Let (X, p) be a PMS, c ⊂ X and φ : C → ℝ^{+} a function on C. Then, the function φ is called a lower semicontinuous (l.s.c) on C whenever
Also, let T : X → X be an arbitrary selfmapping on X such that
where T is called a Caristi map on (X, p).
The following lemma will be used in the proof of the main theorem.
Lemma 3. (see, e.g. [8, 7]) Let (X, p) be a complete PMS. Then

(A)
If p(x, y) = 0 then x = y,

(B)
If x ≠ y, then p(x, y) > 0.
Proof. Proof of (A). Let p(x, y) = 0. By (PM3), we have p(x, x) ≤ p(x, y) = 0 and p(y, y) ≤ p(x, y) = 0. Thus, we have
Hence, by (PM2), we have x = y.
Proof of (B). Suppose x ≠ y. By definition p(x, y) ≥ 0 for all x, y ∈ X. Assume p(x, y) = 0. By part (A), x = y which is a contradiction. Hence, p(x, y) > 0 whenever x ≠ y.
□
Lemma 4. (see, e.g. [8, 7]) Assume x_{
n
} → z as n → ∞ in a PMS (X, p) such that p(z, z) = 0. Then, lim_{n→∞}p(x_{
n
} , y) = p(z, y) for every y ∈ X.
Proof. First, note that lim _{n→∞}p(x_{
n
} , z) = p(z, z) = 0. By the triangle inequality, we have
and
Hence,
Letting n → ∞ we conclude our claim. □
The following theorem is an extension of the result of Caristi ([9]; Theorem 2.1)
Theorem 5. Let (X, p) be a complete PMS, φ : X → ℝ^{+}a lower semicontinuous (l. s.c) function on X. Then, each selfmapping T : X → X satisfying (2.2) has a fixed point in X.
Proof. For each x ∈ X, define
Since x ∈ S(x), then S(x) ≠ ∅. From (2.3), we have 0 ≤ α (x) ≤ φ(x).
Take x ∈ X. We construct a sequence {x_{
n
} } in the following way:
Thus, one can easily observe that
Note that (2.5) implies that {φ(x_{
n
} )} is a decreasing sequence of real numbers, and it is bounded by zero. Therefore, the sequence {φ(x_{
n
} )} is convergent to some positive real number, say L. Thus, regarding (2.5), we have
From (2.5) and (2.6), for each k ∈ ℕ, there exists N_{
k
} ∈ ℕ such that
Regarding the monotonicity of {φ(x_{
n
} )}, for m ≥ n ≥ N_{
k
} , we have
Thus, we obtain
On the other hand, taking (2.5) into account, together with the triangle inequality, we observe that
Analogously,
By induction, we obtain that
and taking (2.9) into account, (2.12) turns into
Since the sequence {φ(x_{
n
} )} is convergent which implies that the righthand side of (2.13) tends to zero. By definition,
Since p(x_{
n
} , x_{
m
} ) tends to zero as n, m → ∞, then (2.14) yields that {x_{
n
} } is Cauchy in (X, d_{
p
} ). Since (X, p) is complete, by Lemma 2, (X, d_{
p
} ) is complete, and thus the sequence {x_{
n
} } is convergent in X, say z ∈ X. Again by Lemma 2,
Since lim_{n,m→∞}p(x_{
n
} , x_{
m
} ) = 0, then by (2.15), we have p(z, z) = 0.
Because φ is l.s.c together with (2.13)
and thus
By definition, z ∈ S(x_{
n
} ) for all n ∈ ℕ and thus α(x_{
n
} ) ≤ φ(z). Taking (2.6) into account, we obtain L ≤ φ (z). Moreover, by l.s.c of φ and (2.6), we have φ (z) lim_{n→∞}φ (x_{
n
} ) = L. Hence, φ (z) = L.
Since z ∈ S(x_{
n
} ) for each n ∈ ℕ and (2.2), then Tz ∈ S(z) and by triangle inequality
is obtained. Hence, Tz ∈ S(x_{
n
} ) for all n ∈ ℕ which yields that α(x_{
n
} ) ≤ φ(Tz) for all n ∈ ℕ.
From (2.6), the inequality φ(Tz) ≥ L is obtained. By φ (Tz) ≤ φ (z), observed by (2.2), and by the observation φ (z) = L, we achieve as follows:
Hence, φ(Tz) = φ (z). Finally, by (2.2), we have p(Tz, z) = 0. Regarding Lemma 3, Tz = z.
