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Strong convergence theorems for quasi-nonexpansive mappings and maximal monotone operators in Hilbert spaces

Abstract

We present the strong convergence theorem for the iterative scheme for finding a common element of the fixed-point set of a quasi-nonexpansive mapping and the zero set of the sums of maximal monotone operators in Hilbert spaces. Our results extend and improve the recent results of Takahashi et al. (J. Optim. Theory Appl. 147:27-41, 2010) and Takahashi and Takahashi (Nonlinear Anal. 69:1025-1033, 2008).

MSC:47H05, 47H09, 47J25.

1 Introduction

Let H be a real Hilbert space and let K be a nonempty closed convex subset of H. Let T:KK be a mapping. We denote by Fix(T) the fixed-point set of T, that is, Fix(T)={xK:Tx=x}. A mapping T:KK is nonexpansive if TxTyxy for all x,yK. Approximation methods for fixed points of nonexpansive mappings have attracted considerable attention (see [15]). A mapping T:KK is quasi-nonexpansive if Fix(T) and Txyxy for all xK and yFix(T). It is well known that the fixed-point set of a quasi-nonexpansive mapping is closed and convex (see [6, 7]). There are some quasi-nonexpansive mappings which are not nonexpansive (see [810]). For example, the level set of a continuous convex function is characterized as the fixed-point set of a nonlinear mapping called the subgradient projection, which is not nonexpansive but quasi-nonexpansive. Quasi-nonexpansive mappings have been discussed in the recent literature (see [911]).

We say that a mapping T:KK is demiclosed at zero if for any sequence { x n }K which converges weakly to x, the strong convergence of the sequence {T x n } to zero implies Tx=0. It is well known that IT is demiclosed whenever T is nonexpansive. In fact, this property is satisfied for more general mappings (see [12, 13]).

Let B be a mapping from H into 2 H . The effective domain of B is denoted by dom(B), namely, dom(B)={xH:Bx}. The graph of B is

Gra(B)= { ( v , r ) H × H : r B v } .

A multi-valued mapping B is said to be monotone if

xy,uv0for all x,ydom(B),uBx, and vBy.

A monotone operator B is said to be maximal if its graph is not properly contained in the graph of any other monotone operator. For a maximal monotone operator B on H and r>0, we define a single-valued operator J r B = ( I + r B ) 1 :Hdom(B), which is called the resolvent of B for r. It is well known that J r B is firmly nonexpansive, that is,

xy, J r B x J r B y J r B x J r B y 2 for any x,yH.

A basic problem for maximal monotone operator B is to

find xHsuch that0Bx.
(1.1)

The classical method for solving problem (1.1) is the proximal point algorithm which was first introduced by Martinet [14]. Rockafellar [15] obtained the weak convergence of the proximal point algorithm for maximal monotone operators. Güler [16] constructed a proximal point iteration that converges weakly but not strongly. Some researchers have devoted their work to modifications of the proximal point algorithm in order to obtain the strong convergence theorem (see [17, 18]). For a positive constant α, a mapping A:KH is said to be α-inverse strongly monotone if

xy,AxAyα A x A y 2 for all x,yK.

We write ( A + B ) 1 0 for the zero set of A+B, that is, ( A + B ) 1 0={xK:0(A+B)x}, where the mapping A:CH is inverse strongly monotone and B is maximal monotone. It is well known that ( A + B ) 1 0=Fix( J λ B (IλA)) for all λ>0 (see [19]). Takahashi et al. [20] presented the following iterative sequence. Let uK, x 1 =xK and let { x n } be a sequence generated by

x n + 1 = α n u+(1 α n ) J λ n B ( x n λ n A x n ).

Under appropriate conditions they proved that the sequence { x n } converges strongly to a point z 0 ( A + B ) 1 0. Lin and Takahashi [21] introduced an iterative sequence that converges strongly to an element of ( A + B ) 1 0 F 1 0, where F is another maximal monotone operator. Takahashi et al. [22] established an iterative scheme for finding a point of ( A + B ) 1 0Fix(T) as follows. Let x 1 =xK and let { x n } be a sequence generated by

x n + 1 = β n x n +(1 β n )T [ α n x + ( 1 α n ) J λ n B ( x n λ n A x n ) ] ,

where T:KK is a nonexpansive mapping.

Motivated by the above results, especially by Chuang et al. [11] and Takahashi et al. [22], we obtain the strong convergence theorem for the iterative scheme for finding a common element of the fixed-point set of a quasi-nonexpansive mapping and the zero set of the sums of maximal monotone operators in Hilbert spaces. Our results extend and improve the recent results of [22] and [23].

