CARISTI AND BANACH FIXED POINT THEOREM ON PARTIAL METRIC SPACE.

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1 CARISTI AND BANACH FIXED POINT THEOREM ON PARTIAL METRIC SPACE. MOHAMAD MUSLIKH Abstract. In this article are shown Caristi and Banach fixed point theorem type in the partial metric space. Both will be proven by using Ekeland s variational principle in partial metric space which also introduced in this article. 1. Introduction Caristi fixed point theorem was generalized by several authors. For example, Bae [1] generalized Caristi s theorem to prove the fixed point theorem for weakly contractive set-valued mapping as well as Banach fixed point theorem in the other way. In recent years many work on domain theory have been made in order to equip semantics domain with a notion of distance, see [2]-[3], [6]-[9]. In particular, Matthews [8] introduced the notion of a partial metric space as a part of the study of donatational semantic of data flow network, showing that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. In this paper we present the Caristi and Banach fixed point theorem type in partial metric spaces. We would also introduce Ekeland variational principle on partial metric spaces and its applications to fixed point. 2. Prelimanaries First, we start with some preliminaries on partial metric spaces. For more details, we refer to reader to [8]. Definition 2.1. Let X be nonempty set. The mapping p : X X R + is said partial metric on X if satisfies (P1) p(x, x) p(x, y) for all x, y X; (P2) x = y if and only if p(x, x) = p(y, y) = p(x, y); (P3) p(x, y) = p(y, x) for all x, y X; (P4) p(x, z) p(x, y) + p(y, z) p(y, y) for all x, y, z X. The pairs (X, p) is called a partial metric space. Note that the self-distance of any point need not be zero. A partial metric is a metric on X if p(x, x) = 0 for any x X Mathematics Subject Classification. Primary 47H10, Secondary 58E30. Key words and phrases. fixed point, partial metric. This work was done while the author was a visiting scholar at a northern university. 1

2 2 MOHAMAD MUSLIKH Example 2.2. Let R be a set real number and the distance function p : R R R be defined by Then p is a partial metric on R. p(x, y) = 1 { x y + x + y }, x, y R. 2 Lemma 2.3. Let (X, p) be a partial metric space, and the function d p : X X [0, ) be defined by Then d p is a metric. d p (x, y) = 2p(x, y) p(x, x) p(y, y), x, y X. Proof. (i) Clear for all x, y X then d p (x, y) 0. (ii) From (P2), we have x = y p(x, x) = p(y, y) = p(x, y) d p (x, y) = 2p(x, y) p(x, x) p(y, y) d p (x, y) = 2p(x, y) 2p(x, y) d p (x, y) = 0. (iii) Clear for all x, y X, d p (x, y) = d p (y, x) (iv) For all x, y, z X and from (P4), we obtain d p (x, z) = 2p(x, z) p(x, x) p(z, z) 2(p(x, y) + p(y, z) p(y, y)) p(x, x) p(z, z) = (2p(x, y) p(x, x) p(y, y)) + (2p(y, z) p(y, y) p(z, z)) = d p (x, y) + d p (y, z). The Lemma 2.3, describe that metric is a special case of partial metric. Therefore the partial matric is a generalization of metric. Definition 2.4. Let (X, p) be a partial metric space, a point x 0 X and > 0. The open ball for a partial metric p are sets of the form B (x 0 ) = {x X p(x 0, x) < }. Since p(x 0, x 0 ) > 0, the open ball are sets of the form B +p(x0,x 0)(x 0 ) = {x X p(x 0, x) < + p(x 0, x 0 )}. Contrary to the metric space case, some open balls may be empty. If > p(x 0, x 0 ), then B (x 0 ) = B p(x0,x 0)(x 0 ). If 0 < p(x 0, x 0 ), we obtain B (x 0 ) = {x X p(x 0, x) < p(x 0, x 0 )} =. This mean the open ball B p(x0,x 0)(x 0 ) be empty set. However may be point x 0 / B p(x0,x 0)(x 0 ). Definition 2.5. A sequence x n in a partial metric space (X, p) converges to x 0 X if, for any > 0 such that x 0 B (x 0 ) there exists N N so that for any n N, x n B (x 0 ). We write lim n x n = x 0. Lemma 2.6. A sequence x n in a partial metric space (X, p). Then x n converges to point x 0 X if and only if lim n p(x n, x 0 ) = p(x 0, x 0 ).

