A Common Fixed Points in Cone and rectangular cone Metric Spaces

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1 Chapter 5 A Common Fixed Points in Cone and rectangular cone Metric Spaces In this chapter we have establish fixed point theorems in cone metric spaces and rectangular cone metric space. In the first part we have proved some generalization of common fixed point on cone metric space with w-distance. Also proved a common fixed point theorem for a pair of weakly compatible self mappings defined on a cone metric space under different contractive conditions. In the second part we have proved some generalization of common fixed point on cone rectangular metric spaces. 5.1 Cone Metric Space Introduction In 2007 Huang and Zhang [72] introduced the notation of cone metric space as natural generalization of the notion of metric space (X, d), where metric d : X X R + is replaced by the metric d : X X E, where E is real Banach space. This theory generalized by see in, [3, 4, 5, 123, 49, 116]. Section (5.1.2) is devoted to the basic operations related to the cone, cone metric space and some basic properties of cone metric spaces with w-distance and some generalization result of H.Lakzian F.Arabyani,[76] are recalled in this section (5.1.3). A fixed 62

2 5.1 Cone Metric Space 63 point theorem of multifunctions proved in cone metric spaces and we generalized some result of S.Rezapour, [124] in section (5.1.4). In section (5.1.5) applying notion of weakly compatible maps we have proved fixed point theorem in cone metric spaces Preliminary Notes Definition [72] Let E always be a real Banach space and P a subset of E. P is called a cone if (i) P is closed, non-empty and P {0}, (ii) ax + by P for all x, y P and non-negative real numbers a, b, (iii) P ( P ) = 0. For a given cone P E, we can define a partial ordering with respect to P by x y if and only if y x P. The notation x < y will stand for x y and x y, while x y will stand for y x intp, where intp denotes the interior of P i.e. P 0. Definition The cone P is called normal if there is a number M > 0 such that for all x, y E, 0 x y implies x M y. The least positive number satisfying above is called the normal constant of P. It is clear that M 1. In the following, let E be a normed linear space, P be a cone in E satisfying int(p ), and denote the partial ordering on E with respect to P. Definition [72] Let X be a non-empty set. Suppose the mapping d : X X E satisfies: (a) 0 d(x, y) for all x, y X and d(x, y) = 0 if and only if x = y, (b) d(x, y) = d(y, x) for all x, y X,

3 5.1 Cone Metric Space 64 (c) d(x, y) d(x, z) + d(z, y) for all x, y, z X. Then d is called a cone metric on X, and (X, d) is called a cone metric space. Example [72] Let E = R 2, P = {(x, y) E x, y 0}, X = R and d : X X E defined by d(x, y) = ( x y, α x y ), where α 0 is constant. Then (X, d) is a cone metric space. Example [123] Let E = R 3, P = {(x, y, z) E x, y, z 0}, X = R and d : X X E defined by d(x, y) = (α x y, β x y, γ x y ), where α, β, γ are positive constants. Then (X, d) is a cone metric space. Example [85] Let E = l p, (0 < p < 1), P = {{x n } n 1 E x n 0, for all n}, (X, ρ) a metric space and d : X X E defined by d(x, y) = { ρ(x,y) 2 n } n 1. Then (X, d) is a cone metric space. Definition [72] Let (X, d) be a cone metric space, x X and {x n } n 1 a sequence in X. Then (a) {x n } n 1 is said to converge to x whenever for every c E with 0 c there is a natural number N such that d(x n, x) c for all n N. We denote this by lim n x n = x or x n x. (b) {x n } n 1 is said to be a Cauchy sequence whenever for every c E with 0 c there is a natural number N such that d(x n, x m ) c, for all n, m N. (c) (X, d) is called a complete cone metric space if every Cauchy sequence is convergent in X. Definition Let (X, d) be a cone metric space and B X. (i) A point b in B is called an interior point of B whenever there exists a point p, 0 p, such that N(b, p) B,

4 5.1 Cone Metric Space 65 where N(b, p) = {y X d(y, b) p}. (ii) A subset A X is called open if each element of A is an interior point of A. Remark [152] Let E be an ordered Banach (normed) space. Then c is an interior point of P, if and only if [ c, c] is neighborhood of 0, where [c, c] = {y E c y c} and N(0, c) = {y E c y c} [c, c]. Remark [158] 1. If u v and v w, then u w. Indeed w u = (w v) + (v u) w v, implies [ (w u), (w u)] [ (w v), (w v)]. 2. If u v and v w, then u w. Indeed w u = (w v) + (v u) w v, implies [ (w u), (w u)] [ (w v), (w v)]. 3. If 0 u c for each c P 0, then u = If a b + c for each c P 0, then a b. 5. If 0 x y in E and 0 a in R, then 0 ax ay. 6. If 0 x n y n for each n N and lim n x n = x, lim n y n = y, then 0 x y. 7. If 0 d(x n, x) b n and b n 0, for any c intp, there exist N N such that n N, then d(x n, x) c, where {x n } is a sequence and x is a given point in X. 8. If E is real Banach space with a cone P and if a λa where a P and 0 < λ < 1, then a = 0. Proof: The condition a λa means λa a P, that is (1 λ)a P. Since a P and 1 λ > 0, then also (1 λ)a P. Thus we have (1 λ)a P ( P ) and a = 0.

5 5.1 Cone Metric Space If c P 0, 0 a n and a n 0, then there exist natural number N such that for all n > N we have a n c. Proof: Let 0 c be given. Choose a symmetric neighborhood V such that c+v P. Since a n 0, there is positive number N such that a n V = V for n > N. This means that c ± a n c + V intp for n > N, that is a n c. Definition The cone P is called regular if every increasing sequence which is bounded from above is convergent. That is if {x n } n 1 is a sequence such that x 1 x 2 y for some y E, then there is x E such that lim n 0 x n x = 0. Equivalently the cone P is regular if and only if every decreasing sequence which is bounded from below is convergent. Lemma [72] Every regular cone is normal. Lemma [72] There is not normal cone with normal constant M < A Fixed Point Theorem in Cone Metric Spaces with w- Distance Huang and Zhang [72] introduced cone metric space without w-distance. The metric space with w-distance was introduced by Osama Kada, Tomonari Suzuki, Wataru Takahashi and Nakoshi Shioji [106, 107]. We compose these concept together and generalized idea of cone metric space with w-distance. Definition [76] Let X be a cone metric space with metric d. Then a mapping p : X X E is called w-distance on X if the following satisfy: (a) 0 p(x, y) for all x, y X, (b) p(x, z) p(x, y) + p(y, z) for all x, y, z X, (c) p(x,.) E is lower semi-continuous for all x X,

