Hybrid and Relaxed Mann Iterations for General Systems of Variational Inequalities and Nonexpansive Mappings

Theorem 1 (see [10].)Let X be a strictly convex and reflexive Banach space with a uniformly Gateaux differentiable norm, C a nonempty closed convex subset of X, $$ a sequence of nonexpansive self-mappings on C such that the common fixed point set $$, and f ∈ Ξ_C with a contractive coefficient ρ ∈ (1/2, 1). For any given x₀ ∈ C, let $$ be the iterative sequence defined by

where $$ and $$ are two sequences in (0,1) with α_n + β_n ≤ 1  (n ≥ 0), $$ is a sequence in [0,1], and W_n is the W-mapping generated by (CY). Assume that

• (i)

  lim _n→∞α_n = 0, $$ and 0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1;

• (ii)

  lim _n→∞|γ_n − γ_n−1| = 0 and limsup _n→∞γ_n < 1.

Then, there hold the following:

• (i)

  lim _n→∞∥x_n+1 − x_n∥ = 0;

• (ii)

  the sequence $$ converges strongly to some p ∈ F which is the unique solution of the variational inequality problem (VIP)

  $$ ()

•  

  provided lim _n→∞γ_n = 0 and β_n ≡ β for some fixed β ∈ (0,1).

Lemma 2 (see [14].)For given $$, $$ is a solution of problem (14) if and only if $$ is a fixed point of the mapping G : C → C defined by

where $$ and P_C is the projection of H onto C.

Theorem 3 (see [38].)Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C. Let A_i : C → X be an α_i-inverse-strongly accretive mapping for each i = 1, …, N. Define the mapping G_i : C → C by G_i = Π_C(I − λ_iA_i) for i = 1, …, N, where λ_i ∈ (0, α_i/κ²) and κ is the 2-uniformly smooth constant of X. Let B : C → C be the K-mapping generated by G₁, …, G_N and ρ₁, …, ρ_N, where ρ_i ∈ (0,1), for all i = 1, …, N − 1, and ρ_N ∈ (0,1]. Let f : C → C a contraction with coefficient ρ ∈ (0,1). Let V : C → C be an η-strictly pseudocontractive mapping and S : C → C be a nonexpansive mapping such that $$. For arbitrarily given x₀ ∈ C, let {x_n} be the sequence generated by

where α ∈ (0, η/κ²). Suppose that {α_n}, {β_n}, {γ_n}, and {δ_n} are the sequences in [0,1], α_n + β_n + γ_n + δ_n = 1 and satisfy the following conditions:

• (i)

  lim _n→∞α_n = 0 and $$;

• (ii)

  {γ_n}, {δ_n}⊂[c, d] for some c, d ∈ (0,1);

• (iii)

  $$;

• (iv)

  0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1.

Then, {x_n} converges strongly to q ∈ F, which solves the following VIP:

Lemma 4 (see [39].)Let X be a 2-uniformly smooth Banach space. Then,

where κ is the 2-uniformly smooth constant of X and J is the normalized duality mapping from X into X^*.

Theorem 5 (see [11].)Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C. Let the mapping B_i : C → X be β_i-inverse-strongly accretive with 0 < μ_i < β_i/κ² for i = 1,2. Let f be a contraction of C into itself with coefficient δ ∈ (0,1). Let $$ be a countable family of nonexpansive mappings of C into itself such that $$, where Ω is the fixed point set of the mapping G defined by (24). For arbitrarily given x₁ ∈ C, let {x_n} be the sequence generated by

Proposition 6 (see [40].)Let X be a real smooth and uniform convex Banach space and let r > 0. Then, there exists a strictly increasing, continuous, and convex function g : [0,2r] → R, g(0) = 0 such that

where B_r = {x ∈ X : ∥x∥ ≤ r}.

Lemma 7 (see [41].)Let {s_n} be a sequence of nonnegative real numbers satisfying

where {α_n}, {β_n}, and {γ_n} satisfy the following conditions:

• (i)

  {α_n}⊂[0,1] and $$;

• (ii)

  limsup _n→∞β_n ≤ 0;

• (iii)

  γ_n ≥ 0, for all n ≥ 0, and $$.

Then, limsup _n→∞s_n = 0.

Lemma 8 (see [42].)Let X be a real Banach space X. Then, for all x, y ∈ X

• (i)

  ∥x+y∥² ≤ ∥x∥² + 2〈y, j(x + y)〉 for all j(x + y) ∈ J(x + y);

• (ii)

  ∥x+y∥² ≥ ∥x∥² + 2〈y, j(x)〉 for all j(x) ∈ J(x).

Lemma 9 (see [43].)Let C be a nonempty closed convex subset of a real smooth Banach space X. Let D be a nonempty subset of C. Let Π be a retraction of C onto D. Then, the following are equivalent:

• (i)

  Π is sunny and nonexpansive;

• (ii)

  ∥Π(x)−Π(y)∥² ≤ 〈x − y, J(Π(x) − Π(y))〉, for all x, y ∈ C;

• (iii)

  〈x − Π(x), J(y − Π(x))〉≤0, for all x ∈ C, y ∈ D.

Lemma 10. Let C be a nonempty closed convex subset of a smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C and let B₁, B₂ : C → X be nonlinear mappings. For given x^*, y^* ∈ C, (x^*, y^*) is a solution of GSVI (13) if and only if x^* = Π_C(y^* − μ₁B₁y^*), where y^* = Π_C(x^* − μ₂B₂x^*).

Proof. We can rewrite GSVI (13) as

which is obviously equivalent to

because of Lemma 9. This completes the proof.

Lemma 11 (see [44].)Let X be a uniformly convex Banach space and B_r = {x ∈ X : ∥x∥ ≤ r}, r > 0. Then, there exists a continuous, strictly increasing, and convex function g : [0, ∞]→[0, ∞], g(0) = 0 such that

for all x, y, z ∈ B_r, and all α, β, γ ∈ [0,1] with α + β + γ = 1.

Lemma 12 (see [45].)Let C be a nonempty closed convex subset of a Banach space X. Let S₀, S₁, … be a sequence of mappings of C into itself. Suppose that $$. Then for each y ∈ C, {S_ny} converges strongly to some point of C. Moreover, let S be a mapping of C into itself defined by Sy = lim _n→∞S_ny for all y ∈ C. Then lim _n→∞sup {∥Sx − S_nx∥ : x ∈ C} = 0.

