text:
\section{Introduction}

Throughout this paper, we assume that
\begin{align*}\text{$X$ is a real Banach space with norm $\|\cdot\|$,}
\end{align*}
that $X^*$ is the continuous dual of $X$,
 and
that $X$ and $X^*$ are paired by $\scal{\cdot}{\cdot}$.
The \emph{closed unit ball} in $X$ is denoted by $B_X=
\menge{x\in X}{\|x\|\leq1}$, and $\ensuremath{\mathbb N}=\{1,2,3,\ldots\}$.

We identify
$X$ with its canonical image in the bidual space $X^{**}$.
As always, $X\times X^*$ and $(X\times X^*)^* = X^*\times X^{**}$
are paired via $$\scal{(x,x^*)}{(y^*,y^{**})} =
\scal{x}{y^*} + \scal{x^*}{y^{**}},$$ where $(x,x^*)\in X\times X^*$
and $(y^*,y^{**}) \in X^*\times X^{**}$.

Let $A\colon X\ensuremath{\rightrightarrows} X^*$
be a \emph{set-valued operator} (also known as multifunction)
from $X$ to $X^*$, i.e., for every $x\in X$, $Ax\subseteq X^*$,
and let
$\ensuremath{\operatorname{gra}} A = \menge{(x,x^*)\in X\times X^*}{x^*\in Ax}$ be
the \emph{graph} of $A$. The \emph{domain} of $A$, written as
 $\ensuremath{\operatorname{dom}} A$,  is $\ensuremath{\operatorname{dom}} A= \menge{x\in X}{Ax\neq\varnothing}$ and
$\ensuremath{\operatorname{ran}} A=A(X)$ is the \emph{range} of $A$. We say $A$ is a
 \emph{linear relation} if $\ensuremath{\operatorname{gra}} A$ is a linear
subspace.
Now let $U\times V\subseteq X\times X^*$.
We say that $A$ is \emph{monotone} with respect to $U\times V$,
if for every $(x,x^*)\in(\ensuremath{\operatorname{gra}} A)\cap (U\times V)$
and $(y,y^*)\in(\ensuremath{\operatorname{gra}} A)\cap (U\times V)$, we have
\begin{equation}
\scal{x-y}{x^*-y^*}\geq 0.
\end{equation}
Of course, by (classical) monotonicity we mean
monotonicity with respect to $X\times X^*$.
Furthermore, we say that
$A$ is \emph{maximally monotone} with respect to $U\times V$
if $A$ is monotone with respect to $U\times V$ and for
every operator $B\colon X\ensuremath{\rightrightarrows} X^*$ that is monotone with respect to $U\times
V$ and such that $(\ensuremath{\operatorname{gra}} A)\cap (U\times V)\subseteq (\ensuremath{\operatorname{gra}} B)\cap (U\times
V)$, we necessarily have $(\ensuremath{\operatorname{gra}} A)\cap (U\times V)=(\ensuremath{\operatorname{gra}} B)\cap(U\times
V)$. Thus, (classical) maximal monotonicity corresponds to
maximal monotonicity with respect to $X\times X^*$.
This slightly unusual presentation is required to state our main results;
moreover,
it yields a more concise formulation of monotone operators of type
(FP).

Now let $A:X\rightrightarrows X^*$ be monotone
and $(x,x^*)\in X\times X^*$.
 We say $(x,x^*)$ is \emph{monotonically related to}
$\ensuremath{\operatorname{gra}} A$ if
\begin{align*}
\langle x-y,x^*-y^*\rangle\geq0,\quad \forall (y,y^*)\in\ensuremath{\operatorname{gra}} A.\end{align*}
If $Z$ is
a real  Banach space with continuous dual $Z^*$ and a subset $S$ of
$Z$, we denote $S^\bot$ by $S^\bot = \menge{z^*\in Z^*}{\langle
z^*,s\rangle= 0,\quad \forall s\in S}$. Given a subset $D$ of  $Z^*$,
we define $D_{\bot}$  by
$D_\bot = \menge{z\in Z}{\langle z,d^*\rangle= 0,
\quad \forall d^*\in D}=D^{\bot}\cap Z$.

The operator \emph{adjoint} of $A$,
written as $A^*$, is defined by
\begin{equation*}
\ensuremath{\operatorname{gra}} A^* =
\menge{(x^{**},x^*)\in X^{**}\times X^*}{(x^*,-x^{**})\in(\ensuremath{\operatorname{gra}} A)^{\bot}}.
\end{equation*}
Note that the adjoint is always a linear relation
with $\ensuremath{\operatorname{gra}} A^*\subseteq X^{**}\times X^* \subseteq X^{**}\times X^{***}$.
These inclusions make it possible to consider
monotonicity properties of $A^*$; however, care is required:
as a linear relation, $\ensuremath{\operatorname{gra}} A^* \subseteq X^{**}\times X^*$ while
as a potential monotone operator we are led to consider
$\ensuremath{\operatorname{gra}} A^* \subseteq X^{**}\times X^{***}$.
Now let $A:X\rightrightarrows X^*$ be a linear relation.
 We say that $A$ is
\emph{skew} if $\ensuremath{\operatorname{gra}} A \subseteq \ensuremath{\operatorname{gra}} (-A^*)$;
equivalently, if $\langle x,x^*\rangle=0,\; \forall (x,x^*)\in\ensuremath{\operatorname{gra}} A$.
Furthermore,
$A$ is \emph{symmetric} if $\ensuremath{\operatorname{gra}} A
\subseteq\ensuremath{\operatorname{gra}} A^*$; equivalently, if $\scal{x}{y^*}=\scal{y}{x^*}$,
$\forall (x,x^*),(y,y^*)\in\ensuremath{\operatorname{gra}} A$.

We now recall three fundamental subclasses of maximally
monotone operators.

 \begin{definition}
 Let $A:X\ensuremath{\rightrightarrows} X^*$ be maximally monotone.
 Then three key types of monotone operators are defined as follows.
 \begin{enumerate}
 \item $A$ is
\emph{of dense type or type (D)} (1971, \cite{Gossez3}) if for every
$(x^{**},x^*)\in X^{**}\times X^*$ with
\begin{align*}
\inf_{(a,a^*)\in\ensuremath{\operatorname{gra}} A}\langle a-x^{**}, a^*-x^*\rangle\geq 0,
\end{align*}
there exist a  bounded net
$(a_{\alpha}, a^*_{\alpha})_{\alpha\in\Gamma}$ in $\ensuremath{\operatorname{gra}} A$
such that
$(a_{\alpha}, a^*_{\alpha})_{\alpha\in\Gamma}$
weak*$\times$strong converges to
$(x^{**},x^*)$.
\item $A$ is
\emph{of type negative infimum (NI)} (1996, \cite{SiNI}) if
\begin{align*}
\sup_{(a,a^*)\in\ensuremath{\operatorname{gra}} A}\big(\langle a,x^*\rangle+\langle a^*,x^{**}\rangle
-\langle a,a^*\rangle\big)
\geq\langle x^{**},x^*\rangle,
\quad \forall(x^{**},x^*)\in X^{**}\times X^*.
\end{align*}
\item
$A$ is \emph{of type Fitzpatrick-Phelps (FP)} (1992, \cite{FP}) if
whenever $V$ is an open convex subset of $X^*$ such that $V\cap \ensuremath{\operatorname{ran}}
A\neq\varnothing$,
it must follow
that $A$ is maximally monotone with respect to $X\times V$.
\end{enumerate}

\end{definition}
\begin{fact}\emph{(See \cite{Si, Si2, BorVan}.)}\label{ITPR}
The following are maximally monotone of type (D), (NI), and (FP).
\begin{enumerate}
\item $\partial f$, where
 $f: X\to\ensuremath{\,\left]-\infty,+\infty\right]}$ is convex, lower semicontinuous, and proper;
\item $A\colon X\ensuremath{\rightrightarrows} X^*$, where $A$ is maximally monotone and $X$ is
reflexive.
\end{enumerate}
\end{fact}
These and other relationships known
amongst these and other monotonicity notions are described in
\cite[Chapter 9]{BorVan}.
As we see in \cite{BBWY2} and \cite{Si4, SiNI, MarSva}, it is now known that the
three classes coincide.

Monotone operators have proven to be a key class of objects in both
modern Optimization and Analysis; see, e.g., \cite{Bor1,Bor2,Bor3},
the books \cite{BC2011,
BorVan,BurIus,ph,Si,Si2,RockWets,Zalinescu,Zeidlerlin,Zeidler}
and the references therein.

Let us now precisely describe the aforementioned Br\'{e}zis-Browder Theorem:

\begin{theorem}[Br\'ezis-Browder in reflexive Banach space \cite{BB76,Brezis-Browder}]
 \label{thm:brbr} Suppose that $X$ is reflexive.
Let $A\colon X \ensuremath{\rightrightarrows} X^*$ be a monotone linear relation
such that $\ensuremath{\operatorname{gra}} A$ is closed.
Then
$A$ is maximally monotone if and only if the adjoint
$A^*$ is monotone.
\end{theorem}

In this paper, we generalize the Br\'{e}zis-Browder Theorem to an arbitrary Banach space.
(See \cite{Si3} for Simons' recent extension of the above result
to {symmetrically self-dual Banach spaces} (SSDB)
spaces as defined in \cite[\S21]{Si2}.)

Our main result is the following.

\begin{theorem}[Br\'ezis-Browder in general Banach space] \label{thm:bbwy}
Let $A\colon X\rightrightarrows X^*$ be a monotone linear
 relation such that $\ensuremath{\operatorname{gra}} A$ is closed.
Then the following are equivalent.
\begin{enumerate}
\item $A$ is maximally monotone of type (D).
\item  $A$ is  maximally monotone  of type (NI).
\item  $A$ is  maximally monotone  of type (FP).
\item  $A^*$ is monotone.
\end{enumerate}
\end{theorem}

In Section~\ref{s:aux}, we collect auxiliary results for future
reference and for the reader's convenience.
In Section~\ref{s:tech}, we provide the key technical step showing that when $A^*$ is monotone then $A$ is of type (D).
Our central result, the
generalized
Br\'ezis-Browder Theorem
(Theorem~\ref{thm:bbwy}), is then proved in Section~\ref{s:main}.
Finally, in Section \ref{s:skew} with the necessary proviso that the domain be closed, we establish further results
such as Theorem~\ref{BrBrTDS:1} relating to the skew part of the operator. This was motivated by and
 extends \cite[Theorem~4.1]{BB} which studied the case of a bounded linear operator.

Finally, let us mention that we adopt standard convex analysis notation.
Given a subset $C$ of $X$,
$\ensuremath{\operatorname{int}} C$ is the \emph{interior} of $C$,
$\overline{C}$ is   the
\emph{norm closure} of $C$.
 For the set $D\subseteq X^*$, $\overline{D}^{\ensuremath{\operatorname{w*}}}$
is the weak$^{*}$ closure of $D$.
If $E\subseteq X^{**}$, $\overline{E}^{\ensuremath{\operatorname{w*}}}$
  is the weak$^{*}$ closure of $E$ in $X^{**}$ with
 the topology induced by $X^*$.
The \emph{indicator function} of $C$, written as $\iota_C$, is defined
at $x\in X$ by
\begin{align}
\iota_C (x)=\begin{cases}0,\,&\text{if $x\in C$;}\\
+\infty,\,&\text{otherwise}.\end{cases}\end{align} For every $x\in
X$, the \emph{normal cone} operator of $C$ at $x$ is defined by
$N_C(x)= \menge{x^*\in X^*}{\sup_{c\in C}\scal{c-x}{x^*}\leq 0}$, if
$x\in C$; and $N_C(x)=\varnothing$, if $x\notin C$.

Let $f\colon X\to \ensuremath{\,\left]-\infty,+\infty\right]}$. Then
$\ensuremath{\operatorname{dom}} f= f^{-1}(\ensuremath{\mathbb R})$ is the \emph{domain} of $f$, and
$f^*\colon X^*\to\ensuremath{\,\left[-\infty,+\infty\right]}\colon x^*\mapsto
\sup_{x\in X}(\scal{x}{x^*}-f(x))$ is
the \emph{Fenchel conjugate} of $f$.
The  \emph{lower semicontinuous hull} of $f$ is denoted by $\overline{f}$.
We say $f$ is proper if $\ensuremath{\operatorname{dom}} f\neq\varnothing$.
Let $f$ be proper. The \emph{subdifferential} of
$f$ is defined by
   $$\partial f\colon X\ensuremath{\rightrightarrows} X^*\colon
   x\mapsto \{x^*\in X^*\mid(\forall y\in
X)\; \scal{y-x}{x^*} + f(x)\leq f(y)\}.$$
For $\varepsilon \geq 0$,
the \emph{$\varepsilon$--subdifferential} of $f$ is defined by
   $\partial_{\varepsilon} f\colon X\ensuremath{\rightrightarrows} X^*\colon
   x\mapsto \menge{x^*\in X^*}{(\forall y\in
X)\; \scal{y-x}{x^*} + f(x)\leq f(y)+\varepsilon}$.
Note that $\partial f = \partial_{0}f$.
We  denote  by $J:=J_X$ the duality map, i.e.,
the subdifferential of the function $\tfrac{1}{2}\|\cdot\|^2$
mapping $X$ to $X^*$. For the properties of $J$, see \cite[Example~2.26]{ph}.