6 5.1 Cone Metric Space 67 (d) for any 0 α, there exists 0 β such that p(z, x) β and p(z, y) β imply d(x, y) α, where α, β E. Definition [76] Let X be a cone metric space with metric d, let p be a w-distance on X, x X and {x n } a sequence in X, then (a) {x n } is called a p-cauchy sequence whenever for every α E, 0 α, there is a positive integer N such that, for all m, n N, p(x m, x n ) α. (b) A sequence {x n } in X is called a p-convergent to a point x X whenever for every α E, 0 α, there is a positive integer N such that, for all n N, p(x, x n ) α. Note that by lower semi-continuous p, for all n N, p(x n, x) α. We denote this by lim n x n = x or x n x. (c) (X, d) is a complete cone metric space with w-distance if every Cauchy sequence is p-convergent. Lemma [76] Let X be a cone metric space with metric d, let p be a w-distance on X and let f be a function from X into E that 0 f(x) for any x X. Then an into function q : X X E given by q(x, y) = f(x) + p(x, y) for each (x, y) X X is also a w-distance. Example [76] (i) Let (X, d) be a metric space. Then p = d is a w-distance on X. (ii) Let X be a norm linear space with Euclidean norm. Then the mapping p : X X [0, ) defined by p(x, y) = x + y, for all x, y X is a w-distance on X. (iii) Let X be a norm linear space with Euclidean norm. Then the mapping p : X X [0, ) defined by p(x, y) = y, for all x, y X is a w-distance on X. Finally, note that the relations intp + intp intp and λintp intp (λ > 0). Example Let E = l 1, P = {{x n } n 1 E x n 0, for all n}, (X, p) a metric space and d : X X E defined by d(x, y) = { p(x,y) 2 n } n 1. Then (X, d) is a cone metric space

7 5.1 Cone Metric Space 68 if set p = d, then (X, d) is cone metric space with w-distance p and the normal constant of P is equal to M = 1. Theorem [76] Let (X, d) be a cone metric space with w-distance p on X and the mapping T : X X there exist r [0, 1/2) such that p(t x, T 2 x) rp(x, T x), for every x X and inf{p(x, y) + p(x, T x) : x X} > 0, for every y X with y T y. Then there is z X such that z = T z. Moreover, if P be a normal cone with normal constant M and v = T (v), then p(v, v) = 0. Theorem Let (X, d) be a complete cone metric space with w-distance p. Let P be a normal cone on X. Suppose a mapping T : X X satisfy the contractive condition p(t x, T y) r[p(t x, x) + p(t y, y)], for all x X, where r [0, 1/2) is a constant. Then, T has a unique fixed point in X. For each x X, the iterative sequence {T n (x)} n 1 converges to the fixed point. Proof. Let x X and define x n+1 = T x n. Then x 1 = T x 0, x 2 = T x 1 = T 2 x 0,, x n = T n x 0. p(t x 0, T 2 x 0 ) = p(t x 0, T (T x 0 )) r{p(t x 0, x 0 ) + p(t 2 x 0, T x 0 )} (1 r)p(t x 0, T 2 x 0 ) rp(t x 0, x 0 ) ( r ) p(t x 0, x 0 ). 1 r

8 5.1 Cone Metric Space 69 Similarly, p(t 2 x 0, T 3 x 0 ) Thus in general, if n is positive integer ( r ) 2p(T x0, x 0 ). 1 r p(t n x 0, T n+1 x 0 ) h n p(t x 0, x 0 ), where h = r, and 0 r < 1. 1 r 2 Let m, n N and m > n, we have to show sequence {T n x 0 } is Cauchy p(t n x 0, T m x 0 ) p(t n x 0, T n+1 x 0 ) + p(t n+1 x 0, T n+2 x 0 ) + + p(t m 1 x 0, T m x 0 ) (h n + h n h m 1 )p(t x 0, x 0 ) < h n (1 + h + h 2 + )p(t x 0, x 0 ) = hn 1 h p(t x 0, x 0 ), Let c E with 0 c. Choose natural number n 1 such that h n 1 h p(t x 0, x 0 ) c 2, for all n n 1. Thus p(t n x 0, T m x 0 ) c for all m > n. Therefore {T n x 0 } is Cauchy in X. Since X is complete space, then there exist x X such that x n x, as n. Choose natural number n 2 such that p(t n x 0, x ) c(1 r) 2 for all n n 2. Hence p(t x, x ) p(t x, T n x 0 ) + p(t n x 0, x ) r{p(t x, x ) + p(t n x 0, T n 1 x 0 )} + p(t n x 0, x ) hp(t n x 0, T n 1 x 0 ) r p(t n x 0, x ) hn 1 r p(t x 0, x 0 ) r p(t n x 0, x ) c 2 + c 2 = c

9 5.1 Cone Metric Space 70 for n n 2. Thus p(t x, x ) c for all m 1. So c p(t m m x, x ) P for all m 1. Since c 0 as m and P is closed so p(t m x, x ) P. But p(t x, x ) P. Therefore p(t x, x ) = 0, so T x = x. Suppose y be another fixed point of T, then p(x, y ) = p(t x, T y ) = r[p(t x, x ) + p(t y, y )] = 0. Hence x = y, thus fixed point of T is unique A Fixed Point Theorem of Multifunctions in Cone Metric Spaces The theory of common fixed points of multifunctions is a generalization of the theory of fixed point of mappings in some sence. The study of fixed point theorems for multivalued mappings was initiated by Kakutani, in 1941, in finite dimensional spaces and was extended to infinite dimensional Banach spaces by Bohnenblust and Karlin, in 1950, and to locally convex spaces by Ky Fan, in There are many articles about fixed point theory and fixed points of multifunctions (for example, [31, 125]). Here we have given generalization of S. Rezapour, [124] result about common fixed points of two multifunctions on the cone metric spaces with normal constant M = 1. Most of familiar cones are normal with normal constant M = 1. But for each k > 1 there are cones with normal constant M > k. First we state the following two lemmas. Lemma [124] Let (X, d) be a cone metric space, P a normal cone with normal constant M = 1 and A a compact set in (X, τ c ), then for every x X there exists a 0 A

