#  $p$-Laplacian operator with potential in generalized Morrey Spaces

René Erlin Castillo1 Departamento de Matemáticas
Universidad Nacional de Colombia,
Bogotá, Colombia<EMAIL_ADDRESS>and  Héctor Camilo Chaparro2
Programa de Matemáticas
Universidad de Cartagena,
Cartagena de Indias, Colombia<EMAIL_ADDRESS>
###### Abstract.

We study some basic properties of generalized Morrey spaces
$\mathcal{M}^{p,\phi}(\mathds{R}^{d})$. Also, the problem $-\mbox{div}(|\nabla
u|^{p-2}\nabla u)+V|u|^{p-2}u=0$ in $\Omega$, where $\Omega$ is a bounded open
set in $\mathds{R}^{d}$, and potential $V$ is assumed to be not equivalent to
zero and lies in $\mathcal{M}^{p,\phi}(\Omega)$, is studied. Finally, we
establish the strong unique continuation for the $p$-Laplace operator in the
case $V\in\mathcal{M}^{p,\phi}(\mathds{R}^{d})$.

###### Key words and phrases:

$p$-Laplacian, generalized Morrey spaces, Strong unique continuation.
2020 Mathematics Subject Classification: 35J05, 46E30, 26D10.

##  1\. Introduction

The so called Morrey space were introduced in 1938 (see [10]) in relation to
regularity problems of solution to partial differential equations.

Although used quite often in problems related to PDE’s, the awareness of such
spaces among mathematicians is not so widespread as in the case of Lebesgue or
Sobolev spaces. Detailed information about Morrey spaces may be found in [1,
11].

The Morrey space $\mathcal{M}^{p,\lambda}(\Omega)$ is defined as

$\mathcal{M}^{p,\lambda}(\Omega)=\left\\{f\in
L_{p}(\Omega):\sup_{\begin{subarray}{c}\lambda\in\Omega\\\
r>0\end{subarray}}\frac{1}{r^{\lambda}}\int\limits_{B(x,r)}|f(y)|^{p}\,dy<\infty\right\\}$

where $1\leq p<\infty$, $\lambda\geq 0$, $\Omega\subset\mathds{R}^{d}$ is a
bounded open set and $B(x,r)$ denotes the open ball centered at $x$ with
radius $r$, i.e. $B(x,r)=\\{y\in\Omega:|y-x|<r\\}$.

This is a Banach space with respect to the norm

$\|f\|_{\mathcal{M}^{p,\lambda}(\Omega)}=\sup_{\begin{subarray}{c}x\in\Omega\\\
r\in(0,\operatorname{diam}(\Omega))\end{subarray}}\left(\frac{1}{r^{\lambda}}\int\limits_{B(x,r)}|f(y)|^{p}\,dy\right)^{\frac{1}{p}}.$

We will explore some properties of spaces $\mathcal{M}^{p,\lambda}(\Omega)$,
aiming to establish some relations with the Fefferman’s inequality, the
Poisson equation, the $p$-Laplacian and the unique continuation principle. The
last is fundamental and of independent interest in the theory of partial
differential equations. Application regarding the vanishing of solutions are
often associated for instance with solvability, stability, geometrical
properties of solutions, and so on.

We are concerned with the following problem

$-\mbox{div}(|\nabla u|^{p-2}\nabla u)+V|u|^{p-2}u=0\quad\mbox{in $\Omega$}$
(1.1)

and the potential $V$ is assumed to be not equivalent to zero and lies in
$\mathcal{M}^{p,\phi}(\Omega)$ for $1<p<\infty$ and $\phi$ is an almost
decreasing function which satisfies the doubling condition.

Specifically, we are interested in studing a family of functions which enjoys
the strong unique continuation property, that is, functions besides the
possible zero functions which have zero infinite order.

By $L_{p,loc}(\Omega)$ we denoted the set of functions $u$ such that

$\int\limits_{K}|u(x)|^{p}\,dx<\infty$

for all compact subset $K\subset\Omega$.

######  Definition 1.1.

We say tha a function $u\in L_{p,loc}(\Omega)$ vanishes of infinite order at
point $x_{0}$ if for any natural number $N$ there exists a constant $C_{N}$
such that

$\int\limits_{B(x_{0},r)}|u(x)|^{p}\,dx\leq C_{N}r^{N}$ (1.2)

for all $n\in\mathds{N}$ and for small positive number $r$.

######  Definition 1.2.

We say that (1.1) has strong unique continuation property if and only if any
solution $u$ of (1.1) in $\Omega$ is identically zero in $\Omega$ provided
that $u$ vanishes of infinite order that a point $x_{0}$ in $\Omega$.

There is an extensive literature on unique continuation property. We refer the
reader to the work of Zamboni on unique continuation for nonnegative solutions
of quasilinear elliptic equation [12], also the work of Jerison-Kenig on
unique continuation for Schrödinger operator [9]. The same work is done by
Chiarenza and Frasca but for linear elliptic operator in the case where $V\in
L^{\frac{n}{2}}$ when $n>2$ [2].

This paper is organized as follows. In section 2, we study a generalization of
Morrey space. Some basic results, such as embeddings, completeness, etc. are
established. In section 3, we shall investigate the relationship between
Fefferman’s inequality and the Poisson equation. Finally, section 4 is devoted
to investigate the strong unique continuation property of the so called
$p$-Laplacian. The $p$-Laplacian has been much studied during the last sixty
eight years, although its theory is by now rather developed, some challenging
problems remain unsolved.

##  2\. Generalized Morrey Space

Generalized Morrey space $\mathcal{M}^{p,\phi}(\mathds{R}^{d})$ are equipped
with the parameter, that is, $1\leq p<\infty$, and the function
$\phi:(0,\infty)\longrightarrow(0,\infty)$. We assume that $\phi$ is in
$\mathscr{G}_{p}$, which is the set of all functions
$\phi:(0,\infty)\longrightarrow(0,\infty)$ such that $\phi$ is almost
decreasing, that is $r\leq s$ implies $\phi(r)\geq C\phi(s)$ and $t\mapsto
t^{\frac{n}{p}}$ is almost increasing, that is $r\leq s$ implies
$r^{\frac{n}{p}}\phi(r)\leq Cs^{\frac{n}{p}}\phi(s)$ for some $C>0$. Note
that, $\phi\in\mathscr{G}_{p}$ implies that $\phi$ satisfies the doubling
condition, that is, there exists $C>0$ such that

$\frac{1}{C}\leq\frac{\phi(r)}{\phi(s)}\leq C$

for every $r$ and $s$ with $\frac{1}{2}\leq\frac{r}{s}\leq 2$.

