1 Introduction

Splitting methods have recently received much attention due to the fact that many nonlinear problems arising in applied areas such as image recovery, signal processing, and machine learning are mathematically modeled as a nonlinear operator equation, and this operator is decomposed as the sum of two nonlinear operators. Study of fixed (zero) point approximation algorithms for computing fixed (zero) points constitutes now a topic of intensive research efforts. Many well-known problems can be studied by using algorithms which are iterative in their nature. As an example, in computer tomography with limited data, each piece of information implies the existence of a convex set in which the required solution lies. The problem of finding a point in the intersection of these convex sets is then of crucial interest, and it cannot be usually solved directly. Therefore, an iterative algorithm must be used to approximate such a point. The well-known convex feasibility problem which captures applications in various disciplines such as image restoration and radiation therapy treatment planning is to find a point in the intersection of common fixed (zero) point sets of a family of nonlinear mappings; see, for example, [116].

In this paper, we will investigate the problem of finding a common solution to inclusion problems and fixed point problems based on an iterative algorithm. Strong convergence of the proposed iterative algorithm has been obtained in the framework of Hilbert spaces.

The organization of this paper is as follows. In Section 2, we provide some necessary preliminaries. In Section 3, an iterative algorithm is proposed and analyzed. Some subresults of the main results are also discussed in this section.

2 Preliminaries

From now on, we always assume that H is a real Hilbert space with the inner product , and the norm , respectively. Let C be a nonempty closed convex subset of H.

Let S:CC be a mapping. F(S) stands for the fixed point set of S; that is, F(S):={xC:x=Sx}.

Recall that S is said to be nonexpansive iff

SxSyxy,x,yC.

S is said to be asymptotically nonexpansive iff there exists a sequence { k n }[1,) with lim n k n =1 such that

S n x S n y k n xy,x,yC.

Recall that S is said to be strictly pseudocontractive iff there exits a positive constant κ such that

S x S y 2 x y 2 +κ ( x S x ) ( y S y ) 2 ,x,yC.

S is said to be asymptotically strictly pseudocontractive iff there exits a positive constant κ and a sequence { k n }[1,) with lim n k n =1 such that

S n x S n y 2 k n x y 2 +κ ( x S n x ) ( y S n y ) 2 ,x,yC.

Let A:CH be a mapping. Recall that A is said to be monotone iff

AxAy,xy0,x,yC.

A is said to be inverse-strongly monotone iff there exists a constant α>0 such that

AxAy,xyα A x A y 2 ,x,yC.

For such a case, A is also said to be α-inverse-strongly monotone. It is not hard to see that inverse-strongly monotone mappings are Lipschitz continuous.

A multivalued operator T:H 2 H with the domain D(T)={xH:Tx} and the range R(T)={Tx:xD(T)} is said to be monotone if for x 1 D(T), x 2 D(T), y 1 T x 1 and y 2 T x 2 , we have x 1 x 2 , y 1 y 2 0. A monotone operator T is said to be maximal if its graph G(T)={(x,y):yTx} is not properly contained in the graph of any other monotone operator. Let I denote the identity operator on H and T:H 2 H be a maximal monotone operator. Then we can define, for each λ>0, a nonexpansive single-valued mapping J λ :HH by J λ = ( I + λ T ) 1 . It is called the resolvent of T. We know that T 1 0=F( J λ ) for all λ>0 and J λ is firmly nonexpansive; see [1723] and the references therein.

Recently, many authors have investigated the solution problems of nonlinear operator equations or inequalities based on iterative methods; see, for instance, [2433] and the references therein. In [19], Kamimura and Takahashi investigated the problem of finding zero points of a maximal monotone operator via the following iterative algorithm:

x 0 H, x n + 1 = α n x n +(1 α n ) J λ n x n ,n=0,1,2,,
(2.1)

where { α n } is a sequence in (0,1), { λ n } is a positive sequence, T:H 2 H is a maximal monotone and J λ n = ( I + λ n T ) 1 . They showed that the sequence { x n } generated in (2.1) converges weakly to some z T 1 (0) provided that the control sequence satisfies some restrictions.

