Unique common fixed point theorems on partial metric spaces

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Research Article

Unique common fixed point theorems on partial metric spaces

Anchalee Kaewcharoen^∗, Tadchai Yuying

Department of Mathematics, Faculty of Science, Naresuan University, Phitsanulok 65000, Thailand.

Communicated by P. Kumam

Abstract

We prove the existence of the unique common fixed point theorems for self mappings which are weakly compatible satisfying some contractive conditions on partial metric spaces. Furthermore, we also prove the result on the continuity in the set of common fixed points for self mappings on partial metric spaces. c2014 All rights reserved.

Keywords: Common fixed points, Weakly compatible mappings, Coincidence points, Partial metric spaces.

2010 MSC: 47H10, 54H25.

1. Introduction and Preliminaries

The common fixed point theorems for mappings satisfying certain contractive conditions in metric spaces have been continually studied for decade (see [1, 3, 5, 6, 8, 9, 10, 11, 12, 14, 15] and references contained therein). In 1976, Jungck [7] proved the existence of common fixed point theorems for commuting mappings in metric spaces where the results require the continuity of one of two such mappings. In 1986, Jungck [8] introduced the concept of compatible mappings and proved that weakly commuting mappings are compatible mappings. After that, Jungck [10], generalized the notion of compatibility by introducing the weakly compatibility.

Recently, Abbas et al. [1] introduced the generalized condition (B) as the following:

∗Corresponding author

Email addresses: [email protected](Anchalee Kaewcharoen),[email protected](Tadchai Yuying)

Received 2013-01-16

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Definition 1.1. LetX be a metric space. A mappingF :X →X is said to satisfy a generalized condition (B) associated with a self mappingf onX if there existsδ∈(0,1) and L≥0 such that

d(F x, F y)≤δM(x, y) +Lmin{d(f x, F x), d(f y, F y), d(f x, F y), d(f y, F x)}, (1.1) for all x, y∈X, where

M(x, y) = max{d(f x, f y), d(f x, F x), d(f y, F y),1

2[d(f x, F y) +d(f y, F x)]}.

Abbas et al. [1] established the existence of a unique common fixed point for two self mappingsF andf onX whereF satisfies a generalized condition (B) associated withf. In this work, we assure the analogous results proved by Abbas et al. [1] for four self mappings in partial metric spaces.

Mathews [13] introduced the notion of partial metric spaces. We now recall some definitions and lemmas that will be used in the sequel.

Definition 1.2. A partial metric on a nonempty set X is a function p : X×X → R⁺ such that for all x, y, 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, z)≤p(x, y) +p(y, z)−p(y, y).

A pair (X, p) is called a partial metric space and pis a partial metric on X.

If p is a partial metric on X, then p generates a T₀ topologyτ_p on X whose base is the family of open p−balls

{B_p(x, ε) :x∈X and ε >0},

whereB_p(x, ε) ={y∈X:p(x, y)< p(x, x) +ε}.For each partial metricponX, the functionp^s:X×X → R⁺ defined by

p^s(x, y) = 2p(x, y)−p(x, x)−p(y, y) (1.2) is a usual metric onX.

Definition 1.3. Let (X, p) be a partial metric space.

(1) A sequence {x_n} in a partial metric space (X, p) converges to a point x ∈ X if limn→∞p(x, x_n) = p(x, x).

(2) A sequence {x_n} in a partial metric space (X, p) is called a Cauchy sequence if limn,m→∞p(x_n, x_m) exists (and is finite).

(3) A partial metric space (X, p) is said to be complete if every Cauchy sequence {x_n} in X converges, with respect toτp,to a point x∈X such that limn,m→∞p(xn, xm) =p(x, x).

Lemma 1.4. [13] Let (X, p) be a partial metric space. Then

(1) A sequence {x_n} in a partial metric space (X, p) is a Cauchy sequence if and only if it is a Cauchy sequence in the metric space (X, p^s).

(2) A partial metric space (X, p) is complete if and only if the metric space (X, p^s) is complete. Moreover,

n→∞lim p^s(x, x_n) = 0 iff lim

n→∞p(x, x_n) = lim

n,m→∞p(x_n, x_m) =p(x, x).

