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# Kannan-type contractions and fixed points in uniform spaces

## Abstract

In uniform spaces, using $J$-families of generalized pseudodistances, we construct four kinds of contractions of Kannan type and, by techniques based on these generalized pseudodistances, we prove fixed point theorems for such contractions. The results are new in uniform and locally convex spaces and even in metric spaces. Examples are given.

MSC: 47H10; 47H09; 54E15; 46A03; 54E35.

## 1 Introduction

Given a space X, Fix(T) denotes the set of fixed points of T : X → X, i.e. Fix(T) = {w X : w = T(w)}. For v0 any point of X, by (vm : m {0} ) we mean the sequence of iteration of T : X → X starting at v0, i.e. $∀ m ∈ { 0 } ∪ ℕ { v m = T [ m ] ( v 0 ) }$.

Recall that maps satisfying the conditions (B) and (K) that are presented in Theorems 1.1 and 1.2 below are called in literature Banach contractions and Kannan contractions, respectively, and first arose in works [1, 2] and [3, 4], respectively.

Theorem 1.1[1, 2]Let (X, d) be a complete metric space. If T : X → X satisfies

$( B ) ∃ λ ∈ [ 0 , 1 ) ∀ x , y ∈ X { d ( T ( x ) , T ( y ) ) ≤ λ d ( x , y ) } ,$

then: (a) T has a unique fixed point w in X; and (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$.

Theorem 1.2[3]Let (X, d) be a complete metric space. If T : XX satisfies (K) η[0,1/2)x, yX{d(T(x), T(y)) ≤ η [d(T(x), x) + d(T(y), y)]},

then: (a) T has a unique fixed point w in X; and (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$.

Theorem 1.3[4]Let (X, d) be a metric space. Assume that: (i) T : X → X satisfies (K); (ii) there exists w X such that T is continuous at a point w; and (iii) there exists a point v0 X such that the sequence (vm : m {0} ) has a subsequence (vmk : k {0} ) satisfying $lim k → ∞ v m k = w$.Then, w is a unique fixed point of T in X.

A great number of applications and extensions of these results have appeared in the literature and plays an important role in nonlinear analysis. The different line of research focuses on the study of the following interesting aspects of fixed point theory in metric spaces and has intensified in the past few decades: (a) the existence and uniqueness of fixed points of various generalizations of Banach and Kannan contractions; (b) the similarity between Banach and Kannan contractions; and (c) the interplay between metric completeness and the existence of fixed points of Banach and Kannan contractions. These aspects have been successfully studied in various papers; see, for example, [521] and references therein.

It is interesting that Theorem 1.2 is independent of Theorem 1.1 that every Banach contraction and every Kannan contraction on a complete metric space has a unique fixed point and that in Theorem 1.3 the completeness of the metric space is omitted. Clearly, Banach contractions are always continuous but Kannan contractions are not necessarily continuous. Next, it is worth noticing that Theorem 1.2 is not an extension of Theorem 1.1. In [5], it is constructed an example of noncomplete metric space X such that each Banach contraction T : XX has a fixed point which implies that Theorem 1.1 does not characterize metric completeness. In [6], it is proved that a metric space X is complete if and only if every Kannan contraction T : XX has a fixed point which implies that Theorem 1.2 characterizes the metric completeness. Similarity between Banach and Kannan contractions may be seen in [7, 8]. In complete metric spaces (X, d), w-distances [9] and τ-distances [10] have found substantial applications in fixed point theory and among others generalizations of Banach and Kannan contractions are introduced, many interesting extensions of Theorems 1.1 and 1.2 to w-distances and τ- distances are obtained, and techniques based on these distances are presented (see, for example, [717]); τ-distances generalize w-distances and metrics d.

The above are some of the reasons why in metric spaces the study of Kannan contractions and generalizations of Kannan contractions plays a particularly important part in fixed point theory.

In this article, in uniform spaces, using $J$-families of generalized pseudodistances, we construct four kinds of contractions of Kannan type (see conditions (C1)-(C4)) and, by techniques based on these generalized pseudodistances, we prove fixed point theorems for such contractions (see Theorems 2.1-2.8). The definitions and the results are new in uniform and locally convex spaces and even in metric spaces. Examples (see Section 12) and some conclusions (see Section 13) are given.

## 2 Statement of main results

Let X be a Hausdorff uniform space with uniformity defined by a saturated family $D = { d α : X 2 → [ 0 , ∞ ) , α ∈ A }$ of pseudometrics d α , $α ∈ A$, uniformly continuous on X2. The notion of $J$-family of generalized pseudodistances on X is as follows:

Definition 2.1 Let X be a uniform space. The family $J = { J α , α ∈ A }$ of maps J α : X2 → [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: $( J 1 ) ∀ α ∈ A ∀ x , y , z ∈ X { J α ( x , z ) ≤ J α ( x , y ) + J α ( y , z ) }$; and

$( J 2 )$For any sequences (x m : m ) and (y m : m ) in X such that

$∀ α ∈ A { lim n → ∞ sup m > n J α ( x n , x m ) = 0 }$
(2.1)

and

$∀ α ∈ A { lim m → ∞ J α ( x m , y m ) = 0 } ,$
(2.2)

the following holds:

$∀ α ∈ A { lim m → ∞ d α ( x m , y m ) = 0 } .$
(2.3)

Remark 2.1 Let X be a uniform space.

(a) Let $J= { J α : X 2 → [ 0 , ∞ ) , α ∈ A )$be a $J$-family. if $∀ α ∈ A ∀ x ∈ x { J α ( x , x ) = 0 }$, then, for each $α∈A$, J α is quasi-pseudometric. Examples of $J$-families such that the maps J α , $α∈A$ are not quasi-pseudometrics are given in Section 12.