10 5.1 Cone Metric Space 71 such that d(x, a 0 ) = inf d(x, a). a A Lemma [124] Let (X, d) be a cone metric space, P a normal cone with normal constant M = 1 and A, B are two compact sets in (X, τ c ), Then sup d (x, A) <, x B where d (x, A) = inf a A d(x, a) for each x in X. Definition [125], Let (X, d) be a cone metric space, P a normal cone with normal constant M = 1, H c (X) the set of all compact subsets of (X, τ c ) and A H c (X). By using above Lemma, we can define h A : H c (X) [0, ) and d H : H c (X) H c (X) [0, ). by h A (B) = sup x A d (x, B) and d H (A, B) = max{h A (B), h B (A)}, respectively. Remark [125], Let (X, d) be a cone metric space with normal constant M = 1. Define ρ : X X [0, ) by ρ(x, y) = d(x, y). Then, (X, ρ) is a metric space. This implies that for each A, B H c (X) and x, y X, we have the following relations (i) d (x, A) d(x, y) + d (y, A), (ii) d (x, A) d (x, B) + h B (A), and (iii) d (x, A) d(x, y) + d (y, B) + h B (A). Theorem Let (X,d) be a complete cone metric space with normal constant M = 1 and T 1, T 2 : X H c (X) two multifunctions satisfying the relation d H (T 1 x, T 2 y) αd (x, T 1 x) + βd (y, T 2 y)

11 5.1 Cone Metric Space 72 for all x, y X, where α + β < 1/2 and α > 0, β > 0 are constants. Then, T 1 and T 2 have a common fixed point, that is, there exists x X such that x T 1 x and x T 2 x. Proof. Let x 0 X be given. By Lemma , there is x 1 T 1 x 0 such that d (x 0, T 1 x 0 ) = d(x 0, x 1 ). Also, there is x 2 T 2 x 1 such that d (x 1, T 2 x 1 ) = d(x 1, x 2 ). Thus, we obtain a sequence {x n } n 1 in X such that x 2n 1 T 1 x 2n 2, x 2n T 2 x 2n 1, d (x 2n 2, T 1 x 2n 2 ) = d(x 2n 2, x 2n 1 ), and d (x 2n 1, T 2 x 2n 1 ) = d(x 2n 1, x 2n ), for all n 1. Thus, for all n 1 we have d(x 2n, x 2n+1 ) = d (x 2n, T 1 x 2n ) h T2 x 2n 1 (T 1 x 2n ) d H (T 2 x 2n 1, T 1 x 2n ) αd (x 2n, T 1 x 2n ) + βd (x 2n 1, T 2 x 2n 1 ) α d(x 2n, x 2n+1 ) + β d(x 2n 1, x 2n ). ( β ) d(x 2n, x 2n+1 ) d(x 2n 1, x 2n ) 1 α c d(x 2n 1, x 2n ), for all n 1, α + β < 1 and c = max{ β, α }. Also 1 α 1 β d(x 2n 1, x 2n ) = d (x 2n 1, T 2 x 2n 1 )

12 5.1 Cone Metric Space 73 h T1 x 2n 2 (T 2 x 2n 1 ) d H (T 1 x 2n 1, T 2 x 2n 1 ) α d(x 2n 2, x 2n 1 ) + β d(x 2n 1, x 2n ) ( α ) d(x 2n 1, x 2n ) d(x 2n 2, x 2n 1 ) 1 β c d(x 2n 2, x 2n 1 ), for all n 1. This implies that for all m 1 d(x m, x m+1 ) c d(x m 1, x m ) c 2 d(x m 2, x m 1 ). c m d(x 0, x 1 ). Then for n > m we have d(x n, x m ) n i=m+1 d(x i, x i 1 ) (c n 1 + c n 2 + c m ) d(x 0, x 1 ) ( c m ) d(x 0, x 1 ) 0, n, m. 1 c Hence, {x n } n 1 is a Cauchy sequence in X. Thus, there exists x X such that x n x. Now by using Remark 5.1.3, we have d (x, T 1 x ) d (x, T 2 x 2n 1 ) + h T2 x 2n 1 (T 1 x ) d (x, T 2 x 2n 1 ) + d H (T 2 x 2n 1, T 1 x ) d(x, x 2n ) + αd (x 2n 1, T 2 x 2n 1 ) + βd (x, T 1 x ), for all n 1. Hence, d (x, T 1 x ) α 1 β d (x 2n 1, T 2 x 2n 1 ) β d(x, x 2n )

13 5.1 Cone Metric Space 74 cd (x 2n 1, T 2 x 2n 1 ) β d(x, x 2n ) = cd(x 2n 1, T 2 x 2n 1 ) β d(x, x 2n ), for all n 1. Therefore, d (x, T 1 x ) = 0. By Lemma , x T 1 x. On the other hand, similarly we have d (x, T 2 x ) d (x, T 1 x 2n ) + h T1 x 2n (T 2 x ) d (x, T 1 x 2n ) + d H (T 1 x 2n, T 2 x ) d(x, x 2n+1 ) + αd (x 2n, T 1 x 2n ) + βd (x, T 2 x ), for all n 1. Hence, d (x, T 2 x ) α 1 β d (x 2n, T 1 x 2n ) β d(x, x 2n+1 ) cd (x 2n, T 1 x 2n ) β d(x, x 2n+1 ) = cd(x 2n, T 1 x 2n ) β d(x, x 2n+1 ), for all n 1. Therefore, d (x, T 2 x ) = 0. By Lemma , x T 2 x. Therefore, x is a common fixed point of T 1 and T 2. Theorem Let (X, d) be a complete cone metric space with normal constant M = 1 and T 1, T 2 : X H c (X) two multifunctions satisfy the relation d H (T 1 x, T 2 y) αd (y, T 1 x) + βd (x, T 2 y) for all x, y X, where α + β < 1/2 and α, β > 0 are a constants. Then, T 1 and T 2 have a common fixed point. Proof. A similar argument to that of the proof of Theorem shows that there exists a Cauchy sequence {x n } n 1 in X such that x 2n 1 T 1 x 2n 2, x 2n T 2 x 2n 1,

