Modified noor iterations for nonexpansive semigroups with generalized contraction in Banach spaces

Abstract

In this article, the modified Noor iterations are considered for the generalized contraction and a nonexpansive semigroup in the framework of a reflexive Banach space which admits a weakly sequentially continuous duality mapping. The strong convergence theorems are obtained under very mild conditions imposed the parameters. The results presented in this article improve and extend the corresponding results announced by Chen and He and Chen et al. and many others.

AMS subject classification: 47H09; 47H10; 47H17.

1. Introduction and preliminaries

Let E be a real Banach space. A mapping T of E into itself is said to be nonexpansive if ||Tx - Ty|| ≤ ||x - y|| for each x,y ∈ E. We denote by Fix(T) the set of fixed points of T. A mapping f : E→E is called α-contraction, if there exists a constant 0 < α < 1 such that ||f(x) - f(y)|| ≤ α||x - y|| for all x, y ∈ E. Throughout this article, we denote by ℕ and ℝ⁺ the sets of positive integers and nonnegative real numbers, respectively. A mapping ψ: ℝ⁺→ℝ⁺ is said to be an L-function if ψ(0) = 0, ψ(t) > 0, for each t > 0 and for every t > 0 and for every s > 0 there exists u > s such that ψ(t) ≤ s, for all t ∈ [s, u], As a consequence, every L-function ψ satisfies ψ(t) < t, for each t > 0.

Definition 1.1. Let (X, d) be a matric space. A mapping f: X→X is said to be :

• (i) a (ψ, L)-function if ψ: ℝ⁺→ℝ⁺ is an L-function and d(f(x), f(y)) < ψ(d(x, y)), for all x, y ∈ X, with x ≠ y:

• (ii) a Meir-Keeler type mapping if for each ε > 0 there exists δ = δ(ε) > 0 such that for each x, y ∈ X, with ε ≤ d(x, y) < ε + δ we have d(f(x),f(y)) < ε.

If, in Definition 1.1 we consider ψ(t) = αt, for each t ∈ ℝ⁺, where α ∈ [0,1), then we get the usual contraction mapping with coefficient α.

Proposition 1.2. [1] Let (X, d) be a matric space and f: X→X be a mapping. The following assertions are equivalent:

• (i) f is a Meir-Keeler type mapping :

• (ii) there exists an L-function ψ: ℝ⁺→ℝ⁺ such that f is a (ψ, L)-contraction.

Lemma 1.3. [2] Let X be a Banach space and C be a convex subset of it. Let T : C→C be a nonexpansive mapping and f is a (ψ, L)-contraction. Then the following assertions hold:

• (i) T ο f is a (ψ, L)-contraction on C and has a unique fixed point in C;

• (ii) for each α ∈ (0, 1) the mapping x ↦ αf(x) + (1 - α)T(x) is a Meir-Keeler type mapping on C.

Lemma 1.4. [3, Proposition 2] Let E be a Banach space and C a convex subset of it. Let f : C→C be a Meir-Keeler type mapping. Then for each ε > 0 there exists r ∈ (0,1) such that for each x,y ∈ C with ||x - y|| ≥ ε we have ||f(x) - f(y)|| ≤ r ||x - y||.

From now on, by a generalized contraction mapping we mean a Meir-Keeler type mapping or (ψ, L)-contraction. In the rest of the article we suppose that the ψ from the definition of the (ψ, L)-contraction is continuous, strictly increasing and η(t) is strictly increasing and onto, where η(t) := t - ψ(t), t ∈ ℝ⁺. As a consequence, we have the η is a bijection on ℝ⁺.

A family $\mathcal{S}=\left\{T\left(t\right):0\le t<\infty \right\}$ of mappings of E into itself is called a nonexpansive semigroup on E if it satisfies the following conditions:

• (i) T(0)x = x for all x ∈ E;

• (ii) T(s +t) = T(s)T(t) for all s, t > 0;

• (iii) ||T(t)x - T(t)y|| ≤ ||x - y|| for all x, y ∈ E and t ≥ 0;

• (iv) for all x ∈ E, the mapping t ↦ T(t)x is continuous.

We denote by $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ the set of all common fixed points of $\mathcal{S}$, that is,

$$\mathsf{\text{Fix}}\left(\mathcal{S}\right):=\left\{x\in E:T\left(t\right)x=x,0\le t<\infty \right\}={\cap }_{t\ge 0}\mathsf{\text{Fix}}\left(T\left(t\right)\right).$$

In [4], Shioji and Takahashi introduced the following implicit iteration in a Hilbert space

$$x_n={\alpha }_{n}x+\left(1-{\alpha }_n\right)\frac{1}{t_n}\underset{0}{\overset{t_n}{\int }}T\left(s\right)x_{n}ds,\forall n\in \mathbb{N}$$

(1.1)

where {α _n } is a sequence in (0,1), {t _n } is a sequence of positive real numbers which diverges to ∞. Under certain restrictions on the sequence {α _n }, Shioji and Takahashi [4] proved strong convergence of the sequence {x _n } to a member of $F\left(\mathcal{S}\right)$. In [5], Shimizu and Takahashi studied the strong convergence of the sequence {x _n } defined by

$$x_{n+1}={\alpha }_{n}x+\left(1-{\alpha }_n\right)\frac{1}{t_n}\underset{0}{\overset{t_n}{\int }}T\left(s\right)x_{n}ds,\forall n\in \mathbb{N}$$

(1.2)

in a real Hilbert space where {T(t) : t ≥ 0} is a strongly continuous semigroup of nonexpansive mappings on a closed convex subset C of a Banach space E and lim _n→∞ t _n = ∞. Using viscosity iterative method, Chen and Song [6] studied the strong convergence of the following iterative method for a nonexpansive semigroup {T(t) : t ≥ 0} with $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$ in a Banach space:

$$x_{n+1}={\alpha }_{n}f\left(x\right)+\left(1-{\alpha }_n\right)\frac{1}{t_n}\underset{0}{\overset{t_n}{\int }}T\left(s\right)x_{n}ds,\forall n\in \mathbb{N},$$

(1.3)

where f is a contraction. Note however that their iterate x _n at step n is constructed through the average of the semigroup over the interval (0, t). Suzuki [7] was the first to introduce again in a Hilbert space the following implicit iteration process:

$$x_n={\alpha }_{n}u+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\forall n\in \mathbb{N},$$

(1.4)

for the nonexpansive semigroup case. In 2002, Benavides et al. [8] in a uniformly smooth Banach space, showed that if $\mathcal{S}$ satisfies an asymptotic regularity condition and {α _n } fulfills the control conditions $\underset{n\to \infty }{\text{lim}}{\alpha }_n=0,{\sum }_{n=1}^{\infty }{\alpha }_n=\infty$, and ${\text{lim}}_{n\to \infty }\frac{{\alpha }_n}{{{\alpha }_n}_{+1}}=0$, then both the implicit iteration process (1.4) and the explicit iteration process (1.5)

$$x_{n+1}={\alpha }_{n}u+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\forall n\in \mathbb{N},$$

(1.5)

converge to a same point of $F\left(\mathcal{S}\right)$. In 2005, Xu [9] studied the strong convergence of the implicit iteration process (1.1) and (1.4) in a uniformly convex Banach space which admits a weakly sequentially continuous duality mapping. Recently Chen and He [10] introduced the viscosity approximation methods:

$$y_n={\alpha }_{n}f\left(y_n\right)+\left(1-{\alpha }_n\right)T\left(t_n\right)y_n,\forall n\in \mathbb{N},$$

(1.6)

and

$$x_{n+1}={\alpha }_{n}f\left(x_n\right)+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\forall n\in \mathbb{N},$$

(1.7)

where f is a contraction, {α _n } is a sequence in (0,1) and a nonexpansive semigroup {T(t) : t ≥ 0}. The strong convergence theorem of {x _n } is proved in a reflexive Banach space which admits a weakly sequentially continuous duality mapping. Very recently, motivated by the above results, Chen et al. [11] proposed the following two modified Mann iterations for nonexpansive semigroups {T(t) : 0 ≤ t < ∞} and obtained the strong convergence theorems in a reflexive Banach space E which admits a weakly sequentially continuous duality mapping:

$$\left\{\begin{array}{c}\hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill x_n={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,\hfill \end{array}\right.$$

(1.8)

and

$$\left\{\begin{array}{c}\hfill x_0\in C,\hfill \\ \hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,\hfill \end{array}\right.$$

(1.9)

where f : C → C is a contraction. They proved that the implicit iterative scheme {x _n } defined by (1.8) converges to an element q of $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$, which solves the following variation inequality problem:

$$⟨\left(f-I\right)q,j\left(x-q\right)⟩\le 0\mathsf{\text{for all }}x\in \mathsf{\text{Fix}}\left(\mathcal{S}\right).$$

Furthermore, Moudafi's viscosity approximation methods have been recently studies by many authors; see the well known results in [12, 13]. However, the involved mapping f is usually considered as a contraction. Note that Suzuki [14] proved the equivalence between Moudafi's viscosity approximation with contractions and Browder-type iterative processes (Halpern-type iterative processes); see [14] for more details.

In this article, inspired by above result, we introduce and study the explicit viscosity iterative scheme for the generalized contraction f and a nonexpansive semigroup {T(t) : t ≥ 0}:

$$\left\{\begin{array}{c}\hfill x_0\in C,\hfill \\ \hfill z_n={\gamma }_n{x}_n+\left(1-{\gamma }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)z_n,\hfill \\ \hfill x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,n\ge 0.\hfill \end{array}\right.$$

(1.10)

The iterative schemes (1.10) are called the three-step(modified Noor) iterations which inspired by three-step(Noor) iterations [15–23]. It is well known that three-step(Noor) iterations, include Mann and two-step iterative methods as special cases. If γ ≡ 1, then (1.10) reduces to (1.9). Furthermore, the implicit iteration (1.8) and explicit iteration (1.10) are considered for the generalized contraction and a nonexpansive semigroup in the framework of a reflexive Banach space which admits a weakly sequentially continuous duality mapping. The strong convergence theorems are obtained under very mild conditions imposed the parameters. The results presented in this article improve and extend the corresponding results announced by Chen and He [10] and Chen et al. [11] and many others.

In order to prove our main results, we need the following lemmas.

