Hybrid Proximal-Point Methods for Zeros of Maximal Monotone Operators, Variational Inequalities and Mixed Equilibrium Problems

Lemma 2.1 (see [36], [37].)If E be a 2-uniformly convex Banach space. Then, for all x, y ∈ E one has

where J is the normalized duality mapping of E and 0 < c ≤ 1.

Lemma 2.2 (see [36], [38].)If E a p-uniformly convex Banach space and let p be a given real number with p ≥ 2. Then, for all x, y ∈ E,J_x ∈ J_p(x) and J_y ∈ J_p(y)

where J_p is the generalized duality mapping of E and 1/c is the p-uniformly convexity constant of E.

Lemma 2.3 (see Xu [37].)Let E be a uniformly convex Banach space. Then, for each r > 0, there exists a strictly increasing, continuous, and convex function K : [0, ∞)→[0, ∞) such that K(0) = 0 and

for all x, y ∈ {z ∈ E : ∥z∥≤r} and λ ∈ [0,1].

Lemma 2.4 (see Kamimura and Takahashi [3].)Let E be a uniformly convex and smooth real Banach space and let {x_n}, {y_n} be two sequences of E. If ϕ(x_n, y_n) → 0 and either {x_n} or {y_n} is bounded, then ∥x_n − y_n∥→0.

Lemma 2.5 (see Alber [39].)Let C be a nonempty, closed, convex subset of a smooth Banach space E and x ∈ E. Then, x₀ = Π_Cx if and only if

Lemma 2.6 (see Alber [39].)Let E be a reflexive, strictly convex, and smooth Banach space, let C be a nonempty closed convex subset of E and let x ∈ E. Then,

Lemma 2.7 (see Kohsaka and Takahashi [4], Lemma 3.2.)Let E be a strictly convex, smooth, and reflexive Banach space, and let V be as in (2.12). Then,

for all x ∈ E and x^*, y^* ∈ E^*.

Lemma 2.8 (see Kohsaka and Takahashi [4], Lemma 3.1.)Let E be a smooth, strictly convex, and reflexive Banach space, T ⊂ E × E^* a maximal monotone operator with T⁻¹0 ≠ ∅, r > 0 and J_r = (J + rT) ⁻¹J. Then,

for all x ∈ T⁻¹0 and y ∈ E.

Theorem 2.9 (see Rockafellar [1].)Let C be a nonempty, closed, convex subset of a Banach space E and A a monotone, hemicontinuous operator of C into E^*. Let T ⊂ E × E^* be an operator defined as follows:

Then, T is maximal monotone and T⁻¹0 = VI(C, A).

Lemma 2.10 (see Tan and Xu [41].)Let {a_n} and {b_n} be two sequence of nonnegative real numbers satisfying the inequality

If $$, then lim _n→∞a_n exists.

Lemma 2.11 (see [29], Lemma 2.7.)Let C be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space E, let θ be a bifunction from C × C to ℝ satisfying (A1)–(A4), let r > 0, and let x ∈ E. Then, there exists z ∈ C such that

Lemma 2.12 (see Takahashi and Zembayashi [29].)Let C be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space E and let Θ be a bifunction from C × C to ℝ satisfying (A1)–(A4). For all r > 0 and x ∈ E, define a mapping T_r : E → C as follows:

for all x ∈ E. Then, the followings hold:

• (1)

  T_r is single-valued,

• (2)

  T_r is a firmly nonexpansive-type mapping, that is, for all x, y ∈ E,

  $$ (2.21)

• (3)

  F(T_r) = EP(Θ),

• (4)

  EP(Θ) is closed and convex.

Lemma 2.13 (see Takahashi and Zembayashi [29].)Let C be a closed, convex subset of a smooth, strictly convex, and reflexive Banach space E, let Θ a bifunction from C × C to ℝ satisfying (A1)–(A4) and let r > 0. Then, for x ∈ E and q ∈ F(T_r),

Lemma 2.14. Let C be a closed convex subset of a smooth, strictly convex and reflexive Banach space E. Let φ : C → ℝ is convex and lower semicontinuous and Θ be a bifunction from C × C to ℝ satisfying (A1)–(A4). For r > 0 and x ∈ E, then there exists u ∈ C such that

Define a mapping K_r : E → C as follows: for all x ∈ E. Then, the followings hold:

• (1)

  K_r is single-valued,

• (2)

  K_r is firmly nonexpansive, that is, for all x, y ∈ E, 〈K_rx − K_ry, JK_rx − JK_ry〉 ≤ 〈K_rx − K_ry, Jx − Jy〉,

• (3)

  F(K_r) = MEP(Θ, φ),

• (4)

  MEP(Θ, φ) is closed and convex.

