# Fixed point theorems for mappings with condition (B)

## Abstract

In this article, a new type of mappings that satisfies condition (B) is introduced. We study Pazy's type fixed point theorems, demiclosed principles, and ergodic theorem for mappings with condition (B). Next, we consider the weak convergence theorems for equilibrium problems and the fixed points of mappings with condition (B).

## 1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : CH be a mapping, and let F(T) denote the set of fixed points of T. A mapping T : CH is said to be nonexpansive if ||Tx - Ty|| ≤ ||x - y|| for all x, y C. A mapping T : CH is said to be quasi-nonexpansive mapping if F(T) ≠ and ||Tx - Ty|| ≤ ||x - y|| for all x C and y F(T).

In 2008, Kohsaka and Takahashi [1] introduced nonspreading mapping, and obtained a fixed point theorem for a single nonspreading mapping, and a common fixed point theorem for a commutative family of nonspreading mappings in Banach spaces. A mapping T : CC is called nonspreading [1] if

$2\parallel Tx-Ty{\parallel }^{2}\le \phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Ty-x{\parallel }^{2}$

for all x, y C. Indeed, T : CC is a nonspreading mapping if and only if

$\parallel Tx-Ty{\parallel }^{2}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}2⟨x-Tx,y-Ty⟩$

for all x, y C[2].

Recently, Takahashi and Yao [3] introduced two nonlinear mappings in Hilbert spaces. A mapping T : CC is called a TY-1 mapping [3] if

$2\parallel Tx-Ty{\parallel }^{2}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}+\parallel Tx-y{\parallel }^{2}$

for all x, y C. A mapping T : CC is called a TY-2 [3] mapping if

$3\parallel Tx-Ty{\parallel }^{2}\le 2\parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Ty-x{\parallel }^{2}$

for all x, y C.

In 2010, Takahashi [4] introduced the hybrid mappings. A mapping T : CC is hybrid [4] if

$\parallel Tx-Ty{\parallel }^{2}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}⟨x-Tx,y-Ty⟩$

for each x, y C. Indeed, T : CC is a hybrid mapping if and only if

$3\parallel Tx-Ty{\parallel }^{2}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}+\parallel Tx-y{\parallel }^{2}+\parallel Ty-x{\parallel }^{2}$

for all x, y C[4].

In 2010, Aoyoma et al. [5] introduced λ-hybrid mappings in a Hilbert space. Note that the class of λ-hybrid mappings contain the classes of nonexpansive mappings, nonspreading mappings, and hybrid mappings. Let λ be a real number. A mapping T : CC is called λ-hybrid [5] if

$\parallel Tx-Ty{\parallel }^{2}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}2\lambda ⟨x-Tx,y-Ty⟩$

for all x, y C.

In 2010, Kocourek et al. [6] introduced (α, β)-generalized hybrid mappings, and studied fixed point theorems and weak convergence theorems for such nonlinear mappings in Hilbert spaces. Let α, β . A mapping T : CH is (α, β)-generalized hybrid [6] if

$\alpha \parallel Tx-Ty{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(1-\alpha \right)\parallel Ty-x{\parallel }^{2}\le \beta \parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(1-\beta \right)\parallel x-y{\parallel }^{2}$

for all x, y C.

In 2011, Aoyama and Kohsaka [7] introduced α-nonexpansive mapping on Banach spaces. Let C be a nonempty closed convex subset of a Banach space E, and let α be a real number such that α < 1. A mapping T : CE is said to be α-nonexpansive if

$\parallel Tx-Ty{\parallel }^{2}\le \alpha \parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\alpha \parallel Ty-x{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(1-2\alpha \right)\parallel x-y{\parallel }^{2}$

for all x, y C.

Furthermore, we observed that Suzuki [8] introduced a new class of nonlinear mappings which satisfy condition (C) in Banach spaces. Let C be a nonempty subset of a Banach space E. Then, T : CE is said to satisfy condition (C) if for all x, y C,

$\frac{1}{2}\parallel x-Tx\parallel \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y\parallel \phantom{\rule{2.77695pt}{0ex}}⇒\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-Ty\parallel \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y\parallel .$

In fact, every nonexpansive mapping satisfies condition (C), but the converse may be false [8, Example 1]. Besides, if T : CE satisfies condition (C) and F(T) ≠ , then T is a quasi-nonexpansive mapping. However, the converse may be false [8, Example 2].

Motivated by the above studies, we introduced Takahashi's $\left(\frac{1}{2},\frac{1}{2}\right)$-generalized hybrid mappings with Suzuki's sense on Hilbert spaces.

Definition 1.1. Let C be a nonempty closed convex subset of a real Hilbert space H, and let T : CH be a mapping. Then, we say T satisfies condition (B) if for all x, y C,

$\frac{1}{2}\parallel x-Tx\parallel \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y\parallel \phantom{\rule{2.77695pt}{0ex}}⇒\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-Ty{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel x-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}.$

Remark 1.1.

1. (i)

In fact, if T is the identity mapping, then T satisfies condition (B).

2. (ii)

Every $\left(\frac{1}{2},\frac{1}{2}\right)$-generalized hybrid mapping satisfies condition (B). But the converse may be false.

3. (iii)

If T : CC satisfies condition (B) and F(T) ≠ , then T is a quasi-nonexpansive mapping, and this implies that F(T) is a closed convex subset of C [9].

Remark 1.2. Let H = , let C be nonempty closed convex subset of H, and let T : CH be a function. In fact, we have

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}\frac{1}{2}\parallel x-Tx\parallel \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y\parallel \hfill \\ ⇔{\left(Tx\right)}^{2}+\phantom{\rule{2.77695pt}{0ex}}{x}^{2}-2xTx\le 4{x}^{2}+4{y}^{2}-8xy\hfill \\ ⇔{\left(Tx\right)}^{2}-2xTx\le 3{x}^{2}+\phantom{\rule{2.77695pt}{0ex}}4{y}^{2}-8xy\hfill \\ ⇔Tx\left(Tx-2x\right)\le \left(3x-2y\right)\left(x-2y\right),\hfill \end{array}$

and

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}|Tx-Ty{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}|x-Ty{|}^{2}\le \phantom{\rule{2.77695pt}{0ex}}|Tx-y{|}^{2}+|x-y{|}^{2}\hfill \\ ⇔{\left(Tx\right)}^{2}+\phantom{\rule{2.77695pt}{0ex}}{\left(Ty\right)}^{2}-2\mathrm{TxTy}+\phantom{\rule{2.77695pt}{0ex}}{x}^{2}+{\left(Ty\right)}^{2}-2xTy\le {\left(Tx\right)}^{2}+{y}^{2}-2\mathrm{yTx}+{x}^{2}+{y}^{2}-2\mathrm{xy}\hfill \\ ⇔2{\left(Ty\right)}^{2}-2Ty\left(Tx+x\right)\le 2{y}^{2}-2y\left(Tx+x\right)\hfill \\ ⇔{2\left(Ty\right)}^{2}-2{y}^{2}\le 2\left(Ty-y\right)\left(Tx+x\right)\hfill \\ ⇔\left(Ty-y\right)\left(Ty+y\right)\le \left(Ty-y\right)\left(Tx+x\right)\hfill \\ ⇔\left(Ty-y\right)\left[\left(Ty+y\right)-\left(Tx+x\right)\right]\le 0.\hfill \end{array}$

Example 1.1. Let H = C = , and let T : CH be defined by Tx : = -x for each x C. Hence, we have the following conditions:

1. (1)

T is $\left(\frac{1}{2},\frac{1}{2}\right)$-generalized hybrid mapping, and T satisfies condition (B).

