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Approximation of common solutions for a fixed point problem of asymptotically nonexpansive mapping and a generalized equilibrium problem in Hilbert space

Abstract

In this paper, we introduce an iterative algorithm to approximate a common solution of a generalized equilibrium problem and a fixed point problem for an asymptotically nonexpansive mapping in a real Hilbert space. We prove that the sequences generated by the iterative algorithm converge strongly to a common solution of the generalized equilibrium problem and the fixed point problem for an asymptotically nonexpansive mapping. The results presented in this paper extend and generalize many previously known results in this research area. Some applications of main results are also provided.

Introduction

Throughout the paper unless otherwise stated, let H be a real Hilbert space with inner product 〈.,.〉 and induced norm .. Let C be a nonempty closed convex subsets of H. Let {xn} be a sequence in H, then xnx (respectively, \(x_{n}\rightharpoonup x\)) denotes strong (respectively, weak) convergence of the sequence {xn} to a point xH. We denote by \(\mathbb {N}\) and \(\mathbb {R}\) the sets of all positive integers and all real numbers, respectively. For every point xH, there exists a unique nearest point of C, denoted by PCx, such that

$$\|x - P_{C}x\|\leq\|x - y\| \textup{ for all } y\in C.$$

Such a PC is called the metric projection from H onto C.

A mapping T:CC is said to be asymptotically nonexpansive [1] if there exists a sequence {kn}[1,) with \(\lim \limits _{n\to \infty }{k_{n}}=1\) such that

$$\begin{array}{*{20}l} \lVert{{T}^{n}x-{T}^{n}y}\rVert\leq{k}_{n}\lVert{x-y}\rVert, \;\forall x,y\in C. \end{array} $$

T is said to be a uniformly k-Lipschitzian for a positive constant k if

$$\begin{array}{*{20}l} \lVert{T^{n}x-T^{n}y}\rVert\leq k\lVert{x-y}\rVert, \forall x,y\in C, \forall n\in\mathbb{N}. \end{array} $$

If \( k_{n}=1, \forall n\in \mathbb {N} \), then T is said to be a nonexpansive mapping. A point xX is called a fixed point for T if Tx=x.

The fixed point problem (in short, FPP) for the mapping T:CC is to find xC such that

$$ Tx = x. $$
(1)

The solution set of FPP (1) is denoted by F(T), that is,

$$F(T)=\{x\in C:Tx=x\}.$$

Let \(F:C\times C \to \mathbb {R}\) be a bifunction and A:CH be a nonlinear mapping. The generalized equilibrium problem is to find zC such that

$$\begin{array}{*{20}l} F(z,y)+\langle{Az,y-z}\rangle\geq0, \forall y\in C. \end{array} $$
(2)

The set of the solution of the problem (2) is denoted by EP(F,A), that is,

$$EP(F,A)=\{z\in C:F(z,y)+\langle{Az,y-z}\rangle\geq0, \forall y\in C\}.$$

If A≡0 in (2), then problem (2) reduces to the equilibrium problem of finding an element zC such that,

$$\begin{array}{*{20}l} F(z,y)\geq 0, \forall y\in C. \end{array} $$
(3)

The set of solutions of problem (3) is denoted by EP(F).

If F≡0 in (2), then the generalized equilibrium problem (2) is reduced to finding a point zC such that,

$$\begin{array}{*{20}l} \langle{Az,y-z}\rangle\geq0,\forall y\in C, \end{array} $$
(4)

which is called the classical variational inequality problem. The set of solution of the problem (4) is denoted by VI(C,A).

If we define F(x,y)=〈Ax,yx〉 for all x,yC, then zEP(F) if and only if 〈Az,yz〉≥0 for all yC and hence zVI(C,A).

The problem (2) is very general in the sense that it includes many special cases such as optimization problems, variational inequalities, minimax problems, and the Nash equilibrium problem in noncooperative games; see Blum and Oettli [2], Kazmi and Rizvi [3], Meche et al.[4], Moudafi and Théra [5], Zegeye et al. [6], and the references therein.

Throughout this paper, let us assume that a bifunction \(F : C\times C\to \mathbb {R}\) satisfies the following conditions:

  1. (A1)

    F(x,x)=0 for all xC;

  2. (A2)

    F is monotone, i.e., F(x,y)+F(y,x)≤0 for all x,yC;

  3. (A3)

    for each x,y,zC;

    $${\lim}_{t\downarrow 0} sup \, F(tz+(1-t)x,y)\leq F(x,y);$$
  4. (A4)

    for each xC, yF(x,y) is convex and lower semi-continuous.

Definition 1

A mapping A:CH is called α-inverse strongly monotone if there exists a positive real number α such that,

$$\begin{array}{*{20}l} \langle{Ax-Ay,x-y}\rangle\geq \alpha\|{Ax-Ay}\|^{2}, \forall x,y\in C. \end{array} $$

Remark 1

Every α-inverse strong monotone mapping is \(\frac {1}{\alpha }\)-Lipschitz mapping; however, the converse may not hold.

Takahashi and Takahashi [7] obtained the following strong convergence theorem to find a common solution of generalized equilibrium problem and the fixed point problem of a nonexpansive mapping in a Hilbert space.

Theorem 1

[7] Let C be a nonempty closed convex subset of a real Hilbert space H. Let \(F:C\times C\to \mathbb {R}\) be a bifunction satisfying (A1), (A2), (A3), and (A4). Let A:CH be an α-inverse strongly monotone mapping, and let T:CC be a nonepansive mapping such that F(T)∩EP(F,A)≠. Let uC and x1C and let {zn}C and {xn}C be sequence generated by

$$\left\{\begin{array}{ll} F(z_{n},y)+\langle{Ax_{n},y-z_{n}}\rangle+\frac{1}{\lambda_{n}}\langle y-z_{n},z_{n}-x_{n}\rangle\geq0, \forall y\in C,\\ x_{n+1}=\beta_{n}x_{n}+(1-\beta_{n})T[\alpha_{n}u+(1-\alpha_{n})z_{n}],\forall n\in\mathbb{N}, \end{array}\right.$$

where {αn}[0,1], {βn}[0,1] and {λn}[0,2α] satisfy

  1. 1.

    \(\lim \limits _{n\to \infty }\alpha _{n}=0\) and \(\sum \limits _{n=1}^{\infty }\alpha _{n}=\infty,\)

  2. 2.

    \(\lim \limits _{n\to \infty }(\lambda _{n}-\lambda _{n+1})=0\), and

  3. 3.

    0<cβnd<1, 0<aλnb<2α.

Then, {xn} converges strongly to z=PF(T)∩EP(F,A)(u).

In this paper, motivated by Takahashi and Takahashi [7], we construct an iterative algorithm for approximating a common solution of a generalized equilibrium problem and the fixed point problem for asymptotically nonexpansive mapping. It is also proved that the proposed algorithm converges strongly to a common solution.

Preliminaries

We now introduce preliminaries which will be used in this paper.

Recall that a mapping f:CC is called a contraction mapping if there exists ρ[0,1) such that

$$\lVert{f(x)-f(y)}\rVert\leq\rho\lVert{x-y}\rVert, \forall x,y\in C.$$

Lemma 1

[2] Let C be a nonempty closed convex subset of a real Hilbert space H. Let \( F:C\times C\to \mathbb {R} \) be a bifunction satisfying (A1), (A2), (A3), and (A4). Let r>0 and xH. Then, there exists zC such that

$$F(z,y)+\frac{1}{r}\langle{y-z,z-x}\rangle\geq0, \forall y\in C.$$

Lemma 2

[8] Let C be a nonempty closed convex subset of H and let \( F:C\times C\to \mathbb {R} \) be a bi-function satisfying (A1), (A2), (A3), and (A4). Then, for any r>0 and xH, there exists zC such that

$$F(z,y)+\frac{1}{r}\langle{y-z,z-x}\rangle\geq 0,~~ \forall y\in C.$$

Furthermore, if

$$T_{r}x=\left\{z\in C:F(z,y)+\frac{1}{r}\langle{y-z,z-x}\rangle\geq 0, \forall y\in C\right\},$$

then the following hold:

  1. (1)

    Tr is single valued,

  2. (2)

    Tr is firmly non-expansive, i.e.,

    $$\lVert{T_{r}x-T_{r}y}\rVert^{2}\leq\langle{T_{r}x-T_{r}y,x-y}\rangle, \forall x,y\in H,$$
  3. (3)

    F(Tr)=EP(F),

  4. (4)

    EP(F) is closed and convex.

