Best proximity points and fixed points for p-summing maps

Abstract

We have found a new type of a contractive condition that ensures the existence and uniqueness of fixed points and best proximity points in uniformly convex Banach spaces. We provide some examples to validate our results. These results generalize some known results from fixed point theory.

AMS Subject Classification: 47H10; 54H25.

1 Introduction

A fundamental result in fixed point theory is the Banach contraction principle. One kind of a generalization of the Banach contraction principle is the notation of cyclical maps [1]. Fixed point theory is an important tool for solving equations Tx = x for mappings T defined on subsets of metric spaces or normed spaces. Because a non-self mapping T : A → B does not necessarily have a fixed point, one often attempts to find an element x which is in some sense closest to Tx. Best proximity point theorems are relevant in this perspective. The notation of best proximity point is introduced in [2]. This definition is more general than the notation of cyclical maps, in sense that if the sets intersect, then every best proximity point is a fixed point. A sufficient condition for the uniqueness of the best proximity points in uniformly convex Banach spaces is given in [2]. It turns out that many of the contractive-type conditions which are investigated for fixed points ensure the existence of best proximity points. Some results of this kind are obtained in [3–6]. It is interesting that in all the investigated conditions for the existence of best proximity the distances between sets are equal. We have found a new type of condition which warrants the existence and the uniqueness of the best proximity points for sets with different distances between them. This new type of a map we have called a p-summing map. We have also shown that this new type of map, the p-summing map, if considered not as a cyclical map, has a unique fixed point.

2 Preliminary results

In this section, we give some basic definitions and concepts which are useful and related to the best proximity points. Let (X, || · ||) be a Banach space. Define a distance between two subsets A, B ⊂ X by dist(A, B) = inf{||x - y|| : x ∈ A, y ∈ B}.

Definition 2.1. [2, 5] Let A₁, A₂,..., A _p be non-empty subsets of a Banach space (X, || · ||) and let $T:\bigcup _{i=1}^p{A}_i\to \bigcup _{i=1}^p{A}_i$. The map T is called a p-cyclic contraction, if it satisfies the following conditions:

• (1) T(A _i ) ⊆ A _{i+ 1}; 1 ≤ i ≤ p, where A _p+i = A _i ;

• (2) For some k ∈ (0,1) the inequality ||Tx - Ty|| ≤ k||x - y|| + (1 - k)dist(A _i , A _{i+ 1}) holds for any x ∈ A _i , y ∈ A _{i+ 1}, 1 ≤ i ≤ p. A point ξ ∈ A _i is said to be a best proximity point of T in A _i if ||ξ - Tξ|| = dist(A _i , A _{i+ 1}).

Definition 2.1 is given for two sets A₁ and A₂ in [2], and for p-sets in [5].

It is proved in [5], that if a map is a p-cyclic contraction, then it has best proximity points for every set A _i , 1 ≤ i ≤ p.

We will use the following two lemmas, established in [2], to proving the uniqueness of the best proximity points.

Lemma 2.1. [2]Let A be a non-empty, closed, convex subset, and B be a non-empty, closed subset of a uniformly convex Banach space. Let${\left\{x_n\right\}}_{n=1}^{\infty }$and${\left\{z_n\right\}}_{n=1}^{\infty }$be sequences in A and${\left\{y_n\right\}}_{n=1}^{\infty }$be a sequence in B satisfying:

(1)$\underset{n\to \infty }{\text{lim}}∥z_n-y_n∥=\text{dist}\left(A,B\right)$;

(2) for every ε > 0 there exists N₀ ∈ ℕ, such that for all m > n ≥ N₀, ||x _n - y _n || ≤ dist(A,B)+ε.

Then for every ε > 0, there exists N₁ ∈ ℕ, such that for all m > n > N₁, holds ||x _m -z _n || ≤ ε.

Lemma 2.2. [2]Let A be a non-empty, closed, convex subset, and B be a non-empty, closed subset of a uniformly convex Banach space. Let${\left\{x_n\right\}}_{n=1}^{\infty }$and${\left\{z_n\right\}}_{n=1}^{\infty }$be sequences in A and${\left\{y_n\right\}}_{n=1}^{\infty }$be a sequence in B satisfying:

(1)$\underset{n\to \infty }{\text{lim}}∥x_n-y_n∥=\text{dist}\left(A,B\right)$;

(2)$\underset{n\to \infty }{\text{lim}}∥z_n-y_n∥=\text{dist}\left(A,B\right)$;

then $\underset{n\to \infty }{\text{lim}}∥x_n-z_n∥=0$

Theorem 2.1. [7]Let (X, || · ||) be a Banach space and F : X → R ∪ {+∞} be a lower semicontinuous function on X that is bounded from below and not identically equal to +∞. Fix ε > 0 and a point x₀ ∈ X, such that

$$F\left(x_0\right)\le \epsilon +\text{inf}\left\{F\left(x\right):x\in X\right\}.$$

Then there exists a point v ∈ X, such that ||x₀ - v|| ≤ 1 and F(v) ≤ F(x₀), and for any w ≠ v there holds the inequality

$$F\left(v\right)\le F\left(w\right)+\epsilon ∥v-w∥.$$

3 Main results

Let (X, || · ||) be a Banach space and A _i ∈ X. We denote P = dist(A, A₂) + dist(A₂, A₃) + dist(A₃,A₁).

