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Strong convergence theorems for equilibrium problems and fixed point problems: A new iterative method, some comments and applications
Fixed Point Theory and Applications volume 2011, Article number: 33 (2011)
Abstract
In this paper, we introduce a new approach method to find a common element in the intersection of the set of the solutions of a finite family of equilibrium problems and the set of fixed points of a nonexpansive mapping in a real Hilbert space. Under appropriate conditions, some strong convergence theorems are established. The results obtained in this paper are new, and a few examples illustrating these results are given. Finally, we point out that some 'socalled' mixed equilibrium problems and generalized equilibrium problems in the literature are still usual equilibrium problems.
2010 Mathematics Subject Classification: 47H09; 47H10, 47J25.
1 Introduction and preliminaries
Throughout this paper, we assume that H is a real Hilbert space with zero vector θ, whose inner product and norm are denoted by 〈·, ·〉 and  · , respectively. The symbols ℕ and ℝ are used to denote the sets of positive integers and real numbers, respectively. Let K be a nonempty closed convex subset of H and T : K → H be a mapping. In this paper, the set of fixed points of T is denoted by F(T). We use symbols → and ⇀ to denote strong and weak convergence, respectively.
For each point x ∈ H, there exists a unique nearest point in K, denoted by P_{ K }x, such that
The mapping P_{ K } is called the metric projection from H onto K. It is well known that P_{ K } satisfies
for every x, y ∈ H. Moreover, P_{ K }x is characterized by the properties: for x ∈ H, and z ∈ K,
Let f be a bifunction from K × K into ℝ. The classical equilibrium problem is to find x ∈ K such that
Let EP(f) denote the set of all solutions of the problem (1.1). Since several problems in physics, optimization, and economics reduce to find a solution of (1.1) (see, e.g., [1, 2]), some authors had proposed some methods to find the solution of equilibrium problem (1.1); for instance, see [1–4]. We know that a mapping S is said to be nonexpansive mapping if for all x, y ∈ K, Sx  Sy ≤ x  y. Recently, some authors used iterative method including composite iterative, CQ iterative, viscosity iterative etc. to find a common element in the intersection of EP(f) and F(S); see, e.g., [5–11].
Let I be an index set. For each i ∈ I, let f_{ i } be a bifunction from K × K into ℝ. The system of equilibrium problem is to find x ∈ K such that
We know that is the set of all solutions of the system of equilibrium problem (1.2).
For each i ∈ I, if f_{ i } (x, y) = 〈A_{ i }x, y  x〉, where A_{ i } : K → K is a nonlinear operator, then the problem (1.2) becomes the following system of variational inequality problem:
It is obvious that the problem (1.3) is a special case of the problem (1.2).
The following Lemmas are crucial to our main results.
Lemma 1.1 (Demicloseness principle[12]) Let H be a real Hilbert space and K a closed convex subset of H. S : K → H is a nonexpansive mapping. Then the mapping I  S is demiclosed on K, where I is the identity mapping, i.e., x_{ n } ⇀ x in K and (I  S)x_{ n } → y implies that × ∈ K and (I  S)x = y.
Lemma 1.2[13] Let {x_{ n } }and {y_{ n } } be bounded sequences in a Banach space E and let {β_{ n } } be a sequence in [0,1] with 0 < lim inf_{n→∞}β_{ n }≤ lim sup_{n→∞}β_{ n }< 1. Suppose x_{n+1}= β_{ n }y_{ n } + (1  β_{ n } )x_{ n } for all integers n ≥ 0 and lim sup_{n→∞}(y_{n+1} y_{ n }   x_{n+1} x_{ n } ) ≤ 0, then lim_{n→∞}y_{ n }  x_{ n }  = 0.
Lemma 1.3[5] Let H be a real Hilbert space. Then the following hold.

