On the Best Proximity Points for p–Cyclic Summing Contractions

Abstract

We present a condition that guarantees the existence and uniqueness of fixed (or best proximity) points in complete metric space (or uniformly convex Banach spaces) for a wide class of cyclic maps, called p–cyclic summing maps. These results generalize some known results from fixed point theory. We find a priori and a posteriori error estimates of the fixed (or best proximity) point for the Picard iteration associated with the investigated class of maps, provided that the modulus of convexity of the underlying space is of power type. We illustrate the results with some applications and examples.

1. Introduction and Preliminaries

Banach contraction principle and its numerous generalizations turn out to be a powerful tool in mathematical research. A direction for a generalization of the Banach contraction principle is the concept of cyclical maps [1]. Fixed point theory is a widely applied technique, when trying to solve $Tx=x$, provided that $T:Z\to Z$, when Z is a metric space. Due to the fact that a non-self mapping $T:Z\to Y$, $Z\cap Y=\varnothing$ do not have a fixed point, an approach can be to search for $x\in Z$ that is as close as possible to its image $Tx$ i.e., to try to solve $min\{\parallel x-Tx\parallel :x\in Z\}$. The last minimization problem, when $min\{\parallel x-Tx\parallel :x\in Z\}=0$, coincides with $x=Tx$. Best proximity point results are applicable in this context. The concept of mentioned above points is initiated by Eldred and Veeramani in [2]. This definition is broader than that of cyclical maps because whenever the sets intersect the best proximity point reduces to a fixed point. A condition that guarantees the existence and uniqueness of best proximity points is presented in [2], provided that the underlying Banach space is uniformly convex. It is well known that a plentiful number of contractive-type maps that are known to have fixed points can be generalized to ensure the existence of best proximity points. The number of such generalizations is enormous and we could not mention even a small part of them. Some results of this kind are obtained in [3,4,5,6,7,8,9] and some very recent investigations [10,11,12,13,14,15,16,17,18]. It is curious that, in all the explored conditions for the presence of best proximity, the distances between the consecutive sets are equal. A condition that is completely different from the known ones and which warrants the existence and uniqueness of the best proximity points and for the cases when the distances between them are not equal is considered in [19]. These new types of maps were named p–cyclic summing contraction maps, but the authors have investigated only the case of $p=3$ there. A further investigation about different classes of p–cyclic summing contraction maps was presented in [20]. We fill the gaps from [19] by proving that the results from [19] can be generalized also for p-cyclic summing contraction maps. Some main tools for the proof are the results from [20].

Error estimates about fixed points for self (or cyclic) maps, starting with the classical Banach contraction principle, some resent results from this year e.g., [21,22] and the approximations of fixed points in [23,24], for example, are one of the greatest advantages in the applications of the fixed points technique. There have been a lack of such results about error estimates for best proximity points. This gap has been filled first for some kind of cyclic maps in [25] and later for other cyclic maps in [26,27,28,29,30].

We have obtained a priori error estimates and a posteriori error estimates for the p–cyclic summing contractions.

The structure of the paper is the following:

Preliminary results—We present the definitions and results, which we will need for the main theorem

Main Result—We define the notion of p–cyclic summing contraction map and we state and prove that any such map has a unique best proximity point and we obtain error estimates, when a sequence of successive iterations is used

Applications—We illustrate the main result, by applying it to the known p–cyclic maps, define in [5], and we extend the results from [5] by getting error estimates. We apply the main result in getting error estimates in the example presented in [19]

Conclusions—We discuss some open problems and possible future generalizations.

2. Preliminary Results

We will recall basic definitions and concepts which are related to our investigation. Let $(X,\rho )$ be a metric space. A distance between two subset $Y,Z\subset X$ is $\mathrm{dist}(Y,Z)=inf\left\{\rho \right(y,z):y\in Y,z\in Z\}$.

Let ${\left\{A_i\right\}}_{i=1}^p$ be non-empty subsets of a metric space $(X,\rho )$. A well-known agreement, just to simplify the notations, is $A_{p+i}=A_i$ for any $i\in \mathbb{N}$. A map $T:{\bigcup }_{i=1}^p{A}_i\to {\bigcup }_{i=1}^p{A}_i$ is called a p–cyclic map if $T\left(A_i\right)\subseteq A_{i+1}$ for every $i=1,2,\dots ,p$. A point $\xi \in A_i$ is called a best proximity point of T in $A_i$ if $\rho (\xi ,T\xi )=\mathrm{dist}(A_i,A_{i+1})$, provided that T is a cyclic map.

