Several common …xed point results for such mappings are proved

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ALMOST RATIONAL CONTRACTION MAPPINGS WITH AN APPLICATION

NAWAB HUSSAIN, HUSEYIN ISIK, AND MUJAHID ABBAS

Abstract. In this paper, we introduce the notion of generalized almost rational contraction with respect to a pair of self mappings on a complete metric space. Several common …xed point results for such mappings are proved. Our results extend and unify various results in the existing literature. An example and application to obtain the existence and uniqueness of common solution of functional equations arising in dynamic programming are also given in order to illustrate the e¤ectiveness of the presented results.

1. Introduction and Preliminaries

Fixed point theory plays a vital role in solving problems arising in various disciplines of mathematical analysis such as split feasibility problems, variational inequality problems, nonlinear optimization problems, equilibrium problems, comple- mentarity problems, selection and matching problems, and problems of proving an existence of solution of integral and di¤erential equations. One of the basic and the most widely applied result in metric …xed point theory is "Banach ( or Banach- Cassioppoli ) Contraction principle" due to Banach [8] . It states that if(X; d)is a complete metric space andf :X !X satis…es

d(f x; f y) kd(x; y);

for allx; y2X;withk2(0;1);thenf has a unique …xed point. The basic idea of this principle rests in the use of successive approximations to establish the existence and uniqueness of solution of an operator equationf(x) =x, particularly it can be employed to prove the existence of solution of di¤erential or integral equations. Due to its applications in mathematics and other related disciplines, Banach contraction principle has been generalized in many directions.

Extensions of Banach contraction principle have been obtained either by gener- alizing the domain of the mapping or by extending the contractive condition on the mappings ( see, [1]-[30] and references therein).

Berinde [10, 11] de…ned the notion of an almost contraction mapping, which is more general than a contraction mapping.

Let(X; d) be a complete metric space. A mapf : X ! X is called an almost contraction if there exist a constant 2(0;1)and some L 0such that

d(f x; f y) d(x; y) +Ld(y; f x); for allx; y2X:

2000Mathematics Subject Classi…cation. Primary 47H10, Secondary 54H25.

Key words and phrases. Common …xed point, admissible mappings, almost contractions, weakly compatible mappings, functional equations.

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Berinde [10] proved some …xed point theorems for almost contractions in a complete metric space which generalized many results in the literature.

In metric …xed point theory, contractive conditions on mappings play vital role in …nding the solution of …xed point problems. It is a common practice to extend and generalize existing contractive contractive conditions and then to employ it to obtain …xed point result in the framework of a metric space. Following this trend, Samet et al. [27] …rst introduced -admissible mappings and then ( - )- contractive type mappings to obtain some interesting generalizations of Banach contraction principle. For more results in this direction, we refer to [14], [22]-[28]

and references mentioned therein. Recently, Alizadeh et al. [6] de…ned the concept of cyclic( ; )-admissible mapping as follows:

De…nition 1 ([6]). Let X be a nonempty set and ; :X ![0;1). A selfmap- ping T onX is called a cyclic( ; )-admissible mapping if

(x) 1for somexin X implies that (T x) 1;

and

(x) 1 for somexinX implies that (T x) 1:

An elementx2X is called a common …xed point of mappings f; g:X !X if f x=gx=x:

Sessa [29] coined the term weakly commuting maps. Jungck [17] generalized the notion of weak commutativity by introducing compatible maps and then weakly compatible maps [18].

De…nition 2([19]). A pair(f; T)of self-mappings on a setX is said to be weakly compatibleiff andT commute at their coincidence point (i.e. f T x=T f x; x2X whenever f x = T x). A point y 2 X is called a point of coincidence of two self- mappingsf andT onX if there exists a point x2Xsuch thaty=T x=f x.

The notationsF(f; g)andC(f; g)stand for the set of all common …xed point and the set of all coincidence points off andg, respectively:In the sequel, the letters R; R⁺, and N will denote the set of all real numbers, the set of all non negative real numbers and the set of all natural numbers, respectively.

To obtain common …xed point results, we extend the de…nition of( ; )-admissible mapping to a pair of two mappings as follows:

De…nition 3. Let f; g; S; T :X !X and ; :X !R⁺: We say that(f; g)is a cyclic ( ; )-admissible with respect to(S; T) if

(i) (Sx) 1for somex2X implies (f x) 1;

(ii) (T x) 1 for somex2X implies (gx) 1:

IfS=T =IX ( identity mapping on X ) in above de…nition, then we have:

De…nition 4. Let f; g :X !X and ; :X !R⁺: We say that a pair (f; g)is a cyclic ( ; )-admissible if

(x) 1 for somex2X implies that (f x) 1;

and

(x) 1 for somex2X implies that (gx) 1:

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On the other hands, Khan et al. [20] introduced and employed the notion of altering distance function to obtain some interesting …xed point results in metric spaces. Note that altering distance functions are continuos whereas Su [30] de…ned generalized altering distance function, not necessarily continuos, as follows:

De…nition 5 ([30]). A mapping : [0;1)![0;1)is called a generalized altering distance function if

(a) is non-decreasing,

(b) (t) = 0 if and only ift= 0.

