Regularization proximal point algorithm for finding a common fixed point of a finite family of nonexpansive mappings in Banach spaces

Abstract

We study the strong convergence of a regularization proximal point algorithm for finding a common fixed point of a finite family of nonexpansive mappings in a uniformly convex and uniformly smooth Banach space.

2010 Mathematics Subject Classification: 47H09; 47J25; 47J30.

1 Introduction

Let E be a Banach space with its dual space E*. For the sake of simplicity, the norms of E and E* are denoted by the symbol || · ||. We write 〈x, x*〉 instead of x*(x) for x* ∈ E* and x ∈ E. We denote as ⇀ and →, the weak convergence and strong convergence, respectively. A Banach space E is reflexive if E = E**.

The problem of finding a fixed point of a nonexpansive mapping is equivalent to the problem of finding a zero of the following operator equation:

$$0\in A\left(x\right)$$

(1.1)

involving the accretive mapping A.

One popular method of solving equation 0 ∈ A(x) is the proximal point algorithm of Rockafellar [1] which is recognized as a powerful and successful algorithm for finding a zero of monotone operators. Starting from any initial guess x₀ ∈ H, this proximal point algorithm generates a sequence {x _n } given by

$$x_{n+1}=J_{c_n}^A\left(x_n+e_n\right),$$

(1.2)

where $J_r^A=(I+rA{)}^{-1}$, ∀r > 0 is the resolvent of A in a Hilbert space H. Rockafellar [1] proved the weak convergence of the algorithm (1.2) provided that the regularization sequence {c _n } remains bounded away from zero, and that the error sequence {e _n } satisfies the condition ${\sum }_{n=0}^{\infty }\parallel e_n\parallel <\infty$. However, Güler's example [2] shows that proximal point algorithm (1.2) has only weak convergence in an infinite-dimensional Hilbert space. Recently, several authors proposed modifications of Rockafellar's proximal point algorithm (1.2) for the strong convergence. For example, Solodov and Svaiter [3] and Kamimura and Takahashi [4] studied a modified proximal point algorithm by an additional projection at each step of iteration. Lehdili and Moudafi [5] obtained the convergence of the sequence {x _n } generated by the algorithm:

$$x_{n+1}=J_{c_n}^{A_n}\left(x_n\right),$$

(1.3)

where A _n = μ _n I + A is viewed as a Tikhonov regularization of A. When A is maximal monotone in a Hilbert space H, Xu [6], Song and Yang [7] used the technique of nonexpansive mappings to get convergence theorems for {x _n } defined by the perturbed version of the algorithm (1.3):

$$x_{n+1}=J_{r_n}^A\left(t_{n}u+\left(1-t_n\right)x_n\right).$$

(1.4)

The equation (1.4) can be written in the following equivalent form:

$$r_{n}A\left(x_{n+1}\right)+x_{n+1}\ni t_{n}u+\left(1-t_n\right)x_n.$$

(1.5)

In this article, we study a regularization proximal point algorithm to solve the problem of finding a common fixed point of a finite family of nonexpansive self-mappings in a uniformly convex and uniformly smooth Banach space E. Moreover, we give some analogue regularization methods for the more general problems, such as: problem of finding a common fixed point of a finite family of nonexpansive mappings T _i , i = 1, 2, ..., N, where T _i is self-mapping or nonself-mapping on a closed convex subset of E.

2 Preliminaries

Definition 2.1. A Banach space E is said to be uniformly convex, if for any ε ∈ (0, 2] the inequalities ||x|| ≤ 1, ||y|| ≤ 1, ||x - y|| ≥ ε imply that there exists a δ = δ(ε) ≥ 0 such that

$$\frac{\parallel x+y\parallel }{2}\le 1-\delta .$$

The function

$${\delta }_E\left(\epsilon \right)=inf\left\{1-2^{-1}\parallel x+y\parallel :\parallel x\parallel =\parallel y\parallel =1,\parallel x-y\parallel =\epsilon \right\}$$

(2.1)

is called the modulus of convexity of the space E. The function δ _E (ε) defined on the interval [0, 2] is continuous, increasing and δ _E (0) = 0. The space E is uniformly convex if and only if δ _E (ε) > 0, ∀ε ∈ (0, 2].