Definition 3.1. Let A _i , i = 1, 2, 3 be subsets of a uniformly convex Banach space (X, || · ||). A map $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ will be called a 3-cyclic summing contraction if it satisfies the following conditions:

• (1) T(A _i ) ⊆ A _{i+ 1}, for every i = 1, 2, 3 and by A ₄ we understand A ₁;

• (2) Let there exists k ∈ (0,1), such that for any x _i ∈ A _i , i = 1, 2, 3 there holds the inequality

  $$\begin{array}{c}∥T{x}_1-T{x}_2∥+∥T{x}_2-T{x}_3∥+∥T{x}_3-T{x}_1∥\\ \le k\left(∥x_1-x_2∥+∥x_2-x_3∥+∥x_3-x_1∥\right)+\left(1-k\right)P\end{array}$$

  (3.1)

Theorem 3.1. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach space X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ is a 3-cyclic summing contraction. Then for any i = 1, 2, 3 there exist unique best proximity points z _i ∈ A _i , such that for any x ∈ A _i the sequence ${\left\{T^{3n}x\right\}}_{n=1}^{\infty }$ converges to z _i . Moreover, T^jz _i = z_i+jis a best proximity point in A _i+j , j = 1, 2 and z _i is a fixed point of the map T³.

Definition 3.2. Let (X, || · ||) be a Banach space. A map T : X → X will be called a 3-summing contraction if there exists k ∈ (0,1), such that for any x ≠ y ≠ z there holds the inequality

$$∥Tx-Ty∥+∥Ty-Tz∥+∥Tz-Tx∥\le k\left(∥x-y∥+∥y-z∥+∥z-x∥\right).$$

(3.2)

Let us mention that any contraction map T : X → X is a 3-summing map, but obviously there are 3-summing maps that are not contractions. The requirement x ≠ y ≠ z in Definition 3.2 is necessary because if we do not impose it, then if we take y = z in (3.2) we will get the classical Banach contraction condition.

Theorem 3.2. Let X be a Banach space and T : X → X be a 3-summing contraction. Then T has a unique fixed point.

It is easy to define a p-summing contraction. Let us mention that all the results in Theorems 3.1 and 3.2 are true for a p-summing contraction. Just for the sake of simplicity we decide to state them and to prove them for a 3-summing contraction.

4 Auxiliary results

Lemma 4.1. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach space X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ be a 3-cyclic summing contraction, then for any x ∈ A _i , i = 1, 2, 3 the iterative sequence ${\left\{T^{n}x\right\}}_{n=1}^{\infty }$ satisfies

$$\underset{n\to \infty }{\text{lim}}\left(∥T^{n+3}x-T^{n+2}x∥+∥T^{n+2}x-T^{n+1}x∥+∥T^{n+1}x-T^{n+3}x∥\right)=P$$

(4.3)

Proof. Let x ∈ A _i . By the chain of inequalities:

$$\begin{array}{c}∥T^{n+3}x-T^{n+2}x∥+∥T^{n+2}x-T^{n+1}x∥+∥T^{n+1}x-T^{n+3}x∥\hfill \\ \le k\left(∥T^{n+2}x-T^{n+1}x∥+∥T^{n+1}x-T^{n}x∥+∥T^{n}x-T^{n+2}x∥\right)\hfill \\ +\left(1-k\right)P\hfill \\ \le k^2\left(∥T^{n+1}x-T^{n}x∥+∥T^{n}x-T^{n-1}x∥+∥T^{n-1}x-T^{n+1}x∥\right)\hfill \\ +\left(1+k\right)\left(1-k\right)P\hfill \\ \le k^3\left(∥T^{n}x-T^{n-1}x∥+∥T^{n-1}x-T^{n-2}x∥+∥T^{n-2}x-T^{n}x∥\right)\hfill \\ +\left(1+k+k^2\right)\left(1-k\right)P\hfill \\ ..........................................................................\hfill \\ \le k^{n+1}\left(∥T^{2}x-Tx∥+∥Tx-x∥+∥x-T^{2}x∥\right)\hfill \\ +\left(1+k+\cdots +k^n\right)\left(1-k\right)P\hfill \\ =k^{n+1}\left(∥T^{2}x-Tx∥+∥Tx-x∥+∥x-T^{2}x∥\right)+\left(1-k^{n+1}\right)P\hfill \end{array}$$

and the fact that T^{n+ 1}x, T^{n+ 2}x and T^{n+ 3}x belong to different sets A _i , i = 1, 2, 3, we get the inequality

$$\begin{array}{ll}\hfill P& =\text{dist}\left(A_1,A_2\right)+\text{dist}\left(A_2,A_3\right)+\text{dist}\left(A_3,A_1\right)\\ \le \left|\right|T^{n+3}x-T^{n+2}x\left|\right|+\left|\right|T^{n+2}x-T^{n+1}x\left|\right|+\left|\right|T^{n+1}x-T^{n+3}x\left|\right|\\ \le k^{n+1}\left(\left|\right|T^{2}x-Tx\left|\right|+\left|\right|Tx-x\left|\right|+\left|\right|x-T^{2}x\left|\right|\right)+\left(1-k^{n+1}\right)P\end{array}$$

and the proof follows because $\underset{n\to \infty }{\text{lim}}{k}^n=0$.

