1 Introduction

The theory of nonexpansive mappings in a Hilbert space is very important because it is applied to convex optimization, the theory of nonlinear evolution equations and others. Browder and Petryshyn [1] introduced a class of nonlinear mappings, called strict pseudo-contractions, which includes the class of nonexpansive mappings. For strict pseudo-contractions, we are interested in finding fixed points of the mappings. We also know the class of inverse-strongly monotone mappings which is related to nonexpansive mappings. For inverse-strongly monotone mappings, we are interested in finding zero points of the mappings. On the other hand, the generalized equilibrium problems which are formulated by the Ky Fan inequality have many important applications in optimization problems, variational inequalities, minimax problems, economics and others. Some methods have been proposed for solving the generalized equilibrium problems in Hilbert spaces; see, for example, [2, 3]. Recently, Liu [4] studied strong convergence theorems for strict pseudo-contractions with equilibrium problems in a Hilbert space. We know from [5] that the solutions of equilibrium problems in [4] are written by using the resolvent of a maximal monotone operator with some domain condition. Furthermore, the class of strict pseudo-contractions is related to the class of inverse-strongly monotone mappings in a Hilbert space.

In this paper, motivated by these results, we prove implicit and explicit strong convergence theorems for finding a common point of the set of zero points of the addition of an inverse-strongly monotone mapping and a maximal monotone operator and the set of zero points of a maximal monotone operator which is related to an equilibrium problem in a real Hilbert space. Such theorems improve and extend the results announced by Liu [4]. The limit point of the implicit strong convergence theorem is simply proved by using an implicit contraction mapping which does not appear in other references. Such a unique fixed point of the mapping is used in the proof of the explicit strong convergence theorem. Using this explicit strong convergence theorem, we obtain well-known and new strong convergence theorems in a Hilbert space.

2 Preliminaries

Throughout this paper, we denote by ℕ and ℝ the sets of positive integers and real numbers, respectively. We also denote by H a real Hilbert space with inner product 〈 ⋅ , ⋅ 〉 and norm \(\lVert \,\cdot\,\rVert \). When {x n } is a sequence in H, we denote the strong convergence of {x n } to xH by x n x and the weak convergence by x n x. We have from [6], for any x,yH and λ∈ℝ,

$$ \|x+y\|^2 \leq\|x\|^2 + 2\langle y, x+y\rangle $$
(1)

and

$$ \big\|\lambda x + (1-\lambda)y\big\|^2 = \lambda\|x \|^2 + (1-\lambda)\|y\|^2 - \lambda(1-\lambda) \|x-y \|^2. $$
(2)

Furthermore we have, for x,y,u,vH,

$$ 2\langle x-y,u-v\rangle= \|x-v\|^2+ \|y-u \|^2- \|x-u\|^2 -\|y-v\|^2. $$
(3)

Let C be a nonempty, closed and convex subset of H. Let T be a mapping of C into H. We denote by F(T) the set of fixed points of T. A mapping T:CH is called nonexpansive iff ∥TxTy∥≤∥xy∥ for all x,yC. If T:CH is nonexpansive, then F(T) is closed and convex; see [6]. For a nonempty, closed and convex subset D of H, the nearest point projection of H onto D is denoted by P D , that is, \(\lVert x-P_{D} x\rVert \leq \lVert x-y\rVert \) for all xH and yD. Such P D is called the metric projection of H onto D. We know that the metric projection P D is firmly nonexpansive, that is,

for all x,yH. Furthermore 〈xP D x,yP D x〉≤0 holds for all xH and yD; see [7].

For a positive number α>0, a mapping A:CH is called α-inverse-strongly monotone iff

$$ \langle x-y, Ax-Ay\rangle\geq\alpha\|Ax-Ay\|^2, \quad \forall x, y\in C. $$
(4)

If A:CH is α-inverse-strongly monotone, then 〈xy,AxAy〉≥0 and \(\lVert Ax-Ay\rVert \leq(1/\alpha)\lVert x-y\rVert \) for all x,yC; see, for example, [8, 9] for inverse-strongly monotone mappings.

