Equivalence of the existence of best proximity points and best proximity pairs for cyclic and noncyclic nonexpansive mappings

Definition 1.1

A point (p, q) ∈ A × B is said to be a best proximity pair for the noncyclic mapping T: A ∪ B → A ∪ B provided that

Also, if T: A ∪ B → A ∪ B is a cyclic mapping, then a point x* ∈ A ∪ B is called a best proximity point for T provided that

Some existence results of best proximity points (pairs) can be found in [1,2,3,4,5].

Definition 1.2

T: A ∪ B → A ∪ B is called a cyclic (noncyclic) nonexpansive mapping, if T is cyclic (noncyclic) and

Also, T is said to be a cyclic (noncyclic) contraction provided that T is cyclic (noncyclic) and there exists α ∈ (0, 1) for which

for all (x, y) ∈ A × B.

It is clear that the class of cyclic (noncyclic) nonexpansive mappings contains the class of cyclic (noncyclic) contractions as a subclass.

In order to state the existence and convergence results of best proximity points (pairs), we need to recall the following concepts and notations.

Definition 1.3

A Banach space X is said to be

(i) uniformly convex if there exists a strictly increasing function δ: [0, 2] → [0, 1] such that the following implication holds for all x, y, p ∈ X, R >0 and r ∈ [0, 2R]:

(ii) strictly convex if the following implication holds for all x, y, p ∈ X and R > 0:

The proximal pair of the pair (A, B) is denoted by (A₀, B₀) and given by

The pair (A, B) is said to be a proximinal pair if A = A₀ and B = B₀.

We shall also adopt the notation

Definition 1.4

[6] A convex pair (K₁, K₂) in a Banach space X is said to have a proximal normal structure (PNS) if for any bounded, closed, convex and proximinal pair (H₁, H₂) ⊆ (K₁,K₂) for which dist(H₁, H₂) = dist(K₁, K₂) and δ(H₁, H₂) > dist(H₁, H₂), there exists (x₁, x₂) ∈ H₁ × H₂ such that

It was announced in [6] that every nonempty, bounded, closed and convex pair in a uniformly convex Banach space X has PNS.

Here, we state the following two existence results which are the main conclusions of [6].

Theorem 1.5

[6, Theorem 2.1] Let (A, B) be a nonempty, weakly compact and convex pair in a Banach space X and suppose (A, B) has PNS. Let T: A ∪ B → A ∪ B be a cyclic nonexpansive mapping. Then, T has a best proximity point.

Theorem 1.6

[6, Theorem 2.2] Let (A, B) be a nonempty, weakly compact and convex pair in a strictly convex Banach space X and suppose (A, B) has PNS. Let T: A ∪ B → A ∪ B be a noncyclic nonexpansive mapping. Then, T has a best proximity pair.

Theorem 1.7

[7, Theorem 3.10] Let (A, B) be a nonempty, closed and convex pair in a uniformly convex Banach space X and let T: A ∪ B → A ∪ B be a cyclic contraction map. For x₀ ∈ A, define x_n+1 := Tx_nfor each n ≥ 0. Then, there exists a unique x* ∈ A such that x_2n → x* and ∥x* − Tx*∥ = dist(A, B).

Proposition 1.8

[8,9] Let (A, B) be a nonempty, bounded, closed and convex pair in a reflexive and strictly convex Banach space X. DefineP:A0∪B0→A0∪B0 as

1. Pis cyclic on A₀ ∪ B₀and∥x−Px∥=dist(A,B)for any x ∈ A₀ ∪ B₀,

2. Pis an isometry, that is,∥Px−Py∥=∥x−y∥for all (x,y) ∈ A₀ × B₀,

3. Pis affine,

4. P2|A0=iA0andP2|B0=iB0, where i_Eis the identity mapping on the subset E of X.

   Moreover, if X is a uniformly convex Banach space, then

5. P|A0andP|B0are continuous.

Theorem 1.9

[8, Theorem 3.2] Let (A, B) be a nonempty, closed and convex pair in a uniformly convex Banach space X and T a noncyclic contraction mapping defined on A ∪ B. Suppose x₀ ∈ A₀and define

Theorem 2.1

Let (A,B) be a nonempty, weakly compact and convex pair in a strictly convex Banach space X. Then, every cyclic nonexpansive mapping defined on A ∪ B has a best proximity point if and only if every noncyclic nonexpansive mapping defined on A ∪ B has a best proximity pair.