We set

z=f : [0;1)![0;1) : is generalized altering distanceg:

=f': [0;1)![0;1): 'is a nondecreasing and right upper semi-continuous and for allt >0; we have (t)> '(t);where is a generalized altering distanceg.

=f : [0;1)![0;1) : is continuous and (t) = 0 if and only ift= 0g: Following the direction in [13], we denote set 1=f 1:R⁺⁶ !R⁺: ₁satis…es (i) (iii)g;where

(i) ₁ is nondecreasing and continuous in each coordinate;

(ii) ₁(t; t; t; t; t; t) t for allt >0;

(iii) ₁(t1; t2; t3; t4; t5; t6) = 0if and only ift1=t2=t3=t4=t5=t6= 0.

2=f 2 :R⁺⁴ !R⁺ : ₂ is continuous in each coordinate and if any one of the argument is zero, then ₂(t1; t2; t3; t4) = 0g.

We now introduced generalized almost rational contraction mappings as follows:

De…nition 6. Let f; g; S and T be selfmaps of a metric space X, and (f; g) be a cyclic ( ; )-admissible with respect to (S; T): We say that (f; g) is generalized almost (S; T)rational contraction if

(Sx) (T y) 1 implies that (d(f x; gy)) '(M(x; y)) +L (N(x; y)) (1.1) for allx; y2X and some L 0; where 2z,'2 ; 2 ;

M(x; y) = ₁ d(Sx; T y); d(Sx; f x); d(T y; gy);d(Sx; gy) +d(f x; T y)

2 ;

d(T y; gy) [1 +d(Sx; f x)]

1 +d(Sx; T y) ;d(f x; T y) [1 +d(Sx; gy)]

1 +d(Sx; T y) ; N(x; y) = ₂(d(Sx; f x); d(T y; gy); d(Sx; gy); d(f x; T y)); with ₁2 ¹;and ₂2 ²:

IfS=T =IX ( identity mapping on X ) in above de…nition, then we have:

De…nition 7. Let f and g be selfmaps of a metric spaceX; and(f; g)be a cyclic ( ; )-admissible:We say that(f; g)is generalized almost rational contraction if

(x) (y) 1implies that (d(f x; gy)) '(M_{f g}(x; y)) +L (N_{f g}(x; y)) for allx; y2X and some L 0; where 2z,'2 ; 2 ;

Mf g(x; y) = ₁ d(x; y); d(x; f x); d(y; gy);d(x; gy) +d(f x; y)

2 ;

d(y; gy) [1 +d(x; f x)]

1 +d(x; y) ;d(f x; y) [1 +d(x; gy)]

1 +d(x; y) ; Nf g(x; y) = ₂(d(x; f x); d(y; gy); d(x; gy); d(f x; y)); with ₁2 1; and ₂2 2:

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In this paper, we obtain several common …xed point results of generalized almost rational contraction pair with respect to a pair(S; T)of selfmappings. Our results extend, generalize and unify comparable results in the existing literature. An example is presented to support the results presented herin. We then employ our results to obtain common …xed points of cyclic mappings on complete metric spaces. As an application of our results, the existence and uniqueness of the common bounded solution of a functional equation arising in dynamic programming are shown.

2. Main Results

In this section we obtain several common …xed point results of almost generalized rational contraction mappings in the framework of complete metric spaces.

We start with the following result.

Theorem 1. Let f; g; S and T be selfmaps of a complete metric space X with f(X) T(X), g(X) S(X) and (f; g) be a generalized almost (S; T) rational contraction pair. Suppose that:

(a) there existsx02X such that (Sx0) 1 and (T x0) 1;

(b) if fx_ng is a sequence inX such that (x_n) 1; (x_n) 1 for alln and x_n!x, then (x) 1 and (x) 1.

Then the pairs (f; S)and(g; T)have a point of coincidence in X. Moreover, if (i) ff; Sgandfg; Tg are weakly compatible and,

(ii) (Su) 1 and (T v) 1 wheneveru2 C(f; S)andv2 C(g; T): Thenf; g; S andT have a common …xed point.

Proof. Let x₀ be a given point in X such that (Sx₀) 1 and (T x₀) 1:

Sincef X T X;we can choose a pointx₁2X such that f x₀ =T x₁: Also, since gX SX;there exists a pointx₂2X such thatgx₁=Sx₂:Continuing this way, we can obtain sequencesfx_ngandfy_ng inX such that

y2n=f x2n =T x2n+1; y2n+1=gx2n+1=Sx2n+2; n2N[ f0g: (2.1) As (f; g) is a cyclic ( ; )-admissible with respect to (S; T) and (Sx0) 1, we have (f x0) = (T x1) 1 which further implies that (gx1) = (Sx2) 1:

Continuing this way, we obtain that (Sx2n) 1 and (T x2n+1) 1 for alln2 N0=N[ f0g:Similarly, by (T x0) 1;we have (T x2n) 1and (Sx2n+1) 1 for alln2N0:Thus

(Sxn) 1and (T xn) 1 for alln2N0: (2.2) Suppose thaty_2n6=y_2n+1 for all n2N0:Now we show that

nlim!1d(yn; yn+1) = 0: (2.3) Putting x= x_2n and y =x_2n+1 in (1.1) and using (2.1) and (2.2), we obtain that