Remark 2

Replacing x with xrAxH in Lemma 1, there exists zC, such that

$$F(z,y)+\langle{Ax,y-z}\rangle+\frac{1}{r}\langle{y-z,z-x}\rangle\geq0, \forall y\in C.$$

Definition 2

[9] Let C be a closed convex subset of a Hilbert space H. A mapping T:CC is called asymptotically regular at x if and only if,

$${\lim}_{n\to\infty}\lVert{T^{n}x-T^{n+1}x}\rVert=0.$$

Lemma 3

[10] Let F:CC be a bifunction satisfying the conditions (A1) and (A2). Let Tr and Ts be defined as in Lemma 2 with r,s>0. For any x,yH, then

$$\lVert{T_{r}y-T_{s}x}\rVert\leq\lVert{y-x}\rVert+\lvert{\frac{r-s}{r}}\rvert\lVert{T_{r}y-y}\rVert.$$

Lemma 4

[11] Let {δn} be a sequence of non negative real numbers, satisfying

$$\begin{array}{*{20}l} \delta_{n+1}\leq(1-s_{n})\delta_{n}+s_{n}\beta_{n}+\gamma_{n}, \forall n\geq0, \end{array} $$

where {sn},{βn} and {γn} satisfies the conditions: (i) {sn}[0,1], \( \sum \limits _{n=1}^{\infty }s_{n}=\infty \) or equivalently, \( \prod \limits _{n=1}^{\infty }(1-s_{n})=0, \) (ii) \( \lim \sup \limits _{n\to \infty }\beta _{n}\leq 0, \) (iii) \( \gamma _{n}\geq 0, \sum \limits _{n=1}^{\infty }\gamma _{n}\leq \infty. \)

Then,

$${\lim}_{n\to\infty}\delta_{n}=0. $$

Lemma 5

[12] Let T be an asymptotically nonexpansive mapping on a closed and convex subset C of a real Hilbert space H. Then, IT is demiclosed at 0. That is, for a sequence {xn} in C, if \( x_{n}\rightharpoonup x \) and xnTxn→0, then xF(T).

Lemma 6

[13] Let H be a real Hilbert space. Then, for any given x,yH, we have the following inequality:

$$\begin{array}{*{20}l} \lVert{x+y}\rVert^{2}\leq\lVert{x}\rVert^{2}+2\langle{y,x+y}\rangle. \end{array} $$

Lemma 7

[14] Let {tn} be a sequence of nonnegative real numbers such that

$$t_{n+1}\leq(1-a_{n})t_{n}+a_{n}\beta_{n},~~n\geq0 $$

where {an} is a sequence in (0,1) and {βn} is a sequence in \( \mathbb {R} \) such that

  1. (C1)

    \( \sum \limits _{n=0}^{\infty }a_{n}=\infty \) or equivalently \( \prod \limits _{n=0}^{\infty }(1-a_{n})=0 \),

  2. (C2)

    \(\limsup \limits _{n\to \infty }\beta _{n}\leq 0 \).

Then

$${\lim}_{n\to\infty}t_{n}=0.$$

Main results

Let \( F:C\times C\to \mathbb {R} \) be a bifunction satisfying (A1), (A2), (A3), and (A4). Let A:CH be α-inverse strongly monotone mapping. Then, it follows from Lemma 2 that for each r>0 and xH there is wC such that

$$T_{r}(x)=\{w\}$$

where \( T_{r}x=\{z\in C:F(z,y)+\frac {1}{r}\langle {y-z,z-x}\rangle \geq 0, \forall y\in C\}=\{w\},\) so that we identify Trx as simply w.

Let f:CC be ρ-contraction mapping and let T:CC be asymptotically nonexpansive mapping. Let {αn}[0,1] and λn(0,2α). For any x1C, we find z1C such that

$$z_{1}=T_{\lambda_{1}}(x_{1}-\lambda_{1}{Ax}_{1}).$$

Then, we can compute x2C by

$$x_{2}=\alpha_{1}f(x_{1})+(1-\alpha_{1}){Tz}_{1}.$$

Also, we can find z2C such that

$$z_{2}=T_{\lambda_{2}}(x_{2}-\lambda_{2}{Ax}_{2}).$$

After that, we can compute x3C by

$$x_{3}=\alpha_{2}f(x_{2})+(1-\alpha_{2})T^{2}z_{2}.$$

Inductively, we can generate the sequence {xn}C as follows:

$$\left\{\begin{array}{ll} x_{1}\in C, \\ z_{n}=T_{\lambda_{n}}(x_{n}-\lambda_{n}{Ax}_{n}),n=1,2,3,....\\ x_{n+1}=\alpha_{n}f(x_{n})+(1-\alpha_{n})T^{n}z_{n}, n=1,2,3,.... \end{array}\right. $$
(5)

Now, we state and prove our convergence theorem as follows:

Theorem 2

Let C be a nonempty closed convex subset of a real Hilbert space H and let \(F:C\times C\to \mathbb {R}\) be a bifunction satisfying (A1), (A2), (A3), and (A4). Let f:CC be ρ-contraction mapping, A:CH be an α-inverse strongly monotone mapping, and T:CC be asymptotically nonexpansive mapping. Assume that T is asymptotically regular on C such that F(T)∩EP(F,A)≠. Let {αn}[0,1] and {λn}[0,2α] satisfy

  1. (i)

    \(\lim \limits _{n\to \infty }\alpha _{n}=0\), \(\sum \limits _{n=1}^{\infty }\alpha _{n}=\infty,\)

  2. (ii)

    0<aλnb<2α,

  3. (iii)

    \(\lim \limits _{n\to \infty }(\lambda _{n}-\lambda _{n+1})=0,\)

  4. (iv)

    \(\lim \limits _{n\to \infty }\frac {k_{n}-1}{\alpha _{n}}=0.\)

For x1C, if {xn} is the sequence defined by the iterative scheme (5), then {xn} converges strongly to z=PF(T)∩EP(F,A)f(z).

Proof

We first show that {xn} is bounded. Let zF(T)∩EP(F,A). Since \( z=T_{\lambda _{n}}(z-\lambda _{n}Az),\)A is α-inverse strongly monotone and 0<λn≤2α for all \(n\in \mathbb {N},\) we have

$$\begin{array}{*{20}l} \lVert{z_{n}-z}\rVert^{2}=&~ \lVert{T_{\lambda_{n}}(x_{n}-\lambda_{n}{Ax}_{n})-T_{\lambda_{n}}(z-\lambda_{n}Az)}\rVert^{2}\\ \leq&~\lVert{(x_{n}-\lambda_{n}{Ax}_{n})-(z-\lambda_{n}Az)}\rVert^{2} \\ =~&\lVert{(x_{n}-z)-\lambda_{n}({Ax}_{n}-Az)}\rVert^{2}\\ =~&\lVert{x_{n}-z}\rVert^{2}-2\lambda_{n}\langle{x_{n}-z,{Ax}_{n}-Az}\rangle+\lambda_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}\\ \leq~&\lVert{x_{n}-z}\rVert^{2}-2\lambda_{n}\alpha\lVert{Ax_{n}-Az}\rVert^{2}+\lambda_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}\\ =~&\lVert{x_{n}-z}\rVert^{2}+\lambda_{n}(\lambda_{n}-2\alpha)\lVert{Ax_{n}-Az}\rVert^{2}\\ \leq~&\lVert{x_{n}-z}\rVert^{2}. \end{array} $$

Hence, we have

$$ \lVert{z_{n}-z}\rVert\leq\lVert{x_{n}-z}\rVert. $$
(6)

