General Iterative Algorithms for Hierarchical Fixed Points Approach to Variational Inequalities

Lemma 2.1 (Browder [10]). Let H be a Hilbert space, C be a closed convex subset of H, and T : C → C be a nonexpansive mapping with F(T) ≠ ∅. If {x_n} is a sequence in C weakly converging to x and if {(I − T)x_n} converges strongly to y, then (I − T)x = y; in particular, if y = 0 then x ∈ F(T).

Lemma 2.2. Let x ∈ H and z ∈ C be any points. Then one has the following:

• (1)

  That z = P_C[x] if and only if there holds the relation:

  $$ ()

• (2)

  That z = P_C[x] if and only if there holds the relation:

  $$ ()

• (3)

  There holds the relation:

  $$ ()
  Consequently, P_C is nonexpansive and monotone.

Lemma 2.3 (Marino and Xu [8]). Let H be a Hilbert space, C be a closed convex subset of H, f : C → H be a contraction with coefficient 0 < ρ < 1, and T : C → C be nonexpansive mapping. Let A be a strongly positive linear bounded operator on a Hilbert space H with coefficient $$. Then, for $$, for x, y ∈ C,

• (1)

  the mapping (I − f) is strongly monotone with coefficient (1 − ρ), that is,

  $$ ()

• (2)

  the mapping (A − γf) is strongly monotone with coefficient $$ that is

  $$ ()

Lemma 2.4 (Xu [4]). Assume that {a_n} is a sequence of nonnegative numbers such that

where {γ_n} is a sequence in (0,1) and {δ_n} is a sequence in ℝ such that

• (1)

  $$,

• (2)

  limsup _n→∞(δ_n/γ_n) ≤ 0 or $$. Then lim _n→∞a_n = 0.

Lemma 2.5 (Marino and Xu [8]). Assume A is a strongly positive linear bounded operator on a Hilbert space H with coefficient $$ and 0 < α ≤ ∥A∥⁻¹. Then $$.

Lemma 2.6 (Acedo and Xu [11]). Let C be a closed convex subset of H. Let {x_n} be a bounded sequence in H. Assume that

• (1)

  The weak ω-limit set ω_w(x_n) ⊂ C,

• (2)

  For each z ∈ C, lim _n→∞∥x_n − z∥ exists. Then {x_n} is weakly convergent to a point in C.

Notation We use → for strong convergence and ⇀ for weak convergence.

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a ρ-contraction with ρ ∈ (0,1). Let S, T : C → C be two nonexpansive mappings with F(T) ≠ ∅. Let A be a strongly positive linear bounded operator on H with coefficient $$. {α_n} and {β_n} are two sequences in (0,1) and $$. Starting with an arbitrary initial guess x₀ ∈ C and {x_n} is a sequence generated by

Suppose that the following conditions are satisfied:

• (C1)

    lim _n→∞α_n = 0 and $$,

• (C2)

    lim _n→∞(β_n/α_n) = τ = 0,

• (C3)

    lim _n→∞(|α_n − α_n−1|/α_n) = 0 and lim _n→∞(|β_n − β_n−1|/β_n) = 0, or

• (C4)

    $$ and $$.

Then the sequence {x_n} converges strongly to a point $$, which is the unique solution of the variational inequality: Equivalently, one has $$.

Proof. We first show the uniqueness of a solution of the variational inequality (3.2), which is indeed a consequence of the strong monotonicity of A − γf. Suppose $$ and $$ both are solutions to (3.2), then $$ and $$. It follows that

The strong monotonicity of A − γf (Lemma 2.3) implies that $$ and the uniqueness is proved.

Next, we prove that the sequence {x_n} is bounded. Since α_n → 0 and lim _n→∞(β_n/α_n) = 0 by condition (C1) and (C2), respectively, we can assume, without loss of generality, that α_n < ∥A∥⁻¹ and β_n < α_n for all n ≥ 1. Take u ∈ F(T) and from (3.1), we have

Since $$ and by Lemma 2.5, we note that By induction, we can obtain which implies that the sequence {x_n} is bounded and so are the sequences {f(x_n)}, {Sx_n}, and {ATy_n}.

