About this Journal Submit a Manuscript Table of Contents
Abstract and Applied Analysis
Volume 2012 (2012), Article ID 736214, 19 pages
http://dx.doi.org/10.1155/2012/736214
Research Article

Common Fixed Point Theorems for a Class of Twice Power Type Contraction Maps in G-Metric Spaces

Institute of Applied Mathematics and Department of Mathematics, Hangzhou Normal University, Hangzhou Zhejiang 310036, China

Received 5 February 2012; Accepted 27 July 2012

Academic Editor: Svatoslav Staněk

Copyright © 2012 Hongqing Ye and Feng Gu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We introduce a new twice power type contractive condition for three mappings in G-metric spaces, and several new common fixed point theorems are established in complete G-metric space. An example is provided to support our result. The results obtained in this paper differ from other comparable results already known.

1. Introduction

The study of fixed points of mappings satisfying certain contractive conditions has been in the center of rigorous research activity. In 2006, a new structure of generalized metric space was introduced by Mustafa and Sims [1] as an appropriate notion of generalized metric space called 𝐺-metric space. Abbas and Rhoades [2] initiated the study of common fixed point in generalized metric space. Recently, many fixed point theorems for certain contractive conditions have been established in 𝐺-metric spaces, and for more details one can refer to [327]. Fixed point problems have also been considered in partially ordered 𝐺-metric spaces [2831], cone metric spaces [32], and generalized cone metric spaces [33].

In 2006, Gu and He [34] introduced a class of twice power type contractive condition in metric space, proving some common fixed point theorems for four self-maps with twice power type Φ-contractive condition.

In this paper, motivated and inspired by the above results, we introduce a new twice power type contractive condition in 𝐺-metric space, and we prove some new common fixed point theorems in complete 𝐺-metric spaces. Our results obtained in this paper differ from other comparable results already known.

Throughout the paper, we mean by the set of all natural numbers. Consistent with Mustafa and Sims [1], the following definitions and results will be needed in the sequel.

Definition 1.1 (see [1]). Let 𝑋 be a nonempty set, and let 𝐺𝑋×𝑋×𝑋𝑅+ be a function satisfying the following axioms:(𝐺1)𝐺(𝑥,𝑦,𝑧=0) if 𝑥=𝑦=𝑧;(𝐺2)0<𝐺(𝑥,𝑥,𝑦), for all 𝑥,𝑦𝑋 with 𝑥𝑦;(𝐺3)  𝐺(𝑥,𝑥,𝑦)𝐺(𝑥,𝑦,𝑧), for all 𝑥,𝑦,𝑧𝑋 with 𝑧 y;(𝐺4)  𝐺(𝑥,𝑦,𝑧)=𝐺(𝑥,𝑧,𝑦)=𝐺(𝑦,𝑧,𝑥)= (symmetry in all three variables);(𝐺5)𝐺(𝑥,𝑦,𝑧)𝐺(𝑥,𝑎,𝑎)+𝐺(𝑎,𝑦,𝑧) for all 𝑥,𝑦,𝑧,𝑎𝑋, (rectangle inequality);then the function 𝐺 is called a generalized metric, or, more specifically, a 𝐺-metric on 𝑋 and the pair (𝑋,𝐺) are called a 𝐺-metric space.

Definition 1.2 (see [1]). Let (𝑋,𝐺) be a 𝐺-metric space, and let {𝑥𝑛} be a sequence of points in 𝑋, a point 𝑥 in 𝑋 is said to be the limit of the sequence {𝑥𝑛} if lim𝑚,𝑛𝐺(𝑥,𝑥𝑛,𝑥𝑚)=0, and one says that sequence {𝑥𝑛} is 𝐺-convergent to 𝑥.

Thus, if 𝑥𝑛𝑥 in a 𝐺-metric space (𝑋,𝐺), then for any 𝜖>0, there exists 𝑁 such that 𝐺(𝑥,𝑥𝑛,𝑥𝑚)<𝜖, for all 𝑛,𝑚𝑁.

Proposition 1.3 (see [1]). Let (𝑋,𝐺) be a 𝐺-metric space, then the followings are equivalent.(1){𝑥𝑛} is 𝐺-convergent to 𝑥.(2)𝐺(𝑥𝑛,𝑥𝑛,𝑥)0 as 𝑛.(3)𝐺(𝑥𝑛,𝑥,𝑥)0 as 𝑛.(4)𝐺(𝑥𝑛,𝑥𝑚,𝑥)0 as 𝑛,𝑚.

Definition 1.4 (see [1]). Let (𝑋,𝐺) be a 𝐺-metric space. A sequence {𝑥𝑛} is called 𝐺-Cauchy sequence if for each 𝜖>0 there exists a positive integer 𝑁 such that 𝐺(𝑥𝑛,𝑥𝑚,𝑥𝑙)<𝜖 for all 𝑛,𝑚,𝑙𝑁; that is, if 𝐺(𝑥𝑛,𝑥𝑚,𝑥𝑙)0 as 𝑛,𝑚,𝑙.

Definition 1.5 (see [1]). A 𝐺-metric space (𝑋,𝐺) is said to be 𝐺-complete if every 𝐺-Cauchy sequence in (𝑋,𝐺) is 𝐺-convergent in 𝑋.

Proposition 1.6 (see [1]). Let (𝑋,𝐺) be a 𝐺-metric space. Then the following are equivalent.(1)The sequence {𝑥𝑛} is 𝐺-Cauchy.(2)For every 𝜖>0, there exists 𝑘 such that 𝐺(𝑥𝑛,𝑥𝑚,𝑥𝑚)<𝜖, for all 𝑛,𝑚𝑘.

Proposition 1.7 (see [1]). Let (𝑋,𝐺) be a 𝐺-metric space. Then the function 𝐺(𝑥,𝑦,𝑧) is jointly continuous in all three of its variables.

Definition 1.8 (see [1]). Let (𝑋,𝐺) and (𝑋,𝐺) be 𝐺-metric space, and 𝑓(𝑋,𝐺)(𝑋,𝐺) be a function. Then 𝑓 is said to be 𝐺-continuous at a point 𝑎𝑋 if and only if for every 𝜀>0, there is 𝛿>0 such that 𝑥,𝑦𝑋 and 𝐺(𝑎,𝑥,𝑦)<𝛿 implies 𝐺(𝑓(𝑎),𝑓(𝑥),𝑓(𝑦))<𝜀. A function 𝑓 is 𝐺-continuous at 𝑋 if and only if it is 𝐺-continuous at all 𝑎𝑋.

Proposition 1.9 (see [1]). Let (𝑋,𝐺) and (𝑋,𝐺) be 𝐺-metric space. Then 𝑓𝑋𝑋 is 𝐺-continuous at 𝑥𝑋 if and only if it is 𝐺-sequentially continuous at 𝑥; that is, whenever {𝑥𝑛} is 𝐺-convergent to 𝑥, {𝑓(𝑥𝑛)} is 𝐺-convergent to 𝑓(𝑥).

