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# Some Common Fixed Point Theorems for Multivalued Mappings in Two Metric Spaces

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In this paper we prove some common fixed point theorems for multivalued mappings in two
complete metric spaces.
AMS Mathematics Subject Classification: 47H10, 54H25

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### Some Common Fixed Point Theorems for Multivalued Mappings in Two Metric Spaces

1. 1. International OPEN ACCESS Journal Of Modern Engineering Research (IJMER) | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 18 | Some Common Fixed Point Theorems for Multivalued Mappings in Two Metric Spaces 1 T. Veerapandi, 2 T. Thiripura Sundari, 3 J. Paulraj Joseph 1 Professor and H.O.D of Science and Humanities, S.Veerasamy Chettiar College of Engineering and Technology, S.V.Nagar, Puliangudi, India 2 Department of Mathematics, Sri K.G.S Arts College, Srivaikuntam, India 3 Professor and H.O.D of Mathematics, Manonmaniam Sundaranar University, Tirunelveli, India I. Introduction In 1981, Fisher [1] initiated the study of fixed points on two metric spaces. In1991, Popa [2] proved some theorems on two metric spaces. After this many authors([3]-[7],[8],[9]-[11]) proved many fixed point theorems in two metric spaces. Using the Banach contraction mapping, Nadler [12] introduced the concept of multi-valued contraction mapping and he showed that a multi-valued contraction mapping gives a fixed point in the complete metric space. Later, some fixed points theorems for multifunctions on two complete metric spaces have been proved in [13], [14] and [15].The purpose of this paper is to give some common fixed point theorems for multi-valued mappings in two metric spaces. Definition1.2 A sequence {xn} in a metric space (X, d) is said to be convergent to a point x  X if given 0 there exists a positive integer n0 such that d(xn,x)   for all n  n0. Definition1.3. A sequence {xn} in a metric space (X, d) is said to be a Cauchy sequence in X if given 0 there exists a positive integer n0 such that d(xm,xn)   for all m,n  n0 . Definition1.4. A metric space (X, d) is said to be complete if every Cauchy sequence in X converges to a point in X. Definition1.5 Let X be a non-empty set and f : X → X be a map. An element x in X is called a fixed point of X if f(x) = x . Definition1.6. Let X be a non-empty set and f , g : X → X be two maps. An element x in X is called a common fixed point of f and g if f(x) = g(x) = x. Definition1.7. Let (X,d1) and (Y,d2) be complete metric spaces and B(X) and B(Y) be two families of all non-empty bounded subsets of X and Y respectively. The function 1 (A,B) for A, B in B(X) and 2 (C,D) for C,D in B (Y) are defined as follow 1 (A,B) = sup{ d1(a,b): a  A, bB } 2 (C,D) = sup{ d2(c,d): c  C, dD } and  (A) = diameter(A) Abstract: In this paper we prove some common fixed point theorems for multivalued mappings in two complete metric spaces. AMS Mathematics Subject Classification: 47H10, 54H25 Keywords: complete metric space, fixed point, multivalued mapping.
2. 2. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 19 | If A consists of a single point a we write 1 (A,B) = 1 (a,B). If B also consists of a single point b we write 1 (A,B) = 1 (a,B) = 1 (a,b). It follows immediately that 1 (A,B) = 1 (B,A)  0, and 1 (A,B)  1 (A,C) + 1 (C,B) for all A,B in B(X). Definition1.8. If {An : n = 1,2,3…} is a sequence of sets in B(X), we say that it converges to the closed set A in B(X) if (i) Each point a in A is the limit of some convergent sequence {an  An : n = 1,2,3…}, (ii) For arbitrary  > 0, there exists an integer N such that An  A for n > N where A is the union of all open spheres with centres in A and radius  . The set A is then said to be the limit of the sequence {An}. Definition1.9. Let f be a multivalued mapping of X into B(X). f is continuous at x in X if whenever {xn} is a sequence of points in X converging to x, the sequence {f(xn)} in B(X) converges to fx in B(X). If f is continuous at each point x  X, then f is continuous mapping of X into B(X). Definition1.10. Let T be a multifunction of X into B(X). z is a fixed point of T if Tz = {z}. Lemma 1.11[16]. If {An} and {Bn} are sequences of bounded subsets of a complete metric space (X, d) which converges to the bounded subsets A and B, respectively, then the sequence {δ(An,Bn)} converges to δ(A,B). II. Main Results Theorem 2.1: Let (X, d1) and (Y, d2) be two complete metric spaces. Let A, B be mappings of X into B(Y) and S, T be mappings of Y into B(X) satisfying the inequalities. 