Some Common Fixed Point Theorems for 𝐹-Contraction Type Mappings in 0-Complete Partial Metric Spaces

Abstract

We prove some common fixed point theorems for 𝐹-contractions in 0-complete partial metric spaces. Our results extend, generalize, and unify several known results in the literature. Some examples are included which show that the generalization is proper.

1. Introduction and Preliminaries

In 1994, Matthews [1] introduced the notion of partial metric spaces, as a part of the study of denotational semantics of dataflow network. In partial metric space, the usual distance was replaced by partial metric, with an interesting property “nonzero self-distance” of points. Also, the convergence of sequences in this space was defined in such a way that the limit of a convergent sequence need not be unique. Matthews showed that the Banach contraction principle is valid in partial metric space and can be applied in program verification. Later, several authors generalized the result of Matthews (see, e.g., [2–29]). O'Neill [22] generalized the concept of partial metric space a bit further by admitting negative distances. The partial metric defined by O'Neill is called dualistic partial metric. Heckmann [21] generalized it by omitting small self-distance axiom. The partial metric defined by Heckmann is called weak partial metric.

Romaguera [23] introduced the notion of 0-Cauchy sequence, 0-complete partial metric spaces and proved some characterizations of partial metric spaces in terms of completeness and 0-completeness. The notion of 0-complete partial metric spaces is more general than the complete partial metric space, as shown by an example in [23].

Recently, Wardowski [30] introduces a new concept of 𝐹-contraction and proves a fixed point theorem which generalizes the Banach contraction principle in a different way than the known results of the literature on complete metric spaces. In this paper, we consider a more generalized type of 𝐹-contraction and prove some common fixed point theorems for such type of mappings in 0-complete partial metric spaces. The results of this paper generalize and extend the results of Wardowski [30], Altun et al. [17], Ćirić [31, 32], and some well-known results in the literature. Some examples are given which show that the results of this paper are proper generalizations of known results.

First we recall some definitions and properties of partial metric space [1, 22, 23, 25, 33].

Definition 1. A partial metric on a nonempty set 𝑋 is a function 𝑝∶𝑋×𝑋→ℝ⁺ (ℝ⁺ stands for nonnegative reals) such that for all 𝑥, 𝑦, and 𝑧∈𝑋(P1)𝑥=𝑦 if and only if 𝑝(𝑥,𝑥)=𝑝(𝑥,𝑦)=𝑝(𝑦,𝑦); (P2)𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦); (P3)𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥); (P4)𝑝(𝑥,𝑦)≤𝑝(𝑥,𝑧)+𝑝(𝑧,𝑦)−𝑝(𝑧,𝑧).A partial metric space is a pair (𝑋,𝑝) such that 𝑋 is a nonempty set, and 𝑝 is a partial metric on 𝑋.

It is clear that, if 𝑝(𝑥,𝑦)=0, then from (P1) and (P2) 𝑥=𝑦. But if 𝑥=𝑦,  𝑝(𝑥,𝑦) may not be 0. Also every metric space is a partial metric space, with zero self-distance.

Each partial metric on 𝑋 generates a 𝑇₀ topology 𝜏_𝑝 on 𝑋 which has a base of the family of open 𝑝-balls {𝐵_𝑝(𝑥,𝜀)∶𝑥∈𝑋,𝜀>0}, where 𝐵_𝑝(𝑥,𝜀)={𝑦∈𝑋∶𝑝(𝑥,𝑦)<𝑝(𝑥,𝑥)+𝜀} for all 𝑥∈𝑋 and 𝜀>0.

Definition 2. A mapping 𝑓∶𝑋→𝑋 is continuous if and only if, whenever a sequence {𝑥_𝑛} in 𝑋 converges with respect to 𝜏_𝑝 to a point 𝑥∈𝑋, the sequence {𝑓𝑥_𝑛} converges with respect to 𝜏_𝑝 to 𝑓𝑥∈𝑋.

Theorem 3 (see [1]). For each partial metric 𝑝∶𝑋×𝑋→ℝ⁺ the pair (𝑋,𝑑), where 𝑑(𝑥,𝑦)=2𝑝(𝑥,𝑦)−𝑝(𝑥,𝑥)−𝑝(𝑦,𝑦) for all 𝑥,𝑦∈𝑋, is a metric space.

Here (𝑋,𝑑) is called induced metric space, and 𝑑 is induced metric. In further discussion until specified,(𝑋,𝑑) will represent induced metric space.

