Open Access

A strong convergence theorem for equilibrium problems and split feasibility problems in Hilbert spaces

Fixed Point Theory and Applications20142014:36

https://doi.org/10.1186/1687-1812-2014-36

Received: 16 June 2013

Accepted: 31 January 2014

Published: 13 February 2014

Abstract

The main purpose of this paper is to introduce an iterative algorithm for equilibrium problems and split feasibility problems in Hilbert spaces. Under suitable conditions we prove that the sequence converges strongly to a common element of the set of solutions of equilibrium problems and the set of solutions of split feasibility problems. Our result extends and improves the corresponding results of some others.

MSC:90C25, 90C30, 47J25, 47H09.

Keywords

equilibrium problems split feasibility problems strong convergence bounded linear operator fixed point

1 Introduction

Let H be a real Hilbert space with inner product , and induced norm , let F : H × H R be a bifunction. Then we consider the following equilibrium problem (EP): find z H such that
F ( z , y ) 0 , y H .
(1.1)
The set of the EP is denoted by Ω, i.e.,
Ω = { z H : F ( z , y ) 0 , y H } .

The problem (1.1) is very general in the sense that it includes, as special cases, optimization problems, variational inequality problems, the Nash equilibrium problems and others, see, for instance, [13]. Some methods have been proposed to solve the EP, see, e.g., [46] and [7, 8].

The split feasibility problem (SFP) was proposed by Censer and Elfving in [9]. It can be formulated as the problem of finding a point x satisfying the property:
x C , A x Q ,
(1.2)

where A is a given M × N real matrix, and C and Q are nonempty, closed and convex subsets in R N and R M , respectively.

Due to its extraordinary utility and broad applicability in many areas of applied mathematics (most notably, fully discretized models of problems in image reconstruction from projections, in image processing, and in intensity-modulated radiation therapy), algorithms for solving convex feasibility problems have been received great attention (see, for instance [1013] and also [1418]).

We assume the SFP (1.2) is consistent, and let Γ be the solution set, i.e.,
Γ = { x C : A x Q } .
It is not hard to see that Γ is closed convex and x Γ if and only if it solves the fixed-point equation
x = P C ( I γ A ( I P Q ) A ) x ,
(1.3)

where P C and P Q are the orthogonal projection onto C and Q, respectively, γ > 0 is any positive constant and A denotes the adjoint of A.

Recently, for the purpose of generality, the SFP (1.2) has been studied in a more general setting. For instance, see [16, 19]. However, the algorithms in these references have only weak convergence in the setting of infinite-dimensional Hilbert spaces. Very recently, He and Zhao [20] introduce a new relaxed CQ algorithm (1.4) such that the strong convergence is guaranteed in infinite-dimensional Hilbert spaces:
x n + 1 = P C n ( α n u + ( 1 α n ) ( x n τ n f n ( x n ) ) ) .
(1.4)

Motivated and inspired by the research going on in the sections of equilibrium problems and split feasibility problems, the purpose of this article is to introduce an iterative algorithm for equilibrium problems and split feasibility problems in Hilbert spaces. Under suitable conditions we prove the sequence converges strongly to a common element of the set of solutions of equilibrium problems and the set of solutions of split feasibility problems. Our result extends and improves the corresponding results of He et al. [20] and some others.

2 Preliminaries and lemmas

Throughout this paper, we assume that H, H 1 or H 2 is a real Hilbert space, A is a bounded linear operator from H 1 to H 2 , and I is the identity operator on H, H 1 or H 2 . If f : H R is a differentiable function, then we denote by f the gradient of the function f. We will also use the notations: → to denote strong convergence, to denote weak convergence and to denote by
w ω ( x n ) = { x | { x n k } { x n }  such that  x n k x }

the weak ω-limit set of { x n } .

Recall that a mapping T : H H is said to be nonexpansive if
T x T y x y , x , y H .
T : H H is said to be firmly nonexpansive if
T x T y 2 x y 2 ( I T ) x ( I T ) y 2 , x , y H .

A mapping T : H H is said to be demi-closed at origin, if for any sequence { x n } H with x n x and lim n ( I T ) x n = 0 , then x = T x .

