Fixed Points of Closed and Compact Composite Sequences of Operators and Projectors in a Class of Banach Spaces

Abstract

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.

1. Introduction

Some results on fixed points related to the contractive compositions of bounded operators in a class of complete metric spaces (𝑋,𝑑), which are Banach spaces if 𝑋 is a vector space on a certain field 𝐹 (usually 𝐑 or 𝐂) and the metric is homogeneous and translation-invariant, are discussed through the paper. In this case, the metric 𝑑 is also a norm and, since the space 𝑋 is a vector space the complete metric space (𝑋,𝑑) is also a Banach space (𝑋,‖‖). 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 {𝑇_𝑘} be a sequence 𝑇_𝑘∶𝑋→𝑋 of self-mappings on 𝑋, where (𝑋,𝑑) is a metric space and consider a sequence {𝑃_𝑘} of (non-necessarily orthogonal) projection operators on 𝑋 of respective ranges 𝑀_𝑘 which are then closed subspaces of 𝑋, [3]. We can then consider a sequence of projection operators {𝑃_𝑀𝑘} with 𝑃_𝑀𝑘∶𝑋→𝑀_𝑘 such that 𝑃_𝑘=𝑃_𝑀𝑘 so that 𝑋=Im𝑃_𝑘⊕Ker𝑃_𝑘and𝑧=𝑃_𝑘𝑥∈Im𝑃_𝑘 is in 𝑀_𝑘forany𝑥∈𝑋and𝑧=𝑥−𝑧=(𝐼−𝑃_𝑘)𝑥∈Ker𝑃_𝑘 for 𝑘∈𝐍₀=𝐍∪{0}. Now, consider sequences {𝑥_𝑘} in 𝑋and{𝑧_𝑘} in 𝑀_𝑘 with  𝑧_𝑘=𝑃_𝑘𝑥_𝑘 such that the identities 𝑥_𝑘+1=𝑇_𝑘𝑥_𝑘=𝑃_𝑘+1𝑥_𝑘+1+𝐼−𝑃_𝑘+1𝑥_𝑘+1=𝑧_𝑘+1+𝐼−𝑃_𝑘+1𝑥_𝑘+1=𝑇_𝑘𝑃_𝑘𝑥_𝑘+𝑇_𝑘𝐼−𝑃_𝑘𝑥_𝑘 (1) hold by construction for 𝑘∈𝐍₀. The subsequent result holds.

Theorem 1. Assume that (𝑋,𝑑) is a complete metric space with the metric 𝑑∶𝑋×𝑋→𝐑₀₊ being homogeneous and translation-invariant and 0∈𝑋;𝑘∈𝐍₀. The following properties hold.(i)If all the self-mappings on 𝑋 in the sequence {𝑇_𝑘} are nonexpansive and the sequence of projection operators {𝑃_𝑘} from 𝑋 to the sequence of subspaces {𝑀_𝑘} is uniformly bounded then 𝑑(𝑧_𝑘+2,𝑧_𝑘+1)<∞;𝑘∈𝐍₀.(ii)Assume that the self-mappings on 𝑋 in the subsequence {𝑇_𝑘}_𝑘≥𝑛0 are contractive for some 𝑛₀∈𝐍₀, that the sequence of operators {𝑇_𝑘} converges to 𝑇∶𝑋→𝑋, and that the projection operator 𝑃∶𝑋→𝑀⊂𝑋 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, ∃lim_𝑘→∞𝑑(𝑧_𝑘+2,𝑧_𝑘+1)=0and{𝑧_𝑘+1=𝑃𝑇_𝑘𝑥_𝑘} is a Cauchy sequence which converges to some unique limit point 𝑧(=𝑃𝑇𝑥)∈𝑀 for any initial iterate 𝑥₀∈𝑋, where 𝑥(=𝑇𝑥)∈𝑋 is the unique fixed point of 𝑇∶𝑋→𝑋.(iii)Assume that there is a strictly sequence of nonnegative integers {𝑗_𝑘} such that the difference sequence {𝑗_𝑘+1−𝑗_𝑘} is uniformly bounded and has a limit 𝐽∈𝐍as𝑘→∞. Assume also that the associate sequence of composite self-mappings 𝑇(𝑗_𝑘+1+1,𝑗_𝑘)={𝑇_𝑗𝑘+1⋯𝑇_𝑗𝑘+1𝑇_𝑗𝑘} is contractive and that the sequence projection operators {𝑃_𝑘} from 𝑋 to the sequence of subspaces {𝑀_𝑘} is uniformly bounded and has a set of subsequences each converging to a set {𝑃_𝑗𝑖}→𝑃_𝑖 of projectors from 𝑋 to {𝑀_𝑗𝑖}. Then, ∃lim_𝑘→∞𝑑(𝑧_𝑗𝑘+1+1,𝑧_𝑗𝑘+1)=0, and there is at most a finite number 𝐽 of distinct Cauchy subsequences {𝑧_𝑗𝑘} with distinct limit points {𝑧₁,…,𝑧_𝐽} in 𝑋.