Example 4 (see [1]). If 𝑝∶ℝ⁺×ℝ⁺→ℝ⁺ is defined by 𝑝(𝑥,𝑦)=max{𝑥,𝑦}, for all 𝑥,𝑦∈ℝ⁺, then (ℝ⁺,𝑝) is a partial metric space.

Example 5. Let (𝑋,𝑝) be a partial metric space; then (𝑋,𝑝^∗) is a partial metric space, where 𝑝^∗(𝑥,𝑦)=𝑑(𝑥,𝑦)+𝑝(𝑥,𝑦) for all 𝑥,𝑦∈𝑋 and 𝑑 is the metric induced by 𝑝.

For some more examples of partial metric spaces, we refer to [14] and the references therein.

Let (𝑋,𝑝) be a partial metric space. (1)A sequence {𝑥_𝑛} in (𝑋,𝑝) converges to a point 𝑥∈𝑋 if and only if 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝(𝑥_𝑛,𝑥).(2)A sequence {𝑥_𝑛} in (𝑋,𝑝) is called the Cauchy sequence if there exists (and is finite) lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).(3)(𝑋,𝑝) is said to be complete if every Cauchy sequence {𝑥_𝑛} in 𝑋 converges with respect to 𝜏_𝑝 to a point 𝑥∈𝑋 such that 𝑝(𝑥,𝑥)=lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).(4)A sequence {𝑥_𝑛} in (𝑋,𝑝) is called 0-Cauchy sequence if lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚)=0. The space (𝑋,𝑝) is said to be 0-complete if every 0-Cauchy sequence in 𝑋 converges with respect to 𝜏_𝑝 to a point 𝑥∈𝑋 such that 𝑝(𝑥,𝑥)=0.

Lemma 6 (see [1, 23, 25, 33]). Let (𝑋,𝑝) be a partial metric space and {𝑥_𝑛} any sequence in 𝑋.(i){𝑥_𝑛} is a Cauchy sequence in (𝑋,𝑝) if and only if it is a Cauchy sequence in metric space (𝑋,𝑑).(ii)(𝑋,𝑝) is complete if and only if the metric space (𝑋,𝑑) is complete. Furthermore, lim_𝑛→∞𝑑(𝑥_𝑛,𝑥)=0 if and only if 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝(𝑥_𝑛,𝑥)=lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).(iii)Every 0-Cauchy sequence in (𝑋,𝑝) is Cauchy in (𝑋,𝑑).(iv)If (𝑋,𝑝) is complete, then it is 0-complete.

The converse assertions of (iii) and (iv) do not hold. Indeed, the partial metric space (ℚ∩ℝ⁺,𝑝), where ℚ denotes the set of rational numbers and the partial metric 𝑝 is given by 𝑝(𝑥,𝑦)=max{𝑥,𝑦}, provides an easy example of a 0-complete partial metric space which is not complete. Also, it is easy to see that every closed subset of a 0-complete partial metric space is 0-complete.

The proof of the following lemma is easy, and for detail we refer to [34] and the references therein.

Lemma 7. Assume 𝑥_𝑛→𝑧 as 𝑛→∞ in a partial metric space (𝑋,𝑝) such that 𝑝(𝑧,𝑧)=0. Then lim_𝑛→∞𝑝(𝑥_𝑛,𝑦)=𝑝(𝑧,𝑦) for all 𝑦∈𝑋.

Analogous to [30] we have following definitions.

Definition 8. Let 𝐹∶ℝ⁺→ℝ be a mapping satisfying: (F1)𝐹 is strictly increasing, that is, for 𝛼,𝛽∈ℝ⁺ such that 𝛼<𝛽 implies 𝐹(𝛼)<𝐹(𝛽); (F2)for each sequence {𝛼_𝑛} of positive numbers lim_𝑛→∞𝛼_𝑛=0 if and only if lim_𝑛→∞𝐹(𝛼_𝑛)=−∞; (F3)there exists 𝑘∈(0,1) such that lim_𝛼→0⁺𝛼^𝑘𝐹(𝛼)=0.

For examples of function 𝐹, we refer to [30]. We denote the set of all functions satisfying properties (F1)–(F3) by ℱ.

Wardowski in [30] defined the 𝐹-contraction as follows.

Let (𝑋,𝜌) be a metric space. A mapping 𝑇∶𝑋→𝑋 is said to be an 𝐹-contraction if there exists 𝐹∈ℱ and 𝜏>0 such that, for all 𝑥,𝑦∈𝑋, 𝜌(𝑇𝑥,𝑇𝑦)>0, we have 𝜏+𝐹𝜌𝑇𝑥,𝑇𝑦≤𝐹𝜌𝑥,𝑦. (1) The following lemma will be useful in proving our main result.

