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 [3–27]. Fixed point problems have also been considered in partially ordered 𝐺-metric spaces [28–31], 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 Ξ¦-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 β„• 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 πΊβˆΆπ‘‹Γ—π‘‹Γ—π‘‹β†’π‘…+ be a function satisfying the following axioms: (𝐺1)𝐺(π‘₯,𝑦,𝑧=0) if π‘₯=𝑦=𝑧; (𝐺2)0<𝐺(π‘₯,π‘₯,𝑦), for all π‘₯,π‘¦βˆˆπ‘‹ with π‘₯≠𝑦; (𝐺3)  𝐺(π‘₯,π‘₯,𝑦)≀𝐺(π‘₯,𝑦,𝑧), for all π‘₯,𝑦,π‘§βˆˆπ‘‹ with 𝑧≠ y; (𝐺4)  𝐺(π‘₯,𝑦,𝑧)=𝐺(π‘₯,𝑧,𝑦)=𝐺(𝑦,𝑧,π‘₯)=β‹― (symmetry in all three variables); (𝐺5)𝐺(π‘₯,𝑦,𝑧)≀𝐺(π‘₯,π‘Ž,π‘Ž)+𝐺(π‘Ž,𝑦,𝑧) for all π‘₯,𝑦,𝑧,π‘Žβˆˆπ‘‹, (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 {π‘₯𝑛} be a sequence of points in 𝑋, a point π‘₯ in 𝑋 is said to be the limit of the sequence {π‘₯𝑛} if limπ‘š,π‘›β†’βˆžπΊ(π‘₯,π‘₯𝑛,π‘₯π‘š)=0, and one says that sequence {π‘₯𝑛} is 𝐺-convergent to π‘₯.

Thus, if π‘₯𝑛→π‘₯ in a 𝐺-metric space (𝑋,𝐺), then for any πœ–>0, there exists π‘βˆˆβ„• such that 𝐺(π‘₯,π‘₯𝑛,π‘₯π‘š)<πœ–, for all 𝑛,π‘šβ‰₯𝑁.

Proposition 1.3 (see [1]). Let (𝑋,𝐺) be a 𝐺-metric space, then the followings are equivalent.(1){π‘₯𝑛} is 𝐺-convergent to π‘₯.(2)𝐺(π‘₯𝑛,π‘₯𝑛,π‘₯)β†’0 as π‘›β†’βˆž.(3)𝐺(π‘₯𝑛,π‘₯,π‘₯)β†’0 as π‘›β†’βˆž.(4)𝐺(π‘₯𝑛,π‘₯π‘š,π‘₯)β†’0 as 𝑛,π‘šβ†’βˆž.

Definition 1.4 (see [1]). Let (𝑋,𝐺) be a 𝐺-metric space. A sequence {π‘₯𝑛} is called 𝐺-Cauchy sequence if for each πœ–>0 there exists a positive integer π‘βˆˆβ„• such that 𝐺(π‘₯𝑛,π‘₯π‘š,π‘₯𝑙)<πœ– for all 𝑛,π‘š,𝑙β‰₯𝑁; that is, if 𝐺(π‘₯𝑛,π‘₯π‘š,π‘₯𝑙)β†’0 as 𝑛,π‘š,π‘™β†’βˆž.

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 {π‘₯𝑛} is 𝐺-Cauchy.(2)For every πœ–>0, there exists π‘˜βˆˆβ„• such that 𝐺(π‘₯𝑛,π‘₯π‘š,π‘₯π‘š)<πœ–, for all 𝑛,π‘šβ‰₯π‘˜.

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

Definition 1.8 (see [1]). Let (𝑋,𝐺) and (π‘‹ξ…ž,πΊξ…ž) be 𝐺-metric space, and π‘“βˆΆ(𝑋,𝐺)β†’(π‘‹ξ…ž,πΊξ…ž) be a function. Then 𝑓 is said to be 𝐺-continuous at a point π‘Žβˆˆπ‘‹ if and only if for every πœ€>0, there is 𝛿>0 such that π‘₯,π‘¦βˆˆπ‘‹ and 𝐺(π‘Ž,π‘₯,𝑦)<𝛿 implies πΊξ…ž(𝑓(π‘Ž),𝑓(π‘₯),𝑓(𝑦))<πœ€. A function 𝑓 is 𝐺-continuous at 𝑋 if and only if it is 𝐺-continuous at all π‘Žβˆˆπ‘‹.

Proposition 1.9 (see [1]). Let (𝑋,𝐺) and (π‘‹ξ…ž,πΊξ…ž) be 𝐺-metric space. Then π‘“βˆΆπ‘‹β†’π‘‹ξ…ž is 𝐺-continuous at π‘₯βˆˆπ‘‹ if and only if it is 𝐺-sequentially continuous at π‘₯; that is, whenever {π‘₯𝑛} is 𝐺-convergent to π‘₯, {𝑓(π‘₯𝑛)} is 𝐺-convergent to 𝑓(π‘₯).

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

2. Main Results

Theorem 2.1. Let (𝑋,𝐺) be a complete 𝐺-metric space. Suppose the three self-mappings 𝑇,𝑆,π‘…βˆΆπ‘‹β†’π‘‹ satisfy the following condition: 𝐺2(𝑇π‘₯,𝑆𝑦,𝑅𝑧)≀𝛼𝐺(π‘₯,𝑇π‘₯,𝑇π‘₯)𝐺(𝑦,𝑆𝑦,𝑆𝑦)+𝛽𝐺(𝑦,𝑆𝑦,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑅𝑧)+𝛾𝐺(π‘₯,𝑇π‘₯,𝑇π‘₯)𝐺(𝑧,𝑅𝑧,𝑅𝑧),(2.1) for all π‘₯,𝑦,π‘§βˆˆπ‘‹, 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 𝑒) and 𝑇,𝑆,𝑅 are all 𝐺-continuous at 𝑒.