Recall that the classical variational inequality is to find an xC such that

Ax,yx0,yC.
(2.2)

In this paper, we use VI(C,A) to denote the solution set of (2.2). It is known that xC is a solution to (2.1) iff x is a fixed point of the mapping P C (IλA), where λ>0 is a constant, I stands for the identity mapping, and P C stands for the metric projection from H onto C. If A is α-inverse-strongly monotone and λ(0,2α], then the mapping P C (IrA) is nonexpansive; see [28] for more details. It follows that VI(C,A) is closed and convex.

In [28], Takahashi an Toyoda investigated the problem of finding a common solution of variational inequality problem (2.1) and a fixed point problem involving nonexpansive mappings by considering the following iterative algorithm:

x 0 C, x n + 1 = α n x n +(1 α n )S P C ( x n λ n A x n ),n0,
(2.3)

where { α n } is a sequence in (0,1), { λ n } is a positive sequence, S:CC is a nonexpansive mapping and A:CH is an inverse-strongly monotone mapping. They proved that the sequence { x n } generated in (2.3) converges weakly to some zVI(C,A)F(S) provided that the control sequence satisfies some restrictions.

In [29], Tada and Takahashi investigated the problem of finding a common solution of an equilibrium problem and a fixed point problem involving nonexpansive mappings by considering the following iterative algorithm:

{ u n C such that F ( u n , u ) + 1 r n u u n , u n x n 0 , u C , x n + 1 = α n x n + ( 1 α n ) S u n
(2.4)

for each n1, where { α n } is a sequence in (0,1), { r n } is a positive sequence, S:CC is a nonexpansive mapping and F:C×CR is a bifunction. They showed that the sequence { x n } generated in (2.4) converges weakly to some zEP(F)F(S), where EP(F) stands for the solution set of the equilibrium problem, provided that the control sequence satisfies some restrictions.

In [30], Manaka and Takahashi introduced the following iteration:

x 1 C, x n + 1 = α n x n +(1 α n )S J λ n (I λ n A) x n ,n1,
(2.5)

where { α n } is a sequence in (0,1), { λ n } is a positive sequence, S:CC is a nonexpansive mapping, A:CH is an inversely-strongly monotone mapping, B:D(B)C 2 H is a maximal monotone operator, J λ n = ( I + λ n B ) 1 is the resolvent of B. They showed that the sequence { x n } generated in (2.5) converges weakly to some z ( A + B ) 1 (0)F(S) provided that the control sequence satisfies some restrictions.

In this paper, motivated by the above results, we consider the problem of finding a common solution to the zero point problems involving two monotone operators and fixed point problems involving asymptotically strictly pseudocontractive mappings based on a one-step iterative method. Weak convergence theorems are established in the framework of Hilbert spaces.

In order to obtain our main results in this paper, we need the following lemmas.

Recall that a space is said to satisfy Opial’s property [34] if, for any sequence { x n }H with x n x, where ⇀ denotes the weak convergence, the inequality

lim inf n x n x< lim inf n x n y

holds for every yH with yx. Indeed, the above inequality is equivalent to the following:

lim sup n x n x< lim sup n x n y.

Lemma 2.1 [20]

Let C be a nonempty, closed, and convex subset of H, A:CH be a mapping, and B:HH be a maximal monotone operator. Then F( J r (IλA))= ( A + B ) 1 (0).

Lemma 2.2 Let H be a real Hilbert space. For any a(0,1) and x,yH, the following holds:

a x + ( 1 a ) y 2 =a x 2 +(1a) y 2 a(1a) x y 2 .

Lemma 2.3 [35]

Let { a n }, { b n }, and { c n } be three nonnegative sequences satisfying the following condition:

a n + 1 (1+ b n ) a n + c n ,n n 0 ,

where n 0 is some nonnegative integer, n = 1 b n < and n = 1 c n <. Then the limit lim n a n exists.

Lemma 2.4 [36]

Let C be a nonempty closed convex subset of H and S be an asymptotically κ-strictly pseudocontractive mapping. Then we have

  1. (a)

    S is uniformly Lipschitz continuous;

  2. (b)

    IS is demiclosed at zero, that is, if { x n } is a sequence in C with x n x and x n S x n 0, then xF(S).

The following lemma can be obtained from [37] immediately.

