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Fixed Points of Closed and Compact Composite Sequences of Operators and Projectors in a Class of Banach Spaces
Some results on fixed points related to the contractive compositions of bounded operators in a class of complete metric spaces which can be also considered as Banach’s spaces are discussed through the paper. The class of composite operators under study can include, in particular, sequences of projection operators under, in general, oblique projective operators. In this paper we are concerned with composite operators which include sequences of pairs of contractive operators involving, in general, oblique projection operators. The results are generalized to sequences of, in general, nonconstant bounded closed operators which can have bounded, closed, and compact limit operators, such that the relevant composite sequences are also compact operators. It is proven that in both cases, Banach contraction principle guarantees the existence of unique fixed points under contractive conditions.
Some results on fixed points related to the contractive compositions of bounded operators in a class of complete metric spaces , which are Banach spaces if is a vector space on a certain field (usually or ) and the metric is homogeneous and translation-invariant, are discussed through the paper. In this case, the metric is also a norm and, since the space is a vector space the complete metric space is also a Banach space . The class of composite operators under study can include, in particular, sequences of projection operators under, in general, oblique projective operators. Section 2 is concerned of composite operators which include sequences of pairs of contractive operators including, in general, oblique projection operators in the operator composite strip. The results are generalized in Section 3 to sequences of, in general, nonconstant bounded closed operators which can have bounded, closed, and compact limits, such that the relevant composite sequences are also compact operators. It is proven in this paper that Banach contraction principle [1–4] guarantees the existence of unique fixed points under contractive conditions fulfilled by some relevant strips of composite operators within in the whole composite sequence of operators.
2. Some Results on Contractive Mappings and Fixed Points under Projection Operators
Let be a sequence of self-mappings on , where is a metric space and consider a sequence of (non-necessarily orthogonal) projection operators on of respective ranges which are then closed subspaces of , . We can then consider a sequence of projection operators with such that so that is in for . Now, consider sequences in in with such that the identities hold by construction for . The subsequent result holds.
Theorem 1. Assume that is a complete metric space with the metric being homogeneous and translation-invariant and . The following properties hold.(i)If all the self-mappings on in the sequence are nonexpansive and the sequence of projection operators from to the sequence of subspaces is uniformly bounded then .(ii)Assume that the self-mappings on in the subsequence are contractive for some , that the sequence of operators converges to , and that the projection operator is constant and bounded (i.e., if it is not orthogonal, i.e., it is oblique, then its norm exceeds one and it is finite) then Property (i) holds. Furthermore, is a Cauchy sequence which converges to some unique limit point for any initial iterate , where is the unique fixed point of .(iii)Assume that there is a strictly sequence of nonnegative integers such that the difference sequence is uniformly bounded and has a limit . Assume also that the associate sequence of composite self-mappings is contractive and that the sequence projection operators from to the sequence of subspaces is uniformly bounded and has a set of subsequences each converging to a set of projectors from to . Then, , and there is at most a finite number of distinct Cauchy subsequences with distinct limit points in .
Proof. Since the metric is homogeneous and translation-invariant then the complete metric space can also be considered as a Banach space under the metric-induced norm defined as . The norm of any projection operator in the considered sequence is defined as . Then, if is the left-composite self-mapping for any , one gets from direct calculations, by using the property that the metric is homogeneous and translation-invariant, the following relations for any iterated sequences in constructed as , with arbitrary :
since the metric is homogeneous and translation-invariant, the norm is an induced-metric norm, then , where , and the self-mappings in the sequence on are all non-expansive; , and the sequence of projection operators from to the sequence of subspaces is uniformly bounded with . Then, one has from (2) where if all the projections are orthogonal and , otherwise. Hence, Property (i). If is a constant bounded projection from to with being constant for and all the self-mappings on of the sequence are contractive then one gets from (2) for the real constant such that that Property (i) holds according to the relation so that from (4) for any initial value of the iteration since . Then, is Cauchy sequence which has a limit in , since is closed, . It is now proven that is the unique limit point in of any sequence of iterates, where is a fixed point of the self-mapping which is unique from Banach contraction principle. It is now proven that is contractive. Assume not so that one has if is not contractive: for nonzero is contractive for . Then, one gets, since , that which is a contradiction since unless converges to zero. If converges to zero then there are and such that for all since and some such that that yields the contradiction Thus, if the subsequence is contractive in then its limit is also contractive. Now, since is contractive then its fixed point is unique since is complete. It is clear that is a limit point in of any iterated sequence. Assume that it is not unique so that there are two limit points for some which is not trivially a fixed point of (since the fixed point of the contractive self-mapping is unique if is complete). Thus, from Banach contraction principle and since is complete, one has as since converges, there is a limit self-mapping on : Thus, . Hence a contradiction to and then in is the unique limit point of even in the event that there is such that . Property (ii) has been proven.
On the other hand, if the sequence of operators is uniformly bounded then and, if furthermore, the sequence of compositie mappings is contractive with some constant given by , where is a strictly increasing sequence of natural numbers such that the sequence is uniformly bounded, one has directly from (1)-(2): for , where is a sequence of finite sets of natural numbers satisfying . Thus, one gets from (10): for any since and. Since is uniformly bounded with existing limit is contractive and the projection sequence from to the sequence of subspaces is uniformly bounded while having a finite set of subsequences converging to a set of projectors from to for then so that is a Cauchy sequence satisfying: for at most distinct points since, by hypothesis, there is a natural number satisfying and since ; . Hence, Property (iii) holds.
Remark 2. The existence of some in the proof of Theorem 1(ii) often happens. For instance if is linear then fulfils the relations for any .
The following auxiliary result to be then used holds.
Lemma 3. Assume that the sequences of linear self-mappings and converge to respective limits and being mappings from to and from to , respectively, in the sense that from any . Let be a Banach space with the norm of any being defined by . For any given such that , where and . If is contractive then the sequence of mappings from to is then asymptotically contractive.
Proof. Direct calculations yield
and for any given , there are such that for all integer for all integer . Thus, if then one has for any :
The last inequality holds if and only if for any given positive real constant and any positive real constant being sufficiently small to satisfy. Since and , there are finite natural numbers such that satisfies that constraint for any integer . If is contractive then for some real constant . Thus, there is such that for any given real and ≤ ≤ and then the sequence is asymptotically contractive.
Note that Lemma 3 holds irrespective of the fact that one of the operators be a projection.
3. Results on Contractive Mappings of Sequences of Composite Bounded Operators
The results of Theorem 1 are now extended to the study of contractive compositions of linear operators belonging to two sequences of bounded operators with so that none of them is necessarily a projection on some subspace of . Some preparatory results are first established. In the following, a Banach space, being equivalent to a complete metric space with a homogeneous and translation-invariant metric induced-norm is considered such that for any real and any . The subsequent result refers to the asymptotic distances in sequences involving a convergent composite sequence of bounded linear operators.
Lemma 4. Consider a Banach space , with , being equivalent to a complete metric space with a homogeneous and translation-invariant metric induced-norm . Consider also a composite sequence of two sequences of bounded linear operators defined by defined by for any , where , provided that and ; for all . The following properties hold.(i)Assume that . Then, ; for all ; for all .(ii)Assume that . Then, ; for all ; for all .(iii)Assume that . Then, ; for all ; for all .(iv)Define the operator composite sequence of operators as with ; for all subject to for and ; for all . Define also the operator composite sequence of operators as for , where if has not a limit as and if . Then Properties (i)–(iii) hold for any .
Proof. Assume that with since , equivalently , is complete so that strongly for since . Then, , = ; for all , and
that leads to
Hence, Property (i). Now, assume that only has a limit. Then,
Hence, Property (ii). Finally, assume that only has a limit. Then,
Hence, Property (iii).
Property (iv) is direct from Properties (i)–(iii) and the associative property of composition of operators since for any , exists in if and for , and then, and ; for all since Then, for any finite , one gets for some positive finite constants and since any linear operator with a limit admits a unique decomposition .
The next result is concerned with the closeness of the limit operator if the sequence of operators is closed.
Lemma 5. Consider a sequence of closed linear operators defined by in a Banach space , such that with , which converge to a limit operator . Then, such a limit is a closed operator which is bounded if all the operators of the sequence are bounded.
Proof. Note the following:(1) for any bounded sequence since ≤ = 0 since . Furthermore, for any is a finite real sequence of constants with for any , since is convergent, and . Again, since is convergent, there is such that for any so that is bounded.(2)If the bounded sequence converges to for any then since is a closed operator for any so that (3)One gets combining the above points (1)-(2) that: → for any where as , and then as for any and as since is a sequence of closed operators which converges. Thus, the limit of bounded converging sequences belongs to the domain of the limit operator. Furthermore, one has for any bounded sequence converging to : as then strongly so that is a closed operator as a result.
The above result can be extended to sequences of operators not all of them being bounded provided that each of such sequences of operators can be decomposed as a composition of subsequences of composite operators such that each of such a composite subsequence is bounded. The above result can be applied to sequences of operators not all of them being bounded. It is well-known that a sequence of linear operators on a Hilbert space [5, 6] is bounded if and only if they are closed and their domain is the whole vector space , [1, 4]. Thus, we have the following result using Lemma 5.
Lemma 6. Consider a sequence of linear bounded operators defined by in a Banach space which converge to a limit operator . Then, such a limit is a bounded linear closed operator.
The subsequent result is concerned with the limit operator of a sequence of linear operators being compact if all the operators in the sequence are bounded and at least one of them is compact.
Lemma 7. The following properties hold.(i)Consider a sequence of bounded compact linear operators defined by in a Banach space , such that with , which converge to a limit operator . Then, such a limit is a compact operator. (ii)Assume that the sequence of bounded operators satisfies that there is at least one compact operator within all subsequences being subject to for some subsequence for any . Then, the composed operator is compact as it is its limit provided that it exists.
Proof. We have to prove that if is bounded then is convergent. Note that for given bounded sequences and ; that and, one gets by taking subsequences , Since , we can find such that for , we have for any given and , since and are bounded subsequences, and converges, so that it is a Cauchy sequence, since contains at least one compact operator. As a result, is arbitrarily small for being sufficiently small. Thus, is convergent. Property (i) has been proven. Property (ii) follows from Property (i) and the fact that any operator composite sequence of bounded operators is a compact operator if there is at least one which is compact.
Now, define the composite operator by
Define also the sequence of composite operators as where replaces each operator in the composite operator by its limit when such a limit exists. A result is now given based on the existence of the following limit:
Theorem 8. Consider the operator composite sequence ; for all of composed linear bounded operators in (31) on a Banach space , subject to for and ; for all , and the sequence of composed linear operators of defined in the same way as as by replacing each operator possessing a limit by such a limit. The following properties hold.(i)Either the sequences and have limits and both limits coincide or none of them has a limit and, furthermore, and → as .(ii)If the limits of Property (i) exist and are finite then the limits of the sequences of operators and as ; for all have the same set of fixed points.(iii)Assume, in addition, that for some , there is at least one compact operator in the composition operator , and that for for and some and that all the operators are closed. If Property (i) holds with as ; for some and , then is contractive and the sequences of composite operators and converge to zero as , and is bounded, closed and compact and has a unique fixed point in to which all sequences with initial conditions in converge.(iv)Assume that there is a (in general, nonunique) strictly increasing sequence of nonnegative integers with and such that for some nonnegative real sequence , and some real constant . Assume, in addition, that for some , there is at least one compact operator in any composite operator and that all the operators are closed. Then, the sequences of composite operators and converge to zero as for any finite . Finally, assume that as . Then, is contractive, continuous, bounded, closed, and compact and has a unique fixed point in to which all sequences with initial conditions in converge.
Proof. Note from the definition of the sequences that for any, since
so that . Then, either both sequences of operators and have the same (finite or infinity) limit or none of them has a limit and as . Hence, Property (i). Property (ii) follows trivially from Property (i) for the case .
To prove Property (iii), note that
so that, since then the limit operator on is continuous, then bounded with for any . Furthermore, for any and the associative property of composition of operators, ; for all , for all