{b},2a\rho^{b})$. By the triangle inequality and a simple
change of variables, we have

$\begin{split}&\int_{\mathbb{R}}|u(x)-u(x+he_{k})|^{p}{\bf
1}_{I_{a}}(|h|)\,\mathrm{d}h\\\ &\leq
N^{p-1}\sum_{j=1}^{N}\int_{\mathbb{R}}\left|u\left(x+\frac{j-1}{N}he_{k}\right)-u\left(x+\frac{j}{N}he_{k}\right)\right|^{p}{\bf
1}_{I_{a}}(|h|)\,\mathrm{d}h\\\
&=N^{p}\sum_{j=1}^{N}\int_{\mathbb{R}}|u(x+(j-1)he_{k})-u(x+jhe_{k})|^{p}{\bf
1}_{I_{a/N}}(|h|)\,\mathrm{d}h.\end{split}$

Since $|h|<\frac{2a}{N}\rho^{b}$, we obtain

$\begin{split}&\int_{\mathbb{R}^{n}}\sup_{\rho>0}\frac{1}{\rho^{(1+ps_{k})b}}\int_{\mathbb{R}}|u(x)-u(x+he_{k})|^{p}{\bf
1}_{I_{a}}(|h|)\,\mathrm{d}h\,\mathrm{d}x\\\ &\leq
N^{p}\left(\frac{2a}{N}\right)^{1+ps_{k}}\sum_{j=1}^{N}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(x+(j-1)he_{k})-u(x+jhe_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{I_{a/N}}(|h|)\,\mathrm{d}h\,\mathrm{d}x.\end{split}$

We change the order of integration by Fubini’s theorem and then use the change
of variables $y=x+(j-1)he_{k}$ to conclude that

$\begin{split}&\sum_{j=1}^{N}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(x+(j-1)he_{k})-u(x+jhe_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{I_{a/N}}(|h|)\,\mathrm{d}h\,\mathrm{d}x\\\
&=\sum_{j=1}^{N}\int_{\mathbb{R}}\int_{\mathbb{R}^{n}}\frac{|u(x+(j-1)he_{k})-u(x+jhe_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{I_{a/N}}(|h|)\,\mathrm{d}x\,\mathrm{d}h\\\
&=\sum_{j=1}^{N}\int_{\mathbb{R}}\int_{\mathbb{R}^{n}}\frac{|u(y)-u(y+he_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{I_{a/N}}(|h|)\,\mathrm{d}y\,\mathrm{d}h\\\
&=N\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(y)-u(y+he_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{I_{a/N}}(|h|)\,\mathrm{d}h\,\mathrm{d}y.\end{split}$

∎

Using the foregoing result for $b=s_{\max}/s_{k}$ allows us to prove a robust
Sobolev-type inequality. Robust in this context means that the appearing
constant in the Sobolev-type inequality is independent of $s_{1},\dots,s_{n}$
and depends on the lower bound $s_{0}$ only.

Before we prove a robust Sobolev-type inequality, we recall the definition of
the Hardy–Littlewood maximal function and sharp maximal function. For $u\in
L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})$,

${\bf
M}u(x)=\sup_{\rho>0}\fint_{M_{\rho}(x)}u(y)\,\textnormal{d}y\quad\text{and}\quad{\bf
M}^{\sharp}u(x)=\sup_{\rho>0}\fint_{M_{\rho}(x)}|u(y)-(u)_{M_{\rho}(x)}|\,\textnormal{d}y,$

where $(u)_{\Omega}=\fint_{\Omega}u(z)\,\textnormal{d}z$. We will use the
maximal function theorem and the sharp maximal function theorem. Note that
$\mathbb{R}^{n}$ is equipped with the metric induced by rectangles of the form
$M_{r}(x)$ and the standard Lebesgue measure. Since
$|M_{2r}|=2^{n}(2r)^{ns_{\max}/\bar{s}}\leq 2^{n/s_{0}}|M_{r}|$, this space is
a doubling space with the doubling constant $2^{n/s_{0}}$.

######  Theorem 2.5.

[32, Theorem 2.2] Let $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some
$s_{0}\in(0,1)$. Then, there is a constant $C_{1}=C_{1}(n,s_{0})>0$ such that

$|\\{x\in\mathbb{R}^{n}:{\bf
M}u(x)>t\\}|\leq\frac{C_{1}}{t}\|u\|_{L^{1}(\mathbb{R}^{n})}$

for all $t>0$ and $u\in L^{1}(\mathbb{R}^{n})$. For $p>0$, there is a constant
$C_{p}=C_{p}(n,p,s_{0})>0$ such that

$\|{\bf M}u(x)\|_{L^{p}(\mathbb{R}^{n})}\leq
C_{p}\|u\|_{L^{p}(\mathbb{R}^{n})}$

for all $u\in L^{p}(\mathbb{R}^{n})$.

We were not able to find a reference for the sharp maximal function theorem
for sets of the type $M_{\rho}$. Actually, we are not sure whether such result
is available in the literature. However, one can follow the ideas of [31,
Section 3.4], where the $L^{p}$ bound is established for the sharp maximal
function with cubes (instead of anisotropic rectangles). In order to prove the
same result for the sharp maximal function with anisotropic rectangles, dyadic
cubes have to be replaced by appropriate anisotropic “dyadic” rectangles. We
construct the anisotropic dyadic rectangles in Appendix B and prove the
following theorem in Appendix C. See also Appendix C for the definition of the
dyadic maximal function ${\bf M}_{d}u$.

######  Theorem 2.6.

Let $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some $s_{0}\in(0,1)$ and let
$0<p_{0}\leq p<\infty$. Then, there is a constant $C=C(n,p,s_{0})>0$ such that
for all $u\in L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})$ with ${\bf M}_{d}u\in
L^{p_{0}}(\mathbb{R}^{n})$,

$\|u\|_{L^{p}(\mathbb{R}^{n})}\leq C\|{\bf
M}^{\sharp}u\|_{L^{p}(\mathbb{R}^{n})}.$

We are now in a position to prove a robust Sobolev-type inequality by using
Lemma 2.4, Theorem 2.5, and Theorem 2.6.

######  Theorem 2.7.

Let $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some $s_{0}\in(0,1)$. Suppose
that $1<p<n/\bar{s}$ and let $p_{\star}=np/(n-p\bar{s})$. Then, there is a
constant $C=C(n,p,p_{\star},s_{0})>0$ such that for every $u\in
V^{p,\mu_{\mathrm{axes}}}(\mathbb{R}^{n}|\mathbb{R}^{n})$

(2.1) $\|u\|_{L^{p_{\star}}(\mathbb{R}^{n})}^{p}\leq
C\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}|u(x)-u(y)|^{p}\mu_{\mathrm{axes}}(x,\mathrm{d}y)\mathrm{d}x.$

###### Proof.

This proof is based on the technique of [47], which uses the maximal and sharp
maximal inequalities. Note that by definition of
$V^{p,\mu_{\mathrm{axes}}}(\mathbb{R}^{n}|\mathbb{R}^{n})$ and Hölder’s
inequality, $V^{p,\mu_{\mathrm{axes}}}(\mathbb{R}^{n}|\mathbb{R}^{n})\subset
L^{p}(\mathbb{R}^{n})\subset L^{p}_{\mathrm{loc}}(\mathbb{R}^{n})\subset
L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})$. Hence, the maximal and sharp maximal
functions are well defined for every function $u\in
V^{p,\mu_{\mathrm{axes}}}(\mathbb{R}^{n}|\mathbb{R}^{n})$. For
$x\in\mathbb{R}^{n}$ and $\rho>0$, we have

(2.2)
$\fint_{M_{\rho}(x)}|u(y)-(u)_{M_{\rho}(x)}|\,\mathrm{d}y\leq\fint_{M_{\rho}(x)}\fint_{M_{\rho}(x)}|u(y)-u(z)|\,\mathrm{d}z\,\mathrm{d}y.$

Let us consider as in [14, Lemma 2.1] a polygonal chain
$\ell=(\ell_{0}(y,z),\cdots,\ell_{n}(y,z))\in\mathbb{R}^{n(n+1)}$ connecting
$y$ and $z$ with

$\ell_{k}(y,z)=(l_{1}^{k},\cdots,l_{n}^{k}),\quad\text{where}~{}l_{j}^{k}=\begin{cases}z_{j},&\text{if}~{}j\leq
k,\\\ y_{j},&\text{if}~{}j>k,\end{cases}$

then $y=\ell_{0}(y,z)$, $z=\ell_{n}(y,z)$, and
$|\ell_{k-1}(y,z)-\ell_{k}(y,z)|=|y_{k}-z_{k}|$ for all $k=1,\cdots,n$. By the
triangle inequality, we have

(2.3)
$\begin{split}&\fint_{M_{\rho}(x)}\fint_{M_{\rho}(x)}|u(y)-u(z)|\,\mathrm{d}z\,\mathrm{d}y\\\
&\leq\sum_{k=1}^{n}\fint_{M_{\rho}(x)}\fint_{M_{\rho}(x)}|u(\ell_{k-1}(y,z))-u(\ell_{k}(y,z))|\,\mathrm{d}z\,\mathrm{d}y.\end{split}$

For a fixed $k$, we set
$w=\ell_{k-1}(y,z)=(z_{1},\cdots,z_{k-1},y_{k},\cdots,y_{n})$ and
$v=y+z-w=(y_{1},\cdots,y_{k-1},z_{k},\cdots,z_{n})$, then
$\ell_{k}(y,z)=w+e_{k}(v_{k}-w_{k})$. By Fubini’s theorem, we obtain

(2.4)
$\begin{split}&\fint_{M_{\rho}(x)}\fint_{M_{\rho}(x)}|u(\ell_{k-1}(y,z))-u(\ell_{k}(y,z))|\,\mathrm{d}z\,\mathrm{d}y\\\
&\leq\fint_{M_{\rho}(x)}\fint_{x_{k}-\rho^{s_{\max}/s_{k}}}^{x_{k}+\rho^{s_{\max}/s_{k}}}|u(w)-u(w+e_{k}(v_{k}-w_{k}))|\,\mathrm{d}v_{k}\,\mathrm{d}w.\end{split}$