2. (2)

T is not a nonspreading mapping. Indeed, if x = 1 and y = -1, then

$2\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}8>0=\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\parallel Ty-x{\parallel }^{2}.$
3. (3)

T is not a TY-1 mapping. Indeed, if x = 1 and y = -1, then

$2\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}8>4=4+0=\phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}.$
4. (4)

T is not a TY-2 mapping. Indeed, if x = 1 and y = -1, then

$3\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=12>0=2\parallel Tx-y{\parallel }^{2}+\parallel Ty-x{\parallel }^{2}.$
5. (5)

T is not a hybrid mapping. Indeed, if x = 1 and y = -1, then

$3\parallel Tx-Ty{\parallel }^{2}=12>4=\phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Ty-x{\parallel }^{2}.$
6. (6)

Now, we want to show that if α ≠ 0, then T is not a α-nonexpansive mapping. For α > 0, let x = 1 and y = -1,

$\parallel Tx-Ty{\parallel }^{2}=4>4-8\alpha =\alpha \parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\alpha \parallel Ty-x{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(1-2\alpha \right)\parallel x-y{\parallel }^{2}.$

For α < 0, let x = y = 1,

$\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=0>8\alpha =\alpha \parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\alpha \parallel Ty-x{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(1-2\alpha \right)\parallel x-y{\parallel }^{2}.$
1. (7)

Similar to (6), if α + β ≠ 1, then T is not a (α, β)-generalized hybrid mapping.

Example 1.2. Let H = , C = [-1, 1], and let T : CC be defined by

$T\left(x\right):=\left\{\begin{array}{c}x\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{1em}{0ex}}x\in \left[-1,0\right],\hfill \\ -x\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{1em}{0ex}}x\in \left(0,1\right],\hfill \end{array}\right\$

for each x C. First, we consider the following conditions:

1. (a)

For x [-1, 0] and $\frac{1}{2}\parallel x-Tx\parallel \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y\parallel$, we know that

(a)1 if y [-1, 0], then Ty = y and (Ty - y)[(Ty + y) - (Tx + x)] = 0;

(a)2 if y [0,1], then Ty = -y and (Ty - y)[(Ty + y) - (Tx + x)] = 4xy ≤ 0.

1. (b)

For x (0, 1] and $\frac{1}{2}\parallel x-Tx\parallel \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y\parallel$, we know that

(b)1 if yx, then xy - x, Tx = -x, and Ty = -y. So, (Ty - y)[(Ty + y) - (Tx + x)] = 0;

(b)2 if y <x, then xx - y and this implies that y ≤ 0. So, (Ty - y)[(Ty + y) - (Tx + x)] = 0.

By these conditions and Remark 1.2, we know that T satisfies condition (B). In fact, T is $\left(\frac{1}{2},\frac{1}{2}\right)$-generalized hybrid mapping. Furthermore, we know that the following conditions:

1. (1)

T is a nonspreading mapping. Indeed, we know that the following conditions hold.

(1)1 If x > 0 and y > 0, then

$2\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=2\parallel x-y{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le 2\parallel x+y{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Ty-x{\parallel }^{2};$

(1)2 If x ≤ 0 and y ≤ 0, then

$2\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=2\parallel x-y{\parallel }^{2}=\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\parallel Ty-x{\parallel }^{2};$

(1)3 If x > 0 and y ≤ 0, then ||Tx - Ty||2 = ||Tx - y||2 = || x+y||2, and ||Ty - x||2 = ||x - y||2. Hence,

$\parallel Tx-y{\parallel }^{2}+\parallel Ty-x{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}2\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=-4xy\ge 0.$
1. (2)

Similar to the above, we know that T is a TY-1 mapping, a TY-2 mapping, a hybrid mapping, (α, β)-generalized hybrid mapping, and T is a α-nonexpansive mapping.

On the other hand, the following iteration process is known as Mann's type iteration process [10] which is defined as

${x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)T{x}_{n},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}n\in ℕ,$

where the initial guess x0 is taken in C arbitrarily and {α n } is a sequence in [0,1].

In 1974, Ishikawa [11] gave an iteration process which is defined recursively by

$\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{2.77695pt}{0ex}}\text{chosen}\phantom{\rule{2.77695pt}{0ex}}\text{arbitrary},\hfill \\ {y}_{n}:=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}T{x}_{n},\hfill \\ {x}_{n+1}:=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}T{y}_{n},\hfill \end{array}\right\$

where {α n } and {β n } are sequences in [0,1].

In 1995, Liu [12] introduced the following modification of the iteration method and he called Ishikawa iteration method with errors: for a normed space E, and T : EE a given mapping, the Ishikawa iteration method with errors is the following sequence

$\left\{\begin{array}{c}{x}_{1}\in E\phantom{\rule{2.77695pt}{0ex}}\text{chosen}\phantom{\rule{2.77695pt}{0ex}}\text{arbitrary},\hfill \\ {y}_{n}:=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}T{x}_{n}+{u}_{n},\hfill \\ {x}_{n+1}:=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}T{y}_{n}+{v}_{n},\hfill \end{array}\right\$

where {α n } and {β n } are sequences in [0,1], and {u n } and {v n } are sequences in E with ${\sum }_{n=1}^{\infty }\parallel {u}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}<\infty$ and ${\sum }_{n=1}^{\infty }\parallel {v}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}<\infty$.

In 1998, Xu [13] introduced an Ishikawa iteration method with errors which appears to be more satisfactory than the one introduced by Liu [12]. For a nonempty convex subset C of E and T : CC a given mapping, the Ishikawa iteration method with errors is generated by

$\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{2.77695pt}{0ex}}\text{chosen}\phantom{\rule{2.77695pt}{0ex}}\text{arbitrary},\hfill \\ {y}_{n}:={a}_{n}{x}_{n}+{b}_{n}T{x}_{n}+{c}_{n}{u}_{n},\hfill \\ {x}_{n+1}:={{a}^{\prime }}_{n}{x}_{n}+{{b}^{\prime }}_{n}T{y}_{n}+{{c}^{\prime }}_{n}{v}_{n},\hfill \end{array}\right\$

where {a n }, {b n }, {c n }, $\left\{{a}_{n}^{\prime }\right\}$, $\left\{{b}_{n}^{\prime }\right\}$, $\left\{{c}_{n}^{\prime }\right\}$ are sequences in [0,1] with a n + b n + c n + = 1 and ${a}_{n}^{\prime }+{b}_{n}^{\prime }+{c}_{n}^{\prime }=1$, and {u n } and {v n } are bounded sequences in C.

Motivated by the above studies, we consider an Ishikawa iteration method with errors for mapping with condition (B).

We also consider the following iteration for mappings with condition (B). Let C be a nonempty closed convex subset of a real Hilbert space H. Let G : C × C be a function. Let T : CH be a mapping. Let {a n }, {b n }, and {θ n } be sequences in [0,1] with a n + b n + θ n = 1. Let {ω n } be a bounded sequence in C. Let {r n } be a sequence of positive real numbers. Let {x n } be defined by u1 H

$\left\{\begin{array}{c}{x}_{n}\in C\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{such}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{that}}\phantom{\rule{2.77695pt}{0ex}}G\left({x}_{n},y\right)+\frac{1}{{r}_{n}}⟨y-{x}_{n},{x}_{n}-{u}_{n}⟩\ge 0\forall y\in C;\hfill \\ {u}_{n+1}:={a}_{n}{x}_{n}+{b}_{n}T{x}_{n}+{\theta }_{n}{\omega }_{n}.\hfill \end{array}\right\$

Furthermore, we observed that Phuengrattana [14] studied approximating fixed points of for a nonlinear mapping T with condition (C) by the Ishikawa iteration method on uniform convex Banach space with Opial property. Here, we also consider the Ishikawa iteration method for a mapping T with condition (C) and improve some conditions of Phuengrattana's result.

