On Banach fixed point theorems for partial metric spaces

@Applied General Topology c Universidad Polit´ecnica de Valencia Volume 6, No. 2, 2005 pp. 229-240 On Banach fixed point theorems for partial metric spaces Oscar Valero∗ Abstract. In this paper we prove several generalizations of the Banach fixed point theorem for partial metric spaces (in the sense of O’Neill) given in [14], obtaining as a particular case of our results the Banach fixed point theorem of Matthews ([12]), and some well-known classical fixed point theorems when the partial metric is, in fact, a metric. 2000 AMS Classification: 54H25, 54E50, 54E99, 68Q55. Keywords: dualistic partial metric, partial metric, complete, quasi-metric, fixed point. 1. Introduction and preliminaries In recent years many works on domain theory have been made in order to equip semantics domain with a notion of distance. In particular, Matthews ([12]) introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks, showing that the Banach contraction mapping theorem can be generalised to the partial metric context for applications in program verification. The existence of several connections between partial metrics and topological aspects of domain theory have been lately pointed by other authors as O’Neill ([15]), Bukatin and Scott ([3], [4]), Waszkiewicz ([23], [24]), Schellekens ([20], [21]), Escardo ([8]), Matthews ([12]) and Romaguera and Schellekens ([19], [18]). Throughout this paper the letters R,R+,Nwill denote the set of real numbers, of nonnegative real numbers and natural numbers, respectively. Le us recall that a partial metric on a (nonempty) set Xis a function p: X×X→R+such that for all x, y, z ∈X: (i)x=y⇔p(x, x) = p(x, y) = p(y, y); ∗The author acknowledges the support of the Spanish Ministry of Science and Technology, Plan Nacional I+D+I, and FEDER, grant BMF2003-02302. 230 O. Valero (ii)p(x, x)≤p(x, y); (iii)p(x, y) = p(y, x); (iv)p(x, z)≤p(x, y) + p(y, z)−p(y, y). Apartial metric space is a pair (X, p) such that Xis a nonempty set and p is a partial metric on X. As we indicated above, O’Neill ([15]) studied several connections between valuations and partial metrics so that he proposed one significant change to Matthews’ definition of the partial metrics, such a change consists of extending their range from R+to R. According to [14], the partial metrics in the O’Neill sense will be called dualistic partial metrics and a pair (X, p) such that Xis a (nonempty) set and pis a dualistic partial metric on Xwill be called a dualistic partial metric space. A paradigmatic example of a dualistic partial metric space is the pair (R, p), where p(x, y) = x∨yfor all x, y ∈R. Each dualistic partial metric pon Xgenerates a T0topology T(p) on X which has as a base the family of open p-balls {Bp(x, ε) : x∈X, ε > 0}, where Bp(x, ε) = {y∈X:p(x, y)< p(x, x) + ε}for all x∈Xand ε > 0. From this fact it immediately follows that a sequence (xn)n∈Nin a dualistic partial metric space (X, p) converges to a point x∈Xif and only if p(x, x) = limn→∞p(x, xn). Following [15] (compare [12]), a sequence (xn)n∈Nin a dualistic partial metric space (X, p) is called a Cauchy sequence if there exists