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 : C → H be a mapping, and let F(T) denote the set of fixed points of T. A mapping T : C → H is said to be nonexpansive if ||Tx - Ty|| ≤ ||x - y|| for all x, y ∈ C. A mapping T : C → H 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 : C → C is called nonspreading [1] if

$$2\parallel Tx-Ty{\parallel }^2\le \parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2$$

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

$$\parallel Tx-Ty{\parallel }^2\le \parallel x-y{\parallel }^2+2⟨x-Tx,y-Ty⟩$$

for all x, y ∈ C[2].

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

$$2\parallel Tx-Ty{\parallel }^2\le \parallel x-y{\parallel }^2+\parallel Tx-y{\parallel }^2$$

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

$$3\parallel Tx-Ty{\parallel }^2\le 2\parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2$$

for all x, y ∈ C.

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

$$\parallel Tx-Ty{\parallel }^2\le \parallel x-y{\parallel }^2+⟨x-Tx,y-Ty⟩$$

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

$$3\parallel Tx-Ty{\parallel }^2\le \parallel x-y{\parallel }^2+\parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2$$

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 : C → C is called λ-hybrid [5] if

$$\parallel Tx-Ty{\parallel }^2\le \parallel x-y{\parallel }^2+2\lambda ⟨x-Tx,y-Ty⟩$$

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 ∈ ℝ. A mapping T : C → H is (α, β)-generalized hybrid [6] if

$$\alpha \parallel Tx-Ty{\parallel }^2+\left(1-\alpha \right)\parallel Ty-x{\parallel }^2\le \beta \parallel Tx-y{\parallel }^2+\left(1-\beta \right)\parallel x-y{\parallel }^2$$

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 : C → E is said to be α-nonexpansive if

$$\parallel Tx-Ty{\parallel }^2\le \alpha \parallel Tx-y{\parallel }^2+\alpha \parallel Ty-x{\parallel }^2+\left(1-2\alpha \right)\parallel x-y{\parallel }^2$$

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 : C → E is said to satisfy condition (C) if for all x, y ∈ C,

$$\frac{1}{2}\parallel x-Tx\parallel \le \parallel x-y\parallel \Rightarrow \parallel Tx-Ty\parallel \le \parallel x-y\parallel .$$

In fact, every nonexpansive mapping satisfies condition (C), but the converse may be false [8, Example 1]. Besides, if T : C → E 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 : C → H be a mapping. Then, we say T satisfies condition (B) if for all x, y ∈ C,

$$\frac{1}{2}\parallel x-Tx\parallel \le \parallel x-y\parallel \Rightarrow \parallel Tx-Ty{\parallel }^2+\parallel x-Ty{\parallel }^2\le \parallel Tx-y{\parallel }^2+\parallel x-y{\parallel }^2.$$

Remark 1.1.

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

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

• (iii) If T : C → C 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 = ℝ, let C be nonempty closed convex subset of H, and let T : C → H be a function. In fact, we have

$$\begin{array}{c}\frac{1}{2}\parallel x-Tx\parallel \le \parallel x-y\parallel \hfill \\ \iff {\left(Tx\right)}^2+x^2-2xTx\le 4x^2+4y^2-8xy\hfill \\ \iff {\left(Tx\right)}^2-2xTx\le 3x^2+4y^2-8xy\hfill \\ \iff Tx\left(Tx-2x\right)\le \left(3x-2y\right)\left(x-2y\right),\hfill \end{array}$$

and

$$\begin{array}{c}|Tx-Ty{|}^2+|x-Ty{|}^2\le |Tx-y{|}^2+|x-y{|}^2\hfill \\ \iff {\left(Tx\right)}^2+{\left(Ty\right)}^2-2\mathrm{TxTy}+x^2+{\left(Ty\right)}^2-2xTy\le {\left(Tx\right)}^2+y^2-2\mathrm{yTx}+x^2+y^2-2\mathrm{xy}\hfill \\ \iff 2{\left(Ty\right)}^2-2Ty\left(Tx+x\right)\le 2y^2-2y\left(Tx+x\right)\hfill \\ \iff {2\left(Ty\right)}^2-2y^2\le 2\left(Ty-y\right)\left(Tx+x\right)\hfill \\ \iff \left(Ty-y\right)\left(Ty+y\right)\le \left(Ty-y\right)\left(Tx+x\right)\hfill \\ \iff \left(Ty-y\right)\left[\left(Ty+y\right)-\left(Tx+x\right)\right]\le 0.\hfill \end{array}$$

