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

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

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

    • 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.
    • 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)
    • 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
    • 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′)}
    • 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)
    • 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 
    • 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),
    • 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′.
    • 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 }
    • 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
    • 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′)}
    • 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}.
    • 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 ,
    • 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) }
    • 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′)
    • 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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