Proposition 1.10 (see, [1]). Let (𝑋,𝐺) be a 𝐺-metric space. Then, for any 𝑥,𝑦,𝑧,𝑎 in 𝑋 it follows that:(i)if 𝐺(𝑥,𝑦,𝑧)=0, then 𝑥=𝑦=𝑧;(ii)𝐺(𝑥,𝑦,𝑧)𝐺(𝑥,𝑥,𝑦)+𝐺(𝑥,𝑥,𝑧); (iii)𝐺(𝑥,𝑦,𝑦)2𝐺(𝑦,𝑥,𝑥); (iv)𝐺(𝑥,𝑦,𝑧)𝐺(𝑥,𝑎,𝑧)+𝐺(𝑎,𝑦,𝑧); (v)𝐺(𝑥,𝑦,𝑧)(2/3)(𝐺(𝑥,𝑦,𝑎)+𝐺(𝑥,𝑎,𝑧)+𝐺(𝑎,𝑦,𝑧));(vi)𝐺(𝑥,𝑦,𝑧)(𝐺(𝑥,𝑎,𝑎)+𝐺(𝑦,𝑎,𝑎)+𝐺(𝑧,𝑎,𝑎)).

2. Main Results

Theorem 2.1. Let (𝑋,𝐺) be a complete 𝐺-metric space. Suppose the three self-mappings 𝑇,𝑆,𝑅𝑋𝑋 satisfy the following condition: 𝐺2(𝑇𝑥,𝑆𝑦,𝑅𝑧)𝛼𝐺(𝑥,𝑇𝑥,𝑇𝑥)𝐺(𝑦,𝑆𝑦,𝑆𝑦)+𝛽𝐺(𝑦,𝑆𝑦,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑅𝑧)+𝛾𝐺(𝑥,𝑇𝑥,𝑇𝑥)𝐺(𝑧,𝑅𝑧,𝑅𝑧),(2.1) for all 𝑥,𝑦,𝑧𝑋, where 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇,𝑆, and 𝑅 have a unique common fixed point (say 𝑢) and 𝑇,𝑆,𝑅 are all 𝐺-continuous at 𝑢.