Lemma 13 (see [35], [46].)Let X be a uniformly smooth Banach space, or a reflexive and strictly convex Banach space with a uniformly Gateaux differentiable norm. Let C be a nonempty closed convex subset of X, T : C → C a nonexpansive mapping with Fix (T) ≠ ∅, and f ∈ Ξ_C. Then, the net {x_t} defined by x_t = tf(x_t)+(1 − t)Tx_t converges strongly to a point in Fix (T). If we define a mapping Q : Ξ_C → Fix (T) by Q(f): = s − lim _t→0x_t, for all f ∈ Ξ_C, then Q(f) solves the VIP:

Lemma 14 (see [47].)Let C be a nonempty closed convex subset of a strictly convex Banach space X. Let $$ be a sequence of nonexpansive mappings on C. Suppose that $$ is nonempty. Let {λ_n} be a sequence of positive numbers with $$. Then, a mapping S on C defined by $$ for x ∈ C is defined well; nonexpansive and $$ holds.

Lemma 15 (see [39].)Given a number r > 0, A real Banach space X is uniformly convex if and only if there exists a continuous strictly increasing function g : [0, ∞)→[0, ∞), g(0) = 0, such that

for all λ ∈ [0,1] and x, y ∈ X such that ∥x∥ ≤ r and ∥y∥ ≤ r.

Lemma 16 (see [48], Lemma  3.2.)Let C be a nonempty closed convex subset of a strictly convex Banach space X. Let $$ be a sequence of nonexpansive self-mappings on C such that $$ and let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1). Then, for every x ∈ C and k ≥ 0, the limit lim _n→∞U_n,kx exists.

Lemma 17 (see [48].)Let C be a nonempty closed convex subset of a strictly convex Banach space X. Let $$ be a sequence of nonexpansive self-mappings on C such that $$ and let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1). Then, $$.

Let μ be a continuous linear functional on l^∞ and s = (a₀, a₁, …) ∈ l^∞. One writes μ_n(a_n) instead of μ(s). μ is called a Banach limit if μ satisfies ∥μ∥ = μ_n(1) = 1 and μ_n(a_n+1) = μ_n(a_n) for all (a₀, a₁, …) ∈ l^∞. If μ is a Banach limit, then, there hold the following:

• (i)

  for all n ≥ 0, a_n ≤ c_n implies μ_n(a_n) ≤ μ_n(c_n);

• (ii)

  μ_n(a_n+r) = μ_n(a_n) for any fixed positive integer r;

• (iii)

  liminf _n→∞a_n ≤ μ_n(a_n) ≤ limsup _n→∞a_n for all (a₀, a₁, …) ∈ l^∞.

Lemma 18 (see [49].)Let a ∈ R be a real number and a sequence {a_n} ∈ l^∞ satisfy the condition μ_n(a_n) ≤ a for all Banach limit μ. If limsup _n→∞(a_n+r − a_n) ≤ 0, then limsup _n→∞a_n ≤ a.

Corollary 19 (see [50].)Let a ∈ R be a real number and a sequence {a_n} ∈ l^∞ satisfy the condition μ_n(a_n) ≤ a for all Banach limit μ. If limsup _n→∞(a_n+1 − a_n) ≤ 0, then, limsup _n→∞a_n ≤ a.

Lemma 20 (see [51].)Let {x_n} and {z_n} be bounded sequences in a Banach space X and let {β_n} be a sequence of nonnegative numbers in [0,1] with 0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1. Suppose that x_n+1 = β_nx_n + (1 − β_n)z_n for all integers n ≥ 0 and limsup _n→∞(∥z_n+1 − z_n∥ − ∥x_n+1 − x_n∥) ≤ 0. Then, lim _n→∞∥x_n − z_n∥ = 0.

Lemma 21 (see [34].)Let C be a nonempty closed convex subset of a smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C and A : C → X an accretive mapping. Then for all λ > 0,

Lemma 22 (see [11].)Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X. Let the mapping B_i : C → X be $$-inverse-strongly accretive. Then, one has

for i = 1,2 where μ_i > 0. In particular, if $$, then I − μ_iB_i is nonexpansive for i = 1,2.

Lemma 23 (see [11].)Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C. Let the mapping B_i : C → X be $$-inverse-strongly accretive for i = 1,2. Let G : C → C be the mapping defined by

If $$ for i = 1,2, then G : C → C is nonexpansive.

Theorem 24. Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C. Let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1) and A_i : C → E an $$-inverse strongly accretive mapping for each i = 0,1, …. Define a mapping G_i : C → C by Π_C(I − λ_iA_i)x = G_ix for all x ∈ C and i = 0,1, …, where $$, κ is the 2-uniformly smooth constant of X. Let B_n : C → C be the W-mapping generated by G_n, G_n−1, …, G₀ and ρ_n, ρ_n−1, …, ρ₀. Let the mapping B_i : C → X be $$-inverse strongly accretive for i = 1,2. Let f : C → C be a contraction with coefficient ρ ∈ (0,1). Let $$ be a countable family of nonexpansive mappings of C into itself such that $$, where Ω is the fixed point set of the mapping G = Π_C(I − μ₁B₁)Π_C(I − μ₂B₂)   with $$ for i = 1,2. For arbitrarily given x₀ ∈ C, let {x_n} be the sequence generated by

where {α_n}, {β_n}, {γ_n}, and {δ_n} are the sequences in [0,1] such that β_n + γ_n + δ_n = 1 for all n ≥ 0. Suppose that the following conditions hold:

• (i)

  $$ and 0 ≤ α_n ≤ 1 − ρ, for all n ≥ n₀ for some integer n₀ ≥ 0;

• (ii)

  liminf _n→∞γ_n > 0 and liminf _n→∞δ_n > 0;

• (iii)

  lim _n→∞(|α_n+1/(1 − (1 − α_n+1)β_n+1) − α_n/(1 − (1 − α_n)β_n)| + |δ_n+1/(1 − β_n+1) − δ_n/(1 − β_n)|) = 0;

• (iv)

  0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1.

Assume that $$ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx = lim _n→∞S_nx for all x ∈ C and suppose that $$. Then, there hold the following:

• (I)

  lim _n→∞∥x_n+1 − x_n∥ = 0;

• (II)

  x_n → q⇔α_n(f(x_n) − x_n) → 0 provided β_n ≡ β for some fixed β ∈ (0,1), where q ∈ F solves the following VIP:

  $$ ()

Proof. First of all, since $$ for i = 0,1, …, it is easy to see that G_i is a nonexpansive mapping for each i = 0,1, …. Since B_n : C → C is the W-mapping generated by G_n, G_n−1, …, G₀ and ρ_n, ρ_n−1, …, ρ₀, by Lemma 16 we know that, for each x ∈ C and k ≥ 0, the limit lim _n→∞U_n,kx exists. Moreover, one can define a mapping B : C → C as follows:

for every x ∈ C. That is, such a B is the W-mapping generated by the sequences $$ and $$. According to Lemma 17, we know that $$. From Lemma 15 and the definition of G_i, we have Fix (G_i) = VI (C, A_i) for each i = 0,1, …. Hence, we have

Next, let us show that the sequence {x_n} is bounded. Indeed, take a fixed p ∈ F arbitrarily. Then, we get p = Gp, p = B_np, and p = S_np for all n ≥ 0. By Lemma 23 we know that G is nonexpansive. Then, from (42), we have

and hence

By induction, we obtain

Thus, {x_n} is bounded, and so are the sequences {y_n}, {Gx_n} and {f(x_n)}.