Lemma 2.5 Let H be a real Hilbert space. The following holds:

i = 1 N a i x i 2 = i = 1 N a i x i 2 i j N a i a j x i x j 2 ,

where N2 denotes some positive integer, a 1 , a 2 ,, a N are real numbers with i = 1 N a i =1 in (0,1) and x 1 , x 2 ,, x N H.

3 Main results

Theorem 3.1 Let C be a nonempty closed convex subset of H. Let N2 be some positive integer and S:CC be an asymptotically strictly pseudocontractive mapping with the constant κ and the sequence { k n }. Let A m :CH be an inverse-strongly monotone mapping with the constant α m and B m be a maximal monotone operator on H such that the domain of B m is included in C for each m{2,3,,N}. Assume F= m = 2 N ( A m + B m ) 1 (0)F(S). Let { α n , 1 },{ α n , 2 },,{ α n , N } and { β n } are real number sequences in (0,1). Let { r n , 2 }, , and { r n , N } be positive real number sequences. Let { x n } be a sequence in C generated in the following iterative process:

{ x 1 C , y n = β n x n + ( 1 β n ) S n x n , x n + 1 = α n , 1 y n + m = 2 N α n , m J r n , m ( x n r n , m A m x n ) , n 1 ,
(3.1)

where J r n , m = ( I + r n , m B m ) 1 is the resolvent of B m . Assume that the sequences { α n , 1 },{ α n , 2 },,{ α n , N }, { β n }, { r n , 2 },,{ r n , N }, and { k n } satisfy the following restrictions:

  1. (a)

    m = 1 N α n , m =1 and 0<a α n , m <1, m{2,,N};

  2. (b)

    0κ β n b<1;

  3. (c)

    0<c r n , m d<2 α m , m{2,,N};

  4. (d)

    n = 1 ( k n 1)<,

where a, b, c, and d are positive real numbers. Then the sequence { x n } generated in (3.1) converges weakly to some point in ℱ.

Proof First, we show I r n , m A m is nonexpansive. In view of the restriction (c), we find that

( I r n , m A m ) x ( I r n , m A m ) y 2 = x y 2 2 r n , m x y , A m x A m y + r n , m 2 A m x A m y 2 x y 2 r n , m ( 2 α m r n , m ) A m x A m y 2 x y 2 .

This proves that I r n , m A m is nonexpansive. Let pF. In view of Lemma 2.1, we find that

p=Sp= J r n , m (p r n , m A m p).

Putting u n , m = J r n , m ( x n r n , m A m x n ), we find that

u n , m p ( x n r n , m A m x n ) ( p r n , m A m p ) x n p .
(3.2)

In view of Lemma 2.2, we find from the restriction (b) that

y n p 2 = β n x n + ( 1 β n ) S n x n p 2 = β n x n p 2 + ( 1 β n ) S n x n p 2 β n ( 1 β n ) x n S n x n 2 β n x n p 2 + ( 1 β n ) k n x n p 2 + ( κ β n ) x n S n x n 2 k n x n p 2 .
(3.3)

From (3.2) and (3.3), we have

x n + 1 p 2 = α n , 1 ( y n p ) + m = 2 N α n , m ( u n , m p ) 2 α n , 1 y n p 2 + m = 2 N α n , m u n , m p 2 α n , 1 k n x n p 2 + m = 2 N α n , m x n p 2 k n x n p 2 .
(3.4)

We draw the conclusion that lim n x n p exists with the aid of Lemma 2.3. This implies that the sequence { x n } is bounded. In view of Lemma 2.5, we find that

x n + 1 p 2 = α n , 1 ( y n p ) + m = 2 N α n , m ( u n , m p ) 2 α n , 1 y n p 2 + m = 2 N α n , m u n , m p 2 α n , 1 α n , r y n u n , r 2 α n , 1 k n x n p 2 + m = 2 N α n , m x n p 2 α n , 1 α n , r y n u n , r 2 k n x n p 2 α n , 1 α n , r y n u n , r 2 , r { 2 , 3 , , N } ,
(3.5)

which yields

α n , 1 α n , r y n u n , r 2 k n x n p 2 x n + 1 p 2 ,r{2,3,,N}.