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 π‘βˆˆπ‘‹ 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 𝑝≠𝑆𝑝 and 𝑝≠𝑅𝑝, 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 𝛽β‰₯1; however, it contradicts with 0≀𝛽≀𝛼+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 π‘₯0βˆˆπ‘‹ be an arbitrary point, and define the sequence {π‘₯𝑛} by π‘₯3𝑛+1=𝑇π‘₯3𝑛,π‘₯3𝑛+2=𝑆π‘₯3𝑛+1,π‘₯3𝑛+3=𝑅π‘₯3𝑛+2, π‘›βˆˆβ„•. If π‘₯𝑛=π‘₯𝑛+1, for some 𝑛, with 𝑛=3π‘š, then 𝑝=π‘₯3π‘š 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=3π‘š+1 or 𝑛=3π‘š+2. Without loss of generality, we can assume that π‘₯𝑛≠π‘₯𝑛+1, for all π‘›βˆˆβ„•.
Next, we prove sequence {π‘₯𝑛} is a 𝐺-Cauchy sequence. In fact, by (2.1) and (𝐺3), we have 𝐺2ξ€·π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ=𝐺2𝑇π‘₯3𝑛,𝑆π‘₯3𝑛+1,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛,𝑇π‘₯3𝑛,𝑇π‘₯3𝑛𝐺π‘₯3𝑛+1,𝑆π‘₯3𝑛+1,𝑆π‘₯3𝑛+1ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,𝑆π‘₯3𝑛+1,𝑆π‘₯3𝑛+1𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛾𝐺3𝑛,𝑇π‘₯3𝑛,𝑇π‘₯3𝑛𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯=𝛼𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+1𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+2𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛾𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+1𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+1ξ€Έξ€·π‘₯+𝛾𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+1ξ€Έ.(2.6) Which gives that 𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3≀π‘₯(𝛼+𝛾)𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ.(2.7) It follows that ξ€·π‘₯(1βˆ’π›½)𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3≀π‘₯(𝛼+𝛾)𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έ.(2.8) From 0≀𝛽<1 we know that 1βˆ’π›½>0. Then, we have 𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3≀𝛼+𝛾𝐺π‘₯1βˆ’π›½3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έ.(2.9) On the other hand, by using (2.1) and (𝐺3), we have 𝐺2ξ€·π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ=𝐺2𝑇π‘₯3𝑛+3,𝑆π‘₯3𝑛+1,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛+3,𝑇π‘₯3𝑛+3,𝑇π‘₯3𝑛+3𝐺π‘₯3𝑛+1,𝑆π‘₯3𝑛+1,𝑆π‘₯3𝑛+1ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,𝑆π‘₯3𝑛+1,𝑆π‘₯3𝑛+1𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+3,𝑇π‘₯3𝑛+3,𝑇π‘₯3𝑛+3𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯=𝛼𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+4𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+2𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+4𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.10) Which implies that 𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4≀π‘₯(𝛼+𝛽)𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.11) It follows that ξ€·π‘₯(1βˆ’π›Ύ)𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4≀π‘₯(𝛼+𝛽)𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ.(2.12) Form the condition 0≀𝛾≀𝛼+𝛽+𝛾<1, we know that 1βˆ’π›Ύ>0. Therefore, we have 𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4≀𝛼+𝛽𝐺π‘₯1βˆ’π›Ύ3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ.(2.13) Again, using (2.1) and (𝐺3), we can get 𝐺2ξ€·π‘₯3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5ξ€Έ=𝐺2𝑇π‘₯3𝑛+3,𝑆π‘₯3𝑛+4,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛+3,𝑇π‘₯3𝑛+3,𝑇π‘₯3𝑛+3𝐺π‘₯3𝑛+4,𝑆π‘₯3𝑛+4,𝑆π‘₯3𝑛+4ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+4,𝑆π‘₯3𝑛+4,𝑆π‘₯3𝑛+4𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+3,𝑇π‘₯3𝑛+3,𝑇π‘₯3𝑛+3𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯=𝛼𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+4𝐺π‘₯3𝑛+4,π‘₯3𝑛+5,π‘₯3𝑛+5ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+4,π‘₯3𝑛+5,π‘₯3𝑛+5𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+4𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5𝐺π‘₯3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+5,π‘₯3𝑛+3,π‘₯3𝑛+4𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.14) Which implies that 