Definition 1.5. [24] A Banach space is said to admit a weakly sequentially continuous normalized duality mapping J from E in E*, if J : E→E* is single-valued and weak to weak* sequentially continuous, that is, if x _n ⇀x in E, then J(x _n ) ⇀* J(x) in E*.

A Banach space E is said to satisfy Opial's condition if for any sequence {x _n } in E, x _n ⇀ x (n→∞) implies

$$\underset{n\to \infty }{\text{lim}\mathsf{\text{sup}}}\parallel x_n-x\parallel <\underset{n\to \infty }{\text{lim}\mathsf{\text{sup}}}\parallel x_n-y\parallel ,\forall y\in E\mathsf{\text{with}}x\ne y.$$

(1.11)

By [25, Theorem 1], it is well known that if E admits a weakly sequentially continuous duality mapping, then E satisfies Opial's condition, and E is smooth.

In order to prove our main result, we need the following lemmas.

Lemma 1.6. Let E be a Banach space and x, y ∈ E, j(x) ∈ J(x), j(x + y) ∈ J(x + y). Then

$$\parallel x{\parallel }^2+2⟨y,j\left(x\right)⟩\le \parallel x+y{\parallel }^2\le \parallel x{\parallel }^2+2⟨y,j\left(x+y\right)⟩.$$

In the following, we also need the following lemma that can be found in the existing literature [13, 26].

Lemma 1.7. Let {a _n } be a sequence of non-negative real numbers satisfying the property

$$a_{n+1}\le \left(1-{\gamma }_n\right)a_n+{\delta }_n,n\ge 0,$$

where {γ _n } ⊆ (0,1) and {δ _n } ⊆ ℝ such that

$${{\displaystyle \sum }}_{n=1}^{\infty }{\gamma }_n=\infty ,andeither\lim {\sup }_{n\to \infty }\frac{{\delta }_n}{{\gamma }_n}\le 0or{{\displaystyle \sum }}_{n=1}^{\infty }\left|{\delta }_n\right|<\infty .$$

Then lim _n→∞ a _n = 0.

Lemma 1.8. [27] Let {x _n } and {y _n } be bounded sequences in a Banach space E and {β _n } a sequence in [0,1] with 0 < lim inf _n→∞ β _n ≤ lim sup _n→∞ β _n < 1. Suppose that x_{n+ 1}= (1 - β _n )y _n + β _n x _n for all n ≥ 0 and

$$\underset{n\to \infty }{\lim \sup }(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Then lim _n→∞ ||y _n - x _n || = 0.

2. Modified Mann iteration for generalized contractions

Now, we are a position to state and prove our main results.

Theorem 2.1. Let E be a reflexive Banach space which admits a weakly sequenctially continuous duality mapping J from E into E*, suppose C is a nonempty closed convex subset of E. Let $\mathcal{S}:=\left\{T\left(t\right):t\ge 0\right\}$ be a nonexpansive semigroup on C such that $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$, and f : C → C a generalized contraction on C. Let {α _n } ⊂ (0,1), {β _n } ⊂ (0,1), and {t _n } ⊂ (0, ∞) be sequences of real numbers satisfying ${\text{lim}}_{n\to \infty }{\alpha }_n={\text{lim}}_{n\to \infty }{t}_n={\text{lim}}_{n\to \infty }\frac{{\beta }_n}{t_n}=0$. Define a sequence {x _n } in C by

$$\left\{\begin{array}{c}\hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill x_n={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,\mathsf{\text{for all }}n\ge 1.\hfill \end{array}\right.$$

(2.1)

Then {x _n } converges strongly to q, as n → ∞; q is the element of $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ such that q is the unique solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the following variational inequality:

$$⟨\left(f-I\right)q,j\left(x-q\right)⟩\le 0\mathsf{\text{for all }}x\in \mathsf{\text{ Fix}}\left(\mathcal{S}\right).$$

(2.2)

Proof. We first show that {x _n } is well defined. For any n ≥ 1, we consider a mapping ${\mathcal{G}}_n$ on C defined by

$${\mathcal{G}}_{n}x={\beta }_{n}f\left(x\right)+\left(1-{\beta }_n\right)U_{n}x,\forall x\in C,$$

where U _n := α _n I + (1 - α _n )T(t _n ). It follows from nonexpansivity of U _n and Lemma 1.3 that ${\mathcal{G}}_n$ is a Meir-Keeler type contraction. Hence ${\mathcal{G}}_n$ has a unique fixed point, denoted as x _n , which uniquely solves the fixed point equation

$$x_n={\alpha }_{n}f\left(x_n\right)+\left(1-{\alpha }_n\right)U_n{x}_n,\forall n\ge 1.$$

Hence {x _n } generated in (2.1) is well defined. Now we show that {x _n } is bounded. Indeed, if we take a fixed point $x\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$, we have

$$\parallel y_n-x\parallel \le {\alpha }_n\parallel x_n-x\parallel +\left(1-{\alpha }_n\right)\parallel T\left(t_n\right)x_n-x\parallel \le \parallel x_n-x\parallel ,$$

(2.3)

and so

$$\begin{array}{ll}\hfill \parallel x_n-x{\parallel }^2& =⟨{\beta }_n\left(f\left(x_n\right)-x\right)+\left(1-{\beta }_n\right)\left(y_n-x\right),j\left(x_n-x\right)⟩\\ ={\beta }_n⟨f\left(x_n\right)-f\left(x\right)+f\left(x\right)-x,j\left(x_n-x\right)⟩+\left(1-{\beta }_n\right)⟨\left(y_n-x\right),j\left(x_n-x\right)⟩\\ \le {\beta }_n\parallel f\left(x_n\right)-f\left(x\right)\parallel \parallel x_n-x\parallel +{\beta }_n⟨f\left(x\right)-x,j\left(x_n-x\right)⟩+\left(1-{\beta }_n\right)\parallel y_n-x\parallel \parallel x_n-x\parallel \\ \le {\beta }_n\psi \left(\parallel x_n-x\parallel \right)\parallel x_n-x\parallel +{\beta }_n\parallel f\left(x\right)-x\parallel \parallel x_n-x\parallel +\left(1-{\beta }_n\right)\parallel x_n-x{\parallel }^2,\end{array}$$

and hence

$$\parallel x_n-x{\parallel }^2\le \psi \left(\parallel x_n-x\parallel \right)\parallel x_n-x\parallel +\parallel f\left(x\right)-x\parallel \parallel x_n-x\parallel .$$

(2.4)

Therefore

$$\eta \left(\parallel x_n-x\parallel \right):=\parallel x_n-x\parallel -\psi \left(\parallel x_n-x\parallel \right)\le \parallel f\left(x\right)-x\parallel ,$$

equivalent to

$$\parallel x_n-x\parallel \le {\eta }^{-1}\left(\parallel f\left(x\right)-x\parallel \right).$$

Thus {x _n } is bounded, and so are {T(t _n )x _n }, {f(x _n )}, and {y _n }. Next, we claim that {x _n } is relatively sequentially compact. Indeed, By reflexivity of E and boundedness of the sequence {x _n } there exists a weakly convergent subsequence $\left\{x_{n_j}\right\}\subset \left\{x_n\right\}$ such that $x_{n_j}\rightharpoonup p$ for some p ∈ C. Now we show that $p\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. Put $x_j=x_{n_j},y_j=y_{n_j},{\alpha }_j={\alpha }_{n_j},{\beta }_j={\beta }_{n_j}$ and $t_j=t_{n_j}$ for j ∈ ℕ, fixed t > 0. Notice that

$$\begin{array}{ll}\hfill \parallel x_j-T\left(t\right)p\parallel & \le \sum _{k=0}^{\left[t/t_j\right]-1}\parallel T\left(\left(k+1\right)t_j\right)x_j-T\left(k{t}_j\right)x_j\parallel \\ +\parallel T\left(\left[t/t_j\right]t_j\right)x_j-T\left(\left[t/t_j\right]t_j\right)p\parallel +\parallel T\left(\left[t/t_j\right]t_j\right)p-T\left(t\right)p\parallel \\ \le \left[t/t_j\right]\parallel T\left(t_j\right)x_j-x_j\parallel +\parallel x_j-p\parallel T\left(t-\left[t/t_j\right]t_j\right)p-p\parallel \\ =\left[t/t_j\right]\frac{{\beta }_j}{1-{\alpha }_j}\parallel x_j-f\left(x_j\right)\parallel +\parallel x_j-p\parallel +\parallel T\left(t-\left[t/t_j\right]t_j\right)p-p\parallel \\ \le \frac{t}{1-{\alpha }_j}\frac{{\beta }_j}{t_j}\parallel x_j-f\left(x_j\right)\parallel +\parallel x_j-p\parallel +\text{max}\left\{\parallel T\left(s\right)p-p\parallel :0\le s\le t_j\right\}.\end{array}$$

For all j ∈ ℕ, we have

$$\underset{j\to \infty }{\lim \sup }\parallel x_j-T(t)p\parallel \le \underset{j\to \infty }{\lim \sup }\parallel x_j-p\parallel .$$

Since Banach space E with a weakly sequentially continuous duality mapping satisfies Opial's condition, T(t)p = p. Therefore $p\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. In Equation (2.4), replace p with x to obtain

$$\parallel x_j-p\parallel \left(\parallel x_j-p\parallel -\psi \left(\parallel x_j-p\parallel \right)\right)\le ⟨f\left(p\right)-p,j\left(x_j-p\right)⟩.$$

Using that the duality map j is single-valued and weakly sequentially continuous from E to E*, we get that

$$\underset{j\to \infty }{\text{lim}}\parallel x_j-p\parallel \left(\parallel x_j-p\parallel -\psi \left(\parallel x_j-p\parallel \right)\right)\le \underset{j\to \infty }{\text{lim}}⟨f\left(p\right)-p,j\left(x_j-p\right)⟩=0.$$

If lim_j→∞||x _j - p|| = 0, then we have done.

If lim _j→∞ (||x _j - p|| - ψ(||x _j -p||)) = 0, then we have lim _j→∞ ||x _j -p|| = lim _j→∞ ψ (||x _j - p||).

