Some Results on the Best Proximity Pair

Abstract

We give some new conditions for existence and uniqueness of best proximity point. We also introduce the concept of strongly proximity pair and give some interesting results.

1. Introduction

Let 𝑋 be a metric space and 𝐴 and 𝐵 nonempty subsets of 𝑋. If there is a pair (𝑥₀,𝑦₀)∈𝐴×𝐵 for which 𝑑(𝑥₀,𝑦₀)=𝑑(𝐴,𝐵), that 𝑑(𝐴,𝐵) is distance of 𝐴 and 𝐵, then the pair (𝑥₀,𝑦₀) is called a best proximity pair for 𝐴 and 𝐵. Best proximity pair evolves as a generalization of the concept of best approximation, and reader can find some important results of it in [1–4].

Now, as in [5] (see also [6–14]), we can find the best proximity points of the sets 𝐴 and 𝐵 by considering a map 𝑇∶𝐴∪𝐵→𝐴∪𝐵 such that 𝑇(𝐴)⊂𝐵 and 𝑇(𝐵)⊂𝐴. We say that the point 𝑥∈𝐴∪𝐵 is a best proximity point of the pair (𝐴,𝐵), if 𝑑(𝑥,𝑇𝑥)=𝑑(𝐴,𝐵), and we denote the set of all best proximity points of (𝐴,𝐵) by 𝑃_𝑇(𝐴,𝐵), that is, 𝑃_𝑇(𝐴,𝐵)∶={𝑥∈𝐴∪𝐵∶𝑑(𝑥,𝑇𝑥)=𝑑(𝐴,𝐵)}. (1.1) Best proximity pair also evolves as a generalization of the concept of fixed point of mappings, because if 𝐴∩𝐵≠∅, every best proximity point is a fixed point of 𝑇.

The concept of approximate best proximity pair on metric space was introduced in [10, Definition 1.1], but it is clear that 𝑃^𝑎_𝑇(𝐴,𝐵)=𝐴×𝐵. Now, in section two of this paper, we give some conditions that guarantee the existence, uniqueness, or compactness of the set 𝑃_𝑇(𝐴,𝐵). Then, in section three, by introducing the concepts of 𝑇-approximatively compact pair and 𝑇-strongly compact pair, we give some characterizations of a subclass of the best proximity points, namely, the strongly proximity pairs of sets.

2. Some Existence Theorems

In this section, we will consider the existence of the best proximity points, by considering some sequences which converge to that best proximity point. At first, we generalize some result of Eldred and Veeramani [6].

Theorem 2.1. Let 𝐴 and 𝐵 be nonempty closed subsets of a complete metric space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 satisfying 𝑇(𝐴)⊂𝐵, 𝑇(𝐵)⊂𝐴, and 𝑑𝑇𝑥,𝑇𝑦≤𝛼𝑑𝑥,𝑦+𝛽𝑑(𝑥,𝑇𝑥)+𝑑𝑦,𝑇𝑦+𝛾𝑑(𝐴,𝐵), (2.1) for all 𝑥,𝑦∈𝐴∪𝐵, where 𝛼,𝛽,𝛾≥0 and 𝛼+2𝛽+𝛾<1. If 𝐴 (or 𝐵) is boundedly compact, then there exists 𝑥∈𝐴∪𝐵 with 𝑑(𝑥,𝑇𝑥)=𝑑(𝐴,𝐵).

