Contractions of Banach, Tarafdar, Meir–Keeler, Ćirić–Jachymski–Matkowski and Suzuki types and fixed points in uniform spaces with generalized pseudodistances

J. Math. Anal. Appl. 404 (2013) 338–350

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Contractions of Banach, Tarafdar, Meir–Keeler, Ćirić–Jachymski–Matkowski and Suzuki types and fixed points in uniform spaces with generalized pseudodistances Kazimierz Włodarczyk ∗ , Robert Plebaniak Department of Nonlinear Analysis, Faculty of Mathematics and Computer Science, University of Łódź, Banacha 22, 90-238 Łódź, Poland

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Article history: Received 5 September 2011 Available online 21 March 2013 Submitted by Tomas Dominguez-Benavides Keywords: Fixed point Iterative approximation Generalized contraction J -family of generalized pseudodistances Uniform space

abstract In the uniform spaces (sequentially complete and not sequentially complete), we introduce the concept of the J -families of generalized pseudodistances, we apply it to construct J -generalized contractions and we provide the conditions guaranteeing the existence and uniqueness of fixed points of these contractions and the convergence to these fixed points of each iterative sequence of these contractions. Our J -generalized contractions essentially extend Banach, Tarafdar, Meir–Keeler, Ćirić–Jachymski–Matkowski and Suzuki contractions to uniform spaces with J -families of generalized pseudodistances. Examples showing a difference between our results and the well-known ones are given. The definitions, the results and the method to investigate uniqueness and iterative approximation of fixed points of the maps presented here are new for maps in uniform and locally convex spaces and even in metric spaces. © 2013 Elsevier Inc. All rights reserved.

1. Introduction In the fixed point theory the study of existence and uniqueness problems by iterative approximation were first proposed by Banach. Theorem 1.1. Let (X , d) be a complete metric space and let T : X → X be a map satisfying the condition: (C1) (Banach [1]) ∃06λ<1 ∀x,y∈X {d(T (x), T (y)) 6 λd(x, y)}. Then the following are true: (i) T has a unique fixed point w in X ; and (ii) ∀w0 ∈X {limm→∞ T [m] (w 0 ) = w}. A number of papers involved its various generalizations, variants and applications. Among numerous papers dealing with such theorems, we choose to refer to the papers Tarafdar [11], Meir–Keeler [7], Ćirić–Jachymski–Matkowski [2,3,6] and Suzuki [8–10]. Theorem 1.2. Let X be a Hausdorff sequentially complete uniform space with uniformity defined by a saturated family D = {dα : α ∈ A} of pseudometrics dα , α ∈ A, uniformly continuous on X 2 and let T : X → X . Assume that: (C2) (Tarafdar [11]) ∀α∈A ∃06λα <1 ∀x,y∈X {dα (T (x), T (y)) 6 λα dα (x, y)}. Then the following are true: (i) T has a unique fixed point w in X ; and (ii) ∀w0 ∈X {limm→∞ T [m] (w 0 ) = w}.

∗

Corresponding author. E-mail addresses: [email protected], [email protected] (K. Włodarczyk), [email protected] (R. Plebaniak).

0022-247X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2013.03.030

K. Włodarczyk, R. Plebaniak / J. Math. Anal. Appl. 404 (2013) 338–350

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Theorem 1.3. Let (X , d) be a complete metric space and let T : X → X be a map satisfying one of the following conditions: (C3) (Meir–Keeler [7]) ∀ε>0 ∃η>0 ∀x,y∈X {d(x, y) < ε + η ⇒ d(T (x), T (y)) < ε}; (C4) (Ćirić–Jachymski–Matkowski [2,3,6]) ∀ε>0 ∃η>0 ∀x,y∈X {d(x, y) < ε + η ⇒ d(T (x), T (y)) 6 ε} and ∀x,y∈X {0 < d(x, y) ⇒ d(T (x), T (y)) < d(x, y)}. Then the following are true: (i) T has a unique fixed point w in X ; and (ii) ∀w0 ∈X {limm→∞ T [m] (w 0 ) = w}. Maps T : X → X satisfying conditions (C1)–(C4) are called in literature Banach, Tarafdar, Meir–Keeler (MK, for short) and Ćirić–Jachymski–Matkowski (CJM, for short) contractions, respectively. In fixed point theory the following definition is important. Definition 1.1 (Suzuki [8]). Let X be a metric space with metric d. A map p : X × X → [0, ∞) is called a τ -distance on X if there exists a map η : X × [0, ∞) → [0, ∞) and the following conditions hold: (S1) ∀x,y,z ∈X {p(x, z ) 6 p(x, y) + p(y, z )}; (S2) ∀x∈X ∀t >0 {η(x, 0) = 0 ∧η(x, t ) > t } and η is concave and continuous in its second variable; (S3) limn→∞ xn = x and limn→∞ supm>n η(zn , p(zn , xm )) = 0 imply ∀w∈X {p(w, x) 6 lim infn→∞ p(w, xn )}; (S4) limn→∞ supm>n p(xn , ym ) = 0 and limn→∞ η(xn , tn ) = 0 imply limn→∞ η(yn , tn ) = 0; and (S5) limn→∞ η(zn , p(zn , xn )) = 0 and limn→∞ η(zn , p(zn , yn )) = 0 imply limn→∞ d(xn , yn ) = 0. Using this definition, Suzuki proved the following Theorem 1.4. Let (X , d) be a complete metric space, let p be a τ -distance on X and let T : X → X . Assume that one of the following conditions holds: (C5) (Suzuki [9, Theorem 3.5]) ∀ε>0 ∃η>0 ∀x,y∈X {p(x, y) < ε + η ⇒ p(T (x), T (y)) < ε}; (C6) (Suzuki [10, Theorem 3.3]) ∀ε>0 ∃η>0 ∀x,y∈X {p(x, y) < ε + η ⇒ p(T (x), T (y)) 6 ε} and ∀x,y∈X {0 < p(x, y) ⇒ p(T (x), T (y)) < p(x, y)}; (C7) (Suzuki [8, Theorem 2]) ∃06λ<1 ∀x,y∈X {p(T (x), T (y)) 6 λp(x, y)}. Then the following are true: (i) T has a unique fixed point w in X ; (ii) ∀w0 ∈X {limm→∞ T [m] (w 0 ) = w}; and (iii) p(w, w) = 0. A map T : X → X satisfying one of the conditions (C5)–(C7) is called a Suzuki τ -contraction. Remark 1.1. Let (X , d) be a metric space. (a) Each Banach or MK contraction satisfies ∀x,y∈X {0 < d(x, y) ⇒ d(T (x), T (y)) < d(x, y)}. (b) Each Tarafdar contraction satisfies ∀α∈A ∀x,y∈X {0 < dα (x, y) ⇒ dα (T (x), T (y)) < dα (x, y)}. (c) Each Banach, Tarafdar, MK or CJM contraction is a continuous map. (d) (C3) includes (C1). (e) Condition (C4) is a slight generalization of condition (C3). (f) In metric spaces (X , d), d is τ -distance. Consequently, Theorem 1.4 includes Theorems 1.1 and 1.3. In this paper, in uniform spaces (sequentially complete and not sequentially complete), we introduce the concept of the J -families of generalized pseudodistances (see Sections 2 and 3), we apply it to construct J -generalized contractions and we provide the conditions guaranteeing the existence and uniqueness of fixed points of these contractions and the convergence to these fixed points of each iterative sequence of these contractions (see Theorems 2.1 and 3.1). Our J generalized contractions essentially extend Banach, Tarafdar, MK, CJM and Suzuki contractions to uniform spaces with J families of generalized pseudodistances (see Sections 4 and 6). Examples showing a difference between our results and the well-known ones are given. The definitions, the results and the method to investigate uniqueness and iterative approximation of fixed points of the maps presented here are new for maps in uniform and locally convex spaces and even in metric spaces. This paper is a continuation of [15–20]. 2. J -families of generalized pseudodistances in uniform spaces and uniqueness and iterative approximation of fixed points of J -generalized contractions Let X be a Hausdorff uniform space with uniformity defined by a saturated family D = {dα : α ∈ A} of pseudometrics dα , α ∈ A, uniformly continuous on X 2 . We denote by Fix(T ) and Per (T ) the sets of all fixed points and periodic points of T : X → X , respectively, i.e., Fix(T ) = {w ∈ X : w = T (w)} and Per (T ) = {w ∈ X : w = T [q] (w) for some q ∈ N}. We start by defining the notions of J -family of generalized pseudodistances on X and J -generalized contractions on X . Definition 2.1. Let X be a Hausdorff uniform space. The family J = {Jα : α ∈ A} of maps Jα : X × X → [0, ∞), α ∈ A, is said to be a J -family of generalized pseudodistances on X (J -family,for short) if the following two conditions hold:

(J 1) ∀α∈A ∀x,y,z ∈X {Jα (x, z ) 6 Jα (x, y) + Jα (y, z )}; and (J 2) For any sequence (xm : m ∈ N) in X such that   ∀α∈A lim sup Jα (xn , xm ) = 0 , n→∞ m>n

