COMMON FIXED POINT THEOREMS FOR FOUR MAPPINGS WITH SOME WEAK CONDITIONS OF COMMUTATIVITY

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1 Novi Sad J. Math. Vol. 36, No. 1, 006, COMMON FIXED POINT THEOREMS FOR FOUR MAPPINGS WITH SOME WEAK CONDITIONS OF COMMUTATIVITY Duran Turkoglu 1, Ishak Altun 1, Brian Fisher Abstract. Some results on the common fied point of two set-valued and two single valued mappings defined on a complete metric space with some weak commutativity conditions have been proved. AMS Mathematics Subject Classification (000): 54H5, 47H10 Key words and phrases: Common fied point, set-valued mapping, weakly commuting mapping, slightly commuting mapping, quasi commuting mapping. 1. Introduction Imdad, Khan and Sessa [3], generalizing the notion of commutativity for set-valued mappings, established the idea of weak commutativity, quasi commutativity, slight commutativity. Under these concepts, Imdad and Ahmad proved Theorems [6], for set-valued mappings. Our work generalizes earlier results due to Pathak, Mishra and Kalinde [5] with the proof techniques of Imdad and Ahmad [6].. Preliminaries Let (X, d) be a metric space, then following [1] we record (i) B(X) = {A : A is a nonempty bounded subset of X} (ii) For A, B B(X) we define δ(a, B) = sup{d(a, b) : a A, b B} If A = {a}, then we write δ(a, B) = δ(a, B) and if B = {b}, then δ(a, B) = d(a, b). One can easily prove that for A, B, C in B(X) δ(a, B) = δ(b, A) 0, δ(a, B) δ(a, C) + δ(c, B) δ(a, A) = sup{d(a, b) : a, b A} = diama and δ(a, B) = 0 implies that A = B = {a}. If {A n } is a sequence in B(X), we say that {A n } converges to A X, and write A n A, iff 1 Department of Mathematics, Faculty of Science and Arts, University of Gazi Teknikokullar, Ankara, TURKEY, s: dturkoglu@gazi.edu.tr, ialtun@gazi.edu.tr Department of Mathematics, Leicester University Leicester, LE1 7RH, ENGLAND, e- mail: fbr@le.ac.uk

2 76 D. Turkoglu, I. Altun, B. Fisher (i) a A implies that a n a for some sequence {a n } with a n A n for n N, and (ii) for any ε > 0 m N such that A n A ε = { X : d(, a) < ε for some a A} for n > m. We need the following lemmas. Lemma 1. [] Suppose {A n } and {B n } are sequences in B(X) and (X, d) is a complete metric space. If A n A B(X) and B n B B(X), then δ(a n, B n ) δ(a, B). Lemma. [3] If {A n } is a sequence of nonempty bounded subsets in the complete metric space (X, d) and if δ(a n, y) 0 for some y X, then A n {y}. Definition 1. [7] The mappings F, S : X X are weakly commuting if for all X, we have d(f S, SF ) d(f, S). Definition. Let F : X B(X) be a set-valued mapping and S : X X a single-valued mapping. Then, following [1, 3], we say that the pair (F, S) is (i) weakly commuting on X if δ(f S, SF ) ma{δ(s, F ), diamsf } for any in X (ii) quasi-commuting on X if SF F S for any in X (iii) slightly commuting on X if δ(f S, SF ) ma{δ(s, F ), diamf } for any in X. Clearly, two commuting mappings satisfy (i) (iii) but the converse may not be true. In [3] it is demonstrated by suitable eamples that the foregoing three concepts are mutually independent and none of them implies the other two. 3. Fied Point Theorems Throughout this section, let R + denote the set of non-negative reals, and let Φ be the family of all mappings φ : (R + ) 5 R + such that φ is upper semi continuous, non-decreasing in each coordinate variable and, for any t > 0, γ(t) = φ(t, t, a 1 t, a t, a 3 t) < t, where γ : R + R + and a 1 + a + a 3 = 8. We need the following lemma. Lemma 3. [4] For any t > 0, γ(t) < t if and only if lim n γn (t) = 0, where γ n denotes the composition of γ n-times with itself. Let F, G be two set-valued mappings of a metric space (X, d) into B(X), and A, B, two self-mappings of X such that (1) F (X) A(X), G(X) B(X),