1 (SAx, TBx′)   max{ 1 (x,x′), 1 ( x,SAx), 1 (x′,TBx′), ½[ 1 (x,TBx′)+ 1 (SAx, x′)], 2 (Ax, Bx′)} --------- (1) 2 (BSy, ATy′)   max{ 2 (y,y′), 2 (y,BSy), 2 (y′,ATy′), ½[ 2 (y,ATy′)+ 2 (BSy,y′)], 2 (Sy,Ty′)} ---------- (2) for all x, x′ in X and y, y′ in Y where 0<  < 1. If one of the mappings A, B, S and T is continuous, then SA and TB have a unique common fixed point z in X and BS and AT have a unique common fixed point w in Y. Further, Az = Bz = {w} and Sw = Tw = {z}. Proof: From (1) and (2) we have 1 (SAC, TBD)   max{ 1 (C,D), 1 (C,SAC) , 1 (D,TBD), ½[ 1 (C,TBD)+ 1 (SAC,D)], 2 (AC,BD)} --------- (3) 2 (BSE, ATF)  max{ 2 (E,F), 2 (E,BSE), 2 (F,ATF), ½[ 2 (E, ATF)+ 2 (BSE,F)], ),(1 TFSE } ---------- (4)  C,D B(X) and E,FB(Y) Let x0 be an arbitrary point in X. Then y1A(x0) since A: X B(Y), x1 S(y1) since S : Y B(X) , y2B(x1) since B: X B(Y) and x2 T(y2) since T : Y B(X). Continuing in this way we get for n 1, y2n-1A(x2n-2), x2n-1S(y2n-1), y2nB(x2n-1) and x2n T(y2n) . We define the sequences {xn} in B(X) and {yn} in B(Y) by choosing a point x2n-1 (SATB)n-1 SAx = X2n-1, x2n (TBSA)n x = X2n, y2n-1A (TBSA)n-1 x = Y2n-1 and y2n B(SATB)n-1 SAx = Y2n  n = 1.2,3, … Now from (3) we have 1 (X2n+1, X2n) = 1 (SAX2n, TBX2n-1)
3. 3. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 20 |   . max{ 1 (X2n, X2n-1 ), 1 (X2n,SAX2n), 1 (X2n-1, TBX2n-1), ½[ 1 (X2n,TBX2n-1)+ 1 (SAX2n, X2n-1)], 2 (AX2n,BX2n-1 )} =  .max { 1 (X2n, X2n-1 ), 1 (X2n, X2n+1), 1 (X2n-1, X2n), ½[ 1 (X2n,X2n)+ 1 ( X2n+1, X2n-1)], 2 (Y2n+1,Y2n) } =  .max{ 1 (X2n, X2n-1 ), 1 (X2n, X2n+1), 1 (X2n-1, X2n), ½. 1 ( X2n+1, X2n-1), 2 (Y2n+1,Y2n) }   .max{ 1 (X2n, X2n-1 ), 1 (X2n, X2n+1), ½. 1 (X2n+1, X2n)+ 1 (X2n, X2n-1), 2 (Y2n+1,Y2n) } Hence 1 (X2n+1, X2n)   . max{ 1 ( X2n, X2n-1) , 2 (Y2n+1,Y2n) } --------------- (5) Now from (4) we have 2 (Y2n+1,Y2n) = 2 (BSY2n-1, ATY2n)   .max{ 2 (Y2n, Y2n-1), 2 (Y2n-1, BSY2n-1), 2 (Y2n, ATY2n), ½[ 2 (Y2n-1, ATY2n)+ 2 (Y2n,BSY2n-1)], 1 (SY2n-1, TY2n)}   . max{ 2 (Y2n-1, Y2n), 1 (X2n-1, X2n)} Similarly we have 1 (X2n, X2n-1)   .max { 1 (X2n-2, X2n-1) , 2 (Y2n-1, Y2n)} ----------------- (6) 2 (Y2n,Y2n-1)   .max { 2 (Y2n-2, Y2n-1), 1 (X2n-1, X2n-2)} from inequalities (5) and (6), we have 1 (Xn+1,Xn)   max { 1 (Xn, Xn-1) , 2 (Yn+1, Yn) }    n .max { 1 (X1, x0) , 2 (Y2, Y1) } → 0 as n→∞ Also 1 (xn+1,xn)  1 (Xn+1,Xn) which implies 1 (xn+1,xn) → 0 as n→∞ Thus {xn} is a Cauchy sequence in X. Since X is complete, it converges to a point z in X. Further 1 (z, Xn)  1 (z, xn)+ 1 (xn, Xn)  1 (z, xn)+ 2 1 (Xn, Xn+1)  1 (z, Xn)  0 as n  Similarly {yn} is a Cauchy sequence in Y and it converges to a point w in Y and 2 (w, Yn)  0 as n . Now 1 (SAx2n,z)  1 ( SAx2n, x2n) + 1 ( x2n, z)  1 ( SAx2n, TBx2n-1) + 1 ( x2n, z)   . max{ 1 (x2n, x2n-1 ), 1 (z,SAx2n), 1 ( x2n, TBz), ½[ 1 (x2n,TBx2n-1 )+ 1 ( x2n-1 ,SAx2n)], 2 (Ax2n, B x2n-1 )} + 1 ( x2n, z)  0 as n . Thus n lim SAx2n = {z} = n lim Sy2n+1 Similarly we prove n lim TBx2n-1 = {z} = n lim Ty2n n lim BSy2n-1 = {w} = n lim Bx2n-1