(b) The family $J=D$ is a $J$-family on X.

It is the purpose of the present paper to prove the following results.

Theorem 2.1 Let X be a Hausdorff uniform space and assume that the map T : X → X and the$J$-family $J= { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$on X satisfy (C 1) $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x , y ∈ X { J α ( T ( x ) , T ( y ) ) ≤ η α [ J α ( T ( x ) , x ) + J α ( T ( y ) , y ) ] }$and, additionally,

(D 1) $∃ v 0 , ω ∈ X ∀ α ∈ A { lim m → ∞ J α ( v m , w ) = 0 }$.

Then: (a) T has a unique fixed point w in X; (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$; and (c) $∀ α ∈ A { J α ( w , w ) = 0 }$.

Theorem 2.2 Let X be a Hausdorff uniform space and assume that the map T : X → X and the$J$-family $J = { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$on X satisfy at least one of the following three conditions:

(C 2) $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x , y ∈ X { J α ( T ( x ) , T ( y ) ) ≤ η α [ J α ( T ( x ) , x ) + J α ( T ( y ) ) , y ] }$,

(C 3) $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x , y ∈ X { J α ( T ( x ) , T ( y ) ) ≤ η α [ J α ( x , T ( x ) ) + J α ( y , T ( y ) ) ] }$,

(c 4) $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x , y ∈ X { J α ( T ( x ) , T ( y ) ) ≤ η α [ J α ( x , T ( x ) ) + J α ( T ( y ) , y ) ] }$,

(D 2) $∃ v 0 , ω ∈ X ∀ α ∈ A { lim m → ∞ J α ( v m , w ) = lim m → ∞ J α ( w , v m ) = 0 }$.

Then: (a) T has a unique fixed point w in X; (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$; and (c) $∀ α ∈ A { J α ( w , w ) = 0 }$.

It is worth noticing that conditions (C1)-(C4) are different and conditions (D1) and (D2) are different since the $J$-family is not symmetric.

Clearly, (D1) include (D2). The following theorem shows that with some additional conditions the converse holds.

Theorem 2.3 Let X be a Hausdorff uniform space and assume that the map T : X → X and the$J$-family$J = { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$on X satisfy at least one of the conditions (C2)-(C4) and, additionally, condition (D1) and at least one of the following conditions (D3)-(D6):

(D 3) $∀ v 0 , w ∈ X { lim m → ∞ v m = w ⇒ ∃ q ∈ ℕ { T [ q ] ( w ) = w } }$,

(D 4) $∀ v 0 , w ∈ X { lim m → ∞ v m = w ⇒ ∃ q ∈ ℕ { T [ q ] i s c o n t i n u o u s a t a p o i n t w } }$,

(D 5) $∀ v 0 , w ∈ X { lim m → ∞ v m = w ⇒ ∃ q ∈ ℕ { lim m → ∞ T [ q ] ( v m ) = T [ q ] ( w ) } }$,

(D 6) $∀ v 0 , w ∈ X { lim m → ∞ v m = w ⇒ ∃ q ∈ ℕ ∀ α ∈ A { lim m → ∞ J α ( T [ q ] ( v m ) , T [ q ] ( w ) ) = 0 } }$.

Then: (a) T has a unique fixed point w in X; (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$; and (c) $∀ α ∈ A { J α ( w , w ) = 0 }$.

The following theorem shows that if we assume that the uniform space is sequentially complete, then the conditions (D1) and (D2) can be omitted.

Theorem 2.4 Let X be a Hausdorff sequentially complete uniform space and assume that the map T : X → X and the$J$-family$J= { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$on X satisfy at least one of the conditions (C1)-(C4) and, additionally, at least one of the conditions (D3)-(D6). Then: (a) T has a unique fixed point w in X; (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$; and (c)$∀ α ∈ A { J α ( w , w ) = 0 }$.

We now introduce the concept of $J$-admissible maps and in the following result these maps will be used to extend Theorem 2.4 to the uniform spaces which are not sequentially complete and without any conditions (D1)-(D6).

Definition 2.2 Let X be a Hausdorff uniform space and let $J = { J α : X 2 → [ 0 , ∞ ) , α ∈ A )$ be a $J$-family on X. We say that T : X → X is $J$-admissible if for each u0 X satisfying $∀ α ∈ A { lim n → ∞ sup m > n J α ( u n , u m ) = 0 }$ there exists w X such that $∀ α ∈ A { lim m → ∞ J α ( u m , w ) = lim m → ∞ J α ( w , u m ) = 0 }$.

Theorem 2.5 Let X be a Hausdorff uniform space and$J= { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$be the$J$-family on X. Let the map T : X → X be$J$-admissible and assume that T and$J$satisfy at least one of the conditions (C1)-(C4). Then: (a) T has a unique fixed point w in X; (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$; and (c) $∀ α ∈ A { J α ( w , w ) = 0 }$.

Also, the following uniqueness results hold.

Theorem 2.6 Let X be a Hausdorff uniform space and assume that the map T : X → X and the$J$- family$J= { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$on X satisfy at least one of the conditions (C1)-(C4) and, additionally, the following conditions (D7) and (D8):

(D 7) There exist q and w X such that T[q]is continuous at a point w;

(D 8) There exists a point v0 X such that the sequence (vm : m {0} ) has a subsequence (vmk : k {0} ) satisfying$lim k → ∞ v m k = w$.

Then: (a) T has a unique fixed point w in X; (b)$lim k → ∞ v m =w$; and (c)$∀ α ∈ A { J α ( w , w ) = 0 }$.

