Existence of Fixed Point Results in 𝐺-Metric Spaces

Abstract

The purpose of this paper is to prove the existence of fixed points of contractive mapping defined on 𝐺-metric space where the completeness is replaced with weaker conditions. Moreover, we showed that these conditions do not guarantee the completeness of 𝐺-metric spaces.

1. Introduction

The study of metric fixed point theory has been researched extensively in the past decades, since fixed point theory plays a major role in mathematics and applied sciences, such as optimization, mathematical models, and economic theories.

Different mathematicians tried to generalize the usual notion of metric space (𝑋,𝑑) such as Gähler [1, 2] and Dhage [3–5] to extend known metric space theorems in more general setting, but different authors proved that these attempts are unvalid (for detail see [6–8]).

In 2005, Mustafa and Sims introduced a new structure of generalized metric spaces (see [9]), which are called 𝐺-metric spaces as generalization of metric space (𝑋,𝑑), to develop and introduce a new fixed point theory for various mappings in this new structure. The 𝐺-metric space is as follows.

Definition 1.1 (see [9]). Let 𝑋 be a nonempty set, and let 𝐺∶𝑋×𝑋×𝑋→ℝ⁺, be a function satisfying the following: (G1)𝐺(𝑥,𝑦,𝑧)=0 if 𝑥=𝑦=𝑧,(G2)0<𝐺(𝑥,𝑥,𝑦); for all 𝑥,𝑦∈𝑋, with 𝑥≠𝑦,(G3)𝐺(𝑥,𝑥,𝑦)≤𝐺(𝑥,𝑦,𝑧), for all 𝑥,𝑦,𝑧∈𝑋 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,𝑎∈𝑋, (rectangle inequality).
Then the function 𝐺 is called a generalized metric , or, more specifically a 𝐺- metric on 𝑋, and the pair (𝑋,𝐺) is a 𝐺-metric space.
Clearly these properties are satisfied when 𝐺(𝑥,𝑦,𝑧) is the perimeter of the triangle with vertices at 𝑥, 𝑦, and 𝑧 in ℝ²; moreover taking 𝑎 in the interior of the triangle shows that (G5) is the best possible.
If (𝑋,𝑑) is an ordinary metric space, then (𝑋,𝑑) can define 𝐺-metrics on 𝑋 by
(E_𝑠)𝐺_𝑠(𝑑)(𝑥,𝑦,𝑧)=𝑑(𝑥,𝑦)+𝑑(𝑦,𝑧)+𝑑(𝑥,𝑧), (E_𝑚)𝐺_𝑚(𝑑)(𝑥,𝑦,𝑧)=max{𝑑(𝑥,𝑦),𝑑(𝑦,𝑧),𝑑(𝑥,𝑧)}.

Proposition 1.2 (see [9]). Let (𝑋,𝐺) be a 𝐺-metric space. Then for any 𝑥,𝑦,𝑧, and 𝑎∈𝑋, it follows that(1)if 𝐺(𝑥,𝑦,𝑧)=0, then 𝑥=𝑦=𝑧,(2)𝐺(𝑥,𝑦,𝑧)≤𝐺(𝑥,𝑥,𝑦)+𝐺(𝑥,𝑥,𝑧),(3)𝐺(𝑥,𝑦,𝑦)≤2𝐺(𝑦,𝑥,𝑥),(4)𝐺(𝑥,𝑦,𝑧)≤𝐺(𝑥,𝑎,𝑧)+𝐺(𝑎,𝑦,𝑧),(5)𝐺(𝑥,𝑦,𝑧)≤(2/3)(𝐺(𝑥,𝑦,𝑎)+𝐺(𝑥,𝑎,𝑧)+𝐺(𝑎,𝑦,𝑧)),(6)𝐺(𝑥,𝑦,𝑧)≤(𝐺(𝑥,𝑎,𝑎)+𝐺(𝑦,𝑎,𝑎)+𝐺(𝑧,𝑎,𝑎)).

