Common Fixed Point Results for Mappings with Rational Expressions

Abstract

We obtain some common fi…xed point results for single as well as set valued mappings involving certain rational expressions in complete partial metric spaces. In the process, we generalize various results of the literature. Two examples are also included to illustrate the fact that our results cannot be obtained from the corresponding results in metric spaces.

1. Introduction and Preliminaries

In 1994, Matthews [1] introduced the concept of a partial metric space and obtained a Banach type fixed point theorem on a complete partial metric space. Later on, several authors (see, e.g., [1–28]) proved fixed point theorems in partial metric spaces. After the definition of the Partial Hausdorff metric, Aydi et al. [9] proved Banach type fixed point result for set valued mappings in complete partial metric space. Here, we prove some common fixed point results for single as well as set valued mappings involving certain rational expressions in complete partial metric spaces and show by examples that the results proved in this paper cannot be deduced from the corresponding results in metric spaces (see Example 10, Remark 13).

We start with recalling some basic definitions and lemmas on partial metric space. The definition of a partial metric space is given by Matthews (see [1]) as follows.

Definition 1. A partial metric on a nonempty set 𝑋 is a function 𝑝∶𝑋×𝑋→[0,∞) such that for all 𝑥,𝑦,𝑧∈𝑋: (P₁)𝑝(𝑥,𝑥)=𝑝(𝑦,𝑦)=𝑝(𝑥,𝑦) if and only if 𝑥=𝑦,(P₂)𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦),(P₃)𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥),(P₄)𝑝(𝑥,𝑧)≤𝑝(𝑥,𝑦)+𝑝(𝑦,𝑧)−𝑝(𝑦,𝑦).

The pair (𝑋,𝑝) is then called a partial metric space.

If (𝑋,𝑝) is a partial metric space, then the function 𝑝^𝑠∶𝑋×𝑋→ℝ⁺ given by 𝑝^𝑠(𝑥,𝑦)=2𝑝(𝑥,𝑦)−𝑝(𝑥,𝑥)−𝑝(𝑦,𝑦), 𝑥,𝑦∈𝑋, is a metric on 𝑋.

A basic example of a partial metric space is the pair (𝑅⁺,𝑝), where 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑅⁺.

Lemma 2 (see [1]). Let (𝑋,𝑝) be a partial metric space; then one has the following.(1)A sequence {𝑥_𝑛} in a partial metric space (𝑋,𝑝) converges to a point 𝑥∈𝑋 if and only if lim_𝑛→∞𝑝(𝑥,𝑥_𝑛)=𝑝(𝑥,𝑥).(2)A sequence {𝑥_𝑛} in a partial metric space (𝑋,𝑝) is called a Cauchy sequence if the lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚) exists and is finite.(3)A partial metric space (𝑋,𝑝) is said to be complete if every Cauchy sequence {𝑥_𝑛} in 𝑋 converges to a point 𝑥∈𝑋; that is, 𝑝(𝑥,𝑥)=lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).(4)A partial metric space (𝑋,𝑝) is complete if and only if the metric space (𝑋,𝑝^𝑠) is complete. Furthermore, lim_𝑛→∞𝑝^𝑠(𝑥_𝑛,𝑧)=0 if and only if 𝑝(𝑧,𝑧)=lim_𝑛→∞𝑝(𝑥_𝑛,𝑧)= lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).

Remark 3 (see [1]). Let (𝑋,𝑝) be a partial metric space and let 𝐴 be a nonempty set in (𝑋,𝑝); then 𝑎∈𝐴 if and only if 𝑝(𝑎,𝐴)=𝑝(𝑎,𝑎), (1) where 𝐴 denotes the closure of 𝐴 with respect to the partial metric 𝑝. Note 𝐴 is closed in (𝑋,𝑝) if and only if 𝐴=𝐴.

Definition 4 (see [24]). Two families of self-mappings {𝑇_𝑖}^𝑚₁ and {𝑆_𝑖}^𝑛₁ are said to be pairwise commuting if(1)𝑇_𝑖𝑇_𝑗=𝑇_𝑗𝑇_𝑖,𝑖,𝑗∈{1,2,…,𝑚};(2)𝑆_𝑘𝑆_𝑙=𝑆_𝑙𝑆_𝑘,𝑘,𝑙∈{1,2,…,𝑛};(3)𝑇_𝑖𝑆_𝑘=𝑆_𝑘𝑇_𝑖,𝑖∈{1,2,…,𝑚}, 𝑘∈{1,2,…,𝑛}.

Now we recall the following definitions and results from [9].

Let 𝐶𝐵^𝑝(𝑋) be the collection of all nonempty, closed, and bounded subsets of 𝑋 with respect to the partial metric 𝑝. For 𝐴∈𝐶𝐵^𝑝(𝑋), we define 𝑝(𝑎,𝐴)=inf𝑝(𝑎,𝑥)∶𝑥∈𝐴. (2) For 𝐴,𝐵∈𝐶𝐵^𝑝(𝑋), 𝛿_𝑝()𝑝(),𝛿𝐴,𝐵=sup𝑎,𝐵∶𝑎∈𝐴_𝑝(𝐵,𝐴)=sup𝑝(𝑏,𝐴)∶𝑏∈𝐵. (3) For 𝐴,𝐵∈𝐶𝐵^𝑝(𝑋), 𝐻_𝑝(𝐴,𝐵)=max𝛿_𝑝(𝐴,𝐵),𝛿_𝑝(𝐵,𝐴). (4)

Proposition 5 (see [9]). Let (𝑋,𝑝) be a partial metric space. For any 𝐴,𝐵,𝐶∈𝐶𝐵^𝑝(𝑋), one has(i)𝛿_𝑝(𝐴,𝐴)=sup{𝑝(𝑎,𝑎)∶𝑎∈𝐴};(ii)𝛿_𝑝(𝐴,𝐴)≤𝛿_𝑝(𝐴,𝐵);(iii)𝛿_𝑝(𝐴,𝐵)=0  implies that  𝐴⊆𝐵; (iv)𝛿_𝑝(𝐴,𝐵)≤𝛿_𝑝(𝐴,𝐶)+𝛿_𝑝(𝐶,𝐵)−inf_𝑐∈𝐶𝑝(𝑐,𝑐).

Proposition 6 (see [9]). Let (𝑋,𝑝) be a partial metric space. For any 𝐴,𝐵,𝐶∈𝐶𝐵^𝑝(𝑋), one has(h₁)𝐻_𝑝(𝐴,𝐴)≤𝐻_𝑝(𝐴,𝐵);(h₂)𝐻_𝑝(𝐴,𝐵)=𝐻_𝑝(𝐵,𝐴);(h₃)𝐻_𝑝(𝐴,𝐵)≤𝐻_𝑝(𝐴,𝐶)+𝐻_𝑝(𝐶,𝐵)−inf_𝑐∈𝐶𝑝(𝑐,𝑐).

Lemma 7 (see [9]). Let 𝐴 and 𝐵 be nonempty, closed, and bounded subsets of a partial metric space (𝑋,𝑝) and ℎ>1. Then, for every 𝑎∈𝐴, there exists 𝑏∈𝐵 such that 𝑝(𝑎,𝑏)≤ℎ𝐻_𝑝(𝐴,𝐵).

Lemma 8 (see [10]). Let 𝐴 and 𝐵 be nonempty, closed, and bounded subsets of a partial metric space (𝑋,𝑝) and 0<ℎ∈ℝ. Then, for every 𝑎∈𝐴, there exists 𝑏∈𝐵 such that 𝑝(𝑎,𝑏)≤𝐻_𝑝(𝐴,𝐵)+ℎ.