In this article, a new type of mappings that satisfies condition (B) is introduced. We study Pazy's type fixed point theorems, demiclosed principles, and ergodic theorem for mappings with condition (B). Next, we consider the weak convergence theorems for equilibrium problems and the fixed points of mappings with condition (B).

## 2 Preliminaries

Throughout this article, let be the set of positive integers and let be the set of real numbers. Let H be a (real) Hilbert space with inner product 〈·, ·〉 and norm || · ||, respectively. We denote the strongly convergence and the weak convergence of {x n } to x H by x n x and x n x, respectively. From [15], for each x, y H and λ [0,1], we have

$\parallel \lambda x+\left(1-\lambda \right)y{\parallel }^{2}=\lambda \parallel x{\parallel }^{2}+\left(1-\lambda \right)\parallel y{\parallel }^{2}-\lambda \left(1-\lambda \right)\parallel x-y{\parallel }^{2}.$

Hence, we also have

$2⟨x-y,u-v⟩=\phantom{\rule{2.77695pt}{0ex}}\parallel x-v{\parallel }^{2}+\parallel y-u{\parallel }^{2}-\parallel x-u{\parallel }^{2}-\parallel y-v{\parallel }^{2}$

for all x, y, u, v H. Furthermore, we know that

$\parallel \alpha x+\beta y+\gamma z{\parallel }^{2}=\alpha \parallel x{\parallel }^{2}+\beta \parallel y{\parallel }^{2}+\gamma \parallel z{\parallel }^{2}-\alpha \beta \parallel x-y{\parallel }^{2}-\alpha \gamma \parallel x-z{\parallel }^{2}-\beta \gamma \parallel y-z{\parallel }^{2}$

for each x, y, z H and α, β, γ [0,1] with α + β + γ = 1 [16].

Let ℓ be the Banach space of bounded sequences with the supremum norm. Let μ be an element of (ℓ)*(the dual space of ℓ). Then, we denote by μ(f) the value of μ at f = (x1, x2, x3, . . .) . Sometimes, we denote by μ n x n the value μ(f). A linear functional μ on ℓ is called a mean if μ(e) = ||μ|| = 1, where e = (1, 1, 1, . . .). For x = (x1, x2, x3, . . .), A Banach limit on ℓ is an invariant mean, that is, μ n x n = μ n x n+1 for any n . If μ is a Banach limit on ℓ, then for f = (x1, x2, x3, . . .) ,

$\underset{n\to \infty }{lim inf}{\phantom{\rule{0.5em}{0ex}}x}_{n}\le {\mu }_{n}{x}_{n}\le \underset{n\to \infty }{lim sup}{\phantom{\rule{0.5em}{0ex}}x}_{n}.$

In particular, if f = (x1, x2, x3, . . .) and x n a , then we have μ(f) = μ n x n = a. For details, we can refer [17].

Lemma 2.1. [17]Let C be a nonempty closed convex subset of a Hilbert space H, {x n } be a bounded sequence in H, and μ be a Banach limit. Let g : C be defined by g(z): = μ n ||x n - z||2for all z C. Then there exists a unique z0 C such that $g\left({z}_{0}\right)=\underset{z\in C}{min}g\left(z\right)$.

Lemma 2.2. [17]Let C be a nonempty closed convex subset of a Hilbert space H. Let P C be the metric projection from H onto C. Then for each x H, we havex - P C x, P C x - y〉 ≥ 0 for all y C.

Lemma 2.3. [17]Let D be a nonempty closed convex subset of a real Hilbert space H. Let P D be the metric projection from H onto D, and let {x n }nbe a sequence in H. If x n x0and P D x n y0, then P D x0= y0.

Lemma 2.4. [18]Let D be a nonempty closed convex subset of a real Hilbert space H. Let P D be the metric projection from H onto D. Let {x n }nbe a sequence in H with ||x n+1 - u||2 ≤ (1 + λ n ) ||x n - u||2+ δ n for all u D and n , where {λ n } and {δ n } are sequences of nonnegative real numbers such that $\sum _{n=1}^{\infty }{\lambda }_{n}<\infty$ and $\sum _{n=1}^{\infty }{\delta }_{n}<\infty$. Then{P D x n } converges strongly to an element of D.

Lemma 2.5. [19]Let {s n } and {t n } be two nonnegative sequences satisfying sn+1s n + t n for each n . If $\sum _{n=1}^{\infty }{t}_{n}<\infty$, then $\underset{n\to \infty }{lim}{s}_{n}$ exists.

The equilibrium problem is to find z C such that

$G\left(z,y\right)\ge 0\phantom{\rule{1em}{0ex}}\mathsf{\text{for each}}\phantom{\rule{1em}{0ex}}y\in C.$
(2.1)

The solution set of equilibrium problem (2.1) is denoted by (EP). For solving the equilibrium problem, let us assume that the bifunction G : C × C satisfies the following conditions:

(A1) G(x, x) = 0 for each x C;

(A2) G is monotone, i.e., G(x, y) + G(y, x) ≤ 0 for any x, y C;

(A3) for each x, y, z C, $\underset{\mathrm{t↓}0}{lim}G\left(\mathrm{tz}+\left(1-t\right)x,y\right)\le G\left(x,y\right)$;

(A4) for each x C, the scalar function yG(x, y) is convex and lower semicontinuous.

Lemma 2.6. [20]Let C be a nonempty closed convex subset of a real Hilbert space H. Let G : C × C be a bifunction which satisfies conditions (A1)-(A4). Let r > 0 and × H. Then there exists z C such that

$G\left(z,y\right)+\frac{1}{r}⟨y-z,z-x⟩\ge 0\phantom{\rule{2.77695pt}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}all\phantom{\rule{2.77695pt}{0ex}}y\in C.$

Furthermore, if

${T}_{r}\left(x\right):=\left\{z\in C:G\left(z,y\right)+\frac{1}{r}⟨y-z,z-x⟩\ge 0\phantom{\rule{2.77695pt}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}all\phantom{\rule{2.77695pt}{0ex}}y\in C\right\},$

then we have:

(i) T r is single-valued;

(ii) T r is firmly nonexpansive, that is, ||T r x - T r y||2 ≤ 〈T r x - T r y, x - yfor each x, y H;

(iii) (EP) is a closed convex subset of C;

(iv) (EP) = F(T r ).

## 3 Fixed point theorems

Proposition 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : CC be a mapping with condition (B). Then for each x, y C, we have:

1. (i)

||Tx - T 2 x||2 + ||x - T 2 x|| ≤ ||x - Tx ||2 ;

2. (ii)

||Tx - T 2 x|| ≤ ||x - Tx || and ||x - T 2 x|| ≤ ||x - Tx ||;

3. (iii)

either $\frac{1}{2}\parallel x-Tx\parallel \le \parallel x-y\parallel$ or $\frac{1}{2}\parallel Tx-{T}^{2}x\parallel \le \parallel Tx-y\parallel$ holds;

4. (iv)

either

$\parallel Tx-Ty{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel x-Ty{\parallel }^{2}\le \parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}$

or

$\parallel {T}^{2}x-Ty\mid {\mid }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-Ty\mid {\mid }^{2}\le \parallel {T}^{2}x-y\mid {\mid }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y\mid {\mid }^{2}$

holds;

1. (v)

$\underset{n\to \infty }{lim}\parallel {T}^{n}x-{T}^{n+2}x\parallel =0$.