The function

$${\rho }_E\left(\tau \right)=sup\left\{2^{-1}\left(\parallel x+y\parallel +\parallel x-y\parallel \right)-1:\parallel x\parallel =1,\parallel y\parallel =\tau \right\},$$

(2.2)

is called the modulus of smoothness of the space E. The function ρ _E (τ) defined on the interval [0, +∞) is convex, continuous, increasing and ρ _E (0) = 0.

Definition 2.2. A Banach space E is said to be uniformly smooth, if

$$\underset{\tau \to 0}{lim}\frac{{\rho }_E\left(\tau \right)}{\tau }=0.$$

(2.3)

It is well known that every uniformly convex and uniformly smooth Banach space is reflexive. In what follows, we denote

$$h_E\left(\tau \right)=\frac{{\rho }_E\left(\tau \right)}{\tau }.$$

(2.4)

The function h _E (τ)is nondecreasing. In addition, it is not difficult to show that the estimate

$$h_E\left(K\tau \right)\le LK{h}_E\left(\tau \right),\forall K>1,\tau >0,$$

(2.5)

is valid, where L is the Figiel's constant [8–10], 1 < L < 1.7. Indeed, we know that the inequality holds ([8])

$$\frac{{\rho }_E\left(\eta \right)}{{\eta }^2}\le L\frac{{\rho }_E\left(\xi \right)}{{\xi }^2},\forall \eta \ge \xi >0.$$

(2.6)

It implies that

$$\xi h_E\left(\eta \right)\le L\eta h_E\left(\xi \right),\forall \eta \ge \xi >0.$$

(2.7)

Taking in (2.7) η = Cτ and ξ = τ, we obtain the inequality:

$$\tau h_E\left(C\tau \right)\le LC\tau h_E\left(\tau \right),$$

(2.8)

which implies that (2.5) holds. Similarly, we have

$${\rho }_E\left(C\tau \right)\le L{C}^2{\rho }_E\left(\tau \right),\forall C>1,\tau >0.$$

(2.9)

Definition 2.3. A mapping j from E onto E* satisfying the condition

$$j\left(x\right)=\left\{f\in E^{*}:⟨x,f⟩=\parallel x{\parallel }^2\mathsf{\text{and}}\parallel f\parallel =\parallel x\parallel \right\}$$

(2.10)

is called the normalized duality mapping of E.

We know that

$$j\left(x\right)=2^{-1}\mathsf{\text{grad}}\parallel x{\parallel }^2.$$

in a smooth Banach space, and the normalized duality mapping J is the identity operator I in a Hilbert space.

Definition 2.4. An operator A : D(A) ⊆ E → E is called accretive, if for all x, y ∈ D(A), there exists j(x - y) ∈ J (x - y) such that

$$⟨A\left(x\right)-A\left(y\right),j\left(x-y\right)⟩\ge 0.$$

(2.11)

Definition 2.5. An operator A : E → E is called m-accretive if it is an accretive operator and the range R(λA + I) = E for all λ > 0, where I is the identity of E.

If A is an m-accretive operator then it is a demiclosed operator, i.e., if the sequence {x _n } ⊂ D(A) satisfies x _n ⇀ x and A(x _n ) → f, then A(x) = f[10, 11].

Definition 2.6. A mapping T : C → E is said to be nonexpansive on a closed convex subset C of Banach space E if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\forall x,y\in C.$$

(2.12)

If T : C → E is a nonexpansive then I - T is an accretive operator. In this case, if the subset C coincides E then I - T is an m-accretive operator.