Proposition 1.3 (see [9]). Every 𝐺-metric space (𝑋,𝐺) will define a metric space (𝑋,𝑑_𝐺) by 𝑑_𝐺(𝑥,𝑦)=𝐺(𝑥,𝑦,𝑦)+𝐺(𝑦,𝑥,𝑥),∀𝑥,𝑦∈𝑋. (1.1)

Definition 1.4 (see [9]). Let (𝑋,𝐺) be a 𝐺-metric space. Then for 𝑥₀∈𝑋, 𝑟>0, the 𝐺-ball with center 𝑥₀and radius 𝑟is 𝐵_𝐺(𝑥₀,𝑟)={𝑦∈𝑋∶𝐺(𝑥₀,𝑦,𝑦)<𝑟}. (1.2)

Proposition 1.5 (see [9]). Let (𝑋,𝐺) be a 𝐺-metric space. Then for any 𝑥₀∈𝑋 and 𝑟>0,one has(1)if 𝐺(𝑥₀,𝑥,𝑦)<𝑟, then 𝑥,𝑦∈𝐵_𝐺(𝑥₀,𝑟),(2)if 𝑦∈𝐵_𝐺(𝑥₀,𝑟), then there exists a 𝛿>0 such that 𝐵_𝐺(𝑦,𝛿)⊆𝐵(𝑥₀,𝑟).

Proof. (1) follows directly from (G3), while (2) follows from (G5) with 𝛿=𝑟−𝐺(𝑥₀,𝑦,𝑦).

It follows from (2) of the above proposition that the family of all 𝐺-balls, ℬ={𝐵_𝐺(𝑥,𝑟)∶𝑥∈𝑋,𝑟>0}, is the base of a topology 𝜏(𝐺) on 𝑋, the 𝐺-metric topology.

Definition 1.6 (see [9]). Let (𝑋,𝐺) be a 𝐺-metric space, let (𝑥_𝑛) be sequence of points of 𝑋, a point 𝑥∈𝑋 is said to be the limit of the sequence (𝑥_𝑛) if lim_{𝑛,𝑚→∞}𝐺(𝑥,𝑥_𝑛,𝑥_𝑚)=0, and we say that the sequence (𝑥_𝑛) is 𝐺-convergent to 𝑥.
Thus, if 𝑥−−−→_𝑛^(𝐺)0, in a 𝐺-metric space (𝑋,𝐺), then for any 𝜖>0, there exists 𝑁∈ℕ such that 𝐺(𝑥,𝑥_𝑛,𝑥_𝑚)<𝜖, for all 𝑛,𝑚≥𝑁, (through this paper we mean by ℕ the set of all natural numbers).

Proposition 1.7 (see [9]). Let (𝑋,𝐺) be a 𝐺-metric space. The sequence (𝑥_𝑛)⊆𝑋 is 𝐺-convergent to 𝑥 if it converges to 𝑥 in the 𝐺-metric topology, 𝜏(𝐺).

Proposition 1.8 (see [9]). Let (𝑋,𝐺) be a 𝐺-metric space. Then for a sequence (𝑥_𝑛)⊆𝑋 and a point 𝑥∈𝑋 the following are equivalent(1)(𝑥_𝑛) is 𝐺-convergent to 𝑥,(2)𝐺(𝑥_𝑛,𝑥_𝑛,𝑥)→0, as 𝑛→∞,(3)𝐺(𝑥_𝑛,𝑥,𝑥)→0, as 𝑛→∞,(4)𝐺(𝑥_𝑚,𝑥_𝑛,𝑥)→0, as 𝑚,𝑛→∞.

Definition 1.9 (see [9]). Let (𝑋,𝐺) and (𝑋^′,𝐺^′) be 𝐺-metric spaces and let 𝑓∶(𝑋,𝐺)→(𝑋^′,𝐺^′) be a function, then 𝑓 is said to be 𝐺-continuous at a point 𝑎∈𝑋 if and only if, given 𝜖>0, there exists 𝛿>0 such that 𝑥,𝑦∈𝑋; 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 𝑎∈𝑋.

Proposition 1.10 (see [9]). Let (𝑋,𝐺), (𝑋^′,𝐺^′) be 𝐺-metric spaces. Then a function 𝑓∶𝑋→𝑋^′ is 𝐺-continuous at a point 𝑥∈𝑋 if and only if it is 𝐺-sequentially continuous at 𝑥; that is, whenever (𝑥_𝑛) is 𝐺-convergent to 𝑥one has(𝑓(𝑥_𝑛)) is 𝐺-convergent to 𝑓(𝑥).

