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\title[A sum theorem for maximal monotone operators]{The sum theorem for maximal monotone operators in reflexive Banach spaces revisited }
\author[M.D. Voisei]{M.D. Voisei}

\address{Baltimore, MD-21208,\\
U.S.A.}
\email{mdvoisei@yahoo.com}


\subjclass{47H05, 46N10}
\keywords{maximal monotone operator, Minkowski sum}
\dedicatory{Dedicated to Gheorghe Moro\c{s}anu on the occasion of his $\it{70}^{th}$ birthday}
\begin{abstract}
Abstract The goal of this note is to present a new shorter proof for the maximal monotonicity of the Minkowski sum of two maximal monotone multi-valued operators defined in a reflexive Banach space under the classical interiority condition involving their domains. 
\end{abstract}
\maketitle

\section{Preliminaries}

Recall the following sum rule for maximal monotone operators:

\begin{theorem} \label{RM} (Rockafellar \cite[Theorem\ 1\ (a),\ p.\ 76]{MR0282272})
	Let $(X,\|\cdot\|)$ be a reflexive Banach space with topological
	dual $X^{*}$ and let $A,\ B:X\rightrightarrows X^{*}$ be multi-valued
	maximal monotone operators from $X$ to $X^{*}$. If $D(A)\cap\operatorname*{int}D(B)\neq\emptyset$
	then $A+B$ is maximal monotone. \end{theorem}

Here $D(T):=\{x\in X\mid T(x)\neq\emptyset\}$ is the \emph{domain}
of $T:X\rightrightarrows X^{*}$, ``$\operatorname*{int}S$'' denotes
the topological interior of $S\subset X$, and $A+B:X\rightrightarrows X^{*}$
is the \emph{Minkowski sum} of $A$ and $B$ defined by
\[
(A+B)(x):=A(x)+B(x):=\{y+v\mid y\in A(x),v\in B(x)\},
\]
for $x\in D(A+B):=D(A)\cap D(B)$. 

The proof of \cite[Theorem\ 1,\ p.\ 76]{MR0282272} relies on the use of the
duality mapping $J$ of $X$ and the (Minty's style) characterization
of maximal monotone operators defined in reflexive Banach spaces.
Similar arguments are used in the presence of an improved qualification
constraint in a second proof of Theorem \ref{RM} (see \cite[Corollary 3.5, p. 286]{MR1398155}).
A third proof of the main theorem involves the exact convolution of
some specially constructed functions based on the Fitzpatrick functions
of $A$ and $B$ (see \cite[Corollary\ 4, p. 1166]{MR2160665}). A
different proof of Theorem \ref{RM} is based on the dual-representability
$A+B$ in the presence of the qualification constraint (see \cite[Remark\ 1,\ p.\ 276]{MR2577332})
and the fact that in a reflexive Banach space dual-representability
is equivalent to maximal monotonicity (see e.g. \cite[Theorem\ 3.1,\ p.\ 2381]{MR1974634}).
All the previously mentioned proofs make use of the duality mapping
$J$ which is characteristic to a normed space.

Our proof relies on the normal cone, is based on full-range characterizations
of maximal monotone operators with bounded domain, and uses the representability
of sums of representable operators, but, avoids the use of $J$ or
the norm. The following intermediary result, is the main ingredient
of our argument. 

\begin{theorem} Let $X$ be a reflexive Banach space, let $T:X\rightrightarrows X^{*}$
	be maximal monotone, and let $C\subset X$ be closed convex and bounded.
	If $D(T)\cap\operatorname*{int}C\neq\emptyset$ then $T+N_{C}$ is
	maximal monotone. \end{theorem}

Here $N_{C}$ denotes the normal cone to $C$ and is defined by $x^{*}\in N_{C}(x)$
if, for every $y\in C$, $\langle y-x,x^{*}\rangle\le0$. Here $\langle\cdot,\cdot\rangle$
denotes the \emph{coupling} or \emph{duality product} of $X\times X^{*}$
and is defined by
\[
c(x,x^{*}):=\langle x,x^{*}\rangle:=x^{*}(x),\ x\in X,\ x^{*}\in X^{*}.
\]

Recall that a multi-valued operator $T:X\rightrightarrows X^{*}$
is 

%\medskip

\emph{$\bullet$} \emph{monotone} if, for every $x_{1}^{*}\in T(x_{1})$,
$x_{2}^{*}\in T(x_{2})$, $\langle x_{1}-x_{2},x_{1}^{*}-x_{2}^{*}\rangle\ge0$. 

%\medskip

\emph{$\bullet$ maximal monotone} if every m.r. to $T$ element $z=(x,x^{*})\in X\times X^{*}$
belongs to $\operatorname*{Graph}T$. 