Definition 2.7. Let G be a nonempty closed convex subset of E. A mapping Q _G : E → G is said to be

• (i) a retraction onto G if $Q_G^2=Q_G$;

• (ii) a nonexpansive retraction if it also satisfies the inequality:

  $$\parallel Q_{G}x-Q_{G}y\parallel \le \parallel x-y\parallel ,\forall x,y\in E;$$

  (2.13)

• (iii) a sunny retraction if for all x ∈ E and for all t ∈ [0, +∞)

  $$Q_G\left(Q_{G}x+t\left(x-Q_{G}x\right)\right)=Q_{G}x.$$

  (2.14)

A closed convex subset C of E is said to be a nonexpansive retract of E, if there exists a nonexpansive retraction from E onto C, and it is said to be a sunny nonexpansive retract of E, if there exists a sunny nonexpansive retraction from E onto C.

Proposition 2.8. [9]Let G be a nonempty closed convex subset of E. A mapping Q _G : E → G is a sunny nonexpansive retraction if and only if

$$⟨x-Q_{G}x,J\left(\xi -Q_{G}x\right)⟩\le 0,\forall x\in E,\forall \xi \in G.$$

(2.15)

Reich [12] showed that if E is uniformly smooth and D is the fixed point set of a nonexpansive mapping from C into itself, then there is a sunny nonexpansive retraction from C onto D, and it can be constructed as follows.

Lemma 2.9. [12]Let E be a uniformly smooth Banach space, and let T : C → C be a nonexpansive mapping with a fixed point. For each u ∈ C and every t ∈ (0, 1), the unique fixed point x _t ∈ C of the contraction C ∋ x ↦ tu + (1 - t)Tx converges strongly as t → 0 to a fixed point of T. Define Q : C → Fix(T) by Qu = lim_t→0x _t . Then, Q is a unique sunny nonexpansive retraction from C onto Fix(T), i.e., Q satisfies the property:

$$⟨u-Qu,j\left(z-Qu\right)⟩\le 0,u\in C,z\in Fix\left(T\right).$$

(2.16)

Definition 2.10. Let C₁ and C₂ be convex subsets of E. The quantity

$$\beta \left(C_1,C_2\right)=\underset{u\in C_1}{sup}\underset{v\in C_2}{inf}\parallel u-v\parallel \left(=\underset{u\in C_1}{sup}d\left(u,C_2\right)\right)$$

is said to be a semideviation of the set C₁ from the set C₂. The function

$$\mathcal{H}\left(C_1,C_2\right)=max\left\{\beta \left(C_1,C_2\right),\beta \left(C_2,C_1\right)\right\}$$

is said to be a Hausdorff distance between C₁ and C₂.

In this article, we will use the following useful lemma:

Lemma 2.11. [7]If E is a uniformly smooth Banach space, C₁and C₂are closed convex subsets of E such that the Hausdorff distance$\mathcal{H}\left(C_1,C_2\right)\le \delta$, and $Q_{C_1}$and$Q_{C_2}$are the sunny nonexpansive retractions onto the subsets C₁and C₂, respectively, then

$$\parallel Q_{C_1}x-Q_{C_2}x{\parallel }^2\le 16R\left(2r+d\right)h_E\left(\frac{16L\delta }{R}\right),$$

(2.17)

where L is Figiel's constant, r = ||x||, d = max{d₁, d₂}, and R = 2(2r + d) + δ. Here d _i = dist(θ, C _i ) = d(θ, C _i ), i = 1, 2, and θ is the origin of the space E.

3 Main results

We need the following lemmas in the proof of our results:

Lemma 3.1. [9]If A = I - T with a nonexpansive mapping T, then for all x, y ∈ D(T), the domain of T

$$⟨Ax-Ay,J\left(x-y\right)⟩\ge L^{-1}{R}^2{\delta }_E\left(\frac{\parallel Ax-Ay\parallel }{4R}\right),$$

(3.1)

where ||x|| ≤ R, ||y|| ≤ R and 1 < L < 1.7 is Figiel's constant.

Lemma 3.2. [13]Let {a _n } be a sequence of nonnegative real numbers satisfying the property:

$$a_{n+1}\le \left(1-{\lambda }_n\right)a_n+{\lambda }_n{\beta }_n+{\sigma }_n,\forall n\ge 0$$

where {λ _n }, {β _n } and {σ _n } satisfy the following conditions.