𝐺π‘₯3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5ξ€Έ+ξ€·π‘₯(𝛽+𝛾)𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.15) It follows that ξ€·π‘₯(1βˆ’π›Ό)𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5≀π‘₯(𝛽+𝛾)𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.16) By the condition 0≀𝛼≀𝛼+𝛽+𝛾<1, we know that 1βˆ’π›Ό>0. Hence, we have 𝐺π‘₯3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5≀𝛽+𝛾𝐺π‘₯1βˆ’π›Ό3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.17) Let π‘ž=max{(𝛼+𝛾)/(1βˆ’π›½),(𝛼+𝛽)/(1βˆ’π›Ύ),(𝛽+𝛾)/(1βˆ’π›Ό)}, then from 0≀𝛼+𝛽+𝛾<1 we know that 0β‰€π‘ž<1. Combining (2.9), (2.13), and (2.17), we have 𝐺π‘₯𝑛,π‘₯𝑛+1,π‘₯𝑛+2ξ€Έξ€·π‘₯β‰€π‘žπΊπ‘›βˆ’1,π‘₯𝑛,π‘₯𝑛+1ξ€Έβ‰€β‹―β‰€π‘žπ‘›πΊξ€·π‘₯0,π‘₯1,π‘₯2ξ€Έ.(2.18) Thus, by (𝐺3) and (𝐺5), for every π‘š,π‘›βˆˆβ„•,π‘š>𝑛, noting that 0β‰€π‘ž<1, we have 𝐺π‘₯𝑛,π‘₯π‘š,π‘₯π‘šξ€Έξ€·π‘₯≀𝐺𝑛,π‘₯𝑛+1,π‘₯𝑛+1ξ€Έξ€·π‘₯+𝐺𝑛+1,π‘₯𝑛+2,π‘₯𝑛+2ξ€Έξ€·π‘₯+β‹―+πΊπ‘šβˆ’1,π‘₯π‘š,π‘₯π‘šξ€Έ,ξ€·π‘₯≀𝐺𝑛,π‘₯𝑛+1,π‘₯𝑛+2ξ€Έξ€·π‘₯+𝐺𝑛+1,π‘₯𝑛+2,π‘₯𝑛+3ξ€Έξ€·π‘₯+β‹―+πΊπ‘šβˆ’1,π‘₯π‘š,π‘₯π‘š+1ξ€Έβ‰€ξ€·π‘žπ‘›+π‘žπ‘›+1+β‹―+π‘žπ‘šβˆ’1𝐺π‘₯0,π‘₯1,π‘₯2ξ€Έβ‰€π‘žπ‘›πΊξ€·π‘₯1βˆ’π‘ž0,π‘₯1,π‘₯2ξ€Έ.(2.19) Which implies that 𝐺(π‘₯𝑛,π‘₯π‘š,π‘₯π‘š)β†’0, as 𝑛,π‘šβ†’βˆž. Thus {π‘₯𝑛} is a 𝐺-Cauchy sequence. Due to the completeness of (𝑋,𝐺), there exists π‘’βˆˆπ‘‹, such that {π‘₯𝑛} is 𝐺-convergent to 𝑒.
Next we prove 𝑒 is a common fixed point of 𝑇,𝑆, and 𝑅. By using (2.1), we have 𝐺2𝑇𝑒,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ=𝐺2𝑇𝑒,𝑆π‘₯3𝑛+1,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺(𝑒,𝑇𝑒,𝑇𝑒)𝐺3𝑛+1,𝑆π‘₯3𝑛+1,𝑆π‘₯3𝑛+1ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,𝑆π‘₯3𝑛+1,𝑆π‘₯3𝑛+1𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛾𝐺(𝑒,𝑇𝑒,𝑇𝑒)𝐺3𝑛+2,𝑅π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯=𝛼𝐺(𝑒,𝑇𝑒,𝑇𝑒)𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+2𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛾𝐺(𝑒,𝑇𝑒,𝑇𝑒)𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έ.(2.20) Letting π‘›β†’βˆž, and using the fact that 𝐺 is continuous on its variables, we can get 𝐺2(𝑇𝑒,𝑒,𝑒)=0.(2.21) Which gives that 𝑇𝑒=𝑒, that is 𝑒 is a fixed point of 𝑇. By using (2.1) again, we have 𝐺2ξ€·π‘₯3𝑛+1,𝑆𝑒,π‘₯3𝑛+3ξ€Έ=𝐺2𝑇π‘₯3𝑛,𝑆𝑒,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+1ξ€Έξ€·π‘₯𝐺(𝑒,𝑆𝑒,𝑆𝑒)+𝛽𝐺(𝑒,𝑆𝑒,𝑆𝑒)𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛾𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+1𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+3ξ€Έ.(2.22) Letting π‘›β†’βˆž at both sides, for 𝐺 is continuous on its variables, it follows that 𝐺2(𝑒,𝑆𝑒,𝑒)=0.(2.23) Therefore, 𝑆𝑒=𝑒; that is, 𝑒 is a fixed point of 𝑆. Similarly, by (2.1), we can also get 𝐺2ξ€·π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έ,𝑅𝑒=𝐺2𝑇π‘₯3𝑛,𝑆π‘₯3𝑛+1ξ€Έξ€·π‘₯,𝑅𝑒≀𝛼𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+1𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+2ξ€Έξ€·π‘₯𝐺(𝑒,𝑅𝑒,𝑅𝑒)+𝛾𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+1𝐺(𝑒,𝑅𝑒,𝑅𝑒).(2.24) On taking π‘›β†’βˆž at both sides, since 𝐺 is continuous on its variables, we get that 𝐺2(𝑒,𝑒,𝑅𝑒)=0.(2.25) Which gives that 𝑒=𝑅𝑒, therefore, 𝑒 is fixed point of 𝑅. Consequently, we have 𝑒=𝑇𝑒=𝑆𝑒=𝑅𝑒, and 𝑒 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(𝑒,𝑒,𝑣)=𝐺2(𝑇𝑒,𝑆𝑒,𝑅𝑣)≀𝛼𝐺(𝑒,𝑇𝑒,𝑇𝑒)𝐺(𝑒,𝑆𝑒,𝑆𝑒)+𝛽𝐺(𝑒,𝑆𝑒,𝑆𝑒)𝐺(𝑣,𝑅𝑣,𝑅𝑣)+𝛾𝐺(𝑒,𝑇𝑒,𝑇𝑒)𝐺(𝑣,𝑅𝑣,𝑅𝑣)=𝛼𝐺(𝑒,𝑒,𝑒)𝐺(𝑒,𝑒,𝑒)+𝛽𝐺(𝑒,𝑒,𝑒)𝐺(𝑣,𝑣,𝑣)+𝛾𝐺(𝑒,𝑒,𝑒)𝐺(𝑣,𝑣,𝑣)=0.(2.26) Which implies that 𝐺2(𝑒,𝑒,𝑣)=0, hence, 𝑒=𝑣. Then we know the common fixed point of 𝑇,𝑆, and 𝑅 is unique.
To show that 𝑇 is 𝐺-continuous at 𝑒, let {𝑦𝑛} be any sequence in 𝑋 such that {𝑦𝑛} is 𝐺-convergent to 𝑒. For π‘›βˆˆβ„•, we have 𝐺2𝑇𝑦𝑛,𝑒,𝑒=𝐺2𝑇𝑦𝑛𝑦,𝑆𝑒,𝑅𝑒≀𝛼𝐺𝑛,𝑇𝑦𝑛,𝑇𝑦𝑛𝑦𝐺(𝑒,𝑆𝑒,𝑆𝑒)+𝛽𝐺(𝑒,𝑆𝑒,𝑆𝑒)𝐺(𝑒,𝑅𝑒,𝑅𝑒)+𝛾𝐺𝑛,𝑇𝑦𝑛,𝑇𝑦𝑛𝑦𝐺(𝑒,𝑅𝑒,𝑅𝑒)=𝛼𝐺𝑛,𝑇𝑦𝑛,𝑇𝑦𝑛𝑦𝐺(𝑒,𝑒,𝑒)+𝛽𝐺(𝑒,𝑒,𝑒)𝐺(𝑒,𝑒,𝑒)+𝛾𝐺𝑛,𝑇𝑦𝑛,𝑇𝑦𝑛𝐺(𝑒,𝑒,𝑒)=0.(2.27) Which implies that limπ‘›β†’βˆžπΊ2(𝑇𝑦𝑛,𝑒,𝑒)=0. Hence {𝑇𝑦𝑛} is 𝐺-convergent to 𝑒=𝑇𝑒. So 𝑇 is 𝐺-continuous at 𝑒. Similarly, we can also prove that 𝑆,𝑅 are 𝐺-continuous at 𝑒. Therefore, we complete the proof.

