#  A strong convergence theorem for solving an equilibrium problem and a fixed
point problem using the Bregman distance

Mostafa Ghadampour1, Ebrahim Soori∗,2, Ravi P. Agarwal3, Donal O′Regan4

###### Abstract.

In this paper, using the Bregman distance, we introduce a new projection-type
algorithm for finding a common element of the set of solutions of an
equilibrium problem and the set of fixed points. Then the strong convergence
of the sequence generated by the algorithm will be established under suitable
conditions. Finally, using MATLAB software, we present a numerical example to
illustrate the convergence performance of our algorithm.

   * Corresponding author
2010 Mathematics Subject Classification: 47H09,47H10
E-mail addresses: m.ghadampour@gmail.com(m. ghadampour<EMAIL_ADDRESS>(E.
Soori), agarwal@tamuk.edu(R. P. Agarwal), donal.oregan@nuigalway.ie(D.
O′Regan).

1 Department of Mathematics, Payame Noor University P. O. Box 19395-3697,
Tehran, Iran.

2 Department of Mathematics, Lorestan University, Lorestan, Khoramabad, Iran.

3 Department of Mathematics Texas A$\&$M University-Kingsville 700 University
Blvd., MSC 172 Kingsville, Texas, USA.

4 School of Mathematics, Statistics, National University of Ireland, Galway,
Ireland.

Keywords: Variational inequality; Bregman nonexpansive mapping; Fixed point
problem; Fréchet differentiable; Asymptotical fixed point.

##  1\.  Introduction

Let $C$ be a nonempty closed and convex subset of a reflexive real Banach
space X. Throughout this paper, $X^{*}$ denotes the dual space of $X$. The
norm and the duality pairing between $X$ and $X^{*}$ are denoted by $\|.\|$
and $\langle.,.\rangle$, respectively. Now $\mathbb{R}$ stands for the set of
real numbers. The equilibrium problem for a bifunction $g:C\times
C\rightarrow\mathbb{R}$ satisfying the condition $g(x,x)=0$ for every $x\in C$
is stated as follows

$\text{Find}\;y^{*}\in C\;\;\text{such}\;\text{that}\;\;g(x,y^{*})\leq 0,$
(1.1)

for all $x\in C$. The set of solutions of (1.1) is denoted by $EP(g)$.

It is well known that variational inequalities arise in optimal control
problems, optimization problems, fixed point problems, partial differential
equations, engineering and equilibrium models and hence they have been
formulated by many authors in recent years(see [33], [28] ).

Tada and Takahashi [26] proposed the hybrid method for finding the common
element of the set of solutions of the monotone equilibrium problem (1.1) and
a set of fixed points of a nonexpansive map represented in their algorithm.
Anh [3] proposed the hybrid extragradient iteration method for finding a
common element of the set of fixed points of a nonexpansive mapping and the
set of solutions of equilibrium problems for a pseudomonotone and Lipschitz-
type continuous bifunction. Eskandani et al.[14] proposed a hybrid
extragradient method and they introduced a new iterative process for
approximating a common element of the set of solutions of equilibrium problems
involving pseudomonotone bifunctions and the set of common fixed points of a
finite family of multi-valued Bregman relatively nonexpansive mappings in
Banach spaces. They proved that for any $x\in C$ the mapping $y\rightarrow
g(x,y)+D_{f}(x,y)$ has a unique minimizer where $g(x,.)$ is proper, convex,
lower semicontinuous and $D_{f}$ is the Bregman distance. Also, Jolaoso et al.
[16] proved that under some suitable conditions, a point $x^{*}\in EP(g)$ if
and only if $x^{*}$ solves the following minimization problem:

$\min\\{\lambda g(x,y)+D_{f}(x,y):y\in C\\}.$

In this paper, motivated by the work of Jolaoso et al. [16], we will present a
new projection-type algorithm for approximating a common solution of a Bregman
nonexpansive mapping which is a solution of (1.1) in the setting of reflexive
Banach spaces. Then using MATLAB software, the main result will be illustrated
with a numerical example.

##  2\. preliminaries

We present some preliminaries and lemmas which will be used in the next
section. Let $f:X\rightarrow(-\infty,\infty]$ be an admissible function, i.e.,
a proper, convex and lower semicontinuous function. The domain of $f$ is the
set $\\{x\in X:f(x)<\infty\\}$ denoted by $\text{dom}f$. The set of minimizers
of f is denoted by Argmin $f$ and its unique element by $argmin_{x\in X}f(x)$,
if Argmin $f$ is a singleton. Let $x\in\text{int}$ $\text{dom}f$, for any
$y\in X$, the directional derivative of f at $x$ is defined by

$f^{\circ}(x,y)=\displaystyle\lim_{t\rightarrow 0}\frac{f(x+ty)-f(x)}{t}.$
(2.1)

If the limit (2.1) as $t\rightarrow 0$ exists for each $y$, then $f$ is said
to be Gâteaux differentiable at $x$. The function $f$ is said to be Gâteaux
differentiable if it is Gâteaux differentiable for all $x\in\text{int}$
$\text{dom}f$. When the limit as $t\rightarrow 0$ in (2.1) is attained
uniformly for any $y\in X$ with $\|y\|=1$, we say that $f$ is Fréchet
differentiable at $x$. Finally, $f$ is said to be uniformly Fréchet
differentiable on a subset $C$ of $X$ if the limit is attained uniformly at
each $x\in C$ and $\|y\|=1$. In this case, the gradient of $f$ at $x$ is the
linear function $\nabla f(x)$ which is defined by $\langle y,\nabla
f(x)\rangle:=f^{\circ}(x,y)$ for all $y\in X$.

Let $x\in\text{int}$ $\text{dom}f$. The subdifferential of $f$ at $x$ is the
convex set defined by

$\partial f(x)=\\{l\in X^{*}:f(x)+\langle y-x,l\rangle\leq f(y),\;\forall y\in
X\\},$ (2.2)

where the Fenchel conjugate of $f$ is the convex function
$f^{*}:X^{*}\rightarrow(-\infty,\infty]$ defined by

$f^{*}(l)=\sup\\{\langle l,x\rangle-f(x):x\in X\\}.$

It is well known that by the Young-Fenchel inequality, if $\partial f(x)$ is
nonempty, then we have

$\langle l,x\rangle\leq f(x)+f^{*}(l),\;\;\forall x\in X,\;l\in X^{*},$

and also

$f(x)+f^{*}(l)=\langle l,x\rangle\Leftrightarrow l\in\partial f(x).$

Let $X$ be a reflexive Banach space. The function $f$ is Legendre if and only
if it satisfies the following two conditions:

  1. (L1)

int $\text{dom}f\neq\emptyset$, $f$ is Gâteaux differentiable and dom $\nabla
f=$ int dom $f$.

  2. (L2)

int $\text{dom}f^{*}\neq\emptyset$, $f^{*}$ is Gâteaux differentiable and dom
$\nabla f^{*}=$ int dom $f^{*}$.

Since $X$ is a reflexive Banach space, $(\partial f)^{-1}=\partial f^{*}$ (see
[7, p 83]). Also, we know that $\nabla f=(\nabla f^{*})^{-1}$, this together
with conditions (L1) and (L2) imply that ran$\nabla f=$ dom $\nabla f^{*}=$
int dom $f^{*}$ and ran$\nabla f^{*}=$ dom $\nabla f=$ int dom $f$. In
addition, if $X$ is reflexive, then $f$ is Legendre if and only if $f^{*}$ is
Legendre (see [5, corollary 5.5]).

Let $f:X\rightarrow(-\infty,\infty]$ be a Gâteaux differentiable function. The
bifunction $D_{f}:domf\times int\;domf\rightarrow[0,+\infty]$ defined by

$D_{f}(y,x):=f(y)-f(x)-\langle y-x,\nabla f(x)\rangle,$ (2.3)

is called the Bregman distance with respect to f (see [20]). The Bregman
distance does not satisfy the well known properties of a metric. Clearly,
$D_{f}(x,x)=0$, but $D_{f}(y,x)=0$ may not imply $x=y$, but when $f$ is
Legendre this indeed holds (see [5, Theorem 7.3(vi)]).

The modulus of total convexity at $x\in\text{int}\;\text{dom}f$ is the
function $\upsilon_{f}(x,.):[0,\infty)\rightarrow[0,\infty]$, defined by

$\upsilon_{f}(x,t):=\inf\\{D_{f}(y,x):y\in\text{dom}f,\;\|y-x\|=t\\}.$

The function $f$ is called totally convex at $x\in\text{int}\;\text{dom}f$ if
$\upsilon_{f}(x,t)$ is positive for any $t>0$ [9]. Let $C$ be a nonempty
subset of $X$. The modulus of total convexity of $f$ on $C$ is defined by

$\upsilon_{f}(C,t)=\inf\\{\upsilon_{f}(x,t):x\in
C\cap\text{int}\;\text{dom}f\\}.$

The function $f$ is called totally convex on bounded subsets if
$\upsilon_{f}(C,t)$ is positive for any nonempty and bounded subset $C$ and
for any $t>0$.

The following result establishes the characteristic continuity properties for
the derivative of a lower semicontinuous convex function.

######  Proposition 1.

[9] Let $f$ be a lower semicontinuous convex function with int
$\text{dom}f\neq\emptyset$. Then the function $f$ is differentiable at the
point $x\in\text{int}$ $\text{dom}f$ if and only if $\partial f(x)$ consists
of a single element.

Let $X$ be a Banach space and $f:X\rightarrow(-\infty,+\infty]$ be a convex
function. The function $f$ is called sequentially consistent if for any two
sequences $\\{x_{n}\\}$ and $\\{y_{n}\\}$ in $X$, such that the first one is
bounded:

$\displaystyle\lim_{n\rightarrow\infty}D_{f}(y_{n},x_{n})=0\Rightarrow\displaystyle\lim_{n\rightarrow\infty}\|y_{n}-x_{n}\|=0.$

######  Lemma 2.1.

[22] If $f:X\rightarrow\mathbb{R}$ is uniformly Fréchet differentiable and
bounded on bounded subsets of $X$, then $\nabla f$ is uniformly continuous on
bounded subsets of $X$ from the strong topology of $X$ to the strong topology
of $X^{*}$.

######  Lemma 2.2.

[9] If $domf$ contains at least two points, then the function $f$ is totally
convex on bounded sets if and only if the function $f$ is sequentially
consistent.

######  Lemma 2.3.

[8] Suppose that $f:X\rightarrow(-\infty,+\infty]$ is a Legendre function. The
function $f$ is totally convex on bounded subsets if and only if $f$ is
uniformly convex on bounded subsets.

######  Lemma 2.4.

[2] Let $f:X\rightarrow Y$ be Gâteaux differentiable at any point of $X$.
Given $u,v\in X$ such that $[u,v]\subset X$, then

$\|f(u)-f(v)\|\leq\sup\\{\|d_{G}f(w)\|:w\in[u,v]\\}\|u-v\|,$

where $d_{G}f(w)$ is called the Gâteaux differential of $f$ at $w$.

######  Lemma 2.5.

[24] Let $f:X\rightarrow\mathbb{R}$ be a Legendre function such that $\nabla
f^{*}$ is bounded on bounded subsets of int $domf^{*}$. Let $x_{0}\in X$. If
$\\{D_{f}(x_{0},x_{n})\\}$ is bounded, then the sequence $\\{x_{n}\\}$ is
bounded too.