2. Results for Single Valued Mappings

The following result, regarding the existence of the common fixed point of the mappings satisfying a contractive condition on the closed ball, is very useful in the sense that it requires the contractiveness of the mappings only on the closed ball instead of the whole space.

Theorem 9. Let 𝑆,𝑇∶𝑋→𝑋 be mappings on a complete PMS (𝑋,𝑝) and 𝑥₀,𝑥,𝑦∈𝑋 and 𝑟>0. Suppose that there exist nonnegative reals 𝛼, 𝛽, and 𝛾 such that 𝛼+𝛽+2𝛾<1. If 𝑆 and 𝑇 satisfy 𝑝+()𝑝𝑝𝑆𝑥,𝑇𝑦≤𝛼𝑝𝑥,𝑦𝛽𝑝𝑥,𝑆𝑥𝑦,𝑇𝑦+𝛾𝑝𝑦,𝑆𝑥𝑥,𝑇𝑦1+𝑝𝑥,𝑦 (5) for all 𝑥,𝑦∈𝐵_𝑝(𝑥₀,𝑟), 𝑝𝑥₀,𝑆𝑥₀≤(1−𝜆)𝑟+𝑝𝑥₀,𝑥₀, (6) where 𝜆=(𝛼+𝛾)/(1−𝛽−𝛾). Then there exists a unique point 𝑢∈𝐵_𝑝(𝑥₀,𝑟) such that 𝑢=𝑆𝑢=𝑇𝑢. Also 𝑝(𝑢,𝑢)=0.

Proof. Let 𝑥₀ be an arbitrary point in 𝑋 and define 𝑥_2𝑘+1=𝑆𝑥_2𝑘,𝑥_2𝑘+2=𝑇𝑥_2𝑘+1,where𝑘=0,1,2,…. (7) We will prove that 𝑥_𝑛∈(𝐵(𝑥₀,𝑟)) for all 𝑛∈ℕ by mathematical induction. Using inequality (6) and the fact that 𝜆=(𝛼+𝛾)/(1−𝛽−𝛾)<1, we have 𝑝𝑥₀,𝑆𝑥₀≤𝑟+𝑝𝑥₀,𝑥₀. (8) It implies that 𝑥₁∈(𝐵(𝑥₀,𝑟)). Let 𝑥₂,…,𝑥_𝑗∈𝐵(𝑥₀,𝑟) for some 𝑗∈𝑁. If 𝑗=2𝑘+1, where 𝑘=0,1,2,…(𝑗−1)/2, so using inequality (5), we obtain 𝑝𝑥_2𝑘+1,𝑥_2𝑘+2=𝑝𝑆𝑥_2𝑘,𝑇𝑥_2𝑘+1≤𝛼𝑝𝑥_2𝑘,𝑥_2𝑘+1+𝛽𝑝𝑥_2𝑘,𝑆𝑥_2𝑘𝑝𝑥_2𝑘+1,𝑇𝑥_2𝑘+1+𝛾𝑝𝑥_2𝑘+1,𝑆𝑥_2𝑘𝑝𝑥_2𝑘,𝑇𝑥_2𝑘+1×1+𝑝𝑥_2𝑘,𝑥_2𝑘+1⁻¹≤𝛼𝑝𝑥_2𝑘,𝑥_2𝑘+1+𝛽𝑝𝑥_2𝑘,𝑥_2𝑘+1𝑝𝑥_2𝑘+1,𝑥_2𝑘+2+𝛾𝑝𝑥_2𝑘+1,𝑥_2𝑘+1𝑝𝑥_2𝑘,𝑥_2𝑘+2×1+𝑝𝑥_2𝑘,𝑥_2𝑘+1⁻¹≤𝛼𝑝𝑥_2𝑘,𝑥_2𝑘+1+𝛽𝑝𝑥_2𝑘,𝑥_2𝑘+1𝑝𝑥_2𝑘+1,𝑥_2𝑘+2+𝛾𝑝𝑥_2𝑘,𝑥_2𝑘+1𝑝𝑥_2𝑘,𝑥_2𝑘+2×1+𝑝𝑥_2𝑘,𝑥_2𝑘+1⁻¹≤𝛼𝑝𝑥_2𝑘,𝑥_2𝑘+1+𝛽𝑝𝑥_2𝑘+1,𝑥_2𝑘+2+𝛾𝑝𝑥_2𝑘,𝑥_2𝑘+2, (9) as 1+𝑝(𝑥_2𝑘,𝑥_2𝑘+1)>𝑝(𝑥_2𝑘,𝑥_2𝑘+1), and so 𝑝𝑥_2𝑘+1,𝑥_2𝑘+2≤𝛼+𝛾𝑝𝑥1−𝛽−𝛾^2𝑘,𝑥_2𝑘+1 (10) which implies that 𝑝𝑥_2𝑘+1,𝑥_2𝑘+2≤𝜆𝑝𝑥_2𝑘,𝑥_2𝑘+1≤⋯≤𝜆^2𝑘+1𝑝𝑥₀,𝑥₁. (11) If 𝑗=2𝑘+2 where 𝑘=0,1,2,…,(𝑗−2)/2, one can easily prove that 𝑝𝑥_2𝑘+2,𝑥_2𝑘+3≤𝜆^2𝑘+2𝑝𝑥₀,𝑥₁. (12) Thus from inequality (11) and (12), we have 𝑝𝑥_𝑗,𝑥_𝑗+1≤𝜆^𝑗𝑝𝑥₀,𝑥₁forsome𝑗∈𝑁. (13) Now 𝑝𝑥₀,𝑥_𝑗+1≤𝑝𝑥₀,𝑥₁+⋯+𝑝𝑥_𝑗,𝑥_𝑗+1−𝑝𝑥₁,𝑥₁+⋯+𝑝𝑥_𝑗,𝑥_𝑗≤𝑝𝑥₀,𝑥₁+⋯+𝜆^𝑗𝑝𝑥₀,𝑥₁=𝑝𝑥₀,𝑥₁1+⋯+𝜆^𝑗−1+𝜆^𝑗≤(1−𝜆)𝑟+𝑝𝑥₀,𝑥₀1−𝜆^𝑗+1𝑥1−𝜆≤𝑟+𝑝₀,𝑥₀ (14) gives 𝑥_𝑗+1∈𝐵(𝑥₀,𝑟). Hence 𝑥_𝑛∈𝐵(𝑥₀,𝑟) for all 𝑛∈ℕ. One can easily prove that 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜆^𝑛𝑝𝑥₀,𝑥₁ 13^∗ for all 𝑛∈ℕ. We now show that {𝑥_𝑛} is a Cauchy sequence. Without loss of generality assume that 𝑚>𝑛. Then, using 13^∗ and the triangle inequality for partial metrics (P₄) we have 𝑝𝑥_𝑛,𝑥_𝑚≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑚−𝑝𝑥_𝑛+1,𝑥_𝑛+1≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑚≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑛+2+𝑝𝑥_𝑛+2,𝑥_𝑚−𝑝𝑥_𝑛+2,𝑥_𝑛+2≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑛+2+𝑝𝑥_𝑛+2,𝑥_𝑚. (15) Inductively, we have 0≤𝑝𝑥_𝑛,𝑥_𝑚≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑛+2+⋯+𝑝𝑥_𝑚−1,𝑥_𝑚≤𝜆^𝑛𝑝𝑥₀,𝑥₁+𝜆^𝑛+1𝑝𝑥₀,𝑥₁+⋯+𝜆^𝑚−1𝑝𝑥₀,𝑥₁≤𝜆^𝑛+𝜆^𝑛+1+⋯+𝜆^𝑚−1𝑝𝑥₀,𝑥₁≤𝜆^𝑛𝑝𝑥1−𝜆⁰,𝑥₁⟶0as𝑛⟶∞(since0<𝜆<1). (16) Thus, lim_𝑛→+∞𝑝𝑥_𝑛,𝑥_𝑚=0. (17) By the definition of 𝑝^𝑠, we get for any 𝑚∈ℕ^∗, 𝑝^𝑠𝑥_𝑛,𝑥_𝑚≤2𝑝𝑥_𝑛,𝑥_𝑚⟶0as𝑛⟶+∞. (18) Hence the sequence {𝑥_𝑛} is a Cauchy sequence in (𝐵(𝑥₀,𝑟),𝑝^𝑠). By Lemma 2(4), {𝑥_𝑛} is a Cauchy sequence in (𝐵(𝑥₀,𝑟),𝑝). Therefore there exists a point 𝑢∈𝐵(𝑥₀,𝑟) with lim_𝑛→∞𝑥_𝑛=𝑢. Also lim_𝑛→∞𝑝^𝑠(𝑥_𝑛,𝑢)=0. Again from Lemma 2(4), we have 𝑝(𝑢,𝑢)=lim_𝑛→+∞𝑝𝑥_𝑛,𝑢=lim_𝑛→+∞𝑝𝑥_𝑛,𝑥_𝑚=0. (19) By the triangle inequality (P₄), we have 𝑝(𝑢,𝑆𝑢)≤𝑝𝑢,𝑥_2𝑛+2+𝑝𝑥_2𝑛+2,𝑆𝑢−𝑝𝑥_2𝑛+2,𝑥_2𝑛+2≤𝑝𝑢,𝑥_2𝑛+2+𝑝𝑇𝑥_2𝑛+1,𝑆𝑢≤𝑝𝑢,𝑥_2𝑛+2+𝛼𝑝𝑥_2𝑛+1+𝑥,𝑢𝛽𝑝_2𝑛+1,𝑇𝑥_2𝑛+1𝑝(𝑢,𝑆𝑢)+𝛾𝑝𝑢,𝑇𝑥_2𝑛+1𝑝𝑥_2𝑛+1×𝑥,𝑆𝑢1+𝑝_2𝑛+1,𝑢⁻¹≤𝑝𝑢,𝑥_2𝑛+2+𝛼𝑝𝑥_2𝑛+1+𝑥,𝑢𝛽𝑝_2𝑛+1,𝑥_2𝑛+2𝑝(𝑢,𝑆𝑢)+𝛾𝑝𝑢,𝑥_2𝑛+2𝑝𝑥_2𝑛+1×𝑥,𝑆𝑢1+𝑝_2𝑛+1,𝑢⁻¹. (20) Letting 𝑛→+∞ and using (19), we obtain 𝑝(𝑢,𝑆𝑢)=0. (21) By (P₁), we concluded that 𝑢=𝑆𝑢. It follows similarly that 𝑢=𝑇𝑢. To prove the uniqueness of common fixed point, let 𝑢^∗∈𝐵(𝑥₀,𝑟) be another common fixed point of 𝑆 and 𝑇, that, is 𝑢^∗=𝑆𝑢^∗=𝑇𝑢^∗. Then 𝑝𝑢,𝑢^∗=𝑝𝑆𝑢,𝑇𝑢^∗≤𝛼𝑝𝑢,𝑢^∗+𝛽𝑝(𝑢,𝑆𝑢)𝑝𝑢^∗,𝑇𝑢^∗+𝛾𝑝𝑢^∗,𝑆𝑢𝑝𝑢,𝑇𝑢^∗1+𝑝(𝑢,𝑢^∗)=𝛼𝑝𝑢,𝑢^∗+𝛾𝑝𝑢^∗,𝑢𝑝𝑢,𝑢^∗1+𝑝(𝑢,𝑢^∗) (22) so that 𝑝(𝑢,𝑢^∗)≤𝛼𝑝(𝑢,𝑢^∗)+𝛾𝑝(𝑢,𝑢^∗) because 1+𝑝(𝑢,𝑢^∗)>𝑝(𝑢,𝑢^∗). Therefore 𝑝(𝑢,𝑢^∗)≤(𝛼+𝛾)𝑝(𝑢,𝑢^∗) which is a contradiction so that 𝑢=𝑢^∗ (as 𝛼+𝛾<1). Hence 𝑆 and 𝑇 have a unique common fixed point in 𝐵(𝑥₀,𝑟).

