# The split common null point problem for generalized resolvents and
nonexpansive mappings in Banach spaces

Bijan Orouji1, Ebrahim Soori2,∗

###### Abstract.

In this paper, the split common null point problem in two Banach spaces is
considered. Then, using the generalized resolvents of maximal monotone
operators and the generalized projections and an infinite family of
nonexpansive mappings, a strong convergence theorem for finding a solution of
the split common null point problem in two Banach spaces in the presence of a
sequence of errors will be proved.

Keywords: Split common null point problem. Maximal monotone operator.
Generalized projection. Generalized resolvent. Nonexpansive mapping.

∗ Corresponding author
2010 Mathematics Subject Classification. 47H10.
bijanorouji@yahoo.com(B. Orouji); sori.e@lu.ac.ir(E. Soori)

##  1\. Introduction

Let $H_{1}$ and $H_{2}$ be two Hilbert spaces and $C$ and $Q$, two nonempty,
closed and convex subsets of $H_{1}$ and $H_{2}$, respectively. Let
$A:H_{1}\rightarrow H_{2}$ be a bounded linear operator. Then the split
feasibility problem (SFP) [8] is: to find $z\in H_{1}$ such that $z\in C\cap
A^{-1}Q$. There exists several generalizations of the SFP: the multiple set
convex sets problem ( MSSFP) [22, 9], the split common fixed point problem
(SCFPP) [10, 23], and the split common null point problem (SCNPP) [7]. (SCNPP)
is as follows: given set-valued mappings $M_{1}:H_{1}\rightarrow 2^{H_{1}}$
and $M_{2}:H_{2}\rightarrow 2^{H_{2}}$, and a bounded linear operator
$A:E\rightarrow F$, find a point $z\in H_{1}$ such that

$z\in M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0),$

where $M_{1}^{-1}0$ and $M_{2}^{-1}0$ are sets of null points of $M_{1}$ and
$M_{2}$, respectively. Many authors have studied the split feasibility problem
and the split common null point problem using nonlinear operators and fixed
points; see, for example, [4, 7, 10, 11, 22, 15, 23, 38]. However, we have not
found many results outside of the framework Hilbert spaces. Note that the
first extension of SFP to Banach spaces is appeared in [28], then this scheme
was later extended to MSSFP in [37]. A very recent generalization for the SFP
is appeared in[29]. The split common null point problem in Banach spaces is
also solved by Takahashi [34, 35, 36].

In this paper, the split common null point problem with generalized resolvents
of maximal monotone operators in two Banach spaces is considered. Then using
the generalized resolvents of maximal monotone operators and the generalized
projections, a strong convergence theorem for finding a solution of the split
null point problem in two banach spaces in the presence of a sequence of
errors is proved.

##  2\. Preliminaries

Let $E$ be a real Banach space with the norm $\|.\|$ and $E^{*}$, the dual
space of $E$. When $\\{x_{n}\\}$ is a sequence in $E$, the strong convergence
of $\\{x_{n}\\}$ to $x\in E$ is denoted by $x_{n}\rightarrow x$ and the weak
convergence to $x\in E$ is denoted by $x_{n}\rightharpoonup x$. A Banach space
$E$ is strictly convex if $\|\frac{x+y}{2}\|<1$, whenever $x,y\in S(E)$,
$x\neq y$ and $S(E)$ is the unite sphere centered at the origin of $E$. $E$ is
said to be uniformly convex if $\delta_{E}(\epsilon)=0$ and
$\delta_{E}(\epsilon)>0$ for all $0<\epsilon\leq 2$ where
$\delta_{E}(\epsilon)$ is the modulus of convexity of $E$ and is defined by

(2.1)
$\displaystyle\delta_{E}(\epsilon)=inf\bigg{\\{}1-\frac{\|x+y\|}{2}:\|x\|,\|y\|\leq
1,\|x-y\|\geq\epsilon\bigg{\\}}$

A uniformly convex Banach space is strictly convex and reflexive. It is also
well known that a uniformly convex Banach space has Kadec Klee property, that
is, $x_{n}\rightharpoonup u$ and $\|x_{n}\|\rightarrow\|u\|$ imply
$x_{n}\rightarrow u$, see [14, 26]. Furthermore, $E$ is called $p$-uniformly
convex if there exists a constant $c>0$ such that $\delta_{E}\geq
c\epsilon^{p}$ for all $\varepsilon\in[0,2]$, where $p$ is a fixed real number
with $p\geq 2$. For example, the $L_{p}$ space is $2$-uniformly convex for
$1<p\leq 2$ and $p$-uniformly convex for $p\geq 2$, see [30]. Let $U=\\{x\in
E:\|x\|=1\\}$. The norm of $E$ is said to be Gateaux differentiable if for
each $x,y\in U$, the limit

(2.2) $\displaystyle\lim_{t_{\rightarrow}0}\frac{\|x+ty\|-\|x\|}{t}$

exists. In this case, $E$ is called smooth. The modulus of smoothness of $E$
is defined by

(2.3) $\displaystyle\rho_{E}(t)=\sup\bigg{\\{}\frac{\|x+y\|+\|x-y\|}{2}-1:x\in
U,\|y\|\leq t\bigg{\\}}.$

If $\displaystyle\lim_{t\rightarrow 0}\frac{\rho_{E}(t)}{t}=0$, then $E$ is
called uniformly smooth. Let $q>1$. If there exists a fixed constant $c>0$
such that $\rho_{E}(t)\leq ct^{q}$, then $E$ is said to be $q$-uniformly
smooth, see [19]. It is well known that a uniformly convex Banach space is
reflexive and strictly convex.

The mapping $J_{E}^{p}$ from $E$ to $2^{E^{*}}$ is defined by

(2.4) $\displaystyle J_{E}^{p}(x)=\\{x^{*}\in E^{*}:\langle
x,x^{*}\rangle=\|x\|\|x^{*}\|,\|x^{*}\|=\|x\|^{p-1}\\},\quad\forall x\in E$

If $p=2$ then $J_{E}^{2}=J_{E}$ is the normalized duality mapping on $E$. Note
that $E$ is smooth if and only if $J_{E}$ is a single-valued mapping of $E$
into $E^{*}$. We also know that $E$ is reflexive if and only if $J_{E}$ is
surjective, and $E$ is strictly convex if and only if $J_{E}$ is one-to-one.
Hence, if $E$ is smooth, strictly convex and reflexive Banach space, then
$J_{E}$ is a single-valued, bijection and in this case, the inverse mapping
$J_{E}^{-1}$ coincides with the duality mapping $J_{{E}^{*}}:E^{*}\rightarrow
2^{E}$, that means $J_{E}^{-1}=J_{{E}^{*}}$. If $E$ is uniformly convex and
uniformly smooth, then is uniformly norm-to-norm continuous on bounded sets of
$E$ and $J_{E}^{-1}=J_{{E}^{*}}$ is also uniformly norm-to-norm continuous on
bounded sets of $E^{*}$. It is known that $E$ is $p$-uniformly convex if and
only if its dual $E^{*}$ is $q$-uniformly smooth where $1<q\leq 2\leq
p<\infty$ with $\frac{1}{p}+\frac{1}{q}=1$. For more details about the mapping
$J_{E}^{p}$ refer to [1, 21, 30, 14, 16, 26, 32, 33].

######  Lemma 2.1.

[12] Let $x,y\in E$ if $E$ is $q$-uniformly smooth, then there is a $c_{q}>0$
so that

$\|x-y\|^{q}\leq\|x\|^{q}-q\langle y,J_{E}^{q}(x)\rangle+c_{q}\|y\|^{q}.$

Suppose that $E$ is a smooth Banach space and $J$ is the duality mapping on
$E$. Define a function $\phi:E\times E\rightarrow\mathbb{R}$ by

(2.5) $\phi_{E}(x,y)=\|x\|^{2}-2\langle x,Jy\rangle+\|y\|^{2}\quad\forall
x,y\in E.$

Observer that, in a Hilbert space H, $\phi(x,y)=\|x-y\|^{2}$ for all $x,y\in
H$. Furthermore, we know that for each $x,y,z,\omega\in E$,

(2.6) $(\|x\|-\|y\|)^{2}\leq\phi(x,y)\leq(\|x\|+\|y\|)^{2};$ (2.7)
$\phi\Big{(}t,J^{-1}\big{(}\lambda
Jx+(1-\lambda)Jy\big{)}\Big{)}\leq\lambda\phi(t,x)+(1-\lambda)\phi(t,y);$
(2.8) $2\langle
x-y,Jz-J\omega\rangle=\phi(x,\omega)+\phi(y,z)-\phi(x,z)-\phi(y,\omega).$

if $E$ is additionally assumed to be a strictly convex Banach space, then

$\phi(x,y)=0\quad\Longleftrightarrow\quad x=y.$

The following lemma is due to Kamimura and Takahashi [20].