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(3) A subset E of a partial metric space (X, p) is closed if whenever {x_n} is a sequence in E such that {x_n}converges to some x∈X, thenx∈E.

Lemma 1.5. [2] Let (X, p) be a partial metric space. Then (1) Ifp(x, y) = 0,then x=y.

(2) Ifx6=y, thenp(x, y)>0.

Definition 1.6. Let (X, p) be a partial metric space. A mapping f :X→X is continuous atx∈X if the sequence {f x_n} converges tof x for every sequence{x_n} inX converging tox.

Definition 1.7. Letf and g are self mappings on a set X. A point x∈X is called a coincidence point of f and g iff x=gx=w wherew is called a point of coincidence of f and g.

Definition 1.8. Two self mappingsf andgon a setXare said to be weakly compatible iff andgcommute at their coincidence points. That is, iff x=gxfor somex∈X, thenf gx=gf x.

In this paper, we prove the uniqueness of a common fixed point of four self mappings on a partial metric space (X, p) satisfying the certain contractive condition and being the weak compatibility. Moreover, we also prove the result on the continuity in the set of common fixed points for self mappings.

2. Main results

We now prove the existence of the unique common fixed point theorems for four self mappings which are weakly compatible on a partial metric space (X, p).The proofs of the mentioned theorems have been taken from the technique used in [1] in the setting of metric spaces.

Theorem 2.1. Let (X, p) be a complete partial metric space. Suppose that f, g, F and Gare self mappings onX satisfying the following conditions:

(a) f(X)⊆g(X) andF(X)⊆G(X).

(b) There exist δ >0 andL≥0 with δ+ 2L <1 such that

p(F x, f y)≤δM(x, y) +Lmin{p(gx, F x), p(Gy, f y), p(gx, f y), p(Gy, F x)}, (2.1) for allx, y∈X, where

M(x, y) = max{p(gx, Gy), p(gx, F x), p(Gy, f y),1

2[p(gx, f y) +p(Gy, F x)]}.

(c) f(X) or g(X) is closed.

If {f, G} and {g, F} are weakly compatible, thenf, g, F andG have a unique common fixed point in X.

Proof. Suppose thatx0 is an arbitrary point inX. Sincef(X)⊆g(X) andF(X)⊆G(X), we can construct a sequence{y_n} inX satisfying

yn=F xn=Gxn+1 and yn+1 =f xn+1=gxn+2 for alln∈N∪ {0}.

By applying (2.1), we have

p(F x_n, f x_n+1) ≤ δM(x_n, x_n+1) +Lmin{p(gx_n, F x_n), p(Gx_n+1, f x_n+1), p(gxn, f xn+1), p(Gxn+1, F xn)}.

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Since

M(xn, xn+1) = max{p(gx_n, Gxn+1), p(gxn, F xn), p(Gxn+1, f xn+1), 1

2[p(gx_n, f x_n+1) +p(Gx_n+1, F x_n)]}

= max{p(y_n−1, yn), p(yn−1, yn), p(yn, yn+1), 1

2[p(yn−1, y_n+1) +p(y_n, y_n)]}

≤ max{p(y_n−1, yn), p(yn, yn+1), 1

2[p(yn−1, yn) +p(yn, yn+1)−p(yn, yn) +p(yn, yn)]}

≤ max{p(y_n−1, y_n), p(y_n, y_n+1)}, and

min{p(gx_n, F xn), p(Gxn+1, f xn+1), p(gxn, f xn+1) +p(Gxn+1, F xn)}

= min{p(y_n−1, yn), p(yn, yn+1), p(yn−1, yn+1), p(yn, yn)}

= min{p(y_n−1, yn+1), p(yn, yn)}, we obtain that

p(y_n, y_n+1) = p(F x_n, f x_n+1)

≤ δmax{p(y_n−1, y_n), p(y_n, y_n+1)}+Lmin{p(y_n−1, y_n+1), p(y_n, y_n)}.

We separate the proof into the following cases.