Example 10. Let 𝑋=[0,+∞) endowed with the usual partial metric 𝑝 defined by 𝑝∶𝑋×𝑋→ℝ⁺ with 𝑝(𝑥,𝑦)=max{𝑥,𝑦}. Clearly, (𝑋,𝑝) is a partial metric space. Now we define 𝑆,𝑇∶𝑋→𝑋 as 𝑆(𝑥)=𝑥116if0≤𝑥≤1𝑥−6𝑇(𝑥)=if𝑥>15𝑥117if0≤𝑥≤1𝑥−7if𝑥>1 (23) for all 𝑥∈𝑋. Taking 𝛼=1/5, 𝛽=1/6, 𝛾=1/8, 𝑥₀=1/2, and 𝑟=1/2, then 𝐵_𝑝(𝑥₀,𝑟)=[0,1]. Also, we have 𝑝(𝑥₀,𝑥₀)=max{1/2,1/2}=1/2, 𝜆=(𝛼+𝛾)/(1−𝛽−𝛾)=39/85 with (1−𝜆)𝑟+𝑝𝑥₀,𝑥₀=46,𝑝𝑥85₀,𝑆𝑥₀1=𝑝2,1=1322<(1−𝜆)𝑟+𝑝𝑥₀,𝑥₀. (24) Also if 𝑥,𝑦∈(1,+∞), then 𝑝1𝑆𝑥,𝑇𝑦=max𝑥−61,𝑥−7≥15+1max𝑥,𝑦61max𝑥,𝑥−61max𝑦,𝑦−7+181max𝑦,𝑥−61max𝑥,𝑦−7×1+max𝑥,𝑦⁻¹+()𝑝𝑝=𝛼𝑝𝑥,𝑦𝛽𝑝𝑥,𝑆𝑥𝑦,𝑇𝑦+𝛾𝑝𝑦,𝑆𝑥𝑥,𝑇𝑦.1+𝑝𝑥,𝑦 (25) So the contractive condition does not hold on whole of 𝑋. Now if 𝑥,𝑦∈𝐵_𝑝(𝑥₀,𝑟), then 𝑝𝑥𝑆𝑥,𝑇𝑦=max,165𝑦≤1175+1max𝑥,𝑦61max𝑥,𝑥−61max𝑦,𝑦−7+181max𝑦,𝑥−61max𝑥,𝑦−7×1+max𝑥,𝑦⁻¹. (26) Therefore, all the conditions of Theorem 9 are satisfied. Thus 0 is the common fixed point of 𝑆 and 𝑇 and 𝑝(0,0)=0. Moreover, note that for any metric 𝑑 on 𝑋𝑑()1𝑆1,𝑇1=𝑑,516>1175𝑑()+11,16𝑑11,𝑑5161,+1178𝑑11,𝑑5161,×((171+𝑑1,1))⁻¹. (27) Therefore common fixed points of 𝑆 and 𝑇 cannot be obtained from a metric fixed point theorem.