Proof Since $\frac{1}{2}\parallel x-Tx\parallel \le \parallel x-Tx\parallel$, it is easy to see (i) and (ii) are satisfied. (iii) Suppose that

$\frac{1}{2}\parallel x-Tx\parallel >\parallel x-y\parallel \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{and}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}\parallel Tx-{T}^{2}x\parallel >\parallel Tx-y\parallel$

holds. So,

$\begin{array}{c}\parallel x-Tx\parallel \le \parallel x-y\parallel \phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\parallel y-Tx\parallel \hfill \\ \phantom{\rule{4em}{0ex}}<\frac{1}{2}\parallel x-Tx\parallel +\phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}\parallel Tx-{T}^{2}x\parallel \hfill \\ \phantom{\rule{4em}{0ex}}\le \frac{1}{2}\parallel x-Tx\parallel +\frac{1}{2}\parallel x-Tx\parallel =\parallel x-Tx\parallel \mathsf{\text{.}}\hfill \end{array}$

This is a contradiction. Therefore, we obtain the desired result. Next, it is easy to get (iv) by (iii).

(v): By (i), we know that

$\parallel {T}^{n+1}x-{T}^{n+2}x{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel {T}^{n}x-{T}^{n+2}x{\parallel }^{2}\le \parallel {T}^{n}x-{T}^{n+1}x{\parallel }^{2}.$

Then {||Tnx - Tn+1x||} is a decreasing sequence, and $\underset{n\to \infty }{lim}\parallel {T}^{n}x-{T}^{n+1}x\parallel$ exists. Furthermore, we have:

$\underset{n\to \infty }{lim}\parallel {T}^{n}x-{T}^{n+2}x{\parallel }^{2}\le \underset{n\to \infty }{lim}\parallel {T}^{n}x-{T}^{n+1}x{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}\underset{n\to \infty }{lim}\parallel {T}^{n+1}x-{T}^{n+2}x{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=0.$

So, $\underset{n\to \infty }{lim}\parallel {T}^{n}x-{T}^{n+2}x\parallel \phantom{\rule{2.77695pt}{0ex}}=0$.

Proposition 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : CC be a mapping with condition (B). Then for each x, y C,

$⟨Tx-Ty,y-Ty⟩\le ⟨x-y,Ty-y⟩\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\parallel {T}^{2}x-x\parallel \cdot \parallel Ty-y\parallel .$

Proof By Proposition 3.1(iv), for each x, y C, either

$\parallel Tx-Ty{\parallel }^{2}+\parallel x-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\parallel x-y{\parallel }^{2}$

or

$\parallel {T}^{2}x-Ty{\parallel }^{2}+\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel {T}^{2}x-y{\parallel }^{2}+\parallel Tx-y{\parallel }^{2}$

holds. In the first case, we have

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}\parallel Tx-Ty{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel x-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}\hfill \\ ⇒\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-Ty{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel x-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}2⟨x-y,y-Ty⟩\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\parallel Ty-y{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel Tx-Ty{\parallel }^{2}+\hfill \\ \phantom{\rule{1em}{0ex}}2⟨Tx-Ty,Ty-y⟩\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\parallel Ty-y{\parallel }^{2}+\parallel x-y{\parallel }^{2}\hfill \\ ⇒⟨x-y,y-Ty⟩\le ⟨Tx-Ty,Ty-y⟩\hfill \\ ⇒⟨Tx-Ty,y-Ty⟩\le ⟨x-y,Ty-y⟩.\hfill \end{array}$

In the second case, we have

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}\parallel {T}^{2}x-Ty{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel Tx-y{\parallel }^{2}+\parallel {T}^{2}x-y{\parallel }^{2}\hfill \\ ⇒\phantom{\rule{2.77695pt}{0ex}}\parallel {T}^{2}x-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}2⟨{T}^{2}x-y,y-Ty⟩\phantom{\rule{2.77695pt}{0ex}}+\parallel y-Ty{\parallel }^{2}+\parallel Tx-Ty{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel Tx-Ty{\parallel }^{2}+\hfill \\ \phantom{\rule{1em}{0ex}}2⟨Tx-Ty,Ty-y⟩\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\parallel y-Ty{\parallel }^{2}+\parallel {T}^{2}x-y{\parallel }^{2}\hfill \\ ⇒⟨Tx-Ty,y-Ty⟩\le ⟨{T}^{2}x-y,Ty-y⟩\hfill \\ \phantom{\rule{8.5em}{0ex}}\le ⟨x-y,Ty-y⟩\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\parallel {T}^{2}x-x\parallel \cdot \parallel Ty-y\parallel .\hfill \end{array}$

Therefore, the proof is completed.

Remark 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : CC be a mapping with condition (B). Then for each x, y C, we have:

1. (a)

||Tx - Ty||2 + ||x - Ty||2 ≤ ||x - y||2 + ||Tx - y||2 + ||T 2 x - x|| · ||Ty - y||.

2. (b)

Tx - Ty, y - Ty〉 ≤ 〈x - y, Ty - y〉 + ||Tx - x|| · ||Ty - y||.

Proof By Proposition 3.2, it is easy to prove Remark 3.1.

The following theorem shows that demiclosed principle is true for mappings with condition (B).

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H, and let T : CC be a mapping with condition (B). Let {x n } be a sequence in C with x n x and $\underset{n\to \infty }{lim}\parallel {x}_{n}-T{x}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$. Then Tx= x.

Proof By Remark 3.1, we get:

$⟨T{x}_{n}-Tx,x-Tx⟩\le ⟨{x}_{n}-x,Tx-x⟩\phantom{\rule{2.77695pt}{0ex}}+\parallel {x}_{n}-T{x}_{n}\parallel \cdot \parallel x-Tx\parallel$

for each n . By assumptions, 〈x - Tx , x - Tx 〉 ≤ 0. So, Tx = x.

Theorem 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H, and let T : CC be a mapping with condition (B). Then {Tnx} is a bounded sequence for some x C if and only if F(T) ≠ .

Proof For each n , let x n := Tnx. Clearly, {x n } is a bounded sequence. By Lemma 2.1, there is a unique z C such that ${\mu }_{n}\parallel {x}_{n}-z{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=\underset{y\in C}{min}{\mu }_{n}\parallel {x}_{n}-y{\parallel }^{2}$. By Proposition 3.2, for each n ,

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}⟨{x}_{n+1}-\mathrm{Tz},z-\mathrm{Tz}⟩\le ⟨{x}_{n}-z,\mathrm{Tz}-z⟩\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\parallel {x}_{n}-{x}_{n+2}\parallel \cdot \parallel z-\mathrm{Tz}\parallel \hfill \\ ⇒\frac{1}{2}\parallel {x}_{n+1}-\mathrm{Tz}{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}\parallel \mathrm{Tz}-z{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}\parallel {x}_{n+1}-z{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}\parallel {x}_{n}-z{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}\parallel z-\mathrm{Tz}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}\parallel {x}_{n}-\mathrm{Tz}{\parallel }^{2}+\parallel {x}_{n}-{x}_{n+2}\parallel \cdot \parallel z-\mathrm{Tz}\parallel \hfill \\ ⇒{\mu }_{n}\parallel {x}_{n}-\mathrm{Tz}{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le {\mu }_{n}\parallel {x}_{n}-z{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\mu }_{n}\parallel {x}_{n}-{x}_{n+2}\parallel \cdot \parallel z-\mathrm{Tz}\parallel .\hfill \end{array}$

By Proposition 3.1(v), μ n ||x n - Tz||2μ n ||x n - z||2. This implies that Tz = z and F(T) ≠ . Conversely, it is easy to see.

Corollary 3.1. Let C be a nonempty bounded closed convex subset of a real Hilbert space H, and let T : CC be a mapping with condition (B). Then F(T) ≠ .