4. 4. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 21 | n lim ATy2n = {w} = n lim Ax2n Suppose A is continuous, then n lim Ax2n = Az = {w}. we prove SAz = {z}. We have 1 (SAz, z) = n lim 1 (SAz, TBx2n-1)  n lim  .max{ 1 (z, x2n-1), 1 (z, SAz), 1 (x2n-1,TBx2n-1), ½[ 1 (z, TBx2n-1) + 1 (x2n-1,SAz)], 2 (Az, Bx2n-1)} < 1 (z, SAz) (since  <1) Thus SAz = {z}. Hence Sw = {z}. (Since Az = {w}) Now we prove BSw = {w}. We have 2 (BSw, w) = n lim 2 (BSw, ATy2n)  n lim  .max{ 2 (w,y2n), 2 (w, BSw), 2 (y2n, ATy2n), ½[ 2 (w,ATy2n)+ 2 (y2n,BSw)], 1 (Sw, Ty2n)} < 2 (w,BSw) (Since  < 1) Thus BSw = {w}. Hence Bz = {w}. (Since Sw = z) Now we prove TBz = {z}. We have 1 (z,TBz) = n lim 1 (SAx2n, TBz)  n lim  .max{ 1 (x2n,z), 1 (x2n,SAx2n), 1 (z,TBz), ½[ 1 (x2n,TBz)+ 1 (z,SAx2n)], 2 (Ax2n, Bz)} < 1 (z,TBz) (Since  < 1) Thus TBz = { z }. Hence Tw = {z}. (Since Bz = {w}) Now we prove ATw = {w}. We have 2 (w, ATw) = n lim 2 (BSy2n-1,ATw)  n lim  .max{ 2 (y2n-1,w), 2 (y2n-1, BSy2n-1), 2 (w, ATw), ½[ 2 (y2n-1,ATw)+ 2 (w,BSy2n-1)], 1 (Sy2n-1, Tw)} < 2 (w,ATw) (Since  < 1) Thus ATw = {w}. The same results hold if one of the mappings B, S and T is continuous. Uniqueness: Let z′ be another common fixed point of SA and TB so that z′ is in SAz′ and TBz′. Using inequalities (1) and (2) we have max{ 1 (SAz′, z′), 1 ( z′, TB z′)}  1 (SAz′, TBz′)   .max{ 1 (z′,z′), 1 (z′,SAz′), 1 (z′,TBz′), ½[ 1 (z′,TBz′)+ 1 (z′,SAz′)], 2 (Az′,Bz′)}
5. 5. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 22 | <  . 2 (Az′,Bz′)   . max{ 2 (ATBz′, Bz′), 2 (Az′, BSAz′)}   . 2 (BSAz′,ATBz′)   2 . max{ 2 (Az′, Bz′), 2 (Az′, BSAz′), 2 ( Bz′,AT Bz′), ½[ 2 (Az′,ATBz′)+ 2 ( BSAz′,Bz′)], 1 (SAz′, TBz′) } <  2 . 1 (SAz′, TBz′)  SAz′ = TBz′ (since  < 1)  SAz′ = TBz′ = {z′} and Az′ = Bz′ = {w′} Thus SAz′ = TBz′ = Sw′ = Tw′ = {z′} and BSw′ = ATw′ = Az′ = Bz′ = {w′} We have 1 (z, z′) = 1 (SAz, TBz′)   .max{ 1 (z,z′), 1 (z,SAz), 1 (z′,TBz′), ½[ 1 (z,TBz′)+ 1 (z′,SAz)], 2 (Az,Bz′)} < 2 (w, w′) 2 (w, w′) = 2 (BSw, ATw′)   .max { 2 (w,w′), 2 (w,BSw), 2 (w′,ATw′), ½[ 2 (w,ATw′)+ 2 (w′,BSw)], 1 (Sw,Tw′) } < 1 (z, z′) Hence 1 (z, z′) < 2 (w, w′) < 1 (z, z′) Thus z = z′. So the point z is the unique common fixed point of SA and TB. Similarly we prove w is a unique common fixed point of BS and AT. Remark :2.2 : If we put A = B, S = T in the above theorem 2.1, we get the following corollary. Corollory 2.3 : Let (X, d1) and (Y, d2) be two complete metric spaces. Let A be a mapping of X into B(Y) and T be a mapping of Y into B(X) satisfying the inequalities. 1 (TAx, TAx′)   max { 1 (x,x′), 1 (x,TAx), 1 (x′,TAx′), ½[ 1 (x,TAx′)+ 1 (x′,TAx)], 2 (Ax,Ax′) } 2 (ATy, ATy′)   max { 2 (y,y′), 2 (y,ATy), 2 (y′,ATy′), ½[ 2 (y,ATy′)+ 2 (y′,ATy)], 1 (Ty,Ty′)} for all x, x′ in X and y, y′ in Y where 0<  < 1. If one of the mappings A and T is continuous, then TA has a unique fixed point z in X and AT has a unique fixed point w in Y. Further, Az = Bz = {w} and Sw = Tw = {z}. Theorem 2.4: Let (X, d1) and (Y, d2) be two complete metric spaces. Let A, B be mappings of X into B(Y) and S, T be mappings of Y into B(X) satisfying the inequalities. 1 (SAx, TBx′)   max{ 1 (x,x′), 1 (x′,TBx′), 2 (Ax,Bx′), ½[ 1 (x,TBx′)+ 1 (SAx,x′)], [ 1 (x,SAx). 1 (x′,TBx′)] / 1 (x,x′)} --------- (1) 2 (BSy, ATy′)   max{ 2 (y,y′), 2 (y′,ATy′), 2 (Sy,Ty′), ½[ 2 (y,ATy′)+ 2 (BSy,y′)], [ 2 (y,BSy). 2 (y′,ATy′)] / 2 (y,y′)} ------- (2)
6. 6. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 23 | for all x, x′ in X and y, y′ in Y where 0<  < 1. If one of the mappings A, B, S and T is continuous, then SA and TB have a unique common fixed point z in X and BS and AT have a unique common fixed point w in Y. Further, Az = Bz = {w} and Sw = Tw = {z}. Proof: From (1) and (2) we have 1 (SAC, TBD)   max{ 1 (C,D), 1 (D,TBD), 2 (AC,BD), ½[ 1 (C,TBD)+ 1 (SAC,D)], [ 1 (C,SAC). 1 (D,TBD)] / 1 (C,D)} --------- (3) 2 (BSE, ATF)  max{ 2 (E,F), 2 (F,ATF), ),(1 TFSE , ½[ 2 (E,ATF)+ 2 (BSE,F)], [ 2 (E,BSE). 2 (F,ATF)] / 2 (E,F)} ---------- (4)  C,D B(X) and E,FB(Y) Let x0 be an arbitrary point in X. Then y1A(x0) since A: X B(Y) , x1 S(y1) since S : Y B(X) , y2B(x1) since B: X B(Y) and x2 T(y2) since T : Y B(X). Continuing in this way we get for n 1, y2n-1A(x2n-2), x2n-1S(y2n-1), y2nB(x2n-1) and x2n T(y2n) . We define the sequences {xn} in B(X) and {yn} in B(Y) by choosing a point x2n-1 (SATB)n-1 SAx = X2n-1, x2n (TBSA)n x = X2n, y2n-1A (TBSA)n-1 x = Y2n-1 and y2n B(SATB)n-1 SAx = Y2n  n = 1.2,3, …. Now from (3) we have 1 (X2n+1, X2n) = 1 (SAX2n, TBX2n-1)   . max{ 1 (X2n, X2n-1), 1 (X2n-1,TBX2n-1), 2 (AX2n,BX2n-1 ), ½[ 1 (X2n,TBX2n-1)+ 1 (SAX2n,X2n-1)], [ 1 (X2n,SAX2n). 