(2.1)

if there exists a sequence (ym : m ∈ N) in X satisfying

∀α∈A





lim Jα (xm , ym ) = 0 ,

m→∞

(2.2)

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then

∀α∈A





lim dα (xm , ym ) = 0 .

m→∞

(2.3)

Remark 2.1. Let X be a Hausdorff uniform space. (a) From (J 1) and (J 2) it follows that if x ̸= y, x, y ∈ X , then we have ∃α∈A {Jα (x, y) ̸= 0 ∨ Jα (y, x) ̸= 0}. Indeed, if ∀α∈A {Jα (x, y) = 0 ∧ Jα (y, x) = 0}, then ∀α∈A {Jα (x, x) = 0}, since, by (J 1), we get ∀α∈A {Jα (x, x) 6 Jα (x, y) + Jα (y, x) = 0}. Now, defining xm = x and ym = y for m ∈ N, we conclude that (2.1) and (2.2) hold. Consequently, by (J 2), we get (2.3) which implies ∀α∈A {dα (x, y) = 0}. However, X is a Hausdorff and hence, since x ̸= y, we have ∃α∈A {dα (x, y) ̸= 0}, a contradiction. (b) Examples of J -families such that the maps Jα , α ∈ A, that are not pseudometrics are given in Section 6. (c) The family J = {Jα : α ∈ A} of maps Jα : X × X → [0, ∞), α ∈ A, such that Jα = dα , α ∈ A, is a J -family on X ; i.e. the family J = D is a J -family on X . (d) Recently, Włodarczyk and Plebaniak in [14] have studied among others the J -families of generalized pseudodistances in cone uniform spaces which generalize distances of Tataru [12], w -distances of Kada et al. [4], τ -distances of Suzuki [8] and τ -functions of Lin and Du [5] in metric spaces and distances of Vályi [13] in uniform spaces. Definition 2.2. Let X be a Hausdorff uniform space and let the family J = {Jα : α ∈ A} of maps Jα : X × X → [0, ∞), α ∈ A, be a J -family on X . We say that T : X → X is a J -generalized contraction on X (J -GC on X , for short) if one of the following conditions (A1)–(A3) holds: (A1)

∀α∈A ∀ε>0 ∃η>0 ∀x,y∈X {Jα (x, y) < ε + η ⇒ Jα (T (x), T (y)) < ε}.

(2.4)

∀α∈A ∀ε>0 ∃η>0 ∀x,y∈X {Jα (x, y) < ε + η ⇒ Jα (T (x), T (y)) 6 ε}

(2.5)

∀α∈A ∀x,y∈X {0 < Jα (x, y) ⇒ Jα (T (x), T (y)) < Jα (x, y)}.

(2.6)

∀α∈A ∃06λα <1 ∀x,y∈X {Jα (T (x), T (y)) 6 λα Jα (x, y)}.

(2.7)

(A2)

and

(A3)

The main result of this paper is the following: Theorem 2.1. Let X be a Hausdorff uniform space and let the family J = {Jα : α ∈ A} of maps Jα : X × X → [0, ∞), α ∈ A, be a J -family on X . Let T : X → X be J -GC on X . Assume that one of the following conditions holds: (A4) The space X is sequentially complete and T [q] is continuous in X for some q ∈ N; (A5) The space X is sequentially complete and if there exist w0 ∈ X and w ∈ X such that limm→∞ T [m] (w 0 ) = w, then T [q] (w) = w for some q ∈ N; (A6) There exist w 0 , w ∈ X such that ∀α∈A {limm→∞ Jα (wm , w) = limm→∞ Jα (w, wm ) = 0}, where ∀m∈N {w m = T [m] (w 0 )}. Then the following are true: (i) T has a unique fixed point w in X ; (ii) ∀w0 ∈X {limm→∞ T [m] (w 0 ) = w}; and (iii) ∀α∈A {Jα (w, w) = 0}. Remark 2.2. (a) If T is D -GC on X and X is a sequentially complete space, then T is continuous in X and, consequently, the assumptions (A4)–(A6) are not necessary. (b) There exist J -GC on X , J ̸= D , which do not satisfy the conditions (A4)–(A6) and for which the assertions (i)–(iii) do not hold; see Section 6. (c) Clearly, Theorem 2.1 holds in locally convex spaces and is new even in metric spaces (see Sections 3, 4 and 6). (d) It is worth noticing that in the proofs of Theorems 1.1–1.4 the assumption that a metric space is complete is essential. Let us observe that this assumption in condition (A6) is not necessary; see Example 6.12. 3. J -generalized pseudodistances in metric spaces and uniqueness and iterative approximation of fixed points of J generalized contractions Let X be a metric space with metric d. We start by defining the notions of J-generalized pseudodistances on X and Jgeneralized contractions on X . Definition 3.1. Let (X , d) be a metric space. The map J : X × X → [0, ∞) is said to be a J-generalized pseudodistance on X if the following two conditions hold: (J1) ∀x,y,z ∈X {J (x, z ) 6 J (x, y) + J (y, z )}; and

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(J2) For any sequence (xm : m ∈ N) in X such that lim sup J (xn , xm ) = 0,

n→∞ m>n

(3.1)

if there exists a sequence (ym : m ∈ N) in X satisfying lim J (xm , ym ) = 0,

(3.2)

lim d(xm , ym ) = 0.

(3.3)

m→∞

then m→∞

Remark 3.1. Let (X , d) be a metric space. (a) From (J1) and (J2) it follows that if x ̸= y, x, y ∈ X , then we have J (x, y) ̸= 0 or J (y, x) ̸= 0. Indeed, if J (x, y) = 0 and J (y, x) = 0, then J (x, x) = 0, since, by (J1), we get J (x, x) 6 J (x, y) + J (y, x) = 0. Now, defining xm = x and ym = y for m ∈ N, we conclude that (3.1) and (3.2) hold. Consequently, by (J2), we get (3.3) which implies d(x, y) = 0. However, X is a Hausdorff and hence, since x ̸= y, we have d(x, y) ̸= 0, a contradiction. (b) Examples of J-generalized pseudodistances which are not metric are given in Section 6. (c) Clearly, each metric d is a J-generalized pseudodistance. Definition 3.2. Let (X , d) be a metric space and let J : X × X → [0, ∞) be a J-generalized pseudodistance on X . We say that T : X → X is a J-generalized contraction on X (J-GC on X , for short) if one of the following conditions (B1)–(B3) holds: (B1)

∀ε>0 ∃η>0 ∀x,y∈X {J (x, y) < ε + η ⇒ J (T (x), T (y)) < ε}.

(3.4)

∀ε>0 ∃η>0 ∀x,y∈X {J (x, y) < ε + η ⇒ J (T (x), T (y)) 6 ε}

(3.5)

∀x,y∈X {0 < J (x, y) ⇒ J (T (x), T (y)) < J (x, y)}.

(3.6)

∃06λ<1 ∀x,y∈X {J (T (x), T (y)) 6 λJ (x, y)}.

(3.7)

(B2) and

(B3)

As a consequence of Theorem 2.1 we derive the following: Theorem 3.1. Let (X , d) be a metric space and let J : X × X → [0, ∞) be a J-generalized pseudodistance on X . Let T : X → X be J-GC on X . Assume that one of the following conditions holds: (B4) The space X is complete and T [q] is continuous in X for some q ∈ N; (B5) The space X is complete and if there exist w 0 ∈ X and w ∈ X such that limm→∞ T [m] (w 0 ) = w , then T [q] (w) = w for some q ∈ N; (B6) There exist w 0 ∈ X and w ∈ X such that limm→∞ J (w m , w) = limm→∞ J (w, w m ) = 0}, where ∀m∈N {wm = T [m] (w 0 )}. Then the following are true: (i) T has a unique fixed point w in X ; (ii) ∀w0 ∈X {limm→∞ T [m] (w 0 ) = w}; and (iii) J (w, w) = 0. Remark 3.2. (a) If T is d-GC on X and X is a complete space, then T is continuous in X and then the assumptions (B4)–(B6) are not necessary. (b) There exist J-GC on X , J ̸= d, which do not satisfy (B4)–(B6) and for which the assertions (i)–(iii) do not hold; see Section 6. 4. Appendix to Theorem 3.1 The following theorem holds: Theorem 4.1. If X is a metric space with metric d and p : X × X → [0, ∞) is a τ -distance in X , then p is a generalized pseudodistance. Proof. It is clear that (S1) implies (J1). For proving that (J2) holds we assume that the sequences (xm : m ∈ N) and (ym : m ∈ N) in X satisfy (3.1) and (3.2), i.e. limn→∞ supm>n p(xn , xm ) = 0 and limm→∞ p(xm , ym ) = 0. Hence, by [8, Lemma 3, p. 450], we obtain that limn→∞ d(xm , ym ) = 0. Therefore, (J2) is satisfied. Consequently, p is a generalized pseudodistance on X .  Remark 4.1. (a) From Remarks 3.1(c) and 3.2(a) it follows that Theorem 3.1 with condition (B4) or (B5) includes Theorems 1.1–1.3. For the fundamental difference between Theorem 3.1 with condition (B4) or (B5) and Theorems 1.1–1.3, see Section 6. (b) Theorem 1.4 is not true if we replace τ -distance by J-generalized pseudodistance; see Examples 6.8–6.12. (c) If Theorem 1.4 holds, then, in particular, the following condition holds: (B5′ ) If there exist w 0 ∈ X and w ∈ X such that limm→∞ T [m] (w 0 ) = w , then T (w) = w . (d) From (c) and Theorem 4.1 it follows that Theorem 3.1 with condition (B4) or (B5) includes Theorem 1.4.

