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Strong convergence theorems for the general split variational inclusion problem in Hilbert spaces

Abstract

The purpose of this paper is to introduce and study a general split variational inclusion problem in the setting of infinite-dimensional Hilbert spaces. Under suitable conditions, we prove that the sequence generated by the proposed new algorithm converges strongly to a solution of the general split variational inclusion problem. As a particular case, we consider the algorithms for a split feasibility problem and a split optimization problem and give some strong convergence theorems for these problems in Hilbert spaces.

1 Introduction

Let C and Q be nonempty closed convex subsets of real Hilbert spaces H 1 and H 2 , respectively. The split feasibility problem (SFP) is formulated as

to find  x C and A x Q,
(1.1)

where A: H 1 H 2 is a bounded linear operator. In 1994, Censor and Elfving [1] first introduced the SFP in finite-dimensional Hilbert spaces for modeling inverse problems which arise from phase retrievals and in medical image reconstruction [2]. It has been found that the SFP can also be used in various disciplines such as image restoration, computer tomography and radiation therapy treatment planning [35]. The SFP in an infinite-dimensional real Hilbert space can be found in [2, 4, 610]. For comprehensive literature, bibliography and a survey on SFP, we refer to [11].

Assuming that the SFP is consistent, it is not hard to see that x C solves SFP if and only if it solves the fixed point equation

x = P C ( I γ A ( I P Q ) A ) x ,

where P C and P Q are the metric projection from H 1 onto C and from H 2 onto Q, respectively, γ>0 is a positive constant, and A is the adjoint of A.

A popular algorithm to be used to solves the SFP (1.1) is due to Byrne’s CQ-algorithm [2]:

x k + 1 = P C ( I γ k A ( I P Q ) A ) x k ,k1,

where γ k (0,2/λ) with λ being the spectral radius of the operator A A.

On the other hand, let H be a real Hilbert space, and B be a set-valued mapping with domain D(B):={xH:B(x)}. Recall that B is called monotone, if uv,x,xy0 for any uBx and vBy; B is maximal monotone, if its graph {(x,y):xD(B),yBx} is not properly contained in the graph of any other monotone mapping. An important problem for set-valued monotone mappings is to find x H such that 0B( x ). Here, x is called a zero point of B. A well-known method for approximating a zero point of a maximal monotone mapping defined in a real Hilbert space H is the proximal point algorithm first introduced by Martinet [12] and generated by Rockafellar [13]. This is an iterative procedure, which generates { x n } by x 1 =xH and

x n + 1 = J β n B x n ,n1,
(1.2)

where { β n }(0,), B is a maximal monotone mapping in a real Hilbert space, and J r B is the resolvent mapping of B defined by J r B = ( I + r B ) 1 for each r>0. Rockafellar [13] proved that if the solution set B 1 (0) is nonempty and lim inf n β n >0, then the sequence { x n } in (1.2) converges weakly to an element of B 1 (0). In particular, if B is the sub-differential ∂f of a proper convex and lower semicontinuous function f:HR, then (1.2) is reduced to

x n + 1 = argmin y H { f ( y ) + 1 2 β n y x n 2 } ,n1.
(1.3)

In this case, { x n } converges weakly to a minimizer of f. Later, many researchers have studied the convergence problems of the proximal point algorithm in Hilbert spaces (see [1421] and the references therein).

Motivated by the works in [1417] and related literature, the purpose of this paper is to introduce and consider the following general split variational inclusion problem.

Let H 1 and H 2 be two real Hilbert spaces, B i : H 1 H 1 and K i : H 2 H 2 , i=1,2, be two families of set-valued maximal monotone mappings, A: H 1 H 2 be a linear and bounded operator, and A be the adjoint of A. The so-called general split variational inclusion problem is

to find  x H 1  such that 0 i = 1 B i ( x )  and 0 i = 1 K i ( A x ) .
(1.4)

The following examples are special cases of (GSVIP) (1.4).