Moreover, using the inequality
$|v_{k}-w_{k}|\leq|v_{k}-x_{k}|+|w_{k}-x_{k}|<2\rho^{s_{\max}/s_{k}}$, we make
the inner integral in the right-hand side of (2.4) independent of $x$. Namely,
we have

(2.5)
$\begin{split}&\fint_{x_{k}-\rho^{s_{\max}/s_{k}}}^{x_{k}+\rho^{s_{\max}/s_{k}}}|u(w)-u(w+e_{k}(v_{k}-w_{k}))|\,\mathrm{d}v_{k}\\\
&\leq
2\fint_{w_{k}-2\rho^{s_{\max}/s_{k}}}^{w_{k}+2\rho^{s_{\max}/s_{k}}}|u(w)-u(w+e_{k}(v_{k}-w_{k}))|\,\mathrm{d}v_{k}\\\
&=2\fint_{-2\rho^{s_{\max}/s_{k}}}^{2\rho^{s_{\max}/s_{k}}}|u(w)-u(w+he_{k})|\,\mathrm{d}h.\end{split}$

Combining (2.2), (2.3), (2.4), and (2.5), we arrive at

(2.6)
$\fint_{M_{\rho}(x)}|u(y)-(u)_{M_{\rho}(x)}|\,\mathrm{d}y\leq\sum_{k=1}^{n}\rho^{s_{\max}}\fint_{M_{\rho}(x)}F_{k}(w)\,\mathrm{d}w,$

where the function $F_{k}$ is defined by

$F_{k}(w):=\sup_{\rho>0}\left(2\rho^{-s_{\max}}\fint_{-2\rho^{s_{\max}/s_{k}}}^{2\rho^{s_{\max}/s_{k}}}|u(w)-u(w+he_{k})|\,\mathrm{d}h\right).$

By Hölder’s inequality,

(2.7)
$\begin{split}\left(\fint_{M_{\rho}(x)}F_{k}(w)\,\mathrm{d}w\right)^{p_{\star}}&\leq\left(\fint_{M_{\rho}(x)}F_{k}^{p}(w)\,\mathrm{d}w\right)^{\frac{{p_{\star}}-p}{p}}\left(\fint_{M_{\rho}(x)}F_{k}(w)\,\mathrm{d}w\right)^{p}\\\
&\leq|M_{\rho}|^{-\frac{{p_{\star}}-p}{p}}\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{{p_{\star}}-p}\left(\fint_{M_{\rho}(x)}F_{k}(w)\,\mathrm{d}w\right)^{p}.\end{split}$

Thus, it follows from (2.6) and (2.7) that

$\begin{split}\Bigg{(}\fint_{M_{\rho}(x)}|u(y)-(u)_{M_{\rho}(x)}|&\,\mathrm{d}y\Bigg{)}^{p_{\star}}\leq
n^{{p_{\star}}-1}\sum_{k=1}^{n}\rho^{{p_{\star}}s_{\max}}\left(\fint_{M_{\rho}(x)}F_{k}(w)\,\mathrm{d}w\right)^{p_{\star}}\\\
&\leq
n^{{p_{\star}}-1}\sum_{k=1}^{n}\rho^{{p_{\star}}s_{\max}}|M_{\rho}|^{-\frac{{p_{\star}}-p}{p}}\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{{p_{\star}}-p}\left(\fint_{M_{\rho}(x)}F_{k}(w)\,\mathrm{d}w\right)^{p}\\\
&\leq\frac{n^{{p_{\star}}-1}}{2^{n\frac{{p_{\star}}-p}{p}}}\sum_{k=1}^{n}\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{{p_{\star}}-p}\left(\fint_{M_{\rho}(x)}F_{k}(w)\,\mathrm{d}w\right)^{p}.\end{split}$

Taking the supremum over $\rho>0$, we obtain

(2.8) $\left({\bf
M}^{\sharp}u(x)\right)^{p_{\star}}\leq\frac{n^{{p_{\star}}-1}}{2^{n\frac{{p_{\star}}-p}{p}}}\sum_{k=1}^{n}\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{{p_{\star}}-p}({\bf
M}F_{k}(x))^{p}.$

We now use Theorem 2.5 and Theorem 2.6. By (C.1) and Theorem 2.5, we know that
${\bf M}_{d}u\in L^{p}(\mathbb{R}^{n})$. Thus, Theorem 2.6 yields that

$\|u\|_{L^{p_{\star}}(\mathbb{R}^{n})}^{p_{\star}}\leq C\|{\bf
M}^{\sharp}u\|_{L^{p_{\star}}(\mathbb{R}^{n})}^{p_{\star}}$

for some $C=C(n,p_{\star},s_{0})>0$. Moreover, assuming $F_{k}\in
L^{p}(\mathbb{R}^{n})$, we have by Theorem 2.5 and (2.8)

$\|{\bf M}^{\sharp}u\|_{L^{p_{\star}}(\mathbb{R}^{n})}^{p_{\star}}\leq
C\sum_{k=1}^{n}\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{{p_{\star}}-p}\|{\bf
M}F_{k}\|_{p}^{p}\leq
C\sum_{k=1}^{n}\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{p_{\star}}$

for some $C=C(n,p,p_{\star},s_{0})>0$. Therefore, it only remains to show that

(2.9) $\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{p}\leq
Cs_{k}(1-s_{k})\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(x)-u(x+he_{k})|^{p}}{|h|^{1+ps_{k}}}\,\mathrm{d}h\,\mathrm{d}x$

for each $k=1,\cdots,n$.

Let us fix $k$. Using Hölder’s inequality, we have

$\begin{split}\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{p}&\leq\int_{\mathbb{R}^{n}}\sup_{\rho>0}\frac{2^{p}}{\rho^{ps_{\max}}}\fint_{-2\rho^{s_{\max}/s_{k}}}^{2\rho^{s_{\max}/s_{k}}}|u(x)-u(x+he_{k})|^{p}\,\mathrm{d}h\,\mathrm{d}x\\\
&=\sum_{i=0}^{\infty}\int_{\mathbb{R}^{n}}\sup_{\rho>0}\frac{2^{p-2}}{\rho^{(1+ps_{k})s_{\max}/s_{k}}}\int_{\mathbb{R}}|u(x)-u(x+he_{k})|^{p}{\bf
1}_{I_{i}}(|h|)\,\mathrm{d}h\,\mathrm{d}x,\end{split}$

where $I_{i}=[2^{-i}\rho^{s_{\max}/s_{k}},2^{-i+1}\rho^{s_{\max}/s_{k}})$. For
each $i$, let $\\{\beta_{j,i}\\}_{j=0}^{\infty}$ be a sequence such that
$\sum_{j}\beta_{j,i}\geq 1$, which will be chosen later. Then,

$\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{p}\leq\sum_{i,j=0}^{\infty}\beta_{j,i}\int_{\mathbb{R}^{n}}\sup_{\rho>0}\frac{2^{p-2}}{\rho^{(1+ps_{k})s_{\max}/s_{k}}}\int_{\mathbb{R}}|u(x)-u(x+he_{k})|^{p}{\bf
1}_{I_{i}}(|h|)\,\mathrm{d}h\,\mathrm{d}x.$

By Lemma 2.4 for $N=2^{j}$, $a=2^{-i}\in(0,1]$ and $b=s_{\max}/s_{k}$, we
obtain

$\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{p}\leq\sum_{i,j=0}^{\infty}2^{p-2+(1+ps_{k})(1-i)+p(1-s_{k})j}\beta_{j,i}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(x)-u(x+he_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{I_{i+j}}(|h|)\,\mathrm{d}h\,\mathrm{d}x.$

We rearrange the double sums to have

$\begin{split}&\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{p}\\\
&\leq\sum_{i=0}^{\infty}\sum_{j=0}^{i}2^{p-2+(1+ps_{k})(1-i+j)+p(1-s_{k})j}\beta_{j,i-j}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(x)-u(x+he_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{I_{i}}(|h|)\,\mathrm{d}h\,\mathrm{d}x.\end{split}$

Let $\beta_{j,i}=p(\log 2)(1-s_{k})2^{-p(1-s_{k})j}$, then

$1\leq\sum_{j=0}^{\infty}\beta_{j,i}=\frac{p(\log
2)(1-s_{k})}{1-2^{-p(1-s_{k})}}\leq 2p<+\infty.$

Since

$\begin{split}\sum_{j=0}^{i}2^{p-2+(1+ps_{k})(1-i+j)+p(1-s_{k})j}\beta_{j,i-j}&=p(\log
2)(1-s_{k})2^{p-2+(1+ps_{k})(1-i)}\sum_{j=0}^{i}2^{(1+ps_{k})j}\\\ &=p(\log
2)(1-s_{k})2^{p-2+(1+ps_{k})(1-i)}\frac{2^{(1+ps_{k})(i+1)}}{2^{1+ps_{k}}-1}\\\
&\leq p(1-s_{k})2^{p(1+s_{k})},\end{split}$

we arrive at

$\begin{split}\|F_{k}\|_{L^{p}(\mathbb{R}^{n})}^{p}&\leq\sum_{i=0}^{\infty}p(1-s_{k})2^{p(1+s_{k})}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(x)-u(x+he_{k})|^{p}}{|h|^{1+ps_{k}}}{\bf
1}_{I_{i}}(|h|)\,\mathrm{d}h\,\mathrm{d}x\\\ &\leq
p2^{2p}\frac{s_{k}}{s_{0}}(1-s_{k})\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}\frac{|u(x)-u(x+he_{k})|^{p}}{|h|^{1+ps_{k}}}\,\mathrm{d}h\,\mathrm{d}x,\end{split}$

which proves (2.9). ∎

Next, we can make use of appropriate cut-off functions to prove a localized
version of the foregoing Sobolev-type inequality.