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Next we show that the following theorem holds: Theorem 4.2. Let (X , d) be a complete metric space, let p be a τ -distance on X and let T : X → X . Assume that one of the conditions (C5)–(C7) holds. Then the following condition holds: (B6′ ) There exist w 0 , w ∈ X such that limm→∞ p(w m , w) = limm→∞ p(w, w m ) = 0}, where ∀m∈N {w m = T [m] (w 0 )}. Proof. By arguments given in the proof of Theorem 1.4 we immediately get that ∀x,y∈X {limm→∞ p(T [m] (x), T [m] (y)) = 0} and, for each w 0 ∈ X , there exists w ∈ X that limn→∞ supm>n p(w n , w m ) = 0, limm→∞ w m = w and T (w) = w , where ∀m∈N {w m = T [m] (w 0 )}. Hence it follows that there exist w 0 ∈ X and w ∈ X satisfying limm→∞ p(w m , w) = limm→∞ p(T [m] (w 0 ), T [m] (w)) = 0 and limm→∞ p(w, w m ) = limm→∞ p(T [m] (w), T [m] (w 0 )) = 0, which yields (B6′ ).  Remark 4.2. From Theorems 4.1 and 4.2 it follows that Theorem 3.1 with condition (B6) includes Theorem 1.4. 5. Proof of Theorem 2.1 The proof will be broken into sixteen steps. Step 1. Assume that (2.4) or (2.7) holds. Then

∀α∈A ∀x,y∈X {Jα (T (x), T (y)) 6 Jα (x, y)}.

(5.1)

Indeed, assume that (2.4) holds and let α0 ∈ A and x0 , y0 ∈ X be arbitrary and fixed. Then, assuming ∀ε>0 {τε = Jα0 (x0 , y0 ) + ε}, from (2.4) we derive ∀ε>0 ∃η>0 ∀u,v∈X {Jα0 (u, v) < τε + η ⇒ Jα0 (T (u), T (v)) < τε }. Of course inequality Jα0 (u0 , v0 ) < τε +η also holds in the case (u0 , v0 ) = (x0 , y0 ) since ∀ε>0 ∀λ>0 {Jα0 (u0 , v0 ) = Jα0 (x0 , y0 ) < Jα0 (x0 , y0 )+ε +λ = τε + λ}. Consequently, ∀ε>0 {Jα0 (T (u0 ), T (v0 )) < τε } holds. Hence, ∀ε>0 {Jα0 (T (x0 ), T (y0 )) < τε = Jα0 (x0 , y0 ) + ε}. This gives Jα0 (T (x0 ), T (y0 )) 6 Jα0 (x0 , y0 ). Therefore, (5.1) holds. Next, if (2.7) holds, then (5.1) is a consequence of (2.7) since ∀α∈A {0 6 λα < 1}. Step 2. Assume that (2.4) or (2.7) holds. Then

∀α∈A ∀x,y∈X {0 < Jα (x, y) ⇒ Jα (T (x), T (y)) < Jα (x, y)}.

(5.2)

Indeed, let (2.4) holds and assume that 0 < Jα0 (x0 , y0 ) for some α0 ∈ A and for some x0 , y0 ∈ X . Then ε = Jα0 (x0 , y0 ) < ε + η for each η > 0 and, by (2.4), Jα0 (T (x0 ), T (y0 )) < ε = Jα0 (x0 , y0 ). Therefore, (5.2) holds. Next, if (2.7) holds, then (5.2) is a consequence of (2.7) since ∀α∈A {0 6 λα < 1}. Step 3. Assume that (2.5) holds. Then

∀α∈A ∀x,y∈X {Jα (x, y) = 0 ⇒ Jα (T (x), T (y)) = 0}.

(5.3)

Indeed, let (2.5) holds and let Jα0 (x0 , y0 ) = 0 for some α0 ∈ A and for some x0 , y0 ∈ X . By (2.5), ∀ε>0 ∃η>0 ∀u,v∈X {Jα0 (u, v) < ε + η ⇒ Jα0 (T (u), T (v)) 6 ε}. Hence, since ∀ε>0 ∀λ>0 {Jα0 (x0 , y0 ) = 0 < ε + λ}, we get ∀ε>0 {Jα0 (T (x0 ), T (y0 )) 6 ε}. This gives Jα0 (T (x0 ), T (y0 )) = 0. Therefore, (5.3) holds. Step 4. Assume that (A1) or (A2) or (A3) holds. Then

∀α∈A ∀x,y∈X





lim Jα (T [m] (x), T [m] (y)) = 0 .

(5.4)

m→∞

(a) Assume that (A1) or (A2) holds and consider the following two cases: Case 1. Suppose first that ∃α0 ∈A ∃x0 ,y0 ∈X ∃m0 ∈{0}∪N {Jα0 (T [m0 ] (x0 ), T [m0 ] (y0 )) = 0}. Then, by (5.1) and (5.3), ∀m>m0 {Jα0 (T [m] (x0 ), T [m] (y0 )) = 0}. Hence we conclude that limm→∞ Jα0 (T [m] (x0 ), T [m] (y0 )) = 0. Case 2. If instead ∀α∈A ∀x,y∈X ∀m∈{0}∪N {Jα (T [m] (x), T [m] (y)) > 0}, then, assuming that α0 ∈ A and x0 , y0 ∈ X are arbitrary and fixed, by (5.2) in the case (A1) or by (2.6) in the case (A2), we get

∀m∈{0}∪N {Jα0 (T [m+1] (x0 ), T [m+1] (y0 )) < Jα0 (T [m] (x0 ), T [m] (y0 ))}. Clearly, the strictly decreasing sequence (Jα0 (T

[m]

(x0 ), T

[m]

(5.5)

(y0 )) : m ∈ {0} ∪ N) is convergent. We show that

lim Jα0 (T [m] (x0 ), T [m] (y0 )) = 0.

(5.6)

m→∞

Assume the contrary, i.e.

∃ε0 >0





lim Jα0 (T [m] (x0 ), T [m] (y0 )) = ε0 .

m→∞

(5.7)

If λ ∈ (0, ε0 ) is arbitrary and fixed, then one can verify that (5.5) and (5.7) imply

∃K =K (λ)∈N ∀m>K {0 < ε0 < Jα0 (T [m+1] (x0 ), T [m+1] (y0 )) < Jα0 (T [m] (x0 ), T [m] (y0 )) < ε0 + λ}.

(5.8)

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From (2.4) in the case (A1) and from (2.5) in the case (A2) it follows that

∃η1 >0 ∀u,v∈X {Jα0 (u, v) < ε0 + η1 ⇒ Jα0 (T (u), T (v)) < ε0 }

(5.9)

∃η2 >0 ∀u,v∈X {Jα0 (u, v) < ε0 + η2 ⇒ Jα0 (T (u), T (v)) 6 ε0 },

(5.10)

and

respectively. Now, by (5.7)–(5.10), if τ = min{η1 , η2 , λ} and η = min{η1 , η2 }, we conclude ∃L=L(τ )∈N;L>K ∀m>L {ε0 < Jα0 (T [m+1] (x0 ), [m+1] T (y0 )) < Jα0 (T [m] (x0 ), T [m] (y0 )) < ε0 + τ 6 ε0 + η} and ∀m>L+1 {Jα0 (T [m] (x0 ), T [m] (y0 )) < ε0 } in the case (A1) and ∀m>L+1 {Jα0 (T [m] (x0 ), T [m] (y0 )) 6 ε0 } in the case (A2), a contradiction. From this we have that (5.6) holds, i.e. limm→∞ Jα0 (T [m] (x0 ), T [m] (y0 )) = 0. Therefore, (5.4) holds. m (b) Assuming now that (A3) holds, we have ∀α∈A ∀x,y∈X {Jα (T [m] (x), T [m] (y)) 6 λm α Jα (x, y)} and ∀α∈A {limm→∞ λα = 0} which gives (5.4). The proof of (5.4) is complete. Step 5. Assume that (A1) or (A2) or (A3) holds. If w 0 ∈ X , then

∀α∈A





lim Jα (w m , w m+1 ) = 0 ∧ lim Jα (w m+1 , w m ) = 0

m→∞

m→∞

(5.11)

where ∀m∈{0}∪N {w m = T [m] (w 0 )}. This is a special case of (5.4) for (x, y) = (w 0 , T (w 0 )) and (x, y) = (T (w 0 ), w 0 ), respectively. Step 6. Assume that (A1) or (A2) holds. Let w 0 ∈ X be arbitrary and fixed and let ∀m∈N {w m = T [m] (w 0 )}. Then

 ∀α∈A



lim sup Jα (w n , w m ) = 0 .

n→∞ m>n

(5.12)

First we see that, by (5.11), the following holds

∀α∈A ∀η>0 ∃M (α,η)∈N ∀s>M (α,η) {Jα (w s , ws+1 ) < η/3 ∧ Jα (w s+1 , ws ) < η/3}.