Proof

Assume that every cyclic nonexpansive mapping defined on A ∪ B has a best proximity point and T: A ∪ B → A ∪ B is a noncyclic nonexpansive mapping. If x ∈ A₀, then there exists a point y ∈ B₀ for which ∥x − y∥ = dist(A, B). By the fact that T is a noncyclic nonexpansive mapping, we obtain ∥Tx − Ty∥ = dist(A,B) and so Tx ∈ A₀ which ensures that T(A₀) ⊆ A₀. Similarly, T(B₀) ⊆ B₀, that is, T is noncyclic on A₀ ∪ B₀. Consider the projection operator P as in (1). It follows from the proof of Theorem 3.2 of [8] that T and P commute on A₀ ∪ B₀. Since P is cyclic and T is noncyclic on A₀ ∪ B₀, we obtain

Therefore, TP is cyclic on A₀ ∪ B₀. In view of the fact that P is an isometry,

for all (x, y) ∈ A₀ × B₀, that is, TP is a cyclic nonexpansive mapping on A₀ ∪ B₀. Now by assumption, there exists a point x* ∈ A₀ such that ∥x⁎−TPx⁎∥=dist(A,B). Moreover,

Strict convexity of X implies that Tx* = x*. Furthermore,

Hence, (x⁎,Px⁎) is a best proximity pair of the mapping T. Conversely, assume that any noncyclic nonexpansive mapping defined on A₀ ∪ B₀ has a best proximity pair and S: A ∪ B → A ∪ B is a cyclic nonexpansive mapping. Thus,

that is, SP is noncyclic on A₀ ∪ B₀ and again since P is an isometry, we conclude that SP is a noncyclic nonexpansive mapping. Now by the assumption SP has a best proximity pair, called (p, q) ∈ A₀ × B₀. Thereby,

We have

and so p is a best proximity point of S. Similarly, we can see that q is a best proximity point of S in B₀ and this completes the proof.□

Corollary 2.2

Theorem 1.6 is a consequence of Theorem 1.5 (see the proof of Theorem 1.6 in [6]).

Theorem 2.3

Theorem 1.9 implies Theorem 1.7 when the initial point of the iterative sequence in Theorem 1.7 is chosen in A₀.

Proof

Let (A, B) be a nonempty, closed and convex pair in a uniformly convex Banach space X and S: A ∪ B → A ∪ B be a cyclic contraction. From Proposition 3.3 of [7], the pair (A₀, B₀) is nonempty. Consider the mapping SP on A₀ ∪ B₀. A proof which is similar to the one of Theorem 2.1 shows that the mapping SP is a noncyclic contraction on A₀ ∪ B₀. Thus, for any x₀ ∈ A₀ if we define

then {(x_n, y_n)} converges to a best proximity pair of the mapping SP. Suppose x_n → p. Continuity of the projection operator P implies that yn→Pp. Since P2 is identity on A₀ and B₀, respectively (Proposition 1.8),

and

Hence, p ∈ A₀ is a best proximity point for the mapping S. We also note that for any n∈ℕ

Theorem 2.4

Theorem 1.7 implies Theorem 1.9 for an even subsequence of the iterative sequence defined in (2).

Proof

Let (A, B) be a nonempty, closed and convex pair in a uniformly convex Banach space X and T: A ∪ B → A ∪ B be a noncyclic contraction. From Proposition 3.4 of [10], the pair (A₀, B₀) is nonempty. Consider the mapping TP on A₀ ∪ B₀. The mapping TP is a cyclic contraction on A₀ ∪ B₀. Thus, for all x₀ ∈ A₀ if we define xn=(TP)nx0, then the sequence {x_2n} converges to a best proximity point of TP, say x* ∈ A₀. In view of the fact that TP is a cyclic nonexpansive mapping by Theorem 2.1, we conclude that (x⁎,Px⁎) is a best proximity pair for T. From Proposition 1.8, since P2 is identity map on A₀, we must have

Continuity of the projection operator on A₀ ensures that y2n:=Px2n→Px⁎ and so the sequence {(x2n,Px2n)} converges to a best proximity pair of T. Notice that for any n∈ℕ we have

Corollary 2.5

The convergence results of Theorem 1.7 and Theorem 1.9 are independent. That is, the convergence result of Theorem 1.7 cannot be implied by the convergence result of Theorem 1.9 and vice versa.

Example 2.1

Consider the Banach space X=ℝ2 and let