Take ε(0,1−ρ). Since \( \frac {k_{n}-1}{\alpha _{n}}\to 0 \) as n, there exists \( N\in \mathbb {N} \) such that

$$(k_{n}-1)<\epsilon\alpha_{n}\ \text{for all}\ n\geq\mathbb{N}.$$

From (5) and (6) it follows that, for all n>N

$$\begin{array}{*{20}l} \lVert{x_{n+1}-z}\rVert=~&\lVert{\alpha_{n}f(x_{n})+(1-\alpha_{n})T^{n}z_{n}-z}\rVert\\ =~&\lVert{\alpha_{n}f(x_{n})-\alpha_{n}f(z)+\alpha_{n}f(z)-\alpha_{n}z+\alpha_{n}z+(1-\alpha_{n})T^{n}z_{n}-z}\rVert\\ =~&\lVert{\alpha_{n}(f(x_{n})-f(z))+\alpha_{n}(f(z)-z)+(1-\alpha_{n})(T^{n}z_{n}-z)}\rVert\\ \leq~&\alpha_{n}\lVert{f(x_{n})-f(z)}\rVert+\alpha_{n}\lVert{f(z)-z}\rVert+(1-\alpha_{n})\lVert{T^{n}z_{n}-z}\rVert\\ \leq~&\alpha_{n}\rho\lVert{x_{n}-z}\rVert+\alpha_{n}\lVert{f(z)-z}\rVert+(1-\alpha_{n})\lVert{T^{n}z_{n}-z}\rVert\\ \leq~&\alpha_{n}\rho\lVert{x_{n}-z}\rVert+\alpha_{n}\lVert{f(z)-z}\rVert+(1-\alpha_{n})k_{n}\lVert{z_{n}-z}\rVert\\ \leq~&\alpha_{n}\rho\lVert{x_{n}-z}\rVert+\alpha_{n}\lVert{f(z)-z}\rVert+(1-\alpha_{n})k_{n}\lVert{x_{n}-z}\rVert\\ =~&(1-\alpha_{n}(1-\rho))\lVert{x_{n}-z}\rVert+\alpha_{n}\lVert{f(z)-z}\rVert+(1-\alpha_{n})(k_{n}-1)\lVert{x_{n}-z}\rVert\\ \leq~&(1-\alpha_{n}(1-\rho))\lVert{x_{n}-z}\rVert+\alpha_{n}\lVert{f(z)-z}\rVert+\alpha_{n}\epsilon\lVert{x_{n}-z}\rVert\\ =~&(1-\alpha_{n}(1-\rho-\epsilon))\lVert{x_{n}-z}\rVert+\alpha_{n}\lVert{f(z)-z}\rVert\\ \leq~&\max\left\{\lVert{x_{n}-z}\rVert,\frac{1}{1-\rho-\epsilon}\lVert{f(z)-z}\rVert\right\}. \end{array} $$

By induction, we see that, for all n≥1

$$\lVert{x_{n}-z}\rVert\leq\max\{\lVert{x_{1}-z}\rVert,\frac{1}{1-\rho-\epsilon}\lVert{f(z)-z}\rVert\}.$$

So {xn} is bounded, hence {Axn},{f(xn)},{zn} and {Tnzn} are bounded.

Next, we have to prove that

$${\lim}_{n\to\infty}\lVert{x_{n+1}-x_{n}}\rVert=0.$$

Since IλnA is non-expansive and by Lemma 3, then we have

$$\begin{array}{*{20}l} \lVert{z_{n+1}-z_{n}}\rVert=~&\lVert T_{\lambda_{n+1}}(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})-T_{\lambda_{n}} (x_{n}-\lambda_{n} {Ax}_{n})\rVert\\ \leq~&\lVert{(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})-(x_{n}-\lambda_{n}{Ax}_{n})}\rVert\\ +~&\left| {\frac{\lambda_{n+1}-\lambda_{n}}{\lambda_{n+1}}}\right| \lVert{T_{\lambda_{n+1}}(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})-(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})}\rVert\\ =~&\lVert{(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})-(x_{n}-\lambda_{n+1}{Ax}_{n})}+(\lambda_{n}-\lambda_{n+1}){Ax}_{n}\rVert\\ +~&\left| {\frac{\lambda_{n+1}-\lambda_{n}}{\lambda_{n+1}}}\right| \lVert{T_{\lambda_{n+1}}(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})-(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})}\rVert\\ \leq~&\lVert{(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})-(x_{n}-\lambda_{n+1}{Ax}_{n})}\rVert+\lvert{\lambda_{n}-\lambda_{n+1}}\rvert\lVert{Ax_{n}}\rVert\\ +~&\left| {\frac{\lambda_{n+1}-\lambda_{n}}{\lambda_{n+1}}}\right| \lVert{T_{\lambda_{n+1}}(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})-(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})}\rVert\\ \leq~&\lVert{x_{n+1}-x_{n}}\rVert+\lvert{\lambda_{n}-\lambda_{n+1}}\rvert\lVert{Ax_{n}}\rVert\\ +~&\left| {\frac{\lambda_{n+1}-\lambda_{n}}{\lambda_{n+1}}}\right| \lVert{T_{\lambda_{n+1}}(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})-(x_{n+1}-\lambda_{n+1}{Ax}_{n+1})}\rVert\\ \leq~&\lVert{x_{n+1}-x_{n}}\rVert+\lvert{\lambda_{n}-\lambda_{n+1}}\rvert\lVert{Ax_{n}}\rVert +\frac{\lvert\lambda_{n+1}-\lambda_{n}\rvert}{\lambda_{n+1}} P_{n+1} \end{array} $$
(7)

where by \( P_{n+1}=\sup \{\lVert {T_{\lambda _{n+1}}(x_{n+1}-\lambda _{n+1}{Ax}_{n+1})-(x_{n+1}-\lambda _{n+1}{Ax}_{n+1})}\rVert \} \).

On the other hand, from \(\phantom {\dot {i}\!} z_{n}=T_{\lambda _{n}}(x_{n}-\lambda _{n}{Ax}_{n}) \) and \(\phantom {\dot {i}\!}z_{n+1}=T_{\lambda _{n+1}}(x_{n+1}-\lambda _{n+1}{Ax}_{n+1}) \), we have

$$\begin{array}{*{20}l} F(z_{n},y)+\langle{Ax_{n},y-z_{n}}\rangle+\frac{1}{\lambda_{n}}\langle{y-z_{n},z_{n}-x_{n}}\rangle\geq0, \forall y\in C. \end{array} $$
(8)

and

$$\begin{array}{*{20}l} F(z_{n+1},y)+\langle{Ax_{n+1},y-z_{n+1}}\rangle+\frac{1}{\lambda_{n+1}}\langle{y-z_{n+1},z_{n+1}-x_{n+1}}\rangle\geq0, \forall y\in C. \end{array} $$
(9)

Putting y=zn+1 in (8) and y=zn in (9), we have

$$\begin{array}{*{20}l} F(z_{n},z_{n+1})+\langle{Ax_{n},z_{n+1}-z_{n}}\rangle+\frac{1}{\lambda_{n}}\langle{z_{n+1}-z_{n},z_{n}-x_{n}}\rangle\geq0. \end{array} $$
(10)

and

$$\begin{array}{*{20}l} F(z_{n+1},z_{n})+\langle{Ax_{n+1},z_{n}-z_{n+1}}\rangle+\frac{1}{\lambda_{n+1}}\langle{z_{n}-z_{n+1},z_{n+1}-x_{n+1}}\rangle\geq0. \end{array} $$
(11)

So, from (A2), we have,

$$\begin{array}{*{20}l} \langle{Ax_{n+1}-{Ax}_{n},z_{n}-z_{n+1}}\rangle+\left\langle {z_{n+1}-z_{n},\frac{z_{n}-x_{n}}{\lambda_{n}}-\frac{z_{n+1}-x_{n+1}}{\lambda_{n+1}}}\right\rangle \geq0. \end{array} $$

And hence,

$$\begin{array}{*{20}l} 0\leq~&\left\langle {z_{n}-z_{n+1},\lambda_{n}({Ax}_{n+1}-{Ax}_{n})+\frac{\lambda_{n}}{\lambda_{n+1}}(z_{n+1}-x_{n+1})-(z_{n}-x_{n})}\right\rangle \\ =&\left\langle {z_{n+1}-z_{n},z_{n}-z_{n+1}+\left(1-\frac{\lambda_{n}}{\lambda_{n+1}}\right)z_{n+1}+(x_{n+1}-\lambda_{n}{Ax}_{n+1})}\right\rangle\\ +~&\left\langle{z_{n+1}-z_{n},(\lambda_{n}{Ax}_{n}-x_{n})-x_{n+1}+ \frac{\lambda_{n}}{\lambda_{n+1}}x_{n+1} }\right\rangle\\ =&\left\langle {z_{n+1}-z_{n},z_{n}-z_{n+1}+\left(1-\frac{\lambda_{n}}{\lambda_{n+1}}\right)(z_{n+1}-x_{n+1})}\right\rangle \\ +~&\left\langle{z_{n+1}-z_{n}, (x_{n+1}-\lambda_{n}{Ax}_{n+1})-(x_{n}-\lambda_{n}{Ax}_{n})}\right\rangle. \end{array} $$