######  Definition 2.1 (Generalized Morrey space).

The generalized Morrey Space
$\mathcal{M}^{p,\phi}(\mathds{R}^{d})=\mathcal{M}^{p,\phi}$ is defined as

$\mathcal{M}^{p,\phi}(\mathds{R}^{d})=\\{f\in
L_{p}(\mathds{R}^{d}):\|f\|_{\mathcal{M}^{p,\phi}}<\infty\\}$

with

$\|f\|_{\mathcal{M}^{p,\phi}(\mathds{R}^{d})}=\sup_{B(a,r)}\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}|f(x)|^{p}\right)$
(2.1)

where $B(a,r)=\\{x\in\mathds{R}^{d}:|x-a|<r\\}$.

Observe that if $1\leq p<\infty$ and $\phi(r)=r^{\frac{\lambda-n}{p}}$ for
$\lambda>0$, then $\mathcal{M}^{p,\phi}=\mathcal{M}^{p,\lambda}$ which means
that we recovered the space defined in (2.1). Now it is propitious to point
out that $\mathds{R}^{d}$ is equipped with a Borel measure $\mu$ satisfying
the growth condition of order $n$, with $0<n\leq d$, that is, there exist a
constant $C>0$ such that

$\mu(B(a,r))\leq Cr^{n}$

for every ball $B(a,r)$ centered at $a\in\mathds{R}^{d}$ with radius $r>0$.

The following lemma indicates that the characteristic function on some balls
is contained in generalized Morrey spaces.

######  Lemma 2.1.

Let $1\leq p<\infty$ and $\phi\in\mathscr{G}_{p}$. If $B_{0}=B(0,r_{0})$ for
some $r_{0}>0$. Then, there exist $C>1$ and $B>0$ such that

$\frac{B}{\phi(r_{0})}\leq\|\chi_{B_{0}}\|_{\mathcal{M}^{p,\phi}}\leq\frac{C}{\phi(r_{0})}.$

###### Proof.

By definition of $\|\cdot\|_{\mathcal{M}^{p,\phi}}$, we have

$\displaystyle\|\chi_{B_{0}}\|_{\mathcal{M}^{p,\phi}}=$
$\displaystyle\sup_{B(a,r)}\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}|\chi_{B_{0}}(x)|^{p}\,d\mu\right)^{\frac{1}{p}}$
$\displaystyle\geq$
$\displaystyle\frac{1}{\phi(r_{0})}\left(\frac{\mu(B_{0})}{r_{0}^{n}}\right)^{\frac{1}{p}}$
$\displaystyle=$ $\displaystyle\frac{B}{\phi(r_{0})}$

where $B=\left(\frac{\mu(B_{0})}{r_{0}^{n}}\right)^{\frac{1}{p}}$, proving the
first inequality.

For the second inequality, we separate the proof into two cases:

First case $r\leq r_{0}$, for this case, we have

$\phi(r)\geq C\phi(r_{0}).$

Thus

$\displaystyle\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}|\chi_{B_{0}}(x)|^{p}\,d\mu\right)^{\frac{1}{p}}\leq$
$\displaystyle\frac{C}{\phi(r_{0})}\left(\frac{\mu(B(a,r)\cap
B_{0})}{r^{n}}\right)^{\frac{1}{p}}$ $\displaystyle\leq$
$\displaystyle\frac{C}{\phi(r_{0})}.$

Second case $r\geq r_{0}$ Since

$r_{0}^{\frac{n}{p}}\phi(r_{0})\leq Cr^{\frac{n}{p}}\phi(r),$

we have

$\displaystyle\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}|\chi_{B_{0}}(x)|^{p}\,d\mu\right)^{\frac{1}{p}}\leq$
$\displaystyle Cr^{\frac{n}{p}}r_{0}^{-\frac{n}{p}}\left(\frac{\mu(B(a,r)\cap
B_{0})}{r^{n}}\right)^{\frac{1}{p}}$ $\displaystyle\leq$
$\displaystyle\frac{Cr^{\frac{n}{p}}r_{0}^{-\frac{n}{p}}}{\phi(r_{0})}\left(\frac{\mu(B_{0})}{r^{n}}\right)^{\frac{1}{p}}$
$\displaystyle\leq$ $\displaystyle\frac{C}{\phi(r_{0})}.$

From these two cases, we can conclude that

$\|\chi_{B_{0}}\|_{\mathcal{M}^{p,\phi}}\leq\frac{C}{\phi(r_{0})}.$

Thus we have proved the inequality. ∎

The following lemma shows us some conditions on the functions
$\phi_{1},\phi_{2}$ for which we can compare the spaces
$\mathcal{M}^{p_{1},\phi_{1}}$ and $\mathcal{M}^{p_{2},\phi_{2}}$. The result
is as follows.

######  Lemma 2.2.

Let $1\leq p_{1}\leq p_{2}<\infty$, $\phi_{1},\phi_{2}\in\mathscr{G}_{p}$.
Then the following statements are equivalent:

  1. (I)

$\phi_{2}\leq C\phi_{1}$

  2. (II)

$\mathcal{M}^{p_{2},\phi_{2}}\subset\mathcal{M}^{p_{1},\phi_{1}}$ with

$\|f\|_{\mathcal{M}^{p_{1},\phi_{1}}}\leq
C\|f\|_{\mathcal{M}^{p_{2},\phi_{2}}}$

for every $f\in\mathcal{M}^{p_{2},\phi_{2}}$.

###### Proof.

Assume that (I) holds and let $f\in\mathcal{M}^{p_{2},\phi_{2}}$ hence by
Hölder’s inequality we have

$\displaystyle\frac{1}{\phi_{1}(r)}\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}|f(x)|^{p_{1}}\right)^{\frac{1}{p_{1}}}$
$\displaystyle\phantom{\frac{1}{\phi_{1}(r)}}\leq\frac{C}{\phi_{2}(r)}\left[\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}(|f(x)|^{p_{1}})^{\frac{p_{2}}{p_{1}}}\,d\mu\right)^{\frac{p_{1}}{p_{2}}}\left(\int\limits_{B(a,r)}d\mu\right)^{1-\frac{p_{1}}{p_{2}}}\right]^{\frac{1}{p_{1}}}$
$\displaystyle\phantom{\frac{1}{\phi_{1}(r)}}\leq\frac{C}{\phi_{2}(r)}\left[\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}|f(x)|^{p_{2}}\,d\mu\right)^{\frac{p_{1}}{p_{2}}}\left(\mu(B(a,r))\right)^{1-\frac{p_{1}}{p_{2}}}\right]^{\frac{1}{p_{1}}}$
$\displaystyle\phantom{\frac{1}{\phi_{1}(r)}}\leq\frac{C}{\phi_{2}(r)}\left[\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}|f(x)|^{p_{2}}\,d\mu\right)^{\frac{p_{1}}{p_{2}}}\left(r^{n}\right)^{1-\frac{p_{1}}{p_{2}}}\right]^{\frac{1}{p_{1}}}$
$\displaystyle\phantom{\frac{1}{\phi_{1}(r)}}\leq\frac{C}{\phi_{2}(r)}\left(\frac{1}{r^{n}}\int\limits_{B(a,r)}|f(x)|^{p_{2}}\,d\mu\right)^{\frac{1}{p_{2}}}$
$\displaystyle\phantom{\frac{1}{\phi_{1}(r)}}\leq
C\|f\|_{\mathcal{M}^{p_{2},\phi_{2}}}.$