Classical split variational inclusion problem. Let B: H 1 H 1 and K: H 2 H 2 be set-valued maximal monotone mappings. The so-called classical split variational inclusion problem (CSVIP) is

to find  x H 1  such that 0B ( x )  and 0K ( A x ) ,
(1.5)

which was introduced by Moudafi [17]. It is obvious that problem (1.5) is a special case of (GSVIP) (1.4). In [17], Moudafi proved that the iteration process

x n + 1 = J λ B ( x n + γ A ( J λ K I ) A x n )

converges weakly to a solution of problem (1.5), where λ and γ are given positive numbers.

Split optimization problem. Let f: H 1 R, g: H 2 R be two proper convex and lower semicontinuous functions. The so-called split optimization problem (SOP) is

to find  x H 1  such that f ( x ) = min y H 1 f(y) and g ( A x ) = min z H 2 g(z).
(1.6)

Denote by B=(f) and K=(g), then B and K both are maximal monotone mappings, and problem (1.6) is equivalent to the following classical split variational inclusion problem, i.e.:

to find  x H 1  such that 0 ( f ( x ) )  and 0 ( g ( A x ) ) .
(1.7)

Split feasibility problem. As in (1.1), let C and Q be two nonempty closed convex subsets of real Hilbert spaces H 1 and H 2 , respectively and A be the same as above. The split feasibility problem is

to find  x C such A x Q.
(1.8)

It is well known that this kind of problems was first introduced by Censor and Elfving [1] for modeling inverse problems arising from phase retrievals and in medical image reconstruction [2]. Also it can be used in various disciplines such as image restoration, computer tomography and radiation therapy treatment planning.

Let i C ( i Q ) be the indicator function of C (Q), i.e.,

i C (x)= { 0 , if  x C , + , if  x C ; i Q (x)= { 0 , if  x Q , + , if  x Q .
(1.9)

Then i C and i Q both are proper convex and lower semicontinuous functions, and its subdifferentials i C and i Q are maximal monotone operators. Consequently problem (1.8) is equivalent to the following ‘split optimization problem’ and ‘Moudafi’s classical split variational inclusion problem’, i.e.,

to find  x H 1  such that  i C ( x ) = min y H 1 i C ( y )  and  i Q ( A x ) = min z H 2 i Q ( z ) to find  x H 1  such that  0 ( i C ( x ) )  and  0 ( i Q ( A x ) ) .
(1.10)

For solving (GSVIP) (1.4), in our paper we propose the following iterative algorithms:

x n + 1 = α n x n + ξ n f( x n )+ i = 1 γ n , i J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] ,n0,
(1.11)

where f: H 1 H 1 is a contraction mapping with a contractive constant k(0,1), { α n }, { ξ n } and { γ n , i } are sequence in [0,1] satisfying some conditions. Under suitable conditions, some strong convergence theorems for the sequence proposed by (1.11) to a solution for (GSVIP) (1.4) in Hilbert spaces are proved. As a particular case, we consider the algorithms for a split feasibility problem and a split optimization problem and give some strong convergence theorems for these problems in Hilbert spaces. Our results extend and improve the related results of Censor and Elfving [1], Byrne [2], Censor et al. [35], Rockafellar [13], Moudafi [14, 17], Eslamian and Latif [15], Eslamian [21], and Chuang [22].

2 Preliminaries

Throughout the paper, we denote by H a real Hilbert space, C be a nonempty closed and convex subset of H. F(T) denote by the set of fixed points of a mapping T. Let { x n } be a sequence in H and xH. Strong convergence of { x n } to x is denoted by x n x, and weak convergence of { x n } to x is denoted by x n x. For every point xH, there exists a unique nearest point in C, denoted by P C x. This point satisfies.

x P C xxy,yC.

The operator P C is called the metric projection. The metric projection P C is characterized by the fact that P C xC and

x P C x, P C xy0,xH,yC.

Recall that a mapping T:CH is said to be nonexpansive, if TxTyxy for every x,yC. T is said to be quasi-nonexpansive, if F(T) and Txpxp for every xC and pF(T). It is easy to see that F(T) is a closed convex subset of C if T is a quasi-nonexpansive mapping. Besides, T is said to be a firmly nonexpansive, if

T x T y 2 x y , T x T y x , y C ; T x T y 2 x y 2 ( I T ) x ( I T ) y 2 x , y C .