Proposition 1.11 (see [9]). 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 1.12 (see [9]). Let (𝑋,𝐺) be a 𝐺-metric space. Then the sequence (𝑥_𝑛)⊆𝑋 is said to be 𝐺-Cauchy if for every 𝜖>0, there exists 𝑁∈ℕ such that 𝐺(𝑥_𝑛,𝑥_𝑚,𝑥_𝑙)<𝜖 for all 𝑛,𝑚,𝑙≥𝑁.

Definition 1.13 (see [9]). A 𝐺-metric space (𝑋,𝐺) is said to be 𝐺-complete (or complete 𝐺-metric space) if every 𝐺-Cauchy sequence in (𝑋,𝐺) is 𝐺-convergent in (𝑋,𝐺).

2. The Main Results

In this section we will prove several theorems in each of which we have omitted the completeness property of 𝐺-metric space and we have obtained the same conclusion as in complete 𝐺-metric space, but with assumed sufficient conditions.

Theorem 2.1. Let (𝑋,𝐺) be a 𝐺-metric space and let 𝑇∶𝑋→𝑋 be a mapping such that 𝑇 satisfies that(A1)𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝑎𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝑏𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝑐𝐺(𝑧,𝑇𝑧,𝑇𝑧) for all 𝑥,𝑦,𝑧∈𝑋 where 0<𝑎+𝑏+𝑐<1,(A2)𝑇 is 𝐺-continuous at a point 𝑢∈𝑋, (A3) there is 𝑥∈𝑋;{𝑇^𝑛(𝑥)} has a subsequence {𝑇^𝑛𝑖(𝑥)}𝐺-converges to 𝑢. Then 𝑢 is a unique fixed point (i.e., 𝑇𝑢=𝑢).

Proof. 𝐺-continuity of 𝑇 at 𝑢 implies that {𝑇^𝑛𝑖+1(𝑥)}𝐺-convergent to 𝑇(𝑢). Suppose 𝑇(𝑢)≠𝑢, consider the two 𝐺-open balls 𝐵₁=𝐵(𝑢,𝜖) and 𝐵₂=𝐵(𝑇𝑢,𝜖) where 𝜖<(1/6)min{𝐺(𝑢,𝑇𝑢,𝑇𝑢),𝐺(𝑇𝑢,𝑢,𝑢)}.
Since 𝑇^𝑛𝑖(𝑥)→𝑢 and 𝑇^𝑛𝑖+1(𝑥)→𝑇𝑢, then there exist 𝑁₁∈ℕ such that if 𝑖>𝑁₁ implies 𝑇^𝑛𝑖(𝑥)∈𝐵₁ and 𝑇^𝑛𝑖+1(𝑥)∈𝐵₂. Hence our assumption implies that we must have 𝐺(𝑇^𝑛𝑖(𝑥),𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+1(𝑥))>𝜖,∀𝑖>𝑁₁. (2.1)
On the other hand we have from (A1), 𝐺(𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+3(𝑥))≤𝑎𝐺(𝑇^𝑛𝑖(𝑥),𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+1(𝑥))+𝑏𝐺(𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+2(𝑥))+𝑐𝐺(𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+3(𝑥),𝑇^𝑛𝑖+3(𝑥)) (2.2) but, by axioms of 𝐺-metric (G3), we have 𝐺(𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+2(𝑥))≤𝐺(𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+3(𝑥)),𝐺(𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+3(𝑥))≤𝐺(𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+3(𝑥)). (2.3)(2.4) So, from (2.3) and (2.4), we see (2.2) becomes 𝐺(𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+3(𝑥))≤𝑞𝐺(𝑇^𝑛𝑖(𝑥),𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+1(𝑥)), (2.5) where 𝑞=𝑎/(1−(𝑏+𝑐)) and 𝑞<1, since 0<𝑎+𝑏+𝑐<1.
Hence (2.3) and (2.5) implies that 𝐺(𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+2(𝑥),𝑇^𝑛𝑖+2(𝑥))≤𝑞𝐺(𝑇^𝑛𝑖(𝑥),𝑇^𝑛𝑖+1(𝑥),𝑇^𝑛𝑖+1(𝑥)). (2.6)
For 𝑙>𝑗>𝑁₁ and by repeated application of (2.6) we have 𝐺(𝑇^𝑛𝑙(𝑥),𝑇^𝑛𝑙+1(𝑥),𝑇^𝑛𝑙+1(𝑥))≤𝑞𝐺(𝑇^𝑛𝑙−1(𝑥),𝑇^𝑛𝑙(𝑥),𝑇^𝑛𝑙(𝑥))≤𝑞²𝐺(𝑇^𝑛𝑙−2(𝑥),𝑇^𝑛𝑙−1(𝑥),𝑇^𝑛𝑙−1(𝑥))≤⋯≤𝑞^{𝑛𝑙−𝑛𝑗}𝐺(𝑇^𝑛𝑗(𝑥),𝑇^𝑛𝑗+1(𝑥),𝑇^𝑛𝑗+1(𝑥)). (2.7) So, as 𝑙→∞ we have lim𝐺(𝑇^𝑛𝑙(𝑥),𝑇^𝑛𝑙+1(𝑥),𝑇^𝑛𝑙+1(𝑥))≤0 which contradict (2.1), hence 𝑇𝑢=𝑢.
Suppose there is 𝑣∈𝑋;𝑇𝑣=𝑣, then from (A1), we have 𝐺(𝑢,𝑣,𝑣)=𝐺(𝑇𝑢,𝑇𝑣,𝑇𝑣)≤𝑎𝐺(𝑢,𝑇𝑢,𝑇𝑢)+(𝑏+𝑐)𝐺(𝑣,𝑇𝑣,𝑇𝑣)=0. (2.8) This prove the uniqueness of 𝑢.