limn,m→∞p(xn, xm). A dualistic partial metric space (X, p) is said to be complete if every Cauchy sequence (xn)n∈Nin Xconverges, with respect to T(p),to a point x∈Xsuch that p(x, x) = limn,m→∞p(xn, xm). As usual a function f:R→Ris monotone non-decreasing (or monotone or increasing)if x≥yimplies f(x)≥f(y). Recently, an extension of the Banach contraction mapping theorem have been proved in the dualistic partial metric context (see [14]) in such a way that the Matthews contraction mapping theorem can be deduced as a special case of such a result. In this paper we extend the notion of contraction between partial metric spaces to the dualistic partial metric context, and present two generalizations and one “local version” of the Banach fixed point theorem given in ([14]), which recuperate the several well-known classical fixed point theorems when the dualistic partial metric converts into a metric. 2. Contractions on dualistic partial metric spaces For the following discussion we recall some basics correspondences between dualistic partial metrics and quasi-metric spaces. Our basic references for quasi-metric spaces are [9] and [11]. In our context by a quasi-metric on a set Xwe mean a nonnegative realvalued function don X×Xsuch that for all x, y, z ∈X: (i)d(x, y) = d(y, x) = 0 ⇔x=y (ii)d(x, y)≤d(x, z) + d(z, y). Fixed point theorems for partial metric spaces 231 Aquasi-metric space is a pair (X, d) such that Xis a (nonempty) set and d is a quasi-metric on X. Each quasi-metric don Xgenerates a T0-topology T(d) on Xwhich has as a base the family of open d-balls {Bd(x, ε) : x∈X, ε > 0}, where Bd(x, ε) = {y∈X:d(x, y)< ε}for all x∈Xand ε > 0. If dis a quasi-metric on X, then the function dsdefined on X×Xby ds(x, y) = max{d(x, y), d(y, x)}, is a metric on X. Next we prove a local version of the Banach fixed point theorem for dualistic partial metric spaces (Theorem 2.3 below) obtaining as a particular case of our result the classical local version of Banach’s fixed point theorem for metric spaces (see [1] and [7]). The proof of the following auxiliary results can be found in [14] (compare [12], [15] and [13]). Lemma 2.1. If (X, p)is a dualistic partial metric space, then the function dp:X×X→R+defined by dp(x, y) = p(x, y)−p(x, x)is a quasi-metric on Xsuch that T(p) = T(dp). As a consequence of Lemma 2.1 a mapping between dualistic partial metric spaces (X, p) and (Y, q) is continuous if it is continuous between the associated quasi-metic spaces. Lemma 2.2. Let (X, p)be a dualistic partial metric space. Then the following assertions are equivalent: (1) (X, p)is complete (2) The induced metric space (X, (dp)s)is complete. Furthermore limn→∞(dp)s(a, xn) = 0 if and only if p(a, a) = limn→∞p(a, xn) = limn,m→∞p(xn, xm). In [14] it was proved the following Banach fixed point theorem which generalies the fixed point theorem for partial metrics of Matthews given in [12]. Theorem 2.3 (Banach fixed point theorem).Let fbe a mapping from a complete dualistic partial metric space (X, p)into itself such that there is a real number cwith 0≤c < 1,satisfying (*) |p(f(x), f(y))| ≤ c|p(x, y)|, for all x, y ∈X. Then fhas a unique fixed point. Corollary 2.4 (Matthews).Let fbe a mapping from a complete partial metric space (X, p)into itself such that there is a real number c with 0≤c < 1, satisfying p(f(x), f(y)) ≤cp(x, y), for all x, y ∈X. Then fhas a unique fixed point. On the other hand, it was showed in [14] that the contractive condition in the statement of Theorem 2.3 can not be replaced by the contractive condition of Corollary 2.4. 