text:
\section{Introduction}\label{Intro}

In this paper, we aim at finding a zero of the sum of a maximally
monotone, a cocoercive, a monotone-Lipschitzian, and a
monotone-continuous operators, which is also in a convex closed
subset $X$ of a Hilbert space.
This inclusion encompasses several problems in partial
differential equations coming from mechanical models
\cite{Gabay1983,Glowinski1975,Goldstein1964}, differential
inclusions \cite{AubinHelene2009,Showalter1997}, game theory
\cite{Nash13},
among other disciplines. When $X$ is the whole Hilbert space,
the algorithms proposed in
\cite{Bot-2019-ACM,Bot2016NA,Bot2013,Bot2015AMC,BotHendrich2013,briceno2011SIAM,CombettesMinh2022,Cevher2020SVVA,Combettes2018MP,CombPes12,Combettes2014Optimization,Csetnek2019AMO,davis2015,Dong2021,DungVu15,Eckstein2017,Eckstein2020,EcksteinJohnstone,Malitsky2020SIAMJO,Raguet-SIAM-2013,Rieger2020AMC,RyuVu2020,Vu13}
can solve this kind of problems  under additional assumptions
or without exploiting the intrinsic properties of the involved
operators.
Indeed, the algorithms in
\cite{Bot2013,Bot2015AMC,BotHendrich2013,briceno2011SIAM,Combettes2018MP,Eckstein2017}
need to compute the resolvents of all the monotone operators, which
are not explicit in general or they can be
numerically expensive. The schemes proposed in
\cite{Bot2016NA,CombPes12,DungVu15,Eckstein2020} take
advantage of the
properties of the monotone-Lipschitzian operator, but the
cocoercivity and the continuity of the others are not leveraged. In
fact,
the algorithms in
\cite{Bot2016NA,CombPes12,DungVu15,Eckstein2020} need to
activate the continuous operator via its resolvent and to explicitly
implement the cocoercive and the monotone-Lipschitzian operators
twice by iteration. In contrast, the algorithms in
\cite{Cevher2020SVVA,Csetnek2019AMO,Malitsky2020SIAMJO,Rieger2020AMC,RyuVu2020}
activate the cocoercive and the monotone-Lipschitzian
operators only once by iteration, but they need to store in
the
memory the two past iterations and the step-size is reduced
significantly. Furthermore, the methods in
\cite{Cevher2020SVVA,Csetnek2019AMO,Malitsky2020SIAMJO} consider only
one maximally monotone operator and, hence, it needs to compute
the resolvent of the sum of the maximally monotone and the
monotone-continuous operator. On the other hand, methods in
\cite{Rieger2020AMC,RyuVu2020} need to calculate the resolvent of
the monotone-continuous operator, which is not simple in general.
In addition, methods proposed in
\cite{Bot-2019-ACM,Combettes2014Optimization,davis2015,Dong2021,EcksteinJohnstone,Raguet-SIAM-2013,Vu13}
take advantage of the cocoercive operator by activating it once by
iteration, but they do not exploit continuity nor the
monotone-Lipschitzian
property of the operators and they need to compute their resolvents.
The method in \cite{CombettesMinh2022} exploits the properties of
the cocoercive and the monotone-Lipschitzian operators, but it does
not take into advantage the monotonicity and continuity of one of the
operators and need to compute its resolvent.
Other methods solving this kind of problems including a normal cone
to a closed vector subspace,
when the continuous operator is zero and either the cocoercive or the
monotone-Lipschitzian operator is zero, are discussed in
\cite{briceno2015Optim,Briceno2015JOTA,Spingran1983AMO}.

In the case when $X$ is a subset of the domain of the maximally
monotone operator,
methods exploiting the monotonicity and continuity property of one of
the operators involved in our problem are proposed in
\cite{BricenoDavis2018,Tseng2000SIAM}. In particular, the algorithm
in  \cite{Tseng2000SIAM} solves our problem by using a line search
procedure involving the sum of the cocoercive, the
monotone-Lipschitzian, and the
monotone-continuous operators. On the other hand, the
\textit{forward-backward-half forward splitting} (FBHF)
method, proposed in \cite{BricenoDavis2018}, solves
our problem via a line search which activates the sum of the
the monotone-Lipschitzian and the monotone-continuous
operators.
As perceived in \cite{BricenoDavis2018}, to reduce the activation of
monotone operators in the line search procedure can reduce the
number of iterations significantly (see, e.g.,
\cite[Table~3]{BricenoDavis2018} in which this reduction is around a
20\%). Moreover, when the explicit implementation of these operators
are expensive (e.g., high dimensional problems), the computational
time can be much larger because line search procedures need to
activate those operators several times by iteration.

In this paper, we propose a fully split method for solving our
monotone
inclusion, which take advantage of the intrinsic properties of each
operator involved in the inclusion. More precisely, the proposed
method activates the cocoercive operator once, the
monotone-Lipschitzian operator twice by iteration, and uses a line
search only implementing the monotone-continuous operator. We
also ensure the weak convergence of our algorithm under
hypotheses on the set $X$ which are
independent of the domain of the maximally monotone operator,
generalizing some results in
\cite{BricenoDavis2018,Tseng2000SIAM}.
We explore an interesting example in optimization, involving
non-linear inequality constraints governed by convex G\^ateaux
differentiable functions. We provide conditions on this function in
order
to guarantee that the saddle operator obtained from the Lagrangian of
the problem is monotone and continuous and satisfies the hypotheses
of our main convergence theorem. Finally, we
provide a numerical experiment which illustrate the efficiency of our
algorithm.

The paper is organized as follows. In Section~\ref{sec:2} we set our
notation. In Section~\ref{sec:3} we provide some technical lemmas,
our splitting  method for solving Problem~\ref{prob:problem1}, and our
convergence result. In Section~\ref{sec:4} we derive an algorithm for
solving a non-linearly constrained composite convex
optimization problem. Finally, in Section~\ref{sec:5} we provide a
numerical experiment illustrating the efficiency of the
method proposed
in Section~\ref{sec:4}.

\section{Preliminaries}\label{sec:2}
Throughout this paper $\H$ and $\mathcal G$ are real Hilbert spaces. We
denote the scalar
product by $\scal{\cdot}{\cdot}$ and the associated norm by
$\|\cdot \|.$
The symbols $\ensuremath{\:\rightharpoonup\:}$ and $\to$ denote the weak and strong
convergence, respectively.
Given a linear bounded
operator $M:\H \to \mathcal G$, we denote its adjoint by $M^*\colon\mathcal G\to\H$.
$\textnormal{Id}$ denotes the identity operator on $\H$.
Let $D\subset \H$ be non-empty, let $T: D \rightarrow \H$, and let
$\beta
\in \left]0,+\infty\right[$.
The operator $T$ is
$\beta$-cocoercive if
\begin{equation} \label{def:coco}
	(\forall x \in D) (\forall y \in D)\quad \langle x-y \mid Tx-Ty
	\rangle \geq
	\beta \|Tx - Ty
	\|^2
\end{equation}
and it is $\beta$-Lipschitzian if
\begin{equation} \label{def:Lips}
	(\forall x \in D) (\forall y \in D)\quad \|Tx-Ty\| \leq  \beta\|x
	- y
	\|.
\end{equation}

Let $A:\H \rightarrow 2^{\H}$ be a set-valued operator. The domain,
range, graph, and the zeros of $A$ are, respectively, $\textnormal{dom}\,\, A =
\menge{x \in \H}{Ax \neq  \varnothing}$, $\textnormal{ran}\,\, A = \menge{u \in
	\H}{(\exists x \in \H)\, u \in Ax}$, $\textnormal{gra}\, A = \menge{(x,u) \in
	\H
	\times \H}{u \in Ax}$, and $\textnormal{zer} A = \menge{x \in \H}{0 \in Ax}$.
The inverse of $A$ is $A^{-1}\colon \H  \to
2^\H \colon u
\mapsto
\menge{x \in \H}{u \in Ax}$ and the resolvent of $A$ is
$J_A=(\textnormal{Id}+A)^{-1}$.
The operator $A$  is monotone if
\begin{equation}\label{def:monotone}
	(\forall (x,u) \in \textnormal{gra}\, A) (\forall (y,v) \in \textnormal{gra}\, A)\quad
	\scal{x-y}{u-v} \geq
	0.
\end{equation}
Moreover, $A$ is maximally monotone if it is monotone and there
exists no
monotone operator $B :\H\to  2^{\H}$ such that $\textnormal{gra}\, B$ properly
contains $\textnormal{gra}\, A$, i.e., for every $(x,u) \in \H \times \H$,
\begin{equation} \label{def:maxmonotone}
	(x,u) \in \textnormal{gra}\, A \quad \Leftrightarrow\quad  (\forall (y,v) \in
	\textnormal{gra}\, A)\ \
	\langle x-y \mid u-v \rangle \geq 0.
\end{equation}
$A$ is locally bounded at $x \in\H$, if there exists $\delta \in
\ensuremath{\left]0,+\infty\right[}$ such that $A(\mathcal{B}(x;\delta))$ is bounded, where
$\mathcal{B}(x;\delta)$ stands for the ball centered at $x$ with
radius $\delta$. Moreover, $A$ is
locally bounded in $\varnothing \neq D \subset \H$ if, for every $x
\in D$, $A$ is locally bounded at $x$.

We denote by $\Gamma_0(\H)$ the class of proper lower
semicontinuous convex functions $f\colon\H\to\ensuremath{\left]-\infty,+\infty\right]}$. Let
$f\in\Gamma_0(\H)$.
The Fenchel conjugate of $f$ is
defined by $f^*\colon u\mapsto \sup_{x\in\H}(\scal{x}{u}-f(x))$,
which
is a function in $\Gamma_0(\H)$. The subdifferential of $f$ is the
maximally monotone operator
$$\partial f\colon \H \to 2^\H \colon x\mapsto
\menge{u\in\H}{(\forall y\in\H)\:\:
	f(x)+\scal{y-x}{u}\le f(y)}.$$
It turns out that $(\partial f)^{-1}=\partial f^*$
and that $\textnormal{zer}\partial f$ is the set of
minimizers of $f$, which is denoted by $\arg\min_{x\in \H}f(x)$.
We denote the proximity operator of $f$ by
\begin{equation}
	\label{e:prox}
	\ensuremath{\text{\rm prox}}_{f}\colon
	x\mapsto\argm{y\in\H}\left(f(y)+\frac{1}{2}\|x-y\|^2\right).
\end{equation}
We have
$\ensuremath{\text{\rm prox}}_f=J_{\partial f}$. Moreover, it follows from \cite[Theorem
14.3]{bauschkebook2017} that
\begin{equation}
	\big(\forall \gamma >0) \quad
	\label{e:Moreau_nonsme}
	\ensuremath{\text{\rm prox}}_{\gamma f}+\gamma \ensuremath{\text{\rm prox}}_{f^*/\gamma} \circ
	\textnormal{Id}/\gamma=\textnormal{Id}.
\end{equation}
We denote by
$\textnormal{lev}_{<0}f=\menge{x\in\H}{f(x)<0}$ and by
$\textnormal{lev}_{\le0}f=\menge{x\in\H}{f(x)\le0}$.
Given a non-empty closed convex set $C\subset\H$, we denote by
$P_C$ the projection onto $C$ and by
$\iota_C\in\Gamma_0(\H)$ the indicator function of $C$, which
takes the value $0$ in $C$ and $\ensuremath{{+\infty}}$ otherwise.
For further properties of monotone operators,
nonexpansive mappings, and convex analysis, the
reader is referred to \cite{bauschkebook2017}.

