Some results in metric fixed point theory

This is a survey of results mainly in metric fixed point theory, including the Darbo–Sadovskĭi theorem using measures of noncompactness. Various different proofs are presented for some of the most important historical results. Furthermore many examples and remarks are added to illustrate the topics of the paper.


Introduction
Fixed point theory is a major branch of nonlinear functional analysis because of its wide applicability. Numerous questions in physics, chemistry, biology, and economics lead to various nonlinear differential and integral equations.
There are two fundamental results, namely Banach's fixed point theorem and Darbos's fixed point theorem The classical Banach contraction principle [2] of Banach's theorem is one of the most useful results in metric fixed point theory. Due to its applications in mathematics and other related disciplines, this principle has been generalized in many directions. Extensions of Banach's contraction principle have been obtained either by generalizing the distance properties of the underlying domain or by modifying the contractive condition on the mappings.
Darbo's fixed point theorem [17] of 1955 uses the condensing principle connected to Kuratowski's measure of noncompactness α [33] of 1930; it is a very important generalization of Schauder's fixed point theorem, and includes the existence part of Banach's fixed point theorem. Other measures of noncompactness were introduced by Goldenštein, Goh'berg and Markus [GGM1], the ball or Hausdorff measure of noncompactness, which was later studied by Goldenštein and Markus [GGM2] in 1968, Istrȃţesku [27] in 1972, and others. Apparently Goldenštein, Goh'berg and Markus were not aware of Kuratowski's and Darbo's work. It is surprising that Darbo's theorem was almost never noticed and applied until in the 1970's mathematicians working in functional analysis, operator theory and differential equations started to apply Darbo's theorem and developed the theory connected with measures of noncompactness. These measures of noncompactness are studied in detail and their use is discussed, for instance, in the monographs [AKP,53,3,28,34,35].

Banach contraction principle
In this section we are going to study the famous Banach fixed point theorem, usually called the Banach contraction principle. This principle from 1922 marks the beginning of the fixed point theory in metric spaces.
We also present several different proofs of Banach's contraction principle Definition 2.1. Let (X, d) be a metric space. A mapping f : X → X is a contraction if there exists some q ∈ [0, 1) such that d(f x, f y) ≤ q · d(x, y), for all x, y ∈ X. (2.1) We observe that every contraction is a continuous mapping. The following theorem shows the existence and uniqueness of a fixed point of an arbitrary contraction on a complete metric space. It is important to mention that there exists a continuous mapping without fixed point property.

Theorem 2.2 (Banach; Banach contraction principle).
If (X, d) is a complete metric space and f : X → X is a contraction, then the mapping f has a unique fixed point in X.
Proof. Let x 0 ∈ X be arbitrary. We define a sequence (x n ) in X such that x n = f (x n−1 ) for (n ∈ N, and prove that (x n ) is a Cauchy sequence, hence convergent in the complete metric space X. We obtain for any n ∈ N, d(x n , x n+1 ) = d(f (x n−1 ), f (x n )) ≤ q · d(x n−1 , x n ) ≤ · · · ≤ q n · d(x 0 , x 1 ), and therefore, if m > n, Since 0 ≤ q < 1, it follows that lim n,m→∞ d(x n , x m ) = 0, hence (x n ) is a Cauchy sequence. Moreover, X is a complete metric space, and so there exists some x ∈ X such that lim n→∞ x n = x. We show f (x) = x by estimating d(x n , f (x)) for n ∈ N; implies lim n→∞ d(x n , f (x)) = 0, and by the uniqueness of the limit of any convergent sequence in a metric space, we conclude f (x) = x.
It remains to prove that such x is uniquely determined. We assume f (y) = y for some y ∈ X, y = x, then d(x, y) = d(f (x), f (y)) ≤ q · d(x, y) and (1 − q)d(x, y) ≤ 0, which contradicts our assumption because 0 < 1 − q ≤ 1. 3. Let f : X → X be a q-contraction on a complete metric space X and z ∈ X be the fixed point of the function f . Then we have (1) the sequence (f n (x)) converges for each x ∈ X and converges to z; (2) d(x, z) ≤ 1/(1 − q) · d(x, f (x)); (3) d(f n (x), z) ≤ q n /(1 − q) · d(x, f (x)); Proof. We only prove the second and third condition, the proofs of the other conditions are analogous.
(3) It follows from f (z) = z, that f n (z) = z, and from the first part of the prove, we have d(f n (x), z) = d(f n (x), f n (z)) ≤ q n d(x, z) ≤ q n 1 − q · d(x, f (x)).
Remark 2.4. There exist various approaches to the Banach fixed point theorem, but the proof above gives a method of how to find the fixed point for a contraction f . It is also known as Picard's iteration method or fixed point iteration. It is based on the idea of defining a sequence of successive iterations. We start with any x 0 ∈ X and define x n = f (x n−1 ) for n ∈ N. The proof presented above guarantees the existence of a limit lim n→∞ x n = x ∈ X such that f (x) = x. If we let m → ∞ in (2.2), then d(x n , x) ≤ q n 1 − q d(x 0 , x 1 ), and this is an estimate for the error made by approximating the solution x by the n−th iteration x n .
We now present a few proofs of Theorem 2.2.
Proof of Theorem 2.2 (Joseph and Kwack [29]). Let c = inf{d(x, f (x)) : x ∈ X}. If c > 0, then c/q > c and there exists x ∈ X such that d(f (x), f (f (x))) ≤ q · d(x, f (x)) < c, which is a contradiction. Hence we must have c = 0. Let (x n ) be a sequence in X such that d(x n , f (x n )) → 0 as n → ∞. We show that (x n ) is a Cauchy sequence, since d(x n , x m ) ≤ d(x n , f (x n )) + d(f (x n ), f (x m )) + d(f (x m , x m ) implies (1 − q)d(x n , x m ) ≤ d(x n , f (x n )) + d(x m , f (x m )).
Hence there exists p ∈ X such that lim n→∞ x n = p, and lim n→∞ d(x n , f (x n )) = 0 implies lim f (x n ) = p. It follows from d(f (x n ), f (p)) ≤ qd(x n , p) that lim n→∞ f (x n ) = f (p), hence f (p) = p. The uniqueness of the fixed point 0f the function f follows from the contractive condition of f .
Proof of Theorem 2.2 (Palais [42]). Let x 1 , x 2 ∈ X. Then we have Hence we obtain the fundamental contraction inequality If x 1 and x 2 are fixed points of the function f , then it follows from (2.3) that x 1 = x 2 , that is, the contraction can have at most one fixed point. Let x ∈ X, n, m ∈ N, and x 1 = f n (x) and x 2 = f m (x). We obtain from (2. 3) ). (2.5) Since 0 ≤ q < 1, it follows that lim n→∞ q n = 0, hence d(f n (x), f m (x)) → 0 as n → ∞ and m → ∞. Therefore the Cauchy sequence (f n (x)) converges, that is, there exists p ∈ X such that lim n f n (x) = p.
Because of the continuity of the function f , we have f (p) = f (lim n f n (x)) = lim n f (f n (x)) = p. We note that letting m → ∞ in (2.4), we obtain d(f n (x), p) ≤ q n 1 − q · d(x, f (x)). (2.6) Proof of Theorem 2.2 (Boyd and Wong [6]). We define ϕ(x) = d(x, f (x)) for x ∈ X. Since f is a contraction, the function ϕ : X → R is continuous and ϕ(f n (x)) → 0 as n → ∞, for each x ∈ X. We put It follows from the conditions above that C m is a closed and nonempty subset of X for each m = 1, 2, . . . . Now we estimate the diameter of the set C m . Let x, y ∈ C m . Then we have .
Since each C m is a closed, nonempty subset of X, C 1 ⊃ C 2 ⊃ C 3 ⊃ . . . and diamC m → 0 as m → ∞, it follows by Cantor's intersection theorem m C m = {ξ}.
Since f (C m ) ⊂ C m for each m, it follows that ξ is a fixed point of the function f , and clearly the fixed point is unique. ( We have for each x ∈ X d(f n (x), ξ) = d(f n (x), f n (ξ)) ≤ q n d(x, ξ) → 0 (n → ∞).
Since d(x, ξ) ≤ d(x, f (x)) + d(f (x), f (ξ)) ≤ d(x, f (x)) + qd(x, ξ), it follows that Hence we again have the estimate Corollary 2.5. Let S be a closed subset of a complete metric space (X, d) and f : S → S be a contraction.
For an arbitrary point x 0 ∈ S, the iterative sequence x n = f (x n−1 ) (n ∈ N) converges to the fixed point of the mapping f .
The following example will show that the statement in Corollary 2.5 does not hold without the assumption that the set S is closed, in general.
Example 2.6. Let d be the natural metric on R defined by d(x, y) = |x − y| for all x, y ∈ R, and S = B 0 (1) = {x ∈ R : |x| < 1}. Then the mapping is a contraction without a fixed point in S.
Banach's fixed point theorem has wide and diverse applications, for instance, in solving various kinds of equations, inclusions, etc.
Example 2.7. If X is a Banach space, A, B ∈ B(X), A is an invertible operator and B − A · A −1 < 1, then the invertibility of B follows from Banach's fixed point theorem.
Proof. It is sufficient to show that, for any y ∈ X, the equation Bx = y has a unique solution x ∈ X. We choose an arbitrary point y in X. If Bx 0 = y for some x 0 ∈ X, then We put z = A −1 y and C = A −1 (B − A). Then we have x 0 = z − Cx 0 . The idea is to show that the function f : X → X defined by f (x) = z − Cx for x ∈ X is a contraction and x 0 is its fixed point. The following inequalities hold for all x, y ∈ X Since A −1 · B − A < 1, f is a contraction and x 0 is the unique fixed point of f . Based on a few elements of an iterative sequence (f n (x)), we may assume, and then easily prove that, because of C < 1, this sequence converges to z − Cx + C 2 x − C 3 x + · · · . We observe that if A = I and C < 1, then I − C + C 2 − C 3 + · · · is an inverse of I + C.
The following corollary shows a relation between f n and f in the case when f n is a contraction.
Corollary 2.8 (Bryant [10]). If (X, d) is a complete metric space and f : X → X is a mapping such that f n is a contraction for some n ≥ 1, then f has a unique fixed point in X.
Since the proof of Banach's theorem is based on an iterative sequence for a point x ∈ X, the next reasonable step in the research was to check local properties and modify this result.
Theorem 2.10. Let (X, d) be a complete metric space and B r (x 0 ) = {x ∈ X : d(x, x 0 ) < r} be the open ball in X for some x 0 ∈ X and r > 0. Also let f : B r (x 0 ) → X be a contraction, that is, and Then the mapping f has a unique fixed point in B r (x 0 ).
Proof. We choose r 0 ∈ [0, r) such that (2.9) holds. Then f : Hence f has a unique fixed point z ∈ B r 0 (x 0 ). It easily follows from (2.8) that z is the unique fixed point of f in B r (x 0 ).