It is easy to prove that if T : H H is a firmly nonexpansive mapping, then T is demi-closed at the origin.

A function f : H R is called convex if
f ( λ x + ( 1 λ ) y ) λ f ( x ) + ( 1 λ ) f ( y ) , λ ( 0 , 1 ) , x , y H .
Lemma 2.1 Let T : H 2 H 2 be a firmly nonexpansive mapping such that ( I T ) x is a convex function from H 2 to R ¯ = [ , + ] . Let A : H 1 H 2 be a bounded linear operator and
f ( x ) : = 1 2 ( I T ) A x 2 , x H 1 .
Then
  1. (i)

    f ( x ) = A ( I T ) A x , x H 1 .

     
  2. (ii)

    f is A 2 -Lipschitz, i.e., f ( x ) f ( y ) A 2 x y , x , y H 1 .

     
Proof (i) From the definition of f, we know that f is convex. First we prove that the limit
f ( x ) , v = lim h 0 + f ( x + h v ) f ( x ) h
exists in R ¯ : = { } R { + } and satisfies
f ( x ) , v f ( x + v ) f ( x ) , v H 1 .
If fact, if 0 < h 1 h 2 , then
f ( x + h 1 v ) f ( x ) = f ( h 1 h 2 ( x + h 2 v ) + ( 1 h 1 h 2 ) x ) f ( x ) .
Since f is convex and h 1 h 2 1 , it follows that
f ( x + h 1 v ) f ( x ) h 1 h 2 f ( x + h 2 v ) + ( 1 h 1 h 2 ) f ( x ) f ( x ) ,
and hence that
f ( x + h 1 v ) f ( x ) h 1 f ( x + h 2 v ) f ( x ) h 2 .
This shows that this difference quotient is increasing, therefore it has a limit in R ¯ as h 0 + :
f ( x ) , v = inf h > 0 f ( x + h v ) f ( x ) h = lim h 0 + f ( x + h v ) f ( x ) h .
(2.1)
This implies that f is differential. Taking h = 1 , (2.1) implies that
f ( x ) , v f ( x + v ) f ( x ) .
Next we prove that
f ( x ) = A ( I T ) A x , x H 1 .
In fact, since
lim h 0 + f ( x + h v ) f ( x ) h = lim h 0 + A x + h A v T A ( x + h v ) 2 ( I T ) A x 2 2 h
(2.2)
and
A x + h A v T A ( x + h v ) 2 ( I T ) A x 2 = A x 2 + h 2 A v 2 + 2 h A A x , v + T A ( x + h v ) 2 A x 2 T A x 2 2 A x , T A ( x + h v ) T A x 2 h A T A ( x + h v ) , v .
(2.3)
Substituting (2.3) into (2.2), simplifying and then letting h 0 + and taking the limit we have
lim h 0 + f ( x + h v ) f ( x ) h = lim h 0 + 2 h { A A x , v A T A ( x + h v ) , v } 2 h = A ( I T ) A x , v , v H 1 .
It follows from (2.1) that
f ( x ) = A ( I T ) A x , x H 1 .
(ii) From (i) we have
f ( x ) f ( y ) = A ( I T ) A x A ( I T ) A y = A [ ( I T ) A x ( I T ) A y ] A A x A y A 2 x y , x , y H 1 .

 □

Lemma 2.2 (See, for example, [21])

Let T : H H be an operator. The following statements are equivalent.
  1. (i)

    T is firmly nonexpansive.

     
  2. (ii)

    T x T y 2 x y , T x T y , x , y H .

     
  3. (iii)

    I T is firmly nonexpansive.

     
Proof (i) (ii): Since T is firmly nonexpansive, for all x , y H we have
T x T y 2 x y 2 ( I T ) x ( I T ) y 2 = x y 2 x y 2 T x T y 2 + 2 x y , T x T y = 2 x y , T x T y T x T y 2 ,
hence
T x T y 2 x y , T x T y , x , y H .
(ii) (iii): From (ii), we know that for all x , y H
( I T ) x ( I T ) y 2 = ( x y ) ( T x T y ) 2 = x y 2 2 x y , T x T y + T x T y 2 x y 2 x y , T x T y = x y , ( I T ) x ( I T ) y .

