Remarks on asymptotic regularity and fixed points
Abstract
Asymptotic regularity allows to provide simple proofs of Banach’s theorem and Kannan’s theorem. Using asymptotic regularity and Kannan’s type conditions we generalize these results, in particular, the Banach contraction principle (see Theorem 2.6 and Corollary 2.10). Further, we discuss the analogous results for monotone mappings on preordered metric spaces, where a preordered binary relation is weaker than a partial order. Next, we will prove a random version of the presented deterministic fixed-point theorems.
Keywords
Asymptotic regularity complete metric space fixed point Banach contraction principle Kannan mapping monotone mapping preordered space measurable space Polish space random operator random variable random fixed pointMathematics Subject Classification
Primary 47H10 Secondary 47H07 47H40 54H25 60H251 Banach and Kannan theorems
Let \(T:X\rightarrow X\) be a mapping. For a initial point \(x_0\in X\), define a sequence of iterates \(x_{n+1}=Tx_n=T^{n+1}x_0\), \(n=0,1,2,\ldots \), and the resulting sequence \(\{x_n\}\) is called the sequence of successive approximations of T.
Hillam [28] proved:
Theorem 1.1
Let T be a continuous map of [0, 1] into [0, 1]. The sequence \(\{x_n=T^nx\}\) of successive approximations of T converges to a fixed point of T if and only if (1) holds.
Smart [44] showed that this result does not extend beyond one-dimensional case:
Example 1.2
There is a continuous mapping T of the closed unit disc in the Euclidean plane such that the origin and points on the unit circle are fixed points and every other point x satisfies \(d(T^nx,T^{n+1}x)\rightarrow 0\) but \(\{T^nx\}\) is not convergent, see [44] for details.
The following observation is trivial:
Lemma 1.3
If T is a continuous map of X into X and if \(d(T^nx,T^{n+1}x)\rightarrow 0\), then any limit point p of the set \(\{T^nx\}\) is a fixed point of T.
Proof
Thus, the continuity of a mapping \(T:X\rightarrow X\) and the fact that the sequence of successive approximations \(\{T^nx\}\) satisfies (1) does not guarantee the existence of a fixed point. For a guarantee that there is a (unique) fixed point, additional assumptions are needed.
Banach’s contraction principle [4] is remarkable in its simplicity, because the contractive condition on the mapping is simple and easy to test, because it requires only a complete metric space for its setting, and because it finds almost canonical applications in the theory of differential and integral equations. In this part, we will give an elementary proof of this result exposing (1), for other proofs see [17, Chapter 2], [18].
Let us recall a few facts.
Definition 1.4
There are many mappings of this type.
Example 1.5
Theorem 1.6
(Banach contraction principle). Let (X, d) be a complete metric space, then each contraction \(T:X\rightarrow X\) has a unique fixed point \(p\in X\), and \(T^nx\rightarrow p\) for each \(x\in X\).
Proof
Remark 1.7
The following trivial fact is noteworthy in that the mapping T is not even assumed to be continuous:
Let (X, d) be a complete metric space and \(T:X\rightarrow X\) be a mapping for which \(T^N\) is contraction for some positive integer \(N>1\), then T has a unique fixed point.
Not only contractions guarantee the existence of a unique fixed point and the possibility of its approximation. In 1968, Kannan [32] established the following theorem, see [25].
Theorem 1.8
Kannan’s theorem is important because Subrahmanyam [46] proved that Kannan’s theorem characterizes the metric completeness. That is, a metric space (X, d) is complete if and only if every mapping satisfying (4) on X with constant \(K<\frac{1}{2}\) has a fixed point. Contractions do not have this property; Connell [11] gave an example of metric space X such that X is not complete and every contraction on X has a fixed point.
Here is an elementary proof of Kannan’s theorem.
Proof
Remark 1.9
Obviously conditions (2) and (4) are independent. Condition (4) is neither stronger nor weaker than the contraction mappings. In particular, the mapping satisfying (4) need not be continuous. In the following examples, the spaces are with the usual metrics.
Example 1.10
Mapping \(Tx=0\) for \(x\leqslant 2\) and \(Tx=-\frac{1}{2}\) for \(x>2\), satisfies (4) with \(K=\frac{1}{5}\), and T is not continuous.
Example 1.11
Contraction \(Tx=\frac{x}{3}\), \(x\in [0,1]\), not satisfied (4) with \(K<\frac{1}{2}\), take \(x=0\) and \(y=1\). If T is a contraction with \(L<\frac{1}{3}\), then T satisfies (4) with \(K<\frac{1}{2}\).