□
The following theorem is a generalization of the result in [10]
Theorem 6. Let φ : X → ℝ^{+}be a l.s.c function on a complete PMS. If φ is bounded below, then there exits z ∈ X such that
Proof. It is enough to show that the point z, obtained in the Theorem 5, satisfies the statement of the theorem. Following the same notation in the proof of Theorem 5, it is needed to show that x ∉ S(z) for x ≠ z. Assume the contrary, that is, for some w ≠ z, we have w ∈ S(z). Then, 0 < p(z, w) ≤ φ(z)  φ (w) implies φ (w) < φ (z) = L. By triangular inequality,
which implies that w ∈ S(x_{
n
} ) and thus α(x_{
n
} ) ≤ φ(w) for all n ∈ ℕ. Taking the limit when n tends to infinity, one can easily obtain L ≤ φ (w), which is in contradiction with φ (w) < φ (z) = L. Thus, for any x ∈ X, x ≠ z implies x ∉ S(z) that is,
□
Theorem 7. Let X and Y be complete partial metric spaces and T : X → X an selfmapping. Assume that R : X → Y is a closed mapping, φ : X → ℝ^{+}is a l.c.s, and a constant k > 0 such that
Then, T has a fixed point.
Proof. For each x ∈ X, we define
For x ∈ X set x_{1} : = x and construct a sequesnce x_{1}, x_{2}, x_{3}, ..., x_{
n
} , ... as in the proof of Theorem 5:
x_{n+1}∈ S(x_{
n
} ) such the for each n ∈ ℕ.
As in Theorem 5, one can easily get that {x_{
n
} } is convergent to z ∈ X. Analogously, {Rx_{
n
} } is Cauchy sequence in Y and convergent to some t. Since R is closed mapping, Rz = t. Then, as in the proof of Theorem 5, we have
As in the proof of Theorem 6, we get that x ≠ z implies x ∉ S(z). From (2.17), Tz ∈ S(z), we have Tz = z.
□
Define p_{
x
} : X → R^{+} such that p_{
x
} (y) = p(x, y).
Theorem 8. Let (X, p) be a complete PMS. Assume for each x ∈ X, the function p_{
x
} defined above is continuous on X, and is a family of mappings f : X → X. If there exists a l.s.c function φ : X → ℝ^{+}such that
then, for each x ∈ X, there is a common fixed point z ofsuch that
Proof. Let S(x): = {y ∈ X : p(x, y) ≤ φ(x)  φ (y)} and α(x): = inf{φ (y): y ∈ S(x)} for all x ∈ X. Note that x ∈ S(x), and so S(x) ≠ ∅ as well as 0 ≤ α (x) ≤ φ(x).
For x ∈ X, set x_{1} := x and construct a sequence x_{1}, x_{2}, x_{3}, ..., x_{
n
} , ... as in the proof of Theorem 5: x_{n+1}∈ S(x_{
n
} ) such that for each n ∈ ℕ. Thus, one can observe that for each n,
(i) p(x_{
n
} , x_{n+1}) ≤ φ(x_{
n
} )  φ(x_{n+1}).
(ii) .
Similar to the proof of Theorem 5, (ii) implies that
Also, using the same method as in the proof of Theorem 5, it can be shown that {x_{
n
} } is a Cauchy sequence and converges to some z ∈ X and φ(z) = L.
We shall show that f(z) = z for all . Assume on the contrary that there is such that f(z) ≠ z. Replace x = z in (2.19); then we get φ(f(z)) < φ (z) = L:
Thus, by definition of L, there is n ∈ ℕ such that φ (f(z)) < α(x_{
n
} ). Since z ∈ S(x_{
n
} ), we have
which implies that f(z) ∈ S(x_{
n
} ). Hence, α(x_{
n
} ) ≤ φ(f(z)) which is in a contradiction with φ (f(z)) < α(x_{
n
} ). Thus, f(z) = z for all .
Since z ∈ S(x_{
n
} ), we have
is obtained. □
The following theorem is a generalization of ([11]; Theorem 2.2).
Theorem 9. Let A be a set, (X, p) as in Theorem 8, g : A → X a surjective mapping anda family of arbitrary mappings f : A → X. If there exists a l.c.s: function φ : X → [0, ∞) such that
and each a ∈ A, then g andhave a common coincidence point, that is, for some b ∈ A; g(b) = f(b) for all .
Proof. Let x be arbitrary and z ∈ X as in Theorem 8. Since g is surjective, for each x ∈ X there is some a = a(x) such that g(a) = x. Let be a fixed mapping. Define by f a mapping h = h(f) of X into itself such that h(x) = f(a), where a = a(x), that is, g(a) = x. Let be a family of all mappings h = h(f). Then, (2.21) yields that
Thus, by Theorem 8, z = h(z) for all . Hence g(b) = f(b) for all , where b = b(z) is such that g(b) = z.
Example 10. Let X = ℝ^{+}and p(x, y) = max{x, y}; then (X, p) is a PMS (see, e.g. [6].) Suppose T : X → X such thatfor all x ∈ X and ϕ(t): [0, ∞) → [0, ∞) such that ϕ (t) = 2t. Then
Thus, it satisfies all conditions of Theorem 5. it guarantees that T has a fixed point; indeed x = 0 is the required point.