• (i) ${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$;

• (ii) lim sup_n→∞ β _n ≤ 0 or ${\sum }_{n=0}^{\infty }\left|{\lambda }_n{\beta }_n\right|<\infty$;

• (iii) σ _n ≥ 0 ∀n ≥ 0 and ${\sum }_{n=0}^{\infty }{\sigma }_n<\infty$.

Then, {a _n } converges to zero.

Lemma 3.3. [9]Let E be a uniformly smooth Banach space. Then, for all x, y ∈ E,

$$\parallel x+y{\parallel }^2\le \parallel x{\parallel }^2+2⟨y,Jx⟩+c{\rho }_E\left(\parallel y\parallel \right),$$

(3.2)

where c = 48 max(L, ||x||, ||y||).

First, we consider the following problem:

$$\mathsf{\text{Finding an element}}{x}^{*}\in S={\cap }_{i=1}^{N}Fix\left(T_i\right),$$

(3.3)

where Fix(T _i ) is the set of fixed points of the nonexpansive mapping T _i : E → E, i = 1, 2, ..., N.

Theorem 3.4. Suppose that E is a uniformly convex and uniformly smooth Banach space which has a weakly sequentially continuous normalized duality mapping j from E to E*. Let T _i : E → E, i = 1, 2, ..., N be nonexpansive mappings with$S={\cap }_{i=1}^{N}Fix\left(T_i\right)\ne \varnothing$. If the sequences {r _n } ⊂ (0, +∞) and {t _n } ⊂ (0, 1) satisfy

• (i) lim_n→∞ t _n = 0; ${\sum }_{n=0}^{\infty }{t}_n=+\infty$;

• (ii) lim_n→∞ r _n = +∞,

then the sequence {x _n } defined by

$$r_n\sum _{i=1}^N{A}_i\left(x_{n+1}\right)+x_{n+1}=t_{n}u+\left(1-t_n\right)x_n,u,x_0\in E,n\ge 0$$

(3.4)

converges strongly to Q _S u, where A _i = I - T _i , i = 1, 2, ..., N and Q _S is a sunny nonexpansive retraction from E onto S.

Proof. First, equation (3.4) defines a unique sequence {x _n } ⊂ E, because for each n, the element x_n+1is a unique fixed point of the contraction mapping f : E → E defined by

$$f\left(x\right)=\frac{r_n}{N{r}_n+1}\sum _{i=1}^N{T}_i\left(x\right)+\frac{1}{N{r}_n+1}\left[t_{n}u+\left(1-t_n\right)x_n\right],x\in E.$$

For every x* ∈ S, we have

$$⟨r_n\sum _{i=1}^N{A}_i\left(x_{n+1}\right),j\left(x_{n+1}-x^{*}\right)⟩\ge 0,\forall n\ge 0.$$

(3.5)

Therefore,

$$⟨t_{n}u+\left(1-t_n\right)x_n-x_{n+1},j\left(x_{n+1}-x^{*}\right)⟩\ge 0,\forall n\ge 0.$$

(3.6)

It gives the inequality as follows:

$$\parallel x_{n+1}-x^{*}{\parallel }^2\le \left\{t_n\parallel u-x^{*}\parallel +\left(1-t_n\right)\parallel x_n-x^{*}\parallel \right\}\times \parallel x_{n+1}-x^{*}\parallel .$$

Consequently, we have

$$\begin{array}{lll}\hfill \parallel x_{n+1}-x^{*}\parallel & \le t_n\parallel u-x^{*}\parallel +\left(1-t_n\right)\parallel x_n-x^{*}\parallel & \hfill \text{(1)}\\ \le max\left(\parallel u-x^{*}\parallel ,\parallel x_n-x^{*}\parallel \right)& \hfill \text{(2)}\\ ⋮& \hfill \text{(3)}\\ \le max\left(\parallel u-x^{*}\parallel ,\parallel x_0-x^{*}\parallel \right),\forall n\ge 0.& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

Therefore, the sequence {x _n } is bounded. Every bounded set in a reflexive Banach space is relatively weakly compact. This means that there exists a subsequence $\left\{x_{n_k}\right\}\subseteq \left\{x_n\right\}$ which converges to a limit $\bar{x}\in E$.