In [10] we have proved the following theorem.

Theorem 2.2 (see [10]). Let (𝑋,𝐺) be a complete 𝐺-metric space and let 𝑇∶𝑋→𝑋 be a mapping satisfies the following condition for all 𝑥,𝑦,𝑧∈𝑋: 𝐺(𝑇(𝑥),𝑇(𝑦),𝑇(𝑧))≤𝑎𝐺(𝑥,𝑇(𝑥),𝑇(𝑥))+𝑏𝐺(𝑦,𝑇(𝑦),𝑇(𝑦))+𝑐𝐺(𝑧,𝑇(𝑧),𝑇(𝑧))+𝑑𝐺(𝑥,𝑦,𝑧), (2.9) where 0≤𝑎+𝑏+𝑐+𝑑<1, then 𝑇 has a unique fixed point, say 𝑢, and 𝑇 is 𝐺-continuous at 𝑢.

We see that if we take 𝑑=0, the following theorem becomes a direct result.

Theorem 2.3. Let (𝑋,𝐺) be complete 𝐺-metric space and let 𝑇∶𝑋→𝑋 be a mapping satisfies for all 𝑥,𝑦,𝑧∈𝑋𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝑎𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝑏𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝑐𝐺(𝑧,𝑇𝑧,𝑇𝑧), (2.10) where 0<𝑎+𝑏+𝑐<1, then 𝑇 has a unique fixed point, say u, and 𝑇 is 𝐺-continuous at 𝑢.

If we compare Theorem 2.3 with Theorem 2.1, we see that in Theorem 2.1 we have omitted the completeness property of the 𝐺-metric space and instead we have assumed conditions (2) and (3). However, the following examples support that conditions (2) and (3) in Theorem 2.1 do not guarantee the completeness of the 𝐺-metric space.

Example 2.4. Let 𝑋=[0,1),𝑇(𝑥)=𝑥/4 and 𝐺(𝑥,𝑦,𝑧)=max{|𝑥−𝑦|,|𝑦−𝑧|,|𝑥−𝑧|}. Then (𝑋,𝐺) is 𝐺-metric space but not complete, since the sequence 𝑥_𝑛=1−1/𝑛 is 𝐺-cauchy which is not 𝐺-convergent in (𝑋,𝐺). However, conditions (2) and (3) in Theorem 2.1 are satisfied.

Theorem 2.5. Let (𝑋,𝐺) be a 𝐺-metric space and let 𝑇∶𝑋→𝑋 be a 𝐺-continuous mapping satisfies the following conditions:(B1)𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝑘{𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝐺(𝑧,𝑇𝑧,𝑇𝑧) for all 𝑥,𝑦,𝑧∈𝑀, where 𝑀 is an every where dense subset of 𝑋 (with respect the topology of 𝐺-metric convergence) and 0<𝑘<1/6,(B2) there is 𝑥∈𝑋 such that {𝑇^𝑛(𝑥)}→𝑥_𝑜. Then 𝑥_𝑜 is unique fixed point.