14 5.1 Cone Metric Space 75 d (x 2n 2, T 1 x 2n 2 ) = d(x 2n 2, x 2n 1 ), and d (x 2n 1, T 2 x 2n 1 ) = d(x 2n 1, x 2n ), for all n 1. Thus, there exists x X such that x n x. Now by using Remark 5.1.3, we have d (x, T 1 x ) d (x, T 2 x 2n 1 ) + h T2 x 2n 1 (T 1 x ) for all n 1. Hence, d (x, T 2 x 2n 1 ) + d H (T 2 x 2n 1, T 1 x ) d(x, x 2n ) + αd (x, T 2 x 2n 1 ) + βd (x 2n 1, T 1 x ) d(x, x 2n ) + α d(x, x 2n ) + βd (x, T 1 x ) + β d(x, x 2n 1 ), d (x, T 1 x ) ( 1 + α ) ( β ) d(x, x 2n ) + d(x, x 2n 1 ), 1 β 1 β for all n 1. Therefore, d (x, T 1 x ) = 0. By Lemma , x T 1 x. Also, we have d (x, T 2 x ) d (x, T 1 x 2n ) + h T1 x 2n (T 2 x ) for all n 1. Hence, d (x, T 1 x 2n ) + d H (T 1 x 2n, T 2 x ) d(x, x 2n+1 ) + αd (x, T 1 x 2n ) + βd (x 2n, T 2 x ) d(x, x 2n+1 ) + α d(x, x 2n+1 ) + βd (x, T 2 x ) + β d(x, x 2n ), d (x, T 2 x ) = ( 1 + α ) ( β ) d(x, x 2n+1 ) + d(x, x 2n ), 1 β 1 β for all n 1. Therefore, d (x, T 2 x ) = 0. By Lemma , x T 2 x. Therefore, x is a common fixed point of T 1 and T A Common Fixed Point For Weakly Compatible Maps In this section, a common fixed point theorem is proved for a pair of weakly compatible self mappings defined on a cone metric space under a contractive condition.

15 5.1 Cone Metric Space 76 Definition [69] Let f and g be self mappings of a set X. If w = fx = gx for some x in X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g. Definition [69] Two self mappings f and g of a set X are said to be weakly compatible if they commute at their coincidence points, that is, if fu = gu for some u X, then fgu = gfu. Proposition [3] Let f and g be weakly compatible self mappings of a set X. If f and g have a unique point of coincidence, that is, w = fx = gx, then w is the unique common fixed point of f and g. Theorem Let (X, d) be a cone metric and P be a normal cone with normal constant K. Let the mappings f, g : X X satisfy the following conditions: (i) f(x) g(x), (ii) g(x) is complete subspace of X, (iii) d(fx, fy) αd(fx, gy) + βd(fy, gx) + γd(fx, gx) + δd(fy, gy), where α, β, γ, δ > 0 and 2α+2β +δ +γ < 1, then f and g have a unique coincidence point in X. Moreover, if f and g are weakly compatible, then f and g have a unique common fixed point. Proof. Let x 0 be an arbitrary point in X. Then, since f(x) g(x), we choose a point x 1 in X such that f(x 0 ) = g(x 1 ). Continuing this process, having chosen x n in X, we obtain x n+1 in X such that f(x n ) = g(x n+1 ). Then d(gx n+1, gx n ) = d(fx n, fx n 1 ) αd(fx n, gx n 1 ) + βd(fx n 1, gx n ) + γd(fx n, gx n ) + δd(fx n 1, gx n 1 ) α{d(fx n, gx n ) + d(gx n, gx n 1 )} + β{d(fx n 1, gx n+1 ) + d(gx n+1, gx n )}

16 5.1 Cone Metric Space 77 + γd(fx n, gx n ) + δd(fx n 1, gx n 1 ) = α{d(gx n+1, gx n ) + d(gx n, gx n 1 )} + β{d(gx n, gx n+1 ) + d(gx n+1, gx n )} + γd(gx n+1, gx n ) + δd(gx n, gx n 1 ) = (α + 2β + γ)d(gx n+1, gx n ) + (α + δ)d(gx n, gx n 1 ). Hence, d(gx n+1, gx n ) hd(gx n, gx n 1 ), where h = α+δ with 2α + 2β + γ + δ < 1. [1 (α+2β+γ)] Now for n > m, we get d(gx n, gx m ) d(gx n, gx n 1 ) + d(gx n 1, gx n 2 ) d(gx m+1, gx m ) (h n 1 + h n h m )d(gx 1, gx 0 ) hm (1 h) d(gx 1, gx 0 ). Using the normality of cone P, implies that d(gx n, gx m ) hm (1 h) K d(gx 1, gx 0 ). Then d(gx n, gx m ) 0 as n, m and so {g(x n )} is a Cauchy sequence in X. Since g(x) is a complete subspace of X, so there exists q in g(x) such that g(x n ) q, as n. Consequently, we can find p in X such that g(p) = q. Thus, d(gx n, fp) = d(fx n 1, fp) αd(fx n 1, gp) + βd(fp, gx n 1 ) + γd(fx n 1, gx n 1 ) + δd(fp, gp) αd(fx n 1, gp) + β[d(fp, gx n ) + d(gx n, gx n 1 )] + γd(fx n 1, gx n 1 )+ δ[d(fp, gx n ) + d(gx n + gp)] (1 β γ)d(gx n, fp) αd(fx n 1, q) + βd(gx n, gx n 1 ) + γd(fx n 1, gx n 1 ) + δd(gx n, q)

17 5.1 Cone Metric Space 78 ( 1 (1 β γ) )[ αd(fx n 1, q) + βd(gx n, gx n 1 ) + γd(fx n 1, gx n 1 )+ ] δd(gx n, q) which, using the normality of cone P, implies that ( K d(gx n, fp) (1 β γ) ( K (1 β γ) = 0, as n. )[ ] αd(fx n 1, q) + βd(gx n, gx n 1 ) + γd(fx n 1, gx n 1 ) + δd(gx n, q) )[ α d(fx n 1, q) + β d(gx n, gx n 1 ) + γ d(fx n 1, gx n 1 ) + ] δ d(gx n, q) Hence, d(gx n, fp) 0, as n. Also, we have d(gx n, gp) 0, as n. The uniqueness of a limit in a cone metric space implies that f(p) = g(p). Again, we show that f and g have a unique point of coincidence. For this, if possible, there exists another point t in X such that f(t) = g(t). Then, we have d(gt, gp) = d(ft, fp) αd(ft, gp) + βd(fp, gt) + γd(ft, gt) + δd(fp, gp), by the normality of cone P, implies that d(gt, gp) = 0, and so, we have gt = gp. Finally, using the Proposition (5.1.15), we conclude that f and g have a unique common fixed point. Example Let E = I 2, for I = [0, 1], P = {(x, y) E x, y 0} I 2, d : I I E such that d(x, y) = ( x y, ξ x y ), where ξ > 0 is a constant. Define fx = ξx 1+ξx, for all x I and gx = ξx for all x I. Then, for ξ = 1, both the mappings f and g are weakly compatible and satisfy all the conditions of the above theorem with x = 0 as a unique common fixed point.