The following theorem shows that Ballion's type Ergodic's theorem is also true for the mapping with condition (B).

Theorem 3.3. Let C be a nonempty closed convex subset of a real Hilbert space H, and let T : CC be a mapping with condition (B). Then the following conditions are equivalent:

1. (i)

for each x C, ${S}_{n}x=\frac{1}{n}\sum _{k=0}^{n-1}{T}^{k}x$ converges weakly to an element of C;

2. (ii)

F(T) ≠ .

In fact, if F(T) ≠ , then for each x C, we know that S n x v, where$v=\underset{n\to \infty }{lim}{P}_{F\left(T\right)}{T}^{n}x$and P F(T) is the metric projection from H onto F(T).

Proof (i) (ii): Take any x C and let x be fixed. Then there exists v C such that S n x v. By Proposition 3.2, for each k , we have:

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}⟨T{T}^{k}x-\mathrm{Tv},v-\mathrm{Tv}⟩\le ⟨{T}^{k}x-v,\mathrm{Tv}-v⟩\phantom{\rule{2.77695pt}{0ex}}+\parallel {T}^{2}{T}^{k}x-{T}^{k}x\parallel \cdot \parallel \mathrm{Tv}-v\parallel \hfill \\ ⇒⟨{T}^{k+1}x-\mathrm{Tv},v-\mathrm{Tv}⟩\le ⟨{T}^{k}x-v,\mathrm{Tv}-v⟩\phantom{\rule{2.77695pt}{0ex}}+\parallel {T}^{k+2}x-{T}^{k}x\parallel \cdot \parallel \mathrm{Tv}-v\parallel \hfill \\ ⇒\sum _{k=0}^{n-2}⟨{T}^{k+1}x-\mathrm{Tv},v-\mathrm{Tv}⟩\hfill \\ \phantom{\rule{1em}{0ex}}\le \sum _{k=0}^{n-2}⟨{T}^{k}x-\mathrm{Tv},\mathrm{Tv}-v⟩\phantom{\rule{2.77695pt}{0ex}}+\sum _{k=0}^{n-2}\parallel {T}^{k+2}x-{T}^{k}x\parallel \cdot \parallel \mathrm{Tv}-v\parallel \hfill \\ ⇒⟨n{S}_{n}x-x-\left(n-1\right)\mathrm{Tv},v-\mathrm{Tv}⟩\le ⟨\left(n-1\right){S}_{n-1}x-\left(n-1\right)\mathrm{Tv},\mathrm{Tv}-v⟩+\hfill \\ \phantom{\rule{1em}{0ex}}\sum _{k=0}^{n-2}\parallel {T}^{k+2}x-{T}^{k}x\parallel \cdot \parallel \mathrm{Tv}-v\parallel \hfill \\ ⇒⟨\frac{n}{n-1}{S}_{n}x-\frac{x}{n-1}-\mathrm{Tv},v-\mathrm{Tv}⟩\hfill \\ \phantom{\rule{1em}{0ex}}\le ⟨{S}_{n-1}x-\mathrm{Tv},\mathrm{Tv}-v⟩+\frac{1}{n-1}\sum _{k=0}^{n-2}\parallel {T}^{k+2}x-{T}^{k}x\parallel \cdot \parallel \mathrm{Tv}-v\parallel .\hfill \end{array}$

By Proposition 3.1(v), $\underset{n\to \infty }{lim}\parallel {T}^{k+2}x-{T}^{k}x\parallel \phantom{\rule{2.77695pt}{0ex}}=0$. This implies that

$\underset{n\to \infty }{lim}\frac{1}{n-1}\sum _{k=0}^{n-2}\parallel {T}^{k+2}x-{T}^{k}x\parallel \phantom{\rule{2.77695pt}{0ex}}=0.$

Since S n x v, we have:

$⟨v-\mathrm{Tv},v-\mathrm{Tv}⟩\le ⟨v-\mathrm{Tv},\mathrm{Tv}-v⟩.$

So, Tv = v.

(ii) (i): Take any x C and u F(T), and let x and u be fixed. Since T satisfies condition (B), ||Tnx - u|| ≤ ||Tn-1x - u|| for each n . Hence, $\underset{n\to \infty }{lim}\parallel {T}^{n}x-u\parallel$ exists and this implies that {Tnx} is a bounded sequence. By Lemma 2.4, there exists z F(T) such that $\underset{n\to \infty }{lim}{P}_{F\left(T\right)}{T}^{n}x=z$. Clearly, z F(T). Besides, we have:

$\parallel {S}_{n}x-u\parallel \phantom{\rule{2.77695pt}{0ex}}\le \frac{1}{n}\sum _{k=0}^{n-1}\parallel {T}^{k}x-u\parallel \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel x-u\parallel .$

So, {S n x} is a bounded sequence. Then there exist a subsequence $\left\{{S}_{{n}_{i}}x\right\}$ of {S n x} and v C such that ${S}_{{n}_{i}}x⇀v$. By the above proof, we have:

$⟨\frac{n}{n-1}{S}_{n}x-\frac{x}{n-1}-\mathrm{Tv},v-\mathrm{Tv}⟩\le ⟨{S}_{n-1}x-\mathrm{Tv},\mathrm{Tv}-v⟩+\frac{1}{n-1}\sum _{k=0}^{n-2}\parallel {T}^{k+2}x-{T}^{k}x\parallel \cdot \parallel \mathrm{Tv}-v\parallel .$

This implies that

$\begin{array}{c}⟨\frac{{n}_{i}}{{n}_{i}-1}{S}_{{n}_{i}}x-\frac{x}{{n}_{i}-1}-Tv,v-Tv⟩\hfill \\ \le ⟨{S}_{{n}_{i}-1}x-\frac{x}{{n}_{i}-1}-Tv,Tv-v⟩+\frac{1}{{n}_{i}-1}\sum _{k=0}^{{n}_{i}-2}\parallel {T}^{k+2}x-{T}^{k}x\parallel \cdot \parallel Tv-v\parallel .\hfill \end{array}$

Since ${S}_{{n}_{i}}x⇀v$, {Tnx} is a bounded sequence, and $\underset{n\to \infty }{lim}\frac{1}{n-1}\sum _{k=0}^{n-2}\parallel {T}^{k+2}x-{T}^{k}x\parallel \phantom{\rule{2.77695pt}{0ex}}=0$, it is easy to see that Tv = v. So, v F(T).

By Lemma 2.2, for each k , 〈Tkx - P F(T) Tkx, P F(T) Tkx - u〉 ≥ 0. This implies that

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}⟨{T}^{k}x-{P}_{F\left(T\right)}{T}^{k}x,u-z⟩\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}⟨{T}^{k}x-{P}_{F\left(T\right)}{T}^{k}x,{P}_{F\left(T\right)}{T}^{k}x-z⟩\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {T}^{k}x-{P}_{F\left(T\right)}{T}^{k}x\parallel \cdot \parallel {P}_{F\left(T\right)}{T}^{k}x-z\parallel \hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {T}^{k}x-z\parallel \cdot \parallel {P}_{F\left(T\right)}{T}^{k}x-z\parallel \hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel x-z\parallel \cdot \parallel {P}_{F\left(T\right)}{T}^{k}x-z\parallel .\hfill \end{array}$

Adding these inequalities from k = 0 to k = n - 1 and dividing by n, we have

$⟨{S}_{n}x-\frac{1}{n}\sum _{k=0}^{n-1}{P}_{F\left(T\right)}{T}^{k}x,u-z⟩\le \frac{\parallel x-z\parallel }{n}\sum _{k=0}^{n-1}\parallel {P}_{F\left(T\right)}{T}^{k}x-z\parallel .$

Since ${S}_{{n}_{k}}x⇀v$ and P F(T) Tkx → z, we get 〈v - z, u - z〉 ≤ 0. Since u is any point of F(T), we know that $v=z=\underset{n\to \infty }{lim}{P}_{F\left(T\right)}{T}^{n}x$.

