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 {x_n} be a sequence in H, then x_n→x (respectively, $x_{n}\rightharpoonup x$) denotes strong (respectively, weak) convergence of the sequence {x_n} to a point x∈H. We denote by $\mathbb {N}$ and $\mathbb {R}$ the sets of all positive integers and all real numbers, respectively. For every point x∈H, there exists a unique nearest point of C, denoted by P_Cx, such that

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

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

A mapping T:C→C is said to be asymptotically nonexpansive [1] if there exists a sequence {k_n}⊂[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 x∈X is called a fixed point for T if Tx=x.

The fixed point problem (in short, FPP) for the mapping T:C→C is to find x∈C 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:C→H be a nonlinear mapping. The generalized equilibrium problem is to find z∈C 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 z∈C 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 z∈C 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,y−x〉 for all x,y∈C, then z∈EP(F) if and only if 〈Az,y−z〉≥0 for all y∈C and hence z∈VI(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:

• (A1) F(x,x)=0 for all x∈C;

• (A2) F is monotone, i.e., F(x,y)+F(y,x)≤0 for all x,y∈C;

• (A3) for each x,y,z∈C;

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

• (A4) for each x∈C, y↦F(x,y) is convex and lower semi-continuous.

Definition 1

A mapping A:C→H 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:C→H be an α-inverse strongly monotone mapping, and let T:C→C be a nonepansive mapping such that F(T)∩EP(F,A)≠∅. Let u∈C and x₁∈C and let {z_n}⊂C and {x_n}⊂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 {α_n}⊂[0,1], {β_n}⊂[0,1] and {λ_n}⊂[0,2α] satisfy

• 1. $\lim \limits _{n\to \infty }\alpha _{n}=0$ and $\sum \limits _{n=1}^{\infty }\alpha _{n}=\infty,$

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

• 3. 0<c≤β_n≤d<1, 0<a≤λ_n≤b<2α.

Then, {x_n} converges strongly to z=P_{F(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:C→C 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 x∈H. Then, there exists z∈C 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 x∈H, there exists z∈C 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) T_r is single valued,

• (2) T_r 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) F(T_r)=EP(F),

• (4) EP(F) is closed and convex.

Remark 2

Replacing x with x−rAx∈H in Lemma 1, there exists z∈C, 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:C→C 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:C→C be a bifunction satisfying the conditions (A1) and (A2). Let T_r and T_s be defined as in Lemma 2 with r,s>0. For any x,y∈H, 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 {δ_n} 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 {s_n},{β_n} and {γ_n} satisfies the conditions: (i) {s_n}⊂[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, I−T is demiclosed at 0. That is, for a sequence {x_n} in C, if $x_{n}\rightharpoonup x$ and x_n−Tx_n→0, then x∈F(T).

Lemma 6

[13] Let H be a real Hilbert space. Then, for any given x,y∈H, 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 {t_n} 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 {a_n} is a sequence in (0,1) and {β_n} is a sequence in $\mathbb {R}$ such that

• (C1) $\sum \limits _{n=0}^{\infty }a_{n}=\infty$ or equivalently $\prod \limits _{n=0}^{\infty }(1-a_{n})=0$,

• (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:C→H be α-inverse strongly monotone mapping. Then, it follows from Lemma 2 that for each r>0 and x∈H there is w∈C 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 T_rx as simply w.

Let f:C→C be ρ-contraction mapping and let T:C→C be asymptotically nonexpansive mapping. Let {α_n}⊂[0,1] and λ_n∈(0,2α). For any x₁∈C, we find z₁∈C such that

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

Then, we can compute x₂∈C by

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

Also, we can find z₂∈C such that

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

After that, we can compute x₃∈C by

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

Inductively, we can generate the sequence {x_n}⊂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:C→C be ρ-contraction mapping, A:C→H be an α-inverse strongly monotone mapping, and T:C→C be asymptotically nonexpansive mapping. Assume that T is asymptotically regular on C such that F(T)∩EP(F,A)≠∅. Let {α_n}⊂[0,1] and {λ_n}⊂[0,2α] satisfy

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

• (ii) 0<a≤λ_n≤b<2α,

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

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

For x₁∈C, if {x_n} is the sequence defined by the iterative scheme (5), then {x_n} converges strongly to z=P_{F(T)∩EP(F,A)}f(z).