It then follows that

$$\begin{array}{*{20}l} \lVert{z_{n+1}-z_{n}}\rVert^{2}\leq~&\lVert{z_{n+1}-z_{n}}\rVert\left\lbrace \left| {1-\frac{\lambda_{n}}{\lambda_{n+1}}}\right|\lVert{z_{n+1}-x_{n+1}}\rVert+\lVert{x_{n+1}-x_{n}}\rVert \right\rbrace \end{array} $$

And so, we have

$$\begin{array}{*{20}l} \lVert{z_{n+1}-z_{n}}\rVert\leq~& \left| {1-\frac{\lambda_{n}}{\lambda_{n+1}}}\right|\lVert{z_{n+1}-x_{n+1}}\rVert+\lVert{x_{n+1}-x_{n}}\rVert. \end{array} $$
(12)

Using condition (ii), we obtain

$$\begin{array}{*{20}l} \lVert{z_{n+1}-z_{n}}\rVert\leq~&\lVert{x_{n+1}-x_{n}}\rVert+\frac{1}{\lambda_{n+1}}\lvert{\lambda_{n+1}-\lambda_{n}}\rvert\lVert{z_{n+1}-x_{n+1}}\rVert\\ \leq~&\lVert{x_{n+1}-x_{n}}\rVert+\frac{1}{a}\lvert{\lambda_{n+1}-\lambda_{n}}\rvert M, \end{array} $$
(13)

where \( M=\sup \limits _{n\geq 1}\lVert {z_{n}-x_{n}}\rVert \). Hence, we have

$$\begin{array}{*{20}l} \lVert{z_{n}-z_{n-1}}\rVert\leq~&\lVert{x_{n}-x_{n-1}}\rVert+\frac{1}{a}\lvert{\lambda_{n}-\lambda_{n-1}}\rvert M. \end{array} $$
(14)

Consider

$$\begin{array}{*{20}l} \lVert{T^{n}z_{n}-T^{n-1}z_{n-1}}\rVert\leq~&\lVert{T^{n}z_{n}-T^{n}z_{n-1}}\rVert+\lVert{T^{n}z_{n-1}-T^{n-1}z_{n-1}}\rVert\\ \leq~& k_{n}\lVert{z_{n}-z_{n-1}}\rVert+\lVert{T^{n}z_{n-1}-T^{n-1}z_{n-1}}\rVert. \end{array} $$
(15)

From (5), (14) and (15), we have that

$$\begin{array}{*{20}l} \lVert{x_{n+1}-x_{n}}\rVert=~&\lVert{\alpha_{n}f(x_{n})+(1-\alpha_{n})T^{n}z_{n}-\alpha_{n-1}f(x_{n-1})-(1-\alpha_{n-1})T^{n-1}z_{n-1}}\rVert\\ \leq~&\alpha_{n}\rho\lVert{x_{n}-x_{n-1}}\rVert+\rvert{\alpha_{n}-\alpha_{n-1}}\lvert\left(\lVert{f(x_{n-1}}\rVert+\lVert{T^{n-1}z_{n-1}}\rVert\right)\\ +~&(1-\alpha_{n})\lVert{T^{n}z_{n}-T^{n-1}z_{n-1}}\rVert\\ \leq~&\alpha_{n}\rho\lVert{x_{n}-x_{n-1}}\rVert+\rvert{\alpha_{n}-\alpha_{n-1}}\lvert K +(1-\alpha_{n})\lVert{T^{n}z_{n}-T^{n-1}z_{n-1}}\rVert \\ \leq~&\alpha_{n}\rho\lVert{x_{n}-x_{n-1}}\rVert+\rvert{\alpha_{n}-\alpha_{n-1}}\lvert K +(1-\alpha_{n})k_{n}\lVert{z_{n}-z_{n-1}}\rVert\\ +~&(1-\alpha_{n})\lVert{T^{n}z_{n-1}-T^{n-1}z_{n-1}}\rVert\\ \leq~&\alpha_{n}\rho\lVert{x_{n}-x_{n-1}}\rVert+\rvert{\alpha_{n}-\alpha_{n-1}}\lvert K +(1-\alpha_{n})(k_{n}-1)\lVert{z_{n}-z_{n-1}}\rVert\\ +~&(1-\alpha_{n})\lVert{T^{n}z_{n-1}-T^{n-1}z_{n-1}}\rVert+(1-\alpha_{n})\lVert{z_{n}-z_{n-1}}\rVert\\ \leq~&\alpha_{n}\rho\lVert{x_{n}-x_{n-1}}\rVert+\rvert{\alpha_{n}-\alpha_{n-1}}\lvert K +(k_{n}-1) \lVert{x_{n}-x_{n-1}}\rVert\\ +~&(k_{n}-1)\frac{1}{a}\lvert{\lambda_{n}-\lambda_{n-1}}\rvert M +\lVert{T^{n}z_{n-1}-T^{n-1}z_{n-1}}\rVert\\ +~&(1-\alpha_{n})\left[ \lVert{x_{n}-x_{n-1}}\rVert+\frac{1}{a}\lvert{\lambda_{n}-\lambda_{n-1}}\rvert M\right]\\ \leq~&\!(1\,-\,\alpha_{n}(1-\rho-\epsilon))\lVert{x_{n}-x_{n-1}}\rVert+\rvert{\alpha_{n}-\alpha_{n-1}}\lvert K +\frac{\epsilon\alpha_{n}}{a}\lvert{\lambda_{n}-\lambda_{n-1}}\rvert M\\ +~&\lVert{T^{n}z_{n-1}-T^{n-1}z_{n-1}}\rVert+\frac{(1-\alpha_{n})}{a}\lvert{\lambda_{n}-\lambda_{n-1}}\rvert M,\\ \leq~&(1-\alpha_{n}(1-\rho-\epsilon))\lVert{x_{n}-x_{n-1}}\rVert+\alpha_{n}(1-\rho-\epsilon)\frac{\lvert{\lambda_{n}-\lambda_{n-1}}\rvert}{a} M\\ +~&(1+\alpha_{n}(2\epsilon+\rho))\frac{\lvert{\lambda_{n}-\lambda_{n-1}}\rvert}{a}M+\rvert{\alpha_{n}-\alpha_{n-1}}\lvert K\\ +~&\lVert{T^{n}z_{n-1}-T^{n-1}z_{n-1}}\rVert, \end{array} $$

where K= sup{f(xn)+Tnzn}. Put sn=αn(1−ρε), \( \beta _{n}=\frac {\lvert {\lambda _{n}-\lambda _{n-1}}\rvert }{a}M \) and \( \gamma _{n}=(1+\alpha _{n}(2\epsilon +\rho))\frac {\lvert {\lambda _{n}-\lambda _{n-1}}\rvert }{a}M+\lvert {\alpha _{n}-\alpha _{n-1}}\rvert K+\lVert {T^{n}z_{n-1}-T^{n-1}z_{n-1}}\rVert \). Then,

$$\begin{array}{*{20}l} \lVert{x_{n+1}-x_{n}}\rVert\leq~&(1-s_{n})\lVert{x_{n}-x_{n-1}}\rVert+s_{n}\beta_{n}+\gamma_{n} \end{array} $$

Using Lemma 4, we have

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\lVert{x_{n+1}-x_{n}}\rVert=0. \end{array} $$
(16)

Further by (13) with the condition that \( {\lim }_{n\to \infty }(\lambda _{n}-\lambda _{n+1})=0, \) we get

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\lVert{z_{n+1}-z_{n}}\lVert=0. \end{array} $$
(17)

Since xn=αn−1f(xn−1)+(1−αn−1)Tn−1zn−1, we have

$$\begin{array}{*{20}l} \lVert{x_{n}-T^{n}z_{n}}\rVert\leq~&\lVert{x_{n}-T^{n-1}z_{n-1}}\rVert+\lVert{T^{n-1}z_{n-1}-T^{n}z_{n}}\rVert\\ \leq~&\lVert{x_{n}-T^{n-1}z_{n-1}}\rVert+\lVert{T^{n-1}z_{n-1}-T^{n}z_{n-1}}\rVert+\lVert{T^{n}z_{n-1}-T^{n}z_{n}}\rVert\\ \leq~&\alpha_{n-1}\lVert{f(x_{n-1})-T^{n-1}z_{n-1}}\rVert\\ +~&\lVert{T^{n-1}z_{n-1}-T^{n}z_{n-1}}\rVert+k_{n}\lVert{z_{n-1}-z_{n}}\rVert. \end{array} $$