Darbo's fixed point theorem
If the contractive condition of f in Theorem 2.2 is relaxed, that is, if we consider so-called nonexpansive mappings f , that is, functions f : X → X satisfying d(f (x), f (y)) ≤ d(x, y) for all x, y ∈ X, then Banach's fixed point theorem need no longer hold.
In 1965, Browder proved a fixed point theorem for nonexpansive maps.
Theorem 3.1 (Browder's fixed point theorem). Let X be a Banach space, C be a convex and bounded subset of X and T : C → C be a nonexpansive map. If X is either a Hilbert space, or a uniformly convex or a reflexive Banach space, then T has a fixed point.
This result uses the convexity hypothesis which is more usual in topological fixed point theory, and the geometric properties of Banach spaces commonly used in linear functional analysis.
The following Brouwer fixed point theorem should be considered in a different setting.
where id is the identity, is closed. It is natural to study what other properties the set F(f ) has. The following theorem shows that no other special features can be inferred, since we will see that for any given non-empty closed subset of the closed unit ball B Obviously this function is continuous. We define the function f : where x 0 is an arbitrary point in F . It is easy to show that f is well defined and continuous. Moreover F(f ) = F and the theorem is proved. An important generalization of Brouwer's fixed point theorem was obtained by Schauder.
Theorem 3.5 (Schauder's fixed point theorem). Every continuous map from a nonempty, compact and convex subset C of a Banach space X into C has a fixed point.
Clearly the conditions in the hypothesis are preserved if the norm of X is replaced by an equivalent norm, so Theorem 3.5 cannot be viewed as a metric fixed point theorem. Schauder's fixed point theorem can be used to prove Peano's existence theorem for the solution of systems of first order ordinary differential equations with initial conditions. The situation is completely different when certain generalizations are considered, in particular those concerning condensing maps, where a condensing map is one under which the image of any set is -in a certain sense -more compact than the set itself. The degree of noncompactness of a set is measured by certain functions called measures of noncompactness.
Darbo's fixed point theorem, which uses Kuratowski's measure of noncompactness α mentioned in the introduction, is a generalization of Schauder's fixed point theorem.
Theorem 3.6 (Darbo's fixed point theorem). Let C be a non-empty bounded, closed and convex subset of a Banach space X and α be the Kuratowski measure of noncompactness on X. If T : C → C is a continuous map such that there exists a constant c ∈ [0, 1) with then T has a fixed point in C.
We will prove a generalization of Theorem 3.6, namely the Darbo-Sadovskiȋ theorem, in the next section.

Measures if noncompactness and the Darbo-Sadovskiȋ theorem
Darbo's fixed point theorem generalizes from compact sets to bounded and closed sets in infinite dimensional Banach spaces, and needs the additional hypothesis of the condensing property in (3.1). As is well known, when we pass from finite to infinite dimensional Banach spaces, bounded and closed subsets need not necessarily be compact. So it is natural to ask if Schauder's fixed point theorem (Theorem 3.5) holds in infinite dimensional Banach spaces for convex, closed and bounded subsets. The following example provides a strong negative answer to this question.
Proof. We consider 2 (Z) with the standard Schauder basis (e (n) ) n∈Z , where for each n ∈ N, e (n) is the sequence with e (n) n = 1 and e (n) k = 0 for k = n, and with the natural norm given by for all x ∈ 2 (Z).
We write B 2 (Z) for the closed unit ball in 2 (Z). Every sequence x = (x n ) n∈Z ∈ 2 (Z) has a unique representation x = n∈Z x n e (n) . We define the left shift operator U : 2 (Z) → 2 (Z) by The relation x implies x n = x 0 for all n > 0 and x n = x 1 for all n < 0. For a sequence in 2 (Z), this is only possible if Finally, T is a fixed point free map. Indeed, if , which is clearly impossible if x = 0, and impossible if x = 0, as we have seen above.
To be able to prove the Darbo-Sadovskiȋ theorem we need to recall the concepts of measures of noncompactness, in particular, the Kuratowski measure of noncompactness, and their most important properties. The results presented here and their proofs can be found, for instance, in [53,35,36].
Since notion of a measure of noncompactness was originally introduced in metric spaces, we are going to give our axiomatic definition in this class of spaces as given in the monograph [53]. In the books [1] and [3], two different patterns are provided for the axiomatic introduction of measures of noncompactness in Banach spaces. The following properties can easily be deduced from the axioms in Definition 4.2. Proposition 4.3. Let φ be a measure of noncompactness on a complete metric space (X, d). Then φ has the following properties Q ⊂Q implies φ(Q) ≤ φ(Q) (monotonicity); (4.1) If Q is finite then φ(Q) = 0 (non-singularity).  Remark 4.4. If X is a Banach space then a measure of noncompactness φ may have some additional properties related to the linear structure of a normed space, for instance φ(λQ) = |λ|φ(Q) for any scalar λ and all Q ∈ M X (homogeneity) (4.5) for any x ∈ X and all Q ∈ M X (translation invariance). For every Q 0 ∈ M X and for all ε > 0 there is δ > 0 such that φ (co(Q)) = φ(Q) for all Q ∈ M X (invariance under the passage to the convex hull).
is called the Kuratowski measure of noncompactness (KMNC), and the real number α(Q) is called the Kuratowski measure of noncompactness of Q.
Now we define the Hausdorff or ball measure of noncompactness which was first introduced by Goldenštein, Goh'berg and Markus in 1957 [GGM1] and later studied by Goldenštein and Markus in 1965 [GGM2].
The definition of the Hausdorff measure of noncompactness is similar to that of the Kuratowski measure of noncompactness and the results are analogous.  In Banach spaces the functions α and χ satisfy some additional properties related to the linear structures of normed spaces. The statements of the following results for the Kuratowski measure of noncompactness are due to Darbo.
Theorem 4.8. Let X be a normed space ψ denote the Kuratowski or Hausdorff measure of noncompactness, and Q, Q 1 , Q 2 ∈ M X . Then we have (subadditivity), (4.10) ψ(λQ) = |λ|ψ(Q) for each scalar λ (homogeneity), (4.12) and ψ(Q) = ψ(co(Q)) (invariance under the passage to the convex hull).  . Let X be a Banach space, φ be a measure of noncompactness which is invariant under passage to the convex hull, C = ∅ be a bounded, closed and convex subset of X and T : C → C be a φ-condensing operator, that is, T is continuous and satisfies φ(T (Q)) < φ(Q) for all bounded non-precompact subsets Q of C. (4.14) Then T has a fixed point.
Proof. We choose a point c ∈ C and denote by Σ the class of all closed and convex subsets K of C such that c ∈ K and T (K) ⊂ K. Furthermore, we put Obviously Σ = ∅, since C ∈ Σ, and B = ∅, since c ∈ B. We also have Since T is φ-condensing, it follows that φ(B) = 0, and so B is compact. Obviously B is also convex. Thus it follows from Schauder's fixed point theorem, Theorem 3.5, that the operator T : C → C has a fixed point.
The following example will show that the theorem of Darbo and Sadovskiȋ fails to be true, if we assume that T is a k-contractive operator with constant k = 1, that is, if we replace the the condensing condition (4.14) by the condition φ(T (Q)) ≤ φ(Q) for all bounded Q of C.
Then we can write T = D + S where D is the one dimensional mapping is an isometry. Hence T is a well-defined, continuous operator, and for every bounded subset Q of B 2 , we have α(T (Q)) ≤ α(D(Q) + S(Q)) ≤ α(D(Q)) + α(S(Q)) = 0 + α(Q).
So T is a k-set-contractive operator with constant k = 1. But T has no fixed points. If T had a fixed point x ∈ B 2 , then we would have x k = x k+1 for all k ∈ N. Since x ∈ 2 , this implies x k = 0 for all k ∈ N, and then T (x) = 1 − x 2 e (0) = e (0) = (0, 0, 0, . . . ), a contradiction.