Corollary 11. Let 𝑆,𝑇∶𝑋→𝑋 be mappings on a complete PMS (𝑋,𝑝). Suppose that there exist nonnegative reals 𝛼, 𝛽, and 𝛾 such that 𝛼+𝛽+2𝛾<1. If 𝑆 and 𝑇 satisfy 𝑝+()𝑝𝑝𝑆𝑥,𝑇𝑦≤𝛼𝑝𝑥,𝑦𝛽𝑝𝑥,𝑆𝑥𝑦,𝑇𝑦+𝛾𝑝𝑦,𝑆𝑥𝑥,𝑇𝑦1+𝑝𝑥,𝑦 (28) for all 𝑥,𝑦∈𝑋. Then there exists a unique point 𝑢∈𝑋 such that 𝑢=𝑆𝑢=𝑇𝑢. Also 𝑝(𝑢,𝑢)=0. Further 𝑆 and 𝑇 have no fixed point other than 𝑢.

By choosing 𝛽=𝛾=0 in Corollary 11, we get the following corollary.

Corollary 12. Let 𝑆, 𝑇∶𝑋→𝑋 be a mappings on complete PMS (𝑋,𝑝). If 𝑆 and 𝑇 satisfy 𝑝𝑆𝑥,𝑇𝑦≤𝛼𝑝𝑥,𝑦 (29) for all 𝑥,𝑦∈𝑋,𝛼<1 (𝛼 is a nonnegative real). Then 𝑆 and 𝑇 have a common fixed point 𝑢∈𝑋 and 𝑝(𝑢,𝑢)=0.

Remark 13. If we impose Banach type contractive condition for a pair 𝑆,𝑇∶𝑋→𝑋 of mappings on a metric space (𝑋,𝑑); that is, 𝑑(𝑆𝑥,𝑇𝑦)≤𝛼𝑑(𝑥,𝑦) for all 𝑥,𝑦∈𝑋, and then it follows that 𝑆𝑥=𝑇𝑥, for all 𝑥∈𝑋 (i.e., 𝑆 and 𝑇 are equal). Therefore the above condition fails to find common fixed points of 𝑆 and 𝑇. This can be seen as 𝑑()1𝑆1,𝑇1=𝑑,516()17>𝛼𝑑1,1=0. (30) However the same condition in partial metric space does not assert that 𝑆=𝑇. This can be seen as by taking the partial metric same as in Example 10, 𝑝()1𝑆1,𝑇1=𝑝,516=517()(1).17≤𝛼𝑝1,1=𝛼 (31) for any 𝛼≥1/3. Hence Corollary 12 cannot be obtained from a metric fixed point theorem.

Remark 14. By equating 𝛼, 𝛽, 𝛾 to 0 in all possible combinations, one can derive a host of corollaries which include Matthews theorem for mappings defined on a complete partial metric space.

By taking 𝑆=𝑇 in the Theorem 9, we get the following corollary.

Corollary 15. Let 𝑇∶𝑋→𝑋 be a mapping on a complete PMS (𝑋,𝑝) and 𝑥₀,𝑥,𝑦∈𝑋 and 𝑟>0. Suppose that there exist nonnegative reals 𝛼,𝛽, and 𝛾 such that 𝛼+𝛽+2𝛾<1. If 𝑇 satisfies 𝑝+()𝑝𝑝𝑇𝑥,𝑇𝑦≤𝛼𝑝𝑥,𝑦𝛽𝑝𝑥,𝑇𝑥𝑦,𝑇𝑦+𝛾𝑝𝑦,𝑇𝑥𝑥,𝑇𝑦1+𝑝𝑥,𝑦 (32) for all 𝑥,𝑦∈𝐵_𝑝(𝑥₀,𝑟), 𝑝𝑥₀,𝑇𝑥₀≤(1−𝜆)𝑟+𝑝𝑥₀,𝑥₀, (33) where 𝜆=(𝛼+𝛾)/(1−𝛽−𝛾). Then there exists a unique point 𝑢∈𝐵_𝑝(𝑥₀,𝑟) such that 𝑢=𝑇𝑢. Also 𝑝(𝑢,𝑢)=0. Further 𝑇 has no fixed point other than 𝑢.

By taking 𝑆=𝑇 in Corollary 11, we get the following corollary.

Corollary 16. Let 𝑇∶𝑋→𝑋 be a mapping on a complete PMS (𝑋,𝑝). Suppose that there exist nonnegative reals 𝛼,𝛽, and 𝛾 such that 𝛼+𝛽+2𝛾<1. If 𝑇 satisfies 𝑝+()𝑝𝑝𝑇𝑥,𝑇𝑦≤𝛼𝑝𝑥,𝑦𝛽𝑝𝑥,𝑇𝑥𝑦,𝑇𝑦+𝛾𝑝𝑦,𝑇𝑥𝑥,𝑇𝑦1+𝑝𝑥,𝑦 (34) for all 𝑥,𝑦∈𝑋. Then there exists a unique point 𝑢∈𝑋 such that 𝑢=𝑇𝑢. Also 𝑝(𝑢,𝑢)=0. Further 𝑇 has no fixed point other than 𝑢.

Now we give an example in favour of Corollary 16.

Example 17. Let 𝑋=[0,4] endowed with the usual partial metric 𝑝 defined by 𝑝(𝑥,𝑦)=max{𝑥,𝑦}. Clearly, (𝑋,𝑝) is a complete partial metric space. Now we define 𝐹∶𝑋→𝑋 as follows: 𝐹(𝑥)=𝑥3𝑥if0≤𝑥<21+𝑥if2≤𝑥≤4 (35) for all 𝑥∈𝑋. Now, let 𝑦≤𝑥. If 𝑥∈[0,2) (and so 𝑦∈[0,2)). Then 𝑝(𝐹𝑥,𝐹𝑦)=𝑥/3, 𝑝(𝑥,𝑦)=𝑥, 𝑝(𝑥,𝐹𝑥)=𝑥, 𝑝(𝑦,𝐹𝑦)=𝑦, 𝑝(𝑦,𝐹𝑥)=𝑥/3, 𝑝(𝑥,𝐹𝑦)=𝑥. Taking 𝛼=1/3, 𝛽=1/15, 𝛾=2/15, we can prove that all the conditions of Corollary 16 are satisfied. Now if 𝑥∈[2,4], then 𝑝(𝐹𝑥,𝐹𝑦)=𝑥/(1+𝑥), 𝑝(𝑥,𝑦)=𝑥, 𝑝(𝑥,𝐹𝑥)=𝑥, 𝑝(𝑦,𝐹𝑦)=𝑦, 𝑝(𝑦,𝐹𝑥)=𝑥/(1+𝑥), 𝑝(𝑥,𝐹𝑦)=𝑥 and taking 𝛼=1/3, 𝛽=1/15, 𝛾=2/15, one can verify the condition of the above corollary. Thus all the conditions of Corollary 16 are satisfied and 𝑢=0 is a fixed point of the mapping 𝐹.

As an application of Theorem 9, we prove the following theorem for two finite families of mappings.

Theorem 18. If {𝑇_𝑖}^𝑚₁ and {𝑆_𝑖}^𝑛₁ are two pairwise commuting finite families of self-mapping defined on a complete partial metric space (𝑋,𝑝) such that the mappings 𝑆 and 𝑇 (with 𝑇=𝑇₁𝑇₂⋯𝑇_𝑚 and 𝑆=𝑆₁𝑆₂⋯𝑆_𝑛) satisfy the contractive condition (5), then the component maps of the two families {𝑇_𝑖}^𝑚₁ and {𝑆_𝑖}^𝑛₁ have a unique common fixed point.