\section{Main Results}\label{sec:3}
We aim at solving the following monotone inclusion
problem.
\begin{pro}\label{prob:problem1}
	Let $X$ be a nonempty closed convex subset of a real Hilbert
	space
	$\H$, let $A: \H \to 2^\H$ be a  maximally monotone operator, let
	$B_1 : \H \to \H$ be a $\beta$-cocoercive operator for some
	$\beta
	>0$, let $B_2 :\H \to \H$ be a monotone and $L$-Lipschitzian
	operator for some $L>0$, and let $B_3 : \H \to 2^\H$ be a
	maximally
	monotone operator such that $B_3$ is single valued and
	continuous
	in $\textnormal{dom}\, A \cup X \subset \textnormal{dom}\, B_3$. Moreover assume that
	$A+B_3$
	is
	maximally
	monotone. The
	problem is to
	\begin{equation}\label{eq:problem1}
		\text{find} \quad x \in X \quad \text{such that} \quad 0 \in
		Ax+B_1x+B_2x+B_3x,
	\end{equation}
	under the assumption that the set of solutions to
	\eqref{prob:problem1} is nonempty.
\end{pro}

We first study some properties of the monotone
operators involved in Problem~\ref{prob:problem1}, which ensure the
finite termination of the backtracking procedure in our method.

\begin{lem}\label{prop:varphiz}
	In the context of Problem~\ref{prob:problem1}, let $z$ and $y$ in
	$\H$, and define
	\begin{equation}\label{eq:varphiz}
		(\forall \gamma > 0) \quad \quad	x_{z,y}(\gamma) =
		J_{\gamma A}
		( z -
		\gamma y) \quad \text{and}
		\quad \varphi_{z,y}( \gamma) = \frac{\| z-
			x_{z,y}(\gamma)\|}{\gamma}.
	\end{equation}
	Then, the following statements holds:
	\begin{enumerate}
		\item\label{prop:varphiz1}
		$\varphi_{z,y}$ is nonincreasing.
		\item\label{prop:varphiz2}
		$(\forall z \in \textnormal{dom}\, A) \displaystyle  \quad \lim_{ \gamma
			\downarrow 0}
		\varphi_{z,y}(\gamma) = \| (A+y)^0 z\| = \min_{w \in
			Az+y}\|w\|
		$.
		\item \label{prop:varphiz23} Set
		\begin{equation}\label{eq:defsetC}
			\mathcal{C}=\menge{z \in \H}{ \lim_{\gamma \downarrow 0}
				\varphi_{z,y}(\gamma)  < +\infty}.
		\end{equation}
		Then, $\textnormal{dom}\, A \subset \mathcal{C}
		\subset \overline{\textnormal{dom}\,} A$.
		\item\label{prop:varphiz3} Suppose that one of the following
		holds:
		\begin{enumerate}
			\item\label{prop:varphiz31} $z \in \mathcal{C}$.

			\item\label{prop:varphiz32} $z \in \textnormal{dom}\, B_3
			\setminus\mathcal{C}$,
			$y=(B_1+B_2+B_3)z$, and $B_3$ is locally bounded at
			$P_{\overline {\textnormal{dom}\,} A} z$.
			\item\label{prop:varphiz33}  $z \in \textnormal{dom}\, B_3
			\setminus\mathcal{C}$,
			$y=(B_1+B_2+B_3)z$, and $\overline{\textnormal{dom}\,} A \subset \ensuremath{\operatorname{int}}
			\textnormal{dom}\,
			B_3$.

		\end{enumerate}
		Then, for every
		$\theta \in \left]0,1\right[$, there exists $\gamma(z) >0$
		such
		that, for
		every  $\gamma \in \left]0, \gamma(z)\right]$,
		\begin{equation}\label{eq:destheta}
			\gamma \| B_3z - B_3 x_{z,y} (\gamma) \| \leq \theta \|z
			- x_{z,y}
			(\gamma)\|.
		\end{equation}

	\end{enumerate}
\end{lem}

\begin{proof}

	Let $z \in \H$. Note that, if $z \in \textnormal{zer} ( A + y )$,

	\begin{align}
		(\forall \gamma > 0 )\quad 0 \in Az+y \quad &\Leftrightarrow
		\quad  z-\gamma y \in \gamma A z + z
		\nonumber \\
		&\Leftrightarrow  \quad  z = x_{z,y} (\gamma)\nonumber \\
		&\Leftrightarrow  \quad  \varphi_{z,y}(\gamma) = 0
		\label{varphiz=0}.
	\end{align}
	In this case, \ref{prop:varphiz1}, \ref{prop:varphiz2}, and
	\ref{prop:varphiz3}
	are clear. Henceforth, assume $z \in \H \setminus \textnormal{zer} ( A + y
	)$.
	It follows from
	\eqref{eq:varphiz}
	that
	\begin{equation}\label{eq:contencionxzy}
		\frac{z-x_{z,y}(\gamma)}{\gamma}  - y \in A
		{x_{z,y}(\gamma)}.
	\end{equation}

	\ref{prop:varphiz1}: For every $\gamma_1$
	and $\gamma_2$ in $\ensuremath{\left]0,+\infty\right[}$, \eqref{eq:contencionxzy}
	and the
	monotonicity of $A$ yield
	\begin{align*}
		0 &\leq \Scal{\frac{z-x_{z,y}(\gamma_1)}{\gamma_1}
			-\frac{z-x_{z,y}(\gamma_2)}{\gamma_2}
		}{x_{z,y}(\gamma_1)-x_{z,y}(\gamma_2)}\\
		& = \Scal{\frac{z-x_{z,y}(\gamma_1)}{\gamma_1}
			-\frac{z-x_{z,y}(\gamma_2)}{\gamma_2}
		}{x_{z,y}(\gamma_1)-z-(x_{z,y}(\gamma_2)-z)}\\
		& = - \frac{1}{\gamma_1} \|z-x_{z,y}(\gamma_1)\|^2 + \left(
		\frac{1}{\gamma_1}+\frac{1}{\gamma_2} \right)
		\scal{z-x_{z,y}(\gamma_1)}{z-x_{z,y}(\gamma_2)}-
		\frac{1}{\gamma_2} \|z-x_{z,y}(\gamma_2)\|^2.
	\end{align*}
	Hence, we obtain

	\begin{align*}
		\gamma_1 \varphi_{z,y}(\gamma_1)^2+\gamma_2
		\varphi_{z,y}(\gamma_2)^2 &\leq ( \gamma_1+\gamma_2) \Scal{
			\frac{z-x_{z,y}(\gamma_1)}{\gamma_1}}{\frac{z-x_{z,y}(\gamma_2)}{\gamma_2}}\\
		&\leq \frac{( \gamma_1+\gamma_2)}{2}
		(\varphi_{z,y}(\gamma_1)^2+\varphi_{z,y}(\gamma_2)^2),
	\end{align*}
	which yields
	$(\gamma_1-\gamma_2)(\varphi_{z,y}(\gamma_1)^2-\varphi_{z,y}(\gamma_2)^2)\leq
	0$ and \ref{prop:varphiz1} follows.

	\ref{prop:varphiz2}: It
	follows
	from the monotonicity of $A$ and \eqref{eq:contencionxzy} that,
	for
	every $w \in  Az +y$ and $\gamma \in \ensuremath{\left]0,+\infty\right[}$,

	\begin{align*}
		0 \leq
		\Scal{\frac{z-x_{z,y}(\gamma)}{\gamma}-w}{x_{z,y}(\gamma)-z},
	\end{align*}
	which yields
	\begin{align}\nonumber
		\frac{1}{\gamma}\|z-x_{z,y}(\gamma)\|^2
		&\leq\scal{w}{z-x_{z,y}(\gamma)}\\
		&\leq \|{w}\|
		\|z-x_{z,y}(\gamma)\|.\label{eq:desinf}
	\end{align}
	Thus, $\varphi_{z,y} (\gamma) \leq \|w\|$. Therefore, since
	\cite[Proposition 20.36]{bauschkebook2017} implies that, for
	every $z
	\in \textnormal{dom}\, A$, $Az+y$ is nonempty, closed, and convex,
	\cite[Theorem
	11.10]{bauschkebook2017} yields
	\begin{align}
		(\forall \gamma \in \ensuremath{\left]0,+\infty\right[})\quad \quad
		\varphi_{z,y}(\gamma)\leq
		\min_{
			w \in Az+y} \|{w}\|. \label{eq:desmin}
	\end{align}
	Hence, since $\varphi_{z,y} \geq 0$, \ref{prop:varphiz1} implies
	that
	$\lim_{\gamma \downarrow 0}
	\varphi_{z,y}(\gamma)$
	exists. In turn, since $z \in \H\setminus \textnormal{zer}
	(A+y)$, it follows from \eqref{eq:desmin} and
	\eqref{varphiz=0} that
	\begin{equation} \label{eq:xtoz}
		0<\varphi_{z,y}(1) \leq \lim_{ \gamma \downarrow 0}
		\varphi_{z,y}(\gamma)\leq  \min_{
			w \in Az+y} \|{w}\|,
	\end{equation}
	which, in view of \eqref{prop:varphiz}, implies
	\begin{equation}\label{eq:xzytoz}
		\lim_{\gamma \downarrow 0} x_{z,y}(\gamma)=z.
	\end{equation}
	Since $(\varphi_{z,y}(\gamma))_{(\gamma >0)}$ is bounded, by
	\cite[Lemma 2.45]{bauschkebook2017}, there exists a sequence
	$(\gamma_k)_{k \in \mathbb N} \subset \ensuremath{\left]0,+\infty\right[}$ and  $\overline{w} \in \H$
	such that $\gamma_k  \downarrow 0$ and $\frac{z-
		x_{z,y}(\gamma_k)}{\gamma_k} \rightharpoonup \overline{w}$ as $k \to
	+\infty$.
	Therefore, \eqref{eq:contencionxzy}, \eqref{eq:xzytoz}, and
	\cite[Proposition~20.38(i)]{bauschkebook2017} imply
	$\overline{w}
	\in
	A z +y $. Hence, noting that
	\begin{align} \nonumber
		(\varphi_{z,y}(\gamma_k))^2&= \left\|\frac{z-
			x_{z,y}(\gamma_k)}{\gamma_k} -
		\overline{w}\right\|^2+\|\overline{w}\|^2+2\Scal{\frac{z-
				x_{z,y}(\gamma_k)}{\gamma_k}-\overline{w}}{\overline{w}}
		\\
		&\geq \|\overline{w}\|^2+2\Scal{\frac{z-
				x_{z,y}(\gamma_k)}{\gamma_k}-\overline{w}}{\overline{w}},
		\label{eq:desparalimite}
	\end{align}
	we deduce
	$$
	\lim_{\gamma \downarrow 0}(\varphi_{z,y}(\gamma))^2 =
	\lim_{k
		\to
		\ensuremath{{+\infty}}}(\varphi_{z,y}(\gamma_k))^2 \geq \|\overline{w}\|^2
	\geq \min_{w \in Az+y} \|w\|^2.$$ Therefore, we obtain from
	\eqref{eq:xtoz} that
	\begin{equation}
		\lim_{ \gamma \downarrow 0} \varphi_{z,y}(\gamma) =  \min_{w
			\in
			Az+y}\|w\|,
	\end{equation}
	and \ref{prop:varphiz2} follows.

	\ref{prop:varphiz23}: It follows from \eqref{eq:defsetC} and
	\ref{prop:varphiz2} that $\textnormal{dom}\, A \subset\mathcal{C}$ . Let $z
	\in \H \setminus \overline{\textnormal{dom}\,} A$ and let $\gamma >0$. The
	firm nonexpansiveness of $J_{\gamma A}$
	\cite[Proposition~23.8(ii)]{bauschkebook2017} implies
	\begin{align}\label{eq:toproj} \nonumber
		\|x_{z,y}(\gamma)-P_{\overline{\textnormal{dom}\,} A} z \| &= \| J_{\gamma
			A}(z-\gamma y) - J_{\gamma A}z + J_{\gamma A}z
		-P_{\overline{\textnormal{dom}\,}
			A} z\|\\ \nonumber
		& \leq \| J_{\gamma A}(z-\gamma y) - J_{\gamma A}z\| +
		\|J_{\gamma A}z -P_{\overline{\textnormal{dom}\,} A} z\|\\
		& \leq \gamma \|y\| + \|J_{\gamma A}z
		-P_{\overline{\textnormal{dom}\,} A}
		z\|.
	\end{align}
	Hence, by taking $\gamma
	\downarrow 0$ in \eqref{eq:toproj} we conclude from
	\cite[Theorem~23.48]{bauschkebook2017} that $
	x_{z,y}(\gamma) \to P_{\overline{\textnormal{dom}\,} A} z$ as $\gamma
	\downarrow
	0$. Then, by the continuity of the norm and $z \notin
	\overline{\textnormal{dom}\,} A$, we deduce
	\begin{equation*}
		\lim_{\gamma \downarrow 0} \| z -x_{z,y}(\gamma) \| = \| z
		-P_{\overline{\textnormal{dom}\,} A} z\|>  0.
	\end{equation*}
	Therefore,
	$\varphi_{z,y}(\gamma)=\| z - x_{z,y}(\gamma) \| / \gamma \to +
	\infty$
	as $\gamma
	\downarrow 0$ and, hence, it follows from \eqref{eq:defsetC}
	that $z \in \H\setminus \mathcal{C}$.

	\ref{prop:varphiz31}: If $z \in  \mathcal{C}$, it
	follows from \ref{prop:varphiz1} that
	\begin{equation} \label{eq:xtoz2}
		0<\varphi_{z,y}(1) \leq \lim_{ \gamma \downarrow 0}
		\varphi_{z,y}(\gamma)< +\infty.
	\end{equation}
	Therefore
	$\lim_{\gamma \downarrow 0}
	x_{z,y}=z$ and the continuity of $B_3$ implies

	\begin{equation} \label{eq:b3to0}
		\lim_{\gamma  \downarrow 0}   B_3 x_{z,y} = B_3 z.
	\end{equation}
	The result follows from \eqref{eq:xtoz2} and \eqref{eq:b3to0}.