######  Lemma 2.6.

[22] Suppose that $f$ is Gâteaux differentiable and totally convex on int
$domf$. Let $x\in$ int $domf$ and $C\subset$ int $domf$ be a nonempty, closed
and convex set. If $\hat{x}\in C$, then the following conditions are
equivalent:

  1. _(i)_

The vector $\hat{x}\in C$ is the Bregman projection of $x$ onto $C$ with
respect to $f$.

  2. _(ii)_

The vector $\hat{x}\in C$ is the unique solution of the variational
inequality:

$\langle z-y,\nabla f(z)-\nabla f(x)\rangle\leq 0,\;\;\forall y\in C.$

  3. _(iii)_

The vector $\hat{x}$ is the unique solution of the inequality:

$D_{f}(y,z)+D_{f}(z,x)\leq D_{f}(y,x),\;\;\forall y\in C.$

The function $V_{f}:X\times X^{*}\rightarrow[0,+\infty]$ is defined by

$V_{f}(x,x^{*})=f(x)-\langle x,x^{*}\rangle+f^{*}(x^{*}),\;\;\;\forall x\in
X,x^{*}\in X^{*}.$ (2.4)

Therefore

$V_{f}(x,x^{*})=D_{f}(x,\nabla f^{*}(x^{*})),\;\;\;\forall x\in X,x^{*}\in
X^{*}.$ (2.5)

Also, by the subdifferential inequality, we obtain

$V_{f}(x,x^{*})+\langle\nabla f^{*}(x^{*})-x,y^{*}\rangle\leq
V_{f}(x,x^{*}+y^{*}),$ (2.6)

for all $x\in X$ and $x^{*},y^{*}\in X^{*}$ [17]. It is known that $V_{f}$ is
convex in the second variable. Hence, for all $z\in X$, we have

$D_{f}\bigg{(}z,\nabla f^{*}\big{(}\sum_{i=1}^{N}t_{i}\nabla
f(x_{i})\big{)}\bigg{)}\leq\sum_{i=1}^{N}t_{i}D_{f}(z,x_{i}),$ (2.7)

where $\\{x_{i}\\}\subset X$ and $\\{t_{i}\\}\subset(0,1)$ with
$\Sigma_{i=1}^{N}t_{i}=1$.

The Bregman projection
$\overleftarrow{Proj}_{C}^{f}:\text{int}(\text{dom}f)\rightarrow C$ is defined
as the necessarily unique vector $\overleftarrow{Proj}_{C}^{f}(x)\in C$ that
satisfies(see [6])

$D_{f}\big{(}\overleftarrow{Proj}_{C}^{f}(x),x\big{)}=\inf\\{D_{f}(y,x):y\in
C\\}.$ (2.8)

Let $X$ be a Banach space. We use $S_{X}$ to denote the unit sphere
$S_{X}=\\{x\in X:\|x\|=1\\}$ and $B_{r}:=\\{y\in X:\|y\|\leq r\\}$ for all
$r>0$, and $B_{r}$ is the closed ball in $X$. Then the function
$f:X\rightarrow\mathbb{R}$ is said to be uniformly convex on bounded subsets
of $X$(see [32]) if $\rho_{r}(t)>0$ for all $r,t>0$, where
$\rho_{r}:[0,\infty)\rightarrow[0,\infty]$ is defined by

$\rho_{r}(t)=\displaystyle\inf_{x,y\in
B_{r},\|x-y\|=t,\alpha\in(0,1)}\frac{\alpha f(x)+(1-\alpha)f(y)-f(\alpha
x+(1-\alpha)y)}{\alpha(1-\alpha)},$

for all $t\geq 0$. The function $\rho_{r}$ is called the gauge of uniform
convexity of $f$.

Now, if $f$ is uniformly convex, then the following lemma is known.

######  Lemma 2.7.

[19] Let $X$ be a Banach space, $r>0$ be a constant and
$f:X\rightarrow\mathbb{R}$ be a uniformly convex function on bounded subsets
of $X$. Then

$f\bigg{(}\sum_{k=0}^{n}\alpha_{k}x_{k}\bigg{)}\leq\sum_{k=0}^{n}\alpha_{k}f(x_{k})-\alpha_{i}\alpha_{j}\rho_{r}(\|x_{i}-x_{j}\|),$

for all $i,j\in\\{0,1,2,...,n\\},x_{k}\in B_{r},\alpha_{k}\in(0,1)$ and
$k=0,1,2,...,n$ with $\sum_{k=0}^{n}\alpha_{k}=1$, where $\rho_{r}$ is the
gauge of uniform convexity of $f$.

A function $f$ on $X$ is said to be coercive[15] if the sublevel set of $f$ is
bounded, equivalently, $\lim_{\|x\|\rightarrow\infty}f(x)=\infty$. A function
$f$ on $X$ is said to be strongly coercive [32] if
$\lim_{\|x\|\rightarrow\infty}\frac{f(x)}{\|x\|}=\infty$. The function $f$ is
also said to be uniformly smooth on bounded subsets([32]) if
$\lim_{t\rightarrow 0}\frac{\sigma_{r}(t)}{t}=0$ for all $r>0$, where
$\sigma_{r}:[0,\infty)\rightarrow[0,\infty]$ is defined by

$\sigma_{r}(t)=\displaystyle\sup_{x,y\in
B_{r},\|x-y\|=t,\alpha\in(0,1)}\frac{\alpha
f(x+(1-\alpha)ty)+(1-\alpha)f(x-\alpha ty)-f(x)}{\alpha(1-\alpha)},$

for all $t\geq 0$. We will need the following Propositions.

######  Proposition 2.

[32] Let $f:X\rightarrow\mathbb{R}$ be a convex function which is strongly
coercive.Then, the following are equivalent:

  1. _(i)_

$f$ is bounded on bounded subsets and uniformly smooth on bounded subsets of
$X$.

  2. _(ii)_

$f$ is Fréchet differentiable and $\nabla f$ is uniformly norm-to-norm
continuous on bounded subsets of $X$.

  3. _(iii)_

dom $f^{*}=X^{*}$, $f^{*}$ is strongly coercive and uniformly convex on
bounded subsets of $X^{*}$.

######  Proposition 3.

[32] Let $f:X\rightarrow\mathbb{\mathbb{R}}$ be a convex function which is
bounded on bounded subsets of $X$. Then the following are equivalent:

  1. _(i)_

$f$ is strongly coercive and uniformly convex on bounded subsets of $X$.

  2. _(ii)_

$dom$ $f^{*}=X^{*}$, $f^{*}$ is bounded on bounded subsets and uniformly
smooth on bounded subsets of $X^{*}$.

  3. _(iii)_

$dom$ $f^{*}=X^{*}$, $f^{*}$ is Fréchet differentiable and $\nabla f^{*}$ is
uniformly norm-to-norm continuous on bounded subsets of $X^{*}$.

######  Lemma 2.8.

[29] Let $C$ be a nonempty convex subset of X and $f:C\rightarrow\mathbb{R}$
be a convex and subdifferentiable function on $C$. Then $f$ attains its
minimum at $x\in C$ if and only if $0\in\partial f(x)+N_{C}(x)$, where
$N_{C}(x)$ is the normal cone of $C$ at $x$, that is

$N_{C}(x):=\\{x^{*}\in X^{*}:\langle x-y,x^{*}\rangle\geq 0,\;\forall y\in
C\\}.$

######  Lemma 2.9.

[12] If f and $g$ are two convex functions on $X$ such that there is a point
$x_{0}\in$ dom $f$ $\cap$ dom $g$ where $f$ is continuous. Then

$\partial(f+g)(x)=\partial f(x)+\partial g(x),\;\;\;\forall x\in X.$

Let $X$ be a real reflexive Banach space and $C$ be a nonempty, closed and
convex subset of $X$. Let $g:C\times C\rightarrow\mathbb{R}$ be a bifunction
such that $g(x,x)=0$ for all $x\in C$. The equilibrium problem ($EP$) with
respect to $g$ on $C$ is stated as follows:

$Find\;y^{*}\in C\;\;such\;that\;\;g(x,y^{*})\leq 0,\;\;for\;all\;x\in C.$
(2.9)

The solution set of equilibrium problem (2.9) is denoted by $EP(g)$.

Throughout this paper, we assume that $g:C\times C\rightarrow\mathbb{R}$ is a
bifunction satisfying the following conditions:

  1. (A1)

$g$ is monotone, i.e., $g(x,y)+g(y,x)\leq 0$, for all $x,y\in C$,

  2. (A2)

$g$ is pseudomonotone, i.e., $g(x,y)\geq 0\Rightarrow g(y,x)\leq 0$, for all
$x,y\in C$,

  3. (A3)

$g$ is a Bregman-Lipschitz-type continuous function, i.e., there exist two
positive constants $c_{1},c_{2}$, such that

$g(x,y)+g(y,z)\geq g(x,z)-c_{1}D_{f}(y,x)-c_{2}D_{f}(z,y),\;\;\forall x,y,z\in
C,$

where $f:X\rightarrow(-\infty,+\infty]$ is a Legendre function. The constants
$c_{1},c_{2}$ are called Bregman-Lipschitz coefficients with respect to $f$,

  4. (A4)

$g$ is weakly continuous on $C\times C$,

  5. (A5)

$g(x,.)$ is convex, lower semicontinuous and subdifferentiable on $C$ for
every fixed $x\in C$,

  6. (A6)

$\limsup_{t\downarrow 0}g(tx+(1-t)y,z)\leq g(y,z)$, for each $x,y,z\in C$.

######  Remark 1.

[31] Every monotone bifunction on $C$ is pseudo-monotone but the converse is
not true. A mapping $A:C\rightarrow X^{*}$ is pseudo-monotone if and only if
the bifunction $g(x,y)=\langle Ax,y-x\rangle$ is pseudo-monotone on $C$.

We denote the set of fixed points of $S$ by $F(S)$, that is $F(S)=\\{x\in
C:x\in Sx\\}$. A point $x\in C$ is called an asymptotic fixed point of $S$ if
$C$ contains a sequence $\\{x_{n}\\}$ which converges weakly to $x$ such that
$\lim_{n\rightarrow\infty}\|x_{n}-Sx_{n}\|=0$. $\hat{F}(S)$ denotes the set of
asymptotic fixed points of $S$ (see [11, 21]). The mapping $T:C\rightarrow C$
is called Bregman nonexpansive if $D_{f}(Tx,Ty)\leq D_{f}(x,y)$, for all
$x,y\in C$.

######  Lemma 2.10.

[14] Let $C$ be a nonempty closed convex subset of a reflexive Banach space
$X$, and $f:X\rightarrow\mathbb{R}$ be a Legendre and strongly coercive
function. Suppose that $g:C\times C\rightarrow\mathbb{R}$ be a bifunction
satisfying A2-A5. For the arbitrary sequences $\\{x_{n}\\}\subset C$ and
$\\{\lambda_{n}\\}\subset(0,\infty)$, let $\\{y_{n}\\}$ and $\\{z_{n}\\}$ be
sequences generated by

$\left\\{\begin{array}[]{lr}y_{n}=argmin\\{\lambda_{n}g(x_{n},y)+D_{f}(y,x_{n}):y\in
C\\},\\\ z_{n}=argmin\\{\lambda_{n}g(y_{n},y)+D_{f}(y,x_{n}):y\in
C\\}.\end{array}\right.$ (2.10)

Then, for all $x^{*}\in EP(g)$
$D_{f}(x^{*},z_{n})\leq
D_{f}(x^{*},x_{n})-(1-\lambda_{n}c_{1})D_{f}(y_{n},x_{n})-(1-\lambda_{n}c_{2})D_{f}(z_{n},y_{n})$.