The rest of this paper is organized as follows. Section 2 contains some important facts and tools. In Section 3, we introduce a new iterative scheme for finding a common element of the fixed-point set of a quasi-nonexpansive mapping and the zero set of the sums of maximal monotone operators, and we prove strong convergence theorem in Hilbert spaces.

2 Preliminaries

Throughout this paper, let H be a real Hilbert space with inner product , and norm , and let K be a nonempty closed convex subset of H. Let be the set of positive integers. We denote the strong convergence and the weak convergence of { x n } to x by x n x and x n x, respectively. For any xH, there exists a unique point P K xK such that

x P K xxy,yK.

P K is called the metric projection of H onto K. Note that P K is a nonexpansive mapping. For xH and zK, we have

z= P K xxz,yz0for every yK.
(2.1)

Let f be a proper lower semicontinuous convex function of H into (,+]. The subdifferential ∂f of f is defined as

f(x)= { z H : f ( y ) f ( x ) z , y x , y H }
(2.2)

for all xH. Rockafellar [24] claimed that ∂f is a maximal monotone operator. Let δ K be the indicator function of K, i.e.,

δ K (x)={ 0 , x K , + , x K .

The subdifferential δ K of δ K is a maximal monotone operator since δ K is a proper lower semicontinuous convex function on H. The resolvent J r δ K of δ K for r is P K (see [21]).

Let A:KH be a nonlinear mapping. The variational inequality problem is to find x ¯ K such that

A x ¯ ,y x ¯ 0for every yK.
(2.3)

The solution set of (2.3) is denoted by VI(K,A). Some methods have been proposed to study the variational inequality problem (see [2528] and the references therein). It is easy to see that VI(K,A)= ( A + δ K ) 1 0, where A is an inverse strongly monotone mapping of K into H (for more details, see [21]).

We collect some useful lemmas.

Lemma 2.1 [29]

Let A:KH be an α-inverse strongly monotone mapping. For all x,yK and λ>0, we have

( I λ A ) x ( I λ A ) y 2 x y 2 +λ(λ2α) A x A y 2 .

In particular, if 0<λ2α, then IλA is a nonexpansive mapping.

Lemma 2.2 [30]

Let { Γ n } be a sequence of real numbers that does not decrease at infinity in the sense that there exists a subsequence { Γ n j } of { Γ n } such that

Γ n j < Γ n j + 1 for alljN.

Define the sequence { τ ( n ) } n n 0 of integers as follows:

τ(n)= max k {kn: Γ k < Γ k + 1 },

where n 0 N such that {k n 0 : Γ k < Γ k + 1 }. Then, for all n n 0 , the following hold:

  1. (1)

    τ(n)τ(n+1) and τ(n);

  2. (2)

    Γ τ ( n ) Γ τ ( n ) + 1 and Γ n Γ τ ( n ) + 1 .

Lemma 2.3 [22]

Let B be a maximal monotone operator on H. Then the following holds:

s t s J s B x J t B x, J s B xx J s B x J t B x 2

for all s,t>0 and xH.

The following lemma is an immediate consequence of the inner product on H.

Lemma 2.4 For all x,yH, the inequality x + y 2 x 2 +2y,x+y holds.

Lemma 2.5 [31]

Let { a n } be a sequence of nonnegative real numbers satisfying a n + 1 (1 α n ) a n + α n β n , where

  1. (i)

    { α n }(0,1), n = 1 α n =;

  2. (ii)

    lim sup n β n 0.

Then lim n a n =0.

3 Strong convergence theorems

In this section, a new iterative scheme for finding a common element of the fixed-point set of a quasi-nonexpansive mapping and the zero set of the sums of maximal monotone operators is presented.

Theorem 3.1 Let K be a nonempty closed convex subset of a real Hilbert space H. Let A:KH and C:KH be α-inverse strongly monotone and γ-inverse strongly monotone, respectively. Suppose that B and D are maximal monotone operators on H such that the domains of B and D are contained in K and that T:KK is a quasi-nonexpansive mapping such that IT is demiclosed at zero. Assume that Ω:=Fix(T) ( A + B ) 1 0 ( C + D ) 1 0. Let { α n } and { β n } be sequences in (0,1) and let { u n } be a sequence in K. Let { x n } be a sequence generated by

{ x 1 K chosen arbitrarily , y n = α n u n + ( 1 α n ) J λ n B ( x n λ n A x n ) , z n = J ψ n D ( y n ψ n C y n ) , x n + 1 = β n x n + ( 1 β n ) T z n .
(3.1)

Suppose the following conditions are satisfied:

(c1) lim n α n =0 and n = 1 α n =;

(c2) lim n u n =u;

(c3) 0< lim inf n β n lim sup n β n <1;

(c4) 0<a λ n b<2α;

(c5) 0<c ψ n d<2γ.