Proof. Suppose 𝑥₀ is an arbitrary point of 𝐴∪𝐵 and define 𝑥_𝑛+1=𝑇𝑥_𝑛. Now, 𝑑𝑥_𝑛+1,𝑥_𝑛+2=𝑑𝑇𝑥_𝑛,𝑇𝑥_𝑛+1≤𝛼𝑑𝑥_𝑛,𝑥_𝑛+1+𝛽𝑑𝑥_𝑛,𝑇𝑥_𝑛+𝑑𝑥_𝑛+1,𝑇𝑥_𝑛+1+𝛾𝑑(𝐴,𝐵). (2.2) So 𝑑𝑥_𝑛+1,𝑥_𝑛+2≤𝛼+𝛽𝑑𝑥1−𝛽^𝑛,𝑥_𝑛+1+𝛾𝑑(),1−𝛽𝐴,𝐵 (2.3) which implies that 𝑑𝑥_𝑛+1,𝑥_𝑛+2≤𝑘𝑑𝑥_𝑛,𝑥_𝑛+1+(1−𝑘)𝑑(𝐴,𝐵), (2.4) where 𝑘=((𝛼+𝛽)/(1−𝛽))<1. Hence, inductively, we have 𝑑𝑥_𝑛+1,𝑥_𝑛≤𝑘^𝑛𝑑𝑥₁,𝑥₀+1−𝑘^𝑛𝑑(𝐴,𝐵), (2.5) and so 𝑑𝑥_𝑛+1,𝑥_𝑛⟶𝑑(𝐴,𝐵). (2.6) Therefore, by Proposition 3.3 of [6], both sequences {𝑥_2𝑛} and {𝑥_2𝑛+1} are bounded. Now, since 𝐴 (or 𝐵) is boundedly compact then {𝑥_2𝑛} has a convergent subsequence, and so, by Proposition 3.2 of [6], there exists 𝑥∈𝐴 such that 𝑑(𝑥,𝑇𝑥)=𝑑(𝐴,𝐵).

Now, we show that the mapping 𝑇, which satisfies (2.1), has a unique best proximity point in the uniformly convex Banach space 𝑋.

Theorem 2.2. Let 𝐴 and 𝐵 be two nonempty closed and convex subsets of a uniformly convex Banach space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 satisfying𝑇(𝐴)⊂𝐵, 𝑇(𝐵)⊂𝐴 and the condition (2.1). Then, there exists a unique element 𝑥∈𝐴 such that ‖𝑥−𝑇𝑥‖=𝑑(𝐴,𝐵). Further, if 𝑥₀∈𝐴 and 𝑥_𝑛+1=𝑇𝑥_𝑛, then {𝑥_2𝑛} converges to the above unique element.

Proof. One can prove this theorem by the method of the Proposition 3.10 of [6].

Note that the uniform convexity of the Banach space 𝑋 is necessary for uniqueness of 𝑃_𝑇(𝐴,𝐵); for instance, let 𝑋=ℝ² with ‖⋅‖_∞, 𝐴=𝑥₀+𝐵_𝑋, and 𝐵=−𝑥₀+𝐵_𝑋 where 𝑥₀=(2,0). If 𝑇∶𝐴∪𝐵→𝐴∪𝐵 by 𝑇(𝑎,𝑏)=(−𝑎,𝑏), then one can easily see that 𝑃_𝑇(𝐴,𝐵) is an infinite set.

Corollary 2.3. Let 𝐴 and 𝐵 be two nonempty closed and convex subsets of a uniformly convex Banach space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 satisfying 𝑇(𝐴)⊂𝐵, 𝑇(𝐵)⊂𝐴, and 𝑇𝑥−𝑇𝑦≤𝑎₁𝑥−𝑦+𝑎₂‖𝑥−𝑇𝑥‖+𝑎₃𝑦−𝑇𝑦+𝑎₄𝑑(𝐴,𝐵) (2.7) for all 𝑥,𝑦∈𝐴∪𝐵, where 𝑎_𝑖≥0,𝑖=1,2,3,4 and ∑⁴_𝑖=1𝑎_𝑖<1. Then, there exists a unique element 𝑥∈𝐴∪𝐵 with ‖𝑥−𝑇𝑥‖=𝑑(𝐴,𝐵).

Proof. Interchange the roles of 𝑥 and 𝑦 in (2.7); then add the new inequality with (2.7).

In the following, we present some new conditions on the mapping 𝑇, such as weak closedness, such that it has a best proximity point in the uniformly convex Banach space 𝑋. We remember that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is said to be weakly closed if 𝑥_𝑛⇀𝑥 weakly in 𝐴∪𝐵 and 𝑇𝑥_𝑛⇀𝑦 weakly, then 𝑇𝑥=𝑦.

