\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{mathrsfs}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 98, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2016/98\hfil Optimization problems]
{Optimization problems involving the \\ fractional Laplacian}
\author[C. Qiu, Y. Huang, Y. Zhou \hfil EJDE-2016/98\hfilneg]
{Chong Qiu, Yisheng Huang, Yuying Zhou}
\address{Chong Qiu \newline
Department of Mathematics,
Taizhou University,
Taizhou 225300, China}
\email{qchsuda@163.com}
\address{Yisheng Huang \newline
Department of Mathematics,
Soochow University,
Suzhou 215006, China}
\email{yishengh@suda.edu.cn}
\address{Yuying Zhou (corresponding author) \newline
Department of Mathematics,
Soochow University,
Suzhou 215006, China}
\email{yuyingz@suda.edu.cn}
\thanks{Submitted January 17, 2016. Published April 19, 2016.}
\subjclass[2010]{49K30, 35A15, 47J30}
\keywords{Optimization; fractional Laplacian; rearrangements}
\begin{abstract}
In this article we study rearrangement optimization problems
related to boundary-value problems involving the fractional Laplacian.
We establish the existence and uniqueness of a solution
under suitable assumptions.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks
\section{Introduction}
A rearrangement optimization problem is referred to an optimization problem
in which the admissible set consists of functions that are rearrangements
of a prescribed function. The theory of rearrangement optimizations has
been established by Burton \cite{G.R87,G.R89}.
Since then, this topic has been widely
studied by many authors in different aspects, see for example
\cite{C.S.C, F.B.G06, F.B09,F.B11, B.R07, M.M10, NE03, QHZ15}.
Burton \cite{G.R89} proved that both the minimization and maximization
problems for the boundary value problem involving the Laplacian have solutions.
However, the results obtained in \cite{G.R89} can not be directly applied
to the optimization problems for the boundary value problem involving
$p$-Laplacian$ (1
1$ and
$r'=\infty$ if $r=1$.
Let $ f\in L^{r}(\Omega)$ and $ g\in L^{r'}(\Omega)$. Suppose that there
exists an increasing (decreasing) function
$\varphi:\mathbb{R}\to\mathbb{R}$ such that
$\varphi\circ g\in\mathscr{R}(f)$. Then $ \varphi\circ g$ is the unique
maximizer (minimizer) of the linear functional $\int_{\Omega} hg \,dx$,
relative to $ h\in\overline{\mathscr{R}(f)}$.
\end{lemma}
\begin{lemma}[{\cite[Lemma 2.3]{B.J09}}] \label{l4}
Suppose that $ f\in L^{r}(\Omega)$ and $ g\in L^{r'}(\Omega)$. Then there
exists $ \hat{f}\in\mathscr{R}(f)$ which maximizes (minimizes) the linear
functional $\int_{\Omega} hgd x$,
relative to $ h\in\overline{\mathscr{R}(f)}$.
\end{lemma}
As in the proof of \cite[Lemma 2.1]{TWY12} we have the following result.
\begin{lemma}\label{l8}
Given $u\in H^{1/2}(\Omega)$, there exists a unique extension
$v \in H^{1}(\mathbb{R}_{+}^{N+1})$ of $u$ such that
\begin{gather*}
-\Delta v(x,y)=0,\quad \text{for } x\in\mathbb{R}^N,y>0,\\
v(x,0)=u(x),\quad \text{for } x\in\mathbb{R}^N.
\end{gather*}
Moreover,
\[
-\partial_{y}v(x,0)=(-\Delta)^{1/2}u(x)
\]
in the sense that
\[
-\int_{\mathbb{R}^N\times\{0\}}\frac{\partial v}{\partial y}\varphi dx
=\int_{\mathbb{R}^N}\varphi(-\Delta)^{1/2}u dx
\]
for every $\varphi\in H^{1/2}(\mathbb{R}^N)$.
\end{lemma}
\section{Minimization related to \eqref{Qs}} \label{s4}
Recall that the energy functional $I:H^s(\Omega)\to \mathbb{R}$ corresponding
to \eqref{Qs} is
\begin{equation}\label{e1}
I(u)=\frac{1}{2}\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
+\int_{\Omega} H(x,u)d x-\int_{\Omega} fud x,
\end{equation}
where $H(x,u)=\int_0^{u}h(x,t)d t$.