Proof. The proof will follow from Theorem 2.1, if we can show that condition (A1) in Theorem 2.1 holds for any 𝑥,𝑦,𝑧∈𝑋.
Let 𝑥,𝑦,𝑧 be any elements of 𝑋.
Case 1. If 𝑥,𝑦,𝑧∈𝑋⧵𝑀, let {𝑥_𝑛}, {𝑦_𝑛}, and {𝑧_𝑛} be a sequences in 𝑀 such that 𝑦_𝑛→𝑦, 𝑥_𝑛→𝑥 and 𝑧_𝑛→𝑧. From (G5) we have 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)+𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦), (2.11) also 𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦)≤𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+𝐺(𝑇𝑧_𝑛,𝑇𝑦_𝑛,𝑇𝑦_𝑛)+𝐺(𝑇𝑦_𝑛,𝑇𝑦,𝑇𝑦) (2.12) and by (B1), we have 𝐺(𝑇𝑧_𝑛,𝑇𝑦_𝑛,𝑇𝑦_𝑛)≤𝑘{𝐺(𝑧_𝑛,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+2𝐺(𝑦_𝑛,𝑇𝑦_𝑛,𝑇𝑦_𝑛)}, (2.13) again by (G5) we have 𝐺(𝑧_𝑛,𝑇𝑧_𝑛,𝑇𝑧_𝑛)≤𝐺(𝑧_𝑛,𝑧,𝑧)+𝐺(𝑧,𝑇𝑧,𝑇𝑧)+𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛),𝐺(𝑦_𝑛,𝑇𝑦_𝑛,𝑇𝑦_𝑛)≤𝐺(𝑦_𝑛,𝑦,𝑦)+𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝐺(𝑇𝑦,𝑇𝑦_𝑛,𝑇𝑦_𝑛). (2.14) So, from (2.13) and (2.14) we see that (2.12) becomes 𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦)≤(1+𝑘)𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+𝐺(𝑇𝑦_𝑛,𝑇𝑦,𝑇𝑦)+𝑘𝐺(𝑧_𝑛,𝑧,𝑧)+2𝑘𝐺(𝑦_𝑛,𝑦,𝑦)+2𝑘𝐺(𝑇𝑦,𝑇𝑦_𝑛,𝑇𝑦_𝑛)+𝑘𝐺(𝑧,𝑇𝑧,𝑇𝑧)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦), (2.15) by the same argument we deduce that 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤(1+𝑘)𝐺(𝑇𝑥,𝑇𝑥_𝑛,𝑇𝑥_𝑛)+𝐺(𝑇𝑦_𝑛,𝑇𝑦,𝑇𝑦)+𝑘𝐺(𝑥_𝑛,𝑥,𝑥)+2𝑘𝐺(𝑦_𝑛,𝑦,𝑦)+2𝑘𝐺(𝑇𝑦,𝑇𝑦_𝑛,𝑇𝑦_𝑛)+𝑘𝐺(𝑥,𝑇𝑥,𝑇𝑥)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦). (2.16)
Hence, by (2.15) and (2.16), we have 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)+𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦)≤{(1+𝑘)𝐺(𝑇𝑥,𝑇𝑥_𝑛,𝑇𝑥_𝑛)+𝐺(𝑇𝑦_𝑛,𝑇𝑦,𝑇𝑦)+𝑘𝐺(𝑥_𝑛,𝑥,𝑥)+2𝑘𝐺(𝑦_𝑛,𝑦,𝑦)+2𝑘𝐺(𝑇𝑦,𝑇𝑦_𝑛,𝑇𝑦_𝑛)+𝑘𝐺(𝑥,𝑇𝑥,𝑇𝑥)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦)}+{(1+𝑘)𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+𝐺(𝑇𝑦_𝑛,𝑇𝑦,𝑇𝑦)+𝑘𝐺(𝑧_𝑛,𝑧,𝑧)+2𝑘𝐺(𝑦_𝑛,𝑦,𝑦)+2𝑘𝐺(𝑇𝑦,𝑇𝑦_𝑛,𝑇𝑦_𝑛)+𝑘𝐺(𝑧,𝑇𝑧,𝑇𝑧)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦)}. (2.17)
Now letting 𝑛→∞ in the above inequality and using the fact that 𝑇 is 𝐺-continuous we get 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝑘{𝐺(𝑥,𝑇𝑥,𝑇𝑥)+4𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝐺(𝑧,𝑇𝑧,𝑇𝑧)}. (2.18)
Case 2. If 𝑥,𝑦∈𝑀 and 𝑧∈𝑋⧵𝑀, let {𝑧_𝑛} be a sequence in 𝑀 such that 𝑧_𝑛→𝑧, then by (G5), we have 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)+𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦) (2.19) but by (B1), we have 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤𝑘{𝐺(𝑥,𝑇𝑥,𝑇𝑥)+2𝐺(𝑦,𝑇𝑦,𝑇𝑦)}, (2.20) and by (G5), we have 𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦)≤𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+𝐺(𝑇𝑧_𝑛,𝑇𝑦,𝑇𝑦). (2.21) Again by (B1), we have 