Furthermore, if $\left\{{S}_{{n}_{i}}x\right\}$ is a subsequence of {S n x} and ${S}_{{n}_{i}}x⇀q$, then q = v by following the same argument as the above proof. Therefore, ${S}_{n}x⇀\upsilon =\underset{n\to \infty }{lim}P{T}^{n}x$, and the proof is completed.

## 4 Weak convergence theorems with errors

Theorem 4.1. Let C be a nonempty closed convex subset of a real Hilbert space H, and let T1, T2 : CC be two mappings with condition (B) and Ω: = F(T1) ∩ F(T2) ≠ . Let {a n }, {b n }, {c n }, {d n }, {θ n }, and {λ n } be sequences in [0,1] with

${a}_{n}+{c}_{n}+{\theta }_{n}={b}_{n}+{d}_{n}+{\lambda }_{n}=1,n\in ℕ.$

Let {u n } and {v n } be bounded sequences in C. Let {x n } and {y n } be defined by

$\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{2.77695pt}{0ex}}chosen\phantom{\rule{2.77695pt}{0ex}}arbitrary,\hfill \\ {y}_{n}:={a}_{n}{x}_{n}+{c}_{n}{T}_{1}{x}_{n}+{\theta }_{n}{u}_{n},\hfill \\ {x}_{n+1}:={b}_{n}{x}_{n}+{d}_{n}{T}_{2}{y}_{n}+{\lambda }_{n}{v}_{n}.\hfill \end{array}\right\$

Assume that:

1. (i)

$\underset{n\to \infty }{lim inf}{a}_{n}{c}_{n}>0$ and $\underset{n\to \infty }{lim inf}{b}_{n}{d}_{n}>0$ ;

2. (ii)

$\sum _{n=1}^{\infty }{\theta }_{n}<\infty$ and $\sum _{n=1}^{\infty }{\lambda }_{n}<\infty$.

Then x n z and y n z, where$z=\underset{n\to \infty }{lim}{P}_{\Omega }{x}_{n}$.

Proof Take any w Ω and let w be fixed. Then for each n , we have:

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}\parallel {y}_{n}-w{\parallel }^{2}\hfill \\ =\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {a}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}{c}_{n}{T}_{1}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}{u}_{n}-w{\parallel }^{2}\hfill \\ =\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{a}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{c}_{n}\parallel {T}_{1}{x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}-{a}_{n}{c}_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{T}_{1}{x}_{n}{\parallel }^{2}-{a}_{n}{\theta }_{n}\parallel {x}_{n}-w{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{c}_{n}{\theta }_{n}\parallel {T}_{1}{x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{u}_{n}{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{a}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{c}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}-{a}_{n}{c}_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{T}_{1}{x}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{a}_{n}{\theta }_{n}\parallel {x}_{n}-w{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{c}_{n}{\theta }_{n}\parallel {T}_{1}{x}_{n}-{u}_{n}{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2},\hfill \end{array}$

and

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}\parallel {x}_{n+1}-w{\parallel }^{2}\hfill \\ =\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {b}_{n}{x}_{n}+{d}_{n}{T}_{2}{y}_{n}+{\lambda }_{n}{v}_{n}-w{\parallel }^{2}\hfill \\ =\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{b}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}\parallel {T}_{2}{y}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\parallel {v}_{n}-w{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}-{b}_{n}{d}_{n}\parallel {x}_{n}-{T}_{2}{y}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{b}_{n}{\lambda }_{n}\parallel {x}_{n}-{v}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\lambda }_{n}\parallel {T}_{2}{y}_{n}-\phantom{\rule{2.77695pt}{0ex}}{v}_{n}{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{b}_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}\parallel {y}_{n}-\phantom{\rule{2.77695pt}{0ex}}w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\parallel {v}_{n}-w{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}-{b}_{n}{d}_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{T}_{2}{y}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{b}_{n}{\lambda }_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{v}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\lambda }_{n}\parallel {T}_{2}{y}_{n}-\phantom{\rule{2.77695pt}{0ex}}{v}_{n}{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{b}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}\left(\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}\right)\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\parallel {v}_{n}-\phantom{\rule{2.77695pt}{0ex}}w{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}-{b}_{n}{d}_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{T}_{2}{y}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{b}_{n}{\lambda }_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{v}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\lambda }_{n}\parallel {T}_{2}{y}_{n}-\phantom{\rule{2.77695pt}{0ex}}{v}_{n}{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\parallel {v}_{n}-w{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}-{b}_{n}{d}_{n}\parallel {x}_{n}-{T}_{2}{y}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{b}_{n}{\lambda }_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{v}_{n}{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\lambda }_{n}\parallel {T}_{2}{y}_{n}-{v}_{n}{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\parallel {v}_{n}-w{\parallel }^{2}.\end{array}$

By Lemma 2.5, $\underset{n\to \infty }{lim}\parallel {x}_{n}-w\parallel$ exists. So, {x n } is a bounded sequence. Now, we set $\underset{n\to \infty }{lim}\parallel {x}_{n}-w\parallel \phantom{\rule{2.77695pt}{0ex}}=t$. Besides,

$\begin{array}{c}\phantom{\rule{1em}{0ex}}{b}_{n}{d}_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{T}_{2}{y}_{n}{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{b}_{n}{\lambda }_{n}\parallel {x}_{n}-\phantom{\rule{2.77695pt}{0ex}}{v}_{n}{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\lambda }_{n}\parallel {T}_{2}{y}_{n}-\phantom{\rule{2.77695pt}{0ex}}{v}_{n}{\parallel }^{2}\\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\parallel {v}_{n}-w{\parallel }^{2}-\parallel {x}_{n+1}-\phantom{\rule{2.77695pt}{0ex}}w{\parallel }^{2}.\end{array}$

This implies that

$\underset{n\to \infty }{lim}{b}_{n}{d}_{n}\parallel {x}_{n}-{T}_{2}{y}_{n}{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=0.$

By assumption, $\underset{n\to \infty }{lim}\parallel {x}_{n}-{T}_{2}{y}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$. Furthermore, we have:

$\parallel {x}_{n+1}-w{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}{b}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}\parallel {y}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\parallel {v}_{n}-w{\parallel }^{2}.$

This implies that

$\begin{array}{c}{b}_{n}{d}_{n}\left(\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}-\parallel {y}_{n}-\phantom{\rule{2.77695pt}{0ex}}w{\parallel }^{2}\right)\hfill \\ \le {d}_{n}\left(\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}-\parallel {y}_{n}-\phantom{\rule{2.77695pt}{0ex}}w{\parallel }^{2}\right)\hfill \\ \le \left(1-{b}_{n}\right)\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{d}_{n}\parallel {y}_{n}-w{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\parallel {x}_{n}-w{\parallel }^{2}-\parallel {x}_{n+1}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\lambda }_{n}\parallel {v}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{d}_{n}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}.\hfill \end{array}$

Hence, $\underset{n\to \infty }{lim}{b}_{n}{d}_{n}\left(\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}-\parallel {y}_{n}-w{\parallel }^{2}\right)=0$. By assumption,

$\underset{n\to \infty }{lim}\left(\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}-\parallel {y}_{n}-w{\parallel }^{2}\right)=0.$

Since $\underset{n\to \infty }{lim}{\theta }_{n}\parallel {u}_{n}-w{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=0$,