Lemma 9. Let (𝑋,𝑝) be a partial metric space and 𝑓,𝑔∶𝑋→𝑋 two mappings. Suppose there exist 𝐹∈ℱ and 𝜏>0 such that, for all 𝑥,𝑦∈𝑋, 𝑝(𝑓𝑥,𝑔𝑦)>0, one has 𝑝𝑝,𝑝𝜏+𝐹𝑓𝑥,𝑔𝑦≤𝐹max𝑥,𝑦,𝑝𝑥,𝑓𝑥,𝑝𝑦,𝑔𝑦𝑥,𝑔𝑦+𝑝𝑦,𝑓𝑥2. (2) If 𝑓 has a fixed point 𝑢∈𝑋, then 𝑢 is a unique common fixed point of 𝑓 and 𝑔, and 𝑝(𝑢,𝑢)=0.

Proof. Let 𝑢∈𝑋 be a fixed point of 𝑓. Suppose 𝑝(𝑓𝑢,𝑔𝑢)>0; then by (2) we obtain 𝑝𝑝(),𝑝𝜏+𝐹𝑓𝑢,𝑔𝑢≤𝐹max𝑢,𝑢,𝑝𝑢,𝑓𝑢,𝑝𝑢,𝑔𝑢𝑢,𝑔𝑢+𝑝𝑢,𝑓𝑢2,𝑝𝑝,𝑝()𝜏+𝐹𝑢,𝑔𝑢≤𝐹max𝑢,𝑔𝑢𝑢,𝑢+𝑝𝑢,𝑔𝑢2,𝜏+𝐹𝑝𝑢,𝑔𝑢≤𝐹𝑝𝑢,𝑔𝑢, (3) a contradiction (as 𝜏>0). Therefore we have 𝑝(𝑓𝑢,𝑔𝑢)=0; that is, 𝑓𝑢=𝑔𝑢=𝑢. Thus 𝑢 is a common fixed point of 𝑓 and 𝑔. Again, if 𝑝(𝑢,𝑢)>0, then by a similar process as above we obtain 𝑝()𝑝𝑝𝑝(),𝜏+𝐹𝑢,𝑢=𝜏+𝐹𝑓𝑢,𝑔𝑢≤𝐹𝑢,𝑔𝑢=𝐹𝑢,𝑢 (4) a contradiction (as 𝜏>0). Therefore, 𝑝(𝑢,𝑢)=0. For uniqueness, let 𝑣 be another common fixed point of 𝑓 and 𝑔; that is, 𝑓𝑣=𝑔𝑣=𝑣. If 𝑝(𝑢,𝑣)>0, then by (2) we obtain 𝑝()𝑝𝑝(),𝑝𝜏+𝐹𝑢,𝑣=𝜏+𝐹𝑓𝑢,𝑔𝑣≤𝐹max𝑢,𝑣,𝑝𝑢,𝑓𝑢,𝑝𝑣,𝑔𝑣𝑢,𝑔𝑣+𝑝𝑣,𝑓𝑢2𝑝()()(),𝑝()()=𝐹max𝑢,𝑣,𝑝𝑢,𝑢,𝑝𝑣,𝑣𝑢,𝑣+𝑝𝑣,𝑢2=𝐹𝑝(𝑢,𝑣), (5) a contradiction. Therefore, 𝑝(𝑢,𝑣)=0; that is, 𝑢=𝑣. Thus a common fixed point is unique.

Now we can state our main results.

2. Main Results

The following theorem extends and generalizes the results of [30] in partial metric spaces.

Theorem 10. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑓,𝑔∶𝑋→𝑋 two mappings. Suppose there exist 𝐹∈ℱ and 𝜏>0 such that, for all 𝑥,𝑦∈𝑋, 𝑝(𝑓𝑥,𝑔𝑦)>0, one has 𝑝𝑝,𝑝𝜏+𝐹𝑓𝑥,𝑔𝑦≤𝐹max𝑥,𝑦,𝑝𝑥,𝑓𝑥,𝑝𝑦,𝑔𝑦𝑥,𝑔𝑦+𝑝𝑦,𝑓𝑥2. (6) If (i) 𝑓 or 𝑔 is continuous or (ii) 𝐹 is continuous, then 𝑓 and 𝑔 have a unique common fixed point 𝑢∈𝑋, and 𝑝(𝑢,𝑢)=0.