Lemma 2.1 (demi-closed principle)

Let C be a nonempty closed convex subset of a real Hilbert space H. Let T:CH be a nonexpansive mapping, and let { x n } be a sequence in C. If x n w and lim n x n T x n =0, then Tw=w.

Lemma 2.2 [23]

Let H be a (real) Hilbert space. Then for all x,yH,

x + y 2 x 2 +2y,x+y.
(2.1)

Lemma 2.3 [24]

Let H be a Hilbert space and let { x n } be a sequence in H. Then, for any given sequence { λ n }(0,1) with n = 1 λ n =1 and for any positive integers i, j with i<j,

n = 1 λ n x n 2 n = 1 λ n x n 2 λ i λ j x i x j 2 .
(2.2)

Lemma 2.4 Let { a n } be a sequence of nonnegative real numbers, { b n } be a sequence of real numbers in (0,1) with n = 1 b n =, { u n } be a sequence of nonnegative real numbers with n = 1 u n <, { t n } be a real numbers with lim sup n t n 0. If

a n + 1 (1 b n ) a n + b n t n + u n ,for each n1,

then lim n a n =0.

Lemma 2.5 [25]

Let { a n } be a sequence of real numbers such that there exists a subsequence { n i } of {n} such that a n i < a n i + 1 for all iN. Then there exists a nondecreasing sequence { m k }N such that m k , a m k a m k + 1 and a k a m k + 1 are satisfied by all (sufficiently large) numbers kN. In fact, m k =max{jk: a j < a j + 1 }.

Lemma 2.6 [22]

Let H be a real Hilbert space, B:H 2 H be a set-valued maximal monotone mapping, β>0, and let J β B be the resolvent mapping of B.

  1. (i)

    For each β>0, J β B is a single-valued and firmly nonexpansive mapping;

  2. (ii)

    D( J β B )=H and F( J β B )= B 1 (0):={xD(B):0Bx};

  3. (iii)

    (I J β B ) is a firmly nonexpansive mapping for each β>0;

  4. (iv)

    suppose that B 1 (0), then for each xH, each x B 1 (0) and each β>0

    x J β B x 2 + J β B x x x x 2 ;
  5. (v)

    suppose that B 1 (0). Then x J β B x, J β B xw0 for each xH and each w B 1 (0), and each β>0.

Lemma 2.7 Let H 1 , H 2 be two real Hilbert spaces, A: H 1 H 2 be a linear bounded operator and A be the adjoint of A. Let B: H 2 2 2 H be a set-valued maximal monotone mapping, β>0, and let J β B be the resolvent mapping of B, then

  1. (i)

    ( I J β B ) A x ( I J β B ) A y 2 (I J β B )Ax(I J β B )Ay,AxAy;

  2. (ii)

    A ( I J β B ) A x A ( I J β B ) A y 2 A 2 (I J β B )Ax(I J β B )Ay,AxAy;

  3. (iii)

    if ρ(0, 2 A 2 ), then (Iρ A (I J β B )A) is a nonexpansive mapping.

Proof By Lemma 2.6(iii), the mapping (I J β B ) is firmly nonexpansive, hence the conclusions (i) and (ii) are obvious.

Now we prove the conclusion (iii).

In fact, for any x,y H 1 , it follows from the conclusions (i) and (ii) that

( I ρ A ( I J β B ) A ) x ( I ρ A ( I J β B ) A ) y 2 = x y 2 2 ρ x y , A ( I J β B ) A x A ( I J β B ) A y + ρ 2 A ( I J β B ) A x A ( I J β B ) A y 2 x y 2 2 ρ A x A y , ( I J β B ) A x ( I J β B ) A y + ρ 2 A 2 ( I J β B ) A x ( I J β B ) A y 2 x y 2 ρ ( 2 ρ A 2 ) ( I J β B ) A x ( I J β B ) A y 2 x y 2 ( since  ρ ( 2 ρ A 2 ) 0 ) .

This completes the proof of Lemma 2.7. □

3 Main results

The following lemma will be used in proving our main results.

