# Maximal operator in Musielak–Orlicz–Sobolev spaces

Piotr Michał Bies, Michał Gaczkowski, Przemysław Górka

###### Abstract.

We study the Hardy-Littlewood maximal operator in the Musielak-Orlicz-Sobolev
space $W^{1,\varphi}(\mathbb{R}^{n})$. Under some natural assumptions on
$\varphi$ we show that the maximal function is bounded and continuous in
$W^{1,\varphi}(\mathbb{R}^{n})$.

Keywords: maximal operator, Musielak–Orlicz–Sobolev spaces

_Mathematics Subject Classification (2020):42B25; 46E35_

##  1\. Introduction

The Hardy-Littlewood maximal operator $M$ plays a central role in mathematics,
e.g. in the theory of function spaces, harmonic analysis and PDE. The
boundedness of $M$ in various types of function spaces is a central issue.
Classical theorem of Hardy, Littlewood and Wiener asserts that for
$1<p\leq\infty$, the maximal operator is bounded on $L^{p}(X,d,\mu)$, where
$(X,d,\mu)$ is a doubling metric measure space (see e.g. [8]). The maximal
operator has been also studied in different function spaces, e.g.: Banach
function spaces [11], Sobolev spaces [9], Lebesgue spaces with variable
exponent [3], Musielak-Orlicz spaces [7]. Moreover, $M$ is bounded in Hölder
spaces $C^{0,s}(X)$ (see [2]), where $(X,d,\mu)$ satisfies the so-called
$\delta$-annular decay property111If no annular decay property is assumed,
then $Mf$ can fail to be continuous, even if $f$ is Lipschitz continuous (see
Example 1.4. in [2]).  and $\mu$ is doubling (see also [4] in the space of
continuous functions $C(X)$ and [1] in the variable exponent Hölder spaces).

Let us observe that since $M$ is sublinear and bounded in $L^{p}$, we get the
continuity of the maximal operator in $L^{p}$. The same holds in the space of
bounded and continuous functions $C(X)$. Iwaniec asked the following question
[5, Question 3]: Is the Hardy–Littlewood maximal operator continuous in
$W^{1,p}(\mathbb{R}^{n})$, $1<p<\infty$? The positive answer to this question
is given in [10]. The main objective of the paper is to study the maximal
operator in the Musielak-Orlicz-Sobolev spaces
$W^{1,\varphi}(\mathbb{R}^{n})$. Under some natural assumptions on $\varphi$
we show that maximal operator is bounded and continuous in
$W^{1,\varphi}(\mathbb{R}^{n})$.

The remainder of the paper is structured as follows. In Section 2, we
introduce the notations and recall the definitions. In Section 3 we prove
boundedness of the maximal operator in the Musielak-Orlicz-Sobolev spaces.
Auxiliary results, which are needed in the proof of the main result, are
contained in Section 4. Our principal assertion, concerning the continuity of
the maximal operator in the Musielak-Orlicz-Sobolev is formulated and proven
in the last section.

##  2\. Preliminaries

Let $A\subset\mathbb{R}^{n}$ be a measurable set, by $|A|$ we denote the
Lebesgue measure of $A$ and by $L^{0}(A)$ the set of measurable functions on
$A$.

Let $\varphi:A\times[0,\infty)\to[0,\infty)$, $p,q>0$. We say that $\varphi$
satisfies $\textnormal{{(aInc)}}_{p}$ if there exists $a\in[1,\infty)$ such
that the inequality holds

$\frac{\varphi(x,s)}{s^{p}}\leq a\frac{\varphi(x,t)}{t^{p}}\quad\text{for
almost all $x\in A$ and for all $0<s<t$},$

and we say that $\varphi$ satisfies $\textnormal{{(aDec)}}_{q}$ if there
exists $a\in[1,\infty)$ such that the inequality holds

$\frac{\varphi(x,t)}{t^{q}}\leq a\frac{\varphi(x,s)}{s^{q}}\quad\text{for
almost all $x\in A$ and for all $0<s<t$}.$

Furthermore, we denote

$\textnormal{{(aInc)}}\
=\bigcup_{p\in(1,\infty)}\textnormal{{(aInc)}}_{p},\,\,\textnormal{{(aDec)}}\
=\bigcup_{p\in(1,\infty)}\textnormal{{(aDec)}}_{p}.$

We say that $\varphi:A\times[0,\infty)\to[0,\infty]$ is a $\Phi$-prefunction
if $\varphi(x,0)=0$, $\varphi(x,\cdot)$ is increasing, $\lim_{t\to
0^{+}}\varphi(x,t)=0$, $\lim_{t\to\infty}\varphi(x,t)=\infty$ for almost every
$x\in A$ and the map $x\mapsto\varphi(x,|f(x)|)$ is measurable for $f\in
L^{0}(A)$. We say that a $\Phi$-prefunction $\varphi$ is a

  * •

weak $\Phi$-function if it satisfies $\textnormal{{(aInc)}}_{1}$,

  * •

convex $\Phi$-function if $\varphi(x,\cdot)$ is left continuous and convex for
almost every $x\in A$,

  * •

strong $\Phi$ function if $\varphi(x,\cdot)$ is continuous and convex for
almost every $x\in A$.

The set of weak $\Phi$-functions, convex $\Phi$-functions and strong
$\Phi$-functions we shall denote by $\Phi_{\textnormal{w}}(A)$,
$\Phi_{\textnormal{c}}(A)$ and $\Phi_{\textnormal{s}}(A)$ respectively. From
the very definition we have
$\Phi_{\textnormal{s}}(A)\subset\Phi_{\textnormal{c}}(A)\subset\Phi_{\textnormal{w}}(A)$.
For $\varphi\in\Phi_{\textnormal{w}}(A)$ and $f\in L^{0}(A)$ we define

$\rho_{\varphi}(f)=\int_{A}\varphi(x,|f(x)|)\;dx,$
$\|f\|_{\varphi,A}=\inf\left\\{\lambda>0\;:\;\rho_{\varphi}\left(\tfrac{f}{\lambda}\right)\leq
1\right\\},$

and a set

$L^{\varphi}(A)=\\{f\in L^{0}(A)\;:\;\exists(\lambda>0)\
\rho_{\varphi}(\lambda f)<\infty\\}.$