Proof . We will proceed in two steps.
Step 1. We prove any fixed point of 𝑇 is a fixed point of 𝑆 and 𝑅 and conversely. Assume that 𝑝𝑋 is such that 𝑇𝑝=𝑝. However, by (2.1), we have 𝐺2(𝑇𝑝,𝑆𝑝,𝑅𝑝)𝛼𝐺(𝑝,𝑇𝑝,𝑇𝑝)𝐺(𝑝,𝑆𝑝,𝑆𝑝)+𝛽𝐺(𝑝,𝑆𝑝,𝑆𝑝)𝐺(𝑝,𝑅𝑝,𝑅𝑝)+𝛾𝐺(𝑝,𝑇𝑝,𝑇𝑝)𝐺(𝑝,𝑅𝑝,𝑅𝑝)=𝛼𝐺(𝑝,𝑝,𝑝)𝐺(𝑝,𝑆𝑝,𝑆𝑝)+𝛽𝐺(𝑝,𝑆𝑝,𝑆𝑝)𝐺(𝑝,𝑅𝑝,𝑅𝑝)+𝛾𝐺(𝑝,𝑝,𝑝)𝐺(𝑝,𝑅𝑝,𝑅𝑝)=𝛽𝐺(𝑝,𝑆𝑝,𝑆𝑝)𝐺(𝑝,𝑅𝑝,𝑅𝑝).(2.2) Now we discuss the above inequality in three cases.
Case (i). If 𝑝𝑆𝑝 and 𝑝𝑅𝑝, then, by (𝐺3), we have 𝐺(𝑝,𝑆𝑝,𝑆𝑝)𝐺(𝑝,𝑆𝑝,𝑅𝑝),𝐺(𝑝,𝑅𝑝,𝑅𝑝)𝐺(𝑝,𝑆𝑝,𝑅𝑝).(2.3) So, the above inequality becomes 𝐺2(𝑝,𝑆𝑝,𝑅𝑝)=𝐺2(𝑇𝑝,𝑆𝑝,𝑅𝑝)𝛽𝐺2(𝑝,𝑆𝑝,𝑅𝑝).(2.4) Since 𝐺2(𝑝,𝑆𝑝,𝑅𝑝)>0, hence we have 𝛽1; however, it contradicts with 0𝛽𝛼+𝛽+𝛾<1, so we get 𝑝=𝑆𝑝=𝑅𝑝.
Case (ii). If 𝑝=𝑅𝑝, then we have 𝐺2(𝑝,𝑆𝑝,𝑅𝑝)=𝐺2(𝑇𝑝,𝑆𝑝,𝑅𝑝)𝛽𝐺(𝑝,𝑆𝑝,𝑆𝑝)𝐺(𝑝,𝑅𝑝,𝑅𝑝)=0.(2.5) Hence we have 𝐺2(𝑝,𝑆𝑝,𝑅𝑝)=0 and so 𝑝=𝑆𝑝=𝑅𝑝.
Case (iii). If 𝑝=𝑆𝑝, we can also get 𝐺2(𝑝,𝑆𝑝,𝑅𝑝)=0. Hence we have 𝑝=𝑆𝑝=𝑅𝑝. Therefore 𝑝 is a common fixed point of 𝑇,𝑆 and 𝑅.
The same conclusion holds if 𝑝=𝑆𝑝 or 𝑝=𝑅𝑝.
Step 2. We prove that 𝑇, 𝑆, and 𝑅 have a unique common fixed point.
Let 𝑥0𝑋 be an arbitrary point, and define the sequence {𝑥𝑛} by 𝑥3𝑛+1=𝑇𝑥3𝑛,𝑥3𝑛+2=𝑆𝑥3𝑛+1,𝑥3𝑛+3=𝑅𝑥3𝑛+2, 𝑛. If 𝑥𝑛=𝑥𝑛+1, for some 𝑛, with 𝑛=3𝑚, then 𝑝=𝑥3𝑚 is a fixed point of 𝑇 and, by the first step, 𝑝 is a common fixed point of 𝑆, 𝑇, and 𝑅. The same holds if 𝑛=3𝑚+1 or 𝑛=3𝑚+2. Without loss of generality, we can assume that 𝑥𝑛𝑥𝑛+1, for all 𝑛.
Next, we prove sequence {𝑥𝑛} is a 𝐺-Cauchy sequence. In fact, by (2.1) and (𝐺3), we have 𝐺2𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3=𝐺2𝑇𝑥3𝑛,𝑆𝑥3𝑛+1,𝑅𝑥3𝑛+2𝑥𝛼𝐺3𝑛,𝑇𝑥3𝑛,𝑇𝑥3𝑛𝐺𝑥3𝑛+1,𝑆𝑥3𝑛+1,𝑆𝑥3𝑛+1𝑥+𝛽𝐺3𝑛+1,𝑆𝑥3𝑛+1,𝑆𝑥3𝑛+1𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑅𝑥3𝑛+2𝑥+𝛾𝐺3𝑛,𝑇𝑥3𝑛,𝑇𝑥3𝑛𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑅𝑥3𝑛+2𝑥=𝛼𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+1𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+2𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+2𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3𝑥+𝛾𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+1𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3𝑥𝛼𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+1𝑥+𝛾𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+1.(2.6) Which gives that 𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥(𝛼+𝛾)𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3.(2.7) It follows that 𝑥(1𝛽)𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥(𝛼+𝛾)𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2.(2.8) From 0𝛽<1 we know that 1𝛽>0. Then, we have 𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝛼+𝛾𝐺𝑥1𝛽3𝑛,𝑥3𝑛+1,𝑥3𝑛+2.(2.9) On the other hand, by using (2.1) and (𝐺3), we have 𝐺2𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4=𝐺2𝑇𝑥3𝑛+3,𝑆𝑥3𝑛+1,𝑅𝑥3𝑛+2𝑥𝛼𝐺3𝑛+3,𝑇𝑥3𝑛+3,𝑇𝑥3𝑛+3𝐺𝑥3𝑛+1,𝑆𝑥3𝑛+1,𝑆𝑥3𝑛+1𝑥+𝛽𝐺3𝑛+1,𝑆𝑥3𝑛+1,𝑆𝑥3𝑛+1𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑅𝑥3𝑛+2𝑥+𝛾𝐺3𝑛+3,𝑇𝑥3𝑛+3,𝑇𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑅𝑥3𝑛+2𝑥=𝛼𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+4𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+2𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+2𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3𝑥+𝛾𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+4𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3𝑥𝛼𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝑥+𝛾𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.10) Which implies that 𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝑥(𝛼+𝛽)𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥+𝛾𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.11) It follows that 𝑥(1𝛾)𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝑥(𝛼+𝛽)𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3.(2.12) Form the condition 0𝛾𝛼+𝛽+𝛾<1, we know that 1𝛾>0. Therefore, we have 𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝛼+𝛽𝐺𝑥1𝛾3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3.