Theorem 2.4. Let 𝐴 and 𝐵 be two nonempty closed and convex subsets of a uniformly convex Banach space 𝑋 such that 𝐴 is bounded. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 satisfying 𝑇(𝐴)⊂𝐵, 𝑇(𝐵)⊂𝐴, and 𝑇𝑥−𝑇𝑦≤𝛼𝑥−𝑦+𝛽‖𝑥−𝑇𝑥‖+𝑦−𝑇𝑦, (2.8) for all 𝑥,𝑦∈𝐴∪𝐵, where 𝛼,𝛽≥0 and 𝛼+2𝛽≤1. If one of the following conditions:(i)𝑇 is weakly closed and 𝑇(𝐴) is bounded,(ii)𝑇 is weakly sequentially continuous,satisfies, then there exists 𝑥∈𝐴 with ‖𝑥−𝑇𝑥‖=𝑑(𝐴,𝐵).

Proof. Let 𝐴₀∶=𝑥∈𝐴∶𝑥−𝑦=𝑑(𝐴,𝐵),forsome𝑦∈𝐵. (2.9) By Lemma 3.2 of [5], 𝐴₀ is nonempty; hence, there are 𝑥₀∈𝐴 and 𝑦₀∈𝐵 such that ‖𝑥₀−𝑦₀‖=𝑑(𝐴,𝐵). For every positive integer 𝑛∈ℕ, define 𝑇_𝑛(𝑥)=1𝑛𝑦₀+11−𝑛1𝑇𝑥,𝑥∈𝐴,𝑛𝑥₀+11−𝑛𝑇𝑥,𝑥∈𝐵. (2.10) Then, for every 𝑥,𝑦∈𝐴∪𝐵, 𝑇_𝑛𝑥−𝑇_𝑛𝑦≤11−𝑛+1𝑇𝑥−𝑇𝑦𝑛𝑑()≤1𝐴,𝐵1−𝑛𝛼+1𝑥−𝑦1−𝑛𝛽‖‖++1𝑥−𝑇𝑥𝑦−𝑇𝑦𝑛𝑑(𝐴,𝐵). (2.11) Therefore, by Theorem 2.2, for every 𝑛∈ℕ, there exists 𝑥_𝑛∈𝐴 such that 𝑥_𝑛−𝑇_𝑛𝑥_𝑛=𝑑(𝐴,𝐵). (2.12) Since 𝐴 is bounded and closed, there exist 𝑥∈𝐴 such that 𝑥_𝑛⇀𝑥 (by passing to a subsequence, if necessary). If (i) holds, then the sequence {𝑇𝑥_𝑛} has a weakly convergent subsequence 𝑇𝑥_𝑛𝑘⇀𝑇𝑥, thanks to the weak closedness of 𝑇. So 𝑥_𝑛𝑘−𝑇𝑥_𝑛𝑘⇀𝑥−𝑇𝑥. On the other hands, since ‖𝑇_𝑛𝑥_𝑛−𝑇𝑥_𝑛‖=(1/𝑛)‖𝑦₀−𝑇𝑥_𝑛‖→0, we have 𝑥_𝑛𝑘−𝑇𝑥_𝑛𝑘≤𝑥_𝑛𝑘−𝑇_𝑛𝑘𝑥_𝑛𝑘+𝑇_𝑛𝑘𝑥_𝑛𝑘−𝑇𝑥_𝑛𝑘⟶𝑑(𝐴,𝐵). (2.13) Therefore, ‖𝑥−𝑇𝑥‖≤liminf_𝑘→∞𝑥_𝑛𝑘−𝑇𝑥_𝑛𝑘≤𝑑(𝐴,𝐵). (2.14) The proof of the statement in the case (ii) is even simpler and is a part of the above proof.