We use the following hypotheses on the functions
$h:\Omega\times\mathbb{R}\to\mathbb{R}$ and $\theta:\mathbb{R}^N\to (0,+\infty)$:
\begin{itemize}
\item [(H1)] $h(x,t)$ is Carath\'{e}odory and is non-decreasing with respect
to the second variable for almost all $x\in\Omega$.
\item [(H2)] There exist $a(x), b(x)\in L^{\infty}(\Omega)$, $00.
\end{equation}
Since $ h(x,\cdot)$ is non-decreasing,
\begin{equation}\label{e109}
\int_{\Omega} (h(x,u_{f})-h(x,w))(u_{f}-w)d x\geq0.
\end{equation}
From \eqref{e67} we obtain that for every $v\in H^s(\Omega)$,
\begin{gather} \label{e7}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u_{f}(x)-u_{f}(y))(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
+\int_{\Omega} h(x,u_{f})vdx
=\int_{\Omega} fv\,dx,\\
\label{e8}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(w(x)-w(y))(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
+\int_{\Omega} h(x,w)vdx=\int_{\Omega} fv\,dx.
\end{gather}
From these two equalities, we obtain that for every $v\in H^s(\Omega)$,
\begin{equation}\nonumber
\begin{aligned}
&\int_{\Omega} (h(x,u_{f})-h(x,w))vdx\\
&= \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(w(x)-u_{f}(x)-w(y)
+u_{f}(y))(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy.
\end{aligned}
\end{equation}
Let $ v=u_{f}-w$ we have
\begin{equation}\label{e14}
\begin{aligned}
&\int_{\Omega} (h(x,u_{f})-h(x,w))(u_{f}-w)dx\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{-(v(x)-v(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
&\leq -\theta_0\|v\|^2 <0,
\end{aligned}
\end{equation}
the last inequality above comes from \eqref{e104}. So, \eqref{e14} contradicts
\eqref{e109}. Therefore, we have proved that $u_f$ is the unique
solution of \eqref{Qs}.
\end{proof}
Let $u_f$ be the unique solution of \eqref{Qs}. Recall that $\Psi(f)$
is defined by \eqref{e0}. considering the optimization problem (Opt*),
we have the following result.
\begin{theorem}\label{thm3}
Suppose that {\rm (H1--(H4)} hold. Then for a fixed nonnegative function
$f_0\in L^{\infty}(\Omega)$ there exists $ f_*\in\mathscr{R}(f_0)$ which
solves the minimization optimization problem {\rm (Opt*)}, i.e.,
$$
\Psi(f_*)=\inf_{f\in\mathscr{R}(f_0)}\Psi(f).
$$
\end{theorem}
\begin{proof}
Let $ A=\inf_{f\in\mathscr{R}(f_0)}\Psi(f)$, then $ A$ is well-defined.
Indeed, for each $f\in\mathscr{R}(f_0)$, we have
\begin{equation}\label{e12}
\begin{aligned}
\Psi(f)
&=\frac{1}{2}\Big(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\Big) \\
&\quad +\int_{\Omega} H(x,u_f)d x-\int_{\Omega} fu_f\,d x\\
&\geq \frac{\theta_0}{2}\|u_{f}\|^{2}-C(\|f\|_{\infty}\|u_{f}\|+\|u_{f}\|^{l+1}).
\end{aligned}
\end{equation}
Noting that $ \|f\|_{\infty}=\|f_0\|_{\infty}$ and $l+1<2$, we deduce
that $A$ must be finite.
Let $\{f_i\}\subset\mathscr{R}(f_0)$ be such that $\Psi(f_i)\to A$
as $i\to\infty$ and we denote $u_i=u_{f_i}$. It follows from
\eqref{e12} that $\{u_i\}$ is bounded in $H^s(\Omega)$, then
it has a subsequence (still denoted $\{u_i\}$) which converges weakly
to $ u\in H^s(\Omega)$ and strongly to $ u$ in $ L^{2}(\Omega)$.
On the other hand, since $
\|f_i\|_{L_{{\infty}}}\equiv\|f_0\|_{L_{{\infty}}}$,
$\{f_i\}$ is bounded in $ L^{2}(\Omega)$, it must contain a subsequence
(still denoted $\{f_i\}$) converging weakly to some
$\bar{f}\in \overline{\mathscr{R}(f_0)}$, the weak closure of
$\mathscr{R}(f_0)$ in $ L^{2}(\Omega)$.
Then
$$
\big|\int_{\Omega} (f_i-\bar{f})ud x\big|\to 0\quad \text{as } i\to\infty,
$$
since $u\in L^{2}(\Omega)$.