232 O. Valero In the sequel a mapping defined from a dualistic partial metric space into itself satisfying the condition (*) will be called a contraction with contraction constant c. The following well-known local version of the Banach fixed point theorem (Proposition 2.5 below) is a useful result with many applications. In particular it allows to show that the property of having a fixed point is invariant by homotopy for contractions and nonexpansive maps in metric spaces and, as consequence, it is applied to solve practice second order homogeneous Dirichlet problems ([1], [7]). Next we extend such this result to the dualistic partial metric case. Proposition 2.5. Let (X, d)be a complete metric space and let x0∈Xand r > 0.Suppose that f:Bd(x0, r)→Xis a contraction with contraction constant L such that d(f(x0), x0)<(1−L)r. Then fhas a unique fixed point in Bd(x0, r). Let (X, d) be a quasi-metric space. From now on by Ydwe denote the clousure of Y⊆Xwith respect to T(d). Remark 2.6. Observe that, in general, the set Bp(0, ε) = {y∈X:p(x, y)≤ ε+p(x, x)}is not closed with respect to T(p). Indeed, consider the dualistic partial metric space (R, p) where the dualistic partial metric pis defined by p(x, y) = x∨y. It is clear that Bp(0,1) = (−∞,1] and R\Bp(0,1) = (1,+∞).So Bp(0,1) is not a closed set in (R, p). Proposition 2.7. Let (X, p)be a complete dualistic partial metric space, x0∈ Xand r > 0.Suppose that f:Bp(x0, r)→Xis a contraction with contraction constant csuch that |p(f(x0), x0)|<(1 −c)r−2|p(x0, x0)| − |p(f(x0), f(x0))|. for all x, y ∈Bp(x0, r).Then fhas a unique fixed point in Bp(x0, r). Proof. It is clear that there exists r0with 0 ≤r0< r such that |p(f(x0), x0)| ≤ (1 −c)r0−2|p(x0, x0)| − |p(f(x0), f(x0))|. Next we show that Bp(x0, r0)(dp)s =Bp(x0, r0).To see this let x∈Bp(x0, r0)(dp)s and (xn)n∈N⊂Bp(x0, r0) such that limn→∞(dp)s(x, xn) = 0.Then, given ε > 0,there exists n0∈Nwith p(x0, xn)−p(x0, x0)≤r0and p(xn, x)−p(xn, xn)< ε whenever n≥n0.Now we only have to show that p(x, x0)−p(x0, x0)≤r0.Indeed, p(x, x0)−p(x0, x0)≤p(x, xn) + p(xn, x0)−p(xn, xn)−p(x0, x0) < ε +r0. Consequently p(x, x0)−p(x0, x0)≤r0and Bp(x0, r0)(dp)s ⊆Bp(x0, r0).Whence Bp(x0, r0)(dp)s =Bp(x0, r0). Fixed point theorems for partial metric spaces 233 We will show that actually f:Bp(x0, r0)→Bp(x0, r0).To this end note that |p(f(x), f(x0))| − |p(x0, x0)| ≤ |p(f(x), f(x0))| − c|p(x0, x0)| ≤c·(|p(x, x0)| − |p(x0, x0)|) ≤c·(p(x, x0)−p(x0, x0)) ≤cr0 for all x∈Bp(x0, r0).Hence p(f(x), x0)−p(x0, x0)≤dp(x0, f(x0)) + dp(f(x0), f(x)) ≤ |p(f(x), f(x0))|+|p(f(x0), x0)|+|p(f(x0), f(x0))|+|p(x0, x0)| ≤cr0+ (1 −c)r0=r0, for all x∈Bp(x0, r0).Therefore f(Bp(x0, r0)) ⊆Bp(x0, r0). Now we can apply Theorem 2.3 to deduce that fhas a fixed point in Bp(x0, r0)⊂Bp(x0, r). Finally we want to show the uniqueness. Suppose that there exist x, y ∈ Bp(x0, r) such that f(x) = x, f(y) = yand x6=y. Since fis a contraction in Bp(x0, r) we have that p(x, y) = p(x, x) = p(y, y) = 0 because of |p(a, b)|=|p(f(a), f(b))| ≤ c|p(a, b)| for all a, b ∈ {x, y},which implies that x=y. This concludes the proof.  