Theorem 2.7 Let X be a Hausdorff sequentially complete uniform space and let the map T : X → X satisfy the condition

(C 5) $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x , y ∈ X { d α ( T ( x ) , T ( y ) ) ≤ η α [ d α ( T ( x ) , x ) + d α ( y , T ( y ) ) ] } .$

Then: (a) T has a unique fixed point w in X; and (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$.

Theorem 2.8 Let X be a Hausdorff uniform space and assume that the map T : X → X satisfies (C5), (D7) and (D8). Then: (a) T has a unique fixed point w in X; and (b) $∀ u 0 ∈ X { lim m → ∞ u m = w }$.

The rest of this article is organized as follows. In Section 3, we prove some auxiliary propositions. In Sections 4-11, we prove Theorems 2.1-2.8, respectively. Section 12 provides examples and comparisons. Section 13 includes some conclusions.

## 3 Auxiliary propositions

In this section, we present some propositions that will be used in Sections 4-11.

Proposition 3.1 Let X be a Hausdorff uniform space and let$J = { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$be a$J$-family. If x ≠ y, x, y X, then$∃ α ∈ A { J α ( x , y ) ≠ 0 ∨ J α ( y , x ) ≠ 0 }$.

Remark 3.1 If x, y X and $∀ α ∈ A { J α ( x , y ) = 0 ∧ J α ( y , x ) = 0 }$, then x = y.

Proof of Proposition 3.1. Suppose that xy and $∀ α ∈ A { J α ( x , y ) = 0 ∧ J α ( y , x ) = 0 }$. Then, $∀ α ∈ A { J α ( x , x ) = 0 }$, since, by ($J 1$), we get $∀ α ∈ A { J α ( x , x ) ≤ J α ( x , y ) + J α ( y , x ) = 0 }$. Defining x m = x and y m = y for m , we conclude that (2.1) and (2.2) hold. Consequently, by ($J2$), we get (2.3) which implies $∀ α ∈ A { d α ( x , y ) = 0 }$. However, X is a Hausdorff and hence, since xy, we have. $∃ α ∈ A { d α ( x , y ) ≠ 0 }$. Contradiction.   □

Proposition 3.2 Let X be a uniform space and let$J = { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$be a$J$-family. Let$Φ= { ϕ α , α ∈ A }$be the family of maps ϕ α : X→ [0, ∞), $α∈A$.

(a) The families$W ( i ) = { W α ( i ) : X 2 → [ 0 , ∞ ) , α ∈ A }$, i = 1, 2, where, for each$α∈A$, $W α ( 1 ) ( x , y ) =max { ϕ α ( x ) , J α ( x , y ) }$and$W α ( 2 ) ( x , y ) =max { ϕ α ( y ) , J α ( x , y ) }$, x, y X, are$J$-families on X.

(b) The families$V ( i ) = { V α ( i ) : X 2 → [ 0 , ∞ ) , α ∈ A }$, i = 1, 2, where for each$α∈A$$V α ( 1 ) ( x , y ) = ϕ α ( x ) + J α ( x , y )$and$V α ( 2 ) ( x , y ) = ϕ α ( y ) + J α ( x , y )$, x, y X, are$J$-families on X.

Remark 3.2$∀ i ∈ { 1 , 2 } ∀ α ∈ A ∀ x , y ∈ X { J α ( x , y ) ≤ W α ( i ) ( x , y ) ∧ J α ( x , y ) ≤ V α ( i ) ( x , y ) }$.

Proof of Proposition 3.2. (a) For each $α ∈ A$ and for each x, y, z X, using $( J 1 )$ for family $J$, we get $W α ( 1 ) ( x , z ) = max { ϕ α ( x ) , J α ( x , z ) } ≤ max { ϕ α ( x ) + ϕ α ( y ) , J α ( x , y ) + J α ( y , z ) } ≤ W α ( 1 ) ( x , y ) + W α ( 1 ) ( y , z ) }$and$W α ( 2 ) ( x , y ) = max { ϕ α ( z ) , J α ( x , z ) } ≤ max { ϕ α ( y ) + ϕ α ( z ) , J α ( x , y ) + J α ( y , z ) } ≤ W α ( 2 ) ( x , y ) + W α ( 2 ) ( y , z ) }$. Therefore, for each i {1, 2}, the condition $( J 1 )$ for family $W ( i )$ holds.

Let i {1, 2} be arbitrary and fixed and let (x m : m ) and (y m : m ) be arbitrary and fixed sequences in X satisfying $∀ α ∈ A { lim n → ∞ sup m > n W α ( i ) ( x n , x m ) = lim m → ∞ W α ( i ) ( x m , y m ) = 0 }$. Then, by Remark 3. 2, we obtain that the. conditions (2.1) and (2.2) for family $J$ hold and, consequently, since $J$ is a $J$-family, by $( J 2 )$, the condition (2.3) is satisfied, i.e. $∀ α ∈ A { lim m → ∞ d α ( x m , y m ) = 0 }$ which gives that $( J 2 )$ for family $W ( i )$ holds.

Therefore, for each i {1, 2},$W ( i )$is $J$- family.

(b) Using $( J 1 )$ for family $J$, we obtain that, for each $α∈A$and for each x, y, z, X, $V α ( 1 ) ( x , z ) = ϕ α ( x ) + J α ( x , z ) ≤ ϕ α ( x ) + J α ( x , y ) + ϕ α ( y ) + J α ( y , z ) = V α ( 1 ) ( x , y ) + V α ( 1 ) ( y , z ) ,$and $V α ( 2 ) ( x , z ) = ϕ α ( z ) + J α ( x , z ) ≤ ϕ α ( y ) + J α ( x , y ) + ϕ α ( z ) + J α ( y , z ) = V α ( 2 ) ( x , y ) + V α ( 2 ) ( y , z )$. Thus, for each i {1, 2}, the condition $( J 1 )$ for family $V ( i )$ holds.

