Implicit Iterative Method for Hierarchical Variational Inequalities

Proposition 2.1. Let C⊆H be a nonempty closed and convex set and A : H → H monotone and hemicontinuous. Then,

• (a)

  [1] VI (C, A) = {x^* ∈ C : 〈Ay, y − x^*〉 ≥ 0,   for  all  y ∈ C},

• (b)

  [1] VI (C, A) ≠ ∅ when C is bounded,

• (c)

  [16, Lemma  2.24] VI (C, A) = Fix (P_C(I − λA)) for all λ > 0, where I stands for the identity mapping on H,

• (d)

  [16, Theorem  2.31] VI (C, A) consists of one point if A is strongly monotone and Lipschitz continuous.

Proposition 2.2. Let C⊆H be a nonempty closed and convex subset and T : C → C a nonexpansive map.

• (a)

  [18, Proposition  5.3] Fix (T) is closed and convex.

• (b)

  [18, Theorem  5.1] Fix (T) ≠ ∅ when C is bounded.

Proposition 2.3 (see [6], Proposition  2.3.)Let C⊆H be a nonempty closed and convex set and A : H → H an α-inverse-strongly monotone operator. Then, for any given λ ∈ (0,2α], the mapping S_λ : H → H, defined by

is nonexpansive and Fix (S_λ) = VI (C, A).

Lemma 2.4 (see [18], Demiclosedness Principle.)Assume that T is a nonexpansive self-mapping on a closed convex subset K of a Hilbert space H. If T has a fixed point, then I − T is demiclosed, that is, whenever {x_n} is a sequence in K weakly converging to some x ∈ K and the sequence {(I − T)x_n} strongly converges to some y, it follows that (I − T)x = y, where I is the identity operator of H.

Lemma 2.5 (see [19], page 80.)Let $$, $$, and $$ be sequences of nonnegative real numbers satisfying the inequality

If $$ and $$, then lim _n→∞a_n exists. If in addition $$ has a subsequence, which converges to zero, then lim _n→∞a_n = 0.

Lemma 2.6 (see [12].)If 0 ≤ λ < 1 and 0 < μ < 2η/κ², then

where $$.

Theorem 3.1. Let H be a real Hilbert space, A an α-inverse-strongly monotone mapping, and F : H → H a κ-Lipschitz continuous and η-strongly monotone mapping for some constants κ, η > 0. Let $$ be N nonexpansive self-maps on H with a nonempty common fixed point set $$. Suppose $$. Denote by T_n : = T_{n  mod   N} for n > N. Let μ ∈ (0,2η/κ²), x₀ ∈ H, $$, $$, and $$ be such that $$, β_n ≤ λ_n and a ≤ α_n ≤ b, for  all  n ≥ 1, for some a, b ∈ (0,1). Then, the sequence $$, defined by

converges weakly to an element of $$.

If, in addition, $$, then {x_n} converges weakly to an element of $$.

Proof. Notice first that the following identity:

holds for all x, y ∈ H and all t ∈ [0,1]. Let $$ be an arbitrary element of $$. Observe that Since A is α-inverse strongly monotone and $$, we have By Lemma 2.6, we have It follows This together with (3.7) yields and so, Since $$ converges, by Lemma 2.5, $$ exists. As a consequence, the sequence {x_n} is bounded. Moreover, we have Therefore, Obviously, it is easy to see that lim _n→∞∥x_n − x_n+i∥ = 0 for each i = 1,2, …, N. Now observe that Also note that the boundedness of {x_n} implies that $$ and $$ are both bounded. Thus, we have This implies Consequently, and hence, for each i = 1,2, …, N, This shows that lim _n→∞∥x_n − T_n+ix_n∥ = 0 for each i = 1,2, …, N. Therefore,

On the other hand, since {x_n} is bounded, it has a subsequence $$, which converges weakly to some $$, and so, we have $$. From Lemma 2.4, it follows that I − T_l is demiclosed at zero. Thus, $$. Since l is an arbitrary element in the finite set {1,2, …, N}, we get $$.

Now, let x^* be an arbitrary element of ω_w(x_n). Then, there exists another subsequence $$ of {x_n}, which converges weakly to x^* ∈ H. Clearly, by repeating the same argument, we get $$. We claim that $$. Indeed, if $$, then by the Opial’s property of H, we conclude that

This leads to a contradiction, and so, we get $$. Therefore, ω_w(x_n) is a singleton set. Hence, {x_n} converges weakly to a common fixed point of the mappings $$, denoted still by x^*.