Most of the results about best proximity points utilize the norm-structure of the underlying space X. Everywhere in the article the distance between the elements of $(X,\parallel ·\parallel )$ will be the classical one $\rho (x,y)=\parallel x-y\parallel$. We will denote by $S_X$ and $B_X$ the unit sphere and the unit ball in $(X,\parallel ·\parallel )$, respectively.

The uniformly convex $(X,\parallel ·\parallel )$ assumption plays a decisive part in the proofs in most of the research about best proximity points.

Definition 1.

([31,32] p. 285) Let $(X,\parallel ·\parallel )$ be a Banach space. For every $\epsilon \in (0,2]$, we define the modulus of convexity of $\parallel ·\parallel$ by

$${\delta }_{(X,\parallel ·\parallel )}\left(\epsilon \right)=inf\left\{1-∥\frac{x+y}{2}∥:x,y\in B_X,\parallel x-y\parallel \ge \epsilon \right\}.$$

The norm is called uniformly convex if ${\delta }_{(X,\parallel ·\parallel )}\left(\epsilon \right)>0$ for all $\epsilon \in (0,2]$. The space $(X,\parallel ·\parallel )$ is then called uniformly convex space.

The modulus of convexity depends both on the space X and its norm $\parallel ·\parallel$. Just to simplify the notations, we will use ${\delta }_{\parallel ·\parallel }$, when there is no risk of confusion.

The next lemmas, proved in [2], are key results that we will need.

Lemma 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) ${lim}_{n\to \infty }\parallel z_n-y_n\parallel =\mathrm{dist}(A,B)$;

(2) for every $\epsilon >0$, there exists $N_0\in \mathbb{N}$, such that, for all $m>n\ge N_0$, $\parallel x_m-y_n\parallel \le \mathrm{dist}(A,B)+\epsilon$,

then, for every $\epsilon >0$, there exists $N_1\in \mathbb{N}$, such that, for all $m>n>N_1$, $\parallel x_m-z_n\parallel \le \epsilon$.

Lemma 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) ${lim}_{n\to \infty }\parallel x_n-y_n\parallel =\mathrm{dist}(A,B)$;

(2) ${lim}_{n\to \infty }\parallel z_n-y_n\parallel =\mathrm{dist}(A,B)$;

then ${lim}_{n\to \infty }\parallel x_n-z_n\parallel =0$.

The inequality

$$∥\frac{x+y}{2}-z∥\le \left(1-{\delta }_X\left(\frac{r}{R}\right)\right)R$$

(1)

for any $x,y,z\in X$, $R>0$, $r\in [0,2R]$, $\parallel x-z\parallel \le R$, $\parallel y-z\parallel \le R$, and $\parallel x-y\parallel \ge r$ holds, provided that the Banach space X is uniformly convex.

The modulus of convexity $\delta$ is a strictly increasing function in any uniformly convex Banach space and consequently there exists its inverse function, which we will denote by ${\delta }^{-1}$. The modulus of convexity $\delta$ is said to be of power type q if the inequality ${\delta }_{\parallel ·\parallel }\left(\epsilon \right)\ge C{\epsilon }^q$ holds for any $\epsilon \in (0,2]$ and for some strictly positive constants C and q ([33], p. 154). It is well known that the inequality ${\delta }_{\parallel ·\parallel }\left(\epsilon \right)\le K{\epsilon }^2$ holds for any Banach space endowed with any norm $\parallel ·\parallel$; thus, if the modulus of convexity is of power type q, then $q\ge 2$.

A comprehensive presentation of the geometry of Banach spaces can be found, for example, in [32,33,34,35].

Let ${\left\{A_i\right\}}_{i=1}^p$ be non-empty subsets of the metric space $(X,\rho )$. We will use the notions $P={\sum }_{i=1}^p\mathrm{dist}(A_i,A_{i+1})$, $d_{i,i+1}=\mathrm{dist}(A_i,A_{i+1})$ and

$$s_p(x_1,x_2,\dots ,x_p)=\sum _{j=1}^{p-1}\rho (x_j,x_{j+1})+\rho (x_p,x_1),$$

(2)

where, if $x_1\in A_i$, then $x_{1+k}\in A_{i+k}$ for every $k=1,2,\dots ,p-1$ (where we use assume that $A_{p+i}=A_i$, for every $i\in \{1,2,\dots ,p\}$). Just for simplicity of the notations, we will denote

$$s_{p,n}\left(x\right)=s_p\left(T^{n}x,T^{n+1}x,T^{n+2}x,\dots ,T^{n+p-1}x\right)$$

for any $x\in {\bigcup }_{i=1}^p{A}_i$, where T is a p–cyclic map.