Proof. From Theorem 9, we can say that the mappings 𝑇 and 𝑆 have a unique common fixed point 𝑧; that is, 𝑇𝑧=𝑆𝑧=𝑧. Now our requirement is to show that 𝑧 is a common fixed point of all the component mappings of both the families. In view of pairwise commutativity of the families {𝑇_𝑖}^𝑚₁ and {𝑆_𝑖}^𝑛₁, (for every 1≤𝑘≤𝑚) we can write 𝑇_𝑘𝑧=𝑇_𝑘𝑇𝑧=𝑇𝑇_𝑘𝑧 and 𝑇_𝑘𝑧=𝑇_𝑘𝑆𝑧=𝑆𝑇_𝑘𝑧 which show that 𝑇_𝑘𝑧 (for every 𝑘) is also a common fixed point of 𝑇 and 𝑆. By using the uniqueness of common fixed point, we can write 𝑇_𝑘𝑧=𝑧 (for every 𝑘) which shows that 𝑧 is a common fixed point of the family {𝑇_𝑖}^𝑚₁. Using the same argument one can also show that (for every 1≤𝑘≤𝑛) 𝑆_𝑘𝑧=𝑧. Thus component maps of the two families {𝑇_𝑖}^𝑚₁ and {𝑆_𝑖}^𝑛₁ have a unique common fixed point.

By setting 𝑇₁=𝑇₂=⋯=𝑇_𝑚=𝐹 and 𝑆₁=𝑆₂=⋯=𝑆_𝑛=𝐺, in Theorem 18, we get the following corollary.

Corollary 19. Let 𝐹, 𝐺∶𝑋→𝑋 be two commuting self-mappings defined on a complete PMS (𝑋,𝑝) satisfying the condition 𝑝𝐹^𝑚𝑥,𝐺^𝑛𝑦+≤𝛼𝑝𝑥,𝑦𝛽𝑝𝑥,𝐹^𝑚𝑥𝑝𝑦,𝐺^𝑛𝑦+𝛾𝑝𝑦,𝐹^𝑚𝑥𝑝𝑥,𝐺^𝑛𝑦1+𝑝𝑥,𝑦 (36) for all 𝑥,𝑦∈𝑋, 𝛼+𝛽+2𝛾<1 (𝛼, 𝛽, and 𝛾 are nonnegative reals). Then F and G have a unique common fixed point.

By setting 𝑚=𝑛 and 𝐹=𝐺=𝑇 in Corollary 19, we deduce the following corollary.

Corollary 20. Let 𝑇∶𝑋→𝑋 be a mapping defined on a complete PMS (𝑋,𝑝) satisfying the condition 𝑝𝑇^𝑛𝑥,𝑇^𝑛𝑦+≤𝛼𝑝𝑥,𝑦𝛽𝑝𝑥,𝑇^𝑛𝑥𝑝𝑦,𝑇^𝑛𝑦+𝛾𝑝𝑦,𝑇^𝑛𝑥𝑝𝑥,𝑇^𝑛𝑦1+𝑝𝑥,𝑦 (37) for all 𝑥,𝑦∈𝑋, 𝛼+𝛽+2𝛾<1 (𝛼, 𝛽, and 𝛾 are nonnegative reals). Then F has a unique fixed point.

By setting 𝛽=𝛾=0, we draw following corollary which can be viewed as an extension of Bryant's theorem [15] for a mapping on a complete PMS (𝑋,𝑝).

Corollary 21. Let 𝐹∶𝑋→𝑋 be a mapping on a complete PMS (𝑋,𝑝). If 𝐹 satisfies 𝑝𝐹^𝑛𝑥,𝐹^𝑛𝑦≤𝛼𝑝𝑥,𝑦 (38) for all 𝑥,𝑦∈𝑋,𝛼<1. Then 𝐹 has a unique fixed point.

The following example demonstrates the superiority of Bryant's theorem over Matthews theorem on complete partial metric space.

Example 22. Let 𝑋=[0,4]. Define the partial metric 𝑝∶𝑋×𝑋→ℝ by 𝑝𝑥,𝑦=max𝑥,𝑦. (39) Then (𝑋,𝑝) is a complete partial metric space. Let 𝐹∶𝑋→𝑋 be defined as follows: 𝐹(𝑥)=𝑥²[[[[[].if𝑥∈0,12if𝑥∈1,20if𝑥∈2,4 (40) Then for 𝑥=0 and 𝑦=1, we get 𝑝(𝐹(0),𝐹(1))=𝑝(0,2)=2>𝛼𝑝(0,1)=𝛼(1), (41) because 0≤𝛼<1. However, 𝐹² satisfies the requirement of Bryant’s theorem and 𝑧= 0 is the unique fixed point of 𝐹.

3. Results for Set Valued Mappings

Theorem 23. Let (𝑋,𝑝) be a complete partial metric space and let 𝑆,𝑇∶𝑋→𝐶𝐵^𝑝(𝑋) be mappings such that 𝐻_𝑝+()𝑝𝑝𝑆𝑥,𝑇𝑦≤𝛼𝑝𝑥,𝑦𝛽𝑝𝑥,𝑆𝑥𝑦,𝑇𝑦+𝛾𝑝𝑦,𝑆𝑥𝑥,𝑇𝑦1+𝑝𝑥,𝑦 (42) for all 𝑥,𝑦∈𝑋, 0≤𝛼,𝛽,𝛾 with 𝛼+𝛽+2𝛾<1. Then 𝑆 and 𝑇 have a common fixed point.