Edelstein's results
For a function f : X → X on a complete metric space (X, d) which satisfies the condition where 0 ≤ λ < 1, the Banach contraction principle yields the existence and uniqueness of fixed points. If we take λ = 1 in the condition in (5.1) then we we obtain a contractive map, that is, a map which satisfies the condition In 1962, Edelstein [20] published a paper in which he studied the fixed points of contractive maps using the next condition and assumption The condition in (5.2) together with the assumption of the existence of x ∈ X such that the iterative sequence (f n (x)) contains a convergent subsequence (f n k (x)) in X, that is, provides the existence of a fixed point of f .
Theorem 5.1 (Edelstein [20]). Let X be a metric space and f : X → X be a contractive map that satisfies the condition in (5.3). Then u = lim k→∞ f n k x is the unique fixed point of f .
The function r is continuous on Y , and there exists a neighborhood U of points (u, f (u)) such that (x, y) ∈ U implies 0 ≤ r(x, y) < R < 1.
) be the open balls with centers in u and f (u), and radius ρ such that and It follows from (5.3) that there exists a natural number N such that k > N implies f n k (x) ∈ B 1 , and (5. 2) and it follows from (5.4) and (5.5) that which is a contradiction to (5.7). This we must have f (u) = u. We assume that v = u is also a fixed point of the function f . Then we have which is a contradiction to (5.2).
The condition in (5.3) is always satisfied for a compact space. Therefore we have Theorem 5.2 (Edelstein [20]). Let (X, d) be a compact metric space and f : X → X be a map. We assume Then the function f has a unique fixed point.
We obtain the following result on the iteration sequence from Theorem 5.1.

Theorem 5.3 (Edelstein [20]).
We assume that the conditions of Theorem 5.1 are satisfied. If the sequence (f n (p)) for p ∈ X contains a convergent subsequence (f n k (p)) then its limit u = lim n→∞ f n (p) in X exists and u is a fixed point of f .
Proof. By Theorem 5.1, we have u = lim k→∞ f n k (p). For given δ > 0, there exists n 0 ∈ N such that k > n 0 implies d(u, f n k (p)) < δ. If m = n k + l > n k , then we have

Rakotch's results
The problem of defining a family of functions F = {α(x, y)} which satisfy the conditions 0 ≤ α < 1 and sup α(x, y) = 1 such that Banach's theorem is satisfied when the constant α is replaced by α(x, y) ∈ F was suggested by H. Hanani, and Rakotch published a result related to this problem in 1962 [43] In this subssection, we present some results from the mentioned paper. Definition 6.1. Let (X, d) be a metric space. We denote by F 1 the family of all functions α(x, y) which satisfy the following conditions: (1) α(x, y) = α(d(x, y)), that is, α depends only on the distance of x and y.
(3) α(d) is a monotone decreasing function of d.
and let f (M ) be a subset of a compact subset of X. Then there exists a unique fixed point of f .
Proof. We assume f (x 0 ) = x 0 and put x n = f n (x 0 ) for n = 1, 2, . . . , that is, By Edelstein's theorem (Theorem 5.1), it suffices to show that x n ∈ M for each n.
Since f is a contractive map, the sequence (d( We obtain from the triangle inequality and (6.1) implies x n ∈ M for all n.
and f (B r (x 0 )) is a subset of a compact subset of X, then the function f has a unique fixed point.
Proof. If we put M = B(x 0 , r) in Theorem 6.2, then by (6.4), the monotony of α(d) and r ≥ 2d that is, we have (6.1).
Theorem 6.4. Let f : X → X be a contractive map on a complete metric space. We assume that there exist M ⊂ X and a point x 0 ∈ M such that Then the function f has a unique fixed point.
Proof. We assume f (x 0 ) = x 0 and define the sequence (x n ) by x n = f n (x 0 ) for n = 1, 2, . . . . As in Theorem 6.2, we have by (6.5) and x n ∈ M for each n.
We are going to prove that the sequence (x n ) is bounded. It follows from (6.6) and the definition of the sequence (x n ) that (6.8) and, by the triangle inequality, we have Hence (6.7) and (6.8) imply So we obtain that is, the sequence (x n ) is bounded. Let p > 0 be an arbitrary natural number. It follows from (6.6) that We prove that (x n ) is a Cauchy sequence. It is enough to show that, for every ε > 0, there exists N which depends only on ε (and not on p) such that, for all . . , n − 1, then we obtain from (6.6) (because of the monotony of the function and then (6.10) implies Since α(ε) < 1 and [α(ε)] n → 0 as n → ∞, there exists a natural number N , independent of p, such that d(x N , x N +p ) < ε for each p > 0. Hence (x n ) is a Cauchy sequence. Since X is a complete metric space, there exists u ∈ X such u = lim n→∞ x n . Because of the continuity of the function f , u is a fixed point of f .
In particular, if M = X, we obtain the next corollary.
where α(x, y) ∈ F 1 . Then the function f has a unique fixed point.
Remark 6.6. The preceding corollary and Theorem 6.4 are generalizations of Banach's fixed point theorem.