Case I : If max{p(y_n−1, yn), p(yn, yn+1)} = p(yn−1, yn) and min{p(y_n−1, yn+1), p(yn, yn)} = p(yn−1, yn+1), then

p(y_n, y_n+1) ≤ δp(yn−1, y_n) +Lp(yn−1, y_n+1)

≤ δp(yn−1, y_n) +L(p(yn−1, y_n) +p(y_n, y_n+1)−p(y_n, y_n))

≤ δp(yn−1, yn) +Lp(yn−1, yn) +Lp(yn, yn+1).

This implies that

p(y_n, y_n+1)≤ δ+L

1−Lp(yn−1, y_n).

Letk1= ^δ+L_1−L.Since δ+ 2L <1,we have k1 <1.Therefore

p(yn, yn+1)≤k1p(yn−1, yn).

Case II : If max{p(yn−1, y_n), p(y_n, y_n+1)}=p(yn−1, y_n) and min{p(yn−1, y_n+1), p(y_n, y_n)}=p(y_n, y_n),then p(yn, yn+1) ≤ δp(yn−1, yn) +Lp(yn, yn)

≤ δp(yn−1, yn) +Lp(yn, yn+1).

This implies that

p(yn, yn+1)≤ δ

1−Lp(yn−1, yn).

Letk₂= _1−L^δ .Since δ+ 2L <1,we have k₂ <1.Therefore

p(y_n, y_n+1)≤k₂p(yn−1, y_n).

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Case III : If max{p(y_n−1, y_n), p(y_n, y_n+1)}=p(y_n, y_n+1) and min{p(y_n−1, y_n+1), p(y_n, y_n)}=p(yn−1, y_n+1), then

p(yn, yn+1) ≤ δp(yn, yn+1) +Lp(yn−1, yn+1)

≤ δp(yn, yn+1) +L(p(yn−1, yn) +p(yn, yn+1)−p(yn, yn))

≤ δp(y_n, y_n+1) +Lp(yn−1, y_n) +Lp(y_n, y_n+1).

This implies that

p(yn, yn+1)≤ L

1−(δ+L)p(yn−1, yn).

Letk3= _1−(δ+L)^L .Since δ+ 2L <1,we have k3 <1.Therefore p(yn, yn+1)≤k3p(yn−1, yn).

Case IV : If max{p(y_n−1, yn), p(yn, yn+1)}=p(yn, yn+1) and min{p(y_n−1, yn+1), p(yn, yn)}=p(yn, yn),then p(yn, yn+1) ≤ δp(yn, yn+1) +Lp(yn, yn)

≤ δp(yn, yn+1) +Lp(yn−1, yn).

This implies that

p(yn, yn+1)≤ L

1−δp(yn−1, yn).

Letk₄= _1−δ^L .Since δ+ 2L <1,we have k₄<1.Therefore

p(y_n, y_n+1)≤k₄p(yn−1, y_n).

Choose k= max{k₁, k2, k3, k4}.Therefore 0< k <1.For eachn∈N,we obtain that

p(yn, yn+1)≤kⁿp(y0, y1). (2.2) We will prove that{y_n} is a Cauchy sequence in (X, p^s). Letm, n ∈N withm > n. By applying (2.2), we have

p(y_m, y_n) ≤ [p(y_n, y_n+1) +p(y_n+1, y_n+2) +· · ·+p(ym−1, y_m)]

−[p(y_n+1, yn+1) +p(yn+2, yn+2) +p(ym−1, ym−1)]

≤ p(yn, yn+1) +p(yn+1, yn+2) +· · ·+p(ym−1, ym)

≤ [kⁿ+kⁿ⁺¹+·+k^m−1]p(y₀, y₁)

≤ kⁿ

1−kp(y0, y1).

It follows that

n,m→∞lim p(ym, yn) = 0. (2.3)

Using (1.2), we have

p^s(y_m, y_n) = 2p(y_m, y_n)−p(y_m, y_m)−p(y_n, y_n)

≤ 2p(y_m, y_n).