18 5.2 Cone Rectangular Metric Space Cone Rectangular Metric Space In this section we introduce fixed point theorems in cone rectangular metric spaces and proved some results on Banach contraction principle in a complete normal cone rectangular metric space also generalize the Kannans fixed point theorem in a cone rectangular metric. At the last we have proved result for expansive onto mappings on cone rectangular metric spaces Introduction Azam, Arshad and Beg, [7] introduced the notion of cone rectangular metric spaces by replacing the triangular inequality of a cone metric space by a rectangular inequality. In this chapter our main results are based on generalization of Banach contraction principle on cone rectangular metric spaces result given by Akbar Azam, Muhammad Arshad and Ismat Beg, [7] and the Kannans fixed point theorem in a cone rectangular metric [101]. Also C.T.Aage and J.N. Salunke, [12] proved fixed point theorem for expansion onto mappings on cone metric spaces and we have extended result for expansive onto mappings on cone rectangular metric Preliminary Notes Definition [101] Let X be a non empty set. Suppose the mapping d : X X E satisfies: (a) 0 d(x, y) for all x, y X and d(x, y) = 0 if and only if x = y; (b) d(x, y) = d(y, x) for all x, y X; (c) d(x, y) d(x, w) + d(w, z) + d(z, y) for all x, y X and for all distinct points w, z X {x, y} [Rectangular Property]. Then d is called a cone rectangular metric on X and d(x, d) is called cone rectangular

19 5.2 Cone Rectangular Metric Space 80 metric space. Example [101] Let X = N, E = R 2 and P = {(x, y) E x, y 0}. Define d : X X E as follow: (0, 0) if x = y, d(x, y) = (3, 9) if x and y are in {1, 2}, (1, 3) if x and y can not both in {1, 2}. Now (X, d) is a cone rectangular metric space but (X, d) is not a cone metric space because it lacks the triangular property: (3, 9) = d(1, 2) > d(1, 3) + d(3, 2) = (1, 3) + (1, 3) = (2, 6), as (3, 9) (2, 6) = (1, 3) P. Example [101] Let E = R 2 and P = {(x, y) E x, y 0} and X = R. Define d : X X E as follow: (0, 0) if x = y, d(x, y) = (3α, 3) if x and y are in {1, 2}, x y, (α, 1) if x and y can not both in {1, 2}, x y, where α > 0 is a constant. Then (X, d) is a cone rectangular metric space but (X, d) is not a cone metric space, since we have (3α, 3) = d(1, 2) > d(1, 3) + d(3, 2) = (2α, 2). Example Let E = R 2 and P = {(x, y) E x, y 0} and X = {a, b, c, e}. Define d : X X E such that d(x, x) = (0, 0) for all x X, d(x, y) = d(y, x) for all x, y X, d(a, b) = (3, α), d(a, c) = d(b, c) = (1, α), d(a, e) = d(b, e) = d(c, e) = (2, α),

20 5.2 Cone Rectangular Metric Space 81 where α > 0 is a constant. Then (X, d) is a cone rectangular metric space but it is not a cone metric space, since we have d(a, b) = (3, α) and d(a, c) + d(c, b) = (2, 2α), but (3, α) and (2, 2α) cannot be compared with respect to. Definition [101] Let (X, d) be a cone rectangular metric space, x X and {x n } n 1 a sequence in X. Then (a) {x n } n 1 is said to converge to x whenever for every c E with 0 c there is a natural number N such that d(x n, x) c for all n N. We denote this by lim n x n = x or x n x. (b) {x n } n 1 is said to be a Cauchy sequence whenever for every c E with 0 c there is a natural number N such that d(x n, x m ) c for all n, m N. (c) (X, d) is called a complete cone rectangular metric space if every Cauchy sequence is convergent in X. Lemma [101] Let (X, d) be a cone rectangular metric space, P be a normal cone. Let {x n } be a sequence in X. Then x n x as n + d(x n, x) 0 as n +. Note that if (X, d) is a cone metric space and {x n } is a convergent sequence in X, then the limit of {x n } is unique ([72]-Lemma 2). In our case uniqueness of the limit is not satisfied in general. We have an example to illustrate this remark. Example [101] We take E = R and P = {x R x 0}. Let {x n } n N be a sequence in Q and a, b R/Q, a b. We put X = {x 1, x 2,, x n, } {a, b} and we

21 5.2 Cone Rectangular Metric Space 82 consider d : X X R defined by d(x, x) = 0 for all x X, d(x, y) = d(y, x) for all x, y X, d(x n, x m ) = 1 for all n, m N, n m, d(x n, b) = 1 n d(x n, a) = 1 n d(a, b) = 1. for all n N, for all n N, We remark (X, d) is not a cone metric space because we have d(x 2, x 3 ) = 1 > d(x 2, a) + (a, x 3 ) = = 5 6. However, (X, d) is a cone rectangular metric space. Now, since d(x n, a) = 1 n n +, we obtain that x n a as n +. Also, we have d(x n, b) = 1 n 0 as 0 as n + and then x n b as n +. Lemma [101] Let (X, d) be a cone rectangular metric space, P be a normal cone. Let {x n } be a sequence in X. Then {x n } is Cauchy sequence if and only if d(x n, x m ) 0 as n, m +. Lemma [101] Let (X, d) be a complete cone rectangular metric space, P be a normal cone with normal constant κ. Let {x n } be a Cauchy sequence in X and suppose that there is positive number N such that (i) x n x m for all n, m > N. (ii) x n, x are distinct points in X, for all n > N. (iii) x n, y are distinct points in X, for all n > N. (iv) x n x and x n y, as n +. Then x = y.