Let i {1, 2} be arbitrary and fixed and let (x m : m ) (y m : m ) be arbitrary and fixed sequences in X satisfying $∀ α ∈ A { lim n → ∞ sup m > n V α ( i ) ( x n , x m ) = lim m → ∞ V α ( i ) ( x m , y m ) = 0 }$. Then, by Remark 3.2, we obtain that the conditions (2.1) and (2.2) for family $J$ hold and, consequently, by $( J 2 )$, $∀ α ∈ A { lim m → ∞ d α ( x m , y m ) = 0 }$.This gives that $( J 2 )$ for family $V ( i )$holds.

We proved that, for each, i {1, 2}, $V ( i )$is a $J$-family.   □

Proposition 3.3 Let X be a uniform space, let$J= { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$be a$J$-family and let T: XX.

(a) If T and$J$satisfy (C1) or (C3), then$∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ X { max J α ( T ( x ) , T [ 2 ] ( x ) ) , J α ( T [ 2 ] ( x ) , T ( x ) ) } ≤ λ α max { J α ( x , T ( x ) ) , J α ( T ( x ) , x ) } }$.

(b) If T and$J$satisfy (C2) or (C4), then$∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) + J α ( T ( x ) , T [ 2 ] ( x ) ) ≤ λ α [ J α ( T ( x ) , x ) + J α ( x , T ( x ) ) ] }$.

Proof. (a) The proof will be broken into two steps.

STEP 1. If (C1) holds, then the assertion holds.

By (C1), $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) ≤ η α [ J α ( T [ 2 ] ( x ) , T ( x ) ) + J α ( T ( x ) , x ) ] ∧ J α ( T ( x ) , T [ 2 ] ( x ) ≤ η α [ J α ( T ( x ) , x ) + J α ( T [ 2 ] ( x ) , T ( x ) ) ] }$ and, since $∀ α ∈ A { η α ∕ ( 1 - η α ) < 1 }$, we see that the first of these inequalities implies $J α ( T [ 2 ] ( x ) , T ( x ) ) ≤ η α ∕ ( 1 - η α ) ] J α ( T ( x ) , x ) ≤ J α ( T ( x ) , x )$. Hence $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x ∈ X { max { J α ( T ( x ) , T [ 2 ] ( x ) ) , J α ( T [ 2 ] ( x ) , T ( x ) ) } ≤ η α [ J α ( T [ 2 ] ( x ) , T ( x ) ) + J α ( T ( x ) , x ) ] ≤ 2 η α J α ( T ( x ) , x ) ≤ 2 η α max { J α ( x , T ( x ) ) , J α ( T ( x ) , x ) } }$. Now, we see that $∀ α ∈ A { λ α = 2 η α < 1 }$.

STEP 2. If (C3) holds, then the assertion holds.

By (C3), $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) ≤ η α [ J α ( T ( x ) , T [ 2 ] ( x ) ) + J α ( x , T ( x ) ) ] ∧ J α ( T ( x ) , T [ 2 ] ( x ) ) ≤ η α [ J α ( x , T ( x ) ) + J α ( T ( x ) , T [ 2 ] ( x ) ) ] }$ and, since $∀ α ∈ A { η α ∕ ( 1 - η α ) < 1 }$, we see that the second from these inequalities implies $J α ( T ( x ) , T [ 2 ] ( x ) ) ≤ [ η α ∕ ( 1 - η α ) ] J α ( x , T ( x ) ) ≤ J α ( x , T ( x ) )$. Hence, we conclude that $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x ∈ X { max { J α ( T [ 2 ] ( x ) , T ( x ) ) , J α ( T ( x ) , T [ 2 ] ( x ) ) } ≤ η α [ J α ( x , T ( x ) ) + J α ( T ( x ) , T [ 2 ] ( x ) ) ] ≤ 2 η α J α ( x , T ( x ) ) ≤ 2 η α max { J α ( x , T ( x ) ) , J α ( T ( x ) , x ) } } .$. It is clear that $∀ α ∈ A { λ α = 2 η α < 1 }$.

(b) The proof will be broken into two steps.

STEP 1. If (C2) holds, then the assertion holds.

By (C2), $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) ≤ η α ( J α ( T [ 2 ] ( x ) , T ( x ) ) + J α ( x , T ( x ) ) ] ∧ J α T ( x ) , T [ 2 ] ( x ) ) ≤ η α [ J α ( T ( x ) , x ) + J α ( T ( x ) , T [ 2 ] ( x ) ) ] }$. Hence, $∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) ≤ λ α J α ( x , T ( x ) ) ∧ J α ( T ( x ) , T [ 2 ] ( x ) ) ≤ λ α J α ( T ( x ) , x ) }$; here $∀ α ∈ A { λ α = η α ∕ ( 1 - η α ) }$. From this, we conclude that $∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) + J α ( T ( x ) , T [ 2 ] ( x ) ) ≤ λ α [ J α ( T ( x ) , x ) + J α ( x , T ( x ) ) ] }$.

STEP 2. If (C4) holds, then the assertion holds.

