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\begin{document}
\title{CHARACTERIZATION OF APPROXIMATE MONOTONE OPERATORS} 




\author[Z. S. Mirsaney]{Zahra Sadat Mirsaney}
\address[Z. S. Mirsaney]{University of Isfahan, Iran.}
\email{\tt z.mirsaney@sci.ui.ac.ir}
\author[M. Rezaei]{Mahboubeh Rezaei}
\address[M. Rezaei]{University of Isfahan, Iran.}
\email{{\tt mrezaie@sci.ui.ac.ir}}

\keywords{$\theta^{'}$-monotone, $\theta$-Fitzpatrick function,  convex,  $\theta$-subdifferential, maximal  $\theta^{'}$-monotone, $\theta$-conjugate. }

\subjclass[2010]{47H05, 49J53, 90C25, 42A50}

\maketitle
\begin{abstract}
Results concerning local boundedness of   operators have a long history. In 1994, Vesel$\acute{\text{y}} $
connected the concept  of    approximate  monotonicity of an operator  with local boundedness of that.  It is our desire in this note to  characterize an approximate monotone operator. Actually, we show that a well-known property of monotone operators, namely representing by convex functions, remains valid for the larger subclass of  operators. In this general framework we establish the similar results by  Fitzpatrick. %\cite{fit}
 Also, celebrated results of   Mart$\acute{\text{i}}$nez-Legaz and  Th$\acute{\text{e}}$ra %  \cite{conv rep},
    inspired  us to  prove that the set of maximal $ \varepsilon$-monotone operators between a normed linear space $  X$ and its continuous dual $ X^{*} $ can be identified as some subset of  convex functions on $ \X $. 
\end{abstract}



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%%%
\section{Introduction}


The concept of monotonicity for multivalued operators defined on a Banach
space and taking values in its dual has imposed itself    \cite{2 teta,3 teta,4 teta,14 teta,15 teta}  play  an  important  role  in  convex  analysis,  optimization  theory,  partial 
differential  equations  and  other  fields  of  mathematics.
 Recently, it was a challenging problem to  represent  maximal monotone operators by convex functions  \cite{ex r1,fit,conv rep,ex r2}.  As we will show the approximate monotone operators are as important as monotone oparators, therefore useful characterization of them is important to the same extent.  Fitzpatrick  developed the representation of monotone operators on $ X$ in terms of the subdifferentials of convex functions on $X\times X^{*}$    \cite{fit}. Our aim here is to extend this idea to  the approximate monotone operators.
Also, in \cite{conv rep}  Mart$\acute{\text{i}}$nez-Legaz and    Th$\acute{\text{e}}$ra proved that the set of maximal monotone operators between a normed linear space $  X$ and its continuous dual
$ X^{*} $ can be identified as some subset of the set of  lower semicontinuous convex proper functions on $X \times X^{*}$, inspired by this idea we will characterize $ \varepsilon$-monotone operator.
 


 According to \cite{teta}, an operator $ T  $ is $ \theta^{'} $-monotone if
 \begin{equation*}
  \langle  u -v, x -y\rangle          \geq \theta^{'}(x, y)\Vert x -y\Vert, ~ \forall (x,u),(y,v) \in \G(T).
\end{equation*}

 

 

  Via some examples  in \cite{teta}, it is shown that the $\theta^{'}$-monotonicity
is more general than most of monotonicity properties known in literature.\\    The $ \theta^{'} $-monotone operators are the key ingredient of  some   branches of mathematics \cite{2 teta,3 teta,9 teta,10 teta,11 teta,12 teta,24 teta}.
As we know, the question about local boundedness of  an operator  is very important in the theory of monotone operators. Classical results due to Rockafellar   \cite{2ves,3ves,1ves},  Borwein and Fitzpatrick \cite{4ves}, answer this question. Using a trick from     \cite{5ves}, Vesel\'{y}  \cite{ves} presented a Banach-Steinhaus type theorem for families of $ \varepsilon $-monotone operators and some useful  consequences.
  Let $ \varepsilon\geq 0$, the  $ \varepsilon $-monotone operator $ T\xxx $ defined as follows
\begin{center}
$  \langle  u -v, x -y\rangle          \geq -\varepsilon,~ \forall (x,u),(y,v) \in \G(T) $.
\end{center}
In this case we can say $ T $ is $ \theta^{'} $-monotone in which  
 \begin{equation}\label{epsilon}
 \theta^{'}(x,y)=\dfrac{-\varepsilon}{\Vert x-y\Vert},~~\theta^{'}(x,x)=0,~~\forall x,y\in D(T):~~x\neq y. 
 \end{equation}
 We see in \cite{ves}, one of the most important result  in the theory of approximate  monotone operators, which establishes a  connection  between $ \varepsilon $-monotonicity of an operator and  local boundedness of that.
Before introducing this proposition, we state  some   basic concepts.

By $ k(M) $ we denote the convex cone generated by a set $  M\subseteq X$, i.e.
\begin{equation*}
k(M)=\co (\underset{t\geq 0}\bigcup tM),
\end{equation*}
in which $\co$ means convex hull of a set. 



We say that $x\in X  $ is a sup-point for a set $  M\subseteq X$ if there exists $ x^{*}\in X^{*}\setminus \lbrace 0\rbrace $ such
that $  \langle x^{*},x \rangle\geq \sup\langle M,x^{*}\rangle$, i.e. if $  M$ is contained in a closed halfspace having $  x$ as a boundary point.





\begin{proposition}\label{cor4ves}
 Let $ \varepsilon\geq 0 $, $  T$ be an $ \varepsilon $-monotone operator on   $  X$ and $ x\in \overline{\D(T)}  $.
Let either $k(\D(T)-X)  $ be a second category set, or $  X$ be barrelled and  $\inte k(\D(T)-X)\neq \emptyset$.
\begin{itemize}
\item[1)] If x is not a sup-point for $ \D(T) $, then $T  $ is locally bounded at $  x$.
\item[2)] If $  T$ is maximal $  \varepsilon$-monotone,
\end{itemize}
 then the following assertions are equivalent:
\begin{itemize}
\item[(i)]
$  x$ is not a sup-point for $ \D(T) $  and $x\in \D(T)  $;
\item[(ii)]
$  x$ is not a sup-point for $ \D(T) $;
 \item[(iii)]
 $  T$ is locally bounded at $  x$;
  \item[(iv)]
  $x\in \D(T)  $ and $T(x)$ is bounded.
\end{itemize}
\end{proposition} 
\begin{proof}
The proof can be found in \cite[Corollary 4]{ves}.
\end{proof}
 As we said   the $ \varepsilon $-monotone operators play an important role in the theory of monotone operators. Consequently, we need the clear recognition of   $ \varepsilon $-monotone operators.\\
 The paper is organized as follows. After preleminaries in  Section 2, we establish in section 3 some useful properties regarding  the convex functions on $ \X $      and their $ \theta $-subdifferential.
In section 4, under reasonable assumptions we show   that 
 the convex functions representing maximal $ \varepsilon$-monotone operators satisfy a minimality condition.
In section 5,  we   restrict our attention  to   prove that the set of maximal $ \varepsilon$-monotone operators between a normed linear space $  X$ and   $  X^{*}$ can be identified as some subset of the set of all  proper convex  functions on $  X\times X^{*}$.\\
\section{Notations and preleminaries}
Throughout  the  paper,  we  use  standard notations  except  special symbols  introduced  when  they  are  defined.  All  spaces  considered  are  Banach.  For  any space  $  X$,  we  consider  its dual  space  $X^{*}  $. \\The norms in $  X$ and $X^{*}  $ will be denoted by $  \Vert.\Vert$.  We
denote  by  $\langle.,.\rangle$  the duality  pairing  between  $X^{*}  $ and  $  X$. The space $  X\times X^{*}$ is also a  Banach space, with the norm given by
\begin{equation*}
\Vert (x,x^{*})\Vert=\Vert x\Vert+\Vert x^{*}\Vert,~~~\forall \x \in \X.
\end{equation*}