22 5.2 Cone Rectangular Metric Space On Banach Contraction Principle in a Complete Normal Cone Rectangular Metric Space In this article we generalized the Banach contraction mapping principle in a complete normal cone rectangular metric space. First we present two theorems whose proofs are similar to Huang and Zhang [[72], Lemmas 1 and 4]. Theorem Let (X, d) be a cone rectangular metric space and P be a normal cone with normal constant κ. Let {x n } be a sequence in X. Then {x n } converges to x if and only if d(x n, x) 0 as n. Theorem Let (X, d) be a cone rectangular metric space, P be a normal cone with normal constant κ. Let {x n } be a sequence in X. Then {x n } is a Cauchy sequence if and only if d(x n, x n+m ) 0 as n. Theorem Let (X, d) be a cone rectangular metric space, P be a normal cone with normal constant κ and the mapping T : X X satisfies: d(t x, T y) αd(x, T x) + βd(y, T y) for all x, y X, where 0 α + β < 1. Then T has a unique fixed point. Proof. Let x 0 be an arbitrary point in X. Define a sequence of points in X as follows: x n+1 = T x n = T n+1 x 0, n = 0, 1, 2,... We can suppose that x 0 is not a periodic point, in fact if x n = x 0, then, d(x 0, T x 0 ) = d(x n, T x n ) = d(t n x 0, T n+1 x 0 ) αd(t n 1 x 0, T n x 0 ) + βd(t n x 0, T n+1 x 0 )

23 5.2 Cone Rectangular Metric Space 84 where 0 α + β < 1 and h =. hd(t n 1 x 0, T n x 0 ) h n d(x 0, T x 0 ), It further implies that [ hn 1 1 h n ]d(x 0, T x 0 ) P. α. It follows that 1 β [hn 1]d(x 0, T x 0 ) P. Hence d(x 0, T x 0 ) P and d(x 0, T x 0 ) = 0, this means x 0 is a fixed point of T. Thus in this sequel of proof we can suppose that x m = x n, for all distinct m, n N. Now by using rectangular property for all y X, we have, d(y, T 4 y) d(y, T y) + d(t y, T 2 y) + d(t 2 y, T 4 y) d(y, T y) + hd(y, T y) + h 2 d(y, T 2 y). Similarly, d(y, T 6 y) d(y, T y) + d(t y, T 2 y) + d(t 2 y, T 3 y) + d(t 3 y, T 4 y) + d(t 4 y, T 6 y) d(y, T y) + hd(y, T y) + h 2 d(y, T y) + h 3 d(y, T y) + h 4 d(y, T 2 y) = 3 h i d(y, T y) + h 4 d(y, T 2 y), for all y X. i=0 Now by induction, we obtain for each k = 2, 3, 4,..., Moreover, for all y X, d(y, T 2k y) 2k 3 i=0 h i d(y, T y) + h 2k 2 d(y, T 2 y). (5.2.1) d(y, T 5 y) d(y, T y) + d(t y, T 2 y) + d(t 2 y, T 3 y) + d(t 3 y, T 4 y) + d(t 4 y, T 5 y) = 4 h i d(y, T y). i=0 By induction, for each k = 0, 1, 2,... we have, d(y, T 2k+1 y) 2k i=0 h i d(y, T y). (5.2.2)

24 5.2 Cone Rectangular Metric Space 85 Using Inequality (5.2.1), for k = 1, 2, 3,... we have, d(t n x 0, T n+2k x 0 ) h n d(x 0, T 2k x 0 ) [ 2k 3 h n i=0 ] h i (d(x 0, T x 0 ) + d(x 0, T 2 x 0 )) + h 2k 2 d(x 0, T x 0 ) + d(x 0, T 2 x 0 ) 2k 2 h n h i [d(x 0, T x 0 ) + d(x 0, T 2 x 0 )] i=0 hn (1 h 2k 1 ) [ ] d(x 0, T x 0 ) + d(x 0, T 2 x 0 ) 1 h [ ] hn d(x 0, T x 0 ) + d(x 0, T 2 x 0 ). 1 h Similarly, for k = 0, 1, 2,..., Inequality (5.2.2) implies that d(t n x 0, T n+2k+1 x 0 ) h n d(x 0, T 2k+1 x 0 ) h n 2k i=0 h i d(x 0, T x 0 ) ( h n )[ ] d(x 0, T x 0 ) + d(x 0, T 2 x 0 ). 1 h ( h d(t n x 0, T n+m n )[ ] x 0 ) d(x 0, T x 0 ) + d(x 0, T 2 x 0 ). 1 h Since P is a normal cone with normal constant κ, therefore, [ h d(t n x 0, T n+m n ] [ ] x 0 ) κ d(x 0, T x 0 ) + d(x 0, T 2 x 0 ). 1 h Therefore, d(t n x 0, T n+m x 0 ) 0 as n. Now Theorem implies that {x n } is a Cauchy sequence in X. Since X is complete, there exists u X such that x n u. By Theorem 5.2.8, we have d(t n x 0, u) 0, as n. By rectangular property, we have d(t u, u) d(u, T n 1 x 0 ) + d(t n 1 x 0, T n+1 x 0 ) + d(t n+1 x 0, u)

25 5.2 Cone Rectangular Metric Space 86 d(u, T n 1 x 0 ) + h n 1 d(x 0, T 2 x 0 ) + d(t n+1 x 0, u). Thus d(t u, u) κ [ d(u, T n 1 x 0 ) + h n 1 d(x 0, T 2 x 0 ) + d(t n+1 x 0, u) ]. (5.2.3) Letting n, we have d(u, T u) = 0. Hence u = T u. Now we show that T have a unique fixed point. For this, assume that there exists another point v in X such that v = T v. Now, d(v, u) = d(t v, T u) αd(v, T v) + βd(u, T u) = 0. Hence, u = v On Kannans Fixed Point Theorem in a Cone Rectangular Metric In this article our main result is On Kannans fixed point theorem in a cone rectangular metric space. Theorem Let (X, d) be a complete cone rectangular metric space, P be a normal cone with normal constant k. Suppose a mapping T : X X satisfies the contractive condition d(t x, T y) αd(x, T x) + βd(x, y) + γd(y, T y), for all x, y X,α, β, γ > 0 and α + 2β + γ < 1. Then, (i) T has a unique fixed point in X. (ii) For any, the iterative sequence converges to the fixed point.