Suppose ||x _n || ≤ R and ||x*|| ≤ R with R > 0. By Lemma 3.1, we have

$$\begin{array}{lll}\hfill {\delta }_E\left(\frac{\parallel A_i\left(x_{n+1}\right)\parallel }{4R}\right)& \le \frac{L}{R^2{r}_n}⟨r_n{A}_i\left(x_{n+1}\right),j\left(x_{n+1}-x^{*}\right)⟩& \hfill \text{(1)}\\ \le \frac{L}{R^2{r}_n}⟨r_n\sum _{k=1}^N{A}_k\left(x_{n+1}\right),j\left(x_{n+1}-x^{*}\right)⟩& \hfill \text{(2)}\\ \le \frac{L}{R^2{r}_n}\parallel t_{n}u+\left(1-t_n\right)x_n-x_{n+1}\parallel .\parallel x_{n+1}-x^{*}\parallel & \hfill \text{(3)}\\ \to 0,n\to \infty ,& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

for every i = 1, 2, ..., N. Since the modulus of convexity δ _E is continuous and E is a uniformly convex Banach space, A _i (x_n+1) → 0, i = 1, 2, ..., N. It is clear that $\bar{x}\in S$ from the demiclosedness of A _i . Hence, noting the inequality (2.15), we obtain

$$\begin{array}{lll}\hfill \underset{n\to \infty }{limsup}⟨u-Q_{S}u,j\left(x_n-Q_{S}u\right)⟩& =\underset{k\to \infty }{lim}⟨u-Q_{S}u,j\left(x_{n_k}-Q_{S}u\right)⟩& \hfill \text{(1)}\\ =⟨u-Q_{S}u,j\left(\bar{x}-Q_{S}u\right)⟩& \hfill \text{(2)}\\ \le 0.& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

(3.7)

Next, we have

$$\begin{array}{lll}\hfill {∥x_{n+1}-Q_{S}u∥}^2& =⟨-r_n\sum _{i=1}^N{A}_i\left(x_{n+1}\right)+t_{n}u+\left(1-t_n\right)x_n-Q_{S}u,J\left(x_{n+1}-Q_{S}u\right)⟩& \hfill \text{(1)}\\ =-⟨r_n\sum _{i=1}^N{A}_i\left(x_{n+1}\right),J\left(x_{n+1}-Q_{S}u\right)⟩& \hfill \text{(2)}\\ +⟨t_{n}u+\left(1-t_n\right)x_n-Q_{S}u,J\left(x_{n+1}-Q_{S}u\right)⟩& \hfill \text{(3)}\\ \le ⟨t_n\left(u-Q_{S}u\right)+\left(1-t_n\right)\left(x_n-Q_{S}u\right),J\left(x_{n+1}-Q_{S}u\right)⟩& \hfill \text{(4)}\\ \le \frac{1}{2}\left\{\parallel t_n\left(u-Q_{S}u\right)+\left(1-t_n\right)\left(x_n-Q_{S}u\right){\parallel }^2+\parallel x_{n+1}-Q_{S}u{\parallel }^2\right\}.& \hfill \text{(5)}\\ \hfill \text{(6) }\end{array}$$

By the Lemma 3.3 and the above inequality, we conclude that

$$\begin{array}{lll}\hfill \left|\right|x_{n+1}-Q_S{u}^2\left|\right|& \le \parallel t_n\left(u-Q_{S}u\right)+\left(1-t_n\right)\left(x_n-Q_{S}u\right){\parallel }^2& \hfill \text{(1)}\\ \le {\left(1-t_n\right)}^2\parallel x_n-Q_{S}u{\parallel }^2+2t_n\left(1-t_n\right)⟨u-Q_{S}u,j\left(x_n-Q_{S}u\right)⟩& \hfill \text{(2)}\\ +c{\rho }_E\left(t_n\parallel u-Q_{S}u\parallel \right).& \hfill \text{(3)}\\ \hfill \text{(4) }\end{array}$$