(2.13) Again, using (2.1) and (𝐺3), we can get 𝐺2𝑥3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5=𝐺2𝑇𝑥3𝑛+3,𝑆𝑥3𝑛+4,𝑅𝑥3𝑛+2𝑥𝛼𝐺3𝑛+3,𝑇𝑥3𝑛+3,𝑇𝑥3𝑛+3𝐺𝑥3𝑛+4,𝑆𝑥3𝑛+4,𝑆𝑥3𝑛+4𝑥+𝛽𝐺3𝑛+4,𝑆𝑥3𝑛+4,𝑆𝑥3𝑛+4𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑅𝑥3𝑛+2𝑥+𝛾𝐺3𝑛+3,𝑇𝑥3𝑛+3,𝑇𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑅𝑥3𝑛+2𝑥=𝛼𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+4𝐺𝑥3𝑛+4,𝑥3𝑛+5,𝑥3𝑛+5𝑥+𝛽𝐺3𝑛+4,𝑥3𝑛+5,𝑥3𝑛+5𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3𝑥+𝛾𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+4𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3𝑥𝛼𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝐺𝑥3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝑥+𝛽𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝑥+𝛾𝐺3𝑛+5,𝑥3𝑛+3,𝑥3𝑛+4𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.14) Which implies that 𝐺𝑥3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝑥𝛼𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5+𝑥(𝛽+𝛾)𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.15) It follows that 𝑥(1𝛼)𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝑥(𝛽+𝛾)𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.16) By the condition 0𝛼𝛼+𝛽+𝛾<1, we know that 1𝛼>0. Hence, we have 𝐺𝑥3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝛽+𝛾𝐺𝑥1𝛼3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.17) Let 𝑞=max{(𝛼+𝛾)/(1𝛽),(𝛼+𝛽)/(1𝛾),(𝛽+𝛾)/(1𝛼)}, then from 0𝛼+𝛽+𝛾<1 we know that 0𝑞<1. Combining (2.9), (2.13), and (2.17), we have 𝐺𝑥𝑛,𝑥𝑛+1,𝑥𝑛+2𝑥𝑞𝐺𝑛1,𝑥𝑛,𝑥𝑛+1𝑞𝑛𝐺𝑥0,𝑥1,𝑥2.(2.18) Thus, by (𝐺3) and (𝐺5), for every 𝑚,𝑛,𝑚>𝑛, noting that 0𝑞<1, we have 𝐺𝑥𝑛,𝑥𝑚,𝑥𝑚𝑥𝐺𝑛,𝑥𝑛+1,𝑥𝑛+1𝑥+𝐺𝑛+1,𝑥𝑛+2,𝑥𝑛+2𝑥++𝐺𝑚1,𝑥𝑚,𝑥𝑚,𝑥𝐺𝑛,𝑥𝑛+1,𝑥𝑛+2𝑥+𝐺𝑛+1,𝑥𝑛+2,𝑥𝑛+3𝑥++𝐺𝑚1,𝑥𝑚,𝑥𝑚+1𝑞𝑛+𝑞𝑛+1++𝑞𝑚1𝐺𝑥0,𝑥1,𝑥2𝑞𝑛𝐺𝑥1𝑞0,𝑥1,𝑥2.(2.19) Which implies that 𝐺(𝑥𝑛,𝑥𝑚,𝑥𝑚)0, as 𝑛,𝑚. Thus {𝑥𝑛} is a 𝐺-Cauchy sequence. Due to the completeness of (𝑋,𝐺), there exists 𝑢𝑋, such that {𝑥𝑛} is 𝐺-convergent to 𝑢.
Next we prove 𝑢 is a common fixed point of 𝑇,𝑆, and 𝑅. By using (2.1), we have 𝐺2𝑇𝑢,𝑥3𝑛+2,𝑥3𝑛+3=𝐺2𝑇𝑢,𝑆𝑥3𝑛+1,𝑅𝑥3𝑛+2𝑥𝛼𝐺(𝑢,𝑇𝑢,𝑇𝑢)𝐺3𝑛+1,𝑆𝑥3𝑛+1,𝑆𝑥3𝑛+1𝑥+𝛽𝐺3𝑛+1,𝑆𝑥3𝑛+1,𝑆𝑥3𝑛+1𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑅𝑥3𝑛+2𝑥+𝛾𝐺(𝑢,𝑇𝑢,𝑇𝑢)𝐺3𝑛+2,𝑅𝑥3𝑛+2,𝑅𝑥3𝑛+2𝑥=𝛼𝐺(𝑢,𝑇𝑢,𝑇𝑢)𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+2𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+2𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3𝑥+𝛾𝐺(𝑢,𝑇𝑢,𝑇𝑢)𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3.(2.20) Letting 𝑛, and using the fact that 𝐺 is continuous on its variables, we can get 𝐺2(𝑇𝑢,𝑢,𝑢)=0.(2.21) Which gives that 𝑇𝑢=𝑢, that is 𝑢 is a fixed point of 𝑇. By using (2.1) again, we have 𝐺2𝑥3𝑛+1,𝑆𝑢,𝑥3𝑛+3=𝐺2𝑇𝑥3𝑛,𝑆𝑢,𝑅𝑥3𝑛+2𝑥𝛼𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+1𝑥𝐺(𝑢,𝑆𝑢,𝑆𝑢)+𝛽𝐺(𝑢,𝑆𝑢,𝑆𝑢)𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3𝑥+𝛾𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+1𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+3.(2.22) Letting 𝑛 at both sides, for 𝐺 is continuous on its variables, it follows that 𝐺2(𝑢,𝑆𝑢,𝑢)=0.(2.23) Therefore, 𝑆𝑢=𝑢; that is, 𝑢 is a fixed point of 𝑆. Similarly, by (2.1), we can also get 𝐺2𝑥3𝑛+1,𝑥3𝑛+2,𝑅𝑢=𝐺2𝑇𝑥3𝑛,𝑆𝑥3𝑛+1𝑥,𝑅𝑢𝛼𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+1𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+2𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+2𝑥𝐺(𝑢,𝑅𝑢,𝑅𝑢)+𝛾𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+1𝐺(𝑢,𝑅𝑢,𝑅𝑢).(2.24) On taking 𝑛 at both sides, since 𝐺 is continuous on its variables, we get that 𝐺2(𝑢,𝑢,𝑅𝑢)=0.(2.25) Which gives that 𝑢=𝑅𝑢, therefore, 𝑢 is fixed point of 𝑅. Consequently, we have 𝑢=𝑇𝑢=𝑆𝑢=𝑅𝑢, and 𝑢 is a common fixed point of 𝑇,𝑆 and 𝑅. Suppose 𝑣 is another common fixed point of 𝑇,𝑆, and 𝑅, and we have 𝑣=𝑇𝑣=𝑆𝑣=𝑅𝑣, then by (2.1), we have 𝐺2(𝑢,𝑢,𝑣)=𝐺2(𝑇𝑢,𝑆𝑢,𝑅𝑣)𝛼𝐺(𝑢,𝑇𝑢,𝑇𝑢)𝐺(𝑢,𝑆𝑢,𝑆𝑢)+𝛽𝐺(𝑢,𝑆𝑢,𝑆𝑢)𝐺(𝑣,𝑅𝑣,𝑅𝑣)+𝛾𝐺(𝑢,𝑇𝑢,𝑇𝑢)𝐺(𝑣,𝑅𝑣,𝑅𝑣)=𝛼𝐺(𝑢,𝑢,𝑢)𝐺(𝑢,𝑢,𝑢)+𝛽𝐺(𝑢,𝑢,𝑢)𝐺(𝑣,𝑣,𝑣)+𝛾𝐺(𝑢,𝑢,𝑢)𝐺(𝑣,𝑣,𝑣)=0.(2.26) Which implies that 𝐺2(𝑢,𝑢,𝑣)=0, hence, 𝑢=𝑣. Then we know the common fixed point of 𝑇,𝑆, and 𝑅 is unique.
To show that 𝑇 is 𝐺-continuous at 𝑢, let {𝑦𝑛} be any sequence in 𝑋 such that {𝑦𝑛} is 𝐺-convergent to 𝑢. For 𝑛, we have 𝐺2𝑇𝑦𝑛,𝑢,𝑢=𝐺2𝑇𝑦𝑛𝑦,𝑆𝑢,𝑅𝑢𝛼𝐺𝑛,𝑇𝑦𝑛,𝑇𝑦𝑛𝑦𝐺(𝑢,𝑆𝑢,𝑆𝑢)+𝛽𝐺(𝑢,𝑆𝑢,𝑆𝑢)𝐺(𝑢,𝑅𝑢,𝑅𝑢)+𝛾𝐺𝑛,𝑇𝑦𝑛,𝑇𝑦𝑛𝑦𝐺(𝑢,𝑅𝑢,𝑅𝑢)=𝛼𝐺𝑛,𝑇𝑦𝑛,𝑇𝑦𝑛𝑦𝐺(𝑢,𝑢,𝑢)+𝛽𝐺(𝑢,𝑢,𝑢)𝐺(𝑢,𝑢,𝑢)+𝛾𝐺𝑛,𝑇𝑦𝑛,𝑇𝑦𝑛𝐺(𝑢,𝑢,𝑢)=0.(2.27) Which implies that lim𝑛𝐺2(𝑇𝑦𝑛,𝑢,𝑢)=0. Hence {𝑇𝑦𝑛} is 𝐺-convergent to 𝑢=𝑇𝑢. So 𝑇 is 𝐺-continuous at 𝑢. Similarly, we can also prove that 𝑆,𝑅 are 𝐺-continuous at 𝑢. Therefore, we complete the proof.