Set w_n : = α_nγf(x_n)+(I − α_nA)Ty_n,   n ≥ 1. We get

It follows that By (3.7) and (3.8), we get From (3.1), we obtain where M is a constant such that Substituting (3.10) into (3.8) to obtain At the same time, we can write (3.12) as From (3.12), (C4), and Lemma 2.5 or from (3.13), (C3), and Lemma 2.5, we can deduce that ∥x_n+1 − x_n∥→0, respectively.

From (3.1), we have

Notice that α_n → 0, β_n → 0, and ∥x_n+1 − x_n∥→0, so we obtain Next, we prove where z = P_F(T)(I − A + γf)z. Since the sequence {x_n} is bounded we can take a subsequence $$ of {x_n} such that and $$. From (3.15) and by Lemma 2.1, it follows that $$. Hence, by Lemma 2.2(1) that Now, by Lemma 2.2(1), we observe that and so Hence, it follows that We observe that Thus, by Lemma 2.4, x_n → z as n → ∞. This is completes.

Corollary 3.2 (Yao et al. [9]). Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a ρ-contraction (possibly nonself) with ρ ∈ (0,1). Let S, T : C → C be two nonexpansive mappings with F(T) ≠ ∅.  {α_n} and {β_n} are two sequences in (0,1). Starting with an arbitrary initial guess x₀ ∈ C and {x_n} is a sequence generated by

Suppose that the following conditions are satisfied:

• (C1)

    lim _n→∞α_n = 0 and $$,

• (C2)

    lim _n→∞(β_n/α_n) = 0,

• (C3)

    lim _n→∞(|α_n − α_n−1|/α_n) = 0 and lim _n→∞(|β_n − β_n−1|/β_n) = 0, or

• (C4)

    $$ and $$.

Then the sequence {x_n} converges strongly to a point $$, which is the unique solution of the variational inequality: Equivalently, one has $$. In particular, if one takes f = 0, then the sequence {x_n} converges in norm to the Minimum norm fixed point $$ of T, namely, the point $$ is the unique solution to the quadratic minimization problem:

Proof. As a matter of fact, if we take A = I and γ = 1 in Theorem 3.1. This completes the proof.

Theorem 3.3. Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a ρ-contraction (possibly nonself) with ρ ∈ (0,1). Let S, T : C → C be two nonexpansive mappings with F(T) ≠ ∅. Let A be a strongly positive linear bounded operator on a Hilbert space H with coefficient $$ and $$. {α_n} and {β_n} are two sequences in (0,1). Starting with an arbitrary initial guess x₀ ∈ C and {x_n} is a sequence generated by

Suppose that the following conditions are satisfied:

• (C1)

    lim _n→∞α_n = 0 and $$,

• (C2)

    lim _n→∞(β_n/α_n) = τ ∈ (0, ∞),

• (C5)

    lim _n→∞((|α_n − α_n−1| + |β_n − β_n−1|)/α_nβ_n) = 0,

• (C6)

    there exists a constant K > 0 such that (1/α_n)|1/β_n − 1/β_n−1| ≤ K.

Then the sequence {x_n} converges strongly to a point $$, which is the unique solution of the variational inequality:

Proof. First of all, we show that (3.27) has the unique solution. Indeed, let $$ and $$ be two solutions. Then

Analogously, we have Adding (3.28) and (3.29), by Lemma 2.3, we obtain and so $$. From (C2), we can assume, without loss of generality, that β_n ≤ (τ + 1)α_n for all n ≥ 1. By a similar argument in Theorem 3.1, we have By induction, we obtain which implies that the sequence {x_n} is bounded. Since (C5) implies (C4) then, from Theorem 3.1, we can deduce ∥x_n+1 − x_n∥→0.