######  Lemma 2.2.

[20] Let $E$ be a uniformly convex and smooth Banach space and let
$\\{y_{n}\\},\\{z_{n}\\}$ be two sequences of $E$, if
$\displaystyle\lim_{n\rightarrow\infty}\phi(y_{n},z_{n})=0$ and either
$\\{y_{n}\\}$ or $\\{z_{n}\\}$ is bounded, then
$\displaystyle\lim_{n\rightarrow\infty}(y_{n}-z_{n})=0$.

Suppose that $C$ is a nonempty, closed and convex subset of a smooth, strictly
convex, and reflexive Banach space $E$. Then for any $x\in E$, there exists a
unique element $z\in C$ such that

(2.9) $\phi(z,x)=\min_{y\in C}\phi(y,x).$

The mapping $\Pi_{C}:E\rightarrow C$ defined by $z=\Pi_{C}x$ is called the
generalized projection of $E$ onto $C$. For example, see [2, 3, 20].

######  Lemma 2.3.

[18] Let $E$ be a smooth, strictly convex and reflexive Banach space and $C$
be a nonempty, closed and convex subset of $E$ and let $x\in E$ and $z\in C$.
Then, the following condition hold:

  * ($1$)

$\phi(z,\Pi_{C}x)+\phi(\Pi_{C}x,x)\leq\phi(z,x)\quad\forall x\in C,y\in E$;

  * ($2$)

$z=\Pi_{C}x\Longleftrightarrow\langle y-z,Jx-Jz\rangle\leq 0\quad\forall y\in
C$.

Suppose that $M$ is a mapping of $E$ into $2^{E^{*}}$ for the Banach space
$E$. The effective domain of $M$ is denote by $dom(M)$, that is,
$dom(M)=\\{x\in E:Mx\neq\emptyset\\}$. A multi-valued mapping $M$ on $E$ is
said to be monotone if $\langle x-y,u^{*}-v^{*}\rangle\geq 0$ for all $x,y\in
dom(M),u^{*}\in Mx$ and $v^{*}\in My$. A monotone operator $M$ on $E$ is said
to be maximal if it’s graph is not property contained in the graph of any
other monotone operator on $E$.

######  Theorem 2.4.

[6, 27] Let $E$ be a uniformly convex and smooth Banach space and let $J$ be
the duality mapping of $E$ into $E^{*}$. Let $M$ be a monotone operator of $E$
into $2^{E^{*}}$. Then $M$ is maximal if and only if for any $r>0$,

$R(J+rM)=E^{*},$

where $R(J+rM)$ is the range of $J+rM$.

Let $E$ be a uniformly convex Banach space with a Gateaux differentiable norm
and let $M$ be a maximal monotone operator of $E$ into $2^{E^{*}}$. For all
$x\in E$ and $r>0$, we consider the following equation

$Jx\in Jx_{r}+rMx_{r}$

This equation has a unique solution $x_{r}$. In fact, it is obvious from
Theorem 3.1 that there exists a solution $x_{r}$ of $Jx\in Jx_{r}+rMx_{r}$.
Assume that $Jx\in Ju+rMu$ and $Jx\in Jv+rMv$. Then there exist $\omega_{1}\in
Mu$ and $\omega_{2}\in Mv$ such that $Jx=Ju+r\omega_{1}$ and
$Jx=Jv+r\omega_{2}$. So, we have that

$\displaystyle 0$ $\displaystyle=\langle u-v,Jx-Jx\rangle$
$\displaystyle=\langle u-v,Ju+r\omega_{1}-(Jv+r\omega_{2})\rangle$
$\displaystyle=\langle u-v,Ju-Jv+r\omega_{1}-r\omega_{2}\rangle$
$\displaystyle=\langle u-v,Ju-Jv\rangle+\langle
u-v,r\omega_{1}-r\omega_{2}\rangle$
$\displaystyle=\phi(u,v)+\phi(v,u)+r\langle u-v,\omega_{1}-\omega_{2}\rangle$
$\displaystyle\geq\phi(u,v)+\phi(v,u)$

and hence $0=\phi(u,v)=\phi(v,u)$. Since $E$ is strictly convex, we have
$u=v$. We defined $J_{r}^{M}$ by $x_{r}=J_{r}^{M}x$, such that $J_{r}^{M},r>0$
are called the generalized resolvents of $M$. The set of null points of $M$ is
defined by $M^{-1}0=\\{z\in E:0\in Mz\\}$. We know that $M^{-1}0$ is closed
and convex; see[26]. Furthermore

(2.10) $\displaystyle\langle J_{r}^{M}x-y,J(x-J_{r}^{M}x)\rangle\geq 0,$

hold for all $x\in E$ and $y\in M^{-1}0$;see [15].

######  Lemma 2.5.

[13] Let $E$ be a real reflexive, strictly convex and smooth Banach space,
$M:E\rightarrow 2^{E^{*}}$ be a maximal monotone with $M^{-1}0\neq\emptyset$,
then for any $x\in E,y\in M^{-1}0$ and $r>0$, we have

$\phi(y,J_{r}^{M}x)+\phi(J_{r}^{M}x,x)\leq\phi(y,x).$

where $J_{r}^{M}:E\rightarrow E$ is defined by $J_{r}^{M}:=(J+rM)^{-1}J$.

######  Definition 2.6.

[24] $M$ is called upper semicontinuous if for any closed subset $C$ of
$E^{*}$, $M^{-1}(C)$ is closed.

######  Theorem 2.7.

[24] Let $M:E\rightarrow 2^{E^{*}}$ be a maximal monotone operator with
$dom(M)=E$. Then, $M$ is upper semicontinuous.

A Banach space $E$ is said to satisfy the Opial condition, if whenever a
sequence $\\{x_{n}\\}$ in $E$ converges weakly to $x_{0}\in E$, then

(2.11)
$\liminf_{n\rightarrow\infty}\|x_{n}-x_{0}\|<\liminf_{n\rightarrow\infty}\|x_{n}-x\|,\quad\forall
x\in E,x\neq x_{0}$

######  Definition 2.8.

Let $C$ be a nonempty convex subset of a Banach space,
$\\{T_{i}\\}_{i\in\mathbb{N}}$ a sequence of nonexpansive mappings of $C$ into
itself and $\\{\lambda_{i}\\}$ a real sequence such that $0\leq\lambda_{i}\leq
1$ for every $i\in\mathbb{N}$. Following [31], for any $n\geq 1$, we define a
mapping $W_{n}$ of $C$ into itself as follows,

$\displaystyle U_{n,n+1}:=I,$ $\displaystyle
U_{n,n}:=\lambda_{n}T_{n}U_{n,n+1}+(1-\lambda_{n})I,$
$\displaystyle\>\;\vdots$ (2.12) $\displaystyle
U_{n,k}:=\lambda_{k}T_{k}U_{n,k+1}+(1-\lambda_{k})I,$
$\displaystyle\>\;\vdots$ $\displaystyle
U_{n,2}:=\lambda_{2}T_{2}U_{n,3}+(1-\lambda_{2})I,$ $\displaystyle W_{n}:=$
$\displaystyle U_{n,1}:=\lambda_{1}T_{1}U_{n,2}+(1-\lambda_{1})I,$

The following results hold for the mappings $W_{n}$.