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

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

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

  $$2\parallel Tx-Ty{\parallel }^2=8>0=\parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2.$$

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

  $$2\parallel Tx-Ty{\parallel }^2=8>4=4+0=\parallel x-y{\parallel }^2+\parallel Tx-y{\parallel }^2.$$

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

  $$3\parallel Tx-Ty{\parallel }^2=12>0=2\parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2.$$

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

  $$3\parallel Tx-Ty{\parallel }^2=12>4=\parallel x-y{\parallel }^2+\parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2.$$

• (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,

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

For α < 0, let x = y = 1,

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

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

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

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

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

• (a) For x ∈ [-1, 0] and $\frac{1}{2}\parallel x-Tx\parallel \le \parallel x-y\parallel$, 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.

• (b) For x ∈ (0, 1] and $\frac{1}{2}\parallel x-Tx\parallel \le \parallel x-y\parallel$, we know that

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

(b)₂ if y <x, then x ≤ x - 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) T is a nonspreading mapping. Indeed, we know that the following conditions hold.

(1)₁ If x > 0 and y > 0, then

$$2\parallel Tx-Ty{\parallel }^2=2\parallel x-y{\parallel }^2\le 2\parallel x+y{\parallel }^2=\parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2;$$

(1)₂ If x ≤ 0 and y ≤ 0, then

$$2\parallel Tx-Ty{\parallel }^2=2\parallel x-y{\parallel }^2=\parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2;$$

(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,

$$\parallel Tx-y{\parallel }^2+\parallel Ty-x{\parallel }^2-2\parallel Tx-Ty{\parallel }^2=-4xy\ge 0.$$

• (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

$$x_{n+1}={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T{x}_n,n\in \mathbb{N},$$

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

$$\left\{\begin{array}{c}{x}_1\in C\text{chosen}\text{arbitrary},\hfill \\ y_n:=\left(1-{\beta }_n\right)x_n+{\beta }_{n}T{x}_n,\hfill \\ x_{n+1}:=\left(1-{\alpha }_n\right)x_n+{\alpha }_{n}T{y}_n,\hfill \end{array}\right.$$

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 : E → E a given mapping, the Ishikawa iteration method with errors is the following sequence

$$\left\{\begin{array}{c}{x}_1\in E\text{chosen}\text{arbitrary},\hfill \\ y_n:=\left(1-{\beta }_n\right)x_n+{\beta }_{n}T{x}_n+u_n,\hfill \\ x_{n+1}:=\left(1-{\alpha }_n\right)x_n+{\alpha }_{n}T{y}_n+v_n,\hfill \end{array}\right.$$

where {α _n } and {β _n } are sequences in [0,1], and {u _n } and {v _n } are sequences in E with ${\sum }_{n=1}^{\infty }\parallel u_n\parallel <\infty$ and ${\sum }_{n=1}^{\infty }\parallel v_n\parallel <\infty$.

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 : C → C a given mapping, the Ishikawa iteration method with errors is generated by

$$\left\{\begin{array}{c}{x}_1\in C\text{chosen}\text{arbitrary},\hfill \\ y_n:=a_n{x}_n+b_{n}T{x}_n+c_n{u}_n,\hfill \\ x_{n+1}:={a^{\prime }}_n{x}_n+{b^{\prime }}_{n}T{y}_n+{c^{\prime }}_n{v}_n,\hfill \end{array}\right.$$

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 → ℝ be a function. Let T : C → H 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

$$\left\{\begin{array}{c}{x}_n\in C\mathsf{\text{such}}\mathsf{\text{that}}G\left(x_n,y\right)+\frac{1}{r_n}⟨y-x_n,x_n-u_n⟩\ge 0\forall y\in C;\hfill \\ u_{n+1}:=a_n{x}_n+b_{n}T{x}_n+{\theta }_n{\omega }_n.\hfill \end{array}\right.$$

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 ℕ be the set of positive integers and let ℝ 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

$$2⟨x-y,u-v⟩=\parallel x-v{\parallel }^2+\parallel y-u{\parallel }^2-\parallel x-u{\parallel }^2-\parallel y-v{\parallel }^2$$