It follows from the H\"{o}lder inequality that
\begin{equation}\label{e72}
\begin{aligned}
\big|\int_{\Omega} (f_iu_i-\bar{f}u)d x\big|
& \leq\big|\int_{\Omega} f_i(u_i-u)d x\big|
+\big|\int_{\Omega} (f_i-\bar{f})ud x\big|\\
&\leq\|f_i\|_{2}\|u_i-u\|_{2}+\big|\int_{\Omega} (f_i-\bar{f})u\,dx\big|\to 0
\end{aligned}
\end{equation}
as $i\to\infty$. By \eqref{e110}, \eqref{e72} and the weak lower
semi-continuity of the norm in $ H^s(\Omega)$, we
obtain
\begin{equation}\label{e0016}
\begin{aligned}
A&=\lim_{i\to\infty}\Psi(f_i)\\
&\geq\frac{1}{2}(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy)\\
&\quad +\int_{\Omega} H(x,u)d x-\int_{\Omega} \bar{f}ud x.
\end{aligned}
\end{equation}
From Lemma \ref{l4} we infer the existence of $\hat{f}\in\mathscr{R}(f_0)$
which maximizes the linear functional $\int_{\Omega} hud x$,
relative to $ h\in\overline{\mathscr{R}(f_0)}$. As a consequence,
\[ %\label{e017}
\int_{\Omega} \bar{f}ud x\leq\int_{\Omega} \hat{f}ud x.
\]
Combining this with \eqref{e0016}, we obtain
\begin{equation}\label{e74}
A\geq \frac{1}{2}(\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy)
+\int_{\Omega} H(x,u)d x-\int_{\Omega} \hat{f}u\,d x.
\end{equation}
By Proposition \ref{prop2},
\begin{equation}\label{e75}
\begin{aligned}
\Psi(\hat{f})&=\inf_{v\in H^s(\Omega)}
\Big(\frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(v(x)-v(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
+\int_{\Omega} (H(x,v)-\hat{f}v)d x\Big)\\
&\leq \frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
+\int_{\Omega}(H(x,u)-\hat{f}u) dx.
\end{aligned}
\end{equation}
It follows from \eqref{e74} and \eqref{e75} that $ \Psi(\hat{f})\leq A$.
On the other hand, recall that $A=\inf_{f\in\mathscr{R}(f_0)}\Psi(f)$ and
$ \hat{f}\in\mathscr{R}(f_0)$,
we must have $ A\leq \Psi(\hat{f})$. So that $ A=\Psi(\hat{f})$. We
complete the proof by letting $ f_{*}= \hat{f}$.
\end{proof}
\section{Maximization and minimization related to \eqref{Ps}} \label{s3}
In this section, we consider two optimization problems (Opt1) and (Opt2)
related to \eqref{Ps}.
The energy functional of \eqref{Ps} is
\[
J(u)=\frac{1}{2}(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy)
-\int_{\Omega} f(x)u(x)d x.
\]
It is easy to see $ J\in C^{1}(H^s(\Omega),\mathbb{R})$ and
$$
J'(u)v=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u(x)-u(y))
(v(x)-v(y))}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy-\int_{\Omega} f(x)v(x)d x.
$$
$u\in H^s(\Omega)$ is a solution of the problem \eqref{Ps}
if and only if $J'(u)v=0,\forall v\in H^s(\Omega)$.
By proposition \ref{prop2}, we have the following result.
\begin{proposition}\label{prop1}
Assume that $\theta $ satisfies {\rm (H3)} and {\rm (H4)}. Then for each
$f$ in $L^{\infty}(\Omega)$, \eqref{Ps} has a unique solution $u_f\in
H^s(\Omega)$, and $ J(u_f)=\inf_{v\in H^s(\Omega)}J(v)$.
\end{proposition}
\begin{remark}\label{rmk1} \rm
Since $J'(u_f)u_f=0$,
\begin{equation}\label{e16}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}
\theta(x-y)\,dx\,dy=\int_{\Omega} f(x)u_f(x)\,d x.
\end{equation}
Recall that $\Phi(f)=\int_{\Omega}fu_f d x$, we have
\begin{equation}\nonumber
\begin{aligned}
&\Phi(f)\\
&= -2\Big(\frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy
-\int_{\Omega} f(x)u_f(x)d x\Big)\\
&= -2\inf_{v\in H^s(\Omega)}\Big(\frac{1}{2}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(v(x)-v(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy-\int_{\Omega} f(x)v(x)d x\Big)\\
&= -2J(u_f).