(5.13)

Now suppose that (5.12) is not true, i.e. there exist α0 ∈ A and ε0 > 0 such that

∀n0 ∈N ∃k>n0 ∃l>k {Jα0 (w k , wl ) > 2ε0 }.

(5.14)

By (2.4) and (2.5), we conclude that

∃η1 >0 ∀x,y∈X {Jα0 (x, y) < ε0 + η1 ⇒ Jα0 (T (x), T (y)) < ε0 }

(5.15)

∃η2 >0 ∀x,y∈X {Jα0 (x, y) < ε0 + η2 ⇒ Jα0 (T (x), T (y)) 6 ε0 },

(5.16)

and respectively, and let now η0 be arbitrary and fixed satisfying 0 < η0 < min{η1 , η2 , ε0 }.

(5.17)

Clearly, by (5.13),

∀s>M (α0 ,η0 ) {Jα0 (w s , ws+1 ) < η0 /3 < ε0 ∧ Jα0 (w s+1 , ws ) < η0 /3 < ε0 }

(5.18)

and, by (5.14) and (5.17), for n0 = M (α0 , η0 ), there exist l > k > n0 such that Jα0 (w k , w l ) > 2ε0 > ε0 + η0 .

(5.19)

Next, for s ∈ N such that k 6 s 6 l, using (5.18), we have Jα0 (w k , w s ) 6 Jα0 (w k , w s+1 ) + Jα0 (w s+1 , w s ) < Jα0 (w k , w s+1 ) + η0 /3 and Jα0 (w k , w s+1 ) 6 Jα0 (w k , w s ) + Jα0 (w s , w s+1 ) < Jα0 (w k , w s ) + η0 /3, which implies

  ∀s∈N,k6s6l Jα0 (wk , ws ) − Jα0 (w k , ws+1 ) < η0 /3.

(5.20)

We see that l ̸= k + 1. Indeed, if l = k + 1, then, by (5.19) and (5.18), we obtain that Jα0 (w k , w l ) > ε0 + η0 and Jα0 (w k , w l ) = Jα0 (w k , w k+1 ) < ε0 , respectively, a contradiction. Therefore, l > k + 2. We see that

∃s0 ∈N,k+16s0 6l {ε0 < ε0 + 2η0 /3 < Jα0 (w k , ws0 ) < ε0 + η0 }.

(5.21)

Otherwise, the following holds

∀s∈N,k+16s6l {ε0 + η0 6 Jα0 (w k , ws ) ∨ Jα0 (w k , ws ) 6 ε0 + 2η0 /3}

(5.22)

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and defining the set S = {s : k + 1 < s 6 l ∧ ε0 + η0 6 Jα0 (w k , w s )}, we first see that S ̸= ∅ since, by (5.19), l ∈ S. Obviously, by the definition of S , k + 1 ̸∈ S. Hence, there exists s∗ = min S satisfying k + 2 6 s∗ 6 l and

ε0 + η0 6 Jα0 (w k , ws∗ ).

(5.23)

However, from (5.23) and the definition of S, we then get ε0 + η0 > Jα0 (w k , w s∗ −1 ) and, by (5.22), since k + 1 6 s∗ − 1 < l,

− (ε0 + 2η0 /3) 6 −Jα0 (w k , ws∗ −1 ).

(5.24)

) 6 |Jα0 (w , ws∗ ) − Next, the consequence of (5.23), (5.24) and (5.20) is η0 /3 6 Jα0 (w , w ) − Jα0 (w , w k s ∗ −1 )| < η0 /3, which is impossible. Therefore, (5.21) is satisfied. Jα0 (w , w By (5.18) and (J 1), we get Jα0 (w k , w s0 ) 6 Jα0 (w k , w k+1 )+Jα0 (w k+1 , w s0 +1 )+Jα0 (w s0 +1 , w s0 ) < η0 /3+Jα0 (w k+1 , w s0 +1 )+ η0 /3. On the other hand, consequence of (5.21) is Jα0 (wk+1 , ws0 +1 ) < ε0 in the case (5.15) and Jα0 (w k+1 , ws0 +1 ) 6 ε0 in the case (5.16). Hence Jα0 (w k , w s0 ) < η0 /3 + ε0 + η0 /3 = ε0 + 2η0 /3, which, by (5.21), is impossible. Therefore, (5.12) is proved. k

s∗

k

s∗ −1

k

Step 7. Assume that (A1) or (A2) holds. Let w 0 ∈ X be arbitrary and fixed and let ∀m∈N {w m = T [m] (w 0 )}. Then

 ∀α∈A



lim sup dα (w n , w m ) = 0 .

(5.25)

n→∞ m>n

Indeed, from (5.12) we claim ∀α∈A ∀ε>0 ∃n1 =n1 (α,ε)∈N ∀n>n1 {sup{Jα (w n , w m ) : m > n} < ε} and, in particular,

∀α∈A ∀ε>0 ∃n1 =n1 (α,ε)∈N ∀n>n1 ∀q∈N {Jα (w n , wq+n ) < ε}.

(5.26)

Let i0 , j0 ∈ N, i0 > j0 , be arbitrary and fixed. If we define um = w i0 +m

and v m = w j0 +m

for m ∈ N,

(5.27)

then (5.26) gives

∀α∈A





lim Jα (w m , um ) = lim Jα (w m , v m ) = 0 .

m→∞

(5.28)

m→∞

Therefore, by (5.12), (5.28) and (J 2),

∀α∈A





lim dα (w m , um ) = lim dα (w m , v m ) = 0 .

m→∞

(5.29)

m→∞

From (5.27) and (5.29) we then claim that

∀α∈A ∀ε>0 ∃n2 =n2 (α,ε)∈N ∀m>n2 {dα (w m , wi0 +m ) < ε/2}

(5.30)

∀α∈A ∀ε>0 ∃n3 =n3 (α,ε)∈N ∀m>n3 {dα (w m , wj0 +m ) < ε/2}.

(5.31)

and

Let now α0 ∈ A and ε0 > 0 be arbitrary and fixed, let n0 = max{n2 (α0 , ε0 ), n3 (α0 , ε0 )} + 1 and let k, l ∈ N be arbitrary and fixed such that k > l > n0 . Then k = i0 + n0 and l = j0 + n0 for some i0 , j0 ∈ N such that i0 > j0 and, using (5.30) and (5.31), we get dα0 (w k , w l ) = dα0 (w i0 +n0 , w j0 +n0 ) 6 dα0 (w n0 , w i0 +n0 ) + dα0 (w n0 , w j0 +n0 ) < ε0 /2 + ε0 /2 = ε0 . Hence, we conclude that ∀α∈A ∀ε>0 ∃n0 =n0 (α,ε)∈N ∀k,l∈N,k>l>n0 {dα (w k , w l ) < ε}. The proof of (5.25) is complete. Step 8. Assume that (A1) or (A2) holds. Let w 0 ∈ X be arbitrary and fixed and let ∀m∈N {w m = T [m] (w 0 )}. If (A4) holds, then there exists a unique w ∈ X such that w = T [q] (w) and limm→∞ w m = w . Indeed, since X is a Hausdorff sequentially complete uniform space and, by Step 7, for each w 0 ∈ X the sequence (w m : m ∈ {0} ∪ N) where ∀m∈N {wm = T [m] (w 0 )}, is a Cauchy sequence, thus there exists a unique w ∈ X such that limm→∞ w m = w . Now we see that if T [q] is continuous for some q ∈ N, then w = T [q] (w). Indeed, we have

wmq+k = T [q] (w (m−1)q+k ) for k = 1, 2, . . . , q and m ∈ N.

(5.32) (m−1)q+k

Clearly, for each k = 1, 2, . . . , q, the sequences (w : m ∈ {0} ∪ N) and (w : m ∈ {0} ∪ N), as subsequences of (w m : m ∈ {0} ∪ N), also converge to w. Since T [q] is continuous, by (5.32), this gives w = T [q] (w). Suppose that T [q] (w) = w and T [q] (v) = v and w ̸= v . Then, by Remark 2.1(a), there exists α0 ∈ A such that Jα0 (w, v) > 0 or Jα0 (v, w) > 0. By (5.2) in the case (A1) and by (2.6) in the case (A2), we get that Jα0 (w, v) = Jα0 (T [q] (w), T [q] (v)) 6 · · · 6 Jα0 (T (w), T (v)) < Jα0 (w, v) if Jα0 (w, v) > 0 and Jα0 (v, w) = Jα0 (T [q] (v), T [q] (w)) 6 · · · 6 Jα0 (T (v), T (w)) < Jα0 (v, w) if Jα0 (v, w) > 0, which is impossible. Consequently, w is a unique fixed point of T [q] and, for each w 0 ∈ X , limm→∞ w m = w . mq+k

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345

Step 9. Assume that (A3) holds. Let w 0 ∈ X be arbitrary and fixed and let ∀m∈N {w m = T [m] (w 0 )}. Then

 ∀α∈A



lim sup Jα (w n , w m ) = 0 .