By (C4), $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) ≤ η α ( J α ( T ( x ) , T [ 2 ] ( x ) ) + J α ( T ( x ) , x ) ] ∧ J α T ( x ) , T [ 2 ] ( x ) ) ≤ η α [ J α ( x , T ( x ) ) + J α ( T [ 2 ] ( x ) , T ( x ) ) ] }$. This gives $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) ≤ η α 2 ∕ ( 1 - η α 2 ) J α ( x , T ( x ) + η α ∕ ( 1 - η α 2 ) J α ( T ( x ) , x ) ∧ J α T ( x ) , T [ 2 ] ( x ) ) ≤ η α ∕ ( 1 - η α 2 ) J α ( x , T ( x ) ) + η α 2 ∕ ( 1 - η α 2 ) J α ( T ( x ) , x ) } .$. Hence, $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) + J α T ( x ) , T [ 2 ] ( x ) ) ≤ ( η α + η α 2 ) ∕ ( 1 - n α 2 ) [ J α ( x , T ( x ) ) + J α ( T ( x ) , x ) ] = η α ∕ ( 1 - η α ) [ J α x , ( T ( x ) ) + J α ( T ( x ) , x ) ]$. Since $∀ α ∈ A { λ α = η α ∕ ( 1 - η α ) < 1 }$, therefore $∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ X { J α ( T [ 2 ] ( x ) , T ( x ) ) + J α T ( x ) , T [ 2 ] ( x ) ) ≤ λ α [ J α ( x , T ( x ) ) + J α ( T ( x ) , x ) ]$.   □

Proposition 3.4 Let X be a uniform space, let$J = { J α : X 2 → [ 0 , ∞ ) , α ∈ A }$be a$J$-family and let T : X → X. Assume that T and$J$satisfy at least one of the conditions (C1)-(C4). Then:

(a) $∀ α ∈ A ∀ u 0 ∈ X { lim n → ∞ sup m > n J α ( u n , u m ) = lim n → ∞ sup m > n J α ( u m , u n ) = 0 }$.

(b) $∀ α ∈ A ∀ u 0 ∈ X { lim m → ∞ J α ( u m , u m + 1 ) = lim m → ∞ J α ( u m + 1 , u m ) = 0 }$.

(c) $∀ α ∈ A ∀ u 0 ∈ X { lim n → ∞ sup m > n d α ( u n , u m ) = 0 }$.

(d) If there exist z X and q such that z = T[q](z), then Fix(T) = {z} and$∀ α ∈ A { J α ( z , z ) = 0 }$.

(e) If v0, w X satisfy (D1), then limm→∞ vm = w.

(a) The proof will be broken into two steps.

STEP 1. If (C1) or (C3) holds, then the assertion holds.

There exist $J$- families $W ( i ) = { W α ( i ) , α ∈ A }$, i {1, 2}, such that

$∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ X { W α ( 1 ) ( T ( x ) , T [ 2 ] ( x ) ) ≤ λ α W α ( 1 ) ( x , T ( x ) ) }$
(3.1)

and

$∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ x { W α ( 2 ) ( T [ 2 ] ( x ) , T ( x ) ) ≤ λ α W α ( 2 ) ( T ( x ) , x ) } .$
(3.2)

Indeed, by Propositions 3.2(a) and 3.3(a), if $∀ α ∈ A ∀ x ∈ X { ϕ α ( 1 ) ( x ) = J α ( T ( x ) , x ) ∧ ϕ α ( 2 ) ( x ) = J α ( x , T ( x ) ) }$, then the maps $W α ( 1 ) ( x , y ) =max { ϕ α ( 1 ) ( x ) , J α ( x , y ) }$and $W α ( 2 ) ( x , y ) =max { ϕ α ( 2 ) ( y ) , J α ( x , y ) }$, x, y X, have properties (3.1)and (3.2), respectively.

Let $α∈A$ and u0 X be arbitrary and fixed. By (3.1), using $( J 1 )$ for $J$-family $W ( 1 )$, if m > n, we get $W α ( 1 ) ( u n , u m ) ≤ ∑ k = n m - 1 W α ( 1 ) ( u k , u k + 1 ) ≤ ∑ k = n m - 1 λ α k W α ( 1 ) ( u 0 , u 1 ) ≤ W α ( 1 ) ( u 0 , u 1 ) λ α n ∕ ( 1 - λ α ) .$. This gives $lim n → ∞ sup m > n W α ( 1 ) ( u n , u m ) =0$ and, by Remark 3.2, we obtain $∀ α ∈ A ∀ u 0 ∈ X { lim n → ∞ sup m > n J α ( u n , u m ) = 0 } .$.

If $α∈A$ and u0 X are arbitrary and fixed, m, n and m > n, then, by (3.2), using $( J 1 )$ for $J$-family $W ( 2 )$, we get $W α ( 2 ) ( u m , u n ) ≤ ∑ k = n m - 1 W α ( 2 ) ( u k + 1 , u k ) ≤ ∑ k = n m - 1 λ α k W α ( 2 ) ( u 1 , u 0 ) ≤ W α ( 2 ) ( u 1 , u 0 ) λ α n ∕ ( 1 - λ α )$ and $lim n → ∞ sup m > n W α ( 2 ) ( u m , u n ) =0.$ Hence, by Remark 3.2, $∀ α ∈ A ∀ u 0 ∈ X { lim n → ∞ sup m > n J α ( u m , u n ) = 0 } .$

STEP 2. If (C2) or (C4) holds, then the assertion holds.