Let B be a mapping of H into 2H. The effective domain of B is denoted by \(\operatorname {dom}(B)\), that is, \(\operatorname {dom}(B)=\{ x\in H: Bx \neq\emptyset\}\). A multi-valued mapping B is said to be a monotone operator on H iff 〈xy,uv〉≥0 for all \(x,y\in \operatorname {dom}(B)\), uBx, and vBy. A monotone operator B on H is said to be maximal iff its graph is not properly contained in the graph of any other monotone operator on H. For a maximal monotone operator B on H and r>0, we may define a single-valued operator \(J_{r}=(I+rB)^{-1}\colon H\to \operatorname {dom}(B)\), which is called the resolvent of B for r. We denote by \(A_{r}=\frac{1}{r}(I-J_{r})\) the Yosida approximation of B for r>0. We know from [10] that

$$ A_rx\in BJ_rx, \quad\forall x\in H, r>0. $$
(5)

Let B be a maximal monotone operator on H and let

$$B^{-1}0 = \{x\in H: 0 \in Bx\}. $$

It is known that B −10=F(J r ) for all r>0 and the resolvent J r is firmly nonexpansive, i.e.,

$$ \|J_rx-J_ry\|^2\leq\langle x-y, J_rx-J_ry\rangle, \quad\forall x,y\in H. $$
(6)

We also know the following lemma from [5].

Lemma 2.1

Let H be a real Hilbert space and let B be a maximal monotone operator on H. For r>0 and xH, define the resolvent J r x. Then the following holds:

for all s,t>0 and xH.

From Lemma 2.1, we have

for all λ,μ>0 and xH; see also [7, 11].

To prove our main results, we need the following lemma [12, 13]:

Lemma 2.2

Let {s n } be a sequence of non-negative real numbers, let {α n } be a sequence of [0,1] with \(\sum_{n=1}^{\infty} \alpha_{n}=\infty\), let {β n } be a sequence of non-negative real numbers with \(\sum_{n=1}^{\infty} \beta_{n}<\infty\), and let {γ n } be a sequence of real numbers with lim sup n→∞ γ n ≤0. Suppose that

$$s_{n+1}\leq(1-\alpha_n)s_n+\alpha_n \gamma_n+\beta_n $$

for all n=1,2,…. Then lim n→∞ s n =0.

3 Strong Convergence Theorems

In this section, we first prove the following implicit strong convergence theorem of Browder’s type [14] in a Hilbert space. Before proving it, we need some definitions. Let H be a Hilbert space. A mapping g:HH is a contraction iff there exists k∈]0,1[ such that∥g(x)−g(y)∥≤kxy∥ for all x,yH. We call such g a k-contraction. A linear bounded operator G:HH is called strongly positive iff there exists \(\overline{\gamma}>0\) such that \(\langle Gx,x\rangle\geq\overline{\gamma} \|x\|^{2}\) for all xH. We call such G a strongly positive operator with coefficient \(\overline{\gamma}>0\). Marino and Xu [15] proved the following result.

Lemma 3.1

Let H be a Hilbert space and let G be a strongly positive bounded linear self-adjoint operator on H with coefficient \(\overline{\gamma}>0\). If 0<γ≤∥G−1, then \(\|I-\gamma G\|\leq1-\gamma\overline{\gamma}\).

Theorem 3.1

Let H be a real Hilbert space and let C be a nonempty, closed and convex subset of H. Let α>0 and let A be an α-inverse-strongly monotone mapping of C into H and let B be a maximal monotone operator on H. Let F be a maximal monotone operator on H such that the domain of F is included in C. Let J λ =(I+λB)−1 and T r =(I+rF)−1 be the resolvents of B and F for λ>0 and r>0, respectively. Let 0<k<1 and let g be a k-contraction of H into itself. Let G be a strongly positive bounded linear self-adjoint operator on H with coefficient \(\overline{\gamma}>0\). Let \(0< \gamma <\frac {\overline{\gamma}}{k}\) and suppose (A+B)−10∩F −10≠∅. Assume that {α n }⊂]0,1[, {λ n }⊂]0,∞[ and {r n }⊂]0,∞[ satisfy

Then, the following hold:

  1. (i)

    For sufficiently large n∈ℕ, define T n :HH by

    Then, T n has a unique fixed point x n in H and {x n } is bounded.