Assume that $$. Let $$ be arbitrary but fixed. Then, it follows from the nonexpansiveness of each T_i and the monotonicity of A that

which implies that where $$. Note that λ_n → 0,   β_n ≤ λ_n,   for  all  n ≥ 1, and $$. Thus, for any ɛ > 0, there exists an integer m₀ ≥ 1 such that $$ for all n ≥ m₀. Consequently, 0 ≤ ɛ + 2〈Ay, y − x_n〉 for all n ≥ m₀. Since x_n⇀x^*, we have ɛ + 2〈Ay, y − x^*〉 ≥ 0 as n → ∞. Therefore, from the arbitrariness of ɛ > 0, we deduce that 〈Ay, y − x^*〉 ≥ 0 for all $$. Proposition 2.1 (a) ensures that that is, $$.

Corollary 3.2 (see [14], Theorem  2.1.)Let H be a real Hilbert space and F : H → Hκ-Lipschitz continuous and η-strongly monotone for some constants κ > 0, η > 0. Let $$ be N nonexpansive self-maps on H such that $$. Let μ ∈ (0,2η/κ²), x₀ ∈ H, $$, and $$ be such that $$ and a ≤ α_n ≤ b,   for  all  n ≥ 1, for some a, b ∈ (0,1). Then, the sequence $$, defined by

converges weakly to a common fixed point of the mappings $$.

Proof. In Theorem 3.1, putting A ≡ 0, we can see readily that for any given positive number α ∈ (0, ∞), A : H → H is an α-inverse-strongly monotone mapping. In this case, we have

Hence, for any given sequence $$ with β_n ≤ λ_n(∀n ≥ 1), the implicit iterative scheme (3.5) reduces to (3.25). Therefore, by Theorem 3.1, we obtain the desired result.

Lemma 3.3. In the setting of Theorem 3.1, we have

• (a)

  $$ exists for each $$,

• (b)

  lim _n→∞d(x_n, C) exists, where d(x_n, C) = inf _p∈C∥x_n − p∥,

• (c)

  liminf _n→∞∥x_n − T_nx_n∥ = 0,

where $$.

Proof. Conclusion (a) follows from (3.12), and conclusion (c) follows from (3.18). We prove conclusion (b). Indeed, for each $$,

This together with (3.12) implies that and hence, where Since $$ and both {Ax_n−1} and {F(x_n−1)} are bounded, it is known that $$. On account of Lemma 2.5, we deduce that lim _n→∞d(x_n, C) exists, that is, conclusion (b) holds.

Theorem 3.4. In the setting of Theorem 3.1, the sequence {x_n} converges strongly to an element of $$if and only if lim  inf _n→∞d(x_n, C) = 0 where $$.

Proof. From (3.28), we derive for each n ≥ 1

where $$. Put $$. Then $$.

Suppose that the sequence {x_n} converges strongly to a common fixed point p of the family $$. Then, lim _n→∞∥x_n − p∥ = 0. Since

we have lim  inf _n→∞ d(x_n, C) = 0.

Conversely, suppose that lim  inf _n→∞d(x_n, C) = 0. Then, by Lemma 3.3 (b), we deduce that lim _n→∞d(x_n, C) = 0. Thus, for arbitrary ɛ > 0, there exists a positive integer N₀ such that

Furthermore, the condition $$ implies that there exists a positive integer N₁ such that Choose N_* = max {N₀, N₁}. Observe that (3.31) yields Note that $$ and $$. Thus, for all n, m ≥ N_* and all $$, we have from (3.35) that Taking the infimum over all $$, we obtain and hence, ∥x_n − x_m∥ ≤ ɛ. This shows that $$ is a Cauchy sequence in H. Let x_n → p ∈ H as n → ∞. Then, we derive from (3.20) that for each l = 1,2, …, N, Therefore, p ∈ Fix (T_l) for each l = 1,2, …, N, and hence, $$.

On the other hand, choose a positive sequence $$ such that ɛ_n → 0 as n → ∞. For each n ≥ 1, from the definition of d(x_n, C), it follows that there exists a point p_n ∈ C such that

Since d(x_n, C) → 0 and ɛ_n → 0 as n → ∞, it is clear that ∥x_n − p_n∥ → 0 as n → ∞. Note that Hence, we get Furthermore, for each β_n ∈ (0,2α], the mapping $$ is defined as follows: From Proposition 2.3, we deduce that $$ is nonexpansive and From Proposition 2.2 (a), we conclude that $$ is closed and convex. Thus, from the condition $$, it is known that C is a nonempty closed and convex set. Since {p_n} lies in C and converges strongly to p, we must have p ∈ C.

Remark 3.5. Setting A = 0 in Lemma 3.3 and Theorem 3.4 above, we shall derive Lemma  2.1 and Theorem  2.2 in [14] as direct consequences, respectively.