A Common Fixed Point Theorem for Two Hybrid Pairs of Mappings in 𝑏-Metric Spaces

Abstract

We introduce a condition (W.C.C) and prove a common fixed point theorem for two hybrid pairs of mappings in b-metric spaces.

1. Introduction and Preliminaries

The study of fixed points for set-valued contraction maps using the Hausdorff metric was initiated by Nadler [1] in 1969, who extended the Banach contraction principle to set-valued mappings. The theory of set-valued maps has many applications in control theory, convex optimization, differential equations, and economics.

Recently Amini-Harandi [2] proved a set-valued version of Ciric's [3] theorem as follows.

Theorem 1. Let (𝑋,𝑑) be a complete metric space. Suppose 𝑇∶𝑋→𝐶𝐵(𝑋) satisfy 𝐻𝑑(),𝑑,𝑇𝑥,𝑇𝑦≤𝑞max𝑥,𝑦,𝑑𝑥,𝑇𝑥,𝑑𝑦,𝑇𝑦𝑥,𝑇𝑦,𝑑𝑦,𝑇𝑥 (1) for all 𝑥,𝑦∈𝑋, where 0≤𝑞<1/2. Then 𝑇 has a fixed point in 𝑋.

Aydi et al. [4] proved a slight version of the above theorem in 𝑏-metric spaces as follows.

Theorem 2. Let (𝑋,𝑑) be a complete 𝑏-metric space. Suppose 𝑇∶𝑋→𝐶𝐵(𝑋) satisfy 𝐻𝑑(),𝑑,𝑇𝑥,𝑇𝑦≤𝑞max𝑥,𝑦,𝑑𝑥,𝑇𝑥,𝑑𝑦,𝑇𝑦𝑥,𝑇𝑦,𝑑𝑦,𝑇𝑥 (2) for all 𝑥,𝑦∈𝑋, where 0≤𝑞<1/(𝑠²+𝑠). Then 𝑇 has a fixed point in 𝑋.

In this paper, we introduce a new condition, namely, “condition (W.C.C)” of mappings which are not continuous and non commuting pairwise. Using this new condition we prove a common fixed point theorem for two pairs of hybrid mappings in 𝑏-metric spaces.

For the sequel, we need the following concepts.

Czerwik [5] presented a generalization of the well-known Banach fixed point theorem in 𝑏-metric spaces by defining the following.

Definition 3. Let 𝑋 be a nonempty set and 𝑠≥1 a given real number. A function 𝑑∶𝑋×𝑋→ℝ⁺ is called a 𝑏-metric if the following axioms are satisfied for all 𝑥,𝑦,𝑧∈𝑋:(𝑏₁)𝑑(𝑥,𝑦)=0 if and only if 𝑥=𝑦, (𝑏₂)𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥), (𝑏₃)𝑑(𝑥,𝑦)≤𝑠[𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)].
Let (𝑋,𝑑) be a 𝑏-metric space. Let CB(𝑋) be the collection of all non-empty closed and bounded subsets of 𝑋. For 𝐴,𝐵∈CB(𝑋), define 𝐻(𝐴,𝐵)=max{𝛿(𝐴,𝐵),𝛿(𝐵,𝐴)}, (3) where 𝛿(){𝑑()},𝛿(){𝑑()},𝐴,𝐵=sup𝑎,𝐵∶𝑎∈𝐴𝐵,𝐴=sup𝑏,𝐴∶𝑏∈𝐵 (4) with 𝑑(𝑎,𝐵)=inf{𝑑(𝑎,𝑏)∶𝑏∈𝐵}. (5) Then 𝐻 is called the Hausdorff metric induced by the 𝑏-metric.
We cite the following lemmas from Czerwik [6]. For more details we refer to [7, 8].