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, . . .) ∈ ℓ^∞. 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 ∈ ℕ. If μ is a Banach limit on ℓ^∞, then for f = (x₁, x₂, x₃, . . .) ∈ ℓ^∞,

$$\underset{n\to \infty }{lim\; inf}{x}_n\le {\mu }_n{x}_n\le \underset{n\to \infty }{lim\; sup}{x}_n.$$

In particular, if f = (x₁, x₂, x₃, . . .) ∈ ℓ^∞ and x _n → a ∈ ℝ, 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 → ℝ 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 have 〈x - 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 }_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 }_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 ∈ 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 s_n+1≤ s _n + t _n for each n ∈ 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

$$G\left(z,y\right)\ge 0\mathsf{\text{for each}}y\in C.$$

(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 → ℝ 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, $\underset{\mathrm{t\downarrow }0}{lim}G\left(\mathrm{tz}+\left(1-t\right)x,y\right)\le G\left(x,y\right)$;

(A4) for each x ∈ C, the scalar function y → G(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 → ℝ be a bifunction which satisfies conditions (A1)-(A4). Let r > 0 and × ∈ H. Then there exists z ∈ C such that

$$G\left(z,y\right)+\frac{1}{r}⟨y-z,z-x⟩\ge 0forally\in C.$$

Furthermore, if

$$T_r\left(x\right):=\left\{z\in C:G\left(z,y\right)+\frac{1}{r}⟨y-z,z-x⟩\ge 0forally\in C\right\},$$

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 - y〉 for 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 : C → C be a mapping with condition (B). Then for each x, y ∈ C, we have:

• (i) ||Tx - T ² x||² + ||x - T ² x|| ≤ ||x - Tx ||² ;

• (ii) ||Tx - T ² x|| ≤ ||x - Tx || and ||x - T ² x|| ≤ ||x - Tx ||;

• (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;

• (iv) either

  $$\parallel Tx-Ty{\parallel }^2+\parallel x-Ty{\parallel }^2\le \parallel Tx-y{\parallel }^2+\parallel x-y{\parallel }^2$$

or

$$\parallel T^{2}x-Ty\mid {\mid }^2+\parallel Tx-Ty\mid {\mid }^2\le \parallel T^{2}x-y\mid {\mid }^2+\parallel Tx-y\mid {\mid }^2$$

holds;

• (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

$$\frac{1}{2}\parallel x-Tx\parallel >\parallel x-y\parallel \mathsf{\text{and}}\frac{1}{2}\parallel Tx-T^{2}x\parallel >\parallel Tx-y\parallel$$

holds. So,

$$\begin{array}{c}\parallel x-Tx\parallel \le \parallel x-y\parallel +\parallel y-Tx\parallel \hfill \\ <\frac{1}{2}\parallel x-Tx\parallel +\frac{1}{2}\parallel Tx-T^{2}x\parallel \hfill \\ \le \frac{1}{2}\parallel x-Tx\parallel +\frac{1}{2}\parallel x-Tx\parallel =\parallel x-Tx\parallel \mathsf{\text{.}}\hfill \end{array}$$

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

$$\parallel T^{n+1}x-T^{n+2}x{\parallel }^2+\parallel T^{n}x-T^{n+2}x{\parallel }^2\le \parallel T^{n}x-T^{n+1}x{\parallel }^2.$$

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:

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

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

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

$$⟨Tx-Ty,y-Ty⟩\le ⟨x-y,Ty-y⟩+\parallel T^{2}x-x\parallel \cdot \parallel Ty-y\parallel .$$

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

$$\parallel Tx-Ty{\parallel }^2+\parallel x-Ty{\parallel }^2\le \parallel Tx-y{\parallel }^2+\parallel x-y{\parallel }^2$$

or

$$\parallel T^{2}x-Ty{\parallel }^2+\parallel Tx-Ty{\parallel }^2\le \parallel T^{2}x-y{\parallel }^2+\parallel Tx-y{\parallel }^2$$

holds. In the first case, we have

$$\begin{array}{c}\parallel Tx-Ty{\parallel }^2+\parallel x-Ty{\parallel }^2\le \parallel Tx-y{\parallel }^2+\parallel x-y{\parallel }^2\hfill \\ \Rightarrow \parallel Tx-Ty{\parallel }^2+\parallel x-y{\parallel }^2+2⟨x-y,y-Ty⟩+\parallel Ty-y{\parallel }^2\le \parallel Tx-Ty{\parallel }^2+\hfill \\ 2⟨Tx-Ty,Ty-y⟩+\parallel Ty-y{\parallel }^2+\parallel x-y{\parallel }^2\hfill \\ \Rightarrow ⟨x-y,y-Ty⟩\le ⟨Tx-Ty,Ty-y⟩\hfill \\ \Rightarrow ⟨Tx-Ty,y-Ty⟩\le ⟨x-y,Ty-y⟩.\hfill \end{array}$$