Let us show that

As a matter of fact, put σ_n = (1 − α_n)β_n, for all n ≥ 0. Then, it follows from (i) and (iv) that

and hence

Define

Observe that

and hence

On the other hand, we note that, for all n ≥ 0,

Furthermore, by (CY), since G_i and U_n,i are nonexpansive, we deduce that for each n ≥ 0

for some constant M₀ > 0. Utilizing (54)–(56), we have

which hence yields

where sup _n≥0{∥f(x_n)∥ + ∥B_nx_n∥ + ∥S_nGx_n∥ + M₀} ≤ M for some M > 0. So, from (58), condition (iii), and the assumption on {S_n}, it follows that (noting that 0 < λ_i ≤ b < 1, for all i ≥ 0)

Consequently, by Lemma 20, we have

It follows from (51) and (52) that

From (42), we have

which hence implies that

Since x_n+1 − x_n → 0 and α_n(f(x_n) − x_n) → 0, we get

Next, we show that ∥x_n − Gx_n∥ → 0 as n → ∞.

Indeed, for simplicity, put q = Π_C(p − μ₂B₂p), u_n = Π_C(x_n − μ₂B₂x_n) and v_n = Π_C(u_n − μ₁B₁u_n). Then, v_n = Gx_n for all n ≥ 0. From Lemma 22, we have

Substituting (65) for (66), we obtain

From (42) and (67), we have

which hence implies that

Since $$ for i = 1,2, and {x_n}, {y_n} are bounded, we obtain from (64), (69), and condition (ii) that

Utilizing Proposition 6 and Lemma 9, we have

which implies that

In the same way, we derive

which implies that

Substituting (72) for (74), we get

By Lemma 8(i), we have from (68) and (75)

which hence leads to

From (70), (77), condition (ii), and the boundedness of {x_n}, {y_n}, {u_n}, and {v_n}, we deduce that

Utilizing the properties of g₁ and g₂, we deduce that

From (79), we get

That is,

Next, let us show that

Indeed, utilizing Lemma 15 and (42), we have

which immediately implies that

So, from (64), the boundedness of {x_n}, {y_n}, and conditions (ii), (iv), it follows that

From the properties of g₃, we have

Taking into account that

we have

From (64), (86), and condition (ii), it follows that

Note that

So, in terms of (81), (89), and Lemma 12, we have

Suppose that β_n ≡ β for some fixed β ∈ (0,1) such that β + γ_n + δ_n = 1 for all n ≥ 0. Define a mapping Vx = (1 − θ₁ − θ₂)Sx + θ₁Bx + θ₂Gx, where θ₁, θ₂ ∈ (0,1) are two constants with θ₁ + θ₂ < 1. Then, by Lemmas 14 and 17, we have that Fix (V) = Fix (S)∩Fix (B)∩Fix (G) = F. For each k ≥ 1, let {p_k} be a unique element of C such that

From Lemma 13, we conclude that p_k → q ∈ Fix (V) = F as k → ∞. Observe that for every n, k

and hence

So, it immediately follows from 0 ≤ α_n ≤ 1 − ρ, for all n ≥ n₀, that

where θ_n = ∥B_np_k − Bp_k∥ + ∥S_nGx_n − B_nx_n∥ + (1/ρ(1 − β))(∥α_n(x_n − f(x_n))∥ + ∥x_n − x_n+1∥). Since lim _n→∞∥B_np_k − Bp_k∥ = lim _n→∞∥S_nGx_n − B_nx_n∥ = lim _n→∞∥α_n(x_n − f(x_n))∥ = lim _n→∞∥x_n − x_n+1∥ = 0, we know that θ_n → 0 as n → ∞.

From (95), we obtain

For any Banach limit μ, from (96), we derive

In addition, note that

It is easy to see from (81) and (91) that

Utilizing (97) and (99), we deduce that

Also, observe that

that is,

It follows from Lemma 8 (ii) and (102) that

So by (100) and (103), we have

and hence

This implies that

Since p_k → q ∈ Fix (V) = F as k → ∞, by the uniform Frechet differentiability of the norm of X we have

On the other hand, from (49) and the norm-to-norm uniform continuity of J on bounded subsets of X, it follows that

So, utilizing Lemma 18 we deduce from (107) and (108) that

which together with (49) and the norm-to-norm uniform continuity of J on bounded subsets of X, implies that

Finally, let us show that x_n → q as n → ∞. Utilizing Lemma 8 (i), from (42) and the convexity of ∥·∥², we get

Applying Lemma 7 to (112), we obtain that x_n → q as n → ∞.

Conversely, if x_n → q ∈ F as n → ∞, then from (42) it follows that

that is, y_n → q. Again from (42) we obtain that

Since x_n → q and y_n → q, we get α_n(f(x_n) − x_n) → 0. This completes the proof.

Corollary 25. Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C. Let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1) and A_i : C → E an $$-inverse strongly accretive mapping for each i = 0,1, …. Define a mapping G_i : C → C by Π_C(I − λ_iA_i)x = G_ix for all x ∈ C and i = 0,1, …, where $$ and κ is the 2-uniformly smooth constant of X. Let B_n : C → C be the W-mapping generated by G_n, G_n−1, …, G₀ and ρ_n, ρ_n−1, …, ρ₀. Let V : C → C be an α-strictly pseudocontractive mapping. Let f : C → C be a contraction with coefficient ρ ∈ (0,1). Let $$ be a countable family of nonexpansive mappings of C into itself such that $$. For arbitrarily given x₀ ∈ C, let {x_n} be the sequence generated by

where 0 < l < α/κ², {α_n}, {β_n}, {γ_n}, and {δ_n} are the sequences in [0,1] such that β_n + γ_n + δ_n = 1 for all n ≥ 0. Suppose that the following conditions hold:

• (i)

  $$ and 0 ≤ α_n ≤ 1 − ρ, for all n ≥ n₀ for some integer n₀ ≥ 0;

• (ii)

  liminf _n→∞γ_n > 0 and liminf _n→∞δ_n > 0;

• (iii)

  lim _n→∞(|α_n+1/(1 − (1 − α_n+1)β_n+1) − α_n/(1 − (1 − α_n)β_n)| + |δ_n+1/(1 − β_n+1) − δ_n/(1 − β_n)|) = 0;

• (iv)

  0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1.