$\underset{n\to \infty }{lim}\left(\parallel {x}_{n}-w{\parallel }^{2}-\parallel {y}_{n}-w{\parallel }^{2}\right)=0.$

Hence, $\underset{n\to \infty }{lim}\parallel {y}_{n}-w\parallel \phantom{\rule{2.77695pt}{0ex}}=\underset{n\to \infty }{lim}\parallel {x}_{n}-w\parallel \phantom{\rule{2.77695pt}{0ex}}=t$. Similar to the above proof, we also get

$\underset{n\to \infty }{lim}\parallel {x}_{n}-{T}_{1}{x}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0.$

Besides,

$\begin{array}{c}∥{y}_{n}-{x}_{n}∥=\phantom{\rule{2.77695pt}{0ex}}\parallel {a}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}{c}_{n}{T}_{1}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}{u}_{n}-{x}_{n}\parallel \hfill \\ \phantom{\rule{4em}{0ex}}\le {c}_{n}\parallel {T}_{1}{x}_{n}-{x}_{n}\parallel +\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {x}_{n}-{u}_{n}\parallel \hfill \\ \phantom{\rule{4em}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel {T}_{1}{x}_{n}-{x}_{n}\parallel +\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {x}_{n}-{u}_{n}\parallel .\hfill \end{array}$

This implies that $\underset{n\to \infty }{lim}\parallel {y}_{n}-{x}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$ and $\underset{n\to \infty }{lim}\parallel {y}_{n}-{T}_{2}{y}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$. Since {x n } is a bounded sequence, there exists a subsequence $\left\{{x}_{{n}_{k}}\right\}$ of {x n } such that ${x}_{{n}_{k}}⇀z$. By Theorem 3.1, z = T1z.

If ${x}_{{n}_{j}}$ is a subsequence of {x n } and ${x}_{{n}_{j}}⇀q$, then T1q = q. Suppose that qz. Then we have:

$\begin{array}{c}\underset{k\to \infty }{lim inf}\parallel {x}_{{n}_{k}}-z\parallel \phantom{\rule{2.77695pt}{0ex}}<\underset{k\to \infty }{lim inf}\parallel {x}_{{n}_{k}}-q\parallel \phantom{\rule{2.77695pt}{0ex}}=\underset{n\to \infty }{lim}\parallel {x}_{n}-q\parallel \phantom{\rule{2.77695pt}{0ex}}=\underset{j\to \infty }{lim}\parallel {x}_{{n}_{j}}-q\parallel \hfill \\ \phantom{\rule{7.5em}{0ex}}<\underset{j\to \infty }{lim inf}\parallel {x}_{{n}_{j}}-z\parallel \phantom{\rule{2.77695pt}{0ex}}=\underset{n\to \infty }{lim}\parallel {x}_{n}-z\parallel \phantom{\rule{2.77695pt}{0ex}}=\underset{k\to \infty }{lim inf}\parallel {x}_{{n}_{k}}-z\parallel .\hfill \end{array}$

And this leads to a contradiction. Then every weakly convergent subsequence of x n has the same limit. So, x n z F(T1). Since x n z and $\underset{n\to \infty }{lim}\parallel {x}_{n}-{y}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$, y n z. By Theorem 3.1, z F(T2). Hence, z Ω.

Next, by Lemma 2.4, PΩx n converges. Then there exists v Ω such that $\underset{n\to \infty }{lim}{P}_{\Omega }{x}_{n}=v$. By Lemma 2.3, PΩz = v. Since z Ω, $z=v=\underset{n\to \infty }{lim}{P}_{\Omega }{x}_{n}$, and the proof is completed.

In Theorem 4.1, if θ n = λ n = 0 for each n , then we have the following result.

Theorem 4.2. Let C be a nonempty closed convex subset of a real Hilbert space H, and let T1, T2 : CC be two mappings with condition (B) and Ω: = F(T1) ∩ F(T2) ≠ . Let {a n } and (b n ) be two sequences in [0,1]. Let {x n } be defined by

$\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{2.77695pt}{0ex}}chosen\phantom{\rule{2.77695pt}{0ex}}arbitrary,\hfill \\ {y}_{n}:={a}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}\left(1-{a}_{n}\right){T}_{1}{x}_{n},\hfill \\ {x}_{n+1}:={b}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}\left(1-{b}_{n}\right){T}_{2}{y}_{n}.\hfill \end{array}\right\$

Assume that$\underset{n\to \infty }{lim inf}{a}_{n}\left(1-{a}_{n}\right)>0$and$\underset{n\to \infty }{lim inf}{b}_{n}\left(1-{b}_{n}\right)>0$. Then x n z and y n z, where$z=\underset{n\to \infty }{lim}{P}_{\Omega }{x}_{n}$.

Furthermore, we also have the following corollaries from Theorem 4.2.

Corollary 4.1. Let C be a nonempty closed convex subset of a real Hilbert space H, and let T : CC be a mapping with condition (B) and F(T) ≠ . Let {a n } and {b n } be two sequences in [0,1]. Let {x n } be defined by

$\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{2.77695pt}{0ex}}chosen\phantom{\rule{2.77695pt}{0ex}}arbitrary,\hfill \\ {y}_{n}:={a}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}\left(1-{a}_{n}\right)T{x}_{n},\hfill \\ {x}_{n+1}:={b}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}\left(1-{b}_{n}\right)T{y}_{n}.\hfill \end{array}\right\$

Assume that$\underset{n\to \infty }{lim inf}{a}_{n}\left(1-{a}_{n}\right)>0$and$\underset{n\to \infty }{lim inf}{b}_{n}\left(1-{b}_{n}\right)>0$. Then x n z and y n z, where$z=\underset{n\to \infty }{lim}{P}_{F\left(T\right)}{x}_{n}$.

Corollary 4.2. Let C be a nonempty closed convex subset of a real Hilbert space H, and let T : CC be a mapping with condition (B) and F(T) ≠ . Let {b n } be a sequence in [0,1]. Let {x n } be defined by

$\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{2.77695pt}{0ex}}chosen\phantom{\rule{2.77695pt}{0ex}}arbitrary,\hfill \\ {x}_{n+1}:={b}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}\left(1-{b}_{n}\right)T{x}_{n}.\hfill \end{array}\right\$

Assume that$\underset{n\to \infty }{lim inf}{b}_{n}\left(1-{b}_{n}\right)>0$. Then x n z, where$z=\underset{n\to \infty }{lim}{P}_{F\left(T\right)}{x}_{n}$.

Theorem 4.3. Let C be a nonempty closed convex subset of a real Hilbert space H. Let G : C × C be a function satisfying (A1)-(A4). Let T : CC be a mapping with condition (B) and Ω: = F(T) ∩ (EP) ≠ . Let {a n }, {b n }, and {θ n } be sequences in [0,1] with a n + b n + θ n = 1. Let {ω n } be a bounded sequence in C. Let {r n } [a, ∞) for some a > 0. Let {x n } be defined by u1 H

$\left\{\begin{array}{c}{x}_{n}\in C\phantom{\rule{2.77695pt}{0ex}}\text{such}\phantom{\rule{2.77695pt}{0ex}}\text{that}\phantom{\rule{2.77695pt}{0ex}}G\left({x}_{n},y\right)+\frac{1}{{r}_{n}}⟨y-{x}_{n},{x}_{n}-{u}_{n}⟩\ge 0\phantom{\rule{1em}{0ex}}\forall y\in C;\hfill \\ {u}_{n+1}:={a}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}{b}_{n}T{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}{\omega }_{n}.\hfill \end{array}\right\$

Assume that:$\underset{n\to \infty }{lim inf}{a}_{n}{b}_{n}>0$, and$\sum _{n=1}^{\infty }{\theta }_{n}<\infty$. Then x n z, where$z=\underset{n\to \infty }{lim}{P}_{\left(EP\right)\cap F\left(T\right)}{x}_{n}$.