Definition 2.

([20]) Let $A_i$, $i=1,2,\dots ,p$ be subsets of a metric space $(X,\rho )$. A map $T:{\bigcup }_{i=1}^p{A}_i\to {\bigcup }_{i=1}^p{A}_i$ is said to be a p–cyclic summing iterated contraction if it satisfies the next two conditions:

(1) T is a p–cyclic map;

(2) there is a constant $k\in (0,1)$, so that, for every $x\in {\bigcup }_{i=1}^p{A}_i$, the inequality

$$s_{p,1}\left(x\right)\le k{s}_{p,0}\left(x\right)+(1-k)P$$

(3)

holds.

We use in the sequel an equivalent form of (3)

$$s_{p,1}\left(x\right)-P\le k(s_{p,0}\left(x\right)-P).$$

(4)

We will need some results from [20].

Definition 3.

[20,36]) Let $A_i$, $i=1,2,\dots ,p$ be non-empty subsets of a metric space and $T:{\bigcup }_{i=1}^p{A}_i\to {\bigcup }_{i=1}^p{A}_i$ be a p–cyclic map. We say that T satisfies the proximal property if whenever ${lim}_{n\to \infty }{x}_n=x\in A_i$, $x_n\in A_i$, and ${lim}_{n\to \infty }\rho (x_n,T{x}_n)=\mathrm{dist}(A_i,A_{i+1})$ hold, it follows that $\rho (x,Tx)=\mathrm{dist}(A_i,A_{i+1})$ for all $i=1,2,\dots p$.

Let us point out that the proximal property for two sets in normed spaces was defined in [36] and for p–sets in [20].

Theorem 1.

([20]) Let $(X,\parallel ·\parallel )$ be a uniformly convex Banach space and $A_i\subset X$, $i=1,2,\dots ,p$ be closed, convex sets and $T:{\cup }_{i=1}^p{A}_i\to {\cup }_{i=1}^p{A}_i$ be a p–cyclic summing iterated contraction.

Then, for every $x\in A_1$, the sequence ${\left\{T^{pn}x\right\}}_{n=1}^{\infty }$ is convergent. If $z={lim}_{n\to \infty }{T}^{pn}x$ and T is continuous at z or T satisfies the proximal property, then $z\in A_1$ is a best proximity point of T in $A_1$, $T^{i}z\in A_{i+1}$ is a best proximity point of T in $A_{i+1}$ for $i=1,2,\dots ,p-1$ and $T^{p}z=z$.

Lemma 3.

([20]) Let $(X,\parallel ·\parallel )$ be a uniformly convex Banach space, $A_i\subset X$, $i=1,2,\dots ,p$ be closed, convex sets and $T:{\cup }_{i=1}^p{A}_i\to {\cup }_{i=1}^p{A}_i$ be a p–cyclic summing iterated contraction. Then, ${lim}_{n\to \infty }\parallel T^{pn+k}x-T^{pn+p+k}x\parallel =0$, $k=0,1,2,\dots ,p-1$.

Lemma 4.

([20]) Let $(X,\rho )$ be a metric space, $A_i\subset X$, $i=1,2,\dots ,p$ be subsets and $T:{\cup }_{i=1}^p{A}_i\to {\cup }_{i=1}^p{A}_i$ be a p–cyclic summing iterated contraction. Then,

$$s_{p,n}\left(x\right)-P\le k^n(s_{p,0}\left(x\right)-P),s_{p,n}\left(x\right)-P\le k^l(s_{p,n-l}\left(x\right)-P),\underset{n\to \infty }{lim}{s}_{p,n}\left(x\right)=P.$$

From Lemma 4, it is easy to observe that there holds the inequality

$$\parallel T^{pn}x-T^{pn+1}x\parallel -d_{i,i+1}\le k^{pn}(s_{p,0}\left(x\right)-P),$$

whenever $x\in A_i$.

3. Main Result

Definition 4.

Let $A_i$, $i=1,2,\dots ,p$ be subsets of a metric space $(X,\rho )$. A map $T:{\bigcup }_{i=1}^p{A}_i\to {\bigcup }_{i=1}^p{A}_i$ will be called a p–cyclic summing contraction if it satisfies the next two assumptions:

(1) T is a p–cyclic map;

(2) there is a constant $k\in (0,1)$, so that the inequality

$$s_p(T{x}_1,T{x}_2,\dots ,T{x}_p)\le k{s}_p(x_1,x_2,\dots ,x_p)+(1-k)P$$

(5)

holds for every $x_i\in A_i$, $i=1,2,\dots p$.