(d(y2n; y2n+1)) = (d(f x2n; gx2n+1))

'(M(x_2n; x_2n+1)) +L (N(x_2n; x_2n+1)); (2.4)

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where

M(x_2n; x_2n+1) = ₁ d(Sx_2n; T x_2n+1); d(Sx_2n; f x_2n); d(T x_2n+1; gx_2n+1); d(Sx2n; gx2n+1) +d(f x2n; T x2n+1)

2 ;d(T x2n+1; gx2n+1) [1 +d(Sx2n; f x2n)]

1 +d(Sx_2n; T x_2n+1) ; d(f x_2n; T x_2n+1) [1 +d(Sx_2n; gx_2n+1)]

1 +d(Sx2n; T x2n+1)

= ₁ d(y_2n ₁; y_2n); d(y_2n ₁; y_2n); d(y_2n; y_2n+1);d(y_2n ₁; y_2n+1) +d(y_2n; y_2n)

2 ;

d(y_2n; y_2n+1) [1 +d(y_2n ₁; y_2n)]

1 +d(y_2n ₁; y_2n) ;d(y_2n; y_2n) [1 +d(y_2n ₁; y_2n+1)]

1 +d(y_2n ₁; y_2n)

= ₁ d(y2n 1; y2n); d(y2n 1; y2n); d(y2n; y2n+1);d(y2n 1; y2n+1)

2 ; d(y2n; y2n+1);0

1 d(y2n 1; y2n); d(y2n 1; y2n); d(y2n; y2n+1);d(y2n 1; y2n) +d(y2n; y2n+1)

2 ;

d(y2n; y2n+1); d(y2n; y2n+1) ; and

N(x_2n; x_2n+1) = ₂(d(Sx_2n; f x_2n); d(T x_2n+1; gx_2n+1); d(Sx_2n; gx_2n+1); d(f x_2n; T x_2n+1))

= ₂(d(y2n 1; y2n); d(y2n; y2n+1); d(y2n 1; y2n+1); d(y2n; y2n)) = 0:

Ifd(y_2n ₁; y_2n) d(y_2n; y_2n+1)for somen2N;then by (2.4);we have (d(y_2n; y_2n+1)) '(M(x_2n; x_2n+1))

' ₁ d(y2n 1; y2n); d(y2n 1; y2n); d(y2n; y2n+1); d(y_2n ₁; y_2n) +d(y_2n; y_2n+1)

2 ; d(y_2n; y_2n+1); d(y_2n; y_2n+1) '( ₁(d(y2n; y2n+1); d(y2n; y2n+1); d(y2n; y2n+1);

d(y_2n; y_2n+1); d(y_2n; y_2n+1); d(y_2n; y_2n+1))) '(d(y2n; y2n+1));

a contradiction to the fact that y_2n 6= y_2n+1. So for all n 2 N; we have d(y_2n; y_2n+1)< d(y_2n ₁; y_2n):

From (2.4);we obtain that

(d(y2n; y2n+1)) '(d(y2n 1; y2n)) (2.5) and hence the sequencefd(y2n; y2n+1)g is decreasing and bounded below. Con- sequently, there existsr 0such thatlimn!1d(y2n; y2n+1) =r:

If r > 0, then taking limit as n ! 1 on both sides of (2.5) and using the properties of and';we have

(r) lim

n!1 (d(y_2n; y_2n+1))

nlim!1'(d(y2n 1; y2n)) '(r);

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a contradiction and hencer= 0; that is,

nlim!1d(y2n; y2n+1) = 0: (2.6)

Puttingx=x2n+1andy=x2n+2in (1.1) and following arguing similar to those given above, we have

nlim!1d(y2n+1; y2n+2) = 0: (2.7)

Therefore lim

n!1d(yn; yn+1) = 0holds.

Now we show thatfy_ng is a Cauchy sequence inX. For this, it is su¢ cient to show thatfy2ng is Cauchy inX. Assume on contrary thatfy2ng is not a Cauchy sequence. Then, there exists some " >0 for which we can …nd two subsequences fy2mkgandfy2nkgoffy2ngsuch thatnk is the smallest index satisfyingnk > mk>

kand

d(y2nk; y2mk) " and d(y2nk 2; y2mk)< ": (2.8) Now

" d(y2nk; y2mk) d(y2nk; y2nk 1) +d(y2nk 1; y2nk 2) +d(y2nk 2; y2mk)

< d(y_2n_k; y_2n_k ₁) +d(y_2n_k ₁; y_2n_k ₂) +":

On taking limit ask! 1on both sides of above inequality, we obtain that

klim!1d(y_2n_k; y_2m_k) =": (2.9)

Also, from triangular inequality, we have

jd(y2nk; y2mk+1) d(y2nk; y2mk)j d(y2mk; y2mk+1):

On taking limit ask! 1on both sides of above inequality and from (2.3) and (2.9), we get

klim!1d(y2nk; y2mk+1) =": (2.10)

Similarly, we obtain that

klim!1d(y2nk 1; y2mk) = lim

k!1d(y2nk 1; y2mk+1) =": (2.11)