Theorem 2.5. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 satisfying 𝑇(𝐴)⊂𝐵, 𝑇(𝐵)⊂𝐴, and 𝑑𝑇𝑥,𝑇²𝑥≤𝑘𝑑(𝑥,𝑇𝑥)+(1−𝑘)𝑑(𝐴,𝐵), (2.15) for all 𝑥∈𝐴∪𝐵, where 0≤𝑘<1. If there are 𝑢∈𝐴∪𝐵 and 𝑛∈ℕ such that 𝑇^𝑛𝑢=𝑢, then 𝑑(𝑢,𝑇𝑢)=𝑑(𝐴,𝐵).

Proof. Suppose there are 𝑢∈𝐴∪𝐵 and 𝑛∈ℕ such that 𝑇^𝑛𝑢=𝑢. If 𝑑(𝐴,𝐵)<𝑑(𝑢,𝑇𝑢), since 𝑇 satisfies (2.15), we have 𝑑(𝑢,𝑇𝑢)=𝑑𝑇𝑇^𝑛−1𝑢,𝑇²𝑇^𝑛−1𝑢≤𝑘𝑑𝑇^𝑛−1𝑢,𝑇𝑇^𝑛−1𝑢+(1−𝑘)𝑑(𝐴,𝐵)≤𝑘²𝑑𝑇^𝑛−2𝑢,𝑇^𝑛−1𝑢+1−𝑘²𝑑()⋮𝐴,𝐵≤𝑘^𝑛𝑑(𝑢,𝑇𝑢)+1−𝑘^𝑛𝑑()(),𝐴,𝐵<𝑑𝑢,𝑇𝑢 (2.16) which is a contradiction, so 𝑑(𝑢,𝑇𝑢)=𝑑(𝐴,𝐵).

Corollary 2.6. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 satisfying 𝑇(𝐴)⊂𝐵, 𝑇(𝐵)⊂𝐴, and (2.1). If there are 𝑢∈𝐴∪𝐵 and 𝑛∈𝑁 such that 𝑇^𝑛𝑢=𝑢, then 𝑑(𝑢,𝑇𝑢)=𝑑(𝐴,𝐵).

Proof. If 𝑦=𝑇𝑥 in (2.1), then 𝑑𝑇𝑥,𝑇²𝑥≤𝛼𝑑(𝑥,𝑇𝑥)+𝛽𝑑(𝑥,𝑇𝑥)+𝑑𝑇𝑥,𝑇²𝑥+𝛾𝑑(𝐴,𝐵). (2.17) So 𝑑𝑇𝑥,𝑇²𝑥≤𝛼+𝛽𝑑()+𝛾1−𝛽𝑥,𝑇𝑥𝑑().1−𝛽𝐴,𝐵 (2.18) Hence, by Theorem 2.5, we have 𝑑(𝑢,𝑇𝑢)=𝑑(𝐴,𝐵).

3. Strongly Proximity Pairs

Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋, 𝛿>0, and 𝑇∶𝐴∪𝐵→𝐴∪𝐵 such that 𝑇(𝐴)⊂𝐵 and 𝑇(𝐵)⊂𝐴. Put 𝑃^𝛿_𝑇(𝐴,𝐵)∶={𝑥∈𝐴∪𝐵∶𝑑(𝑥,𝑇𝑥)<𝑑(𝐴,𝐵)+𝛿}. (3.1) We say that the pair (𝐴,𝐵) is a strongly proximity pair, if it is proximity pair, and, for any neighborhood 𝑉 of 0 in 𝑋 there exists 𝛿>0 such that 𝑃^𝛿_𝑇(𝐴,𝐵)⊆𝑃_𝑇(𝐴,𝐵)+𝑉.

For example, if 𝐴=𝑥,𝑦∶(𝑥−2)²+𝑦²≤1,𝐵=𝑥,𝑦∶(𝑥+2)²+𝑦²≤1, (3.2) and 𝑇(𝑥,𝑦)=(−𝑥,𝑦), then for every 𝜖>0, 𝑃^𝛿_𝑇(𝐴,𝐵)⊆𝑃_𝑇(𝐴,𝐵)+𝑉, where 𝛿=√4+𝜖²−2 and 𝑉 is the sphere with radius 𝜖 and center of zero. Hence the pair (𝐴,𝐵) is a strongly proximity pair.