By taking the supremum of the left side for every $a$ and $r$ we obtain

$\|f\|_{\mathcal{M}^{p_{1},\phi_{1}}}\leq
C\|f\|_{\mathcal{M}^{p_{2},\phi_{2}}}.$

Now, assume that (II) hods. Let $B_{0}=B(0,r_{0})$ as before, then

$\|\chi_{B_{0}}\|_{\mathcal{M}^{p_{1},\phi_{1}}}\leq
C\|\chi_{B_{0}}\|_{\mathcal{M}^{p_{2},\phi_{2}}}.$ (2.2)

By Lemma 2.1 we have

$\frac{1}{\phi_{1}(r_{0})}\leq\|\chi_{B_{0}}\|_{\mathcal{M}^{p_{1},\phi_{1}}}$
(2.3)

and

$\|\chi_{B_{0}}\|_{\mathcal{M}^{p_{1},\phi_{1}}}\leq\frac{C}{\phi_{2}(r_{0})}.$
(2.4)

By (2.2), (2.3) and (2.4) we arrive at

$\frac{1}{\phi_{1}(r_{0})}\leq\frac{C}{\phi_{2}(r_{0})},$

and so

$\phi_{2}(r_{0})\leq C\phi_{1}(r_{0}).$

Since $r_{0}$ is any positive real number, we have

$\phi_{2}\leq\phi_{1}.$

∎

The next result gives us a Hölder’s type inequality.

######  Theorem 2.1.

Let $f\in\mathcal{M}^{p,\phi}(\mathds{R}^{d})$ and
$g\in\mathcal{M}^{q,\phi}(\mathds{R}^{d})$. Then

$\|fg\|_{\mathcal{M}^{1,\phi}}\leq\phi(r)\|f\|_{\mathcal{M}^{p,\phi}}\|g\|_{\mathcal{M}^{q,\phi}}$

with $\frac{1}{p}+\frac{1}{q}=1$.

###### Proof.

By Hölder’s inequality we have

$\displaystyle\frac{1}{\phi(r)r^{n}}\int\limits_{B}|f(x)g(x)|\,d\mu=\frac{\phi(r)}{[\phi(r)]^{2}}\int\limits_{B}\left|\frac{f(x)}{r^{\frac{n}{p}}}\right|\left|\frac{g(x)}{r^{\frac{n}{q}}}\right|\,d\mu$
$\displaystyle\hskip
14.22636pt\leq\phi(r)\left(\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|f(x)|^{p}\,d\mu\right)^{\frac{1}{p}}\left(\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|g(x)|{q}\,d\mu\right)^{\frac{1}{q}}\right)\right)$
$\displaystyle\hskip
14.22636pt\leq\phi(r)\sup_{B}\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|f(x)|^{p}\,d\mu\right)^{\frac{1}{p}}\sup_{B}\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|g(x)|^{q}\,d\mu\right)^{\frac{1}{q}}$
$\displaystyle\hskip
14.22636pt=\phi(r)\|f\|_{\mathcal{M}^{p,\phi}}\|g\|_{\mathcal{M}^{q,\phi}}.$

And so

$\sup_{B}\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|f(x)g(x)|\,d\mu\right)\leq\phi(r)\|f\|_{\mathcal{M}^{p,\phi}}\|g\|_{\mathcal{M}^{q,\phi}}.$

That is

$\|fg\|_{\mathcal{M}^{1,\phi}}\leq\phi(r)\|f\|_{\mathcal{M}^{p,\phi}}\|g\|_{\mathcal{M}^{q,\phi}},$

and so the theorem is proved. ∎

In the coming result we derive the Minkowski inequality, which in this case is
independent of Theorem 2.1.

######  Theorem 2.2.

If $f$ and $g$ belong to $\mathcal{M}^{p,\phi}(\mathds{R}^{d})$, then $f+g$
belongs to $\mathcal{M}^{p,\phi}(\mathds{R}^{d})$. Moreover,

$\|f+g\|_{\mathcal{M}^{p,\phi}}\leq\|f\|_{\mathcal{M}^{p,\phi}}+\|g\|_{\mathcal{M}^{p,\phi}}.$

###### Proof.

Since $f$ and $g$ belong to $\mathcal{M}^{p,\phi}(\mathds{R}^{d})$ by
definition $f$ and $g$ belong to $L_{p}(\mathds{R}^{d})$. Hence the Minkowski
inequality holds. So

$\left(\int\limits_{B}|f+g|^{p}\,d\mu\right)^{\frac{1}{p}}\leq\left(\int\limits_{B}|f|^{p}\,d\mu\right)^{\frac{1}{p}}+\left(\int\limits_{B}|g|^{p}\,d\mu\right)^{\frac{1}{p}}.$

Thus

$\displaystyle\frac{1}{\phi(r)r^{\frac{n}{p}}}\left(\int\limits_{B}|f+g|^{p}\,d\mu\right)^{\frac{1}{p}}$
$\displaystyle\phantom{\frac{1}{\phi(r)r^{\frac{n}{p}}}}\leq\frac{1}{\phi(r)r^{\frac{n}{p}}}\left(\int\limits_{B}|f|^{p}\,d\mu\right)^{\frac{1}{p}}+\frac{1}{\phi(r)r^{\frac{n}{p}}}\left(\int\limits_{B}|g|^{p}\,d\mu\right)^{\frac{1}{p}}$
$\displaystyle\phantom{\frac{1}{\phi(r)r^{\frac{n}{p}}}}=\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|f|^{p}\,d\mu\right)^{\frac{1}{p}}+\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|g|^{p}\,d\mu\right)^{\frac{1}{p}}$
$\displaystyle\phantom{\frac{1}{\phi(r)r^{\frac{n}{p}}}}\leq\sup_{B}\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|f|^{p}\,d\mu\right)^{\frac{1}{p}}+\sup_{B}\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|g|^{p}\,d\mu\right)^{\frac{1}{p}}$
$\displaystyle\phantom{\frac{1}{\phi(r)r^{\frac{n}{p}}}}=\|f\|_{\mathcal{M}^{p,\phi}}+\|g\|_{\mathcal{M}^{p,\phi}}.$