1 (X2n-1,TBX2n-1)] / 1 (X2n, X2n-1)} =  .max{ 1 (X2n, X2n-1), 1 (X2n-1,X2n), 2 (Y2n+1,Y2n ), ½[ 1 (X2n,X2n)+ 1 (X2n+1,X2n-1)], 1 (X2n, X2n+1). 1 (X2n-1, X2n)] / 1 (X2n, X2n-1)} = .max{ 1 (X2n, X2n-1), 2 (Y2n+1,Y2n ), 1/2[ 1 (X2n+1,X2n)+ 1 (X2n, X2n-1)], 1 (X2n,X2n+1)}   .max{ 1 (X2n, X2n-1), 2 (Y2n+1,Y2n )}--------------------------------------- (5) Now from (4) we have 2 (Y2n, Y2n+1) = 2 (BSY2n-1, ATY2n)   . max{ 2 (Y2n-1, Y2n), 2 (Y2n,ATY2n ), 1 (SY2n-1, TY2n), 1/2[ 2 (Y2n-1,ATY2n)+ 2 (BSY2n-1, Y2n)] , 2 (Y2n-1,BSY2n-1). 2 (Y2n,ATY2n) / 2 (Y2n-1, Y2n)}   . max { 2 (Y2n-1, Y2n), 1 (X2n-1, X2n)} Similarly 1 (X2n, X2n-1)   . max{ 1 (X2n-2,X2n-1), 2 (Y2n-1,Y2n)}---------------------------------- (6) 2 (Y2n,Y2n-1)   . max { 2 (Y2n-1, Y2n-2), 1 (X2n-1, X2n-2)} from inequalities (5) and (6), we have 1 (Xn+1,Xn)   max { 1 (Xn, Xn-1) , 2 (Yn+1, Yn) }    n .max { 1 (X1, x0) , 2 (Y2, Y1) } → 0 as n→∞ Also 1 (xn, xn+ 1)  1 (Xn, Xn+ 1)  1 (xn, xn+1)  0 as n 
7. 7. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 24 | Thus {xn} is a Cauchy sequences in X. Since X is complete, {xn} converges to a point z in X. Further 1 (z, Xn)  1 (z, xn)+ 1 (xn, Xn)  1 (z, xn)+ 2 1 (Xn, Xn+1)  1 (z, Xn)  0 as n  Similarly {yn} is a Cauchy sequence in Y and it converges to a point w in Y and 2 (w, Yn)  0 as n . Now 1 (SAx2n,z)  1 ( SAx2n, x2n) + 1 ( x2n, z)  1 ( SAx2n, TBx2n-1) + 1 ( x2n, z)   .max { 1 (x2n, x2n-1), 1 (x2n-1,TBx2n-1), 2 (Ax2n,Bx2n-1 ), ½[ 1 (x2n,TBx2n-1)+ 1 (SAx2n,x2n-1)], [ 1 (x2n,SAx2n). 1 (x2n-1,TBx2n-1)] / 1 (x2n, x2n-1)} + 1 ( x2n, z)  0 as n . Thus n lim SAx2n = {z} = n lim Sy2n+1 Similarly we prove n lim TBx2n-1 = {z} = n lim Ty2n n lim BSy2n-1 = {w} = n lim Bx2n-1 n lim ATy2n = {w} = n lim Ax2n Suppose A is continuous, then n lim Ax2n = Az = {w}. Now we prove SAz = {z} We have 1 (SAz,z)  n lim 1 (SAz, TBx2n-1)  n lim  .max{ 1 (z, x2n-1), 1 ( x2n-1,TB x2n-1), 2 (Az, Bx2n-1), ½[ 1 (z,TBx2n-1)+ 1 (SAz,x2n-1)], 1 (z, SAz), 1 ( x2n-1,TB x2n-1) / 1 (z, x2n-1) } < 1 (z,SAz) (Since  < 1) Thus SAz = {z} = Sw Now we prove BSw = w. We have 2 (BSw, w)  n lim 2 (BSw, ATy2n)  n lim  . max{ 2 (w, y2n), 2 (y2n,ATy2n), 1 (Sw, Ty2n), ½[ 2 (w, ATy2n)+ 2 (BSw,y2n)], 2 (w,BSw). 2 (y2n, ATy2n) / 2 (w, y2n) } < 2 (w,BSw) (Since  < 1) Thus BSw = {w} = Bz. Now we prove TBz = {z}. 1 (z,TBz)  n lim 1 (SAx2n,TBz)  n lim  .max{ 1 (x2n ,z), 1 (z ,TBz), 2 (Ax2n,Bz),
8. 8. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 25 | ½[ 1 (x2n ,TBz)+ 1 (SAx2n ,z)], 1 (x2n ,SAx2n). 1 (z,TBz) / 1 (x2n ,z)} < 1 (z,TBz) (Since  < 1) Thus TBz = {z}= Tw. Now we prove ATw = {w}. 2 (w, ATw)  n lim 2 (BSy2n-1, ATw)  n lim  .max{ 2 (y2n-1, w), 2 (w,ATw), 1 (Sy2n-1, Tw), ½[ 2 (y2n-1, ATw)+ 2 (BSy2n-1, w)], 2 (y2n-1, BSy2n-1). 2 (w,ATw) / 2 (y2n-1, w) } < 2 (w,ATw) (Since  < 1) Thus ATw = {w}. The same results hold if one of the mappings B, S and T is continuous. Uniqueness: Let z′ be another common fixed point of SA and TB so that z′ is in SAz′ and TBz′. Using inequalities (1) and (2) we have max{ 1 (SAz′, z′), 1 ( z′, TB z′)}  1 (SAz′, TBz′)   . max{ 1 (z′, z′), 1 (z′,TBz′), 2 (Az′,Bz′), ½[ 1 (z′,TBz′)+ 1 (SAz′, z′)], [ 1 (z′,SAz′). 1 (z′,TBz′)] / 1 (z′, z′)}   . 2 (Az′,Bz′)   . max{ 2 (ATBz′, Bz′), 2 (Az′,BSAz′)}   . 2 (BSAz′,ATBz′)   2 . max{ 2 (Az′,Bz′) , 2 (Bz′, ATBz′), 1 (SAz′,TBz′) ½[ 2 (z′,BSAz′)+ 2 (ATBz′, z′)], 2 ( z′,BSAz′). 2 ( z′,ATBz′)/ 2 (z′,z′) }   2 1 (SAz′, TBz′)  SAz′ = TBz′ (since  < 1)  SAz′ = TBz′ = {z′} and Az′ = Bz′ = {w′} Thus SAz′ = TBz′ = Sw′ = Tw′ = {z′} and BSw′ = ATw′ = Az′ = Bz′ = {w′} We have 1 (z, z′) = 1 (SAz, TBz′)   . max{ 1 (z, z′), 1 (z′, TBz′), 2 (Az,Bz′), ½[ 1 (z, TBz′)+ 1 (SAz, z′)], 1 (z, SAz) . 1 (z′, TBz′) / 1 (z, z′) } < 2 (w, w′) (Since  < 1) Now 2 (w, w′) = 2 (BSw, ATw′)   . max{ 2 (w, w′), 2 (w′, ATw′), 1 (Sw,Tw′), ½[ 2 (w, ATw′)+ 2 (BSw, w′)], 2 (w, BSw ). 2 (w′, ATw′) / 2 (w, w′)} < 1 (z, z′) Hence 1 (z, z′) < 2 (w, w′) < 1 (z, z′)  z = z′.