Also, if 𝐴=0,𝑦∶−1≤𝑦≤1,𝐵={(𝑥,0)∶1≤𝑥≤2or−2≤𝑥<−1} (3.3) and 𝑇∶𝐴∪𝐵→𝐴∪𝐵 such that𝑇𝑥,𝑦=𝑦+1,𝑥𝑦≥0,𝑥,𝑦∈𝐴,𝑦−1,𝑥𝑦<0,𝑥,𝑦∈𝐴,𝑦,𝑥−1𝑥≥1,𝑥,𝑦∈𝐵,𝑦,𝑥+1𝑥<−1,𝑥,𝑦∈𝐵, (3.4) Therefore 𝑇(𝐴)⊂𝐵, and 𝑇(𝐵)⊂𝐴 and 𝑃_𝑇(𝐴,𝐵)={(0,0),(1,0)}, but the pair (𝐴,𝐵) is not a strongly proximity pair, while it is a proximity pair.

Here, by introducing the concepts of 𝑇-approximatively compact pair and 𝑇-strongly compact pair, we give some characterizations of the strongly proximity pairs of sets.

Definition 3.1. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋 and 𝑇∶𝐴∪𝐵→𝐴∪𝐵 such that 𝑇(𝐴)⊂𝐵 and 𝑇(𝐵)⊂𝐴. We say the following.(i)The sequence {𝑧_𝑛}⊆𝐴∪𝐵 is 𝑇-minimizing if lim_𝑛→∞𝑑𝑧_𝑛,𝑇𝑧_𝑛=𝑑(𝐴,𝐵). (3.5)(ii)The pair (𝐴,𝐵) is 𝑇-approximatively compact pair (𝑇-a.c.p.) if every 𝑇-minimizing sequence {𝑧_𝑛}⊆𝐴∪𝐵 has convergent subsequence.(iii)The pair (𝐴,𝐵) is 𝑇-strongly compact pair (𝑇-s.c.p.) if every 𝑇-minimizing sequence {𝑧_𝑛}⊆𝐴∪𝐵 is convergent.

In the last section, we find some conditions on 𝑇 such that 𝑃_𝑇(𝐴,𝐵)≠∅, and so in this section, we can always suppose that 𝑃_𝑇(𝐴,𝐵)≠∅. At the first, we state an elementary lemma, which can be used in the proof of the main theorems that follow.

Lemma 3.2. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋,𝑇∶𝐴∪𝐵→𝐴∪𝐵 such that 𝑇(𝐴)⊂𝐵 and 𝑇(𝐵)⊂𝐴, and the pair (𝐴,𝐵) is 𝑇-s.c.p. Then, 𝑃_𝑇(𝐴,𝐵) is singleton.

Proof. Let 𝑥,𝑦∈𝑃_𝑇(𝐴,𝐵), hence, 𝑑𝑦,𝑇𝑦=𝑑(𝑥,𝑇𝑥)=𝑑(𝐴,𝐵). (3.6) Now, define 𝑧_𝑛=𝑥,𝑛odd,𝑦,𝑛even. (3.7) Then, the sequence {𝑧_𝑛} is 𝑇-minimizing but is not convergent provided that 𝑥≠𝑦 and so a contradiction.

Now, we can prove the main theorems of this section.

Theorem 3.3. Let 𝐴 and 𝐵 be nonempty closed subsets of a normed space 𝑋, and 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is a continuous function, such that 𝑇(𝐴)⊂𝐵 and 𝑇(𝐵)⊂𝐴 and 𝑃_𝑇(𝐴,𝐵)≠∅. Then, the pair (𝐴,𝐵) is 𝑇-a.c.p. if and only if the pair (𝐴,𝐵) is strongly proximity pair and 𝑃_𝑇(𝐴,𝐵) compact.