Let $(z,z^*)\in X\times X^*$ and $F\colon X\times X^*\to\ensuremath{\,\left]-\infty,+\infty\right]}$.
Then $F_{(z,z^*)}:X\times X^*\rightarrow\ensuremath{\,\left]-\infty,+\infty\right]}$
 \cite{MLegSva,Si2} is defined by
\begin{align}
F_{(z,z^*)}(x,x^*)&=F(z+x,z^*+x^*)-\big(\langle x,z^*\rangle+
\langle z,x^*\rangle+\langle z,z^*\rangle\big)\notag\\
&=F(z+x,z^*+x^*)-\langle z+x,z^*+x^*\rangle+\langle x,x^*\rangle,
\quad \forall(x,x^*)\in X\times X^*.\label{e:timmy}
\end{align}
Let now $Y$ be another real Banach space. We set  $P_X: X\times Y\rightarrow
X\colon (x,y)\mapsto x$.
Let $F_1, F_2\colon X\times Y\rightarrow\ensuremath{\,\left]-\infty,+\infty\right]}$.
Then the \emph{partial inf-convolution}  $F_1\Box_2 F_2$
is the function defined on $X\times Y$ by
\begin{equation*}F_1\Box_2 F_2\colon
(x,y)\mapsto \inf_{v\in Y}
F_1(x,y-v)+F_2(x,v).
\end{equation*}

 \section{Prerequisite results}\label{s:aux}

\begin{fact}\emph{(See \cite[Proposition~2.6.6(c)]{Megg} or
 \cite[Theorem~4.7 and Theorem~3.12]{Rudin}.)}\label{rudin:1}
Let $C$ be a subspace of  $X$, and $D$ be a subspace of $X^*$.
Then \begin{align*}(C^{\bot})_{\bot}=\overline{C}
\quad\text{and}\quad(D_{\bot})^{\bot}=\overline{D}^{\ensuremath{\operatorname{w*}}}.
\end{align*}
\end{fact}

 \begin{fact}[Rockafellar] \label{f:F4}
\emph{(See \cite[Theorem~3(b)]{Rock66}, \cite[Theorem~18.1]{Si2}
or
\cite[Theorem~2.8.7(iii)]{Zalinescu}.)}
Let $f,g:  X\rightarrow\ensuremath{\,\left]-\infty,+\infty\right]}$ be proper convex functions.
Assume that there exists a point $x_0\in\ensuremath{\operatorname{dom}} f \cap \ensuremath{\operatorname{dom}} g$
such that $g$ is continuous at $x_0$. Then
$$\partial (f+g)=\partial f+\partial g.$$
\end{fact}

 \begin{fact}[Br\o ndsted-Rockafellar]\label{f:FJB5}
\emph{(See  \cite[Theorem~3.1.2 or Theorem~3.1.4(ii)]{Zalinescu}.)}
Let $f: X\rightarrow\ensuremath{\,\left]-\infty,+\infty\right]}$ be a proper lower semicontinuous and convex function
and $x^*\in\ensuremath{\operatorname{dom}} f^*$.
Then
there exists   a
sequence $(x_n,x^*_n)_{n\in\ensuremath{\mathbb N}}$ in $\ensuremath{\operatorname{gra}} \partial f$ such that
$x^*_n \to x^*$.
\end{fact}

\begin{fact}[Attouch-Br\'ezis]\emph{(See \cite[Theorem~1.1]{AtBrezis}
 or \cite[Remark~15.2]{Si2}.)}\label{AttBre:1}
Let $f,g: X\rightarrow\ensuremath{\,\left]-\infty,+\infty\right]}$ be proper lower semicontinuous  and convex.
Assume that $
\bigcup_{\lambda>0} \lambda\left[\ensuremath{\operatorname{dom}} f-\ensuremath{\operatorname{dom}} g\right]$
is a closed subspace of $X$.
Then
\begin{equation*}
(f+g)^*(z^*) =\min_{y^*\in X^*} \left[f^*(y^*)+g^*(z^*-y^*)\right],\quad \forall z^*\in X^*.
\end{equation*}
\end{fact}

\begin{fact}[Simons and Z\u{a}linescu]
\emph{(See \cite[Theorem~4.2]{SiZ} or \cite[Theorem~16.4(a)]{Si2}.)}\label{F4}
Let $Y$ be a real Banach space and $F_1, F_2\colon X\times Y \to \ensuremath{\,\left]-\infty,+\infty\right]}$ be proper,
lower semicontinuous, and convex. Assume that
for every $(x,y)\in X\times Y$,
\begin{equation*}(F_1\Box_2 F_2)(x,y)>-\infty
\end{equation*}
and that  $\bigcup_{\lambda>0} \lambda\left[P_X\ensuremath{\operatorname{dom}} F_1-P_X\ensuremath{\operatorname{dom}} F_2\right]$
is a closed subspace of $X$. Then for every $(x^*,y^*)\in X^*\times Y^*$,
\begin{equation*}
(F_1\Box_2 F_2)^*(x^*,y^*)=\min_{u^*\in X^*}
\left[F_1^*(x^*-u^*,y^*)+F_2^*(u^*,y^*)\right].
\end{equation*}\end{fact}

The following result was first established in
\cite[Theorem~7.4]{Borwein}.
Now we give a new proof.
\begin{fact}[Borwein]\label{LinAdSum}
Let $A, B:X\rightrightarrows X^*$ be linear relations such that $\ensuremath{\operatorname{gra}} A$ and $\ensuremath{\operatorname{gra}} B$ are closed.
Assume that
$\ensuremath{\operatorname{dom}} A-\ensuremath{\operatorname{dom}} B$ is closed. Then
\begin{align*}(A+B)^*=A^*+B^*.\end{align*}
\end{fact}
\begin{proof}
We have
\begin{align}
\iota_{\ensuremath{\operatorname{gra}} (A+B)}=\iota_{\ensuremath{\operatorname{gra}} A}\Box_2\iota_{\ensuremath{\operatorname{gra}} B}.\label{LinAdSumP:1}
\end{align}
Let $(x^{**},x^*)\in X^{**}\times X^*$.
Since
$\ensuremath{\operatorname{gra}} A$ and $\ensuremath{\operatorname{gra}} B$ are closed convex,
$\iota_{\ensuremath{\operatorname{gra}} A}$ and  $\iota_{\ensuremath{\operatorname{gra}} B}$ are proper lower semicontinuous and convex.
Then by Fact~\ref{F4} and \eqref{LinAdSumP:1},
 there exists $y^*\in X^*$ such that
\begin{align}
\iota_{\ensuremath{\operatorname{gra}}(A+B)^*}(x^{**},x^*)&=\iota_{\big(\ensuremath{\operatorname{gra}} (A+B)\big)^{\bot}}(-x^*,x^{**})\nonumber\\
&=\iota^*_{\ensuremath{\operatorname{gra}} (A+B)}(-x^*,x^{**})\quad\text{(since $\ensuremath{\operatorname{gra}} (A+B)$ is a subspace)}\nonumber\\
&=
\iota^*_{\ensuremath{\operatorname{gra}} A}(y^*,x^{**})+\iota^*_{\ensuremath{\operatorname{gra}} B}(-x^*-y^*,x^{**})\nonumber\\
&=\iota_{(\ensuremath{\operatorname{gra}} A)^{\bot}}(y^*,x^{**})+\iota_{(\ensuremath{\operatorname{gra}} B)^{\bot}}(-x^*-y^*,x^{**})\nonumber\\
&=\iota_{\ensuremath{\operatorname{gra}} A^*}(x^{**},-y^*)+\iota_{\ensuremath{\operatorname{gra}} B^*}(x^{**},x^*+y^*)\nonumber\\
&=\iota_{\ensuremath{\operatorname{gra}} (A^*+B^*)}(x^{**},x^*).\label{LinAdSumP:2}
\end{align}
Then we have $\ensuremath{\operatorname{gra}} (A+B)^*=\ensuremath{\operatorname{gra}} (A^*+B^*)$ and hence $(A+B)^*=A^*+B^*$.
\end{proof}

\begin{fact}[Simons]
\emph{(See \cite[Lemma~19.7 and Section~22]{Si2}.)}
\label{f:referee}
Let $A:X\rightrightarrows X^*$ be a monotone operator such
 that $\operatorname{gra} A$ is convex with $\operatorname{gra} A
\neq\varnothing$.
Then the function
\begin{equation}
g\colon X\times X^* \rightarrow \left]-\infty,+\infty\right]\colon
(x,x^*)\mapsto \langle x, x^*\rangle + \iota_{\operatorname{gra} A}(x,x^*)
\end{equation}
is proper and convex.
\end{fact}

We also recall the somewhat more precise version of Theorem \ref{thm:brbr}.

\begin{fact}[Br\'ezis and Browder]  \label{Sv:7}\emph{(See \cite[Theorem~2]{Brezis-Browder},
 or \cite{Brezis70,BB76,Si3,Yao}.)}
Suppose that $X$ is reflexive.
Let $A\colon X \ensuremath{\rightrightarrows} X^*$ be a monotone linear relation
such that $\ensuremath{\operatorname{gra}} A$ is closed. Then the following are equivalent.
\begin{enumerate}
\item
$A$ is maximally monotone.
\item
$A^*$ is maximally monotone.
\item
$A^*$ is monotone.
\end{enumerate}
\end{fact}

This has a recent non-reflexive counterpart:

\begin{fact}\emph{(See \cite[Theorem~3.1]{BBWY:1}.)}
\label{TypeD:1}
Let $A:X\rightrightarrows X^*$ be a maximally monotone linear relation.
Then the following are equivalent.
\begin{enumerate}
\item $A$ is of type (D).

\item  $A$ is of type (NI).

\item  $A$ is of type (FP).
\item $A^*$ is monotone

\end{enumerate}
\end{fact}

Comparing of Fact~\ref{TypeD:1} and Fact~\ref{Sv:7},
we observe that the
 hypothesis in the latter (maximality of $A$) is more restrictive than in
  the former (closedness of the graph).
  In \cite[Theorem~3.1]{BBWY:1} we were unable to attack this issue.
The result of the next section redresses our lacuna.

Now let us cite some basic properties of linear relations.

The following result appeared in
 Cross' book \cite{Cross}.  We give new proofs of \ref{Sia:1}--\ref{Th:32}.
 The proof of the \ref{Th:32} below
was adapted from \cite[Remark~2.2]{BWY8}.

\begin{fact}\label{Rea:1}
Let $A:X \rightrightarrows X^*$ be a linear relation.
Then the following hold.
\begin{enumerate}
\item \label{Th:28}$Ax=x^* +A0,\quad\forall x^*\in Ax.$
\item \label{Th:30}
$A(\alpha x+\beta y)=\alpha Ax+\beta Ay, \forall (\alpha,\beta)\in\ensuremath{\mathbb R}^2\smallsetminus\{(0,0)\},\;
\forall x,y\in\ensuremath{\operatorname{dom}} A$.
\item \label{Sia:2b}
$\langle A^*x,y\rangle=\langle x,Ay\rangle$ is a singleton, $\forall x\in \ensuremath{\operatorname{dom}} A^*, \forall y\in\ensuremath{\operatorname{dom}} A$.
\item \label{Sia:1}$(\ensuremath{\operatorname{dom}} A)^{\bot}=A^*0$ is (weak$^*$) closed
 and $\overline{\ensuremath{\operatorname{dom}} A}=(A^*0)_{\perp}$.
\item \label{Sia:2} If $\ensuremath{\operatorname{gra}} A$ is  closed, then $(\ensuremath{\operatorname{dom}} A^*)_{\bot}=A0$ and
$\overline{\ensuremath{\operatorname{dom}} A^*}^{\ensuremath{\operatorname{w*}}}=(A0)^{\perp}$.
\item\label{Th:32} If $\ensuremath{\operatorname{dom}} A$ is closed, then $\ensuremath{\operatorname{dom}} A^*=(\bar{A}0)^{\bot}$
 and thus $\ensuremath{\operatorname{dom}} A^*$ is (weak$^*$) closed, where $\bar{A}$ is the linear
relation whose graph is the closure of the graph of $A$.
\end{enumerate}
\end{fact}
\begin{proof}
\ref{Th:28}: See \cite[Proposition I.2.8(a)]{Cross}.
 \ref{Th:30}: See \cite[Corollary I.2.5]{Cross}.
\ref{Sia:2b}: See \cite[Proposition III.1.2]{Cross}.

\ref{Sia:1}: We have
\begin{align*}
x^*\in A^*0\Leftrightarrow(x^*,0)\in(\ensuremath{\operatorname{gra}} A)^{\bot}
\Leftrightarrow x^*\in(\ensuremath{\operatorname{dom}} A)^{\bot}.
\end{align*}
 Hence $(\ensuremath{\operatorname{dom}} A)^{\bot}= A^*0$ and thus $A^*0$ is weak$^*$ closed.
 By Fact~\ref{rudin:1},
$\overline{\ensuremath{\operatorname{dom}} A}=(A^*0)_{\bot}$.

\ref{Sia:2}: Using Fact~\ref{rudin:1},  \begin{align*}
&x^*\in A0\Leftrightarrow (0,x^*)\in\ensuremath{\operatorname{gra}} A=
 \left[(\ensuremath{\operatorname{gra}} A)^{\bot}\right]_{\bot}=\left[\ensuremath{\operatorname{gra}} -(A^*)^{-1}\right]_{\bot}
 \Leftrightarrow x^*\in(\ensuremath{\operatorname{dom}} A^*)_{\bot}.
\end{align*}
Hence $(\ensuremath{\operatorname{dom}} A^*)_{\bot}=A0$ and thus, by Fact~\ref{rudin:1},
 $\overline{\ensuremath{\operatorname{dom}} A^*}^{\ensuremath{\operatorname{w*}}}=(A0)^{\bot}$.