Finally

$\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|f+g|^{p}\,d\mu\right)^{\frac{1}{p}}\leq\|f\|_{\mathcal{M}^{p,\phi}}+\|g\|_{\mathcal{M}^{p,\phi}}$

holds for any ball $B$. Therefore

$\sup_{B}\frac{1}{\phi(r)}\left(\frac{1}{r^{n}}\int\limits_{B}|f+g|^{p}\,d\mu\right)^{\frac{1}{p}}\leq\|f\|_{\mathcal{M}^{p,\phi}}+\|g\|_{\mathcal{M}^{p,\phi}}$

and so

$\|f+g\|_{\mathcal{M}^{p,\phi}}\leq\|f\|_{\mathcal{M}^{p,\phi}}+\|g\|_{\mathcal{M}^{p,\phi}}.$

∎

Now, we are ready to prove the completeness of $\mathcal{M}^{p,\phi}$ space.

######  Theorem 2.3.

$\mathcal{M}^{p,\phi}$, equipped with the norm (2.1), is a Banach space.

###### Proof.

Let $\\{f_{n}\\}_{n\in\mathds{N}}$ be a Cauchy sequence in
$\mathcal{M}^{p,\phi}(\mathds{R}^{d})$, since
$\mathcal{M}^{p,\phi}(\mathds{R}^{d})\subset L_{p}(\mathds{R}^{d})$, then
$\\{f_{n}\\}_{n\in\mathds{N}}$ is a Cauchy sequence in
$L_{p}(\mathds{R}^{d})$, therefore there exists an $f\in
L_{p}(\mathds{R}^{d})$ such that $f_{n}\to f$ in $L_{p}(\mathds{R}^{d})$.

Thus for $\epsilon>0$ and $r>0$ there exist $n_{r}>0$ such that

$\|f-f_{n}\|_{L_{p}}<\phi(r)r^{\frac{n}{p}}\epsilon\quad\mbox{if}\quad n\geq
n_{r},$

from this latter inequality we arrived at

$\|f-f_{n}\|_{\mathcal{M}^{p,\phi}}<\epsilon\quad\mbox{if}\quad n\geq n_{r}.$
(2.5)

Applying Theorem 2.2 and (2.5) we get

$\displaystyle\|f\|_{\mathcal{M}^{p,\phi}}\leq$
$\displaystyle\|f-f_{n}\|_{\mathcal{M}^{p,\phi}}+\|f_{n_{r}}\|_{\mathcal{M}^{p,\phi}}$
$\displaystyle<$ $\displaystyle\epsilon+\|f_{r_{n}}\|\leq C.$

And so $f\in\mathcal{M}^{p,\phi}(\mathds{R}^{d})$ which end the proof of
Theorem 2.3 ∎

We end this section by proving a Hedberg-type inequality. Before doing so, let
us recall two well-known definitions.

The first one is the maximal operator. Given a function $f\in
L_{1,loc}(\mathds{R}^{d})$ the Hardy-Littlewood maximal function for
$x\in\mathds{R}^{d}$ it is defined as

$Mf(x)=\sup_{r>0}\frac{1}{\mu(B(x,r))}\int\limits_{B(x,r)}|f(y)|\,d\mu.$ (2.6)

The second one is the so-called Riesz potential operator which is defined as

$I_{\alpha}f(x)=\int\limits_{\mathds{R}^{d}}\frac{f(y)}{|x-y|^{n-\alpha}}$
(2.7)

for $0\leq\alpha<n\leq d$. Now, we are ready to establish our announced
result.

######  Theorem 2.4.

Suppose that for some $0\leq\lambda<n-1$ we have

$\int_{r}^{\infty}\phi(t)\,dt\leq Cr^{\lambda+1-n},\quad r>0.$

Then, for any $f\in\mathcal{M}^{p,\phi}$, we have the following pointwise
inequality

$|I_{1}f(x)|\leq
C|Mf(x)|^{1-\frac{1}{n-\lambda}}\|f\|_{\mathcal{M}^{p,\phi}}^{\frac{1}{n-\lambda}}$

for $x\in\mathds{R}^{d}$.

###### Proof.

Let $f\in\mathcal{M}^{p,\phi}$ and $x\in\mathds{R}^{d}$. For every $r>0$ we
have

$\displaystyle|I_{1}f(x)|\leq$
$\displaystyle\int\limits_{|x-y|<r}\frac{|f(y)|d\mu}{|x-y|^{n-1}}+\int\limits_{|x-y|\geq
r}\frac{|f(y)|d\mu}{|x-y|^{n-1}}$ $\displaystyle=$ $\displaystyle A+B.$

Observe that for the first integral we obtain

$\displaystyle A=$
$\displaystyle\int\limits_{|x-y|<r}\frac{|f(y)|d\mu}{|x-y|^{n-1}}$
$\displaystyle=$
$\displaystyle\sum_{j=-\infty}^{-1}\int\limits_{2^{j}r<|x-y|\leq
2^{j+1}r}\frac{|f(y)|d\mu}{|x-y|^{n-1}}$ $\displaystyle\leq$
$\displaystyle\sum_{j=-\infty}^{-1}\frac{1}{(2^{j}r)^{n-1}}\int\limits_{|x-y|\leq
2^{j+1}r}|f(y)|\,d\mu$ $\displaystyle=$
$\displaystyle\sum_{j=-\infty}^{-1}2^{n}(2^{j}r)\frac{1}{(2^{j+1}r)^{n}}\int\limits_{B(x,2^{j+1}r)}|f(y)|\,d\mu$
$\displaystyle\leq$ $\displaystyle 2^{n}rMf(x)\sum_{j=-\infty}^{-1}2^{j}$
$\displaystyle\leq$ $\displaystyle CrMf(f).$ (2.8)