Since ψ is a continuous function, lim _j→∞ ||x _j -p|| = ψ(lim _j→∞ ||x _j -p||). By Definition of ψ, we have lim _j→∞ ||x _j - p|| = 0. Hence {x _n } is relatively sequentially compact, i.e., there exists a subsequence $\left\{x_{n_j}\right\}\subseteq \left\{x_n\right\}$ such that $x_{n_j}\to p$ as j → ∞. Next, we show that p is a solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the variational inequality (2.2). In fact, for any $x\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$,

$$\begin{array}{ll}\hfill \parallel x_n-x{\parallel }^2& =⟨{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n-x,j\left(x_n-x\right)⟩\\ =⟨{\beta }_n\left(f\left(x_n\right)-x_n+x_n-x\right)+\left(1-{\beta }_n\right)\left(y_n-x\right),j\left(x_n-x\right)⟩\\ ={\beta }_n⟨f\left(x_n\right)-x_n,j\left(x_n-x\right)⟩+{\beta }_n⟨x_n-x,j\left(x_n-x\right)⟩+\left(1-{\beta }_n\right)⟨y_n-x,j\left(x_n-x\right)⟩\\ \le {\beta }_n⟨f\left(x_n\right)-x_n,j\left(x_n-x\right)⟩+{\beta }_n\parallel x_n-x{\parallel }^2+\left(1-{\beta }_n\right)\parallel y_n-x\parallel \parallel x_n-x\parallel \\ \le {\beta }_n⟨f\left(x_n\right)-x_n,j\left(x_n-x\right)⟩+{\beta }_n\parallel x_n-x{\parallel }^2+\left(1-{\beta }_n\right)\parallel x_n-x{\parallel }^2.\end{array}$$

Therefore,

$$⟨f\left(x_n\right)-x_n,j\left(x-x_n\right)⟩\le 0.$$

(2.5)

Since the sets {x _n - x} and {x _n - f(x _n )} are bounded and the duality mapping j is singled-valued and weakly sequentially continuous from E into E*, for any fixed $x\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. It follows from (2.5) that

$$⟨f\left(p\right)-p,j\left(x-p\right)⟩-\underset{j\to \infty }{\text{lim}}⟨f\left(x_{n_j}\right)-x_{n_j},j\left(x-x_{n_j}\right)⟩\le 0,\forall x\in \mathsf{\text{ Fix}}\left(\mathcal{S}\right).$$

This is, $p\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$ is a solution of the variational inequality (2.2).

Finally, we show that $p\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$ is the unique solution of the variational inequality (2.2). In fact, supposing $p,q\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$ satisfy the inequality (2.2) with p ≠ q, we get that there exists ε > 0 such that ||p - q|| ≥ ε. By Proposition 1.4 there exists r ∈ (0,1) such that ||f(p)- f(q)|| ≤ r||p - q||. We get that

$$⟨\left(f-I\right)p,j\left(q-p\right)⟩\le 0\mathsf{\text{ and }}⟨\left(f-I\right)q,j\left(p-q\right)⟩\le 0.$$

Adding the two above inequalities, we have that

$$0<\left(1-r\right){\epsilon }^2\le \left(1-r\right)\parallel p-q{\parallel }^2\le ⟨\left(\left(I-f\right)p-\left(I-f\right)q,j\left(p-q\right)\right)⟩\le 0,$$

which is contradiction. We must have p = q, and the uniqueness is proved.

In a similar way, it can be shown that each cluster point of sequence {x _n } is equal to q. Therefore, the entire sequence {x _n } converges to q and the proof is complete.

Setting f is a contraction on C in Theorem 2.1, we have the following results immediately.

Corollary 2.2. [11, Theorem 3.1] Let E be a reflexive Banach space which admits a weakly sequenctially continuous duality mapping J from E into E*, suppose C is a nonempty closed convex subset of E. Let $\mathcal{S}:=\left\{T\left(t\right):t\ge 0\right\}$ be a nonexpansive semigroup on C such that $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$, and f : C → C a contraction on C. Let {α _n } ⊂ (0,1), {β _n } ⊂ (0,1), and {t _n } ⊂ (0, ∞) be sequences of real numbers satisfying ${\text{lim}}_{n\to \infty }{\alpha }_n={\text{lim}}_{n\to \infty }{t}_n={\text{lim}}_{n\to \infty }\frac{{\beta }_n}{t_n}=0$. Define a sequence {x _n } in C by

$$\left\{\begin{array}{c}\hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill x_n={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,foralln\ge 1.\hfill \end{array}\right.$$

Then {x _n } converges strongly to q, as n → ∞ q is the element of $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ such that q is the unique solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the following variational inequality:

$$⟨\left(f-I\right)q,j\left(x-q\right)⟩\le 0forallx\in \mathsf{\text{ Fix}}\left(\mathcal{S}\right).$$

Theorem 2.3. Let E be a reflexive Banach space which admits a weakly sequenctially continuous duality mapping J from E into E*, suppose C is a nonempty closed convex subset of E. Let {T(t) : t ≥ 0}, be a nonexpansive semigroup on C such that $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$, and f : C → C be a generalized contraction on C. Let {α _n } ⊂ (0,1), {β _n } ⊂ (0,1), {γ _n } ⊂ [0,1], and {t _n } ⊂ (0, ∞) be sequences of real numbers satisfying the conditions:

(C1) ${\text{lim}}_{n\to \infty }{\beta }_n=0,{\sum }_{n=0}^{\infty }{\beta }_n=\infty and{\text{lim}}_{n\to \infty }{t}_n=0,$

(C2) lim_n→∞α _n = 0 and lim_n→∞γ _n = 1,

(C3) ${\sum }_{n=0}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty ,{\sum }_{n=0}^{\infty }|{\beta }_{n+1}-{\beta }_n|<\infty ,and{\sum }_{n=0}^{\infty }|{\gamma }_{n+1}-{\gamma }_n|<\infty .$

Define a sequence {x _n } in C by

$$\left\{\begin{array}{c}\hfill x_0\in C,\hfill \\ \hfill z_n={\gamma }_n{x}_n+\left(1-{\gamma }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)z_n,\hfill \\ \hfill x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,n\ge 0.\hfill \end{array}\right.$$

(2.6)

Suppose

$$\sum _{n=0}^{\infty }\underset{x\in \tilde{C}}{\text{sup}}\parallel T\left(t_n\right)x-T\left(t_{n-1}\right)x\parallel <\infty ,$$

where $\tilde{C}$ is any bounded subset of C. Then {x _n } converges strongly to q, as n → ∞ where q is the unique solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the variational inequality (2.2).

Proof. First, we show that {x _n } is bounded. Indeed, if we take a fixed point $x\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. We will prove by induction that

$$\parallel x_n-x\parallel \le M\mathsf{\text{for all }}n\ge 0,$$

where M := {||x₀ - z||, η^-1(||f(x) - x||)}. From Definition of (2.8), notice that

$$\parallel z_n-x\parallel \le {\gamma }_n\parallel x_n-x\parallel +\left(1-{\gamma }_n\right)\parallel T\left(t_n\right)x_n-x\parallel \le \parallel x_n-x\parallel .$$

It follows that

$$\parallel y_n-x\parallel \le {\alpha }_n\parallel x_n-x\parallel +\left(1-{\alpha }_n\right)\parallel T\left(t_n\right)z_n-x\parallel \le {\alpha }_n\parallel x_n-x\parallel +\left(1-{\alpha }_n\right)\parallel x_n-x\parallel \le \parallel x_n-x\parallel .$$

The case n = 0 is obvious.

Suppose that ||x _n - x|| ≤ M, we have

$$\begin{array}{ll}\hfill \parallel x_{n+1}-x\parallel & \le {\beta }_n\parallel f\left(x_n\right)-x\parallel +\left(1-{\beta }_n\right)\parallel y_n-x\parallel \\ \le {\beta }_n\parallel f\left(x_n\right)-f\left(x\right)\parallel +{\beta }_n\parallel f\left(x\right)-x\parallel +\left(1-{\beta }_n\right)\parallel y_n-x\parallel \\ \le {\beta }_n\psi \left(\parallel x_n-x\parallel \right)+{\beta }_n\parallel f\left(x\right)-x\parallel +\left(1-{\beta }_n\right)\parallel x_n-x\parallel \\ ={\beta }_n\psi \left(\parallel x_n-x\parallel \right)+{\beta }_n\eta \left({\eta }^{-1}\left(\parallel f\left(x\right)-x\parallel \right)\right)+\left(1-{\beta }_n\right)\parallel x_n-x\parallel \\ \le {\beta }_n\psi \left(M\right)+{\beta }_n\eta \left(M\right)+\left(1-{\beta }_n\right)M\\ ={\beta }_n\psi \left(M\right)+{\beta }_n\left(M-\psi \left(M\right)\right)+\left(1-{\beta }_n\right)M=M.\end{array}$$

By induction,

$$\parallel x_n-x\parallel \le \text{max}\left\{\parallel x_0-x\parallel ,{\eta }^{-1}\left(\parallel f\left(x\right)-x\parallel \right)\right\},\forall n\ge 0.$$

Thus {x _n } is bounded, and so are {T(t _n )x _n }, {y _n }, {z _n }, {T(t _n )z _n }, and {f(x _n )}. As a result, we obtain by condition (C1),

$$\parallel x_{n+1}-y_n\parallel ={\beta }_n\parallel f\left(x_n\right)-y_n\parallel \to 0.$$

(2.7)

We next show that

$$\parallel x_n-T\left(t_n\right)x_n\parallel \to 0.$$

(2.8)

It suffices to show that

$$\parallel x_{n+1}-x_n\parallel \to 0.$$

(2.9)

Indeed, if (2.9) holds, then noting (2.7), we obtain

$$\begin{array}{ll}\hfill \parallel x_n-T\left(t_n\right)x_n\parallel & \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +\parallel y_n-T\left(t_n\right)z_n\parallel +\parallel T\left(t_n\right)z_n-T\left(t_n\right)x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)z_n\parallel +\parallel z_n-x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel +{\alpha }_n\parallel T\left(t_n\right)x_n-T\left(t_n\right)z_n\parallel \\ +\parallel z_n-x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel +{\alpha }_n\parallel x_n-z_n\parallel +\parallel z_n-x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel +\left(1+{\alpha }_n\right)\parallel x_n-z_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel \\ +\left(1+{\alpha }_n\right)\left(1-{\gamma }_n\right)\parallel x_n-T\left(t_n\right)x_n\parallel .\end{array}$$

