Convergence to Compact Sets of Inexact Orbits of Nonexpansive Mappings in Banach and Metric Spaces

Abstract

We study the influence of computational errors on the convergence to compact sets of orbits of nonexpansive mappings in Banach and metric spaces. We first establish a convergence theorem assuming that the computational errors are summable and then provide examples which show that the summability of errors is necessary for convergence.

1. Introduction

Convergence analysis of iterations of nonexpansive mappings in Banach and metric spaces is a central topic in nonlinear functional analysis. It began with the classical Banach theorem [1] on the existence of a unique fixed point for a strict contraction. Banach's celebrated result also yields convergence of iterates to the unique fixed point. There are several generalizations of Banach's fixed point theorem which show that the convergence of iterates holds for larger classes of nonexpansive mappings. For instance, Rakotch [2] introduced the class of contractive mappings and showed that their iterates also converged to their unique fixed point.

In view of these results and their numerous applications, it is natural to ask if convergence of the iterates of nonexpansive mappings will be preserved in the presence of computational errors. In [3], we provide affirmative answers to this question. Related results can be found, for example, in [4, 5]. More precisely, in [3] we show that if all exact iterates of a given nonexpansive mapping converge (to fixed points), then this convergence continues to hold for inexact orbits with summable errors. In [6], we continued to study the influence of computational errors on the convergence of iterates of nonexpansive mappings in both Banach and metric spaces. We show there that if all the orbits of a nonexpansive self-mapping of a metric space converge to some closed subset of then all inexact orbits with summable errors also converge to this attractor set On the other hand, we also construct examples which show that inexact orbits may fail to converge if the errors are not summable.

Our purpose in the present paper is to consider the case where different exact orbits converge to possibly different compact subsets of In Section 2, we obtain a convergence result (see Theorem 2.1 below) under the assumption that the computational errors are summable. This result is an extension of [3, Theorem 4.2]. In Sections 3 and 4, we provide examples which show that the summability of errors is necessary for convergence (see Proposition 3.1 and Theorem 4.1).

2. Convergence to Compact Sets

Let be a complete metric space. For each and each nonempty and closed subset put

(2.1)

For each mapping set for all

Theorem 2.1.

Let satisfy

(2.2)

Suppose that for each there exists a nonempty compact set such that

(2.3)

Assume that

(2.4)

Then, there exists a nonempty compact subset of such that

(2.5)

Proof.

In order to prove the theorem, it is sufficient to show that any subsequence of has a convergent subsequence.

To see this, it is sufficient to show that for any the following assertion holds:

(P1) any subsequence of has a subsequence which is contained in a ball of radius

Indeed, there is an integer such that

(2.6)

Define a sequence by

(2.7)

(2.8)

There exists a nonempty compact set such that

(2.9)

By (2.4), (2.7), and (2.8),

(2.10)

Assume that is an integer and that for

(2.11)

(Note that in view of (2.10), inequality (2.11) is valid when )

By (2.2) and (2.11),

(2.12)

When combined with (2.4), this implies that

(2.13)

so that (2.11) also holds for Thus, we have shown that for all integers

(2.14)

by (2.6). In view of (2.9), we have for all large enough natural numbers

(2.15)

By (2.15), there exist an integer and a sequence such that

(2.16)

Consider any subsequence of Since the set is compact, the sequence has a convergent subsequence

We may assume without loss of generality that all elements of this convergent subsequence belong to for some

In view of (2.16),

(2.17)

Thus, (P1) holds and this completes the proof of Theorem 2.1.

Note that Theorem 2.1 is an extension of the following result established in [3].

Theorem 2.2.

Let be a complete metric space and let be such that

(2.18)

and for each the sequence converges in

Assume that satisfies and that a sequence satisfies Then, the sequence converges to a fixed point of in

3. First Example of Nonconvergence to Compact Sets

In this section, we show that both Theorems 2.1 and 2.2 cannot, in general, be improved (cf. [6, Proposition 3.1]).

Proposition 3.1.