Then the sequence { x n } converges strongly to P Ω u.

Proof Observe that the set Ω is closed and convex since Fix(T), ( A + B ) 1 0 and ( C + D ) 1 0 are closed and convex.

By Lemma 2.1, for any pΩ, we have

z n p= J ψ n D ( y n ψ n C y n ) J ψ n D ( p ψ n C p ) y n p

and

y n p α n u n p + ( 1 α n ) J λ n B ( I λ n A ) x n p α n u n p + ( 1 α n ) x n p .

It follows that

x n + 1 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 n p max { x n p , u n p } .

The sequence { u n } is bounded due to condition (c2). Hence there exists a positive number L such that sup n { u n p}L. By a simple inductive process, we have

x n + 1 pmax { x 1 p , L } ,

which shows that { x n } is bounded. So are { y n } and { z n }.

Note that

2 x n + 1 x n , x n p= x n + 1 p 2 x n p 2 x n + 1 x n 2

and

x n + 1 x n =(1 β n )(T z n x n ).

Thus we get

x n + 1 p 2 x n p 2 x n + 1 x n 2 = 2 x n + 1 x n , x n p = ( 1 β n ) [ T z n p 2 T z n x n 2 x n p 2 ] ( 1 β n ) [ z n p 2 T z n x n 2 x n p 2 ] α n ( 1 β n ) u n p 2 ( 1 β n ) T z n x n 2

and

x n + 1 p 2 x n p 2 ( 1 β n ) 2 T z n x n 2 α n ( 1 β n ) u n p 2 ( 1 β n ) T z n x n 2 .

This implies that

(1 β n ) β n T z n x n 2 x n p 2 x n + 1 p 2 + α n (1 β n ) u n p 2 .
(3.2)

Set Γ n = x n p 0 2 , where p 0 = P Ω u. We divide the rest proof into two cases.

Case 1. Suppose that Γ n + 1 Γ n for all nN. In this case, the limit lim n Γ n exists and then lim n ( Γ n + 1 Γ n )=0. We obtain

lim n T z n x n =0,
(3.3)

which implies

lim n x n + 1 x n = lim n (1 β n )T z n x n =0.
(3.4)

Note that

J λ n B ( x n λ n A x n ) p 0 2 ( I λ n A ) x n ( I λ n A ) p 0 2 x n p 0 2 + λ n ( λ n 2 α ) A x n A p 0 2 .

It follows that

x n + 1 p 0 2 β n x n p 0 2 + ( 1 β n ) [ α n u n p 0 2 + ( 1 α n ) J λ n B ( x n λ n A x n ) p 0 2 ] β n x n p 0 2 + α n ( 1 β n ) u n p 0 2 + ( 1 α n ) ( 1 β n ) [ x n p 0 2 + λ n ( λ n 2 α ) A x n A p 0 2 ] x n p 0 2 + α n ( 1 β n ) u n p 0 2 + ( 1 α n ) ( 1 β n ) λ n ( λ n 2 α ) A x n A p 0 2 ,

which yields

( 1 α n ) ( 1 β n ) λ n ( 2 α λ n ) A x n A p 0 2 x n p 0 2 x n + 1 p 0 2 + α n ( 1 β n ) u n p 0 2 .

Therefore we get

lim n A x n A p 0 =0.
(3.5)

In a similar way, we have

lim n C y n C p 0 =0.
(3.6)

Letting h n = J λ n B (I λ n A) x n , we have

h n p 0 2 = J λ n B ( I λ n A ) x n J λ n B ( I λ n A ) p 0 2 ( I λ n A ) x n ( I λ n A ) p 0 , h n p 0 1 2 [ x n p 0 2 + h n p 0 2 ( x n h n ) λ n ( A x n A p 0 ) 2 ] ,

from which one deduces that

h n p 0 2 x n p 0 2 ( x n h n ) λ n ( A x n A p 0 ) 2 .