Clearly Proposition 2.5 is a particular case of Proposition 2.7 because of |p(x0, x0)|=|p(f(x0), f(x0))|= 0 when the dualistic partial metric is, in fact, a metric. 3. Generalized Banach’s fixed point theorems for complete dualistic partial metric spaces In the last years many authors have obtained generalizations of the Banach fixed point theorem in complete metric spaces ([2], [5], [6], [10], [16], [17]) Recently, and motivated in part by the applications to Computer Science, some generalizations of a Banach fixed point theorem have been obtained for quasimetric and partial metric spaces (see for instance [12], [14], [22]). In this section our interest is focused on giving two Banach fixed point type theorems where the contractive condition is weakened. One way of extending the Banach theorem arises in a natural way from approximation problems, where the contractive condition depends on a distinguished real function. Proposition 3.1 ([7]).Let (X, d)be a complete metric space and let d(f(x), f(y)) ≤Φ(d(x, y)) for all x, y ∈X, where Φ : [0,∞)→[0,∞)is any monotone non-decresing function with limn→∞Φn(t) = 0 for any fixed t > 0. Then fhas a unique fixed point. The following result generalizes the preceding one. 234 O. Valero Theorem 3.2. Let fbe a mapping of a complete dualistic partial metric space (X, p)into itself such that |p(f(x), f(y))| ≤ Φ(|p(x, y)|) for all x, y ∈X, where Φ : [0,+∞)→[0,+∞)is any monotone non-decreasing function with limn→∞Φn(t) = 0 for each fixed t > 0.Then fhas a unique fixed point. Proof. First note that if t > 0 then Φ(t)< t, because if t≤Φ(t) then Φ(t)≤ Φ(Φ(t)) and therefore t≤Φ2(t).Thus, it is easy to prove, by induction, that t≤Φn(t) for n≥1.It follows that t≤limn→∞Φn(t) = 0,a contradiction. Fix x∈X. It is clear that for each n∈Nwe have |p(fn(x), fn(x))| ≤ Φn(|p(x, x)|) and  p(fn(x), fn+1(x)) ≤Φn(|p(x, f(x))|). Let α:= max{|p(x, x)|,|p(x, f(x))|}.Since, by Lemma 2.1, dp(fn(x), fn+1(x)) = p(fn(x), fn+1(x)) −p(fn(x), fn(x)), we deduce that dp(fn(x), fn+1(x)) ≤ |p(fn(x), fn+1(x))|+|p(fn(x), fn(x))| ≤Φn(|p(x, f(x))|) + Φn(|p(x, x)|) ≤2Φn(α+ 1). Now let n, k ∈N.Then dp(fn(x), fn+k(x)) ≤dp(fn(x), fn+1(x)) + ... +dp(fn+k−1(x), fn+k(x)) ≤Φn(|p(x, f(x))|) + Φn(|p(x, x)|) + ... ... + Φn+k−1(|p(x, f(x))|) + Φn+k−1(|p(x, x)|). Thus dp(fn(x), fn+k(x)) ≤2Φn(α+ 1) + ... + 2Φn+k−1(α+ 1). Similarly we show that dp(fn+1(x), fn(x)) ≤2Φn(α+ 1), and thus dp(fn+k(x), fn(x)) ≤2Φn+k−1(α+ 1) + ... + 2Φn(α+ 1). Therefore (fn(x))n∈Nis a Cauchy sequence in the metric space (X, (dp)s), which is complete by Lemma 2.2. So there is a∈Xsuch that limn→∞(dp)s(a, fn(x)) = 0 and p(a, a) = limn→∞p(a, fn(x)) = limn,m→∞p(fn(x), fm(x)). We want to show that ais the unique fixed point of f. First we claim that ais a fixed point of f. Clearly we have that limn,m→∞dp(fn(x), fm(x)) = 0. Moreover limn→∞p(fn(x), fn(x)) = 0,since |p(fn(x), fn(x))| ≤ Φn(|p(x, x)|)≤Φn(|p(x, x)|+ 