Applying (2.3), we obtain that

n,m→∞lim p^s(ym, yn) = 0. (2.4)

This implies that{y_n}is a Cauchy sequence in (X, p^s).Since X is complete, we have

n→∞lim yn=zfor somez∈X. (2.5)

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By Lemma 1.4 and (2.5), we obtain that p(z, z) = lim

n→∞p(y_n, z) = lim

n,m→∞p(y_m, y_n) (2.6)

From (2.3) and (2.6), we can conclude that p(z, z) = 0.Assume thatg(X) is closed. Therefore there exists a pointu∈X such thatz=gu. Using (2.1), this yields

p(z, F u) ≤ p(z, yn+1) +p(yn+1, F u)−p(yn+1, yn+1)

≤ p(z, y_n+1) +p(F u, f x_n+1)

≤ p(z, y_n+1) +δmax{p(gu, Gx_n+1), p(gu, F u), p(Gx_n+1, f x_n+1), 1

2[p(gu, f xn+1) +p(Gxn+1, F u)]}+Lmin{p(gu, F u), p(Gx_n+1, f xn+1), p(gu, f x_n+1), p(Gx_n+1, F u)}

= p(z, y_n+1) +δmax{p(z, y_n), p(z, F u), p(y_n, y_n+1), 1

2[p(z, yn+1) +p(yn, F u)]}+Lmin{p(z, F u), p(y_n, yn+1), p(z, yn+1), p(yn, F u)}

≤ p(z, y_n+1) +δmax{p(z, y_n), p(z, F u), p(y_n, z) +p(z, y_n+1)−p(z, z), 1

2[p(z, y_n+1) +p(y_n, z) +p(z, F u)−p(z, z)]}+Lmin{p(z, F u), p(y_n, z) +p(z, y_n+1)

−p(z, z), p(z, y_n+1), p(yn, z) +p(z, F u)−p(z, z)}

≤ p(z, y_n+1) +δmax{p(z, y_n), p(z, F u), p(y_n, z) +p(z, y_n+1), 1

2[p(z, y_n+1) +p(y_n, z) +p(z, F u)]}+Lmin{p(z, F u), p(y_n, z) +p(z, y_n+1), p(z, y_n+1), p(y_n, z) +p(z, F u)}.

Taking the limit as n→ ∞ and using the fact thatp(z, z) = 0,we have

p(z, F u)≤δp(z, F u) +Lp(z, F u) = (δ+L)p(z, F u).

It follows that p(z, F u) = 0 and so F u = z = gu. Since F and g are weakly compatible, we obtain that gF u=F gu. Thereforegz=F z.

Since F(X)⊆G(X), there exists a point v∈X such that z=Gv.Applying (2.1), we have p(z, f v) = p(F u, f v)

≤ δmax{p(gu, Gv), p(gu, F u), p(Gv, f v),1

2[p(gu, f v) +p(Gv, F u)]}+ Lmin{p(gu, F u), p(Gv, f v), p(gu, f v), p(Gv, F u)}

= δmax{p(z, z), p(z, z), p(z, f v),1

2[p(z, f v) +p(z, z)]}+ Lmin{p(z, z), p(z, f v), p(z, f v), p(z, z)}

≤ δp(z, f v).

This implies thatp(z, f v) = 0 and so f v =z=Gv. Since G and f are weakly compatible, we obtain that f Gv = Gf v. Therefore f z = Gz. We next prove that z is a common fixed point of f, g, F and G. Using

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(2.1), this yields

p(F z, z) = p(F z, f v)

≤ δmax{p(gz, Gv), p(gz, F z), p(Gv, f v),1

2[p(gz, f v) +p(Gv, F z)]}+ Lmin{p(gz, F z), p(Gv, f v), p(gz, f v), p(Gv, F z)}

= δmax{p(F z, z), p(F z, F z), p(z, z),1

2[p(F z, z) +p(z, F z)]}+ Lmin{p(F z, F z), p(z, z), p(F z, z), p(z, F z)}

≤ δmax{p(F z, z), p(F z, z), p(z, z),1

2[p(F z, z) +p(z, F z)]}+ Lmin{p(F z, F z), p(z, z), p(F z, z), p(z, F z)}

≤ δp(F z, z).