This implies that I T is firmly nonexpansive.

(iii) (i): From (iii) we immediately know that T is firmly nonexpansive.

Let C be a nonempty closed convex subset of H. Recall that for every point x H , there exists a unique nearest point of C, denoted by P C x , such that x P C x x y for all y C . Such a P C is called the metric projection from H onto C. We know that P C is a firmly nonexpansive mapping from H onto C, i.e.,
P C x P C y 2 P C x P C y , x y , x , y H .
Further, for any x H and z C , z = P C x if and only if
x z , z y 0 , y C .
(2.4)

Throughout this paper, let us assume that a bifunction F : H × H R satisfies the following conditions:

(A1) F ( x , x ) = 0 , x H ;

(A2) F is monotone, i.e., F ( x , y ) + F ( y , x ) 0 , x , y H ;

(A3) lim t 0 F ( t z + ( 1 t ) x , y ) F ( x , y ) , x , y , z H ;

(A4) for each x H , y F ( x , y ) is convex and lower semicontinuous.

 □

Lemma 2.3 ([1, 4])

Let H be a Hilbert space and let F : H × H R satisfy (A1), (A2), (A3), and (A4). Then, for any r > 0 and x H , there exists z H such that
F ( z , y ) + 1 r y z , z x 0 , y H .
Furthermore, if
T r x = { z H : F ( z , y ) + 1 r y z , z x 0 , y H } ,
then the following hold:
  1. (1)

    T r is single-valued;

     
  2. (2)

    T r is firmly nonexpansive;

     
  3. (3)

    F ( T r ) = Ω ;

     
  4. (4)

    Ω is closed and convex.

     

The following results play an important role in this paper.

Lemma 2.4 ([22])

Let X be a real Hilbert space, then we have
x + y 2 x 2 + 2 y , x + y , x , y X .

Lemma 2.5 ([23])

Let { x n } and { y n } be bounded sequences in a Banach space X. Let { β n } be a sequence in [ 0 , 1 ] satisfying 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 integer n 0 and
lim sup n ( y n + 1 y n x n + 1 x n ) 0 .

Then lim n y n x n = 0 .

Lemma 2.6 ([24])

Let { a n } be a sequence of nonnegative real numbers such that
a n + 1 ( 1 γ n ) a n + γ n σ n , n = 0 , 1 , 2 , ,
where { γ n } is a sequence in ( 0 , 1 ) , and { σ n } is a sequence in such that
  1. (i)

    n = 0 γ n = ;

     
  2. (ii)

    lim sup n σ n 0 , or n = 0 | γ n σ n | < .

     

Then lim n a n = 0 .

3 Main results

We are now in a position to prove the following theorem.

Theorem 3.1 Let H 1 , H 2 be two real Hilbert spaces, F : H 1 × H 1 R be a bifunction satisfying (A1), (A2), (A3), and (A4). Let A : H 1 H 2 be a bounded linear operator, S : H 1 H 1 be a firmly nonexpansive mapping, and let T : H 2 H 2 be a firmly nonexpansive mapping such that ( I T ) x is a convex function from H 2 to . Assume that C : = F ( S ) Ω and Q : = F ( T ) . Let u H 1 and { x n } be the sequence generated by
{ x 0 H 1  chosen arbitrarily , x n + 1 = β n x n + ( 1 β n ) y n , F ( y n , x ) + 1 λ n x y n , y n z n 0 , x H 1 , z n = S ( α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) ) ,
(3.1)
where
f ( x n ) = 1 2 ( I T ) A x n 2 , f ( x n ) = A ( I T ) A x n 0 n 1 , ξ n = ρ n f ( x n ) f ( x n ) 2 .
If the solution set Γ of SPF (1.2) is not empty, and the sequences { ρ n } ( 0 , 4 ) , { α n } , { β n } ( 0 , 1 ) satisfy the following conditions:
  1. (i)

    lim n α n = 0 , n = 0 α n = ;

     
  2. (ii)

    0 < lim inf n β n lim sup n β n < 1 ;

     
  3. (iii)

    λ n ( a , b ) ( 0 , + ) and lim n ( λ n + 1 λ n ) = 0 ,

     

then the sequence { x n } converges strongly to P Γ u .

