The aim of this manuscript is to discuss some fixed point results for self mappings satisfying certain contractive conditions on Cone Banach space

Dersanambika, K.S.

Fixed points, P.G.

English

Kathmandu University Open Journal Systems

KATHMANDU UNIVERSITY JOURNAL OF SCIENCE, ENGINEERING AND TECHNOLOGY   VOL. 9, No. I, July, 2013, 127-133  127 COMMON FIXED POINT THEOREM ON CONE  BANACH SPACE  1P. G. Varghese *,  2K. S.  Dersanambika  1 Department of Mathematics, Marthoma College, Thiruvalla – 689103, Kerala, India. 2 Department of Mathematics, Fatima Mata National College, Kollam-691001,  Kerala, India * Corresponding address: pg.varghese@ymail.com Received 03 February, 2012; Revised 24 march, 2013  ABSTRACT   The aim of this manuscript is to discus some fixed point results for self mappings satisfying certain contractive conditions on Cone Banach space.  AMS 2000 Subject Classification : 47H10, 54H25 Key words : Cone Banach space, Compatible mappings, Fixed points. INTRODUCTION In 2007, Huang and Zhang [4] introduced the notion of cone metric space, replacing the set of real numbers by ordered Banach space and proved some fixed point theorems for functions satisfying contractive conditions in these spaces. The results in [4] were generalized by Sh. Rezapour and R. Hamlbarani [7] by omitting the normality condition, which is a mile stone in developing fixed point theory in cone metric space. After that several articles on fixed point theorems in cone metric space were obtained by different mathematicians such as M.Abbas, G. Junck [1], D.Ilic [5] etc. Very recently some results on fixed point theorems have been extended to Cone Banach space. E. Karapinar [8] proved some fixed point theorems for self mappings satisfying some contractive condition on a Cone Banach space. Thabet Abdeljawad, E. Karapinar and Kenan Tas [9] have given some generalizations to this theorems. Neeraj Malviya and Sarala Chouhan [6] extended some fixed point theorems to Cone Banach space. DEFINITION 1.1. [4] Let  E, .  be a real Banach space. A subset P E is said to be a cone if and only if (1) P is closed, nonempty and P  {0} (2) a, b R, a,b 0,x, y P   implies ax by P   (3) P ( P) {0}    For a given cone P subset of E, we define a partial ordering   with respect to P by x   y if and only if y – x P. We shall write x < y to indicate that x y  but x y  while x < < y will stand for y x  int P where int P denotes interior of P and is assumed to be nonempty. DEFINITION 1.2. [4] Let X be a nonempty set. Suppose that the mapping d : X × X   E satisfies (1) 0 d(x, y)for everyx, y X,d(x, y) 0    if and only if x = y. (2) d(x, y) d(y,x)for everyx, y X   KATHMANDU UNIVERSITY JOURNAL OF SCIENCE, ENGINEERING AND TECHNOLOGY   VOL. 9, No. I, July, 2013, 127-133  128 (3) d(x, y) d(x,z) d(z, y)foreveryx, y,z X    Then d is a cone metric on X and (X, d) is a cone metric space. DEFINITION 1.3. [8] Let X be a vector space over R. Suppose the mapping .  : X   E satisfies (1) x 0for allx X   (2) x 0if andonlyif x 0   (3) x y x y forallx, y X     (4) kx | k | x forallk R   Then ||. || is called a norm on X, and  X, .  is called a cone normed space. Clearly each cone normed space if a cone metric space with metric defined by d (x, y) = x y . DEFINITION 1.4. [8] Let  X, .  be a cone normed space, x  X and {xn} a sequence in X. Then (1) {xn} converges to x if for every c  E with o << c there is a natural number N such that nx x cfor alln N   . We shall denote it by n n nlim x xor x x   . (2) {xn} is a Cauchy sequence, if for every c E witho c   there is a natural number N such that n mx x cfor alln,m N   . (3)  X, .  