Corollary 2.2. Let (𝑋,𝐺) be a complete 𝐺-metric space. Suppose the three self-mappings 𝑇,𝑆,𝑅𝑋𝑋 satisfy the condition: 𝐺2(𝑇𝑝𝑥,𝑆𝑠𝑦,𝑅𝑟𝑧)𝛼𝐺(𝑥,𝑇𝑝𝑥,𝑇𝑝𝑥)𝐺(𝑦,𝑆𝑠𝑦,𝑆𝑠𝑦)+𝛽𝐺(𝑦,𝑆𝑠𝑦,𝑆𝑠𝑦)𝐺(𝑧,𝑅𝑟𝑧,𝑅𝑟𝑧)+𝛾𝐺(𝑥,𝑇𝑝𝑥,𝑇𝑝𝑥)𝐺(𝑧,𝑅𝑟𝑧,𝑅𝑟𝑧),(2.28) for all 𝑥,𝑦,𝑧𝑋, where 𝑝,𝑠,𝑟,𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇,𝑆, and 𝑅 have a unique common fixed point (say 𝑢) and 𝑇𝑝,𝑆𝑠,𝑅𝑟 are all 𝐺-continuous at 𝑢.

Proof. From Theorem 2.1 we know that 𝑇𝑝,𝑆𝑠,𝑅𝑟 have a unique common fixed point (say 𝑢); that is, 𝑇𝑝𝑢=𝑢,𝑆𝑠𝑢=𝑢,𝑅𝑟𝑢=𝑢, and 𝑇𝑝,𝑆𝑠,𝑅𝑟 are 𝐺-continuous at 𝑢. Since 𝑇𝑢=𝑇𝑇𝑝𝑢=𝑇𝑝+1𝑢=𝑇𝑝𝑇𝑢, so 𝑇𝑢 is another fixed point of 𝑇𝑝, 𝑆𝑢=𝑆𝑆𝑠𝑢=𝑆𝑠+1𝑢=𝑔𝑠𝑔𝑢, so 𝑆𝑢 is another fixed point of 𝑆𝑠, and 𝑅𝑢=𝑅𝑅𝑟𝑢=𝑅𝑟+1𝑢=𝑅𝑟𝑅𝑢, so 𝑅𝑢 is another fixed point of 𝑅𝑟. By (𝐺3) and the condition (2.28) in Corollary 2.2, we have 𝐺2(𝑇𝑢,𝑆𝑠𝑇𝑢,𝑅𝑟𝑇𝑢)=𝐺2(𝑇𝑝𝑇𝑢,𝑆𝑠𝑇𝑢,𝑅𝑟)𝑇𝑢𝛼𝐺(𝑇𝑢,𝑇𝑝𝑇𝑢,𝑇𝑝𝑇𝑢)𝐺(𝑇𝑢,𝑆𝑠𝑇𝑢,𝑆𝑠𝑇𝑢)+𝛽𝐺(𝑇𝑢,𝑆𝑠𝑇𝑢,𝑆𝑠𝑇𝑢)𝐺(𝑇𝑢,𝑅𝑟𝑇𝑢,𝑅𝑟𝑇𝑢)+𝛾𝐺(𝑇𝑢,𝑇𝑝𝑇𝑢,𝑇𝑝𝑇𝑢)𝐺(𝑇𝑢,𝑅𝑟𝑇𝑢,𝑅𝑟𝑇𝑢)=𝛽𝐺(𝑇𝑢,𝑆𝑠𝑇𝑢,𝑆𝑠𝑇𝑢)𝐺(𝑇𝑢,𝑅𝑟𝑇𝑢,𝑅𝑟)𝑇𝑢𝛽𝐺(𝑇𝑢,𝑆𝑠𝑇𝑢,𝑅𝑟𝑇𝑢)𝐺(𝑇𝑢,𝑆𝑠𝑇𝑢,𝑅𝑟𝑇𝑢).(2.29) Since 0𝛽<1, we can get 𝐺2(𝑇𝑢,𝑆𝑠𝑇𝑢,𝑅𝑟𝑇𝑢)=0. That means 𝑇𝑢=𝑇𝑝𝑇𝑢=𝑆𝑠𝑇𝑢=𝑅𝑟𝑇𝑢, hence 𝑇𝑢 is another common fixed point of 𝑇𝑝,𝑆𝑠-and 𝑅𝑟. Since the common fixed point of 𝑇𝑝,𝑆𝑠-and 𝑅𝑟 is unique, we deduce that 𝑢=𝑇𝑢. By the same argument, we can prove 𝑢=𝑆𝑢,𝑢=𝑅𝑢. Thus, we have 𝑢=𝑇𝑢=𝑆𝑢=𝑅𝑢. Suppose 𝑣 is another common fixed point of 𝑇,𝑆, and 𝑅, then 𝑣=𝑇𝑝𝑣=𝑆𝑠𝑣=𝑅𝑟𝑣, and by using the condition (2.28) in Corollary 2.2 again, we have 𝐺2(𝑣,𝑢,𝑢)=𝐺2(𝑇𝑝𝑣,𝑆𝑠𝑢,𝑅𝑟𝑢)𝛼𝐺(𝑣,𝑇𝑝𝑣,𝑇𝑝𝑣)𝐺(𝑢,𝑆𝑠𝑢,𝑆𝑠𝑢)+𝛽𝐺(𝑢,𝑆𝑠𝑢,𝑆𝑠𝑢)𝐺(𝑢,𝑅𝑟𝑢,𝑅𝑟𝑢)+𝛾𝐺(𝑣,𝑇𝑝𝑣,𝑇𝑝𝑣)𝐺(𝑢,𝑅𝑟𝑢,𝑅𝑟𝑢)=𝛼𝐺(𝑣,𝑣,𝑣)𝐺(𝑢,𝑢,𝑢)+𝛽𝐺(𝑢,𝑢,𝑢)𝐺(𝑢,𝑢,𝑢)+𝛾𝐺(𝑣,𝑣,𝑣)𝐺(𝑢,𝑢,𝑢)=0.(2.30) Which implies that 𝐺2(𝑣,𝑢,𝑢)=0, hence 𝑣=𝑢. So the common fixed of 𝑇,𝑆, and 𝑅 is unique. It is obvious that every fixed point of 𝑇 is a fixed point of 𝑆 and 𝑅 and conversely.

Corollary 2.3. Let (𝑋,𝐺) be a complete 𝐺-metric space. Suppose the self-mapping 𝑇𝑋𝑋 satisfies the following condition: 𝐺2(𝑇𝑥,𝑇𝑦,𝑇𝑧)𝛼𝐺(𝑥,𝑇𝑥,𝑇𝑥)𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝛽𝐺(𝑦,𝑇𝑦,𝑇𝑦)𝐺(𝑧,𝑇𝑧,𝑇𝑧)+𝛾𝐺(𝑥,𝑇𝑥,𝑇𝑥)𝐺(𝑧,𝑇𝑧,𝑇𝑧),(2.31) for all 𝑥,𝑦,𝑧𝑋, where 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇 has a unique fixed point (say 𝑢) and 𝑇 is 𝐺-continuous at 𝑢.

Proof. Let 𝑇=𝑆=𝑅 in Theorem 2.1, we can get this conclusion holds.

Corollary 2.4. Let (𝑋,𝐺) be a complete 𝐺-metric space. Suppose the self-mapping 𝑇𝑋𝑋 satisfies the following condition: 𝐺2(𝑇𝑝𝑥,𝑇𝑝𝑦,𝑇𝑝𝑧)𝛼𝐺(𝑥,𝑇𝑝𝑥,𝑇𝑝𝑥)𝐺(𝑦,𝑇𝑝𝑦,𝑇𝑝𝑦)+𝛽𝐺(𝑦,𝑇𝑝𝑦,𝑇𝑝𝑦)𝐺(𝑧,𝑇𝑝𝑧,𝑇𝑝𝑧)+𝛾𝐺(𝑥,𝑇𝑝𝑥,𝑇𝑝𝑥)𝐺(𝑧,𝑇𝑝𝑧,𝑇𝑝𝑧).(2.32) for all 𝑥,𝑦,𝑧𝑋, where 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇 has a unique fixed point (say 𝑢) and 𝑇𝑝 is 𝐺-continuous at 𝑢.

Proof. Let 𝑇=𝑆=𝑅,𝑝=𝑠=𝑟 in Corollary 2.2, we can get this conclusion holds.

Theorem 2.5. Let (𝑋,𝐺) be a complete 𝐺-metric space, and let 𝑇,𝑆,𝑅𝑋𝑋 be three self-mappings in 𝑋, which satisfy the following condition. 𝐺2(𝑇𝑥,𝑆𝑦,𝑅𝑧)𝛼𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇𝑥)+𝛾𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇𝑥).(2.33) for all 𝑥,𝑦,𝑧𝑋, 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇,𝑆 and 𝑅 have a unique common fixed point (say 𝑢) and 𝑇,𝑆,𝑅 are all 𝐺-continuous at 𝑢.