######  Corollary 2.8.

Let $\Lambda\geq 1$ and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some
$s_{0}\in(0,1)$. Suppose that $1<p<n/\bar{s}$. There is
$C=C(n,p,p_{\star},s_{0},\Lambda)>0$ such that for each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $x_{0}\in\mathbb{R}^{n}$,
$r\in(0,1]$, $\lambda\in(1,2]$ and $u\in H^{p,\mu}_{{M_{\lambda
r}(x_{0})}}(\mathbb{R}^{n})$ it holds

$\begin{split}\|u\|_{L^{{p_{\star}}}(M_{r}(x_{0}))}^{p}\
\leq&C\int_{M_{\lambda r}(x_{0})}\int_{M_{\lambda
r}(x_{0})}|u(x)-u(y)|^{p}\mu(x,\mathrm{d}y)\mathrm{d}x\\\
&+C\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-s_{k}p}\right)r^{-ps_{\max}}\|u\|_{L^{p}(M_{\lambda
r}(x_{0}))}^{p},\end{split}$

where ${p_{\star}}$ is defined as in Theorem 2.7.

###### Proof.

Let $\tau:\mathbb{R}^{n}\to\mathbb{R}$ be an admissible cut-off function in
the sense of Definition 2.1. For simplicity of notation we write
$M_{r}=M_{r}(x_{0})$.

By Theorem 2.7 there is a constant $c_{1}=c_{1}(n,p,p_{\star},s_{0})>0$ such
that

$\displaystyle\|u\tau\|_{L^{{p_{\star}}}(\mathbb{R}^{n})}^{p}$
$\displaystyle\leq c_{1}\Bigg{(}\int_{M_{\lambda r}}\int_{M_{\lambda
r}}|u(x)\tau(x)-u(x)\tau(y)|^{p}\,\mu_{\mathrm{axes}}(x,\textnormal{d}y)\,\textnormal{d}x$
$\displaystyle\qquad\qquad+2\int_{M_{\lambda r}}\int_{(M_{\lambda
r})^{c}}|u(x)\tau(x)-u(x)\tau(y)|^{p}\,\mu_{\mathrm{axes}}(x,\textnormal{d}y)\,\textnormal{d}x\Bigg{)}$
$\displaystyle=:c_{1}(I_{1}+2I_{2}).$

We have

$\displaystyle I_{1}$
$\displaystyle\leq\frac{1}{2^{p}}\Bigg{(}\int_{M_{\lambda r}}\int_{M_{\lambda
r}}2^{p-1}\left|(u(y)-u(x))(\tau(x)+\tau(y))\right|^{p}\,\mu_{\mathrm{axes}}(x,\textnormal{d}y)\,\textnormal{d}x$
$\displaystyle\qquad\quad+\int_{M_{\lambda r}}\int_{M_{\lambda
r}}2^{p-1}\left|(u(x)+u(y))(\tau(x)-\tau(y))\right|^{p}\,\mu_{\mathrm{axes}}(x,\textnormal{d}y)\,\textnormal{d}x\Bigg{)}$
$\displaystyle=\frac{1}{2}(J_{1}+J_{2}),$

Since $(\tau(x)+\tau(y))\leq 2$ for all $x,y\in M_{\lambda r}$, we get

$J_{1}\leq 2^{p}\int_{M_{\lambda r}}\int_{M_{\lambda
r}}|u(y)-u(x)|^{p}\,\mu_{\mathrm{axes}}(x,\textnormal{d}y)\,\textnormal{d}x\leq\Lambda
2^{p}\int_{M_{\lambda r}}\int_{M_{\lambda
r}}|u(y)-u(x)|^{p}\,\mu(x,\textnormal{d}y)\,\textnormal{d}x,$

where we used 3 in the second inequality.
Moreover, since $|u(x)+u(y)|^{p}|\tau(x)-\tau(x)|^{p}\leq
2^{p-1}|u(x)|^{p}|\tau(x)-\tau(x)|^{p}+2^{p-1}|u(y)|^{p}|\tau(y)-\tau(x)|^{p}$,
we can again apply 3 and by Lemma 2.2, we get

$\displaystyle J_{2}$ $\displaystyle\leq
2^{p}\left(\sup_{x\in\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}|\tau(y)-\tau(x)|^{p}\mu(x,\textnormal{d}y)\right)\|u\|^{p}_{L^{p}(M_{\lambda
r})}$ $\displaystyle\leq
c_{2}\left(\sum_{k=1}^{n}(\lambda^{\frac{s_{\max}}{s_{k}}}-1)^{-ps_{k}}\right)r^{-ps_{\max}}\|u\|^{p}_{L^{p}(M_{\lambda
r})}\,$

for some $c_{2}>0$, depending on $n$, $p$, $s_{0}$ and $\Lambda$. Moreover, by
Corollary 2.3 there is $c_{3}=c_{3}(n,p,\Lambda)>0$ such that

$I_{2}\leq
c_{3}\left(\sum_{k=1}^{n}(\lambda^{\frac{s_{\max}}{s_{k}}}-1)^{-ps_{k}}\right)r^{-ps_{\max}}\|u\|_{L^{p}(M_{\lambda
r})}^{p}$

Combining these estimates, we find a constant
$C=C(n,p,p_{\star},s_{0},\Lambda)>0$ such that

$\displaystyle\|u\|_{L^{{p_{\star}}}(M_{r})}^{p}$
$\displaystyle\leq\|u\tau\|_{L^{{p_{\star}}}(\mathbb{R}^{n})}^{p}$
$\displaystyle\leq C\Bigg{(}\int_{M_{\lambda r}}\int_{M_{\lambda
r}}|u(x)-u(y)|^{p}\,\mu_{\mathrm{axes}}(x,\textnormal{d}y)\,\,\textnormal{d}x$
$\displaystyle\qquad\qquad\qquad\qquad\qquad+\left(\sum_{k=1}^{n}(\lambda^{\frac{s_{\max}}{s_{k}}}-1)^{-ps_{k}}\right)r^{-ps_{\max}}\|u\|^{p}_{L^{p}(M_{\lambda
r})}\Bigg{)}.$

∎

Applying the same method as in the proof of the Sobolev-type inequality
Theorem 2.7, we can deduce a Poincaré inequality.

######  Theorem 2.9.

Let $p>1$, $\Lambda\geq 1$, and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for
some $s_{0}\in(0,1)$. There is $C=C(n,p,s_{0},\Lambda)>0$ such that for each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $x_{0}\in\mathbb{R}^{n}$,
$r\in(0,1]$ and $u\in L^{p}(M_{r}(x_{0}))$,

$\|u-(u)_{M_{r}(x_{0})}\|_{L^{p}(M_{r}(x_{0}))}^{p}\leq
Cr^{ps_{\max}}\mathcal{E}_{M_{r}(x_{0})}^{\mu}(u,u).$

The proof is analog to the proof of the Poincaré inequality for the case
$p=2$, see [15, Theorem 4.2].

##  3\. Weak Harnack inequality

In this section, we prove Theorem 1.4. The proof is based on Moser’s iteration
technique. We first need to verify a few properties for weak supersolutions to
(1.1).

######  Lemma 3.1.

Let $\Lambda\geq 1$ and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some
$s_{0}\in(0,1)$. Let $1<p\leq n/\bar{s}$, $x_{0}\in\mathbb{R}^{n}$,
$r\in(0,1]$, and $\lambda\in(1,2]$. Set $M_{r}=M_{r}(x_{0})$ and assume $f\in
L^{q/(p\bar{s})}(M_{\lambda r})$ for some $q>n$. There is
$C=C(n,p,s_{0},\Lambda)>0$ such that for each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $u\in V^{p,\mu}(M_{\lambda
r}|\mathbb{R}^{n})$ that satisfies

$\displaystyle\mathcal{E}^{\mu}(u,\varphi)\geq(f,\varphi)\quad\text{for any
nonnegative}~{}\varphi\in H^{p,\mu}_{M_{\lambda r}}(\mathbb{R}^{n}),$
$\displaystyle u(x)\geq\epsilon\quad\text{a.e. in}~{}M_{\lambda r}~{}\text{for
some}~{}\epsilon>0,$

the following holds:

$\begin{split}&\int_{M_{r}}\int_{M_{r}}|\log u(y)-\log
u(x)|^{p}\,\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\ &\leq
C\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-ps_{k}}\right)r^{-ps_{\max}}|M_{\lambda
r}|\\\ &\quad+\epsilon^{1-p}\|f\|_{L^{q/(p\bar{s})}(M_{\lambda r})}|M_{\lambda
r}|^{\frac{q-p\bar{s}}{q}}+2\epsilon^{1-p}|M_{\lambda r}|\sup_{x\in
M_{(\lambda+1)r/2}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}u_{-}^{p-1}(y)\mu(x,\textnormal{d}y).\end{split}$

###### Proof.

Let $\tau$ be an admissible cut-off function in the sense of Definition 2.1
and let $\varphi(x)=\tau^{p}(x)u^{1-p}(x)$, which is well defined since
$\operatorname{supp}(\tau)\subset M_{\lambda r}$. Then, we have

(3.1) $\begin{split}(f,\varphi)&\leq\int_{M_{\lambda r}}\int_{M_{\lambda
r}}|u(x)-u(y)|^{p-2}(u(x)-u(y))\left(\frac{\tau^{p}(x)}{u^{p-1}(x)}-\frac{\tau^{p}(y)}{u^{p-1}(y)}\right)\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\quad+2\int_{M_{\lambda r}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}|u(x)-u(y)|^{p-2}(u(x)-u(y))\frac{\tau^{p}(x)}{u^{p-1}(x)}\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&=:I_{1}+I_{2}.\end{split}$

Similar to the proof of [22, Lemma 1.3], we get the inequality

(3.2)
$\begin{split}&|u(x)-u(y)|^{p-2}(u(x)-u(y))\left(\frac{\tau^{p}(x)}{u^{p-1}(x)}-\frac{\tau^{p}(y)}{u^{p-1}(y)}\right)\\\
&\leq-c_{1}|\log u(x)-\log
u(y)|^{p}\tau^{p}(y)+c_{2}|\tau(x)-\tau(y)|^{p},\end{split}$

where $c_{1},c_{2}>0$ are constants depending only on $p$. Hence, by (3.2) and
Lemma 2.2,