######  Lemma 2.11.

[30] Let $\\{a_{n}\\}$ be a sequence of nonnegative real numbers satisfying
the inequality:

$a_{n+1}\leq(1-\alpha_{n})a_{n}+\alpha_{n}\sigma_{n},\;\;\forall\geq 0,$

where

  1. (a)

$\\{\alpha_{n}\\}\subset[0,1]$ and $\sum_{n=0}^{\infty}\alpha_{n}=\infty$,

  2. (b)

$\limsup_{n\rightarrow\infty}\sigma_{n}\leq
0,or\sum_{n=0}^{\infty}|\alpha_{n}\sigma_{n}|<\infty$.

Then, $\lim_{n\rightarrow\infty}a_{n}=0$.

######  Lemma 2.12.

[18] Let $\\{a_{n}\\}$ be a sequence of real numbers such that there exists a
subsequence $\\{n_{i}\\}$ of $\\{n\\}$ such that $a_{n_{i}}<a_{n_{i}+1}$ for
all $i\in\mathbb{N}$. Then, there exists a subsequence
$\\{m_{k}\\}\subset\mathbb{N}$ such that $m_{k}\rightarrow\infty$ and the
following properties are satisfied by all (sufficiently large) numbers
$k\in\mathbb{N}$ :

$a_{m_{k}}\leq a_{m_{k}+1}\;\;and\;\;a_{k}\leq a_{m_{k}+1}.$

In fact, $m_{k}=\max\\{j\leq k:a_{j}<a_{j+1}\\}$.

We assume that $\varphi:C\times C\rightarrow\mathbb{R}$ is a bifunction
satisfying the following conditions:

  1. (B1)

$\varphi(x,x)=0$, for all $x\in C$.

  2. (B2)

$\varphi$ is monotone, i.e. $\varphi(x,y)+\varphi(y,x)\leq 0$, for all $x,y\in
C$.

  3. (B3)

$\displaystyle\lim_{t\downarrow 0}\varphi(tz+(1-t)x,y)\leq\varphi(x,y)$, for
all $x,y,z\in C$.

  4. (B4)

for all $x\in C$, $y\mapsto\varphi(x,y)$ is convex and lower semicontinuous.

The resolvent of $\varphi$ is the operator $Res^{f}_{\varphi}:X\rightarrow
2^{C}$ defined by [13, 23]

$Res^{f}_{\varphi}x=\\{z\in C:\varphi(z,y)+\langle\nabla f(z)-\nabla
f(x),y-z\rangle\geq 0,\;\;\forall y\in C\\}.$ (2.11)

If $f:X\rightarrow(-\infty,\infty]$ be a Gâteaux differentiable and strongly
coercive function and $\varphi$ satisfies conditions B1 - B4. Then $dom$
$Res^{f}_{\varphi}=X$ (see [23, Lemma 1]).

######  Lemma 2.13.

[23] Suppose that $X$ is a real reflexive Banach space and $C$ a nonempty
closed convex subset of $X$. Let $f:X\rightarrow(-\infty,\infty]$ be a
Legendre function. If $\varphi$ be a bifunction from $C\times C$ to
$\mathbb{R}$ satisfies B1 - B4, then the followings hold:

  1. (i)

$Res^{f}_{\varphi}$ is single-valued.

  2. (ii)

$Res^{f}_{\varphi}$ is a Bregman firmly nonexpansive operator.

  3. (iii)

$F(Res^{f}_{\varphi})=EP(\varphi)$.

  4. (iv)

$EP(\varphi)$ is a closed and convex subset of $C$.

  5. (v)

$D_{f}(u,x)\geq D_{f}(u,Res^{f}_{\varphi}(x))+D_{f}(Res^{f}_{\varphi}(x),x)$,
for all $x\in X$ and $u\in F(Res^{f}_{\varphi})$.

##  3\. Main results

######  Theorem 3.1.

Let $C$ be a nonempty closed convex subset of a real reflexive Banach space
$X$, and let $f:X\rightarrow\mathbb{R}$ be an admissible, strongly coercive
Legendre function which is bounded, uniformly Fréchet differentiable and
totally convex on bounded subsets of $X$. Let $g:C\times
C\rightarrow\mathbb{R}$ be a bifunction satisfying A1-A5. Assume that
$S:C\rightarrow C$ is a Bregman nonexpansive mapping with $\hat{F}(S)=F(S)$.
Let $\varphi:C\times C\rightarrow\mathbb{R}$ be a bifunction satisfying B1-B4
and $\Omega=F(S)\cap EP(g)\cap F(Res^{f}_{\varphi})\neq\emptyset$. Suppose
that $\\{x_{n}\\}$ is a sequence generated by $x_{1}\in C$, $u\in X$ and

$\left\\{\begin{array}[]{lr}y_{n}=argmin\\{\lambda_{n}g(x_{n},y)+D_{f}(y,x_{n}):y\in
C\\},\\\ z_{n}=argmin\\{\lambda_{n}g(y_{n},y)+D_{f}(y,x_{n}):y\in C\\},\\\
v_{n}=\nabla f^{*}(\delta_{n}\nabla
f(Res^{f}_{\varphi}x_{n})+(1-\delta_{n})\nabla f(Res^{f}_{\varphi}z_{n})),\\\
w_{n}=\nabla f^{*}(\gamma_{n,1}\nabla f(v_{n})+\gamma_{n,2}\nabla
f(z_{n})+\gamma_{n,3}\nabla f(Sz_{n})),\\\ u_{n}\in C\;\;such\;that\\\
\varphi(u_{n},y)+\langle\nabla f(u_{n})-\nabla f(w_{n}),y-u_{n}\rangle\geq
0,\;\;\;\forall\;\;y\in C,\\\ k_{n}=\nabla f^{*}(\beta_{n}\nabla
f(w_{n})+(1-\beta_{n})\nabla f(Sw_{n})),\\\ h_{n}=\nabla
f^{*}(\alpha_{n,1}\nabla f(u)+\alpha_{n,2}\nabla f(x_{n})+\alpha_{n,3}\nabla
f(u_{n})+\alpha_{n,4}\nabla f(k_{n})),\\\
x_{n+1}=\overleftarrow{Proj}_{C}^{f}h_{n}.\end{array}\right.$ (3.1)

where $\\{\delta_{n}\\}\subset(0,1)$ and $\\{\alpha_{n,i}\\}^{4}_{i=1}$,
$\\{\beta_{n}\\}$ and $\\{\lambda_{n}\\}$ satisfy the following conditions:

  1. _(i)_

$\\{\alpha_{n,i}\\}\subset(0,1),\sum_{i=1}^{4}\alpha_{n,i}=1,\lim_{n\rightarrow\infty}\alpha_{n,1}=0,$
$\Sigma_{n=1}^{\infty}\alpha_{n,1}=\infty$,
$\liminf_{n\rightarrow\infty}\alpha_{n,i}\alpha_{n,j}>0$ for all $i\neq j$ and
$2\leq i,j\leq 3$.

  2. _(ii)_

$\\{\gamma_{n,i}\\}\subset(0,1)$, $\gamma_{n,1}+\gamma_{n,2}+\gamma_{n,3}=1$,
$\liminf_{n\rightarrow\infty}\gamma_{n,i}\gamma_{n,j}>0$ for all $i\neq j$ and
$1\leq i,j\leq 3$.

  3. _(iii)_

$\\{\lambda_{n}\\}\subset[a,b]\subset(0,p)$, where
$p=\min\\{\frac{1}{c_{1}},\frac{1}{c_{2}}\\}$, $c_{1},c_{2}$ are the Bregman-
Lipschitz coefficients of $g$.

  4. _(iv)_

$\\{\beta_{n}\\}\subset(0,1),\liminf_{n\rightarrow\infty}\beta_{n}(1-\beta_{n})>0$.

Then, the sequence $\\{x_{n}\\}$ converges strongly to
$\overleftarrow{Proj}_{\Omega}^{f}u$.

###### Proof.

First, we show that $\Omega$ is a closed and convex subset of $C$. It follows
from [14, Lemma $2.14$] that $EP(g)$ is a convex and weakly closed (so closed)
subset of $C$. By the conditions (iii) and (iv) of Lemma 2.13,
$F(Res^{f}_{\varphi})$ is a closed and convex subset of $C$. Also, similar to
[25, Proposition $3.1$] $F(S)$ is a closed and convex subset of $C$. Hence,
$\Omega$ is closed and convex subset of $C$.

Let $\hat{u}=\overleftarrow{Proj}_{\Omega}^{f}u$. By (2.7), Lemma 2.10 (v) of
Lemma 2.13, we have

$\displaystyle D_{f}(\hat{u},v_{n})=$ $\displaystyle D_{f}(\hat{u},\nabla
f^{*}(\delta_{n}\nabla f(Res^{f}_{\varphi}x_{n})+(1-\delta_{n})\nabla
f(Res^{f}_{\varphi}z_{n}))$ $\displaystyle\leq$
$\displaystyle\delta_{n}D_{f}(\hat{u},Res^{f}_{\varphi}x_{n})+(1-\delta_{n})D_{f}(\hat{u},Res^{f}_{\varphi}z_{n})$
$\displaystyle\leq$
$\displaystyle\delta_{n}D_{f}(\hat{u},x_{n})+(1-\delta_{n})D_{f}(\hat{u},z_{n})$
$\displaystyle\leq$ $\displaystyle D_{f}(\hat{u},x_{n}).$ (3.2)

Then from (2.7), (3), Lemma 2.10 and the Bregman nonexpansiveness of $S$, we
conclude that

$\displaystyle D_{f}(\hat{u},w_{n})=$ $\displaystyle D_{f}(\hat{u},\nabla
f^{*}(\gamma_{n,1}\nabla f(v_{n})+\gamma_{n,2}\nabla
f(z_{n})+\gamma_{n,3}\nabla f(Sz_{n})))$ $\displaystyle\leq$
$\displaystyle\gamma_{n,1}D_{f}(\hat{u},v_{n})+\gamma_{n,2}D_{f}(\hat{u},z_{n})+\gamma_{n,3}D_{f}(\hat{u},Sz_{n})$
$\displaystyle\leq$
$\displaystyle\gamma_{n,1}D_{f}(\hat{u},v_{n})+\gamma_{n,2}D_{f}(\hat{u},z_{n})+\gamma_{n,3}D_{f}(\hat{u},z_{n})$
$\displaystyle\leq$ $\displaystyle D_{f}(\hat{u},x_{n}).$ (3.3)

It follows from (2.7) and the Bregman nonexpansiveness of $S$ that

$\displaystyle D_{f}(\hat{u},k_{n})=$ $\displaystyle D_{f}(\hat{u},\nabla
f^{*}(\beta_{n}\nabla f(w_{n})+(1-\beta_{n})\nabla f(Sw_{n})))$
$\displaystyle\leq$
$\displaystyle\beta_{n}D_{f}(\hat{u},w_{n})+(1-\beta_{n})D_{f}(\hat{u},Sw_{n})$
$\displaystyle\leq$
$\displaystyle\beta_{n}D_{f}(\hat{u},w_{n})+(1-\beta_{n})D_{f}(\hat{u},w_{n})$
$\displaystyle\leq$ $\displaystyle D_{f}(\hat{u},w_{n}).$ (3.4)

Therefore, from (3),

$D_{f}(\hat{u},k_{n})\leq D_{f}(\hat{u},x_{n}).$ (3.5)

By (2.11) and the algorithm (4.1), we obtain that $u_{n}\in
Res^{f}_{\varphi}w_{n}$. It follows from Lemma 2.13 that $Res^{f}_{\varphi}$
is single valued, hence $u_{n}=Res^{f}_{\varphi}w_{n}$. So from part (v) of
Lemma 2.13, we have