It follows from (C2) that

$$\begin{array}{ll}\hfill \parallel x_n-T\left(t_n\right)x_n\parallel & \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel \\ +\left(1+{\alpha }_n\right)\left(1-{\gamma }_n\right)\parallel x_n-T\left(t_n\right)x_n\parallel \to 0\mathsf{\text{ as }}n\to \infty .\end{array}$$

Suppose that (2.9) is not holds, there exists ε > 0 and subsequence $\parallel x_{n_j+1}-x_{n_j}\parallel$ of ||x_{n + 1}-x _n || such that $\parallel x_{n_j+1}-x_{n_j}\parallel \ge \epsilon$ for all j ∈ ℕ. By Proposition 1.4, there exists r ∈ (0,1) such that $\parallel f\left(x_{n_j+1}\right)-f\left(x_{n_j}\right)\parallel \le r\parallel x_{n_j+1}-x_{n_j}\parallel$ for all j ∈ ℕ. Put $x_j=x_{n_j},{\alpha }_j={\alpha }_{n_j},{\beta }_j={\beta }_{n_j},{\gamma }_j={\gamma }_{n_j}$ and $t_j=t_{n_j}$ for j ∈ ℕ. We calculate x_{j+ 1}- x _j . Observing that

$$x_{j+1}={\beta }_{j}f\left(x_j\right)+\left(1-{\beta }_j\right)y_j\mathsf{\text{and}}{x}_j={\beta }_{j-1}f\left(x_{j-1}\right)+\left(1-{\beta }_{j-1}\right)y_{j-1},$$

we get

$$\begin{array}{ll}\hfill x_{j+1}-x_j& ={\beta }_{j}f\left(x_j\right)+\left(1-{\beta }_j\right)y_j-{\beta }_{j-1}f\left(x_{j-1}\right)-\left(1-{\beta }_{j-1}\right)y_{j-1}\\ ={\beta }_j\left(f\left(x_j\right)-f\left(x_{j-1}\right)\right)+\left({\beta }_j-{\beta }_{j-1}\right)f\left(x_{j-1}\right)+\left(1-{\beta }_j\right)\left(y_j-y_{j-1}\right)\\ -\left({\beta }_j-{\beta }_{j-1}\right)y_{j-1}.\end{array}$$

(2.10)

That is

$$\begin{array}{ll}\hfill \parallel x_{j+1}-x_j\parallel & \le {\beta }_{j}r\parallel x_j-x_{j-1}\parallel +|{\beta }_j-{\beta }_{j-1}|\parallel f\left(x_{j-1}\right)\parallel +\left(1-{\beta }_j\right)\parallel y_j-y_{j-1}\parallel \\ +|{\beta }_j-{\beta }_{j-1}|\parallel y_{j-1}\parallel .\end{array}$$

(2.11)

Noticing that

$$y_j={\alpha }_j{x}_j+\left(1-{\alpha }_j\right)T\left(t_j\right)z_j\mathsf{\text{and}}{y}_{j-1}={\alpha }_{j-1}{x}_{j-1}+\left(1-{\alpha }_{j-1}\right)T\left(t_{j-1}\right)z_{j-1}.$$

We obtain that

$$\begin{array}{ll}\hfill y_j-y_{j-1}& ={\alpha }_j{x}_j+\left(1-{\alpha }_j\right)T\left(t_j\right)z_j-{\alpha }_{j-1}{x}_{j-1}-\left(1-{\alpha }_{j-1}\right)T\left(t_{j-1}\right)z_{j-1}.\\ ={\alpha }_j\left(x_j-x_{j-1}\right)+\left({\alpha }_j-{\alpha }_{j-1}\right)x_{j-1}\\ +\left(1-{\alpha }_{j-1}\right)\left(T\left(t_j\right)z_{j-1}-T\left(t_{j-1}\right)z_{j-1}\right)+\left(1-{\alpha }_j\right)\left(T\left(t_j\right)z_j-T\left(t_j\right)z_{j-1}\right)\\ -\left({\alpha }_j-{\alpha }_{j-1}\right)T\left(t_j\right)z_{j-1}.\end{array}$$

(2.12)

This implied that

$$\begin{array}{ll}\hfill \parallel y_j-y_{j-1}\parallel & \le {\alpha }_j\parallel x_j-x_{j-1}\parallel +|{\alpha }_j-{\alpha }_{j-1}|\parallel x_{j-1}\parallel \\ +\left(1-{\alpha }_{j-1}\right)\parallel T\left(t_j\right)z_{j-1}-T\left(t_{j-1}\right)z_{j-1}\parallel +\left(1-{\alpha }_j\right)\parallel z_j-z_{j-1}\parallel \\ +|{\alpha }_j-{\alpha }_{j-1}|\parallel T\left(t_j\right)z_{j-1}\parallel .\end{array}$$

(2.13)

Again from (2.6) we obtain

$$\begin{array}{ll}\hfill z_j-z_{j-1}& ={\gamma }_j\left(x_j-x_{j-1}\right)+\left(1-{\gamma }_j\right)\left(T\left(t_j\right)x_j-T\left(t_j\right)x_{j-1}\right)\\ +\left(1-{\gamma }_j\right)\left(T\left(t_j\right)x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\right)+\left({\gamma }_j-{\gamma }_{j-1}\right)\left(x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\right),\end{array}$$

that is,

$$\begin{array}{ll}\hfill \parallel z_j-z_{j-1}\parallel & \le {\gamma }_j\parallel x_j-x_{j-1}\parallel +\left(1-{\gamma }_j\right)\parallel T\left(t_j\right)x_j-T\left(t_j\right)x_{j-1}\parallel \\ +\left(1-{\gamma }_j\right)\parallel T\left(t_j\right)x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\parallel +|{\gamma }_j-{\gamma }_{j-1}|\parallel x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\parallel \\ \le \parallel x_j-x_{j-1}\parallel +\left(1-{\gamma }_j\right)\parallel T\left(t_j\right)x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\parallel \\ +|{\gamma }_j-{\gamma }_{j-1}|\parallel x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\parallel .\end{array}$$

(2.14)

Substituting (2.14) into (2.13),

$$\begin{array}{c}\parallel y_j-y_{j-1}\parallel \le {\alpha }_j\parallel x_j-x_{j-1}\parallel +|{\alpha }_j-{\alpha }_{j-1}|\parallel x_{j-1}\parallel \\ +(1-{\alpha }_{j-1})\parallel T(t_j)z_{j-1}-T(t_{j-1})z_{j-1}\parallel +|{\alpha }_j-{\alpha }_{j-1}|\parallel T(t_j)z_{j-1}\parallel \\ +(1-{\alpha }_j)\parallel x_j-x_{j-1}\parallel +(1-{\alpha }_j)(1-{\gamma }_j)\parallel T(t_j)x_{j-1}-T(t_{j-1})x_{j-1}\parallel \\ +(1-{\alpha }_j)|{\gamma }_j-{\gamma }_{j-1}|\parallel x_{j-1}-T(t_{j-1})x_{j-1}\parallel ]\\ \le \parallel x_j-x_{j-1}\parallel +|{\alpha }_j-{\alpha }_{j-1}|\parallel x_{j-1}\parallel \\ +\parallel T(t_j)z_{j-1}-T(t_{j-1})z_{j-1}\parallel +|{\alpha }_j-{\alpha }_{j-1}|\parallel T(t_j)z_{j-1}\parallel \\ +\parallel T(t_j)x_{j-1}-T(t_{j-1})x_{j-1}\parallel \\ +|{\gamma }_j-{\gamma }_{j-1}|\parallel x_{j-1}-T(t_{j-1})x_{j-1}\parallel .\end{array}$$

(2.15)

Substituting (2.15) into (2.11),

$$\begin{array}{ll}\hfill \parallel x_{j+1}-x_j\parallel & \le {\beta }_{j}r\parallel x_j-x_{j-1}\parallel +|{\beta }_j-{\beta }_{j-1}|\parallel f\left(x_{j-1}\right)\parallel +\left(1-{\beta }_j\right)\parallel x_j-x_{j-1}\parallel \\ +\left(1-{\beta }_j\right)|{\alpha }_j-{\alpha }_{j-1}|\parallel x_{j-1}\parallel \\ +\left(1-{\beta }_j\right)\parallel T\left(t_j\right)z_{j-1}-T\left(t_{j-1}\right)z_{j-1}\parallel +\left(1-{\beta }_j\right)|{\alpha }_j-{\alpha }_{j-1}\parallel T\left(t_j\right)z_{j-1}\parallel \\ +\left(1-{\beta }_j\right)\parallel T\left(t_j\right)x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\parallel \\ +\left(1-{\beta }_j\right)|{\gamma }_j-{\gamma }_{j-1}|\parallel x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\parallel \\ +|{\beta }_j-{\beta }_{j-1}|\parallel y_{j-1}\parallel \\ \le \left(1-\left(1-r\right){\beta }_j\right)\parallel x_j-x_{j-1}\parallel +|{\beta }_j-{\beta }_{j-1}|\parallel f\left(x_{j-1}\right)\parallel \\ +|{\alpha }_j-{\alpha }_{j-1}|\parallel x_{j-1}\parallel \\ +\parallel T\left(t_j\right)z_{j-1}-T\left(t_{j-1}\right)z_{j-1}\parallel +|{\alpha }_j-{\alpha }_{j-1}|\parallel T\left(t_j\right)z_{j-1}\parallel \\ +\parallel T\left(t_j\right)x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\parallel \\ +|{\gamma }_j-{\gamma }_{j-1}|\parallel x_{j-1}-T\left(t_{j-1}\right)x_{j-1}\parallel \\ +|{\beta }_j-{\beta }_{j-1}|\parallel y_{j-1}\parallel .\end{array}$$

(2.16)

Hence,

$$\begin{array}{ll}\hfill \parallel x_{j+1}-x_j\parallel & \le \left(1-\left(1-r\right){\beta }_j\right)\parallel x_j-x_{j-1}\parallel +\left(2|{\alpha }_j-{\alpha }_{j-1}|+2|{\beta }_j-{\beta }_{j-1}|+|{\gamma }_j-{\gamma }_{j-1}|\right)M\\ +\underset{x\in \left\{x_n\right\}}{\text{sup}}\parallel T\left(t_j\right)x-T\left(t_{j-1}\right)x\parallel +\underset{z\in \left\{z_n\right\}}{\text{sup}}\parallel T\left(t_j\right)z-T\left(t_{j-1}\right)z\parallel ,\end{array}$$

(2.17)

where M ≥ max{||x_{j- 1}-T(t_{j- 1})x_{j- 1}||,||y_{j- 1}||,||x_{j- 1}||,||T(t _j ) z_{j- 1}||,||f(x_{j- 1})||} for all j.