Lemma 4. Let (𝑋,𝑑) be a 𝑏-metric space. For any 𝐴,𝐵,𝐶∈𝐶𝐵(𝑋) and any 𝑥,𝑦∈𝑋, we have the following: (i)𝑑(𝑥,𝐵)≤𝑑(𝑥,𝑏) for any 𝑏∈𝐵, (ii)𝛿(𝐴,𝐵)≤𝐻(𝐴,𝐵),(iii)𝑑(𝑥,𝐵)≤𝐻(𝐴,𝐵) for any 𝑥∈𝐴,(iv)𝐻(𝐴,𝐴)=0,(v)𝐻(𝐴,𝐵)=𝐻(𝐵,𝐴),(vi)𝐻(𝐴,𝐶)≤𝑠[𝐻(𝐴,𝐵)+𝐻(𝐵,𝐶)],(vii)𝑑(𝑥,𝐴)≤𝑠[𝑑(𝑥,𝑦)+𝑑(𝑦,𝐴)].

Lemma 5. Let (𝑋,𝑑) be a 𝑏-metric space. Let 𝐴,𝐵∈𝐶𝐵(𝑋). Then for each 𝜖>0 and for all 𝑏∈𝐵 there exists 𝑎∈𝐴 such that 𝑑(𝑎,𝑏)≤𝐻(𝐴,𝐵)+𝜖.

Lemma 6. Let (𝑋,𝑑) be a 𝑏-metric space. For 𝐴∈𝐶𝐵(𝑋) and 𝑥∈𝑋, we have 𝑑(𝑥,𝐴)=0⇔𝑥∈𝐴=𝐴.

In this paper, we introduce the following condition on three mappings without continuity and commutativity (𝑊.𝐶.𝐶).

Definition 7. Let (𝑋,𝑑) be a 𝑏-metric space. The mappings (𝑓,𝑔;𝑆) where 𝑓,𝑔∶𝑋→𝑋 and 𝑆∶𝑋→CB(𝑋) are said to satisfy the condition (𝑊.𝐶.𝐶) if 𝑑(𝑓𝑥,𝑔𝑦)≤𝑑(𝑦,𝑆𝑥) for all 𝑥,𝑦∈𝑋.

Remark 8. The mappings (𝑓,𝑔;𝑆) satisfying the condition (𝑊.𝐶.𝐶) need not be continuous and commutative in view of the following examples.

Example 9. Let (𝑋,𝑑) be a 𝑏-metric space, where 𝑋=[0,1] and 𝑑(𝑥,𝑦)=|𝑥−𝑦|².
Define 𝑓,𝑔,𝑆∶𝑋→𝑋 by 𝑥𝑓𝑥=12if[),1𝑥∈0,124if𝑥=1,𝑥𝑔𝑥=6if[),1𝑥∈0,112if𝑥=1,𝑥𝑆𝑥=2if[),1𝑥∈0,14if𝑥=1. (6)(7)(8) We distinguish the following cases.

Case 1. Consider 0≤𝑥<1and0≤𝑦<1Then𝑑=𝑥𝑓𝑥,𝑔𝑦−𝑦126²=1𝑥362−𝑦²=1𝑑.36𝑦,𝑆𝑥 (9)

Case 2. Consider 0≤𝑥<1and𝑦=1Then𝑑=𝑥𝑓𝑥,𝑔𝑦−11212²=11362−𝑥2²<1𝑑.36𝑦,𝑆𝑥 (10)

Case 3. Consider 𝑥=1and0≤𝑦<1Then𝑑=1𝑓𝑥,𝑔𝑦−𝑦246²=1136𝑦−4²=1𝑑.36𝑦,𝑆𝑥 (11)

Case 4. Consider 𝑥=1and𝑦=1Then𝑑=1𝑓𝑥,𝑔𝑦−12412²=1<9576.16=𝑑𝑦,𝑆𝑥 (12) Thus 𝑑(𝑓𝑥,𝑔𝑦)≤𝑑(𝑦,𝑆𝑥) for all 𝑥,𝑦∈𝑋 is satisfied.
In this example all the mappings are discontinuous.