Corollary 2.2. Let (𝑋,𝐺) be a complete 𝐺-metric space. Suppose the three self-mappings 𝑇,𝑆,π‘…βˆΆπ‘‹β†’π‘‹ satisfy the condition: 𝐺2(𝑇𝑝π‘₯,𝑆𝑠𝑦,π‘…π‘Ÿπ‘§)≀𝛼𝐺(π‘₯,𝑇𝑝π‘₯,𝑇𝑝π‘₯)𝐺(𝑦,𝑆𝑠𝑦,𝑆𝑠𝑦)+𝛽𝐺(𝑦,𝑆𝑠𝑦,𝑆𝑠𝑦)𝐺(𝑧,π‘…π‘Ÿπ‘§,π‘…π‘Ÿπ‘§)+𝛾𝐺(π‘₯,𝑇𝑝π‘₯,𝑇𝑝π‘₯)𝐺(𝑧,π‘…π‘Ÿπ‘§,π‘…π‘Ÿπ‘§),(2.28) for all π‘₯,𝑦,π‘§βˆˆπ‘‹, where 𝑝,𝑠,π‘Ÿβˆˆβ„•,𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇,𝑆, and 𝑅 have a unique common fixed point (say 𝑒) and 𝑇𝑝,𝑆𝑠,π‘…π‘Ÿ are all 𝐺-continuous at 𝑒.