Proof. Since the metric is homogeneous and translation-invariant then the complete metric space (𝑋,𝑑) can also be considered as a Banach space (𝑋,‖‖) under the metric-induced norm defined as ‖𝑥−𝑦‖=𝑑(𝑥,𝑦);forall𝑥,𝑦∈𝑋. The norm of any projection operator in the considered sequence is defined as ‖𝑃_𝑘‖=sup_{‖𝑥(∈𝑋)≠0‖≤1}(‖𝑃_𝑘𝑥‖/‖𝑥‖)=sup_{‖𝑥(∈𝑋)‖=1}(‖𝑃_𝑘𝑥‖/‖𝑥‖)=sup_{‖𝑥(∈𝑋)‖=1}‖𝑃_𝑘𝑥‖;𝑘∈𝐍₀. Then, if 𝑇(𝑘+1,𝑗)∶𝑋→𝑋 is the left-composite self-mapping 𝑇(𝑘+1,𝑗)=𝑇_𝑘⋯𝑇_𝑗+1𝑇_𝑗 for any 𝑘(≥𝑗),𝑗∈𝐍₀, one gets from direct calculations, by using the property that the metric is homogeneous and translation-invariant, the following relations for any iterated sequences{𝑥_𝑘}in𝑋and{𝑧_𝑘} in 𝑀_𝑘 constructed as 𝑥_𝑘+1=𝑇_𝑘𝑥_𝑘, 𝑧_𝑘=𝑃_𝑘𝑥_𝑘 with arbitrary 𝑥₀∈𝑋;𝑘∈𝐍: 𝑑𝑧_𝑘+2,𝑧_𝑘+1=𝑑𝑃_𝑘+2𝑥_𝑘+2,𝑃_𝑘+1𝑥_𝑘+1=𝑑𝑃_𝑘+1𝑥_𝑘+2,𝑃_𝑘+1𝑥_𝑘+1+𝑃_𝑘+1𝑥_𝑘+2−𝑃_𝑘+2𝑥_𝑘+2≤𝑑𝑃_𝑘+1𝑥_𝑘+2,𝑃_𝑘+1𝑥_𝑘+1+𝑑𝑃_𝑘+1𝑥_𝑘+1,𝑃_𝑘+1𝑥_𝑘+1+𝑃_𝑘+1𝑥_𝑘+2−𝑃_𝑘+2𝑥_𝑘+2≤𝑑𝑃_𝑘+1𝑥_𝑘+2,𝑃_𝑘+1𝑥_𝑘+1+𝑑𝑃_𝑘+2𝑥_𝑘+2,𝑃_𝑘+1𝑥_𝑘+2=𝑃_𝑘+1𝑥_𝑘+2−𝑃_𝑘+1𝑥_𝑘+1+𝑃_𝑘+2𝑥_𝑘+2−𝑃_𝑘+1𝑥_𝑘+2≤𝑃_𝑘+1𝑥_𝑘+2−𝑥_𝑘+1+𝑃_𝑘+2−𝑃_𝑘+1𝑥_𝑘+2=𝑃_𝑘+1𝑑𝑥_𝑘+1,𝑥_𝑘+2+𝑃_𝑘+2−𝑃_𝑘+1𝑑𝑥_𝑘+2=𝑃,0_𝑘+1𝑑𝑇_𝑘𝑥_𝑘,𝑇_𝑘+1𝑇_𝑘𝑥_𝑘+𝑃_𝑘+2−𝑃_𝑘+1𝑑𝑇(𝑘+2,0)𝑥₀≤𝑃,0_𝑘+1𝑑𝑥₁,𝑥₀+𝑃_𝑘+2−𝑃_𝑘+1𝑑𝑇(𝑘+2,0)𝑥₀≤𝑃,0_𝑘+1𝑑𝑥₁,𝑥₀+𝑃_𝑘+1+𝑃_𝑘+2𝑑𝑥₀≤𝑃,0_𝑘+1𝑑𝑇₀𝑥₀,0+𝑑𝑥₀+𝑃,0_𝑘+1+𝑃_𝑘+2𝑥₀;𝑘∈𝐍₀ (2)
since the metric is homogeneous and translation-invariant, the norm is an induced-metric norm, then ‖𝑥_𝑘‖=𝑑(𝑥_𝑘,0)=𝑑(𝑇(𝑘,0)𝑥₀,0), where 𝑧_𝑘∈𝑀_𝑘, and the self-mappings in the sequence {𝑇_𝑘} on 𝑋 are all non-expansive; 𝑘∈𝐍₀, and the sequence of projection operators {𝑃_𝑘} from 𝑋 to the sequence of subspaces {𝑀_𝑘} is uniformly bounded with sup_𝑘∈𝐍0‖𝑃_𝑘‖≤𝜇<∞. Then, one has from (2) 𝑑𝑧_𝑘+2,𝑧_𝑘+1≤4𝜇𝑥₀<∞;𝑘∈𝐍₀, (3) where 𝜇=1 if all the projections are orthogonal and 𝜇>1, otherwise. Hence, Property (i). If 𝑃_𝑘=𝑃_𝑘+1=𝑃for𝑘∈𝐍₀ is a constant bounded projection from 𝑋 to 𝑀 with 𝑀_𝑘=𝑀⊂𝑋 being constant for 𝑘∈𝐍₀ and all the self-mappings on 𝑋 of the sequence {𝑇_𝑘} are contractive then one gets from (2) for the real constant 𝐾=sup_{𝑘(≥𝑛0)∈𝐍0}𝐾_𝑘 such that 𝐾∈[0,1) that Property (i) holds according to the relation 𝑑𝑧_𝑘+2,𝑧_𝑘+1≤‖𝑃‖𝑑𝑇_𝑘𝑥_𝑘,𝑇_𝑘+1𝑥_𝑘+1≤‖𝑃‖𝑑𝑇_𝑘𝑥_𝑘,𝑇_𝑘𝑥_𝑘+1+𝑇_𝑘+1𝑥_𝑘+1−𝑇_𝑘𝑥_𝑘+1≤‖𝑃‖𝑑𝑇_𝑘𝑥_𝑘,𝑇_𝑘𝑥_𝑘+1+‖𝑃‖𝑑𝑇_𝑘𝑥_𝑘+1,𝑇_𝑘𝑥_𝑘+1+𝑇_𝑘+1𝑥_𝑘+1−𝑇_𝑘𝑥_𝑘+1=‖𝑃‖𝑑𝑇_𝑘𝑥_𝑘,𝑇_𝑘𝑥_𝑘+1+‖𝑃‖𝑑0,𝑇_𝑘+1𝑥_𝑘+1−𝑇_𝑘𝑥_𝑘+1=‖𝑃‖𝑑𝑇_𝑘𝑥_𝑘,𝑇_𝑘𝑥_𝑘+1+𝑑𝑇_𝑘𝑥_𝑘+1,𝑇_𝑘+1𝑥_𝑘+1=‖𝑃‖𝑇_𝑘𝑥_𝑘−𝑇_𝑘𝑥_𝑘+1+𝑇_𝑘𝑥_𝑘+1−𝑇_𝑘+1𝑥_𝑘+1≤‖𝑃‖𝑇_𝑘𝑥_𝑘−𝑥_𝑘+1+‖𝑃‖𝑇_𝑘−𝑇_𝑘+1𝑥_𝑘+1≤‖𝑃‖𝑇_𝑘𝑥_𝑘−𝑥_𝑘+1+𝑇_𝑘−𝑇_𝑘+1×𝑑𝑇(𝑘+1,0)𝑥_𝑛0,𝑇()0≤‖𝑃‖𝐾𝑘+1,0_{𝑘−𝑛0+1}𝑑𝑥_𝑛0+1,𝑥_𝑛0+𝐾1−𝐾^𝑘−𝑛0𝑇1−𝐾^𝑘−𝑇_𝑘+1𝑥_𝑛0𝑘<∞;≥𝑛₀∈𝐍₀ (4) so that ∃lim_𝑘→∞𝑑(𝑧_𝑘+2,𝑧_𝑘+1)=0 from (4) for any initial value 𝑥₀∈𝑋 of the iteration since 𝐾^{𝑘−𝑛0−1}→0and‖𝑇_𝑘−𝑇_𝑘+1‖→0as𝑘→∞since{𝑇_𝑘}→𝑇. Then, {𝑧_𝑘} is Cauchy sequence which has a limit 𝑧 in 𝑀, since 𝑀 is closed, [4]. It is now proven that 𝑧(∈𝑀)=𝑃𝑥 is the unique limit point in 𝑀 of any sequence of iterates, where 𝑥(=𝑇𝑥)∈𝑋 is a fixed point of the self-mapping 𝑇∶𝑋→𝑋 which is unique from Banach contraction principle. It is now proven that 𝑇∶𝑋→𝑋 is contractive. Assume not so that one has if 𝑇∶𝑋→𝑋 is not contractive: 𝑑𝑥_𝑘+1,𝑥_𝑘≤𝑑𝑇𝑥_𝑘+1,𝑇𝑥_𝑘=𝑑𝑇_𝑘𝑥_𝑘+1,𝑇_𝑘𝑥_𝑘+𝑇𝑥_𝑘−𝑇_𝑘𝑥_𝑘+𝑇_𝑘𝑥_𝑘+1−𝑇𝑥_𝑘+1≤𝑑𝑇_𝑘𝑥_𝑘+1,𝑇_𝑘𝑥_𝑘+𝑑𝑇_𝑘𝑥_𝑘,𝑇_𝑘𝑥_𝑘+𝑇𝑥_𝑘−𝑇_𝑘𝑥_𝑘+𝑇_𝑘𝑥_𝑘+1−𝑇𝑥_𝑘+1=𝑑𝑇_𝑘𝑥_𝑘+1,𝑇_𝑘𝑥_𝑘+𝑑0,𝑇𝑥_𝑘−𝑇_𝑘𝑥_𝑘+𝑇_𝑘𝑥_𝑘+1−𝑇𝑥_𝑘+1=𝑑𝑇_𝑘𝑥_𝑘+1,𝑇_𝑘𝑥_𝑘+𝑑𝑇_𝑘𝑥_𝑘+𝑇𝑥_𝑘+1,𝑇𝑥_𝑘+𝑇_𝑘𝑥_𝑘+1≤𝑑𝑇_𝑘𝑥_𝑘+1,𝑇_𝑘𝑥_𝑘+𝑇_𝑘−𝑇𝑥_𝑘+𝑥_𝑘+1≤𝐾_𝑘𝑑𝑥_𝑘+1,𝑥_𝑘+𝑇_𝑘−𝑇𝑥_𝑘+𝑥_𝑘+1;𝑘≥𝑛₀ (5) for nonzero 𝑥_𝑘and𝑥_𝑘+1since𝑇_𝑘∶𝑋→𝑋 is contractive for 𝑘≥𝑛₀. Then, one gets, since ‖𝑇_𝑘−𝑇‖→0as𝑘→∞, that limsup_𝑘→∞1−𝐾_𝑘−𝑇_𝑘−𝑇𝑥_𝑘+𝑥_𝑘+1=limsup_𝑘→∞1−𝐾_𝑘𝑥_𝑘+𝑥_𝑘+1≤0 (6) which is a contradiction since 𝐾_𝑘<1for𝑘≥𝑛₀ unless {𝑥_𝑘} converges to zero. If {𝑥_𝑘} converges to zero then there are 𝑛₁(≥𝑛₀)∈𝐍₀ and 0<𝜆=𝜆(𝑛₁)<1−sup_𝑘≥𝑛1𝐾_𝑘 such that ‖𝑇_𝑘−𝑇‖≤𝜆  for  all  𝑘≥𝑛₁ since ‖𝑇_𝑘−𝑇‖→0as𝑘→∞ and some 𝑘₁≥𝑛₁ such that ‖𝑥_𝑘1‖+‖𝑥_𝑘1+1‖>0 that yields the contradiction 0<1−𝐾_𝑘1−𝜆𝑥_𝑘1+𝑥_𝑘1+1≤0. (7) Thus, if the subsequence {𝑇_𝑘}_𝑘≥𝑛0 is contractive in 𝑋 then its limit 𝑇∶𝑋→𝑋 is also contractive. Now, since 𝑇∶𝑋→𝑋 is contractive then its fixed point is unique since (𝑋,𝑑) is complete. It is clear that 𝑧=𝑃𝑇𝑥 is a limit point in 𝑀 of any iterated sequence. Assume that it is not unique so that there are two limit points 𝑧=𝑃𝑇𝑥,𝑧=𝑃𝑇𝑥(≠𝑧)∈𝑀 for some 𝑥(≠𝑥)∈𝑋 which is not trivially a fixed point of 𝑇∶𝑋→𝑋 (since the fixed point 𝑥∈𝑋 of the contractive self-mapping 𝑇∶𝑋→𝑋 is unique if (𝑋,𝑑) is complete). Thus, from Banach contraction principle and since (𝑋,𝑑) is complete, one has 0⟵𝑑𝑃𝑇_𝑘𝑥,𝑃𝑇_𝑘𝑥=𝑃𝑇_𝑘𝑥−𝑃𝑇_𝑘𝑥≤‖𝑃‖𝑇_𝑘𝑥−𝑇_𝑘𝑥≤‖𝑃‖𝐾^𝑘𝑑(𝑥,𝑥)<∞;𝑘∈𝐍₀ (8) as 𝑘→∞ since {𝑇_𝑘} converges, there is a limit self-mapping 𝑇 on 𝑋: 𝑧⟵𝑃𝑇_𝑘𝑥⟶𝑃𝑇_𝑘𝑥⟶𝑃𝑇𝑥=𝑃𝑥=𝑧. (9) Thus, 𝑃𝑇_𝑘𝑥→𝑧=𝑧as𝑘→∞. Hence a contradiction to 𝑧≠𝑧 and then 𝑧 in 𝑀 is the unique limit point of 𝑃𝑇∶𝑋→𝑀 even in the event that there is 𝑥(≠𝑥=𝑇𝑥)∈𝑋 such that 𝑃𝑇𝑥=𝑃𝑇𝑥=𝑃𝑥=𝑧. Property (ii) has been proven.
On the other hand, if the sequence of operators is uniformly bounded then sup_𝑘∈𝐍0‖𝑃_𝑘‖≤𝜇<∞ and, if furthermore, the sequence of compositie mappings {𝑇(𝑗_𝑘+1,𝑗_𝑘)} is contractive with some constant 𝐾∈[0,1) given by 𝐾=sup_𝑘∈𝐍0(∏^{𝑗𝑘+1−1}_{𝑗=𝑗𝑘}[𝐾_𝑗]),𝑘∈𝐍₀, where {𝑗_𝑘} is a strictly increasing sequence of natural numbers such that the sequence {𝑗_𝑘+1−𝑗_𝑘} is uniformly bounded, one has directly from (1)-(2): 𝑑𝑧_𝑗𝑘+1,𝑧_𝑗𝑘≤𝜇𝑑𝑥_𝑗𝑘,𝑇𝑗_𝑘+1,𝑗_𝑘𝑥_𝑗𝑘≤𝜇𝐾^𝑗𝑘𝑑𝑥₁,𝑥₀+2𝑥₀<∞;𝑘∈𝐍₀,𝑑𝑧_{𝑗𝑘+𝑖𝑘},𝑧_𝑗𝑘≤𝜇𝑑𝑥_𝑗𝑘,𝑇𝑗_𝑘+1,𝑗_𝑘𝑥_𝑗𝑘≤𝜇^{𝑗𝑘+𝑖𝑘−1}_{𝑗=𝑗𝑘}𝐾_𝑗𝐾^𝑗𝑘𝑑𝑥₁,𝑥₀+2𝑥₀<∞ (10) for 𝑘∈𝐍₀, where {{𝑖_𝑘}} is a sequence of finite sets of natural numbers satisfying 𝑗∈{𝑖_𝑘}⇔1≤𝑗<(𝑗_𝑘+1−𝑗_𝑘)+1for𝑘∈𝐍₀. Thus, one gets from (10): ∃lim_{𝑗𝑘→∞}𝑑𝑧_𝑗𝑘+1,𝑧_𝑗𝑘=lim_{𝑗𝑘→∞}𝑑𝑧_{𝑗𝑘+1+𝑗},𝑧_𝑗𝑘=0 (11) for any 𝑗∈{𝑖_𝑘} since 𝑗_𝑘→∞as𝑘→∞ and. Since {𝑗_𝑘+1−𝑗_𝑘} is uniformly bounded with existing limit 𝐽∈𝐍as𝑘→∞,𝑇(𝑗_𝑘+1+1,𝑗_𝑘)={𝑇_𝑗𝑘+1⋯𝑇_𝑗𝑘+1𝑇_𝑗𝑘} is contractive and the projection sequence {𝑃_𝑘} from 𝑋 to the sequence of subspaces {𝑀_𝑘} is uniformly bounded while having a finite set of subsequences {𝑃_𝑗𝑖} converging to a set 𝑃_𝑗𝑖→𝑃_𝑖 of projectors from 𝑋 to {𝑀_𝑘} for 𝑖∈𝐽 then ∃lim_𝑘→∞𝑑𝑧_𝑗𝑘+1+1,𝑧_𝑗𝑘+1=lim_𝑘→∞𝑑𝑃_𝑗𝑘+1+1𝑇𝑗_𝑘+1+1,0𝑥₀,𝑃_𝑗𝑘𝑇𝑗_𝑘+1,0𝑥₀=0 (12) so that {𝑧_𝑗𝑘+1+1=𝑃_𝑗𝑘+1𝑇(𝑗_𝑘+1+1,0)𝑥₀} is a Cauchy sequence satisfying: 𝑧_{𝑗𝑘+𝑖}⟶𝑧_𝑖=𝑃_𝑖lim_𝑘→∞𝑇_{𝑗𝑘+𝑖}⋯𝑇_{𝑗𝑘+𝑖−1}𝑇_𝑗𝑘𝑧_𝑖−1as𝑗_𝑖⟶∞,𝑖∈𝑀;𝑧_𝑖∈𝑋∩^∞_𝑘=0𝑀_{𝑗𝑘+𝑖−1} (13) for at most 𝐽 distinct points {𝑧₁,…,𝑧_𝐽} since, by hypothesis, there is a natural number 𝐽 satisfying ∞>𝐽=lim_𝑘∈𝐍0{𝑗_𝑘+1−𝑗_𝑘}≥1 and since {𝑃_𝑗𝑖}→𝑃_𝑖; 𝑖∈𝐽. Hence, Property (iii) holds.