  2. (ii)

    For any nonempty closed convex subset D of H, P D (IG+γg) has a unique fixed point z 0 in D. This point z 0D is also a unique solution of the variational inequality

    In particular, the set (A+B)−10∩F −10 is a nonempty, closed and convex subset of H and \(P_{(A+B)^{-1}0 \cap F^{-1}0}(I-G+\gamma g)\) has a unique fixed point z 0 in (A+B)−10∩F −10.

  3. (iii)

    The sequence {x n } converges strongly to z 0∈(A+B)−10∩F −10, where {z 0}=VI((A+B)−10∩F −10,Gγg).

Proof

Let us prove (i). For sufficiently large n∈ℕ, define T n :HH by

From α n →0, we have α n ≤∥G−1. Then we have from Lemma 3.1 that for any x,yH,

Since \(0<1-\alpha_{n} (\overline{\gamma}-\gamma k)<1\), T n is a \((1-\alpha_{n} (\overline{\gamma}-\gamma k))\)-contraction of H into itself and hence T n has a unique fixed point x n in H. Next, we show that {x n } is bounded. Let u∈(A+B)−10∩F −10. Using u=α n Gu+uα n Gu, we have, for all n∈ℕ,

Thus we have \(\alpha_{n}(\overline{\gamma}-\gamma k)\|x_{n}-u\|\leq \alpha_{n}\|\gamma g(u)-Gu\|\) and hence

So, we have \(\|x_{n}-u\|\leq\frac{\|\gamma g(u)-Gu\|}{\overline{\gamma }-\gamma k}\). This implies that {x n } is bounded.

Let us prove (ii). Since g:HH is a k-contraction and G is a strongly positive bounded linear self-adjoint operator on H with coefficient \(\overline{\gamma}>0\), we have, for any x,yH,

Then Gγg:HH is a \((\overline{\gamma} -\gamma k)\)-strongly monotone operator. Furthermore, taking a positive number μ with \(\mu(\|G\|+\gamma k)^{2}<2(\overline{\gamma} -\gamma k)\) and \(2\mu(\overline{\gamma} -\gamma k)<1\), we have, for any x,yH,

and

So, Iμ(Gγg) is a contraction of H into itself and hence P D (Iμ(Gγg)) is also a contraction of D into itself. Thus there exists a unique point z 0D such that z 0=P D (Iμ(Gγg))z 0. We also have, for wD,

Thus VI(D,Gγg)={z 0}. Next, we show that (A+B)−10∩F −10 is closed and convex. Since F is a maximal monotone operator, we know from [6] that F −10 is closed and convex. Furthermore, we know from [16] that for any λ>0,

$$ w\in(A+ B)^{-1} 0\quad \Longleftrightarrow\quad w\in F\bigl(J_{\lambda }(I- \lambda A)\bigr). $$

If 0<λ≤2α, then IλA is nonexpansive and then J λ (IλA) is nonexpansive. Thus F(J λ (IλA)) is closed and convex and so is (A+B)−10. Therefore, (A+B)−10∩F −10 is closed and convex. Thus \(P_{(A+B)^{-1}0 \cap F^{-1}0}(I-G+\gamma g)\) has a unique fixed point z 0 in (A+B)−10∩F −10.

Let us prove (iii). Put \(y_{n}=J_{\lambda_{n}}(I-\lambda_{n} A)T_{r_{n}}x_{n}\) and \(u_{n}=T_{r_{n}}x_{n}\) for all n∈ℕ. Since {x n } is bounded, {y n } and {u n } are bounded. Furthermore, {g(x n )} and {Gx n } are also bounded. Let z∈(A+B)−10∩F −10. We note that

$$ \|y_n-z\|= \big\|J_{\lambda_n}(I- \lambda_n A)u_n-z\big\| \leq\|u_n-z\| $$
(7)

and

(8)

Using (6), we have

and hence

$$ \|u_n-z\|^2\leq\|x_n-z \|^2-\|u_n-x_n\|^2. $$
(9)