(3.3) $\begin{split}I_{1}&\leq-c_{1}\int_{M_{r}}\int_{M_{r}}|\log u(x)-\log
u(y)|^{p}\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\quad+C\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-ps_{k}}\right)r^{-ps_{\max}}|M_{\lambda
r}|.\end{split}$

For $I_{2}$, again by Lemma 2.2

(3.4) $\begin{split}I_{2}&\leq 2\int_{M_{\lambda
r}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}(u(x)-u(y))^{p-1}\frac{\tau^{p}(x)}{u^{p-1}(x)}{\bf 1}_{\\{u(x)\geq
u(y)\\}}\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\ &\leq 2\int_{M_{\lambda
r}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}|\tau(x)-\tau(y)|^{p}\mu(x,\textnormal{d}y)\,\textnormal{d}x+2\int_{M_{\lambda
r}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}u_{-}^{p-1}(y)\frac{\tau^{p}(x)}{\epsilon^{p-1}}\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\leq
C\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-ps_{k}}\right)r^{-ps_{\max}}|M_{\lambda
r}|+\frac{2|M_{\lambda r}|}{\epsilon^{p-1}}\sup_{x\in
M_{(\lambda+1)r/2}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}u_{-}^{p-1}(y)\mu(x,\textnormal{d}y),\end{split}$

where we assumed that $\mathrm{supp}(\tau)\subset M_{(\lambda+1)r/2}$.
Combining (3.1), (3.3), and (3.4), and using Hölder’s inequality, we conclude
that

$\begin{split}&\int_{M_{r}}\int_{M_{r}}|\log u(y)-\log
u(x)|^{p}\,\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\ &\leq
C\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-ps_{k}}\right)r^{-ps_{\max}}|M_{\lambda
r}|\\\ &\quad+\epsilon^{1-p}\|f\|_{L^{q/(p\bar{s})}(M_{\lambda r})}|M_{\lambda
r}|^{\frac{q-p\bar{s}}{q}}+2\epsilon^{1-p}|M_{\lambda r}|\sup_{x\in
M_{(\lambda+1)/2}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}u_{-}^{p-1}(y)\mu(x,\textnormal{d}y).\end{split}$

∎

The next theorem is an essential result to prove the weak Harnack inequality.

######  Theorem 3.2.

Let $\Lambda\geq 1$ and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some
$s_{0}\in(0,1)$. Let $1<p<n/\bar{s}$, $x_{0}\in\mathbb{R}^{n}$, and
$r\in(0,1]$. Set $M_{r}=M_{r}(x_{0})$ and assume $f\in
L^{q/(p\bar{s})}(M_{5r/4})$ for some $q>n$. There are
$C=C(n,p,s_{0},q,\Lambda)>0$ and
$\bar{p}=\bar{p}(n,p,s_{0},q,\Lambda)\in(0,1)$ such that for each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $u\in
V^{p,\mu}(M_{5r/4}|\mathbb{R}^{n})$ that satisfies

$\displaystyle\mathcal{E}^{\mu}(u,\varphi)\geq(f,\varphi)\quad\text{for any
nonnegative}~{}\varphi\in H^{p,\mu}_{M_{5r/4}}(\mathbb{R}^{n}),$
$\displaystyle u(x)\geq\epsilon\quad\text{a.e. in}~{}M_{5r/4},$

for

$\epsilon>r^{\delta}\|f\|_{L^{q/(p\bar{s})}(M_{5r/4})}^{\frac{1}{p-1}}+\left(r^{ps_{\max}}\sup_{x\in
M_{9r/8}}\int_{\mathbb{R}^{n}\setminus
M_{5r/4}}u_{-}^{p-1}(y)\mu(x,\textnormal{d}y)\right)^{\frac{1}{p-1}},$

where $\delta=\frac{ps_{\max}}{p-1}\frac{q-n}{q}$, the following holds:

$\left(\fint_{M_{r}}u^{\bar{p}}(x)\,\mathrm{d}x\right)^{1/\bar{p}}\leq
C\left(\fint_{M_{r}}u^{-\bar{p}}(x)\,\mathrm{d}x\right)^{-1/\bar{p}}.$

###### Proof.

We only need to prove that $\log u\in\operatorname*{BMO}(M_{r})$. The rest of
the proof is standard. The Poincaré inequality (see Theorem 2.9) and Lemma 3.1
imply

$\begin{split}&\|\log u-(\log u)_{M_{r}}\|_{L^{p}(M_{r})}^{p}\\\ &\leq
Cr^{ps_{\max}}\mathcal{E}^{\mu}_{M_{r}}(\log u,\log u)\\\ &\leq
C\left(\sum_{k=1}^{n}\left(\left(\frac{5}{4}\right)^{\frac{s_{\max}}{s_{k}}}-1\right)^{-s_{k}p}\right)|M_{5r/4}|+C\epsilon^{1-p}\|f\|_{L^{q/(p\bar{s})}(M_{5r/4})}r^{ps_{\max}}|M_{5r/4}|^{\frac{q-p\bar{s}}{q}}\\\
&\quad+C\epsilon^{1-p}r^{ps_{\max}}|M_{5r/4}|\sup_{x\in
M_{9r/8}}\int_{\mathbb{R}^{n}\setminus
M_{5r/4}}u_{-}^{p-1}(y)\mu(x,\textnormal{d}y)\\\ &\leq C|M_{r}|,\end{split}$

where we used the bound on $\epsilon$ in the last inequality. Finally, by
Hölder’s inequality we obtain

$\|\log u\|_{\mathrm{BMO}(M_{r})}\leq\left(\fint_{M_{r}}|\log u-(\log
u)_{M_{r}}|^{p}\,\textnormal{d}x\right)^{1/p}\leq C,$

which shows that $\log u\in\mathrm{BMO}(B_{r})$. ∎

In order to apply Moser’s iteration for negative exponents, we prove the
following lemma.

######  Lemma 3.3.

Let $\Lambda\geq 1$ and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some
$s_{0}\in(0,1)$. Let $1<p<n/\bar{s},$ $x_{0}\in\mathbb{R}^{n}$, $r\in(0,1]$,
and $\lambda\in(1,2]$. Set $M_{r}=M_{r}(x_{0})$ and assume $f\in
L^{q/(p\bar{s})}(M_{\lambda r})$ for some $q>n$. For each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $u\in V^{p,\mu}(M_{\lambda
r}|\mathbb{R}^{n})$ that satisfies

$\displaystyle\mathcal{E}^{\mu}(u,\varphi)\geq(f,\varphi)\quad\text{for any
nonnegative}~{}\varphi\in H^{p,\mu}_{M_{\lambda r}}(\mathbb{R}^{n}),$
$\displaystyle u(x)\geq\epsilon\quad\text{a.e. in}~{}M_{\lambda r},$

for

$\epsilon>r^{\delta}\|f\|_{L^{q/(p\bar{s})}(M_{\lambda
r/4})}^{\frac{1}{p-1}}+\left(r^{ps_{\max}}\sup_{x\in
M_{(\lambda+1)r/2}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}u_{-}^{p-1}(y)\mu(x,\textnormal{d}y)\right)^{\frac{1}{p-1}},$

the following is true for any $t>p-1$,

$\left\|u^{-1}\right\|_{L^{(t-p+1)\gamma}(M_{r})}^{t-p+1}\leq
C\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-s_{k}p}\right)r^{-s_{\max}p}\left\|u^{-1}\right\|_{L^{t-p+1}(M_{\lambda
r})}^{t-p+1},$

where $\delta=\frac{ps_{\max}}{p-1}\frac{q-n}{q}$, $\gamma=n/(n-p\bar{s})$,
and $C=C(n,p,p_{\star},q,t,s_{0},\Lambda)>0$ is a constant that is bounded
when $t$ is bounded away from $p-1$.

To prove Lemma 3.3, we need the following algebraic inequality.

######  Lemma 3.4.

Let $a,b>0$, $\tau_{1},\tau_{2}\in[0,1]$, and $t>p-1>0$. Then,

$\begin{split}&|b-a|^{p-2}(b-a)(\tau_{1}^{p}a^{-t}-\tau_{2}^{p}b^{-t})\\\
&\geq
c_{1}\left|\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right|^{p}-c_{2}|\tau_{1}-\tau_{2}|^{p}\left(a^{-t+p-1}+b^{-t+p-1}\right),\end{split}$

where $c_{i}=c_{i}(p,t)>0$, $i=1,2$, is bounded when $t$ is bounded away from
$p-1$.

Note that Lemma 3.4 is a discrete version of

$|\nabla v|^{p-2}\nabla v\cdot\nabla(-v^{-t}\tau^{p})\geq
c_{1}\left|\nabla\left(v^{\frac{-t+p-1}{p}}\tau\right)\right|^{p}-c_{2}|\nabla\tau|^{p}v^{-t+p-1}.$

The proof of Lemma 3.4 is provided in Appendix A.