Proof. Assume that 𝑀=((𝛼+𝛾)/(1−𝛽−𝛾)). Let 𝑥₀∈𝑋 be arbitrary but fixed element of 𝑋 and choose 𝑥₁∈𝑆(𝑥₀). By Lemma 8 we can choose 𝑥₂∈𝑇(𝑥₁) such that 𝑝𝑥₁,𝑥₂≤𝐻_𝑝𝑆𝑥₀,𝑇𝑥₁+𝑥𝛼+𝛾≤𝛼𝑝₀,𝑥₁𝑝𝑥+𝛽⁰,𝑆𝑥₀𝑝𝑥₁,𝑇𝑥₁1+𝑝𝑥₀,𝑥₁𝑝𝑥+𝛾¹,𝑆𝑥₀𝑝𝑥₀,𝑇𝑥₁1+𝑝𝑥₀,𝑥₁+𝑥𝛼+𝛾≤𝛼𝑝₀,𝑥₁𝑝𝑥+𝛽⁰,𝑥₁𝑝𝑥₁,𝑥₂1+𝑝𝑥₀,𝑥₁𝑝𝑥+𝛾¹,𝑥₁𝑝𝑥₀,𝑥₂1+𝑝𝑥₀,𝑥₁+𝑥𝛼+𝛾≤𝛼𝑝₀,𝑥₁𝑝𝑥+𝛽⁰,𝑥₁𝑝𝑥₁,𝑥₂1+𝑝𝑥₀,𝑥₁𝑝𝑥+𝛾⁰,𝑥₁𝑝𝑥₀,𝑥₂1+𝑝𝑥₀,𝑥₁+𝑥𝛼+𝛾≤𝛼𝑝₀,𝑥₁+𝛽𝑝𝑥₁,𝑥₂+𝛾𝑝𝑥₀,𝑥₂+𝛼+𝛾. (43) So we get 𝑝𝑥₁,𝑥₂≤𝛼+𝛾𝑝𝑥1−𝛽−𝛾⁰,𝑥₁+𝛼+𝛾.1−𝛽−𝛾 (44) Since 𝑀=((𝛼+𝛾)/(1−𝛽−𝛾)), so it further implies that 𝑝𝑥₁,𝑥₂≤𝑀𝑝𝑥₀,𝑥₁+𝑀. (45) By Lemma 8 we can choose 𝑥₃∈𝑆(𝑥₂) such that 𝑝𝑥₂,𝑥₃≤𝐻_𝑝𝑇𝑥₁,𝑆𝑥₂+𝛼+𝛾²𝑥1−𝛽−𝛾≤𝛼𝑝₁,𝑥₂𝑝𝑥+𝛽¹,𝑇𝑥₁𝑝𝑥₂,𝑆𝑥₂1+𝑝𝑥₁,𝑥₂𝑝𝑥+𝛾²,𝑇𝑥₁𝑝𝑥₁,𝑆𝑥₂1+𝑝𝑥₁,𝑥₂+𝛼+𝛾²𝑥1−𝛽−𝛾≤𝛼𝑝₁,𝑥₂𝑝𝑥+𝛽¹,𝑥₂𝑝𝑥₂,𝑥₃1+𝑝𝑥₁,𝑥₂𝑝𝑥+𝛾²,𝑥₂𝑝𝑥₁,𝑥₃1+𝑝𝑥₁,𝑥₂+𝛼+𝛾²𝑥1−𝛽−𝛾≤𝛼𝑝₁,𝑥₂𝑝𝑥+𝛽¹,𝑥₂𝑝𝑥₂,𝑥₃1+𝑝𝑥₁,𝑥₂𝑝𝑥+𝛾¹,𝑥₂𝑝𝑥₁,𝑥₃1+𝑝𝑥₁,𝑥₂+𝛼+𝛾²𝑥1−𝛽−𝛾≤𝛼𝑝₁,𝑥₂𝑝𝑥+𝛽¹,𝑥₂𝑝𝑥₂,𝑥₃1+𝑝𝑥₁,𝑥₂𝑝𝑥+𝛾¹,𝑥₂𝑝𝑥₁,𝑥₃1+𝑝𝑥₁,𝑥₂+𝛼+𝛾²𝑥1−𝛽−𝛾≤𝛼𝑝₁,𝑥₂+𝛽𝑝𝑥₂,𝑥₃+𝛾𝑝𝑥₁,𝑥₃+𝛼+𝛾².1−𝛽−𝛾 (46) So we get 𝑝𝑥₂,𝑥₃≤𝛼+𝛾1−𝛽−𝛾²𝑝𝑥₀,𝑥₁+2𝛼+𝛾1−𝛽−𝛾². (47) Continuing in this manner, one can obtain a sequence {𝑥_𝑛} in 𝑋 as 𝑥_2𝑛+1∈𝑆(𝑥_2𝑛) and 𝑥_2𝑛+2∈𝑇(𝑥_2𝑛+1) such that 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝑀𝑝𝑥_𝑛−1,𝑥_𝑛+𝑀≤⋯≤𝑀^𝑛𝑝𝑥₀,𝑥₁+𝑛𝑀^𝑛, (48) where 𝑀=((𝛼+𝛽)/(1−𝛼))<1 for all 𝑛≥0. Without loss of generality assume that 𝑛>𝑚. Then, using (48) and the triangle inequality for partial metrics (P₄), we have 𝑝𝑥_𝑛,𝑥_𝑚≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑛+2+⋯+𝑝𝑥_𝑚−1,𝑥_𝑚≤𝑀^𝑛𝑝𝑥₀,𝑥₁+𝑛𝑀^𝑛+𝑀^𝑛+1𝑝𝑥₀,𝑥₁+(𝑛+1)𝑀^𝑛+1+⋯+𝑀^𝑚−1𝑝𝑥₀,𝑥₁+(𝑚−1)𝑀^𝑚−1≤^𝑚−1_𝑖=𝑛𝑀^𝑖𝑝𝑥₀,𝑥₁+^𝑚−1_𝑖=𝑛𝑖𝑀^𝑖⟶0as𝑛⟶+∞(since0<𝑀<1). (49)
By the definition of 𝑝^𝑠, we get, 𝑝^𝑠𝑥_𝑛,𝑥_𝑚≤2𝑝𝑥_𝑛,𝑥_𝑚⟶0as𝑛⟶+∞. (50) This yields that {𝑥_𝑛} is a Cauchy sequence in (𝑋,𝑝^𝑠). Since (𝑋,𝑝) is complete, then from Lemma 2(4), (𝑋,𝑝^𝑠) is a complete metric space. Therefore, the sequence {𝑥_𝑛} converges to some 𝑥^∗∈𝑋 with respect to the metric 𝑝^𝑠; that is, lim_𝑛→+∞𝑝^𝑠(𝑥_𝑛,𝑥^∗)=0. Again, from Lemma 2(4), we get 𝑝𝑥^∗,𝑥^∗=lim_𝑛→+∞𝑝𝑥_𝑛,𝑥^∗=lim_𝑛→+∞𝑝𝑥_𝑛,𝑥_𝑛=0,𝐻_𝑝𝑇𝑥_2𝑛+2,𝑆𝑥^∗≤𝛼𝑝𝑥_2𝑛+2,𝑥^∗𝑝𝑥+𝛽^2𝑛+2,𝑇𝑥_2𝑛+2𝑝𝑥^∗,𝑆𝑥^∗1+𝑝𝑥_2𝑛+2,𝑥^∗𝑝𝑥+𝛾^2𝑛+2,𝑆𝑥^∗𝑝𝑥^∗,𝑇𝑥_2𝑛+21+𝑝𝑥_2𝑛+2,𝑥^∗≤𝛼𝑝𝑥_2𝑛+2,𝑥^∗𝑝𝑥+𝛽^2𝑛+2,𝑥_2𝑛+3𝑝𝑥^∗,𝑆𝑥^∗1+𝑝𝑥_2𝑛+2,𝑥^∗𝑝𝑥+𝛾^2𝑛+2,𝑆𝑥^∗𝑝𝑥^∗,𝑥_2𝑛+11+𝑝𝑥_2𝑛+2,𝑥^∗, (51)(52) taking limit as 𝑛→∞ and using (51), we get lim_𝑛→+∞𝐻_𝑝𝑇𝑥_2𝑛+2,𝑆𝑥^∗=0. (53) Now 𝑥_2𝑛+1∈ 𝑇(𝑥_2𝑛+2) gives that 𝑝𝑥_2𝑛+1,𝑆𝑥^∗≤𝐻_𝑝𝑇𝑥_2𝑛+2,𝑆𝑥^∗, (54) which implies that lim_𝑛→+∞𝑝𝑥_2𝑛+1,𝑆𝑥^∗=0. (55) On the other hand by (P₄), we have 𝑝𝑥^∗,𝑆𝑥^∗≤𝑝𝑥^∗,𝑥_2𝑛+2+𝑝𝑥_2𝑛+2,𝑆𝑥^∗−𝑝𝑥_2𝑛+2,𝑥_2𝑛+2≤𝑝𝑥^∗,𝑥_2𝑛+2+𝑝𝑥_2𝑛+2,𝑆𝑥^∗. (56) Taking limit as 𝑛→+∞ and using (51) and (55), we obtain 𝑝(𝑥^∗,𝑆(𝑥^∗))=0. Therefore, from (51) (𝑝(𝑥^∗,𝑥^∗)=0), we obtain 𝑝𝑥^∗,𝑆𝑥^∗=𝑝𝑥^∗,𝑥^∗ (57) which from Remark 14 implies that 𝑥^∗∈𝑆(𝑥^∗)=𝑆(𝑥^∗). Similarly one can easily prove that 𝑥^∗∈𝑇(𝑥^∗). Thus 𝑆 and 𝑇 have a common fixed point.

Remark 24. For 𝛽=𝛾=0 and 𝑆=𝑇, Theorem 23 reduces to the following result of Aydi et al. [9].