Example 1.12
The condition (2) with \(L=1\), does not imply the existence of a fixed point. The mapping \(Tx=x+1\) for \(x\in {\mathbb {R}}\) is fixed point free. The condition (4) with \(K=\frac{1}{2}\), does not imply the existence of a fixed point. Take the unit circle S on the Euclidean plane and \(Tz=-z\), \(z\in S\).
There are many generalizations of Theorem 1.6 and Theorem 1.8, and unification of conditions (2) and (4), see [5, 14, 31, 39], and references therein. The literature of this subject is extensive.
Conclusion. In this part, we have presented elementary proofs of Banach’s theorem and Kannan’s theorem on a fixed point.
2 Asymptotic regularity, continuity and fixed points
We know many conditions that guarantee the existence of a fixed point, see [2, 17, 26], and references therein. In this part, we present a very simple situation when the mapping T not only satisfies some conditions of Kannan’s type, but it is also continuous and asymptotically regular (as in the Banach theorem).
We recall, asymptotic regularity is a fundamentally important concept in metric fixed point theory, see [2, Chapter IX], and [17, Chapter 9]. It was formally introduced by Browder and Petryshym [7].
Definition 2.1
Obviously, if a mapping \(T:X\rightarrow X\) is a contraction or satisfy (4) or (5) with \(K<\frac{1}{2}\), then T is asymptotically regular. Asymptotic regularity is also satisfied by other mappings. But already the asymptotic regularity and nonexpansiveness (i.e. \(d(Tx,Ty)\leqslant d(x,y)\) for all x, y), more generally, continuity, are independent.
Example 2.2
Let \(B= \{x \in {\mathbb {R}}^2: \Vert x\Vert \leqslant 1\}\) be the closed unit disc in the Euclidean plane and let T be an anticlockwise rotation of \(\frac{\pi }{4}\) about the origin of coordinates. Then T is nonexpansive with the origin as the only fixed point and T is not asymptotically regular. Moreover, the sequence defined by \(\{x_{n+1}=Tx_n,~x_0=(1,0)\}\) does not converge to zero.
Example 2.3
The mapping \(Tx=1-x\), \(0\leqslant x\leqslant 1\), is continuous, is not a contraction and does not satisfy the condition (4), take \(x=0\) and \(y=1\). T has a unique fixed point \(\frac{1}{2}\), and \(d(T^n(0),T^{n+1}(0))\not \rightarrow 0\).
By an averaged mapping we mean one of the form \(T_{\lambda }=(1-\lambda )I+\lambda T\), where \(0<\lambda <1\) and I is the identity operator. When T is nonexpansive, so is \(T_{\lambda }\) and both have the same fixed point set, but \(T_{\lambda }\) has more much felicitous asymptotic behavior than the original mapping.
Ishikawa [30] proved the following theorem with no restrictions on the geometry of the Banach space!
Theorem 2.4
If C is a nonempty bounded closed convex subset of a Banach space X and \(T:C\rightarrow C\) is nonexpansive, then the mapping \(T_{\lambda }\) is asymptotically regular for each \(\lambda \in (0,1)\).
It is known [17] that a nonexpansive mapping \(T:C\rightarrow C\), acting on weakly compact convex subsets of uniformly convex Banach spaces, has a fixed point. Lin [34] gave an example an asymptotically regulate Lipschitzian mapping acting on a weakly compact convex subset of the Hilbert space \(l^2\) which has no fixed point.
Asymptotically regular mappings were studied in many papers, in different contexts, for instance [3, 8, 13, 16, 19, 20, 21, 22, 23, 41, 48].
In 1974, De Blasi [12] proved the following theorem, see [14].
Theorem 2.5
Now, we prove the following new theorem, which is an extension of previous results.
Theorem 2.6
Proof
Remark 2.7
If \(M\geqslant 0\), \(K\geqslant 0\) and \(M+2K<1\), then assumptions of continuity and asymptotic regularity are not necessary for the thesis to hold. If \(0\leqslant M<1\) and \(0\leqslant K<1\), then the continuity assumption is not necessary for the thesis to hold, see [25, 39].
Remark 2.8
Example 2.9
- (a)
T does not satisfy the Banach theorem, take \(x=1\) and \(y=\frac{3}{2}\);
- (b)
T does not satisfy the Kannan theorem, take \(x=0\) and \(y=\frac{3}{2}\);
- (c)
T is asymptotically regular;
- (d)
T is continuous;
- (e)
T satisfies (6) with \(K=2\) and any \(0\leqslant M<1\).