We shall simply write $\|f\|_{\varphi}$ when $A=\mathbb{R}^{n}$. Let us note
that the Musielak-Orlicz space $(L^{\varphi},\|\cdot\|_{\varphi,A})$ is a
quasi-Banach space for $\varphi\in\Phi_{\textnormal{w}}(A)$, and
$(L^{\varphi},\|\cdot\|_{\varphi,A})$ is a Banach space if
$\varphi\in\Phi_{\textnormal{c}}(A)$. It is known (see [6, Proposition
3.2.4]), that if $\varphi,\psi\in\Phi_{\textnormal{w}}(A)$ and
$\varphi\simeq\psi$, then222Let $\varphi,\psi:A\times[0,\infty)\to[0,\infty]$.
We say that $\varphi$ and $\psi$ are equivalent ($\varphi\simeq\psi$) if there
exists $L\geq 1$ such that the inequalities
$\psi(x,t/L)\leq\varphi(x,t)\leq\psi(x,Lt)$ are satisfied for almost every
$x\in A$ and for all $t\in[0,\infty)$. $L^{\varphi}(A)=L^{\psi}(A)$ and
corresponding quasi-norms are equivalent. Moreover, if
$\varphi\in\Phi_{\textnormal{w}}(A)$, then there exists
$\psi\in\Phi_{\textnormal{s}}(A)$ such that $\varphi\simeq\psi$. Thus, even if
$\|\cdot\|_{\varphi,A}$ is not a norm for a certain
$\varphi\in\Phi_{\textnormal{w}}(A)$ it has a Banach space structure. Let us
recall [6, Lemma 3.1.3] that if $\varphi\in\Phi_{\textnormal{w}}(A)$ satisfies
(aDec), then

$L^{\varphi}(A)=\left\\{f\in L^{0}(A)\,:\,\rho_{\varphi}(f)<\infty\right\\}.$

On the other hand [6, Theorem 3.6.6], if $\varphi\in\Phi_{\textnormal{w}}(A)$
satisfies (aInc) and (aDec), then $L^{\varphi}(A)$ is reflexive.

If $\varphi\in\Phi_{\textnormal{w}}(A)$ satisfies $\textnormal{{(aInc)}}_{p}$
with $1\leq p<\infty$, then quantities $\|.\|_{\varphi}$ and
$\rho_{\varphi}(.)$ can be compared using the inequalities (see [6, Corollary
3.2.10])

$\displaystyle\min\\{(\tfrac{1}{a}\rho_{\varphi}(f))^{1/p},1\\}\leq\|f\|_{\varphi}\leq\max\\{(a\rho_{\varphi}(f))^{1/p},1\\},$
(1)

where $a$ is a constant from $\textnormal{{(aInc)}}_{p}$. Let us recall that
if $\varphi\in\Phi_{\textnormal{w}}(A)$, then for $f,g\in L^{0}(A)$ the
following Hölder inequality

$\int_{A}|f(x)||g(x)|\;dx\leq 2\|f\|_{\varphi,A}\|g\|_{\varphi^{*},A},$

holds.333For $\varphi:A\times[0,\infty)\to[0,\infty]$ we define the conjugate
$\varphi^{*}$ as follows $\varphi^{*}(x,t)=\sup_{s\geq 0}(ts-\varphi(x,s))$.

######  Definition 2.1.

Let $A\subset\mathbb{R}^{n}$ be a measurable set. We say that
$\varphi\in\Phi_{\textnormal{w}}(A)$ satisfies (A0) if there exists a constant
$\beta\in(0,1]$ such that $\beta\leq\varphi^{-1}(x,1)\leq{1}/{\beta}$ for
almost everyl $x\in A$, where $\varphi^{-1}$ is left-inverse of $\varphi$ (see
[6, Definition 2.3.1]).

######  Proposition 2.2 ([6, Corollary 3.7.4]).

Let $A\subset\mathbb{R}^{n}$ be a measurable set and let
$\varphi\in\Phi_{\textnormal{w}}(A)$. Then $\varphi$ satisfies (A0) if and
only if there exists $\beta\in(0,1]$ such that $\varphi(x,\beta)\leq
1\leq\varphi(x,1/\beta)$ for almost every $x\in A$.

######  Proposition 2.3 ([6, Corollary 3.7.9]).

Let $A\subset\mathbb{R}^{n}$ be a measurable set with finite measure. If
$\varphi\in\Phi_{\textnormal{w}}(A)$ satisfies (A0)and
$\textnormal{{(aInc)}}_{p}$ with $1\leq p<\infty$, then
$L^{\varphi}(A)\hookrightarrow L^{p}(A)$.

######  Definition 2.4.

Let $\Omega\subset\mathbb{R}^{n}$ be an open set and
$\varphi\in\Phi_{\textnormal{w}}(\Omega)$, then

  1. (1)

$\varphi$ satisfies (A1) if there exists $\beta\in(0,1)$ such that for every
ball $B$ such that $|B|\leq 1$ the following inequality holds

$\beta\varphi^{-1}(x,t)\leq\varphi^{-1}(y,t)$

for every $t\in[1,1/|B|]$ and for almost every $x,y\in B\cap\Omega$.

  2. (2)

$\varphi$ satisfies (A2) if for all $s>0$ there exist $\beta\in(0,1]$ and
$h\in L^{1}(\Omega)\cap L^{\infty}(\Omega)$ such that the following inequality
holds

$\beta\varphi^{-1}(x,t)\leq\varphi^{-1}(y,t)$

for almost every $x,y\in\Omega$ and for all $t\in[h(x)+h(y),s]$.

Next, we define the Musielak–Orlicz–Sobolev spaces. Let $\Omega$ be an open
subset of $\mathbb{R}^{n}$ and let $\varphi\in\Phi_{\textnormal{w}}(\Omega)$.
The Musielak–Orlicz–Sobolev space $W^{1,\varphi}(\Omega)$ is a vector space of
all $f\in L^{\varphi}(\Omega)$ for which the distributional derivatives belong
to $L^{\varphi}(\Omega)$. We equip $W^{1,\varphi}(\Omega)$ with the quasi-norm

$\displaystyle\|u\|_{k,\varphi,\Omega}:=\sum_{|\alpha|\leq
1}\|D_{\alpha}u\|_{\varphi,\Omega}.$

Again, we will write simply $\|u\|_{1,\varphi}$ if $\Omega=\mathbb{R}^{n}$. If
$\varphi\in\Phi_{\textnormal{w}}(\Omega)$ satisfies (A0), (aInc) and (aDec),
then the Musielak–Orlicz–Sobolev space is reflexive [6, Theorem 6.1.4 and
Theorem 3.7.13].

If $f$ is locally integrable and $A$ is a measurable set such that
$0<|A|<\infty$, then we denote the integral average of the function $f$ over
$A$ as

$\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{A}f\,dx=\frac{1}{|A|}\int_{A}f\,dx\,.$

Finally, we formulate and prove the following proposition.

######  Proposition 2.5.

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (aDec) and let
$f\in L^{\varphi}(\mathbb{R}^{n})$. Then

  * (i)

For $\epsilon>0$ there exists $R>0$ such that

$\|f\|_{\varphi,\mathbb{R}^{n}\setminus B(0,R)}<\epsilon.$

  * (ii)

For $\epsilon>0$ there exists $\lambda>0$ such that for any measurable set
$A\subset\mathbb{R}^{n}$ such that $|A|<\lambda$ the following inequality

$\|f\|_{\varphi,A}<\epsilon$

is satisfied.