From (17) with αn→0 as n and T is asymptotically regular on C.It follows that

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\lVert{T^{n}z_{n}-x_{n}}\lVert=0. \end{array} $$
(18)

Now, we have to prove that

$${\lim}_{n\to\infty}\lVert{Tx_{n}-x_{n}}\rVert=0. $$

To show this, we first prove that

$${\lim}_{n\to\infty}\lVert{z_{n}-x_{n}}\rVert= 0. $$

With the fact that A is α-inverse strongly monotone, let us consider the following:

$$\begin{array}{*{20}l} \lVert{z_{n}-z}\rVert^{2}=~&\lVert{T_{\lambda_{n}}(I-\lambda_{n}A)x_{n}-T_{\lambda_{n}}(I-\lambda_{n}A)z}\rVert^{2}\\ \leq~&\lVert{(I-\lambda_{n}A)x_{n}-(I-\lambda_{n}A)z}\rVert^{2}\\ =~&\lVert{(x_{n}-z)-\lambda_{n}({Ax}_{n}-Az}\rVert^{2}\\ =~&\lVert{x_{n}-z}\rVert^{2}-2\lambda_{n}\langle{x_{n}-z,{Ax}_{n}-Az}\rangle+\lambda_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}\\ \leq~&\lVert{x_{n}-z}\rVert^{2}-2\lambda_{n}\alpha\lVert{Ax_{n}-Az}\rVert^{2}+\lambda_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}\\ =~&\lVert{x_{n}-z}\rVert^{2}+\lambda_{n}(\lambda_{n}-2\alpha)\lVert{Ax_{n}-Az}\rVert^{2}. \end{array} $$
(19)

From the convexity of .2, (18), and (19), we have

$$\begin{array}{*{20}l} \lVert{x_{n+1}-z}\rVert^{2}=~&\lVert{\alpha_{n}}f(x_{n})-\alpha_{n}z+\alpha_{n}z+(1-\alpha_{n})T^{n}z_{n}-z\rVert^{2}\\ =~&\lVert{\alpha_{n}(f(x_{n})-z)+(1-\alpha_{n})(T^{n}z_{n}-z)}\rVert^{2}\\ \leq~&\alpha_{n}\lVert{(f(x_{n})-z\rVert^{2}+(1-\alpha_{n})\lVert T^{n}z_{n}-z}\rVert^{2}\\ \leq~&\alpha_{n}\lVert{(f(x_{n})-z\rVert^{2}+(1-\alpha_{n})k_{n}^{2}\lVert z_{n}-z}\rVert^{2} \end{array} $$
(20)
$$\begin{array}{*{20}l} \leq~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})k_{n}^{2}[\lVert{x_{n}-z}\rVert^{2}\\ +~&\lambda_{n}(\lambda_{n}-2\alpha)\lVert{Ax_{n}-Az}\rVert^{2}]\\ =~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})k_{n}^{2}\lVert{x_{n}-z}\rVert^{2}\\ +~&\lambda_{n}(\lambda_{n}-2\alpha)(1-\alpha_{n})k_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2} \end{array} $$
(21)
$$\begin{array}{*{20}l} =~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})\left(k_{n}^{2}-1\right)\lVert{x_{n}-z}\rVert^{2}\\ +~&(1-\alpha_{n})\lVert{x_{n}-z}\rVert^{2} +\lambda_{n}(\lambda_{n}-2\alpha)(1-\alpha_{n})k_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}\\ \leq~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})\left(k_{n}^{2}-1\right)\lVert{x_{n}-z}\rVert^{2}+\lVert{x_{n}-z}\rVert^{2}\\ +~&\lambda_{n}(\lambda_{n}-2\alpha)(1-\alpha_{n})k_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2} \end{array} $$
(22)

which implies that

$$\begin{array}{*{20}l} \lambda_{n}(2\alpha-\lambda_{n})(1-\alpha_{n})k_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}\leq~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})(k_{n}^{2}-1)\lVert{x_{n}-z}\rVert^{2}\\ +~&\lVert{x_{n}-z}\rVert^{2}-\lVert{x_{n+1}-z}\rVert^{2}. \end{array} $$

Since \( \lim \limits _{n\to \infty }\alpha _{n}=0, \lim \limits _{n\to \infty }k_{n}=1 \) and both {f(xn)} and {xn} are bounded by (16), we have

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\lVert{Ax_{n}-Az}\rVert=0. \end{array} $$
(23)

Since (IλnA) is non-expansive and by Lemma 2, we have

$$\begin{array}{*{20}l} \lVert{z_{n}-z}\rVert^{2}=~&\lVert{T_{\lambda_{n}}(x_{n}-\lambda_{n}{Ax}_{n})-T_{\lambda_{n}}(z-\lambda_{n}Az)}\rVert^{2}\\ \leq~&\langle{z_{n}-z,(x_{n}-\lambda_{n}{Ax}_{n})-(z-\lambda_{n}Az)}\rangle\\ =~&\frac{1}{2}\left(\lVert{(x_{n}-\lambda_{n}{Ax}_{n})-(z-\lambda_{n}Az)}\rVert^{2}+\lVert{z_{n}-z}\rVert^{2}\right) \\ ~&-\frac{1}{2}\left(\lVert{(x_{n}-\lambda_{n}{Ax}_{n})-(z-\lambda_{n}Az)-(z_{n}-z)}\rVert^{2}\right) \\ =~&\frac{1}{2}\left(\lVert{(I-\lambda_{n}A)x_{n}-(I-\lambda_{n}A)z}\rVert^{2}+\lVert{z_{n}-z}\rVert^{2}\right) \\ ~&-\frac{1}{2}\left(\lVert{(x_{n}-z_{n})-\lambda_{n}({Ax}_{n}-Az)}\rVert^{2}\right) \\ \leq~&\frac{1}{2}(\lVert{x_{n}-z}\rVert^{2}+\rVert{z_{n}-z}\rVert^{2}-\lVert{(x_{n}-z_{n})-\lambda_{n}({Ax}_{n}-Az)}\rVert^{2})\\ \leq~&\frac{1}{2}\left(\lVert{x_{n}-z}\rVert^{2}+\rVert{z_{n}-z}\rVert^{2}-\lVert{x_{n}-z_{n}}\rVert^{2}+2\lambda_{n}\langle{x_{n}-z_{n},{Ax}_{n}-Az}\rangle\right) \\ -~&\frac{1}{2}\lambda_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2} \end{array} $$

which implies that

$$\begin{array}{*{20}l} \lVert{z_{n}-z}\rVert^{2}\leq~&\lVert{x_{n}-z}\rVert^{2}-\lVert{x_{n}-z_{n}}\rVert^{2}+2\lambda_{n}\langle{x_{n}-z_{n},{Ax}_{n}-Az}\rangle -\lambda_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}. \end{array} $$
(24)

From (20) and (24), we have

$$\begin{array}{*{20}l} \lVert{x_{n+1}-z}\rVert^{2}\leq~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})k_{n}^{2}\left(\lVert{x_{n}-z}\rVert^{2}-\lVert{x_{n}-z_{n}}\rVert^{2}\right) \\ +~& 2\lambda_{n}(1-\alpha)k_{n}^{2}\left(\langle{x_{n}-z_{n},{Ax}_{n}-Az}\rangle-\lambda_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}\right) \\ \leq~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})k_{n}^{2}\lVert{x_{n}-z}\rVert^{2}-(1-\alpha_{n})k_{n}^{2}\lVert{x_{n}-z_{n}}\rVert^{2}\\ +~& 2(1-\alpha_{n})k_{n}^{2}\lambda_{n}\lVert{x_{n}-z_{n}}\rVert\lVert{Ax_{n}-Az}\rVert-(1-\alpha_{n})k_{n}^{2}\lambda_{n}^{2}\lVert{Ax_{n}-Az}\rVert^{2}\\ \leq~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})k_{n}^{2}\lVert{x_{n}-z}\rVert^{2}-(1-\alpha_{n})k_{n}^{2}\lVert{x_{n}-z_{n}}\rVert^{2}\\ +~& 2(1-\alpha_{n})k_{n}^{2}\lambda_{n}\lVert{x_{n}-z_{n}}\rVert\lVert{Ax_{n}-Az}\rVert\\ =~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})(k_{n}^{2}-1)\lVert{x_{n}-z}\rVert^{2}+(1-\alpha_{n})\lVert{x_{n}-z}\rVert^{2}\\ -~&(1-\alpha_{n})k_{n}^{2}\lVert{x_{n}-z_{n}}\rVert^{2} +2(1-\alpha_{n})k_{n}^{2}\lambda_{n}\lVert{x_{n}-z_{n}}\rVert\lVert{Ax_{n}-Az}\rVert)\\ \leq~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})(k_{n}^{2}-1)\lVert{x_{n}-z}\rVert^{2}+\lVert{x_{n}-z}\rVert^{2}\\ -~&(1-\alpha_{n})k_{n}^{2}\lVert{x_{n}-z_{n}}\rVert^{2} + 2(1-\alpha_{n})k_{n}^{2}\lambda_{n}\lVert{x_{n}-z_{n}}\rVert\lVert{Ax_{n}-Az}\rVert. \end{array} $$