Meanwhile, for the second integral, we have the following estimate

$\displaystyle B=$ $\displaystyle\int\limits_{|x-y|\geq
r}\frac{|f(y)|}{|x-y|^{n-1}}\,d\mu$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}\int\limits_{2^{j}r<|x-y|\leq
2^{j+1}r}\frac{|f(y)|d\mu}{|x-y|^{n-1}}$ $\displaystyle\leq$
$\displaystyle\sum_{j=0}^{\infty}\frac{1}{(2^{j}r)^{n-1}}\int\limits_{|x-y|\leq
2^{j+1}r}|f(y)|\,d\mu$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}2^{n}(2^{j}r)\frac{1}{(2^{j+1}r)^{n}}\int\limits_{B(x,2^{j+1}r)}|f(y)|\,d\mu$
$\displaystyle\leq$
$\displaystyle\sum_{j=0}^{\infty}2^{n}(2^{j}r)\frac{1}{(2^{j+1}r)^{n}}\left(\int\limits_{B(x,2^{j+1}r)}|f(y)|^{p}\,d\mu\right)^{\frac{1}{p}}\left(\frac{1}{(2^{j+1}r)^{n}}\mu(B(x,2^{j+1}r))\right)^{\frac{1}{q}}$
$\displaystyle\leq$ $\displaystyle
C\sum_{j=0}^{\infty}2^{n}(2^{j}r)\|f\|_{\mathcal{M}^{p,\phi}}\left(\frac{1}{(2^{j+1}r)^{n}}(2^{j+1}r)^{n}\right)^{\frac{1}{q}}\phi(2^{j+1}r)$
$\displaystyle=$ $\displaystyle
C\left[\sum_{j=0}^{\infty}2^{n}(2^{j}r)\phi(2^{j+1}r)\right]\|f\|_{\mathcal{M}^{p,\phi}}.$

Since $\phi$ is almost decreasing, we observe that for $j=0,1,2,\dots$

$(2^{j}r)\phi(2^{j+1}r)\leq C\int_{2^{j}r}^{2^{j+1}r}\phi(t)\,dt.$

this last inequality and our assumption then lead us to

$\displaystyle B\leq$ $\displaystyle
C\|f\|_{\mathcal{M}^{p,\phi}}\sum_{j=0}^{\infty}\int_{2^{j}r}^{2^{j+1}r}\phi(t)\,dt$
$\displaystyle\leq$ $\displaystyle
C\|f\|_{\mathcal{M}^{p,\phi}}\int_{r}^{\infty}\phi(t)\,dt$ $\displaystyle\leq$
$\displaystyle Cr^{\lambda+1-n}\|f\|_{\mathcal{M}^{p,\phi}}.$

Now, by choosing

$r=\left(\frac{Mf(x)}{\|f\|_{\mathcal{M}^{p,\phi}}}\right)^{\frac{1}{\lambda-n}},$

we obtain

$\displaystyle|I_{1}f(x)|\leq$ $\displaystyle
Cr\left(Mf(x)+r^{\lambda-n}\|f\|_{\mathcal{M}^{p,\phi}}\right)$
$\displaystyle\leq$ $\displaystyle
C[Mf(x)]^{1-\frac{1}{n-\lambda}}\|f\|_{\mathcal{M}^{p,\phi}}^{\frac{1}{\lambda-n}},$

this completes the proof. ∎

##  3\. Fefferman inequality and its realtion with the Poisson equation.

Let us consider the following problem (Dirichlet problem)

$\left\\{\begin{aligned} -\Delta z=&V\quad\mbox{on $B$}\\\ z=&0\quad\mbox{on
$\partial B$.}\end{aligned}\right.$ (3.1)

The equation $-\Delta z=V$ is known as the Poisson equation. It is well known
that the solution of the above problem is given by the convolution

$z(x)=\int\limits_{\mathds{R}^{d}}\phi(x-y)V(y)\,dy,$

where $\phi$ is the fundamental solution of the Laplace equation, and if $V\in
C_{c}^{2}(\mathds{R}^{d})$ it is clear that $z\in C_{c}^{2}(\mathds{R}^{d})$,
(see [8] for details).

Furthermore, $z$ may be written as

$z(x)=\frac{1}{\omega_{n-1}}\int\limits_{\mathds{R}^{d}}\frac{\nabla
z(y)\cdot(x-y)}{|x-y|^{n}}\,dy$

where $\omega_{n-1}$ represents the $(n-1)$-dimensional measure of the sphere
$S^{n-1}$. Then

$\nabla
z(x)=\frac{1}{\omega_{n-1}}\int\limits_{\mathds{R}^{d}}\frac{\nabla^{2}z(y)\cdot(x-y)}{|x-y|^{n}}\,dy.$

Thus

$|\nabla
z(x)|=\frac{1}{\omega_{n-1}}\int\limits_{\mathds{R}^{d}}\frac{|\nabla^{2}z(y)|}{|x-y|^{n-1}}\,dy$

it is known that $\nabla^{2}z=\Delta z$.

######  Definition 3.1.

A function $\omega(x)\geq 1$ is said to be $A_{1}$-class if

$M\omega(x)\leq C_{1}\omega(x)$

for almost $x\in\mathds{R}^{d}$ and for some constant $C_{1}>0$.

Our next task is to state and prove the Fefferman’s inequality, assuming that
$V$ belongs to

$A_{1}\cap\mathcal{M}^{p,\phi}(\mathds{R}^{d})\cap C_{c}^{2}(\mathds{R}^{d})$

with $1\leq p<q\leq\frac{n-\lambda}{n-\lambda-1}$.

Fefferman’s inequality has been shown in different settings. For instance, in
[12], Zamboni proved the Fefferman inequality allowing $V$ to be in a
generalized Kato class. See also [4].

On the other hand, Castillo, Ramos and Rojas proved the Fefferman’s inequality
allowing $V$ to belong to the Kato class with $p=2$, as well as Castillo,
Rafeiro and Rojas proved the Fefferman’s inequality allowing $V$ to be in
$L_{\frac{n}{p}}(\Omega)$ (see [7]). For definitions and details, see [3, 4,
5, 6, 12]

######  Theorem 3.1 (Fefferman’s inequality on $\mathcal{M}^{p,\phi}$).

Suppose that for some $0\leq\lambda<n-1$ we have

$\int_{r}^{\infty}\phi(t)\,dt\leq Cr^{\lambda+1-n}\qquad r>0.$

Let

$1<p\leq q\leq\frac{n-\lambda}{n-\lambda-1}\quad\mbox{and}\quad V\in
A_{1}\cap\mathcal{M}^{p,\phi}\cap C_{c}^{2}(\mathds{R}^{d}).$

Then

$\int\limits_{\mathds{R}^{d}}|u(x)|^{p}V(x)\,d\mu\leq
C(n,p)\|V\|_{\mathcal{M}^{p,\phi}}^{\frac{p}{n-\lambda}}\int\limits_{\mathds{R}^{d}}|\nabla
u(x)|^{p}\,d\mu.$

###### Proof.

For any $u\in C_{c}^{2}(\mathds{R}^{d})$, let us consider a ball such that
$u\in C_{c}^{2}(B)$ and consider the solution $z$ of the Poisson equation

$\left\\{\begin{aligned} -\Delta z=&V\quad\mbox{on $B$}\\\ z=&0\quad\mbox{on
$\partial B$.}\end{aligned}\right.$