9. 9. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 26 | So the point z is the unique common fixed point of SA and TB. Similarly we prove w is a unique common fixed point of BS and AT. Remark: 2.5 : If we put A = B, S = T in the above theorem 2.4, we get the following corollary. Corollory 2.6 : Let (X, d1) and (Y, d2) be two complete metric spaces. Let A be a mapping of X into B(Y) and T be a mapping of Y into B(X) satisfying the inequalities. 1 (TAx, TAx′)   max{ 1 (x,x′), 1 (x′,TAx′), 2 (Ax,Ax′), ½[ 1 (x,TAx′)+ 1 (TAx,x′)], [ 1 (x,TAx). 1 (x′,TAx′)] / 1 (x,x′)} 2 (ATy, ATy′)  max{ 2 (y,y′), 2 (y′,ATy′), 2 (Ty,Ty′),½[ 2 (y,ATy′)+ 2 (ATy,y′)], [ 2 (y,ATy). 2 (y′,ATy′)] / 2 (y,y′)} for all x, x′ in X and y, y′ in Y where 0<  < 1. If one of the mappings A and T is continuous, then TA has a unique fixed point z in X and AT has a unique fixed point w in Y. Further, Az = Bz = {w} and Sw = Tw = {z}. Theorem 2.7: Let (X, d1) and (Y, d2) be complete metric spaces. Let A, B be mappings of X into B(Y) and S, T be mappings of Y into B(X) satisfying the inequalities. 1 (SAx, TBx′) ≤  . max{ 1 (x,x′), 1 (x,SAx), 1 (x′,TBx′), 2 (Ax,Bx′), 2 )',(1 TBxx , 2 )',(1 xSAx } ------- (1) 2 (BSy, ATy′) ≤  . max{ 2 (y,y′), 2 (y,BSy), 2 (y′,ATy′), 1 (Sy,Ty′), 2 )',(2 ATyy , 2 )',(2 yBSy } ---------(2) for all x, x′ in X and y,y′ in Y where 0 ≤  < 1. If one of the mappings A, B, S and T is continuous , then SA and TB have a unique common fixed point z in X and BS and AT have a unique common fixed point w in Y. Further, Az = Bz = {w} and Sw = Tw = {z}. Proof: From (1) and (2) we have 1 (SAC, TBD)   max{ 1 (C,D), 1 (C,SAC), 1 (D,TBD), 2 (AC,BD), 2 ),(1 TBDC , 2 ),(1 DSAC } --------- (3) 2 (BSE, ATF)  max{ 2 (E,F), 2 (E,BSE), 2 (F,ATF), 1 (SE,TF), 2 ),(2 ATFE , 2 ),(2 FBSE } ---------- (4)  C,D B(X) and E,FB(Y) Let x0 be an arbitrary point in X. Then y1A(x0) since A: X B(Y) , x1 S(y1) since S : Y B(X) , y2B(x1) since B: X B(Y) and x2 T(y2) since T : Y B(X). continuing in this way we get for n1, y2n-1A(x2n-2), x2n-1S(y2n-1), y2nB(x2n-1) and x2n T(y2n) . We define the sequences {xn} in B(X) and {yn} in B(Y) by choosing a point x2n-1 (SATB)n-1 SAx = X2n-1, x2n (TBSA)n x = X2n, y2n-1A (TBSA)n-1 x = Y2n-1 and y2n B(SATB)n-1 SAx = Y2n  n = 1.2,3, …. Now from (3) we have 1 (X2n+1, X2n) ≤ 1 (SAX2n, TBX2n-1) ≤  .max{ 1 (X2n, X2n-1), 1 (X2n,SAX2n), 1 (X2n-1, TBX2n-1), 2 (AX2n,BX2n-1 ), , 2 ),( 1221 nn TBXX 2 ),( 1221 nn XSAX }
10. 10. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 27 | =  . max{ 1 (X2n, X2n-1), 1 (X2n, X2n+1), 1 (X2n-1, X2n), 2 (Y2n+1,Y2n), 2 ),( 221 nn XX , 2 ),( 12121  nn XX } ≤  . max{ 1 (X2n, X2n-1), 1 (X2n, X2n+1), 1 (X2n-1, X2n), 2 (Y2n+1,Y2n), 2 )X,(X+)X,(X 1-2n2n12n12n1   } ≤  . max{ 1 (X2n-1, X2n), 2 (Y2n+1, Y2n)} --------------------------- (5) Now from (4) we have 2 ( Y2n, Y2n+1) ≤ 2 (BSY2n-1, ATY2n) ≤  . max{ 2 ( Y2n-1, Y2n), 2 ( Y2n-1, BSY2n-1), 2 ( Y2n, ATY2n), 1 (SY2n-1, TY2n), 2 )ATY,(Y 2n1-2n2 , 2 )Y,(BSY 2n1-2n2 } ≤  . max{ 2 (Y2n-1, Y2n), 1 (X2n-1, X2n)} Similarly 1 (X2n, X2n-1)   . max{ 1 (X2n-2,X2n-1), 2 (Y2n-1,Y2n)}------------------------------- (6) 2 (Y2n,Y2n-1)   . max { 2 (Y2n-1, Y2n-2), 1 (X2n-1, X2n-2)} from inequalities (5) and (6), we have 1 (Xn, Xn+1)   max { 1 (Xn, Xn-1) , 2 (Yn+1, Yn) }    n . max { 1 (x0, X1) , 2 (Y1, Y2) } 0 as n  Also 1 (xn, xn+ 1)  1 (Xn, Xn+ 1)  1 (xn, xn+1)  0 as n  Thus {xn} is a Cauchy sequence in X. Since X is complete, {xn} converges to a point z in X. Further 1 (z, Xn)  1 (z, xn)+ 1 (xn, Xn)  1 (z, xn)+ 2 1 (Xn, Xn+1)  1 (z, Xn)  0 as n  Similarly {yn} is a Cauchy sequence in Y and it converges to a point w in Y and 2 (w, Yn)  0 as n . Now 1 (SAx2n,z)  1 ( SAx2n, x2n) + 1 ( x2n, z)  1 ( SAx2n, TBx2n-1) + 1 ( x2n, z) ≤  .max{ 1 (x2n, x2n-1), 1 (x2n,SAx2n), 1 (x2n-1, TBx2n-1), 2 (Ax2n,Bx2n-1 ), , 2 ),( 1221 nn TBxx 2 ),( 1221 nn xSAx } + 1 ( x2n, z)  0 as n  Thus n lim SAx2n = {z} = n lim Sy2n+1 Similarly we prove n lim TBx2n-1 = {z} = n lim Ty2n n lim BSy2n-1 = {w} = n lim Bx2n-1
