\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 165, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2011/165\hfil Talenti's inequality]
{Consequences of Talenti's inequality becoming equality}

\author[B. Emamizadeh, M. Zivari-Rezapour\hfil EJDE-2011/165\hfilneg]
{Behrouz Emamizadeh, Mohsen Zivari-Rezapour}  % in alphabetical order

\address{Behrouz Emamizadeh \newline
Department of Mathematical Sciences,
Xi'an Jiaotong-Liverpool University\\
111 Ren'ai Road, Suzhou Dushu Lake Higher Education Town \\
Suzhou Industrial Park, Suzhou, Jiangsu  215123 China}
\email{Behrouz.Emamizadeh@xjtlu.edu.cn}

\address{Mohsen Zivari-Rezapour \newline
Faculty of Mathematical  \& Computer Sciences\\
Shahid Chamran University, Ahvaz, Iran}
\email{mzivari@scu.ac.ir}

\thanks{Submitted June 14, 2011. Published December 13, 2011.}
\subjclass[2000]{35J62, 35P30, 35Q74, 35A99, 49J20}
\keywords{Talenti's inequality; $p$-Laplacian operator;
elastic membrane; \hfill\break\indent
rearrangement maximization; unique optimal solution}

\begin{abstract}
 In this article, we consider the case of equality in a well known
 inequality for the  $p$-Laplacian due to Giorgio Talenti.
 Our approach seems to be simpler than the one by Kesavan \cite{kes}.
 We use a result from rearrangement optimization to prove
 the main result in this article. Some physical interpretations
 are also presented.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\newcommand{\Om}{\Omega}
\newcommand{\bd}{\partial\Omega}
\newcommand{\cf}{\cal F}
\newcommand{\intom}{\int_{\Omega}}
\newcommand{\intbom}{\int_{\partial\Omega}}
\newcommand{\phs}{\Phi_{\lambda,s}}

\section{Introduction}

The initial-boundary value problem
\begin{equation}\label{ibvp}
\begin{gathered}
u_{tt}-\Delta u+c(x)u=f(x),\quad\text{in } D\times [0,T]  \\
u=0,\quad\text{on }\partial D\times [0,T]  \\
u=g,\quad u_t=h,\quad\text{on }D\times \{t=0\}
\end{gathered}
\end{equation}
models the vibration of a non-isotropic elastic membrane subject
to an external force. The
steady-state version of \eqref{ibvp} is
\begin{equation}\label{steady}
\begin{gathered}
 -\Delta u+c(x)u=f(x)\quad\text{in }D \\
u=0\quad\text{on }\partial D.
\end{gathered}
\end{equation}
In \eqref{steady} we can suppose $f(x)$ represents a vertical force
on the membrane such as a load distribution, and a non-zero
function $c(x)$ guarantees the membrane is made of several
different materials, hence it is non-isotropic.
Our purpose in this note is related to a common generalization
of \eqref{steady}, namely
\begin{equation}\label{bvp}
\begin{gathered}
-\Delta_p u + c(x)|u|^{p-2}u = f(x), \quad \text{in }D,\\
u=0, \quad\text{on }\partial D,
\end{gathered}
\end{equation}
where $c(x)\geq 0$ is a bounded function, and $f\in L^q(D)$ is
a non-negative function; here $q$ is the conjugate exponent of
$1<p<\infty$; i.e., $\frac{1}{p}+\frac{1}{q}=1$, and $\Delta_p$
denotes the usual $p$-Laplace operator, i.e.
$\Delta_pu=\nabla\cdot (|\nabla u|^{p-2}\nabla u)$. The
``symmetrized" problem corresponding to \eqref{bvp} is the
 boundary value problem
\begin{equation}\label{bvpp}
\begin{gathered}
 -\Delta_p v=f^{\sharp}(x), \quad \text{in }B, \\
v=0, \quad \text{on } \partial B,
\end{gathered}
\end{equation}
where $B$ stands for the ball centered at the origin such that
$\operatorname{Vol}(B)=|D|$. Henceforth, for a measurable
$E\subset\mathbb{R}^n$, $|E|$ indicates the $n$-dimensional Lebesgue
measure of $E$. The notation $f^\sharp$ denotes the standard
Schwarz symmetrization of $f$ on the ball $B$, so $f^\sharp$ is
radial, $f^\sharp (x)=\xi (\|x\|)$, where $\xi$ is a decreasing
function; moreover, $f^\sharp$ and $f$ are equi-measurable:
\[
|\{x\in B: f^\sharp (x)\geq\alpha\}|
=|\{x\in D: f (x)\geq\alpha\}|,
\]
for every $\alpha\geq 0$. The reader may see \cite{ka} for
details. By $u_f\in W^{1,p}_0(D)$ we denote the unique solution of
\eqref{bvp}. It is well known (and easy to prove) that $u_f$ is
the unique minimizer of
\begin{equation}\label{energy}
\Psi (u)=\frac{1}{p}\int_D|\nabla
u|^p\,dx+\frac{1}{p}\int_Dc(x)|u|^p\,dx
-\int_Df(x) u\,dx,
\end{equation}
relative to $u\in W^{1,p}_0(D)$.
Note that \eqref{bvp} satisfies the maximum principle
in the sense that $f\geq 0$ implies $u_f\geq 0$. Indeed,
setting $u\equiv u_f$, and using $-u^-\in W^{1,p}_0(D)$
as a test function,  from \eqref{bvp}, we infer that
\[
\int_D\Delta_p u u^-\,dx
+\int_D c(x)|u|^{p-2}u(-u^-)\,dx=\int_Df(x)(-u^-)\,dx.
\]
Observe that $uu^-=-(u^-)^2$, and $\nabla u\cdot \nabla
u^-=-|\nabla u^-|^2$ on the set $\{x\in D: u(x)\leq 0\}$.
Thus
\begin{align*}
\int_{\{u\leq 0\}}|\nabla u^-|^p\,dx
&\leq \int_{\{u\leq 0\}} |\nabla u^-|^p\,dx
 +\int_{\{u\leq 0\}} c(x)|u^-|^p\,dx\\
&=\int_Df(x) (-u^-)\,dx\leq 0.
\end{align*}
Whence, $u^-$ is  constant. This, in conjunction with the fact
that $u^-$ vanishes on $\partial D$, implies $u^-=0$, hence
$u=u^+\geq 0$. In fact, $u>0$, a result which is implied by the
strong maximum principle, see for example \cite{sabina}.