Proof Since the solution set Ω of EP and the solution set of SPF (1.2) are both closed and convex, Γ (≠) is closed and convex. Thus, the metric projection P Γ is well defined.

Letting p = P Γ u , it follows from Lemma 2.3 that y n = T λ n z n and
y n p = T λ n z n T λ n p z n p .
(3.2)
Observing that S is firmly nonexpansive, we have
z n p = S ( α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) ) p α n ( u p ) + ( 1 α n ) ( x n ξ n f ( x n ) p ) α n u p + ( 1 α n ) x n ξ n f ( x n ) p .
(3.3)
Since p Γ C , f ( p ) = 0 . Observe that I T is firmly nonexpansive, from Lemma 2.2(ii) we have
f ( x n ) , x n p = ( I T ) A x n , A x n A p ( I T ) A x n 2 = 2 f ( x n ) .
(3.4)
This implies that
x n ξ n f ( x n ) p 2 = x n p 2 + ξ n f ( x n ) 2 2 ξ n f ( x n ) , x n p x n p 2 + ξ n 2 f ( x n ) 2 4 ξ n f ( x n ) = x n p 2 ρ n ( 4 ρ n ) f 2 ( x n ) f ( x n ) 2 x n p 2 .
(3.5)
Substituting (3.5) into (3.3), we get
z n p α n u p + ( 1 α n ) x n p .
(3.6)
Thus, from (3.2) and (3.6) we have
x n + 1 p = β n x n + ( 1 β n ) y n p β n x n p + ( 1 β n ) y n p β n x n p + ( 1 β n ) z n p ( 1 α n ( 1 β n ) ) x n p + α n ( 1 β n ) u p .
It turns out that
x n + 1 p max { x n p , u p } .
By induction, we have
x n p max { x 0 p , u p } .

This implies that the sequence { x n } is bounded. From (3.2) and (3.6) we know that { y n } and { z n } both are bounded.

From Lemma 2.4 and (3.5), we have
z n p 2 = S ( α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) ) p 2 α n ( u p ) + ( 1 α n ) ( x n ξ n f ( x n ) p ) 2 ( 1 α n ) x n ξ n f ( x n ) p 2 + 2 α n u p , z n p ( 1 α n ) x n p 2 + 2 α n u p , z n p ( 1 α n ) ρ n ( 4 ρ n ) f 2 ( x n ) f ( x n ) 2 .
(3.7)
Therefore, from Lemma 2.6 and (3.2), (3.7) we have
x n + 1 p 2 = β n x n + ( 1 β n ) y n p 2 β n x n p 2 + ( 1 β n ) z n p 2 β n x n p 2 + ( 1 β n ) ( 1 α n ) x n p 2 + 2 α n ( 1 β n ) u p , z n p ( 1 α n ) ( 1 β n ) ρ n ( 4 ρ n ) f 2 ( x n ) f ( x n ) 2 = x n p 2 α n ( 1 β n ) x n p 2 + 2 α n ( 1 β n ) u p , z n p ( 1 α n ) ( 1 β n ) ρ n ( 4 ρ n ) f 2 ( x n ) f ( x n ) 2 .
(3.8)
On the other hand, without loss of generality, we may assume that there is a constant σ > 0 such that
( 1 α n ) ( 1 β n ) ρ n ( 4 ρ n ) > σ , n 1 .
Setting s n = x n p 2 , we get the following inequality:
s n + 1 s n + α n ( 1 β n ) s n + σ f 2 ( x n ) f ( x n ) 2 2 α n ( 1 β n ) u p , z n p .
(3.9)

Now, we prove s n 0 by employing the technique studied by Maingé [25]. For the purpose we consider two cases.