Proof Take any w Ω and let w be fixed. Putting ${x}_{n}={T}_{{r}_{n}}{u}_{n}$ for each n . Then we have:

$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}\parallel {x}_{n+1}-w{\parallel }^{2}\hfill \\ =\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {T}_{{r}_{n+1}}{u}_{n+1}-w{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {u}_{n+1}-w{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {a}_{n}{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}{b}_{n}T{x}_{n}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}{\omega }_{n}-w{\parallel }^{2}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{a}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{b}_{n}\parallel T{x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {\omega }_{n}-w{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{a}_{n}{b}_{n}\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}\\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{a}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{b}_{n}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {\omega }_{n}-w{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{a}_{n}{b}_{n}\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}\\ \le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\parallel {x}_{n}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {\omega }_{n}-w{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}{a}_{n}{b}_{n}\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}.\end{array}$

By Lemma 2.5, $\underset{n\to \infty }{lim}\parallel {x}_{n}-w\parallel$ exists. So, {x n } is bounded. Furthermore, we have:

1. (a)

$\underset{n\to \infty }{lim}{a}_{n}{b}_{n}\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}=0$;

2. (b)

$\underset{n\to \infty }{lim}\parallel {x}_{n}-T{x}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$;

3. (c)

$\parallel {u}_{n+1}-{x}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}\parallel {b}_{n}T{x}_{n}-{b}_{n}{x}_{n}+{\theta }_{n}{w}_{n}-{\theta }_{n}{x}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel T{x}_{n}-{x}_{n}\parallel +\phantom{\rule{2.77695pt}{0ex}}{\theta }_{n}\parallel {w}_{n}-{x}_{n}\parallel$;

4. (d)

$\underset{n\to \infty }{lim}\parallel {u}_{n+1}-{x}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$;

5. (e)

$\underset{n\to \infty }{lim}\parallel {u}_{n}-w\parallel \phantom{\rule{2.77695pt}{0ex}}=\underset{n\to \infty }{lim}\parallel {x}_{n}-w\parallel$.

Following the same argument as the proof of Theorem 4.2, there exists z C such that x n z and Tz = z. Besides, we also have

$\begin{array}{c}{∥{x}_{n+1}-w∥}^{2}=\phantom{\rule{2.77695pt}{0ex}}\parallel {T}_{{r}_{n+1}}{u}_{n+1}-\phantom{\rule{2.77695pt}{0ex}}w{\parallel }^{2}\hfill \\ \phantom{\rule{5em}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}\parallel {T}_{{r}_{n+1}}{u}_{n+1}-{T}_{{r}_{n+1}}w{\parallel }^{2}\hfill \\ \phantom{\rule{5em}{0ex}}\le ⟨{T}_{{r}_{n+1}}{u}_{n+1}-{T}_{{r}_{n+1}}w,\phantom{\rule{2.77695pt}{0ex}}{u}_{n+1}-w⟩\hfill \\ \phantom{\rule{5em}{0ex}}\le ⟨{x}_{n+1}-w,{u}_{n+1}-w⟩\hfill \\ \phantom{\rule{5em}{0ex}}=\frac{1}{2}\parallel {x}_{n+1}-w{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}\parallel {u}_{n+1}-w{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}\parallel {x}_{n+1}-{u}_{n+1}{\parallel }^{2}.\hfill \end{array}$

This implies that

$\parallel {x}_{n+1}-{u}_{n+1}{\parallel }^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel {u}_{n+1}-w{\parallel }^{2}-\phantom{\rule{2.77695pt}{0ex}}\parallel {x}_{n+1}-w{\parallel }^{2}.$

By (e), $\underset{n\to \infty }{lim}\parallel {x}_{n}-{u}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$. Next, we want to show that z (EP). Since ${x}_{n}={T}_{{r}_{n}}{u}_{n}$,

$G\left({x}_{n},y\right)\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\frac{1}{{r}_{n}}⟨y-{x}_{n},{x}_{n}-{u}_{n}⟩\ge 0\phantom{\rule{1em}{0ex}}\forall y\in C.$

By (A2),

$\frac{1}{{r}_{n}}⟨y-{x}_{n},{x}_{n}-{u}_{n}⟩\ge G\left(y,{x}_{n}\right)\phantom{\rule{1em}{0ex}}\forall y\in C.$

By (A4), (i), and $\underset{n\to \infty }{lim}\parallel {x}_{n}-{u}_{n}\parallel \phantom{\rule{2.77695pt}{0ex}}=0$, we get

$0\ge \underset{n\to \infty }{lim}G\left(y,{x}_{n}\right)\ge G\left(y,z\right)\phantom{\rule{1em}{0ex}}\forall y\in C.$

By (A2), G(z, y) ≥ 0 for all y C. So, z (EP) ∩ F(T) = Ω. By Lemma 2.4, there exists v (EP) ∩ F(T) such that $\underset{n\to \infty }{lim}{P}_{\left(EP\right)\cap F\left(T\right)}{x}_{n}=v$. By Lemma 2.3, z = P(EP)∩F(T)z = v, and the proof is completed.

In Theorem 4.3, if θ n = 0 for each n , then we have the following result.

Theorem 4.4. Let C be a nonempty closed convex subset of a real Hilbert space H. Let G : C × C be a function satisfying (A1)-(A4). Let T : CC be a mapping with condition (B) and Ω: = F(T) ∩ (EP) ≠ . Let {a n } be a sequence in [0,1]. Let {x n } be defined by u1 H

$\left\{\begin{array}{c}{x}_{n}\in C\phantom{\rule{2.77695pt}{0ex}}such\phantom{\rule{2.77695pt}{0ex}}that\phantom{\rule{2.77695pt}{0ex}}G\left({x}_{n},y\right)+\frac{1}{{r}_{n}}⟨y-{x}_{n},{x}_{n}-{u}_{n}⟩\ge 0\phantom{\rule{1em}{0ex}}\forall y\in C;\hfill \\ {u}_{n+1}:={a}_{n}{x}_{n}+\left(1-{a}_{n}\right)T{x}_{n}.\hfill \end{array}\right\$

Assume that: {r n } [a, ∞) for some a > 0 and $\underset{n\to \infty }{lim inf}{a}_{n}\left(1-{a}_{n}\right)>0$. Then x n z, where $z=\underset{n\to \infty }{lim}{P}_{\left(EP\right)\cap F\left(T\right)}{x}_{n}$.

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## Author information

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Correspondence to Lai-Jiu Lin.

### Competing interests

The authors declare that they have no competing interests, except Prof. L.-J. Lin was supported by the National Science Council of Republic of China while he worked on the publish, and C. S. Chuang was supported as postdoctor by the National Science Council of the Republic of China while he worked on this problem.

### Authors' contributions

LJL is responsible for problem resign, coordinator, discussion, revise the important part, and submit. CSC is responsible for the important results of this article, discuss, and draft. ZTY is responsible for giving the examples of this types of problems, discussion. All authors read and approved the final manuscript.

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Lin, LJ., Chuang, CS. & Yu, ZT. Fixed point theorems for mappings with condition (B). Fixed Point Theory Appl 2011, 92 (2011). https://doi.org/10.1186/1687-1812-2011-92

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• DOI: https://doi.org/10.1186/1687-1812-2011-92

### Keywords

• fixed point
• equilibrium problem
• Banach limit
• generalized hybrid mapping
• projection