Boyd and Wong's nonlinear contraction
In this section, we present some results by Boyd and Wong [7] in 1969. In [7], Boyd and Wong studied fixed points for maps of the kind introduced in the next definition.
where Ψ is a function defined on the closure of the range of d, is called a Ψ contraction.
We denote the image of d by P and the closure of P by P . Hence P = {d(x, y) : x, y ∈ X}.
Rakotch [43] proved that if Ψ(t) = α(t)t, where α is a decreasing function with α(t) < 1 for all t > 0, then the map f satisfying (7.1) has a unique fixed point u. It can be shown that if Ψ(t) = α(t)t and α is an increasing function with α(t) < 1 for all t ≥ 0, then the conclusion of Banach's theorem holds true. Boyd and Wong proved that it is enough to assume that Ψ(t) < t for all t > 0 and Ψ is semicontinuous, and if a metric space is convex, then the last condition can be omitted.
We recall that a function ϕ : is said to be upper semi-continuous from the right on X if it is upper semi-continuous from the right at every t ∈ X.
Theorem 7.2. Let (X, d) be a complete metric space and f : X → X be a map satisfying (7.1), where Ψ : P → [0, ∞) is upper semi-continuous from the right on P and satisfies Ψ(t) < t for all t ∈ P \ {0}.
Then the function f has a unique fixed point x 0 and f n (x) → x 0 (→ ∞) for each x ∈ X.
Then, because of (7.1), the sequence (c n ) is monotone decreasing. We put lim n→∞ c n = c ≥ 0, and prove c = 0. If c > 0, then we have which is a contradiction.
We are going to prove that (f n (x)) is a Cauchy sequence for each x ∈ X. Then the limit point of this sequence is the unique fixed point of the function f . We assume that (f n (x)) is not a Cauchy sequence. Then there exist ε > 0 and sequences (m(k)) and (n(k)) of natural numbers with m(k) > n(k) ≥ k such that We may assume that and choose m(k) as the smallest integer greater than n(k) which satisfies (7.5). It follows from (7.2) that Hence d k → ε as k → ∞. Since . This is a contradiction, because we have Ψ(ε) < ε for ε > 0.
The following example will show that the condition of the continuity of the function Ψ in Theorem 7.2 cannot be dropped, in general.
Then X is a closed subset of the real numbers, and so complete. We assume that for each p ∈ P (p = 0), there exists a unique pair (x n , x m ) such that p = d(x n , x m ). We assume that for some integers j, k, m, n with j > k and m > n.
Since the left hand side in (7.9) is irrational or equal to zero and the right hand side is rational, it follows that both sides are equal to zero. Hence we have for which is only possible for n = k. We define the functions f by f (x n ) = x n−1 and Ψ on P by We put Ψ(p) = 0 for p ∈ P \ P . Then we have Ψ(t) < t for all t ∈ P \ {0} and but the function f has no fixed point.
Theorem 7.2 shows that it is not possible to extend the function Ψ from the set P to the set P such that it is upper semi-continuous from the right with Ψ(t) < t for t ∈ P \ {0}. This can directly be seen for the point √ 2 ∈ P \ P . If the condition Ψ(t) < t is replaced by Ψ(t 0 ) = t 0 for some value t 0 , then Theorem 7.2 does not hold. This is shown in the next example.
Now the functions f 1 and f 2 satisfy (7.1), if we define We know that the function Ψ satisfies all the conditions in Theorem 7.2, but Ψ(2) = 2. The function f 1 has two fixed points −1 and 1 and the function f 2 has no fixed points.
Theorem 7.2 is a generalization of Rakotch's theorem. This is shown in the next example.
. . } be the complete metric space with its metric d defined by We define the function f : X → X by and if x ∈ {2, 3, 4, . . . } and x > y, then we have We define the function Ψ by The function Ψ is upper semi-continuous from the right on the set [0, ∞), Ψ(t) < t for all t > 0, and the condition in (7.1) is satisfied. Since there is no decreasing function α with α(t) < 1 for all t > 0 which satisfies (6.11). Furthermore, since there is no increasing function α with α(t) < 1 for all t > 0 which satisfies (6.11).

Theorem of Meir-Keeler
In 1969, Meir and Keeler [39] proved a very interesting theorem and showed that the conclusion of Banach's fixed point theorem can be extended to a more general class of contractions. In this subsection, we present some results of the paper mentioned.
Definition 8.1. Let (X, d) be a metric space. The function f : X → X is said to be a weakly uniformly strict contraction, or a Meir-Keeler contraction (MK contraction) if, for every ε > 0, there exists δ > 0 such that Proof. First we note that (8.1) implies that f is a contractive map, that is, Hence f is a continuous function and has at most one fixed point. We note that if (f n (x)) is a Cauchy sequence for each x ∈ X, then the function has a unique fixed point, and (8.2) is satisfied. This follows from the following consideration. Since X is a complete space, every Cauchy sequence (f n (x)) has a limit u(x). The continuity of f implies Hence u(x) is the unique fixed point of f . The proof of the theorem will be complete if we show that the sequence (f n (x)) = (x n ) of iterations is a Cauchy sequence for each x ∈ X. Let x ∈ X and c n = d(x n , x n+1 ) for n = 1, 2, . . . . It follows from (8.3) that (c n ) is a decreasing sequence. If lim n→∞ c n = ε > 0, then the implication in (8.1) is not true for c m+1 , where c m is chosen such that c m < ε + δ. This implies lim n→∞ c n = 0. We assume that there exists a sequence (x n ) which is not a Cauchy sequence. Then there exists 2ε > 0 such that, for each m 0 ∈ N, there exist n, m ∈ N with n, m > m 0 and d(x m , x n ) > 2ε. It follows from (8.1) that there exists δ > 0 such that The implication in (8.4) remains true if we replace δ by δ = min{δ, ε}. Let m 0 ∈ N be such that c m 0 < δ /3, and let m, n > m 0 such that m < n and d(x m , x n ) > 2ε. We prove that there exists j ∈ {m, m + 1, . . . , n} such that To prove (8.5), we note that d(x n−1 , Let k be the smallest natural number in {m, m + 1, . . . , n}; (clearly m < k ≤ n − 1) such that holds. We prove d(x m , x k ) < ε + δ . If we assume that this is not true, then we have This is a contradiction to the the minimality condition of k in the inequality in (8.7). Therefore the inequality in (8.5) must hold.
This is a contradiction to (8.5). Hence (x n ) is a Cauchy sequence.
It is well known that the Meir-Keeler theorem generalizes Banach's contraction principle [2] and Edelstein's theorem [20]. [2]). Let (X, d) be a complete metric space and f : X → X be a contraction, that is, there exists q ∈ [0, 1) such that Then f has a unique fixed point.
Theorem 8.4 (Edelstein [20]). Let (X, d) be a compact metric space and f : X → X be a map. We assume that Then the function f has a unique fixed point.
Proof. We assume that the function f does not satisfy the condition in (8.1). Then there exist ε > 0 and sequences (x n ) and (y n ) in X such that Since X is a compact set, there exist subsequences (x n k ) and (y m k ) of the sequences (x n ) and (y n ), which converge to some x 0 ∈ X and some y 0 ∈ X, respectively. The continuity of the function f implies This is a contradiction, and consequently the function f must satisfy the condition in (8.1). Now the proof follows from Theorem 8.2.
Rakotch [43], and Boyd and Wong [7] assumed that, among other conditions, the following inequalities are satisfied: d(f (x), f (y)) ≤ ψ(d(x, y)) and ψ(t) < t for all t = 0. (8.11) The next example shows that the Meir-Keeler theorem holds even if the condition in (8.11) is not satisfied.

Theorems by Kannan, Chatterje and Zamfirescu
The first result is by Kannan [30] in 1968.
Theorem 9.1. If (X, d) is a complete metric space, 0 ≤ q < 1/2 and f : X → X be a map such that for all x, y ∈ X, (9.1) then f has a unique fixed point, that is, there exists one and only one u ∈ X such that f (u) = u.
which is a contradiction, and so c = 0. Hence there exists a sequence (x n ) in X such that lim n d(x n , f (x n )) = 0. It follows from that (x n ) is a Cauchy sequence. So there exists p ∈ X such that lim n→∞ x n = p. It follows that lim n→∞ f (x n ) = p. We prove f (p) = p. It follows from and so p = f (p). Now (9.1) implies that the map f has a unique fixed point. Banach's condition (2.1) and Kannan's (9.1) condition are independent. The condition in (2.1) implies the continuity of the map f , but this is not the case for the condition in (9.1). This follows from the following example.
The map f is discontinuous at the point x = 1/2 and so the condition in (2.1) is not satisfied, but the condition in (9.1) is satisfied for q = 4/9. The next theorem was proved by Chatterje [12] in 1972.
then the function f has a unique fixed point.
Proof (Fisher [23]). Let x ∈ X. Then we have So we obtain Since q(1 − q) −1 < 1, it follows that (f n (x)) is a Cauchy sequence in X. Since X is a complete metric space, there exists z ∈ X such that z = lim n f n (x). Now we have Letting n → ∞, we obtain d(z, f (z)) ≤ qd(z, f (z)), and since q < 1/2, we have f (z) = z.
Hence z is a fixed point of the function f . We assume that the function f has one more fixed point z ∈ X. Then we have = 2qd(z, z ).
Since q < 1/2, it follows that z = z , that is, the fixed point of the function f is unique.
Theorem 9.5 (Zamfirescu [54]). Let (X, d) be a complete metric space and f : X → X be a map for which there exist real numbers 0 ≤ α < 1, 0 ≤ β < 1 and γ < 1/2 such that, for each x, y ∈ X, at least one of the following conditions is satisfied: Then the function f has a unique fixed point.
Proof. Let x, y ∈ X. Then at least one of the conditions (z 1 ), (z 2 ) or (z 3 ) is satisfied. If (z 2 ) is satisfied, then we have Similarly, if (z 3 ) is satisfied, we get the following estimate Hence we have We put Then we have 0 ≤ λ < 1, and if (z 2 ) or (z 3 ) is satisfied for each x, y ∈ X, then In a similar way, it can be shown that if (z 2 ) or (z 3 ) is satisfied, then Obviously, (9.2) and (9.3) follow from (z 1 ). It follows from (9.2) that the function f has at least one fixed point. Now we prove the existence of a fixed point of f . Let x 0 ∈ X and x n = f n (x 0 ) for n = 1, 2, . . .
be the Picard iteration of f . If x = x n and y = x n−1 are two successive approximations, then it follows from (9.3) that So (x n ) ∞ n=0 is a Cauchy sequence, and consequently convergent. Let u ∈ X be its limit. Then we have By the triangle inequality and (9.2), it follows that and letting n → ∞, we obtain d(u, f (u)) = 0, hence f (u) = u. (a(x, y) + 2b(x, y) + 2c(x, y)) ≤ λ < 1, such that, for each x, y ∈ X, d(f (x), f (y)) ≤ a(x, y)d(x, y) + b(x, y)(d(x, f (x)) + d(y, f (y))) + c(x, y)(d(x, f (y)) + d(y, f (x))); and (Z ): There exists a constant h with 0 ≤ h < 1 such that, for all x, y ∈ X, It can be proved ( [49]) that the conditions in (Z), (Z ) and (Z ) equivalent. We show that (Z) implies (Z ). If the function f and x, y ∈ X satisfy (z 1 ), then we define a(x, y) = α and b = c = 0. If for x, y ∈ X, for which the function f satisfies (z 2 ), we define b(x, y) = β and a = c = 0, and similarly, in the case of (z 3 ), we define c(x, y) = γ and a = b = 0.
We show that (Z ) implies (Z ). We put and f ∈ (Z ). We show that (Z ) implies (Z).

