In this paper, we introduce a new extragradient algorithm by using generalized metric projection. We prove a strong convergence theorem for finding a common element of the solution set of split feasibility problem and the set of fixed points of relatively nonexpansive mapping and a finite family of resolvent operator and the set of solutions of an equilibrium problem.

1. Introduction

Let C be a nonempty closed convex subset of a real Banach space X with norm ‖.‖ and X^∗ be the dual of X. We consider the following variational inequality problem (VI), which consists in finding a point x ∈ C such that

(1)

where

is a mapping and 〈..〉 denotes the duality pairing. The solution set of the variational inequality problem is denoted by VI(C, A).

The operator

is called

(i)
Monotone if

(2)

(ii)
α-inverse strongly monotone if there exists a constant α > 0 such that

(3)

(iii)
Demiclosed if for all {x_n} ⊂ X with x_n⇀x in X, and y_n ∈ Ax_n with y_n⟶y in X^∗, we have x ∈ X and y ∈ Ax

A monotone mapping B is said to be maximal if its graph G(B) = {(x, Bx): x ∈ D(B)} is not properly contained in the graph of any other monotone mapping. Obviously, the monotone mapping B is maximal if and only if for (x, x^∗) ∈ X × X^∗, 〈x − y, x^∗ − y^∗〉 ≥ 0 for all (y, y^∗) ∈ G(B), then it is implied that x^∗ ∈ Bx.

Assume that

is a nonlinear mapping and f : C × C⟶ℝ is a bifunction. The equilibrium problem (EP) is as follows: find x ∈ C such that

(4)

The solution set of (4) is denoted by EP(f). The equilibrium problem is very general because it includes many well-known problems such as variational inequality problems and saddle point problems (see [1–4]). Several methods have been proposed to solve the equilibrium problem in Hilbert space (see [5]), and some authors obtained weak and strong convergence algorithms for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space (see [6–9]). Then, the authors proved the strong convergence of the algorithms in a uniformly convex and uniformly smooth Banach space (see [10]).

Suppose that C and D are nonempty, closed, and convex subsets of real Banach spaces X₁ and X₂, respectively. The split feasibility problem (SFP) is to find a point

(5)

which A : X₁⟶X₂ is a bounded linear operator. The solution set of (5) is denoted by Ω.

In 1994, the split feasibility problem was first studied by Censor and Elfving [11] in finite dimensional Hilbert spaces. In solving (SFP), Schöpfer et al. [12] proposed the next algorithm in p-uniformly convex real Banach spaces: x₁ ∈ X₁ is chosen arbitrarily and for n ≥ 1,

(6)

where J is the duality mapping, Π_C denotes the Bregman projection, A is a bounded linear operator, and A^∗ is the adjoint of A. Also, they have proven the generated sequence {x_n} by algorithm (6) is weakly convergent under suitable conditions. The split feasibility problems were studied extensively by many authors [13, 14].

In this paper, motivated by Schöpfer et al. [12], we present a new hybrid algorithm using the inverse strongly monotone operation and a finite family of resolvent operator. Then, we show that our generated sequence is strongly converges to a common point, the set of solution set of split feasibility problem, and the fixed point of relatively nonexpansive mapping and the fixed point of resolvent operator.

2. Preliminaries

Let X be a real smooth Banach space with norm ‖.‖ and let X^∗ be the dual space of X. We denote the strong convergence and the weak convergence {x_n} to x in X by x_n⟶x and x_n⇀x, respectively. A function δ : [0, 2]⟶[0, 1] is said to be the modulus of convexity of X as follows:

(7)

for every ε ∈ [0, 2]. A Banach space X is said to be uniformly convex if and only if δ(ε) > 0 for all ε > 0. It is known that a uniformly convex Banach space has the Kadec-Klee property, that is, x_n⇀u and ‖x_n‖⟶‖u‖ imply that x_n⟶u (see [15]). Let p be a fixed real number with p ≥ 2. A Banach space X is called p-uniformly convex [16], if there exists a constant c > 0 such that δ ≥ cϵ^p for all ϵ ∈ [0, 2]. Let S(E) = {x ∈ X : ‖x‖ = 1}. A Banach space X is said to be smooth if for all x ∈ S(X), there exists a unique functional j_x ∈ X^∗ such that 〈x, j_x〉 = ‖x‖ and ‖j_x‖ = 1 (see [17]).