There exist $J$- families $V ( i ) = { V α i , α ∈ A } ,$i {1, 2}, such that

$∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ X { V α ( 1 ) ( T ( x ) , T [ 2 ] ( x ) ) ≤ λ α V α ( 1 ) ( x , T ( x ) ) }$
(3.3)

and

$∀ α ∈ A ∃ λ α ∈ [ 0 , 1 ) ∀ x ∈ X { V α ( 2 ) ( T [ 2 ] ( x ) , T ( x ) ) ≤ λ α V α ( 2 ) ( T ( x ) , x ) } .$
(3.4)

Indeed, by Propositions 3.2(b) and 3.3(b), we have that if $∀ α ∈ A ∀ x ∈ X { ϕ α ( 1 ) ( x ) = J α ( T ( x ) , x ) ∧ ϕ α ( 2 ) ( x ) = J α ( x , T ( x ) ) } ,$ then the maps $V α ( 1 ) ( x , y ) = ϕ α ( 1 ) ( x ) + J α ( x , y )$ and $V α ( 2 ) ( x , y ) = ϕ α ( 2 ) ( y ) + J α ( x , y ) ,x,y∈X,$ have the above properties.

Let $α∈A$ and u0 X are arbitrary and fixed. Then, by (3.3), using $( J 1 )$ for $J$- family $V ( 1 )$, if m > n, we have $V α ( 1 ) ( u n , u m ) ≤ ∑ k = n m - 1 V α ( 1 ) ( u k , u k + 1 ) ≤ ∑ k = n m - 1 λ α k V α ( 1 ) ( u 0 , u 1 ) ≤ V α ( 1 ) ( u 0 , u 1 ) λ α n ∕ ( 1 - λ α ) .$ consequently, $lim n → ∞ sup m > n V α ( 1 ) ( u n , u m ) =0.$ By Remark 3.2, this gives $∀ α ∈ A ∀ u 0 ∈ X { lim n → ∞ sup m > n J α ( u n , u m ) = 0 }$.

If $α∈A$ and u0 X are arbitrary and fixed, m, n and m > n, then, by (3.4), using $( J 1 )$ for $J$- family $V ( 2 )$, we have $V α ( 2 ) ( u m , u n ) ≤ ∑ k = n m - 1 V α ( 2 ) ( u k + 1 , u k ) ≤ ∑ k = n m - 1 λ α k V α ( 2 ) ( u 1 , u 0 ) ≤ V α ( 2 ) ( u 1 , u 0 ) λ α n ( 1 - λ α )$. Hence, we obtain that $lim n → ∞ sup m > n V α ( 2 ) ( u m , u n ) =0$. By Remark 3.2, this gives $∀ α ∈ A ∀ u 0 ∈ X { lim n → ∞ sup m > n J α ( u m , u n ) = 0 }$.

(b) This is a consequence of (a) since $∀ α ∈ A ∀ u 0 ∈ X ∀ n ∈ ℕ { J α ( u n , u n + 1 ) ≤ sup m > n J α ( u n , u m ) ∧ J α ( u n + 1 , u n ) ≤ sup m > n J α ( u m , u n ) }$.

(c) Let u0 X be arbitrary and fixed. By (a),

$∀ α ∈ A { lim n → ∞ sup m > n J α ( u n , u m ) = 0 }$
(3.5)

which implies $∀ α ∈ A ∀ ε > 0 ∃ n 1 = n 1 ( α , ε ) ∈ ℕ ∀ n > n 1 { sup { J α ( u n , u m ) : m > n } < ε }$ and, in particular,

$∀ α ∈ A ∀ ∈ > 0 ∃ n 1 = n 1 ( α , ∈ ) ∈ ℕ ∀ n > n 1 ∀ s ∈ ℕ { J α ( u n , u s + n ) < ε } .$
(3.6)

Let now i0, j0 , i0 > j0, be arbitrary and fixed. If we define

$x m = u i 0 + m and y m = u j 0 + m for m ∈ ℕ ,$
(3.7)

then (3.6) gives

$∀ α ∈ A { lim m → ∞ J α ( u m , x m ) = lim m → ∞ J α ( u m , y m ) = 0 } .$
(3.8)

Therefore, by (3.5), (3.8) and $( J 2 )$,

$∀ α ∈ A { lim m → ∞ d α ( u m , x m ) = lim m → ∞ d α ( u m , y m ) = 0 } .$
(3.9)

From (3.7) and (3.9), we then claim that

$∀ α ∈ A ∀ ε > 0 ∃ n 2 = n 2 ( α , ε ) ∈ ℕ ∀ m > n 2 { d α ( u m , u i 0 + m ) < ε ∕ 2 }$
(3.10)

and

$∀ α ∈ A ∀ ε > 0 ∃ n 3 = n 3 ( α , ε ) ∈ ℕ ∀ m > n 3 { d α ( u m , u j 0 + m ) < ε ∕ 2 } .$
(3.11)

Let now $α 0 ∈A$ and ε0 > 0 be arbitrary and fixed, let n0 = max {n2(α0, ε0), n3(α0, ε0)} + 1 and let k, l be arbitrary and fixed such that k > l > n0. Then, k = i0 + n0 and l = j0 + n0 for some i0, j0 such that i0> j0 and, using (3.10) and (3.11), we get $d α 0 ( u k , u l ) = d α 0 ( u i 0 + n 0 , u j 0 + n 0 ) ≤ d α 0 ( u n 0 , u i 0 + n 0 ) + d α 0 ( u n 0 , u j 0 + n 0 ) < ε 0 2 + ε 0 2 = ε 0$. Hence, we conclude that $∀ α ∈ A ∀ ε > 0 ∃ n 0 = n 0 ( α , ε ) ∈ ℕ ∀ k , l ∈ ℕ , k > l > n 0 { d α ( u k , u l ) < ε }$.

(d) The proof will be broken into two steps.

STEP 1. If (C1) or (C3) holds, then the assertions hold.