In the second case, we have

$$\begin{array}{c}\parallel T^{2}x-Ty{\parallel }^2+\parallel Tx-Ty{\parallel }^2\le \parallel Tx-y{\parallel }^2+\parallel T^{2}x-y{\parallel }^2\hfill \\ \Rightarrow \parallel T^{2}x-y{\parallel }^2+2⟨T^{2}x-y,y-Ty⟩+\parallel y-Ty{\parallel }^2+\parallel Tx-Ty{\parallel }^2\le \parallel Tx-Ty{\parallel }^2+\hfill \\ 2⟨Tx-Ty,Ty-y⟩+\parallel y-Ty{\parallel }^2+\parallel T^{2}x-y{\parallel }^2\hfill \\ \Rightarrow ⟨Tx-Ty,y-Ty⟩\le ⟨T^{2}x-y,Ty-y⟩\hfill \\ \le ⟨x-y,Ty-y⟩+\parallel T^{2}x-x\parallel \cdot \parallel Ty-y\parallel .\hfill \end{array}$$

Therefore, the proof is completed.

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

• (a) ||Tx - Ty||² + ||x - Ty||² ≤ ||x - y||² + ||Tx - y||² + ||T ² x - x|| · ||Ty - y||.

• (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 : C → C be a mapping with condition (B). Let {x _n } be a sequence in C with x _n ⇀ x and $\underset{n\to \infty }{lim}\parallel x_n-T{x}_n\parallel =0$. Then Tx= x.

Proof By Remark 3.1, we get:

$$⟨T{x}_n-Tx,x-Tx⟩\le ⟨x_n-x,Tx-x⟩+\parallel x_n-T{x}_n\parallel \cdot \parallel x-Tx\parallel$$

for each n ∈ 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 : C → C 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 ∈ ℕ, let x _n := Tⁿx. Clearly, {x _n } is a bounded sequence. By Lemma 2.1, there is a unique z ∈ C such that ${\mu }_n\parallel x_n-z{\parallel }^2=\underset{y\in C}{min}{\mu }_n\parallel x_n-y{\parallel }^2$. By Proposition 3.2, for each n ∈ ℕ,

$$\begin{array}{c}⟨x_{n+1}-\mathrm{Tz},z-\mathrm{Tz}⟩\le ⟨x_n-z,\mathrm{Tz}-z⟩+\parallel x_n-x_{n+2}\parallel \cdot \parallel z-\mathrm{Tz}\parallel \hfill \\ \Rightarrow \frac{1}{2}\parallel x_{n+1}-\mathrm{Tz}{\parallel }^2+\frac{1}{2}\parallel \mathrm{Tz}-z{\parallel }^2-\frac{1}{2}\parallel x_{n+1}-z{\parallel }^2\hfill \\ \le \frac{1}{2}\parallel x_n-z{\parallel }^2+\frac{1}{2}\parallel z-\mathrm{Tz}{\parallel }^2-\frac{1}{2}\parallel x_n-\mathrm{Tz}{\parallel }^2+\parallel x_n-x_{n+2}\parallel \cdot \parallel z-\mathrm{Tz}\parallel \hfill \\ \Rightarrow {\mu }_n\parallel x_n-\mathrm{Tz}{\parallel }^2\le {\mu }_n\parallel x_n-z{\parallel }^2+{\mu }_n\parallel x_n-x_{n+2}\parallel \cdot \parallel z-\mathrm{Tz}\parallel .\hfill \end{array}$$

By Proposition 3.1(v), μ _n ||x _n - Tz||² ≤ μ _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 : C → C 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 : C → C be a mapping with condition (B). Then the following conditions are equivalent:

• (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;

• (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 ∈ ℕ, we have:

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

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

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

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|| ≤ ||T^n-1x - u|| for each n ∈ 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:

$$\parallel S_{n}x-u\parallel \le \frac{1}{n}\sum _{k=0}^{n-1}\parallel T^{k}x-u\parallel \le \parallel x-u\parallel .$$

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 $S_{n_i}x\rightharpoonup v$. 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 $S_{n_i}x\rightharpoonup v$, {Tⁿx} is a bounded sequence, and $\underset{n\to \infty }{lim}\frac{1}{n-1}\sum _{k=0}^{n-2}\parallel T^{k+2}x-T^{k}x\parallel =0$, it is easy to see that Tv = v. So, v ∈ F(T).