Hence,

$$\begin{array}{*{20}l} (1-\alpha_{n})k_{n}^{2}\lVert{x_{n}-z_{n}}\rVert^{2}\leq~&\alpha_{n}\lVert{f(x_{n})-z}\rVert^{2}+(1-\alpha_{n})\left(k_{n}^{2}-1\right)\lVert{x_{n}-z}\rVert^{2}+\lVert{x_{n}-z}\rVert^{2}\\ -~&\lVert{x_{n+1}-z}\rVert^{2} +2(1-\alpha_{n})k_{n}^{2}\lambda_{n}\lVert{x_{n}-z_{n}}\rVert\lVert{Ax_{n}-Az}\rVert. \end{array} $$

Since αn→0, kn→1 as n and {xn}, {zn} are bounded with\(\lim \limits _{n\to \infty } \lVert {x_{n}-z}\rVert ^{2}-\lVert {x_{n+1}-z}\rVert ^{2} =0\), we have

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\lVert{x_{n}-z_{n}}\rVert=0. \end{array} $$
(25)

Combining (16) and (25) we have, znxn+1znxn+xnxn+1which implies that

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\lVert{z_{n}-x_{n+1}}\rVert=0. \end{array} $$
(26)

Since TnznznTnznxn+xnzn and from (18) and (25)

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\lVert{T^{n}z_{n}-z_{n}}\rVert=0. \end{array} $$
(27)

Let \( k=\sup _{n\in \mathbb {N}}k_{n}<\infty \). Consequently, by (17) and (27)

$$\begin{array}{*{20}l} \lVert{Tz_{n}-z_{n}}\rVert\leq~&\lVert{Tz_{n}-T^{n+1}z_{n}}\rVert+\lVert{T^{n+1}z_{n}-T^{n+1}z_{n+1}}\rVert\\ +~&\lVert{T^{n+1}z_{n+1}-z_{n+1}}\rVert+\lVert{z_{n+1}-z_{n}}\rVert\\ \leq~& k_{1}\lVert{z_{n}-T^{n}z_{n}}\rVert+k_{n+1}\lVert{z_{n}-z_{n+1}}\rVert\\ +~&\lVert{T^{n+1}z_{n+1}-z_{n+1}}\rVert+\lVert{z_{n+1}-z_{n}}\rVert\\ =~&k_{1}\lVert{z_{n}-T^{n}z_{n}}\rVert+(k_{n+1}+1)\lVert{z_{n}-z_{n+1}}\rVert+\lVert{T^{n+1}z_{n+1}-z_{n+1}}\rVert\\ \leq~&k\lVert{z_{n}-T^{n}z_{n}}\rVert+(k+1)\lVert{z_{n}-z_{n+1}}\rVert+\lVert{T^{n+1}z_{n+1}-z_{n+1}}\rVert \end{array} $$

This implies that

$${\lim}_{n\to\infty}\lVert{Tz_{n}-z_{n}}\rVert=0.$$

Further, we have the following result:

$$\begin{array}{*{20}l} \lVert{Tx_{n}-x_{n}}\rVert\leq~&\lVert{Tx_{n}-{Tz}_{n}}\rVert+\lVert{Tz_{n}-z_{n}}\rVert+\lVert{z_{n}-x_{n}}\rVert\\ \leq~&k_{1}\lVert{x_{n}-z_{n}}\rVert+\lVert{Tz_{n}-z_{n}}\rVert+\lVert{z_{n}-x_{n}}\rVert\\ \leq~&(k_{1}+1)\lVert{x_{n}-z_{n}}\rVert+\lVert{Tz_{n}-z_{n}}\rVert \end{array} $$

which implies that

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\lVert{Tx_{n}-x_{n}}\rVert=0. \end{array} $$
(28)

Since PF(T)∩EP(F,A)f:CC is a ρ-contraction mapping, therefore, by Banach contraction principle, there exists a unique z0F(T)∩EP(F,A) such that z0=PF(T)∩EP(F,A)f(z0). We shall show that

$$\begin{array}{*{20}l} {\lim}_{n\to\infty}\langle{f(z_{0})-z_{0},x_{n}-z_{0}}\rangle\leq0. \end{array} $$
(29)

Since {xn} is bounded sequence, we can choose a subsequence \( \{x_{n_{i}}\} \) of {xn} such that

$$\begin{array}{*{20}l} \limsup_{n\to\infty}\langle{f(z_{0})-z_{0},x_{n}-z_{0}}\rangle={\lim}_{i\to\infty}\langle{f(z_{0})-z_{0},x_{n_{i}}-z_{0}}\rangle. \end{array} $$
(30)

Without loss of generality, we may assume that \( x_{n_{i}}\rightharpoonup \omega \). Since C is closed and convex, C is weakly closed. So, we have ωC. Now, we will show that ωF(T). In fact, since \( x_{n_{i}}\rightharpoonup \omega \) and xnTxn→0 by Lemma 5, we find that ωF(T).

Next, we show that ωEP(F,A). From (25), we have \( z_{n_{i}}\rightharpoonup \omega \).

Since \(\phantom {\dot {i}\!}z_{n}=T_{\lambda _{n}}(x_{n}-\lambda _{n}{Ax}_{n}) \), that is \(F(z_{n},y)+\langle {Ax_{n},y-z_{n}}\rangle +\frac {1}{\lambda _{n}}\langle {y-z_{n},z_{n}-x_{n}}\rangle \geq 0, \forall y\in C\). From (A2), we have \( \langle {Ax_{n},y-z_{n}}\rangle +\frac {1}{\lambda _{n}}\langle {y-z_{n},z_{n}-x_{n}}\rangle \geq F(y,z_{n}), \forall y\in C\).Replacing n with ni in the above inequality, we have,

$$\begin{array}{*{20}l} \langle{Ax_{n_{i}},y-z_{n_{i}}}\rangle+\frac{1}{\lambda_{n_{i}}}\langle{y-z_{n_{i}},z_{n_{i}}-x_{n_{i}}}\rangle\geq F(y,z_{n_{i}}). \end{array} $$
(31)

Put zt=ty+(1−t)ω for all t(0,1] and yC. Then, we have ztC. So, from (31) we have

$$\begin{array}{*{20}l} \langle{z_{t}-z_{n_{i}},{Az}_{t}}\rangle\geq~&\langle{z_{t}-z_{n_{i}},{Az}_{t}}\rangle-\langle{z_{t}-z_{n_{i}},{Ax}_{n_{i}}}\rangle-\langle{z_{t}-z_{n_{i}},\frac{z_{n_{i}}-x_{n_{i}}}{\lambda_{n_{i}}}}\rangle+F(z_{t},z_{n_{i}})\\ =~&\langle{z_{t}-z_{n_{i}},{Az}_{t}-{Az}_{n_{i}}}\rangle+\langle{z_{t}-z_{n_{i}},{Az}_{n_{i}}-{Ax}_{n_{i}}}\rangle\\ -~&\langle{z_{t}-z_{n_{i}},\frac{z_{n_{i}}-x_{n_{i}}}{\lambda_{n_{i}}}}\rangle+F(z_{t},z_{n_{i}}). \end{array} $$

Since \( \lim \limits _{n\to \infty } \lVert {z_{n_{i}}-x_{n_{i}}}\rVert =0\), we have \( \lim \limits _{n\to \infty }\lVert {Az_{n_{i}}-{Ax}_{n_{i}}}\rVert =0 \).