11. 11. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 28 | n lim ATy2n = {w} = n lim Ax2n Suppose A is continuous, then n lim Ax2n = Az = {w}. Now we prove SAz = {z} We have 1 (SAz,z) = n lim 1 (SAz, TBx2n-1) ≤ n lim  . max{ 1 (z, x2n-1), 1 (z,SAz), 1 (x2n-1,TB x2n-1), 2 (Az, Bx2n-1), 2 )TBx(z, 1-2n1 , 2 )x,(SAz 1-2n1 } < 1 (z,SAz) (Since  < 1) Thus SAz = {z}. Hence Sw = {z}. (Since Az = {w}) Now we prove BSw = {w}. We have 2 (BSw, w) = n lim 2 (BSw, ATy2n) ≤ n lim  . max{ 2 (w, y2n), 2 (w, BSw), 2 (y2n, ATy2n), 1 (Sw, Ty2n), 2 )ATy(w, 2n2 , 2 )y(BSw, 2n2 } < 2 (w,BSw) (Since  < 1) Thus BSw = {w}. Hence Bz = {w}. (Since Sw = {z}) Now we prove TBz = {z} 1 (z,TBz) = n lim 1 (SAx2n, TBz) ≤ n lim  . max{ 1 (x2n ,z), 1 (x2n,SAx2n), 1 (z, TBz), 2 (Ax2n, Bz), 2 TBz),(x2n1 , 2 z),(SAx2n1 } < 1 (z, TBz) (Since  < 1) Thus TBz = {z}. Hence Tw = {z}. (Since Bz = {w}) Now we prove ATw = {w}. 2 (w, ATw) = n lim 2 (BSy2n-1, ATw) ≤ n lim  . max{ 2 (y2n-1, w), 2 (y2n-1, BSy2n-1), 2 (w, ATw), 1 (Sy2n-1, Tw), 2 ATw),(y 1-2n2 , 2 w),(BSy 1-2n2 } < 2 (w, ATw) (Since  < 1) Thus ATw = {w}. The same results hold if one of the mappings B, S and T is continuous. Uniqueness: Let z′ be another common fixed point of SA and TB so that z′ is in SAz′ and TBz′. Using inequalities (1) and (2) we have max{ 1 (SAz′, z′), 1 ( z′, TB z′)}
12. 12. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 29 |  1 (SAz′, TBz′) ≤  .max{ 1 (z′, z′), 1 ( z′,SA z′), 1 ( z′, TB z′), 2 (A z′,B z′ ), , 2 )',z'(1 TBz 2 )','(1 zSAz }   . 2 (Az′,Bz′)   . max{ 2 (ATBz′, Bz′), 2 (Az′,BSAz′)}   . 2 (BSAz′,ATBz′) ≤  2 . max{ 2 (Az′, Bz′), 2 ( Az′, BSAz′), 2 (Bz′, ATBz′), 1 (SAz′, TBz′), 2 )ATBz',(Az'2 , 2 )Bz',(BSAz'2 }   2 1 (SAz′, TBz′)  SAz′ = TBz′ (since  < 1)  SAz′ = TBz′ = {z′} and Az′ = Bz′ = {w′} Thus SAz′ = TBz′ = Sw′ = Tw′ = {z′} and BSw′ = ATw′ = Az′ = Bz′ = {w′} We have 1 (z, z′) = 1 (SAz, TBz′) ≤  . max{ 1 (z, z′), 1 (z,SAz), 1 (z′,TBz′), 2 (Az,Bz′), 2 )TBz'(z,1 , 2 )z'(SAz,1 } < 2 (w, w′) (since  < 1) 2 (w, w′) = 2 (BSw,ATw′) ≤  . max{ 2 (w,w′), 2 (w, BSw), 2 (w′,ATw′), 1 (Sw,Tw′), 2 )ATw'(w,2 , 2 )w'(BSw,2 } < 1 (z, z′) (since  < 1) Hence 1 (z, z′) < 2 (w, w′) < 1 (z, z′) Thus z = z′. So the point z is the unique common fixed point of SA and TB. Similarly we prove w is a unique common fixed point of BS and AT. Remark 2.8: If we put A = B, S = T in the above theorem 2.7, we get the following corollary. Corollary 2.9: Let (X,d1) and (Y,d2) be two complete metric spaces. Let A be a mapping of X into Y and T be a mapping of Y into X satisfying the inequalities. 1 (TAx, TAx′) ≤  .max{ 1 (x,x′), 1 (x,TAx), 1 (x′,TAx′), 2 (Ax,Ax′), 2 )',(1 TAxx , 2 )',(1 xTAx } 2 (ATy,ATy′)≤  .max{ 2 (y,y′), 2 (y,ATy), 2 (y′,ATy′), 1 (Ty,Ty′), 2 )',(2 ATyy , 2 )',(2 yATy } for all x, x′ in X and y,y′ in Y where 0 ≤  < 1. If one of the mappings A and T is continuous, then TA has a unique fixed point z in X and AT has a unique fixed point w in Y. Further, Az = Bz = {w} and Sw = Tw = {z}.