The following inequality which is attributed to Giorgio Talenti,
see for example \cite{talenti}, has proven to be an instrumental
tool in partial differential equations,
\begin{equation}\label{talenti}
u_f^\sharp (x)\leq v_{f^\sharp}(x),\quad \forall x\in B,
\end{equation}
where $v_{f^\sharp}$ denotes the solution to \eqref{bvpp}. The
objective of this article is to discuss the consequences of
having equality in \eqref{talenti}. Indeed we are able to
prove the following result.

\begin{theorem}\label{thm1}
Suppose equality holds in \eqref{talenti}. Then
\begin{itemize}
\item[(i)] $c(x)\equiv 0$,
\item[(ii)] $D$ and $f$ are equal to $B$ and $f^\sharp$, respectively,
modulo translations.
\end{itemize}
\end{theorem}

\begin{remark} \label{rmk1.1} \rm
In case $c\equiv 0$, Theorem \ref{thm1} has already been proven, see
for example \cite{kes}. However, our proof is simpler than the known
proof. It relies on a result from rearrangement optimization
theory which itself is intuitively easy to grasp.
\end{remark}

This article is organized as follows. In section 2, we collect known
results from the theory of rearrangements specialized to our
purpose, and in addition recall the well known Polya-Szego
inequality. Section 3 is entirely devoted to the proof of Theorem
\ref{thm1}.
In section 4, a generalization of Theorem \ref{crucial} is
presented.

\section{Preliminaries}

Let us begin by recalling the definition of two functions being
rearrangement of each other.

\begin{definition} \label{def} \rm
Given two measurable functions $f, g: D\to\mathbb{R}$, we say
$f$ and $g$ are rearrangements of each other provided
\[
|\{x\in D: f(x)\geq\alpha\}|
=|\{x\in D: g(x)\geq\alpha\}|,\quad \forall\alpha\in\mathbb{R}.
\]
For $f_0\in L^p(D)$, the set comprising all
rearrangements of $f_0$ is denoted by $\mathcal{R}(f_0)$; i.e.,
\[
\mathcal{R}(f_0)=\{f: \text{$f$ and $f_0$ are
rearrangements of each other}\}.
\]
\end{definition}