Using (3.1), we see that

x n + 1 p 0 2 β n x n p 0 2 + ( 1 β n ) [ α n u n p 0 2 + ( 1 α n ) h n p 0 2 ] β n x n p 0 2 + α n ( 1 β n ) u n p 0 2 + ( 1 α n ) ( 1 β n ) ( x n p 0 2 x n h n 2 + 2 λ n A x n A p 0 , x n h n ) x n p 0 2 + α n ( 1 β n ) u n p 0 2 ( 1 α n ) ( 1 β n ) x n h n 2 + 2 ( 1 α n ) ( 1 β n ) λ n A x n A p 0 , x n h n .

Thus,

( 1 α n ) ( 1 β n ) x n h n 2 x n p 0 2 x n + 1 p 0 2 + α n ( 1 β n ) u n p 0 2 + 2 ( 1 α n ) ( 1 β n ) λ n A x n A p 0 , x n h n .

It follows from (3.5) that

lim n x n h n =0.
(3.7)

This implies that

lim n x n y n =0.
(3.8)

By (3.6), a similar argument shows that

lim n y n z n =0.
(3.9)

As

T z n z n T z n x n + x n y n + y n z n ,

combining (3.3), (3.8) and (3.9) gives

lim n T z n z n =0.
(3.10)

Next we prove that

lim sup n u p 0 , y n p 0 0.
(3.11)

To show this inequality, we choose a subsequence { y n j } of { y n } such that

lim sup n u p 0 , y n p 0 = lim j u p 0 , y n j p 0 .
(3.12)

In view of the boundedness of { y n j }, without loss of generality, we assume that y n j ω. Now we show that ωΩ. According to the fact that { y n } is contained in K and K is a closed convex set, one has ωΩ.

Note that the expressions (3.8) and (3.9) yield x n j ω and z n j ω. By the fact that IT is demiclosed at zero, the expression (3.10) implies ωFix(T).

We prove that ω ( A + B ) 1 0. Due to (c4), there is a subsequence { λ n j i } of { λ n j } such that λ n j i λ 0 [a,b]. Without loss of generality, we assume that λ n j λ 0 . Observe that

x n j J λ 0 B ( I λ 0 A ) x n j x n j y n j + y n j [ α n j u n j + ( 1 α n j ) J λ 0 B ( I λ 0 A ) x n j ] + α n j u n j + ( 1 α n j ) J λ 0 B ( I λ 0 A ) x n j J λ 0 B ( I λ 0 A ) x n j x n j y n j + ( 1 α n j ) J λ n j B ( I λ n j A ) x n j J λ 0 B ( I λ 0 A ) x n j + α n j u n j J λ 0 B ( I λ 0 A ) x n j x n j y n j + ( 1 α n j ) [ J λ n j B ( I λ n j A ) x n j J λ 0 B ( I λ n j A ) x n j + J λ 0 B ( I λ n j A ) x n j J λ 0 B ( I λ 0 A ) x n j ] + α n j u n j J λ 0 B ( I λ 0 A ) x n j x n j y n j + ( 1 α n j ) [ | λ 0 λ n j | λ 0 J λ 0 B ( I λ n j A ) x n j ( I λ n j A ) x n j + | λ n j λ 0 | A x n j ] + α n j u n j J λ 0 B ( I λ 0 A ) x n j .

Hence,

lim j x n j J λ 0 B ( I λ 0 A ) x n j =0.
(3.13)

Since J λ 0 B (I λ 0 A) is nonexpansive, the demiclosedness for a nonexpansive mapping implies that ωFix( J λ 0 B (I λ 0 A)), that is, ω ( A + B ) 1 0.

Note that

y n i J ψ 0 D ( I ψ 0 C ) y n i y n i z n i + J ψ n i D ( I ψ n i C ) y n i J ψ 0 D ( I ψ 0 C ) y n i y n i z n i + J ψ n i D ( I ψ n i C ) y n i J ψ n i D ( I ψ 0 C ) y n i + J ψ n i D ( I ψ 0 C ) y n i J ψ 0 D ( I ψ 0 C ) y n i .

Using a similar argument, we get ω ( C + D ) 1 0. In fact, we have obtained ωΩ.

By (3.12) and (2.1), we have

lim sup n u p 0 , y n p 0 = lim j u p 0 , y n j p 0 = u p 0 , ω p 0 0 .

The inequality (3.11) is obtained.

Finally, we prove that x n p 0 . With the help of Lemma 2.4, we obtain

x n + 1 p 0 2 β n x n p 0 2 + ( 1 β n ) y n p 0 2 β n x n p 0 2 + ( 1 β n ) [ ( 1 α n ) x n p 0 2 + 2 α n u n p 0 , y n p 0 ] [ 1 α n ( 1 β n ) ] x n p 0 2 + 2 α n ( 1 β n ) ( u n u , y n p 0 + u p 0 , y n p 0 ) .

