International Journal of Analysis and Applications ISSN 2291-8639 Volume 5, Number 2 (2014), 154-166 http://www.etamaths.com PREŠIĆ-BOYD-WONG TYPE RESULTS IN ORDERED METRIC SPACES SATISH SHUKLA1,∗ AND STOJAN RADENOVIĆ2 Abstract. The purpose of this paper is to prove some Prešić-Boyd-Wong type fixed point theorems in ordered metric spaces. The results of this paper generalize the famous results of Prešić and Boyd-Wong in ordered metric s- paces. We also initiate the homotopy result in product spaces. Some examples are provided which illustrate the results proved herein. 1. Introduction and preliminaries In 1922 Banach [26] proved the following theorem known as Banach contraction mapping theorem. Theorem 1. Let (X,d) be a complete metric space and f : X → X be a mapping such that (1) d(fx,fy) ≤ λd(x,y) for all x,y ∈ X, where 0 ≤ λ < 1, then there exists a unique x ∈ X such that fx = x. This point x is called the fixed point of mapping f. Due to simplicity and usefulness, several authors generalized the Banach contrac- tion mapping theorem. One such generalization is given by Prešić [24, 25]. Prešić generalized the Banach contraction mapping theorem in product spaces and proved the following theorem. Theorem 2. Let (X,d) be a complete metric space, k a positive integer and f : Xk → X be a mapping satisfying the following contractive type condition: (2) d(f(x1,x2, . . . ,xk),f(x2,x3, . . . ,xk+1)) ≤ k∑ i=1 qid(xi,xi+1), for every x1,x2, . . . ,xk,xk+1 ∈ X, where q1,q2, . . . ,qk are nonnegative constants such that q1 + q2 + · · · + qk < 1. Then there exists a unique point x ∈ X such that f(x,x,. . . ,x) = x. Moreover if x1,x2, . . . ,xk are arbitrary points in X and for n ∈ N, xn+k = f(xn,xn+1, . . . ,xn+k−1), then the sequence {xn} is convergent and lim xn = f(lim xn, lim xn, . . . , lim xn). 2010 Mathematics Subject Classification. 54H25, 47H10. Key words and phrases. Common fixed point; Prešić type mapping; Boyd-Wong fixed point theorem; Partial order. c©2014 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 154 PREŠIĆ-BOYD-WONG TYPE RESULTS 155 Condition (2) in the case k = 1 reduces to the condition (1). So, Theorem 1 is a generalization of the Banach fixed point theorem. The results of Prešić is useful in proving the convergence of some particular sequences and in proving the existence of solutions of differences equations, for example, see [15,24,25,40]. For more on the generalizations of Prešić type operators the reader is referred to [12,16–19,21,28–36]. On the other hand Boyd and Wong [4] generalized the Banach contraction map- ping theorem and proved the following theorem. Theorem 3. Let (X,d) be a complete metric space and f : X → X a mapping that satisfies (3) d(fx,fy) ≤ ψ(d(x,y)) for all x,y ∈ X, where ψ : R+ → R+ is upper semi-continuous function from the right (i.e., λi ↓ λ ≥ 0 ⇒ lim sup i→∞ ψ(λi) ≤ ψ(λ)) such that ψ(t) < t