######  Theorem 2.9.

[31] Let $C$ be a nonempty closed convex subset of a strictly convex Banach
space. Let $\\{T_{i}\\}_{i\in\mathbb{N}}$ be a sequence of nonexpansive
mappings of $C$ into itself such that
$\bigcap_{i=1}^{\infty}\rm{Fix}(T_{i})\neq\emptyset$ and let
$\\{\lambda_{i}\\}$ be a real sequence such that $0\leq\lambda_{i}\leq b<1$
for every $i\in\mathbb{N}$. For any $n\in\mathbb{N}$, let $W_{n}$ be the
$W$-mapping of $C$ into itself generated by $T_{n},T_{n-1},...,T_{1}$ and
$\lambda_{n},\lambda_{n-1},...,\lambda_{1}$. Then

  * (1)

$W_{n}$ is asymptotically regular and nonexpansive and
$\rm{Fix}(W_{n})=\bigcap_{i=1}^{n}\rm{Fix}(T_{i})$, for all $n\in\mathbb{N}$.

  * (2)

for every $x\in C$ and for each positive integer $j$,
$\displaystyle\lim_{n\rightarrow\infty}U_{n,j}x$ exists.

  * (3)

The mapping $W:C\rightarrow C$ defined by
$Wx:=\displaystyle\lim_{n\rightarrow\infty}W_{n}x=\displaystyle\lim_{n\rightarrow\infty}U_{n,1}x$,
for every $x\in C,$ is a nonexpansive mapping satisfying
$\rm{Fix}(W)=\bigcap_{i=1}^{\infty}\rm{Fix}(T_{i})$ and it is called the
$W$-mapping generated by $\\{T_{i}\\}_{i\in\mathbb{N}}$, and
$\\{\lambda_{i}\\}_{i\in\mathbb{N}}$.

######  Theorem 2.10.

[25] Let $\\{T_{i}\\}_{i=1}^{\infty}$ be a sequence of nonexpansive mappings
of $C$ into itself such that
$\bigcap_{i=1}^{\infty}\rm{Fix}(T_{i})\neq\emptyset,\\{\lambda_{i}\\}$ be a
real sequence such that $0<\lambda_{i}\leq b<1,\;(i\geq 1)$. If $D$ is any
bounded subset of $C$, then $\displaystyle\lim_{n\rightarrow\infty}\sup_{x\in
D}\|Wx-W_{n}x\|=0$.

######  Lemma 2.11.

[5] Let $E$ be a strictly convex Banach space and $C\subseteq E$ be a nonempty
and convex subset of $E$. let $T:C\rightarrow E$ be a nonexpansive mapping.
Then $Fix(T)$ is convex.

##  3\. Main results

######  Theorem 3.1.

Let $E$ and $F$ be two $2$-uniformly convex and uniformly smooth real Banach
spaces that satisfy the Opial condition. Let $J_{E}$ and $J_{F}$ be the
duality mappings on $E$ and $F$, respectively. Suppose that $C$ is nonempty,
closed and convex subset of $E$. Let $A:E\longrightarrow F$ be a bounded
linear operator such that $A\neq 0$ with the adjoint operator $A^{*}$. Let
$M_{1}$ be a maximal monotone operator of $E$ into $2^{E^{*}}$ such that
$M_{1}^{-1}0\neq\emptyset$ and $M_{2}$ be a maximal monotone operator of $F$
into $2^{F^{*}}$ such that $M_{2}^{-1}0\neq\emptyset$. Suppose that
$S:C\longrightarrow E$ is a nonexpansive mapping and
$\\{T_{i}\\}_{i=1}^{\infty}:C\longrightarrow C$, a family of nonexpansive
mappings. For every $n\in\mathbb{N}$, let $W_{n}$ be a $W-mapping$ generated
by Definition 2.8. Let $J_{\lambda}^{M_{1}}$ and $Q_{\mu}^{M_{2}}$ be the
generalized resolvents of $M_{1}$ and $M_{2}$ for $\lambda>0$ and $\mu>0$,
respectively. Suppose that $M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq C$.
Let $x_{1}\in C$ and let $\\{x_{n}\\}$, $\\{u_{n}\\}$ and $\\{y_{n}\\}$ be the
sequences generated by

(3.1)
$\displaystyle\begin{cases}u_{n}=J_{E}^{-1}\Big{(}(1-\alpha_{n})J_{E}x_{n}+\alpha_{n}J_{E}\Pi_{C}J_{E}^{-1}(\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n})\Big{)};\\\
z_{n}=J_{\lambda_{n}}^{M_{1}}(u_{n}+e_{n});\quad\quad\omega_{n}=Q_{\mu_{n}}^{M_{2}}(Az_{n});\\\
y_{n}=\Pi_{C}J_{E}^{-1}\Big{(}J_{E}z_{n}-\gamma
A^{*}J_{F}(Az_{n}-\omega_{n})\Big{)};\\\ C_{n}=\\{z\in
C,\big{\langle}\omega_{n}-Az,J_{F}(Az_{n}-\omega_{n})\big{\rangle}\geq
0\\};\\\ D_{n}=\\{z\in E,\phi_{E}(z,z_{n})\leq\phi_{E}(z,u_{n}+e_{n})\\};\\\
Q_{n}=\\{z\in E,\langle x_{n}-z,J_{E}x_{1}-J_{E}x_{n}\rangle\geq 0\\};\\\
x_{n+1}=\Pi_{C_{n}\cap Q_{n}\cap D_{n}}x_{1},\quad\forall
n\in\mathbb{N}.\end{cases}$

where $J_{\lambda_{n}}^{M_{1}}=(J_{E}+\lambda_{n}M_{1})^{-1}J_{E}$ and
$Q_{\mu_{n}}^{M_{2}}=(J_{F}+\mu_{n}M_{2})^{-1}J_{F}$ such that
$\\{\lambda_{n}\\},\\{\mu_{n}\\}\subseteq(0,\infty)$ and $a\in\mathbb{R}$
satisfy in $0<a\leq\lambda_{n},\mu_{n},\forall n\in\mathbb{N}$ and
$0<\gamma<\frac{2}{c\parallel A\parallel^{2}}$ with $c>0$. Let
$\\{\alpha_{n}\\},\\{\sigma_{n}\\}$ be real sequences in $(0,1)$ satisfied in
the conditions:

  * (i)

$\displaystyle\lim_{n\rightarrow\infty}\alpha_{n}=0$;

  * (ii)

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{\|J_{E}x_{n}-J_{E}u_{n}\|}{\alpha_{n}}=0$;

  * (iii)

$\displaystyle\lim_{n\rightarrow\infty}\sigma_{n}=1$.

Consider the error sequence $\\{e_{n}\\}\subseteq E$ such that

  * (iv)

$\displaystyle\lim_{n\rightarrow\infty}\|e_{n}\|=0$.

Let $\Omega=M_{1}^{-1}0\cap
A^{-1}(M_{2}^{-1}0)\cap\big{(}\bigcap_{i=1}^{\infty}Fix(T_{i})\big{)}$.
Suppose that one of the following two conditions holds:

  * (v)

the sequence $\\{x_{n}\\}$ is bounded,
or

  * (vi)

$\Omega\neq\emptyset$.

Then

  1. (a)

$\Omega\neq\emptyset$ if and only if the sequence $\\{x_{n}\\}$ is bounded,

  2. (b)

the sequence $\\{x_{n}\\}$ converges strongly to a point $\omega_{0}\in\Omega$
where $\omega_{0}=\Pi_{\Omega}x_{1}$.