By assumption, we have that ${\sum }_{j=1}^{\infty }{\beta }_j=\infty$, and ${\sum }_{j=1}^{\infty }\left(2|{\alpha }_j-{\alpha }_{j-1}|+2|{\beta }_j-{\beta }_{j-1}|+|{\gamma }_j-{\gamma }_{j-1}|+{\text{sup}}_{x\in \left\{x_n\right\}}\parallel T\left(t_j\right)x-T\left(t_{j-1}\right)x\parallel +{\text{sup}}_{z\in \left\{z_n\right\}}\parallel T\left(t_j\right)z-T\left(t_{j-1}\right)z\parallel \right)<\infty$. Hence, Lemma 1.7 is applicable to (2.17) and we obtain ||x_j+1- x _j ||→0, which is a contradiction. So (2.9) is proved. Applying Theorem 2.1, there is a unique solution $q\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the following variational inequality:

$$⟨f\left(q\right)-q,j\left(x-q\right)⟩\le 0\mathsf{\text{for all }}x\in \mathsf{\text{ Fix}}\left(\mathcal{S}\right).$$

Next, we show that

$$\underset{n\to \infty }{\text{lim}\mathsf{\text{sup}}}⟨f\left(q\right)-q,j\left(x_{n+1}-q\right)⟩\le 0.$$

(2.18)

Indeed, we can take a subsequence $\left\{x_{n_i}\right\}$ of {x _n } such that

$$\underset{n\to \infty }{\text{lim}\mathsf{\text{sup}}}⟨f\left(q\right)-q,j\left(x_{n+1}-q\right)⟩=\underset{i\to \infty }{\text{lim}}⟨f\left(q\right)-q,j\left(x_{n_i+1}-q\right)⟩.$$

By the reflexivity of E and boundedness of the sequence {x _n }, we may assume, without loss of generality, that $x_{n_i}\rightharpoonup p$ for some p ∈ C. Now we show that $p\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. Put $x_i=x_{n_i},{\alpha }_i={\alpha }_{n_i},{\beta }_i={\beta }_{n_i}$ and $t_i=t_{n_i}$ for i ∈ ℕ, let t _i ≥ 0 be such that

$$t_i\to 0\mathsf{\text{and}}\frac{\parallel T\left(t_i\right)x_i-x_i\parallel }{t_i}\to 0,i\to \infty .$$

Fix t > 0. Notice that

$$\begin{array}{ll}\hfill \parallel x_i-T\left(t\right)p\parallel & \le \sum _{k=0}^{\left[t/t_i\right]-1}\parallel T\left(\left(k+1\right)t_i\right)x_i-T\left(k{t}_i\right)x_i\parallel \\ +\parallel T\left(\left[t/t_i\right]t_i\right)x_i-T\left(\left[t/t_i\right]t_i\right)p\parallel +\parallel T\left(\left[t/t_i\right]t_i\right)p-T\left(t\right)p\parallel \\ \le \left[t/t_i\right]\parallel T\left(t_i\right)x_i-x_i\parallel +\parallel x_i-p\parallel +\parallel T\left(t-\left[t/t_i\right]t_i\right)p-p\parallel \\ \le t\frac{\parallel T\left(t_i\right)x_i-x_i\parallel }{t_i}+\parallel x_i-p\parallel +\parallel T\left(t-\left[t/t_i\right]t_i\right)p-p\parallel \\ \le t\frac{\parallel T\left(t_i\right)x_i-x_i\parallel }{t_i}+\parallel x_i-p\parallel +\text{max}\left\{\parallel T\left(s\right)p-p\parallel :0\le s\le t_i\right\}.\end{array}$$

For all i ∈ ℕ, we have

$$\underset{i\to \infty }{\text{lim}\mathsf{\text{sup}}}\parallel x_i-T\left(t\right)p\parallel \le \underset{i\to \infty }{\text{lim}\mathsf{\text{sup}}}\parallel x_i-p\parallel .$$

Since Banach space E with a weakly sequentially continuous duality mapping satisfies Opial's condition, this implies T(t)p = p. Therefore $p\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. In view of the variational inequality (2.2) and the assumption that duality mapping J is weakly sequentially continuous, we conclude

$$\begin{array}{ll}\hfill \underset{n\to \infty }{\text{lim}\mathsf{\text{sup}}}⟨f\left(q\right)-q,j\left(x_{n+1}-q\right)⟩& =\underset{i\to \infty }{\text{lim}}⟨f\left(q\right)-q,j\left(x_{n_i+1}-q\right)⟩\\ =⟨f\left(q\right)-q,j\left(p-q\right)⟩\le 0.\end{array}$$

Then (2.18) is proved. Finally show that x _n → q, i.e. ||x _n - q|| → 0. Suppose that ||x _n - q|| ↛ 0, then there exists ε > 0 and a subsequence $\left\{x_{n_j}\right\}$ of {x _n } such that $\parallel x_{n_j}-q\parallel \ge \epsilon$ for all j ∈ ℕ. Put $x_j=x_{n_j},{\alpha }_j={\alpha }_{n_j},{\beta }_j={\beta }_{n_j}$ and $t_j=t_{n_j}$ for j ∈ ℕ. By Proposition 1.4, there exists r ∈ (0,1) such that ||f(x _j ) - f(q)|| ≤ r||x _j - q|| for all j ∈ ℕ. As a matter of fact, from Lemma 1.6 we have that

$$\begin{array}{l}\parallel x_{j+1}-q{\parallel }^2=\parallel {\beta }_{j}f(x_j)+(1-{\beta }_j)({\alpha }_j{x}_j+(1-{\alpha }_j)T(t_j)z_j)-q{\parallel }^2\\ =\parallel (1-{\beta }_j)({\alpha }_j(x_j-q)+(1-{\alpha }_j(T(t_j)z_j-q))+{\beta }_j(f(x_j)-q){\parallel }^2\\ \le {(1-{\beta }_j)}^2\parallel {\alpha }_j(x_j-q)+(1-{\alpha }_j)(T(t_j)z_j-q){\parallel }^2+2{\beta }_j〈f(x_j)-q,j(x_{j+1}-q)〉\\ \le {(1-{\beta }_j)}^2{({\alpha }_j\parallel x_j-q\parallel +(1-{\alpha }_j)\parallel T(t_j)z_j-q\parallel )}^2\\ +2{\beta }_j〈f(x_j)-f(q),j(x_{j+1}-q)〉+2{\beta }_j〈f(q)-q,j(x_{j+1}-q)〉\\ \le {(1-{\beta }_j)}^2\parallel x_j-q{\parallel }^2+2{\beta }_j\parallel f(x_j)-f(q)\parallel \parallel x_{j+1}-q\parallel +2{\beta }_j〈f(q)-q,j(x_{j+1}-q)〉\\ \le {(1-{\beta }_j)}^2\parallel x_j-q{\parallel }^2+2{\beta }_{j}r\parallel x_j-q\parallel \parallel x_{j+1}-q\parallel +2{\beta }_j〈f(q)-q,j(x_{j+1}-q)〉\\ \le {(1-{\beta }_j)}^2\parallel x_j-q{\parallel }^2+{\beta }_{j}r(\parallel x_j-q{\parallel }^2+\parallel x_{j+1}-q{\parallel }^2)+2{\beta }_j〈f(q)-q,j(x_{j+1}-q)〉\\ \le ({(1-{\beta }_j)}^2+{\beta }_{j}r)\parallel x_j-q{\parallel }^2+{\beta }_{j}r\parallel x_{j+1}-q{\parallel }^2+2{\beta }_j〈f(q)-q,j(x_{j+1}-q)〉.\end{array}$$

It follows that

$$\begin{array}{ll}\hfill \parallel x_{j+1}-q{\parallel }^2& \le \frac{\left(1-\left(2-r\right){\beta }_j+{\beta }_j^2\right)}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+\frac{2{\beta }_j}{1-{\beta }_{j}r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ =\frac{1-\left(2-r\right){\beta }_j}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+\frac{{\beta }_j^2}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+\frac{2{\beta }_j}{1-{\beta }_{j}r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le \frac{1-\left(2-r\right){\beta }_j}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+\frac{{\beta }_j^2}{1-r}\parallel x_j-q{\parallel }^2+\frac{2{\beta }_j}{1-r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le \frac{1-{\beta }_{j}r-2\left(1-r\right){\beta }_j}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+{\beta }_j^{2}M+\frac{2{\beta }_j}{1-r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ =\left(1-\frac{2\left(1-r\right){\beta }_j}{1-{\beta }_{j}r}\right)\parallel x_j-q{\parallel }^2+{\beta }_j^{2}M+\frac{2{\beta }_j}{1-r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le \left(1-2\left(1-r\right){\beta }_j\right)\parallel x_j-q{\parallel }^2+{\beta }_j\left(\frac{2}{1-r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩+{\beta }_{j}M\right),\end{array}$$

where M > 0 such that $M\ge \frac{1}{1-r}\parallel x_j-q{\parallel }^2$. That is,

$$\parallel x_{j+1}-q{\parallel }^2\le \left(1-{\gamma }_j\right)\parallel x_j-q{\parallel }^2+{\delta }_j,$$

(2.19)

where γ _j = 2(1 - r)β _j and $\frac{{\delta }_j}{{\gamma }_j}=\frac{1}{{\left(1-r\right)}^2}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩+\frac{M}{2\left(1-r\right)}{\beta }_j$.

It follows by condition (B1) that γ _j → 0 and ${\sum }_{j=1}^{\infty }{\gamma }_j=\infty$. From (2.18) we have

$$\begin{array}{ll}\hfill \underset{j\to \infty }{\text{lim}\mathsf{\text{sup}}}\frac{{\delta }_j}{{\gamma }_j}& \le \underset{j\to \infty }{\text{lim}\mathsf{\text{sup}}}\frac{1}{{\left(1-r\right)}^2}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩+\underset{j\to \infty }{\text{lim}}\frac{M}{2\left(1-r\right)}{\beta }_j\\ \le \underset{j\to \infty }{\text{lim}\mathsf{\text{sup}}}\frac{1}{{\left(1-r\right)}^2}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\le 0.\end{array}$$

Using Lemma 1.7 onto (2.19), we conclude that ||x _j - q|| → 0. This is a contradiction. Hence x _n → q.