3 CARISTI AND BANACH FIXED POINT THEOREM ON PARTIAL METRIC SPACE 3 Proof. By Definition 2.5, for any > 0, p(x n, x 0 ) < for any n N. Since B (x 0 ) of course p(x 0, x 0 ) this implies p(x n, x 0 ) p(x 0, x 0 ) <, for any n N so that lim n p(x n, x 0 ) = p(x 0, x 0 ). Conversly suppose that p(x 0, x 0 ) = lim n p(x n, x 0 ). If x 0 B (x 0 ), then there exists N N such that for any n N, p(x n, x 0 ) <. This mean x n B (x 0 ) for any n N. By Definition 2.5, x n converges to point x 0 X. If x 0 B (a) with a X, that is p(a, x 0 ) <, then there exists N N such that for any n N, p(x n, x 0 ) p(x 0, x 0 ) < p(x 0, a) so that for any n N we obtain p(x n, a) p(x n, x 0 ) + p(x 0, a) p(x 0, x 0 ) < ( p(x 0, a)) + p(x 0, a) =. This means for any n N, x n B (a). Definition 2.7. A sequence x n in a partial metric space (X, p) is called properly converges to x X if x n converges to x and lim p(x n, x n ) = p(x, x). n In other words, a sequence x n properly converges to x X if lim n p(x n, x) and lim n p(x n, x n ) exists and lim p(x n, x n ) = lim p(x n, x) = p(x, x). n n Notice that every convergent sequence in a metric space converging in partial metric spaces. Definition 2.8. A sequence x n in a partial metric space (X, p) is called a Cauchy sequence if lim m,n p(x n, x m ) exists and is finite. In other words, x n is Cauchy if the numbers sequence p(x n, x m ) converges to some λ R as n and m approach to infinity, that is, if lim n,m p(x n, x m ) = λ <. This means for every > 0 there exists N N such that for all m, n N, p(x n, x m ) λ <. If (X, p) is a metric space then λ = 0. Lemma 2.9. A sequence x n in a partial metric space (X, p). If x n properly converges to x then x n is Cauchy sequence. Proof. By Definition 2.7, p(x, x) = lim n p(x n, x) = lim n p(x n, x n ). By (P1) and (P4) we have p(x, x) = lim p(x n, x n ) lim p(x n, x m ) m,n n,m lim p(x n, x) + lim p(x m, x) p(x, x) n,m n,m = p(x, x). Hence lim n,m p(x n, x m ) = p(x, x). This means, there exists λ R + such that λ = p(x, x) and lim n,m p(x n, x m ) = λ. The sequence x n is Cauchy proved. Theorem A sequence x n in a partial metric space (X, p) is a Cauchy, if and only if for all > 0 there exists N N such that for all n, m N we have p(x n, x m ) p(x m, x m ) < Proof. Since x n is Cauchy, there exists λ R + such that for all > 0 there exists N N and for all n, m N we have p(x n, x m ) λ < 2.