Then we have from (1) and (9)

and hence

From α n →0, we have

$$ \|x_n-u_n\|\to0. $$
(10)

Then we have from (8)

$$ \|y_n-u_n\|\to0. $$
(11)

Take λ 0∈[a,2α]. Putting z n =(Iλ n A)u n , we have from Lemma 2.1

(12)

Furthermore, we have

$$ \big\|J_{\lambda_0}(I-\lambda_0 A)u_n-u_n \big\| \leq\big\|J_{\lambda _0}(I-\lambda_0 A)u_n-y_n \big\| + \|y_n-u_n\|. $$
(13)

We will use these inequalities (12) and (13) later. We know from (ii) that there exists a unique z 0∈(A+B)−10∩F −10 such that

In order to show that x n z 0, it suffices to show that if \(\{ x_{n_{i}}\}\) is any subsequence of {x n }, then we can find a subsequence of \(\{x_{n_{i}}\}\) converging strongly to z 0. Since \(\{x_{n_{i}}\}\) is bounded and \(\{\lambda_{n_{i}}\}\subset [a,2\alpha]\), without loss of generality there exist a subsequence \(\{x_{n_{i_{j}}}\}\) of \(\{x_{n_{i}}\}\) and a subsequence \(\{\lambda_{n_{i_{j}}}\}\) of \(\{\lambda_{n_{i}}\}\) such that \(x_{n_{i_{j}}}\rightharpoonup w\) and \(\lambda_{n_{i_{j}}} \to\lambda_{0}\) for some λ 0∈[a,2α]. From x n u n →0, we have \(u_{n_{i_{j}}}\rightharpoonup w\). Since \(\{u_{n_{i_{j}}}\}\subset C\) and C is closed and convex, we have wC. Using \(\lambda_{n_{i_{j}}} \to\lambda_{0}\) and (12), we have

Furthermore we have from \(\|y_{n_{i_{j}}}-u_{n_{i_{j}}}\|\to0\) and (13) that

Since \(J_{\lambda_{0}}(I-\lambda_{0} A)\) is nonexpansive, we have \(w=J_{\lambda_{0}}(I-\lambda_{0} A)w\) and hence w∈(A+B)−10. We show wF −10. Since F is a maximal monotone operator, we have from (5) that \(A_{r_{n_{i_{j}}}}x_{n_{i_{j}}}\in FT_{r_{n_{i_{j}}}}x_{n_{i_{j}}}\), where A r is the Yosida approximation of F for r>0. Furthermore we have, for any (u,v)∈F,

Since lim inf n→∞ r n >0, \(u_{n_{i_{j}}}\rightharpoonup w\) and \(x_{n_{i_{j}}}-u_{n_{i_{j}}}\to0\), we have

Since F is a maximal monotone operator, we have 0∈Fw and hence wF −10. Thus we have w∈(A+B)−10∩F −10. On the other hand, we have

So we have

Then we have

$$ \alpha_n \overline{\gamma} \|x_n-z_0 \|^2\leq\alpha_{n} \bigl\langle \gamma g(x_n)-Gz_0, x_n-z_0\bigr\rangle $$

and hence

This implies that

In particular we have

Since \(x_{n_{i_{j}}}\rightharpoonup w\), w∈(A+B)−10∩F −10 and

we have

Thus \(x_{n_{i_{j}}}\to z_{0}\). Then {x n } converges strongly to z 0∈(A+B)−10∩F −10 such that

We also know that this z 0 is a unique fixed point of \(P_{(A+B)^{-1}0 \cap F^{-1}0}(I-G+\gamma g)\). This completes the proof. □

Compare the proof of Theorem 3.1(ii) with the proof in [4]. We prove simply that P D (IG+γg) has a unique fixed point by using another contraction mapping which is different from P D (IG+γg). Using this result, we prove Theorem 3.1(iii). Next, we prove a strong convergence theorem of Halpern’s type [17] in a Hilbert space; see also [18].