###### Proof.

(a) Let $\Omega\neq\emptyset$. First, it will be checked that $C_{n}\cap
D_{n}\cap Q_{n}$ is closed and convex for all $n\in\mathbb{N}$. For any $z\in
D_{n}$, it is realized that

$\displaystyle\phi_{E}(z,z_{n})\leq\phi_{E}(z,u_{n}+e_{n})$
$\displaystyle\Leftrightarrow$ $\displaystyle\|z\|^{2}+\|z_{n}\|^{2}-2\langle
z,J_{E}z_{n}\rangle\leq\|z\|^{2}+\|u_{n}+e_{n}\|^{2}-2\langle
z,J_{E}(u_{n}+e_{n})\rangle$ (3.2) $\displaystyle\Leftrightarrow$
$\displaystyle\|u_{n}+e_{n}\|^{2}-\|z_{n}\|^{2}+2\langle
z,J_{E}z_{n}\rangle-2\langle z,J_{E}(u_{n}+e_{n})\rangle\geq 0.$

Because $E$ is a real Banach space, the inner product of $E$ is linear in both
components and jointly continuous. Therefore, it is easily observed from (3)
that $D_{n}$ is closed and convex for all $n\in\mathbb{N}$. Further, since $A$
is a bounded linear operator, it is obvious that $C_{n}$ is closed and convex
for all $n\in\mathbb{N}$. Also, it is evident that $Q_{n}$ is closed and
convex for all $n\in\mathbb{N}$. Consequently, $C_{n}\cap D_{n}\cap Q_{n}$ is
closed and convex for all $n\in\mathbb{N}$.
Now, it will be shown that $M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq
C_{n}$ for all $n\in\mathbb{N}$. In fact, since $Q_{\mu_{n}}^{M_{2}}$ be the
resolvent of $M_{2}$, we have from (2.10) for all $z\in A^{-1}(M_{2}^{-1}0)$
that

$\langle
Q_{\mu_{n}}^{M_{2}}Az_{n}-Az,J_{F}(Az_{n}-Q_{\mu_{n}}^{M_{2}}Az_{n})\rangle\geq
0,\quad\forall n\in\mathbb{N},$

then $M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq C_{n}$ for all
$n\in\mathbb{N}$. Next, it will be demonstrated that $M_{1}^{-1}0\cap
A^{-1}(M_{2}^{-1}0)\subseteq D_{n}$ for all $n\in\mathbb{N}$. Indeed, since
$M_{1}$ is maximal monotone, hence from Lemma 2.5, it is implied for all $z\in
M_{1}^{-1}0$ and $\lambda_{n}>0$ that

$\phi_{E}(z,z_{n})=\phi_{E}\big{(}z,J_{\lambda_{n}}^{M_{1}}(u_{n}+e_{n})\big{)}\leq\phi_{E}(z,u_{n}+e_{n}),\quad\forall
n\in\mathbb{N},$

hence, $M_{1}^{-1}0\subseteq D_{n}$ for all $n\in\mathbb{N}$, and therefore
$M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq D_{n}$ for all $n\in\mathbb{N}$.
Now, it will be shown by induction that $M_{1}^{-1}0\cap
A^{-1}(M_{2}^{-1}0)\subseteq Q_{n}$ for all $n\in\mathbb{N}$. Since $\langle
x_{1}-z,J_{E}x_{1}-J_{E}x_{1}\rangle\geq 0$ for all $z\in E$, it is obvious
that $M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq Q_{1}=E$. Suppose that
$M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq Q_{k}$ for some
$k\in\mathbb{N}$. Then $M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq C_{k}\cap
D_{k}\cap Q_{k}$. From the fact that $x_{k+1}=\Pi_{C_{k}\cap D_{K}\cap
Q_{k}}x_{1}$, it is implied from Lemma 2.3 that

$\langle x_{k+1}-z,J_{E}x_{1}-J_{E}x_{k+1}\rangle\geq 0,\quad\forall z\in
C_{k}\cap D_{k}\cap Q_{k}.$

Since $M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq C_{k}\cap D_{k}\cap
Q_{k}$, it is concluded that

$\langle x_{k+1}-z,J_{E}x_{1}-J_{E}x_{k+1}\rangle\geq 0,\quad\forall z\in
M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0).$

So $M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq Q_{k+1}$. Hence by induction,
$M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\\\ \subseteq Q_{n}$ for all
$n\in\mathbb{N}$. Thus

$M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq C_{n}\cap D_{n}\cap
Q_{n},\quad\forall n\in\mathbb{N}.$

Hence $C_{n}\cap D_{n}\cap Q_{n}\neq\emptyset$. This implies that
$\\{x_{n}\\}$ is well defined.

By Theorem 2.7 and Definition 2.6, the set $M_{1}^{-1}0\cap
A^{-1}(M_{2}^{-1}0)$ is closed. Next, we show that $M_{1}^{-1}0$ and
$A^{-1}(M_{2}^{-1}0)$ are convex subsets of $E$. Indeed, it is observed that
for all $x_{1},x_{2}\in M_{1}^{-1}0$ and for all $t\in[0,1]$

$\displaystyle\langle 0-v,tx_{1}+(1-t)x_{2}-u\rangle=$ $\displaystyle t\langle
0-v,x_{1}-u\rangle+(1-t)\langle 0-v,x_{2}-u\rangle$ $\displaystyle\geq
0,\quad\forall u\in Dom(M_{1}),v\in M_{1}u,$

hence, by the maximal monotonicity of $M_{1}$, it is implied that
$tx_{1}+(1-t)x_{2}\in M_{1}^{-1}0$. Also, for all $x_{1},x_{2}\in
A^{-1}(M_{2}^{-1}0)$ and for all $t\in[0,1]$ we have

$\displaystyle\langle 0-v,tAx_{1}+(1-t)Ax_{2}-u\rangle=$ $\displaystyle
t\langle 0-v,Ax_{1}-u\rangle+(1-t)\langle 0-v,Ax_{2}-u\rangle$
$\displaystyle\geq 0,\quad\forall u\in Dom(M_{2}),v\in M_{2}u,$

and since $A$ is linear then by the maximal monotonicity of $M_{2}$, it is
implied that $tx_{1}+(1-t)x_{2}\in A^{-1}(M_{2}^{-1}0)$. Thus,
$M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)$ is a convex subset of $E$ and therefore,
by Lemma 2.11, $\Omega$ is closed and convex.

Since $\Omega$ is a nonempty, closed and convex subset of $E$, there exists
$\omega_{0}\in\Omega$ such that $\omega_{0}=\Pi_{\Omega}x_{1}$. Since
$x_{n+1}=\Pi_{C_{n}\cap D_{n}\cap Q_{n}}x_{1}$, it is concluded that

$\phi_{E}(x_{n+1},x_{1})\leq\phi_{E}(y,x_{1}),\quad\forall y\in C_{n}\cap
D_{n}\cap Q_{n},$

and since $\omega_{0}\in\Omega\subseteq\ M_{1}^{-1}0\cap
A^{-1}(M_{2}^{-1}0)\subseteq C_{n}\cap D_{n}\cap Q_{n}$, it is observed that

(3.3) $\phi_{E}(x_{n+1},x_{1})\leq\phi_{E}(\omega_{0},x_{1}).$

This means that $\\{x_{n}\\}$ is bounded.

Conversely, suppose $\\{x_{n}\\}$ is bounded. First, we show that
$\displaystyle\lim_{n\rightarrow\infty}\phi_{E}(x_{n+1},x_{n})=0$. Since
$x_{n+1}\in C_{n}\cap D_{n}\cap Q_{n}$ hence

$\displaystyle\phi_{E}(x_{n+1},x_{n})$
$\displaystyle=\phi_{E}(x_{n+1},\Pi_{C_{n-1}\cap D_{n-1}\cap Q_{n-1}}x_{1})$
$\displaystyle\leq\phi_{E}(x_{n+1},x_{1})-\phi_{E}(\Pi_{C_{n-1}\cap
D_{n-1}\cap Q_{n-1}}x_{1},x_{1})$ (3.4)
$\displaystyle=\phi_{E}(x_{n+1},x_{1})-\phi_{E}(x_{n},x_{1}).$

and hence
$\phi_{E}(x_{n},x_{1})+\phi_{E}(x_{n+1},x_{n})\leq\phi_{E}(x_{n+1},x_{1}),$ so

(3.5) $\phi_{E}(x_{n},x_{1})\leq\phi_{E}(x_{n+1},x_{1}),$

therefore, it is concluded from (2.6), (3.5) that
$\\{\phi_{E}(x_{n},x_{1})\\}$ is bounded and nondecreasing. Then, there exists
the limit of $\\{\phi_{E}(x_{n},x_{1})\\}.$ Using (3), it is realized that