Corollary 25 (see [9, Theorem 3.2]). Let (𝑋,𝑝) be a partial metric space. If 𝑇∶𝑋→𝐶𝐵^𝑝(𝑋) is a multivalued mapping such that for all 𝑥,𝑦∈𝑋, one has 𝐻_𝑝𝑇𝑥,𝑇𝑦≤𝑘𝑝(𝑎,𝑏), (58) where 𝑘∈(0,1). Then 𝑇 has a fixed point.

Theorem 26. Let (𝑋,𝑝) be a complete partial metric space and 𝑆,𝑇∶𝑋→𝐶𝐵^𝑝(𝑋) be multivalued mappings such that 𝐻_𝑝𝑝()𝑝𝑝𝑆𝑥,𝑇𝑦≤𝛼𝑥,𝑆𝑥𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑦𝑦,𝑆𝑥𝑝𝑥,𝑇𝑦+𝑝𝑦,𝑆𝑥+𝛽𝑑𝑥,𝑦 (59) for all 𝑥,𝑦∈𝑋, 0≤𝛼,𝛽 with 2𝛼+𝛽<1, and 𝑝(𝑥,𝑇𝑦)+𝑝(𝑦,𝑆𝑥)≠0. Then 𝑆 and 𝑇 have a common fixed point.

Proof. Assume that 𝑙=(𝛼+𝛽)/(1−𝛼). Let 𝑥₀∈𝑋 be arbitrary but fixed element of 𝑋 and choose 𝑥₁∈𝑆(𝑥₀).
When 𝑝(𝑥,𝑇𝑦)+𝑝(𝑦,𝑆𝑥)≠0. By Lemma 8 we can choose 𝑥₂∈𝑇(𝑥₁) such that 𝑝𝑥₁,𝑥₂≤𝐻_𝑝𝑆𝑥₀,𝑇𝑥₁+𝑝𝑥𝛼+𝛽≤𝛼⁰,𝑆𝑥₀𝑝𝑥₀,𝑇𝑥₁+𝑝𝑥₁,𝑇𝑥₁𝑝𝑥₁,𝑆𝑥₀𝑝𝑥₀,𝑇𝑥₁+𝑝𝑥₁,𝑆𝑥₀+𝛽𝑝𝑥₀,𝑥₁+𝑝𝑥𝛼+𝛽≤𝛼⁰,𝑥₁𝑝𝑥₀,𝑥₂+𝑝𝑥₁,𝑥₂𝑝𝑥₁,𝑥₁𝑝𝑥₀,𝑥₂+𝑝𝑥₁,𝑥₁+𝛽𝑝𝑥₀,𝑥₁+𝑝𝑥𝛼+𝛽≤𝛼₀,𝑥₁+𝑝𝑥₁,𝑥₂+𝛽𝑝𝑥₀,𝑥₁+𝛼+𝛽 (60) because 𝑝(𝑥₀,𝑥₂)<𝑝(𝑥₀,𝑥₂)+𝑝(𝑥₁,𝑥₁) and 𝑝(𝑥₁,𝑥₁)<𝑝(𝑥₁,𝑥₁)+𝑝(𝑥₀,𝑥₂). Thus we get 𝑝𝑥₁,𝑥₂≤𝛼+𝛽𝑝𝑥1−𝛼⁰,𝑥₁+𝛼+𝛽.1−𝛼 (61) It further implies that 𝑝𝑥₁,𝑥₂≤𝑙𝑝𝑥₀,𝑥₁+𝑙. (62) By Lemma 8 we can choose 𝑥₃∈𝑆(𝑥₂) such that 𝑝𝑥₂,𝑥₃≤𝐻_𝑝𝑇𝑥₁,𝑆𝑥₂+𝛼+𝛽²𝑝𝑥1−𝛼≤𝛼¹,𝑇𝑥₁𝑝𝑥₁,𝑆𝑥₂+𝑝𝑥₂,𝑆𝑥₂𝑝𝑥₂,𝑇𝑥₁𝑝𝑥₁,𝑆𝑥₂+𝑝𝑥₂,𝑇𝑥₁+𝛽𝑝𝑥₁,𝑥₂+𝛼+𝛽²𝑝𝑥1−𝛼≤𝛼¹,𝑥₂𝑝𝑥₁,𝑥₃+𝑝𝑥₂,𝑥₃𝑝𝑥₂,𝑥₂𝑝𝑥₁,𝑥₃+𝑝𝑥₂,𝑥₂+𝛽𝑝𝑥₁,𝑥₂+𝛼+𝛽²𝑝𝑥1−𝛼≤𝛼₁,𝑥₂+𝑝𝑥₂,𝑥₃+𝛽𝑝𝑥₁,𝑥₂+𝛼+𝛽²1−𝛼 (63) because 𝑝(𝑥₁,𝑥₃)<𝑝(𝑥₁,𝑥₃)+𝑝(𝑥₂,𝑥₂) and 𝑝(𝑥₂,𝑥₂)<𝑝(𝑥₂,𝑥₂)+𝑝(𝑥₁,𝑥₃). Thus we get 𝑝𝑥₂,𝑥₃≤𝛼+𝛽𝑝𝑥1−𝛼¹,𝑥₂+𝛼+𝛽1−𝛼². (64) It further implies that 𝑝𝑥₂,𝑥₃≤𝑙²𝑝𝑥₀,𝑥₀+2𝑙². (65) Continuing in this manner, one can obtain a sequence {𝑥_𝑛} in 𝑋 as 𝑥_2𝑛+1∈𝑆(𝑥_2𝑛) and 𝑥_2𝑛+2∈𝑇(𝑥_2𝑛+1) such that 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝑙𝑝𝑥_𝑛−1,𝑥_𝑛+𝑙≤⋯≤𝑙^𝑛𝑝𝑥₀,𝑥₁+𝑛𝑙^𝑛, (66) where 𝑙=((𝛼+𝛽)/(1−𝛼))<1 for all 𝑛≥0. Without loss of generality assume that 𝑚>𝑛. Then using (66) and the triangle inequality for partial metrics (P₄), one can easily prove that 𝑝𝑥_𝑛,𝑥_𝑚≤^𝑚−1_𝑖=𝑛𝑙^𝑖𝑝𝑥₀,𝑥₁+^𝑚−1_𝑖=𝑛𝑖𝑙^𝑖⟶0as𝑛⟶+∞(since0<𝑙<1). (67)
By the definition of 𝑝^𝑠, 𝑝^𝑠𝑥_𝑛,𝑥_𝑚≤2𝑝𝑥_𝑛,𝑥_𝑚⟶0as𝑛⟶+∞. (68) This yields that {𝑥_𝑛} is a Cauchy sequence in (𝑋,𝑝^𝑠). Since (𝑋,𝑝) is complete, then from Lemma 2(4), (𝑋,𝑝^𝑠) is a complete metric space. Therefore, the sequence {𝑥_𝑛} converges to some 𝑥^′∈𝑋 with respect to the metric 𝑝^𝑠; that is, lim_𝑛→+∞𝑝^𝑠(𝑥_𝑛,𝑥^′)=0. Again, from Lemma 2(4), we get 𝑝𝑥^′,𝑥^′=lim_𝑛→+∞𝑝𝑥_𝑛,𝑥^/=lim_𝑛→+∞𝑝𝑥_𝑛,𝑥_𝑛=0𝐻_𝑝𝑇𝑥_2𝑛+1,𝑆𝑥^′≤𝛼𝑝𝑥_2𝑛+1,𝑇𝑥_2𝑛+1𝑝𝑥_2𝑛+1,𝑆𝑥^′+𝑝𝑥^′,𝑆𝑥^′𝑝𝑥^′,𝑇𝑥_2𝑛+1×𝑝𝑥_2𝑛+1,𝑆𝑥^′+𝑝𝑥^′,𝑇𝑥_2𝑛+1⁻¹+𝛽𝑝𝑥_2𝑛+1,𝑥^/≤𝛼𝑝𝑥_2𝑛+1,𝑥_2𝑛+2𝑝𝑥_2𝑛+1,𝑆𝑥^′+𝑝𝑥^′,𝑆𝑥^′𝑝𝑥^′,𝑥_2𝑛+2×𝑝𝑥_2𝑛+1,𝑆𝑥^′+𝑝𝑥^′,𝑥_2𝑛+2⁻¹+𝛽𝑝𝑥_2𝑛+1,𝑥^/, (69)(70) therefore lim_𝑛→+∞𝐻_𝑝𝑇𝑥_2𝑛+1,𝑆𝑥^′=0. (71) Now 𝑥_2𝑛+2∈ 𝑇(𝑥_2𝑛+1) gives that 𝑝𝑥_2𝑛+2,𝑆𝑥^′≤𝐻_𝑝𝑇𝑥_2𝑛+1,𝑆𝑥^′ (72) which implies that lim_𝑛→+∞𝑝𝑥_2𝑛+2,𝑆𝑥^′=0. (73) On the other hand, we have 𝑝𝑥^′,𝑆𝑥^′≤𝑝𝑥^′,𝑥_2𝑛+2+𝑝𝑥_2𝑛+2,𝑆𝑥^′−𝑝𝑥_2𝑛+2,𝑥_2𝑛+2≤𝑝𝑥^′,𝑥_2𝑛+2+𝑝𝑥_2𝑛+2,𝑆𝑥^′. (74) Taking limit as 𝑛→+∞ and using (69) and (73), we obtain 𝑝(𝑥^′,𝑆(𝑥^′))=0. Therefore, from (69) (𝑝(𝑥^′,𝑥^′)=0), we obtain 𝑝𝑥^′,𝑆𝑥^′=𝑝𝑥^′,𝑥^′ (75) which from Remark 14 implies that 𝑥^′∈𝑆(𝑥^′)=𝑆(𝑥^′). It follows similarly that 𝑥^′∈𝑇(𝑥^′). Thus 𝑆 and 𝑇 have a common fixed point.