###### Proof of Lemma 3.3.

Let $\tau$ be an admissible cut-off function in the sense of Definition 2.1.
Since $\tau=0$ outside $M_{\lambda r}$, the function $\varphi=-\tau^{p}u^{-t}$
is well defined. Using Lemma 3.4, we have

$\begin{split}&(f,-\tau^{p}u^{-t})\geq\mathcal{E}(u,-\tau^{p}u^{-t})\\\
&=\int_{M_{\lambda r}}\int_{M_{\lambda
r}}|u(y)-u(x)|^{p-2}(u(y)-u(x))\left(\tau^{p}(x)u^{-t}(x)-\tau^{p}(y)u^{-t}(y)\right)\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\quad+2\int_{M_{\lambda r}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}|u(y)-u(x)|^{p-2}(u(y)-u(x))\tau^{p}(x)u^{-t}(x)\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\geq c_{1}\int_{M_{\lambda r}}\int_{M_{\lambda
r}}\left|\tau(x)u^{\frac{-t+p-1}{p}}(x)-\tau(y)u^{\frac{-t+p-1}{p}}(y)\right|^{p}\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\quad-c_{2}\int_{M_{\lambda r}}\int_{M_{\lambda
r}}|\tau(x)-\tau(y)|^{p}\left(u^{-t+p-1}(x)+u^{-t+p-1}(y)\right)\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\quad-2\int_{M_{\lambda r}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}(u(x)-u(y))^{p-1}\tau^{p}(x)u^{-t}(x){\bf 1}_{\\{u(y)\leq
u(x)\\}}\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&=:c_{1}I_{1}-c_{2}I_{2}-I_{3},\end{split}$

where $c_{1}$ and $c_{2}$ are constants given in Lemma 3.4. By Corollary 2.8,
we obtain

$\begin{split}I_{1}&=\int_{M_{\lambda r}}\int_{M_{\lambda
r}}\left|\tau(x)u^{\frac{-t+p-1}{p}}(x)-\tau(y)u^{\frac{-t+p-1}{p}}(y)\right|^{p}\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\geq C\left\|\tau
u^{\frac{-t+p-1}{p}}\right\|_{L^{p_{\star}}(M_{r})}^{p}-Cr^{-ps_{\max}}\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-s_{k}p}\right)\left\|\tau
u^{\frac{-t+p-1}{p}}\right\|_{L^{p}(M_{\lambda r})}^{p},\end{split}$

where $p_{\star}=\frac{np}{n-p\bar{s}}$.

For $I_{2}$, we use Lemma 2.2 again to have

$\begin{split}I_{2}&=2\int_{M_{\lambda r}}\int_{M_{\lambda
r}}|\tau(x)-\tau(y)|^{p}u^{-t+p-1}(x)\,\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\leq
Cr^{-ps_{\max}}\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-s_{k}p}\right)\left\|u^{-t+p-1}\right\|_{L^{1}(M_{\lambda
r})}.\end{split}$

For $I_{3}$, assuming that $\mathrm{supp}(\tau)\subset M_{(\lambda+1)r/2}$ and
using Lemma 2.2 we deduce

$\begin{split}I_{3}&\leq C\int_{M_{\lambda r}}\int_{\mathbb{R}^{n}\setminus
M_{\lambda
r}}\left(u^{p-1}(x)+u_{-}^{p-1}(y)\right)\tau^{p}(x)u^{-t}(x)\mu(x,\textnormal{d}y)\,\textnormal{d}x\\\
&\leq
Cr^{-ps_{\max}}\left(\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-s_{k}p}\right)\left\|u^{-t+p-1}\right\|_{L^{1}(M_{\lambda
r})}\\\ &\quad+C\epsilon^{1-p}\left(\sup_{x\in
M_{(\lambda+1)r/2}}\int_{\mathbb{R}^{n}\setminus M_{\lambda
r}}u_{-}^{p-1}(y)\mu(x,\textnormal{d}y)\right)\left\|u^{-t+p-1}\right\|_{L^{1}(M_{\lambda
r})}.\end{split}$

Moreover, we estimate

$\begin{split}|(f,-\tau^{p}u^{-t})|&\leq\epsilon^{1-p}\int_{\mathbb{R}^{n}}|f|\tau^{p}u^{-t+p-1}\,\textnormal{d}x\\\
&\leq\epsilon^{1-p}\|f\|_{L^{q/(p\bar{s})}(M_{\lambda
r})}\left\|\tau^{p}u^{-t+p-1}\right\|_{L^{q/(q-p\bar{s})}(M_{\lambda r})}\\\
&=\epsilon^{1-p}\|f\|_{L^{q/(p\bar{s})}(M_{\lambda r})}\left\|\tau
u^{\frac{-t+p-1}{p}}\right\|_{L^{pq/(q-p\bar{s})}(M_{\lambda
r})}^{p}.\end{split}$

Using Lyapunov’s inequality and Young’s inequality, we have

$\|v\|_{pq/(q-p\bar{s})}^{p}\leq\|v\|_{p_{\star}}^{np/q}\|v\|_{p}^{(qp-
np)/q}\leq\frac{n}{q}\omega\|v\|_{p_{\star}}^{p}+\frac{q-n}{q}\omega^{-n/(q-n)}\|v\|_{p}^{p}$

for any $v\in L^{p_{\star}}\cap L^{p}$ and any $\omega>0$. This yields that

$\begin{split}|(f,-\tau^{p}u^{-t})|&\leq\epsilon^{1-p}\|f\|_{L^{q/(p\bar{s})}(M_{\lambda
r})}\left(\frac{n}{q}\omega\left\|\tau
u^{\frac{-t+p-1}{p}}\right\|_{L^{p_{\star}}}^{p}+\frac{q-n}{q}\omega^{-n/(q-n)}\left\|\tau
u^{\frac{-t+p-1}{p}}\right\|_{L^{p}}^{p}\right)\\\ &\leq
r^{-ps_{\max}\frac{q-n}{q}}\left(\frac{n}{q}\omega\left\|\tau^{p}u^{-t+p-1}\right\|_{L^{\gamma}}+\frac{q-n}{q}\omega^{-n/(q-n)}\left\|\tau^{p}u^{-t+p-1}\right\|_{L^{1}}\right).\end{split}$

Combining all the estimates, we have

$\begin{split}&\left\|\tau^{p}u^{-t+p-1}\right\|_{L^{\gamma}(M_{\lambda
r})}\\\ &\leq
Cr^{-ps_{\max}}\left(1+\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-s_{k}p}\right)\left\|u^{-t+p-1}\right\|_{L^{1}(M_{\lambda
r})}\\\
&\quad+Cr^{-ps_{\max}\frac{q-n}{q}}\left(\frac{n}{q}\omega\left\|\tau^{p}u^{-t+p-1}\right\|_{L^{\gamma}(M_{\lambda
r})}+\frac{q-n}{q}\omega^{-n/(q-n)}\left\|\tau^{p}u^{-t+p-1}\right\|_{L^{1}(M_{\lambda
r})}\right).\end{split}$

Taking $\omega=\varepsilon_{0}r^{-ps_{\max}\frac{q-n}{q}}$ with
$\varepsilon_{0}>0$ small enough, we arrive at

$\begin{split}\left\|u^{-1}\right\|_{L^{(t-p+1)\gamma}(M_{r})}^{t-p+1}&\leq\left\|\tau^{p}u^{-t+p-1}\right\|_{L^{\gamma}(M_{\lambda
r})}\\\ &\leq
C\left(1+\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-s_{k}p}\right)r^{-ps_{\max}}\left\|u^{-1}\right\|_{L^{t-p+1}(M_{\lambda
r})}^{t-p+1},\end{split}$

where $C$ depends on $n$, $p$, $p_{\star}$, $t$, $s_{0}$, $q$ and $\Lambda$,
and is bounded when $t$ is bounded away from $p-1$. Since $\lambda\leq 2$, we
obtain

$\sum_{k=1}^{n}\left(\lambda^{s_{\max}/s_{k}}-1\right)^{-s_{k}p}\geq\sum_{k=1}^{n}\left(\lambda^{1/s_{k}}-1\right)^{-s_{k}p}\geq\sum_{k=1}^{n}2^{-p}=n2^{-p},$

from which the desired result follows. ∎

The standard iteration technique proves the following lemma, see [14, 25].

######  Lemma 3.5.

Under the same assumptions as in Lemma 3.3, for any $p_{0}>0$ there is a
constant $C=C(n,p,p_{\star},q,p_{0},s_{0},\Lambda)>0$ such that

(3.5) $\inf_{M_{r}}u\geq
C\left(\fint_{M_{2r}}u(x)^{-p_{0}}\,\mathrm{d}x\right)^{-1/p_{0}}.$

The proof of Theorem 1.4 follows from Theorem 3.2, Lemma 3.5 and the triangle
inequality.

##  4\. Hölder estimates

This section is devoted to the proof of Theorem 1.5. The general scheme for
the derivation of a priori interior Hölder estimates from the weak Harnack
inequality in the non-local setting has been developed in [25] and applied
successfully to the anisotropic setting [14] when $p=2$. We extend the result
presented in [14] to the general case $p>1$.

Recall that the rectangles in Definition 1.1 satisfy the following property.
For $\lambda>0$ and $\Omega\subset\mathbb{R}^{n}$ open, we have

$\mathcal{E}_{\Omega}^{\mu_{\mathrm{axes}}}(u\circ\Psi,v\circ\Psi)=\lambda^{-(n-\bar{s}p)s_{\max}/\bar{s}}\mathcal{E}_{\Psi(\Omega)}^{\mu_{\mathrm{axes}}}(u,v)\quad\text{for
every}~{}u,v\in V^{p,\mu_{\mathrm{axes}}}(\Omega|\mathbb{R}^{n})$

and

$(f\circ\Psi,\varphi\circ\Psi)=\lambda^{-ns_{\max}/\bar{s}}(f,\varphi)\quad\text{for
every}~{}f\in L^{q/(p\bar{s})}(\Omega),~{}\varphi\in
H^{p,\mu_{\mathrm{axes}}}_{\Omega}(\mathbb{R}^{n}),$

where $\Psi:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is a diffeomorphism given by

(4.1) $\Psi(x)=\begin{pmatrix}\lambda^{s_{\max}/s_{1}}&\cdots&0\\\
\vdots&\ddots&\vdots\\\ 0&\cdots&\lambda^{s_{\max}/s_{n}}\end{pmatrix}x.$

The rectangles from Definition 1.1 are balls in a metric space
$(\mathbb{R}^{n},d)$, where the metric
$d:\mathbb{R}^{n}\times\mathbb{R}^{n}\to[0,\infty)$ is defined as follows:

$d(x,y)=\sup_{k\in\\{1,\dots,n\\}}|x_{k}-y_{k}|^{s_{k}/s_{\max}}.$

By the scaling property and covering arguments provided in [14], it is enough
to show the following theorem.