Lemma 4.2. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach space X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ be a 3-cyclic summing contraction, then for any x ∈ A _i the inequality

$$∥T^{3n+1}x-T^{3n}x∥\le k^{3n-1}\left(\alpha \left(x\right)-P\right)+\text{dist}\left(A_i,A_{i+1}\right)$$

holds, where

$$\alpha \left(x\right)=∥T^{2}x-Tx∥+∥Tx-x∥+∥x-T^{2}x∥$$

Proof. If x ∈ A _i , then T³ⁿx ∈ A _i and T^{3n+ 1}x ∈ A_{i+ 1}. By the proof of Lemma 4.1 we have

$$\begin{array}{c}∥T^{3n+1}x-T^{3n}x∥+P-\text{dist}\left(A_i,A_{i+1}\right)\hfill \\ \le ∥T^{3n+1}x-T^{3n}x∥+∥T^{3n}x-T^{3n-1}x∥+∥T^{3n-1}x-T^{3n+1}x∥\hfill \\ \le k^{3n-1}\left(∥T^{2}x-Tx∥+∥Tx-x∥+∥x-T^{2}x∥\right)+\left(1-k^{3n-1}\right)P\hfill \end{array}$$

thus

$$∥T^{3n+1}x-T^{3n}x∥\le \text{dist}\left(A_i,A_{i+1}\right)+k^{3n-1}\left(\alpha \left(x\right)-P\right).$$

Corollary 4.1. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach space X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ be a 3-cyclic summing contraction, then for any x ∈ A _i there holds

$$\underset{n\to \infty }{\text{lim}}∥T^{3n+1}x-T^{3n}x∥=\text{dist}\left(A_i,A_{i+1}\right).$$

Lemma 4.3. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach space X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ be a 3-cyclic summing contraction, then for any x, y ∈ A _i the inequality

$$∥T^{3n+1}x-T^{3n+3}y∥\le k^{3n-1}\left(\alpha \left(x,y\right)-P\right)+\text{dist}\left(A_i,A_{i+1}\right)$$

holds, where

$$\alpha \left(x,y\right)=∥x-T^{2}y∥+∥T^{2}y-Ty∥+∥Ty-x∥$$

Proof. If x, y ∈ A _i , then T³ⁿy ∈ A _i and T^{3n+ 1}x ∈ A_i+1. By the proof of Lemma 4.1 we have

$$\begin{array}{c}∥T^{3n+1}x-T^{3n+3}y∥+P-\text{dist}\left(A_i,A_{i+1}\right)\hfill \\ \le ∥T^{3n+1}x-T^{3n+3}y∥+∥T^{3n+3}y-T^{3n+2}y∥+∥T^{3n+2}y-T^{3n+1}x∥\hfill \\ \le k^{3n+1}\left(∥x-T^{2}y∥+∥T^{2}y-Ty∥++∥Ty-x∥\right)+\left(1-k^{3n+1}\right)P\hfill \end{array}$$

thus

$$∥T^{3n+1}x-T^{3n+3}y∥\le \text{dist}\left(A_i,A_{i+1}\right)+k^{3n+1}\left(\alpha \left(x,y\right)-P\right).$$

Corollary 4.2. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach pace X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ be a 3-cyclic summing contraction, then for any x, y ∈ A _i there holds

$$\underset{n\to \infty }{\text{lim}}∥T^{3n+1}x-T^{3n+3}y∥=\text{dist}\left(A_i,A_{i+1}\right).$$

The following lemma can be proved in a similar fashion.

Lemma 4.4. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach space X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ be a 3-cyclic summing contraction, then for any x ∈ A _i and for any k ∈ ℕ there hold:

$$\underset{n\to \infty }{\text{lim}}∥T^{3n+1}x-T^{3n\pm 3}x∥=\text{dist}\left(A_i,A_{i+1}\right),$$

(4.4)

$$\underset{n\to \infty }{\text{lim}}∥T^{3n}x-T^{3n\pm 3}x∥=0,$$

(4.5)

$$\underset{n\to \infty }{\text{lim}}∥T^{3n+k+1}x-T^{3n+k\pm 3}x∥=\text{dist}\left(A_{i+k},A_{i+1+k}\right),$$

(4.6)

$$\underset{n\to \infty }{\text{lim}}∥T^{3n+k}x-T^{3n+k\pm 3}x∥=0,$$

(4.7)

Lemma 4.5. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach space X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ be a 3-cyclic summing contraction. If for some x ∈ A _i , i = 1, 2, 3, the iterative sequence ${\left\{T^{3n}x\right\}}_{n=1}^{\infty }$ has a cluster point z, then z is a best proximity point of T in A _i .

Proof. Let $\underset{j\to \infty }{\text{lim}}{T}^{3n_j}x=z$. Then by the continuity of the function f(u) = ||u - v||, for fixed v ∈ X it it follows that $∥z-Tz∥=\underset{j\to \infty }{\text{lim}}∥T^{3n_j}x-Tz∥$. We will prove first that

$$\underset{j\to \infty }{\text{lim}}∥T^{3n_j-1}x-z∥=\underset{j\to \infty }{\text{lim}}∥T^{3n_j-1}x-T^{3n_j}x∥.$$

(4.8)

By the triangle inequality:

$$0\le \left|∥T^{3n_j-1}x-z∥-∥T^{3n_j-1}x-T^{3n_j}x∥\right|\le ∥T^{3n_j}x-z∥$$

(4.9)

it follows that

$$\underset{j\to \infty }{\text{lim}}\left(∥T^{3n_j-1}x-z∥-∥T^{3n_j-1}x-T^{3n_j}x∥\right)=0$$

(4.10)

If we take k = 2 in (4.6) we get

$$\underset{j\to \infty }{\text{lim}}∥T^{3n_j+3}x-T^{3n_j-1}x∥=\text{dist}\left(A_i,A_{i+2}\right).$$

(4.11)

For any x ∈ A _i the inclusions $T^{3n_j}x\in A_i,T^{3n_j-1}x\in A_{i+2}$ hold. Then by the inequality

$$\text{dist}\left(A_i,A_{i+2}\right)\le ∥T^{3n_j-1}x-T^{3n_j}x∥\le ∥T^{3n_j-1}x-T^{3n_j+3}x∥+∥T^{3n_j+3}x-T^{3n_j}x∥,$$

and the equalities (4.5) and (4.11) we get

$$\underset{j\to \infty }{\text{lim}}∥T^{3n_j-1}x-T^{3n_j}x∥=\text{dist}\left(A_i,A_{i+2}\right).$$

(4.12)

Now by (4.10) and (4.12) we found that (4.8) holds true.