Proof. Let 𝑥₀∈𝑋 be arbitrary. Define a sequence {𝑥_𝑛} by 𝑥_2𝑛+1=𝑓𝑥_2𝑛, and 𝑥_2𝑛+2=𝑔𝑥_2𝑛+1 for all 𝑛≥0. If 𝑥_𝑚+1=𝑥_𝑚 for any 𝑚∈ℕ, for example, let 𝑥_2𝑛+1=𝑥_2𝑛, then it follows from (6) that 𝜏+𝐹𝑝𝑥_2𝑛+1,𝑥_2𝑛+2=𝜏+𝐹𝑝𝑓𝑥_2𝑛,𝑔𝑥_2𝑛+1≤𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛,𝑓𝑥_2𝑛,𝑝𝑥_2𝑛+1,𝑔𝑥_2𝑛+1,𝑝𝑥_2𝑛,𝑔𝑥_2𝑛+1+𝑝𝑥_2𝑛+1,𝑓𝑥_2𝑛2=𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛+1,𝑥_2𝑛+2,𝑝𝑥_2𝑛,𝑥_2𝑛+2+𝑝𝑥_2𝑛+1,𝑥_2𝑛+12=𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛,𝑝𝑥_2𝑛,𝑥_2𝑛,𝑝𝑥_2𝑛,𝑥_2𝑛+2,𝑝𝑥_2𝑛,𝑥_2𝑛+2+𝑝𝑥_2𝑛,𝑥_2𝑛2=𝐹𝑝𝑥_2𝑛,𝑥_2𝑛+2, (7) that is, 𝐹(𝑝(𝑥_2𝑛,𝑥_2𝑛+2))≤𝐹(𝑝(𝑥_2𝑛,𝑥_2𝑛+2))−𝜏. As, 𝜏>0, we get a contradiction. Therefore, we must have 𝐹(𝑝(𝑥_2𝑛,𝑥_2𝑛+2))=0; that is, 𝑥_2𝑛=𝑥_2𝑛+1=𝑥_2𝑛+2. Similarly, we obtain 𝑥_2𝑛=𝑥_2𝑛+1=𝑥_2𝑛+2=𝑥_2𝑛+3=⋯, and so 𝑥_2𝑛=𝑓𝑥_2𝑛=𝑔𝑥_2𝑛. Thus, 𝑥_2𝑛 is a common fixed point of 𝑓 and 𝑔. Now, we assume that 𝑥_𝑚≠𝑥_𝑚+1 for all 𝑚∈ℕ; then it follows from (6) that 𝜏+𝐹𝑝𝑥_2𝑛+1,𝑥_2𝑛=𝜏+𝐹𝑝𝑓𝑥_2𝑛,𝑔𝑥_2𝑛−1≤𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛−1,𝑝𝑥_2𝑛,𝑓𝑥_2𝑛,𝑝𝑥_2𝑛−1,𝑔𝑥_2𝑛−1,𝑝𝑥_2𝑛,𝑔𝑥_2𝑛−1+𝑝𝑥_2𝑛−1,𝑓𝑥_2𝑛2=𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛−1,𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛−1,𝑥_2𝑛,𝑝𝑥_2𝑛,𝑥_2𝑛+𝑝𝑥_2𝑛−1,𝑥_2𝑛+12≤𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛−1,𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛−1,𝑥_2𝑛+𝑝𝑥_2𝑛,𝑥_2𝑛+12=𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛−1,𝑝𝑥_2𝑛,𝑥_2𝑛+1. (8) If max{𝑝(𝑥_2𝑛,𝑥_2𝑛−1),𝑝(𝑥_2𝑛,𝑥_2𝑛+1)}=𝑝(𝑥_2𝑛,𝑥_2𝑛+1), then it follows from the above inequality that 𝜏+𝐹(𝑝(𝑥_2𝑛+1,𝑥_2𝑛))≤𝐹(𝑝(𝑥_2𝑛,𝑥_2𝑛+1)), a contradiction (as 𝜏>0). Therefore, we must have max{𝑝(𝑥_2𝑛,𝑥_2𝑛−1),𝑝(𝑥_2𝑛,𝑥_2𝑛+1)}=𝑝(𝑥_2𝑛,𝑥_2𝑛−1). So, setting 𝑝_𝑛=𝑝(𝑥_𝑛,𝑥_𝑛+1) from the above inequality we obtain that 𝐹𝑝_2𝑛≤𝐹𝑝_2𝑛−1−𝜏. (9) Again, using (6) we obtain 