10.Ćirić's generalized contraction
In [13],Ćirić generalized the well-known contractive condition and introduced a concept of a generalized contraction defined as follows.
Definition 10.1 (Ćirić [13]). Let (X, d) be a metric space. A mapping f : X → X is a λ-generalized contraction if, for all x, y ∈ X, there exist some nonnegative numbers q(x, y), r(x, y), s(x, y) and t(x, y) such that sup x,y∈X {q(x, y) + r(x, y) + s(x, y) + 2t(x, y)} = λ < 1, and for all x, y ∈ X, d(f (x), f (y)) ≤ q(x, y)d(x, y) + r(x, y)d(x, f (x)) + s(x, y)d(y, f (y)) + t(x, y)(d(x, f (y)) + d(y, f (x))). If f is given, then the usual notation is O(x). Furthermore, for all n ∈ N, we define the set The space X is said to be an f -orbitally complete metric space if any Cauchy sequence in O(x; f ) for x ∈ X converges in X.
Obviously, every complete metric space is f -orbitally complete, but the converse implication does not hold, in general. It is clear from the proof of Banach's theorem that it is enough to assume that (X, d) is f -orbitally complete instead of complete. The same remark applies for λ-generalized contractions, as is stated in the following theorem.
Theorem 10.5 (Ćirić [13]). If f : X → X is a λ-generalized contraction on an f -orbitally complete metric space X, then, for any x ∈ X, the iterative sequence (f n (x)) converges to the unique fixed point u of f , and Proof. For an arbitrary x ∈ X, we define the sequence (x n ) by x 0 = x and x n = f (x n−1 ) for n ∈ N. Then we obtain from (10.1) and moreover d(x n , x n+1 ) ≤ (q(x n−1 , x n ) + r(x n−1 , x n ))d(x n−1 , x n ) Eberhard Malkowsky and Vladimir Rakočević, Adv. Theory Nonlinear Anal. Appl. 1 (2017), 64-112. 88 Because of q(x, y) + r(x, y) + t(x, y) + λs(x, y) + λt(x, y) ≤ λ, we get q(x, y) + r(x, y) + t(x, y) 1 − s(x, y) − t(x, y) ≤ λ for all x, y ∈ X and, combined with (10.3), it follows that We remark that (10.4) allows us to consider f as a contraction under special assumptions, and Obviously, we have for all m ≥ n ). (10.5) implies that (x n ) is a Cauchy sequence in O(x). Let z ∈ X denote its limit. It remains to show f (z) = z by estimating d(f (z), f (x n )).
that is, z is a fixed point of the function f . The uniqueness easily follows from (10.1) and the estimation inequality is implied by (10.5).
The contractive condition (10.1) for generalized contractions implies many others, thus Theorem 10.5 has numerous consequences among which we will state two analogous to Corollaries 2.8 and 2.10 of Banach's theorem.
Proof. The existence of a unique fixed point directly follows from Theorem 10.5. It remains to estimate d(f n (x), z) for each n ∈ N. Since n = mk + r for m = [n/k] and 0 ≤ r < k, we have As in the case of Banach's theorem, we may consider some local properties of Theorem 10.5. Proof It is clear that x n ∈ B for all n ∈ N, due to (10.6) and mathematical induction. Analogously as in the proof of Theorem 10.5, it follows that (f n (x 0 )) is a Cauchy sequence in B and its limit is a fixed point of f . Inequality (10.1) guarantees uniqueness.

The Reich and Hardy-Rogers theorems
In 1971, Reich [44] proved the following theorem which generalizes Banach's and Kannan's theorems. (We note that for a = b = 0, we obtain Banach's theorem, Theorem 2.2, and for a = b and c = 0, we obtain Kannan's theorem, Theorem 9.1).
Theorem 11.1 (Reich [44]). Let (X, d) be a complete metric space and f : X → X be a map for which there exists nonnegative numbers a, b and c with a + b + c < 1 such that for all x, y ∈ X, d(f (x), f (y)) ≤ ad(x, f (x)) + bd(y, f (y)) + cd(x, y). (11.1) Then the map f has a unique fixed point.
Thus (f n (x)) is a Cauchy sequence, and there exists z ∈ X with z = lim n→∞ f n (x). We are going to show f (z) = z. It suffices to show lim n→∞ f n+1 (x) = f (z). When we choose x = f n (x) and y = z in (11.1), then we have for all n ≥ 1 Thus we obtain for n → ∞ We are going to show that the mapf has a unique fixed point. If we assume that x, y ∈ X with x = y are fixed points of the map f , then we have which implies x = y. But the map f satisfies the condition in(11.1), for instance, for a = 1/6, b = 1/9 and c = 1/3. Corollary 11.3 (Reich [44]). Let (X, d) be a complete metric space and f n : X → X for n = 1, 2, . . . be a sequence of maps satisfying the condition in (11.1) with the same constants a, b and c and with the fixed points u n ∈ X. We define the map f : Then the map f has a unique fixed point u ∈ X and u = lim u n→∞ = u.
Proof. Since the metric d is a continuous function, it follows that the function f satisfies the condition in (11.1), and therefore has a unique fixed point u ∈ X. We note that ≤ ad(u n , f n (u n )) + bd(u, f n (u)) + cd(u n , u) + d(f n (u), f (u)).

Hence we have
Hardy and Rogers [25] improved some of Reich's results [44] including the following theorem.
(i) If (X, d) is a complete metric space and α < 1, then the map f has a unique fixed point. ) is compact, f is continuous and the condition in (11.2) is replaced by for all x = y, and α = 1, then f has a unique fixed point.
The following lemma is essential in the proof of this theorem, but for the reader's convenience, we state it separately.
Lemma 11.5. We assume that (11.2) is satisfied and α < 1. Then there exists β < 1 such that If α = 1 and (11.3) is satisfied, then ). (11.5) Proof. In the first case, for α < 1, we put y = f (x), and observe x) and (11.6), leads to that is, . (11.8) By (11.9), inserting (11.8) in (11.6), we obtain . (11.9) and replacing a and c by b and e (which is permitted because of the symmetry of the metric d), we get If we put (11.11) then (11.4) is satisfied. The remainder of the lemma is shown analogously.
Proof of Theorem 11.4. To prove Part (i), we first observe that, by (11.4), for all m > n, .
Because of (11.5), it follows that y = f (y). The uniqueness is shown as previously discussed.