###### Proof.

Since the proof of (ii) is similar to the proof of (i), we shall give the
proof of (i). Let $\epsilon>0$, then since (aDec) is satisfied, we have

$\int_{\mathbb{R}^{n}}\varphi\left(x,\tfrac{|f(x)|}{\epsilon/2}\right)\,dx<\infty.$

Thus, there exists $R>0$ such that

$\int_{\mathbb{R}^{n}\setminus
B(0,R)}\varphi\left(x,\tfrac{|f(x)|}{\epsilon/2}\right)\,dx\leq 1.$

Therefore, from the very definition of the norm we have
$\|f\|_{\varphi,\mathbb{R}^{n}\setminus B(0,R)}<\epsilon$. ∎

We close this section with couple of examples taken from [6].

######  Example 2.6.

Let $\varphi_{0}(x,t)=\varphi(t)$, where
$\varphi:[0,\infty)\rightarrow[0,\infty)$ is such that
$\varphi_{0}\in\Phi_{\textnormal{w}}$. Then, $\varphi_{0}$ satisifes (A0),
(A1), (A2). Moreover, $\varphi_{0}$ satisifes (aInc), if $\varphi$ satisfies
$\nabla_{2}$ and $\varphi_{0}$ satisifes (aDec), if $\varphi$ satisfies
$\Delta_{2}$.

######  Example 2.7.

Let $\varphi_{1}(x,t)=t^{p(x)}$, where $p:\Omega\rightarrow[1,\infty)$ is a
measurable map, then $\varphi_{1}$ satisifes (A0), $\varphi_{1}$ satisifes
(A1), if $\frac{1}{p}\in C^{\log}$, i.e., there exists $C$ such that for every
distinct $x,y\in\Omega$

$\left|\frac{1}{p(x)}-\frac{1}{p(y)}\right|\leq\frac{C}{\log(e+1/|x-y|)},$

$\varphi_{1}$ satisifes (A2), if $\frac{1}{p}$ satisfies $\log$-Hölder decay
condition, i.e., there exist $C,p_{\infty}$ such that

$\left|\frac{1}{p(x)}-\frac{1}{p_{\infty}}\right|\leq\frac{C}{\log(e+|x|)}.$

Furthermore, $\varphi_{1}$ satisifes (aInc), if $p_{-}>1$ and $\varphi_{1}$
satisifes (aDec), if $p_{+}<\infty$.

######  Example 2.8.

Let $\varphi_{2}(x,t)=t^{p}+a(x)t^{q}$, where $q>p\geq 1$, then $\varphi_{2}$
satisifes (A0), if $a\in L^{\infty}$, $\varphi_{2}$ satisifes (A1), if $a\in
C^{0,\frac{n}{p}(q-p)}$, $\varphi_{2}$ always satisifes (A2), and
$\varphi_{2}$ satisifes (aInc), if $p>1$ and $\varphi_{2}$ satisifes (aDec),
if $q<\infty$.

##  3\. Boundedness of the maximal operator

For $f\in L^{1}_{\textnormal{loc}}(\mathbb{R}^{n})$ we define the maximal
function $Mf\colon\mathbb{R}^{n}\to\mathbb{R}$ in a standard way

$Mf(x)=\sup_{r>0}\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r)}|f(z)|\,dz.$

Let us recall the crucial theorem about boudedness of maximal function in the
Musielak-Orlicz spaces.

######  Theorem 3.1 ([6, Theorem 4.3.4]).

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (A0), (A1),
(A2) and (aInc). Then, the maximal operator

$M\colon L^{\varphi}(\mathbb{R}^{n})\to L^{\varphi}(\mathbb{R}^{n})$

is bounded.

Now, we are in position to formulate and prove the main result of this
section.

######  Theorem 3.2.

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (A0), (A1),
(A2), (aInc) and (aDec). If $f\in W^{1,\varphi}(\mathbb{R}^{n})$, then $Mf\in
W^{1,\varphi}(\mathbb{R}^{n})$ and the inequality

$\displaystyle|D_{i}Mf(x)|\leq MD_{i}f(x),$ (2)

is satisfied for all $i=1,\ldots,n$ and for almost all
$x\in\mathbb{R}^{n}$.444Having in mind Theorem 3.1 we get that $M$ is bunded
from $W^{1,\varphi}(\mathbb{R}^{n})$ to $W^{1,\varphi}(\mathbb{R}^{n})$.

Let us remark that in the setting of $W^{1,p}$ spaces the above theorem has
been proven by Kinnunen [9].

###### Proof.

For $r>0$ we denote by $\chi_{B(0,r)}$ a characteristic function of the ball
$B(0,r)$. Let

$h_{r}=\frac{1}{|B(0,r)|}\chi_{B(0,r)}.$

Then, we have

$|f|*h_{r}(x)=\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r)}|f(y)|\,dy\leq Mf(x).$

This estimate and properties of convolution yields

$|D_{i}(|f|*h_{r})(x)|=|(D_{i}|f|)*h_{r}(x)|\leq MD_{i}|f|(x)$

for almost all $x\in\mathbb{R}^{n}$. By Theorem 3.1 $Mf,MD_{i}|f|\in
L^{\varphi}(\mathbb{R}^{n})$, and therefore $|f|*h_{r}\in
W^{1,\varphi}(\mathbb{R}^{n})$. Let $r_{m}$ be a sequence of all rational
positive numbers, then

$Mf=\sup_{m}|f|*h_{r_{m}}.$

For $g_{k}=\max_{1\leq m\leq k}|f|*h_{r_{m}}$ we have

$\displaystyle|D_{i}g_{k}(x)|\leq\max_{1\leq m\leq
k}\left|D_{i}(|f|*h_{r_{m}})(x)\right|\leq M(D_{i}|f|)(x)=M(D_{i}f)(x)$ (3)

for almost all $x\in\mathbb{R}^{n}$ and all $k\in\mathbb{N}$. Taking into
account (3) and Theorem 3.1, we have

$\displaystyle\|g_{k}\|_{1,\varphi}\leq\|Mf\|_{\varphi}+\sum_{i=1}^{n}\|M(D_{i}f)\|_{\varphi}\leq
C\|f\|_{1,\varphi}.$

In this way we have proven the boudedness of the sequence $g_{k}$ in
$W^{1,\varphi}(\mathbb{R}^{n})$. Since $W^{1,\varphi}(\mathbb{R}^{n})$ is
reflexive, we have a subsequence of $g_{k}$ (still denoted as $g_{k}$) which
is weakly convergent in $W^{1,\varphi}(\mathbb{R}^{n})$. On the other hand, we
have that $g_{k}$ is the increasing sequence converging pointwisely to $Mf$.
Thus, we obtain $g_{k}\rightharpoonup Mf$ in $W^{1,\varphi}(\mathbb{R}^{n})$.
Therefore, from inequality (3) we get