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Now by (2.3), (2.9), (2.10) and (2.11), we have

M(x_2n_k; x_2m_k₊₁) = ₁ d(Sx_2n_k; T x_2m_k₊₁); d(Sx_2n_k; f x_2n_k); d(T x_2m_k₊₁; gx_2m_k₊₁); d(Sx2nk; gx2mk+1) +d(f x2nk; T x2mk+1)

2 ;

d(T x_2m_k₊₁; gx_2m_k₊₁) [1 +d(Sx_2n_k; f x_2n_k)]

1 +d(Sx2nk; T x2mk+1) ; d(f x2nk; T x2mk+1) [1 +d(Sx2nk; gx2mk+1)]

1 +d(Sx_2n_k; T x_2m_k₊₁)

= ₁ d(y2nk 1; y2mk); d(y2nk 1; y2nk); d(y2mk; y2mk+1); d(y_2n_k ₁; y_2m_k₊₁) +d(y_2n_k; y_2m_k)

2 ;

d(y2mk; y2mk+1) [1 +d(y2nk 1; y2nk)]

1 +d(y_2n_k ₁; y_2m_k) ; d(y_2n_k; y_2m_k) [1 +d(y_2n_k ₁; y_2m_k₊₁)]

1 +d(y2nk 1; y2mk) and

N(x_2n_k; x_2m_k₊₁) = ₂(d(Sx_2n_k; f x_2n_k); d(T x_2m_k₊₁; gx_2m_k₊₁); d(Sx_2n_k; gx_2m_k₊₁); d(f x2nk; T x2mk+1))

= ₂(d(y_2n_k ₁; y_2n_k); d(y_2m_k; y_2m_k₊₁); d(y_2n_k ₁; y_2m_k₊₁); d(y_2n_k; y_2m_k)): Thus,

klim!1M(x2nk; x2mk+1) = ₁f";0;0; ";0; "g "; and

klim!1N(x_2n_k; x_2m_k₊₁) = ₂f0;0; "; "g= 0:

From (2.2), we have (Sx2nk) (T x2mk+1) 1. Substituting x = x2nk and y=x2mk+1 in (1.1); we get

(d(y2nk; y2mk+1)) = (d(f x2nk; gx2mk+1))

'(M(x_2n_k; x_2m_k₊₁)) +L (N(x_2n_k; x_2m_k₊₁)): On taking limit ask! 1; we have

(") lim

k!1 (d(y2nk; y2mk+1))

k!1lim'(M(x_2n_k; x_2m_k₊₁)) +L lim

k!1 (N(x_2n_k; x_2m_k₊₁))

= lim

k!1'(M(x2nk; x2mk+1)) '(");

a contradiction and hence fy_2ng is a Cauchy sequence in X. Next, we assume that there existsz2X such that

nlim!1y_n =z: (2.12)

From (2.1) and (2.12), we obtain

nlim!1f x2n = lim

n!1T x2n+1= lim

n!1gx2n+1= lim

n!1Sx2n+2=z: (2.13) We now show thatz is a common …xed point off,g,S andT.

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Sinceg(X) S(X), we can choose a pointuin X such thatz=Su. Suppose thatd(z; f u)6= 0.

By (2.2), (2.13) and the condition(b);we have (Su) (T x_2n+1) 1:Then by usingx=uandy=x_2n+1 in (1.1), we get

(d(f u; gx_2n+1)) '(M(u; x_2n+1)) +L (N(u; x_2n+1)); (2.14) where

M(u; x2n+1) = ₁ d(Su; T x2n+1); d(Su; f u); d(T x2n+1; gx2n+1); d(Su; gx_2n+1) +d(f u; T x_2n+1)

2 ;d(T x_2n+1; gx_2n+1) [1 +d(Su; f u)]

1 +d(Su; T x_2n+1) ; d(f u; T x2n+1) [1 +d(Su; gx2n+1)]

1 +d(Su; T x2n+1)

! 1 0; d(z; f u);0;d(f u; z)

2 ;0; d(f u; z) ; and

N(u; x2n+1) = ₂(d(Su; f u); d(T x2n+1; gx2n+1); d(Su; gx2n+1); d(f u; T x2n+1))

! 2(d(z; f u);0;0; d(f u; z)); asn! 1. From (2.14), we have

(d(f u; z)) lim

n!1 (d(f u; gx_2n+1))

nlim!1'(M(u; x_2n+1)) +Llim

n!1 (N(u; x_2n+1))

= lim

n!1'(M(u; x2n+1))

' ₁ 0; d(z; f u);0;d(f u; z)

2 ;0; d(f u; z) '(d(f u; z)); a contradiction and henced(f u; z) = 0;that isf u=z;and sou2 C(f; S): Similarly, sincef(X) T(X), we can choose a pointv in X such that z=T v.

Suppose thatd(z; gv)6= 0.