Then, by using Theorem 2.4 and Hölder’s inequality, we have

$\displaystyle\int\limits_{\mathds{R}^{d}}|u(x)|^{p}V(x)\,d\mu=-\int\limits_{B}|u(x)|^{p}\Delta
z(x)\,d\mu(x)$ $\displaystyle=\int\limits_{B}\nabla|u(x)|^{p}\nabla
z(z)\,d\mu(x)$ $\displaystyle=p\int\limits_{B}|u(x)|^{p-1}|\nabla u(x)|\nabla
z(x)\,d\mu(x)$ $\displaystyle\leq p\int\limits_{B}|u(x)|^{p-1}|\nabla
u(x)||\nabla z(x)|\,d\mu(x)$
$\displaystyle\leq\frac{p}{\omega_{n-1}}\int\limits_{B}|u(x)|^{p-1}|\nabla
u(x)|\int\limits_{B}\frac{|\nabla^{2}z(y)|}{|x-y|^{n-1}}\,d\mu(y)d\mu(x)$
$\displaystyle=\frac{p}{\omega_{n-1}}\int\limits_{B}|u(x)|^{p-1}|\nabla
u(x)|\int\limits_{B}\frac{|\Delta z(y)|}{|x-y|^{n-1}}\,d\mu(y)d\mu(x)$
$\displaystyle=\frac{p}{\omega_{n-1}}\int\limits_{B}|u(x)|^{p-1}|\nabla
u(x)|\int\limits_{B}\frac{|V(y)|}{|x-y|^{n-1}}\,d\mu(y)d\mu(x)$
$\displaystyle\leq\frac{Cp}{\omega_{n-1}}\int\limits_{B}|u(x)|^{p-1}|\nabla
u(x)||MV(x)|^{1-\frac{1}{n-\lambda}}\|V\|_{\mathcal{M}^{p,\phi}}^{\frac{1}{n-\lambda}}\,d\mu(x)$
$\displaystyle\leq\frac{Cp}{\omega_{n-1}}\|V\|_{\mathcal{M}^{p,\phi}}^{\frac{1}{n-\lambda}}\left[\int\limits_{B}|u(x)|^{p}[MV(x)]^{q\left(1-\frac{1}{n-\lambda}\right)}\,d\mu(x)\right]^{\frac{1}{q}}\left(\int\limits_{B}|\nabla
u(x)|^{p}\,d\mu\right)$
$\displaystyle\leq\frac{Cp}{\omega_{n-1}}\|V\|_{\mathcal{M}^{p,\phi}}^{\frac{1}{n-\lambda}}\left(\int\limits_{B}|u(x)|^{p}V(x)\,d\mu(x)\right)^{\frac{1}{q}}\left(\int\limits_{B}|\nabla
u(x)|^{p}\right)^{\frac{1}{p}}.$

And so

$\int\limits_{B}|u(x)|^{p}V(x)\,d\mu\leq\left(\frac{Cp}{\omega_{n-1}}\right)^{p}\|V\|_{\mathcal{M}^{p,\phi}}^{\frac{p}{n-\lambda}}\int\limits_{B}|\nabla
u(x)|^{p}\,d\mu.$

This completes the proof. ∎

From Fefferman’s inequality we easily derive the following corollary, which
can be obtained through a standard partition of the unity.

######  Corollary 3.1.

Let $V\in\mathcal{M}^{p,\phi}(\mathds{R}^{d})$ and let $\Omega$ be a bounded
subset of $\mathds{R}^{d}$, $\operatorname{Supp}V\subseteq\Omega$. Then, for
any $\sigma>0$, there exists a positive constant $K$ depending on $\sigma$,
such that

$\int\limits_{\Omega}|u(x)|^{p}V(x)\,d\mu\leq\sigma\int\limits_{\Omega}|\nabla
u(x)|^{p}\,d\mu+K(\sigma)\int\limits_{\Omega}\int\limits_{\Omega}|u(x)|^{p}\,d\mu$

for all $u\in C_{0}^{\infty}(\Omega)$.

###### Proof.

Let $\sigma>0$. Let $r$ be a positive number that will be chosen later. Let
$\\{\alpha_{k}^{p}\\}$, $k=1,2,\dots,N(r)$, be a finite partition of the unity
of $\overline{\Omega}$, such that $\operatorname{Supp}V\subseteq B(x_{k},r)$
with $x_{k}\in\overline{\Omega}$. We apply Theorem 3.1 to the functions
$\alpha_{k}$ and we get

$\displaystyle\int\limits_{\Omega}|u(x)|^{p}V(x)\,d\mu$
$\displaystyle=\int\limits_{\Omega}V(x)|u(x)|^{p}\sum_{k=1}^{N(r)}\alpha_{k}^{p}(x)\,d\mu$
$\displaystyle=\sum_{k=1}^{N(r)}\int\limits_{\Omega}V(x)|u(x)\alpha_{k}(x)|^{p}\,d\mu$
$\displaystyle\leq\sum_{k=1}^{N(r)}C\|V\|_{\mathcal{M}^{p,\phi}}^{\frac{p}{n-\lambda}}\left(\int\limits_{\Omega}|\nabla
u(x)|^{p}\alpha_{k}^{p}(x)\,d\mu+\int\limits_{\Omega}|\nabla
u(x)|^{p}||u(x)|^{p}\,d\mu\right)$ $\displaystyle\leq
C\|V\|_{\mathcal{M}^{p,\phi}}^{\frac{p}{n-\lambda}}\left(\int\limits_{\Omega}|\nabla
u(x)|^{p}\,d\mu+\frac{N(r)}{r^{p}}\int\limits_{\Omega}|u(x)|^{p}\,d\mu\right)$
(3.2)

Finally, to obtain the result, it is sufficient to choose $r$ such that

$C\|V\|_{\mathcal{M}^{p,\phi}}^{\frac{p}{n-\lambda}}=\sigma.$

After that, we note that $N(r)\sim r^{-n}$ and the corollary follows. ∎

######  Lemma 3.1.

Let $B_{r}$ and $B_{2r}$ be two concentric balls contained in $\Omega$. Then

$\int\limits_{B_{r}}|\nabla
u(x)|^{p}\,d\mu\leq\frac{C}{r^{p}}\int\limits_{B_{2r}}|u(x)|^{p}\,d\mu,$ (3.3)

where the constant $C$ does not depend on $r$, and $u\in
C_{0}^{\infty}(B_{r})$.