$\int_{A}D_{i}Mf(x)\,dx\leq\int_{A}M(D_{i}f)(x)\,dx\quad\text{ and
}\quad\int_{A}D_{i}Mf(x)\,dx\geq-\int_{A}M(D_{i}f)(x)\,dx,$

for every measurable set $A$ with finite measure. This finishes the proof. ∎

##  4\. Auxiliary results

For $f\in L_{\textnormal{loc}}^{1}(\mathbb{R}^{n})$ and $x\in\mathbb{R}^{n}$
we define (see [10])

$\mathcal{R}f(x)=\left\\{r\geq
0\,:\,\exists(\\{r_{k}\\}\subset(0,\infty))\,r_{k}\to r\wedge
Mf(x)=\lim_{k\to\infty}\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{k})}|f(y)|\,dy\right\\}.$

######  Proposition 4.1.

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (aDec) and
(A0), then for every $f\in L^{\varphi}(\mathbb{R}^{n})$, the following
statements hold.

  * (i)

For all $x\in\mathbb{R}^{n}$ the set $\mathcal{R}f(x)$ is nonempty.

  * (ii)

For all $x\in\mathbb{R}^{n}$ and $r>0$ such that $r\in\mathcal{R}f(x)$ the
equality

$Mf(x)=\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r)}|f(y)|\,dy$

holds.

  * (iii)

For almost all $x\in\mathbb{R}^{n}$ if $0\in\mathcal{R}f(x)$, then

$Mf(x)=|f(x)|.$

###### Proof.

(i) The statement is obvious for $f=0$. Thus, we assume that $f\not=0$. In
this case $Mf(x)>0$ for any $x\in\mathbb{R}^{n}$. Next, for
$x\in\mathbb{R}^{n}$ there exists a sequence $\\{r_{n}\\}\subset(0,\infty)$
such that

$\lim_{n\to\infty}\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{n})}|f(y)|\,dy=Mf(x).$
(4)

Let $p\in(1,\infty)$ be such that $\varphi$ satisfies
$\textnormal{{(aDec)}}_{p}$. We know that $\varphi^{*}$ satisfies
$\textnormal{{(aInc)}}_{p^{\prime}}$ [6, Proposition 2.4.9], and let
$a\in[1,\infty)$ be a constant from the definition of
$\textnormal{{(aInc)}}_{p^{\prime}}$. We also know that $\varphi^{*}$
satisfies (A0)[6, Lemma 3.7.6]. Let $\beta\in(0,1]$ be a constant for
$\varphi^{*}$ from Proposition 2.2.

Let us take $r>0$ such that $|B(x,r)|\geq 1$, then by the Hölder inequality,
inequality (1) and Proposition 2.2 we get the string of inequalities

$\displaystyle\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r)}|f(y)|\,dy$
$\displaystyle\leq\frac{2}{\beta|B(x,r)|}\|f\|_{\varphi}\|\beta\chi_{B(x,r)}\|_{\varphi^{*}}$
$\displaystyle\leq\frac{2}{\beta|B(x,r)|}\|f\|_{\varphi}\max\\{(a\rho_{\varphi^{*}}(\beta\chi_{B(x,r)}))^{\frac{1}{p^{\prime}}},1\\}$
$\displaystyle\leq\frac{2a^{\frac{1}{p^{\prime}}}}{\beta}\|f\|_{\varphi}|B(x,r)|^{\frac{1}{p^{\prime}}-1}.$
(5)

The above inequality yields that $\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r)}|f(y)|\,dy\to 0$ when
$r\to\infty$. Hence, by (4) and since $Mf(x)>0$, we get that $\\{r_{n}\\}$ is
bounded. Thus, there exist a subsequence $\\{r_{n_{k}}\\}$ and
$r\in[0,\infty)$ such that $r_{n_{k}}\to r$. Therefore, we obtain
$r\in\mathcal{R}f(x)$.

(ii) follows from the continouity of integral with respect to measure, and
(iii) is a direct consequence of the Lebesgue differentiation Theorem, since
$f\in L^{1}_{loc}(\mathbb{R}^{n})$. ∎

######  Lemma 4.2.

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (aDec) (A0)
and let $R>0$. Then, for $f\in L^{\varphi}(\mathbb{R}^{n})$ such that
$f\not=0$ we have

$\sup\left\\{r\,:\,\exists(x\in B(0,R))\,r\in\mathcal{R}f(x)\right\\}<\infty.$

###### Proof.

Let us suppose $\sup\left\\{r\,:\,\exists(x\in
B(0,R))\,r\in\mathcal{R}f(x)\right\\}=\infty$, then there exist sequences
$\\{r_{k}\\}\subset(0,\infty)$ and $\\{x_{k}\\}\subset B(0,R)$ such that

$r_{k}\to\infty\quad\textrm{and}\quad r_{k}\in\mathcal{R}f(x_{k}).$

Let $\widetilde{R}>0$ be such that $\int_{B(0,\widetilde{R})}|f|dx>0$ and
define $\widehat{R}=\max\left\\{R,\widetilde{R}\right\\}$. Let us observe that
for any $x\in B(0,R)$ we have

$Mf(x)\geq\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,2\widehat{R})}|f(y)|\,dy\geq\frac{1}{|B(0,2\widehat{R})|}\int_{B(0,\widehat{R})}|f(y)|\,dy>0.$

Therefore, taking into account Proposition 4.1 and inequality (4), we get

$\displaystyle\frac{1}{|B(0,2\widehat{R})|}\int_{B(0,\widehat{R})}|f(y)|\,dy$
$\displaystyle\leq Mf(x_{k})=\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x_{k},r_{k})}|f(y)|\,dy$
$\displaystyle\leq\frac{2a^{\frac{1}{p^{\prime}}}}{\beta}\|f\|_{\varphi}|B(x_{k},r_{k})|^{\frac{1}{p^{\prime}}-1}\to
0,$

and this gives us a contradiction. ∎

######  Corollary 4.3.

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (aDec) (A0)
and let $f\in L^{\varphi}(\mathbb{R}^{n})$ such that $f\not=0$. Then, for
every Lebesgue point $x$ of $f$ the set $\mathcal{R}f(x)$ is compact.

######  Lemma 4.4.

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (aDec) and
(A0). Let $f,f_{m}\in L^{\varphi}(\mathbb{R}^{n})$ be such that $f_{m}\to f$
in $L^{\varphi}(\mathbb{R}^{n})$. If $f\not=0$ and $f_{m}\not=0$, then for all
$R>0$ there exists $\mathfrak{R}>0$ such that

$\sup\left\\{r\,:\,\exists(x\in
B(0,R))\,r\in\mathcal{R}f_{m}(x)\right\\}\leq\mathfrak{R}$

holds for all $m\in\mathbb{N}$.