3 Common fied point theorems for four mappings δ (F, Gy) φ(d (B, Ay), δ q (B, F ) δ q (Ay, Gy), δ r (B, Gy) δ r (Ay, F ), δ s (B, F ) δ s (Ay, F ), δ l (B, Gy) δ l (Ay, Gy)) () for all, y X, where φ Φ, 0 < p, q, q, r, r, s, s, l, l 1 such that = q + q = r + r = s + s = l + l. Then by choosing an arbitrary 0 X and using (1), we can define a sequence {y n } in X by y n+1 = A n+1 F n = X n+1 and (3) y n+ = B n+ G n+1 = X n+, n = 0, 1,,... Let F, G : X B(X) and A, B : X X satisfy conditions (1) and (), and the sequence {y n } is defined by (3), then following the proof techniques of Imdad et al. [6], we can prove the following Lemma 4. If d n = δ(x n, X n+1 ), then lim n d n = 0. Proof. Let us assume that d n+1 > d n, then d n+1 {φ(d n+1, d n+1, 4d n+1, d n+1, d n+1 )} 1 {γ(d n+1 )} 1 < d n+1, which is a contradiction. Hence d n+1 d n. Similarly, one can show that d n+ d n+1. Then {d n } is a decreasing sequence. Now since it follows by induction that d φ(d 1, d 1, 4d 1, d 1, d γ(d 1 ), d n+1 γn (d 1 ) and if d 1 > 0, then Lemma 3 implies that lim d n = 0. If d 1 = 0, then n d n = 0, n = {1,,...}. 1 ) Lemma 5. {y n } is a Cauchy sequence in X. Proof. We show that {y n } is a Cauchy sequence. For this it is sufficient to show that {y n } is a Cauchy sequence. Suppose {y n } is not Cauchy sequence. Then

4 78 D. Turkoglu, I. Altun, B. Fisher there is an ε > 0 such that for an even integer k there eists even integers m(k) > n(k) > k such that (4) d(y n(k), y m(k) ) > ε. For every even integer k, let m(k) be the least positive integer eceeding n(k) satisfying (4) and such that (5) d(y n(k), y m(k) ) < ε. Now ε d(y n(k), y m(k) ) Then by (4) and (5) it follows that d(y n(k), y m(k) ) + d m(k) + d m(k) 1. (6) lim k d(y n(k), y m(k) ) = ε. Also, by the triangle inequality, we have d(y n(k), y m(k) 1 ) d(y n(k), y m(k) ) < d m(k) 1. By using (6) we get d(y n(k), y m(k) 1 ) ε as k. Now by () we get d(y n(k), y m(k) ) d n(k) + δ(f n(k), G m(k) 1 ) which on letting k reduces to d n(k) + {φ(d (y n(k), y m(k) 1 ), d q n(k) dq m(k), [d r (y n(k), y m(k) 1 ) + d r m(k) ] [d r (y m(k) 1, y n(k) ) + d r n(k) ], d s n(k) [ds (y m(k) 1, y n(k) ) + d s n(k) ], [d l (y m(k) 1, y n(k) ) + d l m(k) ]dl m(k) } 1 ε {φ(ε, 0, ε, 0, 0)} 1 {γ(ε } 1 < ε, giving a contradiction. Thus {y n } is a Cauchy sequence. Theorem 1. Let F, G be two set-valued mappings of a complete metric space (X, d) into B(X), and A, B two self-mappings of X satisfying conditions (1), (), (F, B) and (G, A) are slightly commuting and any one of these four mappings is continuous, then F, G, A and B have a unique common fied point in X.