Remark 2. The existence of some 𝑥(≠𝑥=𝑇𝑥)∈𝑋 in the proof of Theorem 1(ii) often happens. For instance if 𝑃𝑇∶𝑋→𝑀 is linear then 𝑥=𝑥+𝑥_𝑎 fulfils the relations 𝑃𝑇𝑥=𝑃𝑥=𝑃𝑇𝑥=𝑧 for any 𝑥_𝑎∈Ker𝑃𝑇.

The following auxiliary result to be then used holds.

Lemma 3. Assume that the sequences of linear self-mappings {𝑇_𝑛} and {𝑃_𝑛} converge to respective limits 𝑃 and 𝑇 being mappings from 𝑋 to 𝑀⊂𝑋 and from 𝑋 to 𝑋, respectively, in the sense that 𝑃_𝑛𝑇_𝑛𝑥→𝑃𝑇𝑥𝑎𝑠𝑛→∞ from any 𝑥∈𝑋. Let (𝑋,‖‖) be a Banach space with the norm of any 𝑄∶𝑋→𝑋 being defined by ‖𝑄‖=sup_{𝑥∈𝑋,‖𝑥‖=1}‖𝑄𝑥‖. For any given 𝛿∈𝐑₊,∃𝑛₀=𝑛₀(𝛿) such that ‖𝑃_𝑛𝑇_𝑛‖≤𝑝𝑡+𝛿, where 𝑝=‖𝑃‖ and 𝑡=‖𝑇‖. If 𝑃𝑇∶𝑋→𝑋 is contractive then the sequence {𝑃_𝑛𝑇_𝑛} of mappings from 𝑋 to 𝑋 is then asymptotically contractive.

Proof. Direct calculations yield 𝑃_𝑛𝑇_𝑛=sup_‖𝑥‖=1𝑃𝑇𝑥+𝑃𝑇_𝑛−𝑇𝑥+𝑃_𝑛−𝑃𝑇_𝑛𝑥≤‖𝑃𝑇‖+‖𝑃‖𝑇_𝑛−𝑇+𝑃_𝑛−𝑃𝑇_𝑛≤‖𝑃𝑇‖+‖𝑃‖𝑇_𝑛+𝑃−𝑇_𝑛−𝑃‖𝑇‖+𝑃_𝑛−𝑃𝑇_𝑛−𝑇 (14) and for any given 𝜀∈𝐑₊, there are 𝑛_𝑖=𝑛_𝑖(𝜀)∈𝐍₀for𝑖=1,2 such that ‖𝑇_𝑛−𝑇‖<𝜀 for all integer 𝑛>𝑛₁and‖𝑃_𝑛−𝑃‖<𝜀 for all integer 𝑛>𝑛₂. Thus, if 𝑛₃=max(𝑛₁,𝑛₂) then one has for any 𝑛>𝑛₃: 𝑃_𝑛𝑇_𝑛≤‖𝑃𝑇‖+(‖𝑃‖+‖𝑇‖+𝜀)𝜀=𝑝𝑡+𝑝+𝑡+𝜀𝜀≤𝑝𝑡+𝛿. (15)
The last inequality holds if and only if 𝜀²+(𝑝+𝑡)𝜀−𝛿≤0 for any given positive real constant 𝛿 and any positive real constant 𝜀=𝜀(𝛿) being sufficiently small to satisfy𝜀≤(1/2)((𝑝+𝑡)²+4𝛿−𝑝−𝑡)≤√𝛿/2. Since ‖𝑃_𝑛‖→𝑝 and ‖𝑇_𝑛‖→𝑡, there are finite natural numbers 𝑛₁=𝑛₁(𝛿,𝜀)=𝑛₁(𝜀),𝑛₂=𝑛₂(𝛿,𝜀)=𝑛₂(𝜀) such that 𝜀 satisfies that constraint ‖𝑃_𝑛𝑇_𝑛‖≤𝑝𝑡+𝛿 for any integer 𝑛>𝑛₃=𝑛₃(𝛿,𝜀)=𝑛₃(𝜀)>max(𝑛₁,𝑛₂). If 𝑃𝑇∶𝑋→𝑋 is contractive then 𝑝𝑡≤𝐾<1 for some real constant 𝐾. Thus, there is 𝑛₃=𝑛₃(𝜀) such that 𝑝𝑡≤𝐾≤‖𝑃_𝑛𝑇_𝑛‖<𝐾₁<1 for any given real 𝛿<𝐾₁−𝐾 and 𝜀≤(1/2)((𝑝+𝑡)²+4𝛿−𝑝−𝑡)  ≤  (1/2)((𝑝+𝑡)²+4(𝐾₁−𝐾)−𝑝−𝑡)  ≤  𝐾₁−𝐾/2 and then the sequence {𝑃_𝑛𝑇_𝑛} 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 {𝑇_𝑘𝑖} with 𝑇_𝑖𝑘∶Dom(𝑇_𝑖𝑘)⊂𝑋→Im(𝑇_𝑖𝑘)⊂𝑋to𝑋;𝑖=1,2 so that none of them is necessarily a projection on some subspace of 𝑋. Some preparatory results are first established. In the following, a Banach space(𝑋,‖‖), being equivalent to a complete metric space (𝑋,𝑑) with a homogeneous and translation-invariant metric induced-norm 𝑑∶𝑋×𝑋→𝐑₀₊ is considered such that ‖𝑥‖=𝑑(𝑥,0)=𝑑(𝑥+𝛼𝑦,𝛼𝑦) for any real 𝛼 and any 𝑥,𝑦∈𝑋. 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  (𝑋,‖‖), with 0∈𝑋, being equivalent to a complete metric space (𝑋,𝑑) with a homogeneous and translation-invariant metric induced-norm 𝑑∶𝑋×𝑋→𝐑₀₊. Consider also a composite sequence of two sequences of bounded linear operators {𝑇_𝑘=𝑇_2𝑘𝑇_1𝑘} defined by 𝑇_𝑘∶Dom(𝑇_𝑘)⊂𝑋→Im(𝑇_𝑘)⊂𝑋 defined by 𝑇_𝑘𝑥=𝑇_2𝑘(𝑇_1𝑘𝑥)=𝑇_2𝑘𝑇_1𝑘𝑥  for  any  𝑥∈Dom(𝑇_𝑘), where 𝑇_𝑖𝑘∶Dom(𝑇_𝑖𝑘)⊂𝑋→Im(𝑇_𝑖𝑘)⊂𝑋;𝑖=1,2, provided that Im(𝑇_1𝑘)(≠∅)⊂Dom(𝑇_2𝑘) and Im(𝑇_2𝑘)∩Dom(𝑇_1𝑘+1)≠∅;  for  all  𝑘∈𝐍₀. The following properties hold.(i)Assume that {𝑇_𝑖𝑘}→𝑇_𝑖(𝑖=1,2). Then, lim_𝑘→∞𝑑(𝑇_𝑘𝑥,𝑇₂𝑇₁𝑥)=0;  for  all  𝑥∈Dom(𝑇_𝑘);  for  all  𝑘∈𝐍₀.(ii)Assume that {𝑇_1𝑘}→𝑇₁(𝑖=1,2). Then, lim_𝑘→∞𝑑(𝑇_𝑘𝑥,𝑇_2𝑘𝑇₁𝑥)=0;  for  all  𝑥∈Dom(𝑇_𝑘);  for  all  𝑘∈𝐍₀.(iii)Assume that {𝑇_2𝑘}→𝑇₂(𝑖=1,2). Then, lim_𝑘→∞𝑑(𝑇_𝑘𝑥,𝑇₂𝑇_1𝑘𝑥)=0;  for  all  𝑥∈Dom(𝑇_𝑘);  for  all  𝑘∈𝐍₀.(iv)Define the operator composite sequence {𝑇(𝑘+𝑖+1,𝑘)} of operators as 𝑇(𝑘+𝑖+1,𝑘)=𝑇_𝑘+𝑖⋯𝑇_𝑘+1𝑇_𝑘 with 𝑇_𝑘=𝑇_𝑗𝑘𝑘𝑇_{𝑗𝑘−1,𝑘}⋯𝑇_2𝑘𝑇_1𝑘;  for  all  𝑖,𝑘∈𝐍₀ subject to Im(𝑇_𝑗𝑘)(≠∅)⊂Dom(𝑇_𝑗+1,𝑘)  for  𝑗∈𝑗_𝑘,Im𝑇_𝑗𝑘𝑘(≠∅)⊂Dom𝑇_1,𝑘+1  and  1≤𝑗_𝑘≤𝐽(∈𝐍₀)<∞;  for  all  𝑘∈𝐍₀. Define also the operator composite sequence of operators {𝑇⁰(𝑘+𝑖+1,𝑘)} as 𝑇⁰(𝑘+𝑖+1,𝑘)=𝑇⁰_𝑘+𝑖⋯𝑇⁰_𝑘+1𝑇⁰_𝑘=𝑇⁰_{𝑗𝑘+𝑖𝑘+𝑖}⋯𝑇⁰_{𝑗𝑘+𝑖−1,𝑘+𝑖}𝑇⁰_1,𝑘+𝑖⋯𝑇⁰_𝑗𝑘𝑘⋯𝑇⁰_2𝑘𝑇⁰_1𝑘 (16) for 𝑖,𝑘∈𝐍₀, where 𝑇⁰_𝑗𝑘=𝑇_𝑗𝑘 if  𝑇_𝑗𝑘 has not a limit as 𝑘→∞ and 𝑇⁰_𝑗𝑘=𝑇_𝑗∞ if 𝑇_𝑗∞=lim_𝑘→∞𝑇_𝑗𝑘. Then lim_𝑘→∞𝑑𝑇(𝑘+𝑖+1,𝑖)𝑥,𝑇⁰()𝑥𝑇𝑘+𝑖+1,𝑖=0;∀𝑥∈Dom₁₀,∀𝑖∈𝐍₀. (17)Properties (i)–(iii) hold for any 𝑥∈Dom(𝑇₁₀).