This implies thatp(F z, z) = 0 and so gz=F z=z.Similarly, applying (2.1), we obtain that p(z, f z) = p(F z, f z)

≤ δmax{p(gz, Gz), p(gz, F z), p(Gz, f z),1

2[p(gz, f z) +p(Gz, F z)]}+ Lmin{p(gz, F z), p(Gz, f z), p(gz, f z), p(Gz, F z)}

= δmax{p(z, f z), p(z, z), p(f z, f z),1

2[p(z, f z) +p(f z, z)]}+ Lmin{p(z, z), p(f z, f z), p(z, f z), p(f z, z)}

≤ δmax{p(z, f z), p(z, z), p(f z, z),1

2[p(z, f z) +p(f z, z)]}+ Lmin{p(z, z), p(f z, f z), p(z, f z), p(f z, z)}

≤ δp(z, f z).

This implies thatp(z, f z) = 0 and so Gz =f z =z. Thereforez is a common fixed point of f, g, F and G.

We will prove the uniqueness of a common fixed point off, g, F and G. Let w be any common fixed point off, g, F and G. By applying (2.1), it follows that

p(z, w) = p(F z, f w)

≤ δmax{p(gz, Gw), p(gz, F z), p(Gw, f w),1

2[p(gz, f w) +p(Gw, F z)]}+ Lmin{p(gz, F z), p(Gw, f w), p(gz, f w), p(Gw, F z)}

= δmax{p(z, w), p(z, z), p(w, w),1

2[p(z, w) +p(w, z)]}+ Lmin{p(z, z), p(w, w), p(z, w), p(w, z)}

≤ δp(z, w).

This implies that p(z, w) = 0 and so z = w. Hence f, g, F and G have a unique common fixed point in X.

Letting F =f andG=g in Theorem 2.1, we immediately obtain the following corollary:

Corollary 2.2. Let(X, p)be a partial metric space. Suppose thatf andg are self mappings onX satisfying the following conditions:

(a) f(X)⊆g(X).

p(f x, f y)≤δM(x, y) +Lmin{p(gx, f x), p(gy, f y), p(gx, f y), p(gy, f x)}, (2.7)

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for allx, y∈X, where

M(x, y) = max{p(gx, gy), p(gx, f x), p(gy, f y),1

2[p(gx, f y) +p(gy, f x)]}.

(c) f(X) or g(X) is complete.

If {f, g} is weakly compatible, then f andg have a unique common fixed point in X.

M(x, y) = max{p(gx, Gy),1

2[p(gx, F x) +p(Gy, f y)],1

2[p(gx, f y) +p(Gy, F x)]}.

Proof. Since the inequality (2.8) implies the inequality (2.1), we have the result obtained from Theorem 2.1.

(b) There exist δ >0 andL≥0 with δ+L < ¹₂ such that

M(x, y) = max{p(gx, Gy), p(gx, F x), p(Gy, f y), p(gx, f y), p(Gy, F x)}.

Proof. Suppose thatx0 is an arbitrary point inX. Sincef(X)⊆g(X) andF(X)⊆G(X), we can construct a sequence{y_n} inX satisfying

yn=F xn=Gxn+1 and yn+1 =f xn+1=gxn+2 for alln∈N∪ {0}.

Applying (2.9), this yields

p(F xn, f xn+1) ≤ δM(xn, xn+1) +Lmin{p(gx_n, F xn), p(Gxn+1, f xn+1), p(gxn, f xn+1), p(Gxn+1, F xn)}.

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Since

M(xn, xn+1) = max{p(gx_n, Gxn+1), p(gxn, F xn), p(Gxn+1, f xn+1), p(gx_n, f x_n+1), p(Gx_n+1, F x_n)}