(5.33)

n→∞ m>n

Indeed, let α0 ∈ A be arbitrary and fixed. For m, n ∈ N such that m > n, using (2.7) and (J 1), we have Jα0 (w n , w m ) 6 (λα0 )n Jα0 (w 0 , w 1 )[1 + λα0 + · · · + (λα0 )m−n−1 ]. Hence supm>n Jα (w n , w m ) = (λα0 )n Jα0 (w 0 , w 1 )/(1 − λα0 ), which implies (5.33). Step 10. Assume that (A3) holds. Let w 0 ∈ X be arbitrary and fixed and let ∀m∈N {w m = T [m] (w 0 )}. Then

 ∀α∈A



lim sup dα (w n , w m ) = 0 .

(5.34)

n→∞ m>n

Arguing as in the proof of Step 7 we obtain that (5.33) implies (5.34). Step 11. Assume that (A3) holds. Let w 0 ∈ X be arbitrary and fixed and let ∀m∈N {w m = T [m] (w 0 )}. If (A4) holds, then there exists a unique w ∈ X such that w = T [q] (w) and limm→∞ w m = w . Since X is a Hausdorff sequentially complete uniform space and, by (5.34), for each w 0 ∈ X , the sequence (w m : m ∈ {0} ∪ N) where ∀m∈N {w m = T [m] (w 0 )}, is a Cauchy sequence, thus there exists a unique w ∈ X such that limm→∞ w m = w and, since (5.32) holds, by continuity of T [q] , we have T [q] (w) = w , i.e. w ∈ Fix(T [q] ). Suppose that T [q] (w) = w and T [q] (v) = v and w ̸= v . Then, by Remark 2.1(a), there exists α0 ∈ A such that Jα0 (w, v) > 0 or Jα0 (v, w) > 0 and, by (5.2), we get that Jα0 (w, v) = Jα0 (T [q] (w), T [q] (v)) 6 λqα0 Jα0 (w, v) < Jα0 (w, v)

or Jα0 (v, w) = Jα0 (T [q] (v), T [q] (w)) 6 λqα0 Jα0 (v, w) < Jα0 (v, w), which is impossible. Consequently, {w} = Fix(T [q] ). Step 12. Assume that (A1) or (A2) or (A3) holds. If (A4) holds, then (i)–(iii) hold. Indeed, from Steps 8 and 11 it follows that there exists a unique w ∈ X such that w = T [q] (w) and limm→∞ w m = w . Hence T (w) = T [q] (T (w)) and, by uniqueness, this gives T (w) = w . Moreover, T [q] (w) = w implies ∀m∈N {w = T [mq] (w)} and ∀α∈A ∀m∈N {Jα (w, w) = Jα (T [mq] (w), T [mq] (w))} hold. Hence, by (5.4), we obtain that ∀α∈A {Jα (w, w) = limm→∞ Jα (T [mq] (w), T [mq] (w)) = 0}.

Step 13. Assume that (A1) or (A2) or (A3) holds. If (A5) holds, then (i)–(iii) hold. Since X is a Hausdorff sequentially complete uniform space, by Steps 7 and 10, for each w 0 ∈ X , the sequence m (w : m ∈ {0} ∪ N) where ∀m∈N {wm = T [m] (w 0 )}, is a Cauchy sequence and thus there exists a unique w ∈ X such that limm→∞ w m = w . Next, by (A5), we get T [q] (w) = w for some q ∈ N. We prove that {w} = Fix(T [q] ). Otherwise, suppose that T [q] (w) = w and T [q] (v) = v and w ̸= v . Then, by Remark 2.1(a), there exists α0 ∈ A such that Jα0 (w, v) > 0 or Jα0 (v, w) > 0. By (5.2) in the case (A1) and by (2.6) in the case (A2), we get that Jα0 (w, v) = Jα0 (T [q] (w), T [q] (v)) 6 · · · 6 Jα0 (T (w), T (v)) < Jα0 (w, v) if Jα0 (w, v) > 0 and Jα0 (v, w) = Jα0 (T [q] (v), T [q] (w)) 6 · · · 6 Jα0 (T (v), T (w)) < Jα0 (v, w) if Jα0 (v, w) > 0, which is impossible. Moreover, by (5.2) in the case (A3), we get that Jα0 (w, v) = Jα0 (T [q] (w), T [q] (v)) 6 λqα0 Jα0 (w, v) < Jα0 (w, v) or Jα0 (v, w) =

Jα0 (T [q] (v), T [q] (w)) 6 λqα0 Jα0 (v, w) < Jα0 (v, w), which is impossible. Consequently, {w} = Fix(T [q] ). We see that {w} = Fix(T ). Otherwise, {w} ̸= Fix(T ) and there exists v ∈ Fix(T ) satisfying w ̸= v . By Remark 2.1(a), there exists α0 ∈ A such that Jα0 (w, v) > 0 or Jα0 (v, w) > 0. Hence, by (5.2) in the case (A1) and by (2.6) in the case (A2), we get Jα0 (w, v) = Jα0 (T (w), T (v)) < Jα0 (w, v) or Jα0 (v, w) = Jα0 (T (v), T (w)) < Jα0 (v, w), respectively, which is impossible. Further, by (5.2) in the case (A3), we get that Jα0 (w, v) = Jα0 (T (w), T (v)) 6 λα0 Jα0 (w, v) < Jα0 (w, v) or Jα0 (v, w) = Jα0 (T (v), T (w)) 6 λα0 Jα0 (v, w) < Jα0 (v, w), which is also impossible. Finally, we see that ∀α∈A {Jα (w, w) = 0}. Indeed, we have ∀α∈A ∀m∈N {Jα (w, w) = Jα (T [m] (w), T [m] (w))}. Consequently, by (5.4), we get ∀α∈A {Jα (w, w) = limm→∞ Jα (T [m] (w), T [m] (w)) = 0}. Step 14. Assume that (A1) or (A2) or (A3) holds and let w 0 , w ∈ X and ∀m∈N {w m = T [m] (w 0 )} be as in condition (A6). Then the following hold: limm→∞ w m = w, T (w) = w, {w} = Fix(T ) and ∀α∈A {Jα (w, w) = 0}. First, we prove limm→∞ w m = w . By (J 1), ∀α∈A {limn→∞ supm>n Jα (w n , w m ) 6 limn→∞ supm>n {Jα (w n , w) + Jα (w, w m )} = 0} and ∀α∈A {limm→∞ Jα (w m , w) = 0}. Thus, defining xm = w m and ym = w for m ∈ N, we conclude that (2.1) and (2.2) hold. Hence, by (J 2), we get (2.3) which implies ∀α∈A {limm→∞ dα (w m , w) = 0}. Next we show that T (w) = w . Indeed, by (J 1) and Steps 1–3, for each α ∈ A and m ∈ N, we have that Jα (T (w), w) 6 Jα (T (w), T (w m )) + Jα (T (w m ), w) 6 Jα (w, w m ) + Jα (w m+1 , w) and Jα (w, T (w)) 6 Jα (w, T (w m )) + Jα (T (w m ), T (w)) 6 Jα (w, w m+1 ) + Jα (w m , w). Hence, using (A6), we obtain ∀α∈A {Jα (T (w), w) = Jα (w, T (w)) = 0} which, by Remark 2.1(a), implies T (w) = w . The proofs of {w} = Fix(T ) and ∀α∈A {Jα (w, w) = 0} are analogous as in Step 13 and will be omitted. Step 15. Assume that (A1) or (A2) or (A3) holds and let w 0 , w ∈ X and ∀m∈N {w m = T [m] (w 0 )} be as in condition (A6). Let v 0 ∈ X , v 0 ̸= w 0 , be arbitrary and fixed and let ∀m∈N {v m = T [m] (v 0 )}. Then

∀α∈A





lim Jα (v m , w) = lim Jα (w, v m ) = 0 .

m→∞

m→∞

(5.35)

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By (J 1), (A6) and Step 4, ∀α∈A {limm→∞ Jα (v m , w) 6 limm→∞ Jα (v m , w m ) + limm→∞ Jα (w m , w) = limm→∞ Jα (T [m] (v 0 ), T (w 0 )) + limm→∞ Jα (w m , w) = 0} and ∀α∈A {limm→∞ Jα (w, v m ) 6 limm→∞ Jα (w, w m ) + limm→∞ Jα (w m , v m ) = limm→∞ Jα (w, w m ) + limm→∞ Jα (T [m] (w 0 ), T [m] (v 0 )) = 0}. Therefore, (5.35) holds. [m]