Consequently, we have

$$\parallel x_{n+1}-Q_{S}u{\parallel }^2\le \left(1-t_n\right)\parallel x_n-Q_{S}u{\parallel }^2+t_n{\beta }_n,$$

(3.8)

where

$${\beta }_n=2\left(1-t_n\right)⟨u-Q_{S}u,j\left(x_n-Q_{S}u\right)⟩+c\frac{{\rho }_E\left(t_n\parallel u-Q_{S}u\parallel \right)}{t_n}.$$

Since E is a uniformly smooth Banach space, $\frac{{\rho }_E\left(t_n\parallel u-Q_{S}u\parallel \right)}{t_n}\to 0,n\to \infty$. By (3.7), we obtain lim sup_n→∞β _n ≤ 0. Hence, an application of Lemma 3.2 on (3.8) yields the desired result. □

Now, we will give a method to solve more generally following problem:

$$\mathsf{\text{Finding an element }}{x}^{*}\in S={\cap }_{i=1}^{N}Fix\left(T_i\right),$$

(3.9)

where T _i : C _i → C _i , i = 1, 2, ..., N is a nonexpansive mapping and C _i is a convex closed nonexpansive retract of E.

Obviously, we have the following lemma:

Lemma 3.5. Let E be a Banach space, and let C be a closed convex retract of E. Let T : C → C be a nonexpansive mapping such that Fix(T) ≠ ∅. Then, Fix(T) = Fix(TQ _C ), where Q _C is a retraction of E onto C.

We consider the iterative sequence {x _n } defined by

$$r_n\sum _{i=1}^N{B}_i\left(x_{n+1}\right)+x_{n+1}=t_{n}u+\left(1-t_n\right)x_n,u,x_0\in E,n\ge 0,$$

(3.10)

where $B_i=I-T_i{Q}_{C_i}$, i = 1, 2, ..., N and $Q_{C_i}$ is a nonexpansive retraction from E onto C _i , i = 1, 2, ..., N.

Theorem 3.6. Suppose that E is a uniformly convex and uniformly smooth Banach space which has a weakly sequentially continuous normalized duality mapping j from E into E*. Let C _i be a convex closed nonexpansive retract of E and let T _i : C _i → C _i , i = 1, 2, ..., N be a nonexpansive mapping such that$S={\cap }_{i=1}^{N}Fix\left(T_i\right)\ne \varnothing$. If the sequences {r _n } ⊂ (0, +∞) and {t _n } ⊂ (0, 1) satisfy

• (i) lim_n→∞ t _n = 0; ${\sum }_{n=0}^{\infty }{t}_n=+\infty$;

• (ii) lim_n→∞ r _n = +∞,

then the sequence {x _n } generated by (3.10) converges strongly to Q _S u, where Q _S is a sunny nonexpansive retraction from E onto S.

Proof. By the Lemma 3.5, we have $S={\cap }_{i=1}^{N}Fix\left(T_i{Q}_{C_i}\right)$ and applying Theorem 3.4, we obtain the proof of this Theorem. □

Next, we study the stability of the regularization algorithm (3.10) in the case that each C _i is a closed convex sunny nonexpansive retract of E with respect to perturbations of operators T _i and constraints C _i , i = 1, 2, ..., N satisfying following conditions:

(P1) Instead of C _i , there is a sequence of closed convex sunny nonexpansive retracts $C_i^n\subset E$, n = 1, 2, 3, ... such that

$$\mathcal{H}\left(C_i^n,C_i\right)\le {\delta }_n,i=1,2,\dots ,N,$$

where {δ _n } is a sequence of positive numbers.