$D_{f}(\hat{u},u_{n})=D_{f}(\hat{u},Res^{f}_{\varphi}w_{n})\leq
D_{f}(\hat{u},w_{n}).$ (3.6)

Then, from (2.7), (3), (3.6) and part (iii) of Lemma 2.6, we obtain

$\displaystyle D_{f}(\hat{u},x_{n+1})\leq$ $\displaystyle
D_{f}(\hat{u},h_{n})$ $\displaystyle=$ $\displaystyle D_{f}(\hat{u},\nabla
f^{*}(\alpha_{n,1}\nabla f(u)+\alpha_{n,2}\nabla f(x_{n})+\alpha_{n,3}\nabla
f(u_{n})+\alpha_{n,4}\nabla f(k_{n})))$ $\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+\alpha_{n,2}D_{f}(\hat{u},x_{n})+(\alpha_{n,3}+\alpha_{n,4})D_{f}(\hat{u},w_{n}).$
(3.7)

Therefore, it follows from (3) that

$\displaystyle D_{f}(\hat{u},x_{n+1})\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+(1-\alpha_{n,1})D_{f}(\hat{u},x_{n})$
$\displaystyle\leq$
$\displaystyle\max\\{D_{f}(\hat{u},u),D_{f}(\hat{u},x_{n})\\},$

hence, by the induction process, we conclude that

$D_{f}(\hat{u},x_{n+1})\leq\max\\{D_{f}(\hat{u},u),D_{f}(\hat{u},x_{1})\\}.$

Now from the above, we have that the sequence $\\{D_{f}(\hat{u},x_{n})\\}$ is
bounded. From Lemma 2.3, $f$ is a uniformly convex function on bounded
subsets. Hence, the condition (i) of Proposition 3 holds, equivalently, the
condition (ii) of the proposition holds, i.e., $f^{*}$ is bounded on bounded
subsets of $X^{*}$. Thus, $\nabla f^{*}$ is also bounded on bounded subsets of
$X^{*}$ (see [4, Proposition 7.8]). From Lemma 2.5, we conclude that the
sequence $\\{x_{n}\\}$ is bounded. It follows from (3), (3), (3.6), Lemma 2.10
and boundednes of $\\{D_{f}(\hat{u},x_{n})\\}$ that
$\\{D_{f}(\hat{u},z_{n})\\}$, $\\{D_{f}(\hat{u},v_{n})\\}$,
$\\{D_{f}(\hat{u},w_{n})\\}$ and $\\{D_{f}(\hat{u},u_{n})\\}$ are bounded,
hence from Lemma 2.5, we conclude that the sequences $\\{z_{n}\\}$,
$\\{v_{n}\\}$, $\\{w_{n}\\}$ and $\\{u_{n}\\}$ are bounded. Hence $\\{\nabla
f(z_{n})\\}$, $\\{\nabla f(v_{n})\\}$ and $\\{\nabla f(w_{n})\\}$ are bounded
(see [9, Proposition 1.1.11]). It follows from algorithm (4.1) that
$\gamma_{n,3}\nabla f(Sz_{n})=\nabla f(w_{n})-\gamma_{n,1}\nabla
f(v_{n})-\gamma_{n,2}\nabla f(z_{n})$. Then from (ii), $\\{\nabla
f(Sz_{n})\\}$ is bounded.

Since $f$ is uniformly Fréchet differentiable and bounded on bounded subsets
of $X$, then from Lemma 2.1, $\nabla f$ is uniformly norm-to-norm continuous
on bounded subsets of $X$. Also from the assumption that $f$ is a convex and
strongly coercive function, we may conclude that the condition (ii) of
Proposition 2 holds, equivalently, the condition (iii) of the proposition
holds, i.e., $f^{*}$ is uniformly convex on bounded subsets of $X^{*}$. Let
$r_{1}=\displaystyle\sup_{n}\\{\|\nabla f(z_{n})\|,\|\nabla
f(Sz_{n})\|,\|\nabla f(v_{n})\|\\}$. Since $\\{\nabla f(z_{n})\\}$, $\\{\nabla
f(Sz_{n})\\}$ and $\\{\nabla f(v_{n})\\}$ are bounded sequences then
$r_{1}<\infty$. Hence, by (2.4), (2.5), (3), Lemmas 2.7, 2.10 and the Bregman
nonexpansiveness of $S$, we have that

$\displaystyle D_{f}(\hat{u},w_{n})=$ $\displaystyle
V_{f}(\hat{u},\gamma_{n,1}\nabla f(v_{n})+\gamma_{n,2}\nabla
f(z_{n})+\gamma_{n,3}\nabla f(Sz_{n}))$ $\displaystyle=$ $\displaystyle
f(\hat{u})-\langle\hat{u},\gamma_{n,1}\nabla f(v_{n})+\gamma_{n,2}\nabla
f(z_{n})+\gamma_{n,3}\nabla f(Sz_{n})\rangle$
$\displaystyle+f^{*}(\gamma_{n,1}\nabla f(v_{n})+\gamma_{n,2}\nabla
f(z_{n})+\gamma_{n,3}\nabla f(Sz_{n}))$ $\displaystyle\leq$ $\displaystyle
f(\hat{u})-\gamma_{n,1}\langle\hat{u},\nabla
f(v_{n})\rangle-\gamma_{n,2}\langle\hat{u},\nabla
f(z_{n})\rangle-\gamma_{n,3}\langle\hat{u},\nabla f(Sz_{n})\rangle$
$\displaystyle+\gamma_{n,1}f^{*}(\nabla f(v_{n}))+\gamma_{n,2}f^{*}(\nabla
f(z_{n}))+\gamma_{n,3}f^{*}(\nabla f(Sz_{n}))$
$\displaystyle-\gamma_{n,2}\gamma_{n,3}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|)$ $\displaystyle=$
$\displaystyle\gamma_{n,1}V_{f}(\hat{u},\nabla
f(v_{n}))+\gamma_{n,2}V_{f}(\hat{u},\nabla
f(z_{n}))+\gamma_{n,3}V_{f}(\hat{u},\nabla f(Sz_{n}))$
$\displaystyle-\gamma_{n,2}\gamma_{n,3}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|)$ $\displaystyle=$
$\displaystyle\gamma_{n,1}D_{f}(\hat{u},v_{n})+\gamma_{n,2}D_{f}(\hat{u},z_{n})+\gamma_{n,3}D_{f}(\hat{u},Sz_{n})$
$\displaystyle-\gamma_{n,2}\gamma_{n,3}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|)$ $\displaystyle\leq$
$\displaystyle\gamma_{n,1}D_{f}(\hat{u},v_{n})+\gamma_{n,2}D_{f}(\hat{u},z_{n})+\gamma_{n,3}D_{f}(\hat{u},z_{n})$
$\displaystyle-\gamma_{n,2}\gamma_{n,3}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|)$ $\displaystyle\leq$ $\displaystyle
D_{f}(\hat{u},x_{n})-\gamma_{n,2}\gamma_{n,3}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|),$ (3.8)

where $\rho^{*}_{r_{1}}$ is the gauge of uniform convexity of $f^{*}$. Also,
by repeating the above process, we obtain

$D_{f}(\hat{u},w_{n})\leq
D_{f}(\hat{u},x_{n})-\gamma_{n,1}\gamma_{n,2}\rho_{r_{1}}^{*}(\|\nabla
f(v_{n})-\nabla f(z_{n})\|).$ (3.9)

It follows from (3) that $\\{D_{f}(\hat{u},k_{n})\\}$ is bounded, hence from
Lemma 2.5, we conclude that $\\{k_{n}\\}$ is bounded. Therefore, $\\{\nabla
f(k_{n})\\}$ is bounded. Hence from $(1-\beta_{n})\nabla f(Sw_{n})=\nabla
f(k_{n})-\beta_{n}\nabla f(w_{n})$, we obtain that $\\{\nabla f(Sw_{n})\\}$ is
bounded. In a similar way as above, from (2.4), (2.5), (3) and Lemma 2.7,
there exists a number $r_{2}=\displaystyle\sup_{n}\\{\|\nabla
f(w_{n})\|,\|\nabla f(Sw_{n})\|\\}$ such that

$D_{f}(\hat{u},k_{n})\leq
D_{f}(\hat{u},x_{n})-\beta_{n}(1-\beta_{n})\rho_{r_{2}}^{*}(\|\nabla
f(w_{n})-\nabla f(Sw_{n})\|),$ (3.10)

Now, from (3) and (3), we have

$\displaystyle D_{f}(\hat{u},x_{n+1})\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+\alpha_{n,2}D_{f}(\hat{u},x_{n})+(\alpha_{n,3}+\alpha_{n,4})D_{f}(\hat{u},x_{n})$
$\displaystyle-(\alpha_{n,3}+\alpha_{n,4})\gamma_{n,2}\gamma_{n,3}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|)$ $\displaystyle=$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+(1-\alpha_{n,1})D_{f}(\hat{u},x_{n})$
$\displaystyle-(\alpha_{n,3}+\alpha_{n,4})\gamma_{n,2}\gamma_{n,3}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|).$ (3.11)

Also, it is implied from (3) and (3.9) that

$\displaystyle D_{f}(\hat{u},x_{n+1})\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+(1-\alpha_{n,1})D_{f}(\hat{u},x_{n})$
$\displaystyle-(\alpha_{n,3}+\alpha_{n,4})\gamma_{n,1}\gamma_{n,2}\rho_{r_{1}}^{*}(\|\nabla
f(v_{n})-\nabla f(z_{n})\|).$ (3.12)

Moreover, from (2.7), (3), (3.6), part (iii) of Lemma 2.6 and (3.10), we
obtain

$\displaystyle D_{f}(\hat{u},x_{n+1})\leq$ $\displaystyle
D_{f}(\hat{u},h_{n})$ $\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+\alpha_{n,2}D_{f}(\hat{u},x_{n})+\alpha_{n,3}D_{f}(\hat{u},u_{n})+\alpha_{n,4}D_{f}(\hat{u},k_{n})$
$\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+(1-\alpha_{n,1})D_{f}(\hat{u},x_{n})$
$\displaystyle-\alpha_{n,4}\beta_{n}(1-\beta_{n})\rho_{r_{2}}^{*}(\|\nabla
f(w_{n})-\nabla f(Sw_{n})\|).$ (3.13)

From (2.7), (3), Lemma 2.10 and the Bregman nonexpansiveness of $S$, we have

$\displaystyle D_{f}(\hat{u},w_{n})\leq$
$\displaystyle\gamma_{n,1}D_{f}(\hat{u},v_{n})+\gamma_{n,2}D_{f}(\hat{u},z_{n})+\gamma_{n,3}D_{f}(\hat{u},Sz_{n})$
$\displaystyle\leq$
$\displaystyle\gamma_{n,1}D_{f}(\hat{u},x_{n})+(\gamma_{n,2}+\gamma_{n,3})D_{f}(\hat{u},z_{n})$
$\displaystyle\leq$
$\displaystyle\gamma_{n,1}D_{f}(\hat{u},x_{n})+(\gamma_{n,2}+\gamma_{n,3})\big{(}D_{f}(\hat{u},x_{n})$
$\displaystyle-(1-\lambda_{n}c_{1})D_{f}(y_{n},x_{n})-(1-\lambda_{n}c_{2})D_{f}(z_{n},y_{n})\big{)}$
$\displaystyle=$ $\displaystyle
D_{f}(\hat{u},x_{n})-(\gamma_{n,2}+\gamma_{n,3})\big{(}(1-\lambda_{n}c_{1})D_{f}(y_{n},x_{n})$
$\displaystyle+(1-\lambda_{n}c_{2})D_{f}(z_{n},y_{n})\big{)}.$ (3.14)