 \ref{Th:32}:
Let $\bar{A}$ be the linear
relation whose graph is the closure of the graph of $A$.
Then $\ensuremath{\operatorname{dom}} A = \ensuremath{\operatorname{dom}}\bar{A}$ and $A^* = \bar{A}^*$.
Then by Fact~\ref{AttBre:1},
\begin{equation*}
\iota_{X^*\times (\bar{A}0)^\perp} =
\iota_{\{0\}\times \bar{A}0}^* =
\big(\iota_{\ensuremath{\operatorname{gra}} \bar{A}} + \iota_{\{0\}\times X^*}\big)^*
= \iota_{\ensuremath{\operatorname{gra}} (-\bar{A}^*)^{-1}} \ensuremath{\mbox{\small$\,\square\,$}} \iota_{X^*\times\{0\}}
= \iota_{X^*\times \ensuremath{\operatorname{dom}} \bar{A}^*}.
\end{equation*}
It is clear that $\ensuremath{\operatorname{dom}} A^*=\ensuremath{\operatorname{dom}} \bar{A}^* = (\bar{A}0)^\perp$ is
weak$^*$ closed, hence closed.
\end{proof}

\section{A key result}\label{s:tech}

The proof of Proposition~\ref{PrTD:1} below  was partially inspired by
 that of \cite[Theorem~3.1]{BBWY:1}.

\begin{proposition}\label{PrTD:1}
Let $A:X\rightrightarrows X^*$ be a monotone linear relation such that $\ensuremath{\operatorname{gra}} A$ is closed
and $A^*$ is monotone. Then $A$ is maximally monotone of type (D).
\end{proposition}

\begin{proof}
By Fact~\ref{TypeD:1}, it suffices to show that $A$ is maximally monotone.
Let $(z,z^*)\in X\times X^*$. Assume that
\begin{align}\text{$(z,z^*)$ is monotonically related to
$\ensuremath{\operatorname{gra}} A$.}\label{TIN:1}
\end{align}
Define
\begin{align*}
F:X\times X^*\to \ensuremath{\,\left]-\infty,+\infty\right]}\colon
(x,x^*)\mapsto \iota_{\ensuremath{\operatorname{gra}} A}(x,x^*)+\langle x,x^*\rangle.
\end{align*}
Fact~\ref{f:referee} implies that
$F$ is convex and since $\ensuremath{\operatorname{gra}} A$ is closed, $F$ is also proper, lower semicontinuous.
Recalling \eqref{e:timmy}, note that
\begin{align}
F_{(z,z^*)} \colon (x,x^*)
\mapsto \iota_{\ensuremath{\operatorname{gra}} A}(z+x,z^*+x^*)+\langle x,x^*\rangle \label{BrB:PSb1}
\end{align}
is proper, lower semicontinuous, and convex.
Set \begin{align}
G(x,x^*) :=
F_{(z,z^*)}(x,x^*)+ \tfrac{1}{2}\|x\|^2
+\tfrac{1}{2}\|x^*\|^2,\quad \forall (x,x^*)\in X\times X^*.\label{BrB:NPSb1}
\end{align}
Then
\begin{align}
\inf G =-G^*(0,0).
\label{BrB:PS1}
\end{align}
By \eqref{BrB:PSb1},  $\inf G\geq 0$. Then $(0,0)\in\ensuremath{\operatorname{dom}} G^*$.
By Fact~\ref{f:FJB5},  there exists a sequence
\begin{align}\big((a_n,a^*_n), (y^*_n, y^{**}_n)\big)_{n\in\ensuremath{\mathbb N}}\;\text{in}\; \ensuremath{\operatorname{gra}} \partial G
\label{BrB:NPSb2}\end{align}
such that
   \begin{align}(y^*_n, y^{**}_n)\rightarrow (0,0).   \label{coer:1}\end{align}
 Thus,
\begin{align}K_n:=\max\big\{\|y^*_n\|, \|y^{**}_n\|\big\}\rightarrow0.
\label{coer:N1}\end{align}
By Fact~\ref{f:F4} and \eqref{BrB:NPSb2}, there exists $(v^*_n, v^{**}_n)\in Ja_n\times J_{X^*} a^*_n$ such that
\begin{align}(y^*_n, y^{**}_n)\in \partial F_{(z,z^*)}(a_n,a^*_n)+(v^*_n,
v^{**}_n),\quad \forall n \in\ensuremath{\mathbb N}.
      \label{coer:N2}\end{align}
By \eqref{coer:N2}, \eqref{BrB:PSb1},
and \cite[Theorem 3.2.4(vi)\&(ii)]{Zalinescu},
 there exists a sequence
 $(z^*_n,z^{**}_n)_{n\in\ensuremath{\mathbb N}}$ in
$(\ensuremath{\operatorname{gra}} A)^{\bot}$ such that
\begin{align}
(y^*_n, y^{**}_n)=({a^{*}_n},
{a_n})+(z^*_n,z^{**}_n)+(v^*_n, v^{**}_n),\quad \forall n\in\ensuremath{\mathbb N}.\label{BrB:PS10}
\end{align}
Since $A^*$ is monotone and $(z^{**}_n,z^{*}_n)\in\ensuremath{\operatorname{gra}}(- A^*)$,
it follows from \eqref{BrB:PS10} that
\begin{align}&\negthinspace\negthinspace\negthinspace\negthinspace\langle
y^*_n, y^{**}_n\rangle+\langle a_n,a^*_n\rangle\nonumber\\ &\quad-\big[\langle y_n^*,
{a_n}\rangle+\langle y_n^{**}, {a^{*}_n}\rangle\big]-\big[\langle
y^*_n, v^{**}_n\rangle+\langle v^*_n,y^{**}_n\rangle\big]\nonumber\\&\quad+\langle a^*_n,v^{**}_n\rangle
+\langle v^*_n, v^{**}_n\rangle+\langle a_n,v^*_n\rangle\nonumber\\
&=\langle y^*_n-a^*_n-v^*_n, y^{**}_n-a_n-v^{**}_n\rangle\nonumber\\
&=\langle z^*_n,z^{**}_n\rangle\leq0, \quad \forall n\in\ensuremath{\mathbb N}.\label{DPNo:1}\end{align}
Since  $(v^*_n, v^{**}_n)\in Ja_n\times J_{X^*} a^*_n$, by \eqref{DPNo:1}, we have
\begin{align}&\negthinspace\negthinspace\negthinspace\negthinspace\langle y^*_n, y^{**}_n\rangle+\langle a_n,a^*_n\rangle\nonumber\\&-\big[\|y_n^*\|\cdot
\|{a_n}\|+\|y_n^{**}\|\cdot \|{a^{*}_n}\|\big]-\big[
\|y^*_n\|\cdot\|a_n^{*}\|+\|a_n\|\cdot\|y^{**}_n\|\big]\nonumber\\
\quad&+\| a^*_n\|^2
-\|a_n\|\cdot\|a^*_n\|+\|a_n\|^2 \leq0, \quad \forall n\in\ensuremath{\mathbb N}.\label{BrB:NPS10}\end{align}
Then by \eqref{BrB:NPS10} and \eqref{coer:N1},
\begin{align}&-K^2_n+\langle a_n,a^*_n\rangle-K_n\big[
\|{a_n}\|+\|{a^{*}_n}\|\big]-
K_n\big[\|a_n^{*}\|+\|a_n\|\big]\nonumber\\&\quad+\tfrac{1}{2}\big[\| a^*_n\|^2
+\|a_n\|^2\big] \leq0, \quad \forall n\in\ensuremath{\mathbb N}.
\label{BrB:NPS11}\end{align}
Hence
\begin{align}\nonumber&-K^2_n+\langle a_n,a^*_n\rangle-2K_n\big[
\|{a_n}\|+\|{a^{*}_n}\|\big]+\tfrac{1}{4}\big[\| a^*_n\|
+\|a_n\|\big]^2 \leq0, \quad \forall n\in\ensuremath{\mathbb N}.
\label{BrB:NPS12}\end{align}
Set $(x_n,x_n^*):=(z+{a_n},z^*+{a^{*}_n}),\quad \forall n\in\ensuremath{\mathbb N}$.
Then by \eqref{BrB:PSb1}, we have
\begin{align}
  F_{(z,z^*)}({a_n},
{a^{*}_n})&=\iota_{\ensuremath{\operatorname{gra}} A}(z+{a_n},z^*
+{a^{*}_n})+\langle {a_n},
{a^{*}_n}\rangle\\&=
  \iota_{\ensuremath{\operatorname{gra}} A}(x_n,x^*_n)+\langle x_n-z,x^*_n-z^*\rangle.
  \label{TINf:5}
\end{align}
By \eqref{coer:N2} and \eqref{TINf:5},
\begin{align}
(x_n, x^*_n)\in\ensuremath{\operatorname{gra}} A,\quad \forall n\in\ensuremath{\mathbb N}.  \label{TINf:6}
\end{align}
Then by \eqref{TINf:6} and   \eqref{TIN:1}, we have
\begin{align}
\langle{a_n},
{a^{*}_n}\rangle=\langle x_n-z,x^*_n-z^*\rangle\geq0,\quad \forall n\in\ensuremath{\mathbb N}.
  \label{TINf:7}
\end{align}
Combining \eqref{BrB:NPS12} and \eqref{TINf:7},
\begin{align}\tfrac{1}{4}\big(\| a^*_n\|
+\|a_n\|\big)^2 \leq K^2_n+2K_n\big(
\|{a_n}\|+\|{a^{*}_n}\|\big), \quad \forall n\in\ensuremath{\mathbb N};
\end{align}
equivalently,
\begin{align}\big(\| a^*_n\|
+\|a_n\|-4K_n\big)^2 \leq 20K^2_n, \quad \forall n\in\ensuremath{\mathbb N}.
\label{BrB:NPS13}\end{align}
In view of \eqref{coer:N1},
\begin{align}
\|a_n\|+\|a^*_n\|\rightarrow0.\label{TINf:3}
\end{align}
Thus
$(a_n,a^*_n)\to (0,0)$ and hence
$(x_n,x^*_n)\to (z,z^*)$.
Finally, by \eqref{TINf:6} and since $\ensuremath{\operatorname{gra}} A$ is closed, we see
$(z,z^*)\in\ensuremath{\operatorname{gra}} A$. Therefore, $A$ is maximally monotone.
\end{proof}

\begin{example}\label{SRE:1}
Let $A:X\rightrightarrows X^*$ be a monotone linear relation such that $\ensuremath{\operatorname{gra}} A$ is closed.
We note that we cannot guarantee
the maximal monotonicity of $A$ even if $A$ is at most single-valued and
densely defined.
To see this, suppose that  $X=\ell^2$,  and that
$A:\ell^2\rightrightarrows \ell^2$ is given by
\begin{align}Ax:=\frac{\bigg(\sum_{i< n}x_{i}-\sum_{i> n}x_{i}\bigg)_{n\in\ensuremath{\mathbb N}}}{2}
=\bigg(\sum_{i< n}x_{i}+\tfrac{1}{2}x_n\bigg)_{n\in\ensuremath{\mathbb N}},
\quad \forall x=(x_n)_{n\in\ensuremath{\mathbb N}}\in\ensuremath{\operatorname{dom}} A,\label{EL:1}\end{align}
where $\ensuremath{\operatorname{dom}} A:=\Big\{ x:=(x_n)_{n\in\ensuremath{\mathbb N}}\in \ell^{2}\mid \sum_{i\geq 1}x_{i}=0,
 \bigg(\sum_{i\leq n}x_{i}\bigg)_{n\in\ensuremath{\mathbb N}}\in\ell^2\Big\}$.
 Then $A$ is an at most single-valued linear relation.
Now \cite[Propositions~3.6]{BWY7} states that
\begin{align}
\label{PF:a2}
A^*x= \bigg(\ensuremath{\tfrac{1}{2}} x_n + \sum_{i> n}x_{i}\bigg)_{n\in\ensuremath{\mathbb N}},
\end{align}
where
\begin{equation*}
x=(x_n)_{n\in\ensuremath{\mathbb N}}\in\ensuremath{\operatorname{dom}} A^*=\bigg\{ x=(x_n)_{n\in\ensuremath{\mathbb N}}\in \ell^{2}\;\; \bigg|\;\;
 \bigg(\sum_{i> n}x_{i}\bigg)_{n\in\ensuremath{\mathbb N}}\in \ell^{2}\bigg\}.
\end{equation*}
 Moreover,
 \cite[Propositions~3.2, 3.5, 3.6 and 3.8]{BWY7}, \cite[Theorem~2.5]{PheSim} and Fact~\ref{Sv:7}
  show
that:
\begin{enumerate}
 \item\label{NEC:1}
 $A$ is maximally monotone and skew;
\item\label{NEC:2} $\ensuremath{\operatorname{dom}} A$ is dense and $\ensuremath{\operatorname{dom}} A\subsetneqq\ensuremath{\operatorname{dom}} A^*$;
\item\label{NEC:3} $A^*$ is  maximally monotone, but not skew;
\item\label{NEC:4} $-A$ is not maximally monotone.
\end{enumerate}
Hence, $-A$ is  monotone with dense domain and
$\ensuremath{\operatorname{gra}} (-A)$ is closed,
but nonetheless $-A$ is not maximally monotone. \ensuremath{\quad \hfill \blacksquare}
\end{example}

\section{The general Br\'ezis-Browder theorem}
\label{s:main}

We may now pack everything together. For ease we repeat Theorem \ref{thm:bbwy}:

\begin{theorem}[Br\'ezis-Browder in general Banach space] \label{BrBrTD:1}
Let $A\colon X\rightrightarrows X^*$ be a monotone linear
 relation  such that $\ensuremath{\operatorname{gra}} A$ is closed.
Then the following are equivalent.
\begin{enumerate}

\item\label{MaT:1} $A$ is maximally monotone of type (D).

\item \label{MaT:2} $A$ is  maximally monotone  of type (NI).

\item \label{MaT:3} $A$ is  maximally monotone  of type (FP).
\item \label{MaT:03} $A^*$ is monotone.

\end{enumerate}
\end{theorem}

\begin{proof}
Directly combine Fact~\ref{TypeD:1} and Proposition~\ref{PrTD:1}.
\end{proof}

The original Br\'ezis and Browder result follows.

\begin{corollary}[Br\'ezis and Browder]
Suppose that $X$ is  reflexive.
Let $A\colon X \ensuremath{\rightrightarrows} X^*$ be a monotone linear relation
such that $\ensuremath{\operatorname{gra}} A$ is closed. Then the following are equivalent.
\begin{enumerate}
\item\label{BRS:1}
$A$ is maximally monotone.
\item\label{BRS:2}
$A^*$ is maximally monotone.
\item\label{BRS:3}
$A^*$ is monotone.
\end{enumerate}
\end{corollary}
\begin{proof}
``\ref{BRS:1}$\Leftrightarrow$\ref{BRS:3}'': Apply Theorem~\ref{BrBrTD:1}
and Fact~\ref{ITPR} directly.

``\ref{BRS:2}$\Rightarrow$\ref{BRS:3}'': Clear.

``\ref{BRS:3}$\Rightarrow$\ref{BRS:2}'': Since $\ensuremath{\operatorname{gra}} A$ is closed, $(A^*)^*=A$.
Apply Theorem~\ref{BrBrTD:1} to $A^*$  .
\end{proof}

In the case of a skew operator we can add maximality of the adjoint and so we prefigure results of the next section:

\begin{corollary}\label{CoSBr:1}
Let $A\colon X \ensuremath{\rightrightarrows} X^*$ be a skew operator
such that $\ensuremath{\operatorname{gra}} A$ is closed. Then the following are equivalent.
\begin{enumerate}
\item\label{MaTS:1} $A$ is maximally monotone of type (D).

\item \label{MaTS:03} $A^*$ is monotone.
\item \label{MaTS:04} $A^*$ is maximally monotone  with respect to $X^{**}\times X^*$.

\end{enumerate}
\end{corollary}
\begin{proof}
By Theorem~\ref{BrBrTD:1}, it only remains to show

``\ref{MaTS:03}$\Rightarrow$\ref{MaTS:04}'':
Let $(z^{**}, z^*)\in X^{**}\times X^*$ be
monotonically related to $\ensuremath{\operatorname{gra}} A^*$. Since $\ensuremath{\operatorname{gra}} (-A)\subseteq\ensuremath{\operatorname{gra}} A^*$,
$(z^{**}, z^*)$ is monotonically related to $\ensuremath{\operatorname{gra}} (-A)$.
Thus $(z^{*}, z^{**})\in\left[\ensuremath{\operatorname{gra}} (-A)\right]^{\bot}$ since $\ensuremath{\operatorname{gra}} A$ is linear. Hence $(z^{**},z^*)\in\ensuremath{\operatorname{gra}} A^*$.
Hence $A^*$ is maximally monotone.
\end{proof}

\begin{remark}
We cannot say $A^*$ is maximally monotone with respect to
$X^{**}\times X^{***}$ in Corollary~\ref{CoSBr:1}\ref{MaTS:04}:
indeed,
let $A$ be defined by \begin{align*}\ensuremath{\operatorname{gra}} A=\{0\}\times X^*.\end{align*}
 Then $\ensuremath{\operatorname{gra}} A^*=\{0\} \times X^*$.
If $X$ is not reflexive, then $X^*\subsetneqq X^{***}$ and so
$\ensuremath{\operatorname{gra}} A^*$ is a proper subset of $\{0\}\times X^{***}$. Hence
$A^*$ is not maximally monotone with respect to $X^{**}\times X^{***}$ although
$A$ is maximally monotone of type (D) (since $A=N_{\{0\}}$
by Fact~\ref{ITPR}).
\end{remark}

In the next section,
we turn to the question of how the skew part of the adjoint behaves.

\section{Decomposition of monotone linear relations}\label{s:skew}
Let us first gather some basic properties about monotone linear relations,
and conditions for them to be maximally monotone.

The next three propositions were
 proven in reflexive spaces in \cite[Proposition~2.2]{BWY3}.
We adjust the proofs to a general Banach space setting.

\begin{proposition}[Monotone linear relations]\label{P:a}
Let $A\colon X \ensuremath{\rightrightarrows} X^*$ be a linear relation.
Then the following hold.
\begin{enumerate}
\item\label{Nov:s1}
Suppose $A$ is monotone. Then
$\ensuremath{\operatorname{dom}} A\subseteq (A0)_{\bot}$ and $A0\subseteq (\ensuremath{\operatorname{dom}} A)^\bot$;
consequently, if $\ensuremath{\operatorname{gra}} A$ is closed, then
$\ensuremath{\operatorname{dom}} A\subseteq\overline{\ensuremath{\operatorname{dom}} A^*}^{\ensuremath{\operatorname{w*}}}\cap X$ and $A0\subseteq A^*0$.
\item \label{Th:032}
$(\forall x\in\ensuremath{\operatorname{dom}} A)(\forall z\in (A0)_{\bot})$
$\langle z,Ax\rangle$ is single-valued.
\item \label{Sia:2c}
$(\forall z \in (A0)_{\bot})$ $\ensuremath{\operatorname{dom}} A \to \ensuremath{\mathbb R} \colon y\mapsto \scal{z}{Ay}$
is linear.
\item\label{Th:031} If $A$ is monotone, then
$(\forall x\in\ensuremath{\operatorname{dom}} A)$ $\langle x,Ax\rangle$ is single-valued.
\item\label{Th:031L}$A$ is monotone $\Leftrightarrow$
$(\forall x\in\ensuremath{\operatorname{dom}} A)$ $\inf\langle x,Ax\rangle$ $\geq 0$.
\item\label{Th:031+} If $(x,x^*)\in (\ensuremath{\operatorname{dom}} A)\times X^*$ is monotonically
related to $\ensuremath{\operatorname{gra}} A$ and $x_{0}^*$ $\in Ax$, then
$x^*-x_{0}^*\in (\ensuremath{\operatorname{dom}} A)^{\perp}.$

\end{enumerate}
\end{proposition}

\begin{proof}
\ref{Nov:s1}: Pick $x\in \ensuremath{\operatorname{dom}} A$.
Then there exists $x^*\in X^*$ such that
  $(x,x^*)\in \ensuremath{\operatorname{gra}} A.$  By
  monotonicity of $A$ and since $\{0\}\times A0\subseteq \ensuremath{\operatorname{gra}} A$,
 we have  $\langle x, x^*\rangle\geq \sup\langle x,A0\rangle$.
Since $A0$ is a linear subspace,
we obtain
$x\bot A0$. This implies
$\ensuremath{\operatorname{dom}} A\subseteq (A0)_{\bot}$ and $A0\subseteq (\ensuremath{\operatorname{dom}} A)^\bot$.
If $\ensuremath{\operatorname{gra}} A$ is closed, then Fact~\ref{Rea:1}\ref{Sia:2}\&\ref{Sia:1}
yields $\ensuremath{\operatorname{dom}} A\subseteq (A0)_{\bot}
\subseteq (A0)^{\bot}=\overline{\ensuremath{\operatorname{dom}} A^*}^{\ensuremath{\operatorname{w*}}}$ and  $A0\subseteq A^*0$.

\ref{Th:032}:
Take $x\in\ensuremath{\operatorname{dom}} A$, $x^*\in Ax$, and $z\in (A0)_{\bot}$.
By Fact~\ref{Rea:1}\ref{Th:28},
$\langle z,Ax\rangle=\langle z,x^*+A0\rangle=\langle z,x^*\rangle$.

\ref{Sia:2c}:
Take $z\in (A0)_{\bot}$. By \ref{Th:032},
$(\forall y\in\ensuremath{\operatorname{dom}} A)$ $\langle z, Ay\rangle$
is single-valued.
Now let $x,y$ be in $\ensuremath{\operatorname{dom}} A$, and let $\alpha,\beta$ be in $\ensuremath{\mathbb R}$.
If $(\alpha,\beta) = (0,0)$, then
$\langle z, A(\alpha x+\beta y)\rangle=\langle z, A0\rangle=0
=\alpha \langle z, Ax\rangle +\beta \langle z, Ay\rangle$.
And if $(\alpha,\beta)\neq (0,0)$, then
Fact~\ref{Rea:1}\ref{Th:30} yields
$\langle z, A(\alpha x+\beta y)\rangle =\langle z, \alpha Ax +\beta Ay\rangle
=\alpha \langle z, Ax\rangle +\beta\langle z, Ay\rangle$.
This verifies  linearity.

\ref{Th:031}: Apply
\ref{Nov:s1}\&\ref{Th:032}.

\ref{Th:031L}: ``$\Rightarrow$'': This  follows from
 the fact that $(0,0)\in\ensuremath{\operatorname{gra}} A$.
``$\Leftarrow$'': If $x$ and $y$ belong to $\ensuremath{\operatorname{dom}} A$, then
Fact~\ref{Rea:1}\ref{Th:30} yields
$\langle x-y,Ax-Ay\rangle=\langle x-y,A(x-y)\rangle\geq 0$.

\ref{Th:031+}: Let $(x,x^*)\in\ensuremath{\operatorname{dom}} A\times X^*$
be monotonically related to $\ensuremath{\operatorname{gra}} A$,
and take $x_0^*\in Ax.$
For every $(v,v^*)\in\ensuremath{\operatorname{gra}} A$, we have $x_0^*+
v^*\in A(x+v)$ (by Fact~\ref{Rea:1}\ref{Th:30}); hence,
$\langle x-(x+v), x^*-(x_0^*+ v^*)\rangle\geq 0$ and thus
$\langle v,  v^*\rangle\geq \langle v, x^*-x_0^*
\rangle$. Now take
$\lambda>0$  and replace $(v,v^*)$ in the last inequality
by $(\lambda v,\lambda v^*)$.
Then divide by $\lambda$ and let $\lambda\rightarrow 0^+$ to see that
$0\geq \sup\langle \ensuremath{\operatorname{dom}} A,\ x^*-x_0^*\rangle$.
Since $\ensuremath{\operatorname{dom}} A$ is linear, it follows that  $x^*-x_0^*\in
 (\ensuremath{\operatorname{dom}} A)^\bot$.
\end{proof}

We define the \emph{symmetric part} and the \emph{skew part} of $A$ via
\begin{equation}
\label{Fee:1}
A_+ := \ensuremath{\tfrac{1}{2}} A + \ensuremath{\tfrac{1}{2}} A^* \quad\text{and}\quad
A_{\mathlarger{\circ}} := \ensuremath{\tfrac{1}{2}} A - \ensuremath{\tfrac{1}{2}} A^*,
\end{equation}
respectively. It is easy to check that $A_+$ is symmetric and that $A_{\mathlarger{\circ}} $
is skew.

\begin{proposition}[Maximally monotone linear relations]\label{linear}
Let $A\colon X \ensuremath{\rightrightarrows} X^*$ be a monotone linear relation.
Then the following hold.
\begin{enumerate}
\item\label{sia:2v} If $A$ is maximally monotone,
 then $(\ensuremath{\operatorname{dom}} A)^{\perp}= A0$ and hence
$\overline{\ensuremath{\operatorname{dom}} A}=(A0)_\bot$.
\item \label{sia:3v}
If $\ensuremath{\operatorname{dom}} A$ is closed, then:
$A$ is maximally monotone $\Leftrightarrow$ $(\ensuremath{\operatorname{dom}} A)^\bot = A0$.
\item \label{sia:3iv}
If $A$ is maximally monotone, then
$\overline{\ensuremath{\operatorname{dom}} A^*}^{\ensuremath{\operatorname{w*}}}\cap X=\overline{\ensuremath{\operatorname{dom}} A}=(A0)_\bot$,
 and
$A0=A^*0=A_+0 = A_{\mathlarger{\circ}}0= (\ensuremath{\operatorname{dom}} A)^{\perp}$ is
(weak$^*$) closed.
\item \label{sia:3vi}
If $A$ is maximally monotone and $\ensuremath{\operatorname{dom}} A$ is closed,
then $\ensuremath{\operatorname{dom}} A^*\cap X=\ensuremath{\operatorname{dom}} A$.

\item \label{sia:3vii}
If $A$ is maximally monotone and $\ensuremath{\operatorname{dom}} A \subseteq \ensuremath{\operatorname{dom}} A^*$,
then
$A=A_{+}+A_{\mathlarger{\circ}}$, $A_{+}=A-A_{\mathlarger{\circ}}$,
 and $A_{\mathlarger{\circ}}=A-A_{+}$.

\item \label{dmns:1}
If $A$ is maximally monotone and $\ensuremath{\operatorname{dom}} A$ is closed,
then both $A_+$ and $A_{\mathlarger{\circ}}$ are maximally monotone.

\item \label{dmns:2}
If $A$ is maximally monotone and $\ensuremath{\operatorname{dom}} A$ is closed,
then $A^*=(A_+)^*+(A_{\mathlarger{\circ}})^*$.
\end{enumerate}
\end{proposition}

\begin{proof}
\ref{sia:2v}: Since
$A+N_{\ensuremath{\operatorname{dom}} A}=A+(\ensuremath{\operatorname{dom}} A)^{\perp}$
is a monotone extension of $A$ and $A$ is
maximally monotone, we must have
$A+(\ensuremath{\operatorname{dom}} A)^{\perp}=A$. Then $A0+(\ensuremath{\operatorname{dom}} A)^{\perp}=A0$. As $0\in A0$,
$(\ensuremath{\operatorname{dom}} A)^{\perp}\subseteq A0.$ Combining with
Proposition~\ref{P:a}\ref{Nov:s1},  we have
$(\ensuremath{\operatorname{dom}} A)^{\bot}=A0$. By Fact~\ref{rudin:1},
$\overline{\ensuremath{\operatorname{dom}} A}=(A0)_{\bot}$.

 \ref{sia:3v}:  ``$\Rightarrow$": Clear from \ref{sia:2v}.
 ``$\Leftarrow$":
 The assumptions and
 Fact~\ref{rudin:1} imply that  $\ensuremath{\operatorname{dom}} A=\overline{\ensuremath{\operatorname{dom}} A}=
\left[ (\ensuremath{\operatorname{dom}} A)^{\bot}\right]_{\bot}=(A0)_{\bot}$.
 Let $(x,x^*)$ be monotonically related to $\ensuremath{\operatorname{gra}} A$.
   We have $\inf\langle x-0, x^*-A0\rangle\geq 0$.
 Then we have $x\in (A0)_{\bot}$ and hence $x\in\ensuremath{\operatorname{dom}} A$.
  Then by Proposition~\ref{P:a}\ref{Th:031+}
 and Fact~\ref{Rea:1}\ref{Th:28}, $x^*\in Ax$. Hence $A$ is maximally monotone.

 \ref{sia:3iv}: By \ref{sia:2v} and  Fact~\ref{Rea:1}\ref{Sia:1},
 $A0=(\ensuremath{\operatorname{dom}} A)^{\perp}=A^*0$ is weak$^*$ closed and thus
 $A_+0 = A_{\mathlarger{\circ}}0=A0=(\ensuremath{\operatorname{dom}} A)^{\perp}$.
 Then by Fact~\ref{Rea:1}\ref{Sia:2} and \ref{sia:2v},
  $\overline{\ensuremath{\operatorname{dom}} A^*}^{\ensuremath{\operatorname{w*}}}\cap X=(A0)_{\bot}=\overline{\ensuremath{\operatorname{dom}} A}$.

 \ref{sia:3vi}: Combine  \ref{sia:3iv} with Fact~\ref{Rea:1}\ref{Th:32}.

 \ref{sia:3vii}:
  We show only the proof of $A=A_{+}+A_{\mathlarger{\circ}}$
as the other two proofs are analogous.
Clearly, $\ensuremath{\operatorname{dom}} A_{+}=\ensuremath{\operatorname{dom}} A_{\mathlarger{\circ}}
=\ensuremath{\operatorname{dom}} A\cap\ensuremath{\operatorname{dom}} A^*=\ensuremath{\operatorname{dom}} A$.
Let $x\in\ensuremath{\operatorname{dom}} A$, and $x^*\in Ax$ and $y^*\in A^*x$. We write
$x^*=\tfrac{x^*+y^*}{2}+\tfrac{x^*-y^*}{2}\in (A_{+}+A_{\mathlarger{\circ}})x$.
Then, by \ref{sia:3iv} and Fact~\ref{Rea:1}\ref{Th:28},
$Ax=x^*+A0=x^*+(A_{+}+A_{\mathlarger{\circ}})0=
(A_{+}+A_{\mathlarger{\circ}})x$.
Therefore, $A=A_{+}+A_{\mathlarger{\circ}}$.

\ref{dmns:1}:
By \ref{sia:3vi},
\begin{align}\ensuremath{\operatorname{dom}} A_+=\ensuremath{\operatorname{dom}}{A_\mathlarger{\circ}}=\ensuremath{\operatorname{dom}} A\ \text{ is closed}.\label{SCK:05}
\end{align}
Hence, by  \ref{sia:3iv},
\begin{align}
A_{\mathlarger{\circ}}0=A_+0=A0=(\ensuremath{\operatorname{dom}} A)^{\bot}=(\ensuremath{\operatorname{dom}} A_+)^{\bot}
= (\ensuremath{\operatorname{dom}} A_{\mathlarger{\circ}})^{\bot}.\label{SCK:M6}
\end{align}
Since $A$ is monotone,
so are  $A_+$ and $A_\mathlarger{\circ}$.
Combining \eqref{SCK:05}, \eqref{SCK:M6}, and  \ref{sia:3v},
we deduce that
$A_+$ and  $A_{\mathlarger{\circ}}$ are maximally monotone.

\ref{dmns:2}: By \ref{sia:3vi}\&\ref{sia:3vii},
\begin{align}
A=A_++ A_{\mathlarger{\circ}}.
\end{align}
Then  by \ref{dmns:1}, \ref{sia:3vi}, and Fact~\ref{LinAdSum},
$A^*=(A_+)^*+ (A_{\mathlarger{\circ}})^*$.
\end{proof}

For a monotone linear relation $A\colon X \ensuremath{\rightrightarrows} X^*$ it will be convenient
to define --- as in, e.g., \cite{BBW} --- a generalized quadratic form
\begin{equation*}
(\forall x\in X)\quad
q_A(x) =  \begin{cases} \tfrac{1}{2}\langle x,Ax\rangle,&\text{if $x\in \ensuremath{\operatorname{dom}} A$};\\
+\infty,&\text{otherwise}.\end{cases}\end{equation*}
We  write $\overline{q_A}$ for the lower semicontinuous hull of $q_A$.

\begin{proposition}\label{Nov:s2} Let $A\colon X \ensuremath{\rightrightarrows} X^*$ be a monotone linear
relation,
let $x$ and $y$ be in $\ensuremath{\operatorname{dom}} A$, and let $\lambda\in\ensuremath{\mathbb R}$.
Then
$q_A$ is single-valued, $q_A\geq0$ and
\begin{align}
&\lambda q_A(x) + (1-\lambda)q_A(y) - q_{A}(\lambda x + (1-\lambda)y) =
\lambda(1-\lambda)q_A(x-y)\nonumber\\& =
\tfrac{1}{2}\lambda(1-\lambda)\scal{x-y}{Ax-Ay}.\label{deli:1}
\end{align}
Consequently, $q_A$ is convex.
\end{proposition}
\begin{proof}
Proposition~\ref{P:a}\ref{Th:031}\&\ref{Th:031L} show
that $q_A$ is single-valued and that $q_A\geq0$.
Combining with Proposition~\ref{P:a}\ref{Nov:s1}\&\ref{Sia:2c}, we obtain
\eqref{deli:1}.
Therefore, $q_A$ is convex.
\end{proof}

As in the classical case, $q_A$ allows us to connect properties of $A_+$ to those of $A$ and $A^*$.

\begin{proposition} \label{f:PheSim} Let  $A:X\rightrightarrows X^*$
 be a monotone  linear relation.
Then the following hold.
\begin{enumerate}
\item \label{f:PheSim:lsc02}  $\overline{q_A}+\iota_{\ensuremath{\operatorname{dom}} A_+}=q_{A_+}$ and thus $q_{A_+}$ is convex.
\item \label{f:PheSim:quad}
$\ensuremath{\operatorname{gra}} A_+ \subseteq\ensuremath{\operatorname{gra}} \partial \overline{q_A}$.
If $A_+$ is maximally monotone, then $A_+=\partial \overline{q_A}$.

\item \label{dms:quad}
If $A$ is maximally monotone and $\ensuremath{\operatorname{dom}} A$ is closed,
then $A_+=\partial \overline{q_A}$.
\item \label{f:PheSim:brah:1}If $A$ is maximally monotone,
then $A^*|_X$ is monotone.
\item \label{f:PheSim:brah:2}If $A$ is maximally
monotone and $\ensuremath{\operatorname{dom}} A$ is closed, then $A^*|_X$ is maximally monotone.
\end{enumerate}
\end{proposition}
\begin{proof}
Let $x\in\ensuremath{\operatorname{dom}} A_+$.

\ref{f:PheSim:lsc02}: By Fact~\ref{Rea:1}\ref{Sia:2b} and Proposition~\ref{P:a}\ref{Th:031},
 $q_{A_+}=q_{A}|_{\ensuremath{\operatorname{dom}} A_+}$.
 Then by Proposition~\ref{Nov:s2},
  $q_{A_+}$ is convex. Let $y\in\ensuremath{\operatorname{dom}} A$. Then by
  Fact~\ref{Rea:1}\ref{Sia:2b},
\begin{align}0\leq\tfrac{1}{2}\langle Ax-Ay,x-y\rangle=
\tfrac{1}{2}\langle Ay,y\rangle+\tfrac{1}{2}\langle Ax,x\rangle-\langle A_+x, y\rangle,
\label{Bor:u2}\end{align}
we have $q_A(y)\geq\langle A_+x, y\rangle-q_A(x)$.
 Take the lower semicontinuous hull of $q_A$ at $y$ to
 deduce that $\overline{q_A}(y)\geq\langle A_+x, y\rangle-q_A(x)$.
  For $y=x$, we have $\overline{q_A}(x)\geq q_A(x)$.
 On the other hand,
  $\overline{q_A}\leq q_A$.
Altogether, $\overline{q_A}(x)=q_A(x)=q_{A_+}(x)$.
Thus \ref{f:PheSim:lsc02} holds.

\ref{f:PheSim:quad}:
Let $y\in\ensuremath{\operatorname{dom}} A$. By \eqref{Bor:u2} and \ref{f:PheSim:lsc02},
\begin{align}q_A(y)\geq
q_A(x)+\langle A_+x,y-x\rangle=\overline{q_A}(x)
+\langle A_+x,y-x\rangle.\label{Bor:u3}
\end{align}
Since $\ensuremath{\operatorname{dom}}\overline{q_A}\subseteq\overline{\ensuremath{\operatorname{dom}} q_A}
=\overline{\ensuremath{\operatorname{dom}} A}$, by \eqref{Bor:u3},
 $\overline{q_A}(z)\geq
\overline{q_A}(x)+\langle A_+x,z-x\rangle,\quad \forall z\in\ensuremath{\operatorname{dom}}\overline{q_A}.$
Hence $A_+x\subseteq\partial \overline{q_A}(x)$.
If $A_+$ is maximally monotone, then  $A_+=\partial \overline{q_A}$.
Thus \ref{f:PheSim:quad} holds.

\ref{dms:quad}: Combine Proposition~\ref{linear}\ref{dmns:1}
with \ref{f:PheSim:quad}.

\ref{f:PheSim:brah:1}:
Suppose to the contrary that $A^*|_X$ is not monotone.
By Proposition~\ref{P:a}\ref{Th:031L},
there exists $(x_0,x_0^*)\in\ensuremath{\operatorname{gra}} A^*$ with $x_0\in X$ such that
 $\langle x_0,x_0^*\rangle<0$.
Now we have
\begin{align}&\langle-x_0-y,x^*_0-y^*\rangle
=-\langle x_0,x^*_0\rangle+\langle y,y^*\rangle+\langle x_0,y^*\rangle
-\langle y,x_0^*\rangle\nonumber\\
&=-\langle x_0,x^*_0\rangle+\langle y,y^*\rangle>0,
\quad \forall (y,y^*)\in\ensuremath{\operatorname{gra}} A.\label{Brezi:1}\end{align}
Thus, $(-x_0,x^*_0)$ is monotonically related to $\ensuremath{\operatorname{gra}} A$.
 By maximal monotonicity of $A$, $(-x_0,x^*_0)\in\ensuremath{\operatorname{gra}} A$.
Then $\langle-x_0-(-x_0),x^*_0-x_0^*\rangle=0$,
 which contradicts \eqref{Brezi:1}. Hence $A^*|_X$ is monotone.

\ref{f:PheSim:brah:2}: By Fact~\ref{Rea:1}\ref{Th:32},
$\ensuremath{\operatorname{dom}} A^*|_X=(A0)_{\bot}$ and thus $\ensuremath{\operatorname{dom}} A^*|_X$ is closed.
By Fact~\ref{rudin:1} and Proposition~\ref{linear}\ref{sia:2v},
$(\ensuremath{\operatorname{dom}} A^*|_X)^{\bot}=((A0)_{\bot})^{\bot}
=\overline{A0}^{\ensuremath{\operatorname{w*}}}=A0$.
Then by Proposition~\ref{linear}\ref{sia:3iv},
$(\ensuremath{\operatorname{dom}} A^*|_X)^{\bot}=A^*0=A^*|_X 0$.
Applying
\ref{f:PheSim:brah:1} and Proposition~\ref{linear}\ref{sia:3v},
we see that
 $A^*|_X$ is maximally monotone.
\end{proof}

The proof of Proposition~\ref{f:PheSim}\ref{f:PheSim:brah:1}
was borrowed from \cite[Theorem~2]{Brezis-Browder}.
Results very similar to
Proposition~\ref{f:PheSim}\ref{f:PheSim:lsc02}\&\ref{f:PheSim:quad}
are verified in \cite[Proposition~18.9]{Yao}.
The proof of the next Theorem~\ref{TSkeD:1}\ref{MaTC:1}$\Rightarrow$\ref{MaTC:02}
was partially inspired by that of \cite[Theorem~4.1(v)$\Rightarrow$(vi)]{BB}.
When the domain of $A$ is closed we can obtain additional information
 about the skew part of $A$.

\begin{theorem}[Monotone relations with closed graph and domain]\label{TSkeD:1}
Let $A:X\rightrightarrows X^*$ be a monotone linear relation
 such that $\ensuremath{\operatorname{gra}} A$ is closed and $\ensuremath{\operatorname{dom}} A$ is closed.
 Then the following are equivalent.\begin{enumerate}
\item\label{MaTC:1} $A$ is maximally monotone of type (D).

\item \label{MaTC:02} $A_{\mathlarger{\circ}}$
is maximally monotone  of type (D)  with respect to $X\times X^*$ and $A^*0=A0$.
\item \label{MaTC:2} $(A_{\mathlarger{\circ}})^*$
is maximally monotone  with respect to $X^{**}\times X^*$ and $A^*0=A0$.
\item \label{MaTC:3} $(A_{\mathlarger{\circ}})^*$ is  monotone and $A^*0=A0$.
\item \label{MaTC:4} $A^*$ is  monotone.
\item \label{MaTC:5} $A^*$ is  maximally monotone with respect to $X^{**}\times X^*$.
\end{enumerate}
\end{theorem}
\begin{proof}
``\ref{MaTC:1}$\Rightarrow$\ref{MaTC:02}": By Fact~\ref{TypeD:1},
\begin{align}
A^*\ \text{is monotone}.\label{SCK:1}
\end{align}
By Proposition~\ref{f:PheSim}\ref{dms:quad} and Fact~\ref{ITPR},
 \begin{align}
A_+\
\text{ is maximally monotone of type (D)}.\label{SCK:04}
\end{align}
By Fact~\ref{TypeD:1},
\begin{align}
(A_+)^* \ \text{is monotone}.\label{SCK:5a}
\end{align}
Now we show that
\begin{align}
(A_{\mathlarger{\circ}})^* \ \text{is monotone}.\label{SCK:5}
\end{align}
Proposition~\ref{linear}\ref{dmns:2} implies
\begin{align}
A^*=(A_+)^*+(A_{\mathlarger{\circ}})^*.\label{SCK:3}
\end{align}
Since $A$ is maximally monotone and $\ensuremath{\operatorname{dom}} A$ is closed,
Proposition~\ref{linear}\ref{dmns:1} implies that
$A_{\mathlarger{\circ}}$ is maximally monotone.
Hence $\ensuremath{\operatorname{gra}} (A_{\mathlarger{\circ}})$ is closed.
On the other hand,
again since $A$ is maximally monotone and $\ensuremath{\operatorname{dom}} A$ is closed,
Proposition~\ref{linear}\ref{sia:3vi} yields
$\ensuremath{\operatorname{dom}} (A_{\mathlarger{\circ}}) = \ensuremath{\operatorname{dom}} A$ is closed.
Altogether, and combining with Fact~\ref{Rea:1}\ref{Th:32}
applied to $A_{\mathlarger{\circ}}$, we obtain
$\ensuremath{\operatorname{dom}} (A_{\mathlarger{\circ}})^* = (A_{\mathlarger{\circ}}0)^{\bot}$.
Furthermore, since $A0 = A_{\mathlarger{\circ}}0$ by
Proposition~\ref{linear}\ref{sia:3iv}, we have
$(A0)^\bot = (A_{\mathlarger{\circ}}0)^\bot$.
Moreover, applying Fact~\ref{Rea:1}\ref{Th:32}
to $A$, we deduce that $\ensuremath{\operatorname{dom}} A^* = (A0)^\bot$.
Therefore,
 \begin{align}\ensuremath{\operatorname{dom}} (A_{\mathlarger{\circ}})^*
 = (A_{\mathlarger{\circ}}0)^{\bot}=(A0)^{\bot}=\ensuremath{\operatorname{dom}} A^*.
 \label{SCK:6}
 \end{align}
 Similarly, we have
  \begin{align}\ensuremath{\operatorname{dom}} (A_+)^*=\ensuremath{\operatorname{dom}} A^*.
 \label{SCK:7}
 \end{align}
Take $(x^{**},x^*)\in\ensuremath{\operatorname{gra}} (A_{\mathlarger{\circ}})^*$.
By \eqref{SCK:3} and \eqref{SCK:6},
there exist $a^*, b^*\in X^*$ such that \begin{align}
(x^{**},a^*)\in \ensuremath{\operatorname{gra}} A^*,
 (x^{**}, b^*)\in \ensuremath{\operatorname{gra}} (A_+)^*\label{SCK:08}\end{align} and
\begin{align}
a^*=b^*+x^*.\label{SCKL:08}
\end{align}
Since $A_+$ is symmetric, $\ensuremath{\operatorname{gra}} A_+\subseteq\ensuremath{\operatorname{gra}} (A_+)^*$.
Thus, by \eqref{SCK:5a},
$(x^{**}, b^*)$ is monotonically related to $\ensuremath{\operatorname{gra}} A_+$.
 By \eqref{SCK:04}, there exist
a  bounded net
$(a_{\alpha}, b^*_{\alpha})_{\alpha\in\Gamma}$ in $\ensuremath{\operatorname{gra}} A_+$
such that
$(a_{\alpha}, b^*_{\alpha})_{\alpha\in\Gamma}$
weak*$\times$strong converges to
$(x^{**},b^*)$.
Thus $(a_{\alpha}, b^*_{\alpha})\in\ensuremath{\operatorname{gra}} (A_+)^*$.
By \eqref{SCK:7} and \eqref{SCK:3},
there exist  $a^*_{\alpha}\in A^* a_{\alpha},
 c^*_{\alpha}\in (A_{\mathlarger{\circ}})^*a_{\alpha}$ such that
\begin{align}
a^*_{\alpha}=b^*_{\alpha}+c^*_{\alpha},\quad\forall \alpha\in\Gamma.
 \label{SCK:8}
\end{align}
Thus by Fact~\ref{Rea:1}\ref{Sia:2b},
\begin{align}
\langle a_{\alpha}, c^*_{\alpha}\rangle
=\langle A_{\mathlarger{\circ}}a_{\alpha}, a_{\alpha}\rangle=0,
\quad  \forall \alpha\in\Gamma.
\label{SCK:9}
\end{align}
Hence for every $\alpha\in\Gamma$,
$(-a_{\alpha}, c^*_{\alpha}) $ is monotonically
 related to $\ensuremath{\operatorname{gra}} A_{\mathlarger{\circ}}$.
By Proposition~\ref{linear}\ref{dmns:1},
\begin{align}
(-a_{\alpha}, c^*_{\alpha})\in\ensuremath{\operatorname{gra}} A_{\mathlarger{\circ}},\quad
\forall \alpha\in\Gamma.\label{SCK:9a}
\end{align}
By \eqref{SCK:1} and \eqref{SCK:08}, we have
\begin{align}
0&\leq\langle x^{**}-a_{\alpha}, a^*-a^*_{\alpha}\rangle
=\langle x^{**}-a_{\alpha}, a^*-b^*_{\alpha}-c^*_{\alpha}\rangle
\quad\text{(by \eqref{SCK:8})} \nonumber\\
&=\langle x^{**}-a_{\alpha}, a^*-b^*_{\alpha}\rangle-\langle x^{**}, c^*_{\alpha}\rangle
+\langle a_{\alpha}, c^*_{\alpha}\rangle\nonumber\\
&=\langle x^{**}-a_{\alpha}, a^*-b^*_{\alpha}\rangle-\langle x^{**}, c^*_{\alpha}\rangle
\quad\text{(by \eqref{SCK:9})}\nonumber\\
&=\langle x^{**}-a_{\alpha}, a^*-b^*_{\alpha}\rangle+\langle x^{*}, a_{\alpha}\rangle
\quad \text{(by \eqref{SCK:9a} and $(x^{**},x^*)\in\ensuremath{\operatorname{gra}} (A_{\mathlarger{\circ}})^*$)}.
\label{SCK:10}
\end{align}
Taking  the limit in \eqref{SCK:10} along with
 $ a_{\alpha} \weakstarly x^{**}$ and $b^*_{\alpha}
\rightarrow b^*$, we have
\begin{align*}
\langle x^{**},x^*\rangle\geq0.
\end{align*}
Hence $(A_{\mathlarger{\circ}})^*$ is monotone
and thus \eqref{SCK:5} holds.  Combining \eqref{SCK:5}, Proposition~\ref{linear}\ref{dmns:1}
and Fact~\ref{TypeD:1}, we see that
$A_{\mathlarger{\circ}}$ is of type (D).

``\ref{MaTC:02}$\Rightarrow$\ref{MaTC:2}$\Rightarrow$\ref{MaTC:3}":
 Apply Corollary~\ref{CoSBr:1} to
$A_{\mathlarger{\circ}}$.

``\ref{MaTC:3}$\Rightarrow$\ref{MaTC:4}":
By Fact~\ref{Rea:1}\ref{Sia:1} and Proposition~\ref{linear}\ref{sia:3v},
$A$ is maximally monotone.
Then by Proposition~\ref{linear}\ref{dmns:2}
and Proposition~\ref{f:PheSim}\ref{dms:quad}, we have
\begin{align}
A^*=(A_+)^*+(A_{\mathlarger{\circ}})^*\ \text{and}
 \ A_+=\partial \overline{q_A}.\label{SCK:11}
\end{align}
Then $A_+$ is of type (D) by Fact~\ref{ITPR}, and hence $(A_+)^*$
 is monotone by Fact~\ref{TypeD:1}.
Thus, by the assumption and \eqref{SCK:11}, we have $A^*$ is monotone.

``\ref{MaTC:4}$\Rightarrow$\ref{MaTC:5}":
By Proposition~\ref{PrTD:1},
$A$ is maximally monotone. Then by Fact~\ref{Rea:1}\ref{Th:32}
and Proposition~\ref{linear}\ref{sia:3iv},
\begin{align}
\ensuremath{\operatorname{dom}} A^*=(A^*0)^{\bot}.\label{MAJo:1}
\end{align}
Then by Fact~\ref{rudin:1} and Fact~\ref{Rea:1}\ref{Sia:1},
\begin{align}
\left[\ensuremath{\operatorname{dom}} A^*\right]_{\bot}=A^*0.\label{MAJo:2}
\end{align}
Let $(x^{**},x^*)\in X^{**}\times X^*$
be monotonically related to $\ensuremath{\operatorname{gra}} A^*$.
Because $\{0\}\times A^*0\subseteq\ensuremath{\operatorname{gra}} A^*$,
we have $\inf\langle x^{**}, x^*-A^*0\rangle\geq 0$.
 Since $A^*0$ is a subspace, $x^{**}\in (A^*0)^{\bot}$.
Then by \eqref{MAJo:1},
\begin{align}x^{**}\in\ensuremath{\operatorname{dom}} A^*.\end{align}
Take $(x^{**},x_0^{**})\in \ensuremath{\operatorname{gra}} A^{*}$ and
$\lambda>0$.
For every $(a^{**},a^*)\in\ensuremath{\operatorname{gra}} A^{*}$, we have $(\lambda a^{**}, \lambda a^*)\in\ensuremath{\operatorname{gra}} A^{*}$ and
hence
 $(x^{**}+\lambda a^{**},x_0^*+
\lambda a^*)\in \ensuremath{\operatorname{gra}} A^{*}$ (since $\ensuremath{\operatorname{gra}} A^{*}$ is a subspace).
Thus
\begin{align*}\lambda\langle a^{**}, x_0^*+\lambda a^*-x^*\rangle=
\langle x^{**}+\lambda a^{**}-x^{**}, x_0^*+\lambda a^*-x^*\rangle\geq 0.\end{align*}
Now divide by $\lambda$ to obtain
$\lambda\langle a^{**},  a^*\rangle\geq \langle a^{**}, x^*-x_0^*
\rangle$.
Then  let $\lambda\rightarrow 0^+$ to see that
$0\geq \sup\langle \ensuremath{\operatorname{dom}} A^*,\ x^*-x_0^*\rangle$.  Thus, $x^*-x_0^*\in (\ensuremath{\operatorname{dom}} A^*)_{\bot}$.
By \eqref{MAJo:2}, $x^*\in x^*_0+A^*0\subseteq A^*x^{**}+A^*0$. Then there exists $(0,z^*)\in \ensuremath{\operatorname{gra}} A^*$ such that $(x^{**}, x^*-z^*)\in\ensuremath{\operatorname{gra}} A^*$. Since $\ensuremath{\operatorname{gra}} A^*$ is s a subspace,
 $(x^{**}, x^*)=(0,z^*)+(x^{**}, x^*-z^*)\in\ensuremath{\operatorname{gra}} A^*$.
Hence
$A^*$ is maximally monotone with respect to $X^{**}\times X^*$.

``\ref{MaTC:5}$\Rightarrow$\ref{MaTC:1}":
  Apply Proposition~\ref{PrTD:1} directly.
\end{proof}

The next three examples show the need for various of our auxiliary hypotheses.

\begin{example}
We cannot remove the condition that $A^*0=A0$ in
 Theorem~\ref{TSkeD:1}\ref{MaTC:3}. For example,
  suppose that $X=\ensuremath{\mathbb R}^2$  and set $e_1=(1,0), e_2=(0,1)$. We define
  $A:X\rightrightarrows X$ by
\begin{equation*}
\ensuremath{\operatorname{gra}} A=\operatorname{span}\{e_1\}\times\{0\}\;\text{ so that }\;
\ensuremath{\operatorname{gra}} A^*=X\times \operatorname{span}\{e_2\}.
\end{equation*}
Then $A$ is monotone, $\ensuremath{\operatorname{dom}} A$ is closed, and $\ensuremath{\operatorname{gra}} A$ is closed.
Thus
\begin{align}
\ensuremath{\operatorname{gra}} A_{\mathlarger{\circ}}=\operatorname{span}\{e_1\}\times\operatorname{span}\{e_2\}
\end{align}
and so
\begin{align*}\ensuremath{\operatorname{gra}} (A_{\mathlarger{\circ}})^*=\operatorname{span}\{e_2\}\times\operatorname{span}\{e_1\}.
\end{align*} Hence  $(A_{\mathlarger{\circ}})^*$ is monotone,
but $A$ is not maximally monotone because $\ensuremath{\operatorname{gra}} A \subsetneqq \ensuremath{\operatorname{gra}} N_X$.
\ensuremath{\quad \hfill \blacksquare}
\end{example}

\begin{example}\label{CDEx:1}
We cannot replace that ``$\ensuremath{\operatorname{dom}} A$ is closed"
 by  that ``$ \ensuremath{\operatorname{dom}} A$ is dense" in the statement of
 Theorem~\ref{TSkeD:1}.
 For example, let $X, A$ be defined as in Example~\ref{SRE:1}
 and consider the operator $A^*$.
Example~\ref{SRE:1}\ref{NEC:3}\&\ref{NEC:2} state that  $A^*$ is maximally
monotone with dense domain; hence, $\ensuremath{\operatorname{gra}} A^*$ is closed.
 Moreover, by Example~\ref{SRE:1}\ref{NEC:1},
 \begin{align}
 (A^*)_{\mathlarger{\circ}}=-A.
 \end{align}
Hence
 \begin{align}
 \left[(A^*)_{\mathlarger{\circ}}\right]^*=-A^*.
 \end{align}
Thus $\left[(A^*)_{\mathlarger{\circ}}\right]^*$ is
 not  monotone by Example~\ref{SRE:1}\ref{NEC:3};
 even though $A^*$ is a classically maximally monotone and densely defined linear operator. \ensuremath{\quad \hfill \blacksquare}
\end{example}

\begin{example}
We cannot remove the condition that
$(A_{\mathlarger{\circ}})^*$ is monotone
in Theorem~\ref{TSkeD:1}\ref{MaTC:3}. For example,
consider the Gossez operator $A$ (see \cite{Gossez72} and \cite{BB}).
It satisfies
$X=\ell^1$, $\ensuremath{\operatorname{dom}} A=X$,
$A_{\mathlarger{\circ}} = A$,
$A0=\{0\}=A^*0$, yet $A^*$ is not monotone.\ensuremath{\quad \hfill \blacksquare}
\end{example}

\begin{remark}
Let $A\colon X\rightrightarrows X^*$
 be a maximally monotone linear relation.
 \begin{enumerate}
 \item
 In general, $(A^*)_{\mathlarger{\circ}}\neq (A_{\mathlarger{\circ}})^*$.
 To see that, let $X, A$ be  as in Example~\ref{SRE:1} again.
  By Example~\ref{SRE:1}\ref{NEC:1},
 we have
\begin{align*}(A^*)_{\mathlarger{\circ}}=-A\;\text{and}\;
 (A_{\mathlarger{\circ}})^*= A^*.
\end{align*}
Hence $(A^*)_{\mathlarger{\circ}}\neq (A_{\mathlarger{\circ}})^*$ by Example~\ref{SRE:1}\ref{NEC:2}.

\item
However, if $X$ is finite-dimensional, we do have $(A^*)_{\mathlarger{\circ}}= (A_{\mathlarger{\circ}})^*$.
Indeed, by Fact~\ref{LinAdSum},
\begin{align*}
(A_{\mathlarger{\circ}})^*=\left(\frac{A-A^{*}}{2}\right)^*
=\frac{A^{*}-A^{**}}{2}=(A^*)_{\mathlarger{\circ}}.
\end{align*}
We expect that  $(A^*)_{\mathlarger{\circ}}= (A_{\mathlarger{\circ}})^*$ for all maximally  monotone linear relations if and only if $X$ is
finite-dimensional.
\end{enumerate}
\end{remark}

We are now able to present our main result relating  monotonicity and adjoint properties of $A$ and those of its skew part
$A_{\mathlarger{\circ}}$.

\begin{theorem} [Adjoint characterizations of type (D)] \label{BrBrTDS:1}
Let $A\colon X\rightrightarrows X^*$
 be a monotone linear relation such that $\ensuremath{\operatorname{gra}} A$ is closed and
$\ensuremath{\operatorname{dom}} A$ is closed.
Then the following are equivalent.
\begin{enumerate}

\item $A$ is maximally monotone of type (D).