###### Proof.

Take $\varphi\in C_{0}^{\infty}(\Omega)$, with
$\operatorname{Supp}\varphi\subset B_{2r}$, $\varphi(x)=1$ for $x\in B_{r}$,
and $|\nabla\varphi|\leq\frac{C}{r}$. By using $\varphi^{p}$ as a test
function in (1.1); we have

$\int\limits_{B_{2r}}-\mbox{div}(|\nabla u|^{p-2}\nabla
u)\varphi^{p}(u)+\int\limits_{B_{2r}}V|u|^{p-2}u\varphi^{p}u=0.$ (3.4)

Thus

$\int\limits_{B_{2r}}|\nabla u|^{p}\varphi^{p}=-p\int\limits_{B_{2r}}|\nabla
u|^{p-2}\varphi^{p-2}\nabla u\cdot\nabla\varphi(\varphi
u)-\int\limits_{B_{2r}}V|\varphi u|^{p}.$ (3.5)

With the help of Young’s inequality for $\frac{p-1}{p}+\frac{1}{p}=1$, we can
estimate the first integral in the right-hand side of (3.5) by

$(p-1)\epsilon^{\frac{p}{p-1}}\int\limits_{B_{2r}}|\nabla
u|^{p}\varphi^{p}+\epsilon^{-p}\int\limits_{B_{2r}}|\nabla\varphi|^{p}|u|^{p}.$
(3.6)

Also, by Corollay 3.1 we can estimate the second integral in the right-hand
side of (3.5) by

$\epsilon\int\limits_{B_{2r}}|\nabla(\varphi
u)|^{p}+C_{\epsilon}\int\limits_{B_{2r}}|\varphi u|^{p}.$ (3.7)

Using this estimates in (3.5) we arrive at

$\displaystyle\int\limits_{B_{2r}}|\nabla u|^{p}\varphi^{p}\leq$
$\displaystyle\left((o-1)\epsilon^{\frac{p}{(p-1)}}+\epsilon\right)\int\limits_{B_{2r}}|\nabla
u|^{p}\varphi^{p}$
$\displaystyle+(\epsilon^{p}+\epsilon)\int\limits_{B_{2r}}|u|^{p}|\nabla
u|^{p}+C_{\epsilon}\int\limits_{B_{2r}}|\nabla u|^{p}|\varphi|^{p}.$

Using the fact that $|\nabla\varphi|\leq\frac{C}{v}$,
$|\varphi|\leq\frac{C}{v}$ and $\varphi=1$ in $B_{r}$, we immediately get
inequality (3.3). ∎

######  Lemma 3.2.

Let $u\in C_{0}^{\infty}(\Omega)$ where $B_{r}$ is the ball of radius $r$ in
$\mathds{R}^{d}$ and let $E=\\{x\in B_{r}:u(x)=0\\}$. Then there exits a
constant $\beta$ (depending only on $n$) such that

$\int\limits_{A}|u|\leq\beta\frac{r^{n}}{\mu(E)}\left(\mathrm{m}(E)\right)^{\frac{1}{n}}\int\limits_{B_{2r}}|\nabla
u|,$ (3.8)

for all ball $B_{r}$, $u$ as above and all measurable sets $A\subset B_{r}$.

###### Proof.

Let us begin by choosing $r$ such that $\mathrm{m}(A)=\mathrm{m}(B)$ where $m$
is the Lebesgue measure, next, for $x\in B_{r}$ and $t\in E$ we obtain

$-u(x)=u(t)-u(x)=\int_{0}^{|x-y|}\frac{du(x-s\omega)}{d\mu(s)}\,d\mu(s)$ (3.9)

where $\omega=\frac{y-x}{|y-x|}$. Now let us integrate equation (3.9) with
respect to $t\in E$, once it is done, we have

$-\mu(E)u(x)=\int\limits_{E}du(t)\int_{0}^{|x-y|}\frac{du}{d\mu(s)}(x+s\omega)\,d\mu(s).$
(3.10)

Now, we need to find a bound for (3.10). By using the Fubini’s Theorem,

$\displaystyle\left|-\mu(E)u(x)\right|=\left|\int\limits_{E}d\mu(t)\int_{0}^{|x-y|}\frac{du}{d\mu(s)}(x+s\omega)\,d\mu(s)\right|$
$\displaystyle\leq\int\limits_{E}du(t)\int_{0}^{|x-y|}\left|\frac{du}{d\mu(s)}(x+s\omega)\right|\,d\mu(s)$
$\displaystyle\leq\int\limits_{B_{r}}du(t)\int_{0}^{|x-y|}\left|\frac{du}{d\mu(s)}(x+s\omega)\right|\,d\mu(s)$
$\displaystyle\leq\int\limits_{B_{2r}}du(t)\int_{0}^{|x-y|}\left|\frac{du}{d\mu(s)}(x+s\omega)\right|\,d\mu(s)$
$\displaystyle=\mu(B_{2r})\int_{0}^{|x-y|}\left|\frac{du}{d\mu(s)}(x+s\omega)\right|\,d\mu(s)$
$\displaystyle\leq\frac{C(2r)^{n}}{\mathrm{m}(B(0,1))}\mathrm{m}(B(0,1))\int_{0}^{|x-y|}\left|\frac{du}{d\mu(s)}(x+s\omega)\right|\,d\mu(s)$
$\displaystyle=\frac{C(2r)^{n}n\mathrm{m}(B(0,1))}{n\mathrm{m}(B(0,1))}\int_{0}^{|x-y|}\left|\frac{du}{d\mu(s)}(x+s\omega)\right|\,d\mu(s)$
$\displaystyle=\frac{C}{\mathrm{m}(\partial B(0,1))}\int\limits_{\partial
B(0,1)}d\omega\int_{0}^{2r}t^{n-1}\,dt\int_{0}^{|x-y|}\left|\frac{du}{d\mu(s)}(x+s\omega)\right|\,d\mu(s)$
$\displaystyle=\frac{C}{\mathrm{m}(\partial
B(0,1))}\int_{0}^{2r}t^{n-1}\,dt\int\limits_{\partial
B(0,1)}d\omega\int_{0}^{|x-y|}\frac{1}{s^{n-1}}\left|\frac{du}{d\mu(s)}(x+s\omega)\right|s^{n}\,d\mu(s)$
$\displaystyle\leq\frac{C}{\mathrm{m}(\partial
B(0,1))}\int_{0}^{2r}t^{n-1}\,dt\int\limits_{\partial
B(0,1)}d\omega\int_{0}^{r}\frac{|\nabla u(y)|}{|x-y|^{n-1}}s^{n-1}\,d\mu(s)$
$\displaystyle=\frac{C(2r)^{n}}{n\mathrm{m}(\partial
B(0,1))}\int\limits_{\partial B(0,1)}\int_{0}^{r}\frac{|\nabla
u(y)|}{|x-y|^{n-1}}s^{n-1}\,d\mu(s)d\omega$
$\displaystyle=\frac{C(2r)^{n}}{n\mathrm{m}(\partial
B(0,1))}\int\limits_{B(x,r)}\frac{|\nabla
u(y)|}{|x-y|^{n-1}}s^{n-1}\,d\mu(y).$

Taking into account that we just write $y=x+r\omega$, thus $r\omega=y-x$ and
$r=|x-y|$ since $|\omega|=1$. Up to now, we have obtained that

$\mu(E)u(x)\leq\frac{C_{1}(2r^{n})}{n}\int\limits_{B(x,r)}\frac{|\nabla
u(y)|}{|x-y|^{n-1}}\,d\mu(y)$ (3.11)

where $C_{1}=\frac{C}{\mathrm{m}(\partial B(0,1))}$.