for each t > 0. Then f has a unique fixed point u ∈ X. Moreover, for each x ∈ X, lim n→∞ fnx = u. Note that the condition (3) in the case ψ(t) = λt reduces to condition (1). So, Theorem 3 is a generalization of the Banach fixed point theorem. Some generaliza- tion of the Boyd-Wong theorem can be found in [9, 10, 13, 14, 20, 23, 38]. The existence of fixed point in partially ordered sets was investigated by Ran and Reurings [1] and then by Nieto and Lopez [7, 8]. Some applications of fixed point theorems in ordered metric spaces to differential equations can be seen in [7, 8]. Several authors generalized the results of these papers in different directions for example, see [2, 3, 5, 6, 22, 27, 37, 39, 41]. The following version of the fixed point theorem was proved, among others, in these papers. Theorem 4. Let (X,�) be a partially ordered set and d be a metric on X such that (X,d) is a complete metric space. Let f : X → X be a nondecreasing map with respect to �. Suppose that the following conditions hold: (i) there exists k ∈ (0, 1) such that d(fx,fy) ≤ kd(x,y) for all x,y ∈ X with y � x; (ii) there exists x0 ∈ X such that x0 � fx0; (iii) if a nondecreasing sequence {xn} in X converges to x ∈ X, then xn ≤ x, for all n ∈ N. Then f has a fixed point x∗ ∈ X. Recently, in [36] Malhotra et al. defined the ordered Prešić type contraction mappings the setting of cone metric spaces (see also [19, 33]) and generalized the result of Prešić in ordered case. In the present paper, we generalize the results of Prešić, Boyd and Wong, Theorem 4 and several known results in and prove some Prešić-Boyd-Wong type fixed point theorems in ordered metric spaces. A homotopy result in the product spaces is also proved. Examples are included which illustrate the results. Following definitions will be needed in sequel. Definition 1. Let X be any nonempty set, k a positive integer and f : Xk → X be a mapping. An element x ∈ X is called a fixed point of f if f(x,x,. . . ,x) = x. 156 SHUKLA AND RADENOVIĆ Definition 2. Let X be a nonempty set, k a positive integer, f : Xk → X and g : X → X be mappings. (i) An element x ∈ X said to be a coincidence point of f and g if gx = f(x,x,. . . ,x). (ii) If w = gx = f(x,x,. . . ,x), then w is called a point of coincidence of f and g. (iii) If x = gx = f(x,x,. . . ,x), then x is called a common fixed point of f and g. (iv) Mappings f and g are said to be commuting if g(f(x,x,. . . ,x)) = f(gx,gx,. . . ,gx) for all x ∈ X. (v) Mappings f and g are said to be weakly compatible if they commute at their coincidence points. Definition 3. Let X be a nonempty set equipped with a partial order relation “ � ”, k a positive integer and f : Xk → X be a mapping. A sequence {xn} in X is said to be nondecreasing with respect to “ � ”, if xn � xn+1 for all n ∈ N. The mapping f is said to be nondecreasing with respect to “ � ” if for any finite