Identifying $  (\X)^{*}$ with $ X^{*}\times X $, we adopt the natural duality between
$ \X $ and $ X^{*}\times X $ given by $\langle (y^{*},y),\x\rangle=\langle y^{*},x\rangle+\langle x^{*},y\rangle  $.


%%%%%%%teta
%%%%%%%teta
%%%%%%%teta
From now on $\theta:(X\times X^{*})\times (X\times X^{*})\longrightarrow\mathbb{R},~\sigma:  X^{*} \times X^{*} \longrightarrow\mathbb{R}$ and $\theta^{'}:X\times X\longrightarrow\mathbb{R}$ be the given functions
 that \begin{center}
$ \theta((x,x^{*}),(y,y^{*}))=\theta^{'}(x,y)+\sigma (x^{*},y^{*}), $
\end{center}
 by the properties that  $\theta^{'}(x, y) = \theta^{'}(y, x),~\sigma(x^{*}, y^{*}) = \sigma(y^{*}, x^{*})$  and $\theta^{'}(x, x) =0=\sigma(x^{*}, x^{*})$ for all $\x, \y \in \X$.\\
 \textbf{ \underline{From now on, the functions   $ \theta,~\theta^{'} $ are defined such as \eqref{epsilon}.} }\label{bas}\\
  In what follows we introduce the concept of $\theta^{'}$-monotonicity for a  multivalued  operator $T\xxx$. \\We denote by $\D(T) = \lbrace x \in X : T(x) \neq \emptyset\rbrace$
 its domain. The graph of the operator $T$ is the set
$\G(T) =\lbrace(x, u)\in X \times X^{*} : u \in T(x)\rbrace$.
We need the following definitions which can be found in \cite{teta}.
\begin{definition}
The operator  $T\xxx$ is $\theta^{'}$-monotone, if  
\begin{equation}
  \langle  u -v, x -y\rangle          \geq \theta^{'}(x, y)\Vert x -y\Vert, ~\forall (x,u),(y,v) \in \G(T).  \label{def 1.1}
\end{equation}

Moreover a $ \theta^{'}$-monotone operator $T$ is maximal $ \theta^{'}$-monotone if for every operator $T^{'}\xxx$ which is $ \theta^{'}$-monotone with $\G(T)\subseteq   \G(T^{'})$, one has $T=T^{'}$.

\end{definition}

\begin{definition}\label{def 1.2}
We say that the pair $(x,x^{*})$ is $ \theta^{'}$-monotonically related to a subset $M $ of $ X \times X^{*} $ if
\begin{equation*}
  \langle  x^{*} -y^{*}, x -y\rangle          \geq \theta^{'}(x, y)\Vert x -y\Vert, ~\forall (y,y^{*}) \in M.
\end{equation*}
\end{definition}

The following result  can be found in    \cite[Proposition 3.4.]{teta}.
\begin{proposition}\label{pro 1.3}
 A $ \theta^{'}$-monotone operator $T\xxx $ is maximal $ \theta^{'}$-monotone
if and only if whenever a pair $(x, u) \in X \times X^{*}$ is $ \theta^{'}$-monotonically related to  $   \G(T)$, it holds that $u \in T(x)$.
\end{proposition}

\begin{example}
Let    $ T:\mathbb{R}\longrightarrow \mathbb{R} $ be  a single-valued operator defined as,  $T(x)=x $   and  suppose that   $\theta^{'}(x,y)=\vert x-y\vert$ for all $ x,y\in \mathbb{R} $.  By Proposition and Definitions above  we conclude that $ T $ is a maximal $\theta^{'}$-monotone.
\end{example}

Let $ f:X\longrightarrow \mathbb{R}\cup\lbrace+\infty\rbrace $ be given. We denote the domain of $f$ by \begin{center}
$ \Dm f:=\lbrace x \in X : f(x)<\infty\rbrace$.
\end{center}
We say  that $f  $
is  proper  if  $  \Dm f$ is  nonempty, also  $ f $ is a convex function if
\begin{center}
$ f(\lambda x +(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y), $
\end{center}
whenever $x, y \in X  $ and $0<\lambda<1  $. The notions of the   subdifferential of a convex function are fundamental in optimization.
The subdifferential of a function $f : X \longrightarrow\mathbb{R}$ is the multivalued operator $\partial f \xxx$ defined by
\begin{center}
$\partial f(x) = \lbrace v \in X^{*} |f(y) \geq f(x) + \langle y - x,v\rangle,~ \forall y \in X\rbrace.$
\end{center}
We recall   a new subdifferential concept, the so-called $\theta^{'}$-subdifferential which is introduced at first in \cite{teta}.
  \begin{definition}\label{def 4.5}
Let  $f :X \longrightarrow \mathbb{R}\cup \lbrace+\infty\rbrace$ be a proper function. One says that $x^{*} \in X^{*}$ is a $\theta^{'}$-subgradient of $f$ in $x \in \Dm f  $, if
\begin{center}
 $  \langle x^{*}, y -x \rangle           \leq f(y) -f(x) -\theta^{'}(x, y)\Vert x -y\Vert, ~\forall y \in X$.
\end{center} The set
 \begin{align}\nonumber
&\partial_{\theta^{'}}f(x)\\
&= \lbrace x^{*} \in X^{*} :   \langle x^{*}, y -x   \rangle           \leq f(y) -f(x) -\theta^{'} (x, y)\Vert x -y\Vert, ~\forall y \in X\rbrace\label{16}
 \end{align}
is called the $\theta^{'}$-subdifferential of $f$ at $x \in \Dm  f $.
\end{definition}
\begin{remark}\label{rem0}
It is easy to check that ${\partial}_{\theta^{'}}f \xxx$ is a $2\theta^{'}$-monotone operator. Let
$x^{*} \in {\partial}_{\theta^{'}}f(x)~\text{and}~y^{*} \in {\partial}_{\theta^{'}}f(y)$, then $  \langle x^{*}, y -x \rangle            \leq f(y) -f(x) -\theta^{'}(x, y)\Vert x-y\Vert$ and
$  \langle y^{*}, x -y \rangle            \leq f(x) -f(y) -\theta^{'}(x, y)\Vert y-x\Vert$, which added give us $  \langle y^{*}-x^{*}, y-x \rangle            \geq
2\theta^{'}(x, y)\Vert x -y\Vert$.
\end{remark}
\begin{example}
Let us consider the function $f : \mathbb{R}\longrightarrow\mathbb{R}$ defined by
\begin{equation*}
f (x)=\left\{
\begin{array}{rl}
0~~~,~~~ &x= 0\\
x^{2}+1, ~~~&x\neq0\\
\end{array} \right.
\end{equation*}
and $ \theta^{'}(x,y)=c\vert x-y\vert $ such that $ c\in \mathbb{R}, c<1 $. By standard calculus we find
\begin{align*}
\partial_{\theta^{'}}f (0)&=\lbrace x^{*}\in \mathbb{R}:\langle x^{*}, y\rangle\leq y^{2}+1-cy^{2},~ \forall y \in \mathbb{R},~y\neq0\rbrace\\
& (\text{if $ y=0 $ all $ x^{*}\in \mathbb{R} $ satisfy \eqref{16}}) \\
&=\lbrace x^{*}\in \mathbb{R}: 0\leq (1-c) y^{2}+1-x^{*} y,~ \forall y \in \mathbb{R},~y\neq0\rbrace\\
&=\lbrace x^{*}\in \mathbb{R}: 0\geq {x^{*}}^{2}-4(1-c)\rbrace\\
&=\lbrace x^{*}\in \mathbb{R}: \vert x^{*}\vert\leq2\sqrt{1-c}\rbrace.
\end{align*}
\end{example}