Step 16. Assume that (A1) or (A2) or (A3) holds. If (A6) holds, then (i)–(iii) hold. This is a consequence of Steps 14, 15, 6 and 9 and condition (J 2). The proof of Theorem 2.1 is now complete.  6. Examples and comparison of J -GC and J -GC with Banach, Tarafdar, MK, CJM and Suzuki contractions In this section we present some examples illustrating the concepts introduced so far. Let X be a Hausdorff uniform space with uniformity defined by a saturated family {dα : α ∈ A} of pseudometrics dα , α ∈ A, uniformly continuous on X 2 . First, we present some examples of J -families and J-generalized pseudodistances. Example 6.1. Let X be a Hausdorff uniform space containing at least three different points, let F and U be sets such that F ⊂ U ⊂ X , let F contain at least one point and let U contain at least two points. Let, for each α ∈ A, the numbers aα , bα and cα be arbitrary and fixed and such that ∀α∈A {0 < aα < bα < cα }. We show that the family J = {Jα : X × X → [0, ∞), α ∈ A}, given by formulae

 0  

aα Jα (x, y) = bα   cα

if x = y ∈ F if x ̸= y and {x, y} ∩ F = {x, y} if {x, y} ∩ (U \ F ) ̸= ∅ and {x, y} ∩ U = {x, y} if {x, y} ∩ U ̸= {x, y},

(6.1)

where x, y ∈ X , α ∈ A, is a J -family on X . To prove (J 1), assume that

∃α0 ∈A ∃x0 ,y0 ,z0 ∈X {Jα0 (x0 , z0 ) > Jα0 (x0 , y0 ) + Jα0 (y0 , z0 )}.

(6.2)

Then we have three cases: Case 1. If Jα0 (x0 , z0 ) = cα0 then, from (6.1), there exists w0 ∈ {x0 , z0 } such that w0 ̸∈ U. By (6.1), Jα0 (w0 , y0 ) = cα0 if w0 = x0 and Jα0 (y0 , w0 ) = cα0 if w0 = z0 , which is in contradiction with (6.2). Case 2. If Jα0 (x0 , z0 ) = bα0 then, from (6.1), x0 , z0 ∈ U and there exists w0 ∈ {x0 , z0 } such that w0 ̸∈ F . By (6.1), Jα0 (w0 , y0 ) > bα0 if w0 = x0 and Jα0 (y0 , w0 ) > bα0 if w0 = z0 , which is in contradiction with (6.2). Case 3. If Jα0 (x0 , z0 ) = aα0 then, from (6.1), x0 , z0 ∈ F and x0 ̸= z0 . By (6.1), Jα0 (x0 , y0 ) > aα0 or Jα0 (y0 , z0 ) > aα0 , which is in contradiction with (6.2). Thus (J 1) holds. To prove that (J 2) holds, we see that if the sequences (xm : m ∈ N) and (ym : m ∈ N) in X satisfy (2.1) and (2.2), then, in particular, (2.2) yields

∀α∈A ∀0<εα m0 {Jα (xm , ym ) < εα }.

(6.3)

Now, by (6.1), denoting m = min{m0 (α, εα ) : α ∈ A}, we conclude that the condition (6.3) gives ′

∀m>m′ {xm = ym ∈ F }.

(6.4)

Using (6.3) and (6.4), we get ∀α∈A ∀0<εα m0 {dα (x , y ) = 0 < εα }. The result is that the sequences (xm : m ∈ N) and (ym : m ∈ N) satisfy (2.3). Hence we find (J 2). m

m

Example 6.2. (a) Let X be a Hausdorff uniform space. The family J = {Jα : α ∈ A} of maps Jα : X × X → [0, ∞), α ∈ A, such that Jα = dα , α ∈ A, is a J -family on X . (b) If X is a metric space with metric d, then J = d is a generalized pseudodistance. Example 6.3. Let X be a metric space with metric d. Let the set E ⊂ X , containing at least two different points, be arbitrary and fixed and let c > 0 satisfies δ(E ) < c where δ(E ) = sup{d(x, y) : x, y ∈ E }. Let J : X × X → [0, ∞) be defined by the formula J (x, y) =



d(x, y) c

if E ∩ {x, y} = {x, y} , if E ∩ {x, y} ̸= {x, y}

x, y ∈ X .

(6.5)

We show that J is a generalized pseudodistance on X . Indeed, it is worth noticing that the condition (J1) does not hold only if there exist some x0 , y0 , z0 ∈ X such that J (x0 , z0 ) > J (x0 , y0 ) + J (y0 , z0 ). This inequality is equivalent to c > d(x0 , y0 ) + d(y0 , z0 ) where J (x0 , z0 ) = c , J (x0 , y0 ) =

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347

d(x0 , y0 ) and J (y0 , z0 ) = d(y0 , z0 ). However, by (6.5), J (x0 , z0 ) = c gives that there exists v ∈ {x0 , z0 } such that v ̸∈ E , J (x0 , y0 ) = d(x0 , y0 ) gives {x0 , y0 } ⊂ E and J (y0 , z0 ) = d(y0 , z0 ) gives {y0 , z0 } ⊂ E. This is impossible. Therefore, ∀x,y,z ∈X {J (x, y) 6 J (x, z ) + J (z , y)}, i.e. the condition (J1) holds. For proving that (J2) holds we assume that the sequences (xm : m ∈ N) and (ym : m ∈ N) in X satisfy (3.1) and (3.2). Then, in particular, (3.2) yields

∀0<εm0 {J (xm , ym ) < ε}.

(6.6)

By (6.6) and (6.5), since ε < c, we conclude that

∀m>m0 {E ∩ {xm , ym } = {xm , ym }}.

(6.7)

From (6.7), (6.5) and (6.6), we get ∀0<εm0 {d(x , y ) < ε}. Therefore, the sequences (x m ∈ N) satisfy (3.3). Consequently, the property (J2) holds. m

m

m

: m ∈ N) and (ym :

The following four examples illustrate the fundamental difference between Theorem 2.1 (with condition (A4) or (A5)) and 3.1 (with condition (B4) or (B5)) and Theorems 1.1–1.3. Example 6.4. Let X = [0, 1], U = [0, 1] \ {1/2, 1}, F = [0, 1/2). Let T : X → X be a continuous map of the form: T (x) = 0 for x ∈ [0, 1/2]; T (x) = (3/2)x − 3/4 for x ∈ (1/2, 3/4]; T (x) = 3/8 for x ∈ (3/4, 1]. Defining

 0   1/2 J (x, y) =  2 3

if x = y ∈ F if x ̸= y and {x, y} ∩ F = {x, y} if {x, y} ∩ (U \ F ) ̸= ∅ and {x, y} ∩ U = {x, y} if {x, y} ∩ U ̸= {x, y},

(6.8)

we see, by Example 6.1 and Definition 3.1, that J is a generalized pseudodistance on X . We prove that T is J-GC on X even for all conditions (B1)–(B3). (F1) T satisfies condition (B1). Indeed, let ε > 0. Consider the following two cases: Case 1. If ε 6 1/2, then there exists η0 = 1/2 such that for each x, y ∈ X satisfying J (x, y) < ε + η0 we have J (x, y) < ε + η0 6 1/2 + 1/2 = 1, so by (6.8), we get J (x, y) = 1/2 or J (x, y) = 0 and, consequently, x ̸= y and {x, y} ⊂ F or x = y and {x, y} ⊂ F , respectively. Hence, by (6.8) and the definition of T , J (T (x), T (y)) = J (0, 0) = 0 < ε. Case 2. If 1/2 < ε , then, for each η > 0, and, for each x, y ∈ X satisfying J (x, y) < ε + η, since T (X ) ⊂ F , by (6.8), we get J (T (x), T (y)) = 0 < ε if T (x) = T (y) and J (T (x), T (y)) = 1/2 < ε if T (x) ̸= T (y). (F2) T satisfies condition (B2). First we prove that (3.5) holds. Indeed, let ε > 0. Consider the following two cases: Case 1. If ε < 1/2, then the argumentation is the same as in the Case 1 of (F1) and will be omitted. Case 2. If 1/2 6 ε , then for each η > 0 and, for each x, y ∈ X satisfying J (x, y) < ε + η, since T (X ) ⊂ F , we have J (T (x), T (y)) = 0 < ε if T (x) = T (y) and J (T (x), T (y)) = 1/2 6 ε if T (x) ̸= T (y). Therefore, (3.5) is satisfied. Now we see that (3.6) holds. Indeed, if 0 < J (x, y), then, by (6.8), 1/2 6 J (x, y) and the following hold: (a) If 1/2 = J (x, y), then x, y ∈ F and x ̸= y and J (T (x), T (y)) = J (0, 0) = 0 < 1/2 = J (x, y); (b) If 1/2 < J (x, y), then, since T (X ) ⊂ F , we have that J (T (x), T (y)) 6 1/2 < J (x, y). (F3) T satisfies condition (B3). Indeed, since T (X ) ⊂ F , thus, by (6.8), we see that J (T (x), T (y)) = 0 if T (x) = T (y) and J (T (x), T (y)) = 1/2 if T (x) ̸= T (y). Therefore, for λ = 1/2 and x, y ∈ X , x ̸= y, we get: Case 1. If T (x) = T (y), then, by the definition of T , it follows that {x, y} ⊂ [0, 1/2] or {x, y} ⊂ [3/4, 1]. For {x, y} ⊂ [0, 1/2], since x ̸= y and 1/2 ̸∈ U, by (6.8), we obtain that J (T (x), T (y)) = 0 < (1/2)λ = λJ (x, y) if {x, y} ⊂ F and J (T (x), T (y)) = 0 < 3λ = λJ (x, y) if {x, y} ∩ {1/2} ̸= ∅. When {x, y} ⊂ [3/4, 1], then {x, y} ∩ (U \ F ) ̸= ∅ and J (T (x), T (y)) = 0 < 2λ = λJ (x, y) if {x, y} ∩ U = {x, y} and J (T (x), T (y)) = 0 < 3λ = λJ (x, y) if {x, y} ∩ U ̸= {x, y}. Case 2. If T (x) ̸= T (y), then J (x, y) = 2 or J (x, y) = 3 since: (a) x ∈ [0, 1/2] implies y ∈ (1/2, 1]; (b) x ∈ (1/2, 3/4) implies y ∈ [0, 1] \ {x}; and (c) x ∈ [3/4, 1] implies y ∈ [0, 3/4). Therefore, J (T (x), T (y)) = 1/2 < λJ (x, y). Consequently, condition (B3) holds. Thus, all assumptions (B1)–(B6) of Theorem 3.1 hold, Fix(T ) = {0}, J (0, 0) = 0 and, for each w 0 , limm→∞ w m = 0. Next we show that the map from Example 6.4 is not d-GC on X . Therefore, the existence of the J-generalized pseudodistance which is different from d is essential. Example 6.5. Let X and the map T : X → X be such as in Example 6.4. (G1) The map T is not a MK contraction. Indeed, if ε0 = 1/4, then for all η > 0 there exist x0 = 1/2 and y0 = 3/4 such that |x0 − y0 | = |1/2 − 3/4| = 1/4 < ε0 + η and |T (x0 ) − T (y0 )| = |0 − 3/8| = 3/8 > 1/4 = ε0 . Consequently, T does not satisfy condition (C3). (G2) The map T is not a CJM contraction. Indeed, let ε0 = 1/4 and let η > 0 be arbitrary and fixed. We consider three cases: Case 1. If 0 < η 6 1/2, x0 = 1/2 and y0 = 3/4 + η/2, then |x0 − y0 | = |1/2 − 3/4 − η/2| = 1/4 + η/2 < ε0 + η and |T (x0 ) − T (y0 )| = |0 − 3/8| = 3/8 > 1/4 = ε0 .