For any normed space there exists an operator such that for all the sequence converges for each and, for any sequence of positive numbers there exists a sequence with for all nonnegative integers which converges to a compact set if and only if the sequence is summable, that is,

Proof.

This is a simple fact because we may take to be the identity operator: Then, we may take to be an arbitrary element of with and define by induction

(3.1)

Evidently, and for all integers so that the convergence of to a compact set is equivalent to the summability of the sequence Proposition 3.1 is proved.

4. Second Example of Nonconvergence to Compact Sets

In Section 3, we have shown that Theorems 2.1 and 2.2 cannot, in general, be improved. However, in Proposition 3.1 every point of the space is a fixed point of the operator and the inexact orbits tend to infinity. In this section, we construct an operator on a certain complete metric space (a bounded, closed, and convex subset of a Banach space) such that all of its orbits converge to its unique fixed point, and for any nonsummable sequence of errors and any initial point, there exists an inexact orbit which does not converge to any compact set (cf. [6, Theorem 4.1]).

Let be the set of all sequences of nonnegative numbers such that For and in set

(4.1)

Clearly, is a complete metric space.

Define a mapping as follows:

(4.2)

In other words, for any

(4.3)

Set for all Clearly,

(4.4)

for all

Theorem 4.1.

Let

(4.5)

and Then, there exists a sequence such that

(4.6)

and the following property holds.

There is no nonempty compact set such that

(4.7)

In the proof of this theorem, we may assume without loss of generality that

(4.8)

We precede the proof of Theorem 4.1 with the following lemma.

Lemma 4.2.

Let let be an integer, and let be a natural number. Then, there exist an integer and a sequence such that

(4.9)

with

Proof.

There is a natural number such that

(4.10)

(4.11)

Set

(4.12)

Then,

(4.13)

By (4.5), there is a natural number such that

(4.14)

By (4.14) and (4.8),

(4.15)

and we may assume without loss of generality that

(4.16)

In view of (4.14) and (4.8),

(4.17)

For define as follows:

(4.18)

(4.19)

Clearly, for is well defined and by (4.18), (4.19), (4.10), and (4.16),

(4.20)

Thus

Let We now estimate If then by (4.2), (4.3), (4.13), and (4.18),

(4.21)

Let We first set

(4.22)

In view of (4.14), (4.2), and (4.3), for all integers When combined with (4.18), this implies that

(4.23)

(4.24)

By (4.18) and (4.23),

(4.25)

for all It now follows from (4.22), (4.25), (4.18), (4.19), and (4.23) that

(4.26)

When combined with (4.12), this implies that

(4.27)

By (4.17) and (4.18),

(4.28)

This completes the proof of Lemma 4.2.

Proof.

In order to prove the theorem, we construct by induction, using Lemma 4.2, a sequence of nonnegative integers and a sequence such that

(4.29)

(4.30)

(4.31)

and for all integers

(4.32)

In the sequel, we use the notation

Set

(4.33)

Assume that is an integer and that we have already defined a (finite) sequence of nonnegative numbers and a (finite) sequence of points such that (4.33) is valid, (4.30) holds for all integers satisfying

(4.34)

and that (4.32) holds for all integers satisfying (Note that for this assumption does hold.)

Now, we show that this assumption also holds for

Indeed, applying Lemma 4.2 with

(4.35)

we obtain that there exist an integer and a sequence such that

(4.36)

Thus, the assumption made for also holds for Therefore, we have constructed by induction a sequence of points and a sequence of nonnegative integers which satisfy (4.30) and (4.31) for all integers respectively, and (4.32) for all integers

Finally, we show that there is no nonempty compact set such that

(4.37)

Assume the contrary. Then, there does exist a nonempty compact set such that

(4.38)

This implies that any subsequence of has a convergent subsequence.

Consider such a subsequence This subsequence has a convergent subsequence There are therefore a point such that

(4.39)

and a natural number such that

(4.40)

Hence we have, for all integers

(4.41)

This, of course, contradicts the inequality The contradiction we have reached completes the proof of Theorem 4.1.