 
We introduce next the very important concept of conjugate functions, originally developed by Fenchel in \cite{84 set}. The Fenchel conjugate of a function $f : X \longrightarrow \mathbb{R}\cup\lbrace +\infty\rbrace$  is the function $f^{*} : X \longrightarrow \mathbb{R}\cup\lbrace +\infty\rbrace$ defined by
\begin{equation*}
 f^{*}(x^{*})=\underset{x\in X}\sup\lbrace   \langle x^{*},x \rangle -f(x)\rbrace.\label{fen}
\end{equation*}

An immediate consequence of the definition of the Fenchel conjugate is the famous Fenchel-Young inequality:
\begin{proposition}[Fenchel-Young Inequality]\label{4.4.1z}
Let   $f : X \longrightarrow \mathbb{R}\cup\lbrace+\infty\rbrace$ be a convex function. Suppose that $  x\in \Dm f$ and $ x^{*}\in X^{*} $. Then satisfy the inequality
\begin{center}
$ f(x)+f^{*}(x^{*})\geq \langle x^{*},x \rangle.$
\end{center}
Equality holds if and only if $x^{*}\in \partial f(x)$.
\end{proposition}
\begin{proof}
The proof can be found in \cite[Proposition 4.4.1]{z}.
\end{proof}

 
Now we are ready to introduce the $\theta$-conjugate function and   inspired by the proof  of Proposition \ref{4.4.1z},   we  extend this  theorem.\\
Let $f$ be a function from $X$ to $\mathbb{R}\cup\lbrace+\infty\rbrace$. Let $y \in X$ be fixed. We define the $\theta$-conjugate function  $f^{*}_{y}(\theta,.) : X^{*} \longrightarrow \mathbb{R}\cup\lbrace+\infty\rbrace$    of  $ f $  at $ y $ by
\begin{equation}
f^{*}_{y}(\theta^{'},\xi)=\underset{x\in X}\sup\lbrace   \langle \xi,x \rangle -f(x)+\theta^{'}(x,y)\Vert x-y\Vert \rbrace\label{17}.
\end{equation}
\begin{proposition}\label{ineq}
Let $f : X \longrightarrow \mathbb{R}\cup\lbrace+\infty\rbrace$  and $ \theta^{'}:X\times X\longrightarrow \mathbb{R} $. Then for all $  x\in \Dm f$ and $ x^{*}\in X^{*} $ we have
\begin{center}
$ f(x)+f^{*}_{x}(\theta^{'},x^{*})= \langle x^{*},x \rangle\Longleftrightarrow x^{*}\in \partial_{\theta^{'}}f(x).$
\end{center}
\end{proposition}
\begin{proof}
For all $  x\in \Dm f$ and $ x^{*}\in X^{*} $ we have following inequality
\begin{align*}
x^{*}\in \partial_{\theta^{'}}f(x)&\Longleftrightarrow \langle x^{*},y-x\rangle\leq f(y)-f(x)-\theta^{'}(x,y)\Vert x-y\Vert,~~\forall y\in X\\
&\Longleftrightarrow  f(x)+f^{*}_{x}(\theta^{'},x^{*})\leq \langle x^{*},x \rangle.
\end{align*}
On the other hand by \eqref{17}, we have
\begin{center}
$f(x)+f^{*}_{x}(\theta^{'},x^{*})\geq \langle x^{*},x \rangle,$
\end{center}
and the proof is complete.
\end{proof}







%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{ convex  functions on $ X\times X^{*} $}
In this section, let us establish some required properties of convex functions.
At first we introduce  a special function  on $ X $,  using the  $ \theta $-subdifferential of function on $ \X $.
\begin{definition}
For each  function $f:X\times X^{*}\longrightarrow\mathbb{R}$ let
\begin{equation*}
T^{\theta}_{f}(x):=\lbrace x^{*}\in X^{*}:(x^{*},x)\in \partial_{\theta}f(x,x^{*}) \rbrace
\end{equation*}
for each $ x\in X $.
\end{definition}
%mybe we must defin teta subdiff

\begin{proposition}\label{2.2}
 $ T^{\theta}_{f} $ is a $ \theta^{'} $-monotone operator on $  X$.
\end{proposition}
\begin{proof}
If $x^{*}\in {T^{\theta}_{f}(x)}$ and $y^{*}\in{T^{\theta}_{f}(y)}$.
For $ x=y  $ we have 
$$\langle x^{*}-y^{*},x-y\rangle\geq\theta^{'}(x,y)\Vert x-y\Vert$$
 On the other hand, by Remark \ref{rem0}   for all $ x\neq y  $, we conclude that 
\begin{align*}
\langle x^{*}-y^{*},x-y\rangle &=\dfrac{1}{2}\langle (x^{*},x)-(y^{*},y),(x,x^{*})-(y,y^{*}) \rangle\\
& \geq \theta((x,x^{*}),(y,y^{*}))\Vert (x,x^{*})-(y,y^{*})\Vert \\
&=\theta^{'}(x,y)\Vert x-y\Vert.
\end{align*}
\end{proof}
This poses a natural question:  is there any convex function  $ f $ on $ \X $ such that  for each  maximal $ \theta^{'} $- monotone operator $ T $,  $ T^{\theta}_{f}=T $? We answer this question in Corollary \ref{answer}, in the next section.