Further from monotonicity of A, we have \(\phantom {\dot {i}\!}\langle {z_{t}-z_{n_{i}},{Az}_{t}-{Az}_{n_{i}}}\rangle \geq 0 \). It follows from (A4) that

$$\begin{array}{*{20}l} F(z_{t},\omega)\leq{\lim}_{n\to\infty}F(z_{t},z_{n_{i}})\leq~&{\lim}_{n\to\infty}\langle{Az_{t},z_{t}-z_{n_{i}}}\rangle=\langle{Az_{t},z_{t}-\omega}\rangle \end{array} $$
(32)

From (A1),(A4), and (32), we have

$$\begin{array}{*{20}l} 0=F(z_{t},z_{t})\leq~& tF(z_{t},y)+(1-t)F(z_{t},\omega)\\ \leq~& tF(z_{t},y)+(1-t)\langle{z_{t}-\omega,{Az}_{t}}\rangle. \end{array} $$

But ztω=ty+(1−t)ωω=t(yω). So, we have the following 0=F(zt,zt)≤tF(zt,y)+(1−t)tyω,Azt〉 and hence 0≤F(zt,y)+(1−t)〈yω,Azt〉. Letting t→0, we have for each yC,

$$\begin{array}{*{20}l} 0\leq~& F(\omega,y)+\langle{y-\omega,A\omega}\rangle.\\ \Longrightarrow~& \omega\in EP(F,A). \end{array} $$
(33)

Since ωF(T)∩EP(F,A), from (30) and the property of metric projection, we have

$$\begin{array}{*{20}l} \limsup_{n\to\infty}\langle{f(z_{0})-z_{0},x_{n}-z_{0}}\rangle=~&{\lim}_{i\to\infty}\langle{f(z_{0})-z_{0},x_{n_{i}}-z_{0}}\rangle\\ =~&\langle{f(z_{0})-z_{0},\omega-z_{0}}\rangle\leq0. \end{array} $$
(34)

Finally, we prove that \( \lim \limits _{n\to \infty }\lVert {x_{n}-z_{0}}\rVert =0 \).

From (5) and Lemma 6, we obtain

$$\begin{array}{*{20}l} \lVert{x_{n+1}-z_{0}}\rVert^{2}=~&\lVert{\alpha_{n}(f(x_{n})-z_{0})+(1-\alpha_{n})(T^{n}z_{n}-z_{0})}\rVert^{2}\\ \leq~&(1-\alpha_{n})^{2}\lVert{T^{n}z_{n}-z_{0}}\rVert^{2}+2\langle{\alpha_{n}(f(x_{n})-z_{0}),x_{n+1}-z_{0}}\rangle\\ =~&(1-\alpha_{n})^{2}\lVert{T^{n}z_{n}-z_{0}}\rVert^{2}+2\alpha_{n}\langle{f(x_{n})-z_{0},x_{n+1}-z_{0}}\rangle\\ \leq~&[(1-\alpha_{n})k_{n}]^{2}\lVert{z_{n}-z_{0}}\rVert^{2}+2\alpha_{n}\langle{f(x_{n})-z_{0},x_{n+1}-z_{0}}\rangle\\ =~&[(1-\alpha_{n})k_{n}]^{2}\lVert{z_{n}-z_{0}}\rVert^{2}+2\alpha_{n}\langle{f(x_{n})-f(z_{0}),x_{n+1}-z_{0}}\rangle\\ +~&2\alpha_{n}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle\\ \leq~&[(1-\alpha_{n})k_{n}]^{2}\lVert{x_{n}-z_{0}}\rVert^{2}+2\alpha_{n}\langle{f(x_{n})-f(z_{0}),x_{n+1}-z_{0}}\rangle\\ +~&2\alpha_{n}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle\\ \leq~&[(1-\alpha_{n})k_{n}]^{2}\lVert{x_{n}-z_{0}}\rVert^{2}+2\alpha_{n}\rho\lVert{x_{n}-z_{0}}\rVert\lVert{x_{n+1}-z_{0}}\rVert\\ +~&2\alpha_{n}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle\\ \leq~&[(1-\alpha_{n})k_{n}]^{2}\lVert{x_{n}-z_{0}}\rVert^{2}+\alpha_{n}\rho(\lVert{x_{n}-z_{0}}\rVert^{2}+\lVert{x_{n+1}-z_{0}}\rVert^{2})\\ +~&2\alpha_{n}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle\\ =~&\left[ (1-\alpha_{n})(k_{n}-1)+(1-\alpha_{n})\right]^{2}\lVert{x_{n}-z_{0}}\rVert^{2}+\alpha_{n}\rho\lVert{x_{n}-z_{0}}\rVert^{2}\\ +~&\alpha_{n}\rho\lVert{x_{n+1}-z_{0}}\rVert^{2} +2\alpha_{n}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle\\ =~&[1-(2-\rho)\alpha_{n}+\alpha_{n}^{2}+(1-\alpha_{n})^{2}(1-k_{n})^{2}+\alpha_{n}\rho\lVert{x_{n+1}-z_{0}}\rVert^{2}\\ +~& 2(1-\alpha_{n})^{2}(k_{n}-1)]\lVert{x_{n}-z_{0}}\rVert^{2} +2\alpha_{n}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle\\ \leq~&[1-(2-\rho)\alpha_{n}+\alpha_{n}^{2}+(1-k_{n})^{2}+2(k_{n}-1)]\lVert{x_{n}-z_{0}}\rVert^{2}\\ +~&\alpha_{n}\rho\lVert{x_{n+1}-z_{0}}\rVert^{2} +2\alpha_{n}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle. \end{array} $$

Let \( P_{n}=\sup \limits _{n\in \mathbb {N}}~\lVert {x_{n}-z_{0}}\rVert ^{2} \), so now we have

$$\begin{array}{*{20}l} \lVert{x_{n+1}-z_{0}}\rVert^{2}\leq~&\frac{1-(2-\rho)\alpha_{n}}{1-\rho\alpha_{n}}\lVert{x_{n}-z_{0}}\rVert^{2}+\frac{\alpha_{n}^{2}+(k_{n}-1)^{2}+2(k_{n}-1)}{1-\rho\alpha_{n}}P_{n}\\ +~&\frac{2\alpha_{n}}{1-\rho\alpha_{n}}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle\\ =~&\left[ 1-\frac{(2-\rho)\alpha_{n}}{1-\rho\alpha_{n}}\right] \lVert{x_{n}-z_{0}}\rVert^{2}+\frac{\alpha_{n}^{2}+(k_{n}-1)^{2}+2(k_{n}-1)}{1-\rho\alpha_{n}}P_{n}\\ +~&\frac{2\alpha_{n}}{1-\rho\alpha_{n}}\langle{f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle\\ =~&(1-a_{n})\lVert{x_{n}-z_{0}}\rVert^{2}+a_{n}\beta_{n}, \end{array} $$

where \( \beta _{n}= \frac {\alpha _{n}^{2}+(k_{n}-1)^{2}+2(k_{n}-1)}{2(1-\rho)\alpha _{n}}P_{n} +\frac {1}{1-\rho }\langle {f(z_{0})-z_{0},x_{n+1}-z_{0}}\rangle \) and \( a_{n}=\frac {2(1-\rho)\alpha _{n}}{1-\rho \alpha _{n}} \). Since \( \lim \limits _{n\to \infty }a_{n}=0 \), \( \sum \limits _{n=0}^{\infty }a_{n}=\infty \), and \( \limsup \limits _{n\to \infty }\beta _{n}\leq 0 \) by (34), then by Lemma 7, we conclude that \(\lim \limits _{n\to \infty }\lVert {x_{n}-z_{0}}\rVert =0\). □

Applications

Using our main theorem (Theorem 2), we obtain strong convergence theorems in Hilbert space.

Theorem 3

Let C be a nonempty closed convex and bounded subset of H. Let \(F:C\times C\to \mathbb {R}\) be a bifunction satisfying (A1), (A2),(A3), and (A4). Let f:CC be ρ-contraction mapping, and let T:CC be asymptotically nonexpansive mapping. Assume that T is asymptotically regular on C such that F(T)∩EP(F)≠. Let {αn}[0,1] and {λn}[0,2α] satisfy

  1. (i)

    \(\lim \limits _{n\to \infty }\alpha _{n}=0\), \(\sum \limits _{n=1}^{\infty }\alpha _{n}=\infty,\)

  2. (ii)

    0<aλnb<2α,

  3. (iii)

    \(\lim \limits _{n\to \infty }(\lambda _{n}-\lambda _{n+1})=0,\)

  4. (iv)

    \(\lim \limits _{n\to \infty }\frac {k_{n}-1}{\alpha _{n}}=0.\)

For x1C, if {xn} is the sequence defined by the iterative scheme (5), then {xn} converges strongly to z=PF(T)∩EP(F)f(z).