Lemma 3.1 Let H 1 and H 2 be two real Hilbert spaces, A: H 1 H 2 be a linear and bounded operator, and A be the adjoint of A. Let B i : H 1 2 H 1 , and K i : H 2 2 H 2 , i=1,2, , be two families of set-valued maximal monotone mappings, and let β>0 and γ>0. If Ω (the solution set of (GSVIP) (1.4)), then x H 1 is a solution of (GSVIP) (1.4) if and only if for each i1, for each γ>0 and for each β>0

x = J β B i ( x γ A ( I J β K i ) A x ) .
(3.1)

Proof Indeed, if x is a solution of (GSVIP) (1.4), then for each i1, γ>0 and β>0,

x B i 1 (0)andA x K i 1 (0),i.e., x = J β B i x  and A x = J β K i A x .

This implies that x = J β B i ( x γA x (I J β K i )A x ).

Conversely, if x solves (3.1), by Lemma 2.6(v), we have

x ( x γ A ( I J β K i ) A x ) , y x 0,y B i 1 (0).

Hence we have

( I J β K i ) A x , A y A x 0,y B i 1 (0).
(3.2)

On the other hand, by Lemma 2.6(v) again

( A x J β K i A x , J β K i A x v 0,v K i 1 (0).
(3.3)

Adding up (3.2) and (3.3), we have

A x J β K i A x , J β K i A x + A y A x v 0,y B i 1 (0), and v K i 1 (0).

Simplifying it, we have

A x J β K i A x 2 A x J β K i A x , A y v 0,y B i 1 (0), and v K i 1 (0).
(3.4)

By the assumption that Ω. Taking wΩ, hence for each i1 w B i 1 (0) and Aw K i 1 (0). In (3.4), taking y=w and v=Aw, then we have

A x J β K i A x 2 =0.

This implies that A x = J β K i A x , and so A x K i 1 (0) for each i1. Hence from (3.1), x = J β B i x , i.e., x B i 1 (0). Hence x is a solution of (GSVIP)(1.4).

This completes the proof of Lemma 3.1. □

We are now in a position to prove the following main result.

Theorem 3.2 Let H 1 , H 2 , A, A , { B i }, { K i }, Ω be the same as in Lemma  3.1. Let f: H 1 H 1 be a contractive mapping with contractive constant k(0,1). Let { α n }, { ξ n }, { γ n , i } be the sequences in (0,1) with α n + ξ n + i = 1 γ n , i =1, for each n0. Let { β i } be a sequence in (0,), and { λ n , i } be a sequence in (0, 2 A 2 ). Let { x n } be the sequence defined by (1.11). If Ω and the following conditions are satisfied:

  1. (i)

    lim n ξ n =0, and n = 0 ξ n =;

  2. (ii)

    lim inf n α n γ n , i >0 for each i1;

  3. (iii)

    0< lim inf n λ n , i lim sup n λ n , i < 2 A 2 ,

then x n x Ω where x = P Ω f( x ), where P Ω is the metric projection from H 1 onto Ω.

Proof (I) First we prove that { x n } is bounded.

In fact, letting zΩ, by Lemma 3.1, for each i1,

z= J β i B i [ z λ n , i A ( I J β i K i ) A z ] .

Hence it follows from Lemma 2.7(iii) that for each i1 and each n1 we have

x n + 1 z = α n x n + ξ n f ( x n ) + i = 1 γ n , i J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] z α n x n z + ξ n f ( x n ) z + i = 1 γ n , i J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] z α n x n z + ξ n f ( x n ) z + i = 1 γ n , i J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] z α n x n z + ξ n f ( x n ) z + i = 1 γ n , i x n z = ( 1 ξ n ) x n z + ξ n f ( x n ) z ( 1 ξ n ) x n z + ξ n f ( x n ) f ( z ) + ξ n f ( z ) z ( 1 ξ n ( 1 k ) ) x n z + ξ n ( 1 k ) 1 k f ( z ) z max { x n z , 1 1 k f ( z ) z } .

By induction, we can prove that

x n zmax { x 0 z , 1 1 k f ( z ) z } ,n0.
(3.5)

This implies that { x n } is bounded, so is {f( x n )}.