Proof. Let the pair (𝐴,𝐵) be 𝑇-a.c.p., and {𝑥_𝑛}⊆𝑃_𝑇(𝐴,𝐵) is an arbitrary sequence. Then for each 𝑛, 𝑑(𝑥_𝑛,𝑇𝑥_𝑛)=𝑑(𝐴,𝐵), and, by hypothesis, the sequence {𝑥_𝑛} has a convergent subsequence to an element of 𝑃_𝑇(𝐴,𝐵). Thus, 𝑃_𝑇(𝐴,𝐵) is compact.
Also, if (𝐴,𝐵) is not strongly proximity pair, then there exist a neighborhood 𝑉 of 0 and a 𝑇-minimizing sequence {𝑧_𝑛}⊆𝐴∪𝐵 with 𝑧_𝑛 not belonging to 𝑃_𝑇(𝐴,𝐵)+𝑉 for all 𝑛≥1. Since (𝐴,𝐵) is 𝑇-a.c.p., there is a subsequence {𝑧_𝑛𝑘} such that 𝑧_𝑛𝑘→𝑧₀. Then, 𝑧₀∈𝑃_𝑇(𝐴,𝐵), and so 𝑧_𝑛∈𝑧₀+𝑉⊆𝑃_𝑇(𝐴,𝐵)+𝑉 for sufficiently large 𝑛, that is a contradiction.
Conversely, suppose that (𝐴,𝐵) is a strongly proximity pair and 𝑃_𝑇(𝐴,𝐵) compact, but (𝐴,𝐵) is not 𝑇-a.c.p. Then, there is a 𝑇-minimizing sequence {𝑧_𝑛}⊆𝐴∪𝐵 without any convergent subsequence. It follows that, for any 𝑥∈𝑃_𝑇(𝐴,𝐵), there is a neighborhood 𝑈_𝑥 of 𝑥 such that, for sufficiently large 𝑛, 𝑧_𝑛 does not belong to 𝑈_𝑥. Since 𝑃_𝑇(𝐴,𝐵) is compact, one can cover 𝑃_𝑇(𝐴,𝐵) by finitely many 𝑈_𝑥𝑖,𝑖=1,2,...,𝑛. So there is a neighborhood 𝑉 of 0 and 𝑛₀∈𝑁 such that for all 𝑛≥𝑛₀, 𝑧_𝑛 does not belong to 𝑃_𝑇(𝐴,𝐵)+𝑉. Since 𝑃_𝑇(𝐴,𝐵) is strongly proximity pair, there exists 𝛿>0 such that 𝑃^𝛿_𝑇(𝐴,𝐵)⊆𝑃_𝑇(𝐴,𝐵)+𝑉. Since {𝑧_𝑛} is a 𝑇-minimizing sequence, 𝑧_𝑛∈𝑃^𝛿_𝑇(𝐴,𝐵) for sufficiently large 𝑛∈𝑁 and this is a contradiction.

Corollary 3.4. Let 𝐴 and 𝐵 be nonempty subsets of normed space 𝑋 such that 𝐴∪𝐵 is compact and 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is continuous such that 𝑇(𝐴)⊂𝐵 and 𝑇(𝐵)⊂𝐴. Then, the pair (𝐴,𝐵) is strongly proximity pair and 𝑃_𝑇(𝐴,𝐵) is compact.

Proof. Since 𝐴∪𝐵 is compact, it is obvious that the pair (𝐴,𝐵) is 𝑇-a.c.p. Now, apply Theorem 3.3.

Theorem 3.5. Let 𝐴 and 𝐵 be nonempty closed subsets of a normed space 𝑋, and 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is continuous such that 𝑇(𝐴)⊂𝐵 and 𝑇(𝐵)⊂𝐴 and 𝑃_𝑇(𝐴,𝐵)≠∅. Then, the pair (𝐴,𝐵) is a 𝑇-s.c.p. if and only if the pair (𝐴,𝐵) is strongly proximity pair and 𝑃_𝑇(𝐴,𝐵) singleton.