𝐺(𝑇𝑧_𝑛,𝑇𝑦,𝑇𝑦)≤𝑘{𝐺(𝑧_𝑛,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+2𝐺(𝑦,𝑇𝑦,𝑇𝑦)}. (2.22) Also, by (G5), we have 𝐺(𝑧_𝑛,𝑇𝑧_𝑛,𝑇𝑧_𝑛)≤𝐺(𝑧_𝑛,𝑧,𝑧)+𝐺(𝑧,𝑇𝑧,𝑇𝑧)+𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛). (2.23) So, from (2.21), (2.22), and (2.23), we see that (2.19) becomes 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝑘{𝐺(𝑥,𝑇𝑥,𝑇𝑥)+2𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝐺(𝑧_𝑛,𝑧,𝑧)+𝐺(𝑧,𝑇𝑧,𝑇𝑧)+𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛)}+𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦). (2.24)
Now letting 𝑛→∞ in the above inequality, we get 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝑘{𝐺(𝑥,𝑇𝑥,𝑇𝑥)+4𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝐺(𝑧,𝑇𝑧,𝑇𝑧)}. (2.25)
Case 3. If 𝑦∈𝑀 and 𝑥,𝑧∈𝑋⧵𝑀, let {𝑥_𝑛} and {𝑧_𝑛} be a sequences in 𝑀 such that 𝑥_𝑛→𝑥 and 𝑧_𝑛→𝑧, but by (G5), we have 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)+𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦),𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤𝐺(𝑇𝑥,𝑇𝑥_𝑛,𝑇𝑥_𝑛)+𝐺(𝑇𝑥_𝑛,𝑇𝑦,𝑇𝑦), (2.26)(2.27) also, from (B1), we have 𝐺(𝑇𝑥_𝑛,𝑇𝑦,𝑇𝑦)≤𝑘{𝐺(𝑥_𝑛,𝑇𝑥_𝑛,𝑇𝑥_𝑛)+2𝐺(𝑦,𝑇𝑦,𝑇𝑦)}, (2.28) and from (G5), we have 𝐺(𝑥_𝑛,𝑇𝑥_𝑛,𝑇𝑥_𝑛)≤𝐺(𝑥_𝑛,𝑥,𝑥)+𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝐺(𝑇𝑥,𝑇𝑥_𝑛,𝑇𝑥_𝑛). (2.29) So, by (2.28) and (2.29), we have 𝐺(𝑇𝑥_𝑛,𝑇𝑦,𝑇𝑦)≤2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝑘𝐺(𝑥_𝑛,𝑥,𝑥)+𝑘𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝑘𝐺(𝑇𝑥,𝑇𝑥_𝑛,𝑇𝑥_𝑛), (2.30) then from (2.27) and (2.30) we have 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤𝑘𝐺(𝑥_𝑛,𝑥,𝑥)+𝑘𝐺(𝑥,𝑇𝑥,𝑇𝑥)+(1+𝑘)𝐺(𝑇𝑥,𝑇𝑥_𝑛,𝑇𝑥_𝑛)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦). (2.31) By the same argument we deduce that 𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦)≤𝑘𝐺(𝑧_𝑛,𝑧,𝑧)+𝑘𝐺(𝑧,𝑇𝑧,𝑇𝑧)+(1+𝑘)𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦). (2.32) Then, from (2.31) and (2.32), we see (2.26) becomes 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)+𝐺(𝑇𝑧,𝑇𝑦,𝑇𝑦)≤𝑘𝐺(𝑥_𝑛,𝑥,𝑥)+𝑘𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝑘𝐺(𝑇𝑥,𝑇𝑥_𝑛,𝑇𝑥_𝑛)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝑘𝐺(𝑧_𝑛,𝑧,𝑧)+𝑘𝐺(𝑧,𝑇𝑧,𝑇𝑧)+𝑘𝐺(𝑇𝑧,𝑇𝑧_𝑛,𝑇𝑧_𝑛)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦). (2.33)
Now letting 𝑛→∞ in the above inequality and using the fact that 𝑇 is 𝐺-continuous we get 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝑘{𝐺(𝑥,𝑇𝑥,𝑇𝑥)+4𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝐺(𝑧,𝑇𝑧,𝑇𝑧)}. (2.34)
So, in all cases we have for all 𝑥,𝑦,𝑧∈𝑋𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑧)≤𝑎𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝑏𝐺(𝑦,𝑇𝑦,𝑇𝑦)+𝑐𝐺(𝑧,𝑇𝑧,𝑇𝑧), (2.35) where 𝑎=𝑘,𝑏=4𝑘,𝑐=𝑘, and 𝑎+𝑏+𝑐<1 since 0<𝑘<1/6, then by Theorem 2.1, 𝑇 has a unique fixed point.