From (3.1), we note that

Hence, it follows that and so Set v_n : = (x_n − x_n+1)/(1 − α_n)β_n. Then, we have From (3.12) in Theorem 3.1 and (C6), we obtain This together with Lemma 2.4 and (C2) imply that From (3.36), for z ∈ F(T), we have By Lemmas 2.2 and 2.3, we obtain Now, we observe that From (C1) and (C2), we have β_n → 0. Hence, from (3.1), we deduce ∥y_n − x_n∥→0 and ∥x_n+1 − Ty_n∥→0. Therefore,

Since v_n → 0, (I − T)y_n → 0, A(Ty_n − x_n) → 0, and $$, every weak cluster point of {x_n} is also a strong cluster point. Note that the sequence {x_n} is bounded, thus there exists a subsequence $$ converging to a point $$. For all z ∈ F(T), it follows from (3.39) that

Letting k → ∞, we obtain By Lemma 2.6 and (3.27) having the unique solution, it follows that $$. Therefore, $$ as n → ∞. This completes the proof.

Corollary 3.4 (Yao et al. [9]). Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a ρ-contraction (possibly nonself) with ρ ∈ (0,1). Let S, T : C → C be two nonexpansive mappings with F(T) ≠ ∅. {α_n} and {β_n} are two sequences in (0,1) Starting with an arbitrary initial guess x₀ ∈ C and {x_n} is a sequence generated by

Suppose that the following conditions are satisfied:

• (C1)

    lim _n→∞α_n = 0 and $$,

• (C2) 

   lim _n→∞(β_n/α_n) = τ ∈ (0, ∞),

• (C5)

    lim _n→∞((|α_n − α_n−1| + |β_n − β_n−1|)/α_nβ_n) = 0,

• (C6)

    there exists a constant K > 0 such that (1/α_n)|1/β_n − 1/β_n−1| ≤ K.

Then the sequence {x_n} converges strongly to a point $$, which is the unique solution of the variational inequality:

Proof. As a matter of fact, if we take A = I and γ = 1 in Theorem 3.3 then this completes the proof.

Corollary 3.5 (Yao et al. [9]). Let C be a nonempty closed convex subset of a real Hilbert space H. Let S, T : C → C be two nonexpansive mappings with F(T) ≠ ∅. {α_n} and {β_n} are two sequences in (0,1). Starting with an arbitrary initial guess x₀ ∈ C and suppose {x_n} is a sequence generated by

Suppose that the following conditions are satisfied:

• (C1)

    lim _n→∞α_n = 0 and $$,

• (C2)

    lim _n→∞(β_n/α_n) = 1,

• (C5)

    lim _n→∞((|α_n − α_n−1| + |β_n − β_n−1|)/α_nβ_n) = 0,

• (C6)

    there exists a constant K > 0 such that (1/α_n)|1/β_n − 1/β_n−1| ≤ K.

Then the sequence {x_n} converges strongly to a point $$, which is the unique solution of the variational inequality:

Proof. As a matter of fact, if we take A = I, f = 0, and γ = 1 in Theorem 3.3 then this is completes the proof.

Remark 3.6. Prototypes for the iterative parameters are, for example, α_n = n^−θ and β_n = n^−ω (with θ, ω > 0). Since |α_n − α_n−1 | ≈ n^−θ and |β_n − β_n−1 | ≈ n^−ω, it is not difficult to prove that (C5) is satisfied for 0 < θ, ω < 1 and (C6) is satisfied if θ + ω ≤ 1.

Remark 3.7. Our results improve and extend the results of Yao et al. [9] by taking A = I and γ = 1 in Theorems 3.1 and 3.3.

Example 3.8. Let H = ℝ,  C = [−1/4, 1/4],  T = I, S = −I,   A = I, f(x) = x², P_C = I,$$, $$ for every n ∈ ℕ, we have τ = 1 and choose $$, ρ = 1/3 and γ = 1. Then {x_n} is the sequence

and $$ as n → ∞, where $$ is the unique solution of the variational inequality