Theorem 3.2

Let H be a real Hilbert space and let C be a nonempty, closed and convex subset of H. Let α>0 and let A be an α-inverse-strongly monotone mapping of C into H and let B be a maximal monotone operator on H. Let F be a maximal monotone operator on H such that the domain of F is included in C. Let J λ =(I+λB)−1 and T r =(I+rF)−1 be the resolvents of B and F for λ>0 and r>0, respectively. Let 0<k<1 and let g be a k-contraction of H into itself. Let G be a strongly positive bounded linear self-adjoint operator on H with coefficient \(\overline{\gamma}>0\). Let \(0< \gamma <\frac {\overline{\gamma}}{k}\) and suppose (A+B)−10∩F −10≠∅. Let x 1=xH and let {x n }⊂H be a sequence generated by

for all n∈ℕ, where {α n }⊂]0,1[, {λ n }⊂]0,∞[ and {r n }⊂]0,∞[ satisfy

Then {x n } converges strongly to a point z 0 of (A+B)−10∩F −10, where z 0∈(A+B)−10∩F −10 is a unique fixed point of \(P_{(A+B)^{-1}0\cap F^{-1}0}(I-G+\gamma g)\). This point z 0∈(A+B)−10∩F −10 is also a unique solution of the variational inequality

Proof

Put \(u_{n}=T_{r_{n}}x_{n}\) and \(y_{n}=J_{\lambda_{n}}(I-\lambda_{n}A)T_{r_{n}}x_{n}\) for all n∈ℕ. Let z∈(A+B)−10∩F −10. Then, we have from \(z=T_{r_{n}}z\) and \(z=J_{\lambda_{n}}(I-\lambda_{n}A)z\) that

(14)

Since x n+1=α n γg(x n )+(Iα n G)y n and z=α n Gz+zα n Gz, we have

Putting

$$K=\max\biggl\{\frac{\lVert \gamma g(z)-Gz\rVert }{\overline{\gamma}- \gamma k},\|x_1-z\|\biggr\},$$

we have ∥x n z∥≤K for all n∈ℕ. Then {x n } is bounded. Furthermore, {u n } and {y n } are bounded. Since

we have

where M 1=sup{γg(x n )∥+∥Gy n ∥:n∈ℕ}. Putting \(z_{n}=(I-\lambda_{n} A)T_{r_{n}}x_{n}\), we have from Lemma 2.1 that

where

$$M_2=\sup\biggl\{\frac{1}{b}\|T_{r_{n+1}}x_{n}-x_{n}\|+\frac{1}{a}\| J_{\lambda_{n+1}}z_{n}-z_n\|+\| AT_{r_{n}}x_{n}\|:n\in\mathbb{N}\biggr\}$$

and 0<br n for all n∈ℕ. Thus we have

where M 3=2M 1+2M 2. Using Lemma 2.2, we obtain

$$ \|x_{n+2}-x_{n+1}\|\to0. $$
(15)

We also have from x n+1=α n γg(x n )+(Iα n G)y n that

Since α n →0 and ∥x n+1x n ∥→0, we get

$$ y_n-x_n\to0. $$
(16)

As in the proof of Theorem 3.1, we have

$$ \|u_n-z\|^2\leq\|x_n-z \|^2-\|u_n-x_n\|^2. $$
(17)

Then we have from (1) and (17) that

and hence

From α n →0 and ∥x n+1x n ∥→0, we have

$$ \|x_n-u_n\|\to0. $$
(18)

Then we have from (16) and (18) that

$$ \|y_n-u_n\|\leq\|y_n-x_n \|+\|x_n-u_n\|\to0. $$
(19)

From \(\sum_{n=1}^{\infty}|\lambda_{n}-\lambda_{n+1}|<\infty\), we find that {λ n } is a Cauchy sequence. So, we have λ n λ 0∈[a,2α]. Since \(u_{n}=T_{r_{n}}x_{n}\), z n =(Iλ n A)u n and \(y_{n}=J_{\lambda_{n}}(I-\lambda_{n}A)T_{r_{n}}x_{n}\), we have from Lemma 2.1 that

(20)

We also have from (19) and (20) that

$$ \big\|u_n-J_{\lambda_{0}}(I-\lambda_{0}A)u_{n} \big\|\leq\|u_n-y_n\|+\big\| y_n-J_{\lambda_{0}}(I- \lambda_{0}A)u_{n}\big\|\to0. $$
(21)