Assume that $$ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx = lim _n→∞S_nx for all x ∈ C and suppose that $$. Then, there hold the following:

• (I)

  lim _n→∞∥x_n+1 − x_n∥ = 0;

• (II)

  x_n → q⇔α_n(f(x_n) − x_n) → 0 provided β_n ≡ β for some fixed β ∈ (0,1), where q ∈ F solves the following VIP

  $$ ()

Proof. In Theorem 24, we put B₁ = I − V, B₂ = 0, and μ₁ = l, where 0 < l < α/κ². Then, GSVI (13) is equivalent to the VIP of finding x^* ∈ C such that

In this case, B₁ : C → X is α-inverse strongly accretive. It is not hard to see that Fix (V) = VI (C, B₁). As a matter of fact, we have, for l > 0,

Accordingly, we know that $$, and

So, the scheme (42) reduces to (115). Therefore, the desired result follows from Theorem 24.

Lemma 26. Let C be a nonempty closed convex subset of a smooth Banach space X and let the mapping B_i : C → X be λ_i-strictly pseudocontractive and α_i-strongly accretive with α_i + λ_i ≥ 1 for i = 1,2. Then, for μ_i ∈ (0,1] one has

for i = 1,2. In particular, if $$, then I − μ_iB_i is nonexpansive for i = 1,2.

Proof. Taking into account the λ_i-strict pseudocontractivity of B_i, we derive for every x, y ∈ C

which implies that

Hence,

Utilizing the α_i-strong accretivity and λ_i-strict pseudocontractivity of B_i, we get

So, we have

Therefore, for μ_i ∈ (0,1] we have

Since $$, it follows immediately that

This implies that I − μ_iB_i is nonexpansive for i = 1,2.

Lemma 27. Let C be a nonempty closed convex subset of a smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C and let the mapping B_i : C → X be λ_i-strictly pseudocontractive and α_i-strongly accretive with α_i + λ_i ≥ 1 for i = 1,2. Let G : C → C be the mapping defined by

If $$, then G : C → C is nonexpansive.

Proof. According to Lemma 26, we know that I − μ_iB_i is nonexpansive for i = 1,2. Hence, for all x, y ∈ C, we have

This shows that G : C → C is nonexpansive. This completes the proof.

Theorem 28. Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let Π_C be a sunny nonexpansive retraction from X onto C. Let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1) and A_i : C → X be ξ_i-strictly pseudocontractive and $$-strongly accretive with $$ for each i = 0,1, …. Define a mapping G_i : C → C by Π_C(I − λ_iA_i)x = G_ix for all x ∈ C and i = 0,1, …, where $$ for all i = 0,1, …. Let B_n : C → C be the W-mapping generated by G_n, G_n−1, …, G₀ and ρ_n, ρ_n−1, …, ρ₀. Let the mapping B_i : C → X  ζ_i-strictly pseudocontractive and $$-strongly accretive with $$ for i = 1,2. Let f : C → C be a contraction with coefficient ρ ∈ (0,1). Let $$ be a countable family of nonexpansive mappings of C into itself such that $$, where Ω is the fixed point set of the mapping G = Π_C(I − μ₁B₁)Π_C(I − μ₂B₂) with $$ for i = 1,2. For arbitrarily given x₀ ∈ C, let {x_n} be the sequence generated by

where {σ_n}, {α_n}, {β_n}, {γ_n}, and {δ_n} are the sequences in [0,1] such that β_n + γ_n + δ_n = 1 and α_n + σ_n ≤ 1 for all n ≥ 0. Suppose that the following conditions hold:

• (i)

  $$ and 0 ≤ α_n + σ_n ≤ 1 − ρ, for all n ≥ n₀ for some integer n₀ ≥ 0;

• (ii)

  liminf _n→∞σ_n > 0, liminf _n→∞γ_n > 0 and liminf _n→∞δ_n > 0;

• (iii)

  lim _n→∞(|α_n+1/(1 − (1 − α_n+1 − σ_n+1)β_n+1) − (α_n/(1 − (1 − α_n − σ_n)β_n))|+  |σ_n+1/(1 − (1 − α_n+1 − σ_n+1)β_n+1) − (σ_n/(1 − (1 − α_n − σ_n)β_n))|+  |δ_n+1/(1 − β_n+1) − δ_n/(1 − β_n)|) = 0;

• (iv)

  0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1.

Assume that $$ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx = lim _n→∞S_nx for all x ∈ C and suppose that $$. Then there hold the following:

• (I)

  lim _n→∞∥x_n+1 − x_n∥ = 0;

• (II)

  x_n → q⇔α_n(f(x_n) − x_n) → 0 provided β_n ≡ β for some fixed β ∈ (0,1), where q ∈ F solves the following VIP

  $$ ()

Proof. First of all, take a fixed p ∈ F arbitrarily. Then we obtain p = Gp, p = B_np and S_np = p for all n ≥ 0. By Lemma 27, we get from (130)

and hence

By induction, we have

which implies that {x_n} is bounded and so are the sequences {y_n}, {Gx_n}, and {f(x_n)}.

Let us show that

As a matter of fact, put θ_n = (1 − α_n − σ_n)β_n, for all n ≥ 0. Then, it follows from (i) and (iv) that

and hence

Define

Observe that

and hence

On the other hand, repeating the same arguments as those of (55) and (56) in the proof of Theorem 24, we can get

for some constant M₀ > 0. Utilizing (140)-(141), we have

where sup _n≥0{∥f(x_n)∥ + ∥Gx_n∥ + ∥B_nx_n∥ + ∥S_nGx_n∥ + M₀} ≤ M for some M > 0. So, from (142), condition (iii), and the assumption on {S_n} it follows that (noting that 0 < λ_i ≤ b < 1, for all i ≥ 0)

Consequently, by Lemma 20, we have

It follows from (137) and (138) that

Next, we show that ∥x_n − Gx_n∥ → 0 as n → ∞.