%\medskip

\emph{$\bullet$ representable} if there is a proper convex $s_{X}\times w^{*}-$lower
semicontinuous $h:X\times X^{*}\to\mathbb{R}\cup\{+\infty\}$ such
that $h\ge c$ and 
\[
\operatorname*{Graph}T=[h=c]:=\{(x,x^{*})\in X\times X^{*}\mid h(x,x^{*})=\langle x,x^{*}\rangle\}.
\]
Here $s_{X}$ denotes the strong topology of $X$ and $w^{*}$ stands
for the weak-star topology of $X^{*}$. 

%\medskip

$\bullet$ \emph{NI} if $\varphi_{T}\ge c$, where $\varphi_{T}$
is the Fitzpatrick function of $T$ which is defined by
\begin{equation}
\varphi_{T}(x,x^{*}):=\sup\{\langle x-a,a^{*}\rangle+\langle a,x^{*}\rangle\mid(a,a^{*})\in\operatorname*{Graph}T\},\ (x,x^{*})\in X\times X^{*}.\label{eq:}
\end{equation}

An element $z=(x,x^{*})\in X\times X^{*}$ is \emph{monotonically
	related} (m.r. for short) \emph{to} $T$ if, for every $(a,a^{*})\in\operatorname*{Graph}T:=\{(u,u^{*})\in X\times X^{*}\mid u\in D(T),\ u^{*}\in T(u)\}$,
$\langle x-a,x^{*}-a^{*}\rangle\ge0$.

\section{Proofs of the main result}

\begin{proof}[Proof of Theorem 1.2] The operator $T+N_{C}$ is representable,
	which follows from the facts that $T$, $N_{C}$ are maximal monotone
	thus representable and $D(T)\cap\operatorname*{int}C\neq\emptyset$
	(see e.g. \cite[Corollary\ 5.6,\ p.\ 470]{MR2453098} or \cite[Theorem 16, p. 818]{MR3958031}).
	
	We prove that $R(T+N_{C})=X^{*}$ which implies that $T+N_{C}$ is
	of NI--type and so it is maximal monotone (see \cite[Theorem 3.4, p. 465]{MR2453098}
	or \cite[Theorem 1 (ii), (7)]{MR2577332}).
	
	It suffices to prove that $0\in R(T+N_{C})$ otherwise we replace
	$T$ by $T-x^{*}$ for an arbitrary $x^{*}\in X^{*}$. 
	
	Consider $F(x,x^{*}):=\varphi_{T}(x,x^{*})+g(x,x^{*})$, with $g(x,x^{*}):=\iota_{C}(x)+\sigma_{C}(-x^{*})$,
	where $\iota_{C}(x)=0$, for $x\in C$; $\iota_{C}(x)=+\infty$, otherwise,
	and $\sigma_{C}(x^{*}):=\sup_{x\in C}\langle x,x^{*}\rangle$, $x^{*}\in X^{*}$.
	
	Then $F\ge0$ due to $\varphi_{T}(x,x^{*})\ge\langle x,x^{*}\rangle$
	and $\iota_{C}(x)+\sigma_{C}(-x^{*})\ge-\langle x,x^{*}\rangle$ (see
	f.i. \cite{MR1009594}). Hence 
	\begin{equation}
	0\le\inf_{X\times X^{*}}F=-(\varphi_{T}+g)^{*}(0,0)=-\min_{(x,x^{*})\in X\times X^{*}}\{\psi_{T}(x,x^{*})+g^{*}(-x^{*},-x)\},\label{conv}
	\end{equation}
	because $C$ is bounded, $g$ is $s_{X}\times s_{X^{*}}-$continuous
	on $\operatorname*{int}C\times X^{*}$, and $X$ is reflexive (see
	f.i. \cite[Theorem 2.8.7, p. 126]{MR1921556}), where $s_{X^{*}}$
	is the strong topology of $X^{*}$. Here ``min'' denotes an infimum
	that is attained when finite, 
	\begin{equation}
	\psi_{T}(x,x^{*})=\varphi_{T}^{*}(x^{*},x),\ (x,x^{*})\in X\times X^{*},\label{eq:-1}
	\end{equation}
	the convex conjugation being taken with respect to the dual system
	$(X\times X^{*},X^{*}\times X^{**})$ and, for every $(x,x^{*})\in X\times X^{*}$,
	$\psi_{T}(x,x^{*})\ge\langle x,x^{*}\rangle$ because $T$ is monotone
	(see e.g. \cite[(12)]{MR2577332}).
	