By the fact that any p–cyclic summing contraction is a p–cyclic summing iterated contraction, it follows that we can apply Theorem 1 for p–cyclic summing contraction.

Theorem 2.

Let $(X,\parallel ·\parallel )$ be a uniformly convex Banach space with modulus of convexity ${\delta }_{\parallel ·\parallel }\left(\epsilon \right)$ and $A_i\subset X$, $i=1,2,\dots p$ be closed, convex sets and $T:{\cup }_{i=1}^p{A}_i\to {\cup }_{i=1}^p{A}_i$ be a p–cyclic summing contraction.

Then, for every $x\in A_1$, the sequence ${\left\{T^{pn}x\right\}}_{n=1}^{\infty }$ is convergent. If $z={lim}_{n\to \infty }{T}^{pn}x$, then $z\in A_1$ is a best proximity point of T in $A_1$, $T^{i}z\in A_{i+1}$ is a best proximity point of T in $A_{i+1}$ for $i=1,2,\dots ,p-1$ and $T^{p}z=z$.

If $(X,\parallel ·\parallel )$ is with modulus of convexity of power type with constants $C>0$ and $q\ge 2$, then

• a priori error estimate holds

  $$∥\xi -T^{pm}x∥\le s_{p,0}\left(x\right)\sqrt[q]{\frac{s_{p,0}\left(x\right)-P}{C{d}_{1,2}}}·\frac{{\left(\sqrt[q]{k}\right)}^{pm}}{1-\sqrt[q]{k^p}};$$

  (6)
• a posteriori error estimate holds

  $$∥T^{pn}x-\xi ∥\le s_{p,pn-1}\left(x\right)\sqrt[q]{\frac{s_{p,pn-1}\left(x\right)-P}{C{d}_{1,2}}}·\frac{\sqrt[q]{k}}{1-\sqrt[q]{k^p}}.$$

  (7)

Proof. 

As far as T is a p–cyclic summing contraction, it follows that it is p–cyclic summing iterated contraction. There, for any arbitrary chosen $x\in A_i$ from Theorem 1, we get that the iterated sequence ${\left\{T^{pn}x\right\}}_{n=0}^{\infty }$ is convergent to a point $z\in X$. From the assumption that $A_i$ are closed subsets, it follows that $z\in A_i$.

Without loss of generality we can assume that $x\in A_1$, indeed, we can enumerate the sets so that $x\in A_1$. This will simplify the notations.

By the continuity of the function $\parallel ·-·\parallel$, it follows that $\parallel z-Tz\parallel ={lim}_{n\to \infty }\parallel T^{pn}x-Tz\parallel$ and

$$\underset{n\to \infty }{lim}\parallel T^{pn-1}x-z\parallel =\underset{n\to \infty }{lim}\parallel T^{pn-1}x-T^{pn}x\parallel .$$

(8)

We apply consecutively (8) and Lemma 4 to obtain the next chain of inequalities:

$$\begin{array}{ccc}{s}_p(z,Tz,T^{2}z,\dots ,T^{p-1}z)-P\hfill & =\hfill & \underset{n\to \infty }{lim}{s}_p(T^{pn}x,Tz,T^{2}z,\dots ,T^{p-1}z)-P\hfill \\ \hfill & \le \hfill & k\left(\underset{n\to \infty }{lim}{s}_p(T^{pn-1}x,z,Tz,\dots ,T^{p-2}z)-P\right)\hfill \\ \hfill & =\hfill & k\left(\underset{n\to \infty }{lim}{s}_p(T^{pn-1}x,T^{pn}x,Tz,\dots ,T^{p-2}z)-P\right)\hfill \\ \hfill & \le \hfill & k^2\left(\underset{n\to \infty }{lim}{s}_p(T^{pn-2}x,T^{pn-1}x,z,\dots ,T^{p-3}z)-P\right)\hfill \\ \hfill & =\hfill & k^2\left(\underset{n\to \infty }{lim}{s}_p(T^{pn-2}x,T^{pn-1}x,T^{pn}x,\dots ,T^{p-3}z)-P\right)\hfill \\ \hfill & \le \hfill & k^3\left(\underset{n\to \infty }{lim}{s}_p(T^{pn-3}x,T^{pn-2}x,T^{pn-1}x,z\dots ,T^{p-4}z)-P\right)\hfill \\ \hfill & =\hfill & \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \hfill \\ \hfill & \le \hfill & k^p\left(\underset{n\to \infty }{lim}{s}_p(T^{pn-p}x,T^{pn-p+1}x,T^{pn-p+2}x,\dots ,T^{pn-1}x)-P\right)\hfill \\ \hfill & \le \hfill & k^p\left(\underset{n\to \infty }{lim}{s}_{p,pn-n}\left(x\right)-P\right)\hfill \\ \hfill & =\hfill & k^p(P-P)=0.\hfill \end{array}$$

