A Continuation Method for Generalized Contractive Mappings and Fixed Point Properties of Generalized Expansive Mappings

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1 Λ46ΦΛ1ff fl Π Vol. 46, No Ω1» ADVANCES IN MATHEMATICS (CHINA) Jan., 2017 doi: /sxjz b A Continuation Method for Generalized Contractive Mappings and Fixed Point Properties of Generalized Expansive Mappings WANG Hongyong (School of Applied Mathematics, Nanjing University of Finance and Economics, Nanjing, Jiangsu, , P. R. China) Abstract: A continuation method for a class of generalized contractive mappings is given, which establishes a homotopy theorem for a class of non-self mappings in a complete metric space. In addition, a type of generalized expansive mappings is introduced and the corresponding fixed point theorem is proved. Finally, the analytical properties of the fixed points of such a generalized expansive mapping are investigated and it is proved that the set of such fixed points is uncountable and closed in a Banach space. Keywords: continuation method; generalized contractive mapping; fixed point; generalized expansive mapping MR(2010) Subject Classification: 47H09; 54H25 / CLC number: O Document code: A Article ID: (2017) Introduction A self-mapping T of a metric space (X, d) is called contractive if there exists α [0, 1) such that d(tx,ty) αd(x, y) for all x, y X. The well-known Banach fixed point theorem says that T has a fixed point if X is complete. For many years, the concept of contractive mapping and the Banach fixed point theorem have been widely generalized. A more general extension of the contractive mapping is presented as follows. Let (X, d) be a complete metric space and T a self-mapping on X with the following property d(tx,ty) α 1 d(x, y)+α 2 d(x, T x)+α 3 d(y, Ty) + α 4 d(x, T y)+α 5 d(y, Tx), x,y X, (0.1) where α i,i=1, 2,, 5, are non-negative real numbers and satisfy 5 i=1 α i < 1. Various types of fixed point theorems concerning this type of mapping T have been established. For instance, Kannan [6] proved that T has a unique fixed point if α 1 = α 4 = α 5 =0andα 2 = α 3. Reich [10] proved that T has a unique fixed point provided that α 4 = α 5 = 0. Hardy and Rogers in [4] proved that T has a unique fixed point if α 2 = α 3 and α 4 = α 5, and further they extended Reich s result to allow α i,i=1, 2,, 5, to be monotonically decreasing functions of d(x, y). Wong [13] proved that T has a unique fixed point under the assumptions that there exist non-negative upper semicontinuous functions α i,i=1, 2,, 5, such that α 2 = α 3,α 4 = α 5, 5 i=1 α i(t) <t for all t>0 and (0.1) is satisfied with α i replaced by αi(d(x,y)) d(x,y). Considering the difficulty to find Received date: Revised date: Foundation item: Supported by NSFC (No ). 98wanghy@163.com

2 112 ADVANCES IN MATHEMATICS (CHINA) Vol. 46 the required α i in applications of the above result, Wong in [14] improved the above conditions by replacing each α i in (0.1) with a symmetric function α i (x, y) ofx X into [0, ) instead of a composite function of d with any function on the real line and gave the corresponding fixed point theorem. Some other variants of the mapping T have also been taken into account and the corresponding fixed point results are presented, see for example, Janos [5], Pathak et al. [8], Agarwal and O Regan [1], etc. In addition, Granas [3] first proposed a continuation method for contractive mappings in the setting of a metric space. Later, Agarwal and O Regan [1] gave general continuation type theorems for a class of generalized contractive homotopies on spaces with two metrics. Their theorems extended previous results of Granas [3], O Regan [7] and Precup [9]. Recently, a continuation method for a class of weakly Kannan mappings was also given by Ariza-Ruiz and Jiménez-Melado [2]. In this paper, we first establish a new local version for a class of