4 4 MOHAMAD MUSLIKH Let n = m N, then p(x m, x m ) λ < 2. Therefore p(x n, x m ) p(x m, x m ) p(x n, x m ) λ + p(x m, x m ) λ <. By (P1), we obtain p(x n, x m ) p(x m, x m ) <. Conversly it is obvious. Definition A partial metric space is complete if every Cauchy sequence properly converges. Definition Let (X,p) be a partial metric space and A be a nonempty subset of X. The diameter of A, denoted by D(A), is given by D(A) = sup{p(x, y) p(x, x) : x, y A} Theorem Let (X, p) be a complete partial metric space and F n be a decreasing sequence (that is, F n F n+1 ) of nonempty closed subsets of X such that D(F n ) 0 as n. Then the intersection n=1f n contains exactly one point. Proof. The first, construct a sequence x n in X by selecting a point x n F n for each n N. Since F n F n+1 for all n, we have x n F n F m for all n > m Let > 0 be given. Since D(F n ) 0, there exists N N such that D(F n ) < for each n N. Since F m, F n F N for each n, m N. Therefore x n, x m F N for each n, m N and thus, we have p(x n, x m ) p(x m, x m ) D(F n ) <. By Theorem 2.10, x n is Cauchy sequence. Since X is complete, there exists x X such that p(x n, x ) p(x, x ) < for each n N. Let n = N be fixed. Then the subsequence {x n, x n+1, } of the sequence x n is contained in F n and still converges to x. F n is closed in complete partial metric space (X, p), it is complete and so x F n for each n N. Hence x n=1f n, that is n=1f n Finally, we show that x is unique in n=1f n. If y n=1f n, then x, y F n for each n N. Therefore 0 p(x, y) p(y, y) D(F n ) 0 as n and 0 p(x, y) p(x, x ) D(F n ) 0 as n. Thus p(x, y) = p(y, y) = p(x, x ). By (P2), x = y. Definition 2.14 ([8]). Let (X, p) be a partial metric space. The mapping f : X X is called a contraction on X if there exists k (0, 1) such that for every x, y X we have (2.1) p(f(x), f(y)) p(f(y), f(y)) k(p(x, y) p(y, y)) Theorem 2.15 ([8]). Each contraction in a complete partial metric space has a unique fixed point. Proof. Suppose f : X X is contraction in a complete partial metric space. Let x n+1 = f(x n ) for n 0. We will first show that x n is a Cauchy sequence. Since f is contraction we obtain p(f(x 0 ), f(x 1 )) p(f(x 1 ), f(x 1 )) k(p(x 0, x 1 ) p(x 1, x 1 )) p(f(x 1 ), f(x 2 )) p(f(x 2 ), f(x 2 )) k 2 (p(x 0, x 1 ) p(x 1, x 1 )). p(f(x n ), f(x n+1 )) p(f(x n+1 ), f(x n+1 )) k n+1 (p(x 0, x 1 ) p(x 1, x 1 )).

5 CARISTI AND BANACH FIXED POINT THEOREM ON PARTIAL METRIC SPACE 5 For all n, m N we obtain p(f(x n ), f(x n+m )) p(f(x n+m ), f(x n+m )) p(f(x n ), f(x n+m 1 )) + p(f(x n+m 1 ), f(x n+m )) p(f(x n+m 1 ), f(x n+m 1 )) p(f(x n+m ), f(x n+m )) k n+m 1 (p(x 0, x 1 ) p(x 1, x 1 )) + p(f(x n+m 1 ), f(x n+m )) p(f(x n+m ), f(x n+m )) k n+m 1 (p(x 0, x 1 ) p(x 1, x 1 )) + p(f(x n+m 1 ), f(x n+m 2 )) + p(f(x n+m 2 ), f(x n+m )) p(f(x n+m 2 ), f(x n+m 2 )) p(f(x n+m ), f(x n+m )) (k n+m 1 + k n+m k n )(p(x 0, x 1 ) p(x 1, x 1 )) = kn 1 k (p(x 0, x 1 ) p(x 1, x 1 )). By Theorem 2.10, x n to be a Cauchy sequence. Since (X, p) is a complete partial metric space, x n properly converges to x X say. We now show that x is a fixed point of f. For all > 0 there exists N N such that for n N, p(x n, x ) p(x n, x n ) < 1 + k and Thus for n N, p(x n, x ) p(x, x ) < 1 + k, p(f(x ), x ) p(x, x ) p(f(x ), f(x n )) + p(f(x n ), x ) p(f(x n ), f(x n )) Thus, as > 0 arbitrary, then p(x, x ) (p(f(x n ), x ) p(f(x n ), f(x n ))) + (p(f(x ), f(x n )) p(x, x )) k(p(x n, x ) p(x n, x n ) + k(p(x, x n ) p(x, x ) < k( 1 + k k ) <. (2.2) p(f(x ), x ) = p(x, x ) Similarly, for n N, p(f(x ), x ) p(f(x ), f(x )) p(f(x ), f(x n )) + p(f(x n ), x ) p(f(x n ), f(x n )) p(f(x ), f(x )) (p(f(x n ), x ) p(f(x n ), f(x n ))) + (p(f(x ), f(x n )) p(f(x ), f(x )) k(p(x n, x ) p(x n, x n ) + k(p(x, x n ) p(x, x ) < k( 1 + k k ) <.