\end{aligned}
\end{equation}
\end{remark}
\begin{theorem}\label{t1}
Assume that $\theta $ satisfies {\rm (H3)} and {\rm (H4)}.
Then for each non-negative function $f_0\in L^{\infty}(\Omega)$,
there exists $ f_1\in\mathscr{R}(f_0)$ which solves the maximization optimization
problem {\rm (Opt1)}, i.e.,
\[
{\Phi(f_1)=}\int_{\Omega} f_1u_{1}\,dx=\sup_{f\in\mathscr{R}(f_0)}
\int_{\Omega} fu_{f}\,dx{
=\sup_{f\in\mathscr{R}(f_0)}\Phi(f)}
\]
where $u_{1}=u_{f_1}$. Moreover,
if $s=\frac{1}{2}$ then there exists an increasing function $\phi$ such that
$f_1=\phi(u_1)$
almost everywhere in $\Omega$.
\end{theorem}
\begin{proof}
Let
\[
M=\sup_{f\in\mathscr{R}(f_0)}\int_{\Omega} fu_{f}\,dx.
\]
We first show that $ M$ is finite. Let $ f\in\mathscr{R}(f_0)$.
From (H4), {\eqref{e16}} and H\"{o}lder's inequality we
find
\begin{equation}\label{e9}
\begin{aligned}
\theta_0\|u_{f}\|^{2}
&\leq\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_f(x)-u_f(y))^2}{| x-y|^{N+2s}}
\theta(x-y)\,dx\,dy \\
&=\int_{\Omega} fu_{f}\,dx
\leq \|f\|_{\infty}\|u_{f}\|_{1}.
\end{aligned}
\end{equation}
Since $\|f\|_{\infty}=\|f_0\|_{\infty}$, it follows from \eqref{e9}
and Lemma \ref{l1} that $ M$ is finite.
Let $\{f_i\}$ be a maximizing sequence and let $ u_i=u_{f_i}$.
From \eqref{e9} it is clear that $\{u_i\}$ is bounded in $ H^{s}(\Omega)$,
hence it has a subsequence (still denoted $\{u_i\}$) that converges weakly
to $ u\in H^{s}(\Omega)$. We also infer that $\{u_i\}$ converges strongly to $ u$
in $ L^{2}(\mathbb{R}^N)$. On the other hand, since $\{f_i\}$ is bounded
in $ L^{\infty}(\Omega)$, it must contain a subsequence(still denoted $\{f_i\}$)
converging weakly to $\bar{f}\in L^{2}(\Omega)$. Note that
$ \bar{f}\in\overline{\mathscr{R}(f_0)}$,
the weak closure of $\mathscr{R}(f_0)$ in $ L^{2}(\Omega)$.
Thus, using the weak lower semi-continuity of the $ H^{s}(\Omega)$
norm and \eqref{e5} we obtain
\begin{equation}\label{e10}
\begin{aligned}
M&=\lim_{i\to\infty}\int_{\Omega} f_iu_i\,dx\\
&=\lim_{i\to\infty}(\int_{\Omega}2f_iu_i\,dx
-\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(u_i(x)-u_i(y))^2}{| x-y|^{N+2s}}
\theta(x-y)\,dx\,dy)\\
&\leq\int_{\Omega}2\bar{f}ud x-\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy.
\end{aligned}
\end{equation}
Note that from Lemma \ref{l4} we infer the existence of
$\hat{f}\in\mathscr{R}(f_0)$ that maximizes the linear functional
$\int_{\Omega} hu\,dx$,
relative to $ h\in\overline{\mathscr{R}(f_0)}$. As a consequence we obtain
\begin{equation}\label{e11}
\int_{\Omega} \bar{f}u\,d x\leq\int_{\Omega} \hat{f}u\,d x.
\end{equation}
Applying \eqref{e16}, \eqref{e10} and \eqref{e11} we find
\begin{align*}
M
&\leq\int_{\Omega} 2\hat{f}ud x-\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(u(x)-u(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
&\leq\int_{\Omega}2\hat{f}\hat{u}\,dx
-\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{(\hat{u}(x)
-\hat{u}(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
&=\int_{\Omega}\hat{f}\hat{u}\,dx\leq M
\end{align*}
where $\hat{u}=u_{\hat{f}}$. Thus let $f_1= \hat{f}$ we complete the
first part of the proof.
We then show that $f_1$ is also a maximizer of the functional
$\int_{\Omega} hu_1d x$,
relative to $ h\in\overline{\mathscr{R}(f_0)}$.