Proof. Assume that {𝑇_𝑖𝑘}→𝑇_𝑖 with 𝑇_𝑖∶𝑋→𝑋(𝑖=1,2) since (𝑋,𝑑), equivalently (𝑋,‖‖), is complete so that 𝑇_𝑖𝑘𝑥→𝑇_𝑖𝑥(∈𝑋) strongly for 𝑖=1,2as𝑘→∞forany𝑥∈Dom(𝑇₁₀) since Im(𝑇_1𝑘)(≠∅)⊂Dom(𝑇_2𝑘)andIm(𝑇_2𝑘)∩Dom(𝑇_1𝑘+1)≠∅;forall𝑘∈𝐍₀. Then, lim_𝑘→∞𝑑(𝑇_𝑘𝑥,  𝑇₂𝑇_1𝑘𝑥)  =  0;  for  all  𝑥∈Dom(𝑇₁₀), and 𝑇_𝑘𝑥=𝑇_2𝑘𝑇_1𝑘𝑥=𝑇_2𝑘𝑇₁𝑥+𝑇_2𝑘𝑇_1𝑘−𝑇₁𝑥=𝑇₂𝑇₁𝑥+𝑇_2𝑘−𝑇₂𝑇₁𝑥+𝑇_2𝑘𝑇_1𝑘−𝑇₁𝑥 (18) that leads to 𝑇_𝑘−𝑇₂𝑇₁𝑥≤𝑇_2𝑘−𝑇₂𝑇₁𝑥+𝑇_2𝑘𝑇_1𝑘−𝑇₁𝑥⟹∃lim_𝑘→∞𝑇_𝑘−𝑇₂𝑇₁𝑥=lim_𝑘→∞𝑑𝑇_𝑘−𝑇₂𝑇₁𝑥,0=lim_𝑘→∞𝑑𝑇_𝑘𝑥,𝑇₂𝑇₁𝑥𝑇=0;∀𝑥∈Dom₁₀. (19) Hence, Property (i). Now, assume that only {𝑇_1𝑘} has a limit. Then, 𝑇_𝑘𝑥=𝑇_2𝑘𝑇_1𝑘𝑥=𝑇_2𝑘𝑇₁𝑥+𝑇_2𝑘𝑇_1𝑘−𝑇₁𝑥⟹𝑇_𝑘−𝑇_2𝑘𝑇₁𝑥≤𝑇_2𝑘𝑇_1𝑘−𝑇₁𝑥;∀𝑥∈Dom𝑇_𝑘,∃lim_𝑘→∞𝑇_𝑘−𝑇_2𝑘𝑇₁𝑥=lim_𝑘→∞𝑑𝑇_𝑘−𝑇_2𝑘𝑇₁𝑥,0=lim_𝑘→∞𝑑𝑇_𝑘𝑥,𝑇_2𝑘𝑇₁𝑥=0;∀𝑥∈Dom𝑇_𝑘. (20) Hence, Property (ii). Finally, assume that only {𝑇_2𝑘} has a limit. Then, 𝑇_𝑘𝑥=𝑇_2𝑘𝑇_1𝑘𝑥=𝑇₂𝑇_1𝑘𝑥+𝑇_2𝑘−𝑇₂𝑇_1𝑘𝑥⟹𝑇_𝑘−𝑇₂𝑇_1𝑘𝑥≤𝑇_2𝑘−𝑇₂𝑇_1𝑘𝑥;∀𝑥∈Dom𝑇_𝑘,∃lim_𝑘→∞𝑇_𝑘−𝑇₂𝑇_1𝑘𝑥=lim_𝑘→∞𝑑𝑇_𝑘−𝑇₂𝑇_1𝑘𝑥,0=lim_𝑘→∞𝑑𝑇_𝑘𝑥,𝑇₂𝑇_1𝑘𝑥𝑇=0;∀𝑥∈Dom_𝑘. (21) Hence, Property (iii).
Property (iv) is direct from Properties (i)–(iii) and the associative property of composition of operators since for any 𝑘∈𝐍₀₊,  𝑇_𝑘𝑥 exists in Im(𝑇_𝑘) if Im(𝑇_1𝑘)⊂Dom(𝑇_2𝑘)  and  Dom(𝑇_1𝑘+1)⊃Im(𝑇_2𝑘)  for  𝑥∈Dom(𝑇₁₀), and then, Dom(𝑇_𝑘)⊂Dom(𝑇_1𝑘)  and  Im(𝑇_𝑘)⊃Im(𝑇_2𝑘);  for  all  𝑘∈𝐍₀ since 𝑇(𝑘+𝑖+1,𝑘)=𝑇_𝑘+𝑖⋯𝑇_𝑘+1𝑇_𝑘=𝑇_{𝑗𝑘+𝑖𝑘+𝑖}⋯𝑇_{𝑗𝑘+𝑖−1,𝑘+𝑖}𝑇_1,𝑘+𝑖⋯𝑇_𝑗𝑘𝑘⋯𝑇_2𝑘𝑇_1𝑘;∀𝑘∈𝐍₀. (22) Then, for any finite 𝑖∈𝐍, one gets 𝑇(𝑘+𝑖+1,𝑘)−𝑇⁰(≤𝑘+𝑖+1,𝑘)_{𝑘+𝑖𝑗=𝑘}𝑐_𝑗𝑜𝑇⁰_𝑗−𝑇_𝑗≤^{𝑘+𝑖𝑗}_𝑗=𝑘^𝑘_ℓ𝑗=1𝑐_𝑗ℓ𝑗𝑜𝑇⁰_ℓ𝑗𝑗−𝑇_ℓ𝑗𝑗⟶0as𝑘⟶∞ (23) for some positive finite constants 𝑐_𝑗 and 𝑐_𝑗ℓ𝑗 since any linear operator 𝑇⁰_ℓ𝑗𝑗 with a limit 𝑇_ℓ∞𝑗 admits a unique decomposition 𝑇⁰_ℓ𝑗𝑗=𝑇_ℓ∞𝑗+𝑇_𝑙𝑗𝑗,with𝑇_𝑙𝑗𝑗→0as𝑘→∞,forallℓ_𝑗∈𝑗_𝑘,𝑗=1,2,…,𝑘+𝑖.