Proof. From Theorem 2.1 we know that 𝑇𝑝,𝑆𝑠,π‘…π‘Ÿ have a unique common fixed point (say 𝑒); that is, 𝑇𝑝𝑒=𝑒,𝑆𝑠𝑒=𝑒,π‘…π‘Ÿπ‘’=𝑒, and 𝑇𝑝,𝑆𝑠,π‘…π‘Ÿ are 𝐺-continuous at 𝑒. Since 𝑇𝑒=𝑇𝑇𝑝𝑒=𝑇𝑝+1𝑒=𝑇𝑝𝑇𝑒, so 𝑇𝑒 is another fixed point of 𝑇𝑝, 𝑆𝑒=𝑆𝑆𝑠𝑒=𝑆𝑠+1𝑒=𝑔𝑠𝑔𝑒, so 𝑆𝑒 is another fixed point of 𝑆𝑠, and 𝑅𝑒=π‘…π‘…π‘Ÿπ‘’=π‘…π‘Ÿ+1𝑒=π‘…π‘Ÿπ‘…π‘’, so 𝑅𝑒 is another fixed point of π‘…π‘Ÿ. By (𝐺3) and the condition (2.28) in Corollary 2.2, we have 𝐺2(𝑇𝑒,𝑆𝑠𝑇𝑒,π‘…π‘Ÿπ‘‡π‘’)=𝐺2(𝑇𝑝𝑇𝑒,𝑆𝑠𝑇𝑒,π‘…π‘Ÿ)𝑇𝑒≀𝛼𝐺(𝑇𝑒,𝑇𝑝𝑇𝑒,𝑇𝑝𝑇𝑒)𝐺(𝑇𝑒,𝑆𝑠𝑇𝑒,𝑆𝑠𝑇𝑒)+𝛽𝐺(𝑇𝑒,𝑆𝑠𝑇𝑒,𝑆𝑠𝑇𝑒)𝐺(𝑇𝑒,π‘…π‘Ÿπ‘‡π‘’,π‘…π‘Ÿπ‘‡π‘’)+𝛾𝐺(𝑇𝑒,𝑇𝑝𝑇𝑒,𝑇𝑝𝑇𝑒)𝐺(𝑇𝑒,π‘…π‘Ÿπ‘‡π‘’,π‘…π‘Ÿπ‘‡π‘’)=𝛽𝐺(𝑇𝑒,𝑆𝑠𝑇𝑒,𝑆𝑠𝑇𝑒)𝐺(𝑇𝑒,π‘…π‘Ÿπ‘‡π‘’,π‘…π‘Ÿ)𝑇𝑒≀𝛽𝐺(𝑇𝑒,𝑆𝑠𝑇𝑒,π‘…π‘Ÿπ‘‡π‘’)𝐺(𝑇𝑒,𝑆𝑠𝑇𝑒,π‘…π‘Ÿπ‘‡π‘’).(2.29) Since 0≀𝛽<1, we can get 𝐺2(𝑇𝑒,𝑆𝑠𝑇𝑒,π‘…π‘Ÿπ‘‡π‘’)=0. That means 𝑇𝑒=𝑇𝑝𝑇𝑒=𝑆𝑠𝑇𝑒=π‘…π‘Ÿπ‘‡π‘’, hence 𝑇𝑒 is another common fixed point of 𝑇𝑝,𝑆𝑠-and π‘…π‘Ÿ. Since the common fixed point of 𝑇𝑝,𝑆𝑠-and π‘…π‘Ÿ is unique, we deduce that 𝑒=𝑇𝑒. By the same argument, we can prove 𝑒=𝑆𝑒,𝑒=𝑅𝑒. Thus, we have 𝑒=𝑇𝑒=𝑆𝑒=𝑅𝑒. Suppose 𝑣 is another common fixed point of 𝑇,𝑆, and 𝑅, then 𝑣=𝑇𝑝𝑣=𝑆𝑠𝑣=π‘…π‘Ÿπ‘£, and by using the condition (2.28) in Corollary 2.2 again, we have 𝐺2(𝑣,𝑒,𝑒)=𝐺2(𝑇𝑝𝑣,𝑆𝑠𝑒,π‘…π‘Ÿπ‘’)≀𝛼𝐺(𝑣,𝑇𝑝𝑣,𝑇𝑝𝑣)𝐺(𝑒,𝑆𝑠𝑒,𝑆𝑠𝑒)+𝛽𝐺(𝑒,𝑆𝑠𝑒,𝑆𝑠𝑒)𝐺(𝑒,π‘…π‘Ÿπ‘’,π‘…π‘Ÿπ‘’)+𝛾𝐺(𝑣,𝑇𝑝𝑣,𝑇𝑝𝑣)𝐺(𝑒,π‘…π‘Ÿπ‘’,π‘…π‘Ÿπ‘’)=𝛼𝐺(𝑣,𝑣,𝑣)𝐺(𝑒,𝑒,𝑒)+𝛽𝐺(𝑒,𝑒,𝑒)𝐺(𝑒,𝑒,𝑒)+𝛾𝐺(𝑣,𝑣,𝑣)𝐺(𝑒,𝑒,𝑒)=0.(2.30) Which implies that 𝐺2(𝑣,𝑒,𝑒)=0, hence 𝑣=𝑒. 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 π‘‡βˆΆπ‘‹β†’π‘‹ satisfies the following condition: 𝐺2(𝑇π‘₯,𝑇𝑦,𝑇𝑧)≀𝛼𝐺(π‘₯,𝑇π‘₯,𝑇π‘₯)𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝛽𝐺(𝑦,𝑇𝑦,𝑇𝑦)𝐺(𝑧,𝑇𝑧,𝑇𝑧)+𝛾𝐺(π‘₯,𝑇π‘₯,𝑇π‘₯)𝐺(𝑧,𝑇𝑧,𝑇𝑧),(2.31) for all π‘₯,𝑦,π‘§βˆˆπ‘‹, where 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇 has a unique fixed point (say 𝑒) and 𝑇 is 𝐺-continuous at 𝑒.

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 π‘‡βˆΆπ‘‹β†’π‘‹ satisfies the following condition: 𝐺2(𝑇𝑝π‘₯,𝑇𝑝𝑦,𝑇𝑝𝑧)≀𝛼𝐺(π‘₯,𝑇𝑝π‘₯,𝑇𝑝π‘₯)𝐺(𝑦,𝑇𝑝𝑦,𝑇𝑝𝑦)+𝛽𝐺(𝑦,𝑇𝑝𝑦,𝑇𝑝𝑦)𝐺(𝑧,𝑇𝑝𝑧,𝑇𝑝𝑧)+𝛾𝐺(π‘₯,𝑇𝑝π‘₯,𝑇𝑝π‘₯)𝐺(𝑧,𝑇𝑝𝑧,𝑇𝑝𝑧).(2.32) for all π‘₯,𝑦,π‘§βˆˆπ‘‹, where 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇 has a unique fixed point (say 𝑒) and 𝑇𝑝 is 𝐺-continuous at 𝑒.