###### Proof.

Let us suppose that the thesis of the lemma dos not hold. Thus, there exist a
sequence $\\{m_{k}\\}\subset\mathbb{N}$, a sequence of points
$\\{x_{k}\\}\subset B(0,R)$ and a sequence $\\{r_{k}\\}$ such that for all $k$
we have

$r_{k}\in\mathcal{R}f_{m_{k}}(x_{k})\quad\textrm{ and }\quad r_{k}\to\infty.$

Let us observe that if $\\{m_{k}\\}$ is bounded, then there exists a
subsequence $\\{m_{k_{l}}\\}$ such that $f_{m_{k_{l}}}$ is constant and
$r_{k_{l}}\in\mathcal{R}f_{m_{k_{l}}}(x_{k})$ with $r_{k_{l}}\to\infty$. But
this contradicts to Lemma 4.2.

Therefore, we assume that $\\{m_{k}\\}$ is unbounded. Hence, there exists an
increasing subsequence $\\{m_{k_{l}}\\}\rightarrow\infty$. Let us denote
$M=\sup_{m}\|f_{m}\|_{\varphi}$ and let $p\in(1,\infty)$ be such $\varphi$
satisfies $\textnormal{{(aDec)}}_{p}$. By inequality (4) we have

$\displaystyle Mf_{m_{k_{l}}}(x_{k_{l}})=\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x_{k_{l}},r_{k_{l}})}|f_{m_{k_{l}}}|(y)dy\leq
C\|f_{m_{k_{l}}}\|_{\varphi}|B(0,r_{k_{l}})|^{-1/p}\leq
CM|B(0,r_{k_{l}})|^{-1/p}\to 0.$ (6)

Let $\widetilde{R}$ be such that $\int_{B(0,\widetilde{R})}|f|dx>0$ and define
$\widehat{R}=\max\left\\{R,\widetilde{R}\right\\}$. For all $m\in\mathbb{N}$
and for all $x\in B(0,R)$ we have

$\displaystyle Mf_{m}(x)$
$\displaystyle\geq\frac{1}{B(0,2\widehat{R})}\int_{B(0,\widehat{R})}|f_{m}(y)|dy$
$\displaystyle\geq\frac{1}{B(0,2\widehat{R})}\int_{B(0,\widehat{R})}|f(y)|dy-\frac{1}{B(0,2\widehat{R})}\int_{B(0,\widehat{R})}|f(y)-f_{m}(y)|\,dy.$
(7)

Since $f_{m}\rightarrow f$ in $L^{\varphi}(\mathbb{R}^{n})$, by Proposition
2.3 there exists $N\in\mathbb{N}$ such that for all $m\geq N$ we have

$\frac{1}{B(0,2\widehat{R})}\int_{B(0,\widehat{R})}|f(y)-f_{m}(y)|\,dy\leq\frac{1}{2B(0,2\widehat{R})}\int_{B(0,\widehat{R})}|f(y)|\,dy.$

Taking into account the above inequality and (4), we obtain for $m\geq N$ the
following bound

$Mf_{m}(x)\geq\frac{1}{2B(0,2\widehat{R})}\int_{B(0,\widehat{R})}|f(y)|\,dy.$

Thus, since $m_{k_{l}}\geq N$ for large $l$, the above inequality contradicts
to (6). ∎

######  Lemma 4.5.

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (A0) and
(aDec), $f\in L^{\varphi}(\mathbb{R}^{n})$ such that $f\neq 0$ and $R>0$.
Moreover, let $\mathfrak{R}>{R_{0}}+R$, where

$R_{0}=\sup\left\\{r\,:\,\exists(x\in B(0,R))\,r\in\mathcal{R}f(x)\right\\},$

and let $g\in L^{1}_{\textnormal{loc}}(\mathbb{R}^{n})$ be such

$g(x)=f(x)\textrm{ for almost all }x\in
B(0,\mathfrak{R})\quad\textrm{and}\quad|g(x)|\leq|f(x)|\textrm{ for almost all
}x\in\mathbb{R}^{n}\setminus B(0,\mathfrak{R}),$

then the following

$\mathcal{R}f(x)=\mathcal{R}g(x)\qquad\textrm{and}\qquad Mf(x)=Mg(x),$

hold for all $x\in B(0,R)$.

###### Proof.

Let $r\in\mathcal{R}f(x)$, then there exists $\\{r_{m}\\}\subset(0,\infty)$
such that

$r_{m}\rightarrow
r\qquad\textrm{and}\qquad\lim_{m\rightarrow\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{m})}|f(y)|\,dy=Mf(x).$

Since $r\leq R_{0}$, we can assume that $r_{m}\leq\mathfrak{R}-R$. Therefore,
we obtain $B(x,r_{m})\subset B(0,\mathfrak{R})$, and we get

$Mg(x)\geq\lim_{m\rightarrow\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{m})}|g(y)|\,dy=\lim_{m\rightarrow\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{m})}|f(y)|\,dy=Mf(x)\geq
Mg(x).$

This yields

$\lim_{m\rightarrow\infty}\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{m})}|g(y)|\,dy=Mg(x).$

Thus, $r\in\mathcal{R}g(x)$ and we finished the proof of the inclusion
$\mathcal{R}f(x)\subset\mathcal{R}g(x)$.

Now, let $\rho\in\mathcal{R}g(x)$ and let us take
$\\{\rho_{m}\\}\subset(0,\infty)$ such that

$\rho_{m}\rightarrow\rho\qquad\textrm{and}\qquad\lim_{m\rightarrow\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,\rho_{m})}|g(y)|\,dy=Mg(x).$

We also take555We can do it since by Proposition 4.1 the set $\mathcal{R}f(x)$
is nonempty. $r\in\mathcal{R}f(x)$ and a sequence $\\{r_{n}\\}$ such that

$r_{m}\rightarrow
r\qquad\textrm{and}\qquad\lim_{m\rightarrow\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{m})}|f(y)|\,dy=Mf(x).$

Since $r\leq R_{0}$, we can also assume that $r_{m}\leq\mathfrak{R}-R$ for all
$m$. We obtain

$\displaystyle Mf(x)$
$\displaystyle\geq\limsup_{m\to\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,\rho_{m})}|f(y)|\,dy\geq\liminf_{m\to\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,\rho_{m})}|f(y)|\,dy\geq\lim_{m\to\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,\rho_{m})}|g(y)|\,dy=Mg(x)$
$\displaystyle\geq\limsup_{m\to\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{m})}|g(y)|\,dy=\lim_{m\to\infty}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r_{m})}|f(y)|\,dy=Mf(x).$