We apply consecutively (4.8) to obtain the next chain of inequalities:

$$\begin{array}{l}\left|\right|z-Tz\left|\right|+\left|\right|Tz-T^{2}z\left|\right|+\left|\right|T^{2}z-z\left|\right|\hfill \\ =\underset{j\to \infty }{\lim }(||T^{3n_j}x-Tz||+||Tz-T^{2}z||+||T^{2}z-T^{3n_j}x||)\hfill \\ \le k\underset{j\to \infty }{\lim }(||T^{3n_j-1}x-z||+||z-Tz||+||Tz-T^{3n_j-1}x||)+(1-k)P\hfill \\ =k\underset{j\to \infty }{\lim }(||T^{3n_j-1}x-T^{3n_j}x||+||T^{3n_j}x-Tz||+||Tz-T^{3n_j-1}x||)\hfill \\ +(1-k)P\hfill \\ \le k^2\underset{j\to \infty }{\lim }(||T^{3n_j-2}x-T^{3n_j-1}x||+||T^{3n_j-1}x-z||+||z-T^{3n_j-2}x||)\hfill \\ +(1+k)(1-k)P\hfill \\ =k^2\underset{j\to \infty }{\lim }(||T^{3n_j-2}x-T^{3n_j-1}x||+||T^{3n_j-1}x-T^{3n_j}x||\hfill \\ +\left|\right|T^{3n_j}x-T^{p{n}_j-2}x\left|\right|)+(1+k)(1-k)P\hfill \\ \le k^3\underset{j\to \infty }{\lim }(||T^{3n_j-3}x-T^{3n_j-2}x||+||T^{3n_j-2}x-T^{3n_j-1}x||\hfill \\ +\left|\right|T^{3n_j-1}x-T^{p{n}_j-3}x\left|\right|)+(1+k+k^2)(1-k)P\hfill \\ =k^{3}P+(1-k^3)P=P.\hfill \end{array}$$

(4.13)

Since z ∈ A _i it follows that Tz ∈ A_i+1, T²z ∈ A_i+2and hence

$$P-\text{dist}\left(A_i,A_{i+1}\right)=\text{dist}\left(A_{i+1},A_{i+2}\right)+\text{dist}\left(A_{i+2},A_i\right)\le ∥Tz-T^{2}z∥+∥T^{2}z-z∥.$$

Consequently by (4.13) we obtain

$$∥z-Tz∥+P-\text{dist}\left(A_i,A_{i+1}\right)\le ∥z-Tz∥+∥Tz-T^{2}z∥+∥T^{2}z-z∥\le P$$

and therefore we get ||z - Tz|| ≤ dist(A _i , A_{i+ 1}). The opposite inequality ||z - Tz|| ≥ dist(A _i , A_{i+ 1}) is obvious and hence we conclude that ||z - Tz|| = dist(A _i , A_{i+ 1}). Thus z is a best proximity point of T in A _i .

Lemma 4.6. Let A _i , i = 1, 2, 3 be closed, convex subsets of a uniformly convex Banach space X and $T:\bigcup _{i=1}^3{A}_i\to \bigcup _{i=1}^3{A}_i$ be a 3-cyclic summing contraction. If for some x ∈ A _i , i = 1, 2, 3, the iterative sequence ${\left\{T^{3n}x\right\}}_{n=1}^{\infty }$ has a cluster point z, then z is a fixed point for T³.

Proof. Let $\underset{j\to \infty }{\text{lim}}{T}^{3n_j}x=z$. Then from the continuity of the function f(u) = ||u-v||, for fixed v ∈ X it follows that $∥z-T^{4}z∥=\underset{j\to \infty }{\text{lim}}∥T^{3n_j}x-T^{4}z∥$ and $∥z-T^{5}z∥=\underset{j\to \infty }{\text{lim}}∥T^{3n_j}x-T^{5}z∥$.

We will prove first that

$$\underset{j\to \infty }{\text{lim}}∥T^{3n_j-4}x-z∥=\underset{j\to \infty }{\text{lim}}∥T^{3n_j-4}x-T^{3n_j}x∥.$$

(4.14)

By the triangle inequality:

$$0\le \left|∥T^{3n_j-4}x-z∥-∥T^{3n_j-4}x-T^{3n_j}x∥\right|\le ∥T^{3n_j}x-z∥$$

(4.15)

it follows that

$$\underset{j\to \infty }{\text{lim}}\left(∥T^{3n_j-4}x-z∥-∥T^{3n_j-4}x-T^{3n_j}x∥\right)=0.$$

(4.16)

For any x ∈ A _i the inclusions $T^{3n_j}x\in A_i,T^{3n_j-4}x\in A_{i+2}$ hold and we can write the inequalities

$$\begin{array}{c}\text{dist}\left(A_i,A_{i+2}\right)\le ∥T^{3n_j-4}x-T^{3n_j}x∥\hfill \\ \le ∥T^{3n_j-4}x-T^{3n_j-1}x∥+∥T^{2n_j-1}x-T^{3n_j+3}x∥+∥T^{3n_j+3}x-T^{3n_j}x∥.\hfill \end{array}$$

(4.17)

From (4.7), (4.5), and (4.11) it follows that

$$\underset{j\to \infty }{\text{lim}}∥T^{3n_j-4}x-T^{3n_j}x∥=\text{dist}\left(A_i,A_{i+2}\right).$$

(4.18)

Now by (4.16) and (4.18) we found that (4.14) holds true. We will omit the proof that

$$\underset{j\to \infty }{\text{lim}}∥T^{3n_j-5}x-z∥=\underset{j\to \infty }{\text{lim}}∥T^{3n_j-5}x-T^{3n_j}x∥,$$

(4.19)

because it is similar to the above one.