Using Fubini's Theorem, it is easy to show that
$\mathcal{R}(f_0)\subset L^p(D)$.
Let us denote by $w_f\in W^{1,p}_0(D)$ the unique solution of
\begin{equation}\label{bvppp}
\begin{gathered}
-\Delta_p w=f, \quad \text{in }D,\\
w=0, \quad \text{on } \partial D,
\end{gathered}
\end{equation}
and define the \emph{energy} functional
$\Phi:L^p(D)\to\mathbb{R}$  by
\begin{equation}
\Phi (f)=\int_Df\, w_f\,dx.
\end{equation}
In \cite{em}, amongst other results, the authors proved the
following result which plays a crucial role in what follows.

\begin{theorem}\label{crucial}
The maximization problem
\begin{equation}\label{maximizationn}
\sup_{f\in\mathcal{R}(f_0)}\Phi (f)
\end{equation}
is solvable; i.e., there exists $\hat f\in\mathcal{R}(f_0)$ such
that
\[
\Phi (\hat f)=\sup_{f\in\mathcal{R}(f_0)}\Phi (f).
\]
Moreover, there exists an increasing function $\eta$, unknown a
priori, such that
\begin{equation}\label{Euler}
\hat f=\eta\circ\hat w,
\end{equation}
almost everywhere in $D$, where $\hat w:=w_{\hat f}$.
\end{theorem}

We are  going to need the following result, which can be found in
\cite{zi}.

\begin{theorem}\label{thm3}
Let $u\in W^{1,p}_0(D)$ be non-negative.
Then $u^\sharp \in W^{1,p}_0(B)$, and
\begin{equation}\label{polya}
\int_B|\nabla u^\sharp |^p\,dx\leq \int_D|\nabla u|^p\,dx.
\end{equation}
Moreover, if equality holds in \eqref{polya}, then
$u^{-1}(\beta,\infty)$ is a translate of
${u^*}^{-1}(\beta,\infty)$, for every $\beta\in [0,M]$, where $M$
is the essential supremum of $u$ over $D$, modulo sets of measure
zero.
\end{theorem}

\begin{lemma}\label{radial}
Suppose $0\leq f_0\in L^p(B)$. Then the maximization problem
\begin{equation}\label{maximization}
\sup_{f\in\mathcal{R}(f_0)}\Phi (f),
\end{equation}
has a unique solution; namely, $f_0^\sharp$, the Schwarz
symmetrization of $f_0$ on $B$. That is,
\begin{itemize}
\item[(a)] $\Phi (f_0^\sharp)=\sup_{f\in\mathcal{R}(f_0)}\Phi (f)$,
and
\item[(b)] $\Phi (f)<\Phi (f_0^\sharp)$, for all
$f\in\mathcal{R}(f_0)\setminus\{f_0^\sharp\}$.
\end{itemize}
\end{lemma}

\begin{proof}
 Part (a) is straightforward. Indeed, for any
$f\in\mathcal{R}(f_0)$, an application of the Hardy-Littlewood
inequality, see \cite{hardy}, \eqref{talenti} yield
\[
\Phi (f)\leq\int_Bf^\sharp w_f^\sharp\,dx\leq\int_Bf^\sharp
w_{f^\sharp}\,dx=\Phi (f^\sharp)=\Phi (f_0^\sharp).
\]
This proves that $f_0^\sharp$ solves \eqref{maximization}, hence
completes the proof of part (a).