5 Common fied point theorems for four mappings Proof. By Lemma 5, the sequence {y n } defined by (3) is a Cauchy sequence in X. Therefore y n z for some z X. Hence the subsequences {y n } = {B n } and {y n+1 } = {A n+1 } of {y n } also converge to z, whereas the sequences of sets {F n } and {G n+1 } converge to the set {z}. Since (F, B) commute slightly, we have δ(bf n, F B n ) ma{δ(b n, F n ), diamf n } which on letting n gives (by Lemma 1) lim δ(bf n, F B n ) = 0. n Now suppose that B is continuous, then we have BB n = By n Bz.Thus d(by n+1, y n+ ) δ(bf n, G n+1 ) δ(bf n, F B n ) + δ(f B n, G n+1 ) δ(bf n, F B n ) + {φ(d (BB n, A n+1 ), [δ q (BB n, BF n ) + δ q (BF n, F B n )] δ q (A n+1, G n+1 ), δ r (BB n, G n+1 ) [δ r (A n+1, BF n ) + δ r (BF n, F B n )], [δ s (BB n, BF n ) + δ s (BF n, F B n )] [δ s (A n+1, BF n ) + δ s (BF n, F B n )], δ l (BB n, G n+1 )δ l (A n+1, G n+1 ))} 1. Suppose Bz z. Then letting n and using Lemma 1 and Lemma we obtain d(bz, z) {φ(d (Bz, z), 0, d (Bz, z), 0, 0)} 1 {γ(d (Bz, z))} 1 < d(bz, z) a contradiction. We must therefore have Bz = z. Similarly, applying the condition () to δ(f Sz, y n+ ) δ(f Sz, GT n+1 ) and letting n, we can prove that F z = {z}, which means that z is in the range of F. Since F (X) A(X), there eist a point z in X such that Az = z. Suppose that Gz z. Then δ(z, Gz ) = δ(f z, Gz ) {φ(0, 0, 0, 0, δ (z, Gz ))} 1 {γ(δ (z, Gz ))} 1 < δ(z, Gz )

6 80 D. Turkoglu, I. Altun, B. Fisher a contradiction. We must therefore have Gz = {z}. Since (G, A) is slightly commuting, we have proving that Gz = Az. If Gz z, then δ(gz, Az) = δ(gaz, AGz ) δ(z, Gz) = δ(f z, Gz) 0, {φ(δ (z, Gz), 0, δ (z, Gz), 0, 0)} 1 {γ(δ (z, Gz))} 1 < δ(z, Gz), a contradiction and so Gz = {z} = Az. Thus we have shown that Bz = Az = F z = Gz = {z}. Hence z is a common fied point of F, G, A and B. Now suppose that F is continuous, then we have {F y n } = {F B n } {F z}. Since By n+1 BF n, the inequality () yields δ(f y n+1, G n+1 ) {φ([δ p (BF n, F B n ) + δ p (F B n, A n+1 )], [δ q (BF n, F B n ) + δ q (F B n, F y n+1 )] δ q (A n+1, G n+1 ), [δ r (BF n, F B n ) + δ r (F B n, G n+1 )] δ r (A n+1, F y n+1 ), [δ s (BF n, F B n ) + δ s (F B n, z) + δ s (z, F y n+1 )] δ s (A n+1, F y n+1 ), [δ l (BF n, F B n ) + δ l (F B n, G n+1 )] δ l (A n+1, G n+1 ))} 1. Suppose F z z. Then letting n, we obtain δ(z, F z) {φ(δ (z, F z), 0, δ (z, F z), δ (z, F z), 0)} 1 {γ(δ (z, F z))} 1 < δ(z, F z) a contradiction and so F z = {z}. Since F (X) A(X), there eists a point z in X such that Az = z. Similarly, using () on δ(gz, F n ) and letting n one can prove that Gz = {z}. Now, by the slight commutativity of (G, A) we find δ(gz, Az) = δ(gaz, AGz ) 0

7 Common fied point theorems for four mappings which gives that Gz = Az. Further, applying () to δ(f n, Gz) and letting n, we can show that Gz = {z} = Az. Since G(X) B(X) there eists a point z in X such that Bz = z. Suppose that F z z. Then δ(f z, z) = δ(f z, Gz) {φ(0, 0, 0, δ (z, F z ), 0)} 1 {γ(δ (z, F z ))} 1 < δ(z, F z ), a contradiction, implying that F z = {z}. By the slight commutativity of (F, B), we have δ(f z, Bz) = δ(f Bz, BF z ) 0, which gives that F z = Bz. Thus we have shown that F z = Gz = Bz = Az = {z}. The other cases, A is continuous and G is continuous, can be disposed of a similar argument as above. For uniqueness, suppose that w is a second distinct fied point of (F, B). Then d(w, z) = δ(f w, Gz) {φ(0, 0, d (w, z), 0, d (w, z))} 1 {γ(d (w, z))} 1 < d(w, z), a contradiction and so the fied point z is unique. Similarly, one can show that z is the unique common fied point of G and A. Theorem. Let F, G be two set-valued mappings of a complete metric space X into B(X), and A, B two self-mappings of (X, d) satisfying conditions (1), (), B is continuous or (1), (), A is continuous. If (F, B) and (G, A) are weakly commuting, then F, G, B and A have a unique common fied point in X. Theorem 3. Let F, G be two set-valued mappings of a complete metric space X into B(X), and A, B two self-mappings of (X, d) satisfying conditions (1), (), F is continuous, (F, B) and (G, A) are quasi-commuting or (1), (), G is G is continuous, (F, B) and (G, A) are quasi-commuting, then F, G, B and A have a unique common fied point in X.