By Lemma 2.2, for each k ∈ ℕ, 〈T^kx - P _F(T) T^kx, P _F(T) T^kx - u〉 ≥ 0. This implies that

$$\begin{array}{c}⟨T^{k}x-P_{F\left(T\right)}{T}^{k}x,u-z⟩\hfill \\ \le ⟨T^{k}x-P_{F\left(T\right)}{T}^{k}x,P_{F\left(T\right)}{T}^{k}x-z⟩\hfill \\ \le \parallel T^{k}x-P_{F\left(T\right)}{T}^{k}x\parallel \cdot \parallel P_{F\left(T\right)}{T}^{k}x-z\parallel \hfill \\ \le \parallel T^{k}x-z\parallel \cdot \parallel P_{F\left(T\right)}{T}^{k}x-z\parallel \hfill \\ \le \parallel x-z\parallel \cdot \parallel P_{F\left(T\right)}{T}^{k}x-z\parallel .\hfill \end{array}$$

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 $S_{n_k}x\rightharpoonup v$ and P _F(T) T^kx → 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 $S_{n_i}x\rightharpoonup q$, then q = v by following the same argument as the above proof. Therefore, $S_{n}x\rightharpoonup \upsilon =\underset{n\to \infty }{lim}P{T}^{n}x$, 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 : C → C 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

$$a_n+c_n+{\theta }_n=b_n+d_n+{\lambda }_n=1,n\in \mathbb{N}.$$

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

$$\left\{\begin{array}{c}{x}_1\in Cchosenarbitrary,\hfill \\ y_n:=a_n{x}_n+c_n{T}_1{x}_n+{\theta }_n{u}_n,\hfill \\ x_{n+1}:=b_n{x}_n+d_n{T}_2{y}_n+{\lambda }_n{v}_n.\hfill \end{array}\right.$$

Assume that:

• (i) $\underset{n\to \infty }{lim\; inf}{a}_n{c}_n>0$ and $\underset{n\to \infty }{lim\; inf}{b}_n{d}_n>0$ ;

• (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 ∈ ℕ, we have:

$$\begin{array}{c}\parallel y_n-w{\parallel }^2\hfill \\ =\parallel a_n{x}_n+c_n{T}_1{x}_n+{\theta }_n{u}_n-w{\parallel }^2\hfill \\ =a_n\parallel x_n-w{\parallel }^2+c_n\parallel T_1{x}_n-w{\parallel }^2+{\theta }_n\parallel u_n-w{\parallel }^2\hfill \\ -a_n{c}_n\parallel x_n-T_1{x}_n{\parallel }^2-a_n{\theta }_n\parallel x_n-w{\parallel }^2-c_n{\theta }_n\parallel T_1{x}_n-u_n{\parallel }^2\hfill \\ \le a_n\parallel x_n-w{\parallel }^2+c_n\parallel x_n-w{\parallel }^2+{\theta }_n\parallel u_n-w{\parallel }^2\hfill \\ -a_n{c}_n\parallel x_n-T_1{x}_n{\parallel }^2-a_n{\theta }_n\parallel x_n-w{\parallel }^2-c_n{\theta }_n\parallel T_1{x}_n-u_n{\parallel }^2\hfill \\ \le \parallel x_n-w{\parallel }^2+{\theta }_n\parallel u_n-w{\parallel }^2,\hfill \end{array}$$