Part (b) is more complicated. By contradiction assume
the assertion is false. Hence, there exists
$f\in\mathcal{R}(f_0)\setminus\{f_0^\sharp\}$ such that
\begin{equation}\label{ali7}
\Phi (f)=\Phi (f_0^\sharp).
\end{equation}
The following inequality follows from the variational formulation
of $w_{f^\sharp}$:
\begin{equation}\label{abbasss}
\frac{1}{p}\int_B |\nabla u|^p\,dx-\int_Bf^\sharp u\,dx
\geq \frac{1}{p}\int_B|\nabla w_{f^\sharp}|^p\,dx
-\int_Bf^\sharp w_{f^\sharp}\,dx,
\end{equation}
for every $u\in W^{1,p}_0(B)$. Thus, by substituting
$u=w_f^\sharp$ in \eqref{abbasss}, we obtain
\[
\frac{1}{p}\int_B|\nabla w_f^\sharp
|^p\,dx+\frac{1}{q}\int_Bf^\sharp w_{f^\sharp}\,dx\geq
\int_Bf^\sharp w_f^\sharp\,dx.
\]
Applying the Hardy-Littlewood inequality to the right hand side
yields
\begin{equation}\label{ali8}
\frac{1}{p}\int_B|\nabla w_f^\sharp
|^p\,dx+\frac{1}{q}\int_Bf^\sharp
w_{f^\sharp}\,dx\geq\int_Bfw_f\,dx=\Phi (f).
\end{equation}
From \eqref{ali7} and \eqref{ali8}, we infer
\begin{equation}\label{ali9}
\int_B|\nabla w_f^\sharp |^p\,dx\geq \Phi (f)
=\int_B|\nabla w_f|^p\,dx.
\end{equation}
Inequality \eqref{ali9} coupled with \eqref{polya} imply that
equality holds in \eqref{ali9}. We now proceed to show that
$w_f=w_f^\sharp$, which follows once we prove the set
$\{x\in B: \nabla w_f=0,\;0<w_f(x)<M\}$ has measure zero, according to
Theorem \ref{thm3}. To this end, let us consider $z\in B$ such that
$0<w(z)<M:=\max_B w(x)$, where we are using $w$ in place of $w_f$
for simplicity. Using the strong maximum principle \cite{sabina}
it is easy to show that $z\in\partial S$, the boundary of the set
$S$, where $S:=\{x\in B: w(x)\geq w(z)\}$. By Theorem 2.2,
$S$ is a ball. Since $z\in\partial S$ we can apply the Hopf
boundary point lemma \cite{vazquez} to conclude that
$\frac{\partial w}{\partial\nu}(z)\neq 0$, where $\nu$ stands for
the unit outward normal vector to $S$ at $z$. Thus, $\nabla
w(z)\neq 0$, hence the set $\{x\in B: \nabla
w_f=0,\;0<w_f(x)<M\}$ has measure zero, as desired. So,
$w=w^\sharp$.

At this stage we utilize Theorem \ref{thm3}. According to
\eqref{Euler}, there exists an increasing function $\eta$ such
that $f=\eta\circ w$. Since $w=w^\sharp$, we infer $f=\eta\circ
w^\sharp$. This, in turn, implies that $f=f^\sharp=f_0^\sharp$,
which is a contradiction.
\end{proof}


\begin{remark} \label{rmk2.1} \rm
There is a nice interpretation of Lemma \ref{radial}, when $p=2$.
Indeed, in this case, the boundary value problem \eqref{bvppp}
models the displacement, $w_f$, of an elastic radial membrane,
fixed around the boundary, and subject to a vertical force, $f$,
which can be taught of as a load distribution. The assertions of
the theorem then are implying that the average displacement, over
the region where the load is positioned, is maximized provided the
load is located at the center of the membrane. It is
intuitively clear that in order to make the membrane stretch as
much as possible we should place the load as far as we can from
the boundary.
\end{remark}

\begin{definition} \label{def2.2} \rm
Suppose $z_i:D_i\to\mathbb{R}$ are measurable functions, $i=1,2$,
and $|D_1|=|D_2|$. We write $z_1\preceq z_2$ provided
\[
z^\sharp_1(x)\leq z^\sharp_2(x),\quad \forall x\in B,
\]
where $B$ is the ball centered at the origin with
$|B|=|D_1|=|D_2|$.
\end{definition}

\begin{remark} \label{rmk2.2} \rm
Note that based on the definition above we observe that Talenti's
inequality \eqref{talenti} can be written as
$u_f\preceq v_{f^\sharp}$, since $v_{f^\sharp}=(v_{f^\sharp})^\sharp$.
\end{remark}

\section{Proof of Theorem \ref{thm1}}

\begin{proof}
From the assumption, $u_f^\sharp =v_{f^\sharp}$, we deduce
\begin{equation}\label{ali1}
\int_Bf^\sharp u^\sharp_f\,dx
=\int_Bf^\sharp\;v_{f^\sharp}\,dx
=\int_B|\nabla v_{f^\sharp}|^p\,dx
=\int_B|\nabla u^\sharp_f|^p\,dx
\leq\int_D|\nabla u_f|^p\,dx,
\end{equation}
where in the last inequality we have applied \eqref{polya}. On the
other hand, by the Hardy-Littlewood inequality, see \cite{hardy},
we have
\begin{equation}\label{ali2}
\int_Df u_f\,dx\leq\int_Bf^\sharp u_f^\sharp\,dx.
\end{equation}
Multiplying the differential equation in \eqref{bvp}, by $u_f$,
integrating over $D$, and remembering the boundary condition,
$u_f=0$, we obtain
\begin{equation}\label{ali3}
\int_Df u_f\,dx=\int_D|\nabla u_f|^p\,dx+\int_Dc(x) u_f^p\,dx.
\end{equation}
the combination of \eqref{ali1}, \eqref{ali2} and \eqref{ali3}
yields
\[
\int_D|\nabla u_f|^p\,dx+\int_Dc(x) u_f^p\,dx\leq\int_D|\nabla
u_f|^p\,dx,
\]
hence $\int_D c(x) u_f^p\,dx\leq 0$. This, in turn, recalling
that $u_f>0$, implies $c\equiv 0$. This completes the proof of
part (i).