Proof

We first show that {x_n} is bounded. Let z∈F(T)∩EP(F,A). Since $z=T_{\lambda _{n}}(z-\lambda _{n}Az),$A is α-inverse strongly monotone and 0<λ_n≤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 {x_n} is bounded, hence {Ax_n},{f(x_n)},{z_n} and {Tⁿz_n} are bounded.

Next, we have to prove that

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

Since I−λ_nA 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=z_n+1 in (8) and y=z_n 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(x_n)∥+∥Tⁿz_n∥}. Put s_n=α_n(1−ρ−ε), $\beta _{n}=\frac {\lvert {\lambda _{n}-\lambda _{n-1}}\rvert }{a}M$ 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 x_n=α_n−1f(x_n−1)+(1−α_n−1)Tⁿ⁻¹z_n−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 α_n→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(x_n)} and {x_n} are bounded by (16), we have

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

(23)

Since (I−λ_nA) 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 α_n→0, k_n→1 as n→∞ and {x_n}, {z_n} 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, ∥z_n−x_n+1∥≤∥z_n−x_n∥+∥x_n−x_n+1∥which implies that

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

(26)

Since ∥Tⁿz_n−z_n∥≤∥Tⁿz_n−x_n∥+∥x_n−z_n∥ 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 P_{F(T)∩EP(F,A)}f:C→C is a ρ-contraction mapping, therefore, by Banach contraction principle, there exists a unique z₀∈F(T)∩EP(F,A) such that z₀=P_{F(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 {x_n} is bounded sequence, we can choose a subsequence $\{x_{n_{i}}\}$ of {x_n} 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 ω∈C. Now, we will show that ω∈F(T). In fact, since $x_{n_{i}}\rightharpoonup \omega$ and x_n−Tx_n→0 by Lemma 5, we find that ω∈F(T).

Next, we show that ω∈EP(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 n_i 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 z_t=ty+(1−t)ω for all t∈(0,1] and y∈C. Then, we have z_t∈C. 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 z_t−ω=ty+(1−t)ω−ω=t(y−ω). So, we have the following 0=F(z_t,z_t)≤tF(z_t,y)+(1−t)t〈y−ω,Az_t〉 and hence 0≤F(z_t,y)+(1−t)〈y−ω,Az_t〉. Letting t→0, we have for each y∈C,

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

(33)

Since ω∈F(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:C→C be ρ-contraction mapping, and let T:C→C be asymptotically nonexpansive mapping. Assume that T is asymptotically regular on C such that F(T)∩EP(F)≠∅. Let {α_n}⊂[0,1] and {λ_n}⊂[0,2α] satisfy

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

• (ii) 0<a≤λ_n≤b<2α,

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

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

For x₁∈C, if {x_n} is the sequence defined by the iterative scheme (5), then {x_n} converges strongly to z=P_F(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 〈x−y,Ax−Ay〉≥α∥Ax−Ay∥²=0,∀x,y∈C. 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:C→C be ρ-contraction mapping, A be an α-inverse strongly monotone mapping of C into H, and T:C→C be asymptotically nonexpansive mapping. Assume that T is asymptotically regular on C such that F(T)∩VI(C,A)≠∅. Let {α_n}⊂[0,1] and {λ_n}⊂[0,2α] satisfy

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

• (ii) 0<a≤λ_n≤b<2α,

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

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

For x₁∈C, if {x_n} is the sequence defined by the iterative scheme (5), then {x_n} converges strongly to z=P_{F(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 〈x_n−λ_nAx_n−z_n,z_n−y〉≥0,∀y∈C. So, we find that z_n=P_C(x_n−λ_nAx_n). 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:C→C 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=I−S, 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:C→C be ρ-contraction mapping, S be a k-strictly pseudo contractive mapping of C into itself, and T:C→C be asymptotically non-expansive mapping. Assume that T is asymptotically regular on C such that F(T)∩EP(F,A)≠∅, where A=I−S. Let {α_n}⊂[0,1] and {λ_n}⊂[0,1−k] satisfy

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

• (ii) 0<a≤λ_n≤b<1−k,

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

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

For x₁∈C, if {x_n} is the sequence defined by the iterative scheme (5), then {x_n} converges strongly to z=P_{F(T)∩EP(F,A)}f(z).

Proof

Since A=I−S 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].