Proof. Define a bifunction F : C × C → ℝ as follows:

It is easily seen that F satisfies (A1)–(A4). Therefore, K_r in Lemma 2.14 can be obtained from Lemma 2.12 immediately.

Theorem 3.1. Let E be a 2-uniformly convex and uniformly smooth Banach space and let C be a nonempty closed convex subset of E. Let Θ be a bifunction from C × C to ℝ satisfying (A1)–(A4) let φ : C → ℝ be a lower semicontinuous and convex function, and let T : E → E^* be a maximal monotone operator. Let J_r = (J + rT) ⁻¹J for r > 0 and let A be an α-inverse-strongly monotone operator of C into E^* with F : = VI(C, A)∩T⁻¹(0)∩MEP(Θ, φ) ≠ ∅ and ∥Ay∥≤∥Ay − Au∥ for all y ∈ C and u ∈ F. Let {x_n} be a sequence generated by x₀ ∈ E with $$ and C₁ = C,

for n ∈ ℕ, where Π_C is the generalized projection from E onto C, J is the duality mapping on E. The coefficient sequence {α_n}, {β_n}⊂(0,1), {r_n}⊂(0, ∞) satisfying lim _n→∞α_n = 0, limsup _n→∞β_n < 1, liminf _n→∞r_n > 0, and {λ_n}⊂[a, b] for some a, b with 0 < a < b < c²α/2, 1/c is the 2-uniformly convexity constant of E. Then, the sequence {x_n} converges strongly to Π_Fx₀.

Proof. We first show that {x_n} is bounded. Put v_n = J⁻¹(Jx_n − λ_nAx_n), let p ∈ F : = VI(C, A)∩T⁻¹(0)∩MEP(Θ, φ), and let $$ be a sequence of mapping define as Lemma 2.14 and $$. By (3.1) and Lemma 2.7, the convexity of the function V in the second variable, we have

Since p ∈ VI(A, C) and A is α-inverse-strongly monotone, we have and by Lemma 2.1, we obtain Substituting (3.3) and (3.4) into (3.2), we get By Lemmas 2.7, 2.8 and (3.5), we have It follows that From (3.1) and (3.7), we obtain So, we have p ∈ C_n+1. This implies that F ⊂ C_n, for all n ∈ ℕ.

From Lemma 2.5 and $$, we have

From Lemma 2.6, one has for all p ∈ F ⊂ C_n and n ≥ 1. Then, the sequence {ϕ(x_n, x₀)} is bounded. Since $$ and $$, we have Therefore, {ϕ(x_n, x₀)} is nondecreasing. Hence, the limit of {ϕ(x_n, x₀)} exists. By the construction of C_n, one has that C_m ⊂ C_n and $$ for any positive integer m ≥ n. It follows that Letting m, n → ∞ in (3.12), we get ϕ(x_m, x_n) → 0. It follows from Lemma 2.4, that ∥x_m − x_n∥→0 as m, n → ∞, that is, {x_n} is a Cauchy sequence. Since E is a Banach space and C is closed and convex, we can assume that x_n → u ∈ C, as n → ∞. Since for all n ∈ ℕ, we also have lim _n→∞ϕ(x_n+1, x_n) = 0. From Lemma 2.4, we get lim _n→∞∥x_n+1 − x_n∥ = 0. Since $$ and by definition of C_n+1, we have Noticing the conditions lim _n→∞α_n = 0 and lim _n→∞ϕ(x_n+1, x_n) = 0, we obtain From again Lemma 2.4, So, by the triangle inequality, we get Since J is uniformly norm-to-norm continuous on bounded sets, we have On the other hand, we observe that It follows that From (3.1), (3.5), (3.6), (3.7), and (3.8), we have and then From conditions lim _n→∞α_n = 0, limsup _n→∞β_n < 1 and (3.20), we obtain By again Lemma 2.4, we have $$.

