Strong Convergence Algorithms for Hierarchical Fixed Points Problems and Variational Inequalities

Lemma 2.1 (see [8].)Let T : C → C be a nonexpansive mapping with F(T) ≠ ∅. If {x_n} is a sequence in C weakly converging to a point x ∈ C and {(I − T)x_n} converges strongly to a point y ∈ C, then (I − T)x = y. In particular, if y = 0, then x ∈ F(T).

Lemma 2.2 (see [9].)Let f : C → H be a contraction with coefficient λ ∈ [0,1) and T : C → C a nonexpansive mapping. Then one has the following.

• (1)

  The mapping (I − f) is strongly monotone with coefficient (1 − λ), that is,

  $$ ()

• (2)

  The mapping I − T is monotone, that is,

  $$ ()

Lemma 2.3 (see [10].)Let {s_n}, {c_n} be the sequences of nonnegative real numbers, and let {a_n}⊂(0,1). Suppose that {b_n} is a real number sequence such that

Assume that $$. Then the following results hold.

• (1)

  If b_n ≤ βa_n, where β ≥ 0, then {s_n} is a bounded sequence.

• (2)

  If one has

  $$ ()

then lim _n→∞s_n = 0.

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a λ-contraction with λ ∈ [0,1). Let S : C → H be a nonexpansive mapping and $$ a countable family of nonexpansive mappings of C into itself such that $$. Set α₀ = 1, and let {α_n}⊂(0,1) be a strictly decreasing sequence and {β_n}⊂(0,1) a sequence satisfying the following conditions:

• (a)

  lim _n→∞α_n = 0 and $$,

• (b)

  lim _n→∞(β_n/α_n) = 0,

• (c)

  $$ and $$.

Then the sequence {x_n} generated by (1.9) converges strongly to a point x^* ∈ F, which is the unique solution of the variational inequality

Proof. First, P_Ff is a contraction from C into itself with a constant λ and C is complete, and there exists a unique x^* ∈ C such that x^* = P_Ff(x^*). From (2.4), it follows that x^* is the unique solution of the problem (3.1).

Now, we prove that {x_n} converges strongly to x^*. To this end, we first prove that {x_n} is bounded. Take p ∈ F. Then it follows from (1.9) that

and hence (note that {α_n} is strictly decreasing) By condition (b), we can assume that β_n ≤ α_n for all n ≥ 1. Hence, from above inequality, we get For each n ≥ 1, let a_n = α_n(1 − λ), b_n = α_n(∥f(p) − p∥+∥Sp − p∥), and c_n = 0. Then {a_n}, {b_n}, and {c_n} satisfy the condition of Lemma 2.3(1). Hence, it follows from Lemma 2.3(1) that {x_n} is bounded and so are {f(x_n)}, {y_n}, {T_ix_n}, and {T_iy_n} for all i ≥ 1. Set $$ for each n ≥ 1. From (1.9), we have where M is a constant such that From (1.9), we have Substituting (3.7) into (3.5), we get that Let a_n = (1 − λ)α_n, b_n = 0, and c_n = (α_n−1 − α_n)+|β_n − β_n−1|. Then conditions (a) and (c) imply that {a_n}, {b_n}, and {c_n} satisfy the condition of Lemma 2.3(2). Thus, by Lemma 2.3(2), we can conclude that Since T_ix_n ∈ C for each i ≥ 1 and $$, we have Now, fixing a z ∈ F, from (1.9) we have Hence it follows that where

Now, from (2.2) and (3.12), we get

From (a), (b), (3.9), and (3.14), we have Since $$ for each i ≥ 1 and {α_n} is strictly decreasing, one has

Next, we prove that

Since {x_n} is bounded, we can take a subsequence $$ of {x_n} such that $$ and From (3.16) and Lemma 2.1, we conclude that x′ ∈ F(T_i) for each i ≥ 1, that is, $$. Then By (2.4), we have Also, Thus, It follows that Let a_n = 2(1 − λ)α_n/(1 + (1 − λ))α_n, b_n = (2(1 − λ)α_n/(1 + (1 − λ)α_n)){(1 − α_n)β_n/(1 − λ)α_n∥Sx^*−x^*∥∥x_n+1 − x^*∥+(1/1 − λ)〈f(x^*) − x^*, x_n+1 − x^*〉}, and c_n = 0 for all n ≥ 1. Since $$, and 2(1 − λ)α_n/(1 + (1 − λ)α_n) ≥ (1 − λ)α_n, we have Therefore, it follows from Lemma 2.3(2) that This completes the proof.

Remark 3.2. In (1.9), if f = 0, then it follows that x_n → x^* = P_F0. In this case, from (3.1), it follows that

that is, Therefore, the point x^* is the unique solution to the quadratic minimization problem

Corollary 3.3. Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a λ-contraction with λ ∈ [0,1). Let S : C → H be a nonexpansive mapping and T : C → C a nonexpansive mapping such that F(T) ≠ ∅. Let x₁ ∈ C and define a sequence {x_n} by

where {α_n}⊂(0,1) and {β_n}⊂(0,1) are two sequences satisfying the following conditions:

• (a)

  lim _n→∞α_n = 0 and $$,

• (b)

  lim _n→∞(β_n/α_n) = 0,

• (c)

  $$ and $$.