Now we give an example which illustrates our Theorem 26.

Example 27. Let 𝑋={1,2,3} be endowed with usual order and let 𝑝 be a partial metric on 𝑋 defined as 𝑝()()()=51,1=𝑝2,2=0,𝑝3,3,𝑝()()=3111,2=𝑝2,1,𝑝()()=9101,3=𝑝3,1,𝑝()()=1202,3=𝑝3,22. (76) Define the mappings 𝑆,𝑇∶𝑋→𝐶𝐵_𝑝(𝑋) by {1}{}{}{1}{}{2}𝑆𝑥=if𝑥,𝑦∈1,21,2otherwise𝑇𝑥=if𝑥,𝑦∈1,2otherwise. (77) Note that 𝑆𝑥 and 𝑇𝑥 are closed and bounded for all 𝑥∈𝑋 with respect to the partial metric 𝑝. To show that for all 𝑥,𝑦 in 𝑋, (59) is satisfied with 𝛼=1/11, 𝛽=4/5, we consider the following cases: if 𝑥,𝑦∈{1,2}, then, 𝐻_𝑝𝑆𝑥,𝑇𝑦=𝐻_𝑝({1},{1})=0 (78) and condition (59) is satisfied obviously.

If 𝑥=𝑦=3, then 𝐻_𝑝𝑆𝑥,𝑇𝑦=𝐻_𝑝},{2=3({1,2}),𝑝()({=910𝑥,𝑆𝑥=𝑝𝑦,𝑆𝑥=𝑝3,1,2}),𝑝({2=20𝑥,𝑇𝑦=𝑝𝑦,𝑇𝑦=𝑝𝑦,𝑆𝑥=𝑝3,})11,𝑝()=524𝑥,𝑦=𝑝3,3.11 (79)

If 𝑥=3, 𝑦=1, then 𝐻_𝑝𝑆𝑥,𝑇𝑦=𝐻_𝑝},{1=3({1,2}),𝑝()({=910𝑥,𝑆𝑥=𝑝3,1,2})({1=920,𝑝𝑥,𝑇𝑦=𝑝3,}),𝑝({1({𝑝()=920𝑦,𝑇𝑦=𝑝1,})=0,𝑝𝑦,𝑆𝑥=𝑝1,1,2})=0,𝑥,𝑦=𝑝3,1.20 (80)

If 𝑥=3, 𝑦=2, then 𝐻_𝑝𝑆𝑥,𝑇𝑦=𝐻_𝑝},{1=3({1,2}),𝑝()({=910𝑥,𝑆𝑥=𝑝3,1,2})({1=920,𝑝𝑥,𝑇𝑦=𝑝3,}),𝑝({1=320𝑦,𝑇𝑦=𝑝2,})({𝑝()=110,𝑝𝑦,𝑆𝑥=𝑝2,1,2})=0,𝑥,𝑦=𝑝3,22. (81)

If 𝑥=1, 𝑦=3, then 𝐻_𝑝𝑆𝑥,𝑇𝑦=𝐻_𝑝1},{2=3({}),𝑝()({1({2=310𝑥,𝑆𝑥=𝑝1,})=0,𝑝𝑥,𝑇𝑦=𝑝1,}),𝑝({2=110𝑦,𝑇𝑦=𝑝3,})2({1=9,𝑝𝑦,𝑆𝑥=𝑝3,}),𝑝()=920𝑥,𝑦=𝑝1,3.20 (82)

If 𝑥=2, 𝑦=3, then 𝐻_𝑝𝑆𝑥,𝑇𝑦=𝐻_𝑝1},{2=3({}),𝑝()({1=310𝑥,𝑆𝑥=𝑝2,})({2𝑝({2=110,𝑝𝑥,𝑇𝑦=𝑝2,})=0,𝑦,𝑇𝑦=𝑝3,})2({1=9,𝑝𝑦,𝑆𝑥=𝑝3,}),𝑝()=120𝑥,𝑦=𝑝2,32. (83) Thus, all the conditions of Theorem 26 are satisfied. Here 𝑥=1 is a common fixed point of 𝑆 and 𝑇.

On the other hand, the metric 𝑝^𝑠 induced by the partial metric 𝑝 is given by 𝑝^𝑠(1,1)=𝑝^𝑠(2,2)=𝑝^𝑠()𝑝3,3=0,_𝑠(2,1)=𝑝^𝑠()=32,14,𝑝^𝑠(3,2)=𝑝^𝑠()=42,37,𝑝^𝑠(3,1)=𝑝^𝑠(1,3)=14.29 (84) Note that in case of ordinary Hausdorff metric, given mapping does not satisfy the condition. Indeed, for 𝑥=1 and 𝑦=3, we have 𝐻1},{2=3𝑆𝑥,𝑇𝑦=𝐻({})4,𝑝^𝑠(𝑥,𝑆𝑥)=𝑝^𝑠({1𝑝1,})=0,_𝑠𝑥,𝑇𝑦=𝑝^𝑠({2=31,})4,𝑝^𝑠𝑦,𝑇𝑦=𝑝^𝑠({2=43,})7,𝑝^𝑠𝑦,𝑆𝑥=𝑝^𝑠(3,{1})=14,29 (85) for the values of 𝛼=1/11, 𝛽=4/5. By a routine calculation one can easily verify that the mapping does not satisfy the condition which involved ordinary Hausdorff metric.

Acknowledgment

This paper was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks DSR technical and financial support.