By (2.2), (2.13) and the condition (b); we have (Sx2n) (T v) 1: Then by usingx=x2n andy=v in (1.1), we deduce

(d(f x2n; gv)) '(M(x2n; v)) +L (N(x2n; v)); (2.15) where

M(x2n; v) = ₁ d(Sx2n; T v); d(Sx2n; f x2n); d(T v; gv); d(Sx_2n; gv) +d(f x_2n; T v)

2 ;d(T v; gv) [1 +d(Sx_2n; f x_2n)]

1 +d(Sx2n; T v) ; d(f x_2n; T v) [1 +d(Sx_2n; gv)]

1 +d(Sx_2n; T v)

! 1 0;0; d(z; gv);d(z; gv)

2 ; d(z; gv);0 ;

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and

N(u; x2n+1) = ₂(d(Sx2n; f x2n); d(T v; gv); d(Sx2n; gv); d(f x2n; T v))

! 2(0; d(z; gv); d(z; gv);0); asn! 1. Now by (2.15), we have

(d(z; gv)) lim

n!1 (d(f x2n; gv))

nlim!1'(M(x_2n; v)) +Llim

n!1 (N(x_2n; v))

= lim

n!1'(M(x2n; v))

' ₁ 0;0; d(z; gv);d(z; gv)

2 ; d(z; gv);0 '(d(z; gv)); a contradiction and henced(z; gv) = 0;that isz=gv;and so v2 C(g; T): Thus, z =f u =Su=gv =T v: By the weak compatibility of the pairs (f; S) and(g; T),f z=Sz andgz=T z:

Sincez2 C(f; S)andv2 C(g; T);we have (Sz) (T v) 1 and so, by (1.1) (d(f z; z)) = (d(f z; gv))

'(M(z; v)) +L (N(z; v)); (2.16) where

M(z; v) = ₁ d(Sz; T v); d(Sz; f z); d(T v; gv); d(Sz; gv) +d(f z; T v)

2 ;d(T v; gv) [1 +d(Sz; f z)]

1 +d(Sz; T v) ; d(f z; T v) [1 +d(Sz; gv)]

1 +d(Sz; T v)

= ₁(d(f z; z);0;0; d(f z; z);0; d(f z; z)) d(f z; z); and

N(z; v) = ₂(d(Sz; f z); d(T v; gv); d(Sz; gv); d(f z; T v))

= ₂(0;0; d(f z; z); d(f z; z)) = 0:

By (2.16), we get

(d(f z; z)) '(d(f z; z));

which implies thatz=f z;and so z=f z=Sz:Similarly, it can be shown that z=gz=T z:This completes the proof.

Corollary 1. Let f; g; S and T be selfmaps of a complete metric space /X with f(X) T(X),g(X) S(X)and(f; g)be a cyclic( ; )-admissible with respect to (S; T) such that

(Sx) (T y) (d(f x; gy)) '(Mmax(x; y)) +L (Nmin(x; y)); (2.17) for allx; y2X and some L 0; where 2z,'2 ; 2 ;and

M_max(x; y) = max d(Sx; T y); d(Sx; f x); d(T y; gy);d(Sx; gy) +d(f x; T y)

2 ;

d(T y; gy) [1 +d(Sx; f x)]

1 +d(Sx; T y) ;d(f x; T y) [1 +d(Sx; gy)]

1 +d(Sx; T y)

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and

Nmin(x; y) = min (d(Sx; f x); d(T y; gy); d(Sx; gy); d(f x; T y)): Assume that the following conditions are satis…ed:

(b) iffxngis a sequence inX such that (xn) 1and (xn) 1for all nand limn!1xn=x, then (x) 1 and (x) 1:

Proof. Let (Sx) (T y) 1 for x; y 2 X: If we take ₁(t1; t2; t3; t4; t5; t6) = maxft1; t2; t3; t4; t5; t6gand ₂(t1; t2; t3; t4) = minft1; t2; t3; t4gin Theorem 1, then by (2.17), we have

(d(f x; gy)) '(M(x; y)) +L (N(x; y)): Hence the result follows.

If we take (Sx) = (T y) = 1;and (t) =t; '(t) = tand (t) =tin Corollary 1, we have a generalized version of Theorem 1 in [10]:

Theorem 2. Let f; g; S and T be selfmaps of a complete metric space /X with f(X) T(X), g(X) S(X). Suppose that there exist a constant 2 (0;1) and someL 0 such that

d(f x; gy) Mmax(x; y) +LNmin(x; y); (2.18) for allx; y2X:Then the pairs(f; S)and(g; T)have a point of coincidence inX.

Moreover, if ff; Sg and fg; Tg are weakly compatible, then f; g; S and T have a common …xed point.

If we takeL= 0in Corollary 1, we have the following result.

Corollary 2. Let f; g; S and T be selfmaps of a complete metric space X with f(X) T(X),g(X) S(X)and(f; g)be a cyclic( ; )-admissible with respect to (S; T) such that

(Sx) (T y) (d(f x; gy)) '(Mmax(x; y)); (2.19) for all x; y2X;where 2z and'2 : Assume that the following conditions are satis…ed:

(b)iffxng is a sequence inX such that (xn) 1and (xn) 1 for allnand limn!1xn=x, then (x) 1 and (x) 1.

If we take'(t) = (t) (t)in Corollary 2, we have the following corollary.