We will use (20) and (21) later. From Theorem 3.1, we can take a unique solution z 0∈(A+B)−10∩F −10 of the variational inequality

We show that lim sup n→∞〈(Gγg)z 0,x n z 0〉≥0. Put

Without loss of generality, there exists a subsequence \(\{x_{n_{i}}\}\) of {x n } such that \(l=\lim_{i\to\infty} \langle (G-\gamma g)z_{0}, x_{n_{i}}-z_{0} \rangle \) and \(\{x_{n_{i}}\}\) converges weakly to some point wH. From ∥x n u n ∥→0, we also find that \(\{u_{n_{i}}\}\) converges weakly to wC. On the other hand, from λ n λ 0∈[a,2α] we have \(\lambda_{n_{i}} \to\lambda_{0}\in[a, 2\alpha]\). Using (20), we have

Furthermore, from (21) we have

Since \(J_{\lambda_{0}}(I-\lambda_{0}A)\) is a nonexpansive mapping of C into H, we have from [19, Lemma 4.1] that \(w=J_{\lambda_{0}}(I-\lambda_{0}A)w\). This means that 0∈Aw+Bw. As in the proof of Theorem 3.1, we can also show wF −10. Thus we have w∈(A+B)−10∩F −10. So, we have

Since x n+1z 0=α n (γg(x n )−Gz 0)+(Iα n G)(y n z 0), we find from (1) that

and hence

(22)

where \(\beta_{n}=\frac{2(\overline{\gamma}-\gamma k)\alpha_{n}}{1-\alpha_{n}\gamma k}\). Since \(\sum^{\infty}_{n=1}\beta_{n}=\infty\), we have from Lemma 2.2 and (22) that x n z 0, where \(z_{0}=P_{(A+B)^{-1}0 \cap F^{-1}0}(I-G+\gamma g)z_{0}\). This completes the proof. □

4 Applications

In this section, using Theorem 3.2, we obtain new strong convergence theorems for in a Hilbert space. Let H be a Hilbert space and let f be a proper lower semicontinuous convex function of H into ]−∞,∞]. Then, the subdifferential ∂f of f is defined as follows:

for all xH. From Rockafellar [20], we know that ∂f is a maximal monotone operator. Let C be a nonempty, closed and convex subset of H and let i C be the indicator function of C. Then i C is a proper lower semicontinuous and convex function on H. So, we can define the resolvent J λ of ∂i C for λ>0, i.e.,

for all xH. We have, for any λ>0, xH and uC,

where N C u is the normal cone to C at u, i.e.,

Let f:C×C→ℝ be a bifunction and let A be a mapping of C into H. A generalized equilibrium problem (with respect to C) is to find \(\hat{x}\in C\) such that

(23)

The set of such solutions \(\hat{x}\) is denoted by EP(f,A), i.e.,

In the case of A=0, EP(f,A) is denoted by EP(f). In the case of f=0, EP(f,A) is also denoted by VI(C,A). This is the set of solutions of the variational inequality for A.

Using Theorem 3.2, we prove a strong convergence theorem for inverse-strongly monotone operators in a Hilbert space.

Theorem 4.1

Let H be a real Hilbert space and let C be a nonempty, closed and convex subset of H. Let α>0 and let A be an α-inverse-strongly monotone mapping of C into H. Let 0<k<1 and let g be a k-contraction of H into itself and let G be a strongly positive bounded linear self-adjoint operator on H with coefficient \(\overline{\gamma}>0\). Take γ with \(0< \gamma <\frac{\overline{\gamma}}{k}\) and suppose VI(C,A)≠∅. Let x 1=xH and let {x n }⊂H be a sequence generated by

for all n∈ℕ, where {α n }⊂]0,1[ and {λ n }⊂]0,∞[ satisfy

Then, the sequence {x n } converges strongly to a point z 0 of VI(C,A), where z 0VI(C,A) is a unique fixed point of P VI(C,A)(IG+γg). This point z 0VI(C,A) is also a unique solution of the variational inequality