Proof. Let 𝑇=𝑆=𝑅,𝑝=𝑠=π‘Ÿ 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,π‘…βˆΆπ‘‹β†’π‘‹ be three self-mappings in 𝑋, which satisfy the following condition. 𝐺2(𝑇π‘₯,𝑆𝑦,𝑅𝑧)≀𝛼𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇π‘₯)+𝛾𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇π‘₯).(2.33) for all π‘₯,𝑦,π‘§βˆˆπ‘‹, 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇,𝑆 and 𝑅 have a unique common fixed point (say 𝑒) and 𝑇,𝑆,𝑅 are all 𝐺-continuous at 𝑒.

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 π‘βˆˆπ‘‹ is such that 𝑇𝑝=𝑝. Now we prove that 𝑝=𝑆𝑝 and 𝑝=𝑅𝑝. If it is not the case, then for 𝑝≠𝑆𝑝 and 𝑝≠𝑅𝑝, 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+𝛾β‰₯1, however it contradicts with the condition 0≀𝛼+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 𝑝≠𝑆𝑝 and 𝑝=𝑅𝑝 or 𝑝=𝑆𝑝 and 𝑝≠𝑅𝑝. 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 π‘₯0βˆˆπ‘‹ be an arbitrary point, and define the sequence {π‘₯𝑛} by π‘₯3𝑛+1=𝑇π‘₯3𝑛,π‘₯3𝑛+2=𝑆π‘₯3𝑛+1,π‘₯3𝑛+3=𝑅π‘₯3𝑛+2, π‘›βˆˆβ„•. If π‘₯𝑛=π‘₯𝑛+1, for some 𝑛, with 𝑛=3π‘š, then 𝑝=π‘₯3π‘š 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=3π‘š+1 or 𝑛=3π‘š+2. Without loss of generality, we can assume that π‘₯𝑛≠π‘₯𝑛+1, for all π‘›βˆˆβ„•. We first prove the sequence {π‘₯𝑛} is a 𝐺-Cauchy sequence. In fact, by using (2.33) and (𝐺3), we have 𝐺2ξ€·π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ=𝐺2𝑇π‘₯3𝑛,𝑆π‘₯3𝑛+1,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+1ξ€Έξ€·π‘₯+𝛾𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+1ξ€Έ.(2.36) Which gives that 𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3≀π‘₯(𝛼+𝛾)𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ.(2.37) It follows that ξ€·π‘₯(1βˆ’π›½)𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3≀π‘₯(𝛼+𝛾)𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έ.(2.38) From 0≀𝛽<1, we know that 1βˆ’π›½>0. Then, we have 𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3≀𝛼+𝛾𝐺π‘₯1βˆ’π›½3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έ.(2.39) On the other hand, by using (2.33) and (𝐺3), we have 𝐺2ξ€·π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ=𝐺2𝑇π‘₯3𝑛+3,𝑆π‘₯3𝑛+1,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+2𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+2𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.40) Which implies that 𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4≀π‘₯(𝛼+𝛽)𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.41) It follows that ξ€·π‘₯(1βˆ’π›Ύ)𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4≀π‘₯(𝛼+𝛽)𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ.(2.42) Since 0≀𝛾<1, we know that 1βˆ’π›Ύ>0. So, we have 𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4≀𝛼+𝛽𝐺π‘₯1βˆ’π›Ύ3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ.(2.43) Again, using (2.33) and (𝐺3), we can get 𝐺2ξ€·π‘₯3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5ξ€Έ=𝐺2𝑇π‘₯3𝑛+3,𝑆π‘₯3𝑛+4,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5𝐺π‘₯3𝑛+4,π‘₯3𝑛+5,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+4,π‘₯3𝑛+5,π‘₯3𝑛+3𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έξ€·π‘₯+𝛾𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4𝐺π‘₯3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5ξ€Έ.(2.44) Which implies that 𝐺π‘₯3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5ξ€Έ+ξ€·π‘₯(𝛽+𝛾)𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.45) It follows that ξ€·π‘₯(1βˆ’π›Ό)𝐺3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5≀π‘₯(𝛽+𝛾)𝐺3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.46) Since 0≀𝛼≀𝛼+𝛽+𝛾<1, we know that 1βˆ’π›Ό>0. So we have 𝐺π‘₯3𝑛+3,π‘₯3𝑛+4,π‘₯3𝑛+5≀𝛽+𝛾𝐺π‘₯1βˆ’π›Ό3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+4ξ€Έ.(2.47) Let π‘ž=max{(𝛼+𝛾)/(1βˆ’π›½),(𝛼+𝛽)/(1βˆ’π›Ύ),(𝛽+𝛾)/(1βˆ’π›Ό)}, then 0β‰€π‘ž<1, and by combining (2.39), (2.43), and (2.47), we have 𝐺π‘₯𝑛,π‘₯𝑛+1,π‘₯𝑛+2ξ€Έξ€·π‘₯β‰€π‘žπΊπ‘›βˆ’1,π‘₯𝑛,π‘₯𝑛+1ξ€Έβ‰€β‹―β‰€π‘žπ‘›πΊξ€·π‘₯0,π‘₯1,π‘₯2ξ€Έ.(2.48) Thus, by (𝐺3) and (𝐺5), for every π‘š,π‘›βˆˆβ„•, if π‘š>𝑛, noting that 0β‰€π‘ž<1, we have 𝐺π‘₯𝑛,π‘₯π‘š,π‘₯π‘šξ€Έξ€·π‘₯≀𝐺𝑛,π‘₯𝑛+1,π‘₯𝑛+1ξ€Έξ€·π‘₯+𝐺𝑛+1,π‘₯𝑛+2,π‘₯𝑛+2ξ€Έξ€·π‘₯+β‹―+πΊπ‘šβˆ’1,π‘₯π‘š,π‘₯π‘šξ€Έξ€·π‘₯≀𝐺𝑛,π‘₯𝑛+1,π‘₯𝑛+2ξ€Έξ€·π‘₯+𝐺𝑛+1,π‘₯𝑛+2,π‘₯𝑛+3ξ€Έξ€·π‘₯+β‹―+πΊπ‘šβˆ’1,π‘₯π‘š,π‘₯π‘š+1ξ€Έβ‰€ξ€·π‘žπ‘›+π‘žπ‘›+1+β‹―+π‘žπ‘šβˆ’1𝐺π‘₯0,π‘₯1,π‘₯2ξ€Έβ‰€π‘žπ‘›πΊξ€·π‘₯1βˆ’π‘ž0,π‘₯1,π‘₯2ξ€Έ.(2.49) Which implies that 𝐺(π‘₯𝑛,π‘₯π‘š,π‘₯π‘š)β†’0, as 𝑛,π‘šβ†’βˆž. Thus {π‘₯𝑛} is a 𝐺-Cauchy sequence. Due to the completeness of (𝑋,𝐺), there exists π‘’βˆˆπ‘‹, such that {π‘₯𝑛} is 𝐺-convergent to 𝑒.