Case 1: { s n } is eventually decreasing, i.e., there exists a sufficient large positive integer k 1 such that s n > s n + 1 holds for all n k . In this case, { s n } must be convergent, and from (3.9) it follows that
σ f 2 ( x n ) f ( x n ) 2 ( s n s n + 1 ) + α n ( 1 β n ) M ,
(3.10)
where M is a constant such that M 2 z n p u p for all n N . Using the condition (i) and (3.10), we have
f 2 ( x n ) f ( x n ) 2 0 ( n ) .
(3.11)
Moreover, it follows from Lemma 2.1(ii) that for all n N
f ( x n ) = f ( x n ) f ( p ) A 2 x n p .
This implies that { f ( x n ) } is bounded. From (3.11) it yields f ( x n ) 0 , namely
( I T ) A x n 0 .
(3.12)
Furthermore, we have
lim n ξ n = 0 .
(3.13)
For any x w ω ( x n ) , and if { x n k } is a subsequence of { x n } such that x n k x H 1 , then
A x n k A x .
(3.14)
On the other hand, from (3.12), we have
( I T ) A x n k 0 .
(3.15)

Since T is demi-closed at origin, from (3.14) and (3.15) we have A x F ( T ) , i.e., A x Q .

In order to prove x C = F ( S ) Ω , we need to prove lim n x n + 1 x n = 0 and lim n x n z n = 0 . In fact, from (3.1) we have
F ( y n , x ) + 1 λ n x y n , y n z n 0 , x H 1 .
Taking x = y n + 1 , we get
F ( y n , y n + 1 ) + 1 λ n y n + 1 y n , y n z n 0 .
Similarly, we also have
F ( y n + 1 , y n ) + 1 λ n + 1 y n y n + 1 , y n + 1 z n + 1 0 .
Adding up the above two inequalities, we get
F ( y n , y n + 1 ) + F ( y n + 1 , y n ) + y n + 1 y n , y n z n λ n y n + 1 z n + 1 λ n + 1 0 .
From (A2), we have
y n + 1 y n , y n z n λ n y n + 1 z n + 1 λ n + 1 0 .
Multiplying the above inequality by λ n and simplifying, we have
y n + 1 y n , y n y n + 1 + y n + 1 z n λ n λ n + 1 ( y n + 1 z n + 1 ) 0 .
Hence we have
y n + 1 y n 2 y n + 1 y n , y n + 1 z n λ n λ n + 1 ( y n + 1 z n + 1 ) = y n + 1 y n , z n + 1 z n + ( 1 λ n λ n + 1 ) ( y n + 1 z n + 1 ) y n + 1 y n ( z n + 1 z n + | 1 λ n λ n + 1 | y n + 1 z n + 1 )
and hence
y n + 1 y n z n + 1 z n + | 1 λ n λ n + 1 | y n + 1 z n + 1 z n + 1 z n + 1 a | λ n + 1 λ n | y n + 1 z n + 1 .
By (3.1) we have
z n + 1 z n = S ( α n + 1 u + ( 1 α n + 1 ) ( x n + 1 ξ n + 1 f ( x n + 1 ) ) ) S ( α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) ) ( α n + 1 α n ) u + ( 1 α n + 1 ) { ( x n + 1 ξ n + 1 f ( x n + 1 ) ) ( x n ξ n f ( x n ) ) } ( α n + 1 α n ) ( x n ξ n f ( x n ) ) ( 1 α n + 1 ) x n + 1 x n + N n x n + 1 x n + N n ,
(3.16)
where
N n = | α n + 1 α n | u + ( 1 α n + 1 ) ( ξ n + 1 f ( x n + 1 ) + ξ n f ( x n ) ) + | α n + 1 α n | x n ξ n f ( x n ) 0 ( n ) .
(3.17)
This implies that
y n + 1 y n x n + 1 x n + 1 a | λ n + 1 λ n | y n + 1 z n + 1 + N n .