(II) Now we prove that for each j1

α n γ n , j x n J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] 2 x n z 2 x n + 1 z 2 + ξ n f ( x n ) z 2 , for each  i 1 .
(3.6)

Indeed, it follows from Lemma 2.3 that for any positive j1

x n + 1 z 2 = α n x n + ξ n f ( x n ) + i = 1 γ n , i J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] z 2 α n x n z 2 + ξ n f ( x n ) z 2 + i = 1 γ n , i J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] z 2 α n γ n , j x n J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] 2 ( 1 ξ n ) x n z 2 + ξ n f ( x n ) z 2 α n γ n , j x n J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] 2 .

Simplifying it, (3.6) is proved.

By the assumption that Ω, and it is easy to prove that Ω is closed and convex. This implies that P Ω is well defined. Again since P Ω f: H 1 Ω is a contraction mapping with contractive constant k(0,1), there exists a unique x Ω such that x = P Ω f x . Since x Ω, it solves (GSVIP) (1.4). By Lemma 3.1,

x = J β j B j ( x λ n , j A ( I J β j K j ) A x ) ,j1,n0.
(3.7)
  1. (III)

    Now we prove that x n x .

In order to prove that x n x (as n), we consider two cases.

Case 1. Assume that { x n x } is a monotone sequence. In other words, for n 0 large enough, { x n x } n n 0 is either nondecreasing or non-increasing. Since { x n x } is bounded, { x n x } is convergence. Again since lim n ξ n =0, and {f( x n )} is bounded, from (3.6) we get

lim n α n γ n , j x n J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] 2 =0.

By condition (ii), we obtain

lim n x n J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] =0.
(3.8)

Now we prove that

lim sup n f ( x ) x , x n x 0.
(3.9)

To show this inequality, we choose a subsequence { x n k } of { x n } such that x n k w, λ n k , i λ i (0, 2 A 2 ) for each i1, and

lim sup n f ( x ) x , x n x = lim n k f ( x ) x , x n k x .
(3.10)

It follows from (3.8) that

J β i B i [ x n λ i A ( I J β i K i ) A x n ] x n J β i B i [ x n λ i A ( I J β i K i ) A x n ] J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] + J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] x n [ x n λ i A ( I J β i K i ) A x n ] [ x n λ n , i A ( I J β i K i ) A x n ] + J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] x n | λ i λ n , i | A ( I J β i K i ) A x n + J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] x n 0 ( as  n ) .

For each i1, J β i B i [I λ i A (I J β i K i )A] is a nonexpansive mapping. Thus from Lemma 2.1, w= J β i B i [I λ i A (I J β i K i )A]w. By Lemma 3.1 wΩ, i.e., w is a solution of (GSVIP) (1.4). Consequently we have

lim sup n f ( x ) x , x n x = lim n k f ( x ) x , x n k x = f ( x ) x , w x 0 .
  1. (IV)

    Finally, we prove that x n P Ω f( x ).

In fact, from Lemma 2.2 we have

x n + 1 x 2 α n ( x n x ) + i = 1 γ n , i J β i B i [ x n λ n , i A ( I J β i K i ) A x n ] x 2 + 2 ξ n f ( x n ) x , x n + 1 x ( 1 ξ n ) 2 x n x 2 + 2 ξ n f ( x n ) f ( x ) , x n + 1 x + 2 ξ n f ( x ) x , x n + 1 x ( 1 ξ n ) 2 x n x 2 + 2 ξ n k x n x x n + 1 x + 2 ξ n f ( x ) x , x n + 1 x ( 1 ξ n ) 2 x n x 2 + ξ n k { x n + 1 x 2 + x n x 2 } + 2 ξ n f ( x ) x , x n + 1 x .

Simplifying it, we have

x n + 1 x 2 ( 1 ξ n ) 2 + ξ n k 1 ξ n k x n x 2 + 2 ξ n 1 ξ n k f ( x ) x , x n + 1 x 1 2 ξ n + ξ n k 1 ξ n k x n x 2 + ξ n 2 1 ξ n k x n x 2 + 2 ξ n 1 ξ n k f ( x ) x , x n + 1 x ( 1 η n ) x n x 2 + η n δ n , n 0 ,

where δ n = ξ n M 2 ( 1 k ) + 1 1 k f( x ) x , x n + 1 x , M= sup n 0 x n x 2 , and η n = 2 ( 1 k ) ξ n 1 ξ n k . It is easy to see that η n 0, n = 1 η n =, and lim sup n δ n 0. Hence by Lemma 2.4, the sequence { x n } converges strongly to x = P Ω f( x ).