	\ref{prop:varphiz32}: Set $B=B_1+B_2+B_3$ and let
	$p=P_{\overline{\textnormal{dom}\,} A} z $. Since $B_3$ is locally bounded at
	$p$,
	there exists
	$\delta_p \in \ensuremath{\left]0,+\infty\right[}$ such that $B_3(\mathcal{B}(p;\delta_p))$ is
	bounded. Now since $y=Bz$ and
	\begin{equation}\label{eq:contencionJz}
		\frac{z-J_{\gamma A} z }{\gamma} \in A J_{\gamma A} z,
	\end{equation} it follows from \eqref{eq:contencionxzy} and
	the
	monotonicity of $A$ that
	\begin{align*} \nonumber
		0 \leq &
		\Scal{\frac{z-x_{z,y}(\gamma)}{\gamma}-Bz-\frac{z-J_{\gamma
					A}z}{\gamma}}{x_{z,y}(\gamma)-J_{\gamma A}z} \\
		=&  - \frac{1}{\gamma} \|x_{z,y}(\gamma)-J_{\gamma A} z\|^2 +
		\scal{Bz}{J_{\gamma A}{z}-x_{z,y}(\gamma)}\\
		\leq& - \frac{1}{\gamma} \|x_{z,y}(\gamma)-J_{\gamma A} z\|^2
		+
		\|Bz\|\|J_{\gamma A}{z} -x_{z,y}(\gamma)\|.
	\end{align*}
	Hence, we obtain
	\begin{equation} \label{eq:cotaBz}
		\|x_{z,y}(\gamma)-J_{\gamma A}z\|\leq	\gamma \|Bz\|  .
	\end{equation}
	Additionally, by \cite[Theorem 23.48]{bauschkebook2017}, there
	exists
	$\gamma_1$ such that, for every $\gamma < \gamma_1$, $\|
	J_{\gamma A} z-p\| \leq \delta_p/2$. By defining
	\begin{equation}
		\overline{\gamma}:=\begin{cases}
			\gamma_1, &\text{ if } Bz=0;\\
			\min \{ \delta_p/(2\|Bz\|),\gamma_1\}, &\text{ if }
			Bz\neq 0,
		\end{cases}
	\end{equation}
	it follows from
	\eqref{eq:cotaBz} that, for every $\gamma <
	\overline{\gamma}$,
	\begin{align*}
		\|x_{z,y}(\gamma) - p \| &\leq \| x_{z,y}(\gamma) -J_{\gamma
			A} z\|
		+ \|J_{\gamma A} z - p \| \\
		& \leq \gamma \|Bz\| + \frac{\delta_p}{2}\\
		&< \delta_p,
	\end{align*}
	which yields $(x_{z,y}(\gamma))_{0<\gamma\leq
		\overline{\gamma}}
	\subset{\mathcal{B}(p,\delta_p)}$ and, thus, $(\|B_3
	z-B_3x_{z,y}(\gamma)\|
	)_{0<\gamma\leq \overline{\gamma}}$ is bounded. Therefore,
	since $z\in \H 	\setminus 	\mathcal{C}$ implies
	$\varphi_{z,\gamma} (\gamma) \to + \infty$ as $\gamma
	\downarrow 0$,  the result follows.

	\ref{prop:varphiz33}: Since $p=P_{\overline{\textnormal{dom}\,} A} z \notin
	\ensuremath{\text{\rm bdry}\,} \textnormal{dom}\, B_3$,  $B_3$ is 	locally bounded at $p$
	\cite[Theorem~21.18]{bauschkebook2017} and
	the result follows from
	\ref{prop:varphiz32}.\qed
\end{proof}

\begin{remark} \begin{enumerate}
		\item In the case $B_2=0$, by setting $y=(B_1+B_3)z$ in
		Lemma~\ref{prop:varphiz}\eqref{prop:varphiz1}\&\eqref{prop:varphiz2},
		we recover \cite[Lemma 2.2(1)]{BricenoDavis2018}.
		\item Realizing that \cite[Lemma 2.2(2)]{BricenoDavis2018} is
		valid for every $z \in \textnormal{dom}\, A$, it is a particular case of
		Lemma~\ref{prop:varphiz}\eqref{prop:varphiz23}\&\eqref{prop:varphiz31}.
	\end{enumerate}
\end{remark}
Now we state our main result.
\begin{teo}\label{teo:convergencia}
	In the context of Problem~\ref{prob:problem1},
	suppose that one of the following holds:
	\begin{enumerate}
		\item \label{teo:hyp1} $X \subset \textnormal{dom}\, A$.

		\item \label{teo:hyp2}  $\overline{\textnormal{dom}\,} A \subset \textnormal{dom}\, B_3$
		and
		$B_3$ is locally bounded in $\textnormal{dom}\, B_3$.

		\item \label{teo:hyp3}  $\overline{\textnormal{dom}\,} A \subset \ensuremath{\operatorname{int}} \textnormal{dom}\,
		B_3$.

	\end{enumerate} Let $ \varepsilon
	\in \left]0, 1\right[$, set $\rho= \min\{2 \beta \varepsilon,
	\sqrt{1-\varepsilon}/L\}$, let $\sigma \in \left]0,1\right[$, let
	$\theta
	\in \left]0,\sqrt{1-\varepsilon} -L\rho\sigma\right[$, let $z_0
	\in \textnormal{dom}\, B_3$, and
	consider the sequence $(z_n)_{n\in \mathbb N}$ defined by the
	recurrence
	\begin{equation}\label{eq:algorithm}
		(\forall n\in\mathbb N)\quad
		\begin{array}{l}
			\left\lfloor
			\begin{array}{l}
				x_n= J_{\gamma_n A}\big(z_n-\gamma_n
				(B_1+B_2+B_3)z_n\big)\\
				z_{n+1}= P_X\big(x_n + \gamma_n
				(B_2+B_3)z_n-\gamma_n
				(B_2+B_3)x_n\big),
			\end{array}
			\right.
		\end{array}
	\end{equation}
	where, for every $n \in \mathbb N$, $\gamma_n$ is the
	largest $\gamma \in \{ \rho \sigma,  \rho \sigma^2,
	\rho \sigma^3, \cdots \}$ satisfying
	\begin{equation}\label{eq:thetateo}
		\gamma \big\|B_3z_n-B_3 J_{\gamma A}\big(z_n - \gamma
		(B_1+B_2+B_3)z_n\big)\big\|
		\leq \theta\big\|z_n-J_{\gamma A} \big(z_n - \gamma
		(B_1+B_2+B_3)z_n\big)\big\|.
	\end{equation}
	Moreover, assume that at least one of the following additional
	statements hold:
	\begin{enumerate}[label=(\roman*)]
		\item \label{teo:hypa} $\displaystyle \liminf_{n \to \infty}
		\gamma_n =
		\delta >0$.
		\item \label{teo:hypb} $B_3$ is uniformly continuous in any
		weakly
		compact subset of $\overline{\ensuremath{\text{\rm conv}\,}} (\textnormal{dom}\, A \cup X)$.
	\end{enumerate}

	Then, $(z_n)_{n \in \mathbb N}$ converges weakly to a solution to
	Problem~\ref{prob:problem1}.

\end{teo}

\begin{proof}

	Set $B=B_1+B_2+B_3$ and fix $n \in \mathbb N$.

	If $z_n \in \mathcal{C}$, where $\mathcal{C}$ is defined in
	\eqref{eq:defsetC}, then $\gamma_n$ is well defined in view of
	Lemma~\ref{prop:varphiz}\eqref{prop:varphiz31}. In particular, if
	\ref{teo:hyp1} holds, $\gamma_n$ is well defined in view of
	Lemma~\ref{prop:varphiz}\eqref{prop:varphiz23}. Now suppose
	that
	$z_n
	\in \H \setminus \mathcal{C}$. If $n=0$, it is clear that
	$z_0
	\in \textnormal{dom}\, B_3 \setminus \mathcal{C}$. Otherwise, since $X
	\subset
	\textnormal{dom}\, B_3$, we have
	$z_n
	\in \textnormal{dom}\, B_3 \setminus \mathcal{C}$. Now, if we assume
	\ref{teo:hyp2}, then $B_3$ is locally bounded in
	$P_{\overline{\textnormal{dom}\,}
		A}z_n\in\overline{\textnormal{dom}\,} A$ and $\gamma_n$ is well defined
	from
	Lemma~\ref{prop:varphiz}\eqref{prop:varphiz32}. Similarly, if we
	assume \ref{teo:hyp3}, $\gamma_n$ is well defined from
	Lemma~\ref{prop:varphiz}\eqref{prop:varphiz33}.

	Now, let $z^* \in \textnormal{zer}(A+B)\cap X$. Note that the maximal
	monotonicity of
	$A+B_3$, the full domain of $B_1$ and $B_2$, and \cite[Corollary
	25.5(i)]{bauschkebook2017} imply that $A+B_2+B_3$ and
	$A+B$
	are
	maximally
	monotone. Then, since
	$B_2+B_3$ is continuous and single valued in $\textnormal{dom}\,
	(B_2+B_3)=\textnormal{dom}\,
	B_3 \supset \textnormal{dom}\, A \cup X$ and $B_1$ is $\beta$-cocoercive, it
	follows from \cite[Proposition
	2.1(1)\&(2)]{BricenoDavis2018} that, for every $n \in \mathbb N$,
	we have
	\begin{align}\label{eq:LemmaDavis}\nonumber
		\|z_{n+1}- z^*\|^2 \leq & \|z_{n}- z^*\|^2-(1-\varepsilon)
		\|z_n-x_n\|^2+\gamma_n^2\|(B_2+B_3)z_n-(B_2+B_3)x_n\|^2\\
		& \hspace*{5cm} -\frac{\gamma_n}{\varepsilon}(2\beta
		\varepsilon
		- \gamma_n)\|B_1z_n-B_1z^*\|^2.
	\end{align}
	Note that the Lipschitz property of $B_2$ and
	\eqref{eq:thetateo} yield
	\begin{align}\label{eq:LemmaDavis2}
		\gamma_n^2\|(B_2+B_3)z_n-(B_2+B_3)x_n\|^2  & \leq
		(L\gamma_n\|z_n-x_n\|+\gamma_n\|B_3z_n-B_3x_n\|)^2
		\nonumber \\
		& \leq (L\gamma_n+\theta)^2\|z_n-x_n\|^2  \nonumber\\
		& \leq  (L\rho \sigma +\theta)^2\|z_n-x_n\|^2.
	\end{align}
	Hence, it follows from \eqref{eq:LemmaDavis} and
	\eqref{eq:LemmaDavis2} that
	\begin{align*}
		(\forall n \in \mathbb N)\quad	\|z_{n+1}- z^*\|^2
		\leq & \|z_n-z^*\|^2-\big((1-\varepsilon)-(L\rho \sigma
		+\theta)^2\big)\|z_n-x_n\|^2.

	\end{align*}
	Therefore, since $(1-\varepsilon-(L \rho \sigma +\theta)^2)>0$,

	\cite[Lemma~3.1(i)]{Comb2001} implies that
	$(\|z_n-z^*\|)_{n \in \mathbb N}$ is a convergent sequence and
	\begin{equation}
		\label{e:zxto0}
		z_n-x_n \to 0.
	\end{equation}
	Let $z \in \H$ be a weak limit point of the
	subsequence
	$(z_n)_{n\in K}$ for some $K \subset \mathbb N$. Then $z$ is also a
	weak
	limit point of
	$(x_n)_{n\in K}$ in view of \eqref{e:zxto0}. Since $X$ is closed
	and convex,
	and
	$(z_n)_{n\in K}$ is a sequence in $X$, we conclude that $z
	\in X$. Let us prove that $z \in \textnormal{zer} (A+B)$.

	\ref{teo:hypa}: Assume that $\liminf_{n\to
		+\infty}\gamma_n=\delta
	>0$.
	Then, there exists $n_0 \in \mathbb N$ such that $\inf_{n \geq n_0}
	\gamma_n
	\geq \delta$. Hence, \eqref{eq:thetateo}, \eqref{eq:algorithm},
	the
	Lipschitz
	continuity of $B_2$, and the
	cocoercivity of $B_1$ yield

	\begin{align}\nonumber
		(\forall n \geq  n_0)	\quad \|B z_n- B x_n\| &\leq \|B_1
		z_n - B_1
		x_n\| +
		\|B_2 z_n - B_2 x_n\|
		+ \|B_3 z_n - B_3x_n\|\\\label{eq:con_z-x}
		&\leq \left(\frac{1}{\beta}+L+\frac{\theta}{\delta}\right)
		\|z_n-x_n\|,
	\end{align}
	which implies $B z_n -B x_n \to 0$ in view of \eqref{e:zxto0}.
	Hence,
	it follows from
	\eqref{eq:algorithm} that
	\begin{equation} \label{eq:contencion1}
		(\forall n \in \mathbb N) \quad
		u_n:=\frac{z_{n}-x_n}{\gamma_n}-Bz_n+Bx_n
		\in
		(A+B)x_n,
	\end{equation}
	and \eqref{e:zxto0} and $\liminf_{n\to +\infty}\gamma_n=\delta
	>0$
	imply that $u_n \to 0$.
	Therefore, since $x_n \rightharpoonup z$, $n \in K$,
	the weak-strong closure of the graph of the maximally monotone
	operator $A+B$ and
	\eqref{eq:contencion1} yield $z \in \textnormal{zer}(A+B)$. The
	convergence follows from
	\cite[Lemma~2.47]{bauschkebook2017}.