######  Theorem 4.1.

Let $\Lambda\geq 1$ and $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some
$s_{0}\in(0,1)$. Let $1<p<n/\bar{s}$. Assume $f\in L^{q/(p\bar{s})}(M_{1})$
for some $q>n$. There are
$\alpha=\alpha(n,p,p_{\star},q,s_{0},\Lambda)\in(0,1)$ and
$C=C(n,p,p_{\star},q,s_{0},\Lambda)>0$ such that for each
$\mu\in\mathcal{K}(p,s_{0},\Lambda)$ and every $u\in
V^{p,\mu}(M_{1}|\mathbb{R}^{n})$ satisfying

$\mathcal{E}^{\mu}(u,\varphi)=(f,\varphi)\quad\text{for any}~{}\varphi\in
H^{p,\mu}_{M_{1}}(\mathbb{R}^{n}),$

we have $u\in C^{\alpha}$ at $0$ and

$|u(x)-u(0)|\leq
C\left(\|u\|_{L^{\infty}(\mathbb{R}^{n})}+\|f\|_{L^{q/(p\bar{s})}(M_{15/16})}\right)d(x,0)^{\alpha}$

for all $x\in M_{1}$.

###### Proof.

We assume that
$2\|u\|_{L^{\infty}(\mathbb{R}^{n})}+\kappa^{-1}\|f\|_{L^{q/(p\bar{s})}(M_{15/16})}\leq
1$ for some $\kappa>0$ which will be chosen later. It is enough to construct
sequences $\\{a_{k}\\}_{k=0}^{\infty}$ and $\\{b_{k}\\}_{k=0}^{\infty}$ such
that $a_{k}\leq u\leq b_{k}$ in $M_{1/4^{k}}$ and $b_{k}-a_{k}=4^{-\alpha k}$
for some $\alpha>0$. For $k=0$, we set $a_{0}=-1/2$ and $b_{0}=1/2$. Assume
that we have constructed such sequences up to $k$ and let us choose $a_{k+1}$
and $b_{k+1}$.

We assume

(4.2) $|\\{u\geq(b_{k}+a_{k})/2\\}\cap
M_{\frac{1}{2}4^{-k}}|\geq|M_{\frac{1}{2}4^{-k}}|/2,$

and then prove that we can choose $a_{k+1}$ and $b_{k+1}$. If (4.2) does not
hold, then we can consider $-u$ instead of $u$. Let $\Psi$ be the
diffeomorphism given by (4.1) with $\lambda=4^{-k}$ and define

$v(x)=\frac{u(\Psi(x))-a_{k}}{(b_{k}-a_{k})/2}\quad\text{and}\quad
g(x)=\frac{\lambda^{ps_{\max}}f(\Psi(x))}{(b_{k}-a_{k})/2}.$

Then, $v\geq 0$ in $M_{1}$ and
$\mathcal{E}^{\mu}_{M_{1}}(v,\varphi)=(g,\varphi)$ for every $\varphi\in
H^{\mu}_{M_{1}}(\mathbb{R}^{n})$. Moreover, it is easy to see that $v\geq
2(1-4^{\alpha j})$ in $M_{4^{j}}$ for every $j\geq 0$ by induction hypothesis.
By applying Theorem 1.4, we obtain

(4.3)
$\begin{split}&\left(\fint_{M_{1/2}}v^{p_{0}}(x)\,\mathrm{d}x\right)^{1/p_{0}}\\\
&\leq C\inf_{M_{1/4}}v+C\sup_{x\in
M_{15/16}}\left(\int_{\mathbb{R}^{n}\setminus
M_{1}}v^{-}(y)^{p-1}\,\mu(x,\textnormal{d}y)\right)^{1/(p-1)}+\|g\|_{L^{q/(p\bar{s})}(M_{15/16})}.\end{split}$

By taking $\alpha<ps_{0}\frac{q-n}{q}$, we have

(4.4) $\|g\|_{L^{q/(p\bar{s})}(M_{15/16})}=2\cdot 4^{(\alpha-
ps_{\max}\frac{q-n}{q})k}\|f\|_{L^{q/(p\bar{s})}(M_{4^{-k}\cdot 15/16})}\leq
2\kappa.$

For $x\in M_{15/16}$ and for each $j\geq 1$, we have $M_{4^{j}}\setminus
M_{4^{j+1}}\subset\mathbb{R}^{n}\setminus M_{4^{j}}(x)$. Hence, by (1.5)

(4.5) $\begin{split}\int_{\mathbb{R}^{n}\setminus
M_{1}}v^{-}(y)^{p-1}\,\mu(x,\textnormal{d}y)&\leq\sum_{j=1}^{\infty}\int_{M_{4^{j}}\setminus
M_{4^{j+1}}}(2(4^{\alpha j}-1))^{p-1}\,\mu(x,\textnormal{d}y)\\\
&\leq\sum_{j=1}^{\infty}(2(4^{\alpha j}-1))^{p-1}\mu(x,\mathbb{R}^{n}\setminus
M_{4^{j}}(x))\\\ &\leq\sum_{j=1}^{l}\Lambda(2(4^{\alpha
j}-1))^{p-1}4^{-ps_{0}j}+2^{p-1}\Lambda\sum_{j=l+1}^{\infty}4^{(\alpha(p-1)-ps_{0})j}.\end{split}$

If we assume that $\alpha<\frac{ps_{0}}{2(p-1)}$, then we can make the last
term in (4.5) as small as we want by taking $l=l(p,s_{0})$ sufficiently large.
Since the first term in (4.5) converges to 0 as $\alpha\to 0$, we have

(4.6) $C\sup_{x\in M_{15/16}}\left(\int_{\mathbb{R}^{n}\setminus
M_{1}}v^{-}(y)^{p-1}\,\mu(x,\textnormal{d}y)\right)^{1/(p-1)}\leq\kappa$

by assuming further that $\alpha=\alpha(n,p,p_{\star},q,s_{0},\Lambda)$ is
sufficiently small.

On the other hand, it follows from (4.2) that

(4.7)
$\left(\fint_{M_{1/2}}v^{p_{0}}(x)\,\mathrm{d}x\right)^{1/p_{0}}\geq\left(\frac{1}{|M_{1/2}|}\int_{M_{1/2}\cap\\{v\geq
1\\}}v^{p_{0}}(x)\,\textnormal{d}x\right)^{1/p_{0}}\geq 2^{-1/p_{0}}.$

Combining (4.3), (4.4), (4.6), and (4.7), and choosing $\kappa>0$ sufficiently
small, we arrive at $\inf_{M_{1/4}}v\geq\kappa_{0}$ for some $\kappa_{0}$. We
take $a_{k+1}=a_{k}+\kappa_{0}(b_{k}-a_{k})/2$ and $b_{k+1}=b_{k}$, and make
$\alpha$ and $\kappa_{0}$ small so that $1-\kappa_{0}/2=4^{-\alpha}$. Then
$a_{k+1}\leq u\leq b_{k+1}$ in $M_{4^{-(k+1)}}$ and
$b_{k+1}-a_{k-1}=4^{-\alpha(k+1)}$, which finishes the proof. ∎

##  Appendix A Algebraic inequalities

In this section we prove Lemma 3.4 using the series of lemmas below.

######  Lemma A.1.

Let $a,b>0$ and $t>p-1>0$. Then,

$|b-a|^{p-2}(b-a)(a^{-t}-b^{-t})\geq
t\left(\frac{p}{t-p+1}\right)^{p}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p}.$

###### Proof.

We may assume that $b>a$. Let $f(x)=-x^{\frac{-t+p-1}{p}}$ and $g(x)=-x^{-t}$,
then by using Jensen’s inequality we have

$\begin{split}\left|\frac{f(b)-f(a)}{b-a}\right|^{p}&=\left|\fint_{a}^{b}f^{\prime}(x)\,\textnormal{d}x\right|^{p}\leq\fint_{a}^{b}(f^{\prime}(x))^{p}\,\textnormal{d}x\\\
&=\frac{1}{t}\left(\frac{t-p+1}{p}\right)^{p}\fint_{a}^{b}g^{\prime}(x)\,\textnormal{d}x=\frac{1}{t}\left(\frac{t-p+1}{p}\right)^{p}\frac{g(b)-g(a)}{b-a},\end{split}$

which proves the lemma. ∎

######  Lemma A.2.

Let $a,b>0$ and $t>p-1>0$. Then,

$|b-a|^{p-1}\min\\{a^{-t},b^{-t}\\}\leq\left(\frac{p}{t-p+1}\right)^{p-1}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p-1}\min\left\\{a^{\frac{-t+p-1}{p}},b^{\frac{-t+p-1}{p}}\right\\}.$

###### Proof.

We may assume that $b>a$. Let $f(x)=-x^{\frac{-t+p-1}{p}}$, then

$\begin{split}\left|\frac{f(b)-f(a)}{b-a}\right|^{p-1}&=\left|\fint_{a}^{b}f^{\prime}(x)\,\textnormal{d}x\right|^{p-1}=\left(\frac{t-p+1}{p}\right)^{p-1}\left(\fint_{a}^{b}x^{\frac{-t-1}{p}}\,\textnormal{d}x\right)^{p-1}\\\
&\geq\left(\frac{t-p+1}{p}\right)^{p-1}\left(\fint_{a}^{b}b^{\frac{-t-1}{p}}\,\textnormal{d}x\right)^{p-1}=\left(\frac{t-p+1}{p}\right)^{p-1}\frac{b^{-t}}{b^{\frac{-t+p-1}{p}}},\end{split}$

which proves the lemma. ∎

######  Lemma A.3.

Let $\tau_{1},\tau_{2}\in[0,1]$ and $p>1$. Then,

$|\tau_{1}^{p}-\tau_{2}^{p}|\leq
p|\tau_{1}-\tau_{2}|\max\\{\tau_{1}^{p-1},\tau_{2}^{p-1}\\}.$