Example 10. Let 𝑑(𝑥,𝑦)=|𝑥−𝑦|² be a 𝑏-metric on a set 𝐴={𝑥,𝑦}. Define 𝑓𝑥=𝑓𝑦=𝑦, 𝑔𝑥=𝑦, 𝑔𝑦=𝑥, and 𝑆𝑥=𝑆𝑦=𝑥. Clearly the mappings (𝑓,𝑔;𝑆) satisfy the condition (𝑊.𝐶.𝐶) and the pairs (𝑓,𝑔), (𝑓,𝑆), and (𝑔,𝑆) are not commuting.

Now, we prove our main result with the condition (𝑊.𝐶.𝐶) for two hybrid pairs of mappings in 𝑏-metric spaces.

2. Main Result

Theorem 11. Let (𝑋,𝑑) be a complete 𝑏-metric space. Let 𝑓,𝑔∶𝑋→𝑋 and 𝑆,𝑇∶𝑋→𝐶𝐵(𝑋) be mappings satisfying the following:(2.1.1) 𝐻(𝑆𝑥,𝑇𝑦)≤𝑞𝑀(𝑥,𝑦), where 𝑀(𝑥,𝑦)=max{𝑑(𝑓𝑥,𝑔𝑦),𝑑(𝑓𝑥,𝑆𝑥), 𝑑(𝑔𝑦,𝑇𝑦), 𝑑(𝑓𝑥,𝑇𝑦), 𝑑(𝑔𝑦,𝑆𝑥)} for all 𝑥,𝑦∈𝑋, where 0≤𝑞<1/(𝑠²+𝑠), (2.1.2)⋃_𝑥∈𝑋𝑆𝑥⊆𝑔(𝑋)and⋃_𝑥∈𝑋𝑇𝑥⊆𝑓(𝑋),(2.1.3)the mappings (𝑓,𝑔;𝑆) or the mappings (𝑓,𝑔;𝑇) satisfy the condition (𝑊.𝐶.𝐶).

Then either the pairs (𝑓,𝑆), and (𝑔,𝑇) have coincidence points or the mappings 𝑆,𝑇,𝑓 and 𝑔 have a common fixed point in 𝑋.