In view of the restriction (a), we find that

lim n y n u n , m =0,r{2,3,,N}.
(3.6)

On the other hand, we have

u n , m p 2 ( x n r n , m A m x n ) ( p r n , m A m p ) 2 = x n p 2 2 r n , m x n p , A m x n A m p + r n , m 2 A m x n A m p 2 x n p 2 r n , m ( 2 α m r n , m ) A m x n A m p 2 .
(3.7)

It follows that

x n + 1 p 2 α n , 1 y n p 2 + m = 2 N α n , m u n , m p 2 α n , 1 k n x n p 2 + m = 2 N α n , m u n , m p 2 k n x n p 2 m = 2 N α n , m r n , m ( 2 α m r n , m ) A m x n A m p 2 .

This in turn implies that

m = 2 N α n , m r n , m (2 α m r n , m ) A m x n A m p 2 k n x n p 2 x n + 1 p 2 .

It follows from the restrictions (b) and (d) that

lim n A m x n A m p=0.
(3.8)

Notice that

u n , m p 2 ( x n r n , m A m x n ) ( p r n , m A m p ) , u n , m p = 1 2 ( ( x n r n A m x n ) ( p r n A m p ) 2 + u n , m p 2 ( x n r n A m x n ) ( p r n A m p ) ( u n , m p ) 2 ) 1 2 ( x n p 2 + u n , m p 2 x n u n , m r n ( A m x n A m p ) 2 ) 1 2 ( x n p 2 + u n , m p 2 x n u n , m 2 r n 2 A m x n A m p 2 + 2 r n x n u n , m A m x n A m p ) 1 2 ( x n p 2 + u n , m p 2 x n u n , m 2 + 2 r n x n u n , m A m x n A m p ) .

It follows that

u n , m p 2 x n p 2 x n u n , m 2 +2 r n , m x n u n , m A m x n A m p.
(3.9)

This implies that

x n + 1 p 2 = α n , 1 ( y n p ) + m = 2 N α n , m ( u n , m p ) 2 α n , 1 y n p 2 + m = 2 N α n , m u n , m p 2 k n x n p 2 m = 2 N α n , m x n u n , m 2 + 2 m = 2 N α n , m r n , m x n u n , m A m x n A m p ,

which finds that

m = 2 N α n , m x n u n , m 2 k n x n p 2 x n + 1 p 2 + 2 m = 2 N α n , m r n , m x n u n , m A m x n A m p .

In view of the restriction (a), we find from (3.8) that

lim n x n u n , m =0.
(3.10)

Notice that

x n y n x n u n , m + u n , m y n .

From (3.6) and (3.10), we obtain that

lim n x n y n =0.
(3.11)

On the other hand, we have

S n x n x n S n x n ( β n x n + ( 1 β n ) S n x n ) + ( β n x n + ( 1 β n ) S n x n ) x n = β n S n x n x n + y n x n ,

which yields

(1 β n ) S n x n x n y n x n .

This implies from the restriction (c) and (3.11) that

lim n S n x n x n =0.
(3.12)

Notice that

x n + 1 x n α n , 1 y n x n + m = 2 N α n , m u n , m x n .

This implies from (3.10) and (3.11) that

lim n x n + 1 x n =0.
(3.13)

On the other hand, we have

x n S x n x n x n + 1 + x n + 1 S n + 1 x n + 1 + S n + 1 x n + 1 S n + 1 x n + S n + 1 x n S x n .

Since S is uniformly continuous, we obtain from (3.12) and (3.13) that

lim n S x n x n =0.
(3.14)

Since { x n } is bounded, there exists a subsequence { x n i } of { x n } such that x n i ωC. We find that ωF(S) with the aid of Lemma 2.4.

Next, we show ω ( A m + B m ) 1 0 for every m{1,2,,N}. In view of (3.10), we can choose a subsequence { u n i , m } of { u n , m } such that u n i , m ω. Notice that

u n , m = J r n , m ( x n r n , m A m x n ).

This implies that

x n r n , m A m x n (I+ r n , m B m ) u n , m .

That is,

x n u n , m r n , m A m x n B m u n , m .

Since B m is monotone, we get for any ( u m , v m )G( B m ) that

u n , m u m , x n u n , m r n , m A m x n v m 0.
(3.15)

Replacing n by n i and letting i, we obtain from (3.10) that

ω u m , A m ω v m 0.