𝜏+𝐹𝑝𝑥_2𝑛+2,𝑥_2𝑛+1=𝜏+𝐹𝑝𝑔𝑥_2𝑛+1,𝑓𝑥_2𝑛=𝜏+𝐹𝑝𝑓𝑥_2𝑛,𝑔𝑥_2𝑛+1≤𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛,𝑓𝑥_2𝑛,𝑝𝑥_2𝑛+1,𝑔𝑥_2𝑛+1,𝑝𝑥_2𝑛,𝑔𝑥_2𝑛+1+𝑝𝑥_2𝑛+1,𝑓𝑥_2𝑛2=𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛+1,𝑥_2𝑛+2,𝑝𝑥_2𝑛,𝑥_2𝑛+2+𝑝𝑥_2𝑛+1,𝑥_2𝑛+12≤𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛+1,𝑥_2𝑛+2,𝑝𝑥_2𝑛,𝑥_2𝑛+1+𝑝𝑥_2𝑛+1,𝑥_2𝑛+22=𝐹max𝑝𝑥_2𝑛,𝑥_2𝑛+1,𝑝𝑥_2𝑛+1,𝑥_2𝑛+2. (10) If max{𝑝(𝑥_2𝑛,𝑥_2𝑛+1),𝑝(𝑥_2𝑛+1,𝑥_2𝑛+2)}=𝑝(𝑥_2𝑛+1,𝑥_2𝑛+2), then it follows from the above inequality that 𝜏+𝐹(𝑝(𝑥_2𝑛+2,𝑥_2𝑛+1))≤𝐹(𝑝(𝑥_2𝑛+2,𝑥_2𝑛+1)), a contradiction (as 𝜏>0). Therefore, we must have max{𝑝(𝑥_2𝑛,𝑥_2𝑛+1),𝑝(𝑥_2𝑛+1,𝑥_2𝑛+2)}=𝑝(𝑥_2𝑛,𝑥_2𝑛+1). So from the above inequality we obtain that 𝐹𝑝_2𝑛+1≤𝐹𝑝_2𝑛−𝜏. (11) Using (9) and (11) we obtain 𝐹𝑝_2𝑛≤𝐹𝑝_2𝑛−1𝑝−𝜏≤𝐹_2𝑛−2−2𝜏≤⋯≤𝐹𝑝₀−2𝑛𝜏. (12) Similarly 𝐹𝑝_2𝑛+1≤𝐹𝑝_2𝑛𝑝−𝜏≤𝐹_2𝑛−1−2𝜏≤⋯≤𝐹𝑝₀−(2𝑛+1)𝜏. (13) It follows from (12) and (13) that lim_𝑛→∞𝐹(𝑝_𝑛)=−∞. As 𝐹∈ℱ by (F2) we have lim_𝑛→∞𝑝_𝑛=0. (14) Again, by (F3), there exists 𝑘∈(0,1) such that lim_𝑛→∞𝑝^𝑘_𝑛𝐹𝑝_𝑛=0. (15) From (12) and (13) we have 𝑝^𝑘_2𝑛𝐹𝑝_2𝑛−𝐹𝑝₀≤−2𝑛𝑝^𝑘_2𝑛𝑝𝜏≤0,_𝑘2𝑛+1𝐹𝑝_2𝑛+1−𝐹𝑝₀≤−(2𝑛+1)𝑝^𝑘_2𝑛+1𝜏≤0. (16) Using (14) and (15) in the above inequalities we obtain lim_𝑛→∞𝑛𝑝_𝑛^𝑘=0. (17) It follows from above that there exists 𝑛₀∈ℕ such that 𝑛(𝑝_𝑛)^𝑘<1 for all 𝑛>𝑛₀; that is, 𝑝_𝑛<1𝑛^1/𝑘,∀𝑛>𝑛₀. (18) Let 𝑚,𝑛∈ℕ with 𝑚>𝑛>𝑛₀; then it follows from (18) that 𝑝𝑥_𝑛,𝑥_𝑚≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑛+2+⋯+𝑝𝑥_𝑚−1,𝑥_𝑚−𝑝𝑥_𝑛+1,𝑥_𝑛+1+𝑝𝑥_𝑛+2,𝑥_𝑛+2+⋯+𝑝𝑥_𝑚−1,𝑥_𝑚−1≤𝑝_𝑛+𝑝_𝑛+1≤1+⋯𝑛^1/𝑘+1(𝑛+1)^1/𝑘=+⋯_∞𝑖=𝑛1𝑖^1/𝑘. (19) As 𝑘∈(0,1), therefore the series ∑^∞_𝑖=𝑛1/𝑖^1/𝑘 converges; so it follows from the above inequality that lim_𝑛→∞𝑝(𝑥_𝑛,𝑥_𝑚)=0; that is, the sequence {𝑥_𝑛} is a 0-Cauchy sequence in 𝑋. Therefore, by 0-completeness of (𝑋,𝑝), there exists 𝑢∈𝑋 such that lim_{𝑛,𝑚→∞}𝑝𝑥_𝑛,𝑥_𝑚=lim_𝑛→∞𝑝𝑥_𝑛,𝑢=𝑝(𝑢,𝑢)=0. (20) We will prove that 𝑢 is a common fixed point of 𝑓 and 𝑔. We consider two cases.