We  prove part (ii). Let us first observe that from
(i) we infer $v_{f^\sharp}=u_{f^\sharp}$. Hence, the hypothesis of
the theorem can be written as $u^\sharp_f=u_{f^\sharp}$. Again,
from the Hardy-Littlewood inequality and \eqref{polya}, we obtain
\begin{equation}\label{ali4}
\begin{split}
\int_D|\nabla u_f|^p\,dx
&=\int_Dfu_f\,dx\leq\int_Bf^\sharp u^\sharp_f\,dx
 =\int_Bf^\sharp u_{f^\sharp}\,dx\\
&= \int_B|\nabla u_{f^\sharp}|^p\,dx
 =\int_B|\nabla u^\sharp_f|^p\,dx  \\
&\leq \int_D|\nabla u_f|^p\,dx.
\end{split}
\end{equation}
Hence, all inequalities in \eqref{ali4} are in fact equalities. In
particular, we obtain
\[
\int_D|\nabla u_f|^p\,dx=\int_B|\nabla u^\sharp_f|^p\,dx.
\]
Thus, we are now in a position to apply Theorem \ref{thm3}, which
implies the sets $\{x\in D: u_f(x)\geq\beta\}$ are translations
of $\{x\in B: u_f^\sharp(x)\geq\beta\}$. So, in particular, we
deduce $D$ is a translation of $B$. Henceforth, without loss of
generality we assume $D=B$. Whence, we have
$\int_B|\nabla u_f|^p\,dx=\int_B|\nabla u_f^\sharp|^p\,dx$.
To complete the proof of (ii), we return to \eqref{ali4},
and recalling that all inequalities are equalities, we obtain
\[
\int_Bfu_f\,dx=\int_Bf^\sharp u_{f^\sharp}\,dx.
\]
Now we can apply Lemma \ref{radial}, which yields $f=f^\sharp$.
\end{proof}

\begin{remark} \label{rmk3.1} \rm
When $p=2$, the boundary value problem \eqref{bvp} reduces
to
\begin{equation}\label{ali6}
\begin{gathered}
-\Delta u+c(x)u=f(x), \quad \text{in } D, \\
u=0, \quad \text{on } \partial D.
\end{gathered}
\end{equation}
The boundary value problem \eqref{ali6} physically models the
displacement of a non-isotropic (assuming $c$ is not identically
zero) elastic membrane, fixed around the boundary, and subject to
a vertical force such as a distribution load. The result of this
paper implies that equality in the \eqref{talenti} is only
possible if the membrane is isotropic; i.e., $c\equiv 0$. In other
words, when the membrane is made of several materials with
different densities it is impossible to have equality in
\eqref{talenti}, hence the best result is
$u_f\preceq v_{f^\sharp}$.
\end{remark}

\section{A special case}
In the maximization problem \eqref{maximizationn}, sometimes the
generator of the rearrangement class, $f_0$, is a ``complicated''
function. We use the term complicated specifically in the
following sense: $f_0=Q(g_0)$, where $Q:\mathbb{R}\to\mathbb{R}^+$
is an increasing and continuous function, and $g_0$ is a non-negative
function that belongs to $L^p(D)$.

In this section we show that if $f_0$ is a complicated function in
the above sense then it is possible to replace
\eqref{maximizationn} with another maximization problem which is
formulated with respect to $g_0$. The main result of this section
is as follows.

\begin{theorem}\label{simplification}
Let $0\leq f_0\in L^p(D)$, and suppose $f_0$ is a complicated
function in the sense described above. Then
\begin{equation}\label{abbas2}
\sup_{f\in\mathcal{R}(f_0)}\int_D fu_f\,dx
=\sup_{g\in\mathcal{R}(g_0)}\int_D Q(g)u_{Q(g)}\,dx.
\end{equation}
\end{theorem}

The proof of Theorem \ref{simplification} relies on the fact that
$\mathcal{R}$ and $T$ commute, as in the following lemma.