	\ref{teo:hypb}: Without loss of generality, suppose
	that
	$\lim_{n \to \infty, n \in K } \gamma_n=0$. Our choice of
	$\gamma_n$
	guarantee that, for every
	$n
	\in
	K$, we have
	\begin{equation} \label{eq:destheta3}
		\widetilde{\gamma}_n \|B_3z_n-B_3 J_{\widetilde{\gamma}_n A}
		(z_n
		-
		\widetilde{\gamma}_n Bz_n)\| >
		\theta\|z_n-J_{\widetilde{\gamma}_n A} (z_n
		- \widetilde{\gamma}_n Bz_n)\|,
	\end{equation}
	where
	$\widetilde{\gamma}_n:=\frac{\gamma_n}{\sigma}>\gamma_n$.
	Now, by the nonincreasing property of $\gamma \mapsto
	\frac{1}{\gamma}\|z-J_{\gamma A} (z - \gamma Bz)\|$ provided
	by
	Lemma~\ref{prop:varphiz}\eqref{prop:varphiz1} with $y =
	B
	z$, we have
	\begin{align}\label{eq:nonincreasinggamma}
		\frac{1}{\widetilde{\gamma}_n}
		\|z_n-J_{\widetilde{\gamma}_n A}
		(z_n - 	\widetilde{\gamma}_n Bz_n)\| \leq
		\frac{1}{\gamma_n}\|z_n-J_{\gamma_n A} (z_n - \gamma_n
		Bz_n)\|,
	\end{align}
	which is equivalent to
	\begin{equation}
		\|z_n-J_{\widetilde{\gamma}_n A} (z_n - \widetilde{\gamma}_n
		Bz_n)\| \leq
		\frac{1}{\sigma}\|z_n-x_n\|.
	\end{equation}
	Thus, by defining
	\begin{equation}\label{eq:xtilde}
		(\forall n \in K) \quad
		\widetilde{x}_n=J_{\widetilde{\gamma}_n	A}
		(z_n-\widetilde{\gamma}_nBz_n),
	\end{equation}
	\eqref{e:zxto0} implies that
	\begin{equation}
		\label{e:auxseqtoz}
		z_n-\widetilde{x}_n \to 0\quad\text{as}\quad n \to\infty, n
		\in
		K.
	\end{equation}
	Therefore, since $z_n \rightharpoonup z$, $n \in K$, we have
	$\widetilde{x}_n \rightharpoonup z$ as $n \to\infty$, $n \in K$.
	Furthermore,
	\eqref{eq:xtilde} yields
	\begin{equation}\label{eq:contencion2}
		\frac{z_n-\widetilde{x}_n}{\widetilde{\gamma}_n}+
		B\widetilde{x}_n-Bz_n \in
		(A+B)\widetilde{x}_n.
	\end{equation}
	Since $\{z\} \cup \bigcup_{n \in K} [\widetilde{x}_n,z_n]$ is a
	weakly
	compact subset of $ \overline{\ensuremath{\text{\rm conv}\,}} (\textnormal{dom}\, A \cup X)$
	\cite[Lemma
	3.2]{salzo2017}, it follows from
	the uniform continuity of $B_3$ and \eqref{e:auxseqtoz} that
	\begin{equation}\label{eq:b3tozero}
		B_3\widetilde{x}_n-B_3z_n \to 0 \text{ as }
		n \to \infty,\ n \in K,
	\end{equation}
	which, combined with \eqref{eq:destheta3}, yields
	$(z_n-\widetilde{x}_n)/\widetilde{\gamma}_n \to 0\text{ as
	} n \to \infty,\ n \in K$. Moreover,
	the Lipschitz continuity of $B_1+B_2$, \eqref{e:auxseqtoz}, and
	\eqref{eq:b3tozero}
	imply
	\begin{equation}
		B\widetilde{x}_n-Bz_n \to 0 \text{ as } n \to \infty,\ n \in
		K.
	\end{equation}
	Altogether, the convergence follows, as in the case
	\ref{teo:hypa},
	from \eqref{eq:contencion2}, the weak-strong closedness of the
	graph
	of
	the maximally
	monotone operator $A+B$, and
	\cite[Lemma~2.47]{bauschkebook2017}.\qed
\end{proof}
\begin{remark}\label{rem:pct}
	\begin{enumerate}
		\item In Theorem~\ref{teo:convergencia}, if $B_3=0$, we have
		$\textnormal{dom}\,
		B_3= \H$ and, for all $n
		\in \mathbb N$, $\gamma_n=\sigma \rho = \sigma
		\min\{2\beta\varepsilon,\sqrt{1-\varepsilon}/L\}$. Since, in
		this
		case $(\gamma_n)_{n\in\mathbb N}$ is constant, the largest step-size
		is
		obtained by taking $\varepsilon=\varepsilon(L,\beta)
		=2/(1+\sqrt{1+16\beta^2L^2})$, which satisfies
		$2\beta\varepsilon =
		\sqrt{1-{\varepsilon}}/L = \chi(L,\beta)$, where
		\begin{equation}
			\chi(L,\beta)= \frac{4\beta}{1+\sqrt{1+16\beta^2L^2}},
		\end{equation}
		and $\gamma_n \equiv \gamma=  \sigma \chi(L,\beta)
		\in \left]0, \chi(L,\beta)
		\right[$. Hence,
		we recover the result in
		\cite[Theorem~2.3(1)]{BricenoDavis2018} for constant
		stepsizes.
		Additionally, if $B_2=0$ and $X=\H$, we have
		$\varepsilon(L,\beta)
		\to 1$ and $\chi(L,\beta)\to 2\beta$ as $ L \to 0$ and
		$\gamma_n \equiv 2\beta \sigma \in \left]0, 2\beta\right[$,
		recovering the the
		forward
		backward algorithm \cite{lionsandmercier1979}.
		On the other hand, if $B_1=0$, we have $\chi(L,\beta) \to
		1/L$
		as $\beta \to \infty$ and
		$\gamma_n \equiv \sigma/L \in \left]0,1/L\right[$, recovering
		the
		result in
		\cite{Tseng2000SIAM} for constant step-sizes.
		\item Suppose that $B_2=0$ and $X\subset \textnormal{dom}\, A$. Then by
		taking $L
		\to 0$, we have $\rho \to 2\beta \varepsilon$ and $\theta \in
		\left]0,\sqrt{1-\varepsilon}\right[$. Hence,
		Theorem~\ref{teo:convergencia}
		recovers \cite[Theorem~2.3(2)]{BricenoDavis2018} noting that
		the
		uniform continuity in weakly compact subsets of
		$\overline{\ensuremath{\text{\rm conv}\,}}(\textnormal{dom}\, A \cup X)=\overline{\textnormal{dom}\,} A$ is needed.
		We
		hence
		generalize
		\cite[Theorem~2.3(1)\&(2)]{BricenoDavis2018} to the case when
		$X\not\subset  \textnormal{dom}\, A$.
	\end{enumerate}
\end{remark}

\section{Application to Convex Optimization with Nonlinear
	Constraints}\label{sec:4}

In this section we consider the following optimization problem.
\begin{pro}\label{prob:opti}
	Let $f\in\Gamma_0 (\H)$, let $g\in \Gamma_0(\mathcal G)$, let $h:\H \to
	\mathbb R$
	be a convex G\^ateaux differentiable
	function such that $\nabla h$ is $\beta^{-1}$-Lipschitzian for
	some $\beta \in \ensuremath{\left]0,+\infty\right[}$, let $M \colon \H \to \mathcal G$ be a bounded
	linear
	operator, and let $e \colon \H \to \ensuremath{\left]-\infty,+\infty\right]}^p \colon x
	\mapsto
	(e_i(x))_{1\le i
		\le p}$ be such that, for every $i\in \{1,\ldots,p\}$, $e_i$
	is
	convex and
	G\^ateaux differentiable in $\ensuremath{\operatorname{int}} \textnormal{dom}\, e_i$, $\textnormal{dom}\, e_i$ is
	closed,
	$\cap_{i=1}^p
	\textnormal{int} \textnormal{dom}\,
	e_{i}\neq \varnothing$, and $\textnormal{dom}\, \partial f \subset
	\cap_{i=1}^n \textnormal{int} \textnormal{dom}\,  e_i $. Assume that $0 \in
	\ensuremath{\text{\rm sri}\,}(\textnormal{dom}\, g - M (\textnormal{dom}\, f))$ and that
	\begin{equation}\label{e:slater}
		\begin{cases}
			(\forall i \in \{1,\ldots,p\}) \quad \quad \textnormal{lev}_{\leq 0}
			e_i \subset
			\ensuremath{\operatorname{int}} \textnormal{dom}\, e_i;\\
			\textnormal{dom}\, (f + g \circ M) \cap \bigcap_{i=1}^n \textnormal{lev}_{<0} e_i
			\neq \varnothing.
		\end{cases}
	\end{equation} The problem is to
	\begin{equation}\label{eq:probopti}
		\min_{e(x)\in \ensuremath{\left]-\infty,0\right]}^p} f(x) + g (Mx) + h(x),
	\end{equation}
	and we assume that solutions exist.
\end{pro}
It follows from  \eqref{e:slater}
and \cite[Proposition~27.21]{bauschkebook2017} that $\hat{x} \in
\H$ is a solution to
Problem~\ref{prob:opti} if and only if there exists $\hat{v} \in
\ensuremath{\left[0,+\infty\right[}^p$ such that
\begin{equation}
	-\sum_{i=1}^p \hat{v}_i\nabla e_i(\hat{x}) \in \partial
	(f+g\circ M + h) \quad
	\text{ and } \quad	(\forall i \in \{1,\ldots,p\}) \quad
	\begin{cases}
		e_i(\hat{x}) \leq 0,\\
		\hat{v}_i e_i(\hat{x}) =0.
	\end{cases}
\end{equation}
Hence, by
\cite[Example~16.13\,\&\,Example~6.42(i)]{bauschkebook2017},
we deduce that

	\begin{equation}\label{eq:inclulag}
		\begin{cases}
			0 \in \partial ( f + g \circ M + h) (\hat{x}) +
			\sum_{i=1}^p
			\hat{v}_i
			\nabla e_i (\hat{x}) \\
			0 \in  N_{\ensuremath{\left[0,+\infty\right[}^p} (\hat{v}) - e(\hat{x}).
		\end{cases}
	\end{equation}
	Then, we deduce from \cite[Theorem
	16.47]{bauschkebook2017}
	that there exists $\hat{u}
	\in \mathcal G$ such that
	\eqref{eq:inclulag} reduces to
	\begin{equation*
		\begin{cases}
			0\in \partial  f (\hat{x}) + M^*\hat{u} + \nabla h
			(\hat{x})  +
			\sum_{i=1}^p \hat{v}_i \nabla e_i (\hat{x})   \\
			0  \in \partial g^*(\hat{u})-M\hat{x}\\
			0 \in  N_{\ensuremath{\left[0,+\infty\right[}^p} (\hat{v}) - e(\hat{x}),
		\end{cases}
	\end{equation*}
	which is equivalent to
	\begin{equation}\label{eq:incluAB}
		\begin{pmatrix}
			0\\
			0\\
			0
		\end{pmatrix} \in \begin{pmatrix}
			\partial f (\hat{x})\\
			\partial g^*(\hat{u})\\
			N_{\ensuremath{\left[0,+\infty\right[}^p} (\hat{v} )
		\end{pmatrix}
		+\begin{pmatrix}
			\nabla h (\hat{x})\\
			0\\
			0
		\end{pmatrix}+
		\begin{pmatrix}
			M^* \hat{u} \\
			-M \hat{x}\\
			0
		\end{pmatrix}
		+ \begin{pmatrix}
			\sum_{i=1}^p\hat{v}_i \nabla e_i (\hat{x})  \\\
			0\\
			-e(\hat{x})
		\end{pmatrix}.
	\end{equation}

	\begin{prop}\label{prop:B3cont}
		In the context of Problem~\ref{prob:opti}, let $X=X_1\times
		X_2 \times
		X_3  \subset \textnormal{dom}\, \partial f \times \textnormal{dom}\, \partial g^* \times
		\ensuremath{\left[0,+\infty\right[}^p$ be
		nonempty, closed, and convex, and define the operator
		\begin{align}\label{eq:defB3prop} \displaystyle
			B_3 \colon \H \times \mathcal G \times \mathbb R &\to 2^{\H \times \mathcal G
				\times \mathbb R}
			\nonumber\\
			(x,u,v) &\mapsto \begin{cases}
				\!\!\left\{\!\left(\displaystyle{\sum_{i=1}^p} v_i
				\nabla  e_i (x) ,
				0,
				-e(x)\right)\!\!\right\},\hspace{-.3cm}& \text{ if }
				v \in
				\ensuremath{\left[0,+\infty\right[}^p \text{ and } x \in \bigcap_{i=1}^p \textnormal{int}
				\textnormal{dom}\,
				e_{i};\\
				\varnothing,& \text{ otherwise.} \\
			\end{cases}
		\end{align}
		Then, the following hold:
		\begin{enumerate}
			\item \label{prop:B3cont1} $B_3$ is maximally monotone.