Proof. We will proceed in two steps.
Step 1. We prove any fixed point of 𝑇 is a fixed point of 𝑆 and 𝑅 and conversely. Assume that 𝑝𝑋 is such that 𝑇𝑝=𝑝. Now we prove that 𝑝=𝑆𝑝 and 𝑝=𝑅𝑝. If it is not the case, then for 𝑝𝑆𝑝 and 𝑝𝑅𝑝, by (2.33) and (𝐺3) we have 𝐺2=(𝑇𝑝,𝑆𝑝,𝑅𝑝)𝛼𝐺(𝑝,𝑇𝑝,𝑆𝑝)𝐺(𝑝,𝑆𝑝,𝑅𝑝)+𝛽𝐺(𝑝,𝑆𝑝,𝑅𝑝)𝐺(𝑝,𝑅𝑝,𝑇𝑝)+𝛾𝐺(𝑝,𝑇𝑝,𝑆𝑝)𝐺(𝑝,𝑅𝑝,𝑇𝑝)=𝛼𝐺(𝑝,𝑝,𝑆𝑝)𝐺(𝑝,𝑆𝑝,𝑅𝑝)+𝛽𝐺(𝑝,𝑆𝑝,𝑅𝑝)𝐺(𝑝,𝑅𝑝,𝑝)+𝛾𝐺(𝑝,𝑝,𝑆𝑝)𝐺(𝑝,𝑅𝑝,𝑝)𝛼𝐺(𝑝,𝑅𝑝,𝑆𝑝)𝐺(𝑝,𝑆𝑝,𝑅𝑝)+𝛽𝐺(𝑝,𝑆𝑝,𝑅𝑝)𝐺(𝑝,𝑅𝑝,𝑆𝑝)+𝛾𝐺(𝑝,𝑅𝑝,𝑆𝑝)𝐺(𝑝,𝑅𝑝,𝑆𝑝)(𝛼+𝛽+𝛾)𝐺2(𝑝,𝑅𝑝,𝑆𝑝).(2.34) It follows that 𝐺2(𝑝,𝑆𝑝,𝑅𝑝)=𝐺2(𝑇𝑝,𝑆𝑝,𝑅𝑝)(𝛼+𝛽+𝛾)𝐺2(𝑝,𝑆𝑝,𝑅𝑝).(2.35) Since 𝐺2(𝑝,𝑆𝑝,𝑅𝑝)>0, hence we have 𝛼+𝛽+𝛾1, however it contradicts with the condition 0𝛼+𝛽+𝛾<1, so we can have 𝑝=𝑆𝑝=𝑅𝑝, hence 𝑝 is a common fixed point of 𝑇,𝑆, and 𝑅.
Analogously, following the similar arguments to those given above, we can obtain a contradiction for 𝑝𝑆𝑝 and 𝑝=𝑅𝑝 or 𝑝=𝑆𝑝 and 𝑝𝑅𝑝. Hence in all the cases, we conclude that 𝑝=𝑆𝑝=𝑅𝑝. The same conclusion holds if 𝑝=𝑆𝑝 or 𝑝=𝑅𝑝.
Step 2. We prove that 𝑇, 𝑆 and 𝑅 have a unique common fixed point. Let 𝑥0𝑋 be an arbitrary point, and define the sequence {𝑥𝑛} by 𝑥3𝑛+1=𝑇𝑥3𝑛,𝑥3𝑛+2=𝑆𝑥3𝑛+1,𝑥3𝑛+3=𝑅𝑥3𝑛+2, 𝑛. If 𝑥𝑛=𝑥𝑛+1, for some 𝑛, with 𝑛=3𝑚, then 𝑝=𝑥3𝑚 is a fixed point of 𝑇 and, by the first step, 𝑝 is a common fixed point of 𝑆, 𝑇, and 𝑅. The same holds if 𝑛=3𝑚+1 or 𝑛=3𝑚+2. Without loss of generality, we can assume that 𝑥𝑛𝑥𝑛+1, for all 𝑛. We first prove the sequence {𝑥𝑛} is a 𝐺-Cauchy sequence. In fact, by using (2.33) and (𝐺3), we have 𝐺2𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3=𝐺2𝑇𝑥3𝑛,𝑆𝑥3𝑛+1,𝑅𝑥3𝑛+2𝑥𝛼𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+1𝑥+𝛾𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+1.(2.36) Which gives that 𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥(𝛼+𝛾)𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3.(2.37) It follows that 𝑥(1𝛽)𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥(𝛼+𝛾)𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2.(2.38) From 0𝛽<1, we know that 1𝛽>0. Then, we have 𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝛼+𝛾𝐺𝑥1𝛽3𝑛,𝑥3𝑛+1,𝑥3𝑛+2.(2.39) On the other hand, by using (2.33) and (𝐺3), we have 𝐺2𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4=𝐺2𝑇𝑥3𝑛+3,𝑆𝑥3𝑛+1,𝑅𝑥3𝑛+2𝑥𝛼𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+2𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝑥+𝛾𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+2𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.40) Which implies that 𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝑥(𝛼+𝛽)𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥+𝛾𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.41) It follows that 𝑥(1𝛾)𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝑥(𝛼+𝛽)𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3.(2.42) Since 0𝛾<1, we know that 1𝛾>0. So, we have 𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝛼+𝛽𝐺𝑥1𝛾3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3.(2.43) Again, using (2.33) and (𝐺3), we can get 𝐺2𝑥3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5=𝐺2𝑇𝑥3𝑛+3,𝑆𝑥3𝑛+4,𝑅𝑥3𝑛+2𝑥𝛼𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝐺𝑥3𝑛+4,𝑥3𝑛+5,𝑥3𝑛+3𝑥+𝛽𝐺3𝑛+4,𝑥3𝑛+5,𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝑥+𝛾𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4𝐺𝑥3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5.(2.44) Which implies that 𝐺𝑥3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝑥𝛼𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5+𝑥(𝛽+𝛾)𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.45) It follows that 𝑥(1𝛼)𝐺3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝑥(𝛽+𝛾)𝐺3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.46) Since 0𝛼𝛼+𝛽+𝛾<1, we know that 1𝛼>0. So we have 𝐺𝑥3𝑛+3,𝑥3𝑛+4,𝑥3𝑛+5𝛽+𝛾𝐺𝑥1𝛼3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+4.