1) Fixed point theorems for partial metric spaces 235 and limn→∞Φn(|p(x, x)|+1) = 0.Thus we deduce that limn,m→∞p(fn(x), fm(x)) = 0.Consequently by Lemma 2.2, p(a, a) = 0.Whence |p(f(a), f(a))| ≤ Φ(|p(a, a)|) = Φ(0) ≤Φ(ε)< ε. It follows that |p(f(a), f(a))|= 0, because the last inequality holds for all ε > 0. On the other hand, since limn→∞p(a, fn(x)) = p(a, a) = 0 we have |p(fn+1(x), f(a))| ≤ Φ(|p(fn(a), a)|)≤Φ(ε)< ε eventually. Thus Lemma 2.2 shows that f(a) is a limit point of (fn(x))n∈Nin (X, (dp)s).Then a=f(a). Finally we show that ais the unique fixed point of f. To this end let b∈X such that f(b) = band b6=a. Then p(a, a) = p(b, b) = p(f(a), f(b)) = 0,since otherwise we have that the following inequality holds |p(x, y)|=|p(f(x), f(y))| ≤ Φ(|p(x, y)|)<|p(x, y)| for any x, y ∈ {a, b}.The proof is concluded.  Remark 3.3. Observe that Theorem 2.3 follows as a special case of the preceding result if we choose Φ(t) = ct with 0 ≤c < 1. We will say that a dualistic partial metric space (X, p) is bounded if there exists k > 0 such that |p(x, y)|< k for every x, y ∈X. Moreover, we define the diameter of a subset Y⊆Xas δp(Y) = sup{|p(x, y)|:x, y ∈Y}if the supremum exists and δp(Y) = +∞otherwise.Note that these notions coincide with the classical notions of bounded metric space and diameter of a metric space, respectively. Furthermore δp(Y)≤δ(dp)s(Y) + δwp(Y),where δwp(Y) = sup{|p(x, x)|:x∈Y}if it exists and δwp(Y) = +∞otherwise. Another way of attempting an extension of the Banach fixed point theorem does not rely on measuring the difference between p(f(x), f(y)) and p(x, y), but, similarly to the metric case, the required condition relies on the behaviour of the induced quasi-metric dp. The below technical results are useful to prove the desired theorem. Lemma 3.4. Let (X, p)be a dualistic partial metric space and Y⊆X. Then δ(dp)s(Y)≤4δp(Y). Proof. It is easily seen that δwp(Y)≤δp(Y).Furthermore (dp)s(x, y)≤dp(x, y) + dp(y, x) = 2p(x, y)−p(x, x)−p(y, y), whence we have (dp)s(x, y)≤2|p(x, y)|+|p(x, x)|+|p(y, y)|. Immediately we deduce that δ(dp)s(Y)≤4δp(Y). Lemma 3.5. Let (X, p)be a complete dualistic partial metric space and let ϕ:X→R+be an arbitrary non-negative function. Assume that inf{ϕ(x) + ϕ(y) : |p(x, y)|+|p(x, x)|+|p(y, y)| ≥ a}=µ(a)>0for all a > 0. 236 O. Valero Then each sequence (xn)n∈Nfor which limn→∞ϕ(xn) = 0 converges with repect to T((dp)s)to the same point of X. Proof. Let An={x∈X:ϕ(x)≤ϕ(xn)}.Clearly Anis a non-empty set and any finite family of such subsets has a non-empty intersection. We show that limn→∞δ(dp)s(An) = 0.To this end, let ε > 0.Then there exists n0∈Nsuch that ϕ(xn)<1 2µ(ε) for all n≥n0.Thus, ϕ(x)+ϕ(y)< µ(ε) whenever x, y ∈Anand n≥n0.Hence, by hypothesis |p(x, y)|< ε. Therefore δp(An)< ε for all n≥n0.Consequently limn→∞δp(An) = 0.By Lemma 3.4 we obtain that limn→∞δ(dp)s(An) = 0, which gives limn→∞δ(dp)s(An (dp)s ) = 0 because of δ(dp)s(An (dp)s ) = δ(dp)s(An).We conclude from Cantor’s intersection theorem that there exists a unique x∈ ∩+∞ n=1An (dp)s .Furthermore, limn→∞(dp)s(x, xn) = 0,since xn∈An (dp)s for each n∈N.Whence we follow that p(x, x) = limn→∞ p(x, xn) = limn→∞ p(xn, xn). Note that we