It yields $\rho\in\mathcal{R}f(x)$, and we get
$\mathcal{R}f(x)=\mathcal{R}g(x)$. The equality $Mf(x)=Mg(x)$ follows from the
proven inequalities. ∎

######  Lemma 4.6.

Let $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies (A0),
(aDec)and (aInc). If $f_{m},f\in L^{\varphi}(\mathbb{R}^{n})$ and
$f_{m}\rightarrow f$ in $L^{\varphi}(\mathbb{R}^{n})$, then for all $R>0$ and
$\lambda>0$ the set $\left\\{x\in
B(0,R)\,:\,\mathcal{R}f_{m}(x)\not\subset\mathcal{R}f(x)_{(\lambda)}\right\\}$
is measurable and666For nonempty set $A\subset\mathbb{R}^{n}$ and $\lambda\geq
0$ we denote $A_{(\lambda)}=\\{x\in\mathbb{R}^{n}:dist(x,A)\leq\lambda\\}$.

$\lim_{m\rightarrow\infty}\left|\left\\{x\in
B(0,R)\,:\,\mathcal{R}f_{m}(x)\not\subset\mathcal{R}f(x)_{(\lambda)}\right\\}\right|=0.$

###### Proof.

If $f=0$, then the lemma is obviously true. It is easy to see that if $f\neq
0$, then $f_{m}\neq 0$ for large $m$. Therefore, we can assume that $f\neq 0$
and $f_{m}\neq 0$. From Lemma 4.2 and Lemma 4.4 there exists $\mathfrak{R}>0$
such that for all $x\in B(0,R)$, $m$ and
$r\in\mathcal{R}f_{m}(x)\cup\mathcal{R}f(x)$ we have $r\leq\mathfrak{R}$. We
define

$g=\chi_{B(0,R+\mathfrak{R})}f\quad\textrm{and}\quad
g_{m}=\chi_{B(0,R+\mathfrak{R})}f_{m}.$

It is obvious that $g_{m}\to g$ in $L^{\varphi}(\mathbb{R}^{n})$ and in
$L^{1}(\mathbb{R}^{n})$ by Proposition 2.3. By Lemma 4.5 we have

$\mathcal{R}g(x)=\mathcal{R}f(x)\quad\textrm{and}\quad\mathcal{R}g_{m}(x)=\mathcal{R}f_{m}(x).$

for any $x\in B(0,R)$. Thus, it yields

$\displaystyle\left\\{x\in
B(0,R)\,:\,\mathcal{R}f_{m}(x)\not\subset\mathcal{R}f(x)_{(\lambda)}\right\\}=\left\\{x\in
B(0,R)\,:\,\mathcal{R}g_{m}(x)\not\subset\mathcal{R}g(x)_{(\lambda)}\right\\}.$
(8)

From Lemma 4.6 for $L^{1}$ -space, which was proven in [10], we obtain

$\left|\left\\{x\in
B(0,R)\,:\,\mathcal{R}g_{m}(x)\not\subset\mathcal{R}g(x)_{(\lambda)}\right\\}\right|\to
0,$

and this finishes the proof. Let us stress that the measurability of the set
$\left\\{x\in
B(0,R)\,:\,\mathcal{R}f_{m}(x)\not\subset\mathcal{R}f(x)_{(\lambda)}\right\\}$
follows from (8) and the measurability of the set $\left\\{x\in
B(0,R)\,:\,\mathcal{R}g_{m}(x)\not\subset\mathcal{R}g(x)_{(\lambda)}\right\\}$,
which was proven in [10]. ∎

##  5\. Continuity of the maximal operator

The following theorem is the main result of the paper.

######  Theorem 5.1.

Let us assume that $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies
(A0), (A1), (A2), (aInc) and (aDec), then the maximal operator

$M\colon W^{1,\varphi}(\mathbb{R}^{n})\to W^{1,\varphi}(\mathbb{R}^{n})$

is continuous.

###### Proof.

First of all we shall prove the formula for weak derivatives of the maximal
function.

######  Lemma 5.2.

Let us assume that $\varphi\in\Phi_{\textnormal{w}}(\mathbb{R}^{n})$ satisfies
(A0), (A1), (A2), (aInc) and (aDec). If $f\in W^{1,\varphi}(\mathbb{R}^{n})$,
then for almost all $x\in\mathbb{R}^{n}$ and for all $i=1,\ldots,n$ we have

$\displaystyle\begin{split}D_{i}Mf(x)&=\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r)}D_{i}|f|(y)\,dy\textrm{
for all $r\in\mathcal{R}f(x)$, $r>0$ and}\\\ D_{i}Mf(x)&=D_{i}|f|(x)\textrm{
if $0\in\mathcal{R}f(x)$.}\end{split}$ (9)

###### Proof.

We can assume that $f\neq 0$. Let $R>0$ and let $\mathfrak{R}=R+R_{0}+1$,
where

$R_{0}=\sup\left\\{r\,:\,\exists(x\in B(0,R))\,r\in\mathcal{R}f(x)\right\\}.$

Let us take $\psi\in C^{\infty}_{\textnormal{c}}(\mathbb{R}^{n})$ such that
$0\leq\psi\leq 1$ and $\psi(x)=1$ for all $x\in B(0,\mathfrak{R})$.

Next, we define $g=f\psi$. Let $p\in(1,\infty)$ be such that $\varphi$
satisfies $\textnormal{{(aInc)}}_{p}$. It is obvious that $g\in
W^{1,\varphi}(\mathbb{R}^{n})$ and $g\in W^{1,p}(\mathbb{R}^{n})$ by
Proposition 2.3. By Lemma 4.5, for $x\in B(0,R)$ we have

$\displaystyle\mathcal{R}f(x)=\mathcal{R}g(x)\quad\textrm{and}\quad
Mf(x)=Mg(x).$ (10)

Since Lemma 5.2 holds for $W^{1,p}$ spaces (see [10]), we have that (9) holds
for $g$. Therefore, having in mind (10) and the form of $g$ we easily get the
thesis. ∎

Let $f_{m}\to f$ in $W^{1,\varphi}(\mathbb{R}^{n})$. We shall show that
$Mf_{m}\to Mf$ in $W^{1,\varphi}(\mathbb{R}^{n})$. Due to Theorem 3.1, it is
enough to show that $D_{i}f_{m}\to D_{i}f$ in $L^{\varphi}(\mathbb{R}^{n})$
for all $1\leq i\leq n$.