(a)
x + y^{2} ≤ y^{2} + 2〈x, x + y〉 for all x, y ∈ H;

(b)
αx + (1  α)y^{2} = αx^{2} + (1  α) y^{2}  α(1  α) x  y^{2} for all x, y ∈ H and α ∈ ℝ;

(c)
x  y^{2} = x^{2} + y^{2}  2 〈x, y〉 for all x, y ∈ H.
Lemma 1.4. [14] Let {a_{ n } } be a sequence of nonnegative real numbers satisfying the following relation:
If

(i)
λ_{ n } ∈ [0,1], or, equivalently, ;

(ii)
or ,
then .
Lemma 1.5[1] Let K be a nonempty closed convex subset of H and F be a bifunction of K × K into ℝ satisfying the following conditions.
(A1) F(x, x) = 0 for all × ∈ K;
(A2) F is monotone, that is, F(x, y) + F(y, x) ≤ 0 for all x, y ∈ K;
(A3) for each x, y, z ∈ K,
(A4) for each × ∈ K, y → F(x, y) is convex and lower semicontinuous.
Let r > 0 and × ∈ H. Then, there exists z ∈ K such that
Lemma 1.6[3] Let K be a nonempty closed convex subset of H and let F be a bifunction of K × K into R satisfying (A1)  (A4). For r > 0 and × ∈ H, define a mapping T_{ r } : H → K as follows:
for all × ∈ H. Then the following hold:

(i)
T _{ r } is singlevalued;

(ii)
T_{ r } is firmly nonexpansive, that is, for any x, y ∈ H,

(iii)
F(T_{ r } ) = EP (F);

(iv)
EP(F) is closed and convex.
2 Main results and their applications
Let I = {1, 2,..., k} be a finite index set, where k ∈ ℕ. For each i ∈ I, let f_{ i } be a bifunctions from K × K into ℝ satisfying the conditions (A1)(A4). Denote by
For each (i, n) ∈ I × ℕ, applying Lemmas 1.5 and 1.6, is a firmly nonexpansive singlevalued mapping such that is closed and convex. For each i ∈ I, let , n ∈ ℕ.
First, let us consider the following example.
Example A Let f_{ i } : [1, 0]×[1,0] → ℝ be defined by f_{ i } (x, y) = (1+x^{2i})(x  y), i = 1, 2, 3. It is easy to see that for any i ∈ {1, 2, 3}, f_{ i } (x, y) satisfies the conditions (A1)(A4) and . Let Sx = x^{3} and , ∀ x ∈ [1, 0] Then g is a contraction from K into itself and S : K → K is a nonexpansive mapping with . Let λ ∈ (0, 1), {r_{ n } } ⊂ [1, + ∞) and {α_{ n } } ⊂ (0,1) satisfy the conditions (i) lim_{n→∞}α_{ n }= 0, and (ii) , or equivalently, ; e.g., let , {α_{ n } } ⊂ (0, 1) and {r_{ n } } ⊂ [1, + ∞) be given by
Define a sequence {x_{ n } } by
Then the sequences {x_{ n } } and , i = 1, 2, 3, defined by (2.1) all strongly converge to 0.
Proof

(a)
By Lemmas 1.5 and 1.6, (2.1) is well defined.

(b)
Let K = [1, 0]. For each i ∈ {1, 2, 3}, define
We claim that for each v ∈ K and any i ∈ {1, 2, 3}, there exists a unique z = 0 ∈ K such that
or, equivalently,
Obviously, z = 0 is a solution of the problem . On the other hand, there does not exist z ∈ [1, 0) such that z  y ≤ 0 and . So z = 0 is the unique solution of the problem .