12.Ćirić's quasi-contraction
In 1971,Ćirić [14] used a concept of generalized contraction to replace the linear combination of distances in (10.1) by their maximum, and defined a new class of contractive mappings called quasi-contractions.
Obviously if a mapping f satisfies condition(2.1), then (12.1) also holds. An example presented byĆirić shows that the converse implication is not true, in general.
Letting n → ∞ in (12.2), we obtain so β(x) = 0 and (f n (x)) is a Cauchy sequence in X. Let z = lim n→∞ f n (x). Because of (12.1), we have that is, f (u) = u. The uniqueness also follows from (12.1). We obtain from (12.6), α(f n (x)) ≤ λ n α(x) and combined with (12.5) If n, m ∈ N and m ≥ n, then and when m → ∞, then

Caristi's Theorem
There are many extensions of Banach's contraction principle, one of the most studied ones is that by Caristi [11], 1976. Caristi's theorem [11] may be motivated by the following consideration. If (X, d) is a metric space and T : X → X is a contraction with a Lipschitz constant k ∈ [0, 1), then we have It is well known that Caristi's theorem (or the Caristi-Kirk, or the Caristi-Kirk-Browder theorem) is equivalent to Ekeland's variation principle [19] which is very important because of its numerous applications. The ordinal proof of the Caristi-Kirk theorem is rather complicated and, in the literature, there are several different proofs of that theorem.
We mention that the map ϕ : X → E (E ⊂ R) is lower semicontinuous at x ∈ X if, for every sequence (x n ), it follows from lim n→∞ x n = x that ϕ(x) ≤ lim inf n→∞ ϕ(x n ). The map ϕ : X → E is lower semicontinuous on X if it is lower semicontinuous at every x ∈ X.
Proof (Ćirić [15]). For each x ∈ X, we put Since x ∈ P (x), P (x) is a nonempty set and 0 ≤ α(x) ≤ φ(x). Let x ∈ X. We define the sequence (x n ) in X such that x 1 = x, and if x 1 , x 2 , . . . , x n are already defined then we define x n+1 ∈ P (x n ) such that φ(x n+1 ) ≤ α(x n ) + 1/n. Hence the sequence (x n ) satisfies the following conditions: Since (φ(x n )) is a decreasing sequence of real numbers, there exists α ≥ 0 such that Let k ∈ N. It follows from (13.2) that there exists N k such that φ(x n ) < α + 1/k for every n ≥ N k . Hence the monotonicity of the sequence (φ(x n )) for m ≥ n ≥ N k implies α ≤ φ(x m ) ≤ φ(x n < α + 1/k, that is, We have from the triangle inequality and the inequality in (13.2) Since (x n ) is a Cauchy sequence and X is a complete metric space the sequence converges to some u ∈ X.
Since u ∈ P (x n ) for each n ∈ N, (13.1) implies T u ∈ P (u), that is, Hence we have T u ∈ P (x n ) for each n ∈ N. It follows that φ(T u) ≥ α(x n ) for each n ∈ N.
Since (13.1) implies φ(T u) ≤ φ(u) and φ(u) = α, we have and so φ(T u) = φ(u). Now (13.1) implies Theorem 13.2 (Ekeland [19], 1972). Let φ : X → R be an upper semicontinuous function on the complete metric space (X, d). If φ is bounded above then there exists u ∈ X such that Proof (Ćirić [15]). We are going to show that u from the proof of Theorem 13.1 is the desired point. Using the same notations for x = u we have to prove x / ∈ P (u). We suppose that this is not the case, that is, for some v = u, we have v ∈ P (u).
it follows that v ∈ P (x n ). Hence we have which is a contradiction to φ(v) < α = φ(u). Hence we have x / ∈ P (u) for x ∈ X with x = u, and so Proof (Brézis and Browder [8]). By Theorem 13.2 there exists u ∈ X which satisfies the condition in (13.6). It follows that T u = u, for T u = u would imply φ(T u) − φ(u) > −d(u, T u), which contradicts (13.1). We note that Theorem 13.2 can be proved by Theorem 13.1. Indeed, if we assume that the conclusion of Theorem 13.2 is not true, then, for each x ∈ X, there exists y ∈ X with y = x such that φ(y) − φ(x) ≤ −d(x, y). Hence we may define a map T : X → X which satisfies (13.1), but does not have a fixed point.
We are going to present a proof of Caristi's theorem given by Kirk and Saliga [31]. First we prove a result by Brézis and Browder [8], the well-known Brézis -Browder [8] principle of ordering.
Let (X, ≤) be a partially ordered set. We denote S(x) = {y ∈ X : x ≤ y} for x ∈ X. A sequence (x n ) in X is said to be increasing if x n ≤ x n+1 for each n ∈ N. Theorem 13.3 (Brézis and Browder [8]). Let the function φ : X → R satisfy the following conditions: (1) x ≤ y implies φ(x) ≤ φ(y); (2) for every increasing sequence (x n ) in X with φ(x n ) ≤ C < ∞ for each n ∈ N, there exists y ∈ X such that x n ≤ y for each n ∈ N; (3) for each x ∈ X there exists u ∈ X such that x ≤ u and φ(x) < φ(u).
Then φ(S(x)) is a bounded set for each x ∈ X.
Proof. For a ∈ X, let p(a) = sup We are going to show p(x) = +∞ for each x ∈ X. We assume that p(x) < ∞ for some x ∈ X. We define a sequence (x n ) by induction such that x 1 = x, x n+1 ∈ S(x n ) and p(x n ) ≤ φ(x n+1 ) + (1/n) for each n ∈ N.
Since φ(x n+1 ) ≤ p(x) < ∞, the condition in (2) implies that there exists y ∈ X such that x n ≤ y for each n.
It follows from the condition in (3) that there exists u ∈ X such that y ≤ u and φ(y) < φ(u). Since x n ≤ u, we have φ(u) ≤ p(x n ) for all n. Furthermore, we have x n+1 ≤ y, so φ(x n+1 ) ≤ φ(y), and so hence φ(u) ≤ φ(y), which is a contradiction.
Theorem 13.4. Let (X, ) be a partially ordered set, x ∈ X and S(x) = {y ∈ X : x y}. We assume that the map ψ : X → R satisfies the following conditions: (a) x y with x = y implies ψ(x) < ψ(y); (b) for each increasing sequence (x n ) in X, for which ψ(x n ) ≤ C < ∞ for each n ∈ N, there exists y ∈ X such that x n y for each n ∈ N; (c) for each x ∈ X, the set ψ(S(x)) is bounded above.
Then, for each x ∈ X there exists x ∈ S(x) such that x is maximal in X, that is, {x } = S(x ).
Proof. We apply Theorem 13.3 to the set X = S(x); since the conditions in (1) and (2) of Theorem 13.3 are satisfied, and the conclusion of the theorem does not hold, it follows that the condition in (3) is not satisfied for some x ∈ S(x). Hence we have S(x ) = {x }.
We remark that the map ϕ : X → R is lower semicontinuous from above if x n ∈ X for n = 1, 2, . . . , lim n→∞ x n = x and (ϕ(x n )) ↓ r imply ϕ(x) ≤ r.
Theorem 13.5 (Kirk and Saliga [31]). We assume that (X, d) is a complete metric space and T : X → X is an arbitrary map such that we have where the map ϕ : X → R is bounded above and lower semicontinuous. Then the map T has a fixed point in X.
Proof. We introduce Brondsted's partial order on X as follows: For each x, y ∈ X, we have x y if and only if d(x, y) ≤ ϕ(x) − ϕ(y), and let ψ = −ϕ. Then the condition in (a) of Theorem 13.4 is satisfied, and the condition in (c) follows from the fact that the map ϕ is bounded below. To show the condition in (b), we assume that (x n ) is an increasing sequence in (X, ) such that ψ(x n ) ≤ C < ∞ for each n. Then (ϕ(x n )) is a decreasing sequence in R, and there exists r ∈ R such that lim n→∞ ϕ(x n ) = r. Since (ϕ(x n )) is a decreasing sequence, we have for each m > n lim n,m→∞ Hence (x n ) is a Cauchy sequence in X. It follows that there exists x ∈ X such that lim n→∞ x n = x. From ϕ(x n ) ↓ r and ϕ(x) ≤ r, it follows that Hence x is an upper bound for the sequence (x n ) in (X, ) and so we have proved the condition in (b). Now it follows by Theorem 13.4 that (X, ) has a maximal element x . Since (13.7) implies x T (x ), we have Siegel [52] proved in 1977 in an original way, a generalized version of Caristi's theorem. Here we present some of his results [52].
Let (X, d) be a complete metric space, φ : X → R + , the set of nonnegative real numbers, and g : X → X be a not necessarily continuous map such that d(x, g(x)) ≤ φ(x) − φ(g(x)) for all x ∈ X.
If a sequence of functions f i for i ≤ 1 < ∞ is given, then we define the product if the limit exists, and call it the countable decomposition of the given sequence of functions.
Lemma 13.7. Let φ be an upper semi continuous function, and (x i ) be a sequence in X such that d( Proof. Since the sequence (φ(x i )) i is not increasing and bounded below by zero, and since d( Lemma 13.8. The sets Φ and Φ g are closed by the composition of functions and if φ is an upper semicontinuous function then the sets Φ and Φ g are closed by the countable composition of sequences of functions.
Proof. We prove that the sets Φ and Φ g are closed by the composition of functions. If f 1 , f 2 ∈ Φ, then we have Hence we have f 2 f 1 ∈ Φ. If f 1 ∈ Φ g , then φ(f 1 (x)) − φ(f 2 (f 1 (x))) ≥ 0 implies φ(f 2 f 1 ) ≤ φ ( g), and so f 2 f 1 ∈ Φ g . The remainder of the proof follows from the fact that, for each x ∈ X, the sequence (x i ) = (f i f i−1 · · · f 1 )(x) satisfies the conditions of Lemma 13.7. Definition 13.9. We introduce the following notations: (1) For A ⊂ X the diameter of A is defined as (2) r(A) = inf x∈A (φ(x)); Proof. We have The main result of Siegel's paper [52] is the following theorem.  such that x = f (x 0 ) and g(x) = x for all g ∈ Φ .
(b) If the elements of Φ are continuous functions, then there exists a sequence of functions f i ∈ Φ and Proof. Let (ε i ) be a sequence of positive real numbers converging to zero and ε > 0. Then there exists We put x 1 = f 1 (x 0 ). Since the set Φ is closed under the composition of functions it follows that S x 1 ⊆ S x 0 and Continuing in this way, we obtain a sequence of function f i such that On the other hand, lim i→∞ δ(S x i ) = 0 implies x = ∩ ∞ i=0 S x i . Now we prove that g(x) = x for each g ∈ Φ . This is a consequence of the fact that g(x) ∈ S x i for each i, and because of g(x) = g( ∞ j=i+1 f j (x i )). We know from the condition in (b) that there exists We are going to show g(x) = x for each g ∈ Φ . We note that g(x i ) ∈ S x i for each i. Since g is a continuous function, for each ε > 0, there exists an i 0 such that Hence if i > i 0 , it follows that d(g(x), x) < ε + ε i . Now ε i → 0 implies d(g(x), x) ≤ ε, and since ε is arbitrary, we have g(x) = x.