8 8 D. Turkoglu, I. Altun, B. Fisher Remark 1. The conclusion of Theorems 1-3 remains valid if the condition () is replaced by δ (F, Gy) αd (B, Ay) + β ma{δ q (B, F ) δ q (Ay, Gy), δ r (B, Gy) δ r (Ay, F ), δ s (B, F ) δ s (Ay, F ), δ l (B, Gy) δ l (Ay, Gy)} ( ) for all, y X, where α > 0, β 0 with α + 4β < 1 and 0 < p, q, q, r, r, s, s, l, l 1 with = q + q = r + r = s + s = l + l. Theorem 4. Let F, G, A and B be self-mappings of a complete metric space (X, d) satisfying (1) and d (F, Gy) φ(d (B, Ay), d q (B, F ) d q (Ay, Gy), d r (B, Gy) d r (Ay, F ), d s (B, F ) d s (Ay, F ), d l (B, Gy) d l (Ay, Gy)) ( ) for all, y X, where φ Φ, 0 < p, q, q, r, r, s, s, l, l 1 such that = q+q = r+r = s+s = l+l and any one of these four mappings is continuous. If (F, B) and (G, A) are weakly commuting, then F, G, B and A have a unique common fied point in X. Remark. By Theorem 4, we get the improved version of Theorem 3.1 of Pathak-Mishra-Kalinde [5]. We now give an eample in which is used Theorem 1. Eample 1. define F = G = Let X be reals with δ induced by the Euclidean metric d and we {0} if 0 [0, 1+3 ] if 0 < 1, A = [0, 1 4 ] if > 1 {0} if 0 [0, 1+ ] if 0 < 1, B = [0, 1 3 ] if > 1 for all in X and let γ : R + R + be given by γ(t) < t 0 if 0 if 0 < 1 1 if > 1 0 if 0 if 0 < 1 1 if > 1

9 Common fied point theorems for four mappings and let φ : (R + ) 5 R + be given by 0 if t i = 0 φ(t 1, t, a 1 t 3, a t 4, a 3 t 5 ) = γ(t) if t i = t and a 1 + a + a 3 = 8 β ma {t i } otherwise for some 0 < β < 1, i = 1,, 3, 4, 5. Then for all in X. Hence F (X) A(X) and G(X) B(X). Now we eamine the following cases case 1 : if 0 and y 0, then δ (F, Gy) = 0 0 = φ(0, 0, 0, 0, 0) case : if 0 and 0 < y 1, then ( ) y δ (F, Gy) = βy = φ(y y, 0, 1 + y 1 + y, 0, y 1 + y ) case 3 : if 0 and y > 1, then δ (F, Gy) = ( ) 1 β = φ(1, 0, 1 3 3, 0, 1 3 ) case 4 : if 0 < 1 and y 0, then ( ) δ (F, Gy) = β = φ(, 0, , 1 + 3, 0) case 5 : if 0 < 1 and y > 1, then δ (F, Gy) = = ( ) 1 3 case 6 : if > 1 and y 0, then δ (F, Gy) = β(1 ) if 1 3 β(1 ) if 1 3 < 3 5 β if 3 5 { φ((1 ),, 1 3,, 1 3 ) if 1 3 φ((1 ),,,, ) if 1 3 < ( ) 1 β = φ(1, 0, 1 4 4, 1 4, 0) case 7 : if > 1 and 0 < y 1, then ( 1 ) δ 4 if y (F, Gy) = ( ) 1 y 1+y if 1 < y = β(1 y) if y 1 4 β(1 y) if 1 4 y < 3 5 βy if 3 5 y { φ((1 y), y, 1 4, 1 4, y) if y 1 4, φ((1 y), y, y, y, y) if 1 4 < y,