generalized contractive mappings considered by Wong [14] and then give a continuation method for this class of mappings. We remark that a condition in our continuation method weakens that of Ariza-Ruiz and Jiménez- Melado [2]. These works are done in Section 1. In Section 2, we first introduce the concept of a class of generalized expansive mappings and then give a fixed point theorem for the class of generalized expansive mappings. Finally, we investigate some properties of the fixed points of such a generalized expansive mapping and prove that the set of such fixed points is an uncountable and closed set in a Banach space. 1 A Continuation Method for Generalized Contractive Mappings In [14], Wong studied a class of generalized contractive mappings and gave the following fixed point result. Proposition 1.1 Let T be a self-mapping on a complete metric space (X, d). Suppose that there exist functions α i, i =1, 2,, 5, of X X into [0, ) such that (a) r sup{ 5 i=1 α i(x, y) :x, y X} < 1; (b) α 2 = α 3, α 4 = α 5 ; (c) for any distinct x, y in X, d(tx,ty) α 1 d(x, y)+α 2 d(x, T x)+α 3 d(y, Ty)+α 4 d(x, T y)+α 5 d(y, Tx), where α i = α i (x, y). Then T has a unique fixed point. On the other hand, for a class of generalized contractions T : X X, where (X, d) isa complete metric space and T satisfies the following properties with constant q (0, 1): d(tx,ty) q max {d(x, y),d(x, T x),d(y, Ty), 12 } (d(x, T y)+d(y, Tx)) for all x, y X, Agarwal and O Regan in [1] gave the following homotopy result. Proposition 1.2 Let (X, d) be a complete metric space. Let Q X be closed and let U X be open and U Q. Suppose that H : Q [0, 1] X satisfies the following three properties: (i) x H(x, λ) forx Q \ U and λ [0, 1];

3 No. 1 Wang H. Y.: A Continuation Method for Generalized Contractive Mappings 113 (ii) There exists q (0, 1) such that for all λ [0, 1] and x, y Q we have { d(h(x, λ),h(y, λ)) q max d(x, y),d(x, H(x, λ)),d(y, H(y, λ)), } 1 (d(x, H(y, λ)) + d(y, H(x, λ))) ; 2 (iii) H(x, λ) is continuous in λ, uniformly for x Q. In addition assume that H(, 0) has a fixed point. Then for each λ [0, 1], H(,λ)alsohas afixedpointinu. In this section, we show that a similar result is true for the class of generalized contractive mappings considered by Wong [14]. For this, we first establish a local version of Proposition 1.1. Let us denote by B(x 0,δ) the open ball centered at x 0 X with radius δ > 0andby B(x 0,δ) the closure of B(x 0,δ). Lemma 1.1 Suppose that (X, d) is a complete metric space. Let x 0 X, δ > 0and T : B(x 0,δ) X be the mapping satisfying the conditions (a), (b) and (c) of Proposition 1.1 with the associated functions α i, i =1, 2,, 5, and all x, y B(x 0,δ). If d(x 0,Tx 0 ) (2 r)(1 r) δ, (1.1) 2+r then T has a fixed point in B(x 0,δ). Proof According to Proposition 1.1, we only need to show that the closed ball B(x 0,δ) is invariant under T. For any x B(x 0,δ), using the condition (c) of Proposition 1.1 with α i = α i (x 0,x), we have d(x 0,Tx) d(x 0,Tx 0 )+d(tx 0,Tx) (1 + α 2 )d(x 0,Tx 0 )+α 1 d(x 0,x)+α 3 d(x, T x)+α 4 d(x 0,Tx)+α 5 d(x, T x 0 ). Using the triangle inequalities that d(x, T x) d(x, x 0 )+d(x 0,Tx)andd(x, T x 0 ) d(x, x 0 )+ d(x 0,Tx 0 ) and simplifying the above inequality, we obtain d(x 0,Tx) (1 + α 2 + α 5 )d(x 0,Tx 0 )+(α 1 + α 3 + α 5 )d(x 0,x)+(α 3 + α 4 )d(x 0,Tx). Hence, d(x 0,Tx) 1+α 2 + α 5 d(x 0,Tx 0 )+ α 1 + α 3 + α 5 d(x 0,x). 1 α 3 α 4 1 α 3 α 4 From the conditions (a) and (b), we deduce that α 2 + α 4 r 2, and hence So we can write 1+α 2 + α 5 = 1+α { 2 + α 4 1+t max 1 α 3 α 4 1 α 2 α 4 α 1 + α 3 + α 5 1 α 3 α 4 r α 2 α 4 1 α 2 α 4 max [0, 1 t : t r ] } 2+r 2 2 r, { r t [0, 1 t : t r ] } r. 2 d(x 0,Tx) 2+r 2 r d(x 0,Tx 0 )+rd(x 0,x). Noting that d(x 0,x) δ and the hypothesis (1.1), it follows that d(x 0,Tx) δ, thus implying that Tx B(x 0,δ). Consequently, by Proposition 1.1, T has a fixed point in B(x 0,δ).