nondecreasing sequence {xn}k+1n=1 we have f(x1,x2, . . . ,xk) � f(x2,x3, . . . ,xk+1). Let g : X → X be a mapping. f is said to be g-nondecreasing with respect to “ � ” if for any finite nondecreasing sequence {gxn}k+1n=1 we have f(x1,x2, . . . ,xk) � f(x2,x3, . . . ,xk+1). Note that for k = 1, above definitions reduce to usual definitions of nondecreasing and g-nondecreasing mappings. Definition 4. Let X be a nonempty set equipped with a partial order relation “ � ”, and g : X → X be a mapping. A nonempty subset A of X is said to be well ordered if every two elements of A are comparable. The elements a,b ∈ A are called g- comparable if ga and gb are comparable. The set A is called g-well ordered if for all a,b ∈A, a and b are g-comparable i.e. ga and gb are comparable. Example 1. Let X = [0,∞), A = [0, 1] and define a relation “ � ” on X by x � y ⇔{(x = y) or (x,y ∈ [0, 1 2 ] with x ≤ y)}. Then � is a partial order relation on X. Define g : X → X by gx = x 2 for all x,y ∈ X. Note that A is not well ordered. Indeed if x,y ∈ ( 1 2 ,∞),x 6= y then neither x � y nor y � x. But g(A) = [0, 1 2 ], therefore A is g-well ordered. Let (X,d) be a metric space equipped with partial order relation “ � ”, then (X,�,d) is called an ordered metric space. Let k be a positive integer and f : Xk → X be a mapping. f is called ordered Prešić contraction if (4) d(f(x1,x2 . . . ,xk),f(x2,x3, . . . ,xk+1)) ≤ k∑ i=1 αid(xi,xi+1), for all x1,x2, . . . ,xk,xk+1 ∈ X with x1 � x2 � ··· � xk � xk+1, where αi are nonnegative constants such that k∑ i=1 αi < 1. If (5) is satisfied for all x1,x2, . . . ,xk,xk+1 ∈ X, then f is called Prešić contraction. Note that in ordered metric spaces a Prešić contraction is necessarily an ordered PREŠIĆ-BOYD-WONG TYPE RESULTS 157 Prešić contraction, but converse may not be true (see examples 3.1 and 3.2 of [36]). Let ψ : Rk+ → R+ be a function satisfying the following conditions: (1) for tn ∈ R+, n ∈ N and tn ↓ t ≥ 0 implies lim sup n→∞ ψ(tn, tn, . . . , tn) ≤ ψ(t,t, . . . , t); (2) ψ(t,t, . . . , t) < t for each t > 0; (3) ψ(t, 0, . . . , 0) + ψ(0, t, 0, . . . , 0) + · · · + ψ(0, . . . , 0, t) ≤ ψ(t,t, . . . , t) for each t ∈ R+. We denote the class of all such functions by Ψ, i.e., ψ ∈ Ψ if and only if ψ satisfies the all above conditions. Example 2. Let ψ : Rk+ → R+ be defined by ψ(t1, t2, . . . , tk) = k∑ i=1 αiti, where αi are nonnegative constants such that k∑ i=1 αi < 1. Then ψ ∈ Ψ. Mapping f : Xk → X is said to be an ordered Prešić-Boyd-Wong contraction if (5) d(f(x1,x2, . . . ,xk),f(x2,x3, . . . ,xk+1)) ≤ ψ(d(x1,x2),d(x2,x3), . . . ,d(xk.xk+1)) for all x1,x2, . . . ,xk,xk+1 ∈ X with x1 � x2 � ··· � xk � xk+1, where ψ ∈ Ψ. If (5) is satisfied for all x1,x2, . . . ,xk,xk+1 ∈ X, then f is called Prešić-Boyd-Wong contraction. Now we can state our main results. 