(c)
We notice that (2.1) is equivalent with (2.2), where
(2.2)
It is easy to see that {x_{ n } } ⊂ [1, 0], so, by (b), for all n ∈ ℕ. We need to prove x_{ n } → 0 as n → ∞. Since z_{ n } = 0 for all n ∈ ℕ, we have y_{ n } = (1 λ)x_{ n } and
for all n ∈ ℕ. For any n ∈ ℕ, from (2.3), we have
Hence {x_{ n }} is a strictly deceasing sequence and x_{ n } ≥ 0 for all n ∈ ℕ. So exists.
On the other hand, for any n, m ∈ ℕ with n > m, using (2.4), we obtain
which implies . Therefore {x_{ n } } strongly converges to 0. □
In this paper, motivated by the preceding Example A, we introduce a new iterative algorithm for the problem of finding a common element in the set of solutions to the system of equilibrium problem and the set of fixed points of a nonexpansive mapping. The following new strong convergence theorem is established in the framework of a real Hilbert space H.
Theorem 2.1 Let K be a nonempty closed convex subset of a real Hilbert space H and I = {1, 2,..., k} be a finite index set. For each i ∈ I, let f_{ i } be a bifunction from K × K into ℝ satisfying (A1)(A4). Let S : K → K be a nonexpansive mapping with . Let λ, ρ ∈ (0, 1) and g : K → K is a ρcontraction. Let {x_{ n } } be a sequence generated in the following manner:
If the above control coefficient sequences {α_{ n } } ⊂ (0, 1) and {r_{ n } } ⊂ (0, +∞) satisfy the following restrictions:
(D1) , and ;
(D2) and.
then the sequences {x_{ n } } and, for all i ∈ I, converge strongly to an element c = P_{Ω}g(c) ∈ Ω. The following conclusion is immediately drawn from Theorem 2.1.
Corollary 2.1 Let K be a nonempty closed convex subset of a real Hilbert space H. Let f be a bifunction from K × K into ℝ satisfying (A1)(A4) and S : K → K be a nonexpansive mapping with Ω = EP(f) ∩F(S) ≠ ∅. Let λ, ρ ∈ (0,1) and g : K → K is a ρcontraction. Let {x_{ n } } be a sequence generated in the following manner:
If the above control coefficient sequences {α_{ n } } ⊂ (0, 1) and {r_{ n } } ⊂ (0, +∞) satisfy all the restrictions in Theorem 2.1, then the sequences {x_{ n } } and {u_{ n } } converge strongly to an element c = P_{Ω}g(c) ∈ Ω, respectively.
If f_{ i } (x, y) ≡ 0 for all (x, y) ∈ K × K in Theorem 2.1 and all i ∈ I, then, from the algorithm (D_{ H } ), we obtain , ∀ i ∈ I. So we have the following result.
Corollary 2.2 Let K be a nonempty closed convex subset of a real Hilbert space H. Let S : K → K be a nonexpansive mapping with F(S) ≠ ∅. Let λ, ρ ∈ (0, 1) and g : K → K is a ρcontraction. Let {x_{ n } } be a sequence generated in the following manner:
If the above control coefficient sequences {α_{ n } } ⊂ (0, 1) satisfy, and, then the sequences {x_{ n } } converge strongly to an element c = P_{Ω}g(c) ∈ F (S).
As some interesting and important applications of Theorem 2.1 for optimization problems and fixed point problems, we have the following.
Application (I) of Theorem 2.1 We will give an iterative algorithm for the following optimization problem with a nonempty common solution set:
where h_{ i } (x), i ∈ {1, 2,..., k}, are convex and lower semicontinuous functions defined on a closed convex subset K of a Hilbert space H (for example, h_{ i } (x) = x^{i} , x ∈ K := [0, 1], i ∈ {1, 2,..., k}).
If we put f_{ i } (x, y) = h_{ i } (y)  h_{ i } (x), i ∈ {1, 2,..., k}, then is the common solution set of the problem (OP), where denote the common solution set of the following equilibrium:
For i ∈ {1, 2,..., k}, it is obvious that the f_{ i } (x, y) satisfies the conditions (A1)(A4). Let S = I (identity mapping), then from (D_{ H } ), we have the following algorithm
where x_{1} ∈ K, λ ∈ (0, 1), g : K → K is a ρcontraction. From Theorem 2.1, we know that {x_{ n } } and , i ∈{1,2,..., k}, generated by (2.5), strongly converge to an element of if the coefficients {α_{ n } } and {r_{ n } } satisfy the conditions of Theorem 2.1.
Application (II) of Theorem 2.1 Let H, K, I, λ, ρ, g be the same as Theorem 2.1. Let A_{1}, A_{2},..., A_{ k } : K → K be k nonlinear mappings with . For any i ∈ I, put f_{ i } (x, y) = 〈x  A_{ i }x, y  x〉, ∀ x, y ∈ K. Since , we have . Let S = I (identity mapping) in the algorithm (D_{ H } ). Then the sequences {x_{ n } } and , defined by the algorithm (D_{ H } ), converge strongly to a common fixed point of {A_{1}, A_{2},..., A_{ k } }, respectively.
The following result is important in this paper.
Lemma 2.1 Let H be a real Hilbert space. Then for any x_{1}, x_{2},... x_{ k } ∈ H and a_{1}, a_{2},..., a_{ k } ∈ [0,1] with, k ∈ ℕ, we have
Proof It is obvious that (2.6) is true if a_{ j } = 1 for some j, so it suffices to show that (2.6) is true for a_{ j } ≠ 1 for all j. The proof is by mathematic induction on k. Clearly, (2.6) is true for k = 1. Let x_{1}, x_{2} ∈ H and a_{1}, a_{2} ∈ [0,1] with a_{1} + a_{2} = 1. By Lemma 1.3, we obtain