Indeed, if \(x,y\in [0,1]\) or \(x,y\in [\frac{3}{2},\frac{5}{3}]\), then \(d(Tx,Ty)=0\), when the condition (6) is obviously satisfied. If \(x\in [0,1]\) and \(y\in [\frac{3}{2},\frac{5}{3}]\), then \(d(Tx,Ty)= 1\) and \(d(x,y)\geqslant \frac{1}{2}\), \(d(x,Tx)+d(y,Ty)\geqslant x+y-1\geqslant \frac{1}{2}\).
Therefore,and any \(0\leqslant M<1\).$$\begin{aligned} \quad d(Tx,Ty)\leqslant M\cdot d(x,y)+2\cdot \{d(x,Tx)+d(y,Ty)\}~\hbox {for all}~x,y\in X, \end{aligned}$$
When \(M=0\), then from Theorem 2.6 we have a significant extension of Banach’s theorem in a new direction:
Corollary 2.10
Let (X, d) be a complete metric space and \(T:X\rightarrow X\) be a continuous and asymptotically regular mapping satisfying (4) with \(0\leqslant K<+\infty \) (especially, \(K\geqslant 1)\), then T has a unique fixed point \(p\in X\) and \(T^nx\rightarrow p\) for each \(x\in X\).
Remark 2.11
Example 2.12
Example 2.13
Remark 2.14
For clarity of this presentation we omit discussion in \(b-\)metric spaces (see [33]) and \(G-\)metric spaces (see [1]) and consideration of semigroups [20].
Conclusion. In this part, we presented a new extension of Banach’s theorem with examples.
3 Fixed point theorems in preordered sets
An interplay between the order and metrical structure of the space turned out to be very fruitful. In Refs. [36, 37], we find an analogue of Banach theorem in partially ordered sets, further extensions are contained in [24, 42, 43]. In all these works, the mapping considered are monotone. For such mappings one of the fundamental results in fixed-point theory is the classical Knaster–Tarski theorem (also known as the Abian–Brown theorem), see [26, 38]. Recently, Espínola and Wiśnicki [15] studied the problem whether the classical Kirk’s theorem for nonexpansive mappings (see [17]) still holds for monotone-nonexpansive mappings. They proved in some partially ordered sets a general theorem which guarantees the existence of a fixed point for monotone mappings (which need not be either monotone-nonexpansive nor continuous), and which does not impose any conditions on the Banach space.
An interesting reference with many applications of the fixed point theory of monotone mappings is [9].
In this section, we extend Corollary 2.10 on preordered metric spaces, where a preordered binary relation is weaker than a partial order. The key feature in this theorem is that the Kannan’s type condition on the map is only assumed to hold on elements that are comparable but not on the entire set on which they are defined, see Example 3.9.
Definition 3.1
- (a)
reflexive if \(x \preccurlyeq x\) for all \(x\in X\),
- (b)
transitive if \(x\preccurlyeq z\) for all \(x,y,z\in X\) such that \(x\preccurlyeq y\) and \(y\preccurlyeq z\).
Example 3.2
Definition 3.3
A preordered metric space is a triple \((X,d,\preccurlyeq )\) where (X, d) is a metric space and \(\preccurlyeq \) is a preordered on X.
One of the most important hypothesis that we shall use in this section is the monotonicity of the involved mappings.
Definition 3.4
Let \(\preccurlyeq \) be a binary relation on X. A map \(T:X\rightarrow X\) is monotone if \(Tx\preccurlyeq Ty\) whenever \(x\preccurlyeq y\).
The following result is the extension of Corollary 2.10 to Kannan’s type mappings on preordered metric spaces.
Theorem 3.5
- (i)
(X, d) is complete,
- (ii)
T is monotone,
- (iii)
T is continuous,
- (iv)
there exists \(x_0\in X\) such that \(x_0\preccurlyeq Tx_0\),
- (v)
T is asymptotically regular, i.e. \(\lim \limits _{n\rightarrow \infty }d(T^nx,T^{n+1}x)=0~\hbox {for all}~x\in X,\)
- (vi)for all \(x,y\in X\) with \(x\preccurlyeq y\),$$\begin{aligned} d(Tx,Ty)\leqslant K\cdot \{d(x,Tx)+d(y,Ty)\}~\hbox {for some}~0\leqslant K<+\infty . \end{aligned}$$(8)
Proof
To prove uniqueness, we assume that \(v\in X\) is another fixed point of T such that \(u\ne v\). By hypothesis, there exists \(w\in X\) such that \(u\preccurlyeq w\) and \(v\preccurlyeq w\).