Case  1. Suppose 𝑓 is continuous. Using continuity of 𝑓, (20), and Lemma 7 we obtain 𝑝𝑢,𝑓𝑢=lim_𝑛→∞𝑝𝑥_2𝑛+1,𝑓𝑢=lim_𝑛→∞𝑝𝑓𝑥_2𝑛,𝑓𝑢=𝑝𝑓𝑢,𝑓𝑢. (21) We claim that 𝑝(𝑓𝑢,𝑓𝑢)=0. Suppose 𝑝(𝑓𝑢,𝑓𝑢)>0. If for each 𝑛∈ℕ there exists 𝑘_𝑛 such that 𝑝(𝑥_𝑘𝑛+1,𝑓𝑢)=0 and 𝑘_𝑛>𝑘_𝑛−1 with 𝑘₀=1, then by Lemma 7 and (20) we have 𝑓𝑢=𝑢, and so, by Lemma 9, 𝑢 is a unique common fixed point of 𝑓 and 𝑔.
Now suppose there exists 𝑛₁∈ℕ such that 𝑝(𝑓𝑢,𝑔𝑥_𝑛)>0 for all 𝑛≥𝑛₁. Then, since 𝑝(𝑓𝑢,𝑓𝑢)>0, there exists 𝜀>0 such that 𝑝(𝑓𝑢,𝑓𝑢)>𝜀.
For any 𝑛≥𝑛₁ we have 𝜏+𝐹𝑝𝑓𝑢,𝑔𝑥_2𝑛+1≤𝐹max𝑝𝑢,𝑥_2𝑛+1,𝑝𝑢,𝑓𝑢,𝑝𝑥_2𝑛+1,𝑔𝑥_2𝑛+1,𝑝𝑢,𝑔𝑥_2𝑛+1+𝑝𝑥_2𝑛+1,𝑓𝑢2=𝐹max𝑝𝑢,𝑥_2𝑛+1,𝑝𝑓𝑢,𝑓𝑢,𝑝𝑥_2𝑛+1,𝑥_2𝑛+2,𝑝𝑢,𝑥_2𝑛+2+𝑝𝑥_2𝑛+1,𝑓𝑢2. (22) Using (20) and (21), there exists 𝑛₂∈ℕ such that 𝑝(𝑢,𝑥_2𝑛+1),𝑝(𝑥_2𝑛+1,𝑥_2𝑛+2)<𝜀/2 and 𝑝(𝑥_2𝑛+1,𝑓𝑢)<𝑝(𝑓𝑢,𝑓𝑢)+𝜀/2 for all 𝑛>𝑛₂; therefore, it follows from the above inequality that 𝜏+𝐹𝑝𝑓𝑢,𝑔𝑥_2𝑛+1𝜀≤𝐹max2,𝜀,𝑝𝑓𝑢,𝑓𝑢2,𝜀/2+𝑝𝑓𝑢,𝑓𝑢+𝜀/22,𝐹𝑝𝑓𝑢,𝑔𝑥_2𝑛+1<𝐹𝑝𝑓𝑢,𝑓𝑢. (23) As 𝐹∈ℱ by (F1) we have 𝑝(𝑓𝑢,𝑔𝑥_2𝑛+1)<𝑝(𝑓𝑢,𝑓𝑢) for all 𝑛>max{𝑛₁,𝑛₂}, a contradiction. Therefore, we must have 𝑝(𝑓𝑢,𝑓𝑢)=𝑝(𝑢,𝑓𝑢)=0; that is, 𝑓𝑢=𝑢, and by Lemma 9, 𝑢 is a unique common fixed point of 𝑓 and 𝑔. Similarly, if 𝑔 is continuous, then 𝑢 is a unique common fixed point of 𝑓 and 𝑔.
Case  2. Now suppose that 𝐹 is continuous. We can assume that there exists 𝑛₃ such that 𝑝(𝑥_𝑛,𝑓𝑢)>0 for all 𝑛≥𝑛₃; otherwise we get 𝑓𝑢=𝑢 (similar as in previous case).
For any 𝑛≥𝑛₃, we obtain from (6) that 𝜏+𝐹𝑝𝑓𝑢,𝑥_2𝑛+2=𝜏+𝐹𝑝𝑓𝑢,𝑔𝑥_2𝑛+1≤𝐹max𝑝𝑢,𝑥_2𝑛+1,𝑝𝑢,𝑓𝑢,𝑝𝑥_2𝑛+1,𝑔𝑥_2𝑛+1,𝑝𝑢,𝑔𝑥_2𝑛+1+𝑝𝑥_2𝑛+1,𝑓𝑢2≤𝐹max𝑝𝑢,𝑥_2𝑛+1,𝑝𝑢,𝑓𝑢,𝑝𝑥_2𝑛+1,𝑥_2𝑛+2,𝑝𝑢,𝑥_2𝑛+2+𝑝𝑥_2𝑛+1,𝑢+𝑝𝑢,𝑓𝑢2. (24) From (20), there exists 𝑛₄∈ℕ such that max𝑝𝑢,𝑥_2𝑛+1,𝑝𝑢,𝑓𝑢,𝑝𝑥_2𝑛+1,𝑥_2𝑛+2,𝑝𝑢,𝑥_2𝑛+2+𝑝𝑥_2𝑛+1,𝑢+𝑝𝑢,𝑓𝑢2=𝑝𝑢,𝑓𝑢, (25) for all 𝑛>𝑛₄. Therefore, it follows from (24) that 𝜏+𝐹𝑝𝑓𝑢,𝑥_2𝑛+2≤𝐹𝑝𝑢,𝑓𝑢,∀𝑛>max𝑛₃,𝑛₄. (26) Using continuity of 𝐹 and letting 𝑛→∞ in the above inequality we obtain 𝜏+𝐹𝑝𝑓𝑢,𝑢≤𝐹𝑝𝑢,𝑓𝑢, (27) a contradiction (as 𝜏>0). Therefore we must have 𝑝(𝑢,𝑓𝑢)=0; that is, 𝑓𝑢=𝑢. Again by Lemma 9  𝑢 is a unique common fixed point of 𝑓 and 𝑔.