Proof

In theorem (2), put A≡0. We obtain that \( F(z_{n},y)+\frac {1}{\lambda _{n}}\langle y-z_{n},z_{n}-x_{n}\rangle \geq 0, \forall y\in C. \) Then, for all α(0,), we have 〈xy,AxAy〉≥αAxAy2=0,x,yC. Thus, we obtain the desired result by Theorem 2. □

Theorem 4

Let C be a nonempty closed convex and bounded subset of a real Hilbert space H. Let f:CC be ρ-contraction mapping, A be an α-inverse strongly monotone mapping of C into H, and T:CC be asymptotically nonexpansive mapping. Assume that T is asymptotically regular on C such that F(T)∩VI(C,A)≠. Let {αn}[0,1] and {λn}[0,2α] satisfy

  1. (i)

    \(\lim \limits _{n\to \infty }\alpha _{n}=0\), \(\sum \limits _{n=1}^{\infty }\alpha _{n}=\infty,\)

  2. (ii)

    0<aλnb<2α,

  3. (iii)

    \(\lim \limits _{n\to \infty }(\lambda _{n}-\lambda _{n+1})=0,\)

  4. (iv)

    \(\lim \limits _{n\to \infty }\frac {k_{n}-1}{\alpha _{n}}=0.\)

For x1C, if {xn} is the sequence defined by the iterative scheme (5), then {xn} converges strongly to z=PF(T)∩VI(C,A)f(z).

Proof

In Theorem 2, put F≡0. Then, we obtain that \(\langle {Ax_{n},y-z_{n}}\rangle +\frac {1}{\lambda _{n}}\langle y-z_{n},z_{n}-x_{n}\rangle \geq 0, \forall y\in C, \forall n\in \mathbb {N}\).

This implies that 〈xnλnAxnzn,zny〉≥0,yC. So, we find that zn=PC(xnλnAxn). Then, we obtain the desired result from Theorem 2. □

Browder and Patryshyn [9] introduced k- strictly pseudocontractive mapping which is as follows:

A mapping S:CC is called k- strictly pseudocontractive if there exists k[0,1) such that

$$\begin{array}{*{20}l} \lVert{Sx-Sy}\rVert^{2}\leq\lVert{x-y}\rVert^{2}+k\lVert{(I-S)x-(I-S)y}\rVert^{2}, \forall x,y\in C. \end{array} $$

Putting A=IS, we know that

$$\begin{array}{*{20}l} \langle{x-y,Ax-Ay}\rangle\geq\frac{1-k}{2}\lVert{Ax-Ay}\rVert^{2}, \forall x,y\in C. \end{array} $$

By using the above definition and Theorem 2, we can obtain the following theorem.

Theorem 5

Let C be a nonempty closed convex and bounded subset of a real Hilbert space H and let \(F:C\times C\to \mathbb {R}\) be a bi-function satisfying (A1−A4). Let f:CC be ρ-contraction mapping, S be a k-strictly pseudo contractive mapping of C into itself, and T:CC be asymptotically non-expansive mapping. Assume that T is asymptotically regular on C such that F(T)∩EP(F,A)≠, where A=IS. Let {αn}[0,1] and {λn}[0,1−k] satisfy

  1. (i)

    \(\lim \limits _{n\to \infty }\alpha _{n}=0\), \(\sum _{n=1}^{\infty }\alpha _{n}=\infty,\)

  2. (ii)

    0<aλnb<1−k,

  3. (iii)

    \(\lim \limits _{n\to \infty }(\lambda _{n}-\lambda _{n+1})=0,\)

  4. (iv)

    \(\lim \limits _{n\to \infty }\frac {k_{n}-1}{\alpha _{n}}=0.\)

For x1C, if {xn} is the sequence defined by the iterative scheme (5), then {xn} converges strongly to z=PF(T)∩EP(F,A)f(z).

Proof

Since A=IS is \( \frac {1-k}{2} \)-inverse strongly monotone mapping. So, by Theorem 2, we obtain the desired result. □

Remark 3

By replacing asymptotically nonexpansive mapping to nonexpansive single valued mapping, it gives an improved version of the main result due to Takahashi and Takahashi [7].

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References

  1. 1

    Goebel, K., Kirk, W. A.: A fixed point theorem for asymptotically nonexpansive mappings. Proc. Amer. Math. Soc. 35, 171–174 (1972).

    MathSciNet  Article  Google Scholar 

  2. 2

    Blum, E., Oettli, W.: From optimization and variational inequalities to equilibrium problems. Math. Stud. 63, 123–145 (1994).

    MathSciNet  MATH  Google Scholar 

  3. 3

    Kazmi, K. R., Rizvi, S. H.: Iterative approximation of a common solution of a split equilibrium problem, a variational inequality problem and a fixed point problem. Egyptian. J. Math. Soc. 21, 44–51 (2013).

    MathSciNet  Article  Google Scholar 

  4. 4

    Meche, T. H, Sangago, M. G, Zegeye, H: Approximating a common solution of a finite family of generalized equilibrium and fixed point problems. SINET: Ethiop. Sci. J. 38(1), 17–28 (2015).

    Google Scholar 

  5. 5

    Moudafi, A., Théra, M.: Proximal and dynamical approaches to equilibrium problems. In: Lecture Notes in Econom. and Math. Systems, Vol. 477, pp. 187–201. Springer (1999). https://doi.org/10.1007/978-3-642-45780-7_12.

    Google Scholar 

  6. 6

    Zegeye, H., Meche, T. H., Sangago, M. G.: Algorithms of common solutions for a fixed point of hemicontractive-type mapping and a generalized equilibrium problem. Int. Adv. J. Math. Sci. 5(1), 20–26 (2017).

    Google Scholar 

  7. 7

    Takahashi, S, Takahashi, W: Strong convergennce theorem for a generalized equilibrium problem and a nonexpansive mapping in a Hilbert space. Nonlinear Anal. TMA. 69, 1025–1033 (2008).

    Article  Google Scholar 

  8. 8

    Combettes, P. L., Hirstoaga, A.: Equilibrium programming in Hilbert spaces. Nonlinear J. Convex Anal. 6, 117–136 (2005).

    MathSciNet  MATH  Google Scholar 

  9. 9

    Browder, F. E., Petryshyn, W. V.: Construction of fixed points of nonlinear mappings in Hilbert space. Math J. Anal. Appl. 20(1967), 197–228.

    MathSciNet  Article  Google Scholar 

  10. 10

    Cianciaruso, F., Marino, G., Muglia, L., Yao, Y.: A hybrid projection algorithm for finding solutions of mixed equilibrium problem and variational inequality problem. Fixed Point Theory Appl. 2010, 19 (2010).

    MathSciNet  Article  Google Scholar 

  11. 11

    Xu, H. K.: Iterative algorithms for nonlinear operators. Lond. J. Math. Soc. 66(2002), 240–256.

    MathSciNet  Article  Google Scholar 

  12. 12

    Cho, Y. J., Zhou, H., Guo, G.: Weak and strong convergence theorems for three-step iterations with errors for asymptotically nonexpansive mappings. Comput. Math. Appl. 47(4-5), 707–717 (2004).

    MathSciNet  Article  Google Scholar 

  13. 13

    Agarwal, R. P., O’Regan, D., Sahu, D. R.: Fixed point theory for Lipschitzian-type mappings with applications. Springer, New York (2009).

    Google Scholar 

  14. 14

    Xu, H. K.: Another control condition in an iterative method for nonexpansive mappings. Bull. Austral. Math. Soc. 65, 109–113 (2002).

    MathSciNet  Article  Google Scholar 

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Acknowledgments

The authors would like to thank and acknowledge Sida-Mathematics Project at the Univeversity of Dar es Salaam for their financial support through out the preparation of this article.

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This project is funded by the Sida-Mathematics project at the University of Dar es Salaam, Tanzania.

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Osward, R., Kumar, S. & Sangago, M.G. Approximation of common solutions for a fixed point problem of asymptotically nonexpansive mapping and a generalized equilibrium problem in Hilbert space. J Egypt Math Soc 27, 43 (2019). https://doi.org/10.1186/s42787-019-0051-8

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Keywords

  • Fixed points
  • Generalized equilibrium problem
  • Asymptotically nonexpansive mapping
  • Iterative algorithm

Mathematics Subject Classification 2010

  • 47J05
  • 47J20
  • 47H09