A General Iterative Algorithm with Strongly Positive Operators for Strict Pseudo-Contractions
Abstract
This paper deals with a new iterative algorithm {xn} with a strongly positive operator A for a k-strict pseudo-contraction T and a non-self-Lipschitzian mapping S in Hilbert spaces. Under certain appropriate conditions, the sequence {xn} converges strongly to a fixed point of T, which solves some variational inequality. The results here improve and extend some recent related results.
1. Introduction
Let C be a closed convex subset of Hilbert space H with inner product 〈·, ·〉 and norm ∥·∥, T : C → H be a nonlinear mapping. The fixed point set of T is denoted by Fix (T); that is, Fix (T) = {x ∈ C, Tx = x}. Fixed point problem is very general in the sense that it includes, as spacial cases, optimization problems, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games, and others.
Recently the problems of the approximation of the common fixed points of nonexpansive mappings were extended to the case of a family of finite or infinite pseudo-contractions; see, for example, [9–11].
2. Preliminaries
In this section, we recall some useful definitions and lemmas for the proof of the main results.
Definition 1. A mapping T : C ↦ C is said to be L-Lipschitzian, if there exists a constant L > 0 such that
It is clear that a Lipschitzian map is a contractive map when 0 < L < 1 and is a nonexpansive map when L = 1. If k = 0; then a k-strict pseudo-contraction map is a nonexpansive map.
Definition 2. A mapping PC : H ↦ C is said to be the metric projection, if for any x ∈ H, there exists a unique nearest point in C denoted by PCx such that
Lemma 3 (see [13].)Let x ∈ H and z ∈ C be any points. There holds
Lemma 4 (see [9], Demiclosedness princple.)Let C be a nonempty closed convex subset of a real Hilbert space H and let T : C ↦ C be a nonexpansive mapping with F(T) ≠ ∅. If {xn} is a sequence in C weakly converging to x and if {(I − T)xn} converges strongly to y, then (I − T)x = y; in particular if y = 0, then x ∈ F(T).
Lemma 5 (see [14].)Let λ be a number in [0,1] and μ ≥ 0. Let F : H ↦ H be a t-Lipschitzian and η-strongly monotone operator on a Hilbert space. Associate with a nonexpansive mapping T : H ↦ H and define the mapping Tλ : H ↦ H by
Lemma 6 (see [4].)Assume that A is a strongly positive bounded linear operator on a Hilbert space H with coefficient and 0 < ρ ≤ ∥A∥−1; then .
Lemma 7 (see [15].)Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : C ↦ C be a k-strict pseudo-contractive mapping. Let γ and δ be two nonnegative real numbers such that (γ + δ)k ≤ γ; then
Lemma 8 (see [16].)Let H be a Hilbert space and C a nonempty convex subset of H. Let T : C ↦ H be a k-strict pseudo-contractive mapping. Define a mapping Jx = δx + (1 − δ)Tx for all x ∈ C. Then as δ ∈ [k, 1), J is a nonexpansive mapping such that F(J) = F(T).
Lemma 9 (see [17].)Let {αn} be a sequence of nonnegative real numbers satisfying the following relation: αn+1 ≤ (1 − γn)αn + δn, where (i) {γn}⊂(0,1), ; (ii) limsup n→∞(δn/γn) = 0 or ; then lim n→∞αn = 0.
3. Main Results
In this section, we prove the strong convergence results on the iterative algorithm for k-strict pseudo-contractions.
Theorem 10. Let C be a nonempty closed convex subset of a real Hilbert space H, S : C ↦ H a non-self-L-Lipschitzian mapping, and T : C ↦ C a k-strict pseudo-contractive mapping such that Fix (T) ≠ ∅. Let F : C ↦ H be a t-Lipschitzian and η-strongly monotone mapping and A : C ↦ C a -strongly positive bounded linear operator. For a given x0 ∈ C, let the sequences {xn} and {yn} generated by (11), where {αn}, {γn}, {δn}∈[0,1], satisfy the following conditions:
- (i)
[1 − μ(η − μt2/2)]((1 + k)/(1 − k)) ≤ 1, μ(η − μt2/2) − τL > 0, ;
- (ii)
lim n→∞γn = 0, lim n→∞δn = 0, , , (γn + δn)k ≤ γn;
- (iii)
lim n→∞(αn/(γn + δn)) = 0, , , .
Proof. The proof is divided into five steps.
Step 1. We first show that the sequences {xn}, {yn} are bounded. Take p ∈ Fix (T), own to T : C ↦ C be a k-strict pseudo-contractive mapping, and define Jx = kx + (1 − k)Tx. By Lemma 8 J is nonexpansive and Fix (J) = Fix (T); therefore Tx = (1/(1 − k))(Jx − kx):
Thus we immediately get that T is a (1 + k)/(1 − k)-Lipschitzian mapping. Then we estimate ∥yn − p∥:
On the other hand, notice that lim n→∞γn = 0, lim n→∞δn = 0; without loss of generality, we may assume that γn + δn ≤ ∥A∥−1; thus
Step 2. Now we prove that ∥xn+1 − xn∥→0 as n → ∞. Denote ξ = μ(η − (μt2/2)):
Step 3. Now we prove that ∥xn+1 − Txn∥→0 as n → ∞:
Step 4. Now we show that limsup n→∞〈(I − A)x*, xn − x*〉≤0, where x* ∈ Fix (T) is the unique solution of the variational inequality. Take a subsequence of {xn} such that
Observe that the sequence {xn} is bounded; without loss of generality we may assume that . By Lemma 4, we get x′ ∈ Fix (T). Therefore by Lemma 3, we have
Step 5. Next we prove that ∥xn+1 − x*∥→0 as n → ∞:
Notice that
Remark 11. The iterative algorithm in Theorem 10 here is a new approximating method, and Lemma 7 plays a key role in the proof of the main results which makes the proof simple.
Acknowledgments
The authors would like to thank editors and referees for many useful comments and suggestions for the improvement of the paper. This work is partially supported by the Natural Science Foundation of Zhejiang Province (Y6100696, Y6110270) and the National Natural Science Foundation (11071169, 11271330).