We apply consecutively (4.14), (4.8), and (4.19) to obtain the next chain of inequalities:

$$\begin{array}{l}\left|\right|z-T^{4}z\left|\right|+\left|\right|T^{4}z-T^{5}z\left|\right|+\left|\right|T^{5}z-z\left|\right|\hfill \\ =\underset{j\to \infty }{\lim }(||T^{3n_j}x-T^{4}z||+||T^{4}z-T^{5}z||+||T^{5}z-T^{3n_j}x||)\hfill \\ \le k\underset{j\to \infty }{\lim }(||T^{3n_j-1}x-T^{3}z||+||T^{3}z-T^{4}z||+||T^{4}z-T^{3n_j-1}x||)\hfill \\ +(1-k)P\hfill \\ \le k^4\underset{j\to \infty }{\lim }(||T^{3n_j-4}x-z||+||z-Tz||+||Tz-T^{3n_j-4}x||)\hfill \\ +(1-k^4)P\hfill \\ =k^4\underset{j\to \infty }{\lim }(||T^{3n_j-4}x-T^{3n_j}x||+||T^{3n_j}x-Tz||+||Tz-T^{3n_j-4}x||)\hfill \\ +(1-k^4)P\hfill \\ \le k^5\underset{j\to \infty }{\lim }(||T^{3n_j-5}x-T^{3n_j-1}x||+||T^{3n_j-1}x-z||+||z-T^{3n_j-5}x||)\hfill \\ +(1-k^5)P\hfill \\ =k^5\underset{j\to \infty }{\lim }(||T^{3n_j-5}x-T^{3n_j-1}x||+||T^{3n_j-1}x-T^{3n_j}x||\hfill \\ +\left|\right|T^{3n_j}x-T^{3n_j-5}x\left|\right|)+(1-k^4)P\hfill \\ \le k^6\underset{j\to \infty }{\lim }(||T^{3n_j-6}x-T^{3n_j-2}x||+||T^{3n_j-2}x-T^{3n_j-1}x||\hfill \\ +\left|\right|T^{3n_j-1}x-T^{3n_j-6}x\left|\right|)+(1-k^6)P\hfill \\ =k^{6}P+(1-k^6)P=P.\hfill \end{array}$$

(4.20)

By z ∈ A _i it follows that T⁴z ∈ A_i+1, T⁵z ∈ A_{i+ 2}and hence

$$P-\text{dist}\left(A_i,A_{i+1}\right)=\text{dist}\left(A_{i+1},A_{i+2}\right)+\text{dist}\left(A_{i+2},A_i\right)\le ∥T^{4}z-T^{5}z∥+∥T^{5}z-z∥.$$

Consequently by (4.20) we obtain

$$∥z-T^{4}z∥+P-\text{dist}\left(A_i,A_{i+1}\right)\le ∥z-T^{4}z∥+∥T^{4}z-T^{5}z∥+∥T^{5}z-z∥\le P.$$

and therefore we get ||z - T⁴z|| ≤ dist(A _i , A_{i+ 1}). The opposite inequality ||z - T⁴z|| ≥ dist(A _i ,A_{i+ 1}) is obvious and therefore it follows that ||z - Tz|| = dist(A _i , A_i+1). Now by Lemma 4.5 we get that

$$∥z-T^{4}z∥=∥z-Tz∥=\text{dist}\left(A_i,A_{i+1}\right)$$

and from the uniform convexity of X it follows that T⁴x = Tx.

By the inequality

$$\begin{array}{c}∥T^{4}z-T^{3}z∥+∥T^{3}z-T^{2}z∥+∥T^{2}z-Tz∥\hfill \\ =∥T^{4}z-T^{3}z∥+∥T^{3}z-T^{2}z∥+∥T^{2}z-T^{4}z∥\hfill \\ \le k\left(∥T^{3}z-T^{2}z∥+∥T^{2}z-Tz∥+∥Tz-T^{3}z∥\right)+\left(1-k\right)P\hfill \\ =k\left(∥T^{3}z-T^{2}z∥+∥T^{2}z-Tz∥+∥T^{4}z-T^{3}z∥\right)+\left(1-k\right)P\hfill \end{array}$$

we get

$$\left(1-k\right)∥T^{4}z-T^{3}z∥+\left(1-k\right)\left(P-\text{dist}\left(A_i,A_{i+1}\right)\right)\le \left(1-k\right)P,$$

i.e. ||T⁴z - T³z|| ≤ dist(A _i , A_{i+ 1}). By the obvious inequality ||T⁴z - T³z|| ≥ dist(A _i , A_{i+ 1}) it follows that ||T⁴z - T³z|| = dist(A _i , A_{i+ 1}). Now from

$$∥z-T^{4}z∥=∥T^{3}z-T^{4}z∥=\text{dist}\left(A_i,A_{i+1}\right)$$

and the uniform convexity of X it follows that T³z = z.

Lemma 4.7. If T is a 3-summing contraction then T is continuous.