\begin{lemma} \label{simple}
Let $0\leq h_0\in L^p(D)$. Let $T:\mathbb{R}\to\mathbb{R}^+$
be an increasing and continuous function.
Then
\begin{equation}\label{abbas1}
\mathcal{R}(T(h_0))=T(\mathcal{R}(h_0)).
\end{equation}
\end{lemma}

\begin{proof}
 Let us first prove the inclusion
$T(\mathcal{R}(h_0))\subseteq \mathcal{R}(T(h_0))$. To this end,
let $l\in T(\mathcal{R}(h_0))$. So, there exists
$h\in\mathcal{R}(h_0)$ such that $l=T(h)$. Note that for every
$\alpha\geq 0$, there exists $\beta\geq 0$ such that
$T^{-1}([\alpha,\infty))=[\beta,\infty)$, where $T^{-1}$ denotes
the inverse image, since $T$ is increasing and continuous. Hence
\[
\{x\in D:  l(x)\geq\alpha\}=\{x\in D: h(x)\geq\beta\}.
\]
This, in turn, implies
\begin{align*}
|\{x\in D: l(x)\geq\alpha\}|
&= |\{x\in D: h(x)\geq\beta\}|\\
&= \{x\in D: h_0(x)\geq\beta\}| \\
&= |\{x\in D: T(h_0)(x)\geq\alpha\}|.
\end{align*}
Thus, $l\in\mathcal{R}(T(h_0))$, as desired.

Now we prove that $\mathcal{R}(T(h_0))\subset T(\mathcal{R}(h_0))$.
Consider $l\in\mathcal{R}(T(h_0))$, so
$l^*=T(h_0)^*$. Here, ``*'' stands for the decreasing rearrangement
operator. So, for example,
\[
l^*(s)=\inf \{\alpha\in\mathbb{R}:  |\{x\in D: l(x)\geq\alpha\}|\leq
s\},
\]
see \cite{ryff} for details. Since $T$ is decreasing, we infer
$l^*=T(h_0)^*=T(h_0^*)$. At this stage, we use another result from
\cite{ryff}; namely, that there exists a measure preserving map
$\psi: D\to D$ such that $l=l^*\circ\psi$. Therefore,
$l=T(h_0^*)\circ\psi= T(h_0^*\circ\psi)$. Since $\psi$ is measure
preserving, we infer $h_0^*\circ \psi\in\mathcal{R}(h_0)$, hence
$l\in T(\mathcal{R}(h_0))$, which finishes the proof of the
lemma.
\end{proof}

\begin{proof}[Proof of Theorem \ref{simplification}]
 Let us set
$$
L=\sup_{f\in\mathcal{R}(f_0)}\int_Dfu_f\,dx,\quad
R=\sup_{g\in\mathcal{R}(g_0)}\int_D Q(g)u_{Q(g)}\,dx.
$$
We prove only $L\leq R$, since the proof of
$L\geq R$ is similar.
Consider $f\in\mathcal{R}(f_0)$. Then
$f\in\mathcal{R}(Q(g_0))=Q(\mathcal{R}(g_0))$, by Lemma
\ref{simple}. Thus, $f=Q(g)$, for some $g\in\mathcal{R}(g_0)$.
Therefore
\[
\int_Dfu_f\,dx=\int_DQ(g)u_{Q(g)}\,dx\leq R.
\]
Since $f\in \mathcal{R}(f_0)$ is arbitrary we deduce
$L\leq R$.
\end{proof}


To illustrate the advantages of Theorem \ref{simplification},
we consider the following example.

 Let $P:\mathbb{R}\to\mathbb{R}^+$ be a bounded function.
Consider the boundary value problem
\begin{equation}\label{example}
\begin{gathered}
-\Delta_p u= \int_0^{g(x)}P(s)\,ds, \quad \text{in }D,\\
u=0, \quad \text{on }\partial D.
\end{gathered}
\end{equation}
Setting $f(x)= \int_0^{g(x)}P(s)\; ds$, Theorem
\ref{simplification} implies that
\[
\sup_{f\in\mathcal{R}(f_0)}\int_Dfu_f\,dx
=\sup_{g\in\mathcal{R}(g_0)}\int_D\Big(\int_0^{g(x)}P(s)\,ds\Big)
u_{(\int_0^{g(x)}P(s)\,ds)}\,dx.
\]

\subsection*{Acknowledgements}
The authors wish to thank the anonymous referee for his/her
valuable comments regarding the proof of Lemma 2.3.
The first author would like to thank Dr. Stephen J. Shaw,
the head of the Mathematical Sciences Department
of Xi'an Jiaotong-Liverpool University, for his support
during the course of this project.


\begin{thebibliography}{00}

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