2. Main results Theorem 5. Let (X,�,d) be a complete ordered metric space, k a positive integer. Let f : Xk → X, g : X → X be two mappings such that f(Xk) ⊂ g(X) and g(X) is a closed subspace of X. Suppose following conditions hold: (I) (6) d(f(x1,x2, . . . ,xk),f(x2,x3, . . . ,xk+1)) ≤ ψ(d(gx1,gx2),d(gx2,gx3), . . . ,d(gxk,gxk+1)), for all x1,x2, . . . ,xk,xk+1 ∈ X with gx1 � gx2 � ···� gxk � gxk+1, where ψ ∈ Ψ; (II) there exist x1 ∈ X such that gx1 � f(x1,x1, . . . ,x1); (III) f is g-nondecreasing; (IV) if a nondecreasing sequence {gxn} converges to gu ∈ X, then gxn � gu for all n ∈ N and gu � ggu. Then f and g have a point of coincidence. If in addition f and g are weakly compatible, then f and g have a common fixed point v ∈ X. Moreover, the set of common fixed points of f and g is g-well ordered if and only if f and g have a unique common fixed point. Proof. Starting with given x1 ∈ X, we define a sequence {yn} as follows: let y1 = gx1, as f(X k) ⊂ g(X) and gx1 � f(x1,x1, . . . ,x1), define yn+1 = gxn+1 = f(xn,xn, . . . ,xn), n ∈ N. Then gx1 � gx2 i.e. y1 � y2 and f is g-nondecreasing, so y2 = f(x1,x1, . . . ,x1) � f(x1,x1, . . . ,x1,x2) � f(x1,x1, . . . ,x1,x2,x2) � ···� f(x1,x2, . . . ,x2) � f(x2,x2, . . . ,x2) = gx3 = y3 158 SHUKLA AND RADENOVIĆ i.e. y2 = gx2 � y3 = gx3. Continuing this procedure, we obtain gx1 � gx2 � ···� gxn � gxn+1 � ··· , i.e. y1 � y2 � ···� yn � yn+1 � ··· . Thus {yn} = {gxn} is a nondecreasing sequence with respect to “ � ”. For simplicity set dn = d(yn,yn+1),n ∈ N. We may assume that dn > 0 for all n ∈ N, otherwise coincidence point and point of coincidence of f and g exist trivially. We shall show that lim n→∞ dn = 0. Note that dn+1 = d(yn+1,yn+2) = d(f(xn,xn, . . . ,xn),f(xn+1,xn+1, . . . ,xn+1)) ≤ d(f(xn,xn, . . . ,xn),f(xn, . . . ,xn,xn+1)) +d(f(xn, . . . ,xn,xn+1),f(xn, . . . ,xn,xn+1,xn+1)) + · · · + d(f(xn,xn+1, . . . ,xn+1),f(xn+1, . . . ,xn+1)) and gxn � gxn+1, ψ ∈ Ψ, so it follows from (6) that dn+1 ≤ ψ(0, . . . , 0,d(gxn,gxn+1)) + ψ(0, . . . , 0,d(gxn,gxn+1), 0) + · · · + ψ(d(gxn,gxn+1), 0, . . . , 0) = ψ(0, . . . , 0,dn) + ψ(0, . . . , 0,dn, 0) + · · · + ψ(dn, 0, . . . , 0) ≤ ψ(dn,dn, . . . ,dn) < dn, for all n ∈ N. Therefore {dn} is a monotonic nondecreasing sequence and bounded below, so lim n→∞ dn exists. Let lim n→∞ dn = δ ≥ 0. Assume δ > 0, then as ψ ∈ Ψ we obtain δ = lim n→∞ dn+1 ≤ lim n→∞ ψ(dn,dn, . . . ,dn) ≤ ψ(δ,δ, . . . ,δ) < δ, a contradiction, so δ = 0. We shall show that {yn} is Cauchy sequence. Assume that {yn} is not Cauchy, then there exists � > 0 and integers ml,nl, l ∈ N such that ml > nl ≥ l and d(ynl,yml ) ≥ � for l ∈ N. Also, choosing ml as small as possible, it may be assumed that d(yml−1,ynl ) < �. So for each l ∈ N, we have � ≤ d(yml,ynl ) ≤ d(yml,yml−1) + d(yml−1,ynl ) ≤ dml−1 + � PREŠIĆ-BOYD-WONG TYPE RESULTS 159 and it follows from the fact lim n→∞ dn = 0 that lim l→∞ d(yml,ynl ) = �. Observe that � ≤ d(yml,ynl ) ≤ d(yml,yml+1) + d(yml+1,ynl+1) + d(ynl+1,ynl ) = dml + dnl + d(f(xnl, . . . ,xnl ),f(xml, . . . ,xml )) ≤ dml + dnl + d(f(xnl, . . . ,xnl ),f(xnl, . . . ,xnl,xml )) +d(f(xnl, . . . ,xnl,xml ),f(xnl, . . . ,xnl,xml,xml )) + · · · + d(f(xnl,xml, . . . ,xml ),f(xml, . . . ,xml )). As ml > nl and {yn} is nondecreasing with respect to “ � ”, so ynl � yml i.e., gxnl � gxml, therefore it follows from (6) and the above inequality that � ≤ d(yml,ynl ) ≤ dml + dnl + ψ(0, . . . , 0,d(ynl,yml )) + ψ(0, . . . , 0,d(ynl,yml ), 0) + · · · + ψ(d(ynl,yml ), 0, . . . , 0) ≤ dml + dnl + ψ(d(ynl,yml ), . . . ,d(ynl,yml )). Letting l →∞ and using the facts that lim n→∞ dn = 0 and ψ ∈ Ψ, we have � = lim l→∞ d(yml,ynl ) ≤ lim l→∞ ψ(d(yml,ynl ), . . . ,d(yml,ynl )) ≤ ψ(�, . . . ,�) < �, which is a contradiction. Therefore {yn} = {gxn} is a Cauchy sequence in g(X). As g(X) is closed, there exist u,v ∈ X such that v = gu and (7) lim n→∞ yn = lim n→∞ gxn = gu = v. We shall show that u is a coincidence point and v is a point of coincidence of f and g. Note that d(v,f(u,u,. . . ,u)) ≤ d(v,yn+1) + d(yn+1,f(u,u,. . . ,u)) = d(v,yn+1) + d(f(xn,xn, . . . ,xn),f(u,u,. . . ,u)) ≤ d(v,yn+1) + d(f(xn,xn, . . . ,xn),f(xn, . . . ,xn,u)) +d(f(xn, . . . ,xn,u)),f(xn, . . . ,xn,u,u)) + · · · + d(f(xn,u, . . . ,u)),f(u,. . . ,u)). If v 6= f(u,u,. . . ,u), then by (IV) we have gxn � gu,gu � ggu, so using (6) it follows from the above inequality that d(v,f(u,u,. . . ,u)) ≤ d(v,yn+1) + ψ(0, . . . , 0,d(gxn,gu)) + ψ(0, . . . , 0,d(gxn,gu), 0) + · · · + ψ(d(gxn,gu), 0, . . . , 0) ≤ d(v,yn+1) + ψ(d(gxn,gu), . . . ,d(gxn,gu)) < d(v,yn+1) + d(gxn,gu), letting n →∞ and using (7) we obtain d(v,f(u,u,. . . ,u)) = 0 i.e. f(u,u,. . . ,u) = gu = v. Thus u is a coincidence point and v is a point of coincidence of f and g. Suppose f and g are weakly compatible, so f(v,v, . . . ,v) = f(gu,gu,. . . ,gu) = g(f(u,u,. . . ,u)) = gv. 160 SHUKLA AND RADENOVIĆ Again if d(gu,gv) > 0 then as gu � ggu = gv we obtain from (6) that d(v,f(v,v, . . . ,v)) = d(f(u,u,. . . ,u),f(v,v, . . . ,v)) ≤ d(f(u,u,. . . ,u),f(u,. . . ,u,v)) + d(f(u,. . . ,u,v),f(u,. . . ,u,v,v)) + · · · + d(f(u,v, . . . ,v),f(v, . . . ,v)) ≤ ψ(0, . . . , 0,d(gu,gv)) + ψ(0, . . . , 0,d(gu,gv), 0) + · · · + ψ(d(gu,gv), 0, . . . , 0) ≤ ψ(d(gu,gv), . . . ,d(gu,gv)) < d(gu,gv) = d(v,f(v,v, . . . ,v)), a contradiction, therefore v = gv = f(v,v, . . . ,v). Thus v is the common fixed point of f and g. Suppose the set of common fixed points of f and g is g-well ordered. We shall show that common fixed point is unique. Assume on contrary that v′ is another common fixed point of f and g i.e. v′ = gv′ = f(v′,v′, . . . ,v′) and v 6= v′. As v and v′ are g-comparable, let e.g. gv � gv′. From (6), it follows that d(v,v′) = d(f(v,v, . . . ,v),f(v′,v′, . . . ,v′)) ≤ d(f(v,v, . . . ,v),f(v, . . . ,v,v′)) + d(f(v, . . . ,v,v′),f(v, . . . ,v,v′,v′)) + · · · + d(f(v,v′, . . . ,v′),f(v′,v′, . . . ,v′)) ≤ ψ(0, . . . , 0,d(gv,gv′)) + ψ(0, . . . , 0,d(gv,gv′), 0) + · · · + ψ(d(gv,gv′), 0, . . . , 0) ≤ ψ(d(gv,gv′), . . . ,d(gv,gv′)) < d(v,v′) a contradiction. Therefore, v = v′, i.e., the common fixed point is unique. For converse, if common fixed point of f and g is unique then the set of common fixed points of f and g being singleton therefore g-well ordered. � Remark 1. For k = 1 the above theorem is a generalization and extension of result of Boyd and Wong in ordered metric spaces. Following is a simple example which illustrate the above result. Example 3. Let X = [0,∞) with the usual metric and partial order �= {(x,y) : x,y ∈ X,y ≤ x}. For k = 2, define f : X2 → X and g : X → X by f(x1,x2) = x1 + x2 3 + x1 + x2 for all x1,x2 ∈ X and gx = x for all x ∈ X. Define ψ : R2+ → R by ψ(t1, t2) = t1 + t2 3 + |t1 − t2| for all t1, t2 ∈ R+. Then it easy to see that all the conditions of Theorem 5 are satisfied and 0 is the unique common fixed point of f and g in X. Taking g = IX i.e. identity mapping of X in Theorem 5, we get the following fixed point result for ordered Prešić-Boyd-Wong contraction. Corollary 6. Let (X,�,d) be a complete ordered metric space, k a positive integer. Let f : Xk → X be a mapping such that the following conditions hold: (I) f is ordered Prešić-Boyd-Wong contraction; (II) there exist x1 ∈ X such that x1 � f(x1,x1, . . . ,x1); (III) f is nondecreasing; PREŠIĆ-BOYD-WONG TYPE RESULTS 161 (IV) if a nondecreasing sequence {xn} converges to u ∈ X, then xn � u for all n ∈ N. Then f has a fixed point v ∈ X. Moreover, the set of fixed points of f is well ordered if and only if f has a unique fixed point. The following example illustrate the case when the known results are not appli- cable but the Corollary 6 of this paper is applicable. Example 4. Let X = [0, 2] and d is the usual metric on X, then (X,d) is a complete metric space. For k = 2, define a mapping f : X2 → X by f(x,y) =   x 1 + x , if (x,y) ∈ [0, 1) × [0, 1) ∪ [1, 2] × [0, 1); y 1 + y , if (x,y) ∈ [0, 1) × [1, 2]; 0, otherwise, and a function ψ : Rk+ → R+ by ψ(t1, t2) = t1 1 + |t1/2 − t2| for all t1, t2 ∈ R+. Let � be a partial order define on X by � = {(x,y) : (x,y) ∈ [0, 1) × [0, 1) with y ≤ x}∪{(x,y) : (x,y) ∈ [1, 2] × (0, 1)} ∪{(x,x) : x ∈ X}, then ψ ∈ Ψ. Now by careful calculations one can see that all the conditions of Corollary 6 are satisfied and 0 is the unique fixed point of f. Note that, f is not an ordered Prešić type contraction, therefore it is not a Prešić type contraction. To see this, take arbitrary values x,y = z ∈ [0, 1) and then condition (5) is not satisfied. Following theorem is a generalization of the result of Prešić and Boyd and Wong in metric spaces. Theorem 7. Let (X,d) be a complete metric space, k a positive integer. Let f : Xk → X, g : X → X be two mappings such that f(Xk) ⊂ g(X) and g(X) is a closed subspace of X. Suppose following