Indeed, in terms of Lemma 11, from (130), we have

Then, it immediately follows from 0 ≤ α_n + σ_n ≤ 1 − ρ, for all n ≥ n₀ that

for all n ≥ n₀. Since ∥α_n(f(x_n) − x_n)∥ → 0 and {x_n} is bounded, we deduce from (145) and condition (ii) that

Utilizing the properties of g, we have

Also, from (130) we have

which hence leads to

So, it is easy to see from (145), (149), and ∥α_n(f(x_n) − x_n)∥ → 0 that

We note that

Therefore, from (149) and (152) it follows that

Repeating the same arguments as those of (86), (89), and (91) in the proof of Theorem 24, we can obtain

Suppose that β_n ≡ β for some fixed β ∈ (0,1) such that β + γ_n + δ_n = 1 for all n ≥ 0. Define a mapping Vx = (1 − θ₁ − θ₂)Sx + θ₁Bx + θ₂Gx, where θ₁, θ₂ ∈ (0,1) are two constants with θ₁ + θ₂ < 1. Then, by Lemmas 14 and 17, we have that Fix (V) = Fix (S)∩Fix (B)∩Fix (G) = F. For each k ≥ 1, let {p_k} be a unique element of C such that

From Lemma 13, we conclude that p_k → q ∈ Fix (V) = F as k → ∞. Observe that for every n, k

and hence

So, it immediately follows from 0 ≤ α_n ≤ 1 − ρ, for all n ≥ n₀ that

where θ_n = ∥B_np_k − Bp_k∥+  ∥S_nGx_n − B_nx_n∥+  1/(ρ(1 − β))(∥α_n(x_n − f(x_n))∥+  ∥Gx_n − x_n∥+  ∥x_n − x_n+1∥). Since lim _n→∞∥B_np_k − Bp_k∥ =   lim _n→∞∥S_nGx_n − B_nx_n∥ =   lim _n→∞∥α_n(x_n − f(x_n))∥ =   lim _n→∞∥Gx_n − x_n∥ =   lim _n→∞∥x_n − x_n+1∥ =   0, we know that τ_n → 0 as n → ∞.

From (159), we obtain

For any Banach limit μ, from (160) we derive

Repeating the same arguments as those of (99), in the proof of Theorem 24, we can get

Utilizing (161) and (162), we deduce that

Also, observe that

Repeating the same arguments as those of (106) in the proof of Theorem 24, we can get

Since p_k → q ∈ Fix (V) = F as k → ∞, by the uniform Gateaux differentiability of the norm of X, we have

On the other hand, from (135) and the norm-to-weak* uniform continuity of J on bounded subsets of X, it follows that

So, utilizing Lemma 18, we deduce from (166) and (167) that

which, together with (135) and the norm-to-norm uniform continuity of J on bounded subsets of X, implies that

Finally, let us show that x_n → q as n → ∞. Utilizing Lemma 8 (i), from (130) and the convexity of ∥·∥², we get

Applying Lemma 7 to (171), we obtain that x_n → q as n → ∞.

Conversely, if x_n → q ∈ F as n → ∞, then from (130) it follows that

as n → ∞; that is, y_n → q. Again from (130) we obtain that

Since x_n → q and y_n → q, we get α_n(f(x_n) − x_n) → 0. This completes the proof.

Corollary 29. Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let Π_C be a sunny nonexpansive retraction from X onto C. Let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1) and A_i : C → Xξ_i-strictly pseudocontractive and $$-strongly accretive with $$ for each i = 0,1, …. Define a mapping G_i : C → C by Π_C(I − λ_iA_i)x = G_ix for all x ∈ C and i = 0,1, …, where $$ for all i = 0,1, …. Let B_n : C → C be the W-mapping generated by G_n, G_n−1, …, G₀ and ρ_n, ρ_n−1, …, ρ₀. Let V : C → C be a self-mapping such that I − V : C → X is λ-strictly pseudocontractive and α-strongly accretive with α + λ ≥ 1. Let f : C → C be a contraction with coefficient ρ ∈ (0,1). Let $$ be a countable family of nonexpansive mappings of C into itself such that $$. For arbitrarily given x₀ ∈ C, let {x_n} be the sequence generated by

where $$ and {σ_n}, {α_n}, {β_n}, {γ_n}, and {δ_n} are the sequences in [0,1] such that β_n + γ_n + δ_n = 1 and α_n + σ_n ≤ 1 for all n ≥ 0. Suppose that the following conditions hold:

• (i)

  $$ and 0 ≤ α_n + σ_n ≤ 1 − ρ, for all n ≥ n₀ for some integer n₀ ≥ 0;

• (ii)

  liminf _n→∞σ_n > 0, liminf _n→∞γ_n > 0 and liminf _n→∞δ_n > 0;

• (iii)

  lim _n→∞(|α_n+1/(1 − (1 − α_n+1 − σ_n+1)β_n+1) − α_n/(1 − (1 − α_n − σ_n)β_n)|+  |σ_n+1/(1 − (1 − α_n+1 − σ_n+1)β_n+1) − σ_n/(1 − (1 − α_n − σ_n)β_n)|+  |δ_n+1/(1 − β_n+1) − δ_n/(1 − β_n)|) =   0;

• (iv)

  0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1.

Assume that $$ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx = lim _n→∞S_nx for all x ∈ C and suppose that $$. Then there hold the following:

• (I)

  lim _n→∞∥x_n+1 − x_n∥ = 0;

• (II)

  x_n → q⇔α_n(f(x_n) − x_n) → 0 provided β_n ≡ β for some fixed β ∈ (0,1), where q ∈ F solves the following VIP

  $$ ()

Proof. In Theorem 28, we put B₁ = I − V, B₂ = 0, and μ₁ = l, where $$. Then, GSVI (13) is equivalent to the VIP of finding x^* ∈ C such that

In this case, B₁ : C → X is λ-strictly pseudocontractive and α-strongly accretive. Repeating the same arguments as those in the proof of Corollary 25, we can infer that Fix (V) = VI (C, B₁). Accordingly, $$, and

So, scheme (130) reduces to (174). Therefore, the desired result follows from Theorem 31.

Remark 30. Our Theorems 24 and 28 improve, extend, supplement and develop Ceng and Yao’s [10, Theorem  3.2], Cai and Bu’s [11, Theorem  3.1], Kangtunyakarn’s [38, Theorem  3.1], and Ceng and Yao’s [8, Theorem  3.1], in the following aspects.

• (i)

  The problem of finding a point $$ in our Theorems 24 and 28 is more general and more subtle than every one of the problem of finding a point $$ in [10, Theorem  3.2], the problem of finding a point $$ in [11, Theorem  3.1], the problem of finding a point $$ in [38, Theorem  3.1], and the problem of finding a point q ∈ Fix (T) in [8, Theorem  3.1].

• (ii)

  The iterative scheme in [8, Theorem  3.1] is extended to develop the iterative schemes (42) and (130) in our Theorems 24 and 28 by virtue of the iterative schemes of [11, Theorem  3.1] and [10, Theorems  3.2]. The iterative schemes (42) and (130) in our Theorems 24 and 28 are more advantageous and more flexible than the iterative scheme of [8, Theorem  3.1] because they can be applied to solving three problems (i.e., GSVI (13), fixed point problem and infinitely many VIPs), and involve several parameter sequences {α_n}, {β_n}, {γ_n}, {δ_n}, (and {σ_n}).