(9)

Since $z\in A_1$, it follows that $Tz\in A_2$, $T^{k}z\in A_{1+k}$. From the inequalities $\parallel T^{k}z-T^{k+1}z\parallel \ge d_{1+k,2+k}$, $k=0,\dots ,p-2$ and $\parallel T^{p-1}z-z\parallel \ge d_{p,1}$ and (9), it follows that

$$\parallel T^{k}z-T^{k+1}z\parallel -d_{1+k,2+k}\le s_p(z,Tz,T^{2}z,\dots ,T^{p-1}z)-P=0$$

for $k=0,\dots ,p-2$ and

$$\parallel T^{p-1}z-z\parallel -d_{p,1}\le s_p(z,Tz,T^{2}z,\dots ,T^{p-1}z)-P=0.$$

Thus, $\parallel T^{k}z-T^{k+1}z\parallel =d_{1+k,2+k}=\mathrm{dist}(A_{i+k},A_{i+k+1})$ for $k=0,\dots ,p-2$ and $\parallel T^{p-1}z-z\parallel =d_{p,1}=\mathrm{dist}(A_p,A_1)$. Therefore, z is a best proximity point of T in $A_1$ and $T^{k}z$ is a best proximity point of T in $A_{1+k}$, for $k=1,2,\dots ,p-1$.

From the inequality $s_p(T^{p}z,Tz,T^{2}z,\dots ,T^{p-1}z)\le k{s}_p(T^{p-1}z,z,Tz,\dots ,T^{p-2}z)=s_{p,0}\left(z\right)=P$, it follows that $\parallel T^{p}z-Tz\parallel -d_{1,2}\le s_{p,0}\left(z\right)-P=0$ and thus $\parallel T^{p}z-Tz\parallel =d_{1,2}$. From the equality $\parallel z-Tz\parallel =d_{1,2}$ and Lemma 1, it follows that $T^{p}z=z$.

Now, we will prove the a priori error estimate. Let us assume now that $(X,\parallel ·\parallel )$ is uniformly convex with a modulus of convexity of power type with constants $C>0$ and $q\ge 2$.

For any $x\in A$, $n\in \mathbb{N}$ and $l\le 2n$ there holds the inequality

$${\delta }_{\parallel ·\parallel }\left(\frac{\parallel T^{2n}x-T^{2n+2}x\parallel }{d_{1,2}+k^l{S}_{2n-l,2n+1-l}\left(x\right)}\right)\le \frac{k^l{S}_{2n-l,2n+1-l}\left(x\right)}{d_{1,2}+k^l{S}_{2n-l,2n+1-l}\left(x\right)}.$$

Indeed, let $x\in A_1$ be arbitrarily chosen. Let us denote $S_{p,pn-l}\left(x\right)=s_{p,pn-l}\left(x\right)-P$. From Lemma 4, we have the inequalities

$$\parallel T^{pn}x-T^{pn+1}x\parallel \le d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P),$$

$$\parallel T^{pn+p}x-T^{pn+1}x\parallel \le d_{1,2}+k^{l+1}(s_{p,pn-l}\left(x\right)-P)<d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)$$

and

$$\parallel T^{pn+p}x-T^{pn}x\parallel \le \parallel T^{pn+p}x-T^{pn+1}x\parallel +\parallel T^{pn+1}x-T^{pn}x\parallel \le 2\left(d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)\right).$$

After a substitution in (1) with $x=T^{pn}x$, $y=T^{pn+p}x$, $z=T^{pn+1}x$, $r=\parallel T^{pn+p}x-T^{pn}x\parallel$ and $R=d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)$ and, using the convexity of the set A, we get the chain of inequalities

$$\begin{array}{ccc}{d}_{1,2}\hfill & \le \hfill & {\displaystyle ∥\frac{T^{pn}x+T^{pn+p}x}{2}-T^{pn+1}x∥}\hfill \\ \hfill & \le \hfill & {\displaystyle \left(1-{\delta }_{\parallel ·\parallel }\left(\frac{\parallel T^{pn}x-T^{pn+p}x\parallel }{d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)}\right)\right)\left(d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)\right).}\hfill \end{array}$$