Proof. Let fix x₀ ∈ X and let ${\left\{y_n\right\}}_{n=1}^{\infty }$ and ${\left\{z_n\right\}}_{n=1}^{\infty }$ be two sequences, that are convergent to x₀. Then for any ε > 0 there is n₀ ∈ ℕ, such that for every n ≥ n₀ there holds ||x - y _n || + ||y _n - z _n || + ||z _n - y _n || < ε. By the inequalities

$$\begin{array}{c}∥T{x}_0-T{y}_n∥\le ∥T{x}_0-T{y}_n∥+∥T{y}_n-T{z}_n∥+∥T{z}_n-T{x}_0∥\\ \le k\left(∥x_0-y_n∥+∥y_n-z_n∥+∥z_n-x_0∥\right)<\epsilon \end{array}$$

it follows that T is continuous at x₀.

5 Proof of main results

Proof of Theorem 3.1. First we will show that for any x ∈ A _i the sequence ${\left\{T^{3n}x\right\}}_{n=1}^{\infty }$ is convergent, i.e., it is enough to prove that the sequence ${\left\{T^{3n}x\right\}}_{n=1}^{\infty }$ is a Cauchy sequence.

Claim 5.1. For any ε > 0 there exists n₀, such that for any m > n ≥ n₀ there holds the inequality

$$∥T^{3m}x-T^{3n+1}x∥+∥T^{3n+1}x-T^{3n+2}x∥+∥T^{3n+2}x-T^{3m}x∥<P+\epsilon$$

(5.21)

Proof of Claim 5.1. Suppose the contrary, i.e., there is ε > 0 such that for every k ∈ ℕ there are m _k > n _k ≥ k so that

$$∥T^{3m_k}x-T^{3n_k+1}x∥+∥T^{3n_k+1}x-T^{3n_k+2}x∥+∥T^{3n_k+2}x-T^{3m_k}x∥\ge P+\epsilon .$$

(5.22)

Choose m _k to be the smallest integer satisfying (5.22). Now from Lemma 4.5 we have that

$$\underset{k\to \infty }{\text{lim}}∥T^{3m_k}x-T^{3m_k-3}x∥=0$$

and by

$$\begin{array}{c}P+\epsilon \le ∥T^{3m_k}x-T^{3n_k+1}x∥+∥T^{3n_k+1}x-T^{3n_k+2}x∥+∥T^{3n_k+2}x-T^{3m_k}x∥\hfill \\ \le ∥T^{3m_k}x-T^{3m_k-3}x∥+∥T^{3m_k-3}x-T^{3n_k+1}x∥\hfill \\ +∥T^{3n_k+1}x-T^{3n_k+2}x∥+∥T^{3n_k+2}x-T^{3m_k}x∥\hfill \end{array}$$

we get

$$P+\epsilon \le \underset{k\to \infty }{\text{lim}}\left(∥T^{3m_k}x-T^{3n_k+1}x∥+∥T^{3n_k+1}x-T^{3n_k+2}x∥+∥T^{3n_k+2}x-T^{3m_k}x∥\right)\le P+\epsilon .$$

Thus

$$\underset{k\to \infty }{\text{lim}}\left(∥T^{3m_k}x-T^{3n_k+1}x∥+∥T^{3n_k+1}x-T^{3n_k+2}x∥+∥T^{3n_k+2}x-T^{3m_k}x∥\right)=P+\epsilon$$

Now from the triangular inequality we have

$$\begin{array}{c}\left|\right|T^{3m_k}x-T^{3n_k+1}x\left|\right|+\left|\right|T^{3n_k+1}x-T^{3n_k+2}x\left|\right|+\left|\right|T^{3n_k+2}x-T^{3m_k}x\left|\right|\hfill \\ \le \left|\right|T^{3m_k}x-T^{3m_k+3}x\left|\right|+\left|\right|T^{3m_k+3}x-T^{3n_k+4}x\left|\right|\hfill \\ +\left|\right|T^{3n_k+4}x-T^{3n_k+1}x\left|\right|\hfill \\ +\left|\right|T^{3n_k+1}x-T^{3n_k+4}x\left|\right|+\left|\right|T^{3n_k+4}x-T^{3n_k+5}x\left|\right|\hfill \\ +\left|\right|T^{3n_k+5}x-T^{3n_k+2}x\left|\right|\hfill \\ +\left|\right|T^{3n_k+2}x-T^{3n_k+5}x\left|\right|+\left|\right|T^{3n_k+5}x-T^{3m_k+3}x\left|\right|\hfill \\ +\left|\right|T^{3m_k+3}x-T^{3m_k}x\left|\right|\hfill \end{array}$$

(5.23)

and by Lemma 4.4, taking a limit in (5.23) and applying condition (3.1) three times we get we get

$$\begin{array}{c}\underset{k\to \infty }{\lim }(||T^{3m_k}x-T^{3n_k+1}x||+||T^{3n_k+1}x-T^{3n_k+2}x||+||T^{3n_k+2}x-T^{3m_k}x||)\hfill \\ \le \underset{k\to \infty }{\lim }(||T^{3m_k+3}x-T^{3n_k+4}x||+||T^{3n_k+4}x-T^{3n_k+5}x||\hfill \\ +\left|\right|T^{3n_k+5}x-T^{3m_k+3}x\left|\right|)\hfill \\ \le k^3\underset{k\to \infty }{\lim }(||T^{3m_k}x-T^{3n_k+1}x||+||T^{3n_k+1}x-T^{3n_k+2}x|\hfill \\ +\left|\right|T^{3n_k+2}x-T^{3m_k}x\left|\right|)+(1-k^3)P\hfill \end{array}$$

i.e.

$$P+\epsilon \le k^3\left(P+\epsilon \right)+\left(1-k^3\right)P,$$

which is a contradiction and Claim 5.1 is proved.