(3.6) $\lim_{n\longrightarrow\infty}\phi_{E}(x_{n+1},x_{n})=0.$

From Lemma 2.2, it is implied that

(3.7) $\lim_{n\longrightarrow\infty}\|x_{n+1}-x_{n}\|=0.$

Next, we show that
$\displaystyle\lim_{n\rightarrow\infty}\phi_{E}(x_{n+1},u_{n})=0$. From the
inequality (2.7), it is implied that

$\displaystyle\phi_{E}(x_{n+1},u_{n})=$
$\displaystyle\phi_{E}\big{(}x_{n+1},J_{E}^{-1}\big{(}(1-\alpha_{n})J_{E}x_{n}$
$\displaystyle+\alpha_{n}J_{E}\Pi_{C}J_{E}^{-1}(\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n})\big{)}\big{)}$
$\displaystyle\leq$ $\displaystyle(1-\alpha_{n})\phi_{E}(x_{n+1},x_{n})$
$\displaystyle+\alpha_{n}\phi_{E}\big{(}x_{n+1},\Pi_{C}J_{E}^{-1}(\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n})\big{)}$
$\displaystyle\leq$ $\displaystyle(1-\alpha_{n})\phi_{E}(x_{n+1},x_{n})$
$\displaystyle+\alpha_{n}\phi_{E}\big{(}x_{n+1},J_{E}^{-1}(\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n})\big{)}$
$\displaystyle\leq$ $\displaystyle(1-\alpha_{n})\phi_{E}(x_{n+1},x_{n})$
$\displaystyle+\alpha_{n}\sigma_{n}\phi_{E}(x_{n+1},W_{n}x_{n})+\alpha_{n}(1-\sigma_{n})\phi_{E}(x_{n+1},Sx_{n}),$

and since from (2.6), $\\{\phi_{E}(x_{n+1},W_{n}x_{n})\\}$ and
$\\{\phi_{E}(x_{n+1},Sx_{n})\\}$ are bounded. hence, using $(i)$ and (3.6), it
is concluded that

(3.8) $\lim_{n\rightarrow\infty}\phi_{E}(x_{n+1},u_{n})=0.$

Therefore, it is realized from Lemma 2.2 that

(3.9) $\lim_{n\rightarrow\infty}\|x_{n+1}-u_{n}\|=0.$

Hence, it is followed from (3.7) and (3.9) that

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

Furthermore, since $E$ is uniformly smooth and $J_{E}$ is uniformly
continuous, it is implied from (3.10) that

(3.11) $\lim_{n\rightarrow\infty}\|J_{E}x_{n}-J_{E}u_{n}\|=0.$

Also, since $x_{n+1}\in D_{n}$ then

(3.12) $\phi_{E}(x_{n+1},z_{n})\leq\phi_{E}(x_{n+1},u_{n}+e_{n})$

Notice that

$\displaystyle\phi_{E}(x_{n+1},u_{n}+e_{n})-\phi_{E}(x_{n+1},u_{n})=$
$\displaystyle\|u_{n}+e_{n}\|-\|u_{n}\|$ (3.13) $\displaystyle+2\langle
x_{n+1},J_{E}u_{n}-J_{E}(u_{n}+e_{n})\rangle$

Since $J_{E}$ is uniformly continuous on each bounded subset of $E$ and
$\displaystyle\lim_{n\rightarrow\infty}\|e_{n}\|=0$, we know from (3.8) and
(3.12) that
$\displaystyle\lim_{n\rightarrow\infty}\phi_{E}(x_{n+1},u_{n}+e_{n})=0$, which
implies that

(3.14) $\lim_{n\rightarrow\infty}\phi_{E}(x_{n+1},z_{n})=0.$

Therefore, it is implied from Lemma 2.2 that

(3.15) $\lim_{n\rightarrow\infty}\|x_{n+1}-z_{n}\|=0.$

Hence, it is followed from (3.7) and (3.15) that

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

From (3.10) and (3.16) it is implied that

(3.17) $\lim_{n\rightarrow\infty}\|u_{n}-z_{n}\|=0.$

Let $z\in C_{n}\cap D_{n}\cap Q_{n}$. Using Lemma 2.1, it is concluded that

$\displaystyle\phi_{E}(z,y_{n})=$
$\displaystyle\phi_{E}\big{(}z,\Pi_{C}J_{E}^{-1}\big{(}J_{E}z_{n}-\gamma
A^{*}J_{F}(Az_{n}-\omega_{n})\big{)}\big{)}$ $\displaystyle\leq$
$\displaystyle\phi_{E}\big{(}z,J_{E}^{-1}\big{(}J_{E}z_{n}-\gamma
A^{*}J_{F}(Az_{n}-\omega_{n})\big{)}\big{)}$ $\displaystyle=$
$\displaystyle\|z\|^{2}+\|J_{E}z_{n}-\gamma
A^{*}J_{F}(Az_{n}-\omega_{n})\|^{2}$ $\displaystyle-2\langle
z,J_{E}z_{n}-\gamma A^{*}J_{F}(Az_{n}-\omega_{n})\rangle$ $\displaystyle=$
$\displaystyle\|z\|^{2}+\|J_{E}z_{n}-\gamma
A^{*}J_{F}(Az_{n}-\omega_{n})\|^{2}-2\langle z,J_{E}z_{n}\rangle$
$\displaystyle+2\gamma\langle z,A^{*}J_{F}(Az_{n}-\omega_{n})\rangle$
$\displaystyle\leq$ $\displaystyle\|z\|^{2}+\|J_{E}z_{n}\|^{2}-2\gamma\langle
Az_{n},J_{F}(Az_{n}-\omega_{n})\rangle$
$\displaystyle+c\gamma^{2}\|A\|^{2}\|Az_{n}-\omega_{n}\|^{2}-2\langle
z,J_{E}z_{n}\rangle+2\gamma\langle Az,J_{F}(Az_{n}-\omega_{n})\rangle$
$\displaystyle=$ $\displaystyle\phi_{E}(z,z_{n})-2\gamma\langle
Az_{n}-Az,J_{F}(Az_{n}-\omega_{n})\rangle$
$\displaystyle+c\gamma^{2}\|A\|^{2}\|Az_{n}-\omega_{n}\|^{2}$ $\displaystyle=$
$\displaystyle\phi_{E}(z,z_{n})-2\gamma\langle
Az_{n}-\omega_{n},J_{F}(Az_{n}-\omega_{n})\rangle$
$\displaystyle-2\gamma\langle\omega_{n}-Az,J_{F}(Az_{n}-\omega_{n})\rangle+c\gamma^{2}\|A\|^{2}\|Az_{n}-\omega_{n}\|^{2}$
$\displaystyle=$
$\displaystyle\phi_{E}(z,z_{n})-2\gamma\|Az_{n}-\omega_{n}\|^{2}-2\gamma\langle\omega_{n}-Az,J_{F}(Az_{n}-\omega_{n})\rangle$
$\displaystyle+c\gamma^{2}\|A\|^{2}\|Az_{n}-\omega_{n}\|^{2}$ (3.18)
$\displaystyle\leq$
$\displaystyle\phi_{E}(z,z_{n})-\gamma(2-c\gamma\|A\|^{2})\|Az_{n}-\omega_{n}\|^{2},$

hence, it is implied from the condition $0<\gamma<\dfrac{2}{c\|A\|^{2}}$ that

(3.19) $\phi_{E}(z,y_{n})\leq\phi_{E}(z,z_{n}).$

Also, using (3.19) and (3.14), it is concluded that

(3.20) $\lim_{n\rightarrow\infty}\phi_{E}(x_{n+1},y_{n})=0.$

Therefore, it is followed from (3.7), (3.20) and Lemma 2.2 that

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

Next, it is evaluated that
$\displaystyle\lim_{n\rightarrow\infty}\|Az_{n}-\omega_{n}\|=0.$ Indeed,
putting $z=x_{n+1}$, from (3), it is observed that

(3.22)
$\gamma(2-c\gamma\|A\|^{2})\|Az_{n}-\omega_{n}\|^{2}\leq\phi_{E}(x_{n+1},z_{n})-\phi_{E}(x_{n+1},y_{n})\leq\phi_{E}(x_{n+1},z_{n}),$

for all $n\in\mathbb{N}$. Since $0<\gamma<\frac{2}{c\parallel
A\parallel^{2}}$, it is concluded from (3.14) that

(3.23) $\lim_{n\rightarrow\infty}\|Az_{n}-\omega_{n}\|=0.$

Since $\\{x_{n}\\}$ is bounded, from (3.10), (3.16), (3.21) and (3.23), the
sequences $\\{u_{n}\\},$ $\\{z_{n}\\},$ $\\{y_{n}\\}$ and $\\{\omega_{n}\\}$
are bounded.