Let z Fix (T[q]) for some z X and q . First, we prove that if $W ( 1 )$ is $J$-family defined in the proof of (a), then

$∀ α ∈ A { W α ( 1 ) ( z , T ( z ) ) = 0 } .$
(3.12)

Otherwise, we have $∃ α 0 ∈ A { W α 0 ( 1 ) ( z , T ( z ) ) > 0 }$. Then, by (3.1), since z = T[q](z) = Τ[2q]= T[2q](z), there exists $λ α 0 ∈ [ 0 , 1 )$, such that $W α 0 ( 1 ) ( z , T ( z ) ) = W α 0 ( 1 ) ( T [ 2 q ] ( z )$, $T [ 2 q ] ( T ( z ) ) )= W α 0 ( 1 ) ( T ( T [ 2 q - 1 ] ( z ) )$, $T [ 2 ] ( T [ 2 q - 1 ] ( z ) ) )≤ λ α 0 W α 0 ( 1 ) ( T ( T [ 2 q - 2 ] ( z ) )$¸ $T [ 2 ] ( T [ 2 q - 1 ] ( z ) ) ) ≤ λ α 0 2 W α 0 ( 1 ) ( T [ 2 q - 1 ] ( z )$, $T ( T [ 2 q - 2 ] ( z ) ) )≤⋯≤ λ α 0 2 q W α 0 ( 1 ) ( z , T ( z ) ) < W α 0 ( 1 ) ( z , T ( z ) )$, which is absurd. Therefore, (3.12) is satisfied.

Next, we see that

$∀ α ∈ A { W α ( 1 ) ( T ( z ) , z ) = 0 } .$
(3.13)

Otherwise, $∃ α 0 ∈ A { W α 0 ( 1 ) ( T ( z ) , z ) > 0 }$ and, since z = T[q](z) = T[2q](z) and if q + 1 < 2q, then by (3.1) and (3.12), for some , we get 0 < $0< W α 0 ( 1 ) ( T ( z ) , z ) = W α 0 ( 1 ) ( T ( T [ q ] ( z ) )$, $T [ 2 q ] ( z ) ) = W α 0 ( 1 ) ( T [ q + 1 ] ( z )$, $T [ 2 q ] ( z ) ) ≤ ∑ i = q + 1 2 q - 1 λ α 0 i W α 0 ( 1 ) ( z , T ( z ) ) ≤ [ λ α 0 q + 1 ∕ ( 1 - λ α 0 ) ] W α 0 ( 1 ) ( z , T ( z ) ) =0$, which is absurd. If q + 1 = 2q, i.e. q = 1, then z = T(z) = T2(z) and, by (3.1) and (3.12), $0< W α 0 ( 1 ) ( T ( z ) , z ) = W α 0 ( 1 ) ( T ( z ) , T [ 2 ] ( z ) ) ≤ λ α 0 W α 0 ( 1 ) ( z , T ( z ) ) =0$, which is absurd. Therefore, (3.13) holds.

Now, using Remark 3.2, we see that (3.12) and (3.13) gives

$∀ α ∈ A { J α ( z , T ( z ) ) = J α ( T ( z ) , z ) = 0 } ,$
(3.14)

which, by Remark 3.1, implies that z Fix (T).

By $( J1 )$and (3.14), we obtain $∀ α ∈ A { J α ( z , z ) ≤ J α ( z , T ( z ) ) + J α ( T ( z ) , z ) = 0 }$.

We show that z is a unique fixed point of T. Otherwise, there exist a, b Fix (T) such that ab. Then, using above for q = 1, we obtain $∀ α ∈ A { J α ( a , T ( a ) ) = J α ( T ( a ) , a ) = 0 }$ and $∀ α ∈ A { J α ( b , T ( b ) ) = J α ( T ( b ) , b ) = 0 }$. Hence, if (C1) holds, then for each $α∈A$, by (C1), J α (a, b) = J α (T (a), T (b)) ≤ η α [J α (T (a), a) + J α (T (b), b)] = 0 and J α (b, a) = J α (T (b), T (a)) ≤ η α [J α (T (b), b) + J α (T (a), a)] = 0 where η α [0, 1/ 2). Hence, $∀ α ∈ A { J α ( a , b ) = J α ( b , a ) = 0 }$. By Remark 3.1, this implies a = b. Contradiction. Similarly, if (C3) holds, then, for each $α∈A$, by (C3), J α (a, b) = J α (T (a), T (b)) ≤ η α [J α (a, T (a)) + J α (b, T (b))] = 0 and J α (b, a) = J α (T (b), T (a)) ≤ η α [J α (b, T (b)) + J α (a, T (a))] = 0 where η α [0, 1/ 2). Thus, $∀ α ∈ A { J α ( a , b ) = J α ( b , a ) = 0 }$ which, by Remark 3.1, implies a = b. Contradiction.

STEP 2. If (C2) or (C4) holds, then the assertions hold.

Let z Fix(T[q]) for some z X and q . We prove that if $V ( 1 )$ is $J$-family defined in the proof of (a), then

$∀ α ∈ A { V α ( 1 ) ( z , T ( z ) ) = 0 } .$
(3.15)

Otherwise, we have $∃ α 0 ∈ A { V α 0 ( 1 ) ( z , T ( z ) ) > 0 }$ and, consequently, by (3.3), since z = T[q](z) = T[2q](z), we get, for some $λ α 0 ∈ [ 0 , 1 ) , V α 0 ( 1 ) ( z , T ( z ) ) = V α 0 ( 1 ) ( T [ 2 q ] ( z )$, $T [ 2 q ] ( T ( z ) ) ) = V α 0 ( 1 ) ( T ( T [ 2 q - 1 ] ( z ) )$, $T [ 2 ] ( T [ 2 q - 1 ] ( z ) ) ) ≤ λ α 0 V α 0 ( 1 ) ( T ( T [ 2 q - 2 ] ( z ) )$, $T [ 2 ] ( T [ 2 q - 2 ] ( z ) ) ) ≤ λ α 0 2 V α 0 ( 1 ) ( T [ 2 q - 2 ] ( z )$, $T ( T [ 2 q - 2 ] ( z ) ) ) ≤⋯≤ λ α 0 2 q V α 0 ( 1 ) ( z , T ( z ) ) < V α 0 ( 1 ) ( z , T ( z ) )$, which is absurd. Therefore, (3.15) holds.