10 84 D. Turkoglu, I. Altun, B. Fisher case 8 : if > 1 and y > 1, then δ (F, Gy) = case 9 : if 0 < 1 and 0 < y 1, then ( 1+3) if δ (F, Gy) = ( ) y 1+y if subcase 9 1 : if φ(( y), y, y 1+y < y < 1+3 <, then 1 + 3, 1 + 3, y) = y 1+y < ( ) 1 β = φ(0, 1, 1, 1, 1) 3 { ( y 1+y ( 1+3 subcase 9 : if 1+3 < y <, then { φ(( y) β ( y), y, y, y, y) = βy y 1+y < subcase 9 3 : if 1+3 < < y, then { φ(( y) β ( y), y, y, y, y) = βy 1+3 < y subcase 9 4 : if 1+y < < y, then { φ(( y) β ( y), y, y, y, y) = βy subcase 9 5 : if φ(( y), y, 1+3 < < y 1+y < y, then y 1 + y, y, y 1 + y ) = 1+3 < y ) ( ) 1+3 ) ) < ( y 1+y β ( y) if 1+3 ( y) β 1+3 if ( y) < 1+3 { β ( y) subcase 9 6 : if 1+y < y <, then { φ(( y) β ( y), y, y, y, y) = βy and δ (F, Gy) φ(d (B, Ay), if y ( y) if ( y) < y if y ( y), if ( y) < y, if y ( y) if ( y) < y y if 1+y ( y) β y 1+y if ( y) < y 1+y if y ( y) if ( y) < y δ q (B, F ) δ q (Ay, Gy), δ r (B, Gy) δ r (Ay, F ), δ s (B, F ) δ s (Ay, F ), δ l (B, Gy) δ l (Ay, Gy))

11 Common fied point theorems for four mappings for 0 < p = q = q = r = r = s = s = l = l = 1 and = q + q = r + r = s + s = l + l. Also (F, B) and (G, A) are slightly commuting. Really, F B = GA = and i) if 0, then {0} if 0 [0, 1+3 ] if 0 < 1, BF = [0, 1 4 ] if > 1 {0} if 0 [0, 1+ ] if 0 < 1, AG = [0, 1 3 ] if > 1 0 if 0 [0, 1+3 ] if 0 < 1 [0, 1 4 ] if > 1 0 if 0 [0, 1+ ] if 0 < 1 [0, 1 3 ] if > 1 δ(f B, BF ) = 0 0 = δ(f, B) ma{δ(f, B), diamf }, δ(ga, AG) = 0 0 = δ(g, A) ma{δ(g, A), diamg} ii) if 0 < 1, then δ(f B, BF ) = δ(ga, AG) = = δ(f, B) ma{δ(f, B), diamf }, = δ(g, A) ma{δ(g, A), diamg}. 1 + iii) if > 1, then δ(f B, BF ) = 1 1 = δ(f, B) ma{δ(f, B), diamf }, 4 δ(ga, AG) = 1 1 = δ(g, A) ma{δ(g, A), diamg}. 3 Further, B and T are continuous. Then F, G, B and A have a unique common fied point in X by Theorem 1. References [1] Sessa, S., Khan, M. S., Imdad, M., A common fied point theorem with a weak commutativity condition. Glas. Math. 1 (41) (1986), [] Fisher, B., Common fied point of mappings and set-valued mappings. Rostock Math Kolloq. 18 (1981), [3] Imdad, M., Khan, M. S., Sessa, S., On some weak conditions of commutativity in common fied point theorems. Int. J. Math. Math. Sci. 11 () (1988), [4] Singh, S. P., Meade, B. A., A common fied point theorems. Bull. Austral. Math. Soc. 16 (1977), [5] Pathak, H. K., Mishra, S. N., Kalinde, A. K., Common fied point theorems with applications to nonlinear integral equations. Demonstratio Math. 3 (1999),

12 86 D. Turkoglu, I. Altun, B. Fisher [6] Imdad, M., Ahmad, A., On common fied point of mappings and set-valued mappings with some weak conditions of commutativity. Publ. Math. Debrecen. 44 (1994), [7] Sessa, S., On a weak commutativity condition of mappings in fied point considerations. Publ. Inst. Math. (Beograd) 3 (46) (198), Received by the editors August 3, 005