Now we prove 𝑒 is a common fixed point of 𝑇,𝑆, and 𝑅. By using (2.33), we have 𝐺2𝑇𝑒,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ=𝐺2𝑇𝑒,𝑆π‘₯3𝑛+1,𝑅π‘₯3𝑛+2≀𝛼𝐺𝑒,𝑇𝑒,𝑆π‘₯3𝑛+1𝐺π‘₯3𝑛+1,𝑆π‘₯3𝑛+1,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,𝑆π‘₯3𝑛+1,𝑅π‘₯3𝑛+2𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·,𝑇𝑒+𝛾𝐺𝑒,𝑇𝑒,𝑆π‘₯3𝑛+1𝐺π‘₯3𝑛+2,𝑅π‘₯3𝑛+2ξ€Έξ€·,𝑇𝑒=𝛼𝐺𝑒,𝑇𝑒,π‘₯3𝑛+2𝐺π‘₯3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έξ€·π‘₯+𝛽𝐺3𝑛+1,π‘₯3𝑛+2,π‘₯3𝑛+3𝐺π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έξ€·,𝑇𝑒+𝛾𝐺𝑒,𝑇𝑒,π‘₯3𝑛+2𝐺π‘₯3𝑛+2,π‘₯3𝑛+3ξ€Έ.,𝑇𝑒(2.50) Letting π‘›β†’βˆž, and using the fact that 𝐺 is continuous on its variables and 𝛾<1, we can get 𝐺2(𝑇𝑒,𝑒,𝑒)≀𝛾𝐺2(𝑒,𝑒,𝑇𝑒).(2.51) Which gives that 𝑇𝑒=𝑒, hence 𝑒 is a fixed point of 𝑇. By using (2.33) again, we have 𝐺2ξ€·π‘₯3𝑛+1,𝑆𝑒,π‘₯3𝑛+3ξ€Έ=𝐺2𝑇π‘₯3𝑛,𝑆𝑒,𝑅π‘₯3𝑛+2ξ€Έξ€·π‘₯≀𝛼𝐺3𝑛,π‘₯3𝑛+1𝐺,𝑆𝑒𝑒,𝑆𝑒,π‘₯3𝑛+3ξ€Έξ€·+𝛽𝐺𝑒,𝑆𝑒,π‘₯3𝑛+3𝐺π‘₯3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+1ξ€Έξ€·π‘₯+𝛾𝐺3𝑛,π‘₯3𝑛+1𝐺π‘₯,𝑆𝑒3𝑛+2,π‘₯3𝑛+3,π‘₯3𝑛+1ξ€Έ.(2.52) Letting π‘›β†’βˆž at both sides, for 𝐺 is continuous in its variables, it follows that 𝐺2(𝑒,𝑆𝑒,𝑒)≀𝛼𝐺2(𝑒,𝑆𝑒,𝑒).(2.53) For 0≀𝛼<1, Therefore, we can get 𝐺2(𝑒,𝑆𝑒,𝑒)=0, hence 𝑆𝑒=𝑒, hence 𝑒 is a fixed point of 𝑆. Similarly, by (2.33), we can also get 𝐺2ξ€·π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έ,𝑅𝑒=𝐺2𝑇π‘₯3𝑛,𝑆π‘₯3𝑛+1ξ€Έξ€·π‘₯,𝑅𝑒≀𝛼𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2𝐺π‘₯3𝑛+1,π‘₯3𝑛+2ξ€Έξ€·π‘₯,𝑅𝑒+𝛽𝐺3𝑛+1,π‘₯3𝑛+2𝐺,𝑅𝑒𝑒,𝑅𝑒,π‘₯3𝑛+1ξ€Έξ€·π‘₯+𝛾𝐺3𝑛,π‘₯3𝑛+1,π‘₯3𝑛+2𝐺𝑒,𝑅𝑒,π‘₯3𝑛+1ξ€Έ.(2.54) On taking π‘›β†’βˆž at both sides, since 𝐺 is continuous in its variables, we get that 𝐺2(𝑒,𝑒,𝑅𝑒)≀𝛽𝐺2(𝑒,𝑒,𝑅𝑒).(2.55) Since 0≀𝛽<1, so we get 𝐺2(𝑒,𝑒,𝑅𝑒)=0, hence 𝑒=𝑅𝑒, therefore, 𝑒 is a fixed point of 𝑅. Consequently, we have 𝑒=𝑇𝑒=𝑆𝑒=𝑅𝑒, and 𝑒 is a common fixed point of 𝑇,𝑆, and 𝑅. Suppose 𝑣≠𝑒 is another common fixed point of 𝑇,𝑆, and 𝑅, and we have 𝑣=𝑇𝑣=𝑆𝑣=𝑅𝑣, then by (2.33), we have 𝐺2(𝑒,𝑒,𝑣)=𝐺2(𝑇𝑒,𝑆𝑒,𝑅𝑣)≀𝛼𝐺(𝑒,𝑇𝑒,𝑆𝑒)𝐺(𝑒,𝑆𝑒,𝑅𝑣)+𝛽𝐺(𝑒,𝑆𝑒,𝑅𝑣)𝐺(𝑣,𝑅𝑣,𝑇𝑒)+𝛾𝐺(𝑒,𝑇𝑒,𝑆𝑒)𝐺(𝑣,𝑅𝑣,𝑇𝑒)=𝛼𝐺(𝑒,𝑒,𝑒)𝐺(𝑒,𝑒,𝑣)+𝛽𝐺(𝑒,𝑒,𝑣)𝐺(𝑣,𝑣,𝑒)+𝛾𝐺(𝑒,𝑒,𝑒)𝐺(𝑣,𝑣,𝑒).(2.56) Which gives that 𝐺2(𝑒,𝑒,𝑣)≀𝛽𝐺(𝑒,𝑒,𝑣)𝐺(𝑣,𝑣,𝑒).(2.57) Hence, we can get 𝐺(𝑒,𝑒,𝑣)≀𝛽𝐺(𝑣,𝑣,𝑒). By using (2.33) again, we get 𝐺2(𝑒,𝑣,𝑣)=𝐺2(𝑇𝑒,𝑆𝑣,𝑅𝑣)≀𝛼𝐺(𝑒,𝑇𝑒,𝑆𝑣)𝐺(𝑣,𝑆𝑣,𝑅𝑣)+𝛽𝐺(𝑣,𝑆𝑣,𝑅𝑣)𝐺(𝑣,𝑅𝑣,𝑇𝑒)+𝛾𝐺(𝑒,𝑇𝑒,𝑆𝑣)𝐺(𝑣,𝑅𝑣,𝑇𝑒)=𝛼𝐺(𝑒,𝑒,𝑣)𝐺(𝑣,𝑣,𝑣)+𝛽𝐺(𝑣,𝑣,𝑣)𝐺(𝑣,𝑣,𝑒)+𝛾𝐺(𝑒,𝑒,𝑣)𝐺(𝑣,𝑣,𝑒).(2.58) Which implies that 𝐺2(𝑒,𝑣,𝑣)≀𝛾𝐺(𝑒,𝑒,𝑣)𝐺(𝑣,𝑣,𝑒).(2.59) Hence, we can get 𝐺(𝑒,𝑣,𝑣)≀𝛾𝐺(𝑒,𝑒,𝑣).(2.60) By combining 𝐺(𝑒,𝑒,𝑣)≀𝛽𝐺(𝑣,𝑣,𝑒), we can have 𝐺(𝑒,𝑣,𝑣)≀𝛾𝐺(𝑒,𝑒,𝑣)≀𝛽𝛾𝐺(𝑣,𝑣,𝑒).(2.61) Since 𝑣≠𝑒,𝐺(𝑒,𝑣,𝑣)>0, so we have that 𝛽𝛾β‰₯1. Since 0≀𝛽,𝛾<1, we know 0≀𝛽𝛾<1, so it’s a contradiction. Hence, we get 𝑒=𝑣. Then we know the common fixed point of 𝑇,𝑆, and 𝑅 is unique.
To show that 𝑇 is 𝐺-continuous at 𝑒, let {𝑦𝑛} be any sequence in 𝑋 such that {𝑦𝑛} is 𝐺-convergent to 𝑒. For π‘›βˆˆβ„•, we have 𝐺2𝑇𝑦𝑛,𝑒,𝑒=𝐺2𝑇𝑦𝑛𝑦,𝑆𝑒,𝑅𝑒≀𝛼𝐺𝑛,𝑇𝑦𝑛,𝑆𝑒𝐺(𝑒,𝑆𝑒,𝑅𝑒)+𝛽𝐺(𝑒,𝑆𝑒,𝑅𝑒)𝐺𝑒,𝑅𝑒,𝑇𝑦𝑛𝑦+𝛾𝐺𝑛,𝑇𝑦𝑛𝐺,𝑆𝑒𝑒,𝑅𝑒,𝑇𝑦𝑛𝑦=𝛼𝐺𝑛,𝑇𝑦𝑛,𝑒𝐺(𝑒,𝑒,𝑒)+𝛽𝐺(𝑒,𝑒,𝑒)𝐺𝑒,𝑒,𝑇𝑦𝑛𝑦+𝛾𝐺𝑛,𝑇𝑦𝑛𝐺,𝑒𝑒,𝑒,𝑇𝑦𝑛𝑦=𝛾𝐺𝑛,𝑇𝑦𝑛𝐺,𝑒𝑒,𝑒,𝑇𝑦𝑛.(2.62) Which implies that 𝐺𝑇𝑦𝑛𝑦,𝑒,𝑒≀𝛾𝐺𝑛,𝑇𝑦𝑛,𝑒.(2.63) On taking π‘›β†’βˆž at both sides, considering 𝛾<1, we get limπ‘›β†’βˆžπΊ(𝑇𝑦𝑛,𝑒,𝑒)=0. Hence {𝑇𝑦𝑛} is 𝐺-convergent to 𝑒=𝑇𝑒. So 𝑇 is 𝐺-continuous at 𝑒. Similarly, we can also prove that 𝑆,𝑅 are 𝐺-continuous at 𝑒. 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(π‘₯,𝑦,𝑧)=|π‘₯βˆ’π‘¦|+|π‘¦βˆ’π‘§|+|π‘§βˆ’π‘₯|, for all π‘₯,𝑦,𝑧 in 𝑋. Let 𝑇, 𝑆, and 𝑅 be three self-mappings defined by ⎧βŽͺ⎨βŽͺβŽ©ξ‚ƒ1𝑇π‘₯=1,π‘₯∈0,2ξ‚„67ξ‚€1,π‘₯∈2ξ‚„βŽ§βŽͺ⎨βŽͺ⎩7,1,𝑆π‘₯=81,π‘₯∈0,2ξ‚„67ξ‚€1,π‘₯∈2ξ‚„6,1,𝑅π‘₯=7[],π‘₯∈0,1.(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 π‘₯,π‘¦βˆˆ[0,1/2], π‘§βˆˆ[0,1], then 𝐺2(𝑇π‘₯,𝑆𝑦,𝑅𝑧)=𝐺2ξ‚€71,8,67=4,ξ‚€749𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)=𝐺π‘₯,1,8=||||+|||7π‘₯βˆ’1π‘₯βˆ’8|||+18β‰₯12+38+18ξ‚€7=1,𝐺(𝑦,𝑆𝑦,𝑅𝑧)=𝐺𝑦,8,67=|||7π‘¦βˆ’8|||+|||6π‘¦βˆ’7|||+1β‰₯3568+5+114=3564,ξ‚€6𝐺(𝑧,𝑅𝑧,𝑇π‘₯)=𝐺𝑧,7=|||6,1π‘§βˆ’7|||+17+||||1π‘§βˆ’1β‰₯0+71+0=7.(2.65) Thus, we have 𝐺24(𝑇π‘₯,𝑆𝑦,𝑅𝑧)=349≀𝛼⋅1β‹…43+𝛽⋅4β‹…171+𝛾⋅1β‹…7≀𝛼𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇π‘₯)+𝛾𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇π‘₯).(2.66)