\begin{example}
Let $ g:X\longrightarrow \mathbb{R} $ be a function. If for arbitrary $ z\in X $, the function $ f^{z} $ on $ \X $  defined as follows
\begin{center}
 $f^{z}(x,x^{*})=g(x)+g^{*}_{z}(\theta^{'},x^{*}),~\forall \x \in  \X$.
\end{center}
Then $T^{\theta}_{f^{x}}(x)= \partial_{\theta^{'}}g(x)$ for all $ x\in X $.
\end{example}
\begin{proof}
If $x^{*}\in \partial_{\theta^{'}}g(x)$, then by \eqref{16} and \eqref{17}, we have
\begin{align*}
\langle x, y^{*}-x^{*}\rangle&\leq -g(x)-g^{*}_{x}(\theta^{'},x^{*})+\langle y^{*},x\rangle\\
&\leq -g^{*}_{x}(\theta^{'},x^{*})+\underset{z\in X}\sup\lbrace\langle y^{*},z\rangle-g(z) +\theta^{'}(x,z)\Vert x-z\Vert\rbrace\\
&=-g^{*}_{x}(\theta^{'},x^{*})+g^{*}_{x}(\theta^{'},y^{*}),
\end{align*}
so that, $x\in \partial g^{*}_{x}(\theta^{'},x^{*})$. Hence by an elementary
computation, we have
\begin{align*}
&\langle (x^{*},x),\y -\x \rangle\leq g(y)+g^{*}_{x}(\theta^{'},y^{*})\\
&-[g^{*}_{x}(\theta^{'},x^{*})+g(x)]-\theta^{'}(x,y)\Vert x-y\Vert \\
&=f^{x}(y,y^{*})-f^{x}(x,x^{*}) -\theta^{'}(x,y)\Vert x-y\Vert \\
&\leq f^{x}(y,y^{*})-f^{x}(x,x^{*}) -\theta(\x,\y)\Vert \x-\y\Vert ,
\end{align*}
hence  $(x^{*},x)\in \partial_{\theta}f^{x}(x,x^{*})$. On the other hand,  if $(x^{*},x)\in \partial_{\theta}f^{x}(x,x^{*})$, then for $ u\in X $ we have
\begin{align*}
\langle x^{*},u\rangle &= \langle (x^{*},x),(u,0)\rangle\leq f^{x}(x+u,x^{*})-f^{x}(x,x^{*})\\
&-\theta((x+u,x^{*}),(x,x^{*}))\Vert(u,0)\Vert\\
&=g(x+u)-g(x)-\theta^{'}(x+u,x)\Vert u\Vert+\underset{0}{\underbrace{ g^{*}_{x}(\theta^{'},x^{*})-g^{*}_{x}(\theta^{'},x^{*})}},
\end{align*}
so that $x^{*}\in \partial_{\theta^{'}}g(x)$ as required.
\end{proof}

We will be interested in the case when $f(x,x^{*})\geq \langle x^{*},x\rangle$ for all $  (x,x^{*})\in X\times X^{*}$. This allows a simple way to guarantee $x^{*}\in T^{\theta}_{f}(x)$.
\begin{theorem}\label{2.4}
Suppose that  $f$ is a  convex function on $X\times X^{*}$. If for some $(y,y^{*})\in X\times X^{*}$ we have   $f(y,y^{*}) =\langle y^{*},y\rangle$   and $f(x,x^{*})\geq \langle x^{*},x\rangle$
for all $(x,x^{*})$ in some neighbourhood $U$ of $(y,y^{*})$, then $y^{*}\in T^{\theta}_{f}(y)$.
\end{theorem}
\begin{proof}
Let $(z,z^{*})\in X\times X^{*}$ and $1>s > 0$, so that $(y+sz,y^{*}+sz^{*})\in U$. Then
%u is absorbing
\begin{align*}
f(y+z,y^{*}+&z^{*})-f(y,y^{*})\geq \dfrac{f(y+sz,y^{*}+sz^{*})-f(y,y^{*})}{s}\\
&+\dfrac{s(1-s)\theta((y+z,y^{*}+z^{*}), (y,y^{*}))\Vert (y+z,y^{*}+z^{*})- (y,y^{*})\Vert}{s}\\
%by assumption
&\geq\dfrac{ \langle y+sz,y^{*}+sz^{*}\rangle-\langle y^{*},y\rangle}{s}\\
&+(1-s)\theta((y+z,y^{*}+z^{*}), (y,y^{*}))\Vert (y+z,y^{*}+z^{*})- (y,y^{*})\Vert\\%-teta is positiv
&= \langle z^{*},y\rangle+\langle y^{*},z\rangle+ s \langle z^{*},z\rangle\\
&+(1-s)\theta((y+z,y^{*}+z^{*}), (y,y^{*}))\Vert (y+z,y^{*}+z^{*})- (y,y^{*})\Vert\\
&\text{Letting $ s\longrightarrow 0^{+} $   we have}\\
&=\langle (y^{*},y),(z,z^{*})\rangle\\
&+\theta((y+z,y^{*}+z^{*}), (y,y^{*}))\Vert (y+z,y^{*}+z^{*})- (y,y^{*})\Vert,
\end{align*}
%by convexity f

so that $ (y^{*},y)\in \partial_{\theta}f(y,y^{*}) $, hence  $y^{*}\in T^{\theta}_{f}(y)$ as required.
\end{proof}

Because of interesting applications of  the previous theorem in the following sections, we are
concerned   with obtaining functions satisfying the hypotheses of this theorem. Now denote the $x$-section of $f$ by $f_{x}(x^{*}) := f(x,x^{*})$ for $x\in X$ and $x^{*}\in X^{*}$.
\begin{theorem}\label{2.5}
If for some $ \x \in \X $ we have $x\in \partial_{\sigma}f_{x}(x^{*})$ and $ (f_{x})^{*}_{x^{*}}(\sigma,x)=0 $, then $f(x,x^{*}) = \langle x^{*},x \rangle$.

% such that $ \sigma:X^{*}\times X^{*}\longrightarrow \mathbb{R}$. If
%
\end{theorem}
\begin{proof}
By $ (f_{x})^{*}_{x^{*}}(\sigma,x)=0 $,  we have
\begin{equation}\label{sup1}
\underset{z^{*}\in X^{*}}\sup\lbrace \langle x,z^{*}\rangle+\sigma(x^{*},z^{*})\Vert x^{*}-z^{*}\Vert-f_{x}(z^{*})\rangle\rbrace=0,
\end{equation}
and  $x\in \partial_{\sigma}f_{x}(x^{*})$ conclude that  \begin{center}
 $\langle u^{*},x\rangle \leq f_{x}(x^{*}+u^{*})-f_{x}(x^{*})-\sigma(x^{*},x^{*}+u^{*})\Vert u^{*}\Vert,$
\end{center}
 for all $  u^{*}\in X^{*}$, so we have
\begin{equation*}
\langle x,x^{*}+u^{*} \rangle -f(x,x^{*}+u^{*})+\sigma(x^{*},x^{*}+u^{*})\Vert u^{*}\Vert\leq \langle x,x^{*}\rangle-f(x,x^{*}).
\end{equation*}
Taking the supremum over $u^{*}$, by  \eqref{sup1} we have $ f(x,x^{*})\leq  \langle x,x^{*}\rangle$.
However putting $z^{*}=x^{*}$ in \eqref{sup1} we get $ f(x,x^{*})\geq  \langle x,x^{*}\rangle$, so $ f(x,x^{*})= \langle x,x^{*}\rangle$.
\end{proof}
\begin{corollary}
Suppose that  $f$ is a  convex function on $X\times X^{*}$.  If  $ (f_{x})^{*}_{x^{*}}(\sigma,x)=0 $ for some $ (x,x^{*})\in X\times X^{*}$ and  $f(y,y^{*})\geq \langle y^{*},y\rangle$
for all $(y,y^{*}) \in X\times X^{*}$. Then $ x\in \partial_{\sigma}{f_{x}}(x^{*}) $ if and only if $x^{*}\in T^{\theta}_{f}(x)$.

 
\end{corollary}
\begin{proof}

Combine Theorems \ref{2.4} and \ref{2.5}, we conclude that  $x^{*}\in T^{\theta}_{f}(x)$.
Conversly, take $x^{*}\in T^{\theta}_{f}(x)$, then for all $ u^{*}\in   X^{*} $
\begin{align*}
\langle (x^{*},x),(x,x^{*}+ u^{*})-&(x,x^{*})\rangle \leq f_{x}(x^{*}+ u^{*})-f_{x}(x^{*})\\
&-\theta((x,x^{*}+ u^{*}),(x,x^{*}))\Vert (x,x^{*}+ u^{*})-(x,x^{*})\Vert\\
&\Downarrow\\
 \langle  x, u^{*} \rangle &\leq f_{x}(x^{*}+ u^{*})-f_{x}(x^{*})-\theta((x,x^{*}+ u^{*}),(x,x^{*}))\Vert u^{*}\Vert\\
 &=f_{x}(x^{*}+ u^{*})-f_{x}(x^{*})-\sigma(x^{*}+ u^{*},x^{*})\Vert u^{*}\Vert\\
\end{align*}
This completes the proof. 
\end{proof}

\section{convex functions from $ \varepsilon $-monotone operators}
 
Fitzpatrick   \cite{fit}, showed that the convex functions representing maximal monotone operators satisfy a minimality condition. It is natural to ask whether this result  is still valid for the  larger class of operators.
In this section we use  a  convex function on $  X\times X^{*}$ for  reperesenting  an $ \varepsilon $-monotone operator on $  X$ and show that this functions satisfy a minimality condition.

\begin{definition}
Let $T$ be a $ \theta^{'} $-monotone operator.  The $ \theta$-Fitzpatrick function is defined as 
\begin{equation*}
L^{\theta }_{T}(x,x^{*})=\underset{(y,y^{*})\in \G(T)}\sup\lbrace \langle x^{*},y\rangle+\langle y^{*},x-y\rangle+\theta(\x,\y)\Vert \x-\y\Vert\rbrace.
\end{equation*}
for $x\in X$ and $x^{*}\in X^{*}$.
\end{definition}
The first result is immediate from the definition.
\begin{proposition}\label{conv}
Let  $ T $ be  an $ \varepsilon $-monotone operator as in \eqref{epsilon} and $ \G(T) $   is  a subset of $ \X  $ satisfies following  property, 
\begin{equation}
\lambda \x+(1-\lambda)\y \in  \G(T) \Longleftrightarrow \x=\y\in \G(T),
\end{equation}\label{pdt}
 for all $\lambda \in (0,1)$, then the function $ L^{\theta}_{T}$ is convex on $  X\times X^{*}$.
\end{proposition}
%\begin{proof}
%Let  $ \x,~\y \in \X $ and $ (z,z^{*})=\lambda \x+(1-\lambda)\y $ for an arbitrary $ \lambda\in \mathbb{R}_{+} $. We claim that for all $ (u,u^{*})\in \G(T) $
%\begin{align*}
%&\langle z^{*},u\rangle+\langle u^{*},z-u\rangle+\theta^{'}(z,u)\Vert z-u\Vert\leq\\
%&\lambda(\langle x^{*},u\rangle+\langle u^{*},x-u\rangle+\theta^{'}(x,u)\Vert x-u\Vert)+\\
%&(1-\lambda)(\langle y^{*},u\rangle+\langle u^{*},y-u\rangle+\theta^{'}(x,u)\Vert x-u\Vert).
%\end{align*}
%If $z\in \D(T) $, then $ x=y=z $ and $\theta^{'}(z,u)\Vert z-u\Vert\leq \lambda \theta^{'}(x,u)\Vert x-u\Vert+(1-\lambda)\theta^{'}(y,u)\Vert y-u\Vert$.
%On the other hand, if  $z\notin D(T) $ we have two cases:\\
%one case is, $ x\neq y \in D(T) $. Another case is $ x=y  \notin D(T)   $.

%\end{proof}
As a start to examining $  \partial_{\theta}{L^{\theta}_{T}}$ we have the following definition and  result.

\begin{lemma}\label{3.3}
Let $ T $ be   an $ \varepsilon $-monotone operator.
 If  there exists $ (y,y^{*})\in \G(T) $ such that  for $  (x,x^{*})\in X\times X^{*}$ we have 
\begin{equation*}
L^{\theta^{'}}_{T}(x,x^{*})=\langle x^{*},y\rangle+\langle y^{*},x-y\rangle+\theta(\x,\y)\Vert \x-\y\Vert,
\end{equation*}

then $ (y,y^{*})\in  \partial_{\theta}{L^{\theta}_{T}}(x,x^{*}) $.
\end{lemma}
\begin {proof}
For each $ u\in X $ and $ u^{*}\in X^{*} $ we have
\begin{align*}
&L^{\theta}_{T}(x+u,x^{*}+u^{*})-L^{\theta }_{T}(x,x^{*})\\
&=\underset{(v,v^{*})\in \G(T)}\sup\lbrace \langle x^{*}+u^{*},v\rangle+\langle v^{*},x+u\rangle-\langle v,v^{*}\rangle+\\
&\theta ((x+u,x^{*}+u^{*}),(v,v^{*}))\Vert(x+u,x^{*}+u^{*})-(v,v^{*})\Vert\rbrace-L_{T}(x,x^{*})\\
%(v,v*)=(y,y*)
&\geq  \langle y,x^{*}+u^{*}\rangle+\langle x+u, y^{*}\rangle-\langle y,y^{*}\rangle\\
&+\theta ((x+u,x^{*}+u^{*}),(y,y^{*}))\Vert(x+u,x^{*}+u^{*})-(y,y^{*})\Vert-\langle x^{*},y\rangle-\\ &\langle y^{*},x-y\rangle-\theta ((x^{*},x),(y,y^{*}))\Vert(x^{*},x)-(y,y^{*})\Vert\\
&\geq \langle (y^{*},y),(u,u^{*})\rangle+\theta((x+u,x^{*}+u^{*}),(x,x^{*}))\Vert (u,u^{*})\Vert
\end {align*}
so we have $(y^{*},y)\in  \partial_{\theta}{L^{\theta}_{T}}(x,x^{*}) $. 
\end{proof}
\begin{theorem}\label{3.4}
If $ T $ is a $ \theta^{'} $-monotone operator   and $ (x,x^{*})\in \G(T) $, then $ L^{\theta}_{T}(x,x^{*})=\langle x^{*},x \rangle $. 
\end{theorem}
\begin{proof}
By $ \theta^{'} $-monotonicity of $T  $, for all  $(y,y^{*})\in \G(T)$ we have
\begin{align*}
\langle x,x^{*}\rangle&\geq \langle x^{*},y\rangle+\langle y^{*},x-y\rangle+\theta^{'}(x,y)\Vert x-y\Vert\\
&\geq \langle x^{*},y\rangle+\langle y^{*},x-y\rangle+\theta(\x,\y)\Vert \x-\y\Vert,
 \end{align*}
 so that $ L^{\theta}_{T}(x,x^{*})\leq\langle x,x^{*}\rangle $. On the other hand,
  \begin{equation*}
 L^{\theta}_{T}(x,x^{*})\geq\langle x,x^{*}\rangle+\langle x-x,x^{*}\rangle+\theta(\x,\x)\Vert \x-\x\Vert=\langle x,x^{*}\rangle.
 \end{equation*}

\end{proof}


\begin{theorem}\label{3.6}
If  $ f(x,x^{*})\geq \langle x,x^{*}\rangle$ for all $ (x,x^{*}) \in X\times X^{*} $,  then $ L^{\theta}_{T^{\theta}_{f}}\leq f.$
\end{theorem}
\begin{proof}
 By our assumptions for all  $ (y,y^{*})\in \X $ we have
 \begin{align*}
 &L^{\theta}_{T^{\theta}_{f}}(y,y^{*})- f(y,y^{*})\\
&=\underset{x^{*}\in T^{\theta}_{f}x}\sup\lbrace \langle y^{*},x\rangle+\langle x^{*},y-x\rangle+\theta(\x,\y)\Vert \x-\y\Vert-f(y,y^{*})\rbrace\\
&=\underset{(x^{*},x)\in \partial_{\theta}f(x,x^{*}) } \sup\lbrace\langle y^{*}-x^{*},x\rangle+\langle x^{*},y-x\rangle\\
&+\langle x^{*},x\rangle+\theta(\x,\y)\Vert \x-\y\Vert-f(y,y^{*})\rbrace\\
&=\underset{(x^{*},x)\in \partial_{\theta}f(x,x^{*}) } \sup\lbrace f(y,y^{*})-f(x,x^{*})-\theta((x,x^{*}),(y,y^{*}))\Vert (x,x^{*})-(y,y^{*})\Vert\\
&+\langle x^{*},x\rangle+\theta(\x,\y)\Vert \x-\y\Vert-f(y,y^{*})\rbrace \\
&\leq 0.
 \end{align*}
Consequently, $ L^{\theta}_{T^{\theta}_{f}}\leq f.$
\end{proof}

\begin{theorem}\label{3.7}
If $T$ be a $ \theta^{'} $-monotone operator and $ f $ be a  convex function such that $ f(y,y^{*})=\langle y^{*},y \rangle $ for all $(y,y^{*})\in \G(T)$ and $ f(x,x^{*})\geq\langle x^{*},x \rangle $ for all $(x,x^{*})\in X\times X^{*}$, then $ L^{\theta}_{T}\leq f.$
\end{theorem}
\begin{proof}
By Theorem \ref{2.4}, if $y^{*}\in Ty$, then $y^{*}\in T^{\theta}_{f}y$.
Thus for all $x\in X$ and $x^{*}\in X^{*}$ we  have
\begin{align*}
&L^{\theta^{'}}_{T}(x,x^{*})\\
&=\underset{(y,y^{*})\in \G(T)}\sup\lbrace \langle x^{*},y\rangle+\langle y^{*},x\rangle-\langle y^{*},y\rangle+ \theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
&\leq \underset{(y,y^{*})\in G(T^{\theta}_{f})}\sup\lbrace \langle x^{*},y\rangle+\langle y^{*},x\rangle-\langle y^{*},y\rangle+ \theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
&=L^{\theta}_{T^{\theta}_{f}}(x,x^{*})\leq f(x,x^{*}),
\end{align*}
by Theorem \ref{3.6}.
\end{proof}

\begin{theorem}\label{3.8}
If $T$ is a $ \theta^{'} $-monotone operator.   Then $T$ is maximal $ \theta^{'} $-monotone  if 
and only if $ L^{\theta}_{T}(x,x^{*})>\langle x^{*},x \rangle$  whenever $x\in X$ and $x^{*}\in X^{*}\setminus T(x)$.   
\end{theorem}
\begin{proof}
If $ L^{\theta}_{T}(x,x^{*})\leq \langle x^{*},x \rangle $, then  we have  \begin{center}
$ \langle x^{*},y\rangle+\langle y^{*},x-y\rangle+\theta(\x,\y)\Vert \x-\y\Vert \leq \langle x^{*},x \rangle$
\end{center} for all 
$(y,y^{*})\in \G(T)$ so \begin{center}
$\langle x^{*}-y^{*},x-y \rangle\geq \theta(\x,\y)\Vert \x-\y\Vert.$
\end{center}
Generally, if $x=y$, then $\langle x^{*}-y^{*},x-y \rangle= \theta^{'}(x,y)\Vert x-y\Vert=0.$
On the other hand, for all $ y\neq x $ we have 
\begin{center}
$\langle x^{*}-y^{*},x-y \rangle\geq \theta(\x,\y)\Vert \x-\y\Vert= \theta^{'}(x,y)\Vert x-y\Vert.$
\end{center}
When $  T$ is maximal $ \theta^{'} $-monotone that implies $x^{*}\in Tx$.\\ Conversely,  if $T$ is not maximal $ \theta^{'} $-monotone, then there are $x\in X$ and $x^{*}\in X^{*}\setminus T(x)$ such that \begin{center}
$\langle x^{*}-y^{*},x-y \rangle\geq \theta^{'}(x,y)\Vert x-y\Vert\geq\theta(\x,\y)\Vert \x-\y\Vert$
\end{center}
 for all  $(y,y^{*})\in \G(T)$. It follows that  $ L^{\theta}_{T}(x,x^{*})\leq \langle x^{*},x \rangle$.%-teta positiv 
\end{proof}
\begin{corollary}\label{3.9}
If  $T$ be a maximal $ \theta^{'} $-monotone  operator,  Then 
$ L^{\theta}_{T}(x,x^{*})\geq \langle x,x^{*}\rangle $ for all $x\in  X$ and $x^{*}\in X^{*}$, and  $L^{\theta}_{T}(x,x^{*})=\langle x^{*},x \rangle$  if and only if 
$x^{*}\in Tx  $.
\end{corollary}
\begin{proof}
Use Theorems \ref{3.4} and \ref{3.8}.
\end{proof}
\begin{theorem}\label{final}
If $T$ be a maximal $ \theta^{'} $-monotone such that $ \G(T) $ satisfying \eqref{pdt},  then $L^{\theta}_{T}$ is minimal element of following family
\begin{align*}
&{\mathcal{H}}^{\theta^{'}}(T)=\\
&\lbrace f:X\times X^{*}\longrightarrow \mathbb{R}: f ~\text{is}~\text{convex}~,~ f(x,x^{*})\left\{\begin{array}{rl}=\langle x^{*},x\rangle & \text{if}~(x,x^{*})\in \G(T),\\ 
 \geq\langle x^{*},x\rangle &   ~\text{otherwise}
 \end{array}\right.  \rbrace.
\end{align*}

 
\end{theorem}
\begin{proof}
we have $ L^{\theta^{'}}_{T}\leq f$  for any such function  $  f$  by Theorem \ref{3.7}.  However,  $ L^{\theta^{'}}_{T}$ has 
the  required properties by Proposition \ref{conv} and Corollary \ref{3.9}. 
\end{proof}
\begin{corollary}\label{answer}
 Under assumptions of the previous theorem, $ T^{\theta}_{L^{\theta}_{T}}=T $.
\end{corollary}
\begin{proof}
The result follows from Proposition \ref{2.2}, Theorems \ref{2.4} and \ref{final}.
\end{proof}