			\item \label{prop:B3cont2} Suppose that one of the
			following
			holds:

			\begin{enumerate}
				\item\label{prop:hip1} $(\nabla e_i)_{1\leq i \leq
					p}$ are
				bounded and uniformly
				continuous in every weakly compact subset of
				$\overline{\textnormal{dom}\,} \partial f$.
				\item\label{prop:hip2} $\H$ is finite dimensional
				and
				$(\nabla e_i)_{1\leq i \leq p}$ are
				continuous in every compact subset of
				$\overline{\textnormal{dom}\,} \partial
				f$.
			\end{enumerate}
			Then,
			$B_3$ is uniformly continuous in every compact subset of
			$\overline{\textnormal{dom}\,} \partial f \times \overline{\textnormal{dom}\,}
			\partial
			g^*
			\times
			\ensuremath{\left[0,+\infty\right[}^p$.
		\end{enumerate}

	\end{prop}

	\begin{proof} \ref{prop:B3cont1}: Consider the saddle-function
		\begin{align*}
			\ell \colon \H \times \mathcal G \times  \mathbb R^p & \to \ensuremath{\left[-\infty,+\infty\right]}\\
			(x,u,v) &\mapsto\begin{cases} e(x)\cdot v, & \text{
					if } v \in 	\ensuremath{\left[0,+\infty\right[}^p \text{ and } x \in
				\bigcap_{i=1}^p
				\textnormal{dom}\, e_{i};\\
				+\infty, &\text{ if } v \in 	\ensuremath{\left[0,+\infty\right[}^p \text{ and } x
				\notin
				\bigcap_{i=1}^p \textnormal{dom}\,
				e_{i};		\\
				-\infty, &\text{ if } v \notin \ensuremath{\left[0,+\infty\right[}^p.
			\end{cases}
		\end{align*}
		Note that, if $v \in \ensuremath{\left[0,+\infty\right[}^p$,
		\begin{equation}
			\ell\colon(x,u,v) \mapsto \begin{cases}
				e(x)\cdot v,
				&
				\text{
					if } x \in \bigcap_{i=1}^p
				\textnormal{dom}\, e_{i};\\
				+\infty, &\text{ otherwise}
			\end{cases}
		\end{equation}
		and, if  $v \notin \ensuremath{\left[0,+\infty\right[}^p$,
		$\ell(\cdot,\cdot,v)\equiv-\infty$. Hence,
		for every $v
		\in \mathbb R^p$, $\ell(\cdot, \cdot , v)$ is lower-semicontinuous.
		Additionally,
		if $(x,u) \in \bigcap_{i=1}^p \textnormal{dom}\, e_i\times\mathcal G$, we have
		\begin{equation}
			-\ell\colon(x,u,v)\mapsto \begin{cases}
				-e(x)\cdot v,
				&
				\text{
					if } v \in 	\ensuremath{\left[0,+\infty\right[}^p ;\\
				+\infty, &\text{ otherwise}
			\end{cases}
		\end{equation}
		and, if $x \notin \bigcap_{i=1}^p \textnormal{dom}\, e_i$ and $u\in\mathcal G$,
		\begin{equation}
			-\ell\colon(x,u,v)\mapsto\begin{cases}
				-\infty &
				\text{ if } v \in 	\ensuremath{\left[0,+\infty\right[}^p ;\\
				+\infty, &\text{ otherwise}.
			\end{cases}
		\end{equation}
		Therefore, for every $(x,u) \in \H\times \mathcal G$,
		$-\ell(x,u,\cdot)$ is lower-semicontinuous. Furthermore,
		\begin{equation}
			(\forall (x,u) \in \H\times \mathcal G) (\forall v \in \mathbb R^p)
			\quad B_3(x,u,v) =
			\partial \ell(\cdot,\cdot,v)(x,u) \times  \partial
			(-\ell(x,u,\cdot))(v).
		\end{equation} The result
		follows from \cite[Corollary 1]{Rockafellar70}.

		\ref{prop:B3cont2}: First, assume \ref{prop:hip1}.
		Let $Y=Y_1\times Y_2 \times Y_3 \subset \overline{\textnormal{dom}\,}
		\partial f \times  \overline{\textnormal{dom}\,}  \partial g^* \times
		\ensuremath{\left[0,+\infty\right[}^p$ be a weakly compact set. Let
		$\boldsymbol{x}= (x_1,u_1,v_1)$ and
		$\boldsymbol{y}=(x_2,u_2,v_2)$ in $Y$, fix $i \in
		\{1\ldots,p\}$, define
		$\rho_i \colon \left[0,1\right] \to \mathbb R \colon t
		\mapsto e_i(x_1+t(x_2-x_1))$,
		which is differentiable in $\left]0,1\right[$. Since $Y_1$ is
		weakly compact, by
		\cite[Theorem 3.37]{bauschkebook2017}, $\ensuremath{\text{\rm conv}\,} Y_1$ is also
		weakly
		compact. Moreover, we deduce from the boundedness of
		$\nabla e_i$ in $\ensuremath{\text{\rm conv}\,} Y_1
		\subset
		\overline{\textnormal{dom}\,} \partial f$ that there
		exists $K_i>0$ such that $\sup_{x \in \ensuremath{\text{\rm conv}\,} Y_1}\|\nabla
		e_i(x)\| \leq
		K_i$. Therefore, since ${\rho_i}'\colon t \mapsto
		\scal{\nabla
			e_i(x_1+t(x_2-x_1))}{x_2-x_1}$, we obtain
		\begin{align*}
			|e_i(x_2)-e_i(x_1)| &=|\rho_i(1)-\rho_i(0)|\\
			&= \left|\int_0^1 {\rho_i}'(t)dt\right|\\
			&= \left|\int_0^1\scal{\nabla
				e_i(x_1+t(x_2-x_1))}{x_2-x_1} dt\right|\\
			&\leq \int_0^1 \|\nabla
			e_i(x_1+t(x_2-x_1))\|\|x_2-x_1\|dt\\
			&\leq  K_i \|x_2-x_1\|.
		\end{align*}
		Thus,  we conclude $|e_i(x_1)-e_i(x_2)| \leq K_i \|x_2-x_1\|$
		and therefore
		\begin{equation}\label{eq:eKlip}
			\|e(x_2)-e(x_1)\| \leq K \|x_2-x_1\|,
		\end{equation}
		where $K=\max_{i\in \{1,\ldots,p\}} K_i$.  Since $Y_3$ is
		weakly compact, there
		exists $V >0$ such that $\sup_{v \in Y_3} \|v\| \leq V$
		\cite[Lemma~2.36]{bauschkebook2017}.
		Let $\varepsilon > 0 $. The uniform continuity of $(\nabla
		e_i)_{1\leq i
			\leq p}$ implies
		the existence of $\delta >0$ such that
		\begin{equation}\label{eq:geaco}
			(\forall i \in \{1,\ldots,p\}) (\forall (z_1,z_2) \in
			Y_1^2)\quad
			\|z_1-z_2\|< \delta  \Rightarrow \|
			\nabla e_i (z_1) - \nabla e_i ( z_2)  \|^2 \leq
			\frac{\varepsilon^2}{4 p
				V^2}.
		\end{equation}
		Now suppose that
		\begin{equation*}
			\| \boldsymbol{x} - \boldsymbol{y}  \|^2 \leq \min
			\left\{
			\frac{\varepsilon^2}{4p  K^2}   , \delta^2  \right\}.
		\end{equation*}
		Then, \eqref{eq:eKlip}, the convexity of $\|\cdot\|^2$, and
		\eqref{eq:geaco} imply
		\begin{align*}
			\|B_3 &\boldsymbol{x} - B_3 \boldsymbol{y}\|^2 = \left\|
			\sum_{i=1}^p
			v_{1,i} \nabla e_i(x_1) - v_{2,i} \nabla e_i(x_2)
			\right\|^2 +
			\|e(x_1)-e(x_2) \|^2\\
			&\leq 2  \left\|\sum_{i=1}^p (v_{1,i} -v_{2,i}) \nabla
			e_i(x_1)
			\right\|^2 +
			2  \left\|\sum_{i=1}^p v_{2,i} (\nabla e_i(x_1)-\nabla
			e_i(x_2))
			\right\|^2  +  K^2 \|x_1-x_2\|^2 \\
			&\leq 2p  \sum_{i=1}^p |v_{1,i} -v_{2,i}|^2 \left\|\nabla
			e_i(x_1)
			\right\|^2 +
			2p   \sum_{i=1}^p |v_{2,i}|^2 \left\|\nabla
			e_i(x_1)-\nabla
			e_i(x_2)
			\right\|^2  +  K^2 \|x_1-x_2\|^2 \\
			&\leq   2p  K^2  \left\| v_{1} - v_{2}\right\|^2 + 2p
			V^2
			\sum_{i=1}^p \left\|
			\nabla e_i (x_1) -  \nabla e_i (x_2)\right\|^2    +
			K^2\|x_1-x_2\|^2\\
			&\leq   2p K^2 \left\| \boldsymbol{x} - \boldsymbol{y}
			\right\|^2
			+\frac{\varepsilon^2}{2}\\
			& \leq \varepsilon^2.
		\end{align*}
		Therefore $B_3$ is uniformly continuous in $Y$.

		Now, assume \ref{prop:hip2}. Since $\H$ is finite dimensional
		the
		weak and strong topologies coincide \cite[Fact
		2.33]{bauschkebook2017}. Hence, since  $\nabla e_i$ is
		continuous, it
		is bounded and uniformly continuous in every compact subset
		of $X$.
		The result follows from \ref{prop:hip1}.\qed
	\end{proof}
	\begin{remark}
		Note that if, for every $i \in \{1,\ldots,p\}$, $\nabla e_i$
		is bounded
		and uniformly continuous in every weakly compact subset of
		$\textnormal{dom}\, f$, since
		$\overline{\textnormal{dom}\,}\partial f \subset \textnormal{dom}\,
		f$, $B_3$ is uniformly continuous
		in every compact
		subset of $\overline{\textnormal{dom}\,} \partial f \times \overline{\textnormal{dom}\,}
		\partial g^* \times \ensuremath{\left[0,+\infty\right[}^p$ in view of
		Proposition~\ref{prop:B3cont}.
	\end{remark}
	\begin{prop} In Problem~\ref{prob:opti}, let
		$X=X_1\times X_2 \times X_3  \subset
		\textnormal{dom}\,
		\partial f \times \textnormal{dom}\, \partial g^* \times \ensuremath{\left[0,+\infty\right[}^p$ be
		nonempty,
		closed,
		and convex, let $ \varepsilon
		\in\left]0, 1\right[$, set $\rho= \min\{2 \beta \varepsilon,
		\sqrt{1-\varepsilon}/\|M\|\}$, let $\sigma \in
		\left]0,1\right[$, let
		$\theta
		\in\left]0,\sqrt{1-\varepsilon} -\|M\|\rho\sigma\right[$.

		For every $\boldsymbol{z}=(z^1,z^2,z^3) \in \H \times \mathcal G
		\times \mathbb R^p$
		define
		$\boldsymbol{\Phi}_{\boldsymbol{z}}\colon \gamma \mapsto
		(\Phi_{\boldsymbol{z}}^1
		(\gamma),\Phi_{\boldsymbol{z}}^2(\gamma),\Phi_{\boldsymbol{z}}^3(\gamma))$,
		where
		\begin{align}\label{eq:defx123}\nonumber
			\Phi_{\boldsymbol{z}}^1\colon \gamma&\mapsto
			\ensuremath{\text{\rm prox}}_{\gamma
				f}\left(z^1-\gamma \left( \nabla h (z^1) + M^*z^2 +
			\sum_{i=1}^p
			z_i^3 \nabla e_i (z^1)\right)\right)\\ \nonumber
			\Phi_{\boldsymbol{z}}^2\colon
			\gamma&\mapsto\ensuremath{\text{\rm prox}}_{\gamma
				g^*}
			(z^2+\gamma
			M z^1)\\
			\Phi_{\boldsymbol{z}}^3\colon \gamma&\mapsto
			P_{\ensuremath{\left[0,+\infty\right[}^p}\big(z^3+\gamma e
			(z^1)\big).
		\end{align}
		Let $\boldsymbol{z}_0=(z_0^1,z_0^2,z_0^3)\in \H\times \mathcal G
		\times
		\mathbb R^p$ and consider the 	recurrence, for every $n\in\mathbb N$,
		\begin{equation}\label{eq:algorithmop}
			\begin{array}{l}
				\left\lfloor
				\begin{array}{l}
					x_n^1=\Phi_{\boldsymbol{z}_n}^1(\gamma_n)\\
					x_n^2=\Phi_{\boldsymbol{z}_n}^2(\gamma_n) \\
					x_n^3=\Phi_{\boldsymbol{z}_n}^3(\gamma_n)\\
					z_{n+1}^1= P_{X_1}\big(x_n^1 + \gamma_n (M^*
					z_n^2 +
					\sum_{i=1}^p z_{n,i}^3 \nabla e_i(z_n^1)
					)-\gamma_n (M^*
					x_n^2 + \sum_{i=1}^p  x_{n,i}^3 \nabla e_i(x_n^1)
					)\big)\\
					z_{n+1}^2= P_{X_2} (x_n^2 - \gamma_n M z_n^1 +
					\gamma_n
					M x_n^1) \\
					z_{n+1}^3= P_{X_3} (x_n^3 - \gamma_n e(z_n^1) +
					\gamma_n e  (x_n^1))\\
					\boldsymbol{z}_{n+1}=(z_{n+1}^1,z_{n+1}^2,z_{n+1}^3),
				\end{array}
				\right.
			\end{array}
		\end{equation}
		where $\gamma_n$ is the largest $\gamma \in \{ \rho\sigma,
		\rho\sigma^2, \rho\sigma^3,  \dots \}$ satisfying
		\begin{multline}\label{eq:thetateoop}
			\gamma^2 \left( \left\|\sum_{i=1}^p  z_{n,i}^3\nabla e_i
			(z_n^1)  -
			\Phi_{\boldsymbol{z}_n,i}^3(\gamma) \nabla e
			\big(\Phi_{\boldsymbol{z}_n}^1(\gamma)\big)\right\|^2+
			\left\|
			e(z_{n}^1) -
			e\big(\Phi_{\boldsymbol{z}_n}^1(\gamma)\big)
			\right\|^2\right)\\
			\leq \theta^2
			\pnorm{\boldsymbol{z}_n -
				\boldsymbol{\Phi}_{\boldsymbol{z}_n}(\gamma)}^2.
		\end{multline}
		Moreover, assume that at least one of the following
		additional
		statements hold:
		\begin{enumerate}[label=(\roman*)]
			\item \label{teo:ophypa} $\displaystyle \liminf_{n \to
				\infty}
			\gamma_n = \delta >0$.
			\item \label{teo:ophypb} For every $i \in \{1,
			\ldots,p\}$, $\nabla e_i$
			is bounded and uniformly continuous in every weakly
			compact
			subset of $\overline{\textnormal{dom}\,}\partial f$.
		\end{enumerate}
		Then, $(z_n^1)_{n \in \mathbb N}$ converges weakly to a solution to
		Problem~\ref{prob:opti}.
	\end{prop}

	\begin{proof}
		Let ${\bm{\mathcal{H}}} = \H \times \mathcal G \times \mathbb R^p$, define
		\begin{equation}\label{eq:defoperators}
			\begin{cases}
				A \colon {\bm{\mathcal{H}}}\to 2^{{\bm{\mathcal{H}}}} \colon (x,u,v) \mapsto
				\partial f (x) \times
				\partial g^*(u) \times N_{\ensuremath{\left[0,+\infty\right[}^p}(v),\\
				B_1 \colon {\bm{\mathcal{H}}}\to {\bm{\mathcal{H}}} \colon (x,u,v) \mapsto (\nabla
				h (x) , 0 , 0),\\
				B_2 \colon {\bm{\mathcal{H}}}\to {\bm{\mathcal{H}}} \colon (x,u,v) \mapsto
				(M^*u,-Mx,0),\\
			\end{cases}
		\end{equation}
		and consider the operator $B_3$ defined in
		\eqref{eq:defB3prop}.
		Note that $A$ is maximally monotone \cite[Proposition 20.23
		\&
		Proposition 20.25]{bauschkebook2017}, $B_1$ is
		$\beta$-cocoercive
		\cite[Corollary 18.17]{bauschkebook2017}, $B_2$ is
		$\|M\|$-Lipschitzian
		\cite[Proposition~2.7(ii)]{briceno2011SIAM} \&
		\cite[Fact~2.20]{bauschkebook2017}, and the operator  $B_3$
		is
		maximally
		monotone by Proposition~\ref{prop:B3cont}. Furthermore, note
		that
		$\textnormal{dom}\, A= \textnormal{dom}\, (\partial f)\times \textnormal{dom}\,(\partial g^*) \times
		\ensuremath{\left[0,+\infty\right[}^p$ and
		$\textnormal{dom}\, B_3=\left(\cap_{i=1}^n \textnormal{int}\textnormal{dom}\,  e_i\right)\times
		\mathcal G
		\times \ensuremath{\left[0,+\infty\right[}^p$. Hence, it follows from $\textnormal{dom}\, (\partial f)
		\subset
		\cap_{i=1}^n \textnormal{int} \textnormal{dom}\,  e_i $ that $\textnormal{dom}\, A \cup
		X \subset \textnormal{dom}\, B_3$ and $0 \in \textnormal{int}
		\left(\textnormal{dom}\, A-\textnormal{dom}\, B_3\right)$.  Therefore, $A+B_3$ is
		maximally monotone
		\cite[Corollary~25.5(ii)]{bauschkebook2017}.
		Altogether, the inclusion in \eqref{eq:incluAB} is a
		particular
		instance of Problem~\ref{prob:problem1}. Define, for every $n
		\in
		\mathbb N$, $\boldsymbol{x}_n=(x_n^1,x_n^2,x_n^3)$. Hence,
		\eqref{eq:defx123}, \eqref{eq:defB3prop}, and
		\eqref{eq:defoperators}
		yield
		\begin{equation}\label{eq:algorithmnegrita}
			(\forall n\in\mathbb N)\quad
			\begin{array}{l}
				\left\lfloor
				\begin{array}{l}
					\boldsymbol{x}_n= J_{\gamma_n
						A}(\boldsymbol{z}_n-\gamma_n
					(B_1+B_2+B_3)\boldsymbol{z}_n)\\
					\boldsymbol{z}_{n+1}= P_X (\boldsymbol{x}_n +
					\gamma_n
					(B_2+B_3)\boldsymbol{z}_n-\gamma_n
					(B_2+B_3)\boldsymbol{x}_n),
				\end{array}
				\right.
			\end{array}
		\end{equation}
		where $\gamma_n$, by \eqref{eq:thetateoop} and
		\eqref{eq:defoperators}, satisfies
		\begin{equation}\label{eq:thetateodem}
			\gamma \|B_3\boldsymbol{z}_n-B_3 J_{\gamma A}
			(\boldsymbol{z}_n
			- \gamma (B_1+B_2+B_3)\boldsymbol{z}_n)\| \leq
			\theta\|\boldsymbol{z}_n-J_{\gamma A} (\boldsymbol{z}_n -
			\gamma
			(B_1+B_2+B_3)\boldsymbol{z}_n)\|.
		\end{equation}
		Note that, if we assume \ref{teo:ophypb}, by
		Proposition~\ref{prop:B3cont}, $B_3$ is uniformly continuous
		in
		every weak compact subset of $\overline{\textnormal{dom}\,}\partial f
		\times \overline{\textnormal{dom}\,} \partial g^* \times \ensuremath{\left[0,+\infty\right[}^p =
		\overline{\ensuremath{\text{\rm conv}\,}}(\textnormal{dom}\, A ) =
		\overline{\ensuremath{\text{\rm conv}\,}}(\textnormal{dom}\, A\cup X)$ \cite[Corollary~21.14 \&
		Exercise~3.2]{bauschkebook2017}. In view of $0 \in \ensuremath{\text{\rm sri}\,} (\textnormal{dom}\,
		g - M(\textnormal{dom}\, f))$,
		\eqref{e:slater}, and \cite[Proposition
		27.21]{bauschkebook2017}, there exists a solution
		$\boldsymbol{\hat{z}}=(\hat{x},\hat{u},\hat{v})
		\in \H\times \mathcal G \times \mathbb R^p$ to
		\eqref{eq:incluAB} such that $\hat{x}$ is a solution to
		Problem~\ref{prob:opti}. Altogether, since $X
		\subset \textnormal{dom}\, A$,
		Theorem~\ref{teo:convergencia} implies that
		$\boldsymbol{z}_n
		\rightharpoonup \boldsymbol{\hat{z}}$ and the result follows.\qed
	\end{proof}
	\section{Numerical Experiments}\label{sec:5}
	In this section we consider the following optimization problem

	\begin{equation}\label{eq:numericproblem}
		\min_{\substack{ {y}^0 \leq {x}\leq {y}^1\\
				x_i(\ln (x_i/a_i) -1 ) -r_i \le 0, \\
				i \in \{1, \ldots, n\}}}  \alpha \|M x\|_1 +
		\frac{1}{2}\|Ax-z\|^2,
	\end{equation}
	where $M \in \mathbb R	^{{r}\times {n}}$, $A\in \mathbb R^{m\times n}$,
	$z \in \mathbb R^m$, ${y}^0=(\eta_i^0)_{1\le i\le n} \in \mathbb R^n$,
	${y}^1=(\eta_i^1)_{1\le i\le N} \in \mathbb R^N$, and, for every $i \in
	\{1,
	\ldots, p\}$, $r_i \in
	\left]-a_i
	,
	0\right[$ and $a_i \in \ensuremath{\left]0,+\infty\right[}$. Set
	\begin{equation}
		\begin{cases}
			C=\times_{i=1}^n[\eta_i^0,\eta_i^1],\\
			f=\iota_{C},\\
			g=\alpha \|\cdot\|_1,\\
			h=\|A\cdot -z\|^2/2,\\
			e=(e_i(\cdot))_{i=1}^n,
		\end{cases}
	\end{equation}
	where
	\begin{equation}
		(\forall i=1,\ldots,n) \quad e_i \colon \mathbb R^n \to
		\ensuremath{\left]-\infty,+\infty\right]} \colon
		x \mapsto \begin{cases}
			x_i(\ln (x_i/a_i)-1)-r_i,& \text{ if } x_i > 0;\\
			-r_i,  &\text{ if } x_i = 0 ;\\
			+\infty, &\text{ otherwise}.
		\end{cases}
	\end{equation}
	Then, we have $f \in \Gamma_0(\mathbb R^n)$, $g \in
	\Gamma_0(\mathbb R^r)$, and $\nabla h$ is $\|A\|^2$-Lipschitizian.
	Additionally, for every $i \in \{1,\ldots,n\}$, $e_i$ is
	G\^ateaux
	differentiable in $\ensuremath{\left]0,+\infty\right[}^n$,
	\begin{equation}\label{eq:nablaei}
		(\nabla e_i (x))_k=	\begin{cases}
			\ln x_k ,& \text{ if } k =i; \\
			0, & \text{ otherwise},
		\end{cases}
	\end{equation}
	$\textnormal{dom}\, e_i$ is closed,
	$\cap_{i=1}^n \textnormal{dom}\, e_i =
	\ensuremath{\left[0,+\infty\right[}^n$, \eqref{e:slater} holds, and
	\begin{equation*}
		0 \in
		\textnormal{int}\left(\textnormal{dom}\,(\partial f)-\cap_{i=1}^n \textnormal{dom}\,
		e_i\right)=\times_{i=1}^N\left]-\infty,\eta_i^1\right[.
	\end{equation*} Hence,
	the
	optimization
	problem in
	\eqref{eq:numericproblem} is a particular instance of
	Problem~\ref{prob:opti}. In this setting, since
	$g^*=\iota_{[-\alpha,\alpha]^r}$
	\cite[Example~13.32(v) \& Proposition
	13.23(i)]{bauschkebook2017}, we
	consider $X_1\times
	X_2 \times X_3 = C\times [-\alpha,\alpha]^r \times \ensuremath{\left[0,+\infty\right[}^p$ in
	order to
	write
	the recurrence in \eqref{eq:algorithmop} as
	Algorithm~\ref{alg:opti}
	below.