(Sx) (T y) (d(f x; gy)) (M_max(x; y)) (M_max(x; y)); (2.20)

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for all x; y2X;where 2zand 2 : Assume that the following conditions are satis…ed:

(a) there existsx₀2X such that (Sx₀) 1 and (T x₀) 1;

(b)iffx_ng is a sequence inX such that (x_n) 1and (x_n) 1 for allnand lim_n_!1x_n=x, then (x) 1 and (x) 1.

If we take (Sx) = (T y) = 1 in Corollary 3, we have a generalized version of Theorem 2.1 in [3]:

Theorem 3. Let f; g; S and T be selfmaps of a complete metric space /X with f(X) T(X), g(X) S(X). Suppose that for any x; y2X; 2 z; and 2 ; we have

(d(f x; gy)) (Mmax(x; y)) (Mmax(x; y)); (2.21) for allx; y2X:Then the pairs(f; S)and(g; T)have a point of coincidence inX.

Moreover, if ff; Sg and fg; Tg are weakly compatible, then f; g; S and T have a common …xed point.

If we take (t) =tin Corollary 2, we have the following result.

(Sx) (T y)d(f x; gy) '(M_max(x; y)); (2.22) for allx; y2X;where'2 :Assume that the following conditions are satis…ed:

(a) there existsx₀2X such that (Sx₀) 1 and (T x₀) 1;

(b)iffx_ng is a sequence inX such that (x_n) 1and (x_n) 1 for allnand lim_n!1x_n=x, then (x) 1 and (x) 1.

For the uniqueness of the …xed point of a generalized almost (S; T) rational contraction, we will consider the following hypothesis.

(H) For allx; y2 F(f; g; S; T);we have (Sx) 1 and (T y) 1:

Theorem 4. Adding condition(H)to the hypotheses of Theorem 1, we obtain the uniqueness of the common …xed point off,g,S andT:

Proof. Suppose thatx=f x=gx=Sx=T xandy=f y=gy=Sy=T y:Then, from(H); (Sx) (T y) 1:

Applying (1.1), we obtain

(d(x; y)) = (d(f x; gy))

'(M(x; y)) +L (N(x; y)); (2.23)

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where

M(x; y) = ₁ d(Sx; T y); d(Sx; f x); d(T y; gy); d(Sx; gy) +d(f x; T y)

2 ;d(T y; gy) [1 +d(Sx; f x)]

1 +d(Sx; T y) ; d(f x; T y) [1 +d(Sx; gy)]

1 +d(Sx; T y)

= ₁(d(x; y);0;0; d(x; y);0; d(x; y)) d(x; y); and

N(x; y) = ₂(d(Sx; f x); d(T y; gy); d(Sx; gy); d(f x; T y))

= ₂(0;0; d(x; y); d(x; y)) = 0:

From (2.23), we have

(d(x; y)) '(d(x; y)); which implies thatd(x; y) = 0;that is,x=y:

Remark 1. Adding condition (H) to the hypotheses of Corollaries 1-4, we obtain the uniqueness of the common …xed point off,g,S andT:

If we chooseS=T =IX in Theorem 1, we have the following corollary.

Corollary 5. Letf andgbe selfmaps of a complete metric spaceX and(f; g)be a generalized almost rational contraction pair. Assume that the following conditions hold:

(a) there existsx₀2X such that (x₀) 1 and (x₀) 1;

(b) iffx_ngis a sequence inX such that (x_n) 1for allnandlim_n_!1x_n =x;

then (x) 1.

Thenf andg have a common …xed point. Moreover, if (x) 1 and (y) 1 wheneverx; y2 F(f; g);thenf andg have a unique common …xed point.

Now, we furnish the following example which illustrates Theorem 1 as well as Theorem 4.

Example 1. LetX =Rbe endowed with the usual metric, (t) =t,'(t) =¹₂tfor allt 0, and ₁(t₁; t₂; t₃; t₄; t₅; t₆) = maxft₁; t₂; t₃; t₄; t₅; t₆gfor allt₁; t₂; t₃; t₄; t₅; t₆ 0:

De…ne the self-mappingsf; g; S andT on X by f x=

( x

4 if x2 ¹₂;0 ;

4x

5 if x2Rn ¹2;0 and gx= ( 2x

15 if x2 0;¹₂ ;

7x

10 if Rn 0;¹₂ Sx=

( 4x

5 if x2 0;¹₂ ;

3x if Rn 0;¹₂ and T x= ( 4x

5 ifx2 ¹2;0 ; 2x ifRn ¹2;0 :

Note thatf(X) T(X) andg(X) S(X), ff; Sg and fg; Tg are weakly compatible.

De…ne ; :X ![0;1)as (x) =

(1 if x2 ²₅;0 ;

0 otherwise and (x) =

(1 if x2 0;²₅ ; 0 otherwise.

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If x2X is such that (Sx) 1, then Sx2 ²₅;0 and hencex2 0;¹₂ : By the de…nitions of f and ;we have f x2 0;²₅ and so (f x) 1:If for some x2X, we have (T x) 1 then T x2 0;²₅ and hence x2 ¹₂;0 :By the de…nitions of g and ; we have gx 2 ²5;0 and so (gx) 1: Thus (f; g) is a cyclic ( ; )- admissible with respect to (S; T). Moreover, (Sx0) 1 and (T x0) 1 holds forx0= 0:

Iffxngis any sequence inX such that (xn) 1and (xn) 1 for alln2Nand x_n!x;then by the de…nition of and ;we havex_n2 ²₅;0 \ 0;²₅ =f0gfor alln2Nand sox2 f0gwhich implies that (x) 1 and (x) 1:

Next, we show that (f; g) is a generalized almost(S; T) rational contraction pair.