Corollary 2.6. Let (𝑋,𝐺) be 𝐺-metric space and let 𝑇∶𝑋→𝑋 be a mapping such that 𝑇 satisfies that(C1)𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤𝑎𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝑏𝐺(𝑦,𝑇𝑦,𝑇𝑦) for all 𝑥,𝑦∈𝑋 where 0<𝑎+𝑏<1,(C2)𝑇 is 𝐺-continuous at a point 𝑢∈𝑋, (C3) there is 𝑥∈𝑋;{𝑇^𝑛(𝑥)} has a subsequence {𝑇^𝑛𝑖(𝑥)}𝐺-converges to 𝑢. Then 𝑢 is a unique fixed point.

Proof. Let 𝑧=𝑦 in condition (A1), then we see that every mapping satisfies condition (C1) will satisfy condition (A1), so the proof follows from Theorem 2.1.

Corollary 2.7. Let (𝑋,𝐺) be 𝐺-metric space and let 𝑇∶𝑋→𝑋 be a 𝐺-continuous mapping satisfies that(D1)𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤𝑘{𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝐺(𝑦,𝑇𝑦,𝑇𝑦)} for all 𝑥,𝑦∈𝑀 where 𝑀 is an every where dense subset of 𝑋 (with respect the topology of 𝐺-metric convergence) and 0<𝑘<1/6,(D2) there is 𝑥∈𝑋 such that {𝑇^𝑛(𝑥)}→𝑥_𝑜. Then 𝑥_𝑜 is unique fixed point.

Proof. Let 𝑧=𝑦 in condition (B1), then we see that every mapping satisfies condition (D1) will satisfy condition (B1), so the proof follows from Theorem 2.5.

Corollary 2.8. Let (𝑋,𝐺) be 𝐺-metric space and let 𝑇∶𝑋→𝑋 be a mapping such that 𝑇 satisfies that(E1)𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤𝑘𝐺(𝑥,𝑦,𝑦) for all 𝑥,𝑦∈𝑋 where 0<𝑘<1/4,(E2)𝑇 is 𝐺-continuous at a point 𝑢∈𝑋, (E3) there is 𝑥∈𝑋;{𝑇^𝑛(𝑥)} has a subsequence {𝑇^𝑛𝑖(𝑥)}𝐺-converges to 𝑢. Then 𝑢 is a unique fixed point.

Proof. By axioms of 𝐺-metric (G5), we have 𝐺(𝑥,𝑦,𝑦)≤𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)+𝐺(𝑇𝑦,𝑦,𝑦),𝐺(𝑇𝑦,𝑦,𝑦)≤2𝐺(𝑦,𝑇𝑦,𝑇𝑦), (2.36) so, from (2.36), we see that (E1) becomes 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤𝑘𝐺(𝑥,𝑦,𝑦)≤𝑘𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝑘𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)+2𝑘𝐺(𝑦,𝑇𝑦,𝑇𝑦), (2.37) then 𝑇 will satisfy the following condition 𝐺(𝑇𝑥,𝑇𝑦,𝑇𝑦)≤𝑎𝐺(𝑥,𝑇𝑥,𝑇𝑥)+𝑏𝐺(𝑦,𝑇𝑦,𝑇𝑦) (2.38) for all 𝑥,𝑦∈𝑋, where 𝑎=𝑘/(1−𝑘), 𝑏=2𝑘/(1−𝑘), and 𝑎+𝑏<1, since 𝑘<1/4.
So, condition (C1) is satisfied and the proof follows from Corollary 2.6.

Acknowledgment

The authors are very grateful to the referees for their helpful suggestions.