13. 13. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 30 | Theorem 2.10: Let (X, d1) and (Y, d2) be complete metric spaces. Let A, B be mappings of X into B(Y) and S, T be mappings of Y into B(X) satisfying the inequalities. 1 (SAx, TBx′) ≤  . max{ 1 (x,x′), 1 (x,SAx), 2 (Ax,Bx′), 2 )',(1 TBxx , 2 )',(1 xSAx [ 1 (x,SAx). 1 (x′,TBx′)] / 1 (x,x′) } ------- (1) 2 (BSy, ATy′) ≤  . max{ 2 (y,y′), 2 (y,BSy), 1 (Sy,Ty′), 2 )',(2 ATyy , 2 )',(2 yBSy [ 2 (y,BSy). 2 (y′,ATy′)] / 2 (y,y′) } ---------(2) for all x, x′ in X and y,y′ in Y where 0 ≤  < 1. If one of the mappings A, B, S and T is continuous , then SA and TB have a unique common fixed point z in X and BS and AT have a unique common fixed point w in Y. Further, Az = Bz = {w} and Sw = Tw = {z}. Proof: From (1) and (2) we have 1 (SAC, TBD)   max{ 1 (C,D), 1 (C,SAC), 2 (AC,BD), 2 ),(1 TBDC , 2 ),(1 DSAC [ 1 (C,SAC). 1 (D,TBD)] / 1 (C,D) } --------- (3) 2 (BSE, ATF)  max{ 2 (E,F), 2 (E,BSE), 1 (SE,TF), 2 ),(2 ATFE , 2 ),(2 FBSE [ 2 (E,BSE). 2 (F′,ATF)] / 2 (E,F)} ---------- (4)  C,D B(X) and E,FB(Y) Let x0 be an arbitrary point in X. Then y1A(x0) since A: X B(Y), x1 S(y1) since S : Y B(X), y2B(x1) since B: X B(Y) and x2 T(y2) since T : Y B(X).Continuing in this way we get for n1, y2n-1A(x2n-2), x2n-1S(y2n-1), y2nB(x2n-1) and x2n T(y2n) . We define the sequences {xn} in B(X) and {yn} in B(Y) by choosing a point x2n-1 (SATB)n-1 SAx = X2n-1, x2n (TBSA)n x = X2n, , y2n-1A (TBSA)n-1 x = Y2n-1 and y2n B(SATB)n-1 SAx = Y2n  n = 1.2,3, …. Now from (3) we have 1 (X2n+1, X2n) ≤ 1 (SAX2n, TBX2n-1) ≤  .max{ 1 (X2n, X2n-1), 1 (X2n,SAX2n), 2 (AX2n,BX2n-1 ), , 2 ),( 1221 nn TBXX 2 ),( 1221 nn XSAX , [ 1 (X2n,SAX2n). 1 ( X2n-1, TBX2n-1)] / 1 (X2n, X2n-1)} =  . max{ 1 (X2n, X2n-1), 1 (X2n, X2n+1), 2 (Y2n+1,Y2n), 2 ),( 221 nn XX , 2 ),( 12121  nn XX ,[ 1 (X2n,X2n+1). 1 (X2n-1,X2n)] / 1 (X2n, X2n-1) } ≤  . max{ 1 (X2n, X2n-1), 1 (X2n, X2n+1), 2 (Y2n+1,Y2n), 0, 2 )X,(X+)X,(X 1-2n2n12n12n1   , 1 (X2n,X2n+1) } ≤  . max{ 1 (X2n-1, X2n), 2 (Y2n+1, Y2n)} --------------------------- (5) Now from (4) we have 2 ( Y2n, Y2n+1) ≤ 2 (BSY2n-1, ATY2n) ≤  max{ 2 (Y2n-1, Y2n), 2 ( Y2n-1, BSY2n-1), 1 (SY2n-1, TY2n), 2 )ATY,(Y 2n1-2n2 ,
14. 14. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 31 | 2 )Y,(BSY 2n1-2n2 ,[ 2 (Y2n-1,BSY2n-1). 2 (Y2n,ATY2n) / 2 (Y2n-1, Y2n) } ≤  . max{ 2 (Y2n-1, Y2n), 1 (X2n-1, X2n)} Similarly 1 (X2n, X2n-1)   . max{ 1 (X2n-2,X2n-1), 2 (Y2n-1,Y2n)}------------------------------- (6) 2 (Y2n,Y2n-1)   . max { 2 (Y2n-1, Y2n-2), 1 (X2n-1, X2n-2)} from inequalities (5) and (6) we have 1 (Xn, Xn+1)   max { 1 (Xn, Xn-1) , 2 (Yn+1, Yn) }    n . max { 1 (x0, X1) , 2 (Y1, Y2) } 0 as n  Also 1 (xn, xn+ 1)  1 (Xn, Xn+ 1)  1 (xn, xn+1)  0 as n  Thus {xn} is a Cauchy sequence in X. Since X is complete, {xn} converges to a point z in X. Further 1 (z, Xn)  1 (z, xn)+ 1 (xn, Xn)  1 (z, xn)+ 2 1 (Xn, Xn+1)  1 (z, Xn)  0 as n  Similarly {yn} is a Cauchy sequence in Y and it converges to a point w in Y and 2 (w, Yn)  0 as n . Now 1 (SAx2n,z)  1 ( SAx2n, x2n) + 1 ( x2n, z)  1 ( SAx2n, TBx2n-1) + 1 ( x2n, z) ≤ c.max{ 1 (x2n, x2n-1), 1 (x2n,SAx2n), 2 (Ax2n,Bx2n-1 ), , 2 ),( 1221 nn TBxx 2 ),( 1221 nn xSAx ,[ 1 (x2n,SAx2n). 1 (x2n-1,TBx2n-1)] / 1 (x2n, x2n-1) } + 1 ( x2n, z)  0 as n  Thus n lim SAx2n = {z} = n lim Sy2n+1 Similarly we prove n lim TBx2n-1 = {z} = n lim Ty2n n lim BSy2n-1 = {w} = n lim Bx2n-1 n lim ATy2n = {w} = n lim Ax2n Suppose A is continuous, then n lim Ax2n = Az = {w}. Now we prove SAz = {z} We have 1 (SAz,z) = n lim 1 (SAz, TBx2n-1) ≤ n lim  . max{ 1 (z, x2n-1), 1 (z,SAz), 2 (Az, Bx2n-1), 2 )TBx(z, 1-2n1 , 2 )x,(SAz 1-2n1 , 1 (z, SAz), 1 ( x2n-1,TB x2n-1) / 1 (z, x2n-1) }