The norm of X is said to be

differentiable if for all x, y ∈ S(X), the limit

(8)

exists. In this case, X is said to be smooth, and X is called uniformly smooth if the limit (8) is attained uniformly for all x, y ∈ S(X) [18]. If a Banach space X is uniformly convex, then X is reflexive and strictly convex, and X^∗ is uniformly smooth [17]. The duality mapping

on X is defined by

(9)

for every x ∈ X. If X is a p-uniformly convex and uniformly smooth, then

is single valued, one-to-one and satisfies

, where

is the duality mapping of X (see [19]). If p = 2, then J_X = J₂ = J is the normalized duality mapping. It is well known that if X is a reflexive, strictly convex, and smooth Banach space and

is the duality mapping on X^∗, then

. If X is a uniformly smooth and uniformly convex Banach space, then J_X is uniformly norm to norm continuous on bounded sets of X, and

is also uniformly norm to norm continuous on bounded sets of X^∗. Let X be a smooth Banach space and let J_X be the duality mapping on X. The function ϕ : X × X⟶ℝ is defined by

(10)

Clearly, from (10), we can conclude that

(11)

If X is a reflexive, strictly convex, and smooth Banach space, then for all x, y ∈ X

(12)

Also, it is clear from the definition of the function ϕ that the following condition holds for all x, y ∈ X,

(13)

Now, the function V : X × X^∗⟶ℝ is defined as follows:

(14)

for all x ∈ X and x^∗ ∈ X^∗. Moreover,

for all x ∈ X and x^∗ ∈ X^∗. If X is a reflexive strictly convex and smooth Banach space with X^∗ as its dual, then

(15)

for all x ∈ X and all x^∗, y^∗ ∈ X^∗ [20].

An operator A : C⟶X^∗ is hemicontinuous at x₀ ∈ C, if for any sequence {x_n} converging to x₀ along a line implies that the sequence {Ax_n} is weakly convergent to Ax₀, i.e., Ax_n = A(x₀ + t_nx)⇀Ax₀ as t_n⟶0 for all x ∈ C.

The generalized projection Π_C : X⟶C is a mapping that assigns to an arbitrary point x ∈ X, the minimum point of the functional ϕ(y, x); that is, Π_Cx = x₀, where x₀ is the solution of the minimization problem

(16)

The existence and uniqueness of the operator Π_C follows from the properties of the functional ϕ(x, y) and strict monotonicity of the mapping J [21]. Suppose that C is a nonempty closed convex subset of X and T is a self mapping on C. We denote the set of fixed points of T by F(T), that is F(T) = {x ∈ C : x ∈ Tx}. A point p ∈ C is called an asymptotically fixed point of T if C contains a sequence {x_n} which converges weakly to p such that Tx_n − x_n⟶0 [17]. The set of asymptotical fixed points of T will be denoted by . A mapping T from C into itself is said to be relatively nonexpansive if and ϕ(p, Tx) ≤ ϕ(p, x) for all x ∈ C and p ∈ F(T). The asymptotic behavior of a relatively nonexpansive mapping was studied in [22, 23].

We need the following lemmas for proving our main results.

Lemma 1. (see [24].)Let X be a smooth and uniformly convex Banach space and let {x_n} and {y_n} be two sequences of X. If ϕ(x_n, y_n)⟶0 and either {x_n} or {y_n} is bounded, then x_n − y_n⟶0.

Lemma 2. (see [21].)Let C be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space X and let y ∈ X. Then,

(17)

Lemma 3. (see [21].)Let C be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space X, let x ∈ X, and let z ∈ C. Then,

(18)

Lemma 4. (see [25].)Let X be a 2-uniformly convex and smooth Banach space. Then, for all x, y ∈ X, we have that

(19)

where 1/c(0 ≤ c ≤ 1) is the 2-uniformly convex constant of X.

Lemma 5. (see [25].)Let X be a uniformly convex Banach space and r > 0. Then, there exists a continuous strictly increasing convex function g : [0, 2r]⟶[0, ∞) such that g(0) = 0 and

(20)

for all x, y ∈ B_r(0) = {z ∈ X : ‖z‖ ≤ r} and t ∈ [0, 1].

Lemma 6. (see [24].)Let X be a uniformly convex Banach space and r > 0. Then, there exists a continuous strictly increasing convex function g : [0, 2r]⟶[0, ∞) such that g(0) = 0 and

(21)

for all x, y ∈ B_r(0) = {z ∈ X : ‖z‖ ≤ r}.

Lemma 7. (see [25].)Let x, y ∈ X. If X is p-uniformly smooth, then there is a c > 0 so that

(22)

Throughout this paper, we assume that f : C × C⟶ℝ is a bifunction satisfying the following conditions

(A1) f(x, x) = 0 for all x ∈ C

(A2) f is monotone, i.e., f(x, y) + f(y, x) ≤ 0, for all x, y ∈ C

(A3) lim_t↓0f(tz + (1 − t)x, y) ≤ f(x, y), for all x, y, z ∈ C

(A4) For each x ∈ C, y ↦ f(x, y) is convex and lower semicontinuous.