Let (Sx) (T y) 1: Then Sx2 ²₅;0 and T y2 0;²₅ and so x2 0;¹₂ and y2 ¹2;0 : Thus, we have

(d(f x; gy)) = jf x gyj= 4x 5

7y 10

= 8x 7y

10

8x 10 =1

2jSx f xj 1

2M(x; y) ='(M(x; y)) '(M(x; y)) +L (N(x; y));

for some L 0 and 2 : Note that assumption (ii) of Theorem 1 and the condition(H)also hold. Thus, by Theorems 1 and 4,f; g; S andT have a unique common …xed point.

3. Common fixed points of cyclic mappings

Let A andB be two nonempty subsets of a set X. A mapping f : X ! X is said to be cyclic (with respect toAandB) iff(A) B andf(B) A.

The …xed point theory of cyclic contractive mappings is a recent development.

Kirk et al. [21] in 2003 introduced a class of mappings which satisfy contraction condition for points x and y where x 2 A and y 2 B. For more work in this direction, we refer to [4, 25].

De…nition 8. The mappingsf; g; S; T :A[B !A[Bare called cyclic iff A T B andgB SA, whereA; B are nonempty subsets of a metric space(X; d).

As an application of our results in Section 2, we obtain some …xed point results of cyclic mappings in the setup of complete metric spaces.

Theorem 5. Let A andB be two closed subsets of complete metric spaceX such that A\B 6=; and f; g; S; T : A[B ! A[B with f A T B andgB SA:

Assume that

(d(f x; gy)) '(M(x; y)) +L (N(x; y)); (3.1) for any x2 A and y 2B and someL 0; where 2 z, '2 ; 2 : If S andT are one to one then there existsz2A\B such thatf z=gz=Sz=T z: If ff; Sgandfg; Tg are weakly compatible, thenf; g; SandT have a unique common

…xed pointz2A\B:

Proof. De…ne ; :X ![0;1)by (x) = 1; x2SA;

0; otherwise and (x) = 1; x2T B;

0; otherwise :

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Let (Sx) (T y) 1:ThenSx2SAandT y2T B:SinceSandT are one to one, we havex2Aandy 2B:From (3.1), we obtain that

(d(f x; gy)) '(M(x; y)) +L (N(x; y)):

Let (Sx) 1for some x2X;so Sx2SA and thenx2A: Hence,f x2T B and so (f x) 1:Again, let (T x) 1for somex2X:ThenT x2T B and sox2B:

Hence,gx2SAand (gx) 1:Therefore,(f; g)is a cyclic( ; )-admissible with respect to(S; T).

There exists anx₀2A\B;asA\B is nonempty. This implies thatSx₀2SA andT x₀2T B and so (Sx₀) 1and (T x₀) 1:

Letfx_ng be a sequence inX such that (x_n) 1 and (x_n) 1for alln2N and x_n !xas n! 1: Thenx_n 2SA\T B for alln2N and sox2SA\T B:

This implies that (x) 1 and (x) 1:

Thus all the conditions(a)and(b)of Theorem 1 hold. So there existu; z2A[B such thatu=f z =gz =Sz =T z:On the other hand, sinceS andT are one to one, there existz12A; z22Bsuch thatSz1=Sz2=uandT z1=T z2=uwhich implies thatz1=z2=z:Therefore,u=Sz=T zforz2A\B:

Finally, suppose thatff; Sgand fg; Tg are weakly compatible. Following argu- ments similar to those in proof of Theorem 1, we haveu=f u=gu=Su=T u:

The uniqueness of the common …xed point follows from (3.1).

4. An application in dynamic programming

The existence and uniqueness of solutions of functional equations and system of functional equations arising in dynamic programming have been studied by using di¤erent …xed point results (see, [2, 7, 16]).

Throughout this section, we assume thatU andV are Banach spaces,W U is a state space,D V is a decision space. We now prove the existence and uniqueness of the common solution of the following a system of functional equations:

p_i(x) = sup

y2Dfq(x; y) +Q_i(x; y; p_i( (x; y)))g; x2W (4.1) where :W D!W; q:W D!RandQi:W D R!R; i2 f1;2;3;4g. It is well known that equation of the type (4.1) provides useful tools for mathematical optimization, computer and dynamic programming (see, [9, 12]).

LetB(W)denote the space of all bounded real-valued functions de…ned on the setW, whereB(W)is endowed with the metricd(h; k) =supx2Wjhx kxjfor all h; k2B(W):Note thatB(W)is a complete metric space.

We consider the operatorsfi:B(W)!B(W)given by f_ih(x) = sup

y2Dfq(x; y) +Q_i(x; y; h_i( (x; y)))g;

forx2W; hi 2B(W);where i2 f1;2;3;4g;these operators are well-de…ned if the functionsqi andGi are bounded.