It follows from (3.11) and Lemma 2.5 that { x n } converges strongly to p 0 .

Case 2. Suppose that there exists a subsequence { Γ n i } of { Γ n } such that

Γ n i < Γ n i + 1 for all iN.

We define τ:NN by

τ(n)=max{kn: Γ k < Γ k + 1 }.

Lemma 2.2 shows that Γ τ ( n ) Γ τ ( n ) + 1 . Therefore we have from (3.2)

lim n T z τ ( n ) x τ ( n ) =0
(3.14)

and

lim n x τ ( n ) + 1 x τ ( n ) =0.
(3.15)

As in the proof of Case 1, we obtain

lim sup n u p 0 , y τ ( n ) p 0 0.
(3.16)

Observe that

x τ ( n ) + 1 p 0 2 [ 1 α τ ( n ) ( 1 β τ ( n ) ) ] x τ ( n ) p 0 2 + 2 α τ ( n ) ( 1 β τ ( n ) ) ( u τ ( n ) u , y τ ( n ) p 0 + u p 0 , y τ ( n ) p 0 ) .

It follows that

x τ ( n ) p 0 2 2 ( u τ ( n ) u , y τ ( n ) p 0 + u p 0 , y τ ( n ) p 0 ) ,
(3.17)

which implies that

lim sup n x τ ( n ) p 0 2 0.

Thus we get

lim n x τ ( n ) p 0 =0.
(3.18)

It follows from (3.15) and (3.18) that

lim n x τ ( n ) + 1 p 0 =0.
(3.19)

Lemma 2.2 implies that

lim n x n p 0 =0.

The proof is completed. □

The following result is a direct consequence of Theorem 3.1.

Corollary 3.2 Let K be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone operator of K into H and let B be a maximal monotone operator on H such that the domain of B is contained in K. Let T:KK be a quasi-nonexpansive mapping such that IT is demiclosed at zero. Assume that Fix(T) ( A + B ) 1 0. Let { α n } and { β n } be sequences in (0,1) and let { u n } be a sequence in K. Let { x n } be a sequence generated by

{ x 1 K chosen arbitrarily , y n = α n u n + ( 1 α n ) J λ n B ( x n λ n A x n ) , x n + 1 = β n x n + ( 1 β n ) T y n .
(3.20)

If conditions (c1)-(c4) are satisfied, then the sequence { x n } converges strongly to the element P Fix ( T ) ( A + B ) 1 0 u.

Proof Letting C=0 and D= δ K in Theorem 3.1, the desired result follows. □

Let us consider the variational inequality problem. Recall that the subdifferential δ K of δ K is a maximal monotone operator and VI(K,A)= ( A + δ K ) 1 0, where A is an inverse strongly monotone mapping. We obtain the following result.

Corollary 3.3 Let K be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone operator of K into H and let T:KK be a quasi-nonexpansive mapping such that IT is demiclosed at zero. Assume that Fix(T)VI(K,A). Let { α n } and { β n } be sequences in (0,1) and let { u n } be a sequence in K. Let { x n } be a sequence generated by

{ x 1 K chosen arbitrarily , y n = α n u n + ( 1 α n ) P K ( x n λ n A x n ) , x n + 1 = β n x n + ( 1 β n ) T y n .
(3.21)

If conditions (c1)-(c4) are satisfied, then the sequence { x n } converges strongly to the element P Fix ( T ) VI ( K , A ) u.

Proof Corollary 3.2 easily yields the desired result. □

Remark Corollaries 3.2 and 3.3 improve and extend Theorem 3.1 of Takahashi et al. [22] and Theorem 4.2 of Takahashi and Takahashi [23] in the following aspects, respectively.

  1. (1)

    The nonexpansive mapping is extended to the quasi-nonexpansive mapping.

  2. (2)

    The constant vector u is replaced by the variables u n with lim n u n =u.

  3. (3)

    The condition lim n ( λ n + 1 λ n )=0 is removed.

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Wu, Hc., Cheng, Cz. & Qu, Dn. Strong convergence theorems for quasi-nonexpansive mappings and maximal monotone operators in Hilbert spaces. J Inequal Appl 2014, 318 (2014). https://doi.org/10.1186/1029-242X-2014-318

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Keywords

  • quasi-nonexpansive mapping
  • maximal monotone operator
  • variational inequality