26 5.2 Cone Rectangular Metric Space 87 Proof. Let x 0 X and T x n = x n+1. Now d(t x 0, T 2 x 0 ) = d(t x 0, T (T (x 0 )) (1 γ)d(t x 0, T 2 x 0 ) (α + β)d(x 0, T x 0 ) αd(x 0, T x 0 ) + βd(x 0, T x 0 ) + γd(t x 0, T 2 x 0 ) d(t x 0, T 2 x 0 ) ( α + β ) d(x 0, T x 0 ), (5.2.4) 1 γ ( α + β ) d(t 2 x 0, T 3 x 0 ) d(t x 0, T 2 x 0 ) 1 γ ( α + β )( α + β ) d(x 0, T x 0 ) 1 γ 1 γ ( α + β ) 2d(x0 =, T x 0 ). (5.2.5) 1 γ Thus in general, if n is positive integer, then where h = α+β 1 γ and 0 < h < 1. We divide the proof into two cases. d(t n x 0, T n+1 x 0 ) h n d(x 0, T x 0 ), Case I: First assume that T m x 0 = T n x 0 for m, n N, m n. Let m > n, then T m n (T n x 0 ) = T n x 0, i.e. T p y = y, where p = m n, y = T n x. Now since p > 1, we have d(y, T y) = d(t p y, T p+1 y) h p d(y, T y), where 0 < h < 1. We obtain d(y, T y) P and d(y, T y) P which implies that d(y, T y) = 0, i.e. T y = y. Case II: Assume that T m x 0 T n x 0 for m, n N, m n. Clearly we have d(t n x 0, T n+1 x 0 ) h n d(x 0, T x 0 ),

27 5.2 Cone Rectangular Metric Space 88 and d(t n x 0, T n+2 x 0 ) =d(t (T n 1 x 0, T (T n+1 x 0 )) αd(t n 1 x 0, T n x 0 ) + βd(t n 1 x 0, T n+1 x 0 )+ γd(t n+1 x 0, T n+2 x 0 ) [ αh n 1 d(x 0, T x 0 ) + β d(t n 1 x 0, T n x 0 ) + d(t n x 0, T n+2 x 0 )+ ] d(t n+2 x 0, T n+1 x 0 ) + γh n+1 d(x 0, T x 0 ) (1 β)d(t n x 0, T n+2 x 0 ) αh n 1 d(x 0, T x 0 ) + βh n 1 d(x 0, T x 0 )+ βh n+1 d(x 0, T x 0 ) + γh n+1 d(x 0, T x 0 ) = h n 1 (α + β)d(x 0, T x 0 ) + h n+1 (β + γ)d(x 0, T x 0 ) d(t n x 0, T n+2 x 0 ) h n 1 d(x 0, T x 0 ). Now if m > 2 is odd then writing m = 2l + 1, l 1 and using the fact that T p x 0 T r x 0 for p, r N, p r, we get d(t n x 0, T n+m x 0 ) d(t n x 0, T n+1 x 0 ) + d(t n+1 x 0, T n+2 x 0 ) d(t n+2l x 0, T n+2l+1 x 0 ) h n d(x 0, T x 0 ) + h n+1 d(x 0, T x 0 ) h n+2l d(x 0, T x 0 ) < hn 1 h d(x 0, T x 0 ). Again if m > 2 is even then writing m = 2l, l 2 and using the same arguments as before, we get d(t n x 0, T n+m x 0 ) d(t n x 0, T n+2 x 0 ) + d(t n+2 x 0, T n+3 x 0 ) d(t n+2l 1 x 0, T n+2l x 0 ) h n 1 d(x 0, T x 0 ) + h n+2 d(x 0, T x 0 ) h n+2l 1 d(x 0, T x 0 ) [ h n 1 + h n + h n+1 + ]d(x 0, T x 0 ) < hn 1 1 h d(x 0, T x 0 ). Thus combining all the cases we have d(t n x 0, T n+m x 0 ) < hn 1 1 h d(x 0, T x 0 ),

28 5.2 Cone Rectangular Metric Space 89 for all m, n N. Hence we get ( h d(t n x 0, T n+m n 1 ) x 0 ) k d(x 0, T x 0 ), 1 h for all m, n N. ( ) Since k d(x 0, T x 0 ) 0, as n +, {T n x 0 } is a Cauchy sequence. By the h n 1 1 h completeness of X, there is x X such that T n x 0 x, as n +. We shall now show that T x = x. Without any loss of generality, we can assume that T r x x, for any r N. We have d(x, T x ) d(x, T n x ) + d(t n x, T n+1 x ) + d(t n+1 x, T x ) d(x, T n x ) + d(t n x, T n+1 x ) + αd(t n x, T n+1 x ) + βd(t n x, x ) + γd(x, T x ) (1 γ)d(x, T x ) (1 + α)d(t n+1 x, T n x ) + (1 + β)d(t n x, x ) ( 1 + α ) ( 1 + β ) d(x, T x ) d(t n+1 x, T n x ) + d(t n x, x ) 1 γ 1 γ ( 1 + α ) h n ( 1 + β ) h 1 γ 1 h d(t x, x n 1 ) + 1 γ 1 h d(t x, x ). Hence ( 1 + α ) h d(x, T x n ( 1 + β ) h ) 1 γ 1 h d(t x, x n 1 ) + 1 γ 1 h d(t x, x ) 0, as n +. So we obtain d(x, T x ) = 0, i.e., T x = x. Now, if y is another fixed point of T, then d(x, y ) = d(t x, T y ) αd(t x, x ) + βd(x, y ) + γd(y, T y ) < d(x, y ), since 0 < β < 1. which implies that d(x, y ) = 0, i.e., x = y.