= max{p(y_n−1, y_n), p(yn−1, y_n), p(y_n, y_n+1), p(yn−1, y_n+1), p(y_n, y_n)}

= max{p(y_n−1, yn), p(yn, yn+1), p(yn−1, yn+1)}

≤ max{p(y_n−1, yn), p(yn, yn+1), p(yn−1, yn) +p(yn, yn+1)−p(yn, yn)}

≤ max{p(yn−1, yn), p(yn, yn+1), p(yn−1, yn) +p(yn, yn+1)}

= p(yn−1, y_n) +p(y_n, y_n+1), and

min{p(gx_n, F x_n), p(Gx_n+1, f x_n+1), p(gx_n, f x_n+1) +p(Gx_n+1, F x_n)}

= min{p(y_n−1, y_n), p(y_n, y_n+1), p(yn−1, y_n+1), p(y_n, y_n)}

= min{p(y_n−1, y_n+1), p(y_n, y_n)}, we obtain that

p(yn, yn+1) = p(F xn, f xn+1)

≤ δ(p(yn−1, y_n) +p(y_n, y_n+1)) +Lmin{p(yn−1, y_n+1), p(y_n, y_n)}.

We separate the proof into the following cases.

Case I : If min{p(y_n−1, y_n+1), p(y_n, y_n)}=p(yn−1, y_n+1),then p(y_n, y_n+1) ≤ δ(p(yn−1, y_n) +p(y_n, y_n+1)) +Lp(yn−1, y_n+1)

≤ δp(yn−1, y_n) +δp(y_n, y_n+1) +L(p(yn−1, y_n) +p(y_n, y_n+1)−p(y_n, y_n))

≤ δp(yn−1, y_n) +δp(y_n, y_n+1) +Lp(yn−1, y_n) +Lp(y_n, y_n+1).

This implies that

p(y_n, y_n+1)≤ δ+L

1−(δ+L)p(yn−1, y_n).

Letk1= _1−(δ+L)^δ+L .Since δ+L < ¹₂,we have k1<1.Therefore p(y_n, y_n+1)≤k₁p(yn−1, y_n).

Case II : If min{p(y_n−1, yn+1), p(yn, yn)}=p(yn, yn),then p(y_n, y_n+1) ≤ δ(p(yn−1, y_n) +p(y_n, y_n+1)) +Lp(y_n, y_n)

≤ δp(yn−1, yn) +δp(yn, yn+1) +Lp(yn−1, yn).

This implies that

p(y_n, y_n+1)≤ δ+L

1−δp(yn−1, y_n).

Letk2= ^δ+L_1−δ.Sinceδ+L < ¹₂,we have k2<1.Therefore

p(y_n, y_n+1)≤k₂p(yn−1, y_n).

Choose k= max{k₁, k2}.Therefore 0< k <1.For eachn∈N,we obtain that

p(y_n, y_n+1)≤kⁿp(y₀, y₁). (2.10) We can complete the proof by the same arguments appeared in Theorem 2.1.

Letting F =f andG=g in Theorem 2.4, we immediately have the following result:

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(a) f(X)⊆g(X).

(b) There exist δ >0 andL≥0 with δ+L < ¹₂ such that

p(f x, f y)≤δM(x, y) +Lmin{p(gx, f x), p(gy, f y), p(gx, f y), p(gy, f x)}, (2.11) for allx, y∈X, where

M(x, y) = max{p(gx, gy), p(gx, f x), p(gy, f y), p(gx, f y), p(gy, f x)}.

(c) f(X) or g(X) is complete.

If {f, G} is weakly compatible, then f and g have a unique common fixed point in X.

We finally prove the result on the continuity in the set of common fixed points for self mappings in partial metric spaces.

Theorem 2.6. Let (X, p) be a partial metric space. Suppose that f, g and T are self mappings on X satisfying the following conditions:

(a) There exist δ∈(0,1)and L≥0 such that

p(T x, f y)≤δM(x, y) +Lmin{p(gx, T x), p(gy, f y), p(gx, f y), p(gy, T x)}, (2.12) for allx, y∈X, where

M(x, y) = max{p(gx, gy), p(gx, T x), p(gy, f y),1

2[p(gx, f y) +p(gy, T x)]}.

(b) The setF(f, g, T) ={z∈X:f z =gz=T z=z, p(z, z) = 0} of all common fixed points of f, g andT is nonempty.

If g is continuous at z∈F(f, g, T), then f and T are continuous atz.