Proof. Suppose 𝑀(𝑥,𝑦)=0 for some 𝑥,𝑦∈𝑋.
Then 𝑓𝑥∈𝑆𝑥 and 𝑔𝑦∈𝑇𝑦. Thus 𝑥 is a coincidence point of 𝑓 and 𝑆, and 𝑦 is a coincidence point of 𝑔 and 𝑇.
Assume that 𝑀(𝑥,𝑦)>0 for all 𝑥,𝑦∈𝑋.
Let 𝜖=(1/2)(1/(𝑠²+𝑠)−𝑞) and 𝛽=𝑞+𝜖=(1/2)(1/(𝑠²+𝑠)+𝑞).
Then 𝜖>0 and 0<𝛽<1.
Let 𝑥₀∈𝑋. From (2.1.2), there exist 𝑥₁,𝑦₁∈𝑋 such that 𝑦₁=𝑔𝑥₁∈𝑆𝑥₀.
From (2.1.2), there exist 𝑥₂,𝑦₂∈𝑋 such that 𝑦₂=𝑓𝑥₂∈𝑇𝑥₁ and from Lemma 5, we have 𝑑𝑦₁,𝑦₂≤𝐻𝑆𝑥₀,𝑇𝑥₁+𝜖𝑀𝑥₀,𝑥₁≤𝑞𝑀𝑥₀,𝑥₁+𝜖𝑀𝑥₀,𝑥₁,from()𝑥2.1.1=𝛽𝑀₀,𝑥₁. (13) Again from (2.1.2), there exist 𝑥₃,𝑦₃∈𝑋 such that 𝑦₃=𝑔𝑥₃∈𝑆𝑥₂ and from Lemma 5, we have 𝑑𝑦₂,𝑦₃≤𝐻𝑆𝑥₂,𝑇𝑥₁+𝜖𝑀𝑥₂,𝑥₁≤𝑞𝑀𝑥₂,𝑥₁+𝜖𝑀𝑥₂,𝑥₁,from()𝑥2.1.1=𝛽𝑀₂,𝑥₁. (14) Continuing in this way, we get the sequences {𝑥_𝑛} and {𝑦_𝑛} in 𝑋 such that 𝑦_2𝑛+1=𝑔𝑥_2𝑛+1∈𝑆𝑥_2𝑛, 𝑦_2𝑛+2=𝑓𝑥_2𝑛+2∈𝑇𝑥_2𝑛+1 for all 𝑛=0,1,2,3,… and 𝑑𝑦_2𝑛+1,𝑦_2𝑛≤𝛽𝑀𝑥_2𝑛,𝑥_2𝑛−1,𝑑𝑦_2𝑛+1,𝑦_2𝑛+2≤𝛽𝑀𝑥_2𝑛,𝑥_2𝑛+1for𝑛=1,2,3,…. (15)
For simplification, write 𝑑_𝑛=𝑑(𝑦_𝑛,𝑦_𝑛+1).
Now 𝑑_2𝑛+1=𝑑𝑦_2𝑛+1,𝑦_2𝑛+2≤𝛽𝑀𝑥_2𝑛,𝑥_2𝑛+1=𝛽max𝑑𝑦_2𝑛,𝑦_2𝑛+1,𝑑𝑦_2𝑛,𝑦_2𝑛+1,𝑑𝑦_2𝑛+1,𝑦_2𝑛+2,𝑑𝑦_2𝑛,𝑦_2𝑛+2𝑑,0≤𝛽max_2𝑛,𝑑_2𝑛+1,𝑠𝑑_2𝑛+𝑑_2𝑛+1 (16) which in turn yields 𝑑_2𝑛+1≤𝛾𝑑_2𝑛, (17) where 𝛾=max{𝛽,𝑠𝛽/(1−𝑠𝛽)}.
Similarly, we can show that 𝑑_2𝑛≤𝛾𝑑_2𝑛−1. (18) Thus, from (17) and (18), we have 𝑑_𝑛≤𝛾𝑑_𝑛−1 for all 𝑛=1,2,3,…, so that 𝑑_𝑛≤𝛾^𝑛𝑑₀∀𝑛=1,2,3,…. (19) If 𝛾=𝛽, then 𝛾𝑠=𝛽𝑠=(𝑠/2)(1/(𝑠²+𝑠)+𝑞)<𝑠/(𝑠²+𝑠)=1/(𝑠+1)≤1/2<1.
If 𝛾=𝑠𝛽/(1−𝑠𝛽), then 𝛾𝑠=𝑠(𝑠𝛽/(1−𝑠𝛽))<𝑠[1/(𝑠+1)/1−1/(𝑠+1)]=1.