\section{corresponding approximate monotone operators and subset of functions on $ \X $}
As we see in \cite{conv rep},    the set of maximal monotone
operators between a normed linear space $  X$ and its continuous dual
$ X^{*} $ can be identified as some subset of the set $\Gamma(\X)  $ of all lower
semicontinuous convex proper functions on $\X  $.\\
Using   the identification mentioned above 
, we want to identify  a  family of maximal $ \varepsilon $-monotone  operators to a  subset  of convex functions on $ \X $.
  The set of all $\theta^{'}$-maximal monotone operators is  denoted by $\mathfrak{M}_{\theta^{'}}(X)$,
in fact,  all of    $  \varepsilon$-monotone operators.\\
Considering $\varphi: X\times X^{*}\longrightarrow \overline{\mathbb{R}}=\left[ -\infty,+\infty\right]$, which is convex and proper, i.e., $\varphi$ is not identically equal to $+\infty$ and  define
\begin{equation*}
b(\varphi) = \lbrace (x, x^{*}) \in X\times X^{*}: \varphi(x,x^{*})\leq   \langle x,x^{*} \rangle           \rbrace.
\end{equation*}
Note $\delta_{b(\varphi)}$ the indicator function of $b(\varphi)$, i.e.

\begin{flalign*}
\delta_{b(\varphi)}(x,x^{*})= \left\{
\begin{array}{cc}
0&\text{if}~(x,x^{*})\in b(\varphi), \\
\\
+\infty &\text{if}~ (x,x^{*}) \notin b(\varphi). 
\end{array} \right.
\end{flalign*}
  By \eqref{17} we define   $\varphi^{*}_{(x,x^{*})}(\theta,(x^{*},x))$  as follow  
\begin{equation*}
\underset{(y,y^{*})\in X\times X^{*}}\sup\lbrace  \langle (x^{*},x),\y \rangle         -\varphi(y,y^{*}) +\theta(\x,\y)\Vert \x-\y\Vert\rbrace. 
\end{equation*}
Let $\Phi_{\theta}(X)$  be the set of all convex  proper function $\varphi: X\times X^{*}\longrightarrow \overline{\mathbb{R}}=\left[ -\infty,+\infty\right]$ such that
 \begin{equation}
 \varphi(x,x^{*})=(\varphi +\delta_{b(\varphi)} )^ {*}_{(x,x^{*})}(\theta,(x^{*},x)),~~~\forall (x,x^{*})\in X\times X^{*}.\label{21}
 \end{equation}
 