and

$$\begin{array}{c}\parallel x_{n+1}-w{\parallel }^2\hfill \\ =\parallel b_n{x}_n+d_n{T}_2{y}_n+{\lambda }_n{v}_n-w{\parallel }^2\hfill \\ =b_n\parallel x_n-w{\parallel }^2+d_n\parallel T_2{y}_n-w{\parallel }^2+{\lambda }_n\parallel v_n-w{\parallel }^2\hfill \\ -b_n{d}_n\parallel x_n-T_2{y}_n{\parallel }^2-b_n{\lambda }_n\parallel x_n-v_n{\parallel }^2-d_n{\lambda }_n\parallel T_2{y}_n-v_n{\parallel }^2\hfill \\ \le b_n\parallel x_n-w{\parallel }^2+d_n\parallel y_n-w{\parallel }^2+{\lambda }_n\parallel v_n-w{\parallel }^2\hfill \\ -b_n{d}_n\parallel x_n-T_2{y}_n{\parallel }^2-b_n{\lambda }_n\parallel x_n-v_n{\parallel }^2-d_n{\lambda }_n\parallel T_2{y}_n-v_n{\parallel }^2\hfill \\ \le b_n\parallel x_n-w{\parallel }^2+d_n\left(\parallel x_n-w{\parallel }^2+{\theta }_n\parallel u_n-w{\parallel }^2\right)+{\lambda }_n\parallel v_n-w{\parallel }^2\hfill \\ -b_n{d}_n\parallel x_n-T_2{y}_n{\parallel }^2-b_n{\lambda }_n\parallel x_n-v_n{\parallel }^2-d_n{\lambda }_n\parallel T_2{y}_n-v_n{\parallel }^2\hfill \\ \le \parallel x_n-w{\parallel }^2+d_n{\theta }_n\parallel u_n-w{\parallel }^2+{\lambda }_n\parallel v_n-w{\parallel }^2\hfill \\ -b_n{d}_n\parallel x_n-T_2{y}_n{\parallel }^2-b_n{\lambda }_n\parallel x_n-v_n{\parallel }^2-d_n{\lambda }_n\parallel T_2{y}_n-v_n{\parallel }^2\hfill \\ \le \parallel x_n-w{\parallel }^2+d_n{\theta }_n\parallel u_n-w{\parallel }^2+{\lambda }_n\parallel v_n-w{\parallel }^2.\end{array}$$

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 $\underset{n\to \infty }{lim}\parallel x_n-w\parallel =t$. Besides,

$$\begin{array}{c}{b}_n{d}_n\parallel x_n-T_2{y}_n{\parallel }^2+b_n{\lambda }_n\parallel x_n-v_n{\parallel }^2+d_n{\lambda }_n\parallel T_2{y}_n-v_n{\parallel }^2\\ \le \parallel x_n-w{\parallel }^2+d_n{\theta }_n\parallel u_n-w{\parallel }^2+{\lambda }_n\parallel v_n-w{\parallel }^2-\parallel x_{n+1}-w{\parallel }^2.\end{array}$$

This implies that

$$\underset{n\to \infty }{lim}{b}_n{d}_n\parallel x_n-T_2{y}_n{\parallel }^2=0.$$

By assumption, $\underset{n\to \infty }{lim}\parallel x_n-T_2{y}_n\parallel =0$. Furthermore, we have:

$$\parallel x_{n+1}-w{\parallel }^2\le b_n\parallel x_n-w{\parallel }^2+d_n\parallel y_n-w{\parallel }^2+{\lambda }_n\parallel v_n-w{\parallel }^2.$$

This implies that

$$\begin{array}{c}{b}_n{d}_n\left(\parallel x_n-w{\parallel }^2+{\theta }_n\parallel u_n-w{\parallel }^2-\parallel y_n-w{\parallel }^2\right)\hfill \\ \le d_n\left(\parallel x_n-w{\parallel }^2+{\theta }_n\parallel u_n-w{\parallel }^2-\parallel y_n-w{\parallel }^2\right)\hfill \\ \le \left(1-b_n\right)\parallel x_n-w{\parallel }^2+d_n{\theta }_n\parallel u_n-w{\parallel }^2-d_n\parallel y_n-w{\parallel }^2\hfill \\ \le \parallel x_n-w{\parallel }^2-\parallel x_{n+1}-w{\parallel }^2+{\lambda }_n\parallel v_n-w{\parallel }^2+d_n{\theta }_n\parallel u_n-w{\parallel }^2.\hfill \end{array}$$

Hence, $\underset{n\to \infty }{lim}{b}_n{d}_n\left(\parallel x_n-w{\parallel }^2+{\theta }_n\parallel u_n-w{\parallel }^2-\parallel y_n-w{\parallel }^2\right)=0$. By assumption,

$$\underset{n\to \infty }{lim}\left(\parallel x_n-w{\parallel }^2+{\theta }_n\parallel u_n-w{\parallel }^2-\parallel y_n-w{\parallel }^2\right)=0.$$