Now by Claim 5.1 we have that for any ε > 0 there is n₀, such that

$$∥T^{3m}x-T^{3n+1}x∥<\text{dist}\left(A_i,A_{i+1}\right)+\epsilon$$

for every m > n ≥ n₀ and by Corollary 4.1 and Lemma 2.1 we have that the sequence ${\left\{T^{3n}x\right\}}_{n=1}^{\infty }$ is a Cauchy sequence. Thus lim_n→∞T³ⁿx = z and z is a best proximity point of T in A _i .

For the next proof we will follow the idea in [8], how to use a variational principle to prove a fixed point theorem.

Proof of Theorem 3.2. Let us define the function F : X → ℝ by

$$F\left(x\right)=∥Tx-x∥+∥T^{2}x-Tx∥+∥T^{2}x-x∥.$$

Since by Lemma 4.7 the function T is continuous and so is F. It is easy to see that F is bounded form below and not identically +∞. Choose ε₀ > 0, such that k + ε < 1. There exists x₀ ∈ X, such that F(x₀) < ε₀ + inf{F(x) : x ∈ X}, because F is continuous and bounded from below. Therefore, we can apply Theorem 2.1. By Theorem 2.1 there is v ∈ X, such that ||x₀- v|| ≤ 1 and for every w ∈ X there holds the inequality

$$F\left(v\right)\le F\left(w\right)+\epsilon ∥v-w∥.$$

Suppose that T does not have a fixed point then F(v) > 0 for every v ∈ X. Put w = Tv. Then we get the inequality F(v) ≤ F(Tv) + ε||v - Tv||, i.e.,

$$\begin{array}{ll}\hfill F\left(v\right)& =\left|\right|Tv-v\left|\right|+\left|\right|T^{2}v-Tv\left|\right|+\left|\right|T^{2}v-v\left|\right|\\ \le \left|\right|T^{2}v-Tv\left|\right|+\left|\right|T^{3}v-T^{2}v\left|\right|+\left|\right|T^{3}v-Tv\left|\right|+\epsilon \left|\right|v-Tv\left|\right|\\ \le k\left(\left|\right|Tv-v\left|\right|+\left|\right|T^{2}v-Tv\left|\right|+\left|\right|T^{2}v-v\left|\right|\right)+\epsilon \left|\right|v-Tv\left|\right|\\ =\left(k+\epsilon \right)\left|\right|Tv-v\left|\right|+k\left(\left|\right|T^{2}v-Tv\left|\right|+\left|\right|T^{2}v-v\left|\right|\right).\end{array}$$

By the last chain of inequalities we get

$$\left(1-k-\epsilon \right)∥Tv-v∥+\left(1-k\right)\left(∥T^{2}v-Tv∥+∥T^{2}v-v∥\right)\le 0,$$

which is a contradiction and therefore Tv = v.

Let us suppose that T has two fixed points x ≠ y. Let z ∈ X, be fixed and different from x, y. There is s₀ ∈ ℕ, such that $k^{s_0}<\frac{∥x-y∥}{∥x-y∥+∥y-z∥+∥z-x∥}$. Then for any s ≥ s₀ by 3.2 we get

$$\begin{array}{ll}\hfill \left|\right|x-y\left|\right|& =\left|\right|T^{s}x-T^{s}y\left|\right|\le \left|\right|T^{s}x-T^{s}y\left|\right|+\left|\right|T^{s}y-T^{s}z\left|\right|+\left|\right|T^{s}z-T^{s}x\left|\right|\\ \le k\left(\left|\right|T^{s-1}x-T^{s-1}y\left|\right|+\left|\right|T^{s-1}y-T^{s-1}z\left|\right|+\left|\right|T^{s-1}z-T^{s-1}x\left|\right|\right)\\ \le k^2\left(\left|\right|x-y\left|\right|+\left|\right|y-z\left|\right|+\left|\right|z-x\left|\right|\right),\end{array}$$

which is a contradiction and thus T has a unique fixed point.

We would like to illustrate Theorem 3.1 with two example:

Example 5.1. Consider the Euclidian space $\left({ℝ}_2^3,{∥\cdot ∥}_2\right)$, endowed with the Euclidian norm ${∥\left(x,y,z\right)∥}_2=\sqrt{x^2+y^2+z^2}$. Let $X\subset {ℝ}_2^3$ be $X=\left\{\left(x,y,z\right):x\in \left[4,5\right],y,z=0\right\},Y\subset {ℝ}_2^3$ be $Y=\left\{\left(x,y,z\right):y\in \left[1,2\right],x,z=0\right\},Z\subset {ℝ}_2^3$ be Z = {(x, y, z): z ∈ [1, 2], x, y = 0}. Define the 3-cyclic map T: X → Y, T :Y → Z, T : Z → X by

$$\begin{array}{ll}\hfill T\left(x,0,0\right)& =\left(0,\frac{x}{8}+\frac{1}{2},0\right),x\in \left[4,5\right]\\ \hfill T\left(0,y,0\right)& =\left(0,0,\frac{y}{8}+\frac{7}{8}\right),y\in \left[1,2\right]\\ \hfill T\left(0,0,z\right)& =\left(\frac{z}{8}+\frac{31}{8},0,0\right),z\in \left[1,2\right].\end{array}$$