Hence, let us integrate both sides of (3.11) with respect to $x$ on the set
$A$. Indeed

$\displaystyle\mu(E)\int\limits_{A}|u(x)|\,d\mu(x)\leq\frac{C(2r)^{n}}{n}\int\limits_{A}\int\limits_{B(x,r)}\frac{|\nabla
u(y)|}{|x-y|^{n-1}}\,d\mu(y)d\mu(A)$
$\displaystyle\phantom{\mu(E)}\leq\frac{C(2r)^{n}}{n}\int\limits_{B(x,r)}\frac{d\mu(x)}{|x-y|^{n-1}}\int\limits_{B(x,r)}|\nabla
u(y)|d\mu(y)$
$\displaystyle\phantom{\mu(E)}=\frac{C(2r)^{n}}{n}n\mathrm{m}(B(0,1))\int_{0}^{r}t^{1-r}t^{n-1}\,dt\int\limits_{B(x,r)}|\nabla
u(y)|d\mu(y)$
$\displaystyle\phantom{\mu(E)}=C(2r^{n})\mathrm{m}(B(0,1))r\int\limits_{B(x,r)}|\nabla
u(y)|\,d\mu(y)$
$\displaystyle\phantom{\mu(E)}=C(2r)^{n}(\mathrm{m}(B(0,1)))^{1-\frac{1}{n}}(r^{n}\mathrm{m}(B(0,1)))^{\frac{1}{n}}\int\limits_{B(x,r)}|\nabla
u(y)|d\mu(y)$
$\displaystyle\phantom{\mu(E)}=C(2r)^{n}(\mathrm{m}(B(0,1)))^{1-\frac{1}{n}}(r^{n}\mathrm{m}(B))^{\frac{1}{n}}.$

Finally

$\int\limits_{A}|u(x)|\,d\mu(x)\leq\frac{\beta
r^{n}(\mathrm{m}(A))^{\frac{1}{n}}}{\mu(E)}\int\limits_{B(x,2r)}|\nabla
u(y)|\,d\mu(y)$

where

$\beta=2^{n}(\mathrm{m}(B(0,1)))^{1-\frac{1}{n}}.$

∎

##  4\. Strong Unique Continuation

In this section, we proceed to establish the strong unique continuation
property of the eigenfunction for the $p$-Laplacian operator in the case
$V\in\mathcal{M}^{p,\phi}(\mathds{R}^{d})$.

######  Theorem 4.1.

Let $u\in C_{0}^{\infty}(\Omega)$ be a solution of $\eqref{eq1.2}$. If $u=0$
on a set $E$ of positive measure, then $u$ has zero of infinite orden in
$p$-mean.

###### Proof.

Let $E=\\{x\in B_{r}:u(x)=0\\}$ with $0<r<1$ and let $E^{c}$ the complement of
$E$.

Taking $r_{0}$ smaller if necessary, we can assume that
$B_{r_{0}}\subset\Omega$. Since $u=0$ on $E$ by Lemma 3.2 we have

$\displaystyle\int\limits_{B_{r}}|u|^{p}=$
$\displaystyle\int\limits_{B_{r}\cap E^{c}}|u|^{p}\leq\frac{\beta
r^{n}(\mathrm{m}(E^{c}\cap B_{r}))}{\mu(E\cap
B_{r})}\int\limits_{B_{r}}|\nabla|u|^{p}|$ $\displaystyle=$ $\displaystyle
C_{n}\int\limits_{B_{r}}|u|^{p-1}|\nabla u|$

where

$C_{n}=\frac{p\beta(\mathrm{m}(B(0,1)))^{\frac{1}{n}}}{\mu(E)}.$

By Hölder’s inequality

$\int\limits_{B_{r}}|u|^{p}\leq C_{n}\left(\int\limits_{B_{r}}|\nabla
u|^{p}\right)^{\frac{1}{p}}\left(\int\limits_{B_{r}}|u|^{p}\right)^{\frac{p-1}{p}}$
(4.1)

and by using Young’s inequality, we get

$\int\limits_{B_{r}}|u|^{p}\leq
C_{n}^{r}\left(r^{p-1}\int\limits_{B_{r}}|\nabla
u|^{p}+\frac{p-1}{r}\int\limits_{B_{r}}|u|^{p}\right).$ (4.2)

Finally by Lemma (3.1), we have

$\int\limits_{B_{r}}|u|^{p}\leq C_{n}\int\limits_{B_{2r}}|u|^{p}.$ (4.3)

Now, let us introduce the following function

$f(r)=\int\limits_{B_{r}}|u|^{p}.$ (4.4)

Observe that consequently $r_{0}$ depends on $n$. Then (4.3) can be written as

$f(r)\leq M_{n}2^{-n}f(2r)\quad\mbox{for $r\leq r_{0}$}$ (4.5)

where $M_{n}=2^{-n}C_{n}$.

Integrating (4.5), we get

$-f(\rho)\leq M_{n}2^{-kn}f(2^{n}\rho)\quad\mbox{if $2^{k-1}\rho\leq r_{0}$}$
(4.6)

Now given that $0<r<r_{0}(n)$, choose $k\in\mathds{N}$ such that

$2^{-k}r_{0}\leq r\leq 2^{-k+1}r_{0}.$ (4.7)

From (4.6), we obtain

$f(r)\leq M_{n}2^{-kn}f(2^{k}r)\leq M_{n}2^{-kn}f(2r).$ (4.8)

Since $2^{-k}\leq\frac{r}{r_{0}}$, we finally obtain

$f(r)\leq M_{n}\left(\frac{r}{r_{0}}\right)^{n}f(2r_{0}).$ (4.9)

And thus, we have

$\int\limits_{B_{r}(x_{0})}|u(y)|^{p}\,d\mu(y)\leq
M_{n}\left(\frac{r}{r_{0}}\right)^{n}f(2r_{0})$ (4.10)

and this shows that (1.2) holds, which means that $u$ has zero of infinite
order in $p$-mean at $x_{0}$. ∎

######  Corollary 4.1.

Equation (1.1) has strong unique continuation property.

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