Then the sequence {x_n} converges strongly to a point x^* ∈ F(T), which is the unique solution of the variational inequality

Corollary 3.4. Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a λ-contraction with λ ∈ [0,1). Let S : C → C be a nonexpansive mapping and T : C → C a nonexpansive mapping such that F(T) ≠ ∅. Let x₁ ∈ C and define a sequence {x_n} by

where the sequences {α_n}⊂(0,1) and {β_n}⊂(0,1) are two sequences satisfying the following conditions:

• (a)

  lim _n→∞α_n = 0 and $$,

• (b)

  lim _n→∞(β_n/α_n) = 0,

• (c)

  $$ and $$.

Then the sequence {x_n} converges strongly to a point x^* ∈ F(T), which is the unique solution of the variational inequality

Theorem 3.5. Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a λ-contraction with λ ∈ [0,1). Let S : C → C be a nonexpansive mapping and $$ a countable family of nonexpansive mappings such that $$. Set α₀ = 1. Let x₁ ∈ C and define a sequence {x_n} by

where {α_n}⊂(0,1) is a strictly decreasing sequence and {β_n}⊂(0,1) is a sequence satisfying the following conditions:

• (a)

  lim _n→∞α_n = 0, and $$,

• (b)

  lim _n→∞(β_n/α_n) = τ ∈ (0, ∞),

• (c)

  $$ and $$,

• (d)

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

• (e)

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

Then the sequence {x_n} generated by (3.34) converges strongly to a point x^* ∈ F, which is the unique solution of the variational inequality

Proof. First, the proof of Theorem  3.2 of [1] shows that (3.35) has the unique solution. By a similar argument as in that of Theorem 3.1, we can conclude that {x_n} is bounded, ∥x_n+1 − x_n∥→0 and ∥x_n − T_ix_n∥→0 as n → ∞. Note that conditions (a) and (b) imply that β_n → 0 as n → ∞. Hence we have

It follows that, for all i ≥ 1, Now, it follows from (3.36) and (3.37) that, for all i ≥ 1, From (3.8), we get Note that Hence, from (3.39), we have Let a_n = (1 − λ)α_n and b_n = α_nK∥x_n − x_n−1∥+M[|β_n − β_n−1|/β_n + (α_n−1 − α_n)/β_n]. From conditions (a) and (d), we have By Lemma 2.3(2), we get From (3.34), we have Hence it follows that and hence Let v_n = (x_n − x_n+1)/(1 − α_n)β_n. For any z ∈ F, we have By Lemma 2.2, we have By (2.4), we have Now, from (3.47)–(3.51) it follows that Observe that (3.52) implies that Since v_n → 0, y_n − T_iy_n → 0 for all i ≥ 1, and ∥u_n−1 − u_n∥/α_n → 0 as n → ∞, every weak cluster point of {x_n} is also a strong cluster point. Since {x_n} is bounded, there exists a subsequence $$ of {x_n} converging to a point x^* ∈ C. Note that x_n − T_ix_n → 0 as n → ∞ for all i ≥ 1. By the demiclosed principle for a nonexpansive mapping, we have x^* ∈ F(T_i) for all i ≥ 1 and so $$. From (3.47), (3.48), (3.50), and (3.51), it follows that, for all z ∈ F, Since v_n → 0, (I − T_i)y_n → 0 for all i ≥ 1, and ∥u_n − u_n−1∥/α_n = 0, letting k → ∞ in (3.54), we obtain Since (3.35) has the unique solution, it follows that $$. Since every weak cluster point of {x_n} is also a strong cluster point, we conclude that $$ as n → ∞. This completes the proof.

Corollary 3.6. Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C → H be a λ-contraction with λ ∈ [0,1). Let S : C → C be a nonexpansive mapping and T : C → C a nonexpansive mapping such that F(T) ≠ ∅. Let x₁ ∈ C and define a sequence {x_n} by

where {α_n}⊂(0,1) and {β_n}⊂(0,1) are the sequences satisfying the following conditions:

• (a)

  lim _n→∞α_n = 0 and $$,

• (b)

  lim _n→∞(β_n/α_n) = τ ∈ (0, ∞),

• (c)

  $$ and $$,

• (d)

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

• (e)

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

Then the sequence {x_n} generated by (3.56) converges strongly to a point x^* ∈ F(T), which is the unique solution of the variational inequality

Remark 3.7. In (1.9), if S = I and f is a self-contraction of C, then we get

which is well known as the viscosity method studied by Moudafi [6] and Xu [7]. If S and f are both self-mappings of C in (1.9), then we get the algorithm of Cianciaruso et al. [5].