The proof is completed.

If γ _n ≡ 1, then we have the following Corollary.

Corollary 2.4. Let E be a reflexive Banach space which admits a weakly sequenctially continuous duality mapping J from E into E*, suppose C is a nonempty closed convex subset of E. Let {T(t) : t ≥ 0} be a nonexpansive semigroup on C such that $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$, and f : C → C be a generalized contraction on C. Let {α _n } ⊂ (0,1), {β _n } ⊂ (0,1), and {t _n } ⊂ (0, ∞) be sequences of real numbers satisfying the conditions:

(C1) ${\text{lim}}_{n\to \infty }{\beta }_n=0,{\sum }_{n=0}^{\infty }{\beta }_n=\infty and{\text{lim}}_{n\to \infty }{t}_n=0,$

(C2) lim _n→∞ α _n = 0,

(C3) ${\sum }_{n=0}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty ,{\sum }_{n=0}^{\infty }|{\beta }_{n+1}-{\beta }_n|<\infty .$

Define a sequence {x _n } in C by

$$\left\{\begin{array}{c}\hfill x_0\in C,\hfill \\ \hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,n\ge 0.\hfill \end{array}\right.$$

(2.20)

Suppose ${\sum }_{n=0}^{\infty }\parallel T\left(t_n\right)x_{n-1}-T\left(t_{n-1}\right)x_{n-1}\parallel <\infty$. Then {x _n } converges strongly to q, as n → ∞ where q is the unique solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the variational inequality (2.2).

Setting f is a contraction on C in Corollary 2.4, we have the following results immediately.

Corollary 2.5. [11, Theorem 3.2] Let E be a reflexive Banach space which admits a weakly sequenctially continuous duality mapping J from E into E*, suppose C is a nonempty closed convex subset of E. Let {T(t) : t ≥ 0} be a nonexpansive semigroup on C such that $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$, and f : C → C be a contraction on C. Let {α _n } ⊂ (0,1), {β _n } ⊂ (0,1), and {t _n } ⊂ (0, ∞) be sequences of real numbers satisfying the conditions:

(C1) ${\text{lim}}_{n\to \infty }{\beta }_n=0,{\sum }_{n=0}^{\infty }{\beta }_n=\infty and{\text{lim}}_{n\to \infty }{t}_n=0,$

(C2) lim _n→∞ α _n = 0,

(C3) ${\sum }_{n=0}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty ,{\sum }_{n=0}^{\infty }|{\beta }_{n+1}-{\beta }_n|<\infty .$

Define a sequence {x _n } in C by

$$\left\{\begin{array}{c}\hfill x_0\in C,\hfill \\ \hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,n\ge 0.\hfill \end{array}\right.$$

(2.21)

Suppose ${\sum }_{n=0}^{\infty }\parallel T\left(t_n\right)x_{n-1}-T\left(t_{n-1}\right)x_{n-1}\parallel <\infty$. Then {x _n } converges strongly to q, as n → ∞; where q is the unique solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the variational inequality (2.2).

Questions

• (i) Could we obtain Theorem 2.3 with other control conditions which are different from (C2) and (C3)?

• (ii) Could we weaken the control condition (*) by the strictly weaker condition (**):

  $${\text{lim}}_{n\to \infty }{\text{sup}}_{x\in \tilde{C}}\parallel T\left(t_n\right)x-T\left(t_{n-1}\right)x\parallel =0?$$

The following theorem gives the affirmative answers to these question mentioned above.

Theorem 2.6. Let E be a reflexive Banach space which admits a weakly sequenctially continuous duality mapping J from E into E*, suppose C is a nonempty closed convex subset of E. Let {T(t) : t ≥ 0}, be a nonexpansive semigroup on C such that $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$, and f : C → C be a generalized contraction on C. Let {α _n } ⊂ (0,1), {β _n } ⊂ (0,1), {γ _n } ⊂ [0,1] and {t _n } ⊂ (0, ∞) be sequences of real numbers satisfying the conditions:

(B1) ${\text{lim}}_{n\to \infty }{\beta }_n=0,{\sum }_{n=0}^{\infty }{\beta }_n=\infty and{\text{lim}}_{n\to \infty }{t}_n=0,$

(B2) α _n + (1 + α _n )(1 - γ _n ) ∈ [0,a) for some a ∈ (0,1),

(B3) 0 < lim inf_n→∞α _n ≤ lim sup_n→∞α _n < 1,

(B4) lim_n→∞(γ_n+1- γ _n ) = 0.

Define a sequence {x _n } in C by

$$\left\{\begin{array}{c}\hfill x_0\in C,\hfill \\ \hfill z_n={\gamma }_n{x}_n+\left(1-{\gamma }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)z_n,\hfill \\ \hfill x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,n\ge 0.\hfill \end{array}\right.$$

(2.22)

Suppose

$$\underset{n\to \infty }{\text{lim}}\underset{x\in \tilde{C}}{\text{sup}}\parallel T\left(t_n\right)x-T\left(t_{n-1}\right)x\parallel =0,$$

where $\tilde{C}$ is any bounded subset of C. Then {x _n } converges strongly to q, as n → ∞; where q is the unique solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the variational inequality (2.2).

Proof. First, we show that {x _n } is bounded. Indeed, if we take a fixed point $x\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. We will prove by induction that

$$\parallel x_n-x\parallel \le M\mathsf{\text{for all }}n\ge 0,$$

where M := {||x₀ - z||,η^-1(||f(x) - x||)}. From Definition of (2.22), notice that

$$\parallel z_n-x\parallel \le {\gamma }_n\parallel x_n-x\parallel +\left(1-{\gamma }_n\right)\parallel T\left(t_n\right)x_n-x\parallel \le \parallel x_n-x\parallel .$$

It follows that

$$\parallel y_n-x\parallel \le {\alpha }_n\parallel x_n-x\parallel +\left(1-{\alpha }_n\right)\parallel T\left(t_n\right)z_n-x\parallel \le {\alpha }_n\parallel x_n-x\parallel +\left(1-{\alpha }_n\right)\parallel x_n-x\parallel \le \parallel x_n-x\parallel .$$

The case n = 0 is obvious.

Suppose that ||x _n - x|| ≤ M, we have

$$\begin{array}{ll}\hfill \parallel x_{n+1}-x\parallel & \le {\beta }_n\parallel f\left(x_n\right)-x\parallel +\left(1-{\beta }_n\right)\parallel y_n-x\parallel \\ \le {\beta }_n\parallel f\left(x_n\right)-f\left(x\right)\parallel +{\beta }_n\parallel f\left(x\right)-x\parallel +\left(1-{\beta }_n\right)\parallel y_n-x\parallel \\ \le {\beta }_n\psi \left(\parallel x_n-x\parallel \right)+{\beta }_n\parallel f\left(x\right)-x\parallel +\left(1-{\beta }_n\right)\parallel x_n-x\parallel \\ ={\beta }_n\psi \left(\parallel x_n-x\parallel \right)+{\beta }_n\eta \left({\eta }^{-1}\left(\parallel f\left(x\right)-x\parallel \right)\right)+\left(1-{\beta }_n\right)\parallel x_n-x\parallel \\ \le {\beta }_n\psi \left(M\right)+{\beta }_n\eta \left(M\right)+\left(1-{\beta }_n\right)M\\ ={\beta }_n\psi \left(M\right)+{\beta }_n\left(M-\psi \left(M\right)\right)+\left(1-{\beta }_n\right)M=M.\end{array}$$

By induction,

$$\parallel x_n-x\parallel \le \text{max}\left\{\parallel x_0-x\parallel ,{\eta }^{-1}\left(\parallel f\left(x\right)-x\parallel \right)\right\},\forall n\ge 0.$$

Thus {x _n } is bounded, and so are {T(t _n )x _n }, {y _n }, {z _n } and {f(x _n )}. As a result, we obtain by condition (B1),

$$\parallel x_{n+1}-y_n\parallel ={\beta }_n\parallel f\left(x_n\right)-y_n\parallel \to 0.$$

(2.23)

We next show that

$$\parallel x_n-T\left(t_n\right)x_n\parallel \to 0.$$

(2.24)

It suffices to show that

$$\parallel x_{n+1}-x_n\parallel \to 0.$$

(2.25)

Indeed, if (2.25) holds, then noting (2.23), we obtain

$$\begin{array}{ll}\hfill \parallel x_n-T\left(t_n\right)x_n\parallel & \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +\parallel y_n-T\left(t_n\right)z_n\parallel +\parallel T\left(t_n\right)z_n-T\left(t_n\right)x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)z_n\parallel +\parallel z_n-x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel +{\alpha }_n\parallel T\left(t_n\right)x_n-T\left(t_n\right)z_n\parallel \\ +\parallel z_n-x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel +{\alpha }_n\parallel x_n-z_n\parallel +\parallel z_n-x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel +\left(1+{\alpha }_n\right)\parallel x_n-z_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel +{\alpha }_n\parallel x_n-T\left(t_n\right)x_n\parallel \\ +\left(1+{\alpha }_n\right)\left(1-{\gamma }_n\right)\parallel x_n-T\left(t_n\right)x_n\parallel \end{array}$$

and hence (1 - (α _n +(1+ α _n )(1 - γ _n )))||x _n -T(t _n )x _n || ≤ ||x _n - x_n+1|| + ||x_n+1- y _n || → 0 as n → ∞. Using (B2), we conclude that (2.24) holds. Define the sequence {u _n } by

$$u_n=\frac{x_{n+1}-{\sigma }_n{x}_n}{\left(1-{\sigma }_n\right)},$$

where σ _n = (1 - β _n )α _n . Then x_{n+ 1}= σ _n x _n + (1 - σ _n )u _n . Next multiplication gives us that