(P2) For each set $C_i^n$, there is a nonexpansive self-mapping $T_i^n:C_i^n\to C_i^n$, i = 1, 2, ..., N satisfying the conditions: if for all t > 0, there exists the increasing positive functions g(t) and ξ(t) such that g(0) ≥ 0, ξ(0) = 0 and x ∈ C _i , $y\in C_i^m$, ||x - y|| ≤ δ, then

$$\parallel T_{i}x-T_i^{m}y\parallel \le g\left(max\left\{\parallel x\parallel ,\parallel y\parallel \right\}\right)\xi \left(\delta \right).$$

(3.11)

We establish the convergence and stability of the regularization method (3.10) in the form:

$$r_n\sum _{i=1}^N{B}_i^n\left(z_{n+1}\right)+z_{n+1}=t_{n}u+\left(1-t_n\right)z_n,u,z_0\in E,n\ge 0,$$

(3.12)

where $B_i^n=I-T_i^n{Q}_{C_i^n}$, i = 1, 2, ..., N and $Q_{C_i^n}$ is a sunny nonexpansive retraction from E onto $C_i^n$, i = 1, 2, ..., N.

Theorem 3.7. Suppose that E is a uniformly convex and uniformly smooth Banach space which has a weakly sequentially continuous normalized duality mapping j from E into E*. Let C _i be a convex closed sunny nonexpansive retract of E and let T _i : C _i → C _i , i = 1, 2, ..., N be nonexpansive mappings such that$S={\cap }_{i=1}^{N}Fix\left(T_i\right)\ne \varnothing$. If the conditions (P1) and (P2) are fulfilled, and the sequences {r _n }, {δ _n } and {t _n } satisfy

• (i) lim_n→∞ t _n = 0; ${\sum }_{n=0}^{\infty }{t}_n=+\infty$;

• (ii) lim_n→∞ r _n = +∞;

• (iii) ${\sum }_{n=0}^{\infty }{r}_n\xi \left(a\sqrt{h_E\left({\delta }_n\right)}\right)<+\infty$ for each a > 0,

then the sequence {z _n } generated by (3.12) converges strongly to Q _S u, where Q _S is a sunny nonexpansive retraction from E onto S.

Proof. For each n, ${\sum }_{i=1}^N{B}_i^n$ is an m-accretive operator on E, so the equation (3.12) defines a unique element z_n+1∈ E. From the equations (3.10) and (3.12), we have

$$\begin{array}{c}{r}_n⟨\sum _{i=1}^N{B}_i^n\left(z_{n+1}\right)-B_i^n\left(x_{n+1}\right),j\left(z_{n+1}-x_{n+1}\right)⟩\\ +r_n⟨\sum _{i=1}^N{B}_i^n\left(x_{n+1}\right)-B_i\left(x_{n+1}\right),j\left(z_{n+1}-x_{n+1}\right)⟩+\parallel z_{n+1}-x_{n+1}{\parallel }^2\\ =\left(1-t_n\right)⟨z_n-x_n,j\left(z_{n+1}-x_{n+1}\right)⟩.\end{array}$$

(3.13)

By the accretivity of ${\sum }_{i=1}^N{B}_i^n$ and the equation (3.13), we deduce

$$\parallel z_{n+1}-x_{n+1}\parallel \le \left(1-t_n\right)\parallel z_n-x_n\parallel +r_n\sum _{i=1}^N\parallel B_i^n\left(x_{n+1}\right)-B_i\left(x_{n+1}\right)\parallel .$$

(3.14)

For each i ∈ {1, 2, ..., N},

$$\parallel B_i^n\left(x_{n+1}\right)-B_i\left(x_{n+1}\right)\parallel =\parallel T_i^n{Q}_{C_i^n}{x}_{n+1}-T_i{Q}_{C_i}{x}_{n+1}\parallel .$$

(3.15)

Since {x _n } is bounded and $\mathcal{H}\left(C_i,C_i^n\right)\le {\delta }_n$, there exist constants K_1,i> 0 and K_2,i> 1 such that

$$\parallel Q_{C_i^n}{x}_{n+1}-Q_{C_i}{x}_{n+1}\parallel \le K_{1,i}\sqrt{h_E\left(K_{2,i}{\delta }_n\right)}\le K_{1,i}\sqrt{K_{2,i}L}\sqrt{h_E\left({\delta }_n\right)}.$$