Since $\underset{n\to \infty }{lim}{\theta }_n\parallel u_n-w{\parallel }^2=0$,

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

Hence, $\underset{n\to \infty }{lim}\parallel y_n-w\parallel =\underset{n\to \infty }{lim}\parallel x_n-w\parallel =t$. Similar to the above proof, we also get

$$\underset{n\to \infty }{lim}\parallel x_n-T_1{x}_n\parallel =0.$$

Besides,

$$\begin{array}{c}∥y_n-x_n∥=\parallel a_n{x}_n+c_n{T}_1{x}_n+{\theta }_n{u}_n-x_n\parallel \hfill \\ \le c_n\parallel T_1{x}_n-x_n\parallel +{\theta }_n\parallel x_n-u_n\parallel \hfill \\ \le \parallel T_1{x}_n-x_n\parallel +{\theta }_n\parallel x_n-u_n\parallel .\hfill \end{array}$$

This implies that $\underset{n\to \infty }{lim}\parallel y_n-x_n\parallel =0$ and $\underset{n\to \infty }{lim}\parallel y_n-T_2{y}_n\parallel =0$. Since {x _n } is a bounded sequence, there exists a subsequence $\left\{x_{n_k}\right\}$ of {x _n } such that $x_{n_k}\rightharpoonup z$. By Theorem 3.1, z = T₁z.

If $x_{n_j}$ is a subsequence of {x _n } and $x_{n_j}\rightharpoonup q$, then T₁q = q. Suppose that q ≠ z. Then we have:

$$\begin{array}{c}\underset{k\to \infty }{lim\; inf}\parallel x_{n_k}-z\parallel <\underset{k\to \infty }{lim\; inf}\parallel x_{n_k}-q\parallel =\underset{n\to \infty }{lim}\parallel x_n-q\parallel =\underset{j\to \infty }{lim}\parallel x_{n_j}-q\parallel \hfill \\ <\underset{j\to \infty }{lim\; inf}\parallel x_{n_j}-z\parallel =\underset{n\to \infty }{lim}\parallel x_n-z\parallel =\underset{k\to \infty }{lim\; inf}\parallel x_{n_k}-z\parallel .\hfill \end{array}$$

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 $\underset{n\to \infty }{lim}\parallel x_n-y_n\parallel =0$, y _n ⇀ z. By Theorem 3.1, z ∈ F(T₂). Hence, z ∈ Ω.

Next, by Lemma 2.4, P_Ωx _n converges. Then there exists v ∈ Ω such that $\underset{n\to \infty }{lim}{P}_{\Omega }{x}_n=v$. By Lemma 2.3, P_Ωz = 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 ∈ 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 : C → C 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

$$\left\{\begin{array}{c}{x}_1\in Cchosenarbitrary,\hfill \\ y_n:=a_n{x}_n+\left(1-a_n\right)T_1{x}_n,\hfill \\ x_{n+1}:=b_n{x}_n+\left(1-b_n\right)T_2{y}_n.\hfill \end{array}\right.$$

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 : C → C 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

$$\left\{\begin{array}{c}{x}_1\in Cchosenarbitrary,\hfill \\ y_n:=a_n{x}_n+\left(1-a_n\right)T{x}_n,\hfill \\ x_{n+1}:=b_n{x}_n+\left(1-b_n\right)T{y}_n.\hfill \end{array}\right.$$

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 : C → C be a mapping with condition (B) and F(T) ≠ ∅. Let {b _n } be a sequence in [0,1]. Let {x _n } be defined by

$$\left\{\begin{array}{c}{x}_1\in Cchosenarbitrary,\hfill \\ x_{n+1}:=b_n{x}_n+\left(1-b_n\right)T{x}_n.\hfill \end{array}\right.$$

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 → ℝ be a function satisfying (A1)-(A4). Let T : C → C 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

$$\left\{\begin{array}{c}{x}_n\in C\text{such}\text{that}G\left(x_n,y\right)+\frac{1}{r_n}⟨y-x_n,x_n-u_n⟩\ge 0\forall y\in C;\hfill \\ u_{n+1}:=a_n{x}_n+b_{n}T{x}_n+{\theta }_n{\omega }_n.\hfill \end{array}\right.$$