• (iii)

  Our Theorems 24 and 28 extend and generalize Ceng and Yao [8, Theorem  3.1] from a nonexpansive mapping to a countable family of nonexpansive mappings, and Ceng and Yao’s [10, Theorems  3.2], to the setting of the GSVI (13) and infinitely many VIPs, Kangtunyakarn [38, Theorem  3.1], from finitely many VIPs to infinitely many VIPs, from a nonexpansive mapping to a countable family of nonexpansive mappings and from a strict pseudocontraction to the GSVI (13). In the meantime, our Theorems 24 and 28 extend and generalize Cai and Bu’s [11, Theorem  3.1], to the setting of infinitely many VIPs.

• (iv)

  The iterative schemes (42) and (130) in our Theorems 24 and 28 are very different from every one in [10, Theorem  3.2], [11, Theorem  3.1], [38, Theorem  3.1], and [8, Theorem  3.1] because the mappings G and T_n in [11, Theorem  3.1] and the mapping T in [8, Theorem  3.1] are replaced with the same composite mapping S_nG in the iterative schemes (42) and (130) and the mapping W_n in [10, Theorem  3.2] is replaced with B_n.

• (v)

  Cai and Bu’s proof in [11, Theorem  3.1] depends on the argument techniques in [14], the inequality in 2-uniformly smooth Banach spaces (see Lemma 4), and the inequality in smooth and uniform convex Banach spaces (see Proposition 6). Because the composite mapping S_nG appears in the iterative scheme (42) of our Theorem 24, the proof of our Theorem 24 depends on the argument techniques in [14], the inequality in 2-uniformly smooth Banach spaces (see Lemma 4), the inequality in smooth and uniform convex Banach spaces (see Proposition 6), the inequality in uniform convex Banach spaces (see Lemma 15 in Section 2 of this paper), and the properties of the W-mapping and the Banach limit (see Lemmas 16–18 in Section 2 of this paper). However, the proof of our Theorem 28 does not depend on the argument techniques in [14], the inequality in 2-uniformly smooth Banach spaces (see Lemma 4), and the inequality in smooth and uniform convex Banach spaces (see Proposition 6). It depends on only the inequality in uniform convex Banach spaces (see Lemma 15 in Section 2 of this paper) and the properties of the W-mapping and the Banach limit (see Lemmas 16–18 in Section 2 of this paper).

• (vi)

  The assumption of the uniformly convex and 2-uniformly smooth Banach space X in [11, Theorem  3.1] is weakened to the one of the uniformly convex Banach space X having a uniformly Gateaux differentiable norm in our Theorem 28. Moreover, the assumption of the uniformly smooth Banach space X in [8, Theorem  3.1] is replaced with the one of the uniformly convex Banach space X having a uniformly Gateaux differentiable norm in our Theorem 28. It is worth emphasizing that there is no assumption on the convergence of parameter sequences {α_n}, {β_n}, {γ_n}, and {δ_n} (and {σ_n}) to zero in our Theorems 24 and 28.

Theorem 31. Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C. Let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1) and A_i : C → X an $$-inverse strongly accretive mapping for each i = 0,1, …. Define a mapping G_i : C → C by Π_C(I − λ_iA_i)x = G_ix for all x ∈ C and i = 0,1, …, where $$ and κ is the 2-uniformly smooth constant of X. Let B_n : C → C be the W-mapping generated by G_n, G_n−1, …, G₀ and ρ_n, ρ_n−1, …, ρ₀. Let the mapping B_i : C → X be $$-inverse strongly accretive for i = 1,2. Let f : C → C be a contraction with coefficient ρ ∈ (0,1). Let $$ be a countable family of nonexpansive mappings of C into itself such that $$, where Ω is the fixed point set of the mapping G = Π_C(I − μ₁B₁)Π_C(I − μ₂B₂) with $$ for i = 1,2. For arbitrarily given x₀ ∈ C, let {x_n} be the sequence generated by

where {α_n}, {β_n}, {γ_n}, and {δ_n} are the sequences in (0,1) such that α_n + β_n + γ_n + δ_n = 1 for all n ≥ 0. Suppose that the following conditions hold:

• (i)

  lim _n→∞α_n = 0 and $$;

• (ii)

  {γ_n}, {δ_n}⊂[c, d] for some c, d ∈ (0,1);

• (iii)

  lim _n→∞(|β_n − β_n−1| + |γ_n − γ_n−1| + |δ_n − δ_n−1|) = 0;

• (iv)

  0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1.

Assume that $$ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx = lim _n→∞S_nx for all x ∈ C and suppose that $$. Then, there hold the following:

• (I)

  lim _n→∞∥x_n+1 − x_n∥ = 0;

• (II)

  the sequence $$ converges strongly to some q ∈ F which is the unique solution of the variational inequality problem (VIP)

  $$ ()

Proof. First of all, since $$ for i = 0,1, …, it is easy to see that G_i is a nonexpansive mapping for each i = 0,1, …. Since B_n : C → C is the W-mapping generated by G_n, G_n−1, …, G₀, and ρ_n, ρ_n−1, …, ρ₀, by Lemma 16 we know that, for each x ∈ C and k ≥ 0, the limit lim _n→∞U_n,kx exists. Moreover, one can define a mapping B : C → C as follows:

for every x ∈ C. That is, such a B is the W-mapping generated by the sequences $$ and $$. According to Lemma 17, we know that $$. From Lemma 21 and the definition of G_i, we have Fix (G_i) = VI (C, A_i) for each i = 0,1, …. Hence, we have

Next, let us show that the sequence {x_n} is bounded. Indeed, take a fixed p ∈ F arbitrarily. Then, we get p = Gp, p = B_np, and p = S_np for all n ≥ 0. By Lemma 23, we know that G is nonexpansive. Then, from (178), we have

By induction, we obtain

Hence, {x_n} is bounded, and so are the sequences {Gx_n} and {f(x_n)}.

Let us show that

As a matter of fact, observe that x_n+1 can be rewritten as follows:

where z_n = (α_nf(x_n) + γ_nB_nx_n + δ_nS_nGx_n)/(1 − β_n). Observe that

On the other hand, we note that, for all n ≥ 1,

Furthermore, by (CY), since G_i and U_n,i are nonexpansive, we deduce that for each n ≥ 1

for some constant M > 0. Taking into account 0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1, we may assume, without loss of generality, that $$. Utilizing (186)–(188), we have

where $$ for some M₁ > 0. Thus, from (189), conditions (i), (iii) and the assumption on {S_n}, it follows that (noting that 0 < λ_i ≤ b < 1, for all i ≥ 0)

Since 0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1, by Lemma 20 we get

Consequently,

Next we show that ∥x_n − Gx_n∥ → 0 as n → ∞.

