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