$$\begin{array}{ll}\hfill u_{n+1}-u_n& =\frac{\left(x_{n+2}-{\sigma }_{n+1}{x}_{n+1}\right)}{1-{\sigma }_{n+1}}-\frac{x_{n+1}-{\sigma }_n{x}_n}{1-{\sigma }_n}\\ =\frac{{\beta }_{n+1}f\left(x_{n+1}\right)+\left(1-{\beta }_{n+1}\right)y_{n+1}-{\sigma }_{n+1}{x}_{n+1}}{1-{\sigma }_{n+1}}-\frac{{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n-{\sigma }_n{x}_n}{1-{\sigma }_n}\\ =\left(\frac{{\beta }_{n+1}f\left(x_{n+1}\right)}{1-{\alpha }_{n+1}}-\frac{{\beta }_{n}f\left(x_n\right)}{1-{\alpha }_n}\right)\\ +\frac{\left(1+{\beta }_{n+1}\right)\left({\alpha }_{n+1}{x}_{n+1}+\left(1-{\alpha }_{n+1}\right)T\left(t_{n+1}\right)z_{n+1}\right)-{\alpha }_{n+1}{x}_{n+1}}{1-{\sigma }_{n+1}}\\ -\frac{\left(1-{\beta }_n\right)\left({\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)z_n\right)-{\alpha }_n{x}_n}{1-{\sigma }_n}\\ =\left(\frac{{\beta }_{n+1}f\left(x_{n+1}\right)}{1-{\sigma }_{n+1}}-\frac{{\beta }_{n}f\left(x_n\right)}{1-{\sigma }_n}\right)\\ +\left(T\left(t_{n+1}\right)z_{n+1}-\frac{{\alpha }_{n+1}}{1-{\sigma }_{n+1}}T\left(t_{n+1}\right)z_{n+1}\right)-\left(T\left(t_n\right)z_n-\frac{{\alpha }_n}{1-{\sigma }_n}T\left(t_n\right)z_n\right)\\ =\frac{{\beta }_{n+1}}{1-{\sigma }_{n+1}}\left(f\left(x_{n+1}\right)-T\left(t_{n+1}\right)z_{n+1}\right)-\frac{{\beta }_n}{1-{\sigma }_n}\left(f\left(x_n\right)-T\left(t_n\right)z_n\right)\\ +\left(T\left(t_{n+1}\right)z_{n+1}-T\left(t_{n+1}\right)z_n\right)-\left(T\left(t_{n+1}\right)z_n-T\left(t_n\right)z_n\right).\end{array}$$

Then we have

$$\begin{array}{ll}\hfill \parallel u_{n+1}-u_n\parallel & \le \frac{{\beta }_{n+1}}{1-{\sigma }_{n+1}}\parallel f\left(x_{n+1}\right)-T\left(t_{n+1}\right)z_{n+1}\parallel -\frac{{\beta }_n}{1-{\sigma }_n}\parallel f\left(x_n\right)-T\left(t_n\right)z_n\parallel \\ +\parallel z_{n+1}-z_n\parallel +\parallel T\left(t_{n+1}\right)z_n-T\left(t_n\right)z_n\parallel .\end{array}$$

(2.26)

From (2.22) we have

$$\begin{array}{ll}\hfill z_{n+1}-z_n& ={\gamma }_{n+1}{x}_{n+1}+\left(1-{\gamma }_{n+1}\right)T\left(t_{n+1}\right)x_{n+1}-{\gamma }_n{x}_n-\left(1-{\gamma }_n\right)T\left(t_n\right)x_n\\ ={\gamma }_{n+1}\left(x_{n+1}-x_n\right)+\left({\gamma }_{n+1}-{\gamma }_n\right)x_n+\left(1-{\gamma }_{n+1}\right)\left(T\left(t_{n+1}\right)x_{n+1}-T\left(t_{n+1}\right)x_n\right)\\ +\left(1-{\gamma }_{n+1}\right)\left(T\left(t_{n+1}\right)x_n-T\left(t_n\right)x_n\right)-\left({\gamma }_{n+1}-{\gamma }_n\right)T\left(t_n\right)x_n\\ ={\gamma }_{n+1}\left(x_{n+1}-x_n\right)+\left({\gamma }_{n+1}-{\gamma }_n\right)\left(x_n-T\left(t_n\right)x_n\right)\\ +\left(1-{\gamma }_{n+1}\right)\left(T\left(t_{n+1}\right)x_{n+1}-T\left(t_{n+1}\right)x_n\right)+\left(1-{\gamma }_{n+1}\right)\left(T\left(t_{n+1}\right)x_n-T\left(t_n\right)x_n\right),\end{array}$$

that is,

$$\begin{array}{ll}\hfill \parallel z_{n+1}-z_n\parallel & \le {\gamma }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\gamma }_{n+1}-{\gamma }_n|\parallel x_n-T\left(t_n\right)x_n\parallel +\left(1-{\gamma }_{n+1}\right)\parallel x_{n+1}-x_n\parallel \\ +\left(1-{\gamma }_{n+1}\right)\parallel T\left(t_{n+1}\right)x_n-T\left(t_n\right)x_n\parallel \\ \le \parallel x_{n+1}-x_n\parallel +|{\gamma }_{n+1}-{\gamma }_n|\parallel x_n-T\left(t_n\right)x_n\parallel T\left(t_{n+1}\right)x_n-T\left(t_n\right)x_n\parallel .\end{array}$$

(2.27)

Substituting (2.27) into (2.26) that

$$\begin{array}{ll}\hfill \parallel u_{n+1}-u_n\parallel -\parallel x_{n+1}-x_n\parallel & \le \frac{{\beta }_{n+1}}{1-{\sigma }_{n+1}}\parallel f\left(x_{n+1}\right)-T\left(t_{n+1}\right)z_{n+1}\parallel -\frac{{\beta }_n}{1-{\sigma }_n}\parallel f\left(x_n\right)-T\left(t_n\right)z_n\parallel \\ +|{\gamma }_{n+1}-{\gamma }_n|\parallel x_n-T\left(t_n\right)x_n\parallel +\parallel T\left(t_{n+1}\right)x_n-T\left(t_n\right)x_n\parallel \\ +\parallel T\left(t_{n+1}\right)z_n-T\left(t_n\right)z_n\parallel \\ \le \frac{{\beta }_{n+1}}{1-{\sigma }_{n+1}}\parallel f\left(x_{n+1}\right)-T\left(t_{n+1}\right)z_{n+1}\parallel -\frac{{\beta }_n}{1-{\sigma }_n}\parallel f\left(x_n\right)-T\left(t_n\right)z_n\parallel \\ +|{\gamma }_{n+1}-{\gamma }_n|\parallel x_n-T\left(t_n\right)x_n\parallel +\underset{x\in \left\{x_n\right\}}{\text{sup}}\parallel T\left(t_{n+1}\right)x-T\left(t_n\right)x\parallel \\ +\underset{z\in \left\{z_n\right\}}{\text{sup}}\parallel T\left(t_{n+1}\right)z-T\left(t_n\right)z\parallel .\end{array}$$

(2.28)

By (B1), (B4), $\underset{n\to \infty }{\text{lim}}\underset{x\in \left\{x_n\right\}}{\text{sup}}\parallel T\left(t_{n+1}\right)x-T\left(t_n\right)x\parallel =0$, $\underset{n\to \infty }{\text{lim}}\underset{z\in \left\{z_n\right\}}{\text{sup}}\parallel T\left(t_{n+1}\right)z-T\left(t_n\right)z\parallel =0$ and (2.28), we obtain that

$$\underset{n\to \infty }{\text{lim}\mathsf{\text{sup}}}\left(\parallel u_{n+1}-u_n\parallel -\parallel x_{n+1}-x_n\parallel \right)\le 0.$$

Hence by Lemma 1.8, we have lim _n→∞ ||u _n - x _n || = 0. It follows from (B3) that

$$\underset{n\to \infty }{\text{lim}}\parallel x_{n+1}-x_n\parallel =\underset{n\to \infty }{\text{lim}}\left(1-{\sigma }_n\right)\parallel u_n-x_n\parallel =0.$$

Hence (2.24) holds. Applying Theorem 2.1, there is a unique solution $q\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the following variational inequality:

$$⟨f\left(q\right)-q,j\left(x-q\right)⟩\le 0\mathsf{\text{for all }}x\in \mathsf{\text{ Fix}}\left(\mathcal{S}\right).$$

Next, we show that

$$\underset{n\to \infty }{\text{lim}\mathsf{\text{sup}}}⟨f\left(q\right)-q,j\left(x_{n+1}-q\right)⟩\le 0.$$

(2.29)

Indeed, we can take a subsequence $\left\{x_{n_i}\right\}$ of {x _n } such that

$$\underset{n\to \infty }{\text{lim}\mathsf{\text{sup}}}⟨f\left(q\right)-q,j\left(x_{n+1}-q\right)⟩=\underset{i\to \infty }{\text{lim}}⟨f\left(q\right)-q,j\left(x_{n+1}-q\right)⟩.$$

By the reflexivity of E and boundedness of the sequence {x _n }, we may assume, without loss of generality, that $x_{n_i}\rightharpoonup p$ for some p ∈ C. Now we show that $p\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. Put $x_i=x_{n_i},{\alpha }_i={\alpha }_{n_i},{\beta }_i={\beta }_{n_i}$ and $t_i=t_{n_i}$ for i ∈ ℕ, let t _i ≥ 0 be such that

$$t_i\to 0\mathsf{\text{ and }}\frac{\parallel T\left(t_i\right)x_i-x_i\parallel }{t_i}\to 0,i\to \infty .$$

Fix t > 0. Notice that

$$\begin{array}{ll}\hfill \parallel x_i-T\left(t\right)p\parallel & \le \sum _{k=0}^{\left[t/t_i\right]-1}\parallel T\left(\left(k+1\right)t_i\right)x_i-T\left(k{t}_i\right)x_i\parallel \\ +\parallel T\left(\left[t/t_i\right]t_i\right)x_i-T\left(\left[t/t_i\right]t_i\right)p\parallel +\parallel T\left(\left[t/t_i\right]t_i\right)p-T\left(t\right)p\parallel \\ \le \left[t/t_i\right]\parallel T\left(t_i\right)x_i-x_i\parallel +\parallel x_i-p\parallel +\parallel T\left(t-\left[t/t_i\right]t_i\right)p-p\parallel \\ \le t\frac{\parallel T\left(t_i\right)x_i-x_i\parallel }{t_i}+\parallel x_i-p\parallel +\parallel T\left(t-\left[t/t_i\right]t_i\right)p-p\parallel \\ \le t\frac{\parallel T\left(t_i\right)x_i-x_i\parallel }{t_i}+\parallel x_i-p\parallel +\text{max}\left\{\parallel T\left(s\right)p-p\parallel :0\le s\le t_i\right\}.\end{array}$$