29 5.2 Cone Rectangular Metric Space An Expansion Mappings on Cone Rectangular Metric Spaces C.T.Aage and J.N. Salunke, [12] proved fixed point theorem for expansion onto mappings on cone metric spaces in this article we have proved fixed point theorems using same expansion onto mappings on cone rectangular metric spaces. Theorem Let (X, d) be a complete cone rectangular metric space. Suppose the mapping T : X X is onto and satisfies the contractive condition d(t x, T y) Kd(x, y), (5.2.6) for all x, y X, where K > 1 is a constant. Then T has a unique fixed point in X. Proof. If T x = T y, then 0 Kd(x, y) 0 = d(x, y) x = y. Thus, T is one to one. Define G = T 1 d(x, y) = d(t T 1 x, T T 1 y) Kd(T 1 x, T 1 y) = Kd(Gx, Gy), So d(gx, Gy) Md(x, y), where M = 1 K < 1. By Theorem (2.3) in [123], G has unique fixed point u in X i.e. Gu = u T 1 u = u u = T u. Therefore, u is a unique fixed point of T. Corollary Let (X, d) be a complete cone rectangular metric space. Suppose a mapping T : X X is onto and satisfies for some positive integer n d(t n x, T n y) Kd(x, y), for all x, y X, where K > 1 is a constant. Then T has a unique fixed point in X. Theorem Let (X, d) be a complete cone rectangular metric space. Suppose the mapping T : X X is continuous, onto and satisfies the contractive condition d(t x, T y) K(d(T x, x) + d(t y, y)), for all x, y X, where 1 < K 1 is a constant. Then T has a fixed point in X. 2

30 5.2 Cone Rectangular Metric Space 91 Proof. For each x 0 X, since T is onto, there exist x 1 X such that T x 1 = x 0. Similarly, for each n 1 there exist x n+1 X such that x n = T x n+1. If x n 1 = x n, then x n is a fixed point of T. Thus, Suppose that x n 1 x n for all n 1. Then d(x n, x n 1 ) = d(t x n+1, T x n ) K(d(T x n+1, x n+1 ) + d(t x n, x n )) = K(d(x n, x n+1 ) + d(x n 1, x n )), so, ( 1 K d(x n, x n+1 ) K = hd(x n 1, x n ). ) d(x n 1, x n ) From this we get d(x n, x n+1 ) h n d(x 0, x 1 ), where h = 1 K K n < m we have and 0 h < 1. Now for d(x n, x m ) d(x n, x n+1 ) + d(x n+1, x n+2 ) + + d(x m 1, x m ) (h n + h n h m 1 )d(x 0, x 1 ) hn 1 h d(x 0, x 1 ). Let 0 c be given. Choose a natural number N 1 such that h n 1 h d(x 0, x 1 ) c, for all n N 1. Thus d(x n, x m ) c, for n < m. Therefore, {x n } n 1 is Cauchy sequence in (X, d). Since (X, d) is a complete cone rectangular metric space, there exist x X such that x n x, as n. If T is continuous, then d(t x, x ) d(t x, T x n ) + d(t x n, T x n+1 ) + d(t x n+1, x ) 0, as n, since x n x and T x n T x as n. Therefore d(t x, x ) = 0 and so T x = x. Thus T having fixed point in X.

31 5.2 Cone Rectangular Metric Space 92 Theorem Let (X, d) be a complete cone rectangular metric space. Suppose the mapping T : X X is continuous, onto and satisfies the contractive condition d(t x, T y) kd(x, y) + ld(t x, y), for all x, y X, where k > 1, l 0 are constants. Then T has a fixed point in X. Proof. For each x 0 X, since T is onto, there exist x 1 X such that T x 1 = x 0. Similarly, for each n 1 there exist x n+1 X such that x n = T x n+1. If x n 1 = x n, then x n is a fixed point of T. Thus, Suppose that x n 1 x n for all n 1. Then d(x n, x n 1 ) = d(t x n+1, T x n ) kd(x n+1, x n ) + ld(t x n+1, x n ) = kd(x n+1, x n ) + ld(x n, x n ) = kd(x n+1, x n ), so, d(x n, x n+1 ) 1 k d(x n 1, x n ) = hd(x n 1, x n ), where h = 1. Now for n < m we have k d(x n, x m ) d(x n, x n+1 ) + d(x n+1, x n+2 ) + + d(x m 1, x m ) (h n + h n h m 1 )d(x 0, x 1 ) hn 1 h d(x 0, x 1 ). Let 0 c be given. Choose a natural number N 1 such that h n 1 h d(x 0, x 1 ) c, for all n N 1. Thus d(x n, x m ) c, for N 1 < n < m. Therefore, {x n } n 1 is Cauchy sequence in (X, d). Since (X, d) is a complete cone rectangular metric space, there exist x X such that x n x, as n.

32 5.2 Cone Rectangular Metric Space 93 If T is continuous, then d(t x, x ) d(t x, T x n ) + d(t x n, T x n+1 ) + d(t x n+1, x ) 0, as n, since x n x and T x n T x as n. Therefore d(t x, x ) = 0 and so T x = x. Thus T having fixed point in X. Theorem Let (X, d) be a complete cone rectangular metric space. Suppose the mapping T : X X is continuous, onto and satisfies the contractive condition d(t x, T y) kd(x, y) + ld(x, T x) + pd(y, T y), for all x, y X, where k 1, l > 1 and p < 1 are constants with k + l + p > 1. Then T has a fixed point in X. Proof. For each x 0 X, since T is onto, there exist x 1 X such that T x 1 = x 0. Similarly, for each n 1 there exist x n+1 X such that x n = T x n+1. If x n 1 = x n, then x n is a fixed point of T. Thus, Suppose that x n 1 x n for all n 1. Then d(x n, x n 1 ) = d(t x n+1, T x n ) kd(x n+1, x n ) + ld(x n+1, T x n+1 ) + pd(x n, T x n ) = kd(x n+1, x n ) + ld(x n+1, x n ) + pd(x n, x n 1 ), so, d(x n, x n+1 ) ( 1 p ) d(x n, x n 1 ) k + l = hd(x n, x n 1 ), where h = 1 p k+l with 0 < h < 1. Now for n < m we have d(x n, x m ) d(x n, x n+1 ) + d(x n+1, x n+2 ) + + d(x m 1, x m ) (h n + h n h m 1 )d(x 0, x 1 )

33 5.2 Cone Rectangular Metric Space 94 hn 1 h d(x 0, x 1 ). Let 0 c be given. Choose a natural number N 1 such that h n 1 h d(x 0, x 1 ) c, for all n N 1. Thus d(x n, x m ) c, for N 1 < n < m. Therefore, {x n } n 1 is Cauchy sequence in (X, d). Since (X, d) is a complete cone rectangular metric space, there exist x X such that x n x, as n. If T is continuous, then d(t x, x ) d(t x, T x n ) + d(t x n, T x n+1 ) + d(t x n+1, x ) 0, as n, since x n x and T x n T x as n. Therefore d(t x, x ) = 0 and so T x = x. Thus T having fixed point in X.