Lemma 8. (see [26].)Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space X. Let A : C⟶X^∗ be an α-inverse-strongly monotone operator and f be a bifunction from C × C to ℝ satisfying (A₁) − (A₄). Then, for all r > 0 the following hold

(i)
For x ∈ X, there exists u ∈ C such that

(23)

(ii)
If X is additionally uniformly smooth and K_r : X⟶C is defined as

(24)

then, the following conditions hold:

K_r is single-valued

K_r is firmly nonexpansive, i.e., for all x, y ∈ X,

(25)

EP is a closed convex subset of C.

(26)

Definition 9.

Let X be a real smooth and uniformly convex Banach space and let be a maximal monotone operator. For all ι > 0, define the operator by for all x ∈ X.

Lemma 10. (see [18].)Let X be a real smooth and uniformly convex Banach space, and let be a maximal monotone operator. Then, M⁻¹0 is a closed and convex subset of X, and the graph G(M) of M is demiclosed.

Lemma 11. Let X be a real reflexive, strictly convex, and let smooth Banach space and be a maximal monotone operator with M⁻¹0 ≠ ∅. Then, for all x ∈ X, y ∈ M⁻¹0 and ι > 0, then .

3. Main Results

In this section, we introduce our new extragradient algorithm.

Theorem 12. Let X₁ and X₂ are real 2-uniformly convex and uniformly smooth Banach spaces. Suppose that C and D are nonempty closed and convex subsets of X₁ and X₂, respectively. Suppose that g is a bifunction from C × C to ℝ which satisfies the conditions A1-A4, A : X₁⟶X₂ is a bounded linear operator and is the adjoint of A. Let be a maximal monotone operator with for all i = 1, 2, ⋯, k. Assume that B : C⟶X^∗ is an α-inverse strongly monotone operator, and f is a relatively nonexpansive mappings from C into itself and . Let {x_n} is a sequence generated by v₁ ∈ C and

(27)

where r_n ∈ [a, ∞) for some a > 0, {s_n} and {β_n} are real sequences in [a, b] ⊂ (0, 1), and τ and

satisfy the following conditions:

(i)
, , liminf_n⟶∞α_n,1α_n,2 > 0, and liminf_n⟶∞α_n,3 > 0
(ii)
τ is real number such that 0 < τ < 2/c‖A‖², where c depends on 2-uniformly smoothness of

Then, {x_n} converges strongly to .

Proof. Let . By (10), Lemma 2 and the convexity of ‖.‖², we have

(28)

Now, it follows from Lemma 11 and the above that

(29)

(30)

(31)

Let k_n = Au_n − P_DAu_n. From (10) and Lemmas 2 and 7, we have that

(32)

Since

for each y ∈ D and for each x ∈ X₂, we have that

(33)

From (32), our assumptions, and the above, we conclude that

(34)

In a similar way as above, we obtain that

(35)

It follows from (10), (29), (34), (35), Lemma 11, and the convexity of ‖.‖² that

(36)

(37)

(38)

(39)

By (10), (29), (37), Lemmas 2, 8, the condition (i), the convexity of ‖.‖², and the relatively nonexpansiveness of f, we have that

(40)

Therefore, is bounded, and exists. Now, by (11), we conclude that {x_n} is bounded. It follows from (29), (34), (35), (37), (40), and relatively nonexpansiveness of f that the sequences {u_n}, {z_n}, {y_n}, {w_n}, {v_n}, and {f(x_n)} are bounded.

Next, by (28), (37), (40), and Lemma 11, we conclude that

(41)

(42)

(43)

Now, from (11) and (41), we have the following inequalities:

(44)

Now, since

is convergent, it follows from (44), the conditions (i), and our assumptions that

(45)

Therefore, from Lemma 1, we have that

(46)

Then,

(47)

From (13), (47), the boundedness of the sequences {x_n},

, and using uniformly norm-to-norm continuity of J on bounded sets, it is clear that

(48)

By (29), (34), (35), (36), (40), and Lemma 11, we conclude that

(49)

Hence, from (11), the above, and our assumptions, we obtain the following results

(50)

(51)

Since

is convergent, we conclude from (i), (50), (51), and our assumptions that

(52)

(53)

Now, from (29), (34), (35), (37), and (40), we have

(54)

Hence, it follows from (54) that

(55)

Then, it follows from (i) and our assumptions that

(56)

From (29), (34), (35), (37), and (40), we have

(57)

So,

(58)

Therefore, it follows from (i) and our assumptions that

(59)

Suppose that r₁ = sup_n{‖f(x_n)‖, ‖u_n‖}. Therefore, from Lemma 5, there exists a continuous strictly increasing convex function g₁ : [0, 2r₁]⟶[0, ∞) such that g₁(0) = 0 and using (29), (37), Lemmas 2 and 8, the convexity of ‖.‖², and the condition relatively nonexpansiveness of f, we have that

(60)

So,

(61)

Since

exists. Therefore, it follows from the condition (i) that

(62)

