1 Introduction

Let H be a real Hilbert space with inner product , and induced norm , let F:H×HR be a bifunction. Then we consider the following equilibrium problem (EP): find zH such that

F(z,y)0,yH.
(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:

xC,AxQ,
(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={xC:AxQ}.

It is not hard to see that Γ is closed convex and xΓ 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:HR 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:HH is said to be nonexpansive if

TxTyxy,x,yH.

T:HH is said to be firmly nonexpansive if

T x T y 2 x y 2 ( I T ) x ( I T ) y 2 ,x,yH.

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

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

A function f:HR is called convex if

f ( λ x + ( 1 λ ) y ) λf(x)+(1λ)f(y),λ(0,1),x,yH.

Lemma 2.1 Let T: H 2 H 2 be a firmly nonexpansive mapping such that (IT)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 (IT)Ax, x H 1 .

  2. (ii)

    f is A 2 -Lipschitz, i.e., f(x)f(y) A 2 xy, 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 h0+:

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 (IT)Ax,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 h0+ 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 (IT)Ax,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:HH be an operator. The following statements are equivalent.

  1. (i)

    T is firmly nonexpansive.

  2. (ii)

    T x T y 2 xy,TxTy, x,yH.

  3. (iii)

    IT is firmly nonexpansive.

Proof (i) ⇒ (ii): Since T is firmly nonexpansive, for all x,yH 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 xy,TxTy,x,yH.

(ii) ⇒ (iii): From (ii), we know that for all x,yH

( 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 IT 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 xH, there exists a unique nearest point of C, denoted by P C x, such that x P C xxy for all yC. 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,xy,x,yH.

Further, for any xH and zC, z= P C x if and only if

xz,zy0,yC.
(2.4)

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

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

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

(A3) lim t 0 F(tz+(1t)x,y)F(x,y), x,y,zH;

(A4) for each xH, yF(x,y) is convex and lower semicontinuous.

 □

Lemma 2.3 ([1, 4])

Let H be a Hilbert space and let F:H×HR satisfy (A1), (A2), (A3), and (A4). Then, for any r>0 and xH, there exists zH such that

F(z,y)+ 1 r yz,zx0,yH.

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 +2y,x+y,x,yX.

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 n0 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 insuch 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 (IT)x is a convex function from H 2 to ℝ. 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 pΓC, f(p)=0. Observe that IT 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 up+(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 pmax { x n p , u p } .

By induction, we have

x n pmax { 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,n1.

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 )up, 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 k1 such that s n > s n + 1 holds for all nk. 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 M2 z n pup for all nN. 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 nN

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 (IS) 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 =tx+(1t) 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 t0, 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 )up, 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 kn: 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 iN 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 ) )M0.

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 ) 2up, 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. □