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Generalizations of Caristi Kirk's Theorem on Partial Metric Spaces

Fixed Point Theory and Applications20112011:4

  • Received: 27 January 2011
  • Accepted: 21 June 2011
  • Published:


In this article, lower semi-continuous maps are used to generalize Cristi-Kirk's fixed point theorem on partial metric spaces. First, we prove such a type of fixed point theorem in compact partial metric spaces, and then generalize to complete partial metric spaces. Some more general results are also obtained in partial metric spaces.

2000 Mathematics Subject Classification 47H10,54H25


  • Partial metric space
  • Lower semi-continuous
  • Fixed point theory

1. Introduction and preliminaries

In 1992, Matthews [1, 2] introduced the notion of a partial metric space which is a generalization of usual metric spaces in which d(x, x) are no longer necessarily zero. After this remarkable contribution, many authors focused on partial metric spaces and its topological properties (see, e.g. [3]-[8])

Let X be a nonempty set. The mapping p : X × X → [0, ∞) is said to be a partial metric on X if for any x, y, z X the following conditions hold true:

(PM1) p(x, y) = p(y, x) (symmetry)

(PM2) If p(x, x) = p(x, y) = p(y, y) then x = y (equality)

(PM3) p(x, x) ≤ p(x, y) (small self-distances)

(PM4) p(x, z) + p(y, y) ≤ p(x, y) + p(y, z) (triangularity)

for all x, y, z X. The pair (X, p) is then called a partial metric space(see, e.g. [1, 2]). We use the abbreviation PMS for the partial metric space (X, p).

Notice that for a partial metric p on X, the function d p : X × X → [0, ∞) given by

is a (usual) metric on X. Observe that each partial metric p on X generates a T0 topology τp on X with 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. Similarly, closed p-ball is defined as B p [x, ε] = {y X : p(x, y) ≤ p(x, x) + ε}

Definition 1. (see, e.g. [1, 2, 6])

(i) A sequence {x n } in a PMS (X, p) converges to x X if and only if p(x, x) = limn→∞p(x, x n ),

(ii) a sequence {x n } in a PMS (X, p) is called Cauchy if and only if limn,m→∞p(x n , x m ) exists (and finite),

(iii) A PMS (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) = limn,m→∞p(x n , x m ).

(iv) A mapping f : XX is said to be continuous at x0 X, if for every ε > 0, there exists δ > 0 such that f(B(x0, δ)) B(f(x0), ε).

Lemma 2. (see, e.g. [1, 2, 6])

(A) A sequence {x n } is Cauchy in a PMS (X, p) if and only if {x n } is Cauchy in a metric space (X, d p ),

(B) A PMS (X, p) is complete if and only if a metric space (X, d p ) is complete. Moreover,

2. Main Results

Let (X, p) be a PMS, c X and φ : C+ a function on C. Then, the function φ is called a lower semi-continuous (l.s.c) on C whenever
Also, let T : XX be an arbitrary self-mapping 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
  1. (A)

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

  2. (B)

    If xy, 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 xy. 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 xy.

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, limn→∞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

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+a lower semi-continuous (l. s.c) function on X. Then, each self-mapping T : XX 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 , there exists N k such that
Regarding the monotonicity of {φ(x n )}, for mnN k , we have
Thus, we obtain
On the other hand, taking (2.5) into account, together with the triangle inequality, we observe that
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 right-hand 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 limn,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 and thus α(x n ) ≤ φ(z). Taking (2.6) into account, we obtain Lφ (z). Moreover, by l.s.c of φ and (2.6), we have φ (z) limn→∞φ (x n ) = L. Hence, φ (z) = L.

Since z S(x n ) for each n and (2.2), then Tz S(z) and by triangle inequality

is obtained. Hence, Tz S(x n ) for all n which yields that α(x n ) ≤ φ(Tz) for all 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+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 xz. Assume the contrary, that is, for some wz, 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 . Taking the limit when n tends to infinity, one can easily obtain Lφ (w), which is in contradiction with φ (w) < φ (z) = L. Thus, for any x X, xz implies x S(z) that is,

Theorem 7. Let X and Y be complete partial metric spaces and T : XX an self-mapping. Assume that R : XY is a closed mapping, φ : X+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 x1 : = x and construct a sequesnce x1, x2, x3, ..., x n , ... as in the proof of Theorem 5:

xn+1 S(x n ) such the for each 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 xz implies x S(z). From (2.17), Tz S(z), we have Tz = z.

Define p x : XR+ 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 : XX. If there exists a l.s.c function φ : X+such that
then, for each x X, there is a common fixed point z of such 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 x1 := x and construct a sequence x1, x2, x3, ..., x n , ... as in the proof of Theorem 5: xn+1 S(x n ) such that for each n . Thus, one can observe that for each n,

(i) p(x n , xn+1) ≤ φ(x n ) - φ(xn+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 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 : AX a surjective mapping and a family of arbitrary mappings f : AX. If there exists a l.c.s: function φ : X → [0, ∞) such that

and each a A, then g and have 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 = +and p(x, y) = max{x, y}; then (X, p) is a PMS (see, e.g. [6].) Suppose T : XX such that for 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.


Authors’ Affiliations

Department of Mathematics, Atilim University, 06836 Incek, Ankara, Turkey


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© Karapinar; licensee Springer. 2011

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