\begin{proposition}\label{pro 6.1}
Suppose that $\varphi\in \Phi_{\theta}(X)$, then
 \begin{equation}
 \varphi(x,x^{*})\geq  \langle x,x^{*} \rangle,~~~~~           \forall~(x,x^{*})\in X\times X^{*}.\label{star}
 \end{equation}
 \end{proposition}
\begin{proof}
If $(x, x^{*}) \notin b(\varphi)$, then obviously inequality \eqref{star} holds in this case.
If $(x, x^{*}) \in b(\varphi)$. Then
\begin{align*}
\varphi(x,x^{*})&=\dfrac{\varphi(x,x^{*})+(\varphi +\delta_{b(\varphi)} )^{*}_{(x,x^{*})}(\theta,(x^{*},x))}{2}\\
&\geq\dfrac{\langle (x^{*},x),(x,x^{*})\rangle}{2}\\
&\geq \langle x^{*},x\rangle
\end{align*}
\end{proof}
%(x,x^{*})\notin {b(\varphi)} clear
%(x,x^{*})\in {b(\varphi)}\Longrightarrow  \varphi(x,x^{*})=1/2(\varphi(x,x^{*})+\varphi(x,x^{*}))
%=1/2(\varphi(x,x^{*})+(\varphi -\theta(x,.)\Vert x-.\Vert)^{*}(x^{*},x))=1/2((\varphi-\theta(x,.)\Vert x-.\Vert)(x,x^{*})+(\varphi(.,.) -\theta(x,.)\Vert x-.\Vert)^{*}(x^{*},x))\geq(x,x^{*})
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem} If    $T \in \mathfrak{M}_{\theta^{'}}(X)$ and $ \G(T) $ satisfy \eqref{pdt}.  
Then $L^{\theta}_{T}\in \Phi_{\theta}(X)$. Moreover, the mapping
\begin{equation*}
\mathfrak{M}_{\theta^{'}}(X) \ni T\longmapsto L^{\theta}_{T}\in \Phi_{\theta}(X)
\end{equation*}
is a bijection, with inverse
\begin{equation*}
\Phi_{\theta}(X)\ni \varphi\longmapsto T_{\varphi}\in \mathfrak{M}_{\theta^{'}}(X)
\end{equation*}
given by $ T{\varphi}(x) = \lbrace x^{*} \in  X^{*}:\varphi(x, x^{*}) =   \langle x, x^{*} \rangle           \rbrace$.
\end{theorem}
\begin{proof}
Clearly, by Theorem \ref{final},  we conclude that   $b(L^{\theta}_{T})=\G(T)$.
 For each $(x, x^{*}) \in X\times X^{*}$, one has
\begin{align*}
&L^{\theta}_{T}(x, x^{*})= \\
&\underset{(y,y^{*})\in b(L^{\theta}_{T})}\sup\lbrace    \langle y, x^{*} \rangle           +   \langle x,y^{*} \rangle           -  \langle y,y^{*} \rangle           +\theta(\x,\y)\Vert \x  - \y \Vert\rbrace=\\
&\underset{(y,y^{*})\in b(L^{\theta}_{T})}\sup\lbrace    \langle (x^{*},x),(y,y^{*}) \rangle           -L^{\theta}_{T}(y, y^{*})+\theta(\x,\y)\Vert \x-\y\Vert \rbrace\\
%=\underset{(y,y^{*})\in b(\varphi^{\theta}_{T})}\sup\lbrace    \langle (x,x^{*}),(y,y^{*}) \rangle           -\varphi^{\theta}_{T}(y, y^{*})-\delta_{b(\varphi^{\theta}_{T})}(y, y^{*})+\theta(\x , \y)\Vert \x - \y\Vert \rbrace\\
&=(L^{\theta}_{T} +\delta_{b(L^{\theta}_{T})})^{*}_{\x}(\theta,(x^{*},x))
 \end{align*}
Hence $L^{\theta}_{T}\in \Phi_{\theta}(X)$.
Now, we shall prove that, for $\varphi\in \Phi_{\theta}(X)$, one has $T_{\varphi}\in \mathfrak{M}_{\theta^{' }}(X)$. Let
$x^{*}\in T_{\varphi}(x)$ and $y^{*}\in T_{\varphi}(y)$. Clearly for $ x=y $
\begin{equation*}
 \langle x -y, x^{*} - y^{*} \rangle   \geq
 \theta^{'}(x,y)\Vert x-y\Vert.
\end{equation*}
On the other hand, for all $ x\neq y $  we have
\begin{align*}
&\langle x -y, x^{*} - y^{*} \rangle =   \langle x, x^{*} \rangle  +   \langle y, y^{*} \rangle -  \langle x, y^{*} \rangle  -  \langle y, x^{*} \rangle \\
&=\varphi(x, x^{*})+\varphi(y, y^{*})-  \langle x, y^{*} \rangle            -  \langle y, x^{*} \rangle           \\
&= \varphi(x, x^{*})+ (\varphi+\delta_{b(\varphi)})^{*}_{(y,y^{*})}(\theta,(y^{*},y)) -  \langle x, y^{*} \rangle            -  \langle y, x^{*} \rangle           \\
&=(\delta_{b(\varphi)}+\varphi)(x,x^{*})+(\delta_{b(\varphi)}+\varphi)^{*}_{(y,y^{*})}(\theta,(y^{*},y))-  \langle x, y^{*} \rangle            -  \langle y, x^{*} \rangle         
\end{align*}
Hence, by \eqref{ineq} we have \begin{center}
$  \langle x -y, x^{*} - y^{*} \rangle           \geq \theta(\x,\y)\Vert \x-\y\Vert=\theta^{'}(x,y)\Vert x-y\Vert,$
\end{center}
establishing the $\theta^{'}$-monotonicity of $T_{\varphi}$.\\
In order to prove that it is maximal, fix $(x, x^{*}) \in X \times X^{*}$ and suppose that
\begin{equation*}
 \langle x -y, x^{*} - y^{*} \rangle   \geq
 \theta^{'}(x,y)\Vert x-y\Vert\geq\theta(\x,\y)\Vert \x-\y\Vert,
\end{equation*}
for all  $(y, y^{*})\in G(T_{\varphi})$, then we have
\begin{align*}
&\langle x,x^{*} \rangle \geq\\
& \underset{(y, y^{*})\in G(T^{\theta}_{\varphi})}\sup\lbrace   \langle x, y^{*} \rangle            +  \langle y, x^{*} \rangle           - \langle y, y^{*} \rangle           +\theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
&=\underset{\varphi(y, y^{*}) =   \langle y, y^{*} \rangle           }\sup\lbrace   \langle x, y^{*} \rangle            +  \langle y, x^{*} \rangle           -\varphi(y, y^{*})+\theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
%%%(y, y^{*})\inG(T^{\theta}_{\varphi})\Longrightarrow\varphi(y, y^{*}) =   \langle y, y^{*}~\varphi(y, y^{*}) \leq   \langle y, y^{*}
&=\underset{\varphi(y, y^{*}) \leq   \langle y, y^{*} \rangle           }\sup\lbrace   \langle x, y^{*} \rangle            +  \langle y, x^{*} \rangle           -\varphi(y, y^{*}) +\theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
&=\underset{(y, y^{*}) \in b(\varphi) }\sup\lbrace   \langle x, y^{*} \rangle            +  \langle y, x^{*} \rangle           -\varphi(y, y^{*}) +\theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
%&=\underset{ (y, y^{*}) \in X\times X^{*} }\sup\lbrace   \langle x, y^{*} \rangle            +  \langle y, x^{*} \rangle           -\varphi(y, y^{*})-\delta_{\varphi(b)}(y) +\theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
%\varphi\in\varphi\in \Phi_{\theta}(X)
&=(\varphi +\delta_{b(\varphi)})^{*}_{(x,x^{*})}(\theta,(x^{*},x))=\varphi(x,x^{*})\\
&\geq   \langle x,x^{*} \rangle    ~~~\text{by Proposition \ref{pro 6.1}}.
\end{align*}
Hence, $\langle x^{*},x\rangle=\varphi(x,x^{*})$ i.e. $(x,x^{*})\in \G(T_{\varphi})$. This proves that $\G(T_{\varphi})$ is
maximal $ \theta^{'} $-monotone according to the Proposition \ref{pro 1.3}.
It only remains to prove that, for $T \in \mathfrak{M}_{\theta^{'}}(X)$, one has $T_{L^{\theta}_{T}}=T$ and  for $\varphi\in \Phi_{\theta}(X)$, one has $L^{\theta}_{T_{\varphi}}=\varphi$.\\
To prove the first equality, let $x \in X$. We have
\begin{align*}
T_{L^{\theta}_{T}}(x)&=\lbrace x^{*}\in X^{*}:L^{\theta}_{T}(x,x^{*})=   \langle x, x^{*} \rangle\rbrace\\
%%by 6.1
&=\lbrace x^{*}\in X^{*}:L^{\theta}_{T}(x,x^{*})\leq  \langle x, x^{*} \rangle \rbrace          \\
&=\lbrace x^{*}\in X^{*}: (x,x^{*})\in b(L^{\theta}_{T})\rbrace\\
&=\lbrace x^{*}\in X^{*}: (x,x^{*})\in \G(T)\rbrace\\
&=T(x).
\end{align*}
To prove the second equality, let $(x,x^{*})\in X\times X^{*}$. We have
\begin{align*}
&L^{\theta}_{T_{\varphi}}(x,x^{*})\\
&=\underset{(y,y^{*})\in G(T_{\varphi})}\sup\lbrace  \langle x,y^{*} \rangle           +  \langle y,x^{*} \rangle           -  \langle y,y^{*} \rangle +\theta(\x,\y)\Vert \x-\y\Vert\rbrace \\
&=\underset{\varphi(y, y^{*}) =   \langle y, y^{*} \rangle           }\sup\lbrace  \langle x,y^{*} \rangle           +  \langle y,x^{*} \rangle           -  \langle y,y^{*} \rangle           +\theta(\x,\y)\Vert \x-\y\Vert\rbrace \\
&=\underset{\varphi(y,y^{*}) \leq   \langle y,y^{*} \rangle           }\sup\lbrace  \langle x,y^{*} \rangle           +  \langle y,x^{*} \rangle           -\varphi(y,y^{*})+\theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
&=\underset{(y,y^{*}) \in b(\varphi)}\sup\lbrace  \langle x,y^{*} \rangle           +  \langle y,x^{*} \rangle           -\varphi(y,y^{*})+\theta(\x,\y)\Vert \x-\y\Vert\rbrace\\
%&=\underset{(y,y^{*}) \in X\times X^{*}}\sup\lbrace  \langle x,y^{*} \rangle           +  \langle y,x^{*} \rangle           -\varphi(y,y^{*})+\theta(\x,\y)\Vert \x-\y\Vert-\delta_{b(\varphi)}(y,y^{*})\rbrace\\ 
&=(\varphi+\delta_{b(\varphi)})^{*}_{(x,x^{*})}(\theta,(x^{*},x))=\varphi(x,x^{*})
\end{align*}
\end{proof}

\begin{corollary}
  Let  $T$ be a $\theta^{'}$-monotone operator and  $ \G(T) $ satisfying \eqref{pdt}. $T$ is maximal $\theta^{'}$-monotone if and only if there exists (unique) $\varphi\in \Phi_{\theta}(X)$ such that
\begin{center}
$T(x)= \lbrace x^{*}\in X^{*}: (x^{*},x)=\varphi(x,x^{*})\rbrace,~~ \forall x\in X$.
\end{center}
\end{corollary}


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