A Theorem by Bollenbacher and Hicks
The following result is related to Caristi's theorem 13.1.
Let (X, d) be a metric space and f : X → X be a map. Then there exists a map φ : if and only if the series converges for each x ∈ X.
Proof. We assume that the condition in (14.1) is satisfied. We show that the series in (14.2) converges. This follows from the fact that, for each n ∈ N, If the series (14.2) converges for each x ∈ X, the we define a map φ : We note that if the condition in (14.1) is satisfied for each y ∈ (x, ∞), then the series (14.2) converges for x, since the sequence of partial sums is nondecreasing and bounded by φ(x).
In 1988, Bollenbacher and Hicks [5] proved the following very interesting theorem, the corollaries of which include many generalizations of Banach's fixed point theorem.
Theorem 14.2 (Bollenbacher and Hicks [5]). Let (X, d) be a metric space, and f : X → X and φ : X → [0, ∞). We assume that there exists x such that and that each Cauchy sequence in O(x, ∞) converges to some point in X. Then we have: (1) lim n→∞ f n (x) = x exists; (2) d(f n (x), x) ≤ φ(f n (x)); (3) f (x) = x if and only if G(x) = d(x, f (x)) is f -orbitally upper semicontinuous at x; Proof. It follows from Theorem 14.1 that the series converges.
We prove that (f n (x)) is a Cauchy sequence. This follows since, for each m > n, and from the fact that the series ∞ k=n d(f k (x), f k+1 (x)) converges. Hence there exist x ∈ X such that the condition in (1) is satisfied. Now from the condition in (2) follows as m → ∞. To prove the condition in (3), we assume that x n = f n (x) → x as (n → ∞). If G is f -orbitally upper semicontinuous at x, then we have and so f (x) = x. Now we assume f (x) = x and that (x n ) is a sequence in O(x, ∞) such that lim n→∞ x n = x. Then we have and so G is an f -orbitally upper semicontinuous function at x. The condition in (4) follows from and since as n → ∞, we get d(x, x) ≤ φ(x). Corollary 14.3. ( [26]) Let (X, d) be a complete metric space and 0 < k < 1. We assume that, for f : X → X, there exists x such that d(f (y), f 2 (y)) ≤ kd(y, f (y)) for each y ∈ O(x, ∞). (2) d(f n (x), x) ≤ k n (1 − k) −1 d(x, f (x));
The Mann iterations are more general than the Picard iterations, that is, the Picard iterations are special cases of the Mann iterations which Mann introduced in his paper [37] in 1953.
Let E be a convex compact subset of a Banach space X, and T : E → E be a continuous map. By Schauder's fixed point theorem [51], there exists at least one fixed point of the function T , that is, there exists p ∈ E such that T (p) = p.
In 1953, Mann ([37]) studied the problem of constructing a sequence (x n ) in E which converges to a fixed point of T . Usually an arbitrary initial value x 1 ∈ E is chosen, and then the sequence of successive iterations (x n ) of x 1 defined by x n+1 = T (x n ) for n = 1, 2, . . .
is considered. If this sequences converges, then its limit is a fixed point of the function T .
Definition 15.1 (Dotson [18]). Let E be a vector space, C be a convex subset of E, f : C → C be a map and x 1 ∈ C. We assume that the infinite matrix A = [a nj ] satisfies the conditions (A 1 ) a nj ≥ 0 for all j ≤ n and a nj = 0 for j > n; (A 2 ) n j=1 a nj = 1 for each n ≥ 1; (A 3 ) lim n→∞ a nj = 0 for each j ≥ 1.
We define the sequence (x n ) by a nj x j .  37]). If one of the sequences (x n ) or (v n ) is convergent, then they both converge. In this case, they converge to the same limit point which is a fixed point of the function T .
Proof. Let lim n→∞ x n = p. Since A is a regular matrix, it follows that lim n→∞ v n = p. The continuity of the function T implies lim n→∞ T (v n ) = T (p), and from T (v n ) = x n+1 , it follows that T (p) = p. If we assume lim n→∞ v n = q, then lim n→∞ x n+1 = T (q), and the regularity of the matrix A implies lim n→∞ v n = T (q). Hence we have T (q) = q.
If the sequences (x n ) and (v n ) are not convergent, then, since E is a compact set, each of the two sequences has at least two distinct accumulation points.
Let X and V be the sets of accumulation points of the sequences (x n ) and (v n ), respectively. |a n+1,k − a n,k | = 0, (15.2) then X and V are closed and connected sets.
Proof. The set V is closed and compact, and by (15.2), lim n→∞ (v n+1 − v n ) = 0. Hence the set V is connected. Since the function T is continuous and X = T (V ), it follows that X is a closed and connected set.
Theorem 15.4 ([37]). The set V is a subset of co(X), where co(X) denotes the convex hull of the set X.
Proof. By Mazur's theorem [38], co(X) is a closed set. All but finitely many terms of the sequence (x n ) are elements of each open set that contains the set co(X). Hence for all sufficiently large n, v n are arbitrarily close to the the set X. Hence the limit point of each convergent subsequence of the sequence (v n ) is an element of the set co(X).
Example 15.5. Let A be the Cesàro matrix of order 1, that is, The matrix A satisfies all the assumptions for a matrix in this subsection. In this case, the Mann method M (x 1 , A, T ) is usually referred to as the mean value method, where the initial value is x 1 ∈ E and x n+1 = T (v n ) and v n = 1 n n k=1 x k for all n = 1, 2, . . . .