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 {𝑇_𝑛} defined by 𝑇_𝑛∶Dom(𝑇_𝑛)⊂𝑋→Im(𝑇_𝑛)⊂𝑋 in a Banach space (𝑋,‖‖), such that Im(𝑇_𝑛)⊂Dom(𝑇_𝑛+1) with Im(𝑇_𝑛)∩Dom(𝑇_𝑛+1)≠∅, which converge to a limit operator 𝑇∶Dom(𝑇)⊂𝑋→Im(𝑇)⊂𝑋. 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){𝑇_𝑛𝑥_𝑚^(𝑛)}→𝑇𝑥_𝑚^(𝑛)as𝑛,𝑚→∞ for any bounded sequence {𝑥_𝑚^(𝑛)}⊂Dom(𝑇_𝑛)  since  ∃lim_{𝑛,𝑚→∞}‖(𝑇_𝑛−𝑇)𝑥_𝑚^(𝑛)‖  ≤  lim_{𝑛,𝑚→∞}‖𝑇_𝑛−𝑇‖‖𝑥_𝑚^(𝑛)‖  =  0  since  {𝑇_𝑛}→𝑇. Furthermore, ‖𝑇𝑥‖≤𝑇_𝑛𝑥+𝑇_𝑛−𝑇𝑥≤𝐾_𝑛+𝐾_𝑛𝑥‖≤𝐾_𝑛‖‖𝑥‖+𝐾_𝑛‖𝑥‖;∀𝑥∈𝑋 (24) for any 𝑛∈𝐍₀,𝐾_𝑛≥‖𝑇_𝑛‖ is a finite real sequence of constants {𝐾_𝑛} with 𝐾_𝑛≤𝐾 for any 𝑛∈𝐍₀, since {‖𝑇_𝑛‖} is convergent, and 𝐾_𝑛≥‖𝑇_𝑛−𝑇‖;forall𝑛∈𝐍₀. Again, since {‖𝑇_𝑛‖} is convergent, there is 𝑛₀∈𝐍 such that 𝐾≥‖𝐾_𝑛‖  for  any  𝑛(≥𝑛₀)∈𝐍₀,‖𝑇𝑥‖≤‖𝐾_𝑛‖‖𝑥‖+‖𝐾_𝑛‖‖𝑥‖≤(𝐾+𝐾)‖𝑥‖ so that 𝑇∶𝑋→𝑋 is bounded.(2)If the bounded sequence {𝑥_𝑚^(𝑛)}⊂Dom(𝑇_𝑛) converges to 𝑥^(𝑛)⊂Dom(𝑇_𝑛) for any 𝑛∈𝐍₀ then {𝑇_𝑛𝑥_𝑚^(𝑛)}→𝑇_𝑛𝑥^(𝑛)forany𝑛∈𝐍₀as𝑚→∞ since {𝑇_𝑛} is a closed operator for any 𝑛∈𝐍₀ so that ∃lim_𝑚→∞𝑇_𝑛𝑥_𝑚^(𝑛)−𝑇_𝑛𝑥^(𝑛)≤lim_𝑚→∞𝑇_𝑛𝑥_𝑚^(𝑛)−𝑥^(𝑛)≤𝑇_𝑛lim_𝑚→∞𝑥_𝑚^(𝑛)−𝑥^(𝑛)=0. (25)(3)One gets combining the above points (1)-(2) that: {𝑇_𝑛𝑥_𝑚^(𝑛)}→𝑇_𝑛𝑥^(𝑛)→𝑇𝑥_𝑚^(𝑛)→𝑇𝑥as𝑛,𝑚  →  ∞if{𝑥_𝑚^(𝑛)}→𝑥^(𝑛)→𝑥 for any 𝑛∈𝐍₀ where {𝑥_𝑚^(𝑛)}→{𝑥^(𝑛)}⊂Dom(𝑇_𝑛)  as  𝑚→∞, and then {𝑥_𝑚^(𝑛)}→{𝑥^(𝑛)}∈Dom(𝑇_𝑛)  as  𝑚→∞  for  any  𝑛∈𝐍₀  and  {𝑥_𝑚^(𝑛)}→{𝑥^(𝑛)}→𝑥∈Dom(𝑇)  as  𝑛,𝑚→∞ since {𝑇_𝑛} 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 {𝑥_𝑛} converging to 𝑥∈Dom(𝑇): 𝑇_𝑛𝑥_𝑛−𝑇𝑥=𝑇𝑥_𝑛+𝑇_𝑛−𝑇𝑥_𝑛≤‖𝑇𝑥−𝑇𝑥‖_𝑛−𝑥+𝑇_𝑛−𝑇𝑥_𝑛⟶0 (26)as 𝑘→∞ then 𝑇_𝑛𝑥_𝑛→𝑇𝑥 strongly so that 𝑇∶Dom(𝑇)⊂𝑋→Im(𝑇)⊂𝑋 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 𝑋, [1, 4]. Thus, we have the following result using Lemma 5.

Lemma 6. Consider a sequence of linear bounded operators {𝑇_𝑛} defined by 𝑇_𝑛∶𝑋→𝑋 in a Banach space (𝑋,‖‖) which converge to a limit operator 𝑇∶𝑋→𝑋. Then, such a limit is a bounded linear closed operator.

Proof. Since the operators are all bounded then their domain is 𝑋, their range is in 𝑋 and are all closed. The conditions of Lemma 5 hold with 𝑋≡Im𝑇_𝑛⊂𝑋≡Dom𝑇_𝑛+1withIm𝑇_𝑛∩Dom𝑇_𝑛+1=𝑋≠∅. (27) Then, the limit 𝑇∶𝑋→𝑋 of {𝑇_𝑛} is also bounded and closed from Lemma 5.

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 {𝑇_𝑛} defined by 𝑇_𝑛∶Dom(𝑇_𝑘)⊂𝑋→Im(𝑇_𝑛)⊂𝑋 in a Banach space (𝑋,‖‖), such that Im(𝑇_𝑛)⊂Dom(𝑇_𝑛+1) with Im(𝑇_𝑛)∩Dom(𝑇_𝑛+1)≠∅, which converge to a limit operator 𝑇∶Dom(𝑇)⊂𝑋→Im(𝑇)⊂𝑋. Then, such a limit is a compact operator. (ii)Assume that the sequence {𝑇_𝑛} of bounded operators satisfies that there is at least one compact operator within all subsequences {𝑇_𝑗𝑛,𝑇_𝑗𝑛+1,…,𝑇_{𝑗𝑛+1−1}} being subject to max_𝑛∈𝐍0(𝑗_𝑛+1−𝑗_𝑛)≤𝑐_𝑗<∞  for some subsequence {𝑗_𝑛}⊂𝐍₀  for any  𝑛∈𝐍₀. Then, the composed operator 𝑇(𝑛,𝑚) is compact as it is its limit provided that it exists.

Proof. We have to prove that if {𝑥_𝑛} is bounded then {𝑇𝑥_𝑛} is convergent. Note that for given bounded sequences {𝑥_𝑛^(𝑖)} and {𝑥_𝑛^(𝑗)};  𝑖,𝑗,𝑛∈𝐍₀ that 𝑇𝑥_𝑛^(𝑖)−𝑇𝑥_𝑛^(𝑗)=𝑇−𝑇_𝑛𝑥_𝑛^(𝑖)+𝑇_𝑛𝑥_𝑛^(𝑖)−𝑇_𝑛𝑥_𝑛^(𝑗)−𝑇−𝑇_𝑛𝑥_𝑛^(𝑗)≤𝑇−𝑇_𝑛𝑥_𝑛^(𝑖)+𝑇−𝑇_𝑛𝑥_𝑛^(𝑗)+𝑇_𝑛𝑥_𝑛^(𝑖)−𝑥_𝑛^(𝑗) (28) and, one gets by taking subsequences {𝑧_𝑖}⊂{𝑥_𝑛^(𝑖)}, {𝑧_𝑗}⊂{𝑥_𝑛^(𝑗)}𝑇𝑧_𝑖−𝑇𝑧_𝑗≤𝑇−𝑇_𝑛𝑧_𝑖+𝑧_𝑗+𝑇_𝑛𝑧_𝑖−𝑇_𝑛𝑧_𝑗. (29) Since {𝑇_𝑛}→𝑇, we can find 𝑛₀,𝑖₀∈𝐍₀ such that for 𝑛(≥𝑛₀)∈𝐍₀,min(𝑖,𝑗)>𝑖₀, we have 𝑇−𝑇_𝑛<𝜀,𝑧4𝑐max_𝑖,𝑧_𝑗≤𝐾_𝑧𝑇≤2𝑐<∞,_𝑛𝑧_𝑖−𝑇_𝑛𝑧_𝑗<𝜀2 (30) for any given 𝑐 and 𝜀=𝜀(𝑐)∈𝐑₊, since {𝑧_𝑖} and {𝑧_𝑗} are bounded subsequences, and {𝑇_𝑛𝑧_𝑖} converges, so that it is a Cauchy sequence, since {𝑇_𝑛} contains at least one compact operator. As a result, ‖𝑇𝑧_𝑖−𝑇𝑧_𝑗‖≤𝜀/2+𝜀/2=𝜀 is arbitrarily small for 𝜀 being sufficiently small. Thus, {𝑇𝑧_𝑖} 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 𝑇(𝑘+𝑖+1,𝑘)∶𝑋→𝑋;forall𝑖,𝑘(≥𝑖)∈𝐍₀ by 𝑇(𝑘+𝑖+1,𝑘)=𝑇_𝑘+𝑖⋯𝑇_𝑘+1𝑇_𝑘=𝑇_{𝑗𝑘+𝑖𝑘+𝑖}⋯𝑇_{𝑗𝑘+𝑖−1,𝑘+𝑖}𝑇_1,𝑘+𝑖⋯𝑇_𝑗𝑘𝑘⋯𝑇_2𝑘𝑇_1𝑘;∀𝑥∈Dom𝑇₁₀;∀𝑘∈𝐍₀. (31)

Define also the sequence {𝑇⁰(𝑘+𝑖+1,𝑘)} of composite operators as 𝑇⁰(𝑘+𝑖+1,𝑘)=𝑇⁰_𝑘+𝑖⋯𝑇⁰_𝑘+1𝑇⁰_𝑘;forall𝑘∈𝐍₀ where 𝑇⁰_𝑘+𝑖 replaces each operator in the composite operator 𝑇_𝑘+𝑖 by its limit when such a limit exists. A result is now given based on the existence of the following limit: lim_𝑘→∞𝑑𝑇(𝑘+𝑖+1,𝑖)𝑥,𝑇⁰()𝑥𝑇𝑘+𝑖+1,𝑖=0;∀𝑥∈Dom₁₀. (32)

The following result is obtained from Lemmas 4–7.

Theorem 8. Consider the operator composite sequence {𝑇(𝑘+𝑖+1,𝑘)};  for  all  𝑘,𝑖∈𝐍₀ of composed linear bounded operators in (31) on a Banach space (𝑋,‖‖), subject to Im(𝑇_𝑗𝑘)(≠∅)⊂Dom(𝑇_𝑗+1,𝑘)  for  𝑗∈𝑗_𝑘,Im𝑇_𝑗𝑘𝑘(≠∅)⊂Dom𝑇_1,𝑘+1  and  1≤𝑗_𝑘≤𝐽(∈𝐍₀)<∞;  for  all  𝑘∈𝐍₀, and the sequence of composed linear operators {𝑇⁰(𝑘+𝑖+1,𝑘)} of defined in the same way as {𝑇(𝑘+𝑖+1,𝑘)}  as  𝑘→∞ by replacing each operator possessing a limit by such a limit. The following properties hold.(i)Either the sequences {𝑇(𝑘+𝑖+1,𝑘)} and {𝑇⁰(𝑘+𝑖+1,𝑘)} have limits and both limits coincide or none of them has a limit and, furthermore, and 𝑇(𝑘+𝑖+1,𝑘)  →  𝑇⁰(𝑘+𝑖+1,𝑘)  as  𝑘→∞.(ii)If the limits of Property (i) exist and are finite then the limits of the sequences of operators {𝑇(𝑘+𝑖+1,𝑘)}  and  {𝑇⁰(𝑘+𝑖+1,𝑘)}  as  𝑘→∞;  for  all  𝑖∈𝐍₀ have the same set of fixed points.(iii)Assume, in addition, that for some 𝑘∈𝐍₀, there is at least one compact operator in the composition operator 𝑇(𝑘+𝑖+1,𝑘), and that Im(𝑇_𝑗𝑘)(≠∅)⊂Im(𝑇_𝑗𝑘)⊂Dom(𝑇_𝑗+1,𝑘)  for  𝑗∈𝑗_𝑘,Im𝑇_{𝑗𝑘+ℓ𝑘}(≠∅)⊂Im𝑇_{𝑗𝑘+ℓ𝑘}(≠∅)⊂Dom𝑇_1,𝑘+1  for  0≤ℓ≤𝑖 and some 𝑖∈𝐍₀ and that all the operators are closed. If Property (i) holds with 𝑇(𝑘+𝑖+1,𝑘)→𝑇⁰(𝑘+𝑖+1,𝑘)→𝑇^∗  as  𝑘→∞; for some 𝑖∈𝐍₀  and ‖𝑇^∗‖≤𝐾<1, then 𝑇^∗ is contractive and lim_{𝑘,𝑛→∞}𝑇(𝑘+𝑛(𝑖+1),𝑘)𝑥−𝑇((𝑘+𝑛(𝑖+1),𝑘),𝑘)𝑦=lim_𝑘→∞𝑇⁰(𝑘+𝑛(𝑖+1),𝑘)𝑥−𝑇⁰(𝑘+𝑛(𝑖+1),𝑘)𝑦=lim_{𝑘,𝑛→∞}𝑇^{∗(𝑖+1)𝑛}𝑥−𝑇^{∗(𝑖+1)𝑛}𝑦=0;∀𝑥,𝑦∈Dom𝑇₁₀, (33) the sequences of composite operators {𝑇(𝑘+𝑗,𝑗)} and {𝑇⁰(𝑘+𝑗,𝑗)} converge to zero as 𝑘→∞,  and  𝑇^∗∶Dom(𝑇^∗)⊂𝑋→Im(𝑇^∗)⊂𝑋 is bounded, closed and compact and has a unique fixed point in Dom(𝑇^∗)∩Im(𝑇^∗) to which all sequences with initial conditions in Im(𝑇^∗) converge.(iv)Assume that there is a (in general, nonunique) strictly increasing sequence of nonnegative integers {𝑗_𝑘} with 𝑗₀=0 and 0<𝑗_𝑘+1−𝑗_𝑘≤𝑚<∞ such that 𝑇𝑗_𝑘+2,𝑗_𝑘≤𝐾𝑗_𝑘+2,𝑗_𝑘+1𝑇𝑗_𝑘+1,𝑗_𝑘≤𝐾𝑇𝑗_𝑘+1,𝑗_𝑘;∀𝑘∈𝐍₀ (34) for some nonnegative real sequence {𝐾_𝑘(𝑗_𝑘+1,𝑗_𝑘)};𝑘∈𝐍₀, and some real constant 𝐾∈[0,1). Assume, in addition, that for some 𝑘∈𝐍₀, there is at least one compact operator in any composite operator 𝑇(𝑗_𝑘+1,𝑗_𝑘) and that all the operators are closed. Then, the sequences of composite operators {𝑇(𝑘+𝑗,𝑗)} and {𝑇⁰(𝑘+𝑗,𝑗)} converge to zero as 𝑘→∞ for any finite 𝑗∈𝐍₀. Finally, assume that 𝑇(𝑗_𝑘+1,𝑗_𝑘)→𝑇^∗_𝑔  as  𝑘→∞. Then, 𝑇^∗_𝑔 is contractive, continuous, bounded, closed, and compact and has a unique fixed point in Dom(𝑇^∗)∩Im(𝑇^∗) to which all sequences with initial conditions in Im(𝑇^∗) converge.