Since J is uniformly norm-to-norm continuous on bounded sets, we obtain

Applying (3.5) and (3.6), we observe that and, hence, for all n ∈ ℕ. Since 0 < a ≤ λ_n ≤ b < c²α/2, lim _n→∞α_n = 0, limsup _n→∞β_n < 1 and (3.20), we have From Lemmas 2.6, 2.7, and (3.4), we get From Lemma 2.4 and (3.27), we have Since J is uniformly norm-to-norm continuous on bounded sets, we obtain Since {x_n} is bounded, there exists a subsequence $$ of {x_n} such that $$. Since x_n − w_n → 0, then we get $$ as i → ∞.

Now, we claim that u ∈ F. First, we show that u ∈ T⁻¹0. Indeed, since lim  inf _n→∞r_n > 0, it follows from (3.24) that

If (z, z^*) ∈ T, then it holds from the monotonicity of T that for all i ∈ ℕ. Letting i → ∞, we get 〈z − u, z^*〉≥0. Then, the maximality of T implies u ∈ T⁻¹0.

Next, we show that u ∈ VI(C, A). Let B ⊂ E × E^* be an operator as follows:

By Theorem 2.9, B is maximal monotone and B⁻¹0 = VI(A, C). Let (v, w) ∈ G(B). Since w ∈ Bv = Av + N_C(v), we get w − Av ∈ N_C(v). From w_n ∈ C, we have On the other hand, since w_n = Π_CJ⁻¹(Jx_n − λ_nAx_n), then by Lemma 2.5, we have Thus, It follows from (3.34) and (3.36) that where M = sup _n≥1{∥v − w_n∥}. From (3.29) and (3.30), we obtain 〈v − u, w〉≥0. By the maximality of B, we have u ∈ B⁻¹0 and, hence, u ∈ VI(C, A).

Next, we show that u ∈ MEP(Θ, φ). Since $$. From Lemmas 2.13 and 2.14, we have

Similarly by (3.20), and so Since J is uniformly norm-to-norm continuous on bounded sets, we obtain From (3.1) and (A2), we also have Hence, From ∥x_n − u_n∥→0, ∥x_n − w_n∥→0, we get $$. Since $$, it follows by (A4) and the weak, lower semicontinuous of φ that For t with 0 < t ≤ 1 and y ∈ C, let y_t = ty + (1 − t)u. Since y ∈ C and u ∈ C, we have y_t ∈ C and hence Θ(y_t, u) + φ(u) − φ(y_t) ≤ 0. So, from (A1), (A4), and the convexity of φ, we have Dividing by t, we get Θ(y_t, y) + φ(y) − φ(y_t) ≥ 0. From (A3) and the weakly lower semicontinuity of φ, we have Θ(u, y) + φ(y) − φ(u) ≥ 0 for all y ∈ C implies u ∈ MEP(Θ, φ). Hence, u ∈ F : = VI(C, A)∩T⁻¹(0)∩MEP(Θ, φ).

Finally, we show that u = Π_Fx. Indeed, from $$ and Lemma 2.5, we have

Since F ⊂ C_n, we also have Taking limit n → ∞, we have By again Lemma 2.5, we can conclude that u = Π_Fx₀. This completes the proof.

Corollary 3.2. Let E be a 2-uniformly convex and uniformly smooth Banach space, let C be a nonempty, closed, convex subset of E. Let Θ be a bifunction from C × C to ℝ satisfying (A1)–(A4) let φ : C → ℝ be a lower semicontinuous and convex function, and let T : E → E^* be a maximal monotone operator. Let J_r = (J + rT) ⁻¹J for r > 0 with F : = T⁻¹(0)∩MEP(Θ, φ) ≠ ∅. Let {x_n} be a sequence generated by x₀ ∈ E with $$ and C₁ = C,

for n ∈ ℕ, where Π_C is the generalized projection from E onto C, J is the duality mapping on E. The coefficient sequence {α_n}, {β_n}⊂(0,1), {r_n}⊂(0, ∞) satisfying lim _n→∞α_n = 0, limsup _n→∞β_n < 1 and lim  inf _n→∞r_n > 0. Then, the sequence {x_n} converges strongly to Π_Fx₀.