This means A m ω m B m ω, that is, 0( A m + B m )(ω). Hence we get ω ( A m + B m ) 1 (0) for every m{1,2,,N}. This completes the proof that ωF.

Suppose there is another subsequence { x n j } of { x n } such that x n j ω . Then we can show that ω F in the same way. Assume ω ω . Since lim n x n p exits for any pF. Put lim n x n ω=d. Since the space satisfies Opial’s condition, we see that

d = lim inf i x n i ω < lim inf i x n i ω = lim n x n ω = lim inf j x n j ω < lim inf j x n j ω = d .

This is a contradiction. This shows that ω= ω . This proves that the sequence { x n } converges weakly to ωF. This completes the proof. □

If N=2, then we have the following.

Corollary 3.2 Let C be a nonempty closed convex subset of H. Let S:CC be an asymptotically strictly pseudocontractive mapping with the constant κ and the sequence { k n }. Let A:CH be an inverse-strongly monotone mapping with the constant α, and B be a maximal monotone operator on H such that the domain of B is included in C. Assume F= ( A + B ) 1 (0)F(S). Let { α n , 1 }, { α n , 2 }, and { β n } be real number sequences in (0,1). Let { r n } be a positive real number sequence. Let { x n } be a sequence in C generated in the following iterative process:

{ x 1 C , y n = β n x n + ( 1 β n ) S n x n , x n + 1 = α n , 1 y n + α n , 2 J r n ( x n r n A 2 x n ) , n 1 ,

where J r n = ( I + r n B ) 1 is the resolvent of B. Assume that the sequences { α n , 1 }, { α n , 2 }, { β n }, { r n }, and { k n } satisfy the following restrictions:

  1. (a)

    m = 1 2 α n , m =1 and 0<a α n , m <1, m{1,2};

  2. (b)

    0κ β n b<1;

  3. (c)

    0<c r n d<2α;

  4. (d)

    n = 1 ( k n 1)<,

where a, b, c, and d are positive real numbers. Then the sequence { x n } converges weakly to some point in ℱ.

If S is asymptotically nonexpansive, then we find from Theorem 3.1 the following by letting β n =0.

Corollary 3.3 Let C be a nonempty closed convex subset of H. Let N2 be some positive integer and S:CC be an asymptotically nonexpansive mapping with the sequence { k n }. Let A m :CH be an inverse-strongly monotone mapping with the constant α m and let B m be a maximal monotone operator on H such that the domain of B m is included in C for each m{2,3,,N}. Assume F= m = 2 N ( A m + B m ) 1 (0)F(S). Let { α n , 1 },{ α n , 2 },,{ α n , N }, and { β n } be real number sequences in (0,1). Let { r n , 2 }, , and { r n , N } be positive real number sequences. Let { x n } be a sequence in C generated in the following iterative process:

x 1 C, x n + 1 = α n , 1 S n x n + m = 2 N α n , m J r n , m ( x n r n , m A m x n ),n1,

where J r n , m = ( I + r n , m B m ) 1 is the resolvent of B m . Assume that the sequences { α n , 1 },{ α n , 2 },,{ α n , N }, { β n }, { r n , 2 },,{ r n , N }, and { k n } satisfy the following restrictions:

  1. (a)

    m = 1 N α n , m =1 and 0<a α n , m <1, m{2,,N};

  2. (b)

    0<b r n , m c<2 α m , m{2,,N};

  3. (c)

    n = 1 ( k n 1)<,

where a, b and c are positive real numbers. Then the sequence { x n } converges weakly to some point in ℱ.

If S is the identity mapping, then we draw from Theorem 3.1 the following.