Case 2. If π‘₯∈[0,1/2],π‘¦βˆˆ(1/2,1],π‘§βˆˆ[0,1], then we can get 𝐺2(𝑇π‘₯,𝑆𝑦,𝑅𝑧)=𝐺2ξ‚€61,7,67=4,ξ‚€649𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)=𝐺π‘₯,1,7=||||+|||6π‘₯βˆ’1π‘₯βˆ’7|||+17β‰₯12+5+1147ξ‚€6=1,𝐺(𝑦,𝑆𝑦,𝑅𝑧)=𝐺𝑦,7,67=|||6π‘¦βˆ’7|||+|||6π‘¦βˆ’7|||ξ‚€6β‰₯0+0=0,𝐺(𝑧,𝑅𝑧,𝑇π‘₯)=𝐺𝑧,7=|||6,1π‘§βˆ’7|||+17+||||1π‘§βˆ’1β‰₯0+71+0=7.(2.67) Thus, we have 𝐺24(𝑇π‘₯,𝑆𝑦,𝑅𝑧)=149≀𝛼⋅1β‹…0+𝛽⋅0β‹…71+𝛾⋅1β‹…7≀𝛼𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇π‘₯)+𝛾𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇π‘₯).(2.68)

Case 3. If π‘₯∈(1/2,1],π‘¦βˆˆ[0,1/2],π‘§βˆˆ[0,1], then we have 𝐺2(𝑇π‘₯,𝑆𝑦,𝑅𝑧)=𝐺2ξ‚€67,78,67=1,ξ‚€6784𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)=𝐺π‘₯,7,78=|||6π‘₯βˆ’7|||+|||7π‘₯βˆ’8|||+1156β‰₯0+0+=156,ξ‚€756𝐺(𝑦,𝑆𝑦,𝑅𝑧)=𝐺𝑦,8,67=|||7π‘¦βˆ’8|||+|||6π‘¦βˆ’7|||+1β‰₯3568+5+114=3564,ξ‚€6𝐺(𝑧,𝑅𝑧,𝑇π‘₯)=𝐺𝑧,7,67=|||6π‘§βˆ’7|||+|||6π‘§βˆ’7|||β‰₯0+0=0.(2.69) Thus, we have 𝐺21(𝑇π‘₯,𝑆𝑦,𝑅𝑧)=1784≀𝛼⋅⋅35643+𝛽⋅41β‹…0+𝛾⋅56β‹…0≀𝛼𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇π‘₯)+𝛾𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇π‘₯).(2.70)

Case 4. If π‘₯,π‘¦βˆˆ(1/2,1],π‘§βˆˆ[0,1], then we have 𝐺2(𝑇π‘₯,𝑆𝑦,𝑅𝑧)=𝐺2ξ‚€67,67,67=0.(2.71) Thus, we have 𝐺2(𝑇π‘₯,𝑆𝑦,𝑅𝑧)=0≀𝛼𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑦,𝑆𝑦,𝑅𝑧)+𝛽𝐺(𝑦,𝑆𝑦,𝑅𝑧)𝐺(𝑧,𝑅𝑧,𝑇π‘₯)+𝛾𝐺(π‘₯,𝑇π‘₯,𝑆𝑦)𝐺(𝑧,𝑅𝑧,𝑇π‘₯).(2.72)

Then in all of the above cases, the mappings 𝑇,𝑆, and 𝑅 satisfy the contractive condition (2.33) of Theorem 2.5 with 𝛼=1/7,𝛽=1/7,𝛾=4/7. So that all the conditions of Theorem 2.5 are satisfied. Moreover, 6/7 is the unique common fixed point for all of the three mappings 𝑇,𝑆, and 𝑅.

At last, we prove 𝑇,𝑆, and 𝑅 are all 𝐺-continuous at the common fixed point 6/7. Since 6/7∈(1/2,1], and let the sequence {𝑦𝑛}βŠ‚(0,1] and 𝑦𝑛→(6/7)(π‘›β†’βˆž), then there exists π‘βˆˆβ„• such that {𝑦𝑛}βŠ‚(1/2,1], for all 𝑛>𝑁. Without loss of generality, we can assume that {𝑦𝑛}βŠ‚(1/2,1], and so 𝑇𝑦𝑛=6/7,𝑆𝑦𝑛=6/7 and 𝑅𝑦𝑛=6/7. Therefore, limπ‘›β†’βˆžπ‘‡π‘¦π‘›=limπ‘›β†’βˆžπ‘†π‘¦π‘›=limπ‘›β†’βˆžπ‘…π‘¦π‘›=67.(2.73) Which implies that 𝑇,𝑆, and 𝑅 are all 𝐺-continuous at the common fixed point 6/7.

Corollary 2.7. Let (𝑋,𝐺) be a complete 𝐺-metric space. Suppose the three self-mappings 𝑇,𝑆,π‘…βˆΆπ‘‹β†’π‘‹ satisfy the condition: 𝐺2(𝑇𝑝π‘₯,𝑆𝑠𝑦,π‘…π‘Ÿπ‘§)≀𝛼𝐺(π‘₯,𝑇𝑝π‘₯,𝑆𝑠𝑦)𝐺(𝑦,𝑆𝑠𝑦,π‘…π‘Ÿπ‘§)+𝛽𝐺(𝑦,𝑆𝑠𝑦,π‘…π‘Ÿπ‘§)𝐺(𝑧,π‘…π‘Ÿπ‘§,𝑇𝑝π‘₯)+𝛾𝐺(π‘₯,𝑇𝑝π‘₯,𝑆𝑠𝑦)𝐺(𝑧,π‘…π‘Ÿπ‘§,𝑇𝑝π‘₯),(2.74) for all π‘₯,𝑦,π‘§βˆˆπ‘‹, where 𝑝,𝑠,π‘Ÿβˆˆβ„•, 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇,𝑆, and 𝑅 have a unique common fixed point (say 𝑒) and 𝑇𝑝,𝑆𝑠,π‘…π‘Ÿ are all 𝐺-continuous at 𝑒.

Proof. Since the proof of Corollary 2.7 is very similar to that of Corollary 2.2, so we delete it.

Corollary 2.8. Let (𝑋,𝐺) be a complete 𝐺-metric space, and let π‘‡βˆΆπ‘‹β†’π‘‹ be a self-mapping in 𝑋, which satisfies the following condition: 𝐺2(𝑇π‘₯,𝑇𝑦,𝑇𝑧)≀𝛼𝐺(π‘₯,𝑇π‘₯,𝑇𝑦)𝐺(𝑦,𝑇𝑦,𝑇𝑧)+𝛽𝐺(𝑦,𝑇𝑦,𝑇𝑧)𝐺(𝑧,𝑇𝑧,𝑇π‘₯)+𝛾𝐺(π‘₯,𝑇π‘₯,𝑇𝑦)𝐺(𝑧,𝑇𝑧,𝑇π‘₯).(2.75) for all π‘₯,𝑦,π‘§βˆˆπ‘‹, where 𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇 has a unique fixed point (say 𝑒) and 𝑇 is 𝐺-continuous at 𝑒.

Corollary 2.9. Let (𝑋,𝐺) be a complete 𝐺-metric space, and let π‘‡βˆΆπ‘‹β†’π‘‹ be a self-mapping in 𝑋, which satisfies the following condition: 𝐺2(𝑇𝑝π‘₯,𝑇𝑝𝑦,𝑇𝑝𝑧)≀𝛼𝐺(π‘₯,𝑇𝑝π‘₯,𝑇𝑝𝑦)𝐺(𝑦,𝑇𝑝𝑦,𝑇𝑝𝑧)+𝛽𝐺(𝑦,𝑇𝑝𝑦,𝑇𝑝𝑧)𝐺(𝑧,𝑇𝑝𝑧,𝑇𝑝π‘₯)+𝛾𝐺(π‘₯,𝑇𝑝π‘₯,𝑇𝑝𝑦)𝐺(𝑧,𝑇𝑝𝑧,𝑇𝑝π‘₯).(2.76) for all π‘₯,𝑦,π‘§βˆˆπ‘‹, where π‘βˆˆπ‘,𝛼,𝛽,𝛾 are nonnegative real numbers and 𝛼+𝛽+𝛾<1. Then 𝑇 has a unique fixed point (say 𝑒) and 𝑇𝑝 is 𝐺-continuous at 𝑒.

Acknowledgments

The present studies are supported by the National Natural Science Foundation of China (11071169), the Natural Science Foundation of Zhejiang Province (Y6110287, Y12A010095).