	\begin{algorithm}
		\caption{}
		\label{alg:opti}
		\begin{algorithmic}[1]
			\STATE{Fix $\boldsymbol{z}_0=(z_0^1, z_0^2,z_0^3) \in
				\mathbb R^n
				\times \mathbb R^r \times \mathbb R^n $. Let $\sigma \in
				\left]0,1\right[$, let
				$\varepsilon
				=\|A\|^4
				\frac{\sqrt{1+16\|M\|^2/\|A\|^4}-1}{8\|M\|^2}$, let
				$\theta =
				2\varepsilon\|M\|(1-\sigma)/\|A\|^2$, and let $\epsilon
				>0$.}
			\WHILE{ $r_n > \epsilon $}
			\STATE{$\gamma=2\varepsilon \|M\|$}
			\STATE{{\tt V} $=0$}
			\WHILE{ {\tt V} $=0$}
			\STATE{$\gamma \to \gamma \cdot \sigma$}
			\STATE{$\Phi^1(\gamma) =
				P_{[\eta_1,\eta_2]^n}\Big(z^1_n-\gamma
				\big( A^* (A z^1_n- z) + M^*z^2_n + \sum_{i=1}^p
				z_{n,i}^3
				\ln(z_{n,i}^1)\big)\Big)$}
			\STATE{$\Phi^2(\gamma)=\gamma (\textnormal{Id}-\ensuremath{\text{\rm prox}}_{ \alpha \|\cdot
					\|_1/\gamma}) (z_n^2 / \gamma + M z^1_n)$}
			\STATE{$\Phi^3(\gamma)=P_{\ensuremath{\left[0,+\infty\right[}^n}\big(z^3_n+\gamma e
				(z^1_n)\big)$}
			\STATE{$\boldsymbol{\Phi}(\gamma)=(\Phi^1(\gamma),\Phi^2(\gamma),\Phi^3(\gamma))$}

			\IF{$ \sum_{i=1}^p  |z_{n,i}^3\ln  (z_{n,i}^1)  -
				\Phi_{i}^3(\gamma)
				\ln
				(\Phi_{i}^1(\gamma))|^2+ \left\| e(z_{n}^1) -
				e(\Phi^1(\gamma))
				\right\|^2\leq \frac{\theta^2}{\gamma^2}
				\pnorm{\boldsymbol{z}_n -
					\boldsymbol{\Phi}(\gamma)}^2$ }
			\STATE{{\tt V} $=1$}
			\ENDIF
			\ENDWHILE
			\STATE{$\gamma_n=\gamma$}
			\STATE{$(x_n^1,x_n^2,x_n^3)=(\Phi^1(\gamma_n),\Phi^2(\gamma_n),\Phi^3(\gamma_n))$}
			\STATE{	$z_{n+1}^1= P_{[\eta_1,\eta_2]^N} \big(x_n^1 +
				\gamma_n (M^* z_n^2 + \sum_{i=1}^p z_{n,i}^3
				\ln(z_{n,i}^1)
				)-\gamma_n (M^* x_n^2 + \sum_{i=1}^p  x_{n,i}^3
				\ln(x_{n,i}^1)
				)\big)$ }
			\STATE{	$z_{n+1}^2= P_{[-\alpha,\alpha]^r}(x_n^2 -
				\gamma_n M
				z_n^1 + \gamma_n M x_n^1)$ }
			\STATE{	$z_{n+1}^3=P_{\ensuremath{\left[0,+\infty\right[}^p} (x_n^3 - \gamma_n e(z_n^1) +
				\gamma_n e  (x_n^1))$ }
			\STATE{
				$\boldsymbol{z}_{n+1}=(z_{n+1}^1,z_{n+1}^2,z_{n+1}^3)$
			}
			\STATE{
				$r_n=\mathcal{R}\big(\boldsymbol{z}_{n+1},
				\boldsymbol{z}_{n}\big)$ }
			\STATE{$n \to n+1$}
			\ENDWHILE
			\RETURN{$\boldsymbol{z}_{n+1}$}
		\end{algorithmic}
	\end{algorithm}
	We compare Algorithm~\ref{alg:opti} with the algorithm propose
	in
	\cite{BricenoDavis2018} called FBHF and with the MATLAB's
	\texttt{fmincon} (interior point).

	To solve problem in \eqref{eq:probopti} with FBHF algorithm, we
	consider $X =X_1\times X_2 \times X_3$ and the
	operators (see \eqref{eq:incluAB} and \cite[Theorem
	2.3]{BricenoDavis2018})

	\begin{equation}\label{eq:opBricenoDavis2018}
		A=\begin{pmatrix}
			\partial f (\hat{x})\\
			\partial g^*(\hat{u})\\
			N_{\ensuremath{\left[0,+\infty\right[}^p} (\hat{v} )
		\end{pmatrix}, \quad B_1 = \begin{pmatrix}
			\nabla h (\hat{x})\\
			0\\
			0
		\end{pmatrix}, \quad B_2+B_3 =
		\begin{pmatrix}
			M^* \hat{u} +\sum_{i=1}^p\hat{v}_i \nabla e_i (\hat{x})
			\\
			-M \hat{x}\\
			-e(\hat{x})
		\end{pmatrix}.
	\end{equation}
	In our numerical experiments, we generate $20$ random
	realizations of
	$A$, $M$, $z$, and $r_1,\ldots, r_n$ for dimensions  $n=m \in
	\{600,900,1200\}$ and $r \in \{n/3, n/2,2n/3\}$. In each
	realization we define $a_i=9$ for $i\in \{ 1,\ldots,n\}$, $\alpha
	=
	0.05$, and
	$y^0 = \hat{y}_0$ and $y^1=\hat{y}_1 + \text{rand}(n)$, where
	$\hat{y}_1$ and $\hat{y}_2$ satisfies
	$e(\hat{y}_0)=e(\hat{y}_1)=0$.
	For Algorithm~\ref{alg:opti} we consider $\sigma = 0.99$. For
	FBHF we
	consider $\varepsilon = 0.8$, $\theta= \sqrt{1-\varepsilon}/2$,
	$\sigma= 0.99$, the maximally monotone operator $A$, the
	cocoercive
	operator $B_1$, and the continuous operator $B_1+B_2$ on
	\eqref{eq:opBricenoDavis2018} (see \cite[Theorem
	2.3]{BricenoDavis2018}).

	In Table~\ref{t:1} we provide the
	average time and iterations to achieve a tolerance
	$\epsilon=10^{-6}$ for the instances mentioned above. We can
	observe
	that, for each instance, Algorithm~\ref{alg:opti} is more
	efficient
	than the method FBHF and \texttt{fmincon}.
	Algorithm~\ref{alg:opti}
	and FBHF are similar in number of iterations, but each iteration
	of
	FBHF is more expensive in time than Algorithm~\ref{alg:opti}.
	This is
	because FBHF needs, additionally, to activate the operators
	$M^*$
	and
	$M$ in each line search. This difference is larger as the
	dimension
	of the problem increases. Although \texttt{fmincon} needs less
	iterations than Algorithm~\ref{alg:opti} and FBHF to reach the
	stop
	criterion, each iteration is very expensive in CPU time. Indeed,
	Algorithm~\ref{alg:opti} reaches the stop criterion in 20\% of
	the
	time that \texttt{fmincon} takes.

	\begin{table}
		{ \caption{ Average time and average number of iterations of
				20
				random realizations of problem in
				\eqref{eq:numericproblem} for
				Algorithm~\ref{alg:opti}, FBHF, and \texttt{fmincon}.
				\label{t:1}}
			\begin{center}
				\begin{tabular}{|c|c|c|c|c|c|c|c|}\cline{2-8}
					\multicolumn{1}{c}{ }	&
					\multicolumn{1}{|c}{$\epsilon=10^{-6}$} &
					\multicolumn{2}{|c|}{$
						N_2=N_1/3$}&\multicolumn{2}{c|}{$N_2=N_1/2$}
					&\multicolumn{2}{c|}{ $N_2=2N_1/3$}\\  \hline
					$N_1$ & Algorithm &  Av. Time (s)& Av. Iter & Av.
					Time
					(s)& Av.
					Iter &
					A. Time (s)& A. Iter  \\ \hline  \hline
					\multirow{3}{*}{$600$}& Alg.~\ref{alg:opti} &
					6.82 &
					7845  & 9.61
					& 10384 & 15.79 & 16136  \\ \cline{2-8}
					&FBHF  &  10.47 & 8280  & 14.22 & 10885 & 23.28 &
					16772
					\\
					\cline{2-8}
					&\texttt{fmincon}  &  52.52 & 238  & 66.25 & 276
					& 69.78
					&
					251	 \\ \hline
					\hline
					\multirow{3}{*}{$900$} & Alg.~\ref{alg:opti}
					& 19.26 & 8185 & 28.89  & 11932 & 52.69 & 20653
					\\
					\cline{2-8}
					& FBHF  & 31.28 & 8568 &46.06& 12375 & 84.17 &
					21757\\
					\cline{2-8}
					& \texttt{fmincon} &  256.71 & 350  & 309.33  &
					408 &
					292.21
					& 368	 \\ \hline
					\hline
					\multirow{3}{*}{$1200$}& Alg.~\ref{alg:opti}  &
					36.01 &
					8809 &
					62.82 & 14490& 110.76 & 24778   \\
					\cline{2-8}
					&FBHF  &  59.41 & 9231 &98.60 & 14783 & 174.08 &
					25633\\
					\cline{2-8}
					&\texttt{fmincon}  & 694.86 & 457& 839.06 & 528
					& 790.66
					&
					462	 \\ \hline
				\end{tabular}
		\end{center}}
	\end{table}

	\section{Conclusions}
	We provide an algorithm for finding a zero of the sum of a
	maximally
	monotone, a cocoercive, a monotone-Lipschitzian, and a
	monotone-continuous operators in a Hilbertian setting which
	splits
	their influence in its implementation. The proposed method
	exploits the intrinsic properties of each operator activating
	implicitly the maximally monotone operator via its resolvent,
	explicitly the cocoercive and the monotone-Lipschitzian operator,
	and for the continuous operator activation a line search
	procedure is
	needed. This method generalizes previous results in
	\cite{BricenoDavis2018,lionsandmercier1979,Tseng2000SIAM} and it
	is more efficient than other algorithms in the literature when
	applied to
	non-linearly constrained convex optimization problems.

	\section*{acknowledgements}
		The first author thanks the support of  Centro de
		Modelamiento
		Matemático (CMM), ACE210010 and FB210005, BASAL funds for
		centers of excellence, grant Redes 180032, and grant
		FONDECYT 1190871
		from ANID-Chile. The second author thanks the support
		of ANID-Subdirección de Capital Humano/Doctorado
		Nacional/2018-21181024 and by the Direcci\'on de Postgrado y
		Programas from
		UTFSM through Programa de Incentivos a la Iniciaci\'on
		Cient\'ifica
		(PIIC).