\item $A$ is  maximally monotone  of type (NI).

\item $A$ is  maximally monotone  of type (FP).
\item  $A^*$ is monotone.
\item  $A^*$  is maximally monotone with respect to $X^{**}\times X^*$.
\item $A_{\mathlarger{\circ}}$
is maximally monotone  of type (D) and $A^*0=A0$.
\item $(A_{\mathlarger{\circ}})^*$
is maximally monotone with respect to $X^{**}\times X^*$ and $A^*0=A0$.
\item $(A_{\mathlarger{\circ}})^*$ is  monotone and $A^*0=A0$.

\end{enumerate}
\end{theorem}

\begin{proof}
Apply Theorem~\ref{TSkeD:1} and Theorem~\ref{BrBrTD:1}.
\end{proof}

The work in \cite{BBWY3} suggests that in every nonreflexive Banach space there is a maximally monotone linear relation which is not of type (D).

When  $A$   is linear and continuous, Theorem~\ref{BrBrTDS:1}
can also be deduced from \cite[Theorem~4.1]{BB}.
When $X$ is reflexive and $\ensuremath{\operatorname{dom}} A$ is closed,
Theorem~\ref{BrBrTDS:1} turns into the following
refined version of Fact~\ref{Sv:7}:

\begin{corollary} \label{BrBrTDS:2}
Suppose that $X$ is reflexive and let
$A\colon X\rightrightarrows X^*$
 be a monotone linear relation.
 such that $\ensuremath{\operatorname{gra}} A$ is closed and $\ensuremath{\operatorname{dom}} A$ is closed.
Then the following are equivalent.
\begin{enumerate}
\item
$A$ is  maximally monotone.
\item
$A^*$ is monotone.
\item
$A^*$ is maximally monotone.
\item
$A0=A^*0$.
\end{enumerate}
\end{corollary}
\begin{proof}
``(i)$\Leftrightarrow$(ii)$\Leftrightarrow$(iii)$\Rightarrow$(iv)'':
This follows from
Theorem~\ref{BrBrTDS:1} and
Fact~\ref{ITPR}(ii).

``(iv)$\Rightarrow$(i)'':
Fact~\ref{Rea:1}\ref{Sia:1} implies that
$(\ensuremath{\operatorname{dom}} A)^\perp = A^*0 = A0$.
By Proposition~\ref{linear}\ref{sia:3v}, $A$ is maximally monotone.
\end{proof}

When $X$ is finite-dimensional, the closure assumptions in
the previous result are automatically satisfied
and we thus obtain the following:

\begin{corollary}
Suppose that $X$ is finite-dimensional.
Let $A\colon X\rightrightarrows X^*$
 be a monotone linear relation.
Then the following are equivalent.
\begin{enumerate}
\item
$A$ is  maximally monotone.
\item
$A^*$ is monotone.
\item
$A^*$ is maximally monotone.
\item
$A0=A^*0$.
\end{enumerate}
\end{corollary}

\section*{Acknowledgments}
Heinz Bauschke was partially supported by the Natural Sciences and
Engineering Research Council of Canada and
by the Canada Research Chair Program.
Jonathan  Borwein was partially supported by the Australian Research  Council.
Xianfu Wang was partially supported by the Natural
Sciences and Engineering Research Council of Canada.

redpajama_set_name: RedPajamaArXiv


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Men at Otero Raise Money for Victims of the El Paso Shooting
The weather was dreary and wet, but their spirits were high.
"It sounded like a really good way for us to do something to give back to the community. I think there can often be some misconceptions about us as inmates: that we don't really care; that we don't have compassion."
More than 100 inmates at MTC's Otero County Prison Facility participated in a walk/run to raise money for victims of the deadly August shooting in nearby El Paso, Texas.
"I decided it was time to try and help out people outside of our little community here."
Justin Trout was touched when he heard about the tragedy.
"So, I sat down and thought, 'what better way to do it than to come out and run and walk.' I contacted our rec specialist, and she helped put it together."
"I knew that was coming from their hearts, you know? And I thought that was a great idea: giving something back to the community.
Officer Castile opened some doors but says it was really the inmates who made it happen.
"They just ran with it. They did the stop watches, they did the timing. All I did was stand around and oversee them."
"I think we're fortunate to have the capability to do this here. We're really fortunate to have the people we have that supported this all the way to the top."
"I hope that the money goes to a good cause and helps some people who are really struggling, going through some grieving right now. And I just want to let them know that we do care. And we hope that there's some healing and some mending."
About Otero County Prison Facility
Otero County Prison Facility is accredited by the American Correctional Association—
achieving 100% compliance for both mandatory and non-mandatory standards in
its most recent audit. Other accreditations and certifications include the Correctional
Education Association and the Prison Rape Elimination Act (PREA).
Learn More See Jobs in Corrections
See How MTC Supports the Small City of Diboll, Texas
Vocational Training, Certifications, Caring Staff and Much More at the Cleveland Correctional Center
County Judge, Warden, Instructors, and Inmates All Agree—Change Can Happen in Prison
NEWS: Otero County inmates raise money for El Paso shooting victims
U.S. District Court Judge Recent Ruling Confirms MTC's Commitment to Providing Quality Prison Services
NEWS: Fathers Receive Early Christmas Present
MTC Rolls Out Staff Culture of Caring Initiative
A Committed Staff and Exceptional Programming at the Dalby Facility Result in Success
Men at MTC's Taft Facility Are Having a Big Impact on People Throughout the World
He Admits He Wasted Years of Life, But Now He's Turning the Corner

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Android : References to a Context and memory leaks I've read that it is a mistake and a source of memory leaks in Android application to keep a long-lived references to a Context.
But I don't understand if it is ok to create a class that looks like this one:
public class HelperClass {
    private Context context;

    public HelperClass(Context context) {
        this.context = context;
    }
    public void myHelperMethod() {
        // uses this.context
    }
}

And call it from an Activity:
public class MyActivity extends Activity {
    public void onCreate(Bundle savedInstanceState) {
        HelperClass h = new HelperClass(this);
        h.myHelperMethod();
    }

    ...
}

A: The scope of the HelperClass is only within your onCreate function, so once onCreate executed, your "h" object is no longer needed and subject to garbage collection.
It would be a different story if "h" was a static member - THAT would be a great way to leak memory.

A: This is fine, and will not cause a memory leak.
As soon as onCreate finishes executing, h will be out of scope and become eligible for garbage collection.  If h was static, then you would run into problems.  Only when the reference to the context outlives the lifecycle of the context itself will a memory leak occur.  A few helpful hints:

*

*Use Context.getApplicationContext() when possible.  This context will live as long as your application is alive.

*Be careful when using static fields and inner classes.

*Run your application through a profiler to check for leaks.

redpajama_set_name: RedPajamaStackExchange


text:
Book Review: 'Clark' by Clark Terry
John Scott G Clark Terry, Count Basie, Dizzy Gillespie, Duke Ellington, John Scott G, Miles Davis, Quincy Jones, University of California Press Nov 12, 2011
MuseWire COLUMN: Storytellers are to be cherished and Clark Terry should be on a pedestal for his thoroughly entertaining autobiography. Brimming with life, love, music, and great characters, this book is as much a history of the twentieth century as it is a history of his ninety years (and counting!).
The opening line of Clark Terry's book tells you a few things about the man: "I made my first trumpet with scraps from the junkyard." This brilliantly sets the stage for a story of a man with an indomitable will and an intense dedication to music. It also shows a guy who is just the right amount of crazy.
"Clark" (written with his wife, Gwen Terry) is a book that is chock full of sights, sounds, smells and situations involving big characters and big emotions, all played out against nine decades of history. There are pimps, whores, con artists, racists, and a full complement of good people and bad.
Jazz giants like Duke Ellington and Dizzy Gillespie are a part of his tale and we get to eavesdrop on some of the conversations. His meet-ups with Miles Davis, scattered through his life, are interesting or acerbic, or both. Through it all, his lifelong friendships are touching and his dedication to jazz (and the teaching of it) is inspirational.
Okay, you were warned about what's coming up. There are exciting interactions with the fringes of society and some of the power brokers with whom he had to deal, including the Purple Gang, one of whom would stroll through the club in a "sharp suit, escorted by his bodyguards." This man's "signature style was a white cap and a real pigeon roosting on his shoulder." And there are passages like this:
…one of the acts was a short little piano player named Peanuts. He rolled the piano from table to table while a sexy girl named Mary Calendar sang and did risque dancing. It was customary for her to do little private dances at a particular table. After her dance, she'd pick up any coins given to her with her "other set of lips."
What happens next is even more outrageous but I'll let you discover all the wild scenes for yourself.
Terry may not be a great writer in the classic sense of the term, but he is a great storyteller in any sense of the term. And his life is one big, bountiful, beautiful story. While his style gets to the nitty-gritty in scene after scene, it also has some of the positives and negatives of verbal history. Like the turnaround. Here is the conclusion of a section dealing with his having to pass a swimming test while he served in the Navy during World War II:
I got out and walked away hoping that they had accepted what I'd done. When the instructor blew the whistle, I glanced back and saw the next guy jump in. No one called me back, so I walked away calmly, while my insides were jumping for joy.
A few weeks later I got an unwelcome letter from home.
That type of thing happens a lot and it rather charmingly propels you forward.
On His Own at 12
Kicked out of the house before his thirteenth birthday, Terry learned how to make his way through life the hard way. There was scrapin' and scrappin' galore as he struggled to survive while always returning to music, whether playing with others or spending hours and hours practicing. Until there were sores. Until there was blood.
To enter this world is to be swept away with him, as in his first trip into New York:
Billy met me at Grand Central Station. I was overwhelmed by all the people. All colors, foreign accents, moving, pushing, talking. Billy was grinning like crazy, telling me all about the things he had planned for us. I didn't hear much of what he was saying while we were walking to the subway, 'cause I was totally wowed! Tall buildings, a sea of yellow taxi cabs, restaurants, stores, and honking horns.
Billy said, "Don't look up. They'll know you're a mark."
While the book is a story of a man's journey through life, love, and music, it is also about racism. The tales of indignities and violence are numerous and incendiary. Some of them resulted in flight; some in calls for first aid; some were deadly.
On the other hand, the colors of people are noted frequently by Terry, almost always in a descriptive and friendly manner. This guy or that gal might be the "color of nutmeg," "Dark-skinned," "Medium-brown skinned," "Olive complexion," "caramel-skinned," "light complexion," "color of bittersweet chocolate," and so on.
At one point, he talks about joining the colored musicians' union: "The whites had theirs and we had ours, but we all read the same black notes on white manuscript." It's hard to top a sentence like that.
While his love for playing knew no boundaries, there were some performing situations that had a bit more tension than others:
All the cats from St. Louis carried a shank — in common language, a knife. So did I. When I'd come to work, before I took my music out of the book, I'd push the switchblade button. Bling! The sharp metal would flash in the lights. Then I'd throw the knife so that the point of the blade would stick in the fiberglass music stand. Boing! …"Good evening, everybody."
I have nearly a page of notes leading to cool quotes in the book. Like when Terry mentions that "Jazz saved my sanity. Which made me love it even more." Or when he is talking about a trumpet player who was waiting tables "because it paid more money than gigging." Some things never change.
Throughout the book, the teaching of jazz becomes an all-powerful force after working with his first student, "a skinny young kid" trumpet player of about twelve years old named Quincy Jones. For Terry, passing along the insights and traditions of jazz became almost an obsession:
It was the greatest feeling in the world to think that I had something to do with the careers of my students. The greatest feeling. I loved the awards that came from here and there, but the thing I loved most was to see those young people grow. Watch their dreams come true.
Jazz is a constant throughout the journey, but so are pretty ladies, all kinds of food, good deeds, wrong turns, and more food. Every type of meat and vegetable seems to be simmering in the background somewhere. Because of the segregation the bands endured on the road, they often had to cook for themselves. "We cooked beans in a pot on a hotplate or on the back of an iron."
Time and again, Terry provides delicious descriptions of great, hot jazz. He refers to train rhythms, horse hooves on the street, the tap dancing of a childhood friend who affixed PET Milk can bottoms on the soles of his shoes, "raindrops on the coal shed in the old chicken yard," and much more. As a writer, I love these sections, especially here where he gets inside the genius of Count Basie:
During his easygoing renditions, there was a soul-stirring feeling. His four-four rhythm swayed a head-nodding and ear-bending groove, an easy foot-tapping bounce. Then during his up-tempo renditions he created a finger-popping force that made you move.
Recording History
In the back of the book is a nineteen-page discography compiled by Tom Lord, editor of the Jazz Discography Online. It lists 371 releases. After reading Terry's stories, I now want to hear every single one of them.
BLOG: http://keepswinging.blogspot.com/2008/01/clark-terry-bob-brookmeyer-quintet.html
"Clark: The Autobiography of Clark Terry"
by Clark Terry with Gwen Terry
Preface by Quincy Jones
Forward by Bill Cosby
Introduction by David Demsey
University of California Press, Hardbound, 344 pages, ISBN: 9780520268463, $34.95; Adobe PDF E-Book, ISBN: 9780520949782, $34.95; ePUB Format, ISBN: 9780520949782, $34.95
http://www.ucpress.edu/book.php?isbn=9780520268463
Article is Copr. © 2011 by John Scott G, and originally published on MusicIndustryNewswire-dot-com before the site was revamped as MuseWire.com in March 2015 – all commercial rights reserved. Disclosure: no fee or other consideration was provided to JSG, this site or its publisher, by the book author, publisher or agent(s).
John Scott G
John Scott G, an admitted word nerd, writes books, plays, screenplays, and political commentary. Author of "Area Code 666," "Secret Sex," and "Ambient Deviant Speedmetal Polka," Mr. G also writes under the pseudonym Gerald Laurence. Every day he happily rubs a few phrases up against each other to create sparks in your brain. You're welcome.
→ John Scott G
Feature Articles /
New Album 'Battle Cry: Songs of America's Heroes' to Help Veterans
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Book Review: 'Love is the Cure' by Elton John