It follows that
y n + 1 y n x n + 1 x n 1 a | λ n + 1 λ n | y n + 1 z n + 1 + N n .
In view of condition (iii) and (3.17) we get
lim sup n ( y n + 1 y n x n + 1 x n ) 0 .
By Lemma 2.5, we obtain
lim n y n x n = 0 .
(3.18)
Consequently
x n + 1 x n = β n x n + ( 1 β n ) y n x n = ( 1 β n ) y n x n 0 ( n ) .
(3.19)
Since S is firmly nonexpansive, it follows from (3.1) that
2 z n p 2 = 2 S ( α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) ) S p 2 2 α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) p , z n p = α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) p 2 + z n p 2 α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) p z n + p 2 = α n ( u p ) + ( 1 α n ) ( x n ξ n f ( x n ) p ) 2 + z n p 2 α n ( u z n ) + ( 1 α n ) ( x n ξ n f ( x n ) z n ) 2 ( 1 α n ) x n p 2 + z n p 2 x n z n 2 + M n ,
where
M n : = α n u p 2 + ( 1 α n ) ξ n f ( x n ) 2 2 ( 1 α n ) ξ n x n p , f ( x n ) α n u z n 2 ( 1 α n ) { ξ n f ( x n ) 2 2 x n z n , ξ n f ( x n ) } + α n x n z n 2 + α n ( 1 α n ) x n u ξ n f ( x n ) 2 0 ( as  n ) .
Therefore we have
z n p 2 x n p 2 x n z n 2 + M n .
This together with (3.8) shows that
x n + 1 p 2 β n x n p 2 + ( 1 β n ) z n p 2 x n p 2 ( 1 β n ) x n z n 2 + ( 1 β n ) M n .
Then we obtain
( 1 β n ) x n z n 2 x n p 2 x n + 1 p 2 + ( 1 β n ) M n = s n s n + 1 + ( 1 β n ) M n .
Therefore, we get
lim n x n z n = 0 .
(3.20)
By virtue of (3.18), we have
lim n y n z n = 0 .
(3.21)
Now, we turn to a proof that x C = F ( S ) Ω . For this purpose, we denote
v n : = α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) .
In view of condition (i) and (3.13) we have
v n x n = α n u + ( 1 α n ) ( x n ξ n f ( x n ) ) x n = α n ( u x n ) ( 1 α n ) ξ n f ( x n ) α n u x n + ( 1 α n ) ξ n f ( x n ) 0 .
(3.22)
Since S is firmly nonexpansive (and so it is also nonexpansive), it follows from Lemma 2.4 that
z n + 1 p 2 = S v n + 1 S x n + S x n S p 2 S x n S p 2 + 2 S v n + 1 S x n + 1 + S x n + 1 S x n , z n + 1 p x n p 2 ( I S ) x n 2 + 2 ( v n + 1 x n + 1 + x n + 1 x n ) z n + 1 p .
Thus, we have
( I S ) x n 2 x n p 2 z n + 1 p 2 + 2 ( v n + 1 x n + 1 + x n + 1 x n ) z n + 1 p x n p 2 ( z n + 1 x n + 1 x n + 1 p ) 2 + 2 ( v n + 1 x n + 1 + x n + 1 x n ) z n + 1 p x n p 2 x n + 1 p 2 z n + 1 x n + 1 2 + 2 x n + 1 p z n + 1 x n + 1 + 2 ( v n + 1 x n + 1 + x n + 1 x n ) z n + 1 p = s n s n + 1 z n + 1 x n + 1 2 + 2 x n + 1 p z n + 1 x n + 1 + 2 ( v n + 1 x n + 1 + x n + 1 x n ) z n + 1 p .
(3.23)

It follows from (3.19), (3.20), and (3.22) that ( I S ) x n 0 . In view of x n k x and that S is demi-closed at origin, we get x F ( S ) .