In fact, we notice that for each $g\in\mathscr{R}(f_0)$,
\begin{align*}
\int_{\Omega} f_1u_1dx{\geq}& \int_{\Omega} gu_gdx\\
&\geq \int_{\Omega}2gu_1d x-\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{(u_1(x)-u_1(y))^2}{| x-y|^{N+2s}}\theta(x-y)\,dx\,dy\\
&= \int_{\Omega}2gu_1d x-\int_{\Omega} f_1u_1dx,
\end{align*}
which implies that
\begin{equation}\label{e113}
\int_{\Omega} f_1u_1 dx\geq \int_{\Omega} gu_1 dx,\quad \forall g\in\mathscr{R}(f_0).
\end{equation}
If $g\in\overline{\mathscr{R}(f_0)}$ then we may choose a sequence
$\{g_n\}\subset\mathscr{R}(f_0)$ such that $\{g_n\}$ converge weakly to $g$
in $L^{2}(\Omega)$.
By \eqref{e113}, we obtain
$$
\int_{\Omega} f_1u_1\, dx\geq \int_{\Omega} g_nu_{1} dx\to \int_{\Omega} gu_1\,dx
$$
as $n\to\infty$.
So that
\begin{equation}\label{e114}
\int_{\Omega} f_1u_1 \,dx\geq \int_{\Omega} gu_1 \,dx,\quad
\forall g\in\overline{\mathscr{R}(f_0)}
\end{equation}
and our claim is valid.
Let $E=\{x\in \Omega: f_1(x)=0\}$ and define $S=sup_{x\in E}(u_1(x))$,
we claim that $u_1(x)\geq S$ on $E^c$ almost everywhere. Arguing by
contradiction suppose the claim is false. Therefore there exists a
number $S_10$ such that $u_1(x)S_2$ on $D$
almost everywhere. We can assume $|A|=|D|$. Next, consider a measure-preserving
map $T:A\to D$. Using $T$ we define a particular rearrangement of $f_1$,
denoted by $\overline{f}$, as follows:
$$
\overline{f}=\begin{cases}
f_1(Tx) & x\in A,\\
f_1(T^{-1}x) & x\in D,\\
f_1(x) & x\in \Omega\backslash(A\cup D).
\end{cases}
$$
Thus
\begin{align*}
& \int_{\Omega} \overline{f}u_1dx-\int_{\Omega} f_1u_1dx\\
&=\int_{A\bigcup D}\overline{f}u_1dx-\int_{A\cup D}f_1u_1dx\\
&=\int_{A}\overline{f}u_1dx+\int_{A}(u_1\circ T)f_1 dx
-\int_{A}f_1u_1dx-\int_{A}(u_1\circ T)\overline{f}dx\\
&=\int_{A}(u_1\circ T-u_1)(f_1-\overline{f})dx>(S_2-S_1)\int_{A}f_1dx>0.
\end{align*}
Therefore $\int_{\Omega} \overline{f}u_1dx>\int_{\Omega} f_1u_1dx$,
which contradicts \eqref{e114}.
Since $s=\frac{1}{2}$, by Lemma \ref{l8} we know that $u_1$ has no level
set of positive measure on $E^c$, then by Lemma \ref{l2} we infer
the existence of an increasing function $\phi_1$ such that
$\phi_1(u_1)$ is a rearrangement of $f_1$ on $E^c$. Now we define an
increasing function $\phi_2$ as
$$
\phi_2(t)=\begin{cases}
0 &t0$ a.e. $x\in \Omega$. By Lemma \ref{l8} we have that
each level set of $u_{\underline{g}}$ has zero measure. By Lemmas \ref{l2}
and \ref{l3}, there exists a decreasing function $\phi$ such that
$\phi\circ u_{\underline{g}}$ is a rearrangement of $\underline{g}$ and
the unique minimizer of the linear functional $\int_{\Omega}gu_{\underline{g}}dx$,
related to $g\in\overline{\mathscr{R}(f_0)}$.
So that $\underline{g}=\phi\circ u_{\underline{g}}\in \mathscr{R}(f_0)$.
We complete the proof by letting $f_2=\underline{g}$.
\end{proof}
\subsection*{Acknowledgements}
This work was supported by Natural Science Foundation of China
(11171247, 11371273, 11471235,11501437) and the scientific research
fund project of Taizhou University (TZXY2015QDXM033).
The authors are grateful to the anonymous referee for
the useful comments.
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