4 114 ADVANCES IN MATHEMATICS (CHINA) Vol. 46 In the following we give a continuation type theorem for the generalized contractive mapping T described in Proposition 1.1. Theorem 1.1 Let (X, d) be a complete metric space and U be an open subset of X. Suppose that H : U [0, 1] X satisfies the following conditions: (i) x H(x, λ) for all x U and all λ [0, 1]; (ii) There exist functions α i = α i (x, y), i=1, 2,, 5, from U U into [0, 1) with α 2 = α 3 and α 4 = α 5 such that r sup{ 5 i=1 α i(x, y) :x, y U} < 1andforallx, y U, λ [0, 1], d(h(x, λ),h(y, λ)) α 1 d(x, y)+α 2 d(x, H(x, λ)) + α 3 d(y, H(y, λ)) (1.2) + α 4 d(x, H(y, λ)) + α 5 d(y, H(x, λ)); (iii) H(x, λ) is continuous in λ, uniformly for x U. If H(, 0) has a fixed point in U, then H(,λ)alsohasafixedpointinU for all λ [0, 1]. Proof We follow the ideas from the proof of [1, Theorem 3.1]. Consider the set A = {λ [0, 1] : x = H(x, λ) forsomex U}. Since H(, 0) has a fixed point in U, it means that 0 A, soa is nonempty. We will prove that A =[0, 1], and for this we only need to show that A is both closed and open in [0, 1]. Now we proceed to show that A is closed in [0, 1]. Let {λ n } be a sequence in A that converges to λ [0, 1]. By the definition of A, there exists a sequence {x n } in U such that x n = H(x n,λ n ). We will prove that {x n } converges to a point x 0 U with x 0 = H(x 0,λ), thus showing that λ A. Noting the fact that x n = H(x n,λ n ) for all n 1 and using (1.2) with α i = α i (x n,x m ) and λ = λ m,wehave d(x n,x m )=d(h(x n,λ n ),H(x m,λ m )) d(h(x n,λ n ),H(x n,λ m )) + d(h(x n,λ m ),H(x m,λ m )) d(h(x n,λ n ),H(x n,λ m )) + α 1 d(x n,x m )+α 2 d(x n,h(x n,λ m )) + α 4 d(x n,h(x m,λ m )) + α 5 d(x m,h(x n,λ m )). (1.3) Substituting the inequalities d(x n,h(x m,λ m )) d(x n,x m )+d(x m,h(x m,λ m )) = d(x n,x m ) and d(x m,h(x n,λ m )) d(x m,x n )+d(x n,h(x n,λ m )) into the right-hand side of (1.3) and simplifying it, we obtain Hence, d(x n,x m ) (1 + α 2 + α 5 )d(h(x n,λ n ),H(x n,λ m )) + (α 1 + α 4 + α 5 )d(x n,x m ). d(x n,x m ) 1+α 2 + α 5 d(h(x n,λ n ),H(x n,λ m )) 1 α 1 α 4 α 5 1+r 1 r [d(h(x n,λ n ),H(x n,λ)) + d(h(x n,λ),h(x n,λ m ))], from which, Condition (iii) can ensure that {x n } is a Cauchy sequence. So there exists an x 0 U such that x n x 0 as n. We claim that x 0 U and x 0 = H(x 0,λ). In fact, having in mind that x n = H(x n,λ n ) for all n 1, we have d(x 0,H(x 0,λ)) d(x 0,x n )+d(h(x n,λ n ),H(x 0,λ)) d(x 0,x n )+d(h(x n,λ n ),H(x 0,λ n )) + d(h(x 0,λ n ),H(x 0,λ)).

5 No. 1 Wang H. Y.: A Continuation Method for Generalized Contractive Mappings 115 Using (1.2) with α i = α i (x n,x 0 )gives d(x 0,H(x 0,λ)) (1 + α 1 )d(x 0,x n )+α 3 d(x 0,H(x 0,λ n )) + α 4 d(x n,h(x 0,λ n )) + α 5 d(x 0,H(x n,λ n )) + d(h(x 0,λ n ),H(x 0,λ)) (1 + α 1 )d(x 0,x n )+α 3 d(x 0,H(x 0,λ n )) + α 4 d(x n,x 0 ) + α 4 d(x 0,H(x 0,λ n )) + α 5 d(x 0,x n )+d(h(x 0,λ n ),H(x 0,λ)) =(1+α 1 + α 4 + α 5 )d(x 0,x n )+(α 3 + α 4 )d(x 0,H(x 0,λ n )) + d(h(x 0,λ n ),H(x 0,λ)) (1 + r)d(x 0,x n )+rd(x 0,H(x 0,λ n )) + d(h(x 0,λ n ),H(x 0,λ)). Letting n and by (iii), we obtain that d(x 0,H(x 0,λ)) rd(x 0,H(x 0,λ)). Since r<1, it follows that d(x 0,H(x 0,λ)) = 0 and thus shows that x 0 = H(x 0,λ). It is a direct result that x 0 U from (i). Consequently, we have proven λ A. ItmeansthatA is closed in [0, 1]. Next we show that A is open in [0, 1]. Let λ 0 A. Then there exists an x 0 U such that x 0 = H(x 0,λ 0 ). Since U is open, there exists a δ>0 such that the open ball B(x 0,δ) U. Consider the mapping H(,λ):B(x 0,δ) X for λ [0, 1]. Condition (iii) guarantees that there exists η = η(δ) > 0 such that for all λ (λ 0 η, λ 0 + η) [0, 1], d(x 0,H(x 0,λ)) = d(h(x 0,λ 0 ),H(x 0,λ)) < (2 r)(1 r) δ. 2+r Thus, by Lemma 1.1, H(,λ) has a fixed point in B(x 0,δ), that is, there exists an x B(x 0,δ) with x = H(x, λ) for all λ (λ 0 η, λ 0 + η) [0, 1]. This means that (λ 0 η, λ 0 + η) [0, 1] A, and hence A is open in [0, 1]. Remark 1.1 Ariza-Ruiz and Jiménez-Melado [2] proved a similar result to Theorem 1.1 for weakly Kannan mappings. In their theorem (see [2, Theorem 3.1]), it was assumed that, in place of the inequality (1.2), the following conditions hold for all x, y U and λ [0, 1], α(x, y) d(h(x, λ),h(y, λ)) [d(x, H(x, λ)) + d(y, H(y, λ))] 2 and θ(a, b) =sup{α(x, y) :a d(x, y) b} < 1 for all 0 <a b. It is easy to see that our hypothesis (1.2) relaxes the above constraints on the mappings H(,λ). Example Consider the metric space (X, d), where X =[ 1, 1] and d(x, y) = x y. Let T : X X be the map defined by 5 [ Tx= 36 x, x 1, 9 ), x 29 [ 9 ] (1.4) 20, x 10, 1. Then T is a nonlinear and continuous increasing self-mapping on X. For x, y in [ 1, 9 10 ), we define α 1 (x, y) = 5 36, α 2(x, y) =α 3 (x, y) =α 4 (x, y) =α 5 (x, y) =0. Forx, y in X with x or y in [ 9 10, 1], we take α 1(x, y) =α 4 (x, y) =α 5 (x, y) =0,α 2 (x, y) =α 3 (x, y) = 3 7. Then each α i is a function of X X into [0, 1) and 5 i=1 α i(x, y) < 1. In the following we first check that the map T defined in (1.4) satisfies the conditions of Proposition 1.1. Obviously, we only need to check that Condition (c) is satisfied. Suppose that

6 116 ADVANCES IN MATHEMATICS (CHINA) Vol. 46 x, y [ 1, 9 10 ). Then d(tx,ty)= 5 36 d(x, y) =α 1(x, y)d(x, y). Now suppose that x, y [ 9 10, 1]. Then Hence, it follows that d(tx,ty)= 7 7 d(x, y) 4 40, α 2 (x, y)d(x, T x)+α 3 (x, y)d(y, Ty) = 3 7 [ ] (x + y) > d(tx,ty) α 2 (x, y)d(x, T x)+α 3 (x, y)d(y, Ty). Finally, let us suppose that x [ 1, 9 10 ), y [ 9 10, 1]. Then d(tx,ty)=ty Tx= 7 4 y x. We need to consider two cases. Case (i): x [0, 9 10 ), y [ 9 10, 1]. In this case we have α 2 (x, y)d(x, T x)+α 3 (x, y)d(y, Ty) = 3 ( x 3 4 y + 29 ). 20 Since 7 4 y x 3 ( x 3 4 y + 29 ) 20 Condition (c) is satisfied. Case (ii): x [ 1, 0), y [ 9 10, 1]. Then α 2 (x, y)d(x, T x)+α 3 (x, y)d(y, Ty) = 3 7 Noting that x<0 in Case (ii) and the relation 7 4 y x 3 ( x 3 4 y + 29 ) 20 = (y 1) x 0, ( x 3 4 y + 29 ). 20 = (y 1) x 0, it follows that Condition (c) is satisfied. So T satisfies the conditions of Proposition 1.1. Next we define H :[ 1, 1] [0, 1] [ 1, 1] by H(x, λ) =λt x and check that H satisfies Conditions (i), (ii) and (iii) of Theorem 1.1. It is obvious that H satisfies (i). To check Condition (ii), note that for all x, y [ 1, 1] and λ [0, 1], d(h(x, λ),h(y, λ)) = λ Tx Ty = λd(tx,ty). Using a similar method as stated above for the verification of T, it is not hard to check that H satisfies Condition (ii). Finally, Condition (iii) is clearly satisfied with the definition of H. 2 Fixed Points of Generalized Expansive Mappings In 1982, Wang et al. [12] introduced the concepts of three kinds of expansive type mappings corresponding to three types of the contractive type mappings summarized by Rhoades [11].One of the three expansive type mappings is stated as follows.

7 No. 1 Wang H. Y.: A Continuation Method for Generalized Contractive Mappings 117 A self-mapping f on a complete metric space (X, d) is called a second type expansive mapping if there exist non-negative real numbers a, b and c with a + b + c>1such that for all x, y X, x y, d(f(x),f(y)) ad(x, f(x)) + bd(y, f(y)) + cd(x, y). Wang et al. [12] proved the existence of fixed points of the expansive mapping f and discussed some properties of the fixed points on certain conditions. In this section we first generalize the concept of the above expansive mapping to a more general case and get a class of generalized expansive mappings. Then we give a fixed point theorem for the class of generalized expansive mappings. Finally, we prove that the set of fixed points of such a mapping is uncountable and closed in a Banach space under some hypotheses. Definition 2.1 Let (X, d) be a metric space. A self-mapping T on X is called generalized expansive if there exist real functions β i = β i (x, y), i=1, 2, 3, from X X into [0, ) such that inf{ 3 i=1 β i(x, y) :x, y X} > 1andforallx, y X, d(tx,ty) β 1 d(x, y)+β 2 d(x, T x)+β 3 d(y, Ty). (2.1) Remark 2.1 If we replace the inequality (2.1) in Definition 2.1 by the relation d(tx,ty) β 1 d(x, y)+β 2 d(x, T x)+β 3 d(y, Ty)+β 4 d(x, T y)+β 5 d(y, Tx) withinf{ 5 i=1 β i(x, y) :x, y X} > 1, then we may obtain another class of generalized expansive mappings. However, the set of these mappings in this case is only a subset of that in Definition 2.1. Theorem 2.1 Let (X, d) be a complete metric space and T be the generalized expansive mapping described in Definition 2.1. Suppose that either p 1 =inf{β 1 (x, y) :x, y X} > 0or p 3 =inf{β 3 (x, y) :x, y X} > 0, and T is surjective on X. ThenT has at least a fixed point in X. In particular, if p 1 =inf{β 1 (x, y) :x, y X} > 1, then T has a unique fixed point in X. Proof For any x 0 X, thereexistsanx 1 X such that Tx 1 = x 0 since T is a surjective self-mapping on X. For the point x 1, likewise, there exists an x 2 X such that Tx 2 = x 1. Repeating this procedure, we can obtain a sequence {x n } X with Tx n = x n 1 for all n = 1, 2,. Using (2.1) with β i = β i (x n,x n+1 ), we obtain d(x n 1,x n )=d(tx n,tx n+1 ) β 1 d(x n,x n+1 )+β 2 d(x n,tx n )+β 3 d(x n+1,tx n+1 ) = β 1 d(x n,x n+1 )+β 2 d(x n,x n 1 )+β 3 d(x n+1,x n ) =(β 1 + β 3 )d(x n,x n+1 )+β 2 d(x n,x n 1 ). By the assumption on p 1 or p 3,wehavethatβ 1 + β 3 > 0. Hence, d(x n+1,x n ) 1 β 2 d(x n,x n 1 ). (2.2) β 1 + β 3 Let p inf{ 3 i=1 β i(x, y) :x, y X} > 1andq = 1 p. Applying a well-known inequality that for a b and t 0, we obtain from (2.2) a b a+t b+t 1 d(x n+1,x n ) d(x n,x n 1 ) 1 β 1 + β 2 + β 3 p d(x n,x n 1 ). By induction, we obtain for n =1, 2,, d(x n,x n 1 ) 1 p n 1 d(x 1,x 0 )=q n 1 d(x 1,x 0 ).

8 118 ADVANCES IN MATHEMATICS (CHINA) Vol. 46 Let m and n be two positive integers with m>n.thenwehave d(x m,x n ) d(x m,x m 1 )+d(x m 1,x m 2 )+ + d(x n+1,x n ) (q m 1 + q m q n )d(x 1,x 0 ). Since q<1, it follows that d(x m,x n ) qn 1 q d(x 1,x 0 ). So {x n } is a Cauchy sequence, and hence it converges to a point x in X due to the completeness of X. Next we consider two cases. Case (i): There exists a positive integer N such that x n = x for all n>n.inthiscasewe claim that x is a fixed point of T in X. In fact, we have x n+1 = x n = x for all n>n. This implies that T x = Tx n+1 = x n = x, thus showing that x is a fixed point of T. Case (ii): For an arbitrary positive integer N, thereexistsn>nsuch that x n x. Let ȳ X be such that T ȳ = x. We will prove that ȳ is a fixed point of T and further ȳ = x. Indeed, using (2.1) with β i = β i (x n+1, ȳ), we have d(x n, x) =d(tx n+1,tȳ) β 1 d(x n+1, ȳ)+β 2 d(x n+1,x n )+β 3 d(ȳ, x). If p 1 > 0, we have from the above inequality that d(x n, x) β 1 d(x n+1, ȳ) p 1 d(x n+1, ȳ). If p 3 > 0, we obtain that d(x n, x) β 3 d(ȳ, x) p 3 d(ȳ, x). In either case, letting n and noting that x n x as n,wehavethatd( x, ȳ) = 0. This means that ȳ = x = T ȳ. In the following we consider the special case when p 1 =inf{β 1 (x, y) :x, y X} > 1. In this case we first prove that T has an inverse mapping, denoted as T 1.Infact,foranyx y, if Tx = Ty, then we have from (2.1) that 0 β 1 d(x, y) +β 2 d(x, T x) +β 3 d(y, Ty) β 1 d(x, y), which contradicts the fact that β 1 d(x, y) > 0. Thus T is an injection. Note that T is also a surjection, then we say that T is a one-to-one mapping on X, thus implying that T has inverse mapping T 1. We now prove that T has a unique fixed point. For any x, y X, replacing x and y in (2.1) by T 1 x and T 1 y, respectively, we obtain d(x, y) β 1 d(t 1 x, T 1 y)+β 2 d(t 1 x, x)+β 3 d(t 1 y, y) β 1 d(t 1 x, T 1 y), where β i = β i (T 1 x, T 1 y), i=1, 2, 3. Hence d(t 1 x, T 1 y) 1 β 1 d(x, y) 1 p 1 d(x, y), where 0 < 1 p 1 < 1. Therefore, T 1 agrees with the Banach fixed point theorem, and thus T 1 has a unique fixed point x X, thatis,t 1 x = x. This means that T x = x, and shows that x is also the unique fixed point of T. Remark 2.2 Theorem 2.1 gives the fixed point results of the generalized expansive mapping T defined in (2.1). Apart from the case when p 1 > 1, we find that it is hard to prove the uniqueness of the fixed point for the mapping T unless some additional conditions are given. A simple example of this class of generalized expansive mappings is the identity mapping. It is easy to check that there exist real functions β i = β i (x, y)(i =1, 2, 3) satisfying the desired condition and p 1 =inf{β 1 (x, y) :x, y X} 1 such that (2.1) holds for the identity mapping. Obviously, every point x X is the fixed point of the identity mapping. In order to discuss the properties of fixed points of the generalized expansive mapping T, we introduce a set A x and describe its characteristics in the following.

9 No. 1 Wang H. Y.: A Continuation Method for Generalized Contractive Mappings 119 For x X, wewrite A x = {y X : T n y x, n }, where T n denotes the n-fold composition of the inverse mapping T 1 with itself, defined as T n = T 1 (T (n 1) ) for all positive integers n 2. Lemma 2.1 Let (X, d) be a complete metric space and T be the generalized expansive mapping given in Definition 2.1. Suppose that p 1 =inf{β 1 (x, y) :x, y X} > 0andp 3 = inf{β 3 (x, y) :x, y X} > 0. Let T be a one-to-one mapping with inverse mapping T 1 and x be any fixed point of T. If the set of accumulation points of A x is nonempty, then the accumulation point set consists only of a singleton and the sole accumulation point is identical to x. Proof Since x is a fixed point of T, it is easy to see that x A x.soa x is nonempty. Let ỹ be any accumulation point of A x. We will prove that ỹ = x. By the properties of accumulation point, there exists a sequence {y k } A x such that y k ỹ for all positive integers k and y k ỹ as k. Since both T and T 1 are one-to-one and y k ỹ, wehavet 1 y k T 1 ỹ. From (2.1) with β i = β i (T 1 y k,t 1 ỹ), we obtain d(y k, ỹ) =d(t (T 1 y k ),T(T 1 ỹ)) β 1 d(t 1 y k,t 1 ỹ)+β 2 d(t 1 y k,y k )+β 3 d(t 1 ỹ, ỹ). Hence, we have that d(y k, ỹ) β 1 d(t 1 y k,t 1 ỹ) p 1 d(t 1 y k,t 1 ỹ). Letting k and noting that y k ỹ as k and p 1 > 0, we obtain that d(t 1 y k,t 1 ỹ) 0ask. Similarly, we deduce from p 3 > 0thatd(T 1 ỹ, ỹ) = 0. Since d(t 1 y k, ỹ) d(t 1 y k,t 1 ỹ)+ d(t 1 ỹ, ỹ) =d(t 1 y k,t 1 ỹ), it follows that T 1 y k ỹ as k. Repeating the above process, we have that T n y k ỹ as k for all positive integers n. For every fixed positive integer k, using (2.1) with β i = β i (T (m+1) y k,t m y k ), we have for all m 1, d(t m y k,t (m 1) y k )=d(t(t (m+1) y k ),T(T m y k )) (β 1 + β 2 )d(t (m+1) y k,t m y k )+β 3 d(t m y k,t (m 1) y k ), from which, it follows that d(t (m+1) y k,t m y k ) 1 β 3 β 1 + β 2 d(t m y k,t (m 1) y k ) 1 d(t m y k,t (m 1) y k ) β 1 + β 2 + β 3 1 p d(t m y k,t (m 1) y k ), (2.3) where p =inf{ 3 i=1 β i(x, y) :x, y X} > 1. From (2.3), using induction and the notation q = 1 p, we deduce d(t (m+1) y k,t m y k ) q m d(t 1 y k,y k ). (2.4) For two arbitrary positive integers m and n with m>n, from (2.4) we obtain d(t m y k,t n y k ) d(t m y k,t (m 1) y k )+d(t (m 1) y k,t (m 2) y k ) + + d(t (n+1) y k,t n y k ) (q m 1 + q m q n )d(t 1 y k,y k )

10 120 ADVANCES IN MATHEMATICS (CHINA) Vol. 46 qn 1 q d(t 1 y k,y k ). Since d(t 1 y k,y k ) d(t 1 y k, ỹ)+d(ỹ, y k ) 0ask, there exists a constant M>0such that d(t 1 y k,y k ) M for all k 1. Thus we obtain for all k 1, d(t m y k,t n y k ) qn M. (2.5) 1 q Noting that each y k belongs to A x,wehavethatt m y k x as m for every fixed k. Letting m in (2.5) gives d(x, T n y k ) qn 1 q M. For arbitrary ε>0, we take sufficiently large n such that qn 1 q M<εso as to obtain d(x, T n y k ) <ε. (2.6) Noting that T n y k ỹ as k for all n 1, it follows that d(x, ỹ) ε by letting k in (2.6). In view of the arbitrariness of ε, we conclude that ỹ = x. The proof of Lemma 2.1 is complete. The following theorem answers the question concerning the number of the fixed points of T in a Banach space on certain conditions. Theorem 2.2 Let (X, ) be a (nonempty) Banach space and d be a distance on X induced by the norm. Suppose that T : X X is the generalized expansive mappings described in Definition 2.1 satisfying the additional conditions 0 <p 1 =inf{β 1 (x, y) :x, y X} 1andp 3 =inf{β 3 (x, y) :x, y X} > 0. If T is a one-to-one mapping, then the set of all fixed points of T is an uncountable and closed set. Proof We first prove the uncountability of the set F (T )offixedpointsoft by contradiction. For this we suppose that the set F (T ) is at most countable, then F (T ) can be represented as { x k }, in which, the number of all fixed points x k is either finitely many or countable. For every x k, we define a set A k = {y X : T n y x k,n }. From the proof of Theorem 2.1, we can see that for any x 0 X, the sequence {x n } produced by the iteration process x n+1 = T 1 x n for all n =0, 1, 2, converges to some fixed point of T, that is, the sequence {T n x 0 } converges to a point in F (T ). Thus we say that x 0 must belong to some A k. Hence, we have X = k A k. The completeness of X and the well-known Baire s theorem of category imply that X is a set of the second category. Hence, there is at least a set A k such that A k is not a nowhere dense set. Consequently, there exists a ball S = B(a, δ) of radius δ>0 and center at some a X such that A k is dense in S, thatis,s A k,wherea k denotes the closure of A k. We will show that S A k. For this, let y be any point in S. Ify is not in A k,theny must be an accumulation point of A k since y A k. By Lemma 2.1, we have y = x k A k, which is a contradiction. So we assert that S A k. Since X is a normed linear space, every point y S A k ( X) is an accumulation point of S, and hence, an accumulation point of A k. Therefore, according to Lemma 2.1, all points of S are identical to x k. This means that the ball S with positive radius contains only a single point, which obviously is impossible to a normed linear space. Thus we conclude that the set F (T )is uncountable.

11 No. 1 Wang H. Y.: A Continuation Method for Generalized Contractive Mappings 121 Next we proceed to prove that F (T ) is closed. Let y 0 be an accumulation point of F (T ), then there exists a sequence { x n } F (T )with x n y 0 for all n 1 such that x n y 0 as n.letȳ be the point in X such that T ȳ = y 0. From (2.1) with β i = β i ( x n, ȳ), we have d( x n,y 0 )=d(t x n,tȳ) β 1 d( x n, ȳ)+β 3 d(ȳ, Tȳ) β 1 d( x n, ȳ) p 1 d( x n, ȳ). Noting that x n y 0 as n and p 1 > 0, by letting n,weobtaind(y 0, ȳ) = 0, and hence y 0 =ȳ. This implies that Ty 0 = T ȳ = y 0, and thus shows that y 0 F (T ). Consequently, F (T ) is closed. This finishes the proof of Theorem 2.2. Acknowledgements The author thanks the reviewers for valuable suggestions and useful comments. References [1] Agarwal, R.P. and O Regan, D., Fixed point theory for generalized contractions on spaces with two metrics, J. Math. Anal. Appl., 2000, 248(2): [2] Ariza-Ruiz, D. and Jiménez-Melado, A., A continuation method for weakly Kannan maps, Fixed Point Theory Appl., 2010, 2010: Article ID , 12 pages. [3] Granas, A., Continuation method for contractive maps, Topol. Methods Nonlinear Anal., 1994, 3(2): [4] Hardy, G.E. and Rogers, T.D., A generalization of a fixed point theorem of Reich, Canad. Math. Bull., 1973, 16(2): [5] Janos, L., On mappings contractive in the sense of Kannan, Proc. Amer. Math. Soc., 1976, 61(1): [6] Kannan, R., Some results on fixed points, II, Amer. Math. Monthly, 1969, 76(4): [7] O Regan, D., Fixed point theorems for nonlinear operators, J. Math. Anal. Appl., 1996, 202(2): [8] Pathak, H.K., Kang, S.M. and Cho, Y.J., Coincidence and fixed point theorems for nonlinear hybrid generalized contractions, Czechoslovak Math. J., 1998, 48(2): [9] Precup, R., Discrete continuation method for boundary value problems on bounded sets in Banach spaces, J. Comput. Appl. Math., 2000, 113(1/2): [10] Reich, S., Some remarks concerning contraction mappings, Canad. Math. Bull., 1971, 14(1): [11] Rhoades, B.E., A comparison of various definitions of contractive mappings, Trans. Amer. Math. Soc., 1977, 226: [12] Wang, S.Z., Li, B.Y. and Gao, Z.M., Expansive operators and their fixed point theorems, Adv. Math. (China), 1982, 11(2): (in Chinese). [13] Wong, C.S., Generalized contractions and fixed point theorems, Proc. Amer. Math. Soc., 1974, 42(2): [14] Wong, C.S., Fixed point theorems for generalized nonexpansive mappings, J. Aust. Math. Soc., 1974, 18(3): ffi)&#*"ο(/!%fiff+ffi)ψ-*"ομρß$. 102 (Ψ± ffl fifflffl, Ψ±, Ξffi, ),' 4^C9RcMEdbYfW:clPa@?, g]3>qkgkiorcmam fw:cb S<N. evrcmedlhfw, C9RZT:7=;<N, 6gc<:[H_, juri`7=;d8r Banach KGkcB7JX:5F. flχν Pa@?; EdbYfW; 7=;; EdLhfW