15. 15. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 32 | < 1 (z,SAz) (Since  < 1) Thus SAz = {z}. Hence Sw = {z}. (Since Az = {w}) Now we prove BSw = {w}. We have 2 (BSw, w) = n lim 2 (BSw, ATy2n) ≤ n lim  max{ 2 (w, y2n), 2 (w, BSw), 1 (Sw, Ty2n), 2 )ATy(w, 2n2 , 2 )y(BSw, 2n2 , 2 (w,BSw). 2 (y2n, ATy2n) / 2 (w, y2n)} < 2 (w,BSw) (Since  < 1) Thus BSw = {w}. Hence Bz = {w}. (Since Sw = {z}) Now we prove TBz = {z} 1 (z,TBz) = n lim 1 (SAx2n, TBz) ≤ n lim  . max{ 1 (x2n ,z), 1 (x2n,SAx2n), 2 (Ax2n, Bz), 2 TBz),(x2n1 , 2 z),(SAx2n1 , 1 (x2n ,SAx2n). 1 (z,TBz) / 1 (x2n ,z)} < 1 (z, TBz) (Since  < 1) Thus TBz = {z}. Hence Tw = {z}. (Since Bz = {w}) Now we prove ATw = {w}. 2 (w, ATw) = n lim 2 (BSy2n-1, ATw) ≤ n lim  .max{ 2 (y2n-1, w), 2 (y2n-1, BSy2n-1), 1 (Sy2n-1, Tw), 2 ATw),(y 1-2n2 , 2 w),(BSy 1-2n2 , 2 (y2n-1, BSy2n-1). 2 (w,ATw) / 2 (y2n-1, w)} < 2 (w, ATw) (Since  < 1) Thus ATw = {w}. The same results hold if one of the mappings B, S and T is continuous. Uniqueness: Let z′ be another common fixed point of SA and TB so that z′ is in SAz′ and TBz′. Using inequalities (1) and (2) we have max{ 1 (SAz′, z′), 1 ( z′, TBz′)}  1 (SAz′, TBz′) ≤  .max{ 1 (z′, z′), 1 ( z′,SA z′), 2 (A z′,B z′ ), , 2 )',z'(1 TBz 2 )','(1 zSAz ,[ 1 (z′,SAz′). 1 (z′,TBz′)] / 1 (z′, z′) }   . 2 (Az′,Bz′)   . max{ 2 (ATBz′, Bz′), 2 (Az′,BSAz′)}   . 2 (BSAz′,ATBz′)
16. 16. Some Common Fixed Point Theorems for Multi valued Mappings in Two Metric Spaces | IJMER | ISSN: 2249–6645 | www.ijmer.com | Vol. 4 | Iss. 3 | Mar. 2014 | 33 | ≤  2 .max{ 2 (Az′, Bz′), 2 ( Az′, BSAz′), 1 (SAz′, TBz′), 2 )ATBz',(Az'2 , 2 )Bz',(BSAz'2 , 2 ( z′,BSAz′). 2 ( z′,ATBz′)/ 2 (z′,z′)}   2 1 (SAz′, TBz′)  SAz′ = TBz′ (since  < 1)  SAz′ = TBz′ = {z′} and Az′ = Bz′ = {w′} Thus SAz′ = TBz′ = Sw′ = Tw′ = {z′} and BSw′ = ATw′ = Az′ = Bz′ = {w′} We have 1 (z, z′) = 1 (SAz, TBz′) ≤  . max{ 1 (z, z′), 1 (z,SAz), 2 (Az,Bz′), 2 )TBz'(z,1 , 2 )z'(SAz,1 , 1 (z, SAz) . 1 (z′, TBz′) / 1 (z, z′)} < 2 (w, w′) (since  < 1) 2 (w, w′) = 2 (BSw,ATw′) ≤  . max{ 2 (w,w′), 2 (w, BSw), 1 (Sw,Tw′), 2 )ATw'(w,2 , 2 )w'(BSw,2 , 2 (w, BSw ). 2 (w′, ATw′) / 2 (w, w′)} < 1 (z, z′) (since  < 1) Hence 1 (z, z′) < 2 (w, w′) < 1 (z, z′) Thus z = z′. So the point z is the unique common fixed point of SA and TB. Similarly we prove w is a unique common fixed point of BS and AT. Remark :2.11 : If we put A = B, S = T in the above theorem 2.10, we get the following corollary. Corollary 2.12: Let (X,d1) and (Y,d2) be two complete metric spaces. Let A, B be mappings of X into B(Y) and S, T be mappings of Y into B(X) satisfying the inequalities. 1 (TAx, TAx′) ≤  . max{ 1 (x,x′), 1 (x,TAx), 2 (Ax,Ax′), 2 )',(1 TAxx , 2 )',(1 xTAx [ 1 (x,TAx). 1 (x′,TAx′)] / 1 (x,x′) } 2 (ATy, ATy′) ≤  . max{ 2 (y,y′), 2 (y,ATy), 1 (Ty,Ty′), 2 )',(2 ATyy , 2 )',(2 yATy [ 2 (y,ATy). 2 (y′,ATy′)] / 2 (y,y′) } for all x, x′ in X and y, y′ in Y where 0 ≤  < 1. If one of the mappings A and T is continuous, then TA has a unique fixed point z in X and AT has a unique fixed point w in Y. Further, Az = Bz = {w} and Sw = Tw = {z}.
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