is a complete cone normed space if every Cauchy sequence is convergent. A complete cone normed space is called a Cone Banach space. DEFINITION 1.5. Let f and g be self mappings on a cone normed space  X, . , they are said to be compatible if n n nlim ||fgx gfx || 0    for every sequence {xn} in X with n n n nlim fx lim gx y    for some point y in X. PROPOSITION 1.6. [8] Let  X, .  be a cone normed space. P be a normal cone with constant K. Let {xn} be a sequence in X. Then (1) {xn} converges to x if and only if n|| x x || 0asn   . (2) {xn} is a Cauchy sequence if and only is n m|| x x || 0asn, m    (3) if the {xn} converges to x and {yn} converges to y then n n|| x y || || x y ||    PROPOSITION 1.7. Let f and g be compatible mappings on a cone normed space  X,|| .||  such that n n n nlim fx lim gx y    for some point y in X and for every sequence {xn} in X. Then n nlim gfx fy   if f is continuous.  RESULTS THEOREM 2.1. Let (X, ||.||) be a Cone Banach space, P is a cone in E, with ||x|| = d(x, 0), f and g be self mappings on X satisfying  || fx fy ||    a || gx gy || b || fx gx || c || fy gy || d || fx gy || e || fy gx ||          (1) KATHMANDU UNIVERSITY JOURNAL OF SCIENCE, ENGINEERING AND TECHNOLOGY   VOL. 9, No. I, July, 2013, 127-133  129 where a + b + c + 2d + e  [0, 1), x, y,  X. If the range of g contains the range of f, g(X) is complete, f; g are continuous and compatible. Then f and g have a unique common fixed point. PROOF : Let 0x X,  since 0f (x ) f (X) g(X)   there is x1  X such that f (x0) = g(x1). Let y1 = f(x0) = g(x1). Now 1x X , since  1f x f (X) g(X)   there is an x2 such that                    f(x1) = g(x2). Let 2 1 2y f (x ) g(x )   continuing like this we get a sequence {yn} in X such that n n 1 ny f (x ) g(x ),n 0,1,2........    We shall prove   n n 1 n 1 n 2|| y y || || y y ||       Where  < 1. We have  n n 1|| y y ||   =  n 1 n|| g(x ) g(x ) ||      =  n n 1|| f (x ) f (x ) ||      n n 1 n na || g(x ) g(x ) || b || f (x ) g(x ) ||          n 1 n 1 n n 1c || f (x )g(x ) || d || f x g x ||      + n 1 ne|| fx g(x ) ||     =  n 1 n 2 n n 1 n 1 n 2a || y y || b || y y || c || y y ||             + n n 2 n 1 n 1d || y y || e || y y ||                                            =  n 1 n 2 n n 1 n 1 n 2a || y y || b || y y || c || y y ||             n n 1 n 1 n 2d || y y || e || y y ||       (1-  b - d) n n 1|| y y ||    (a + c + d) n 1 n 2|| y y ||    n n 1 n 1 n 2a c d|| y y || || y y ||1 b d       Let a c d1 b d   , then 1   Therefore  n n 1 n 1 n 2|| y y || || y y ||     , for every n Hence  n 1n n 1 1 0|| y y || || y y ||     Let n > m  n m|| y y ||     n n 1 n 1 n 2 n 1 m|| y y || || y y || ......... || y y ||                                  n 1 n 2 m 1 0........ || y y ||       KATHMANDU UNIVERSITY JOURNAL OF SCIENCE, ENGINEERING AND TECHNOLOGY   VOL. 9, No. I, July, 2013, 127-133  130                        m1 0|| y y ||1     (2) Let c   0, then there is a 0   such that c N (0) P   where  N (0) y :|| y ||    .  Since 1  , for 0   there is a positive integer N such that m1 0|| y y || for m N.1     Hence m1 0|| y y || N (0)1 . So m1 0|| y y || N (0)1  .      Thereforem1 0c || y y || c N (0) P1    . This implies m1 0|| y y || c1 .           Therefore (2) becomes n m|| y y || c   for n, m   N. Hence {yn} is a Cauchy sequence in       g(X). But g(X) is complete therefore there is an z g(X)  such that n n n n n n 1lim y lim f (x ) lim g(x ) z.       