Case 2. Assume that { x n x } is not a monotone sequence. Then, by Lemma 2.3, we can define a sequence of positive integers: {τ(n)}, n n 0 (where n 0 large enough) by

τ(n)=max { k n : x k x x k + 1 x } .
(3.11)

Clearly {τ(n)} is a nondecreasing sequence such that τ(n) as n, and for all n n 0

x τ ( n ) x x τ ( n ) + 1 x .
(3.12)

Therefore { x τ ( n ) x } is a nondecreasing sequence. According to Case (1), lim n x τ ( n ) x =0 and lim n x τ ( n ) + 1 x =0. Hence we have

0 x n x max { x n x , x τ ( n ) x } x τ ( n ) + 1 x 0,as n.

This implies that x n x and x = P Ω f( x ) is a solution of (GSVIP) (1.4).

This completes the proof of Theorem 3.2. □

In Theorem 3.2, if B i =B and K i =K, for each i1, where B: H 1 2 H 1 and K: H 2 2 H 2 are two set-valued maximal monotone mappings, then from Theorem 3.2 we have the following.

Theorem 3.3 Let H 1 , H 2 , A, A , B, K, Ω, f be the same as in Theorem  3.2. Let { α n }, { ξ n }, { γ n } be the sequence in (0,1) with α n + ξ n + γ n =1 for each n0. Let β>0 be any given positive number, and { λ n } be a sequence in (0, 2 A 2 ). Let { x n } be the sequence defined by

x n + 1 = α n x n + ξ n f( x n )+ γ n J β B [ x n λ n A ( I J β K ) A x n ] ,n0.
(3.13)

If Ω and the following conditions are satisfied:

  1. (i)

    lim n ξ n =0, and n = 0 ξ n =;

  2. (ii)

    lim inf n α n γ n >0;

  3. (iii)

    0< lim inf n λ n lim sup n λ n < 2 A 2 ,

then x n x Ω where x = P Ω f( x ).

4 Applications

In this section we shall utilize the results presented in Theorem 3.2 and Theorem 3.3 to study some problems.

4.1 Application to split optimization problem

Let H 1 and H 2 be two real Hilbert spaces. Let h: H 1 R and g: H 2 R be two proper, convex and lower semicontinuous functions, and A: H 1 H 2 be a linear and bounded operators. The so-called split optimization problem (SOP) is

to find  x H 1  such that h ( x ) = min y H 1 h(y) and g ( A x ) = min z H 2 g(z).
(4.1)

Denote by h=B and g=K. It is know that B: H 1 2 H 1 (resp. K: H 2 2 H 2 ) is a maximal monotone mapping, so we can define the resolvent J β B = ( I + β B ) 1 and J β K = ( I + β K ) 1 , where β>0. Since x and A x is a minimum of h on H 1 and g on H 2 , respectively, for any given β>0, we have

x B 1 (0)=F ( J β B ) ,andA x K 1 (0)=F ( J β K ) .
(4.2)

This implies that the (SOP) (4.1) is equivalent to the split variational inclusion problem (SVIP) (4.2). From Theorem 3.3 we have the following.

Theorem 4.1 Let H 1 , H 2 , A, B, K, h, g be the same as above. Let f, { α n }, { ξ n }, { γ n } be the same as in Theorem  3.3. Let β>0 be any given positive number, and { λ n } be a sequence in (0, 2 A 2 ). Let { x n } be a sequence generated by x 0 H 1

{ y n = argmin z H 2 { g ( z ) + 1 2 β z A x n 2 } , z n = x n λ n A ( A x n y n ) , w n = argmin y H 1 { h ( y ) + 1 2 β y z n 2 } , x n + 1 = α n x n + ξ n f ( x n ) + γ n w n , n 0 .
(4.3)

If Ω 1 , the solution set of the split optimization problem (4.1), and the following conditions are satisfied:

  1. (i)

    lim n ξ n =0, and n = 0 ξ n =;

  2. (ii)

    lim inf n α n γ n >0;

  3. (iii)

    0< lim inf n λ n lim sup n λ n < 2 A 2 ,

then x n x Ω 1 where x = P Ω 1 f( x ).