Let us fix $\epsilon>0$, then by Proposition 2.5 there exist $R>0$ and
$\lambda>0$ such that $\|MD_{i}f\|_{{p(\cdot)},\mathbb{R}^{n}\setminus
B(0,R)}<\epsilon/2$ and $\|MD_{i}f\|_{{p(\cdot)},A}<\epsilon/2$ whenever
$|A|<\lambda$. For every $x\in\mathbb{R}^{n}$ we define functions
$u_{x},u^{m}_{x}\colon[0,\infty)\to\mathbb{R}$ as follows

$\displaystyle u_{x}(r)=\begin{cases}\mathop{\mathchoice{\kern
1.28333pt\vbox{\hrule width=3.50002pt}\kern-4.78336pt}{\kern
1.16664pt\vbox{\hrule width=3.50002pt}\kern-4.66666pt}{\kern
0.91664pt\vbox{\hrule width=2.75pt}\kern-3.66664pt}{\kern
0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r)}D_{i}f&\textrm{ if
}r>0,\\\ D_{i}f(x)&\textrm{ if }r=0\end{cases}\quad\quad
u^{m}_{x}(r)=\begin{cases}\mathop{\mathchoice{\kern 1.28333pt\vbox{\hrule
width=3.50002pt}\kern-4.78336pt}{\kern 1.16664pt\vbox{\hrule
width=3.50002pt}\kern-4.66666pt}{\kern 0.91664pt\vbox{\hrule
width=2.75pt}\kern-3.66664pt}{\kern 0.74997pt\vbox{\hrule
width=2.25pt}\kern-2.99998pt}}\mathopen{}\int_{B(x,r)}D_{i}f_{m}&\textrm{ if
}r>0,\\\ D_{i}f_{m}(x)&\textrm{ if }r=0.\end{cases}$

By the Lebesgue differentiation Theorem and since
$\lim_{r\rightarrow\infty}u_{x}(r)=0$, we have that $u_{x}$ is uniformly
continuous for almost all $x\in\mathbb{R}^{n}$ and $u_{x}$ is continuous on
$(0,\infty)$ for all $x\in\mathbb{R}^{n}$. In other words the set

$\mathcal{N}=\left\\{x\in B(0,R)\,:\,u_{x}\text{ is not uniformly
continuous}\right\\}$

is a null set. Next, for $\delta>0$ we define

$C_{\delta}=\left\\{x\in B(0,R)\,:\,\exists(r_{1},r_{2}\in[0,\infty))\
|r_{1}-r_{2}|\leq\delta\textrm{ and
}|u_{x}(r_{1})-u_{x}(r_{2})|>\epsilon/\|\chi_{B(0,R)}\|_{\varphi}\right\\}.$

Since $\bigcap\limits_{\delta>0}C_{\delta}\subset\mathcal{N}$ and
$C_{\delta_{1}}\subset C_{\delta_{2}}$ when $\delta_{1}<\delta_{2}$, we can
choose777Since
$C_{\delta}=\bigcup_{\begin{subarray}{c}r_{1},r_{2}\in[0,\infty)\cap\mathbb{Q}\\\
|r_{1}-r_{2}|\leq\delta\end{subarray}}\left\\{x\in
B(0,R)\,:\,|u_{x}(r_{1})-u_{x}(r_{2})|>\epsilon/\|\chi_{B(0,R)}\|_{\varphi}\right\\}$,
we have that $C_{\delta}$ is measurable.  $\delta_{0}$ such that
$|C_{\delta_{0}}|<\frac{\lambda}{2}$. Due to Lemma 4.6 there exists $m_{0}$
such that

$\left|\left\\{x\in
B(0,R)\,:\,\mathcal{R}f_{m}(x)\not\subset\mathcal{R}f(x)_{(\delta_{0})}\right\\}\right|<\frac{\lambda}{2},\quad\text{
for }m\geq m_{0}.$

We denote the set in the above formula by $C^{m}$. Let us also denote by
$\Psi$ a set of $x\in\mathbb{R}^{n}$ such that (9) is satisfied for $f$ and
$f_{m}$ for all $m$ and $x$ is the Lebesgue point of $f,f_{m},D_{i}f$ and
$D_{i}f_{m}$ for all $m$. From Lemma 5.2 and the Lebesgue differentiation
Theorem we have $|\mathbb{R}^{n}\setminus\Psi|=0$.

Now, let us take $m\geq m_{0}$, $x\in\Psi$, then for
$r_{1}\in\mathcal{R}f_{m}(x)$ and $r_{2}\in\mathcal{R}f(x)$ we have

$\displaystyle\left|D_{i}Mf_{m}(x)-D_{i}Mf(x)\right|$
$\displaystyle=\left|u_{x}^{m}(r_{1})-u_{x}(r_{2})\right|$
$\displaystyle\leq\left|u_{x}^{m}(r_{1})-u_{x}(r_{1})\right|+\left|u_{x}(r_{1})-u_{x}(r_{2})\right|$
$\displaystyle\leq
M\left(D_{i}f_{m}-D_{i}f\right)(x)+\left|u_{x}(r_{1})-u_{x}(r_{2})\right|.$
(11)

Next, if $x\in(\mathbb{R}^{n}\setminus(C_{\delta_{0}}\cup
C^{m}\cup(\mathbb{R}^{n}\setminus B(0,R)))\cap\Psi$ we can find
$r_{1}\in\mathcal{R}f_{m}(x)$ and $r_{2}\in\mathcal{R}f(x)$ such, that888The
compactness of $\mathcal{R}f(x)$ is used in this place.
$|r_{1}-r_{2}|\leq\delta_{0}$. Therefore, by the definition of
$C_{\delta_{0}}$, we obtain

$\left|u_{x}(r_{1})-u_{x}(r_{2})\right|\leq\epsilon/\|\chi_{B(0,R)}\|_{\varphi}.$

If $x\in(C_{\delta_{0}}\cup C^{m}\cup(\mathbb{R}^{n}\setminus
B(0,R))\cap\Psi$, then we use the following bound

$\left|u_{x}(r_{1})-u_{x}(r_{2})\right|\leq 2MD_{i}f(x).$

Combing those inequalities with (5) we get

$\displaystyle\left\|D_{i}Mf_{m}-D_{i}Mf\right\|_{\varphi,\mathbb{R}^{n}}$
$\displaystyle\leq\left\|M(D_{i}f_{m}-D_{i}f)\right\|_{\varphi,\mathbb{R}^{n}}+\|\tfrac{\epsilon}{\|\chi_{B(0,R)}\|_{\varphi}}\|_{\varphi,B(0,R)}$
$\displaystyle+\left\|2MDf\right\|_{\varphi,C_{\delta_{0}}\cup
C^{m}}+\left\|2MDf\right\|_{\varphi,\mathbb{R}^{n}\setminus B(0,R)}$
$\displaystyle\leq
C\left\|D_{i}f_{m}-D_{i}f\right\|_{\varphi,\mathbb{R}^{n}}+3\epsilon.$