It is easy to check that

$$\begin{array}{c}\text{max}\left\{\sqrt{{\left(\frac{x}{8}+\frac{1}{2}\right)}^2+{\left(\frac{y}{8}+\frac{7}{8}\right)}^2}-\frac{1}{2}\sqrt{\left(x^2+y^2\right)}:x\in \left[4,5\right],y\in \left[1,2\right]\right\}=\sqrt{2}-\frac{\sqrt{17}}{2},\\ \text{max}\left\{\sqrt{{\left(\frac{x}{8}+\frac{1}{2}\right)}^2+{\left(\frac{z}{8}+\frac{31}{8}\right)}^2}-\frac{1}{2}\sqrt{{\left(x\right)}^2+z^2}:x\in \left[4,5\right],z\in \left[1,2\right]\right\}=\frac{\sqrt{17}}{2},\\ \text{max}\left\{\sqrt{{\left(\frac{y}{8}+\frac{7}{8}\right)}^2+{\left(\frac{z}{8}+\frac{31}{8}\right)}^2}-\frac{1}{2}\sqrt{\left(y^2+z^2\right)}:x\in \left[1,2\right],z\in \left[1,2\right]\right\}=\sqrt{17}-\frac{\sqrt{2}}{2}.\end{array}$$

Thus we get that for every x ∈ X, y ∈ Y, z ∈ Z there holds the inequality:

$${∥Tx-Ty∥}_2+{∥Ty-Tz∥}_2+{∥Tz-Tx∥}_2\le \frac{1}{2}\left({∥x-y∥}_2+{∥y-z∥}_2+{∥z-x∥}_2\right)+\frac{1}{2}P,$$

because $P=\text{dist}\left(X,Y\right)+\text{dist}\left(Y,Z\right)+\text{dist}\left(Z,X\right)=2\sqrt{17}+\sqrt{2}$. The distances between the three sets are different. The map T is not a cyclical contraction. Indeed, there exists ε₀, δ₀ > 0, such that for any z, y ∈ [1,1 + δ₀] we have

$${∥Tz-Ty∥}_2\ge \sqrt{17}-{\epsilon }_0>\sqrt{2}+{\epsilon }_0\ge k{∥z-y∥}_2+\left(1-k\right)\text{dist}\left(Z,X\right).$$

Example 5.2. Consider the Banach space $\left({ℝ}_2^3,\left|∥\cdot ∥\right|\right)$, where ||| · ||| = || · ||₂ + || · ||₁ and ||(x, y, z)||₁ = |x| + |y| + |z|. Let $X\subset {ℝ}_2^3$ be X = {(x, y, z): x ∈ [4, 5], y, z = 0}, $Y\subset {ℝ}_2^3$ be Y = {(x, y, z) : y ∈ [1, 2], x, z = 0}, $Z\subset {ℝ}_2^3$ be Z = {(x, y, z): z ∈ [1, 2], x, y = 0}. Define the 3-cyclic map T: X → Y, T: Y → Z, T : Z → X by

$$\begin{array}{ll}\hfill T\left(x,0,0\right)& =\left(0,\frac{x}{8}+\frac{1}{2},0\right),x\in \left[4,5\right]\\ \hfill T\left(0,y,0\right)& =\left(0,0,\frac{y}{8}+\frac{7}{8}\right),y\in \left[1,2\right]\\ \hfill T\left(0,0,z\right)& =\left(\frac{z}{8}+\frac{31}{8},0,0\right),z\in \left[1,2\right].\end{array}$$

It is easy to check for every x ∈ X, y ∈ Y, z ∈ Z that

$$\begin{array}{c}\text{max}\left\{∥\left|Tx-Ty\right|∥-\frac{1}{2}∥\left|x-y\right|∥:x\in \left[4,5\right],y\in \left[1,2\right]\right\}=-\frac{1}{2}+\sqrt{2}-\frac{\sqrt{17}}{2},\\ \text{max}\left\{∥\left|Tx-Tz\right|∥-\frac{1}{2}∥\left|x-z\right|∥:x\in \left[4,5\right],z\in \left[1,2\right]\right\}=\frac{5}{4}+\frac{\sqrt{17}}{2},\\ \text{max}\left\{∥\left|Ty-Tz\right|∥-\frac{1}{2}∥\left|y-z\right|∥:y\in \left[1,2\right],z\in \left[1,2\right]\right\}=4+\sqrt{17}-\frac{\sqrt{2}}{2}\end{array}$$

and $P=2\sqrt{17}+\sqrt{2}+12$. Therefore, there holds the inequality

$$∥\left|Tx-Ty\right|∥+∥\left|Ty-Tz\right|∥+∥\left|Tz-Tx\right|∥\le \frac{1}{2}\left(∥\left|x-y\right|∥+∥\left|y-z\right|∥+∥\left|z-x\right|∥\right)+\frac{1}{2}P.$$

It remains to show that the space (R³, ||| · |||) is uniformly convex. Let us consider its dual space (R³,||| · |||*). The norm ||| · ||| is strictly convex, then ||| · |||* is Geteaux differentiable [9, 10]. The space (R³,||| · |||*) is finite dimensional and therefore ||| · |||* is uniformly Frechet differentiable and consequently ||| · ||| is uniformly convex [9, 10].

The distances between the three sets are different. The map T is not a cyclical contraction. Indeed, there exist ε₀, δ₀ > 0, such that for any z, y ∈ [1,1 + δ₀] we have

$$∥\left|Tz-Ty\right|∥\ge \sqrt{17}+5-{\epsilon }_0>\sqrt{2}+2+{\epsilon }_0\ge k∥z-y∥+\left(1-k\right)\text{dist}\left(Z,X\right).$$