which means that (2.6) hold for k = 2. Suppose that (2.6) is true for k = l ∈ ℕ. Let x_{1}, x_{2},..., x_{ l } , x_{l+1}∈ H and a_{1}, a_{2},..., a_{ l }, a_{l+1}∈ [0, 1) with . Let . Then applying the induction hypothesis we have
Hence, the equality (2.6) is also true for k = l + 1. This completes the induction. □
3 Proof of Theorem 2.1
We will proceed with the following steps.
Step 1: There exists a unique c ∈ Ω ⊂ H such that P_{Ω}g(c) = c.
Since P_{Ω}g is a ρcontraction on H, Banach contraction principle ensures that there exists a unique c ∈ H such that c = P_{Ω}g(c) ∈ Ω.
Step 2: We prove that the sequences {x_{ n } }, {y_{ n } }, {z_{ n } } and , ∀i ∈ I, are all bounded.
First, we notice that (D_{ H } ) is equivalent with (Z_{ H } ), where
For each i ∈ I, we have
For any n ∈ ℕ, from (Z_{ H } ) we have
and
Since g is a ρcontraction, it follows from (3.2) that
By induction, we obtain
which shows that {x_{ n } } is bounded. Also, we know that {y_{ n } }, {z_{ n } } and , ∀i ∈ I, are all
bounded.
Step 3: We prove lim_{n→∞}x_{n+1} x_{ n }  = 0.
For each i ∈ I, since , , from (Z_{ H } ), we have
and
By (3.3) and (3.4) and (A2),
which implies
It follows from (3.5) that
Let . For any n ∈ ℕ, since , by (3.6), we have
Set
where β_{ n } = 1  (1  λ)(1  α_{ n } ), n ∈ ℕ. Then for each n ∈ ℕ,
and
For any n ∈ ℕ, since
by (3.7), it follows that
From this and (D1), (D2), we get
By Lemma 1.2 and (3.11),
Owing to (3.9) and (3.12), we obtain
Step 4: We show .
By (3.6), (3.13) and (D2), we have
From (Z_{ H } ), we get
Since x_{ n }  y_{ n }  ≤ x_{ n }  x_{n+1} + x_{n+1} y_{ n } , by (3.13) and (3.14),
which implies that
By Lemma 1.6,
which yields that
From (3.15) and Lemma 2.1,
Since
where
We have
Letting n → ∞ in the inequality (3.16), we obtain
Furthermore, it is easy to prove that
For any i ∈ I, since
it implies
Step 5: Prove lim sup_{n→∞}〈g(c)  q, x_{ n }  c〉 ≤ 0.
Take a subsequence of {x_{ n } } such that
Since is bounded, there exists a subsequence of which is still denoted by such that as ℓ → ∞. Notice that for each i ∈ I, by (3.17), so we also have as ℓ → ∞, ∀ i ∈ I.
We want to show z ∈ Ω. First, we show that z ∈ F(S). In fact, since and as ℓ → ∞, by Lemma 1.1, we have (I  S)z = θ or, equivalently, z ∈ F(S).
For each i ∈ I, since , ∀ y ∈ K, it follows from (A2) that
and hence
Applying (3.17) and (A4),
Let y ∈ K be given. Put y_{ t } = ty + (1  t)z, t ∈ (0, 1). Then y_{ t } ∈ K and f_{ i } (y_{ t } , z) ≤ 0 for all i ∈ I. By (A1) and (A4), we get
For any i ∈ I, by (A3), we have
Hence, from (3.21), . Therefore, we proved . On the other hand, by (3.19), we obtain
Step 6: Finally, we prove {x_{ n } } and , for all i ∈ I, converge strongly to c = P_{Ω}g(c) ∈ Ω.
From (Z_{ H } ) and (a) of Lemma 1.3, we have
For any n ∈ ℤ, let
and
From (3.23), we have
It is easy to verify that all conditions of Lemma 1.4 are satisfied. Hence, applying Lemma 1.4, we obtain lim_{n→∞}a_{ n }= 0 which implies
or equivalence, {x_{ n } } strongly converges to c. By (3.17), we can prove that for any i ∈ I, strongly converges to c. The proof of Theorem 2.1 is completed. □
4 Further remarks
Let K be a nonempty closed convex subset of H and f be a bifunction of K × K into ℝ.
Remark 4.1 Recently, some authors introduced the following mixed equilibrium problem (MEP, for short) (see [15–17] and references therein) and generalized equilibrium problem (GEP, for short) (see [18–20] and references therein):
where φ : C → ℝ is a realvalued function.
where A : C → H is a nonlinear operator.
In [15–17], the authors gave some iterative methods for finding the solution of MEP when the bifunction f(x, y) admits the conditions (A1)(A4) and the realvalued function φ satisfies the following condition:
(A5) φ : C → ℝ is a proper lower semicontinuous and convex function.
However, in this case, we argue that the problem MEP is still the equilibrium problem (1.1). In fact, if we put f_{1}(x, y) = f(x, y), f_{2}(x, y) = φ(y)  φ(x) and F(x, y) = f_{1}(x, y) + f_{2}(x, y) for each (x, y) ∈ C × C, then f_{1}(x, y) satisfies the conditions (A1)(A4), f_{2}(x, y) satisfies the condition (A5) and the function φ must satisfy the conditions (A1)(A4). This shows that for each (x, y) ∈ C × C, F(x, y) satisfies the conditions (A1)(A4). So, when we study the solution of MEP, we only need to study the solution of the equilibrium (1.1). This also shows that some "socalled" mixed equilibrium problem studied in [15–17] is still the equilibrium problem (1.1).
Remark 4. 2 Let us recall some wellknown definitions. A mapping T : C → C is said to be