(10)

From (10), we obtain the inequality

$${\delta }_{\parallel ·\parallel }\left(\frac{\parallel T^{pn}x-T^{pn+p}x\parallel }{d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)}\right)\le \frac{k^l(s_{p,pn-l}\left(x\right)-P)}{d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)}.$$

(11)

From the assumption that X is uniform convexity of X, it follows that both ${\delta }_{\parallel ·\parallel }$ and its inverse function ${\delta }_{\parallel ·\parallel }^{-1}$ are strictly increasing functions. From (11), we get

$$\parallel T^{pn}x-T^{pn+p}x\parallel \le \left(d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)\right){\delta }_{\parallel ·\parallel }^{-1}\left(\frac{k^l(s_{p,pn-l}\left(x\right)-P)}{d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)}\right).$$

(12)

It is easy to observe that

$$k^l(s_{p,pn-l}\left(x\right)-P)\le s_{p,pn-l}\left(x\right)-P\le s_{p,pn-l}\left(x\right)-d_{1,2}$$

i.e.,

$$d_{1,2}\le d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)\le s_{p,pn-l}\left(x\right).$$

From the inequality ${\delta }_{\parallel ·\parallel }\left(t\right)\ge C{t}^q$, we get the inequality ${\delta }_{\parallel ·\parallel }^{-1}\left(t\right)\le {\left(\frac{t}{C}\right)}^{1/q}$ and by the last inequality and (12), we obtain

$$\begin{array}{ccc}\parallel T^{pn}x-T^{pn+p}x\parallel \hfill & \le \hfill & \left(d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)\right)\sqrt[q]{\frac{k^l(s_{p,pn-l}\left(x\right)-P)}{C.\left(d_{1,2}+k^l(s_{p,pn-l}\left(x\right)-P)\right)}}\hfill \\ \hfill & \le \hfill & s_{p,pn-l}\left(x\right)\sqrt[q]{\frac{s_{p,pn-l}\left(x\right)-P}{C{d}_{1,2}}}{\left(\sqrt[q]{k}\right)}^l.\hfill \end{array}$$

(13)

We have proven in the first part that there exists a unique $\xi \in A_i$, such that $\parallel \xi -T\xi \parallel =\mathrm{dist}(A_i,A_{i+1})$, $T^p\xi =\xi$ and $\xi$ is a limit of the sequence ${\left\{T^{pn}x\right\}}_{n=1}^{\infty }$ for any $x\in A_i$.

After substituting in (13) l with $pn$, we obtain the inequality

$$\sum _{n=1}^{+\infty }∥T^{pn}x-T^{pn+p}x∥\le s_{p,0}\left(x\right)\sqrt[q]{\frac{s_{p,0}\left(x\right)-P}{C{d}_{1,2}}}\sum _{n=1}^{+\infty }{\left(\sqrt[q]{k}\right)}^{pn}=s_{p,0}\left(x\right)\sqrt[q]{\frac{s_{p,0}\left(x\right)-P}{C{d}_{1,2}}}·\frac{\sqrt[q]{k^p}}{1-\sqrt[q]{k^p}}$$

and, consequently, the series ${\sum }_{n=1}^{+\infty }(T^{pn}x-T^{pn+p}x)$ is absolutely convergent. Therefore, for any $m\in \mathbb{N}$, $\xi =T^{pm}x-{\sum }_{n=m}^{+\infty }\left(T^{pn}x-T^{pn+p}x\right)$ holds and consequently we get the inequality

$$∥\xi -T^{pm}x∥\le \sum _{n=m}^{+\infty }∥T^{pn}x-T^{pn+p}x∥\le s_{p,0}\left(x\right)\sqrt[q]{\frac{s_{p,0}\left(x\right)-P}{C{d}_{1,2}}}·\frac{{\left(\sqrt[q]{k}\right)}^{pm}}{1-\sqrt[q]{k^p}}.$$

It remains to prove the ”a posteriori“ error estimate. After a substitution with $l=1+pi$ in (13), we obtain

$$\parallel T^{pn+pi}x-T^{pn+p(i+1)}x\parallel \le s_{p,pn-1}\left(x\right)\sqrt[q]{\frac{s_{p,pn-1}\left(x\right)-P}{C{d}_{1,2}}}{\left(\sqrt[q]{k}\right)}^{1+pi}.$$