Proof. In Theorem 3.1 if A ≡ 0, then (3.1) reduced to (3.49).

Theorem 4.1. Let E be a 2-uniformly convex and uniformly smooth Banach space whose duality mapping J is weak sequentially continuous. Let T : E → E^* be a maximal monotone operator and let J_r = (J + rT) ⁻¹J for r > 0. Let C be a nonempty, closed, convex subset of E such that D(T) ⊂ C ⊂ J⁻¹(⋂_r>0R(J + rT)), let Θ be a bifunction from C × C to ℝ satisfying (A1)–(A4), let φ : C → ℝ be a lower semicontinuous and convex function, and let A be an α-inverse-strongly monotone operator of C into E^* with F : = VI(C, A)∩T⁻¹(0)∩MEP(Θ, φ) ≠ ∅ and ∥Ay∥≤∥Ay − Au∥ for all y ∈ C and u ∈ F. Let {x_n} be a sequence generated by x₁ = x ∈ C and

for n ∈ ℕ ∪ {0}, where Π_C is the generalized projection from E onto C, J is the duality mapping on E. The coefficient sequence {α_n}, {β_n}⊂[0,1], {r_n}⊂(0, ∞) satisfying $$, limsup _n→∞β_n < 1  liminf _n→∞r_n > 0 and {λ_n}⊂[a, b] for some a, b with 0 < a < b < c²α/2, 1/c is the 2-uniformly convexity constant of E. Then, the sequence {Π_Fx_n} converges strongly to an element of F, which is a unique element v ∈ F such that where Π_F is the generalized projection from C onto F.

Proof. Put v_n = J⁻¹(Ju_n − λ_nAu_n). Let p ∈ F : = VI(C, A)∩T⁻¹(0)∩MEP(Θ, φ), by Lemma 2.14 and nonexpansiveness of K_r, we have

By (4.1) and Lemma 2.7, the convexity of the function V in the second variable, we obtain Since p ∈ VI(A, C) and A is α-inverse-strongly monotone, we also have Substituting (4.5) and (4.6) into (4.4) and (4.3), we get By Lemmas 2.7, 2.8, (4.7), and using the same argument in Theorem 3.1, (3.6), we obtain and hence by Lemma 2.6 and (4.7), we note that for all n ≥ 0. So, from $$ and Lemma 2.10, we deduce that lim _n→∞ϕ(p, x_n) exists. This implies that {ϕ(p, x_n)} is bounded. It implies that {x_n}, {y_n}, {z_n}, and $$ are bounded. Define a function g : F → [0, ∞) as follows: Then, by the same argument as in proof of [43, Theorem 3.1], we obtain g is a continuous convex function and if ∥z_n∥→∞, then g(z_n) → ∞. Hence, by [34, Theorem 1.3.11], there exists a point v ∈ F such that Put w_n = Π_Fx_n for all n ≥ 0. We next prove that w_n → v as n → ∞. Suppose on the contrary that there exists ϵ₀ > 0 such that, for each n ∈ ℕ, there is n′ ≥ n satisfying ∥w_n′ − v∥≥ϵ₀. Since v ∈ F, we have for all n ≥ 0. This implies that Since (∥v∥−∥Π_Fx_n∥) ² ≤ ϕ(v, w_n) ≤ ϕ(v, x_n) for all n ≥ 0 and {x_n} is bounded, {w_n} is bounded. By Lemma 2.3, there exists a stricly increasing, continuous, and convex function K : [0, ∞)→[0, ∞) such that K(0) = 0 and for all n ≥ 0. Now, choose σ satisfying 0 < σ < (1/4)K(ϵ₀). Hence, there exists n₀ ∈ ℕ such that for all n ≥ 0. Thus, there exists k ≥ n₀ satisfying the following: From (4.9), (4.14), and (4.16), we obtain for all n ≥ 0. Hence, This is a contradiction. So, {w_n} converges strongly to v ∈ F : = VI(C, A) ∩ T⁻¹(0) ∩ MEP(Θ, φ). Consequently, v ∈ F is the unique element of F such that This completes the proof.