redpajama_set_name: RedPajamaArXiv


text:
Park Won-soon, outspoken mayor of Seoul, found dead
Police said Park's body was found near a restaurant nestled in wooded hills stretching across northern Seoul after a more than seven-hour search.
By KIM TONG-HYUNGAssociated Press
SEOUL, South Korea — Park Won-soon, the three-term mayor of the South Korean capital of Seoul, a fierce critic of economic inequality, and a potential presidential candidate in 2022, was found dead early Friday. He was 64.
Police said Park's body was found near a restaurant nestled in wooded hills stretching across northern Seoul after a more than seven-hour search involving hundreds of police officers, firefighters, drones and dogs. They gave no further details, including the cause of death.
The Seoul Metropolitan Government earlier said Park did not come to work on Thursday and had canceled his schedule, including a meeting with a presidential official.
In this June 5, 2014, file photo, Park Won-soon, then candidate for Seoul city mayor of the main opposition party New Politics Alliance for Democracy celebrates his victory in the Seoul mayoral election at his office in Seoul, South Korea. Police say on Thursday, July 9. AP Photo/Lee Jin-man, File
His daughter reported him missing Thursday afternoon, saying he had given her a "will-like" verbal message and left home. He was last seen on security video entering a park at the mouth of the hills late Thursday morning.
It wasn't immediately clear what caused Park's disappearance and death. Police didn't confirm local media reports that one of Park's secretaries had lodged a complaint over alleged sexual harassment.
Park, a liberal human rights lawyer who once led two of South Korea's most influential civic groups, was credited with winning the country's first sexual harassment conviction as an attorney. He was elected Seoul mayor in 2011, upsetting his conservative opponent as an independent candidate backed by opposition liberals. He became the city's first mayor elected to a third term in June 2018 and had been considered a potential presidential candidate in the next election in 2022.
Park mostly maintained his activism as mayor, lamenting the country's growing gap between rich and poor, gender inequality, and corrupt ties between large businesses and politicians.
He was also a vocal critic of Japan, which ruled the Korean Peninsula as a colony from 1910 to the end of World War II, over what he described as Tokyo's refusal to sincerely repent for atrocities such as forced labor and a system of sexual slavery for Japanese troops.
Despite positioning himself as a champion of the poor and powerless, Park was criticized for pushing ahead with aggressive redevelopment projects that razed old commercial and housing districts and drove out tenants who couldn't afford the spike in rents.
During his first terms, Park established himself as a fierce opponent of former conservative President Park Geun-hye and openly supported the millions of protesters who flooded the streets of his city in late 2016 and 2017 calling for her ouster over a corruption scandal. Months after her impeachment, Park Geun-hye was formally removed from office by a court ruling in March 2017 and is currently serving a decades-long prison term for bribery, abuse of power and other charges.
In recent months, Park Won-soon led an active campaign against the coronavirus as it spread in the city, shutting down thousands of nightspots and issuing an administrative order banning rallies in major downtown streets.
South Korea has seen the sudden deaths of key political figures before.
Former liberal President Roh Moo-hyun, who was a close friend and mentor of current President Moon Jae-in, leapt to his death in 2009, a year after leaving office, amid allegations that family members had taken bribes from a businessman during his presidency. Former military dictator Park Chung-hee, the father of Park Geun-hye, was assassinated by his spy chief during a late-night drinking session in 1979.
Joe Biden proposes $700 billion 'Buy American' campaign
Trump faced racial issues with Asian Americans even before virus
Central Maine Obituaries
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Mais de um tipo de layout em um ContentPage É possível ter mais de uma Tipo de layout, dentro de um ContentPage?
Estou desenvolvendo um app,e ele precisa de uma barra superior com alguns botões e uma listagem de Nomes.
Como pode ser feito?

redpajama_set_name: RedPajamaStackExchange


text:
Hypocrisy or Decency? The Left's Dirty Little Secret By Matt Sanchez
In case you haven't heard, another gay sex scandal has hit the news cycle as Washington state Representative, Republican Richard Curtis, announced his resignation after the details of his sexual encounter with a male escort were aired.
During a segment on MSNBC, Dan Abrams, accused the representative of "hypocrisy" for voting against "legislation for gay rights." Of course, we heard the same whining after the Senator Larry Craig and Mark Foley incidents.
With so many liberals crying hypocrisy, it's obvious the charge has much less to do with reason than it does with desperation.
Cody Castagna, a self-described male-escort and pornstar, claimed Washington Representative Rick Curtis was a sexual "freak". Abrams warned the encounter between the legislator and the callboy could not be described on TV, "but let's just say it involved lingerie, rope and a stethoscope."
Another segment commentator, Laura Flanders, a British born mouth-piece for Air America–yes, apparently they still are on the air–accused Representative Curtis and closeted Republicans of being "chickens" for not standing up for "who they are."
"Being chicken that disqualifies you from holding office," said Flanders, so proud that someone was actually listening to what she had to say.
For this liberal loon, Larry Craig's bathroom tap dance, Mark Foley's love texts to teenage interns and Representative Curtis' fetish for female nightwear is "who they are" and therefore these legislators should vote for "gay rights" or else they're hypocrites.
Let's stop and consider how ridiculous this statement is.
When a chain smoker afflicted with cancer votes for anti-smoking legislation, he probably has good reason for at least attempting to shield the public from what he is already suffering. Why would Foley, Craig or the latest, Curtis, be so proud of their wayward behavior that they would want to pass legislation to validate it? Were these men showing hypocrisy and denial or just insight and concern?
On television, radio, print and now even in the public schools, the purveyors of perversion have forced sexual issues onto prime time airwaves like raw sewage into a city water reserve–completely bypassing the filters and waste treatment plant. Things that have always existed, bondage, sadism, public and group sex have steadily become matters for public consumption. It gets to the point were it all seems, common, normal and harmless.
My generation of Americans are numb to the "freak factor". Experts agree that kids who have been sexually abused are more promiscuous and often have an unhealthy sexual relationship. The sexual revolution, as exemplified through the gay agenda, has been molesting the American public for nearly three decades.
During the whole Abrams piece no one, besides old-school conservative, Pat Buchanan, even questioned if hiring a self-described male-escort for sex or cheating on one's wife was shameful.
For liberals, sex in public bathrooms and inter-office relations with interns is just being open and honest. Freak sex behavior is fine, as long as you tow the homosexual party line for the Eldorado of legitimacy: Marriage.
"When you're trying to tell people how to live their lives you'd better be sure you're walking the straight line." Warned Laura Flanders, who probably would be tolerant if her husband had been caught having sex in her black sequin dress. After all, no one is perfect.
And here is what is at the core of the syphilitic liberal brain.
Conservatives realize human beings are imperfect, but that shouldn't stop anyone from doing the right thing. Of course, the term "right thing" leaves liberals confused and wondering if this is discrimination against those who can only use their left hand.
Most Americans have a sense of fairness about personal preferences, and this extends to sexual preference. The libertarian branch of conservatism considers sexuality a private matter intimately linked to the pursuit of happiness, or as liberals call it "Girls gone Wild, without the guilt trip."
"As long as you're not hurting anyone and it's between two consenting adults, the government should have no say in the matter," is what many people on both the right and left will say, but this gesture of civility is wasted on a pick-and-choose tolerance driven campaign for radical social change.
The dirty little secret is that the gay lifestyle is marginal and most Americans know it.
Go to San Francisco, the Mecca of the gay rights movement and you'll find a city where public sex is not only common, but encouraged by a city council that when convened, looks like stand-ins for the statuettes that represent the Seven Deadly Sins.
In a city where civil unions have been accepted for over a generation, the City by the Bay, San Francisco is anything but a gay paradise. It boasts the highest HIV infection rate in the country; so much for the pipe dream the official recognition of gay relationships would somehow encourage monogamy and stability.
Folsom Street hosts the homosexual version of a county fair where there are public displays of leather bondage and a wide-array of anal plugs for sale. You'll probably want to skip the bobbing for apples, horseback ride and the kiss the girl for $1 booth, but the fisting demonstration is proof of diversity.
Oh sure, supporters of homosexual marriage and gay rights will insist on love and commitment as the grounds for their "right to equality." This is window dressing for the pink Trojan Horse, but like the rear ends of the senior citizens wearing leather chaps at the fair, there's too much hanging out for the public not to notice the hideous truth.
While producing and filming adult gay films, I thought sex on film was edgy, counter-culture and most of all a black eye for prudish American hypocrisy. This was my way of acting out and showing how little respect I had for the public, friends and most of all family. I felt like a rebel, an outsider that truly enjoyed pretending I was indifferent to criticism, but I was just one person on the fringe.
Boy, look how much "progress" we have made. The marginal lifestyle I espoused is quickly being packaged as a human rights campaign. No homosexual rights advocate will ever denounce lewd public behavior, pornography or promiscuity. They are too busy trying to get the terms "mother" and "father" banned from school textbooks and insisting junior high school kids benefit from live demonstrations on how to properly put on a condom.
For a relatively small interest group, the gay lobbyist and their enablers in high places have come a long way, nevertheless, Americans reject the homosexual agenda and it's not just for conservatives anymore. Representative Curtis, who opposed same-sex marriage and civil unions, was elected with over a 10 point lead in hopelessly liberal Washington State.
Oregon, New Jersey and New York, also liberal states have just said no to "gay rights". If given the chance to vote, the citizens of Massachusetts would repeal gay marriage, because they realize same-sex wedlock would be reduced to the status of Representative Barney Frank's roommate.
The keyword tolerance can trace it's Latin roots back to the verb for endurance, so how much will the American public take? We'll have the answer to that question, when no one dares to object to what should obviously be objectionable behavior.
"I sincerely apologize for any pain my actions may have caused." Representative Curtis said, after stepping down. It's nice to know someone has a sense of decency, or as liberals call it, hypocrisy.
Columnist Matt Sanchez is currently working on his first book, Gay Jihad: What the radical homosexual movement has in store for you and your family
The 50 Most Popular Conservative Websites
I thought it might be interesting to see which right-of-center websites actually receive the most traffic. So, I went to Alexa.com and ranked the 50 ...


text:
Home — Essay Samples — History — Winston Churchill — Churchill and Cyber Attacks in Russia
Churchill and Cyber Attacks in Russia
Subject: Law, Crime & Punishment, Geography & Travel, History
Category: Crime, Asia, Historical Figures
Essay Topic: Cyber Crimes, Russia, Winston Churchill
Cyber-attacks have become a common phenomenon in the modern society. Three factors contribute towards launching a cyber-attack to an individual or a state. They are; vulnerability factor, fear factor and spectacular factor. The infrastructure that is affected by cyber-attacks includes gas plants, power stations, water plants, and transportation control systems. The outcomes of these attacks include water shortages, power outages, traffic jams and stealing of money among others. This paper looks to analyze views on whether Winston Churchill would support cyber-attacks in Russia. This is basing on the values, and the past life of Churchill and his relationship with Stalin since this event started occurring after he died.
Infrastructure affected by cyber attacks
In the modern society thinking cyber-attacks main target essential infrastructures such as water and gas plants, power stations and transportation control systems. This provides the necessary needs for building an economy of a country therefore when disrupted it leads to an economic downfall hence it can be used to bring down Russia. However, this view cannot be upheld by Churchill who had this belief that Russia was a riddle that was enfolded in a mystery inside an enigma (Herbert G. n.p). Churchill believed it was hard for one to break into the national interests of Russia since they were very secretive even to their allies (Churchill, Winston S. pg.7). He never wanted any war or attack to Russia this is despite having misunderstandings with Dwight D. Eisenhower during the last year of World War II. After assuming power, they wrote to each other and ensured open communication to bury differences for the better relationship development between their countries (John Lukacs, n.p).
One of the quotes of Churchill is "Russia must be saved by Russian exertions, and it must be from the heart of the Russian people" (Churchill, pg.29). He wanted what is best for the Russian government and encourage the local citizens to cooperate with another to solve their internal political challenges (Churchill, Winston, and Harry pg.5). This means he is against the raid of Russia by any external forces hence he would not support cyber-attacks since it may lead to instability in the country. Also, in one of his quotes, Churchill says that continuous efforts if the key to unlocking an individual potential (Churchill, pg.14). He disregarded strength or intelligence as a way of uncovering potentiality (Douglas, pg.50). Cyber-attack is as a result of intelligence of people for which he never advocated for hence Churchill would not support the attack.
Expected outcomes
In the modern day, the expected results of cyber-attack to a country is power outages, water shortages and traffic jams among others. This may cause civil wars since the nation being attacked would want to retaliate but also using the same mechanism. According to Churchill having enemies is a means of standing up for something (National Churchill Museum, n.p). Therefore the attacks on Russia would mean it had a stance that was differing from other nations hence leading into the cyber-attacks. Churchill believed that it was difficult for Russia to be invaded because of their secrets. When the invasion occurs then it will be difficult to forecast the revenge actions Russia would take thus the outcomes will only explain the steps that Russia will pursue. The Russians themselves will contain the damages that will be caused by the attacks.
Churchill and Stalin had a professional working relationship and seem to agree on various things. For example, the worked together to end the Second World War through holding several meeting for which Churchill traveled over 10,000 miles to attend (James Taylor, n.p). This is because they both believed in protecting the sovereignty and ensuring protecting the interests of their countries for the future. When they were making agreements on how to divide colony state they agreed that Great Britain would take 90% of Greece while USSR will receive 90% of Poland Churchill (Yergin, pg.1). These common interests would their beliefs of protecting the interests of their countries. Therefore, because of their professional relationship, Churchill would not want to support the cyber-attacks in Russia. However, Stalin and Churchill differed at the end of the Second World War when Russia failed to honor the Yalta agreement (Buhite, pg. 354). Churchill then planned a total military war to deal with the betrayal of Russia. The agreement was made between Churchill, Stalin, and Roosevelt. At this time of the of Churchill time, he would have supported the cyber-attacks. Churchill during his political career would have considered defense of liberty from Russian cyber aggressions because he believed in protecting the interests of Britain first (Churchill, Winston, and Harry, pg.5).
Churchill would not have supported the cyber-attacks in Russia because he believed there as a need to have an excellent relationship between the two countries. Also, Churchill believed that Russia could with the issue of national interests and when attacked he had no idea of how they respond hence a threat to the world when attacked. He had a professional working relationship with Stalin since they agreed on protecting the interests of their countries. The bond failed when Russia trough Stalin was unable to honor the agreement they had signed before the Second World War. At this time Churchill could have supported the cyber-attacks because he was planning a military war to deal with the betrayal of Russia.
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