(1)
vexpansive if Tx  Ty ≥ vx  y for all x, y ∈ C. In particular, if v = 1, then T is called expansive.

(2)
vstrongly monotone if there exists a constant v > 0 such that
Clearly, any vstrongly monotone mapping is vexpansive.

(3)
uinverse strongly monotone if there exists a constant u > 0 such that

(4)
LLipschitz continuous if Tx  Ty ≤ Lx  y for all x, y ∈ C. In particular, if L = 1, then T is called nonexpansive.
It is easy to see that a uinverse strongly monotone operator is Lipschitz continuous.
For the problem GEP, if the nonlinear operator A : C → H is a uinverse strongly monotone operator and the bifunction f(x, y) admits the conditions (A1)(A4), we argue that the problem GEP is still the problem (1.1) and so it is indeed not a generalization. In fact, if A is a uinverse strongly monotone operator from C into H, then A is a continuous operator. So, we obtain easily that the function (x, y) → <Ax, y  x〉, ∀x, y ∈ C, satisfies the conditions (A1)(A4). Hence, if we put F(x, y) = f(x, y) + 〈Ax, y  x〉 ≥ 0, then the problem GEP studied in [18–20] is still the problem (1.1).
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Acknowledgements
Zhenhua He was supported by the Natural Science Foundation of Yunnan Province (2010ZC152) and the Scientific Research Foundation from Yunnan Province Education Committee (08Y0338); WeiShih Du was supported by the National Science Council of the Republic of China.
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He, Z., Du, WS. Strong convergence theorems for equilibrium problems and fixed point problems: A new iterative method, some comments and applications. Fixed Point Theory Appl 2011, 33 (2011). https://doi.org/10.1186/16871812201133
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DOI: https://doi.org/10.1186/16871812201133