On the other hand, from x n k x and (3.18), we obtain y n k x . From (3.1), for any x H 1 , we have
F ( y n , x ) + 1 λ n x y n , y n z n 0 .
From (A2), we have
1 λ n x y n , y n z n F ( x , y n ) , x H 1 .
Replacing n by n k , we have
x y n k , y n k z n k λ n k F ( x , y n k ) , x H 1 .
Since y n k z n k λ n k 0 and y n k x , from (A4) we have
F ( x , x ) 0 , x H 1 .
(3.24)
Put w t = t x + ( 1 t ) x for all t ( 0 , 1 ] and x H 1 . Then we get w t H 1 . So, from (3.24) we have
F ( w t , x ) 0 , x H 1 .
From (A4), we have
0 = F ( w t , w t ) t F ( w t , x ) + ( 1 t ) F ( w t , x ) t F ( w t , x ) ,
and hence F ( w t , x ) 0 . Letting t 0 , we have
F ( x , x ) 0 , x H 1 .
This implies x Ω . Consequently, x C , and hence w w ( x n ) Γ . Furthermore, in view of (3.20) we have
lim sup n u p , z n p = lim sup n u p , x n p = max w w w ( x n ) u P Γ u , w P Γ u 0 .
On the other hand, from (3.9), we have
s n + 1 ( 1 α n ( 1 β n ) ) s n + 2 α n ( 1 β n ) u p , z n p .
(3.25)

Applying Lemma 2.6 to (3.25), from the condition (i) we obtain s n 0 , that is, x n p .

Case 2: { s n } is not eventually decreasing, that is, we can find a positive integer n 0 such that s n 0 s n 0 + 1 . Now we define
U n : = { n 0 k n : s k s k + 1 } , n > n 0 .
It easy to see that U n is nonempty and satisfies U n U n + 1 . Let
ψ ( n ) : = max U n , n > n 0 .
It is clear that ψ ( n ) as n (otherwise, { s n } is eventually decreasing). It is also clear that s ψ ( n ) s ψ ( n ) + 1 for all n > n 0 . Moreover, we prove that
s n s ψ ( n ) + 1 , n > n 0 .
(3.26)
In fact, if ψ ( n ) = n , then the inequality (3.26) is trivial; if ψ ( n ) < n , from the definition of ψ ( n ) , there exists some i N such that ψ ( n ) + i = n , we deduce that
s ψ ( n ) + 1 > s ψ ( n ) + 2 > > s ψ ( n ) + i = s n ,
and the inequality (3.26) holds again. Since s ψ ( n ) s ψ ( n ) + 1 for all n > n 0 , it follows from (3.10) that
σ f 2 ( x ψ ( n ) ) f ( x ψ ( n ) ) 2 α ψ ( n ) ( 1 β ψ ( n ) ) M 0 .
Noting that { f ( x ψ ( n ) ) } is bounded, we get f ( x ψ ( n ) ) 0 . By the same argument to the proof in case 1, we have w w ( x ψ ( n ) ) Γ . From (3.19) we have
lim n x ψ ( n ) x ψ ( n ) + 1 = 0 .
(3.27)
Furthermore, in view of (3.20), we can deduce that
lim sup n u p , z ψ ( n ) p = lim sup n u p , x ψ ( n ) p = max w w w ( x ψ ( n ) ) u P Γ u , w P Γ u 0 .
(3.28)
Since s ψ ( n ) s ψ ( n ) + 1 , it follows from (3.9) that
s ψ ( n ) 2 u p , z ψ ( n ) p , n > n 0 .
(3.29)
Combining (3.28) and (3.29) we have
lim sup n s ψ ( n ) 0 ,
(3.30)
and hence s ψ ( n ) 0 , which together with (3.27) implies that
s ψ ( n ) + 1 ( x ψ ( n ) p ) + ( x ψ ( n ) + 1 x ψ ( n ) ) s ψ ( n ) + x ψ ( n ) + 1 x ψ ( n ) 0 .

Noting the inequality (3.26), this shows that s n 0 , that is, x n p . This completes the proof of Theorem 3.1. □

Declarations

Acknowledgements

This study was supported by the Scientific Research Fund of Sichuan Provincial Education Department (13ZA0199) and the Scientific Research Fund of Sichuan Provincial Department of Science and Technology (2012JYZ011) and by the National Natural Science Foundation of China (Grant No. 11361070).

Authors’ Affiliations

(1)
Department of Mathematics, Yibin University
(2)
College of Statistics and Mathematics, Yunnan University of Finance and Economics
(3)
Faculty of Computing, Engineering and Technology, Staffordshire University

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© Tang et al.; licensee Springer. 2014

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