Since f and g are continuous 2n nlim f (x ) f (z), 2n nlim g (x )  = g(z), n nlim gf (x ) g(z),   n nlim fg(x ) f (z)  . Now we shall prove f(z) = g(z). We have 2n n|| f (x ) gf (x ) ||  = 2n n|| f (x ) fg(x ) ||      2 2n n n na ||gf (x ) g (x ) || b || f (x ) gf (x ) ||        2 2 2n n n nc ||fg(x ) g (x ) || d || f (x ) g (x ) ||   n ne ||fg(x ) gf (x ) ||   Letting n  . We get  || f (z) g(z) ||    b ||f (z) g(z) || c|| f (z) g(z) ||   +     d ||f (z) g(z) || e|| f (z) g(z) ||    (1 b c d e) || f (z) g(z) ||        0 But (1- b – c – d - e) > 0. Therefore || f (z) g(z) || 0   which implies f(z) = g(z). Hence z is a point of coincidence of f and g. To prove uniqueness if possible z1 is another point of coincidence of f and g. Then y1 = f(z1) = g(z1).  Now  1|| f (z) f (z ) ||    1a || g(z) g(z ) || b || f (z) g(z) ||           1 1 1c || f (z ) g(z ) || d || f (z) g(z ) ||    + 1e || f (z ) g(z) ||    =   1 1a || f (z) f (z ) || d || f (z) f (z ) ||   1e || f (z ) f (z) ||   (1- a – d - e) || f(z) – f(z1)||   0 But (1 - a - d - e) > 0. Hence f (z) = f (z1). So f (z) = g(z) = f(z1) = g(z1), the uniqueness of z. Since f and g are compatible by definition 2.5, f and g have a unique common fixed point. THEOREM 2.2 Let f, g, h, l be mappings on Cone Banach space  X,|| . ||  into itself, with ||x|| = d(x, 0) satisfying the conditions. KATHMANDU UNIVERSITY JOURNAL OF SCIENCE, ENGINEERING AND TECHNOLOGY   VOL. 9, No. I, July, 2013, 127-133  131 (1) || hx – ly||   a || fx – hx|| + b ||fx-ly|| + c || gy – ly|| for all x, y   X, a, b, c  0,           a+2b+c <1. (2) f and g are onto mapping (3) f is continuous (4) f and h; g and l commute Then f, g, h and l have a unique common fixed point. PROOF:  Let 0x X be arbitrary, then 0l(x ) X , since f is onto there is an 1x X  such that f(x1) = l(x0). Let y0 = f(x1)= l(x0). Again 1x X  since g is onto there is a 2x X  such that g (x2) = h (x1). Let y2 = g(x2) = h(x1). Continuing like this we get a sequence {yn} such that y2n = f(x2n+1) = l (x2n) and y2n+1 = g(x2n+2) = h (x2n+1). We have   2n 1 2n|| y y ||   =  2n 1 2n|| h(x ) l(x ) ||      2n 1 2n 1 2n 1 2na || f (x ) h(x ) || b || f (x ) l(x ) ||     2n 2nc || g(x ) h(x ) ||    2n 2n 1|| y y ||     2n 2 2n 1 2n 2 2na || y y || b || y y ||     2n 1 2nc || y y ||         2n 2 2n 1 2n 2 2n 1a || y y || b || y y ||              2n 1 2n 2n 1 2nb || y y || c || y y ||      (1 – b - c) ||y2n-1- y2n ||    2n 2 2n 1(a b) || y y ||     2n 1 2n|| y y ||       a b1 b c  2n 2 2n 1|| y y ||   Let 1a b1 b c  , 1 1  . Therefore  2n 1 2n|| y y ||     1 2n 2 2n 1|| y y ||      (3)  2n 2n 1|| y y ||  =  2n 1 2n|| hx lx ||        2n 1 2n 1 2n 1 2na || fx hx || b || fx lx ||     2n 2nc || gx lx ||    2n 2n 1|| y y ||     2n 2n 1 2n 2na || y y || b || y y ||   2n 1 2nc || y y ||   2n 2n 1(1 a) || y y ||       2n 1 2nc || y y ||   Let 2c1 a , then 2 1  . Therefore  2n 2n 1|| y y ||    2 2n 1 2n|| y y ||     (4) Let 1 2max ( , )     then 1  , (6) and (7) becomes  2n 1 2n 2n 2 2n 1|| y y || || y y ||         (5) And  2n 2n 1 2n 1 2n|| y y || || y y ||        (6) KATHMANDU UNIVERSITY JOURNAL OF SCIENCE, ENGINEERING AND TECHNOLOGY   VOL. 9, No. I, July, 2013, 127-133  132 From (5) and (6) we get  nn n 1 1 0|| y y || || y y ||        (7) For every n. Let n  >  m  n m|| y y ||     m m 1 m 1 m 2|| y y || || y y ||     n 1 n....... || y y ||           m m 1 n 1 1 0....... || y y ||         <   m1 0|| y y ||1   (8) Let c > 0, then there is a 0   such that c N (0) P   where N (0) {y E :|| y || }     .Since 1   there exist a positive integer N such that m1 0|| y y ||1   for every m N . Hence m1 0|| y y || N (0)1 , which implies m1 0|| y y || N (0)1  . Therefore m1 0c || y y || c N (0) P1    . This implies m1 0|| y y || c1  for n, m   N. Hence by (8) n m|| y y || c   for every n, m N  so by definition 2.4 {yn} is a Cauchy sequence in X. Since X is complete there is an z   X such that ny z . Therefore 2n 1 2n 2n 1 2n{f (x )},{g(x )},{h(x )},{l(x )}   converges to z. Continuity of f implies 22n 1f (x ) f (z).   