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Case 2. If 1/2 < η 6 3/4, x0 = 1/4 and y0 = 1/4 + η, then |x0 − y0 | = |1/4 − 1/4 − η| = η < ε0 + η and, since 3/4 < 1/4 + η = y0 , by definition of T we obtain that |T (x0 ) − T (y0 )| = |0 − 3/8| = 3/8 > 1/4 = ε0 . Case 3. If η > 3/4, x0 = 1/2 and y0 = 1, then |x0 − y0 | = |1/2 − 1| = 1/2 < ε0 + η and |T (x0 ) − T (y0 )| = |0 − 3/8| = 3/8 > 1/4 = ε0 . Consequently, T does not satisfy condition (C4). (G3) The map T is not a Banach contraction. Indeed, (C3) includes (C1). Now we show that Theorem 2.1 essentially generalized Theorem 1.2 of Tarafdar. Example 6.6. Let Y = RN = R × R × R × · · · be non-normable real Hausdorff locally convex and sequentially complete space with calibration C = {c1 , c2 , c3 , . . .}, cα (x) = |xα | , α ∈ N, x = (x1 , x2 , x3 , . . .) ∈ Y, and let X = [0, 1]N be a Hausdorff sequentially complete uniform space with uniformity defined by a saturated family D = {dα : α ∈ A} of pseudometrics dα , α ∈ A, dα (x, y) = cα (x − y), x, y ∈ X. Let U = [0, 1] \ {1/2, 1} and let F = [0, 1/2). Defining J = {Jα : X × X → [0, ∞), α ∈ N}, where, for α ∈ N and (x, y) ∈ X × X,

 0  

1/2 Jα (x, y) = 2  3

if xα = yα ∈ F if xα ̸= yα and {xα , yα } ∩ F = {xα , yα } if {xα , yα } ∩ (U \ F ) ̸= ∅ and {xα , yα } ∩ U = {xα , yα } if {xα , yα } ∩ U ̸= {xα , yα },

we see, by Example 6.1, that the family J is a J -family on X. Let T = (T1 , T2 , . . .) : X → X be a continuous map of the form 0 (3/2)xα − 3/4 3/8

 Tα (x) =

for xα ∈ [0, 1/2] for xα ∈ (1/2, 3/4], for xα ∈ (3/4, 1]

α ∈ N, x ∈ X.

Then ∀w0 ∈X ∀m>2 {wm = T [m] (w0 ) = (T1 (w0 ), T2 (w0 ), . . .) = 0 = (0, 0, . . .)}, 0 is a unique fixed point of T in X and, by Example 6.4(F3), T satisfies condition (A3) and, by Example 6.5(G3), T is not Tarafdar contraction (C2). Clearly, by Example 6.2, Theorem 2.1 (with condition (A4) or (A5)) includes Theorem 1.2. [m]

[m]

Now, we show that condition (3.6) in Theorem 3.1 with condition (B2) is necessary. Example 6.7. Let (X , d) be a metric space, where X = [0, 1] ∪ {2}, d : X × X → (0, ∞], d(x, y) = |x − y| , x, y ∈ X . Let T : X → X be a continuous map of the form T (x) =

for x ∈ [0, 1] for x = 2.



2 1

(6.9)

Define J = d. By Example 6.2(b), J is a generalized pseudodistance. First we show that T satisfies the condition (3.5):

∀ε>0 ∃η>0 ∀x,y∈X {d(x, y) < ε + η ⇒ d(T (x), T (y)) 6 ε}.

(6.10)

We need to consider two different cases: Case 1. Let 0 < ε < 1. For η = 1 − ε > 0, by (6.9), we have that A = {(x, y) ∈ X × X : d(x, y) < ε + η = ε + (1 − ε) = 1} ⊂ [0, 1]. Consequently, ∀x,y∈A {d(T (x), T (y)) = d(2, 2) = 0 6 ε}. Case 2. Let 1 6 ε . In this case, for η = 2 > 0, we have that 3 6 ε + η and, by (6.9), we conclude the following: B = {(x, y) ∈ X × X : d(x, y) < ε + η} = X × X ; ∀(x,y)∈B {0 < d(T (x), T (y)) 6 d(2, 1) = 1 6 ε}. This gives that (6.10) holds. Now let us observe that T is not contractive. Indeed, for (1, 2) ∈ X × X we have d(T (1), T (2)) = d(2, 1) = 1, i.e. ∀x,y∈X {0 < d(x, y) ⇒ d(T (x), T (y)) < d(x, y)} does not hold. It is easy to see that Fix(T ) = ∅. The following examples illustrate the difference between Theorems 3.1 and 1.4. First, we show that J defined in Example 6.3 is not a τ -distance. Example 6.8. Let (X , d) be a metric space where X = [0, 1] and d : X × X → [0, ∞) is of the form d(x, y) = |x − y|, x, y ∈ X . Let E = (0, 1/2] and, for each γ > 0, define the map J γ : X × X → [0, ∞) by J γ (x, y) =



d(x, y) 1/2 + γ

if E ∩ {x, y} = {x, y} , if E ∩ {x, y} ̸= {x, y}

x, y ∈ X .

Using Example 6.3 we find that, for each γ > 0, the J γ is a generalized pseudodistance on X .

(6.11)

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349

We show that, for each γ > 0, the map pγ : X × X → [0, ∞) of the form pγ (x, y) = J γ (x, y), x, y ∈ X , is not a τ -distance on X . Indeed, suppose that there exists γ0 > 0, such that pγ0 is a τ -distance on X and consider a sequence (xm : m ∈ N) defined by the formula: xm = 1/m, m ∈ N. Of course, (xm : m ∈ N) converges in X and lim xm = 0.

(6.12)

m→∞

Consequently, (xm : n ∈ N) is a Cauchy sequence on X , i.e. lim sup d(xn , xm ) = 0.

(6.13)

n→∞ m>n

Next, since ∀m>2 {xm ∈ E }, by (6.11), we get ∀n,m>2 {pγ0 (xn , xm ) = J γ0 (xn , xm ) = d(xn , xm )}. This and (6.13) give lim sup pγ0 (xn , xm ) = lim sup d(xn , xm ) = 0.

n→∞ m>n

n→∞ m>n

(6.14)

Using now (6.14) and [8, Lemma 3, p. 450], we obtain that (xm : m ∈ N) is pγ0 -Cauchy (recall that if X is a metric space with metric d and p is a τ -distance on X , then a sequence (xm : m ∈ N) in X is called p-Cauchy if there exists a function η : X × [0, ∞) → [0, ∞) satisfying (S2)–(S5) and a sequence (z m : m ∈ N) in X such that limn→∞ supm>n η(z n , p(z n , xm )) = 0; see [8, p. 449]). Consequently, there exists a map η : X ×[0, ∞) → [0, ∞) satisfying (S2)–(S5) and a sequence (z m : m ∈ N) in X such that lim sup η(z n , pγ0 (z n , xm )) = 0.