conditions hold: (8) d(f(x1,x2, . . . ,xk),f(x2,x3, . . . ,xk+1)) ≤ ψ(d(gx1,gx2),d(gx2,gx3), . . . ,d(gxk,gxk+1)), for all x1,x2, . . . ,xk,xk+1 ∈ X, where ψ ∈ Ψ. Then f and g have a point of coincidence. If in addition f and g are weakly compatible, then f and g have a unique common fixed point v ∈ X. Proof. We note that the inequality (8) is true for all x1,x2, . . . ,xk,xk+1 ∈ X, therefore the proof of theorem follows from similar process as used in the proof of Theorem 5. � Taking g = IX i.e. identity mapping of X in Theorem 7, we get the following fixed point result for Prešić-Boyd-Wong contraction. Corollary 8. Let (X,d) be a complete metric space, k a positive integer. Let f : Xk → X be a Prešić-Boyd-Wong contraction. Then f has a unique fixed point v ∈ X. 162 SHUKLA AND RADENOVIĆ Remark 2. Note that, for ψ(t1, t2, . . . , tk) = k∑ i=1 αiti, where αi are nonnegative constants such that k∑ i=1 αi < 1, Corollary 8 reduces to the Prešić theorem. 3. A homotopy result In this section we prove a homotopy result for Prešić type mapping on product space. Theorem 9. Let (X,d) be any complete metric space, U an open subset of X. Suppose H : (U)k × [0, 1] → X be a function such that the following conditions hold: (i) for every x ∈ ∂U(here ∂U is the boundary of U) and λ ∈ [0, 1], x 6= H(x,x,. . . ,x,λ); (ii) for all x1,x2, . . . ,xk,xk+1 ∈ U and λ ∈ [0, 1] (9) d(H(x1,x2, . . . ,xk,λ),H(x2,x3, . . . ,xk+1,λ)) ≤ k∑ i=1 αid(xi,xi+1), where αi are nonnegative constants such that k∑ i=1 αi < 1 k ; (iii) for all x1,x2, . . . ,xk ∈ U and λ,µ ∈ [0, 1] there exists M ≥ 0 such that (10) d(H(x1,x2, . . . ,xk,λ),H(x1,x2, . . . ,xk,µ)) ≤ M|λ−µ|. If Hλ=λ′ has a fixed point in U for at least one λ ′ ∈ [0, 1], then Hλ has a fixed point in U for all λ ∈ [0, 1]. Furthermore, for any fixed λ ∈ [0, 1], the fixed point of Hλ is unique. Proof. Define F = {λ ∈ [0, 1] : x = H(x,x,. . . ,x,λ) for some x ∈ U}. As Hλ=λ′ for at least one λ ′ ∈ [0, 1], has a fixed point in U, i.e., there exists x ∈ U such that H(x,x,. . . ,x,λ′) = x, so λ′ ∈ F and Fx 6= ∅. We shall show that F is both open and closed in [0, 1] and therefore by connectedness F = [0, 1]. (I) F is closed: Let {λn} be any sequence in F and lim n→∞ λn = λ ∈ [0, 1]. As λn ∈ F for all n ∈ N so there exists xn ∈ U such that xn = H(xn,xn, . . . ,xn,λn) for all n ∈ N. Note that, for all n,m ∈ N with m > n we have d(xn,xm) = d(H(xn,xn, . . . ,xn,λn),H(xm,xm, . . . ,xm,λm)) ≤ d(H(xn, . . . ,xn,λn),H(xn, . . . ,xn,xm,λn)) +d(H(xn, . . . ,xn,xm,λn),H(xn, . . . ,xn,xm,xm,λn)) + · · · + d(H(xn,xm, . . . ,xm,λn),H(xm, . . . ,xm,λn)) +d(H(xm, . . . ,xm,λn),H(xm, . . . ,xm,λm)). Using (9) and (10) it follows that d(xn,xm) ≤ αkd(xn,xm) + αk−1d(xn,xm) + · · · + α1d(xn,xm) + M|λn −λm| = [ k∑ i=1 αi]d(xn,xm) + M|λn −λm| PREŠIĆ-BOYD-WONG TYPE RESULTS 163 i.e. d(xn,xm) ≤ M 1 − k∑ i=1 αi |λn −λm|. Letting n → ∞ and using the fact that