Proof. Assume that z ∈ F(f, g, T) and {x_n} is a sequence in X converging to z. Using (2.12), we obtain that

p(T z, f xn)≤δM(z, xn) +Lmin{p(gz, T z), p(gx_n, f xn), p(gz, f xn), p(gxn, T z)}, where

M(z, x_n) = max{p(gz, gx_n), p(gz, T z), p(gx_n, f x_n),1

2[p(gz, f x_n) +p(gx_n, T z)]}.

This implies that

p(T z, f x_n) ≤ δmax{p(gz, gx_n), p(z, z), p(gx_n, f x_n),1

2[p(f z, f x_n) +p(gx_n, gz)]}+ Lmin{p(z, z), p(gx_n, f xn), p(f z, f xn), p(gxn, gz)}

≤ δmax{p(gz, gx_n), p(gx_n, gz) +p(f z, f x_n)−p(z, z),1

2[p(f z, f x_n) +p(gx_n, gz)]}

≤ δmax{p(gz, gx_n), p(gx_n, gz) +p(f z, f x_n),1

2[p(f z, f x_n) +p(gx_n, gz)]}

= δ(p(gxn, gz) +p(f z, f xn)).

It follows that

p(f z, f x_n)≤δ(p(gx_n, gz) +p(f z, f x_n)).

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Therefore

p(f z, f x_n)≤ δ

1−δp(gx_n, gz). (2.13)

By continuity ofg, we obtain that

n→∞lim p(gx_n, gz) =p(gz, gz) =p(z, z) = 0.

Using (2.13), this yields

n→∞lim p(f z, f x_n) = 0.

This implies thatf is continuous at z. Similarly, by applying (2.12), we have

p(T xn, f z)≤δM(xn, z) +Lmin{p(gx_n, T xn), p(gz, f z), p(gxn, f z), p(gz, T xn)}, where

M(xn, z) = max{p(gx_n, gz), p(gxn, T xn), p(gz, f z),1

2[p(gxn, f z) +p(gz, T xn)]}.

This implies that

p(T xn, f z) ≤ δmax{p(gx_n, gz), p(gxn, T xn), p(z, z),1

2[p(gxn, gz) +p(T z, T xn)]}+ Lmin{p(gx_n, T x_n), p(z, z), p(gx_n, gz), p(T z, T x_n)}

≤ δmax{p(gx_n, gz), p(gxn, gz) +p(T z, T xn)−p(z, z),1

2[p(gxn, gz) +p(T z, T xn)]}

≤ δmax{p(gx_n, gz), p(gxn, gz) +p(T z, T xn),1

2[p(gxn, gz) +p(T z, T xn)]}

= δ(p(gx_n, gz) +p(T z, T x_n)).

Therefore

p(T x_n, T z)≤ δ

1−δp(gx_n, gz). (2.14)

By continuity ofg, we obtain that

n→∞lim p(T xn, T z) = 0.

This implies thatT is continuous at z.

If T =f in Theorem 2.6, then we obtain the following results:

p(f x, f y)≤δM(x, y) +Lmin{p(gx, f x), p(gy, f y), p(gx, f y), p(gy, f x)}, (2.15) for allx, y∈X, where

M(x, y) = max{p(gx, gy), p(gx, f x), p(gy, f y),1

2[p(gx, f y) +p(gy, f x)]}.

(b) The set F(f, g) = {z ∈ X : f z = gz = z, p(z, z) = 0} of all common fixed points of f and g is nonempty.

If g is continuous at z∈F(f, g), then f is continuous at z.

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Corollary 2.8. (Theorem 2.7, [1]) Let (X, d) be a metric space. Suppose that f and g are self mappings onX satisfying the following conditions:

d(f x, f y)≤δM(x, y) +Lmin{d(gx, f x), d(gy, f y), d(gx, f y), d(gy, f x)}, (2.16) for allx, y∈X, where

M(x, y) = max{d(gx, gy), d(gx, f x), d(gy, f y),1

2[d(gx, f y) +d(gy, f x)]}.

(b) The set F(f, g) ={z∈X:f z=gz =z} of all common fixed points of f and g is nonempty.

If g is continuous at z∈F(f, g), then f is continuous at z.

Acknowledgements:

This research is supported by Naresuan University under grant R2557B055. . References

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