Let \(\{w_n=Tw_{n-1}\}\) be the sequence of successive approximations of T based on \(w_0=w\). As T is monotone, \(v=Tv\preccurlyeq Tw=w_1\) and \(u=Tu\preccurlyeq Tw=w_1\). By induction, \(v\preccurlyeq w_n\) and \(u\preccurlyeq w_n\) for all \(n\geqslant 0\).
Case 1. If \(v=w_{n_0}\) for some \(n_0\geqslant 0\), then \(v=Tv=Tw_{n_0}=w_{n_0+1}\) and by induction, \(w_n=v\) for all \(n\geqslant n_0\), so \(w_n\rightarrow v\).
Thus \(w_n\rightarrow v\) and \(w_n\rightarrow u\). The uniqueness of the limit concludes that \(u=v\), so T has a unique fixed point. \(\square \)
Remark 3.6
After the appearance of the Ran and Reurings’ result [37], Nieto and Rodríguez-López [36] changed the continuity of the mapping T with the condition nondecreasing regularity (Definition 3.7). Now, we exchanged the continuity of the mapping T with the condition nondecreasing regularity and we obtain in preordered metric spaces an analogue of [25, Theorem 3.1].
Definition 3.7
Let (X, d) be a metric space, let \(A\subset X\) be a nonempty subset and let \(\preccurlyeq \) be a binary relation on X. Then triple \((A,d,\preccurlyeq )\) is said to be nondecreasing regular if for all sequence \(\{x_n\}\subset A\) such that \(\{x_n\}\rightarrow x\in A\) and \(x_n\preccurlyeq x_{n+1}\) for all \(n\in {\mathbb {N}}\), we have that \(x_n\preccurlyeq x\) for all \(n\in {\mathbb {N}}\).
Theorem 3.8
- (i)
(X, d) is complete,
- (ii)
T is monotone,
- (iii)
\((X,d,\preccurlyeq )\) is nondecreasing regular,
- (iv)
there exists \(x_0\in X\) such that \(x_0\preccurlyeq Tx_0\),
- (v)
T is asymptotically regular, i.e. \(\lim \limits _{n\rightarrow \infty }d(T^nx,T^{n+1}x)=0~\hbox {for all}~x\in X,\)
- (vi)for all \(x,y\in X\) with \(x\preccurlyeq y\),$$\begin{aligned} d(Tx,Ty)\leqslant K\cdot \{d(x,Tx)+d(y,Ty)\}~\hbox {for some}~0\leqslant K<1. \end{aligned}$$
Proof
Observe that condition (4) with \(0\leqslant K<1\), see [25, Theorem 3.1], is slightly stronger than condition (vi) of Theorem 3.8, which only requires the inequality for comparable points, that is, for all \(x,y\in X\) such that \(x\preccurlyeq y\) or \(y\preccurlyeq x\).
Example 3.9
In the next example, we have shown that if condition (9) in Theorems 3.5 and 3.8 fails, it is possible to find examples of functions T with more than one fixed point.
Example 3.10
This shows that conditions in Theorems 3.5 and 3.8 do not imply uniqueness of the fixed point.
In this example, condition (9) does not hold, since given two different elements in X, there is no upper bound of them. In this case, T may have more than one fixed point.
Conclusion. In this section, we discussed the extension of Banach’s theorem in preordered metric spaces.
4 Random fixed-point theorems
An interesting aspect of the nonlinear analysis is to randomize deterministic fixed-point theorems of nonlinear mappings. The study of fixed-point theorems for random operators was initiated by the Prague school of probability research. The first results were studied in 1955–1956 by Špaček and Hanš in the context of Fredholm integral equations with random kernel, see for instance [45]. In a separable metric space, random fixed-point theorems for contraction mappings were proved by Hanš [27] (for some set-valued mappings see [40]).
In many cases, the mathematical models or equations used to describe phenomena in biology, physics, engineering contain certain parameters whose values are unknown. Then, it is more realistic to consider such equations as random operator equations. These equations are much more difficult to handle mathematically than deterministic equations [6].
It has been shown that when the underlying measurable space \((\Omega ,\Sigma )\) is a Suslin family (see [47] for definitions), a deterministic fixed-point theorem may, in general, correspond to a random fixed-point theorem. However, it is unknown if the same is true when the measurable space \((\Omega ,\Sigma )\) is not a Suslin family.
Nieto et al. [35] proved the random version in partially ordered metric spaces of the classical Banach contraction principle. In this section, we will prove some random fixed point theorems for single-valued operators which are asymptotically regular and satisfies some Kannan’s type conditions.
Let \((\Omega ,\Sigma )\) be a measurable space with \(\Sigma \) a \(\sigma \)-algebra of subsets of \(\Omega \). For a metric space (X, d), we denote by CL(X) the family of all nonempty closed subsets of X.