have actually proved that for any sequence (xn)n∈Nsuch that limn→∞ϕ(xn) = 0 there exists a limit point x∈Xwith respect to T((dp)s). Let (yn)n∈Nbe any other sequence such that limn→∞ϕ(yn) = 0 and let y∈Xits limit point. Since limn→∞ϕ(xn) = limn→∞ϕ(yn) = 0 we get that, given ε > 0,there exists n1∈Nsuch that ϕ(xn)<1 2µ(ε), ϕ(yn)<1 2µ(ε) whenever n≥n1.In consequence |p(xn, yn)|+|p(xn, xn)|+|p(yn, yn)|< ε whenever n≥n1,because otherwise µ(ε) = inf{ϕ(x) + ϕ(y) : |p(x, y)|+|p(x, x)|+|p(y, y)| ≥ ε} ≤ϕ(xn) + ϕ(yn)< µ(ε), which is a contradiction. We show that x=y. Indeed, (dp)s(x, y)≤(dp)s(x, xn) + (dp)s(xn, yn) + (dp)s(yn, y) ≤2ε+ (dp)s(xn, yn) ≤2ε+ 2|p(xn, yn)|+|p(xn, xn)|+|p(yn, yn)|<4ε. So x=y, since the preceding inequality is true for every ε > 0.This completes the proof.  Theorem 3.6. Let (X, p)be a complete dualistic partial metric space (X, p) and let f:X→Xbe a continuous mapping from (X, (dp)s)to (X, (dp)s) such that the functions ϕ(x) = dp(x, f(x)) and ψ(x) = dp(f(x), x)satisfy the following conditions: (1) inf{ϕ(x) + ϕ(y) + ψ(x) + ψ(y) : |p(x, y)|+|p(x, x)|+|p(y, y)| ≥ a}= µ(a)>0for all a > 0 (2) infx∈X(ϕ(x) + ψ(x)) = 0. Then fhas a unique fixed point. Proof. First we construct a sequence (xn)n∈N⊂Xsuch that limn→∞ϕ(xn) + ψ(xn) = 0.Let ε > 0 be given then there exists xε∈Xsuch that ϕ(xε) + Fixed point theorems for partial metric spaces 237 ψ(xε)< ε. Put xn=x1/n for each n∈N.Thus the sequence (xn)n∈Nsatisfies the following: For every ε≥1, ϕ(xn) + ψ(xn)<1≤εwhenever n > 2. For every ε < 1,there always exists n0∈Nsuch that ϕ(xn)+ψ(xn)<1 n0< ε for all n≥n0. It follows that limn→∞ϕ(xn) + ψ(xn) = 0 as we claim. Since limn→∞ϕ(xn)+ψ(xn) = 0,by Lemma 3.5 there exists a unique x∈X such that limn→∞(dp)s(x, xn) = 0.Then ϕ(x)−ϕ(xn) = p(x, f(x)) −p(x, x)−p(xn, f(xn)) + p(xn, xn) < ε +p(x, f(x)) −p(xn, f(xn)) <2ε+p(xn, f(x)) −p(xn, f(xn)) ≤2ε+p(f(xn), f(x)) −p(f(xn), f(xn)) <3ε. Similarly is showed that ϕ(xn)−ϕ(x)<3εeventually. Hence, we conclude that ϕ(x) = limn→∞ϕ(xn). On the other hand, limn→∞ϕ(xn)≤limn→∞(ϕ(xn) + ψ(xn)) = 0. Thus we have ϕ(x) = 0 and p(x, f(x)) = p(x, x). Furthermore, ψ(x)−ψ(xn) = p(x, f(x)) −p(f(x), f(x)) −p(xn, f(xn)) + p(f(xn), f(xn)) < ε +p(x, f(x)) −p(xn, f(xn)) <2ε+p(xn, f(x)) −p(xn, f(xn)) ≤2ε+p(f(xn), f(x)) −p(f(xn), f(xn)) <3ε. Again we similarly show that ψ(xn)−ψ(x)<3εeventually. So ψ(x) = limn→∞ψ(xn)≤limn→∞(ϕ(xn) + ψ(xn)) = 0.Whence we deduce that p(x, f(x)) = p(f(x), f(x)) = p(x, x).By condition (i) of the definition of a dualistic partial metric x=f(x),and fhas a fixed point. Next we show the uniqueness. To this end we consider y∈Xsuch that y6=xand f(y) = y. Then we would have α:= |p(x, y)|+|p(x, x)|+|p(y, y)| 6= 0, because otherwise p(x, y) = p(x, x) = p(y, y) = 0 and x=y. It follows that 0< µ(α) = inf{ϕ(x) + ϕ(y) + ψ(x) + ψ(y) : |p(x, y)|+|p(x, x)|+|p(y, y)| ≥ α} ≤ϕ(x) + ϕ(y) + ψ(x) + ψ(y) = 0, a contradiction. So that x=y.  As a consequence we obtain the following classical result which can be found in [7]. Corollary 3.7. Let fbe a mapping from a complete metric space (X, d)into itself and ϕ:X→R+the non-negative function defined by ϕ(x) = d(x, f(x)). Assume that inf{ϕ(x) + ϕ(y) : d(x, y)≥a}=µ(a)>0 for all a > 0