6 6 MOHAMAD MUSLIKH Thus, as > 0 arbitrary, then (2.3) p(f(x ), x ) = p(f(x ), f(x )) Thus from (2.2),(2.3) and by (P2), f(x ) = x and so f has been that shown to have a fixed point. It just remains to show that x is unique. Suppose y X and y = f(y ), then, p(x, y ) p(y, y ) = p(f(x ), f(y )) p(f(y ), p(y )) k(p(x, y ) p(y, y )). it follow p(x, y ) p(y, y ) = 0 as 0 k < 1. Similarly, Suppose x X and x = f(x ), then, p(x, y ) p(x, x ) = p(f(x ), f(y )) p(f(x ), f(x )) k(p(x, y ) p(x, x )). it follow p(x, y ) p(x, x ) = 0 as 0 k < 1. According By axiom (P2), p(x, y ) = p(x, x ) = p(y, y ), thus y = x is unique. 3. Main results In this main result will be shown a fixed point theorem of both Caristi [4] and Banach [8] on partial metric space. In addition it shall be shown the prove Caristi fixed point theorem with two methods, that is, without and use Ekeland s variational principle. Similarly, for the Banach fixed point theorem. We start with the following lemma needed to prove our maint result. Lemma 3.1. Let (X, p) be a partial metric space and the function ϕ : X [0, ) is lower semicontinuous. For any x, y X we define relation p on X by (3.1) x p y p(x, y) p(x, x) ϕ(x) ϕ(y) Then the relation p is partial ordered on X. Proof. (i) It is clear that p(x, x) p(x, x) = 0 = ϕ(x) ϕ(x) so that x p x is reflexsif. (ii) If x p y then p(x, y) p(x, x) ϕ(x) ϕ(y) and if y p x then p(y, x) p(y, y) ϕ(y) ϕ(x). This implies 2p(x, y) p(x, x) p(y, y) = 0. Of course p(x, y) = p(x, x) = p(y, y). By P2, we obtain x = y. (iii) If x p y then p(x, y) p(x, x) ϕ(x) ϕ(y) and if y p z then p(y, z) p(y, y) ϕ(y) ϕ(x). This implies 2p(x, z) p(x, x) p(x, y) + p(y, z) p(y, y) p(x, x) ϕ(x) ϕ(z) and hence x p z. Lemma 3.2. (Zorn s Lemma). Let X be nonempty partially ordered set in which every totally set has a upper bound. Then X has at least one maximal element. The following is Caristi fixed point theorem type on partial metric space.