Proof. Suppose that (𝐴,𝐵) is 𝑇-s.c.p. By Theorem 3.3, (𝐴,𝐵) is strongly proximity pair, and, by Lemma 3.2, 𝑃_𝑇(𝐴,𝐵) is singleton.
Conversely, suppose (𝐴,𝐵) is strongly proximity pair and 𝑃_𝑇(𝐴,𝐵)={𝑧₀}. Let 𝑉 be a neighborhood of 0. Since (𝐴,𝐵) is strongly proximity pair, there exists 𝛿>0 such that 𝑃^𝛿_𝑇(𝐴,𝐵)⊆𝑧₀+𝑉. Thus, for any 𝑇-minimizing sequence {𝑧_𝑛}⊆𝐴∪𝐵, 𝑧_𝑛∈𝑃^𝛿_𝑇(𝐴,𝐵)⊆𝑧₀+𝑉 for sufficiently large 𝑛. Hence, 𝑧_𝑛→𝑧₀.

Theorem 3.6. Let 𝐴 and 𝐵 be nonempty closed and convex subsets of a uniformly convex Banach space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 satisfying 𝑇(𝐴)⊂𝐵, 𝑇(𝐵)⊂𝐴, and, for every 𝑥,𝑦∈𝐴𝑇𝑥−𝑇𝑦≤𝛼𝑥−𝑦(0<𝛼<1). (3.8) Then, the pair (𝐴,𝐵) is a 𝑇-s.c.p. if and only if 𝑃_𝑇(𝐴,𝐵) is singleton.

Proof. The necessary condition follows from Theorem 3.5.
For the proof of sufficient condition, suppose that 𝑃_𝑇(𝐴,𝐵)={𝑧₀} but (𝐴,𝐵) is not a 𝑇-s.c.p. Then, there is a 𝑇-minimizing sequence {𝑧_𝑛}⊆𝐴∪𝐵 that is not convergent. It follows that there exists a subsequence {𝑧_𝑛𝑘} of {𝑧_𝑛} and a scaler 𝑑>0 such that for all integer 𝑘, 𝑧_𝑛𝑘−𝑧≥𝑑. (3.9) By uniform convexity of 𝑋, there exists 𝜖>0 such that (𝑑(𝐴,𝐵)+𝜖)1−𝛿_𝑋𝑑(1−𝛼)𝑑()().𝐴,𝐵+𝜖<𝑑𝐴,𝐵 (3.10) Since lim_𝑛→∞‖𝑧_𝑛−𝑇𝑧_𝑛‖=𝑑(𝐴,𝐵), there exists 𝑘 such that 𝑧_𝑛𝑘−𝑇𝑧_𝑛𝑘<𝑑(𝐴,𝐵)+𝜖. (3.11) Also, 𝑧₀−𝑇𝑧₀=𝑑(𝐴,𝐵)<𝑑(𝐴,𝐵)+𝜖. (3.12) But 𝑧₀−𝑇𝑧₀−𝑧_𝑛𝑘−𝑇𝑧_𝑛𝑘≥𝑧₀−𝑧_𝑛𝑘−𝑇𝑧₀−𝑇𝑧_𝑛𝑘≥𝑧₀−𝑧_𝑛𝑘−𝑧₀−𝑦_𝑛𝑘𝛼=𝑧₀−𝑧_𝑛𝑘)(),(1−𝛼>𝑑1−𝛼 (3.13) Because 𝐴 and 𝐵 are convex, ((𝑧+𝑧_𝑛𝑘)/2)∈𝐴, and 𝑇𝑧+𝑇𝑧_𝑛𝑘/2∈𝐵, the following inequality leads to a contradiction: 𝑧+𝑧_𝑛𝑘2−𝑇𝑧+𝑇𝑧_𝑛𝑘2≤(𝑑(𝐴,𝐵)+𝜖)1−𝛿_𝑋𝑑(1−𝛼)𝑑()().𝐴,𝐵+𝜖<𝑑𝐴,𝐵 (3.14)