(14)

From (14), we get that there holds the inequality

$$\begin{array}{ccc}\parallel T^{pn}x-T^{p(n+m)}x\parallel \hfill & \le \hfill & {\displaystyle \sum _{j=0}^{m-1}\parallel T^{pn+pj}x-T^{pn+p(j+1)}x\parallel }\hfill \\ \hfill & \le \hfill & {\displaystyle \sum _{j=0}^{m-1}{s}_{p,pn-1}\left(x\right)\sqrt[q]{\frac{s_{p,pn-1}\left(x\right)-P}{C{d}_{1,2}}}{\left(\sqrt[q]{k}\right)}^{1+pj}}\hfill \\ \hfill & =\hfill & {\displaystyle s_{p,pn-1}\left(x\right)\sqrt[q]{\frac{s_{p,pn-1}\left(x\right)-P}{C{d}_{1,2}}}\sum _{j=0}^{m-1}{\left(\sqrt[q]{k}\right)}^{1+pj}}\hfill \\ \hfill & =\hfill & {\displaystyle s_{p,pn-1}\left(x\right)\sqrt[q]{\frac{s_{p,pn-1}\left(x\right)-P}{C{d}_{1,2}}}·\frac{1-{\left(\sqrt[q]{k}\right)}^{pm}}{1-\sqrt[q]{k^p}}\sqrt[q]{k}}\hfill \end{array}$$

(15)

and, after letting $m\to \infty$ in (15), we obtain the inequality

$$∥T^{pn}x-\xi ∥\le s_{p,pn-1}\left(x\right)\sqrt[q]{\frac{s_{p,pn-1}\left(x\right)-P}{C{d}_{1,2}}}·\frac{\sqrt[q]{k}}{1-\sqrt[q]{k^p}}.$$

□

4. Applications

Let us recall the definition of p–cyclic contractions.

Definition 5.

([5]) Let $A_i$, $i=1,2,\dots ,p$ be subsets of a metric space $(X,\rho )$. A map $T:{\bigcup }_{i=1}^p{A}_i\to {\bigcup }_{i=1}^p{A}_i$ will be called a p–cyclic contraction if it satisfies the next two assumptions:

(1) T is a p–cyclic map;

(2) there is a constant $k\in (0,1)$, so that the inequality

$$\rho (Tx,Ty)\le k\rho (x,y)+(1-k)\mathrm{dist}(A_i,A_{i+1})$$

(16)

holds for every $x\in A_i$ and every $y\in A_{i+1}$, $i=1,2,\dots ,p$

Theorem 3.

Let $(X,\parallel ·\parallel )$ be a uniformly convex Banach space with a modulus of convexity ${\delta }_{\parallel ·\parallel }$ and $A_i\subset X$, $i=1,2,\dots ,p$ be closed, convex sets and $T:{\cup }_{i=1}^p{A}_i\to {\cup }_{i=1}^p{A}_i$ be a p–cyclic contraction.

Then, for every $x\in A_1$, the sequence ${\left\{T^{pn}x\right\}}_{n=1}^{\infty }$ is convergent. If $z={lim}_{n\to \infty }{T}^{pn}x$, then $z\in A_1$ is a best proximity point of T in $A_1$, $T^{i}z\in A_{i+1}$ is a best proximity point of T in $A_{i+1}$ for $i=1,2,\dots ,p-1$ and $T^{p}z=z$;

If there exist $C>0$ and $q\ge 2$, such that ${\delta }_{\parallel ·\parallel }\left(\epsilon \right)\ge C{\epsilon }^q$, then

• a priori error estimate holds

  $$∥\xi -T^{pm}x∥\le s_{p,0}\left(x\right)\sqrt[q]{\frac{s_{p,0}\left(x\right)-P}{C{d}_{1,2}}}·\frac{{\left(\sqrt[q]{k}\right)}^{pm}}{1-\sqrt[q]{k^p}};$$

  (17)
• a posteriori error estimate holds

  $$∥T^{pn}x-\xi ∥\le s_{p,pn-1}\left(x\right)\sqrt[q]{\frac{s_{p,pn-1}\left(x\right)-P}{C{d}_{1,2}}}·\frac{\sqrt[q]{k}}{1-\sqrt[q]{k^p}}.$$

  (18)

The first part of the theorem is proven in [5].

Proof. 