Proof. Note from the definition of the sequences {𝑇(𝑘+𝑖+1,𝑘)}and{𝑇⁰(𝑘+𝑖+1,𝑘)} that for any, since 𝑇⁰(≤𝑇(+𝑇𝑘+𝑖+1,𝑘)𝑘+𝑖+1,𝑘)₀()(,𝑇(≤𝑇𝑘+𝑖+1,𝑘−𝑇𝑘+𝑖+1,𝑘)𝑘+𝑖+1,𝑘)₀(+𝑇𝑘+𝑖+1,𝑘)₀(𝑘+𝑖+1,𝑘)−𝑇(𝑘+𝑖+1,𝑘) (35) so that limsup_𝑘→∞|‖𝑇(𝑘+𝑖+1,𝑘)‖−‖𝑇⁰(𝑘+𝑖+1,𝑘)‖|≤0. Then, either both sequences of operators {𝑇(𝑘+𝑖+1,𝑘)} and {𝑇⁰(𝑘+𝑖+1,𝑘)} have the same (finite or infinity) limit or none of them has a limit and 𝑇(𝑘+𝑖+1,𝑘)→𝑇⁰(𝑘+𝑖+1,𝑘) as 𝑘→∞. Hence, Property (i). Property (ii) follows trivially from Property (i) for the case lim_𝑘→∞𝑇⁰(𝑘+𝑖+1,𝑘)=lim_𝑘→∞𝑇(𝑘+𝑖+1,𝑘)=𝑇^∗;forall𝑖∈𝐍₀.
To prove Property (iii), note that 𝑇()𝑇()𝑦,𝑇()𝑇()𝑥−𝑇()𝑇()𝑦𝑘+2𝑖+1,𝑘𝑥−𝑘+2𝑖+1,𝑘≤𝐾𝑥−𝑦=𝐾𝑑𝑥,𝑦𝑘+2𝑖+1,𝑘+2𝑖+1𝑘+2𝑖+1,𝑘𝑘+2𝑖+1,𝑘+2𝑖+1𝑘+2𝑖+1,𝑘≤𝐾²,⋮𝑥−𝑦lim_𝑛→∞𝑇()𝑇()𝑥−𝑇()𝑇()𝑦𝑘+𝑛𝑖+1,𝑘+𝑛𝑖+1𝑘+𝑛𝑖+1,𝑘𝑘+𝑛𝑖+1,𝑘+𝑛𝑖+1𝑘+𝑛𝑖+1,𝑘≤𝐾^𝑛,𝑑𝑇𝑥−𝑦_∗𝑛𝑥−𝑇^∗𝑛𝑦=𝑑lim_𝑘→∞𝑇(𝑘+𝑛𝑖+1,𝑘)𝑥−lim_𝑘→∞𝑇(𝑘+𝑛𝑖+1,𝑘)𝑦=lim_𝑘→∞𝑑𝑇(𝑘+𝑛𝑖+1,𝑘)𝑥−𝑇(𝑘+𝑛𝑖+1,𝑘)𝑦=lim_𝑘→∞𝑑𝑇(𝑘+𝑛𝑖+1,𝑘)𝑥,𝑇(𝑘+𝑛𝑖+1,𝑘)𝑦=lim_𝑘→∞𝑑𝑇⁰(𝑘+𝑛𝑖+1,𝑘)𝑥,𝑇⁰(𝑘+𝑛𝑖+1,𝑘)𝑦=lim_𝑘→∞𝑑𝑇(𝑘+𝑛𝑖+1,𝑘)𝑥,𝑇⁰(𝑘+𝑛𝑖+1,𝑘)𝑦=lim_𝑘→∞𝑇(𝑘+𝑛𝑖+1,𝑘)𝑥−𝑇(𝑘+𝑛𝑖+1,𝑘)𝑦≤𝐾^𝑛𝑑𝑥,𝑦 (36)
so that, since 𝐾∈[0,1) then the limit operator 𝑇^∗ on 𝑋 is continuous, then bounded with lim_𝑛→∞𝑑𝑇^∗𝑛𝑥−𝑇^∗𝑛𝑦=𝑑lim_𝑛→∞𝑇^∗𝑛𝑥−lim_𝑛→∞𝑇^∗𝑛𝑦=0 (37) for any 𝑥,𝑦∈Dom(𝑇₁₀). Furthermore, for any 𝑘∈𝐍₀ and the associative property of composition of operators, ∃𝑛=𝑛(𝑗,𝑘)=max(𝑖∈𝐙₀₊∶𝑇(𝑘+𝑗,𝑗)=𝑇(𝑘+𝑗,𝑘₁)𝑇^∗𝑛(𝑘₂,𝑗);  for  all  𝑗∈𝐍₀,  for  all  𝑘(>𝑗)∈𝐍) with 𝑛=𝑛(𝑗,𝑘) being unique for each given 𝑗∈𝐍₀,𝑘(>𝑗)∈𝐍 and the given operator decomposition being also trivially unique. Note that ‖𝑇(𝑘+𝑗,𝑗)‖→0 as 𝑘→∞ for any 𝑗∈𝐍₀ since 𝑛=𝑛(𝑗,𝑘)→∞ if (𝑘−𝑗)→∞ since ‖𝑇^∗𝑛‖≤𝐾^𝑛→0as𝑛→∞ since 𝐾∈[0,1). Similarly, it is proven that ‖𝑇⁰(𝑘+𝑗,𝑗)‖→0as𝑘→∞.
On the other hand, note the following.(1)Any convergent sequence {𝑥_𝑘}for𝑘∈𝐍₀ constructed from the composed operators 𝑇(𝑘+𝑖+1,𝑘)=𝑇_𝑘+𝑖⋯𝑇_𝑘+1𝑇_𝑘=𝑇_{𝑗𝑘+𝑖𝑘+𝑖}⋯𝑇_{𝑗𝑘+𝑖−1,𝑘+𝑖}𝑇_1,𝑘+𝑖⋯𝑇_𝑗𝑘𝑘⋯𝑇_2𝑘𝑇_1𝑘 (38) for any 𝑘∈𝐍₀ as follows 𝑥_1𝑘∈Dom(𝑇_1𝑘),𝑥_2𝑘=𝑇_2𝑘𝑥_1𝑘,…,𝑥_𝑗𝑘𝑘=𝑇_𝑗𝑘𝑘𝑥_{𝑗𝑘−1,𝑘},𝑥_1,𝑘+1=𝑇_1,𝑘+1𝑥_𝑗𝑘𝑘,…  converges to a point 𝑥 in Dom(𝑇^∗), since all the operators in the above composite sequence of operators are closed and then the limit operator 𝑇^∗∶Dom(𝑇^∗)⊂𝑋→Im(T^∗)⊂Im(𝑇^∗)⊂Dom(𝑇^∗)⊂𝑋 is also bounded and closed (from Lemma 6 and the associative property of operator compositions), and 𝑇^∗𝑥 in Im(𝑇^∗)⊂Im(𝑇^∗) with Im(𝑇^∗) being closed (i.e., Im(𝑇^∗) is relatively compact) since all composite sequences of operators 𝑇(𝑘+𝑗,𝑗) are compact for any given 𝑗∈𝐍₀,𝑘(>𝑗)∈𝐍 since at least one of the operators within any of such sequences is compact and all of them are bounded, [1, 3, 4].(2)Any convergent sequence {𝑇^∗𝑛𝑥} of elements in Dom(𝑇^∗) with 𝑥∈Dom(𝑇₁₀) converges to some point 𝑥^∗ in Dom(𝑇^∗), which maps to 𝑇^∗𝑥^∗ in Im(𝑇^∗)⊂Im(𝑇^∗)⊂Dom(𝑇^∗) which is also the limit of the same convergent sequence. Such a limit {𝑇^∗𝑛𝑥} has a limit in Dom(𝑇^∗)∩Im(𝑇^∗) which is also the unique fixed point of 𝑇^∗. Otherwise, if there were two distinct fixed points 𝑥^∗ and 𝑦^∗ then it would are 𝑥,𝑦∈Dom(𝑇₁₀) such that lim_𝑛→∞𝑑(𝑇^∗𝑛𝑥−𝑇^∗𝑛𝑦)=𝑑(𝑥^∗,𝑦^∗)>0, then a contradiction and hence Property (iii).
To prove Property (iv), note that strictly increasing sequence of nonnegative integers {𝑗_𝑘} with 𝑗₀=0 and 0<𝑗_𝑘+1−𝑗_𝑘≤𝑚<∞ such that 𝑇𝑗_𝑘