From (3) and (3), we have

$\displaystyle D_{f}(\hat{u},x_{n+1})\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+\alpha_{n,2}D_{f}(\hat{u},x_{n})+(\alpha_{n,3}+\alpha_{n,4})D_{f}(\hat{u},w_{n})$
$\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+\alpha_{n,2}D_{f}(\hat{u},x_{n})+(\alpha_{n,3}+\alpha_{n,4})\big{[}D_{f}(\hat{u},x_{n})$
$\displaystyle-(\gamma_{n,2}+\gamma_{n,3})\big{(}(1-\lambda_{n}c_{1})D_{f}(y_{n},x_{n})+(1-\lambda_{n}c_{2})D_{f}(z_{n},y_{n})\big{)}\big{]}$
$\displaystyle=$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+(1-\alpha_{n,1})D_{f}(\hat{u},x_{n})-(\alpha_{n,3}+\alpha_{n,4})(\gamma_{n,2}$
$\displaystyle+\gamma_{n,3})\big{(}(1-\lambda_{n}c_{1})D_{f}(y_{n},x_{n})+(1-\lambda_{n}c_{2})D_{f}(z_{n},y_{n})\big{)}.$

Thus it follows that

$\displaystyle(\alpha_{n,3}+\alpha_{n,4})(\gamma_{n,2}+\gamma_{n,3})(1-$
$\displaystyle\lambda_{n}c_{1})D_{f}(y_{n},x_{n})$ $\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+D_{f}(\hat{u},x_{n})-D_{f}(\hat{u},x_{n+1})$
$\displaystyle-(\alpha_{n,3}+\alpha_{n,4})(\gamma_{n,2}+\gamma_{n,3})(1-\lambda_{n}c_{2})D_{f}(z_{n},y_{n}).$
(3.15)

Now by (2.5), (2.6), (2.7), (3), (3.5), (3.6) and the part (iii) of Lemma 2.6,
we conclude that

$\displaystyle D_{f}(\hat{u},x_{n+1})\leq$ $\displaystyle D_{f}(\hat{u},\nabla
f^{*}(\alpha_{n,1}\nabla f(u)+\alpha_{n,2}\nabla f(x_{n})+\alpha_{n,3}\nabla
f(u_{n})+\alpha_{n,4}\nabla f(k_{n}))$ $\displaystyle=$ $\displaystyle
V_{f}(\hat{u},\alpha_{n,1}\nabla f(u)+\alpha_{n,2}\nabla
f(x_{n})+\alpha_{n,3}\nabla f(u_{n})+\alpha_{n,4}\nabla f(k_{n}))$
$\displaystyle\leq$ $\displaystyle V_{f}(\hat{u},\alpha_{n,1}\nabla
f(u)+\alpha_{n,2}\nabla f(x_{n})+\alpha_{n,3}\nabla
f(u_{n})+\alpha_{n,4}\nabla f(k_{n})$ $\displaystyle-\alpha_{n,1}(\nabla
f(u)-\nabla f(\hat{u})))+\alpha_{n,1}\langle h_{n}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle$ $\displaystyle=$ $\displaystyle
V_{f}(\hat{u},\alpha_{n,2}\nabla f(x_{n})+\alpha_{n,3}\nabla
f(u_{n})+\alpha_{n,4}\nabla f(k_{n})+\alpha_{n,1}\nabla f(\hat{u}))$
$\displaystyle+\alpha_{n,1}\langle h_{n}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle$ $\displaystyle=$ $\displaystyle D_{f}(\hat{u},\nabla
f^{*}(\alpha_{n,1}\nabla f(\hat{u})+\alpha_{n,2}\nabla
f(x_{n})+\alpha_{n,3}\nabla f(u_{n})+\alpha_{n,4}\nabla f(k_{n})))$
$\displaystyle+\alpha_{n,1}\langle h_{n}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle$ $\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},\hat{u})+\alpha_{n,2}D_{f}(\hat{u},x_{n})+\alpha_{n,3}D_{f}(\hat{u},u_{n})+\alpha_{n,4}D_{f}(\hat{u},k_{n})$
$\displaystyle+\alpha_{n,1}\langle h_{n}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle$ $\displaystyle\leq$
$\displaystyle\alpha_{n,2}D_{f}(\hat{u},x_{n})+\alpha_{n,3}D_{f}(\hat{u},x_{n})+\alpha_{n,4}D_{f}(\hat{u},x_{n})$
$\displaystyle+\alpha_{n,1}\langle h_{n}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle$ $\displaystyle=$
$\displaystyle(1-\alpha_{n,1})D_{f}(\hat{u},x_{n})+\alpha_{n,1}\langle
h_{n}-\hat{u},\nabla f(u)-\nabla f(\hat{u})\rangle.$ (3.16)

Also, from (2.7), we have

$\displaystyle D_{f}(w_{n},h_{n})\leq$
$\displaystyle\alpha_{n,1}D_{f}(w_{n},u)+\alpha_{n,2}D_{f}(w_{n},x_{n})+\alpha_{n,3}D_{f}(w_{n},u_{n})+\alpha_{n,4}D_{f}(w_{n},k_{n})$
$\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(w_{n},u)+\alpha_{n,2}D_{f}(w_{n},x_{n})+\alpha_{n,3}D_{f}(w_{n},u_{n})$
$\displaystyle+\alpha_{n,4}\big{(}\beta_{n}D_{f}(w_{n},w_{n})+(1-\beta_{n})D_{f}(w_{n},Sw_{n})\big{)}$
$\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(w_{n},u)+\alpha_{n,2}D_{f}(w_{n},x_{n})+\alpha_{n,3}D_{f}(w_{n},u_{n})$
$\displaystyle+\alpha_{n,4}(1-\beta_{n})D_{f}(w_{n},Sw_{n}).$ (3.17)

We now consider the following two possible cases.

## Case1

There exists some $n_{0}\in\mathbb{N}$ such that $\\{D_{f}(\hat{u},x_{n})\\}$
is nonincreasing for all $n\geq n_{0}$. Therefore,
$\displaystyle\lim_{n\rightarrow\infty}D_{f}(\hat{u},x_{n})$ exists and
$D_{f}(\hat{u},x_{n})-D_{f}(\hat{u},x_{n+1})\rightarrow 0$ as
$n\rightarrow\infty$. From (3), the conditions (i), (ii) and (iii), it follows
that $D_{f}(y_{n},x_{n})\rightarrow 0$ as $n\rightarrow\infty$, then by Lemma
2.2 and boundedness of $\\{x_{n}\\}$, we obtain

$\displaystyle\lim_{n\rightarrow\infty}\|y_{n}-x_{n}\|=0.$ (3.18)

In a similar way, from (3), we have

$\displaystyle\lim_{n\rightarrow\infty}\|z_{n}-y_{n}\|=0.$ (3.19)

By (3.18) and (3.19), we have

$\displaystyle\lim_{n\rightarrow\infty}\|z_{n}-x_{n}\|=0.$ (3.20)

It follows from (3.18) that the sequence $\\{y_{n}\\}$ is bounded. Next, from
(3) and the condition (i), we have

$\displaystyle(\alpha_{n,3}+\alpha_{n,4})\gamma_{n,2}\gamma_{n,3}$
$\displaystyle\rho_{r_{1}}^{*}(\|\nabla f(z_{n})-\nabla f(Sz_{n})\|)$
$\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+(1-\alpha_{n,1})D_{f}(\hat{u},x_{n})-D_{f}(\hat{u},x_{n+1})$
$\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+D_{f}(\hat{u},x_{n})-D_{f}(\hat{u},x_{n+1})\rightarrow
0,\;\;as\;\;n\rightarrow\infty.$ (3.21)

Hence, it follows from the conditions (i) and (ii) that
$\displaystyle\lim_{n\rightarrow\infty}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|)=0$. Furthermore, we claim that
$\displaystyle\lim_{n\rightarrow\infty}\|\nabla f(z_{n})-\nabla
f(Sz_{n})\|=0$. If not, there exist a subsequence $\\{n_{m}\\}$ of $\\{n\\}$
and a positive number $\epsilon_{1}$ such that $\|\nabla f(z_{n_{m}})-\nabla
f(Sz_{n_{m}})\|\geq\epsilon_{1}$. Therefore $\rho_{r_{1}}^{*}(\|\nabla
f(z_{n_{m}})-\nabla f(Sz_{n_{m}})\|)\geq\rho_{r_{1}}^{*}(\epsilon_{1})$ for
all $m\in\mathbb{N}$, because $\rho_{r_{1}}^{*}$ is nondecreasing. Letting
$m\rightarrow\infty$, we conclude that $\rho_{r_{1}}^{*}(\epsilon_{1})\leq 0$
which is a contradiction to the uniform convexity of $f^{*}$ on bounded
subsets of $X^{*}$. Therefore $\displaystyle\lim_{n\rightarrow\infty}\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|=0$. By Proposition 3 and Lemma 2.3, $\nabla f^{*}$
is uniformly norm-to-norm continuous on bounded subsets of $X^{*}$. Then

$\displaystyle\lim_{n\rightarrow\infty}\|z_{n}-Sz_{n}\|=0.$ (3.22)

From (3) and (3), the conditions (ii), (iv) and a similar technique as in the
above, we have

$\displaystyle\lim_{n\rightarrow\infty}\|v_{n}-z_{n}\|=0\;\;\&\;\;\displaystyle\lim_{n\rightarrow\infty}\|w_{n}-Sw_{n}\|=0.$
(3.23)

Now, since $\\{x_{n}\\}$ is bounded and $X$ is reflexive, then by [27, Theorem
1.2.14], there exists a subsequence $\\{x_{n_{m}}\\}$ of $\\{x_{n}\\}$ and a
point $q\in X$ such that $x_{n_{m}}\rightharpoonup q$. Therefore by (3.20),
$z_{n_{m}}\rightharpoonup q$. Using (3.22), we obtain $q\in\hat{F}(S)=F(S)$.

Next, we show that $q\in EP(g)$. For this purpose, let $x,y\in\text{int}$
$\text{dom}f$. We prove that $\nabla f(x)-\nabla
f(y)\in\partial_{1}D_{f}(x,y)$ which $\partial_{1}D_{f}(x,y)$ is the
subdifferential of $D_{f}(x,y)$ at the first component. We let
$\vartheta:=\nabla f(x)-\nabla f(y)$. Note that

$\displaystyle 0\geq-D_{f}(z,x)=$ $\displaystyle f(x)-f(z)-\langle x-z,\nabla
f(x)\rangle$ $\displaystyle=$ $\displaystyle f(x)-f(z)-\langle x-z,\nabla
f(y)\rangle-\langle x-z,\vartheta\rangle$ $\displaystyle=$ $\displaystyle
f(x)-f(z)-\langle x-y,\nabla f(y)\rangle-\langle y-z,\nabla
f(y)\rangle-\langle x-z,\vartheta\rangle$ $\displaystyle=$ $\displaystyle
f(x)-f(y)-\langle x-y,\nabla f(y)\rangle+f(y)-f(z)$ $\displaystyle-\langle
y-z,\nabla f(y)\rangle-\langle x-z,\vartheta\rangle$ $\displaystyle=$
$\displaystyle D_{f}(x,y)-D_{f}(z,y)-\langle x-z,\vartheta\rangle,$

for all $z\in\text{dom}f$. Then

$D_{f}(x,y)+\langle z-x,\vartheta\rangle\leq D_{f}(z,y),$

for all $z\in\text{dom}f$. Therefore, $\nabla f(x)-\nabla
f(y)=\vartheta\in\partial_{1}D_{f}(x,y)$. On the other hand, using the fact
that $D_{f}(.,x)$ is convex, differentiable and lower semicontinuous on int
dom$f$, it follows from Proposition 1 that $\partial_{1}D_{f}(x,y)=\\{\nabla
f(x)-\nabla f(y)\\}$. Since
$y_{n}=argmin\\{\lambda_{n}g(x_{n},y)+D_{f}(y,x_{n}):y\in C\\}$, from the
condition $A5$, Lemmas 2.8 and 2.9, it follows that

$\displaystyle 0\in\partial\\{$
$\displaystyle\lambda_{n}g(x_{n},y_{n})+D_{f}(y_{n},x_{n})\\}+N_{C}(y_{n})$
$\displaystyle=$
$\displaystyle\lambda_{n}\partial_{2}g(x_{n},y_{n})+\partial_{1}D_{f}(y_{n},x_{n})+N_{C}(y_{n}).$

Hence, there exist $\zeta_{n}\in\partial_{2}g(x_{n},y_{n})$ and $\eta_{n}\in
N_{C}(y_{n})$ such that

$\lambda_{n}\zeta_{n}+\nabla f(y_{n})-\nabla f(x_{n})+\eta_{n}=0.$ (3.24)

Since $\eta_{n}\in N_{C}(y_{n})$, then

$\langle y_{n}-z,\eta_{n}\rangle\geq 0,\;\;\forall z\in C,$ (3.25)

from (3.24) and (3.25), we have

$\displaystyle 0\leq\langle y_{n}-z,$
$\displaystyle-\lambda_{n}\zeta_{n}-\nabla f(y_{n})+\nabla f(x_{n})\rangle\ $
$\displaystyle=$ $\displaystyle\lambda_{n}\langle
z-y_{n},\zeta_{n}\rangle-\langle y_{n}-z,\nabla f(y_{n})-\nabla
f(x_{n})\rangle,$

for all $z\in C$. Therefore

$\langle y_{n}-z,\nabla f(y_{n})-\nabla f(x_{n})\rangle\leq\lambda_{n}\langle
z-y_{n},\zeta_{n}\rangle.$ (3.26)

Moreover, since $\zeta_{n}\in\partial_{2}g(x_{n},y_{n})$, then

$g(x_{n},y_{n})-g(x_{n},z)\leq\langle y_{n}-z,\zeta_{n}\rangle,$ (3.27)

for all $z\in C$. Now, from (3.26) and (3.27), we have

$\frac{1}{\lambda_{n}}[\langle y_{n}-z,\nabla f(y_{n})-\nabla
f(x_{n})\rangle]\leq g(x_{n},z)-g(x_{n},y_{n}),$

for all $z\in C$. Now, substituting $n_{m}$ instead of $n$ in the above, we
conclude that

$\frac{1}{\lambda_{n_{m}}}[\langle y_{n_{m}}-z,\nabla f(y_{n_{m}})-\nabla
f(x_{n_{m}})\rangle]\leq g(x_{n_{m}},z)-g(x_{n_{m}},y_{n_{m}}),$ (3.28)

for all $z\in C$. From (3.18), (3.28), Lemma 2.1, the conditions (iii), $A4$
and the fact that $x_{n_{m}}\rightharpoonup q$ as $m\rightarrow\infty$, we
have that $g(q,z)\geq 0$ for all $z\in C$, then from $A2$, we have $g(z,q)\leq
0$, for all $z\in C$, i.e., $q\in EP(g)$.

Now, we show that $q\in F(Res^{f}_{\varphi})$. Since $\\{u_{n}\\}$ is bounded,
hence $\\{\nabla f(u_{n})\\}$ is bounded. Let
$r_{3}=\displaystyle\sup_{n}\\{\|\nabla f(u)\|,\|\nabla f(x_{n})\|,\|\nabla
f(u_{n})\|,\|\nabla f(k_{n})\|\\}$. Then by (2.4), (2.5), (3), (3.5), (3.6)
and Lemmas 2.6, 2.7, we have

$\displaystyle D_{f}(\hat{u},x_{n+1})\leq$ $\displaystyle
D_{f}(\hat{u},h_{n})$ $\displaystyle=$ $\displaystyle
V_{f}(\hat{u},\alpha_{n,1}\nabla f(u)+\alpha_{n,2}\nabla
f(x_{n})+\alpha_{n,3}\nabla f(u_{n})+\alpha_{n,4}f(k_{n}))$ $\displaystyle=$
$\displaystyle f(\hat{u})-\langle\hat{u},\alpha_{n,1}\nabla
f(u)+\alpha_{n,2}\nabla f(x_{n})+\alpha_{n,3}\nabla
f(u_{n})+\alpha_{n,4}\nabla f(k_{n})\rangle$
$\displaystyle+f^{*}(\alpha_{n,1}\nabla f(u)+\alpha_{n,2}\nabla
f(x_{n})+\alpha_{n,3}\nabla f(u_{n})+\alpha_{n,3}\nabla f(k_{n}))$
$\displaystyle\leq$ $\displaystyle
f(\hat{u})-\alpha_{n,1}\langle\hat{u},\nabla
f(u)\rangle-\alpha_{n,2}\langle\hat{u},\nabla
f(x_{n})\rangle-\alpha_{n,3}\langle\hat{u},\nabla f(u_{n})\rangle$
$\displaystyle-\alpha_{n,4}\langle\hat{u},\nabla
f(k_{n})\rangle+\alpha_{n,1}f^{*}(\nabla f(u))+\alpha_{n,2}f^{*}(\nabla
f(x_{n}))$ $\displaystyle+\alpha_{n,3}f^{*}(\nabla
f(u_{n}))+\alpha_{n,4}f^{*}(\nabla f(k_{n}))$
$\displaystyle-\alpha_{n,2}\alpha_{n,3}\rho^{*}_{r_{3}}(\|\nabla
f(x_{n})-\nabla f(u_{n})\|)$ $\displaystyle=$
$\displaystyle\alpha_{n,1}V_{f}(\hat{u},\nabla
f(u))+\alpha_{n,2}V_{f}(\hat{u},\nabla
f(x_{n}))+\alpha_{n,3}V_{f}(\hat{u},\nabla f(u_{n}))$
$\displaystyle+\alpha_{n,4}V_{f}(\hat{u},\nabla
f(k_{n}))-\alpha_{n,2}\alpha_{n,3}\rho^{*}_{r_{3}}(\|\nabla f(x_{n})-\nabla
f(u_{n})\|)$ $\displaystyle=$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+\alpha_{n,2}D_{f}(\hat{u},x_{n})+\alpha_{n,3}D_{f}(\hat{u},u_{n})+\alpha_{n,4}D_{f}(\hat{u},k_{n})$
$\displaystyle-\gamma_{n,2}\gamma_{n,3}\rho_{r_{1}}^{*}(\|\nabla
f(z_{n})-\nabla f(Sz_{n})\|)$ $\displaystyle\leq$
$\displaystyle\alpha_{n,1}D_{f}(\hat{u},u)+(1-\alpha_{n,1})D_{f}(\hat{u},x_{n})$
$\displaystyle-\alpha_{n,2}\alpha_{n,3}\rho^{*}_{r_{3}}(\|\nabla
f(x_{n})-\nabla f(u_{n})\|).$ (3.29)

Hence, it follows from the condition (i) and a similar technique as (3.22)
that

$\displaystyle\lim_{n\rightarrow\infty}\|x_{n}-u_{n}\|=0.$ (3.30)

Note $\\{x_{n}\\}$, $\\{z_{n}\\}$ are bounded and $\nabla f$ is bounded on
bounded subsets of $X$ and also $\nabla f^{*}$ is bounded on bounded subsets
of $X^{*}$. Then consequently $\\{w_{n}\\}$ is bounded. In a similar way, it
follows that $\\{k_{n}\\}$ and $\\{h_{n}\\}$ are bounded.

From Lemma 2.4, the inequalities (2.3), (3.22) and (3.23), we have

$\displaystyle\lim_{n\rightarrow\infty}D_{f}(z_{n},v_{n})=0,\displaystyle\lim_{n\rightarrow\infty}D_{f}(z_{n},Sz_{n})=0,\displaystyle\lim_{n\rightarrow\infty}D_{f}(w_{n},Sw_{n})=0,$
(3.31)

also from (2.7), we have

$D_{f}(z_{n},w_{n})\leq\gamma_{n,1}D_{f}(z_{n},v_{n})+\gamma_{n,2}D_{f}(z_{n},z_{n})+\gamma_{n,3}D_{f}(z_{n},Sz_{n}),$

so from (3.31), we obtain that

$\displaystyle\lim_{n\rightarrow\infty}D_{f}(z_{n},w_{n})=0.$

Now, from Lemma 2.2, we conclude that

$\displaystyle\lim_{n\rightarrow\infty}\|z_{n}-w_{n}\|=0.$ (3.32)

By Lemma 2.4, the inequalities (3.20), (3.30) and (3.32), we obtain that

$\displaystyle\lim_{n\rightarrow\infty}D_{f}(w_{n},x_{n})=0,\;\;\;\displaystyle\lim_{n\rightarrow\infty}D_{f}(w_{n},u_{n})=0.$
(3.33)

Then, as the above, we have

$\displaystyle\lim_{n\rightarrow\infty}\|u_{n}-w_{n}\|=0.$

Therefore, from Lemma 2.1, it is implied that

$\displaystyle\lim_{n\rightarrow\infty}\|\nabla f(u_{n})-\nabla f(w_{n})\|=0.$
(3.34)

From (3.30) and $x_{n_{m}}\rightharpoonup q$, we obtain that
$u_{n_{m}}\rightharpoonup q$. Since $u_{n}=Res^{f}_{\varphi}w_{n}$, then we
have

$\varphi(u_{n},y)+\langle\nabla f(u_{n})-\nabla f(w_{n}),y-u_{n}\rangle\geq
0,\;\;\;\forall\;y\in C.$

Now, substituting $n_{m}$ instead of $n$ in the above, we conclude that

$\varphi(u_{n_{m}},y)+\langle\nabla f(u_{n_{m}})-\nabla
f(w_{n_{m}}),y-u_{n_{m}}\rangle\geq 0,\;\;\;\forall\;y\in C.$

By (B2), we obtain

$\langle\nabla f(u_{n_{m}})-\nabla
f(w_{n_{m}}),y-u_{n_{m}}\rangle\geq-\varphi(u_{n_{m}},y)\geq\varphi(y,u_{n_{m}}),\;\;\;\forall\;y\in
C.$ (3.35)

Now, we know from (B4) that $\varphi(x,.)$ is convex and lower semicontinuous.
Then from [1, Propositions 1.9.13 and 2.5.2], it is also weakly lower
semicontinuous. Hence, letting $m\rightarrow\infty$, we conclude from (3.34)
and (3.35) that $\varphi(y,q)\leq 0$ for all $y\in C$. Then from (2.9), $q\in
EP(\varphi)$. So by the condition (iii) of Lemma 2.13, we have that $q\in
F(Res^{f}_{\varphi})$. Therefore $q\in\Omega$.

It follows from (3), (3.31), (3.33), the boundedness of $\\{D_{f}(w_{n},u)\\}$
and the conditions (i), (iv) that
$\lim_{n\rightarrow\infty}D_{f}(w_{n},h_{n})=0$. Hence, from Lemma 2.2 and the
boundedness of $\\{h_{n}\\}$, we have

$\displaystyle\lim_{n\rightarrow\infty}\|w_{n}-h_{n}\|=0.$

Then from (3.20), (3.32) and the above, we obtain

$\displaystyle\lim_{n\rightarrow\infty}\|x_{n}-h_{n}\|=0.$ (3.36)

Now, we show that $\limsup_{n\rightarrow\infty}\langle h_{n}-\hat{u},\nabla
f(u)-\nabla f(\hat{u})\rangle\leq 0$. Since $\\{x_{n}\\}$ is bounded, we
conclude that there exists a subsequence $\\{x_{n_{k}}\\}$ of $\\{x_{n}\\}$
which

$\limsup_{n\rightarrow\infty}\langle x_{n}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle=\lim_{k\rightarrow\infty}\langle x_{n_{k}}-\hat{u},\nabla
f(u)-\nabla f(\hat{u})\rangle.$

Since $\\{x_{n_{k}}\\}$ is bounded then there exists a subsequence
$\\{x_{n_{k_{i}}}\\}$ of $\\{x_{n_{k}}\\}$ that converges weakly to some
$\kappa\in\Omega$. Without loss of generality, we can assume that
$x_{n_{k}}\rightharpoonup\kappa$. Then from Lemma 2.6, we have that

$\limsup_{n\rightarrow\infty}\langle x_{n}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle=\langle\kappa-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle\leq 0.$ (3.37)

Next, It follows from (3.36) and (3.37) that

$\limsup_{n\rightarrow\infty}\langle h_{n}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle=\limsup_{n\rightarrow\infty}\langle x_{n}-\hat{u},\nabla
f(u)-\nabla f(\hat{u})\rangle\leq 0.$ (3.38)

Therefore, from (3), (3.38) and Lemma 2.11, we have
$\displaystyle\lim_{n\rightarrow\infty}D_{f}(\hat{u},x_{n})=0$. Therefore from
Lemma 2.2, $x_{n}\rightarrow\hat{u}$ as $n\rightarrow\infty$.

## Case2

There exists a subsequence $\\{n_{j}\\}$ of $\\{n\\}$, such that

$D_{f}(\hat{u},x_{n_{j}})<D_{f}(\hat{u},x_{n_{j}+1}),$

for all $j\in\mathbb{N}$. Hence, from Lemma 2.12, there exists a subsequence
$\\{m_{k}\\}\subset\mathbb{N}$, such that $m_{k}\rightarrow\infty$ and the
following properties are satisfied:

$D_{f}(\hat{u},x_{m_{k}})\leq
D_{f}(\hat{u},x_{m_{k}+1})\;\;\text{and}\;\;D_{f}(\hat{u},x_{k})\leq
D_{f}(\hat{u},x_{m_{k}+1}),$ (3.39)

for all $k\in\mathbb{N}$. From (3), we have

$\displaystyle(\alpha_{m_{k},3}+\alpha_{m_{k},4})$
$\displaystyle(\gamma_{m_{k},2}+\gamma_{m_{k},3})(1-\lambda_{m_{k}}c_{1})D_{f}(y_{m_{k}},x_{m_{k}})$
$\displaystyle\leq$
$\displaystyle\alpha_{m_{k},1}D_{f}(\hat{u},u)+D_{f}(\hat{u},x_{m_{k}})-D_{f}(\hat{u},x_{{m_{k}}+1})$
$\displaystyle-(\alpha_{m_{k},3}+\alpha_{m_{k},4})(\gamma_{m_{k},2}+\gamma_{m_{k},3})(1-\lambda_{m_{k}}c_{2})D_{f}(z_{m_{k}},y_{m_{k}}),$

for all $k\in\mathbb{N}$. Then using (3.39), the conditions (i), (ii) and
(iii), it follows that $D_{f}(y_{m_{k}},x_{m_{k}})\rightarrow 0$ as
$k\rightarrow\infty$. So by Lemma 2.2 and the boundedness of
$\\{x_{m_{k}}\\}$, we obtain

$\displaystyle\lim_{n\rightarrow\infty}\|y_{m_{k}}-x_{m_{k}}\|=0.$

In a similar way, from (3), we have

$\displaystyle\lim_{k\rightarrow\infty}\|y_{m_{k}}-z_{m_{k}}\|=0.$

Now proceeding with the same argument as in Case 1, we conclude that

$\displaystyle\limsup_{k\rightarrow\infty}\langle h_{m_{k}}-\hat{u},$
$\displaystyle\nabla f(u)-\nabla f(\hat{u})\rangle$
$\displaystyle=\limsup_{k\rightarrow\infty}\langle x_{m_{k}}-\hat{u},\nabla
f(u)-\nabla f(\hat{u})\rangle\leq 0.$ (3.40)

Also, from (3) and (3.39), we obtain

$\displaystyle D_{f}(\hat{u},x_{{m_{k}}+1})$ $\displaystyle\leq$
$\displaystyle(1-\alpha_{m_{k},1})D_{f}(\hat{u},x_{m_{k}})+\alpha_{m_{k},1}\langle
h_{m_{k}}-\hat{u},\nabla f(u)-\nabla f(\hat{u})\rangle$ $\displaystyle\leq$
$\displaystyle(1-\alpha_{m_{k},1})D_{f}(\hat{u},x_{{m_{k}}+1})+\alpha_{m_{k},1}\langle
h_{m_{k}}-\hat{u},\nabla f(u)-\nabla f(\hat{u})\rangle,$

hence,

$D_{f}(\hat{u},x_{{m_{k}}+1})\leq\langle h_{m_{k}}-\hat{u},\nabla f(u)-\nabla
f(\hat{u})\rangle.$

Then, it follows from (3.39) that

$D_{f}(\hat{u},x_{k})\leq D_{f}(\hat{u},x_{m_{k}+1})\leq\langle
h_{m_{k}}-\hat{u},\nabla f(u)-\nabla f(\hat{u})\rangle.$ (3.41)

From (Case2) and (3.41), we conclude that
$\displaystyle\lim_{k\rightarrow\infty}D_{f}(\hat{u},x_{k})=0$. Therefore from
Lemma 2.2, $x_{k}\rightarrow\hat{u}$ as $k\rightarrow\infty$. This completes
the proof. ∎

##  4\. Applications and numerical examples

The following theorem will be concluded from theorem 3.1.

######  Theorem 4.1.

Let $C$ be a nonempty closed convex subset of a real reflexive Banach space
$X$, and let $f:X\rightarrow\mathbb{R}$ be an admissible, strongly coercive
Legendre function which is bounded, uniformly Fréchet differentiable and
totally convex on bounded subsets of $X$. Let $g:C\times
C\rightarrow\mathbb{R}$ be a bifunction satisfying A1-A5. Assume that
$S:C\rightarrow C$ is a Bregman nonexpansive mapping with $\hat{F}(S)=F(S)$.
Let $\Omega=F(S)\cap EP(g)\neq\emptyset$. Suppose that $\\{x_{n}\\}$ is a
sequence generated by $x_{1}\in C$, $u\in X$ and

$\left\\{\begin{array}[]{lr}y_{n}=argmin\\{\lambda_{n}g(x_{n},y)+D_{f}(y,x_{n}):y\in
C\\},\\\ z_{n}=argmin\\{\lambda_{n}g(y_{n},y)+D_{f}(y,x_{n}):y\in C\\},\\\
v_{n}=\nabla f^{*}(\delta_{n}\nabla
f(\overleftarrow{Proj}_{C}^{f}x_{n})+(1-\delta_{n})\nabla
f(\overleftarrow{Proj}_{C}^{f}z_{n})),\\\ w_{n}=\nabla
f^{*}(\gamma_{n,1}\nabla f(v_{n})+\gamma_{n,2}\nabla
f(z_{n})+\gamma_{n,3}\nabla f(Sz_{n})),\\\
u_{n}=\overleftarrow{Proj}_{C}^{f}w_{n},\\\ k_{n}=\nabla f^{*}(\beta_{n}\nabla
f(w_{n})+(1-\beta_{n})\nabla f(Sw_{n})),\\\ h_{n}=\nabla
f^{*}(\alpha_{n,1}\nabla f(u)+\alpha_{n,2}\nabla f(x_{n})+\alpha_{n,3}\nabla
f(u_{n})+\alpha_{n,4}\nabla f(k_{n})),\\\
x_{n+1}=\overleftarrow{Proj}_{C}^{f}h_{n}.\end{array}\right.$ (4.1)

where $\\{\delta_{n}\\}\subset(0,1)$ and $\\{\alpha_{n,i}\\}^{4}_{i=1}$,
$\\{\beta_{n}\\}$ and $\\{\lambda_{n}\\}$ satisfy the following conditions:

  1. _(i)_

$\\{\alpha_{n,i}\\}\subset(0,1),\sum_{i=1}^{4}\alpha_{n,i}=1,\lim_{n\rightarrow\infty}\alpha_{n,1}=0,$
$\Sigma_{n=1}^{\infty}\alpha_{n,1}=\infty$,
$\liminf_{n\rightarrow\infty}\alpha_{n,i}\alpha_{n,j}>0$ for all $i\neq j$ and
$2\leq i,j\leq 3$.

  2. _(ii)_

$\\{\gamma_{n,i}\\}\subset(0,1)$, $\gamma_{n,1}+\gamma_{n,2}+\gamma_{n,3}=1$,
$\liminf_{n\rightarrow\infty}\gamma_{n,i}\gamma_{n,j}>0$ for all $i\neq j$ and
$1\leq i,j\leq 3$.

  3. _(iii)_

$\\{\lambda_{n}\\}\subset[a,b]\subset(0,p)$, where
$p=\min\\{\frac{1}{c_{1}},\frac{1}{c_{2}}\\}$, $c_{1},c_{2}$ are the Bregman-
Lipschitz coefficients of $g$.

  4. _(iv)_

$\\{\beta_{n}\\}\subset(0,1),\liminf_{n\rightarrow\infty}\beta_{n}(1-\beta_{n})>0$.

Then, the sequence $\\{x_{n}\\}$ converges strongly to
$\overleftarrow{Proj}_{\Omega}^{f}u$.

###### Proof.

Putting $\varphi(x,y)=0$ for all $x,y\in C$ in Theorem 3.1, by Lemma 2.6, we
have $Res^{f}_{\varphi}=\overleftarrow{Proj}_{C}^{f}$,
$u_{n}=\overleftarrow{Proj}_{C}^{f}w_{n}$ and $F(Res^{f}_{\varphi})=C$. ∎

######  Example 4.2.

Let $X=\mathbb{R}$, $C=[0,2]$, $f(.)=\frac{1}{2}\|.\|^{2}$ and define the
bifunction $g:C\times C\rightarrow\mathbb{R}$ by

$g(x,y):=16y^{2}+9xy-25x^{2},$

for all $x$, $y\in C$. Next, $g$ satisfies the conditions A1 - A6 as follows:
(A1) $g$ is monotone:

$g(x,y)+g(y,x)=16y^{2}+9xy-25x^{2}+16x^{2}+9xy-25y^{2}=-9(x-y)^{2}\leq 0,$

for all $x,y\in C$.
(A2) $g$ is pseudomonotone, for all $x$, $y\in C$, because $g$ is monotone.
(A3) $g$ is Bregman-Lipschitz-type continuous with $c_{1}=c_{2}=9$:

$\displaystyle g(x,y)+g(y,z)-g(x,z)$ $\displaystyle=16y^{2}+9xy-25x^{2}$
$\displaystyle+16z^{2}+9yz-25y^{2}-16z^{2}-9xz+25x^{2}$ $\displaystyle=$
$\displaystyle-9(y^{2}-xy-yz+xz)$ $\displaystyle=$
$\displaystyle-\frac{9}{2}(y^{2}-2xy+x^{2}+y^{2}-2yz+z^{2}-x^{2}+2xz-z^{2})$
$\displaystyle=$ $\displaystyle-9D_{f}(x,y)-9D_{f}(y,z)+9D_{f}(x,z)$
$\displaystyle\geq$ $\displaystyle-9D_{f}(x,y)-9D_{f}(y,z).$ (4.2)

(A4) Note since $C\subseteq\mathbb{R}$ is weakly compact we need only show
that $g:C\times C\to\mathbb{R}$ is weakly sequentially continuous. Let $x$,
$y\in C$ and let $\\{x_{n}\\}$ and $\\{y_{n}\\}$ be two sequences in C
converging weakly to x and y, respectively (note the weak and strong
convergence in $\mathbb{R}$ are equivalent), and then

$\displaystyle\lim_{n\rightarrow\infty}g(x_{n},y_{n})=\displaystyle\lim_{n\rightarrow\infty}16y_{n}^{2}+9x_{n}y_{n}-25x_{n}^{2}=16y^{2}+9xy-25x^{2}=g(x,y).$

(A5) Note $g(x,.)$ is convex, lower semicontinuous and subdifferentiable on
$C$ for every fixed $x\in C$.
(A6) Note

$\displaystyle\limsup_{t\rightarrow 0}g(tx+(1-t)y,z)=$
$\displaystyle\displaystyle\limsup_{t\rightarrow
0}[16z^{2}+9(tx+(1-t)y)z-25(tx+(1-t)y)^{2}]$ $\displaystyle=g(y,z),$ (4.3)

for each $x$, $y$, $z\in C$. Define $S:C\rightarrow C$ by $S(x)=\frac{x}{3}$,
for all $x\in C$. Hence, $F(S)=\\{0\\}$ and

$\displaystyle D_{f}(Sx,Sy)$ $\displaystyle=D_{f}(\frac{x}{3},\frac{y}{3})$
$\displaystyle=$ $\displaystyle
f(\frac{x}{3})-f(\frac{y}{3})-\langle\frac{x}{3}-\frac{y}{3},\frac{y}{3}\rangle$
$\displaystyle=$
$\displaystyle\frac{x^{2}}{18}-\frac{y^{2}}{18}-\frac{xy}{9}+\frac{y^{2}}{9}$
$\displaystyle=$
$\displaystyle\frac{1}{18}(x-y)^{2}\leq\frac{1}{2}(x-y)^{2}=D_{f}(x,y),$

for all $x,y\in C$. Therefore $S$ is Bregman nonexpansive. Let
$p\in\hat{F(S)}$ then $C$ contains a sequence $\\{x_{n}\\}$ such that
$x_{n}\rightharpoonup p$ and
$\displaystyle\lim_{n\rightarrow\infty}(Sx_{n}-x_{n}))=0$, then $p=0$. So
$\hat{F}(S)=\\{0\\}=F(S)$. Now, define the bifunction $\varphi:C\times
C\rightarrow\mathbb{R}$ by $\varphi(x,y)=0$ for all $x,y\in C$. It is clear
that $\varphi$ satisfies the conditions $B1-B4$. By Lemma 2.6, we conclude
that $Res^{f}_{\varphi}=Proj^{f}_{C}$. Now, if $\lambda_{n}=\frac{1}{32}$ by
definition of $y_{n}$ in our algorithm, we have

$y_{n}=argmin\\{\frac{1}{32}g(x_{n},y)+D_{f}(y,x_{n}):y\in C\\},$

therefore, $y_{n}=\frac{23}{64}x_{n}$. Similarly

$z_{n}=argmin\\{\frac{1}{32}g(y_{n},y)+D_{f}(y,x_{n}):y\in C\\},$

hence, $z_{n}=\frac{1841}{(64)^{2}}x_{n}$. Also
$v_{n}=\delta_{n}P_{C}x_{n}+(1-\delta_{n})P_{C}z_{n}$,
$w_{n}=\gamma_{n,1}v_{n}+\gamma_{n,2}z_{n}+\frac{1}{3}\gamma_{n,3}z_{n}$,
$u_{n}=P_{C}w_{n}$,
$k_{n}=\beta_{n}w_{n}+(1-\beta_{n})\frac{1}{3}w_{n}=(\frac{1}{3}+\frac{2}{3}\beta_{n})w_{n}$,
$h_{n}=\alpha_{n,1}u+\alpha_{n,2}x_{n}+\alpha_{n,3}u_{n}+\alpha_{n,4}k_{n}$
and $x_{n+1}=P_{C}h_{n}$. We choose $\alpha_{n,1}=\frac{1}{4n}$,
$\alpha_{n,2}=\alpha_{n,3}=\alpha_{n,4}=\frac{1}{3}-\frac{3}{4n}$,
$\beta_{n}=\frac{1}{2}+\frac{1}{n}$, $\delta_{n}=\frac{1}{2}$ and
$\gamma_{n,1}=\gamma_{n,2}=\gamma_{n,3}=\frac{1}{3}$. See the table
LABEL:tableexample1 and Figure LABEL:pp1 with the initial point $x_{1}=5$ of
the sequence $\\{x_{n}\\}$.

## References

  * [1] Agarwal, R. P., O’Regan, D. and Sahu, D. R.: Fixed point theory for Lipschitzian-type mappings with applications, vol. 6, Springer, New York, (2009)
  * [2] Ambrosetti, A., Prodi, G.: A Primer of Nonlinear Analysis. Cambridge University Press, Cambridge (1993)
  * [3] Anh, P.N.: A hybrid extragradient method for pseudomonotone equilibrium problems and fixed point problems. Bull. Malays. Math. Sci. Soc. 36(1), 107-116 (2013)
  * [4] Bauschke, H. H., Borwein, J. M.: On projection algorithms for solving convex feasibility problems, SIAM Rev. 38 367-426 (1996)
  * [5] Bauschke, H.H., Borwein, J. M., Combettes, P. L.: Essential smoothness, essential strict convexity, and Legendre functions in Banach spaces, Communications in Contemporary Mathematics, 3, 615-647 (2001)
  * [6] Bregman, L.M.: A relaxation method for finding the common point of convex sets and its application to the solution of problems in convex programming. USSR Comput. Math. Math. Phys. 7, 200-217 (1967)
  * [7] Bonnans, J.F., Shapiro, A.: Perturbation Analysis of Optimization Problems. Springer, New York (2000)
  * [8] Butnariu, D., Iusem, A.N., Z˘alinescu, C.: On uniform convexity, total convexity and convergence of the proximal point and outer Bregman projection algorithms in Banach spaces. J. Convex Anal. 10, 35-61 (2003)
  * [9] Butnariu, D., Iusem, A.N.: Totally Convex Functions for Fixed Points Computation and Infinite Dimensional Optimization. Kluwer Academic Publishers, Dordrecht (2000)
  * [10] Butnariu, D., Resmerita, E.: Bregman distances, totally convex functions and a method for solving operator equations in Banach spaces. Abstr. Appl. Anal. Art. ID 84919, 1-39 (2006)
  * [11] Censor, Y., Reich, S.: Iterations of paracontractions and firmly nonexpansive operators with applications to feasibility and optimization. Optimization. 37, 323-339, (1996)
  * [12] Cioranescu, I.: Geometry of Banach Spaces, Duality Mappings and Nonlinear Problems. Kluwer Academic Publishers, Dordrecht (1990)
  * [13] Combettes, P. L., Hirstoaga S. A.: Equilibrium programming in Hilbert spaces. Journal of Nonlinear and Convex Analysis, 6, 117-136, (2005)
  * [14] Eskandani, G.Z., Raeisi, M., Rassias, T.M.: A hybrid extragradient method for solving pseudomonotone equilibrium problems using Bregman distance. J. Fixed Point Theory Appl. 20, 132 (2018). https://doi.org/10.1007/s11784-018-0611-9
  * [15] Hiriart-Urruty J-B, Lemarchal C: Grundlehren der mathematischen Wissenschaften 306. In Convex Analysis and Minimization Algorithms II. Springer, Berlin; (1993)
  * [16] Jolaoso, L.O., Taiwo, A., Alakoya, T.O.: A Strong Convergence Theorem for Solving Pseudo-monotone Variational Inequalities Using Projection Methods. J Optim Theory Appl 185, 744-766 (2020) https://doi.org/10.1007/s10957-020-01672-3
  * [17] Kohsaka, F., Takahashi,W.: Proximal point algorithm with Bregman functions in Banach spaces. J. Nonlinear Convex Anal. 6, 505-523 (2005)
  * [18] Maing´e, P.E.: Strong convergence of projected subgradient methods for nonsmooth and nonstrictly convex minimization. Set-valued Anal. 16, 899-912 (2008)
  * [19] Naraghirad, E., Yao, J.C.: Bregman weak relatively nonexpansive mappings in Banach spaces. Fixed Point Theory Appl. (2013). https://doi.org/10.1186/ 1687-1812-2013-141
  * [20] Reem, D., Reich, S., De Pierro, A.: Re-examination of Bregman functions and new properties of their divergences. Optimization 68, 279-348 (2019)
  * [21] Reich, S.: A weak convergence theorem for the alternating method with Bregman distances. in Theory and applications of Nonlinear Operators of Accretive and Monotone Type, pp. 313- 318,Marcel Dekker, NewYork, NY,USA, (1996)
  * [22] Reich, S., Sabach, S.: A strong convergence theorem for proximal type- algorithm in reflexive Banach spaces. J. Nonlinear Convex Anal. 10, 471-485 (2009)
  * [23] Reich, S., Sabach, S.: Two strong convergence theorems for Bregman strongly nonexpansive operators in reflexive Banach spaces.Nonlinear Analysis: Theory,Methods $\&$ Applications. 73, 122-135 (2010)
  * [24] Sabach, S.: Products of finitely many resolvents of maximal monotone mappings in reflexive banach spaces. SIAM J. Optim. 21, 1289-1308 (2011)
  * [25] Shahzad, N., Zegeye, H.: Convergence theorem for common fixed points of a finite family of multi-valued Bregman relatively nonexpansive mappings. Fixed Point Theory Appl. (2014). https://doi.org/10.1186/1687-1812-2014-152
  * [26] Tada, A., Takahashi,W.: Strong convergence theorem for an equilibrium problem and a nonexpansive mapping. In: Takahashi, W., Tanaka, T. (eds.) Nonlinear Analysis and Convex Analysis. Yokohama Publishers, Yokohama (2006)
  * [27] Takahashi, W.: Nonlinear Functional Analysis. Yokohama Publishers, Yokohama (2000)
  * [28] Thong, D.V., Dung, V.T. Cho, Y.J.: A new strong convergence for solving split variational inclusion problems. Numer Algor 86, 565-591 (2021). https://doi.org/10.1007/s11075-020-00901-0
  * [29] Tiel, J.V.: Convex Analysis: An introductory text. Wiley, New York (1984)
  * [30] Xu, H.K.: Another control condition in an iterative method for nonexpansive mappings. Bull. Austral. Math. Soc. 65, 109-113 (2002)
  * [31] Yao, J.C.: Variational inequalities with generalized monotone operators. Math. Oper. Res. 19, 691-705 (1994)
  * [32] Zălinescu, C.: Convex analysis in general vector spaces. World Scientific Publishing, Singapore (2002)
  * [33] Zhao, X., K$\ddot{\text{o}}$bis, M.A., Yao, Y.: A Projected Subgradient Method for Nondifferentiable Quasiconvex Multiobjective Optimization Problems. J Optim Theory Appl (2021). https://doi.org/10.1007/s10957-021-01872-5