Theorem 4.2. Let E be a 2-uniformly convex and uniformly smooth Banach space whose duality mapping J is weakly sequentially continuous. Let T : E → E^* be a maximal monotone operator and let J_r = (J + rT) ⁻¹J for r > 0. Let C be a nonempty closed convex subset of E such that D(T) ⊂ C ⊂ J⁻¹(⋂_r>0R(J + rT)), let Θ be a bifunction from C × C to ℝ satisfying (A1)–(A4), let φ : C → ℝ be a lower semicontinuous and convex function, and let A be an α-inverse-strongly monotone operator of C into E^* with F : = VI(C, A)∩T⁻¹(0)∩MEP(Θ, φ) ≠ ∅ and ∥Ay∥≤∥Ay − Au∥ for all y ∈ C and u ∈ F. Let {x_n} be a sequence generated by x₁ = x ∈ C and

for n ∈ ℕ ∪ {0}, where Π_C is the generalized projection from E onto C, J is the duality mapping on E. The coefficient sequence {α_n}, {β_n}⊂[0,1], {r_n}⊂(0, ∞) satisfying $$, limsup _n→∞β_n < 1  liminf _n→∞r_n > 0 and {λ_n}⊂[a, b] for some a, b with 0 < a < b < c²α/2, 1/c is the 2-uniformly convexity constant of E. Then, the sequence {x_n} converges weakly to an element v of F, where v = lim _n→∞Π_Fx_n.

Proof. By Theorem 4.1, we have {x_n} is bounded and so are {z_n}, $$.

From (4.9), we obtain

and then Since lim _n→∞α_n = 0, limsup _n→∞β_n < 1 and {ϕ(p, x_n)} exists, then we have By again Lemma 2.4, we have $$. Since J is uniformly norm-to-norm continuous on bounded sets, we obtain Apply (4.7), (4.8), and (4.9), we observe that and hence for all n ∈ ℕ. Since 0 < a ≤ λ_n ≤ b < c²α/2, lim _n→∞α_n = 0 and limsup _n→∞β_n < 1, we have

From Lemmas 2.6, 2.7, and (4.7), we get

From Lemma 2.4 and (4.27), we have Since J is uniformly norm-to-norm continuous on bounded sets, we obtain

Since {z_n} is bounded, there exists a subsequence $$ of {z_n} such that $$. It follows that $$ and $$ as i → ∞.

Now, we claim that u ∈ F. First, we show that u ∈ T⁻¹0. Indeed, since liminf _n→∞r_n > 0, it follows that

If (z, z^*) ∈ T, then it holds from the monotonicity of T that for all i ∈ ℕ. Letting i → ∞, we get 〈z − u, z^*〉≥0. Then, the maximality of T implies u ∈ T⁻¹0.

Next, we show that u ∈ VI(C, A). Let B ⊂ E × E^* be an operator as follows:

By Theorem 2.9, B is maximal monotone and B⁻¹0 = VI(A, C). Let (v, w) ∈ G(B). Since w ∈ Bv = Av + N_C(v), we get w − Av ∈ N_C(v). From z_n ∈ C, we have On the other hand, since z_n = Π_CJ⁻¹(Ju_n − λ_nAu_n). Then, by Lemma 2.5, we have Thus, It follows from (4.34) and (4.36) that where M = sup _n≥1{∥v − z_n∥}. From (4.29) and (4.30), we obtain 〈v − u, w〉≥0. By the maximality of B, we have u ∈ B⁻¹0 and hence u ∈ VI(C, A).

Next, we show u ∈ MEP(f) = F(K_r). From $$. It follows from (4.7), (4.8), and (4.9) that

or, equivalently, with lim _n→∞α_n = 0 and limsup _n→∞β_n < 1, yield that lim _n→∞ϕ(p, u_n) = lim _n→∞ϕ(p, x_n).