(2.47) Let 𝑞=max{(𝛼+𝛾)/(1𝛽),(𝛼+𝛽)/(1𝛾),(𝛽+𝛾)/(1𝛼)}, then 0𝑞<1, and by combining (2.39), (2.43), and (2.47), we have 𝐺𝑥𝑛,𝑥𝑛+1,𝑥𝑛+2𝑥𝑞𝐺𝑛1,𝑥𝑛,𝑥𝑛+1𝑞𝑛𝐺𝑥0,𝑥1,𝑥2.(2.48) Thus, by (𝐺3) and (𝐺5), for every 𝑚,𝑛, if 𝑚>𝑛, noting that 0𝑞<1, we have 𝐺𝑥𝑛,𝑥𝑚,𝑥𝑚𝑥𝐺𝑛,𝑥𝑛+1,𝑥𝑛+1𝑥+𝐺𝑛+1,𝑥𝑛+2,𝑥𝑛+2𝑥++𝐺𝑚1,𝑥𝑚,𝑥𝑚𝑥𝐺𝑛,𝑥𝑛+1,𝑥𝑛+2𝑥+𝐺𝑛+1,𝑥𝑛+2,𝑥𝑛+3𝑥++𝐺𝑚1,𝑥𝑚,𝑥𝑚+1𝑞𝑛+𝑞𝑛+1++𝑞𝑚1𝐺𝑥0,𝑥1,𝑥2𝑞𝑛𝐺𝑥1𝑞0,𝑥1,𝑥2.(2.49) Which implies that 𝐺(𝑥𝑛,𝑥𝑚,𝑥𝑚)0, as 𝑛,𝑚. Thus {𝑥𝑛} is a 𝐺-Cauchy sequence. Due to the completeness of (𝑋,𝐺), there exists 𝑢𝑋, such that {𝑥𝑛} is 𝐺-convergent to 𝑢.
Now we prove 𝑢 is a common fixed point of 𝑇,𝑆, and 𝑅. By using (2.33), we have 𝐺2𝑇𝑢,𝑥3𝑛+2,𝑥3𝑛+3=𝐺2𝑇𝑢,𝑆𝑥3𝑛+1,𝑅𝑥3𝑛+2𝛼𝐺𝑢,𝑇𝑢,𝑆𝑥3𝑛+1𝐺𝑥3𝑛+1,𝑆𝑥3𝑛+1,𝑅𝑥3𝑛+2𝑥+𝛽𝐺3𝑛+1,𝑆𝑥3𝑛+1,𝑅𝑥3𝑛+2𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑇𝑢+𝛾𝐺𝑢,𝑇𝑢,𝑆𝑥3𝑛+1𝐺𝑥3𝑛+2,𝑅𝑥3𝑛+2,𝑇𝑢=𝛼𝐺𝑢,𝑇𝑢,𝑥3𝑛+2𝐺𝑥3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝑥+𝛽𝐺3𝑛+1,𝑥3𝑛+2,𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑇𝑢+𝛾𝐺𝑢,𝑇𝑢,𝑥3𝑛+2𝐺𝑥3𝑛+2,𝑥3𝑛+3.,𝑇𝑢(2.50) Letting 𝑛, and using the fact that 𝐺 is continuous on its variables and 𝛾<1, we can get 𝐺2(𝑇𝑢,𝑢,𝑢)𝛾𝐺2(𝑢,𝑢,𝑇𝑢).(2.51) Which gives that 𝑇𝑢=𝑢, hence 𝑢 is a fixed point of 𝑇. By using (2.33) again, we have 𝐺2𝑥3𝑛+1,𝑆𝑢,𝑥3𝑛+3=𝐺2𝑇𝑥3𝑛,𝑆𝑢,𝑅𝑥3𝑛+2𝑥𝛼𝐺3𝑛,𝑥3𝑛+1𝐺,𝑆𝑢𝑢,𝑆𝑢,𝑥3𝑛+3+𝛽𝐺𝑢,𝑆𝑢,𝑥3𝑛+3𝐺𝑥3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+1𝑥+𝛾𝐺3𝑛,𝑥3𝑛+1𝐺𝑥,𝑆𝑢3𝑛+2,𝑥3𝑛+3,𝑥3𝑛+1.(2.52) Letting 𝑛 at both sides, for 𝐺 is continuous in its variables, it follows that 𝐺2(𝑢,𝑆𝑢,𝑢)𝛼𝐺2(𝑢,𝑆𝑢,𝑢).(2.53) For 0𝛼<1, Therefore, we can get 𝐺2(𝑢,𝑆𝑢,𝑢)=0, hence 𝑆𝑢=𝑢, hence 𝑢 is a fixed point of 𝑆. Similarly, by (2.33), we can also get 𝐺2𝑥3𝑛+1,𝑥3𝑛+2,𝑅𝑢=𝐺2𝑇𝑥3𝑛,𝑆𝑥3𝑛+1𝑥,𝑅𝑢𝛼𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2𝐺𝑥3𝑛+1,𝑥3𝑛+2𝑥,𝑅𝑢+𝛽𝐺3𝑛+1,𝑥3𝑛+2𝐺,𝑅𝑢𝑢,𝑅𝑢,𝑥3𝑛+1𝑥+𝛾𝐺3𝑛,𝑥3𝑛+1,𝑥3𝑛+2𝐺𝑢,𝑅𝑢,𝑥3𝑛+1.(2.54) On taking 𝑛 at both sides, since 𝐺 is continuous in its variables, we get that 𝐺2(𝑢,𝑢,𝑅𝑢)𝛽𝐺2(𝑢,𝑢,𝑅𝑢).(2.55) Since 0𝛽<1, so we get 𝐺2(𝑢,𝑢,𝑅𝑢)=0, hence 𝑢=𝑅𝑢, therefore, 𝑢 is a fixed point of 𝑅. Consequently, we have 𝑢=𝑇𝑢=𝑆𝑢=𝑅𝑢, and 𝑢 is a common fixed point of 𝑇,𝑆, and 𝑅. Suppose 𝑣𝑢 is another common fixed point of 𝑇,𝑆, and 𝑅, and we have 𝑣=𝑇𝑣=𝑆𝑣=𝑅𝑣, then by (2.33), we have 𝐺2(𝑢,𝑢,𝑣)=𝐺2(𝑇𝑢,𝑆𝑢,𝑅𝑣)𝛼𝐺(𝑢,𝑇𝑢,𝑆𝑢)𝐺(𝑢,𝑆𝑢,𝑅𝑣)+𝛽𝐺(𝑢,𝑆𝑢,𝑅𝑣)𝐺(𝑣,𝑅𝑣,𝑇𝑢)+𝛾𝐺(𝑢,𝑇𝑢,𝑆𝑢)𝐺(𝑣,𝑅𝑣,𝑇𝑢)=𝛼𝐺(𝑢,𝑢,𝑢)𝐺(𝑢,𝑢,𝑣)+𝛽𝐺(𝑢,𝑢,𝑣)𝐺(𝑣,𝑣,𝑢)+𝛾𝐺(𝑢,𝑢,𝑢)𝐺(𝑣,𝑣,𝑢).(2.56) Which gives that 𝐺2(𝑢,𝑢,𝑣)𝛽𝐺(𝑢,𝑢,𝑣)𝐺(𝑣,𝑣,𝑢).(2.57) Hence, we can get 𝐺(𝑢,𝑢,𝑣)𝛽𝐺(𝑣,𝑣,𝑢). By using (2.33) again, we get 𝐺2(𝑢,𝑣,𝑣)=𝐺2(𝑇𝑢,𝑆𝑣,𝑅𝑣)𝛼𝐺(𝑢,𝑇𝑢,𝑆𝑣)𝐺(𝑣,𝑆𝑣,𝑅𝑣)+𝛽𝐺(𝑣,𝑆𝑣,𝑅𝑣)𝐺(𝑣,𝑅𝑣,𝑇𝑢)+𝛾𝐺(𝑢,𝑇𝑢,𝑆𝑣)𝐺(𝑣,𝑅𝑣,𝑇𝑢)=𝛼𝐺(𝑢,𝑢,𝑣)𝐺(𝑣,𝑣,𝑣)+𝛽𝐺(𝑣,𝑣,𝑣)𝐺(𝑣,𝑣,𝑢)+𝛾𝐺(𝑢,𝑢,𝑣)𝐺(𝑣,𝑣,𝑢).(2.58) Which implies that 𝐺2(𝑢,𝑣,𝑣)𝛾𝐺(𝑢,𝑢,𝑣)𝐺(𝑣,𝑣,𝑢).(2.59) Hence, we can get 𝐺(𝑢,𝑣,𝑣)𝛾𝐺(𝑢,𝑢,𝑣).(2.60) By combining 𝐺(𝑢,𝑢,𝑣)𝛽𝐺(𝑣,𝑣,𝑢), we can have 𝐺(𝑢,𝑣,𝑣)𝛾𝐺(𝑢,𝑢,𝑣)𝛽𝛾𝐺(𝑣,𝑣,𝑢).(2.61) Since 𝑣𝑢,𝐺(𝑢,𝑣,𝑣)>0, so we have that 𝛽𝛾1. Since 0𝛽,𝛾<1, we know 0𝛽𝛾<1, so it’s a contradiction. Hence, we get 𝑢=𝑣. Then we know the common fixed point of 𝑇,𝑆, and 𝑅 is unique.
To show that 𝑇 is 𝐺-continuous at 𝑢, let {𝑦𝑛} be any sequence in 𝑋 such that {𝑦𝑛} is 𝐺-convergent to 𝑢. For 𝑛, we have 𝐺2𝑇𝑦𝑛,𝑢,𝑢=𝐺2𝑇𝑦𝑛𝑦,𝑆𝑢,𝑅𝑢𝛼𝐺𝑛,𝑇𝑦𝑛,𝑆𝑢𝐺(𝑢,𝑆𝑢,𝑅𝑢)+𝛽𝐺(𝑢,𝑆𝑢,𝑅𝑢)𝐺𝑢,𝑅𝑢,𝑇𝑦𝑛𝑦+𝛾𝐺𝑛,𝑇𝑦𝑛𝐺,𝑆𝑢𝑢,𝑅𝑢,𝑇𝑦𝑛𝑦=𝛼𝐺𝑛,𝑇𝑦𝑛,𝑢𝐺(𝑢,𝑢,𝑢)+𝛽𝐺(𝑢,𝑢,𝑢)𝐺𝑢,𝑢,𝑇𝑦𝑛𝑦+𝛾𝐺𝑛,𝑇𝑦𝑛𝐺,𝑢𝑢,𝑢,𝑇𝑦𝑛𝑦=𝛾𝐺𝑛,𝑇𝑦𝑛𝐺,𝑢𝑢,𝑢,𝑇𝑦𝑛.(2.62) Which implies that 𝐺𝑇𝑦𝑛𝑦,𝑢,𝑢𝛾𝐺𝑛,𝑇𝑦𝑛,𝑢.(2.63) On taking 𝑛 at both sides, considering 𝛾<1, we get lim𝑛𝐺(𝑇𝑦𝑛,𝑢,𝑢)=0. Hence {𝑇𝑦𝑛} is 𝐺-convergent to 𝑢=𝑇𝑢. So 𝑇 is 𝐺-continuous at 𝑢. Similarly, we can also prove that 𝑆,𝑅 are 𝐺-continuous at 𝑢. Therefore, we complete the proof.