From the condition (iii) and the fact that
$\\{\|J_{E}Sx_{n}-J_{E}W_{n}x_{n}\|\\}$ is bounded, it is observed that

$\displaystyle\lim_{n\rightarrow\infty}\|$
$\displaystyle\big{(}\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n}\big{)}-J_{E}W_{n}x_{n}\|$
$\displaystyle=\lim_{n\rightarrow\infty}\|(1-\sigma_{n})J_{E}Sx_{n}-(1-\sigma_{n})J_{E}W_{n}x_{n}\|$
$\displaystyle=\lim_{n\rightarrow\infty}(1-\sigma_{n})\|J_{E}Sx_{n}-J_{E}W_{n}x_{n}\|=0,$

hence, from the continuity of $J_{E}\Pi_{C}J_{E}^{-1}$, it is understood that

$\displaystyle\lim_{n\rightarrow\infty}\|J_{E}\Pi_{C}J_{E}^{-1}\big{(}\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n}\big{)}-J_{E}\Pi_{C}J_{E}^{-1}(J_{E}W_{n}x_{n})\|$
(3.24)
$\displaystyle=\lim_{n\rightarrow\infty}\|J_{E}\Pi_{C}J_{E}^{-1}\big{(}\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n}\big{)}-J_{E}(W_{n}x_{n})\|=0.$

Now, it is shown that
$\displaystyle\lim_{n\rightarrow\infty}\|J_{E}u_{n}-J_{E}W_{n}x_{n}\|=0$,

$\displaystyle\|J_{E}u_{n}-J_{E}W_{n}x_{n}\|=$
$\displaystyle\big{\|}\big{(}(1-\alpha_{n})J_{E}x_{n}$
$\displaystyle+\alpha_{n}J_{E}\Pi_{C}J_{E}^{-1}\big{(}\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n})\big{)}$
$\displaystyle-J_{E}W_{n}x_{n}\big{\|}$ $\displaystyle=$
$\displaystyle\big{\|}\big{(}(1-\alpha_{n})J_{E}x_{n}-(1-\alpha_{n})J_{E}W_{n}x_{n}\big{)}$
$\displaystyle+\alpha_{n}\big{(}J_{E}\Pi_{C}J_{E}^{-1}\big{(}\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n}\big{)}$
$\displaystyle-J_{E}W_{n}x_{n}\big{)}\big{\|}$ $\displaystyle\leq$
$\displaystyle(1-\alpha_{n})\|J_{E}x_{n}-J_{E}u_{n}\|$
$\displaystyle+(1-\alpha_{n})\|J_{E}u_{n}-J_{E}W_{n}x_{n}\|$
$\displaystyle+\alpha_{n}\big{\|}\big{(}J_{E}\Pi_{C}J_{E}^{-1}\big{(}\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n}\big{)}$
$\displaystyle-J_{E}W_{n}x_{n}\big{)}\big{\|},$

which implies that

$\displaystyle\alpha_{n}\|J_{E}u_{n}-J_{E}W_{n}x_{n}\|\leq$
$\displaystyle(1-\alpha_{n})\|J_{E}x_{n}-J_{E}u_{n}\|$
$\displaystyle+\alpha_{n}\big{\|}\big{(}J_{E}\Pi_{C}J_{E}^{-1}\big{(}\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n}\big{)}$
$\displaystyle-J_{E}W_{n}x_{n}\big{)}\big{\|},$

therefore,

$\displaystyle\|J_{E}u_{n}-J_{E}W_{n}x_{n}\|\leq$
$\displaystyle(1-\alpha_{n})\frac{\|J_{E}x_{n}-J_{E}u_{n}\|}{\alpha_{n}}+\big{\|}\big{(}J_{E}\Pi_{C}J_{E}^{-1}\big{(}\sigma_{n}J_{E}W_{n}x_{n}$
$\displaystyle+(1-\sigma_{n})J_{E}Sx_{n}\big{)}-J_{E}W_{n}x_{n}\big{)}\big{\|}$

now, using (3) and the condition (ii), it is concluded that

$\lim_{n\rightarrow\infty}\|J_{E}u_{n}-J_{E}W_{n}x_{n}\|=0,$

hence, because $E^{*}$ is uniformly smooth, it is induced that

$\lim_{n\rightarrow\infty}\|u_{n}-W_{n}x_{n}\|=0,$

therefore, it is deduced from (3.10) that

(3.25) $\lim_{n\rightarrow\infty}\|x_{n}-W_{n}x_{n}\|=0.$

Since $\\{x_{n}\\}$ is bounded, there exists a subsequence $\\{x_{n_{k}}\\}$
which converges weakly to a point $x^{*}\in E$. First, we show that
$x^{*}\in\bigcap_{i=1}^{\infty}Fix(T_{i})$. To see that, by Theorems 2.9 and
2.10, the mapping $W:C\rightarrow C$ satisfies

(3.26) $\lim_{n\rightarrow\infty}\|W_{n}x^{*}-Wx^{*}\|=0.$

Moreover, from Theorem 2.9, it is followed that
$Fix(W)=\cap_{i=1}^{\infty}Fix(T_{i})$. Assume that
$x^{*}\notin\cap_{i=1}^{\infty}Fix(T_{i})$ then $x^{*}\neq Wx^{*}$ and using
(3.25), (3.26) and Opial’s property of Banach space, it is concluded that

$\displaystyle\liminf_{k\rightarrow\infty}\|x_{n_{k}}-x^{*}\|<$
$\displaystyle\liminf_{k\rightarrow\infty}\|x_{n_{k}}-Wx^{*}\|$
$\displaystyle\leq$
$\displaystyle\liminf_{k\rightarrow\infty}\big{(}\|x_{n_{k}}-W_{n_{k}}x_{n_{k}}\|+\|W_{n_{k}}x_{n_{k}}-W_{n_{k}}x^{*}\|$
$\displaystyle+\|W_{n_{k}}x^{*}-Wx^{*}\|\big{)}$ $\displaystyle\leq$
$\displaystyle\liminf_{k\rightarrow\infty}\|x_{n_{k}}-x^{*}\|.$

which is a contradiction. Therefore, $x^{*}\in\cap_{i=1}^{\infty}Fix(T_{i})$.

Next, it will be checked that $x^{*}\in M_{1}^{-1}0$. From (3.10) and the fact
that $\\{x_{n_{k}}\\}$ converges weakly to $x^{*}$, there exists a subsequence
$\\{u_{n_{k}}\\}$ of $\\{u_{n}\\}$ converging weakly to $x^{*}$ and therefore
from (3.17), it is induced that
$\\{J_{\lambda_{n_{k}}}^{M_{1}}(u_{n_{k}}+e_{n_{k}})\\}$ converges weakly to
$x^{*}$. Also, from $(iv)$ and (3.17), it is implied that

$\lim_{n\rightarrow\infty}\|(u_{n}+e_{n})-J_{\lambda_{n}}^{M_{1}}(u_{n}+e_{n})\|=0,$

hence, since $E$ is uniformly smooth, it is understood that

(3.27)
$\lim_{n\rightarrow\infty}\|J_{E}(u_{n}+e_{n})-J_{E}J_{\lambda_{n}}^{M_{1}}(u_{n}+e_{n})\|=0.$

Since $J_{\lambda_{n}}^{M_{1}}$ is the generalized resolvent of $M_{1}$, it is
observed that

$\frac{J_{E}(u_{n}+e_{n})-J_{E}J_{\lambda_{n}}^{M_{1}}(u_{n}+e_{n})}{{\lambda_{n}}}\in
M_{1}J_{\lambda_{n}}^{M_{1}}(u_{n}+e_{n}),\quad\forall n\in\mathbb{N}.$

From the monotonicity of $M_{1}$, it is deduced that

(3.28)
$\Big{\langle}r-J_{\lambda_{n_{k}}}^{M_{1}}(u_{n_{k}}+e_{n_{k}}),t^{*}-\frac{J_{E}(u_{n_{k}}+e_{n_{k}})-J_{E}J_{\lambda_{n_{k}}}^{M_{1}}(u_{n_{k}}+e_{n_{k}})}{{\lambda_{n_{k}}}}\Big{\rangle}\geq
0,$

for all $(r,t^{*})\in M_{1}$. From (3.27) and the condition
$0<a\leq\lambda_{n_{k}}$, it is followed that $\langle
r-x^{*},t^{*}-0\rangle\geq 0,$ for all $(r,t^{*})\in M_{1}.$ Since $M_{1}$ is
maximal monotone, we have $x^{*}\in M_{1}^{-1}0$.

Next, we show that $x^{*}\in A^{-1}(M_{2}^{-1}0)$. From (3.16) and the fact
that $\\{x_{n_{k}}\\}$ converges weakly to $x^{*}$, there exists a subsequence
$\\{z_{n_{k}}\\}$ of $\\{z_{n}\\}$ converging weakly to $x^{*}$ and since $A$
is bounded and linear, we also have that $\\{Az_{n_{k}}\\}$ converges weakly
to $Ax^{*}$. Therefore, from (3.23), we have
$\\{Q_{\mu_{n_{k}}}^{M_{2}}Az_{n_{k}}\\}$ converges weakly to $Ax^{*}$. Also,
since $F$ is uniformly smooth, it is induced from (3.23) that

(3.29) $\lim_{n\rightarrow\infty}\|J_{F}Az_{n}-J_{F}\omega_{n}\|=0.$

Since $Q_{\mu_{n}}^{M_{2}}$ is the generalized resolvent of $M_{2}$, it is
understood that

$\frac{J_{F}Az_{n}-J_{F}Q_{\mu_{n}}^{M_{2}}Az_{n}}{{\mu_{n}}}\in
M_{2}Q_{\mu_{n}}^{M_{2}}Az_{n},\quad\forall n\in\mathbb{N}.$

From the monotonicity of $M_{2}$, it follows that

(3.30)
$\Big{\langle}b-Q_{\mu_{n_{k}}}^{M_{2}}Az_{n_{k}},f^{*}-\frac{J_{F}Az_{n_{k}}-J_{F}Q_{\mu_{n_{k}}}^{M_{2}}Az_{n_{k}}}{{\mu_{n_{k}}}}\Big{\rangle}\geq
0,\quad\forall(b,f^{*})\in M_{2}.$

From (3.29) and the condition $0<a\leq\mu_{n_{k}}$, it is concluded that
$\langle b-Ax^{*},f^{*}-0\rangle\geq 0,$ for all $(b,f^{*})\in M_{2}$. Since
$M_{2}$ is maximal monotone, it is implied that $x^{*}\in
A^{-1}(M_{2}^{-1}0)$.
Therefore, $x^{*}\in\Omega=M_{1}^{-1}0\cap
A^{-1}(M_{2}^{-1}0)\cap\big{(}\bigcap_{i=1}^{\infty}Fix(T_{i})\big{)}$ hence,
$\Omega\neq\emptyset$.

(b) Now, let $\bar{x}$ be an arbitrary element of $\omega_{\omega}(x_{n})$
(the set of all weak limit point of the sequence $\\{x_{n}\\}$). Then there
exists another subsequence $\\{x_{n_{i}}\\}$ of $\\{x_{n}\\}$ which converges
weakly to $\bar{x}$. Clearly, repeating the same argument, it is implied that
$\bar{x}\in\Omega$. It is claimed that $\bar{x}=x^{*}$. Indeed, suppose that
$\bar{x}\neq x^{*}$. Obviously, from (3.7), the sequences $\\{x_{n}\\}$ is
cauchy and hence the sequences $\\{\|x_{n}-\bar{x}\|\\}$ and
$\\{\|x_{n}-x^{*}\|\\}$ are convergent. Again, using Opial’s property, it is
concluded that

$\displaystyle\lim_{n\rightarrow\infty}\|x_{n}-\bar{x}\|$
$\displaystyle=\liminf_{i\rightarrow\infty}\|x_{n_{i}}-\bar{x}\|<\liminf_{i\rightarrow\infty}\|x_{n_{i}}-x^{*}\|=\lim_{n\rightarrow\infty}\|x_{n}-x^{*}\|$
$\displaystyle=\liminf_{k\rightarrow\infty}\|x_{n_{k}}-x^{*}\|<\liminf_{k\rightarrow\infty}\|x_{n_{k}}-\bar{x}\|=\lim_{n\rightarrow\infty}\|x_{n}-\bar{x}\|,$

this is a contradiction and thus $\bar{x}=x^{*}$. Therefore,
$\omega_{\omega}(x_{n})$ is singleton. Thus $\\{x_{n}\\}$ converges weakly to
$x^{*}\in\Omega$. Since norm is weakly lower semicontinuous, it is implied
from (3.3) that

$\displaystyle\phi_{E}(\omega_{0},x_{1})$
$\displaystyle=\phi_{E}(\Pi_{\Omega}x_{1},x_{1})\leq\phi_{E}(x^{*},x_{1})\qquad\qquad$
$\displaystyle=\|x^{*}\|^{2}-2\langle x^{*},J_{E}x_{1}\rangle+\|x_{1}\|^{2}$
$\displaystyle\leq\liminf_{k\rightarrow\infty}(\|x_{n_{k}}\|^{2}-2\langle
x_{n_{k}},J_{E}x_{1}\rangle+\|x_{1}\|^{2})$
$\displaystyle=\liminf_{k\rightarrow\infty}\phi_{E}(x_{n_{k}},x_{1})$
$\displaystyle\leq\limsup_{k\rightarrow\infty}\phi_{E}(x_{n_{k}},x_{1})\leq\phi_{E}(\omega_{0},x_{1}),$

hence, from the definition of $\Pi_{\Omega}x_{1}$, it is understood that
$\omega_{0}=x^{*}$ and

$\lim_{k\rightarrow\infty}\phi_{E}(x_{n_{k}},x_{1})=\phi_{E}(x^{*},x_{1})=\phi_{E}(\omega_{0},x_{1}).$

So, it is deduced that
$\displaystyle\lim_{k\rightarrow\infty}\|x_{n_{k}}\|=\|\omega_{0}\|$. From the
Kadec-Klee property of $E$, it is concluded that
$\displaystyle\lim_{k\rightarrow\infty}x_{n_{k}}=\omega_{0}$. Therefore,
$\displaystyle\lim_{k\rightarrow\infty}x_{n}=\omega_{0}$. Thus $\\{x_{n}\\}$
converges strongly to $x^{*}$ where $x^{*}=\Pi_{\Omega}x_{1}$. ∎

##  4\. Applications and numerical example

In this section, using Theorem 3.1, a new strong convergence theorem in Banach
spaces will be demonstrated. Let $E$ be a Banach space and $f$ be a proper,
lower semicontinuous and convex function of $E$ into $(-\infty,\infty]$.
Recall the definition of the subdifferential $\partial f$ of $f$ as follows:

$\partial f(x)=\\{z^{*}\in E^{*}:f(x)+\langle y-x,z^{*}\rangle\leq
f(y),\,\,\forall y\in E\\}$

for all $x\in E$. It is known that $\partial f$ is a maximal monotone operator
by Rocfellar [21]. Let $C$ be a nonempty, closed and convex subset of $E$ and
$i_{C}$ be the indicator function of $C$, i.e.,

$i_{C}(x)=\left\\{\begin{array}[]{r}0\quad x\in C,\\\ \,\infty\quad x\notin
C.\end{array}\right.$

Then $i_{C}$ is a proper, lower semicontinuous and convex function on $E$ and
hence, the subdifferential $\partial i_{C}$ of $i_{C}$ is a maximal monotone
operator. Therefore, the generalized resolvent $j_{\lambda}$ of $\partial
i_{C}$ for $\lambda>0$ is defined as follows:

(4.1) $J_{\lambda}x=(J+\lambda\partial i_{C})^{-1}Jx,\quad\forall x\in E.$

For any $x\in E$ and $u\in C$, the following relations are hold:

$\displaystyle u=$ $\displaystyle J_{\lambda}x\Longleftrightarrow Jx\in
Ju+\lambda\partial i_{C}u$
$\displaystyle\Longleftrightarrow\frac{1}{\lambda}(Jx-Ju)\in\partial i_{C}u$
$\displaystyle\Longleftrightarrow i_{C}y\geq\langle y-u,\frac{1}{\lambda}(Jx-
Ju)\rangle+i_{C}u,\,\,\,\forall y\in E$ $\displaystyle\Longleftrightarrow
0\geq\langle y-u,\frac{1}{\lambda}(Jx-Ju)\rangle,\,\,\,\forall y\in C$
$\displaystyle\Longleftrightarrow\langle y-u,Jx-Ju\rangle\leq 0,\,\,\,\forall
y\in C$ $\displaystyle\Longleftrightarrow u=\Pi_{C}x.$

Next, using Theorem (3.1), a strong convergence theorem for finding minimizers
of convex functions in two Banach spaces is demonstrated.

######  Theorem 4.1.

Let $E$ and $F$ be two $2$-uniformly convex and uniformly smooth real Banach
spaces that satisfies the Opial condition and let $J_{E}$ and $J_{F}$ be the
duality mappings on $E$ and $F$, respectively. Let $C$ and $Q$ be nonempty,
closed and convex subsets of $E$ and $F$ respectively. Let $A:E\longrightarrow
F$ be a bounded linear operator such that $A\neq 0$ and with the adjoint
operator $A^{*}$. Suppose that $S:C\longrightarrow E$ be a nonexpansive
mapping and $\\{T_{i}\\}_{i=1}^{\infty}:C\longrightarrow C$ a family of
nonexpansive mappings. For every $n\in\mathbb{N}$, let $W_{n}$ be a
$W-mapping$ generated by Definition 2.8. Let $x_{1}\in C$ and let
$\\{x_{n}\\}$, $\\{u_{n}\\}$ and $\\{y_{n}\\}$ be the sequences generated by

$\displaystyle\begin{cases}u_{n}=J_{E}^{-1}\Big{(}(1-\alpha_{n})J_{E}x_{n}+\alpha_{n}J_{E}\Pi_{C}J_{E}^{-1}(\sigma_{n}J_{E}W_{n}x_{n}+(1-\sigma_{n})J_{E}Sx_{n})\Big{)};\\\
z_{n}=\Pi_{C}(u_{n}+e_{n});\quad\quad\omega_{n}=\Pi_{Q}(Az_{n});\\\
y_{n}=\Pi_{C}J_{E}^{-1}\Big{(}J_{E}z_{n}-\gamma
A^{*}J_{F}(Az_{n}-\omega_{n})\Big{)};\\\ C_{n}=\\{z\in
C,\big{\langle}\omega_{n}-Az,J_{F}(Az_{n}-\omega_{n})\big{\rangle}\geq
0\\};\\\ D_{n}=\\{z\in E,\phi_{E}(z,z_{n})\leq\phi_{E}(z,u_{n}+e_{n})\\};\\\
Q_{n}=\\{z\in E,\langle x_{n}-z,J_{E}x_{1}-J_{E}x_{n}\rangle\geq 0\\};\\\
x_{n+1}=\Pi_{C_{n}\cap Q_{n}\cap D_{n}}x_{1},\quad\forall n\in\mathbb{N}.\\\
\end{cases}$

where $0<\gamma<\frac{2}{c\parallel A\parallel^{2}}$ with $c>0$. Let
$\\{\alpha_{n}\\},\\{\sigma_{n}\\}$ be real sequences in $(0,1)$ satisfying
the conditions

  * ($i$)

$\displaystyle\lim_{n\rightarrow\infty}\alpha_{n}=0$;

  * ($ii$)

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{\|J_{E}x_{n}-J_{E}u_{n}\|}{\alpha_{n}}=0$;

  * (iii)

$\displaystyle\lim_{n\rightarrow\infty}\sigma_{n}=1$;

and the error sequence $\\{e_{n}\\}\subseteq E$ such that

  * ($iv$)

$\displaystyle\lim_{n\rightarrow\infty}\|e_{n}\|=0$.

Let $\Omega=C\cap A^{-1}Q\cap\big{(}\bigcap_{i=1}^{\infty}Fix(T_{i})\big{)}$.
Suppose that one of the following two conditions is hold:

  * (v)

the sequence $\\{x_{n}\\}$ is bounded,
or

  * (vi)

$\Omega\neq\emptyset$.

Then

  1. (a)

$\Omega\neq\emptyset$ if and only if the sequence $\\{x_{n}\\}$ is bounded,

  2. (b)

The sequence $\\{x_{n}\\}$ converges strongly to a point $\omega_{0}\in\Omega$
where $\omega_{0}=\Pi_{\Omega}x_{1}$.

Next, Theorem 3.1 will be illustrated by an example:
A numerical example Let $E=F=\mathbb{R}$, the set of real numbers, with the
inner product defined by $\langle x,y\rangle=xy,\forall x,y\in\mathbb{R}$, and
usual norm $|\cdot|$. Suppose that $C=[0,1]$ and the mapping
$A:\mathbb{R}\rightarrow\mathbb{R}$ is defined by $A(x)=-2x,\forall
y\in\mathbb{R}$. Let $T_{i}:C\rightarrow C$ be the identity function for each
$i\in\mathbb{N}$ and hence the mapping $W_{n}:C\rightarrow C$ is the identity
function for each $n\in\mathbb{N}$. Also suppose that
$S:C\rightarrow\mathbb{R}$ is the identity function. Let
$M_{1},M_{2}:\mathbb{R}\rightarrow 2^{\mathbb{R}}$ be defined by
$M_{1}(x)=\\{2x\\},\forall x\in\mathbb{R}$ and $M_{2}(y)=\\{3y\\},\forall
y\in\mathbb{R}$. Then $M_{1}^{-1}0\cap A^{-1}(M_{2}^{-1}0)\subseteq C$ and
$\Omega\neq\emptyset$. Let $\\{\alpha_{n}\\}$ and $\\{\sigma_{n}\\}$ be
arbitrary real sequences in $(0,1)$ such that
$\displaystyle\lim_{n\rightarrow\infty}\alpha_{n}=0$ and
$\displaystyle\lim_{n\rightarrow\infty}\sigma_{n}=1$. Let
$\lambda_{n}=\mu_{n}=0.25,\>e_{n}=n^{-1}$ for each $n\in\mathbb{N}$, and
$\gamma=0.1$. Then the sequences
$\\{x_{n}\\},\\{u_{n}\\},\\{z_{n}\\},\\{\omega_{n}\\}$ and $\\{y_{n}\\}$
generated by (3.1) az follows: given initial value $x_{1}\in C$

$\displaystyle\begin{cases}u_{n}=x_{n}\\\
z_{n}=\dfrac{2}{3}(x_{n}+n^{-1});\quad\quad\omega_{n}=\dfrac{-16}{21}(x_{n}+n^{-1});\\\
y_{n}=\Pi_{C}\big{(}\dfrac{116}{210}(x_{n}+n^{-1})\big{)};\\\
C_{n}=\big{\\{}z\in C,z\leq\dfrac{16(x_{n}+n^{-1})}{42}\big{\\}};\\\
D_{n}=\big{\\{}z\in E,z\leq\dfrac{5x_{n}+2n^{-1}}{6}\big{\\}};\\\
Q_{n}=\\{z\in E,(x_{n}-z)(x_{1}-x_{n})\geq 0\\};\\\ x_{n+1}=\Pi_{C_{n}\cap
Q_{n}\cap D_{n}}x_{1},\quad\forall n\in\mathbb{N}.\\\ \end{cases}$

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