Proof Since h=B, g:=K, and y n = argmin z H 2 {g(z)+ 1 2 β z A x n 2 }, we have

0 [ K ( z ) + 1 β ( z A x n ) ] z = y n ,i.e.,A x n (βK+I)( y n ).

This implies that

y n = J β K (A x n ).
(4.4)

Similarly, from (4.3), we have

w n = J β B ( z n ).
(4.5)

From (4.3)-(4.5), we have

w n = J β B ( x n λ n A ( I J β K ) A x n ) .
(4.6)

Therefore (4.3) can be rewritten as

x n + 1 = α n x n + ξ n f( x n )+ γ n J β B ( x n λ n A ( I J β K ) A x n ) ,n0.
(4.7)

The conclusion of Theorem 4.1 can be obtained from Theorem 3.3 immediately. □

4.2 Application to split feasibility problem

Let C H 1 and Q H 2 be two nonempty closed convex subsets and A: H 1 H 2 be a bounded linear operator. Now we consider the following split feasibility problem, i.e.: to find

x C such that A x Q.
(4.8)

Let i C and i Q be the indicator functions of C and Q defined by (1.9). Let N C (u) be the normal cone at u H 1 defined by

N C (u)= { z H 1 : z , v u 0 , v C } .

Since i C and i Q both are proper convex and lower semicontinuous functions on H 1 and H 2 , respectively, and the subdifferential i C of i C (resp. i Q of i Q ) is a maximal monotone operator, we can define the resolvents J β i C of i C and J β i Q of i Q by

J β i C ( x ) = ( I + β i C ) 1 ( x ) , x H 1 , J β i Q ( x ) = ( I + β i Q ) 1 ( x ) , x H 2 ,

where β>0. By definition, we know that

i C ( x ) = { z H 1 : i C ( x ) + z , y x i C ( y ) , y H 1 } = { z H 1 : z , y x 0 , y C } = N C ( x ) , x C .

Hence, for each β>0, we have

u = J β i C ( x ) x u β N C ( u ) x u , y u 0 , y C u = P C ( x ) .

This implies that J β i C = P C . Similarly J β i Q = P Q . Taking h(x)= i C (x) and g(x)= i Q (x) in (4.1), then the (SFP) (4.8) is equivalent to the following split optimization problem:

to find  x H 1  such that  i C ( x ) = min y H 1 i C (y) and  i Q ( A x ) = min z H 2 i Q (z).
(4.9)

Hence, the following result can be obtained from Theorem 4.1 immediately.

Theorem 4.2 Let H 1 , H 2 , A, A , i C , i Q be the same as above. Let f, { α n }, { ξ n }, { γ n } be the same as in Theorem  4.1. Let { λ n } be a sequence in (0, 2 A 2 ). Let { x n } be the sequence defined by

x n + 1 = α n x n + ξ n f( x n )+ γ n P C [ x n λ n A ( I P Q ) A x n ] ,n0.
(4.10)

If the solution set of the split optimization problem (4.4) Ω 2 , and the following conditions are satisfied:

  1. (i)

    lim n ξ n =0, and n = 0 ξ n =;

  2. (ii)

    lim inf n α n γ n >0;

  3. (iii)

    0< lim inf n λ n lim sup n λ n < 2 A 2 ,

then x n x Ω 2 where x = P Ω 2 f( x ).

Remark 4.3 Theorem 4.2 extends and improves the main results in Censor and Elfving [1] and Byrne [2].

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Acknowledgements

The authors would like to express their thanks to the referees and the editors for their kind and helpful comments and advice. This work was supported by the National Natural Science Foundation of China (Grant No. 11361070).

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Chang, Ss., Wang, L. Strong convergence theorems for the general split variational inclusion problem in Hilbert spaces. Fixed Point Theory Appl 2014, 171 (2014). https://doi.org/10.1186/1687-1812-2014-171

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Keywords

  • general split variational inclusion problem
  • split feasibility problem
  • split optimization problem
  • quasi-nonexpansive mapping
  • zero point
  • resolvent mapping