Proof

Put B=F=∂i C in Theorem 3.2. Then, we have, for λ n >0 and r n >0,

Furthermore, we have (∂i C )−10=C and (A+∂i C )−10=VI(C,A). In fact, we have, for zC,

Thus we obtain the desired result by Theorem 3.2. □

Let C be a nonempty, closed and convex subset of H. Then, U:CH is called a widely strict pseudo-contraction iff there exists r∈ℝ with r<1 such that

We call such U a widely r-strict pseudo-contraction. If 0≤r<1, then U is a strict pseudo-contraction [1]. Furthermore, if r=0, then U is nonexpansive. Conversely, let T:CH be a nonexpansive mapping and define U:CH by \(U=\frac{1}{1+n}T+\frac{n}{1+n}I\) for all xC and n∈ℕ. Then U is a widely (−n)-strict pseudo-contraction. In fact, from the definition of U, it follows that T=(1+n)UnI. Since T is nonexpansive, we have, for any x,yC,

and hence

Using Theorem 3.2, we obtain an extension of Zhou’s strong convergence theorem [21] in a Hilbert space.

Theorem 4.2

Let H be a real Hilbert space and let C be a nonempty, closed and convex subset of H. Let r∈ℝ with r<1 and let U be a widely r-strict pseudo-contraction of C into H such that F(U)≠∅. Let uC, x 1=xC and let {x n }⊂C be a sequence generated by

for all n∈ℕ, where {α n }⊂]0,1[ and {t n }⊂]−∞,1[ satisfy

Then, the sequence {x n } converges strongly to a point z 0 of F(U), where z 0=P F(U) u.

Proof

Put B=F=∂i C and A=IU in Theorem 3.2. Furthermore, put g(x)=u and G(x)=x for all xH. Then, we can take \(\overline{\gamma}=\frac{1}{2}\). Since \(\| g(x)-g(y)\|=0\leq\frac{1}{3}\|x-y\|\) for all x,yH, we can take \(k=\frac{1}{3}\) and hence set γ=1. Putting a=1−b,λ n =1−t n and 2α=1−r in Theorem 3.2, we get from rt n b<1 that 0<aλ n ≤2α,

and

Furthermore, we have, for zC,

Since F(U)≠∅, we find, as in the proof of [21, Fact 3], that F(P C U)=F(U). We also have z 0=P F(U)(IG+γg)z 0=P F(U)(z 0z 0+1⋅u)=P F(U) u. Thus, we obtain the desired result by Theorem 3.2. □

For solving the equilibrium problem, let us assume that the bifunction f:C×C→ℝ satisfies the following conditions:

  1. (A1)

    f(x,x)=0 for all xC;

  2. (A2)

    f is monotone, i.e., f(x,y)+f(y,x)≤0 for all x,yC;

  3. (A3)

    for all x,y,zC,

    $$ \limsup_{t\downarrow0}f\bigl(tz+(1-t)x,y\bigr)\leq f(x,y); $$
  4. (A4)

    for all xC, f(x,⋅) is convex and lower semicontinuous.

Then, we know the following lemma which appears implicitly in Blum and Oettli [22].

Lemma 4.1

(Blum and Oettli)

Let C be a nonempty, closed and convex subset of H and let f be a bifunction of C×C intosatisfying (A1)(A4). Let r>0 and xH. Then, there exists zC such that

$$ f(z,y)+\frac{1}{r}\langle y-z,z-x\rangle\geq0,\quad \forall y\in C. $$

The following lemma was also given in Combettes and Hirstoaga [23].

Lemma 4.2

Assume that f:C×C→ℝ satisfies (A1)(A4). For r>0 and xH, define a mapping T r :HC as follows:

$$ T_rx:= \biggl\{z\in C:f(z,y)+\frac{1}{r}\langle y-z, z-x \rangle\geq 0,\ \forall y\in C \biggr\} $$

for all xH. Then, the following hold:

  1. (1)

    T r is single-valued;

  2. (2)

    T r is a firmly nonexpansive mapping, i.e., for all x,yH,

    $$ \| T_rx-T_ry\|^2\leq\langle T_rx-T_ry,x-y\rangle; $$
  3. (3)

    F(T r )=EP(f);

  4. (4)

    EP(f) is closed and convex.