The following corollaries are immediate consequences of Theorem 10.

Corollary 11. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑓,𝑔∶𝑋→𝑋 two mappings. Suppose there exist 𝐹∈ℱ and 𝜏>0 such that, for all 𝑥,𝑦∈𝑋, 𝑝(𝑓𝑥,𝑔𝑦)>0, one has 𝜏+𝐹𝑝𝑓𝑥,𝑔𝑦≤𝐹max𝑝𝑥,𝑦,𝑝𝑥,𝑓𝑥,𝑝𝑦,𝑔𝑦. (28) If (i) 𝑓 or 𝑔 is continuous or (ii) 𝐹 is continuous, then 𝑓 and 𝑔 have a unique common fixed point 𝑢∈𝑋, and 𝑝(𝑢,𝑢)=0.

Corollary 12. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑓,𝑔∶𝑋→𝑋 two mappings. Suppose there exist 𝐹∈ℱ and 𝜏>0 such that, for all 𝑥,𝑦∈𝑋, 𝑝(𝑓𝑥,𝑔𝑦)>0, one has 𝜏+𝐹𝑝𝑓𝑥,𝑔𝑦≤𝐹𝑝𝑥,𝑦. (29) If (i) 𝑓 or 𝑔 is continuous or (ii) 𝐹 is continuous, then 𝑓 and 𝑔 have a unique common fixed point 𝑢∈𝑋, and 𝑝(𝑢,𝑢)=0.

Taking 𝑓=𝑔 in Theorem 10 we obtain the following corollary.