Finally, since $D_{i}f_{m}\to D_{i}f$ in $W^{1,\varphi}(\mathbb{R}^{n})$, the
proof follows. ∎

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$\displaystyle-\theta(P-1)\dfrac{\beta^{{}^{\prime}}\gamma}{4}(\bar{q}_{2}^{P/2}\bar{p}_{2}-\bar{q}_{1}^{P/2}\bar{p}_{1})+\frac{1}{4}{\beta^{{}^{\prime}}}^{2}\gamma.$
(187)

Similar to RS assumption, a must satisfy the condition $a=P-1$ and the
previous expression for even $P\geq 4$ reads as

$\displaystyle\mathcal{A}^{(P)}=\ln
2+\frac{1}{\theta}\mathbb{E}_{1}\ln\mathbb{E}_{2}\cosh^{\theta}g(\bm{J},\bar{m})-\dfrac{\gamma\beta^{{}^{\prime}}}{4}\bar{q}_{2}^{P/2-1}\bar{p}_{2}\Big{(}P-(P-1)\bar{q}_{2}\Big{)}$
$\displaystyle+\frac{\beta^{{}^{\prime}}}{2}\bar{m}^{P}(1-P)-\theta(P-1)\dfrac{\beta^{{}^{\prime}}\gamma}{4}(\bar{q}_{2}^{P/2}\bar{p}_{2}-\bar{q}_{1}^{P/2}\bar{p}_{1})+\frac{1}{4}{\beta^{{}^{\prime}}}^{2}\gamma+\mathcal{O}(N^{-1})$
(188)

in the thermodynamics limit ($N\to\infty$) the correction terms will vanish. ∎

##  Appendix D Proof of Lemma 2

We prove only (51) regarding Lemma 2, being the proofs for the others obtained
in the same way. First of all, using (50) we see that

$\displaystyle\partial_{t}\mathcal{A}_{N}=\frac{\beta^{{}^{\prime}}}{2}\langle
m_{1}^{P}\rangle+\sqrt{\frac{\beta^{{}^{\prime}}}{N^{P-1}}}\frac{1}{2N\sqrt{t}}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\left[\mathcal{W}_{2}\sum_{\bm{i},\mu}\xi_{i}^{\mu}\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})\right]$
(189)

Now, using Stein’s lemma (24), we may rewrite the second member of (189) as

$\displaystyle\sqrt{\frac{\beta^{{}^{\prime}}}{N^{P-1}}}\frac{1}{2N\sqrt{t}}\sum_{\bm{i},\mu}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\left[\partial_{\eta_{\bm{i}}^{\mu}}\bigg{(}\mathcal{W}_{2}\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})\bigg{)}\right]=D_{1}+D_{2}+D_{3}$
(190)

Let’s investigate those three terms:

$\displaystyle D_{1}=$
$\displaystyle\sqrt{\frac{\beta^{{}^{\prime}}}{N^{P-1}}}\frac{1}{2N\sqrt{t}}\sum_{\bm{i},\mu}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\bigg{[}\mathcal{W}_{2}\partial_{\xi_{i}^{\mu}}\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})\bigg{]}=$
$\displaystyle=$
$\displaystyle\frac{\beta^{{}^{\prime}}}{2N^{P}}\sum_{\bm{i},\mu}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\left[\mathcal{W}_{2}\omega((\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})^{2})\right]-\frac{\beta^{{}^{\prime}}}{2N^{P}}\sum_{\bm{i},\mu}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\left[\mathcal{W}_{2}\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})^{2}\right]=$
$\displaystyle=$
$\displaystyle\frac{\beta^{{}^{\prime}}K}{2N^{P/2}}\left[\langle
p_{11}\rangle-\langle p_{12}q_{12}^{P/2}\rangle_{2}\right]$ (191)
$\displaystyle D_{2}=$
$\displaystyle\sqrt{\frac{\beta^{{}^{\prime}}}{N^{P-1}}}\frac{1}{2N\sqrt{t}}\sum_{\bm{i},\mu}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\left[\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})\frac{\partial_{\eta_{\bm{i}}^{\mu}}\mathcal{Z}_{2}^{\theta}}{\mathbb{E}_{2}\left(\mathcal{Z}_{2}^{\theta}\right)}\right]=$
$\displaystyle=$
$\displaystyle\frac{\beta^{{}^{\prime}}\theta}{2N^{P}}\sum_{\bm{i},\mu}\mathbb{E}_{0}\left\\{\mathbb{E}_{1}\left[\mathbb{E}_{2}\left(\mathcal{W}_{2}\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})^{2}\right)\right]\right\\}=\frac{\beta^{{}^{\prime}}K}{2N^{P/2}}\theta\langle
p_{12}q_{12}^{P/2}\rangle_{2}$ (192) $\displaystyle D_{3}=$
$\displaystyle\sqrt{\frac{\beta^{{}^{\prime}}}{N^{P-1}}}\frac{1}{2N\sqrt{t}}\sum_{\bm{i},\mu}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\left[\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu}){\mathcal{Z}_{2}^{(P)}}^{\theta}\partial_{\eta_{\bm{i}}^{\mu}}\frac{1}{\mathbb{E}_{2}\left({\mathcal{Z}_{2}^{(P)}}^{\theta}\right)}\right]=$
$\displaystyle=$
$\displaystyle-\frac{\beta^{{}^{\prime}}\theta}{2N^{P}}\sum_{\bm{i},\mu}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\left[\omega(\sigma_{i}\tau_{\mu})\mathcal{W}_{2}\mathbb{E}_{2}\left(\mathcal{W}_{2}\frac{\partial_{\eta_{\bm{i}}^{\mu}}\mathcal{Z}_{2}^{(P)}}{\mathcal{Z}_{2}^{(P)}}\right)\right]=$
$\displaystyle=$
$\displaystyle-\frac{\beta^{{}^{\prime}}\theta}{2N^{P}}\sum_{i,\mu}\mathbb{E}_{0}\mathbb{E}_{1}\mathbb{E}_{2}\left[\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})\mathcal{W}_{2}\mathbb{E}_{2}\left(\omega(\sigma_{i_{1}}\ldots\sigma_{i_{P/2}}\tau_{\mu})\mathcal{W}_{2}\right)\right]=-\frac{\beta^{{}^{\prime}}K}{2N^{P/2}}\theta\langle
p_{12}q_{12}^{P/2}\rangle_{1}$ (193)

Putting (D), (192) and (193) inside (190), and (190) inside (189) we find
(51).

## Acknowledgments

The Authors acknowledge INFN (FIELDTURB) and INdAM (GNFM) for providing
computational facilities and the grants by MUR (PRIN 2017, Project no.
2017JFFHSH, Stochastic Models for Complex Systems) and by MEACI (Project
”BULBUL”: Scientific, technological and industrial cooperation between Italy
and Israel).

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