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Revised and expanded, A Performer's Guide to Seventeenth Century Music is a comprehensive reference guide for students and professional musicians. The book contains useful material on vocal and choral music and style; instrumentation; performance practice; ornamentation, tuning, temperament; meter and tempo; basso continuo; dance; theatrical production; and much more. The volume includes new chapters on the violin, the violoncello and violone, and the trombone—as well as updated and expanded reference materials, internet resources, and other newly available material. This highly accessible handbook will prove a welcome reference for any musician or singer interested in historically informed performance.
Jeffery Kite-Powell is Professor Emeritus at the Florida State University College of Music. He is translator of Michael Praetorius's Syntagma Musicum III and editor of A Performer's Guide to Renaissance Music, Second Edition (IUP, 2007).
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The year is winding down.
I'm down to just TWO photoshoots left and then 2017 will be wrapped up. I snapped this newborn yesterday and it was one of my favorite shoots ever. I've done a few extended family sessions (here and here) this past week and they've been wonderful...even with having to do them mostly inside.
I have a 2017 year in review post I've been working on, a 2017 favorites from Pics & Paws post, Annie's 11 month post, and my 365 matte project post ALLLLL in the works that I'll be finishing up come Sunday and publishing for you...so get ready for a gigantic post dump that day! HAHA!
Anyways, I snapped a few pics of the girls having breakfast this morning and gosh I just love these. Lifestyle, real life photography is where my heart is at lately and I love it so. It'll actually be the theme for my 2018 project! 365 days of real life candids, all 4x4 formatted. I.cannot.wait.
Anyways, I can't believe we only have TWO days left of 2017! I'm determined to send it out with a bang (I loved 2017. It was a great year!) yet welcome 2018 and all of the things to come with open arms.
Hope everyone has a great one!
In a nutshell, I'm a photographer (alexandriamooneyphotography.com), birth stories documenter, Google Certified Innovator, tech nerd, Apple lover, history buff, root beer enthusiast, avid reader, unofficial movie critic, world traveler, micro-blogger (@AlexRMooney), insta-overgrammer (@alexandria.mooney), reality TV addict, former division-one softball player, former #googleglass explorer, GoPro picture-taker, self-proclaimed meteorologist, Tula wearer, cloth diaper addict, dog lover, devoted wife to a crime scene investigator, and, most importantly, a mama to Patrick, Rosie, Annie, our angel Clark, and this blog's namesake, our 15-year-old dachshund/terrier mix named Scooter who went across the rainbow bridge in December of 2018 (@Scooter_the_Dog).
First Trip to the Park!
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Savage Universal, the worldwide leader in backgrounds and photographic products, is pleased to announce that it has become a sponsor of acclaimed international photographer Zhang Jingna.
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Jingna's relationship with Savage began when she used the company's backdrops for her famous Motherland Chronicles series. She will be a featured photographer on Savage's blog, showcasing her photographs and providing reviews on select Savage products. "I have been using Savage backdrops for many years and am very excited about the new possibilities and directions my work can take with the support of Savage Universal. I very much look forward to working together!" said Jingna.


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Previous PostPrevious 4th ANNUAL FOUNDATION DAY KICKS OFF THE 2018 SCI CONVENTION!


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Q: Logging a "persons" activity in Linux So I've read many of the posts related to this and have just come up more confused than before.  There are recommendations for various tools including ttyrec, snoopy, acct, rootsh, sudosh, ttyrpld, unix auditing and more.
In my case, I want to be able to log all commands executed on a system (like history with timestamps enabled), but I also want to know who did what? However, we all log in via ssh to the same small subset of user accounts (depending on what we're doing). How can I get a log of commands including the information a "who" will give me (regarding connection) so I can trace the action to a specific person as a posed to just a generic "user"?

A: The auditd package is designed for this, you can use PAM to assign each login a session ID so you can track activity on an account back to the original login (and then check the login log to determine where that person logged in from), and use the ausearch command to locate the events you're interested in.  You can read a pretty comprehensive guide here though some of the finer details will need to be changed to match your distribution.  Redhat has a site with a FAQ and some other docs here.
Since this doesn't rely on trying to log commands typed into shells or record the users' command history, it should log things like opening vi (which could maybe be done over X or in screen), writing a script (maybe one character at a time with vi's cut/paste commands and a few runs of :.!/usr/games/fortune to get some source data that you might not be able to log) then running the :%!/bin/bash command.

A: There are a variety of tools at your disposal already, which your question (and these answers) hint at like history, snoopy, auditd, sudo logs, etc...but if you have a "Subset of accounts" people are using, there's no way at the server to tell who did what. The only way you can specifically tell who did it is if users have their own computers that they use specifically and use keyloggers to tell what they were physically typing at that keyboard.
Whenever you share accounts you can't tell what was actually going on, since you'd need further proof who was using your root account or bob account or whatever your people were doing. If you're trying to investigate what happened in a specific incident, you may need to review your access policies and procedures, your recovery procedures, and evaluate your employees and/or engage in retraining if necessary (or their trustworthiness with sensitive accounts) rather than directly focus on hunting down who did something, as it may suck up more resources than you stand to gain in finding the person who did it.
Otherwise you might want to look into forensic investigation techniques for tracking down what happened (drive imaging, log tracing, etc.) If you're not investigating an incident, look into reviewing your policies and instituting better tracking and account verification (only you have root, only Bob uses his account using sudo to get access to higher privileges, install and monitor auditd, etc.) and be careful not to make your trusted circle feel like they're being held under a microscope or you might alienate the people trying to get their jobs done (or hinder them from getting their jobs done).

A: Linux auditd (http://people.redhat.com/sgrubb/audit/) will give you most power in watching who did what. You have more about it in DerfK's answer.
Nothing, however, is going to tell you, who logged in as webadmin, if there are n people who have access to the webadmin account. I would suggest to use named accounts for each user and then either use su - or sudo to run commands from the "function" account.


Most of this tutorial is for the Almost all Linux Operating Systems, primarily Ubuntu, Linux Mint, etc.
Big Kudos for Matt who made on his own a Linux guide for OpenVPN.
First, let's get started by mentioning this guide is mostly for Ubuntu and Ubuntu-based distributions, the most popular one and Linux Mint is very similar. For those trying to install openvpn on distributions such as CentOS/Redhat, Fedora, etc, a lot of the syntax is *similar* to what's being written now, and it can be found on Google. This is the Client side of OpenVPN we're installing, not the Server part of it.
1) Use opening a terminal (keyboard shortcut: Ctrl + Alt + T, there's also using the "Start Menu" or "Command Menu" and then typing "Terminal" and clicking on it to get to it, but Ctrl + Alt + T is the easiest/quickest. Some distributions of Linux can be quite user-friendly, and even have some similarities, the terminal is the Command Prompt for Windows ; but do not confuse them; they're entirely different. Mac OSX has a terminal itself with similar commands to Linux—but again, do NOT confuse the them.. most commands are different.). It is recommended to keep the Linux terminal open for all steps, but you might need to open another to verify a few things.
2) Install OpenVPN client by entering sudo apt-get -y install openvpn and you'll enter your Ubuntu Password—same goes for Linux Mint. .. both of those usually use the same thing.
3) Head back to the tigerVPN website. Go into Dashboard, click on the "Geeks" tab Remember: Your username, password—ALL VPN Credentials are stored on this part of the website. Now click on "Download" next to OpenVPN Configuration File under where it says "OpenVPN Configuration" – a file named "config-256bit.zip" will asked to be downloaded. Save it under the Downloads folder.. if you didn't try to save it anywhere else, that's the default folder for a file to be downloaded, and that's where you should leave it.
4) Moving on, you will now need to extract/unzip config-256bit.zip in the terminal. To do that, type unzip config-256bit.zip -d ~/tiger (config-256bit.zip is the file name that just downloaded previously) Now you've extracted all OpenVPN files provided by TigerVPN into the "~/tiger" directory.
5) Now that you've extracted the OpenVPN files into the ~/tiger through the terminal, type cd ~/tiger/config to get to the folder with the extracted files to use for OpenVPN.
6) Make sure you are in the right directory through a terminal, simply enter "ls -la" command and it will print full configuration file list.
7) This first example will just be entering sudo openvpn --config "US - New York @tigervpn.com.ovpn" – Then just hit enter. This is just to get you familiar with your footing regarding the openvpn. This will be when your credentials will be asked for.
This error occurs because you cannot use spaced file names in a Linux terminal without using quotes around them. Guess what, though? I've got good news for you. There's a MUCH shorter way to do this AND you avoid this error! :) Just type sudo openvpn --config " and press Tab; As you can see, you'll see every single one you have been provided with.. so as a reminder, just use that and press tab so you know what to type next. See how this gets easier? You can use a single quote or a double quote. Both will work. Try sudo openvpn --config "US and then press Tab. As you can see, it narrows it down to just the US locations. You can technically keep typing one character at a time and pressing tab.. it'll just keep giving you the information to type, in case you forget or if you don't want to type the rest. Try just typing this: sudo openvpn --config "US - New York and then press Tab. It'll finish the rest for you. Let's try it in the next step.
9) Type sudo openvpn --config "US and then, of course, press Tab. (this is when you press tab a couple of times to make sure that it makes your choices shorter by typing one more thing each time; instead of sudo openvpn --config 'US [press tab] and seeing the same locations (from the US obviously—let's use New York as an example), type sudo openvpn --config "US - New and press Tab. This method goes unknown by many new users and will get more until you get just exactly what you want and helps you type as little as you need for ANY location.
10) Try just typing out sudo openvpn --config 'US - New York and then pressing Tab. It'd be the same if you typed out sudo openvpn --config 'US - New York @tigervpn.com.ovpn' and pressing Tab because then it shows up as the only selection as it is typed out completeley. At that point, all you have to do is press enter. With all of this, You'll have the full listing(s), ability to reduce typing time/capacity, unlike this guide (sorry, I also try to teach with a guide to help people understand why things are different), and then all you'd need is to do press enter, type your sudo password (your Ubuntu/password, if it asks. It depends on how much time you spent in between browsing/working and actually using the guide. It's hard for anyone with even an a normal attention span to keep up with this, including myself.) – last step is to enter your credentials for tigerVPN. Let's also remember that your tigerVPN username and password are located at the Dashboard under "Geeks" on the tigerVPN website. Make sure you log in on to the website to retrieve your information!
11) After doing all of this, under Ubuntu Linux, Linux Mint, and others similar to it at least, run sudo openvpn --config 'US - New York @tigervpn.com.ovpn' either by typing it out or by pressing Tab to make it easier, press enter, type in your passwords and you should successfully connect to tigerVPN with OpenVPN!
12) To disconnect from the OpenVPN connection simply go back to your terminal and hold the Ctrl key down and press C (Usually known as Ctrl + C – You just hold the two keys at the same time. It basically means to close or and terminate a process). If that doesn't work, just close the terminal and say yes if it asks if you're sure. Just remember, you'll need to redo this process again, but you get used to it, and using the tab method in the guide can really helps :) Hopefully the shortcuts with the Tab button will help. Also, a word of consideration, once you've connected, for the love of your VPN connection, don't close the terminal window or the connection will definitely close.


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text:
Home » News » Alzheimer's Signs Reversed in Mouse Study
Alzheimer's Signs Reversed in Mouse Study
Researchers have successfully reversed the formation of amyloid plaques in the brains of mice with Alzheimer's disease, thereby improving the animals' cognitive function.
Investigators from the Cleveland Clinic Lerner Research Institute discovered that gradually depleting an enzyme called BACE1 eliminates the plaques.
The study, which appears in the Journal of Experimental Medicine, raises hopes that drugs targeting this enzyme will be able to successfully treat Alzheimer's disease in humans. The vast majority of experimental treatments using a rodent model — whether for diabetes or cancer or Alzheimer's — fail to work in humans.
Researchers explain that one of the earliest events in Alzheimer's disease is an abnormal buildup of beta-amyloid peptide, which can form large, amyloid plaques in the brain and disrupt the function of neuronal synapses.
Also known as beta-secretase, BACE1 helps produce beta-amyloid peptide by cleaving amyloid precursor protein (APP). Drugs that inhibit BACE1 are therefore being developed as potential Alzheimer's disease treatments. However, the drugs may have serious side-effects because BACE1 controls many important processes.
Mice completely lacking BACE1 suffer severe neurodevelopmental defects. To investigate whether inhibiting BACE1 in adults might be less harmful, researcher Dr. Riqiang Yan and colleagues generated mice that gradually lose this enzyme as they grow older. These mice developed normally and appeared to remain perfectly healthy over time.
The researchers then bred these rodents with mice that start to develop amyloid plaques and Alzheimer's disease when they are 75 days old. The resulting offspring also formed plaques at this age, even though their BACE1 levels were approximately 50 percent lower than normal.
Remarkably, however, the plaques began to disappear as the mice continued to age and lose BACE1 activity, until, at 10 months old, the mice had no plaques in their brains at all.
"To our knowledge, this is the first observation of such a dramatic reversal of amyloid deposition in any study of Alzheimer's disease mouse models," Yan said.
Decreasing BACE1 activity also resulted in lower beta-amyloid peptide levels and reversed other hallmarks of Alzheimer's disease, such as the activation of microglial cells and the formation of abnormal neuronal processes.
Loss of BACE1 also improved the learning and memory of mice with Alzheimer's disease. However, when the researchers made electrophysiological recordings of neurons from these animals, they found that depletion of BACE1 only partially restored synaptic function, suggesting that BACE1 may be required for optimal synaptic activity and cognition.
"Our study provides genetic evidence that preformed amyloid deposition can be completely reversed after sequential and increased deletion of BACE1 in the adult," Yan said.
"Our data show that BACE1 inhibitors have the potential to treat Alzheimer's disease patients without unwanted toxicity. Future studies should develop strategies to minimize the synaptic impairments arising from significant inhibition of BACE1 to achieve maximal and optimal benefits for Alzheimer's patients."
Source: Rockefeller University Press/Science Daily
Nauert PhD, R. (2018). Alzheimer's Signs Reversed in Mouse Study. Psych Central. Retrieved on January 20, 2020, from https://psychcentral.com/news/2018/02/19/alzheimers-signs-reversed-in-mouse-study/132747.html
Last updated: 8 Aug 2018 (Originally: 19 Feb 2018)
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