(6.15)

n→∞ m>n

Now (6.12), (6.15) and condition (S3) imply that

∀w∈X {pγ0 (w, 0) 6 lim inf pγ0 (w, xm )}.

(6.16)

m→∞

However, for w0 = 1/2 ∈ X , since ∀m>2 {xm ∈ E } and w0 = 1/2 ∈ E, according to (6.11), we calculate, since 0 ̸∈ E, that pγ0 (w0 , 0) = pγ0 (1/2, 0) = J γ0 (1/2, 0) = 1/2 + γ0 > 1/2 = lim infm→∞ d(1/2, xm ) = lim infm→∞ J γ0 (1/2, xm ) = lim infm→∞ pγ0 (1/2, xm ) = lim infm→∞ pγ0 (w0 , xm ) which, by (6.16), is impossible. Next, we construct T and J such that T is J-GC on X , conditions (B1)–(B6) hold and J is not τ -distance on X . Example 6.9. Let (X , d) be a metric space where X = [0, 1] and d : X × X → [0, ∞) is of the form d(x, y) = |x − y|, x, y ∈ X . Let T : X → X be a map given by the formula: T (x) =

1/4 1/2



if x = 0 if x ∈ X \ {0}.

(6.17)

Clearly, T [2] is continuous. Let E = (0, 1/2] and define the map J : X × X → [0, ∞) by J (x, y) =

d(x, y) 2



if E ∩ {x, y} = {x, y} , if E ∩ {x, y} ̸= {x, y}

x, y ∈ X .

(6.18)

Using Example 6.8 (for γ = 3/2 > 0) we find that the map J is a generalized pseudodistance on X and the map p : X × X → [0, ∞) of the form p(x, y) = J (x, y), x, y ∈ X , is not a τ -distance on X . We prove that T is J-GC on X even for all conditions (B1)–(B3). Moreover, all conditions (B4)–(B6) hold. (H1) T satisfies condition (B1). Indeed, let ε > 0 be arbitrary and fixed. We consider the following two cases. Case 1. If ε 6 1/4, then there exists η = 1 > 0 such that for each x, y ∈ X , if J (x, y) < ε + η < 2, then, by (6.18), E ∩ {x, y} = {x, y} and since x, y ∈ E = (0, 1/2], by (6.17), T (x) = T (y) = 1/2 ∈ E and this, by (6.18), gives that J (T (x), T (y)) = d(T (x), T (y)) = d(1/2, 1/2) = 0 < ε . Case 2. If ε > 1/4, then there exists η = 2 > 0 such that, for each x, y ∈ X , if J (x, y) < ε + η we have that J (T (x), T (y)) = d(T (x), T (y)) 6 1/4 < ε since T (x), T (y) ∈ {1/4, 1/2} ⊂ E. As a consequence, T satisfies condition (B1). (H2) T satisfies condition (B2). By the same argumentation as in (H1) we have that the condition (3.5) holds. We show that (3.6) also holds. With the aim of this, let x, y ∈ X , such that J (x, y) > 0, be arbitrary and fixed. Since T (x), T (y) ∈ E, so: J (T (x), T (y)) = d(T (x), T (y)) = 1/4 < 2 = J (x, y) if 0 ∈ {x, y}; J (T (x), T (y)) = d(T (x), T (y)) = 0 < J (x, y) if 0 ̸∈ {x, y}. Thus (3.6) holds, and as a consequence, T satisfies condition (B2). (H3) satisfies condition (B3). Then there exists λ = 1/2 such that, for each x, y ∈ X , J (T (x), T (y)) = d(T (x), T (y)) = 1/4 6 1 = (1/2)J (x, y) = λJ (x, y) if 0 ∈ {x, y} and J (T (x), T (y)) = d(T (x), T (y)) = 0 6 (1/2)J (x, y) = λJ (x, y) if 0 ̸∈ {x, y}. Therefore, T satisfies condition (B3). Thus, all assumptions (B1)–(B6) of Theorem 3.1 hold, w = 1/2 is a unique fixed point of T in X , for each w 0 ∈ X , we have ∀m>2 {wm = 1/2} and, by (6.18), J (1/2, 1/2) = 0.

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K. Włodarczyk, R. Plebaniak / J. Math. Anal. Appl. 404 (2013) 338–350

Now, we prove that the map T defined in above example is not d-GC on X . Example 6.10. We see that, for the map T defined in Example 6.9, the conditions (B1)–(B3) do not hold when J = d. We have: (W1) T does not satisfy condition (B1). Indeed, let x = 0, y = 1/4 and ε = d(0, 1/4) = 1/4 > 0. Then, for each η > 0, we get that d(0, 1/4) < ε + η, but d(T (x), T (y)) = d(1/4, 1/2) = 1/4 = d(0, 1/4) = d(x, y). (W2) T does not satisfy condition (B2). We see that, for T , condition (3.6) does not hold. Indeed, if x = 0, y = 1/4, then d(0, 1/4) = 1/4 > 0, but d(T (x), T (y)) = d(1/4, 1/2) = 1/4 = d(0, 1/4) = d(x, y). (W3) T does not satisfy condition (B3). Indeed, suppose that there exists λ ∈ [0, 1), such that ∀x,y∈X {d(T (x), T (y)) 6 λd(x, y)}. Then, for x = 0, y = 1/4, we have 1/4 = d(1/4, 1/2) = d(T (x), T (y)) 6 λd(x, y) < d(x, y) = d(0, 1/4) = 1/4, which is absurd. We construct T and J such that T is J-GC on X for condition (B1), assumptions (B4)–(B6) and assertions (i)–(iii) do not hold and J is not a τ -distance on X . This gives that at last one from the assumptions (B4) or (B5) or (B6) is necessary. Example 6.11. Let (X , d) be a metric space where X = [0, 1] and d : X ×X → [0, ∞) is of the form d(x, y) = |x−y|, x, y ∈ X . Let T : X → X be a map given by the formula: T (x) = 1/2 if x ∈ {0, 1} and T (x) = x/4 + 3/4 if x ∈ (0, 1). Let E = (0, 1) and define the map J : X × X → [0, ∞) by (6.18). Using analogous considerations as in Example 6.8 we find that the map J is a generalized pseudodistance on X and not a τ -distance on X . (Y1) T is J-GC on X for condition (B1). Indeed, first, we see that, ∀x∈X {T (x) ∈ {1/2} ∪ (3/4, 1) ⊂ E } and, consequently, J (T (0), T (1)) = |T (0) − T (1)| = |1/2 − 1/2| = 0 < 10/8 = (5/8)J (0, 1), ∀y∈(0,1) {J (T (0), T (y)) = |T (0) − T (y)| = |y/2 + 1/4| 6 3/4 < 10/8 = (5/8)J (0, y)}, ∀y∈(0,1) {J (T (1), T (y)) = |T (1) − T (y)| = |y/2 + 1/4| 6 3/4 < 10/8 = (5/8)J (1, y)}, ∀x,y∈(0,1) {J (T (x), T (y)) = |T (x) − T (y)| = (1/4) |x − y| 6 (5/8) |x − y| = (5/8)J (x, y). Therefore, the condition (B1) of the form ∀x,t ∈X {J (T (x), T (y)) 6 (5/8)J (x, y)} holds. (Y2) Assumptions (B4)–(B6) and all assertions (i)–(iii) do not hold. Indeed, if we denote ∀m∈N {w m = T [m] (w 0 )} for w 0 ∈ X , then we have: (a) for each q ∈ N, T [q] is not continuous in X ; (b) there exists exactly one w = 1 ∈ X such that, for each w 0 ∈ X , limm→∞ w m = 1 and Fix(T ) = ∅; (c) ∀w0 ∈X {limm→∞ J (w m , 1) = limm→∞ J (1, wm ) = 2}; (d) J (1, 1) = 2. Finally, we illustrate Theorem 3.1 with condition (B6). Example 6.12. Let (X , d) be a metric space where X = [0, 1) and d : X ×X → [0, ∞) is of the form d(x, y) = |x−y|, x, y ∈ X . Let T : X → X be a map given by the formula: T (x) = 1/4 if x = 0 and T (x) = 1/2 if x ∈ X \ {0}. Let E = (0, 1/2] and define the map J : X × X → [0, ∞) by (6.18). The map J is a generalized pseudodistance on X and not a τ -distance on X , X is not complete, T is J-GC on X with assumption (B1) (see (H1) in Example 6.9), condition (B6) holds since, even for each w 0 ∈ X , there exists w = 1/2 ∈ X such that limm→∞ J (w m , w) = limm→∞ J (w, w m ) = 0, Fix(T ) = {1/2}, ∀x∈X ∀q>2 {T [q] (x) = 1/2} and J (1/2, 1/2) = 0. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]

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