We will show how the above theorem follows from Theorem 2. Let us choose an arbitrary $x_i\in A_i$, $i=1,2,\dots ,p$. Then, after summing the inequalities

$$\parallel T{x}_i-T{x}_{i+1}\parallel \le k\parallel x_i-x_{i+1}\parallel +(1-k)\mathrm{dist}(A_i,A_{i+1}),i=1,2,\dots ,p-1.$$

and

$$\parallel T{x}_p-T{x}_1\parallel \le k\parallel x_p-x_1\parallel +(1-k)\mathrm{dist}(A_p,A_1),$$

we get

$$s_p(T{x}_1,T{x}_2,\dots ,T{x}_p)\le k{s}_p(x_1,x_2,\dots ,x_p)+(1-k)P.$$

(19)

The proofs of the error estimates follow directly from (19). □

We believe that similar results about the error estimates can be obtained, for example, for the classical p–cyclic Kannan maps investigated in [6] or for the proximal contractions; see, e.g., [37].

We will illustrate Theorem 2 with an example from [19].

It is well known that any Hilbert space with a norm generated by the scalar product there holds the inequality ${\delta }_{{\parallel ·\parallel }_p}\left(\epsilon \right)\ge \frac{{\epsilon }^2}{8}$ [38].

Example ([19]) Let the underlying space X be three-dimensional space $({\mathbb{R}}_2^3,\parallel ·{\parallel }_2)$, endowed with the Euclidian norm ${\parallel (x,y,z)\parallel }_2=\sqrt{x^2+y^2+z^2}$. Let $A_1\subset {\mathbb{R}}_2^3$ be $A_1=\{(x,y,z):x\in [4,5],y,z=0\}$, $A_2\subset {\mathbb{R}}_2^3$ be $A_2=\{(x,y,z):y\in [1,2],x,z=0\}$, $A_3\subset {\mathbb{R}}_2^3$ be $A_3=\{(x,y,z):z\in [1,2],x,y=0\}$. Define the 3-cyclic map $T:A_i\to A_{i+1}$, for $i=1,2,3$ and $A_4=A_1$ by

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

It is shown in [20] that, for every $x\in A_1$, $y\in A_2$, $z\in A_3$, there holds the inequality:

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

where $P=\mathrm{dist}(A_1,A_2)+\mathrm{dist}(A_2,A_3)+\mathrm{dist}(A_3,A_1)=2\sqrt{17}+\sqrt{2}$. The distances between the three sets are different and $\mathrm{dist}(X,Y)=\sqrt{17}$. The map T is not a cyclical contraction in the sense of [5].

Let us start with an initial guess $x=5$ in X, see

Table 1. The values of the iterated sequence $T^{n}x$ if started with $x=5$ in X.

n	0	1	2	5
$T^{3n}x$	5.00	4.00195	4.00000381	4.000000000
$T^{3n+1}x$	1.12	1.00024	1.00000047	1.000000000
$T^{3n+2}x$	1.01	1.00003	1.00000006	1.000000000

,

Table 2. Number of the necessary iterations needed if started with $x=5$ in X to get an $\epsilon$ a priori error estimate.

$\mathit{\epsilon }$	0.1	0.01	0.0001	0.000001
n	6	9	13	17

and

Table 3. Number of the necessary iterations needed if started with $x=5$ in X to get an $\epsilon$ a posteriori error estimate.

$\mathit{\epsilon }$	0.1	0.01	0.0001	0.000001
n	2	3	5	7

.

5. Conclusions

Let us mention that we get a larger number of the iterations that are needed to get the desired error. It happens because we use the modulus of convexity, which is the infinum of $∥1-\frac{x+y}{2}∥$ among all $x,y\in S_X$, such that $\parallel x-y\parallel \ge \epsilon$. A reason for this may be that the modulus of convexity is greater in the direction of the best proximity point $\xi$ than in the other directions, but, for the estimation of the error, we do not use it. We would like to pose the following question of whether it possible to get better estimates if we use the directional modulus of convexity ${\delta }_{\parallel ·\parallel }(x,\epsilon )$ [39]? For the estimations, we use geometric progression and that is why we impose the condition for the modulus of convexity to be of power type ([33], p. 154). Is it possible to obtain error estimates if the modulus of convexity is not of power type? Results about best proximity points in modular function spaces are obtained in [40,41]. Is it possible to generalize the notion of best proximity points in modular function spaces for p–cyclic summing contractions and to get error estimates? Sufficient conditions for the existence of best proximity points for weak p–cyclic Kannan contraction is obtained in [42]. It seems that the technique of obtaining error estimates could be possible to be applied for these class of maps.