We note
In many special problems, the iterative method M (x 1 , A, T ) converges even when the method T n x 1 diverges.
Example 15.6. Let E = {x ∈ R 2 : x ≤ 1}, where · is the Euclidean norm. Furthermore, let A be the Cesàro matrix of order 1 and the function T : E → E be the rotation about the center by the angle π/4. Then the Picard iteration T n (x 1 ) does not converge for any x 1 ∈ E \ {0}. Using Mann's method the M (x 1 , A, T ), the sequences (x n ) and (v n ) always converge (on a spiral) to the center, independently of the choice of the initial value x 1 . a n+1,j = (1 − a n+1,n+1 )a nj for (j = 1, 2, . . . , n; n = 1, 2, . . . ); (A 5 ) either a nn = 1 for all n, or a nn < 1 for all n > 1.
In his paper [18], Dotson proved the following theorem.
In the literature, mainly the normal Mann iterative method is studied.
Proof. Without loss of generality, we may assume a = 0 and b = 1. By Brouwer's fixed point theorem and our assumption, there exists a unique fixed point x ∈ [0, 1] of the function f . Now we have for all y ∈ [0, 1] with y < x it follows that f (y) − y > 0; (16.5) for all y ∈ [0, 1] with (y > x) it follows that f (y) − y < 0. (16.6) If x = 0, then we obviously have (16.5). If x > 0 and if there exists which is a contradiction to the uniqueness of the fixed point. The case (16.6) is proved analogously. There are two alternatives I. and II. for the sequence (x n ): I. There exists n 1 ∈ N such x n 1 = x. Then x n = x for all n ≥ n 1 and the theorem is proved. II. For each n ∈ N, we have x n = x.
In this case, we have the following three possibilities: 1. There exists n 0 ∈ N such that x n < x for all n > n 0 . Then we have and (16.5) implies that (x n ) is a monotone increasing sequence; so the sequence converges, since x n ≤ 1 for all n. By Theorem 15.2, and since the function f has only one fixed point x ∈ [0, 1], it follows that lim n x n = x.
2. There exists m 0 ∈ N such that x n > x for all n ≥ m 0 . In this case, it follows by (16.6) that lim n→∞ x n = x, as in Case 1..

3.
We assume that the possibilities 1. and 2. are not true. Let ε > 0 be given. We choose n 0 ∈ N such that |x n+1 − x n | < ε for all n ≥ n 0 . This is possible, since |x n+1 − x n | ≤ 2c n and lim n→∞ c n = 0.
We are going to prove that there exists n 1 ∈ N with n 1 ≥ n 0 such that |x n 1 − x| < ε, that is, there exists n 1 ≥ n 0 such that − ε < x n 1 − x < ε. (16.7) If (16.7) is not true, then x n ≤ x − ε or x n ≥ x + ε for each n ≥ n 0 .
Now, if x n 0 ≤ x − ε, then x n ≤ x − ε for all n ≥ n 0 (because of |x n+1 − x n | < ε), that is, the condition in 1. is satisfied. Analogously, if x n 0 ≥ x + ε, then x n ≥ x + ε for all n ≥ n 0 (again because of |x n+1 − x n | < ε), that is, the condition in 2. is satisfied. Hence, in all cases, the conditions in 1. or 2. are satisfied. So we have shown (16.7).
We are going to prove that we have |x n − x| < ε for all n ≥ n 1 . This is true for n = n 1 . If n ≥ n 1 and if |x n − x| < ε, then we have the following possibilities A. and B.: A. x − ε < x n < x. Then we have (a) or (b) for x n+1 : (a) x n+1 < x. In this case, x n+1 − x n = c n (f (x n ) − x n ) and (16.5) imply x n+1 − x n > 0, Hence we have |x n+1 − x| = x − x n+1 < x − x n = |x n − x| < ε.
Hence |x n+1 − x| < ε. It follows by mathematical induction that |x n − x| < ε for all n ≥ n 1 , thus lim n x n = x.
We note that if we put c n := 1/(n + 1) for all n, then Theorem 16.3 implies Theorem 16.1.
In 1971, Franks and Marzec [24] showed that the condition of the uniqueness of the fixed point p in Theorem 16.1 is not necessary.

1.
We prove that the sequence (x n ) is convergent. The sequence (x n ) is in [0, 1], and so has at least one accumulation point. We assume that the sequence (x n ) has two distinct accumulation points ξ 1 and ξ 2 with ξ 1 < ξ 2 .
Thus ξ 1 is not an accumulation point of the sequence (x n ), which contradicts our assumption. If f (x * ) < x * , then, similarly as above, we obtain that ξ 2 is not an accumulation point of the sequence (x n ), which again is a contradiction. Hence f (x * ) = x * for each x * ∈ (ξ 1 , ξ 2 ).
b. Let us prove that ξ 1 and ξ 2 are not accumulation points of the sequence (x n ). We note that x n / ∈ (ξ 1 , ξ 2 ) for n = 1, 2, . . . . (16.16) If f (x n ) =x n , then (16.12) impliesx m =x n for all m > n. So neither ξ 1 nor ξ 2 can be an accumulation point of the sequence (x n ). Furthermore, (16.13) and (16.16) imply that there exists a natural number M such thatx M ≥ ξ 2 for all n > M . Hence ξ is not an accumulation point of the sequence (x n ). It follows fromx M ≤ ξ 1 thatx n < ξ 1 < ξ 2 for all n > M . Hence ξ 2 is not an accumulation point of the sequence (x n ). Consequently the sequence (x n ) cannot have two distinct accumulation points, and so this sequence is convergent. We put lim nxn = ξ ∈ [0, 1].
2. We show f (ξ) = ξ. We assume f (ξ) > ξ. Let Since the sequence (x n ) converges to ξ and the function f is continuous, there exists a natural number N such that f (x n ) −x n > ε for each n > N . It follows from (16.12) that x n+1 −x n = f (x n ) −x n n + 1 > ε n + 1 . Sox n → ∞ as n → ∞, which contradicts the fact thatx m ∈ [0, 1] for all m. If f (ξ) < ξ, then it can be shown thatx n → −∞ as n → ∞, which again is a contradiction. So we have f (ξ) = ξ.
Rhoades ([48], [47] and [16]), among other things, generalized many results presented in this section. He noted the importance of the condition in (15.2).
Let X be a normed space, E be a nonempty, closed, bounded and convex, subset of X and f : E → E be a map which has at least one fixed point in E, and let A be an infinite matrix. We consider the iterative scheme x 0 = x 0 ∈ E (16.17) x n+1 = f (x n ) for n = 0, 1, 2, . . . (16.18) x n = n k=0 a nk x k for n = 1, 2, 3, . . . . (16.19) The question is which are the necessary and sufficient conditions for the matrix A such that the above iterative scheme converges to a fixed point of the function f ? Many results were obtained by the use of the iterative scheme of the form above (16.17)-(16.19) for various classes of infinite matrices.
An infinite matrix A is said to be regular if x ∈ c and x n → l as n → ∞ implies A n (x) = ∞ k=0 a nk x k → l as n → ∞. The matrix A is triangular if all entries below the main diagonal are equal to zero. We consider regular triangular matrices A which satisfy 0 ≤ a nk ≤ 1 for all n, k = 0, 1, 2, . . . The conditions in (16.20) and (16.21) are necessary for x n , x n ∈ E. The scheme (16.17)- (16.19) is a Mann method [37]. Barone proved in [4] (see (15.2)) that a necessary condition that a regular matrix A maps all bounded sequence into sequences with the property that the set of their accumulation points is connected is the following lim n ∞ k=0 |a nk − a n−1,k | = 0. (16.22) In [46], Rhoades made the following assumption. Rhoades showed that the statement above is true for the large class of weighted means matrices. (For the definition and properties of these matrices see [25, p. 57].) The weighted means method is a triangular method of the matrix A = (a nk ) defined by a nk = p k /P n , where p 0 > 0, p n ≥ 0 for n > 0, P n = n k=0 p k and P n → ∞ as n → ∞. Then the matrix A satisfies the condition in (16.22) if and only if p n /P n → 0 as n → ∞.
Theorem 16.6 (Rhoades [48]).Let A be the matrix of a regular weighted means method which satisfies the condition in (16.22). Let E = [a, b] and f : E → E be a continuous map. Then the iterative scheme (16.17)- (16.19) converges to a fixed point of the function f .
Now, by the proof of Theorem 16.4, the sequence (x n ) is convergent. We have to show that the sequence converges to a fixed point of the function f . Let z = lim n→∞ x n . Then we have lim n→∞ f (x n ) = f (z). It follows from x n+1 = f (x n ) for each n ∈ N that lim n→∞ x n = f (z). Since A is a regular matrix, we obtain z = lim n→∞ x n = lim n→∞ A n (x) = f (z).
If we choose c n = (n + 1) −1 , the previous statement is Theorem 16.1.