Next, we prove that

$∀ α ∈ A { V α ( 1 ) ( T ( z ) , z ) = 0 } .$
(3.16)

Otherwise, $∃ α 0 ∈ A { V α 0 ( 1 ) ( T ( z ) , z ) > 0 }$ and, since z = T[q](z) = T[2q](z), if q + 1 < 2q, then, by (3.3) and (3.15), for some , we have $0< V α 0 ( 1 ) ( T ( z ) , z ) = V α 0 ( 1 ) ( T ( T [ q ] ( z ) )$, $T [ 2 q ] ( z ) ) = V α 0 ( 1 ) ( T [ q + 1 ] ( z )$, $T [ 2 q ] ( z ) ) ≤ ∑ i = q + 1 2 q - 1 λ α 0 i V α 0 ( 1 ) ( z , T ( z ) ) ≤ [ λ α 0 q + 1 ∕ ( 1 - λ α 0 ) ] V α 0 ( 1 ) ( z , T ( z ) ) = 0$, which is absurd. If q + 1 = 2q, i.e. q = 1, then z T (z) = T2 (z) and, by (3.3) and (3.15), $0< V α 0 ( 1 ) ( T ( z ) , z ) = V α 0 ( 1 ) ( T ( z )$, $T [ 2 ] ( z ) ) ≤ λ α 0 V α 0 ( 1 ) ( z , T ( z ) ) =0$, which is absurd. Therefore, (3.16) holds.

By Remark 3.2, (3.15) and (3.16) implies

$∀ α ∈ A { J α ( z , T ( z ) ) = J α ( T ( z ) , z ) = 0 } ,$
(3.17)

which, by Remark 3.1, gives z Fix (T).

Now, by $( J 1 )$ and (3.17), we obtain $∀ α ∈ A { J α ( z , z ) ≤ J α ( z , T ( z ) ) + J α ( T ( z ) , z ) = 0 }$.

We show that z is a unique fixed point of T. Otherwise, there exist a, b Fix(T) such that ab. Then, by above considerations for q = 1, we get $∀ α ∈ A { J α ( a , T ( a ) ) = J α ( T ( a ) , a ) = 0 }$ and $∀ α ∈ A { J α ( b , T ( b ) ) = J α ( T ( b ) , b ) = 0 }$. Hence, if (C2) holds, then for each $α ∈ A$, by (C2), J α (a, b) = J α (T (a), T (b)) ≤ α [J α (T (a), a) + J α (b, T (b))] = 0 and J α (b, a) = J α (T (b), T (a)) ≤ η α [J α (T (b), b) + J α (a, T (a))] = 0 where η α [0, 1/ 2). Therefore, $∀ α ∈ A { J α ( a , b ) = J α ( b , a ) = 0 }$. Hence, by Remark 3.1, we get a = b, which is impossible. Similarly, if (C4) holds, then, for each $α∈A$, by (C4), J α (a, b) = J α (T (a), T (b)) ≤ η α [J α (a, T (a)) + J α (T (b), b)] = 0 and J α (b, a) = J α (T (b), T (a)) ≤ η α [J α (b, T (b)) + J α (T (a), a)] = 0 where η α [0, 1/ 2). Therefore $∀ α ∈ A { J α ( a , b ) = J α ( b , a ) = 0 }$ and, by Remark 3.1, we get a = b, which is impossible.

(e) Indeed, by (a), we have $∀ α ∈ A { lim m → ∞ sup m > n J α ( v n , v m ) = 0 }$. Next, by (D1), $∀ α ∈ A { lim m → ∞ J α ( v m , w ) = 0 }$. Hence, defining x m = vm and y m = w for m , we conclude that (2.1) and (2.2) hold for sequences (x m : m ) and (y m : m ) in X. Therefore, by $( J 2 )$, we get (2.3) which implies $∀ α ∈ A { lim m → ∞ d α ( v m , w ) = 0 }$.   □

## 4 Proof of Theorem 2.1

The proof will be broken into 11 steps.

STEP 1. If v0, w X satisfy (D1), then $∀ α ∈ A { lim m → ∞ J α ( v m , v m + 1 ) = lim m → ∞ J α ( v m + 1 , v m ) = 0 }$. This follows from Proposition 3.4(b).

STEP 2. If v0, w X satisfy (D1), then lim m→∞ vm = w. This follows from Proposition 3.4(e).

STEP 3. If v0, w X satisfy (D1), then$∀ α ∈ A { J α ( T ( w ) , w ) = 0 }$.

Indeed, by $( J 1 )$ and (C1), $∀ α ∈ A ∃ η α ∈ [ 0 , 1 ∕ 2 ) ∀ m ∈ ℕ { J α ( T ( w ) , w ) ≤ J α ( T ( w )$, $T ( v m ) ) + J α ( T ( v m ) , w ) ≤ η α [ J α ( T ( w ) , w ) + J α ( v m + 1$