Since h and f commute 2n 1 2n 1hf (x ) fh(x ) f (z)   . We have  2n 1 2n|| hf (x ) l(x ) ||      22n 1 2n 1a || f (x ) hf (x ) ||  +       22n 1 2n 2n 2nb || f (x ) l(x ) || c || g(x ) l(x ) ||     Letting n   we get  || f (z) z ||       a || f (z) f (z) || b || f (z) z || c || z z ||          (1 b) || f (z) z || 0    But 1- b > 0 so || f(z) – z || = 0 which implies f(z) = z. Again 2n 2n 2n 2n|| h(z) l(x ) || a || f (z) h(z) || || bf (z) l(x ) || c || g(x ) l(x ) ||        Letting n   we get  || h(z) z || a || z h(z) || b || z z || c || z z ||                            (1 a) || h(z) z || 0    But 1-a > 0, hence h(z) = z. Since g is onto there is u   X, such that z = g(u). We have  2n 1|| hf (x ) l(u)     2 22n 1 2n 1 2n 1a || f (x ) hf (x ) || b || f (x ) l(u) ||                  c || g(u) l(u) ||  Letting n   we get KATHMANDU UNIVERSITY JOURNAL OF SCIENCE, ENGINEERING AND TECHNOLOGY   VOL. 9, No. I, July, 2013, 127-133  133  || z l(u) ||     a || z z || b || z l(u) || c || z l(u) ||        (1 b c) || z l(u) || 0     Since 1-b-c > 0, this implies z = l(u). Therefore g(u) = l(u) = z. Since g and l commute    l(z)  = lg (u) = gl (u) = g(z). We have  2n 1|| h(x l(z) ||     2n 1 2n 1 2n 1a || f (x ) h(x ) || b || f (x ) l(z) ||      c || g(u) l(u) ||   Letting n   we get  || z l(z) ||    a || z z || b || z l(z) || c || z l(z) ||        (1 b c) || z l(z) || 0     But 1- b - c > 0, hence z = l(z). This implies g(z) = l(z) = z. Therefore                                      f (z) = g(z) = h(z) = l(z) = z. Hence z is a fixed point of f, g, h and l. To prove the uniqueness let z  be another fixed point of f, g, h and l.  || z z ||  = || h(z) l(z) ||                   a || f (z) h(z) || b || f (z) l(z ) || c || g(z ) l(z ) ||                        = a || z z || b || z z || c || z z ||       (1-b - c ) || z z || 0   But 1- b - c > 0 thus z z . Therefore f, g, h and l has a unique common fixed point.  REFERENCES [1]  Abbas M, Jungck G, Common fixed point results for noncommuting mappings without continuity in cone metric space, J. Math. Anal. Appl. 341 (2008)416.       [2]        Deimling K, Non linear functional analysis, Springer- Veriag, (1985). [3]  Erdal Karapinar, Fixed point Theorems in Cone Banach Space, Hindawi Publishing Corporation, Fixed point Theory and Applications, Article ID 609281, (2009)9. [4]  Huang L G , Zhang X, Cone metric space and fixed point theorems, Math. Anal. Appl. 332 (2) (2007)1468. [5]  Ilic D and Rakoevic V, Common fixed points for maps on cone metric space, J.Math. Anal. Appl. 341 (2) (2008) 876. [6]    Kang S M and Rhoades B, Fixed points for four mappings, Math. Japonica, 37(6) (1992), 1053. [7]  Neeraj Malviya and Sarla Chouhan, Proving Fixed point Theorems Using General   Principles in Cone Banach Space, International Mathematical Forum, 6(3)(2011)  115. [8]  Sh. Rezapour, R Hamlbarani, Some notes on the paper “Cone Metric space and Fixed point Theorems of contractive mappings”, J. Math, Anal. Appl. 345 (2008) 719.  [9]  Thabet Abdeljawad, Erdal Karapinar and Kenan Tas, Common fixed point Theorems in Cone Banach spaces, Hacettpe Journal of Mathematics and Statistics, 40(2) (2011) 211. 

Common Fixed Point Theorem on Cone Banach Space

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Common Fixed Point Theorem on Cone Banach Space

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