Definition 4.1
A set-valued operator \(F:\Omega \rightarrow 2^X\) is called \(\Sigma \)-measurable if for any open subset B of X, the set \(F^{-1}(B)=\{\omega \in \Omega :F(\omega )\cap B\ne \emptyset \}\) belongs to \(\Sigma \).
Definition 4.2
A measurable (single-valued) operator \(x:\Omega \rightarrow X\) is called a selector for a measurable set-valued operator \(F:\Omega \rightarrow 2^X\) if \(x(\omega )\in F(\omega )\) for all \(\omega \in \Omega \).
Definition 4.3
A mapping \(T:\Omega \times X\rightarrow X\) is called a random operator if for each \(x\in X\), the map \(T(\cdot ,x):\Omega \rightarrow X\) is measurable.
Definition 4.4
A measurable operator \(x:\Omega \rightarrow X\) is said to be a random fixed point of random operator \(T:\Omega \times X\rightarrow X\) if \(T(\omega ,x(\omega ))=x(\omega )\) for all \(\omega \in \Omega \).
Equivalently, it is a measurable selection for the set-valued map \(Fix ~T:\Omega \rightarrow 2^X\) defined by \(Fix~ T(\omega )=\{x\in X:T(\omega ,x)=x\}\).
We recall, a random mapping \(T:\Omega \times X\rightarrow X\) is said to be continuous if for each fixed \(\omega \in \Omega \), the map \(T(\omega ,\cdot ):X\rightarrow X\) has this particular property.
We will list the following results related to the concept of measurability.
Theorem 4.5
[29]. Let \((\Omega ,\Sigma )\) be a measurable space, X be a separable metric space and Y a metric space. If \(T:\Omega \times X\rightarrow Y\) is measurable in \(\omega \in \Omega \) and continuous in \(x\in X\), respectively, and if \(x:\Omega \rightarrow X\) is measurable, then \(T(\cdot ,x(\cdot )):\Omega \rightarrow X\) is a measurable.
Theorem 4.6
[47]. Let \((\Omega ,\Sigma )\) be a measurable space, (Y, d) be a Polish space (i.e. complete and separable metric space) and \(F:\Omega \rightarrow CL(Y)\) a measurable map. Then F has a measurable selection.
Definition 4.7
The following result is the randomization of Theorem 2.6.
Theorem 4.8
(We do not assume the measurability of the functions \(M(\cdot )\) and \(K(\cdot )\).)
Proof
Corollary 4.9
Corollary 4.10
Now, we establish a random version of some fixed-point theorem in preordered metric spaces.
Theorem 4.11
- (i)
T is a continuous random operator,
- (ii)for each \(\omega \in \Omega \), the function \(T(\omega ,\cdot )\) is monotone operator, i.e.$$\begin{aligned} (x,y\in X~\hbox {and}~x\preccurlyeq y)~~\Longrightarrow ~~T(\omega ,x)\preccurlyeq T(\omega ,y), \end{aligned}$$
- (iii)there exists a random variable \(x_0:\Omega \rightarrow X\) with$$\begin{aligned} x_0(\omega )\preccurlyeq T(\omega ,x_0(\omega ))~\hbox {or}~x_0(\omega )\succcurlyeq T(\omega ,x_0(\omega ))~\hbox {for each}~\omega \in \Omega , \end{aligned}$$
- (iv)
T is asymptotically regular,
- (v)there exists a function \(K:\Omega \rightarrow [0,\infty )\) such that for each \(\omega \in \Omega \),for every comparable \(x,y\in X\), i.e. \(x\preccurlyeq y\) or \(y\preccurlyeq x\).$$\begin{aligned} d(T(\omega ,x),T(\omega ,y))\leqslant K(\omega )\cdot \{d(x,T(\omega ,x))+d(y,T(\omega ,y))\} \end{aligned}$$
for every \(x,y\in X\), there exists \(z\in X\) that is comparable to x and y.
Proof
Remark 4.12
- (v\('\))
- there exists functions \(M:\Omega \rightarrow [0,1)\) and \(K:\Omega \rightarrow [0,\infty )\) such that for each \(\omega \in \Omega \),for every comparable \(x,y\in X\), i.e. \(x\preccurlyeq y\) or \(y\preccurlyeq x\).$$\begin{aligned}&d(T(\omega ,x),T(\omega ,y))\\&\quad \leqslant M(\omega )\cdot d(x,y)+K(\omega )\cdot \{d(x,T(\omega ,x))+d(y,T(\omega ,y))\} \end{aligned}$$
Notes
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