Indeed, for simplicity, put q = Π_C(p − μ₂B₂p), u_n = Π_C(x_n − μ₂B₂x_n) and v_n = Π_C(u_n − μ₁B₁u_n). Then, v_n = Gx_n for all n ≥ 0. From Lemma 26 we have

Substituting (193) for (194), we obtain

By Lemma 8, we have from (178) and (195)

which hence implies that

Since ∥x_n − x_n+1∥ → 0, $$ for i = 1,2, and {x_n} is bounded, we obtain from conditions (i), (ii) that

Utilizing Proposition 6 and Lemma 9, we have

which implies that

In the same way, we derive

which implies that

Substituting (200) for (202), we get

By Lemma 8, we have from (196) and (203)

which hence leads to

From (198), (205), conditions (i), (ii) and the boundedness of {x_n}, {u_n}, and {v_n}, we deduce that

Utilizing the properties of g₁ and g₂, we deduce that

From (207), we get

That is,

Next, let us show that

Indeed, observe that x_n+1 can be rewritten as follows:

where e_n = γ_n + δ_n and $$. Utilizing Lemma 11 and (211), we have

which hence implies that

Utilizing (184), conditions (i), (ii), (iv), and the boundedness of {x_n} and {f(x_n)}, we get

From the properties of g₃, we have

Utilizing Lemma 15 and the definition of $$, we have

which hence yields

Since {x_n} and $$ are bounded and $$ as n → ∞, we deduce from condition (ii) that

From the properties of g₄, we have

On the other hand, x_n+1 can also be rewritten as follows:

where d_n = α_n + δ_n and $$. Utilizing Lemma 11 and the convexity of ∥·∥², we have

which hence implies that

From (184), conditions (i), (ii), (iv), and the boundedness of {x_n} and {f(x_n)}, we have

Utilizing the properties of g₅, we have

which, together with (219), implies that

That is,

We note that

So, in terms of (209), (226), and Lemma 12, we have

Suppose that β_n ≡ β for some fixed β ∈ (0,1) such that α_n + β + γ_n + δ_n = 1 for all n ≥ 0. Define a mapping Vx = (1 − θ₁ − θ₂)Sx + θ₁Bx + θ₂Gx, where θ₁, θ₂ ∈ (0,1) are two constants with θ₁ + θ₂ < 1. Then by Lemmas 14 and 17, we have that Fix (V) = Fix (S)∩Fix (B)∩Fix (G) = F. For each k ≥ 1, let {p_k} be a unique element of C such that

From Lemma 13, we conclude that p_k → q ∈ Fix (V) = F as k → ∞. Observe that for every n, k

where θ_n = α_n∥f(x_n) − Bp_k∥ + (1 − β)∥B_np_k − Bp_k∥ + δ_n∥S_nGx_n − B_nx_n∥. Since lim _n→∞α_n = lim _n→∞∥B_np_k − Bp_k∥ = lim _n→∞∥S_nGx_n − B_nx_n∥ = 0, we know that θ_n → 0 as n → ∞.

From (230), we obtain

where τ_n = θ_n[2(β∥x_n − Bp_k∥ + (1 − β)∥x_n − p_k∥) + θ_n] → 0 as n → ∞.

For any Banach limit μ, from (231) we derive

In addition, note that

It is easy to see from (209) and (228) that

Utilizing (232) and (234), we deduce that

Also, observe that

that is,

It follows from Lemma 8(ii) and (237) that

So by (235) and (238), we have

and hence

This implies that

Since p_k → q ∈ Fix (V) = F as k → ∞, by the uniform Frechet differentiability of the norm of X, we have

On the other hand, from (184) and the norm-to-norm uniform continuity of J on bounded subsets of X, it follows that

So, utilizing Lemma 18 we deduce from (242) and (243) that

which, together with (184) and the norm-to-norm uniform continuity of J on bounded subsets of X, implies that

Finally, let us show that x_n → q as n → ∞. Utilizing Lemma 8 (i), from (178) and the convexity of ∥·∥², we get

Applying Lemma 7 to (246), we obtain that x_n → q as n → ∞. This completes the proof.

Corollary 32. Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π_C be a sunny nonexpansive retraction from X onto C. Let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1) and A_i : C → X an $$-inverse strongly accretive mapping for each i = 0,1, …. Define a mapping G_i : C → C by Π_C(I − λ_iA_i)x = G_ix for all x ∈ C and i = 0,1, …, where $$ and κ is the 2-uniformly smooth constant of X. Let B_n : C → C be the W-mapping generated by G_n, G_n−1, …, G₀, and ρ_n, ρ_n−1, …, ρ₀. Let V : C → C be an α-strictly pseudocontractive mapping. Let f : C → C be a contraction with coefficient ρ ∈ (0,1). Let $$ be a countable family of nonexpansive mappings of C into itself such that $$. For arbitrarily given x₀ ∈ C, let {x_n} be the sequence generated by

where 0 < l < α/κ² and {α_n}, {β_n}, {γ_n}, and {δ_n} are the sequences in (0,1) such that α_n + β_n + γ_n + δ_n = 1 for all n ≥ 0. Suppose that the following conditions hold:

• (i)

  lim _n→∞α_n = 0 and $$;

• (ii)

  {γ_n}, {δ_n}⊂[c, d] for some c, d ∈ (0,1);

• (iii)

  lim _n→∞(|β_n − β_n−1 | +|γ_n − γ_n−1 | +|δ_n − δ_n−1|) = 0;

• (iv)

  0 < lim  inf _n→∞β_n ≤ lim  sup _n→∞β_n < 1.

Assume that $$ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx = lim _n→∞S_nx for all x ∈ C and suppose that $$. Then, there hold the following:

• (I)

  lim _n→∞∥x_n+1 − x_n∥ = 0;

• (II)

  the sequence $$ converges strongly to some q ∈ F which is the unique solution of the variational inequality problem (VIP)

  $$ ()

•  

  provided β_n ≡ β for some fixed β ∈ (0,1).

Proof. In Theorem 31, we put B₁ = I − V, B₂ = 0 and μ₁ = l where 0 < l < α/κ². Then GSVI (13) is equivalent to the VIP of finding x^* ∈ C such that

In this case, B₁ : C → X is α-inverse strongly accretive. Repeating the same arguments as those in the proof of Corollary 25, we can infer that Fix (V) = VI (C, B₁). Accordingly, we know that $$, and

So, scheme (178) reduces to (247). Therefore, the desired result follows from Theorem 31.