(3.16)

By the condition (P2),

$$\parallel T_i^n{Q}_{C_i^n}{x}_{n+1}-T_i{Q}_{C_i}{x}_{n+1}\parallel \le g\left(M_i\right)\xi \left(K_{1,i}\sqrt{K_{2,i}L}\sqrt{h_E\left({\delta }_n\right)}\right),$$

(3.17)

where $M_i=max\left\{sup\parallel Q_{\underset{i}{\overset{n}{C}}}{x}_{n+1}\parallel ,sup\parallel Q_{C_i}{x}_{n+1}\parallel \right\}<+\infty$.

From (3.14), (3.15) and (3.17), we obtain

$$\parallel z_{n+1}-x_{n+1}\parallel \le \left(1-t_n\right)\parallel z_n-x_n\parallel +Ng\left(M\right)r_n\xi \left({\gamma }_{1,2}\sqrt{h_E\left({\delta }_n\right)}\right),$$

(3.18)

where M = max{M₁, M₂, ..., M _N } < +∞ and ${\gamma }_{1,2}=\underset{i=1,2,\dots ,N}{max}\left\{K_{1,i}\sqrt{K_{2,i}L}\right\}$.

By the above assumption and Lemma 3.2, we conclude that ||z _n - x _n || → 0. In addition, by Theorem 3.6,

$$\parallel z_n-Q_{S}u\parallel \le \parallel z_n-x_n\parallel +\parallel x_n-Q_{S}u\parallel \to 0,\mathsf{\text{as}}n\to \infty ,$$

(3.19)

which implies that {z _n } converges strongly to Q _S u. □

Finally, in this article we give a method to solve the following problem:

$$\mathsf{\text{Finding an element }}{x}^{*}\in S={\cap }_{i=1}^{N}Fix\left(T_i\right),$$

(3.20)

where T _i : C _i → E, i = 1, 2, ..., N is nonexpansive nonself-mapping and C _i is a closed convex sunny nonexpansive retract of E.

Lemma 3.8. [14]Let C be a closed convex subset of a strictly convex Banach space E and let T be a nonexpansive mapping from C into E. Suppose that C is a sunny nonexpansive retract of E. If Fix(T) ≠ ∅, then Fix(T) = Fix(Q _C T), where Q _C is a sunny nonexpansive retraction from E onto C.

We have the following result:

Theorem 3.9. Suppose that E is a uniformly convex and uniformly smooth Banach space which has a weakly sequentially continuous normalized duality mapping j from E into E*. Let C _i be a convex closed sunny nonexpansive retract of E and let T _i : C _i → E, i = 1, 2, ..., N be nonexpansive mappings such that$S={\cap }_{i=1}^{N}Fix\left(T_i\right)\ne \varnothing$. If the sequences {r _n } ⊂ (0, +∞) and {t _n } ⊂ (0, 1) satisfy

• (i) lim_n→∞ t _n = 0; ${\sum }_{n=0}^{\infty }{t}_n=+\infty$;

• (ii) lim_n→∞ r _n = +∞,

then the sequence {u _n } defined by

$$r_n\sum _{i=1}^N{f}_i\left(u_{n+1}\right)+u_{n+1}=t_{n}u+\left(1-t_n\right)u_n,u,u_0\in E,n\ge 0,$$

(3.21)

converges strongly to Q _S u, where Q _S is a sunny nonexpansive retraction from E onto S and$f_i=I-Q_{C_i}{T}_i{Q}_{C_i}$, i = 1, 2, ..., N.

Proof. By the Lemma 3.5 and Lemma 3.8, $S={\cap }_{i=1}^{N}Fix\left(T_i\right)={\cap }_{i=1}^{N}Fix\left(f_i\right)$. Applying Theorem 3.4, we obtain the proof of this Theorem. □