###### Proof.

The desired inequality follows from the convexity of the function
$f(\tau)=\tau^{p}$. ∎

######  Lemma A.4.

Let $a,b>0$, $\tau_{1},\tau_{2}\in[0,1]$, and $t>p-1>0$. Then,

(A.1)
$\begin{split}&\min\\{\tau_{1}^{p},\tau_{2}^{p}\\}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p}\\\
&\geq
2^{1-p}\left|\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right|^{p}-|\tau_{1}-\tau_{2}|^{p}\max\\{a^{-t+p-1},b^{-t+p-1}\\}\end{split}$

and

(A.2)
$\begin{split}&\max\\{\tau_{1}^{p},\tau_{2}^{p}\\}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p}\\\
&\leq
2^{p-1}\left|\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right|^{p}+2^{p-1}|\tau_{1}-\tau_{2}|^{p}\max\\{a^{-t+p-1},b^{-t+p-1}\\}.\end{split}$

###### Proof.

For (A.1), we assume that $\tau_{1}\geq\tau_{2}$. Then, we obtain from

$\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}=\tau_{2}\left(a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right)+(\tau_{1}-\tau_{2})a^{\frac{-t+p-1}{p}}$

that

$\begin{split}2^{1-p}\left|\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right|^{p}&\leq\tau_{2}^{p}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p}+|\tau_{1}-\tau_{2}|^{p}a^{-t+p-1}\\\
&\leq\tau_{2}^{p}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p}+|\tau_{1}-\tau_{2}|^{p}\max\\{a^{-t+p-1},b^{-t+p-1}\\},\end{split}$

from which (A.1) follows. The other case $\tau_{1}<\tau_{2}$ can be proved in
the same way.

For (A.2), we assume that $\tau_{1}\geq\tau_{2}$. Then

$\begin{split}\max\\{\tau_{1}^{p},\tau_{2}^{p}\\}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p}&=\left|\left(\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right)+(\tau_{2}-\tau_{1})b^{\frac{-t+p-1}{p}}\right|^{p}\\\
&\leq
2^{p-1}\left|\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right|^{p}+2^{p-1}|\tau_{2}-\tau_{1}|^{p}b^{-t+p-1}.\end{split}$

The proof for the case $\tau_{1}<\tau_{2}$ is the same. ∎

###### Proof of Lemma 3.4.

We may assume that $b>a$. We begin with the equality

(A.3)
$\begin{split}&|b-a|^{p-2}(b-a)(\tau_{1}^{p}a^{-t}-\tau_{2}^{p}b^{-t})\\\
&=(b-a)^{p-1}(a^{-t}-b^{-t})\tau_{1}^{p}+(b-a)^{p-1}b^{-t}(\tau_{1}^{p}-\tau_{2}^{p})=:A+B.\end{split}$

By Lemma A.1 and (A.1), we have

(A.4) $\begin{split}A&\geq
t\left(\frac{p}{t-p+1}\right)^{p}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p}\min\\{\tau_{1}^{p},\tau_{2}^{p}\\}\\\
&\geq
t\left(\frac{p}{t-p+1}\right)^{p}\left(2^{1-p}\left|\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right|^{p}-|\tau_{1}-\tau_{2}|^{p}\max\\{a^{-t+p-1},b^{-t+p-1}\\}\right).\end{split}$

For $B$, we use Lemma A.2, Lemma A.3, and Young’s inequality to obtain

$\begin{split}B&\geq-p\left(\frac{p}{t-p+1}\right)^{p-1}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p-1}b^{\frac{-t+p-1}{p}}|\tau_{1}-\tau_{2}|\max\\{\tau_{1}^{p-1},\tau_{2}^{p-1}\\}\\\
&\geq-(p-1)\left(\frac{p}{t-p+1}\right)^{p}\varepsilon^{p/(p-1)}\left|a^{\frac{-t+p-1}{p}}-b^{\frac{-t+p-1}{p}}\right|^{p}\max\\{\tau_{1}^{p},\tau_{2}^{p}\\}-\frac{1}{\varepsilon^{p}}b^{-t+p-1}|\tau_{1}-\tau_{2}|^{p}\end{split}$

for any $\varepsilon>0$. Using (A.2), we have

(A.5)
$\begin{split}B&\geq-2^{p-1}(p-1)\left(\frac{p}{t-p+1}\right)^{p}\varepsilon^{p/(p-1)}\left|\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right|^{p}\\\
&\quad-\left(2^{p-1}(p-1)\left(\frac{p}{t-p+1}\right)^{p}\varepsilon^{p/(p-1)}+\frac{1}{\varepsilon^{p}}\right)|\tau_{1}-\tau_{2}|^{p}\max\\{a^{-t+p-1},b^{-t+p-1}\\}.\end{split}$

Combining (A.3), (A.4), and (A.5), and then taking $\varepsilon$ so that
$2^{p-1}\varepsilon^{p/(p-1)}=2^{1-p}$, we arrive at

$\begin{split}&|b-a|^{p-2}(b-a)(\tau_{1}^{p}a^{-t}-\tau_{2}^{p}b^{-t})\\\
&\geq
c_{1}\left|\tau_{1}a^{\frac{-t+p-1}{p}}-\tau_{2}b^{\frac{-t+p-1}{p}}\right|^{p}-c_{2}|\tau_{1}-\tau_{2}|^{p}\left(a^{-t+p-1}+b^{-t+p-1}\right),\end{split}$

where

$c_{1}=\frac{2^{1-p}p^{p}}{(t-p+1)^{p-1}}\quad\text{and}\quad
c_{2}=\left(t+2^{1-p}(p-1)\right)\left(\frac{p}{t-p+1}\right)^{p}+2^{2(p-1)^{2}}.$

Note that $c_{1}$ and $c_{2}$ are bounded when $t$ is bounded away from $p-1$.
∎

##  Appendix B Anisotropic dyadic rectangles

Let us briefly sketch the construction of anisotropic “dyadic” rectangles.
These objects can be used to prove the lower bound in $L^{p}$ for the sharp
maximal function ${\bf M}^{\sharp}u$.

We construct anisotropic dyadic rectangles having the following properties:

  1. (i)

For each integer $k\in\mathbb{Z}$, a countable collection
$\\{Q_{k,\alpha}\\}_{\alpha}$ covers the whole space $\mathbb{R}^{n}$.

  2. (ii)

Each $Q_{k}$ ($=Q_{k,\alpha}$ for some $\alpha$) has an interior of the form
$\mathrm{Int}(Q_{k})=M_{2^{-k}}(x)$. We call $Q_{k}$ an anisotropic dyadic
rectangle of generation $k$.

  3. (iii)

Every $Q_{k,\alpha}$ is contained in $Q_{k-1,\beta}$ for some $\beta$. We call
$Q_{k-1,\beta}$ a predecessor of $Q_{k,\alpha}$.

  4. (iv)

If $Q_{k,0},\dots,Q_{k,2^{n}}$ are $2^{n}+1$ different anisotropic dyadic
rectangles of generation $k$, then $\cap_{i=0}^{2^{n}}Q_{k,i}=\emptyset$.

  5. (v)

If $2^{n}+1$ different anisotropic dyadic rectangles
$Q_{k_{0}},\dots,Q_{k_{2^{n}}}$, $k_{0}\leq\dots\leq k_{2^{n}}$, have a non-
empty intersection, then $Q_{j}\subset Q_{i}$ for some $0\leq i<j\leq 2^{n}$.

######  Remark B.1.

  1. (1)

A predecessor may not be unique.

  2. (2)

$2^{n}$ different anisotropic dyadic rectangles from the same generation may
have a non-empty intersection.

Such a family of anisotropic dyadic rectangles can be easily constructed.
Since the sets are rectangles, it is sufficient to exemplify the construction
in one dimension. Let $Q_{0}=[0,1)$. Then, a countable collection
$\\{Q_{0}+z\\}_{z\in\mathbb{Z}}$ constitutes the zeroth generation. Let
$N=\lfloor 2^{s_{\max}/s_{1}}\rfloor$. In order to construct the first
generation, we take a disjoint family of (left-closed and right-opened) $N$
intervals in $Q_{0}$ starting from 0 with length $2^{-s_{\max}/s_{1}}$ such
that the following interval starts at the endpoint of the previous interval.
If the right-endpoint of the last interval is 1, then these intervals
constitute the first generation and there is nothing to do. Thus, we assume
from now on that $2^{s_{\max}/s_{1}}\notin\mathbb{Z}$. In this case, we add an
interval $[1-2^{-s_{\max}/s_{i}},1)$ so that

$Q_{0}=\left(\bigcup_{i=0}^{N-1}Q_{1,i}\right)\cup Q_{1,N},$

where

$Q_{1,i}=[i2^{-s_{\max}/s_{1}},(i+1)2^{-s_{\max}/s_{1}})\quad\text{for}~{}i=0,\dots,N-1,\quad
Q_{1,N}=[1-2^{-s_{\max}/s_{i}},1),$

and $N=\lfloor 2^{s_{\max}/s_{1}}\rfloor$. Then, the collection
$\\{Q_{1,i}+z\\}_{0\leq i\leq N,z\in\mathbb{Z}}$ forms the first generation of
intervals satisfying (i)-(iv).

Figure 1. This figure shows the construction of the family $Q_{1,i}$.