Now we introduce an example to support Theorem 2.5.

Example 2.6. Let 𝑋=[0,1], and let (𝑋,𝐺) be a 𝐺-metric space defined by 𝐺(𝑥,𝑦,𝑧)=|𝑥𝑦|+|𝑦𝑧|+|𝑧𝑥|, for all 𝑥,𝑦,𝑧 in 𝑋. Let 𝑇, 𝑆, and 𝑅 be three self-mappings defined by 1𝑇𝑥=1,𝑥0,2671,𝑥27,1,𝑆𝑥=81,𝑥0,2671,𝑥26,1,𝑅𝑥=7[],𝑥0,1.(2.64)
Next we proof the mappings 𝑇, 𝑆, and 𝑅 are satisfying Condition (2.33) of Theorem 2.5 with 𝛼=1/7, 𝛽=1/7 and 𝛾=4/7.

Case 1. If 𝑥,𝑦[0,1/2], 𝑧[0,1], then 𝐺2(𝑇𝑥,𝑆𝑦,𝑅𝑧)=𝐺271,8,67=4,749𝐺(𝑥,𝑇𝑥,𝑆𝑦)=𝐺𝑥,1,8=||||+|||7𝑥1𝑥8|||+1812+38+187=1,𝐺(𝑦,𝑆𝑦,𝑅𝑧)=𝐺𝑦,8,67=|||7𝑦8|||+|||6𝑦7|||+13568+5+114=3564,6𝐺(𝑧,𝑅𝑧,𝑇𝑥)=𝐺𝑧,7=|||6,1𝑧7|||+17+||||1𝑧10+71+0=7.(2.65) Thus, we have 𝐺24(𝑇𝑥,𝑆𝑦,𝑅𝑧)=349𝛼143+𝛽4171+𝛾17𝛼𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇𝑥)+𝛾𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇𝑥).(2.66)

Case 2. If 𝑥[0,1/2],𝑦(1/2,1],𝑧[0,1], then we can get 𝐺2(𝑇𝑥,𝑆𝑦,𝑅𝑧)=𝐺261,7,67=4,649𝐺(𝑥,𝑇𝑥,𝑆𝑦)=𝐺𝑥,1,7=||||+|||6𝑥1𝑥7|||+1712+5+11476=1,𝐺(𝑦,𝑆𝑦,𝑅𝑧)=𝐺𝑦,7,67=|||6𝑦7|||+|||6𝑦7|||60+0=0,𝐺(𝑧,𝑅𝑧,𝑇𝑥)=𝐺𝑧,7=|||6,1𝑧7|||+17+||||1𝑧10+71+0=7.(2.67) Thus, we have 𝐺24(𝑇𝑥,𝑆𝑦,𝑅𝑧)=149𝛼10+𝛽071+𝛾17𝛼𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇𝑥)+𝛾𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇𝑥).(2.68)

Case 3. If 𝑥(1/2,1],𝑦[0,1/2],𝑧[0,1], then we have 𝐺2(𝑇𝑥,𝑆𝑦,𝑅𝑧)=𝐺267,78,67=1,6784𝐺(𝑥,𝑇𝑥,𝑆𝑦)=𝐺𝑥,7,78=|||6𝑥7|||+|||7𝑥8|||+11560+0+=156,756𝐺(𝑦,𝑆𝑦,𝑅𝑧)=𝐺𝑦,8,67=|||7𝑦8|||+|||6𝑦7|||+13568+5+114=3564,6𝐺(𝑧,𝑅𝑧,𝑇𝑥)=𝐺𝑧,7,67=|||6𝑧7|||+|||6𝑧7|||0+0=0.(2.69) Thus, we have 𝐺21(𝑇𝑥,𝑆𝑦,𝑅𝑧)=1784𝛼35643+𝛽410+𝛾560𝛼𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇𝑥)+𝛾𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇𝑥).(2.70)

Case 4. If 𝑥,𝑦(1/2,1],𝑧[0,1], then we have 𝐺2(𝑇𝑥,𝑆𝑦,𝑅𝑧)=𝐺267,67,67=0.(2.71) Thus, we have 𝐺2(𝑇𝑥,𝑆𝑦,𝑅𝑧)=0𝛼𝐺(𝑥,𝑇𝑥,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺<