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SEVERAL FIXED POINT THEOREMS CONCERNING τ-DISTANCE TOMONARI SUZUKI Received 21 October 2003 and in revised form 10 March 2004 Using the notion of τ-distance, we prove several fixed point theorems, which are general- izations of fixed point theorems by Kannan, Meir-Keeler, Edelstein, and Nadler. We also discuss the properties of τ-distance. 1. Introduction In 1922, Banach proved the following famous fixed point theorem [1]. Let (X,d)bea complete metric space. Let T be a contractive mapping on X, that is, there exists r ∈ [0,1) satisfying d(Tx,Ty) ≤ rd(x, y) (1.1) for all x, y ∈ X. Then there exists a unique fixed point x 0 ∈ X of T. This theorem, called the Banach contraction principle, is a forceful tool in nonlinear analysis. This princi- ple has many applications and is extended by several authors: Caristi [2], Edelstein [5], Ekeland [6, 7], Meir and Keeler [14], Nadler [15], and others. These t heorems are also extended; see [4, 9, 10, 13, 23, 25, 26, 27] and others. In [20], the author introduced the notion of τ-distance and extended the Banach contraction principle, Caristi’s fixed point theorem, and Ekeland’s ε-variational principle. In 1969, Kannan proved the follow- ing fixed point theorem [12]. Let (X,d) be a complete metric space. Let T be a Kannan mapping on X, that is, there exists α ∈ [0,1/2) such that d(Tx,Ty) ≤ α  d(Tx,x)+d(Ty, y)  (1.2) for all x, y ∈ X. Then there exists a unique fixed point x 0 ∈ X of T. We note that Kan- nan’s fixed point theorem is not an extension of the Banach contraction principle. We also know that a metric space X is complete if and only if every Kannan mapping has a fixed point, while there exists a metric space X such that X is not complete and every contractive mapping on X has a fixed point; see [3, 17]. Copyright © 2004 Hindawi Publishing Corporation Fixed Point Theory and Applications 2004:3 (2004) 195–209 2000 Mathematics Subject Classification: 54H25, 54E50 URL: http://dx.doi.org/10.1155/S168718200431003X 196 Fixed point theorems concerning τ-distance In this paper, using the notion of τ-distance, we prove several fixed point theorems, which are generalizations of fixed point theorems by Kannan, Meir-Keeler, Edelstein, and Nadler. We also discuss the properties of τ-distance. 2. τ-distance Throughout this paper, we denote by N the set of all positive integers. In this section, we discuss some properties of τ-distance. Let (X,d) be a metric space. Then a function p from X × X into [0,∞)iscalledaτ-distance on X [20] if there exists a function η from X × [0,∞)into[0,∞) and the following are satisfied: (τ1) p(x,z) ≤ p(x, y)+p(y,z)forallx, y, z ∈ X; (τ2) η(x,0) = 0andη(x,t) ≥ t for all x ∈ X and t ∈ [0,∞), and η is concave and con- tinuous in its second variable; (τ3) lim n x n = x and lim n sup{η(z n , p(z n ,x m )) : m ≥ n}=0implyp(w,x) ≤ liminf n p(w,x n )forallw ∈ X; (τ4) lim n sup{p(x n , y m ):m ≥ n}=0andlim n η(x n ,t n ) = 0 imply lim n η(y n ,t n ) = 0; (τ5) lim n η(z n , p(z n ,x n )) = 0andlim n η(z n , p(z n , y n )) = 0 imply lim n d(x n , y n ) = 0. We may replace (τ2) by the following (τ2)  (see [20]): (τ2)  inf{η(x, t):t>0}=0forallx ∈ X,andη is nondecreasing in its second variable. The metric d is a τ-distance on X. Many useful examples are stated in [11, 16, 18, 19, 20, 21, 22, 24]. It is very meaningful that one τ-distance generates other τ-distances. In the sequel, we discuss this fact. Proposition 2.1. Let (X,d) be a metric space. Let p be a τ-distance on X and let η be a function satisfying (τ2)  ,(τ3), (τ4), and (τ5). Let q be a function from X × X into [0,∞) satisfying (τ1) q .Supposethat (i) there exists c>0 such that min{p(x, y),c}≤q(x, y) for x, y ∈ X, (ii) lim n x n =x and lim n sup{η(z n ,q(z n ,x m )) : m≥n}=0 imply q(w, x)≤ liminf n q(w,x n ) for w ∈ X. Then q is also a τ-distance on X. Proof. We p ut θ(x,t) = t + η(x, t) (2.1) for x ∈ X and t ∈ [0,∞). Note that η(x,t) ≤ θ(x,t)forallx ∈ X and t ∈ [0,∞). Then, by the assumption, (τ1) q ,(τ2)’ θ ,and(τ3) q,θ hold. We assume that lim n sup{q(x n , y m ): m ≥ n}=0andlim n θ(x n ,t n ) = 0. Then lim n sup{p(x n , y m ):m ≥ n}=0andlim n t n = lim n η(x n ,t n ) = 0 clearly hold. From (τ4), we have lim n η(y n ,t n ) = 0 and hence lim n θ(y n ,t n ) = 0. Therefore, we have shown (τ4) q,θ . We assume that lim n θ(z n ,q(z n ,x n ))= 0andlim n θ(z n ,q(z n , y n )) = 0. By the definition of θ,wehavelim n η(z n ,q(z n ,x n )) = 0and lim n q(z n ,x n ) = 0. So, by the assumption, lim n η(z n , p(z n ,x n )) = 0 holds. We can similarly prove lim n η(z n , p(z n , y n )) = 0. Therefore, from (τ5), lim n d(x n , y n ) = 0. Hence, we have shown (τ5) q,θ . This completes the proof.  As a direct consequence of Proposition 2.1, we obtain the following proposition. Tomonari Suzuki 197 Proposition 2.2. Let p be a τ-distance on a metric space X.Letq beafunctionfromX × X into [0,∞) satisfying (τ1) q .Supposethat (i) there exists c>0 such that min{p(x, y),c}≤q(x, y) for x, y ∈ X, (ii) for every convergent sequence {x n } with limit x satisfying p(w,x) ≤ liminf n p(w,x n ) for all w ∈ X, q(w,x) ≤ liminf n q(w,x n ) holds for all w ∈ X. Then q is also a τ-distance on X. Using the above proposition, we obtain the following one w hich is used in the proof of generalized Kannan’s fixed point theorem. Proposition 2.3. Let p be a τ-distance on a metric space X and let α be a function from X into [0,∞).Thentwofunctionsq 1 and q 2 from X × X into [0,∞),definedby (i) q 1 (x, y) = max{α(x), p(x, y)} for x, y ∈ X, (ii) q 2 (x, y) = α(x)+p(x, y) for x, y ∈ X, are τ-distances on X. Proof. We h ave q 1 (x, z) = max  α(x), p(x,z)  ≤ max  α(x)+α(y), p(x, y)+p(y,z)  ≤ q 1 (x, y)+q 1 (y,z), q 2 (x, z) = α(x)+p(x,z) ≤ α(x)+α(y)+p(x, y)+p(y,z) = q 2 (x, y)+q 2 (y,z), (2.2) for all x, y,z ∈ X. We note that p(x, y) ≤ q 1 (x, y) ≤ q 2 (x, y) (2.3) for all x, y ∈ X.Weassumethatasequence{x n } satisfies lim n x n = x and p(w,x) ≤ liminf n p(w,x n )forallw ∈ X. Then it is clear that q 1 (w,x) ≤ liminf n q 1 (w,x n )and q 2 (w,x) ≤ liminf n q 2 (w,x n )forallw ∈ X.ByProposition 2.2, q 1 and q 2 are τ-distances on X. This completes the proof.  Let (X,d) be a metric space and let p be a τ-distance on X. Then a sequence {x n } in X is called p-Cauchy [20] if there exist a function η from X × [0,∞)into[0,∞) satisfying (τ2)–(τ5) and a sequence {z n } in X such that lim n sup{η(z n , p(z n ,x m )) : m ≥ n}=0. The following lemmas are very useful in the proofs of fixed point theorems in Section 3. Lemma 2.4 [20]. Let (X,d) be a metric space and let p be a τ-distance on X.If{x n } is a p-Cauchy sequence, then {x n } is a Cauchy sequence. Moreover, if {y n } is a sequence satisfying lim n sup{p(x n , y m ):m ≥ n}=0, then {y n } is also a p-Cauchy sequence and lim n d(x n , y n ) = 0. Lemma 2.5 [20]. Let (X,d) be a metric space and let p be a τ-distance on X.Ifasequence {x n } in X satisfies lim n p(z,x n ) = 0 for some z ∈ X, then {x n } is a p-Cauchy sequence. 198 Fixed point theorems concerning τ-distance Moreover, if a sequence {y n } in X also satisfies lim n p(z, y n ) = 0, then lim n d(x n , y n ) = 0.In particular, for x, y,z ∈ X, p(z,x) = 0 and p(z, y) = 0 imply x = y. Lemma 2.6 [20]. Let (X,d) be a metric space and let p be a τ-distance on X.Ifase- quence {x n } in X satisfies lim n sup{p(x n ,x m ):m>n}=0, then {x n } is a p-Cauchy se- quence. Moreover, if a sequence {y n } in X satisfies lim n p(x n , y n ) = 0, then {y n } is also a p-Cauchy sequence and lim n d(x n , y n ) = 0. 3. Fixed point theorems In this section, we prove several fixed point theorems in complete metric spaces. In [20], the following theorem connected with Hicks-Rhoades theorem [8] was proved and used in the proofs of generalizations of the Banach contraction pr inciple, Caristi’s fixed point theorem, and so on. In this paper, this theorem is used in the proof of a generalization of Kannan’s fixed point theorem. Theorem 3.1 [20]. Let X be a complete metric space and let T be a mapping on X.Suppose that there exist a τ-distance p on X and r ∈ [0,1) such that p(Tx,T 2 x) ≤ rp(x,Tx) for all x ∈ X. Assume that either of the following holds: (i) if lim n sup{p(x n ,x m ):m>n}=0, lim n p(x n ,Tx n ) = 0,andlim n p(x n , y) = 0, then Ty= y; (ii) if {x n } and {Tx n } converge to y, then Ty= y; (iii) T is continuous. Then there exists x 0 ∈ X such that Tx 0 = x 0 .Moreover,ifTz = z, then p(z, z) = 0. As a direct consequence, we obtain the following theorem. Theorem 3.2. Let X be a complete metric space and let p be a τ-distance on X.LetT be a mapping on X. Suppose that there exists r ∈ [0,1) such that either (a) or (b) holds: (a) max{p(T 2 x, Tx), p(Tx,T 2 x)}≤r max{p(Tx,x), p(x,Tx)} for all x ∈ X; (b) p(T 2 x, Tx)+p(Tx,T 2 x) ≤ rp(Tx,x)+rp(x, Tx) for all x ∈ X. Further, assume that either of the following holds: (i) if lim n sup{p(x n ,x m ):m>n}=0, lim n p(Tx n ,x n ) = 0, lim n p(x n ,Tx n ) = 0,and lim n p(x n , y) = 0, then Ty= y; (ii) if {x n } and {Tx n } converge to y, then Ty= y; (iii) T is continuous. Then there exists x 0 ∈ X such that Tx 0 = x 0 .Moreover,ifTz = z, then p(z, z) = 0. Proof. In the case of (a), we define a function q by q(x, y) = max{p(Tx,x), p(x, y)}.In the case of (b), we define a function q by q(x, y) = p(Tx,x)+p(x, y). By Proposition 2.3, q is a τ-distance on X. In both cases, we have q  Tx,T 2 x  ≤ rq(x,Tx) (3.1) Tomonari Suzuki 199 for all x ∈ X. Conditions (ii) and (iii) are not connected with τ-distance p. In the case of (i), since p(x, y) ≤ q(x, y), p(Tx,x) ≤ q(x,Tx), (3.2) for all x, y ∈ X, T has a fixed point in X by Theorem 3.1.IfTz = z,thenq(z,z) = 0, and hence p(z,z) = 0. This completes the proof.  We now prove a generalization of Kannan’s fixed point theorem [12]. Let X be a metric space, let p be a τ-distance on X,andletT be a mapping on X.ThenT is called a Kannan mapping with respect to p if there exists α ∈ [0,1/2) such that either (a) or (b) holds: (a) p(Tx,Ty) ≤ αp(Tx,x)+αp(Ty, y)forallx, y ∈ X; (b) p(Tx,Ty) ≤ αp(Tx,x)+αp(y,Ty)forallx, y ∈ X. Theorem 3.3. Let (X,d) be a complete metric space, let p be a τ-distance on X,andletT be a Kannan mapping on X with respect to p. Then T has a unique fixed point x 0 ∈ X.Further, such x 0 satisfies p(x 0 ,x 0 ) = 0. Proof. In the case of (a), there exists α ∈ [0,1/2) such that p(Tx,Ty) ≤ αp(Tx,x)+ αp(Ty, y)forx, y ∈ X.Since p  T 2 x, Tx  ≤ αp  T 2 x, Tx  + αp(Tx,x), (3.3) we have p  T 2 x, Tx  ≤ α 1 − α p(Tx,x) ≤ p(Tx,x) (3.4) for x ∈ X.Puttingr = 2α ∈ [0,1), we have max  p  T 2 x, Tx  , p  Tx,T 2 x  ≤ αp  T 2 x, Tx  + αp(Tx,x) ≤ rp(Tx,x) ≤ r max  p(Tx,x), p(x,Tx)  (3.5) for all x ∈ X.Weassumelim n sup{p(x n ,x m ):m>n}=0, lim n p(Tx n ,x n ) = 0, lim n p(x n ,Tx n ) = 0, and lim n p(x n , y) = 0. Then, by Lemma 2.6, {x n } and {Tx n } are p- Cauchy and lim n→∞ d  x n , y  = lim n→∞ d  Tx n , y  = 0. (3.6) Now we have p(Ty, y) ≤ liminf n→∞ p  Ty,Tx n  ≤ liminf n→∞  αp(Ty, y)+αp  Tx n ,x n  = αp(Ty, y), (3.7) 200 Fixed point theorems concerning τ-distance and hence p(Ty, y) = 0. Since p(T 2 y,Ty) ≤ p(Ty, y) = 0andp(T 2 y, y) ≤ p(T 2 y,Ty)+ p(Ty, y) = 0, we have Ty = y by Lemma 2.5. Therefore, by Theorem 3.2, there exists x 0 ∈ X such that Tx 0 = x 0 and p(x 0 ,x 0 ) = 0. Further, a fixed point of T is unique. In fact, if Tz = z,thenp(z,z) = 0byTheorem 3.2.Sowehave p  x 0 ,z  = p  Tx 0 ,Tz  ≤ αp  Tx 0 ,x 0  + αp(Tz,z) = αp  x 0 ,x 0  + αp(z,z) = 0. (3.8) By Lemma 2.5 again, we have x 0 = z. In the case of (b), there exists α ∈ [0,1/2) such that p(Tx,Ty) ≤ αp(Tx,x)+αp(y,Ty)forx, y ∈ X. Then, putting r = α/(1 − α) ∈ [0,1), we have p(Tx,T 2 x) ≤ rp(Tx,x)andp(T 2 x, Tx) ≤ rp(x,Tx)forallx ∈ X.So, p  T 2 x, Tx  + p  Tx,T 2 x  ≤ rp(Tx,x)+rp(x,Tx) (3.9) for all x ∈ X.Weassumelim n sup{p(x n ,x m ):m>n}=0, lim n p(Tx n ,x n ) = 0, lim n p(x n , Tx n ) = 0, and lim n p(x n , y) = 0. Then {x n } and {Tx n } are p-Cauchy and lim n d(x n , y) = lim n d(Tx n , y) = 0. So we have p(Ty, y) ≤ liminf n→∞ p  Ty,Tx n  ≤ liminf n→∞  αp(Ty, y)+αp  x n ,Tx n  = αp(Ty, y), (3.10) and hence p(Ty, y) = 0. Since p(Ty,T 2 y) ≤ rp(Ty, y) = 0, we have y = T 2 y by Lemma 2.5.So,p(y,Ty)= p(T 2 y,Ty)≤rp(y, Ty), and hence p(y,Ty)=0. We also have p(y, y)≤ p(y, Ty)+p(Ty, y) = 0. So we have Ty = y by Lemma 2.5. Therefore, by Theorem 3.2, there exists x 0 ∈ X such that Tx 0 = x 0 and p(x 0 ,x 0 ) = 0. As in the case of (a), we obtain that a fixed point of T is unique.  In general, τ-distance p does not satisfy p(x, y) = p(y,x). So conditions (a) and (b) in the definition of Kannan mappings differ from conditions (c) and (d) in the following theorem. Indeed, there exists a mapping T on a complete metric space X such that (c) and (d) hold, and T has no fixed points; see [19]. However, under the assumption that T is continuous, T has a fixed point. Theorem 3.4. Let X be a complete metric space and let T be a continuous mapping on X. Suppose that there exist a τ-distance p on X and α ∈ [0,1/2) such that either (c) or (d) holds: (c) p(Tx,Ty) ≤ αp(x,Tx)+αp(Ty, y) for all x, y ∈ X; (d) p(Tx,Ty) ≤ αp(x,Tx)+αp(y,Ty) for all x, y ∈ X. Then there exists a unique fixed point x 0 ∈ X of T.Moreover,suchx 0 satisfies p(x 0 ,x 0 ) = 0. Proof. In the case of (c), putting r = α/(1 − α) ∈ [0,1), from p(Tx,T 2 x) ≤ αp(x,Tx)+ αp(T 2 x, Tx)andp(T 2 x, Tx) ≤ αp(Tx,T 2 x)+αp(Tx,x), we have p  T 2 x, Tx  + p  Tx,T 2 x  ≤ rp(Tx,x)+rp(x,Tx) (3.11) Tomonari Suzuki 201 for all x ∈ X.So,byTheorem 3.2, we prove the desired result. In the case of (d), we have p(Tx,T 2 x) ≤ rp(x,Tx)forallx ∈ X. Therefore, by Theorem 3.1, we prove the desired result. This completes the proof.  We next prove a generalization of Meir and Keeler’s fixed point theorem [14]. Theorem 3.5. Let X be a complete metric space, let p be a τ-distance on X,andletT be a mapping on X.Supposethatforanyε>0,thereexistsδ>0 such that for every x, y ∈ X, p(x, y) <ε+ δ implies p(Tx,Ty) <ε. Then T has a unique fixed point x 0 in X. Further, such x 0 satisfies p(x 0 ,x 0 ) = 0. Proof. We first show p(Tx,Ty) ≤ p(x, y)forallx, y ∈ X. For an arbitrary λ>0, there exists δ>0suchthatforeveryz,w ∈ X, p(z,w) <p(x, y)+λ + δ implies p(Tz,Tw) < p(x, y)+λ.Sincep(x, y) <p(x, y)+λ + δ,wehavep(Tx,Ty) <p(x, y)+λ.Sinceλ>0is arbitrary, we obtain p(Tx,Ty) ≤ p(x, y). We next show lim n→∞ p  T n x, T n y  = 0 ∀x, y ∈ X. (3.12) In fact, {p(T n x, T n y)} is nonincreasing and hence converges to some real number r.We assume r>0. Then there exists δ>0suchthatforeveryz,w ∈ X, p(z,w) <r+ δ implies p(Tz,Tw) <r.Forsuchδ,wecanchoosem ∈ N such that p(T m x, T m y) <r+ δ.Sowe have p(T m+1 x, T m+1 y) <r. This is a contradiction, and hence (3.12)holds.Letu ∈ X and put u n = T n u for every n ∈ N.From(3.12), we have lim n p(u n ,u n+1 ) = 0. We will show that lim n→∞ sup m>n p  u n ,u m  = 0. (3.13) Let ε>0 be arbitrary. Then, without loss of generality, there exists δ ∈ (0,ε)suchthat for every z,w ∈ X, p(z,w) <ε+ δ implies p(Tz,Tw) <ε.Forsuchδ, there exists n 0 ∈ N such that p(u n ,u n+1 ) <δfor every n ≥ n 0 . Assume that there exists m>≥ n 0 such that p(u  ,u m ) > 2ε.Since p  u  ,u +1  <ε+ δ<p  u  ,u m  , (3.14) there exists k ∈ N with <k<msuch that p  u  ,u k  <ε+ δ ≤ p  u  ,u k+1  . (3.15) Then, since p(u  ,u k ) <ε+ δ,wehavep(u +1 ,u k+1 ) <ε. On the other hand, we have p  u  ,u k+1  ≤ p  u  ,u +1  + p  u +1 ,u k+1  <δ+ ε. (3.16) This is a contradiction. Therefore, m>n≥ n 0 implies p(u n ,u m ) ≤ 2ε, and hence (3.13) holds. By Lemma 2.6, {u n } is p-Cauchy. So, {u n } is also a Cauchy sequence by Lemma 2.4. 202 Fixed point theorems concerning τ-distance Hence there exists x 0 ∈ X such that {u n } converges to x 0 .From(τ3), we have limsup n→∞ p  u n ,Tx 0  ≤ limsup n→∞ p  u n−1 ,x 0  = limsup n→∞ p  u n ,x 0  ≤ limsup n→∞ liminf m→∞ p  u n ,u m  ≤ lim n→∞ sup m>n p  u n ,u m  = 0. (3.17) By Lemma 2.6 again, {u n } converges to Tx 0 , and hence Tx 0 = x 0 .From(3.12), we obtain p  x 0 ,x 0  = lim n→∞ p  T n x 0 ,T n x 0  = 0. (3.18) If z = Tz,then p  x 0 ,z  = lim n→∞ p  T n x 0 ,T n z  = 0. (3.19) So, from Lemma 2.5, x 0 = z. Therefore, a fixed point of T is unique. This completes the proof.  Let X be a metric space and let p be a τ-distance on X.Forε ∈ (0, ∞], X is called ε-chainable with respect to p if, for each (x, y) ∈ X × X, there exists a finite sequence {u 0 ,u 1 ,u 2 , ,u  } in X such that u 0 = x, u  = y,andp(u i−1 ,u i ) <εfor i = 1,2, ,.We will prove a generalization of Edelstein’s fixed point theorem [5]. Theorem 3.6. Let X be a complete metric space. Suppose that X is ε-chainable with respect to p for some ε ∈ (0,∞] and for some τ-distance p on X.LetT be a mapping on X.Suppose that there exists r ∈ [0,1) such that p(Tx,Ty) ≤ rp(x, y) for all x, y ∈ X with p(x, y) <ε. Then there exists a unique fixed point x 0 ∈ X of T. Further, such x 0 satisfies p(x 0 ,x 0 ) = 0. Proof. We first show lim n→∞ p  T n x, T n y  = 0 (3.20) for all x, y ∈ X.Letx, y ∈ X be fixed. Then there exist u 0 ,u 1 ,u 2 , ,u  ∈ X such that u 0 = x, u  = y,andp(u i−1 ,u i )<εfor i = 1,2, ,.Sincep(u i−1 ,u i ) <ε,wehavep(Tu i−1 ,Tu i ) ≤ rp(u i−1 ,u i ) <ε.Thus p  T n u i−1 ,T n u i  ≤ rp  T n−1 u i−1 ,T n−1 u i  ≤··· ≤r n p  u i−1 ,u i  . (3.21) Therefore limsup n→∞ p  T n x, T n y  ≤ limsup n→∞   i=1 p  T n u i−1 ,T n u i  ≤ lim n→∞   i=1 r n p  u i−1 ,u i  = 0. (3.22) Tomonari Suzuki 203 We have shown (3.20). Let x ∈ X be fixed. From (3.20), there exists n 0 ∈ N such that p  T n x, T n+1 x  <ε (3.23) for n ≥ n 0 .Then,form>n≥ n 0 ,wehave p  T n x, T m x  ≤ m−1  k=n p  T k x, T k+1 x  ≤ m−1  k=n r k−n 0 p  T n 0 x, T n 0 +1 x  ≤ r n−n 0 1 − r p  T n 0 x, T n 0 +1 x  . (3.24) Hence, lim n sup{p(T n x, T m x):m>n}=0. By Lemma 2.6, {T n x} is p-Cauchy. By Lemma 2.4, {T n x} is a Cauchy sequence. So, {T n x} converges to some x 0 ∈ X.Since limsup n→∞ p  T n x, x 0  ≤ limsup n→∞ liminf m→∞ p  T n x, T m x  ≤ lim n→∞ sup m>n p  T n x, T m x  = 0, (3.25) we have limsup n→∞ p  T n x, Tx 0  ≤ lim n→∞ rp  T n−1 x, x 0  = 0. (3.26) By Lemma 2.6,weobtainTx 0 = x 0 .Ifz is a fixed point of T,thenwehave p  x 0 ,z  = lim n→∞ p  T n x 0 ,T n z  = 0 (3.27) from (3.20). We also have p(x 0 ,x 0 ) = 0. Therefore, z = x 0 by Lemma 2.5. This completes the proof.  Let X be a metric space and let p be a τ-distance on X. Then, a set-valued mapping T from X into itself is called p-contractive if Tx is nonempty for each x ∈ X and there exists r ∈ [0,1) such that Q(Tx,Ty) ≤ rp(x, y) (3.28) for all x, y ∈ X,where Q(A,B) = sup a∈A inf b∈B p(a, b). (3.29) The following theorem is a generalization of Nadler’s fixed point theorem [15]. Theorem 3.7. Let (X,d) be a complete metric space and let p be a τ-distance on X.LetT be a set-valued p-contractive mapping from X into itself such that for any x ∈ X, Tx is a nonempty closed subset of X. Then there exists x 0 ∈ X such that x 0 ∈ Tx 0 and p(x 0 ,x 0 ) = 0. 204 Fixed point theorems concerning τ-distance Remark 3.8. z ∈ Tz does not necessarily imply p(z,z) = 0; see Example 3.9. Proof. By the assumption, there exists r  ∈ [0,1) such that Q(Tx,Ty) ≤ r  p(x, y)forall x, y ∈ X.Putr = (1 + r  )/2 ∈ [0,1) and fix x, y ∈ X and u ∈ Tx. Then, in the case of p(x, y) > 0, there is v ∈ Ty satisfying p(u,v) ≤ rp(x, y). In the case of p(x, y) = 0, we have Q(Tx,Ty) = 0. Then there exists a sequence {v n } in Ty satisfying lim n p(u, v n ) = 0. By Lemma 2.5, {v n } is p-Cauchy, and hence {v n } is Cauchy. Since X is complete and Ty is closed, {v n } converges to some point v ∈ Ty.Thenwehave p(u, v) ≤ lim n→∞ p  u,v n  = 0 = rp(x, y). (3.30) Hence, we have shown that for any x, y ∈ X and u ∈ Tx, there is v ∈ Ty with p(u,v) ≤ rp(x, y). Fix u 0 ∈ X and u 1 ∈ Tu 0 . Then there exists u 2 ∈ Tu 1 such that p(u 1 ,u 2 ) ≤ rp(u 0 ,u 1 ). Thus, we have a sequence {u n } in X such that u n+1 ∈ Tu n and p(u n ,u n+1 ) ≤ rp(u n−1 ,u n )foralln ∈ N.Foranyn ∈ N,wehave p  u n ,u n+1  ≤ rp  u n−1 ,u n  ≤ r 2 p  u n−2 ,u n−1  ≤··· ≤r n p  u 0 ,u 1  , (3.31) and hence, for any m,n ∈ N with m>n, p  u n ,u m  ≤ p  u n ,u n+1  + p  u n+1 ,u n+2  + ···+ p  u m−1 ,u m  ≤ r n p  u 0 ,u 1  + r n+1 p  u 0 ,u 1  + ···+ r m−1 p  u 0 ,u 1  ≤ r n 1 − r p  u 0 ,u 1  . (3.32) By Lemma 2.6, {u n } is a p-Cauchy sequence. Hence, by Lemma 2.4, {u n } is a Cauchy sequence. So, {u n } converges to some point v 0 ∈ X.Forn ∈ N,from(τ3), we have p  u n ,v 0  ≤ liminf m→∞ p  u n ,u m  ≤ r n 1 − r p  u 0 ,u 1  . (3.33) By hypothesis, we also have w n ∈ Tv 0 such that p(u n ,w n ) ≤ rp(u n−1 ,v 0 )forn ∈ N.Sowe have limsup n→∞ p  u n ,w n  ≤ limsup n→∞ rp  u n−1 ,v 0  ≤ lim n→∞ r n 1 − r p  u 0 ,u 1  = 0. (3.34) By Lemma 2.6, {w n } converges to v 0 .SinceTv 0 is closed, we have v 0 ∈ Tv 0 .Forsuchv 0 , there exists v 1 ∈ Tv 0 such that p(v 0 ,v 1 ) ≤ rp(v 0 ,v 0 ). Thus, we also have a sequence {v n } in X such that v n+1 ∈ Tv n and p(v 0 ,v n+1 ) ≤ rp(v 0 ,v n )foralln ∈ N.Sowehave p  v 0 ,v n  ≤ rp  v 0 ,v n−1  ≤··· ≤r n p  v 0 ,v 0  . (3.35) Hence limsup n→∞ p  u n ,v n  ≤ lim n→∞  p  u n ,v 0  + p  v 0 ,v n  = 0. 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(4.21) n=1 we have q6 (x,z) ≤ q6 (x, y) + q6 (y,z) for x, y,z ∈ X We note that q5 (x, y) ≤ q6 (x, y) for x, y ∈ X We suppose limn xn = x and limn sup{θ(zn , q6 (zn ,xm )) : m ≥ n} = 0 Then we 208 Fixed point theorems concerning τ-distance have limn sup{θ(zn , q5 (zn ,xm )) : m ≥ n} = 0 In such case, we have already shown that pk (w,x) ≤ liminf n pk (w,xn ) for w ∈ X and k ∈ N Fix λ with ∞ λ< liminf pk w,xn... 1 (1979), no 3, 443– 474 T L Hicks and B E Rhoades, A Banach type fixed -point theorem, Math Japon 24 (1979), no 3, 327–330 J R Jachymski, Caristi’s fixed point theorem and selections of set-valued contractions, J Math Anal Appl 227 (1998), no 1, 55–67 J S Jung, Y J Cho, B S Lee, G M Lee, and H K Pathak, Some nonconvex minimization theorems in complete metric spaces and applications, Panamer Math J 7 (1997),... Publishers, Yokohama, 2000 D Tataru, Viscosity solutions of Hamilton-Jacobi equations with unbounded nonlinear terms, J Math Anal Appl 163 (1992), no 2, 345–392 J.-S Ume, Some existence theorems generalizing fixed point theorems on complete metric spaces, Math Japon 40 (1994), no 1, 109–114 C.-K Zhong, On Ekeland’s variational principle and a minimax theorem, J Math Anal Appl 205 (1997), no 1, 239–250... proof Example 3.9 Put X = {0,1} and define a τ-distance p on X by p(x, y) = y for all x, y ∈ X, and a set-valued p-contractive mapping T from X into itself by T(x) = X for all x ∈ X Then 1 ∈ X is a fixed point of T and p(1,1) = 0 4 Other examples of τ-distances In this section, we give other examples of τ-distances generated by either some τ-distance p or a family of τ-distances Proposition 4.1 Let p be . SEVERAL FIXED POINT THEOREMS CONCERNING τ-DISTANCE TOMONARI SUZUKI Received 21 October 2003 and in revised form 10 March 2004 Using the notion of τ-distance, we prove several fixed point theorems, . {y n } is also a p-Cauchy sequence and lim n d(x n , y n ) = 0. 3. Fixed point theorems In this section, we prove several fixed point theorems in complete metric spaces. In [20], the following theorem. Takahashi, Fixed point theorems and characterizations of metric completeness, Topol. Methods Nonlinear Anal. 8 (1996), no. 2, 371–382. [23] W. Takahashi, Existence theorems generalizing fixed point theorems