Suppose that the following conditions hold.

(A1) p; q:W D!RandG; K :W D R!Rare bounded, (A2)

for anyh2B(W); there existsk2B(W)such thatf1h=f4k;

and

for anyh^p2B(W); there exists k^p2B(W)such thatf₂h^p=f₃k^p;

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(A3)

f1f3(h) =f3f1(h); wheneverf1(h) =f3(h) for some h2B(W);

and

f2f4(k) =f4f2(k); wheneverf2(k) =f4(k) for somek2B(W);

(A4) there exist ; : X !R such that if (f₃h) 0 and (f₄k) 0 for all h; k2B(W);then for every(x; y)2W Dandt2W;we have

jQ1(x; y; h(x)) Q2(x; y; k(x))j ln (1 +M(h; k)); (4.2) where

M(h; k) = max d(f₃h(t); f₄k(t)); d(f₃h(t); f₁h(t));

d(f4k(t); f2k(t));d(f₃h(t); f₂k(t)) +d(f₁h(t); f₄k(t))

2 ;

d(f4k(t); f2k(t)) [1 +d(f3h(t); f1h(t))]

1 +d(f₃h(t); f₄k(t)) ; d(f1h(t); f4k(t)) [1 +d(f3h(t); f2k(t))]

1 +d(f3h(t); f4k(t)) ; (A5)

(f₃h) 0 for someh2X implies that (f₁h) 0;

and

(f4h) 0for some h2X implies that (f2h) 0;

(A6) iffh_ng is a sequence inB(W)such that (h_n) 0 and (h_n) 0for all n2N[ f0gandh_n!h asn! 1;then (h ) 0and (h ) 0;

(A7) there exists h₀2B(W)such that (f₃h₀) 0and (f₄h₀) 0;

(A8) (f₃h) 0 and (f₄h^p) 0wheneverf₁h=f₃handf₂h^p=f₄h^p:

(A9) (f3k) 0 and (f4k^p) 0 whenever k =f1k =f2k =f3k =f4k and k^p=f1k^p=f2k^p=f3k^p=f4k^p:

Theorem 6. Assume that conditions(A1)-(A9)are satis…ed:Then functional equation (4.1) has a unique common bounded solution inW.

Proof. Let be an arbitrary positive number,x2W andh1; h22B(W)such that (f3h1) 0and (f4h2) 0:Then there existy1; y22D such that

f₁h₁(x)< q(x; y₁) +Q₁(x; y₁; h₁( (x; y₁))) + ; (4.3) f₂h₂(x)< q(x; y₂) +Q₂(x; y₂; h₂( (x; y₂))) + ; (4.4) f1h1(x) q(x; y2) +Q1(x; y2; h1( (x; y2))); (4.5) f2h2(x) q(x; y1) +Q2(x; y1; h2( (x; y1))): (4.6) From (4.3) and (4.6), we have

f₁h₁(x) f₂h₂(x) < Q₁(x; y₁; h₁( (x; y₁))) Q₂(x; y₁; h₂( (x; y₁))) + jQ1(x; y1; h1( (x; y1))) Q2(x; y1; h2( (x; y1)))j+

ln (1 +M(h₁; h₂)) + : (4.7)

Similarly, from (4.4) and (4.5), we obtain that

f₂h₂(x) f₁h₁(x)<ln (1 +M(h₁; h₂)) + : (4.8)

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By (4.7) and (4.8), we have

jf1h1(x) f2h2(x)j<ln (1 +M(h1; h2)) + or, equivalently,

d(f₁h₁; f₂h₂) ln (1 +M(h₁; h₂)) + : Since >0is arbitrary, we get

d(f₁h₁; f₂h₂) ln (1 +M(h₁; h₂)): De…ne ; :B(W)![0;1)by

(h) = 1; if (h) 0whereh2B(W);

0; otherwise.

and

(h) = 1; if (h) 0whereh2B(W);

0; otherwise.

Also, de…ne'; : [0;1)![0;1)by (t) =t and '(t) = ln (1 +t):Thus, we have

(f₃h₁) (f₄h₂) (d(f₁h₁; f₂h₂)) '(M(h₁; h₂))

'(M(h1; h2)) +L (N(h1; h2)); whereL 0; 2 and

N(h1; h2) = (d(f3h(t); f1h(t)); d(f4k(t); f2k(t)); d(f3h(t); f2k(t)); d(f1h(t); f4k(t))): If f = f1; g = f2, S = f3 and T = f4; it is easy to observe that all the

hypotheses of Corollary 2 are satis…ed. Thereforef1; f2; f3 andf4 have a unique common …xed point and hence the system of functional equations (4.1) has a unique bounded common solution.

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Nawab Hussain, Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia

E-mail address: [email protected]

Huseyin Isik, Department of Mathematics, Faculty of Science, University of Gazi, 06500-Teknikokullar, Ankara, TURKEY, Department of Mathematics, Mu^¸^S Alparslan University, Mu^¸^S49100, Turkey

Mujahid Abbas, Department of Mathematics and Applied Mathematics, University Pretoria, Lynnwood Road, Pretoria 0002, South Africa