Corollary 3.4 Let C be a nonempty closed convex subset of H. Let N2 be some positive integer. Let A m :CH be an inverse-strongly monotone mapping with the constant α m and let B m be a maximal monotone operator on H such that the domain of B m is included in C for each m{2,3,,N}. Assume F= m = 2 N ( A m + B m ) 1 (0). Let { α n , 1 },{ α n , 2 }, , and { α n , N } be real number sequences in (0,1). Let { r n , 2 }, , and { r n , N } be positive real number sequences. Let { x n } be a sequence in C generated in the following iterative process:

x 1 C, x n + 1 = α n , 1 x n + m = 2 N α n , m J r n , m ( x n r n , m A m x n ),n1,

where J r n , m = ( I + r n , m B m ) 1 is the resolvent of B m . Assume that the sequences { α n , 1 },{ α n , 2 },,{ α n , N }, { r n , 2 }, , and { r n , N } satisfy the following restrictions:

  1. (a)

    m = 1 N α n , m =1 and 0<a α n , m <1, m{2,,N};

  2. (b)

    0<b r n , m c<2 α m , m{2,,N},

where a, b, and c are positive real numbers. Then the sequence { x n } converges weakly to some point in ℱ.

Let f:H(,] be a proper lower semicontinuous convex function. Define the subdifferential

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

for all xH. Then ∂f is a maximal monotone operator of H into itself; see [38] for more details. Let C be a nonempty closed convex subset of H and i C be the indicator function of C, that is,

i C x={ 0 , x C , , x C .

Furthermore, we define the normal cone N C (v) of C at v as follows:

N C v= { z H : z , y v 0 , y H }

for any vC. Then i C :H(,] is a proper lower semicontinuous convex function on H and i C is a maximal monotone operator. Let J r x= ( I + r i C ) 1 x for any r>0 and xH. From i C x= N C x and xC, we get

v = J r x x v + r N C v x v , y v 0 , y C , v = P C x ,

where P C is the metric projection from H into C. Similarly, we can get that x ( A + i C ) 1 (0)xVI(A,C).

Corollary 3.5 Let C be a nonempty closed convex subset of H. Let N2 be some positive integer and S:CC be an asymptotically strictly pseudocontractive mapping with the constant κ and the sequence { k n }. Let A m :CH be an inverse-strongly monotone mapping with the constant α m for each m{2,3,,N}. Assume F= m = 2 N VI(C, A m )F(S). Let { α n , 1 },{ α n , 2 },,{ α n , N }, and { β n } be real number sequences in (0,1). Let { r n , 2 }, , and { r n , N } be positive real number sequences. Let { x n } be a sequence in C generated in the following iterative process:

{ x 1 C , y n = β n x n + ( 1 β n ) S n x n , x n + 1 = α n , 1 y n + m = 2 N α n , m P C ( x n r n , m A m x n ) , n 1 .

Assume that the sequences { α n , 1 },{ α n , 2 },,{ α n , N }, { β n }, { r n , 2 },,{ r n , N }, and { k n } satisfy the following restrictions:

  1. (a)

    m = 1 N α n , m =1 and 0<a α n , m <1, m{2,,N};

  2. (b)

    0κ β n b<1;

  3. (c)

    0<c r n , m d<2 α m , m{2,,N};

  4. (d)

    n = 1 ( k n 1)<,

where a, b, c, and d are positive real numbers. Then the sequence { x n } converges weakly to some point in ℱ.

Proof Putting B m = i C for every m{2,3,,N}, we see J r n , m = P C . We can immediately draw from Theorem 3.1 the desired conclusion. □

If S is the identity mapping, then we find from Corollary 3.5 the following.

Corollary 3.6 Let C be a nonempty closed convex subset of H. Let N2 be some positive integer. Let A m :CH be an inverse-strongly monotone mapping with the constant α m for each m{2,3,,N}. Assume F= m = 2 N VI(C, A m ). Let { α n , 1 },{ α n , 2 }, , and { α n , N } be real number sequences in (0,1). Let { r n , 2 }, , and { r n , N } be positive real number sequences. Let { x n } be a sequence in C generated in the following iterative process:

x 1 C, x n + 1 = α n , 1 x n + m = 2 N α n , m P C ( x n r n , m A m x n ),n1.

Assume that the sequences { α n , 1 },{ α n , 2 },,{ α n , N }, { β n }, { r n , 2 },,{ r n , N }, and { k n } satisfy the following restrictions:

  1. (a)

    m = 1 N α n , m =1 and 0<a α n , m <1, m{2,,N};

  2. (b)

    0<b r n , m c<2 α m , m{2,,N};

  3. (c)

    n = 1 ( k n 1)<,

where a, b, and c are positive real numbers. Then the sequence { x n } converges weakly to some point in ℱ.