	From $g^{*}(x^{*},x)=\iota_{C}(-x)+\sigma_{C}(x^{*})$, $(x,x^{*})\in X\times X^{*}$
	and (\ref{conv}) there exists $(\bar{x},\bar{x}^{*})\in X\times X^{*}$
	such that $\psi_{T}(\bar{x},\bar{x}^{*})+\iota_{C}(\bar{x})+\sigma_{C}(-\bar{x}^{*})\le0$
	which implies that $\iota_{C}(\bar{x})+\sigma_{C}(-\bar{x}^{*})=-\langle\bar{x},\bar{x}^{*}\rangle$,
	i.e., $-\bar{x}^{*}\in N_{C}(\bar{x})$ and $\psi_{T}(\bar{x},\bar{x}^{*})=\langle\bar{x},\bar{x}^{*}\rangle$,
	that is, $\bar{x}^{*}\in T(\bar{x})$ since $T$ is representable
	(see \cite[Theorem 1, p. 270]{MR2577332}). Therefore $0\in(T+N_{C})(\bar{x},\bar{x}^{*})$
	and so $0\in R(T+N_{C})$. \end{proof}

\begin{proof}[Proof of Theorem 1.1] 
	First we prove that we can assume
	without loss of generality that $D(B)$ is bounded. Indeed, assume
	that the result is true for that case. Let $z=(x,x^{*})$ be m.r.
	to $A+B$. Take $C\subset X$ closed convex and bounded with $x\in\operatorname*{int}C$
	and $D(A)\cap\operatorname*{int}D(B)\cap\operatorname*{int}C\neq\emptyset$
	e.g. $C:=[x_{0},x]+U$, where $[x_{0},x]:=\{tx_{0}+(1-t)x\mid0\le t\le1\}$
	and $U$ is a closed convex bounded neighborhood of $0$, and $x_{0}\in D(A)\cap\operatorname*{int}D(B)$.
	Note that $z$ is m.r. to $A+B+N_{C}=A+(B+N_{C})$ which is maximal
	monotone since, according to Theorem 2, $B+N_{C}$ is maximal monotone,
	$D(B+N_{C})$ is bounded, and $x_{0}\in D(A)\cap\operatorname*{int}D(B+N_{C})\neq\emptyset$.
	Hence $z\in\operatorname*{Graph}(A+B+N_{C})$ or $x^{*}\in(A+B)(x)$
	because $N_{C}(x)=\{0\}$. Therefore $A+B$ is maximal monotone.
	
	It remains to prove that, whenever $D(B)$ is bounded, $R(A+B)=X^{*}$
	or sufficiently $0\in R(A+B)$ (since $A+B$ is representable, see
	again \cite[Corollary\ 5.6]{MR2453098}).
	
	Let $F(x,x^{*}):=\varphi_{A}(x,x^{*})+\varphi_{B}(x,-x^{*})$, $g(x,x^{*}):=\varphi_{B}(x,-x^{*})$,
	$(x,x^{*})\in X\times X^{*}$. Since $A$, $B$ are maximal monotone,
	for every $(x,x^{*})\in X\times X^{*}$, $\min\{\varphi_{A}(x,x^{*}),\varphi_{B}(x,x^{*})\}\ge\langle x,x^{*}\rangle$
	which imply $F\ge0$ and so 
	\begin{equation}
	0\le\inf_{X\times X^{*}}F=-(\varphi_{A}+g)^{*}(0,0)=-\min_{(x,x^{*})\in X\times X^{*}}\{\psi_{A}(x,x)+\psi_{B}(x,-x^{*})\},\label{eq:-2}
	\end{equation}
	because $D(B)$ bounded provides $D(B)\times X^{*}\subset\operatorname*{dom}g$,
	$g$ is $s_{X}\times s_{X^{*}}-$continuous on $\operatorname*{int}D(B)\times X^{*}$,
	and $X$ is reflexive (see again \cite[Theorem 2.8.7, p. 126]{MR1921556}).
	More precisely, for every $(x,x^{*})\in D(B)\times X^{*}$ there is
	$\overline{x}^{*}\in B(x)$ and so
	\[
	\begin{aligned}\varphi_{B}(x,x^{*}) & :=\sup\{\langle x-b,b^{*}\rangle+\langle b,x^{*}\rangle\mid(b,b^{*})\in\operatorname*{Graph}B\}\\
	& \le\sup\{\langle x-b,\overline{x}^{*}\rangle+\langle b,x^{*}\rangle\mid(b,b^{*})\in\operatorname*{Graph}B\}\\
	& \le\langle x,\overline{x}^{*}\rangle+\|x^{*}-\overline{x}^{*}\|\sup_{b\in D(B)}\|b\|<+\infty.
	\end{aligned}
	\]
	
	There exists $(\bar{x},\bar{x}^{*})\in X\times X^{*}$ such that $\psi_{A}(\bar{x},\bar{x}^{*})+\psi_{B}(\bar{x},-\bar{x}^{*})\le0$
	which implies that $\psi_{A}(\bar{x},\bar{x}^{*})=\langle\bar{x},\bar{x}^{*}\rangle$,
	$\psi_{B}(\bar{x},-\bar{x}^{*})=-\langle\bar{x},\bar{x}^{*}\rangle$,
	i.e., $\bar{x}^{*}\in A(\bar{x})$ and $-\bar{x}^{*}\in B(\bar{x})$
	from which $0\in R(A+B)$. \end{proof}

\begin{remark}
Theorem 2 still holds if we replace the assumption $C$ bounded with
$D(T)$ bounded. In this case an alternate proof of Theorem 1 can
be performed with $A+N_{C}$ instead of $A$ and a similar argument
as in the current proof. 
\end{remark}


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