lim n→∞ λn = λ it follows from the above inequality that lim n→∞ d(xn,xm) = 0, therefore {xn} is a Cauchy sequence. As X is complete, there exists u ∈ U such that lim n→∞ xn = u. Now for any n ∈ N we obtain d(xn,H(u,u,. . . ,u,λ)) = d(H(xn,xn, . . . ,xn,λn),H(u,u,. . . ,u,λ)) ≤ d(H(xn, . . . ,xn,λn),H(xn, . . . ,xn,u,λn)) +d(H(xn, . . . ,xn,u,λn),H(xn, . . . ,xn,u,u,λn)) + · · · + d(H(xn,u, . . . ,u,λn),H(u,. . . ,u,λn)) +d(H(u,u,. . . ,u,λn),H(u,u,. . . ,u,λ)), using (9) and (10) it follows that d(xn,H(u,u,. . . ,u,λ)) ≤ αkd(xn,u) + αk−1d(xn,u) + · · · + α1d(xn,u) + M|λn −λ| = [ k∑ i=1 αi]d(xn,u) + M|λn −λ|. As lim n→∞ λn = λ and lim n→∞ xn = u, we obtain lim n→∞ d(xn,H(u,u,. . . ,u,λ)) = d(u,H(u,u,. . . ,u,λ)) = 0 i.e. u = H(u,u,. . . ,u,λ) and u ∈ U. As (i) holds, therefore u ∈ U so λ ∈F. Thus F is closed. (II) F is open: Let λ0 ∈F, then there exists u0 ∈ U such that u0 = H(u0, . . . ,u0,λ0). As U is open, there exists δ > 0 such that B(u0,δ) = {x ∈ X : d(x,u0) < δ} ⊂ U. Fix � > 0 with (11) � < 1 −k k∑ i=1 αi M δ. Let λ ∈ (λ0 −�,λ0 + �), then for all x1,x2, . . . ,xk ∈ B(u0,δ) = {x ∈ X : d(x,u0) ≤ δ}, we have d(u0,H(x1,x2, . . . ,xk,λ)) = d(H(u0,u0, . . . ,u0,λ0),H(x1,x1, . . . ,xk,λ)) ≤ d(H(u0,u0, . . . ,u0,λ0),H(u0, . . . ,u0,x1,λ0)) +d(H(u0, . . . ,u0,x1,λ0),H(u0, . . . ,u0,x1,x2,λ0)) + · · · + d(H(u0,x1, . . . ,xk−1,λ0),H(x1, . . . ,xk,λ0)) +d(H(x1,x2, . . . ,xk,λ0),H(x1,x2, . . . ,xk,λ)). 164 SHUKLA AND RADENOVIĆ It follows from (9), (10) and the above inequality that d(u0,H(x1,x2, . . . ,xk,λ)) ≤ [ k∑ i=1 αi]d(u0,x1) + [ k∑ i=2 αi]d(x1,x2) + · · · +[ k∑ i=k−1 αi]d(xk−2,xk−1) + αkd(xk−1,xk) + M|λ0 −λ| < [k k∑ i=1 αi]δ + M�. Using (11) in the above inequality we obtain d(u0,H(x1,x2, . . . ,xk,λ)) < [k k∑ i=1 αi]δ + [1 −k k∑ i=1 αi]δ = δ. Therefore d(u0,H(x1,x2, . . . ,xk,λ)) < δ i.e. H(x1,x2, . . . ,xk,λ) ∈ B(u0,δ). Thus, for each fixed λ ∈ (λ0 − �,λ0 + �), Hλ is a self map of B(u0,δ), so we can apply Corollary 8 and Remark 2, to deduce that Hλ has a fixed point in U, and as (i) holds, this fixed point must be in U. Thus λ ∈F for all λ ∈ (λ0 − �,λ0 + �). Therefore F is open in [0, 1] i.e. F = [0, 1]. Thus Hλ has a fixed point in U for all λ ∈ [0, 1]. For uniqueness, let λ ∈ [0, 1] be fixed and for this fixed λ, u and v be two fixed points of Hλ in U i.e. u = H(u,u,. . . ,u,λ) and v = H(v,v, . . . ,v,λ) and u 6= v. Then it follows from (9) that d(u,v) = d(H(u,u,. . . ,u,λ),H(v,v, . . . ,v,λ)) ≤ d(H(u,. . . ,u,λ),H(u,. . . ,u,v,λ)) + d(H(u,. . . ,u,v,λ),H(u,. . . ,u,v,v,λ)) + · · · + d(H(u,v, . . . ,v,λ),H(v,v, . . . ,v,λ)) ≤ αkd(u,v) + αk−1d(u,v) + · · · + α1d(u,v) = [ k∑ i=1 αi]d(u,v) < d(u,v), a contradiction. Thus fixed point is unique. � For k = 1 in the above theorem, we obtain following Homotopy result. Corollary 10. Let (X,d) be any complete metric space, U an open subset of X. 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