From Lemmas 2.13 and 2.14, for p ∈ F,

This implies that lim _n→∞ϕ(u_n, x_n) = 0. Noticing Lemma 2.4, we get Since J is uniformly norm-to-norm continuous on bounded sets, we obtain

From (4.20) and (A2), we also have

Hence, From ∥u_n − z_n∥→0, we get $$. Since $$, it follows by (A4) and the weakly lower semicontinuous of φ that For t with 0 < t ≤ 1 and y ∈ C, let y_t = ty + (1 − t)u. Since y ∈ C and u ∈ C, we have y_t ∈ C and hence Θ(y_t, u) + φ(u) − φ(y_t) ≤ 0. So, from (A1), (A4), and the convexity of φ, we have Dividing by t, we get Θ(y_t, y) + φ(y) − φ(y_t) ≥ 0. From (A3) and the weakly lower semicontinuity of φ, we have Θ(u, y) + φ(y) − φ(u) ≥ 0 for all y ∈ C implies u ∈ MEP(Θ, φ). Hence, u ∈ F : = VI(C, A)∩T⁻¹(0)∩MEP(Θ, φ).

By Theorem 4.1, the {Π_Fx_n} converges strongly to a point v ∈ F which is a unique element of F such that

By the uniform smoothness of E, we also have $$.

Finally, we prove u = v. From Lemma 2.5 and u ∈ F, we have

Since J is weakly sequentially continuous, $$ and u_n − x_n → 0. Then, On the other hand, since J is monotone, we have Hence, Since E is strict convexity, it follows that u = v. Therefore, the sequence {x_n} converges weakly to v = lim _n→∞Π_Fx_n. This completes the proof.

Theorem 5.1. Let E be a 2-uniformly convex and uniformly smooth Banach space and let K be a nonempty closed convex subset of E. Let Θ be a bifunction from K × K to ℝ satisfying (A1)–(A4) let φ : K → ℝ be a lower semicontinuous and convex function, and let T : E → E^* be a maximal monotone operator. Let J_r = (J + rT) ⁻¹J for r > 0 and let A be an α-inverse-strongly monotone operator of K into E^* with F : = T⁻¹(0)∩CP(K, A)∩MEP(Θ, φ) ≠ ∅ and ∥Ay∥≤∥Ay − Au∥ for all y ∈ K and u ∈ F. For an initial point x₀ ∈ E with $$ and K₁ = K,

for n ∈ ℕ, where Π_K is the generalized projection from E onto K and J is the duality mapping on E. The coefficient sequence {α_n}, {β_n}⊂(0,1), {r_n}⊂(0, ∞) satisfying lim _n→∞α_n = 0, limsup _n→∞β_n < 1, lim  inf _n→∞r_n > 0 and {λ_n}⊂[a, b] for some a, b with 0 < a < b < c²α/2, 1/c is the 2-uniformly convexity constant of E. Then, the sequence {x_n} converges strongly to Π_Fx₀.

Proof. As in the proof Lemma 7.1.1 of Takahashi in [44], we have VI(C, A) = CP(K, A). So, we obtain the desired result.

Theorem 5.2. Let E be a 2-uniformly convex and uniformly smooth Banach space whose duality mapping J is weakly sequentially continuous. Let T : E → E^* be a maximal monotone operator and let J_r = (J + rT) ⁻¹J for r > 0. Let K be a nonempty closed convex subset of E such that D(T) ⊂ K ⊂ J⁻¹(⋂_r>0R(J + rT)), let Θ be a bifunction from K × K to ℝ satisfying (A1)–(A4), let φ : K → ℝ be a proper lower semicontinuous and convex function, and let A be an α-inverse-strongly monotone operator of K into E^* with F : = CP(K, A)∩T⁻¹(0)∩MEP(Θ, φ) ≠ ∅ and ∥Ay∥≤∥Ay − Au∥ for all y ∈ K and u ∈ F. Let {x_n} be a sequence generated by x₁ = x ∈ K and

for n ∈ ℕ ∪ {0}, where Π_K is the generalized projection from E onto K, J is the duality mapping on E. The coefficient sequence {α_n}, {β_n}⊂[0,1], {r_n}⊂(0, ∞) satisfying $$, limsup _n→∞β_n < 1  liminf _n→∞r_n > 0 and {λ_n}⊂[a, b] for some a, b with 0 < a < b < c²α/2, 1/c is the 2-uniformly convexity constant of E. Then, the sequence {x_n} converges weakly to an element v of F, where v = lim _n→∞Π_Fx_n.

Proof. It follows by Lemma 7.1.1 of Takahashi in [44], we have VI(C, A) = CP(K, A). Hence, Theorem 4.2, {x_n} converges weakly to an element v of F, where v = lim _n→∞Π_Fx_n.