7 CARISTI AND BANACH FIXED POINT THEOREM ON PARTIAL METRIC SPACE 7 Theorem 3.3. Let (X, p) be a complete partial metric space and f : X X be a mapping on X. Suppose there exists a lower semicontinuous function ϕ : X [0, ) such that (3.2) p(x, f(x)) p(f(x), f(x)) ϕ(x) ϕ(f(x)) for all x X. Then f has a fixed point. Proof. For any x, y X we define the relation p on X by (3.3) x p y p(x, y) p(y, y) ϕ(x) ϕ(y). By Lemma 3.1, (X, p ) is a partial ordered. Let x 0 X be an arbitrary but fixed element of X. Then by Zorn s Lemma, we obtain totally ordered subset M of X containing x 0. Let M = {x α } α I, where I is totally ordered and (3.4) x α p x β α p β for all α, β I. Since {ϕ(x α )} is decreasing net in R +, thre exits r 0 such that ϕ(x α ) r as α increases. Let > 0 be given. Then there exists α 0 I such that for α p α 0 we have (3.5) r ϕ(x α ) ϕ(x α0 ) r +. Let β p α p α 0, then by (4) we obtain (3.6) p(x α, x β ) p(x β, x β ) ϕ(x α ) ϕ(x β ) r + r =. which implies that {x α } is a Cauchy net in X by Theorem Since X is complete, there exixts x X such that x α x as α increases. From the lower semicontinuity of ϕ we deduce that ϕ(x α ) r. If β p α then p(x α, x β ) p(x β, x β ) ϕ(x α ) ϕ(x β ). Letting β as increses we obtain (3.7) p(x α, x) p(x, x) ϕ(x α ) r ϕ(x α ) ϕ(x). which gives is x α p x for α I. In particular, x 0 p x. Since M is maximal, x M. Moreover, the condition (3) implies that (3.8) x α p x p f(x) for all α I. Again by maximality, f(x) M. Since x M, f(x) p x and hence f(x) = x. The mapping f satisfying (3.2) is called Caristi s maps. Again we writen self map f : X X contraction on a partial metric space, the following p(f(x), f(y)) p(f(x), f(x)) kk[p(x, y) p(y, y)] for all x, y X and for some k (0, 1) For y = f(x) will deduce the following thus p(f(x), f 2 (x)) p(f(x), f(x)) k[p(x, f(x)) p(f(x), f(x))] [p(x, f(x)) p(x, x)] k[p(x, f(x)) p(x, x)] [p(x, f(x)) p(x, x)] [p(f(x), f 2 (x)) p(f(x), f(x))].

8 8 MOHAMAD MUSLIKH Hence or (1 k)[p(x, f(x)) p(x, x)] p(x, f(x)) p(x, x) [p(x, f(x)) p(x, x)] [p(f(x), f 2 (x)) p(f(x), f(x))] 1 1 [p(x, f(x)) p(x, x)] 1 k 1 k [p(f(x), f 2 (x)) p(f(x), f(x))] If the function ϕ : X [0, ) defined by we obtained ϕ(x) = 1 [p(x, f(x)) p(x, x)], 1 k p(x, f(x)) p(x, x) ϕ(x) ϕ(f(x)). It appears that f is a Caristi s mapping on partial metric space. Thus the contraction mapping is a special case of Caristi s mapping. The following will be given the Ekeland s Variational Principle on partial metric spaces. Theorem 3.4. Let (X, p) be a complete partial metric space and ϕ : X [0, ) be a lower semicontinuous function. Let > 0 and x X be given such that ϕ( x) inf ϕ(x) +. x X Then for a given δ > 0 there exists x X such that (a) ϕ(x ) ϕ( x) (b) p( x, x ) δ + p(x, x ) (c) ϕ(x ) ϕ(x) + δ (p(x, x ) p(x, x )) for all x X \ {x }. Proof. For δ > 0, we set p δ (x, y) p δ (y, y) = 1 δ (p(x, y) p(y, y)). Then p δ is equivalent to p and (X, p δ ) is complete. Let us define a partial ordering on X by (3.9) x y ϕ(x) ϕ(y) (p δ (x, y) p δ (y, y)). It is easy to see that this ordering is (i) reflexsive, that is, for all x X, x x;(ii) antisymmetric, that is,for all x, y X, x y and y x imply x = y; (iii) transitive, that is, for all x, y, z X, x y and y z imply x z. We define a sequence E n of subsets of X as follow: We start x 1 = x and define E 1 = {x X : x x 1 } ; x 2 E 1 such that ϕ(x 2 ) inf x E1 ϕ(x) + 2, E 2 = {x X : x x 2 } ; x 3 E 2 such that ϕ(x 3 ) inf x E2 ϕ(x) + 2 2, and inductively E n = {x X : x x n } ; x n+1 E n such that ϕ(x n+1 ) inf x En ϕ(x) + 2 n. Clearly, E 1 E 2 E 3 Let u m E n with u m u X. Then u m x n and