We continue to construct the intervals of generation 2 that fill in $Q_{1,i}$
for each $0\leq i\leq N-2$. However, we have to be careful in filling in
$Q_{1,N-1}$ and $Q_{1,N}$ since $Q_{1,N-1}\cap Q_{1,N}\neq\emptyset$. Suppose
that we filled in $Q_{1,N-1}$ and $Q_{1,N}$ as above, i.e.,

$Q_{1,N-1}=\left(\bigcup_{i=0}^{N-1}Q_{2,i}\right)\cup
Q_{2,N}\quad\text{and}\quad
Q_{1,N}=\left(\bigcup_{i=0}^{N-1}\tilde{Q}_{2,i}\right)\cup\tilde{Q}_{2,N}$

for some intervals $Q_{2,i}$ and $\tilde{Q}_{2,i}$, $0\leq i\leq N$, of length
$4^{-s_{\max}/s_{1}}$. Let $K$ be the smallest integer such that
$\overline{Q_{2,K}}\cap Q_{1,N}\neq\emptyset$. Then, we have

$Q_{1,N-1}\cup
Q_{1,N}=\left(\bigcup_{i=0}^{K}Q_{2,i}\right)\cup\left(\bigcup_{i=0}^{N-1}\tilde{Q}_{2,i}\right)\cup\tilde{Q}_{2,N}$

and at most two different intervals among
$\\{Q_{2,0},\dots,Q_{2,K},\tilde{Q}_{2,0},\dots,\tilde{Q}_{2,N}\\}$ can
intersect. Therefore, these intervals constitute the second generation
satisfying (i)-(iv).

Figure 2. This figure shows the construction of generation 2.

In this way, we construct intervals of generation $k$ for all $k\geq 0$.

Let us now construct intervals of generation $k<0$. It is easy to observe that
a collection $\\{Q_{-1}+Nz\\}_{z\in\mathbb{Z}}$ of intervals of generation
$-1$ satisfies (i)-(iv), where

$Q_{-1}=[0,2^{s_{\max}/s_{1}}).$

For the generation of $-2$, let $K$ be the largest integer such that
$Q_{-1}+NK\subset Q_{-2}$, where $Q_{-2}=[0,4^{s_{\max}/s_{1}})$. Then, the
intervals $Q_{-2}+N(K+1)z$, $z\in\mathbb{Z}$, form the generation of $-2$,
which satisfying (i)-(iv). We continue this process to construct intervals of
all generations $k<0$.

We show that the intervals constructed in this way satisfy the property (v) as
well. Suppose that three different intervals $Q_{k_{0}},Q_{k_{1}}$, and
$Q_{k_{2}}$, $k_{0}\leq k_{1}\leq k_{2}$, have a non-empty intersection. If
$k_{1}=k_{2}$, then $Q_{k_{1}}\subset Q_{k_{0}}$ or $Q_{k_{2}}\subset
Q_{k_{0}}$. If $k_{1}<k_{2}$, then either $Q_{k_{2}}\subset Q_{k_{1}}$ or not.
In the former case, we are done. In the latter case,
$Q_{k_{2}}\subset\tilde{Q}_{k_{1}}$ for some $\tilde{Q}_{k_{1}}\neq
Q_{k_{1}}$, which reduces to the case $k_{1}=k_{2}$.

##  Appendix C Sharp maximal function theorem

In this section we prove Theorem 2.6 by using the anisotropic dyadic
rectangles. For $u\in L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})$, we define a
dyadic maximal function ${\bf M}_{d}u$ by

${\bf M}_{d}u(x)=\sup_{x\in Q}\fint_{Q}|u(y)|\,\mathrm{d}y,$

where the supremum is taken over all anisotropic dyadic rectangles $Q$. Since

(C.1) ${\bf M}_{d}u\leq{\bf M}u,$

Theorem 2.5 also holds for the dyadic maximal function ${\bf M}_{d}u$. We
first prove a good-lambda estimate using the dyadic maximal function. See [31,
Theorem 3.4.4].

######  Theorem C.1.

Let $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some $s_{0}\in(0,1)$. There
exists a constant $C=C(n,s_{0})>0$ such that

$|\\{x\in\mathbb{R}^{n}:{\bf M}_{d}u(x)>2\lambda,{\bf
M}^{\sharp}u(x)\leq\gamma\lambda\\}|\leq C\gamma|\\{x\in\mathbb{R}^{n}:{\bf
M}_{d}u(x)>\lambda\\}|$

for all $\gamma>0$, $\lambda>0$, and $u\in
L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})$.

###### Proof.

Let $\Omega_{\lambda}=\\{x\in\mathbb{R}^{n}:M_{d}(f)(x)>\lambda\\}$. We may
assume that $|\Omega_{\lambda}|<+\infty$ since otherwise there is nothing to
prove. For each $x\in\Omega_{\lambda}$, we find a maximal anisotropic dyadic
rectangle $Q^{x}$ such that

(C.2) $x\in
Q^{x}\subset\Omega_{\lambda}\quad\text{and}\quad\fint_{Q^{x}}|f|>\lambda.$

There are at most $2^{n}$ different maximal anisotropic dyadic rectangles of
the same generation satisfying (C.2), but we can still choose anyone of them.
Let $Q_{j}$ be the collection of all such rectangles $Q^{x}$ for all
$x\in\Omega_{\lambda}$. Then, we have $\Omega_{\lambda}=\cup_{j}Q_{j}$. Note
that different rectangles $Q_{j}$ may have an intersection, but the
intersection is contained in at most $2^{n}$ different maximal rectangles of
the same generation. This is a consequence of the properties (iv) and (v) of
anisotropic dyadic rectangles. Hence,

$\sum_{j}|Q_{j}|\leq 2^{n}|\Omega_{\lambda}|.$

Therefore, the desired result follows once we have

(C.3) $|\\{x\in Q_{j}:{\bf M}_{d}u(x)>2\lambda,{\bf
M}^{\sharp}u(x)\leq\gamma\lambda\\}|\leq C\gamma|Q_{j}|$

for some $C=C(n,s_{0})$. Indeed, one can prove (C.3) by following the second
paragraph of the proof of [31, Theorem 3.4.4], using Theorem 2.5 for ${\bf
M}_{d}$, and replacing [31, Equation (3.4.8)] by

$\frac{1}{\lambda}\int_{Q_{j}}|u(y)-(u)_{Q_{j}^{\prime}}|\,\mathrm{d}y\leq\frac{2^{ns_{\max}/\bar{s}}}{\lambda}\frac{|Q_{j}|}{|Q_{j}^{\prime}|}\int_{Q_{j}^{\prime}}|u(y)-(u)_{Q_{j}^{\prime}}|\,\mathrm{d}y\leq\frac{2^{n/s_{0}}}{\lambda}|Q_{j}|{\bf
M}^{\sharp}u(\xi_{j})$

for all $\xi_{j}\in Q_{j}$, where $Q_{j}^{\prime}$ is anyone of predecessors
of $Q_{j}$. ∎

######  Theorem C.2.

Let $s_{1},\dots,s_{n}\in[s_{0},1)$ be given for some $s_{0}\in(0,1)$, and let
$0<p_{0}\leq p<\infty$. Then, there is a constant $C=C(n,p,s_{0})>0$ such that
for all functions $u\in L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})$ with ${\bf
M}_{d}u\in L^{p_{0}}(\mathbb{R}^{n})$ we have

$\|{\bf M}_{d}u\|_{L^{p}(\mathbb{R}^{n})}\leq C\|{\bf
M}^{\sharp}u\|_{L^{p}(\mathbb{R}^{n})}.$

Theorem C.2 can be proved in the same way as in the proof of [31, Theorem
3.4.5] except that we use Theorem C.1 instead of [31, Theorem 3.4.4]. Finally,
we combine the inequality

$\|u\|_{L^{p}(\mathbb{R}^{n})}\leq\|{\bf M}_{d}u\|_{L^{p}(\mathbb{R}^{n})},$

which comes from the Lebesgue differentiation theorem, and Theorem C.2 to
conclude Theorem 2.6. See [31, Corollary 3.4.6].

##  Appendix D Pointwise convergence of the fractional orthotropic
$p$-Laplacian

This section provides the proof of pointwise convergence of the fractional
orthotropic $p$-Laplacian as $s\nearrow 1$.

######  Proposition D.1.

Let $u\in C^{2}(\mathbb{R}^{n})\cap L^{\infty}(\mathbb{R}^{n})$ and
$x\in\mathbb{R}^{n}$ be such that $\partial_{i}u(x)\neq 0$ for all
$i=1,\dots,n$. Let $s_{i}=s$ for all $i=1,\dots,n$. Let $L$ be the operator in
(1.2) with $\mu=\mu_{\mathrm{axes}}$ and $A^{p}_{\mathrm{loc}}$ be as in
(1.3). Then, $Lu(x)\to A^{p}_{\mathrm{loc}}u(x)$ as $s\nearrow 1$ up to a
constant.

###### Proof.

Let us fix a point $x\in\mathbb{R}^{n}$ with $\partial_{i}u(x)\neq 0$. For
each $i=1,\dots,n$, let us define $u_{i}:\mathbb{R}\to\mathbb{R}$ by
$u_{i}(x_{i})=u(x_{1},\dots,x_{i},\dots,x_{n})$ as a function of one variable.
Then $u_{i}\in C^{2}(\mathbb{R})\cap L^{\infty}(\mathbb{R})$ and
$u_{i}^{\prime}(x_{i})\neq 0$. We write

$\begin{split}Lu(x)=\sum_{i=1}^{n}s(1-s)\int_{\mathbb{R}}\frac{|u_{i}(y_{i})-u(x_{i})|^{p-2}(u_{i}(y_{i})-u_{i}(x_{i}))}{|x_{i}-y_{i}|^{1+sp}}\,\textnormal{d}y_{i}=-\sum_{i=1}^{n}(-\partial^{2})^{s}_{p}u_{i}(x_{i}),\end{split}$

which is the sum of one-dimensional fractional $p$-Laplacians. By [11, Theorem
2.8], we have

$-(-\partial^{2})^{s}_{p}u_{i}(x_{i})\to\frac{\textnormal{d}}{\textnormal{d}x_{i}}\left(\left|\frac{\textnormal{d}u_{i}}{\textnormal{d}x_{i}}(x_{i})\right|^{p-2}\frac{\textnormal{d}u_{i}}{\textnormal{d}x_{i}}(x_{i})\right)$

as $s\nearrow 1$, for each $i=1,\dots,n$, up to a constant depending on $p$
only. Consequently, by summing up $Lu(x)\to A^{p}_{\mathrm{loc}}u(x)$ as
$s\nearrow 1$. ∎

### Conflict of Interest

Authors state no conflict of interest

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