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 ∈ ℕ. Then we have:

$$\begin{array}{c}\parallel x_{n+1}-w{\parallel }^2\hfill \\ =\parallel T_{r_{n+1}}{u}_{n+1}-w{\parallel }^2\hfill \\ \le \parallel u_{n+1}-w{\parallel }^2\hfill \\ \le \parallel a_n{x}_n+b_{n}T{x}_n+{\theta }_n{\omega }_n-w{\parallel }^2\hfill \\ \le a_n\parallel x_n-w{\parallel }^2+b_n\parallel T{x}_n-w{\parallel }^2+{\theta }_n\parallel {\omega }_n-w{\parallel }^2-a_n{b}_n\parallel x_n-T{x}_n{\parallel }^2\\ \le a_n\parallel x_n-w{\parallel }^2+b_n\parallel x_n-w{\parallel }^2+{\theta }_n\parallel {\omega }_n-w{\parallel }^2-a_n{b}_n\parallel x_n-T{x}_n{\parallel }^2\\ \le \parallel x_n-w{\parallel }^2+{\theta }_n\parallel {\omega }_n-w{\parallel }^2-a_n{b}_n\parallel x_n-T{x}_n{\parallel }^2.\end{array}$$

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

• (a) $\underset{n\to \infty }{lim}{a}_n{b}_n\parallel x_n-T{x}_n{\parallel }^2=0$;

• (b) $\underset{n\to \infty }{lim}\parallel x_n-T{x}_n\parallel =0$;

• (c) $\parallel u_{n+1}-x_n\parallel =\parallel b_{n}T{x}_n-b_n{x}_n+{\theta }_n{w}_n-{\theta }_n{x}_n\parallel \le \parallel T{x}_n-x_n\parallel +{\theta }_n\parallel w_n-x_n\parallel$;

• (d) $\underset{n\to \infty }{lim}\parallel u_{n+1}-x_n\parallel =0$;

• (e) $\underset{n\to \infty }{lim}\parallel u_n-w\parallel =\underset{n\to \infty }{lim}\parallel x_n-w\parallel$.

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

$$\begin{array}{c}{∥x_{n+1}-w∥}^2=\parallel T_{r_{n+1}}{u}_{n+1}-w{\parallel }^2\hfill \\ =\parallel T_{r_{n+1}}{u}_{n+1}-T_{r_{n+1}}w{\parallel }^2\hfill \\ \le ⟨T_{r_{n+1}}{u}_{n+1}-T_{r_{n+1}}w,u_{n+1}-w⟩\hfill \\ \le ⟨x_{n+1}-w,u_{n+1}-w⟩\hfill \\ =\frac{1}{2}\parallel x_{n+1}-w{\parallel }^2+\frac{1}{2}\parallel u_{n+1}-w{\parallel }^2-\frac{1}{2}\parallel x_{n+1}-u_{n+1}{\parallel }^2.\hfill \end{array}$$

This implies that

$$\parallel x_{n+1}-u_{n+1}{\parallel }^2\le \parallel u_{n+1}-w{\parallel }^2-\parallel x_{n+1}-w{\parallel }^2.$$

By (e), $\underset{n\to \infty }{lim}\parallel x_n-u_n\parallel =0$. Next, we want to show that z ∈ (EP). Since $x_n=T_{r_n}{u}_n$,

$$G\left(x_n,y\right)+\frac{1}{r_n}⟨y-x_n,x_n-u_n⟩\ge 0\forall y\in C.$$

By (A2),

$$\frac{1}{r_n}⟨y-x_n,x_n-u_n⟩\ge G\left(y,x_n\right)\forall y\in C.$$

By (A4), (i), and $\underset{n\to \infty }{lim}\parallel x_n-u_n\parallel =0$, we get

$$0\ge \underset{n\to \infty }{lim}G\left(y,x_n\right)\ge G\left(y,z\right)\forall y\in C.$$

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 ∈ 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 → ℝ be a function satisfying (A1)-(A4). Let T : C → C 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

$$\left\{\begin{array}{c}{x}_n\in CsuchthatG\left(x_n,y\right)+\frac{1}{r_n}⟨y-x_n,x_n-u_n⟩\ge 0\forall y\in C;\hfill \\ u_{n+1}:=a_n{x}_n+\left(1-a_n\right)T{x}_n.\hfill \end{array}\right.$$

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 $z=\underset{n\to \infty }{lim}{P}_{\left(EP\right)\cap F\left(T\right)}{x}_n$.