For all i ∈ ℕ, we have

$$\underset{i\to \infty }{\text{lim}\mathsf{\text{sup}}}\parallel x_i-T\left(t\right)p\parallel \le \underset{i\to \infty }{\text{lim}\mathsf{\text{sup}}}\parallel x_i-p\parallel .$$

Since Banach space E with a weakly sequentially continuous duality mapping satisfies Opial's condition, this implies T(t)p = p. Therefore $p\in \mathsf{\text{Fix}}\left(\mathcal{S}\right)$. In view of the variational inequality (2.2) and the assumption that duality mapping J is weakly sequentially continuous, we conclude

$$\begin{array}{c}\underset{n\to \infty }{\lim \sup }〈f(q)-q,j(x_{n+1}-q)〉=\underset{i\to \infty }{\lim }〈f(q)-q,j(x_{ni+1}-q)〉\\ =〈f(q)-q,j(p-q)〉\le 0.\end{array}$$

Then (2.29) is proved. Finally, show that x _n → q, i.e., ||x _n -q|| → 0. Suppose that ||x _n -q|| ↛ 0, then there exists ε > 0 and a subsequence $\left\{x_{n_j}\right\}$ of {x _n } such that $\parallel x_{n_j}-q\parallel \ge \epsilon$ for all j ∈ ℕ. Put $x_j=x_{n_j},{\alpha }_j={\alpha }_{n_j},{\beta }_j={\beta }_{n_j}$ and $t_j=t_{n_j}$ for j ∈ ℕ. By Proposition 1.4, there exists r ∈ (0,1) such that ||f(x _j ) - f(q)|| ≤ r||x _j - q|| for all j ∈ ℕ. As a matter of fact, from 1.6 we have that

$$\begin{array}{ll}\hfill \parallel x_{j+1}-q{\parallel }^2& =\parallel {\beta }_{j}f\left(x_j\right)+\left(1-{\beta }_j\right)\left({\alpha }_j{x}_j+\left(1-{\alpha }_j\right)T\left(t_j\right)z_j\right)-q{\parallel }^2\\ =\parallel \left(1-{\beta }_j\right)\left({\alpha }_j\left(x_j-q\right)+\left(1-{\alpha }_j\right)\left(T\left(t_j\right)z_j-q\right)\right)+{\beta }_j\left(f\left(x_j\right)-q\right){\parallel }^2\\ \le {\left(1-{\beta }_j\right)}^2\parallel {\alpha }_j\left(x_j-q\right)+\left(1-{\alpha }_j\right)\left(T\left(t_j\right)z_j-q\right){\parallel }^2+2{\beta }_j⟨f\left(x_j\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le {\left(1-{\beta }_j\right)}^2{\left({\alpha }_j\parallel x_j-q\parallel +\left(1-{\alpha }_j\right)\parallel T\left(t_j\right)z_j-q\parallel \right)}^2\\ +2{\beta }_j⟨f\left(x_j\right)-f\left(q\right),j\left(x_{j+1}-q\right)⟩+2{\beta }_j⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le {\left(1-{\beta }_j\right)}^2\parallel x_j-q{\parallel }^2+2{\beta }_j\parallel f\left(x_j\right)-f\left(q\right)\parallel \parallel x_{j+1}-q\parallel +2{\beta }_j⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le {\left(1-{\beta }_j\right)}^2\parallel x_j-q{\parallel }^2+2{\beta }_{j}r\parallel x_j-q\parallel \parallel x_{j+1}-q\parallel +2{\beta }_j⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le {\left(1-{\beta }_j\right)}^2\parallel x_j-q{\parallel }^2+{\beta }_{j}r\left(\parallel x_j-q{\parallel }^2+\parallel x_{j+1}-q{\parallel }^2\right)+2{\beta }_j⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le \left({\left(1-{\beta }_j\right)}^2+{\beta }_{j}r\right)\parallel x_j-q{\parallel }^2+{\beta }_{j}r\parallel x_{j+1}-q{\parallel }^2+2{\beta }_j⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩.\end{array}$$

It follows that

$$\begin{array}{ll}\hfill \parallel x_{j+1}-q{\parallel }^2& \le \frac{\left(1-\left(2-r\right){\beta }_j+{\beta }_j^2\right)}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+\frac{2{\beta }_j}{1-{\beta }_{j}r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ =\frac{1-\left(2-r\right){\beta }_j}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+\frac{{\beta }_j^2}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+\frac{2{\beta }_j}{1-{\beta }_{j}r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le \frac{1-\left(2-r\right){\beta }_j}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+\frac{{\beta }_j^2}{1-r}\parallel x_j-q{\parallel }^2+\frac{2{\beta }_j}{1-r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le \frac{1-{\beta }_{j}r-2\left(1-r\right){\beta }_j}{1-{\beta }_{j}r}\parallel x_j-q{\parallel }^2+{\beta }_j^{2}M+\frac{2{\beta }_j}{1-r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ =\left(1-\frac{2\left(1-r\right){\beta }_j}{1-{\beta }_{j}r}\right)\parallel x_j-q{\parallel }^2+{\beta }_j^{2}M+\frac{2{\beta }_j}{1-r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\\ \le \left(1-2\left(1-r\right){\beta }_j\right)\parallel x_j-q{\parallel }^2+{\beta }_j\left(\frac{2}{1-r}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩+{\beta }_{j}M\right),\end{array}$$

where M > 0 such that $M\ge \frac{1}{1-r}\parallel x_j-q{\parallel }^2$. That is,

$$\parallel x_{j+1}-q{\parallel }^2\le \left(1-{\gamma }_j\right)\parallel x_j-q{\parallel }^2+{\delta }_j,$$

(2.30)

Where γ _j = 2(1 - r)β _j and $\frac{{\delta }_j}{{\gamma }_j}=\frac{1}{{\left(1-r\right)}^2}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩+\frac{M}{2\left(1-r\right)}{\beta }_j$.

It follows by condition (B1) that γ _j → 0 and ${\sum }_{j=1}^{\infty }{\gamma }_j=\infty$. From (2.29) we have

$$\begin{array}{ll}\hfill \underset{j\to \infty }{\text{lim}\mathsf{\text{sup}}}\frac{{\delta }_j}{{\gamma }_j}& \le \underset{j\to \infty }{\text{lim}\mathsf{\text{sup}}}\frac{1}{{\left(1-r\right)}^2}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩+\underset{j\to \infty }{\text{lim}}\frac{M}{2\left(1-r\right)}{\beta }_j\\ \le \underset{j\to \infty }{\text{lim}\mathsf{\text{sup}}}\frac{1}{{\left(1-r\right)}^2}⟨f\left(q\right)-q,j\left(x_{j+1}-q\right)⟩\le 0.\end{array}$$

Using Lemma 1.7 onto (2.30), we conclude that ||x _j -q|| → 0. This is a contradiction. Hence x _n →q.

The proof is completed.

If γ _n ≡ 1, then we have the following Corollary.

Corollary 2.7. Let E be a reflexive Banach space which admits a weakly sequenctially continuous duality mapping J from E into E*, suppose C is a nonempty closed convex subset of E. Let {T(t) : t ≥ 0}, be a nonexpansive semigroup on C such that $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$, and f : C → C be a generalized contraction on C. Let {α _n } ⊂ (0,1), {β _n } ⊂ (0,1), and {t _n } ⊂ (0, ∞) be sequences of real numbers satisfying the conditions:

(B1) $\underset{n\to \infty }{\text{lim}}{\beta }_n=0,{\sum }_{n=0}^{\infty }{\beta }_n=\infty ,$

(B2) 0 < lim inf_n→∞α _n ≤ lim sup _n→∞ α _n < 1,

(B3) lim_n→∞t _n = 0.

Define a sequence {x _n } in C by

$$\left\{\begin{array}{c}\hfill x_0\in C,\hfill \\ \hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,n\ge 0.\hfill \end{array}\right.$$

Suppose lim_n→∞||T(t _n )x_{n- 1}- T(t_{n- 1})x_{n- 1}|| = 0. Then {x _n } converges strongly to q, as n → ∞ where q is the unique solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the variational inequality (2.2).

Setting f is a contraction on C in Corollary 2.7, we have the following results immediately.

Corollary 2.8. Let E be a reflexive Banach space which admits a weakly sequenctially continuous duality mapping J from E into E*, suppose C is a nonempty closed convex subset of E. Let {T(t) : t ≥ 0}, be a nonexpansive semigroup on C such that $\mathsf{\text{Fix}}\left(\mathcal{S}\right)\ne \varnothing$, and f : C → C be a contraction on C. Let {α _n } ⊂ (0,1), {β _n } ⊂ (0,1), and {t _n } ⊂ (0, ∞) be sequences of real numbers satisfying the conditions:

(B1) $\underset{n\to \infty }{\text{lim}}{\beta }_n=0,{\sum }_{n=0}^{\infty }{\beta }_n=\infty ,$

(B2) 0 < lim inf_n→∞α _n ≤ lim sup_n→∞α _n < 1,

(B3) lim_n→∞t _n = 0.

Define a sequence {x _n } in C by

$$\left\{\begin{array}{c}\hfill x_0\in C,\hfill \\ \hfill y_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T\left(t_n\right)x_n,\hfill \\ \hfill x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)y_n,n\ge 0.\hfill \end{array}\right.$$

Suppose lim _n→∞ ||T(t _n )x_{n- 1}- T(t_{n- 1})x_{n- 1}|| = 0. Then {x _n } converges strongly to q, as n → ∞ where q is the unique solution in $\mathsf{\text{Fix}}\left(\mathcal{S}\right)$ to the variational inequality (2.2).

Remark 2.9. Theorem 2.3 generalize and improve [11, Theorem 3.2]. In fact,

• (i) The iterations (1.10) can reduce to (1.9).

• (ii) The contraction is replaced by the generalized contraction in both modified Mann iterations (1.8) and (1.9).

• (iii) We can obtain the Theorem 2.3 with control conditions (B2), (B3), and (B4) which are different from (C2) and (C3).