Because g₁ is continues function, we conclude that

(63)

Therefore,

(64)

Since

is uniformly norm-to-norm continuous on bounded sets, it imply that

(65)

Using (13), (65), the uniformly norm-to-norm continuity of

on bounded sets, and the boundedness of the sequences {f(x_n)} and {u_n}, we conclude that

(66)

By (48) and using our assumptions, we obtain that

(67)

Then, it follows from Lemma 1 that

(68)

Now, by (15), (56), and Lemmas 2 and 4, we conclude that

(69)

Then, using Lemma 1, we obtain

(70)

Also, from (15), (59), and Lemma 2, and the same way used for proving (70), we can conclude that

(71)

Then, using Lemma 1, we get

(72)

Now, it follows from (13), (52), and Lemma 1 that

(73)

Similarly, from (13), (53), and Lemma 1, we have

(74)

then, by (72), we obtain that

(75)

Now, by (13), (73), (75), and using uniformly norm-to-norm continuity of

on bounded sets, it is implied that

(76)

It follows from (10), (76), Lemma 2, and the convexity of ‖.‖² that

(77)

Now, by Lemma 1, we have lim_n⟶∞‖z_n − w_n‖ = 0. Therefore, we obtain from (70) that lim_n⟶∞‖u_n − w_n‖ = 0, then by (13), we conclude that

(78)

From (66), (78), Lemma 2, and our assumptions, it implied that

(79)

Therefore, by Lemma 1, we have

(80)

Let r₂ = sup_n{‖v_n‖, ‖x_n‖}. Therefore, by Lemma 6, there exists a continuous, convex, and strictly increasing function g₂ : [0, 2r₂]⟶[0, ∞) such that g₂(0) = 0 and

(81)

It follows from (40), (81), Lemma 8, and the fact that

, we conclude that

(82)

Therefore,

(83)

because g₂ is a continuous strictly increasing convex function. Now, by (80) and (83), we have

(84)

From (68) and (84), we obtain that

(85)

This shows that {x_n} is a Cauchy sequence, so {x_n} converges strongly to a point q ∈ C. Therefore, by (68), (70), and (72), we imply that {u_n}, {y_n}, and {z_n} converge strongly to q.

Next, we prove that

. It follows from (46) and uniformly continuity of

on bounded subset of X₁ that

as n⟶∞. Get

; hence, by Definition 9, we have

. Therefore, there exists h_n ∈ M₁η_n such that

(86)

So, by the above observation, h_n⟶0 as n⟶∞. On the other hand, since x_n⇀q, we can conclude from (47) that η_n⇀q. Then, from Lemma 10, 0 ∈ M₁q, i.e., . Similar to the above, by using (46), we can also prove that for all i = 2, 3, ⋯k. Hence, .

Next, we show that q ∈ F(f). From (65), (68), and the triangle inequality, we conclude that

(87)

Hence, q is an asymptotic fixed point of f. Then, because f is a relatively nonexpansive mapping. Hence, q ∈ F(f).

Now, we prove that q ∈ EP(g). Since

is uniformly norm-to-norm continuous on bounded sets, it follows from (83) that

(88)

, we conclude that

for all y ∈ C. Moreover, by the condition A2, g(y, x_n) ≤ −g(x_n, y) for all y ∈ C. Therefore,

(89)

for all y ∈ C. Using (88), the condition A4, and letting n⟶∞, we have that

(90)

for all y ∈ C. Let y_λ = λy + (1 − λ)q for all y ∈ C and λ ∈ (0, 1). It follows from (90), the conditions A1, A4, and the monotonicity of B that

(91)

for all y ∈ C. Therefore, 0 ≤ g(y_λ, y) + 〈By_λ, y − y_λ〉. Using the condition A3 and letting λ⟶0, we obtain that 0 ≤ g(q, y) + 〈Bq, y − q〉 for all y ∈ C. Then, q ∈ EP(g).

Finally, we prove that q ∈ Ω. From (56), we have that ‖P_DAq − Aq‖ = lim_n⟶∞‖P_DAu_n − Au_n‖ = 0. Therefore, Aq ∈ D, i.e., q ∈ Ω. Hence, , and this completed the proof.

Conflicts of Interest

This work does not have any conflicts of interest.

Open Research

Data Availability

No data were used to support the study.

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All articles

Extragradient Methods for Solving Split Feasibility Problem and General Equilibrium Problem and Resolvent Operators in Banach Spaces

Abstract

1. Introduction

2. Preliminaries

3. Main Results

Conflicts of Interest

Open Research

Data Availability

References

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Extragradient Methods for Solving Split Feasibility Problem and General Equilibrium Problem and Resolvent Operators in Banach Spaces

Abstract

1. Introduction

2. Preliminaries

3. Main Results

Conflicts of Interest

Open Research

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