Corollary 13. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑓∶𝑋→𝑋 a mapping. Suppose there exist 𝐹∈ℱ and 𝜏>0 such that, for all 𝑥,𝑦∈𝑋, 𝑝(𝑓𝑥,𝑓𝑦)>0, one has 𝑝𝑝,𝑝𝜏+𝐹𝑓𝑥,𝑓𝑦≤𝐹max𝑥,𝑦,𝑝𝑥,𝑓𝑥,𝑝𝑦,𝑓𝑦𝑥,𝑓𝑦+𝑝𝑦,𝑓𝑥2. (30) If (i) 𝑓 is continuous or (ii) 𝐹 is continuous, then 𝑓 has a unique fixed point 𝑢∈𝑋, and 𝑝(𝑢,𝑢)=0.

The following is a fixed point result for 𝐹-contraction in 0-complete partial metric space and follows from Corollary 13.

Corollary 14. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑓∶𝑋→𝑋 a mapping. Suppose there exist 𝐹∈ℱ and 𝜏>0 such that, for all 𝑥,𝑦∈𝑋,  𝑝(𝑓𝑥,𝑓𝑦)>0, one has 𝜏+𝐹𝑝𝑓𝑥,𝑓𝑦≤𝐹𝑝𝑥,𝑦. (31) If (i) 𝑓 is continuous or (ii) 𝐹 is continuous, then 𝑓 has a unique fixed point 𝑢∈𝑋, and 𝑝(𝑢,𝑢)=0.

The following are some examples which illustrate the above results and that the generalizations are proper.

Example 15. Let 𝑋=[0,1]∩ℚ, and let 𝑝∶𝑋×𝑋→ℝ⁺ be defined by 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋. Then (𝑋,𝑝) is a 0-complete partial metric space, but it is not complete partial metric space. Indeed, the induced metric space (𝑋,𝑑), where 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈𝑋, is not a complete metric space. Define 𝑓∶𝑋→𝑋 by 𝑥𝑓𝑥=4,if[);1𝑥∈0,18,if𝑥=1. (32) We note that 𝑓 satisfies the condition (31) of Corollary 14 with 𝐹(𝑡)=log𝑡 for all 𝑡∈ℝ⁺, 𝜏∈(0,log4], and 0∈𝑋 is unique fixed point of 𝑓 with 𝑝(0,0)=0. On the other hand, the metric version of Corollary 14 is not applicable because (𝑋,𝑑) is not complete metric space. Also, this example shows that the class of 𝐹-contraction in partial metric spaces is wider than that in metric spaces. Indeed, for 𝑥=1, 𝑦=9/10, there is no 𝐹∈ℱ and 𝜏>0 such that 𝜏+𝐹(𝑑(𝑓𝑥,𝑓𝑦))≤𝐹(𝑑(𝑥,𝑦)), where 𝑑 is usual as well as the metric induced by 𝑝.

The following example illustrates the case when Corollary 11 is applicable, while Corollary 12 is not, as well as the metric versions of Corollary 11 are not applicable.

Example 16. Let 𝑋={0,1,2,3}, and let 𝑝∶𝑋×𝑋→ℝ⁺ be defined by 𝑝𝑥,𝑦=𝑥−𝑦+max𝑥,𝑦,∀𝑥,𝑦∈𝑋. (33) Then, (𝑋,𝑝) is a 0-complete partial metric space. Note that the metric induced by 𝑝 is given by 𝑑(𝑥,𝑦)=3|𝑥−𝑦| for all 𝑥,𝑦∈𝑋. Define 𝑓,𝑔∶𝑋→𝑋 by 𝑓0=0,𝑓1=0,𝑓2=1,𝑓3=1,𝑔0=0,𝑔1=0,𝑔2=0,𝑔3=1. (34) Now, by a careful observation one can see that 𝑓 and 𝑔 satisfy the condition (28) of Corollary 11, with 𝐹(𝑡)=log𝑡 for all 𝑡∈ℝ⁺, 𝜏∈(0,log(3/2)], and 𝑓 and 𝑔 have a unique common fixed point, namely, 0∈𝑋 with 𝑝(0,0)=0. While, 𝑓 and 𝑔 do not satisfy the condition (29) of Corollary 12. Indeed, for 𝑥=2, 𝑦=2, there are no 𝐹∈ℱ and 𝜏>0 such that 𝜏+𝐹(𝑝(𝑓𝑥,𝑔𝑦))≤𝐹(𝑝(𝑥,𝑦)). Again, 𝑓 and 𝑔 do not satisfy the metric versions of Condition (28) of Corollary 11. Indeed, 𝑥=2, 𝑦=1 are the points where the induced metric and usual metric versions of condition (28) of Corollary 11 are not satisfied.