Theorem 33. Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let Π_C be a sunny nonexpansive retraction from X onto C. Let $$ be a sequence of positive numbers in (0, b] for some b ∈ (0,1) and A_i : C → Xξ_i-strictly pseudocontractive and $$-strongly accretive with $$ for each i = 0,1, …. Define a mapping G_i : C → C by Π_C(I − λ_iA_i)x = G_ix for all x ∈ C and i = 0,1, …, where $$ for all i = 0,1, …. Let B_n : C → C be the W-mapping generated by G_n, G_n−1, …, G₀ and ρ_n, ρ_n−1, …, ρ₀. Let the mapping B_i : C → X be ζ_i-strictly pseudocontractive and $$-strongly accretive with $$ for i = 1,2. Let f : C → C be a contraction with coefficient ρ ∈ (0,1). Let $$ be a countable family of nonexpansive mappings of C into itself such that $$, where Ω is the fixed point set of the mapping G = Π_C(I − μ₁B₁)Π_C(I − μ₂B₂) with $$ for i = 1,2. For arbitrarily given x₀ ∈ C, let {x_n} be the sequence generated by

where {α_n}, {β_n}, {γ_n}, {δ_n}, and {σ_n} are the sequences in (0,1) such that α_n + β_n + γ_n + δ_n = 1 for all n ≥ 0. Suppose that the following conditions hold:

• (i)

  lim _n→∞α_n = 0 and $$;

• (ii)

  {γ_n}, {δ_n}⊂[c, d] for some c, d ∈ (0,1);

• (iii)

  $$;

• (iv)

  0 < lim  inf _n→∞β_n ≤ lim  sup _n→∞β_n < 1 and 0 < lim  inf _n→∞σ_n ≤ lim  sup _n→∞σ_n < 1.

Assume that $$ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx = lim _n→∞S_nx for all x ∈ C and suppose that $$. Then, there hold the following:

• (I)

  lim _n→∞∥x_n+1 − x_n∥ = 0;

• (II)

  the sequence $$ converges strongly to some q ∈ F which is the unique solution of the variational inequality problem (VIP)

  $$ ()

Proof. First of all, it is easy to see that (251) can be rewritten as follows:

Take a fixed p ∈ F arbitrarily. Then, we obtain p = Gp, p = B_np and S_np = p for all n ≥ 0. Thus, we get from (253)

and hence

By induction, we have

which implies that {x_n} is bounded and so are the sequences {y_n}, {Gx_n} and {f(x_n)}.

Let us show that

As a matter of fact, observe that y_n can be rewritten as follows:

where z_n =  (α_nf(x_n) + γ_nB_nx_n + δ_nS_nGx_n)/ (1 − β_n) . Observe that

On the other hand, repeating the same arguments as those of (52) and (54) in the proof of Theorem 24, we can deduce that for all n ≥ 1

for some constant M > 0. Taking into account 0 < liminf _n→∞β_n ≤ limsup _n→∞β_n < 1, we may assume, without loss of generality, that $$. Utilizing (259)-(260) we have

where $$ for some M₁ > 0. In the meantime, observe that

This together with (261), implies that

where sup _n≥0{M₁ + ∥Gx_n − z_n∥} ≤ M₂ for some M₂ > 0. Since $$ and (1 − σ_n)α_n(1 − ρ)/(1 − β_n) ≥ (1 − σ_n)α_n(1 − ρ), we obtain from conditions (i) and (iv) that $$. Thus, applying Lemma 7 to (263), we deduce from condition (iii) and the assumption on {S_n} that (noting that 0 < λ_i ≤ b < 1, for all i ≥ 0)

Next, we show that ∥x_n − Gx_n∥ → 0 as n → ∞.

Indeed, according to Lemma 8 we have from (253)

Utilizing Lemma 15 we get from (253) and (265)

which hence yields

Since α_n → 0 and ∥x_n+1 − x_n∥ → 0, from condition (iv) and the boundedness of {x_n}, it follows that

Utilizing the properties of g, we have

which, together with (253) and (257), implies that

That is,

Since

it immediately follows from (269) and (271) that

On the other hand, observe that y_n can be rewritten as follows:

where e_n = γ_n + δ_n and $$. Utilizing Lemma 11, we have

which hence implies that

Utilizing (271), conditions (i), (ii), (iv), and the boundedness of {x_n}, {y_n} and {f(x_n)}, we get

From the properties of g₁, we have

Utilizing Lemma 15 and the definition of $$, we have

which leads to

Since {x_n} and $$ are bounded, we deduce from (278) and condition (ii) that

From the properties of g₂, we have

Furthermore, y_n can also be rewritten as follows:

where d_n = α_n + δ_n and $$. Utilizing Lemma 11 and the convexity of ∥·∥², we have

which hence implies that

Utilizing (271), conditions (i), (ii), (iv), and the boundedness of {x_n}, {y_n} and {f(x_n)}, we get

From the properties of g₃, we have

Thus, from (282) and (287), we get

That is,

Therefore, from Lemma 12, (273), and (289), it follows that

That is,

Suppose that β_n ≡ β for some fixed β ∈ (0,1) such that α_n + β + γ_n + δ_n = 1 for all n ≥ 0. Define a mapping Vx = (1 − θ₁ − θ₂)Sx + θ₁Bx + θ₂Gx, where θ₁, θ₂ ∈ (0,1) are two constants with θ₁ + θ₂ < 1. Then, by Lemmas 14 and 17, we have that Fix (V) = Fix (S)∩Fix (B)∩Fix (G) = F. For each k ≥ 1, let {p_k} be a unique element of C such that

From Lemma 13, we conclude that p_k → q ∈ Fix (V) = F as k → ∞. Repeating the same arguments as those of (81) in the proof of Theorem 24, we can conclude that for every n, k

where θ_n = α_n∥f(x_n) − Bp_k∥ + (1 − β)∥B_np_k − Bp_k∥ + δ_n∥S_nGx_n − B_nx_n∥. Since lim _n→∞α_n = lim _n→∞∥B_np_k − Bp_k∥ = lim _n→∞∥S_nGx_n − B_nx_n∥ = 0, we know that θ_n → 0 as n → ∞. So, it immediately follows that

where τ_n = θ_n[2(β∥x_n − Bp_k∥ + (1 − β)∥x_n − p_k∥) + θ_n] + ∥x_n+1 − y_n∥[2∥y_n − Bp_k∥ + ∥x_n+1 − y_n∥] → 0 as n → ∞.

For any Banach limit μ, from (294), we derive

In addition, note that

It is easy to see from (273) and (291) that

Utilizing (295) and (297), we deduce that

Repeating the same arguments as those of (99) in the proof of Theorem 24, we can obtain that

Since p_k → q ∈ Fix (V) = F as k → ∞, by the uniform Gateaux differentiability of the norm of X we have

On the other hand, from (257) and the norm-to-weak* uniform continuity of J on bounded subsets of X, it follows that

So, utilizing Lemma 18, we deduce from (300) and (301) that

which together with (271) and the norm-to-weak* uniform continuity of J on bounded subsets of X, implies that

Finally, let us show that x_n → q as n → ∞. Utilizing Lemma 8 (i), from (253) and the convexity of ∥·∥, we get

and hence

From conditions (i) and (iv), it is easy to see that $$. Applying Lemma 7 to (305), we infer that x_n → q as n → ∞. This completes the proof.