9 CARISTI AND BANACH FIXED POINT THEOREM ON PARTIAL METRIC SPACE 9 so ϕ(u m ) ϕ(x n ) (p δ (u m, x n ) p δ (x n, x n )). therefore ϕ(u) lim m inf ϕ(u m) ϕ(x n ) lim m inf(p δ(u m, x n ) p δ (x n, x n )) ϕ(x n ) (p δ (u, x n ) p δ (x n, x n )). Thus u E n. We conclude that each E n is closed. Take any point x E n, one on hand x x n, implies that (3.10) ϕ(x) ϕ(x n ) (p δ (x, x n ) p δ (x n, x n )). We observe that x also belongs to E n 1 E n. so it is one of the points which entered in the competition when we picked x n. Therefore, (3.11) ϕ(x n ) inf y E n 1 ϕ(y) + From (3.10) and (3.11), we obtain It is follow that ϕ(x) + 2n 1 2 n 1. ϕ(x) + (p δ (x, x n ) p δ (x n, x n )) ϕ(x) + 2 p δ (x, x n ) p δ (x n, x n ) 2 n+1. n 1.. for all x E n. Which resulted D(E n ) 2 n and hence D(E n ) 0 (n ). Since (X, p δ ) is complete and E n is a decreasing sequence of closed sets, by Theorem 2.13, we infer that E n = {x } Since x E 1, we have n=1 x x 1 = x ϕ(x ) ϕ( x) (p δ (x, x) p δ ( x, x)) ϕ( x). Hence (a) is proved. Now we write p δ ( x, x n ) p δ (x n, x n ) = p δ (x 1, x n ) p δ (x n, x n ) and taking limit as n, we obtain n 1 [p δ (x i, x i+1 ) p δ (x i, x i )] i=1 n 1 2 i+1. i=1 1 δ (p( x, x ) p(x, x )) = p δ ( x, x ) p δ (x, x ) 1

10 10 MOHAMAD MUSLIKH and so (p( x, x ) δ + p(x, x )). This proves (b). Finally, let x x, of course x / n=1 E n, so x x, which means that ϕ(x) > ϕ(x ) [p δ (x, x ) p δ (x, x ) = ϕ(x ) δ [p(x, x ) p(x, x )] and hence (c) proved. We now present, so called the weak formulation of Ekeland s Variational Principle. Corollary 3.5. Let (X, p) be a complete partial metric space and ϕ : X [0, ) be a lower semicontinuous function. Then for any given > 0 there exists x X be such that ϕ(x ) inf ϕ(x) +. x X and ϕ(x ) < ϕ(x) + [p(x, x ) p(x, x )]. for all x x X. Definition 3.6. Let (X, p) be a partial metric space. The function f : X X is called continuous at the point x 0 if, for any sequence x n in X converges to x 0, then a sequence f(x n ) converges to f(x 0 ). Lemma 3.7. Let (X, p) be a partial metric space and the function f : X X. Then for each x X, the function ϕ x : X [0, ) defined by ϕ x (y) = p(x, f(y)). If f is contraction, then the function ϕ x is continuous on X.. Proof. Assume that a sequence y n converges to y in X, then lim n p(y n, y) p(y, y) = 0. For each x X and k (0, 1) and by (P4),we have ϕ x (y n ) ϕ x (y) = p(x, f(y n )) p(x, f(y)) p(f(y), f(y n )) p(f(y), f(y)) = p(f(y), f(y n )) p(f(y), f(y)) < k(p(y n, y) p(y, y)). This yields lim n ϕ x (y n ) = ϕ x (y) because lim n p(y n, y) p(y, y) = 0. Now, we will present the proves fixed point theorem which using Ekeland s variational principle. As first applications of Ekeland s variational principle, we prove Caristi s fixed point theorem version. Theorem 3.8. Let (X, p) be a complete partial metric space and f : X X be a mapping on X. Suppose there exists a lower semicontinuous function ϕ : X [0, ) such that (3.12) p(x, f(x)) p(f(x), f(x)) ϕ(x) ϕ(f(x)) for all x X. Then f has a fixed point.