Thus 𝛾𝑠<1. (20) Since 𝛾<1, from (19), we have 𝑑_𝑛⟶0as𝑛⟶∞. (21) Now for 𝑚,𝑛∈ℕ with 𝑚>𝑛, we have 𝑑𝑦_𝑛,𝑦_𝑚≤𝑠𝑑𝑦_𝑛,𝑦_𝑛+1+𝑠²𝑑𝑦_𝑛+1,𝑦_𝑛+2+⋯+𝑠^𝑚−𝑛𝑑𝑦_𝑚−1,𝑦_𝑚≤𝑠𝛾^𝑛+𝑠²𝛾^𝑛+1+⋯+𝑠^𝑚−𝑛𝛾^𝑚−1𝑑₀,from(19)≤𝑠𝛾^𝑛1+𝑠𝛾+⋯+𝑠^{𝑚−𝑛−1}𝛾^{𝑚−𝑛−1}+⋯𝑑₀=𝛾^𝑛𝑠𝑑1−𝛾𝑠⁰→0as𝑛,𝑚→∞. (22) Hence {𝑦_𝑛} is a Cauchy sequence in 𝑋. Since 𝑋 is complete, there exists 𝑧∈𝑋 such that 𝑦_𝑛→𝑧 which in turn yields 𝑓𝑥_2𝑛→𝑧 and 𝑔𝑥_2𝑛+1→𝑧 as 𝑛→∞.
Suppose the mappings (𝑓,𝑔;𝑆) satisfy the condition (𝑊.𝐶.𝐶). Then 𝑑𝑓𝑥,𝑔𝑦≤𝑑𝑦,𝑆𝑥∀𝑥,𝑦∈𝑋. (23) From (23), we have 𝑑𝑓𝑥_2𝑛,𝑔𝑧≤𝑑𝑧,𝑆𝑥_2𝑛≤𝑑𝑧,𝑔𝑥_2𝑛+1. (24) Letting 𝑛→∞, we get 𝑑𝑧,𝑔𝑧≤0sothat𝑔𝑧=𝑧. (25)
Now 1𝑠𝑑(𝑧,𝑇𝑧)≤𝑑𝑧,𝑔𝑥_2𝑛+1+𝑑𝑔𝑥_2𝑛+1,𝑇𝑧≤𝑑𝑧,𝑔𝑥_2𝑛+1+𝐻𝑆𝑥_2𝑛,𝑇𝑧≤𝑑𝑧,𝑦_2𝑛+1𝑑+𝑞max𝑓𝑥_2𝑛,𝑔𝑧,𝑑𝑓𝑥_2𝑛,𝑆𝑥_2𝑛,𝑑,𝑑𝑔𝑧,𝑇𝑧𝑓𝑥_2𝑛,𝑇𝑧,𝑑𝑔𝑧,𝑆𝑥_2𝑛≤𝑑𝑧,𝑦_2𝑛+1𝑑𝑦+𝑞max^2𝑛,𝑧,𝑑𝑦_2𝑛,𝑦_2𝑛+1(),𝑑𝑦,𝑑𝑧,𝑇𝑧_2𝑛,𝑇𝑧,𝑑𝑧,𝑦_2𝑛+1. (26) Using (vii) of Lemma 4 and letting 𝑛→∞ and using (21), we get 1𝑠𝑑(𝑧,𝑇𝑧)≤0+𝑞𝑠𝑑(𝑧,𝑇𝑧). (27) Thus we have 𝑑(𝑧,𝑇𝑧)≤𝑞𝑠²𝑑(𝑧,𝑇𝑧), (28) so that 𝑑(𝑧,𝑇𝑧)=0since𝑞𝑠²<1. (29) From Lemma 6, we have 𝑧∈𝑇𝑧. Thus 𝑔𝑧=𝑧∈𝑇𝑧. (30) From (23), we get 𝑑𝑓𝑧,𝑧=𝑑𝑓𝑧,𝑔𝑧≤𝑑(𝑧,𝑆𝑧). (31) Now from (2.1.1), we have 𝑑()()𝑑,𝑑{𝑑()[𝑑()(,()(𝑧,𝑆𝑧≤𝐻𝑆𝑧,𝑇𝑧≤𝑞max𝑓𝑧,𝑔𝑧,𝑑𝑓𝑧,𝑆𝑧,𝑑𝑔𝑧,𝑇𝑧𝑓𝑧,𝑇𝑧,𝑑𝑔𝑧,𝑆𝑧≤𝑞max𝑧,𝑆𝑧,𝑠𝑧,𝑆𝑧+𝑑𝑧,𝑆𝑧)]0,𝑠𝑑𝑧,𝑆𝑧,𝑑𝑧,𝑆𝑧)}from(30)and()().31=2𝑞𝑠𝑑𝑧,𝑆𝑧 (32)
Hence 𝑧∈𝑆𝑧, since 2𝑞𝑠<1.
From (31), we have 𝑓𝑧=𝑧. Thus 𝑓𝑧=𝑧∈𝑆𝑧. (33) From (30) and (33), it follows that 𝑧 is a common fixed point of 𝑓,𝑔,𝑆, and 𝑇.

Acknowledgment

The authors are thankful to the referee for his valuable suggestions.