We call such T r the resolvent of f for r>0. Using Lemmas 4.1 and 4.2, Takahashi, Takahashi and Toyoda [5] obtained the following lemma. See [24] for a more general result.

Lemma 4.3

Let H be a Hilbert space and let C be a nonempty, closed and convex subset of H. Let f:C×C→ℝ satisfy (A1)(A4). Let A f be a set-valued mapping of H into itself defined by

Then, \(\mathit{EP}(f)=A_{f}^{-1}0\) and A f is a maximal monotone operator such that \(\operatorname {dom}(A_{f}) \subset C\). Furthermore, for any xH and r>0, the resolvent T r of f coincides with the resolvent of A f , i.e.,

Using Lemma 4.3 and Theorem 3.2, we obtain the following result which generalizes Liu’s strong convergence theorem [4].

Theorem 4.3

Let H be a real Hilbert space and let C be a nonempty, closed and convex subset of H. Let r∈ℝ with r<1 and let U be a widely r-strict pseudo-contraction of C into H and let f be a bifunction of C×C intosatisfying (A1)(A4). Let T r be the resolvent of f for r>0. Let 0<k<1 and let g be a k-contraction of H into itself. Let G be a strongly positive bounded linear self-adjoint operator on H with coefficient \(\overline{\gamma}>0\). Let \(0< \gamma <\frac {\overline{\gamma}}{k}\) and suppose F(U)∩EP(f)≠∅. Let x 1=xH and let {x n }⊂H be a sequence generated by

for all n∈ℕ, where {α n }⊂]0,1[, {t n }⊂]−∞,1[ and {r n }⊂]0,∞[ satisfy

Then, the sequence {x n } converges strongly to a point z 0 of F(U)∩EP(f), where z 0F(U)∩EP(f) is a unique fixed point of P F(U)∩EP(f)(IG+γg). This point z 0F(U)∩EP(f) is also a unique solution of the variational inequality

Proof

For the bifunction f:C×C→ℝ, we can define A f in Lemma 4.3. Putting A=IU, Bx=0 for all ∈H and F=A f in Theorem 3.2, we obtain from Lemma 4.3 that \(J_{\lambda_{n}} = I\) for all λ n >0 and \(T_{r_{n}} = (I+r_{n}A_{f})^{-1}\) for all r n >0. As in the proof of Theorem 4.2, the sequence {t n } and U are changed in {λ n } and A. We have also from Lemma 4.3 that

Furthermore, we have, for zC,

$$ z \in(A+ B)^{-1}0 \quad \Longleftrightarrow\quad z\in F(U). $$

So, we obtain the desired result by Theorem 3.2. □

Remark 4.1

We note that two assumptions 0≤r and lim n→∞ t n =b in Liu’s theorem [4] do not appear in Theorem 4.3.

5 Concluding Remarks

(1) We cannot directly prove that the mapping P D (IG+rg) in Theorem 3.1 is a contraction. We proved that the mapping has a unique fixed point by using another contraction which is different from the mapping. Then we showed two strong convergence theorems (Theorems 3.1 and 3.2) by using this result. It seems that such methods are new.

(2) The domain of the maximal monotone operator A f in Lemma 4.3, which is deduced from an equilibrium problem, is included in C. Thus the maximal monotone operator F in Theorems 3.1 and 3.2 covers the equilibrium problem. Our methods for the resolvents of the maximal monotone operator F, which are used in the proofs of Theorems 3.1 and 3.2, are more general than methods for solving the equilibrium problem.

(3) Since the class of inverse-strongly monotone mappings contains strict pseudo-contractions, our two theorems are general and useful.

(4) For the fixed point problem of nonself-mappings, we use generally the metric projections. For such a problem, we used the resolvents of a maximal monotone operator B in Theorems 3.1 and 3.2. Consequently, we solve the problem of finding a zero point of the addition of an inverse-strongly monotone mapping and a maximal monotone operator.

(5) Our results (Theorems 3.1 and 3.2) are also used for finding a common fixed point of two commuting nonexpansive mappings defined on a bounded, closed and convex subset of a Hilbert space.