11 CARISTI AND BANACH FIXED POINT THEOREM ON PARTIAL METRIC SPACE 11 Proof. By using Cororally 3.5 with = 1, we obtain x X such that (3.13) ϕ(x ) < ϕ(x) + [p(x, x ) p(x, x)] for all x X \ {x }. We assumed for all y = f(x ) X are such that y x. Then from (3.12) and (3.13), we have p(x, y) p(y, y) ϕ(x ) ϕ(y) and ϕ(x ) < ϕ(y) + [p(y, x ) p(y, y)] which cannot hold simultaneously. Hence, x = f(x ) As second applications of Ekeland s variational principle, we prove the wellknown Banach contraction theorem. Theorem 3.9. Let (X, p) be a complete partial metric space and f : X X be a contraction mapping. Then f has a unique fixed point in X. Proof. Consider the function ϕ : X [0, ) defined by ϕ(x) = p(x, f(x)), for all x X. By Lemma 3.7, ϕ is a continuous on X. Choose > 0 such that 0 < < 1 k, where k (0, 1). By Lemma 3.5, there exists x X such that for all x X. Putting x = f(x ), we have ϕ(x ) < ϕ(x) + [p(x, x ) p(x, x)] p(x, f(x )) p(x, f(x)) + [p(x, x ) p(x, x)] = p(f(x ), f(f(x ))) + [p(f(x ), x ) p(f(x ), f(x ))] k[p(x, f(x )) p(f(x ), f(x )] + [p(f(x ), x ) p(f(x ), f(x ))] = (k + )[p(x, f(x )) p(f(x ), f(x ))] (k + )p(x, f(x )) If x f(x ), then we obtain 1 (k + ), which contradict to our assumption that 1 > (k + ). Therefore, we have x = f(x ). The uniqueness of x can be proved as in the Theorem References [1] Bae J.S, Fixed point theorem for weakly contractive multivalued maps. J. Math. Anal. Appl.284 (2003), [2] Berinde V,Vetro F, Common fixed points of mappings satisfying implicit contractive conditions. Fixed Point Theory Appl.2012 (2012), [3] Chatterjea SK, Fixed point theorem. C.R. Acad Bulgare Sci.25(1972), [4] Caristi J,, Fixed point theorem for mapping stisfying inwardness condition. Trans. Amer. Math. Soc.215(1976), [5] Ekeland I, Remarques sur les problemes variationnels I. C. R. Acad Sci. Paris SPr. A-B 275,(1972),, [6] Kada O, Suzuki T, Takahashi, W Nonconvex minimization theorem an fixed points in metric spaces. Math. Jepon 44(1996), [7] Kannan R, Some Result on fixed points-ii. Am. Math. Mon. 76(1969), [8] Matthews S.G, Partial metric topology. In Proc. 8th Summer Conference on General Topology Appl. Ann. New York Acad. Sci.vol.78 (1994),

12 12 MOHAMAD MUSLIKH [9] Mizoguchi N, Takahashi W. Fixed point theorem for multivalued mapping on complete metric spaces. J. Math. Anal. Appl.141 (1989), Mathematics Department, University Brawijaya, Malang, Indonesia address: