# Applications of Measure of Noncompactness in Matrix Operators on Some Sequence Spaces

Authors: M. Mursaleen; A. Latif
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We determine the conditions for some matrix transformations fromn(ϕ), where the sequence space n(ϕ), which is related to the ℓp spaces, was introduced by Sargent (1960). We also obtain estimates for the norms of the bounded linear operators defined by these matrix transformations and find conditions to obtain the corresponding subclasses of compact matrix operators by using the Hausdorff measure of
noncompactness.


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## Body

## 1. Introduction and Preliminaries

We shall writew for the set of all complex sequences x=(xk)k=0∞. Let φ,ℓ∞,c, and c0 denote the sets of all finite, bounded, convergent, and null sequences, respectively. We write ℓp:={x∈w:∑k=0∞|xk|p<∞} for 1≤p<∞. By e and e(n)(n∈ℕ), we denote the sequences such that ek=1 for k=0,1,…, and en(n)=1 and ek(n)=0(k≠n). For any sequence x=(xk)k=0∞, let x[n]=∑k=0nxke(k) be its n-section. Moreover, we write bs and cs for the sets of sequences with bounded and convergent partial sums, respectively.A sequence(b(n))n=0∞in a linear metric space X is called Schauder  basis if for every x∈X, there is a unique sequence (λn)n=0∞ of scalars such that x=∑n=0∞λnb(n). A sequence space X with a linear topology is called a K-space if each of the maps pi:X→C defined by pi(x)=xi is continuous for all i∈ℕ. A K-space is called an FK-space if X is complete linear metric space; a BK-space is a normed FK-space. An FK-space X⊃ϕ is said to have AK if every sequence x=(xk)k=0∞∈X  has a unique representation x=∑k=0∞xke(k), that is, x[n]=∑k=0nxke(k)→x as n→∞.The spacesc0, c and ℓp(1≤p<∞) all have Schauder bases but the space ℓ∞ has no Schauder basis. Among the other classical sequence spaces, the spaces c0 and ℓp(1≤p<∞) have AK.Let(X,∥·∥)  be a normed space. Then the unit sphere and closed unit ball in X are denoted by SX:={x∈X:∥x∥=1}  and B̅X:={x∈X:∥x∥≤1}. If X⊃φ is a BK-space and a=(ak)∈w, then we define(1.1)‖a‖X*=sup⁡x∈SX⁡|∑k=0∞akxk|
provided the expression on the right-hand side exists and is finite.Theα-, β-, and γ-duals of a subset X of w are, respectively, defined by (1.2)Xα={a=(ak)∈w:ax=(akxk)∈l1∀x=(xk)∈X},Xβ={a=(ak)∈w:ax=(akxk)∈cs∀x=(xk)∈X},Xγ={a=(ak)∈w:ax=(akxk)∈bs∀x=(xk)∈X}.Throughout this paper, the matrices are infinite matrices of complex numbers. IfA is an infinite matrix with complex entries ank(n,k∈ℕ), then we write A=(ank) instead of A=(ank)n,k=0∞. Also, we write An for the sequence in the nth row of A, that is, An=(ank)n,k=0∞ for every n∈ℕ. In addition, if x=(xk)∈w, then we define the A-transform of x as the sequence Ax=(An(x))n=0∞, where (1.3)An(x)=∑k=0∞ankxk;(n∈N)
provided the series on the right converges for each n∈ℕ.LetX and Y be subsets of w and A=(ank) an infinite matrix. Then, we say that A defines a matrix mapping from X into Y, and we denote it by writing A:X→Y if Ax exists and is in Y for all x∈X. By (X,Y), we denote the class of all infinite matrices that map X into Y. Thus A∈(X,Y) if and only if An∈Xβ for all n∈ℕ and Ax∈Y for all x∈X.Lemma 1.1 (see[1]).
Let† denote any of the symbols α, β, or γ. Then, we have c0†=c†=ℓ∞†=ℓ1, ℓ1†=ℓ∞ and ℓp†=ℓq, where 1<p<∞ and q=p/(p-1).Lemma 1.2 (see[1, 2]).
LetX be any of the spaces c0, c, ℓ∞, or ℓp(1≤p<∞). Then, we have ∥·∥X*=∥·∥Xβ on Xβ, where ∥·∥Xβ  denotes the natural norm on the dual space Xβ.Lemma 1.3 (see[1, 2]).
LetX⊃φ and Y be BK-spaces. Then, we have(a)
(X,Y)⊂ℬ(X,Y), that is, every matrix A∈(X,Y) defines an operator LA∈ℬ(X,Y) by LA(x)=Ax for all x∈X;(b)
ifX has AK, then ℬ(X,Y)⊂(X,Y), that is, for every operator L∈ℬ(X,Y) there exists a matrix A∈(X,Y) such that L(x)=Ax for all x∈X.Furthermore, we have the following results on the operator norms.Lemma 1.4 (see[2]).
LetX⊃φ be a BK-space and Y any of the spaces c0, c, or ℓ∞. If A∈(X,Y), then
(1.4)‖LA‖=‖A‖(X,l∞)=sup⁡n⁡‖An‖X*<∞,
where ∥A∥(X,ℓ∞) denotes the operator norm for the matrix A∈(X,ℓ∞).Sargent [3] defined the following sequence spaces.LetC denote the space whose elements are finite sets of distinct positive integers. Given any element σ of C, we denote by c(σ) the sequence {cn(σ)}  such that cn(σ)=1 for n∈σ and cn(σ)=0 otherwise. Further (1.5)Cs={σ∈C:∑n=1∞cn(σ)≤s},that is,Cs is the set of those σ whose support has cardinality at most s, and we get (1.6)Φ={ϕ=(ϕk)∈w:0<ϕ1≤ϕn≤ϕn+1and(n+1)ϕn≥nϕn+1}.Forϕ∈Φ, the following sequence spaces were introduced by Sargent [3] and further studied in [4] (1.7)m(ϕ)={x=(xk)∈w:‖x‖m(ϕ)=sup⁡s≥1⁡sup⁡σ∈Cs(1ϕs∑k∈σ|xk|)<∞},n(ϕ)={x=(xk)∈w:‖x‖n(ϕ)=sup⁡u∈S(x)(∑k=1∞|uk|Δϕk)<∞},
where Δϕk=ϕk-ϕk-1 and S(x) denotes the set of all sequences that are rearrangements of x.Remark 1.5 ([3]).
(i) The spacesm(ϕ) and n(ϕ) are BK spaces with their respective norms. (ii) If ϕn=1 for all n∈ℕ, then m(ϕ)=ℓ1,n(ϕ)=ℓ∞;if ϕn=n for all n∈ℕ, then m(ϕ)=ℓ∞,  n(ϕ)=ℓ1. (iii)ℓ1⊆m(ϕ)⊆ℓ∞[ℓ∞⊇n(ϕ)⊇ℓ1] for all ϕ of Φ.(iv)(m(ϕ))†=n(ϕ) and (n(ϕ))†=m(ϕ), where † is any of the symbols α, β, or γ.Recently, Makowsky and Mursaleen [5] have characterized the classes of compact operators on some BK-spaces, namely, 𝒞(m(ϕ),ℓp)(1≤p≤∞),  𝒞(n(ϕ),ℓp)(1≤p<∞),𝒞(ℓp,m(ϕ))(1<p≤∞), and 𝒞(ℓp,n(ϕ))(1≤p≤∞). In this paper, we determine the conditions for the classes of matrix transformations (n(ϕ),c0),(n(ϕ),c), and (n(ϕ),ℓ∞), and establish estimates for the norms of the bounded linear operators defined by these matrix transformations. Further, we obtain the necessary and sufficient (or only sufficient) conditions for the corresponding subclasses of compact matrix operators C(n(ϕ),c0),C(n(ϕ),c), and C(n(ϕ),ℓ∞) by using the Hausdorff measure of noncompactness.


## 2. The Hausdorff Measure of Noncompactness

Let(X,∥·∥) be a normed space. Then the unit sphere and closed unit ball in X are denoted by SX:={x∈X:∥x∥=1} and B̅X:={x∈X:∥x∥≤1}. If X and Y are Banach spaces then B(X,Y) is the set of all bounded linear operators L:X→Y;B(X,Y) is a Banach space with the operator norm given by ∥L∥=supx∈SX∥L(x)∥ for all L∈ℬ(X,Y). A linear operator L:X→Y is said to be compact if the domain of L is all of X, and for every bounded sequence (xn) in X, the sequence (L(xn))  has a subsequence which converges in Y. We denote the class of all compact operators in ℬ(X,Y) by 𝒞(X,Y). An operator L∈ℬ(X,Y) is said to be of finite rank if dimR(L)<∞, where R(L) is the range space of L. An operator of finite rank is clearly compact. In particular, if Y=ℂ then we write X* for the set of all continuous linear functionals on X with the norm ∥f∥=supx∈SX|f(x)|.The Hausdorff measure of noncompactness was defined byGoldenšteinet al. in 1957 [6].LetS and M be subsets of a metric space (X,d) and ε>0. Then, S is called an ε-net of M in X if for every x∈M there exists s∈S such that d(x,s)<ε. Further, if the set S is finite, then the ε-net S of M is called a finiteε-net of M, and we say that M has a finite ε-net in X. A subset of a metric space is said to be totally bounded if it has a finite ε-net for every ε>0.ByℳX, we denote the collection of all bounded subsets of a metric space (X,d). If Q∈ℳX, then the Hausdorff measure of noncompactness of the set Q, denoted by χ(Q), is defined by (2.1)χ(Q):=inf⁡{ϵ>0:Q⊂∪i=1nB(xi,ri),xi∈X,ri<ϵ(i=1,2,…),n∈N}.The functionχ:ℳX→[0,∞) is called the Hausdorff measure of noncompactness.The basic properties of the Hausdorff measure of noncompactness can be found in [2, 7–9] and for recent developments, see [10–18]. If Q, Q1, and Q2 are bounded subsets of a metric space (X,d), then (2.2)χ(Q)=0iffQistotallybounded,Q1⊂Q2impliesχ(Q1)≤χ(Q2).Further, ifX is a normed space, then the function χ has some additional properties connected with the linear structure, for example,(2.3)χ(Q1+Q2)≤χ(Q1)+χ(Q2),χ(αQ)=|α|χ(Q)∀α∈C.LetX and Y be Banach spaces and χ1 and χ2 the Hausdorff measures of noncompactness on X and Y, respectively. An operator L:X→Y is said to be (χ1,χ2)-bounded if L(Q)∈ℳY for all Q∈ℳX and there exist a constant C≥0 such that χ2(L(Q))≤Cχ1(Q) for all Q∈ℳX. If an operator L is (χ1,χ2)-bounded, then the number ∥L∥(χ1,χ2):=inf{C≥0:χ2(L(Q))≤Cχ1(Q)forallQ∈ℳX} is called the (χ1,χ2)-measure of noncompactnessofL. If χ1=χ2=χ, then we write ∥L∥(χ1,χ2)=∥L∥χ.The most effective way in the characterization of compact operators between the Banach spaces is by applying the Hausdorff measure of noncompactness. This can be achieved as follows: LetX and Y be Banach spaces and L∈B(X,Y). Then, the Hausdorff measure of noncompactness of L, denoted by ∥L∥χ, can be determined by(2.4)‖L‖χ=χ(L(SX)),
and we have that L is compact if and only if (2.5)‖L‖χ=0.Now, the following result gives an estimate for the Hausdorff measure of noncompactness in Banach spaces with Schauder bases. It is known that if(bk)k=0∞ is a Schauder basis for a Banach space X, then every element x∈X has a unique representation x=∑k=0∞αk(x)bk, where αk(k∈ℕ) are called the basis functionals. Moreover, for each n∈ℕ, the operator Pn:X→X defined by Pn(x)=∑k=0nαk(x)bk(x∈X) is called the projector onto the linear span of {b0,b1,…,bn}. Besides, all operators Pn and I-Pn are equibounded, where I denotes the identity operator on X.Theorem 2.1 (see[7]).
LetX be a Banach space with a Schauder basis (bk)k=0∞, Q∈ℳX, and Pn:X→X(n∈ℕ) the projector onto the linear span of {b0,b1,…,bn}. Then, we have
(2.6)1a⋅lim⁡n→∞⁡sup⁡(sup⁡x∈Q⁡‖(I-Pn)(x)‖)≤χ(Q)≤lim⁡n→∞sup⁡(sup⁡x∈Q‖(I-Pn)(x)‖),
where a=limsupn→∞∥I-Pn∥.In particular, the following result shows how to compute the Hausdorff measure of noncompactness in the spacesc0 and ℓp(1≤p<∞), which are BK-spaces with AK.Theorem 2.2 (see[7]).
LetQ be a bounded subset of the normed space X, where X is ℓp for 1≤p<∞ or c0. If Pn:X→X(n∈ℕ) is the operator defined by Pn(x)=x[n]=(x0,x1,…,xn,0,0,…) for all x=(xk)k=0∞∈X, then we have
(2.7)χ(Q)=lim⁡n→∞(sup⁡x∈Q‖(I-Pn)(x)‖).The Haudorff measure of noncompactness forn(ϕ) has recently been determined in [19] as follows.Theorem 2.3.
LetQ be a bounded subset of n(ϕ). Then
(2.8)χ(Q)=lim⁡k→∞sup⁡x∈Q(sup⁡u∈S(x)(∑n=k∞|un|Δϕn)).


## 3. Main Results

First we prove the following basic lemma.Lemma 3.1.
IfA∈(n(ϕ),c), then the following hold
(3.1)αk=lim⁡n→∞ankexistsforeveryk∈N,(3.2)α=(αk)∈m(ϕ),(3.3)sup⁡n‖An-α‖n(ϕ)*<∞,(3.4)lim⁡n→∞An(x)=∑k=0∞αkxk∀x=(xk)∈n(ϕ).
Proof.
We write∥x∥=∥x∥n(ϕ), for short. Since A∈(n(ϕ),c), we have
(*)∥LA∥=sup⁡n‖An‖n(ϕ)*<∞.
Further, since e(k)∈n(ϕ) and hence Ae(k)∈c for all k∈ℕ. Consequently, the limits αk in (3.1) exist for all k∈ℕ.  Now, let x∈n(ϕ) be given. Then there is a positive constant K such that ∥x[p]∥≤K∥x∥ for all p∈ℕ. Thus we have
(3.5)|∑k=0pankxk|=|An(x[p])|≤‖Ax[p]‖l∞=‖LA(x[p])‖l∞≤K‖LA‖‖x‖
for all p,n∈ℕ. Hence, we obtain from (3.1) that
(3.6)|∑k=0pαkxk|=lim⁡n→∞|∑k=0pankxk|≤K‖LA‖‖x‖;(p∈N).
This implies that αx=(αkxk)∈bs, and since x∈X was arbitrary, we deduce that α∈n(ϕ)γ. But n(ϕ)γ=n(ϕ)β and hence (3.2) holds. Moreover, since n(ϕ)⊃φ is a BK space, (3.2) implies ∥α∥n(ϕ)*<∞ by (Wilansky [19, Theorem  7.2.9]). Therefore, we get (3.3) from  (15) by using (1.3).  Now, define the matrix B=(bnk) by bnk=ank-αk for all n,k∈ℕ. Then, it is obvious that Bn∈n(ϕ)β=m(ϕ) for all n∈ℕ. Also, it follows by (3.3) that
(3.7)sup⁡n∥Bn∥n(ϕ)*=sup⁡n‖An-α‖n(ϕ)*<∞.
Furthermore, we have from (3.1) that
(3.8)lim⁡n→∞Bn(e(k))=lim⁡n→∞⁡bnk=0(k∈N),
that is, Be(k)∈c0 for all k∈ℕ. This leads us to the consequence that B∈(n(ϕ),c0) by (Malkowsky-Rakocevic [2, Theorem  1.23(c)]). Hence,  limn→∞Bn(x)=0 for all x∈n(ϕ), which yields (3.4).This completes the proof of the lemma.Theorem 3.2.
(a) IfA∈(n(ϕ),c0), then
(3.9)‖LA‖χ=lim⁡r→∞⁡(sup⁡n>r⁡‖An‖n(ϕ)*)(b) If A∈(n(ϕ),c), then
(3.10)12⋅lim⁡r→∞⁡(sup⁡n≥r⁡‖An-α‖n(ϕ)*)≤‖LA‖χ≤lim⁡r→∞(sup⁡n≥r‖An-α‖n(ϕ)*),
where α=(αk) with αk=limr→∞ankfor all k∈ℕ.
(c) IfA∈(n(ϕ),ℓ∞), then
(3.11)0≤‖LA‖χ≤lim⁡r→∞(sup⁡n≥r‖An‖n(ϕ)*).Proof.
We writeS=Sn(ϕ), for short. Then, we have by (2.4) and Lemma 1.3 (a) that(3.12)∥LA∥χ=χ(AS).
For (a), we have AS∈ℳc0. Thus, it follows by Theorem 2.2 that
(3.13)∥LA∥χ=χ(AS)=lim⁡r→∞(sup⁡x∈S∥(I-Pr)(Ax)∥l∞),
where Pr:c0→c0(r∈ℕ) is the operator defined by Pr(x)=(x0,x1,…,xr,0,0,…) for all x=(xk)∈c0. This yields that ∥(I-Pr)(Ax)∥ℓ∞=supn>r|An(x)| for all x∈n(ϕ) and every r∈ℕ. Thus, by combining (1.1) and (1.3), we have for every r∈ℕ that
(3.14)sup⁡x∈S⁡‖(I-Pr)(Ax)‖l∞=sup⁡n>r⁡‖An‖n(ϕ)*.
Hence, by (3.13) we get (3.9).
To prove(b), we have AS∈ℳc. Thus, we are going to apply Theorem 2.1 to get an estimate for the value of χ(AS) in (3.12). For this, we know that every z=(zn)∈c has a unique representation z=z̅e+∑n=0∞(zn-z̅)e(n), where z̅=limn→∞zn. Thus, we define the projectors Pr:c→c(r∈ℕ) by P0(z)=z̅e and Pr(z)=z̅e+∑n=0r-1(zn-z̅)e(n) for r≥1. Then, we have for every r∈ℕ that (I-Pr)(z)=∑n=r∞(zn-z̅)e(n) and hence
(3.15)‖(I-Pr)(z)‖l∞=sup⁡n>r|zn-z̅|
for allz∈c and every r∈ℕ. Obviously ∥(I-Pr)(z)∥ℓ∞≤2∥z∥ℓ∞, hence ∥I-Pr∥≤2 for all r∈ℕ. Further, for each r∈ℕ, we define the sequence z(r)=(zn(r))∈c by zr(r)=-1 and zn(r)=1 for n≠r. Then ∥z(r)∥ℓ∞=1 and limn→∞zn(r)=1. Therefore,  ∥I-Pr∥≥∥(I-Pr)(z(r))∥ℓ∞=2  by (3.15). Consequently, we have ∥I-Pr∥=2 for all r∈ℕ. Hence, from (3.12) we obtain by applying Theorem 2.1 that
(3.16)12⋅μ(A)≤‖LA‖χ≤μ(A),
where
(3.17)μ(A)=lim⁡r→∞sup⁡(sup⁡x∈S‖(I-Pr)(Ax)‖l∞).
Now, it is given that A∈(n(ϕ),c). Thus, it follows from Lemma 3.1 that the limits αk=limn→∞ank exist for all k, α=(αk)∈n(ϕ)β=m(ϕ) and
(3.18)lim⁡n→∞An(x)=∑k=0∞αkxk
for all x=(xk)∈n(ϕ). Therefore, we derive from (3.15) that
(3.19)‖(I-Pr)(Ax)‖l∞=sup⁡n≥r|An(x)-∑k=0∞αkxk|=sup⁡n≥r|∑k=0∞(ank-αk)xk|
for allx=(xk)∈n(ϕ) and every r∈ℕ. Consequently, we obtain by (1.3) that
(3.20)sup⁡x∈S‖(I-Pr)(Ax)‖l∞=sup⁡n≥r‖An-α‖n(ϕ)*(r∈N).
Hence, we get (3.10) from (3.16).
Finally, to prove(c) we define Pr:ℓ∞→ℓ∞(r∈ℕ) as in the proof of part (a) for all x=(xk)∈ℓ∞.   Then, it is clear that
(3.21)AS⊂Pr(AS)+(I-Pr)(AS)(r∈N).
Thus, it follows by the elementary properties of the functionχ that
(3.22)0≤χ(AS)≤χ(Pr(AS))+χ((I-Pr)(AS))=χ((I-Pr)(AS))≤sup⁡x∈S‖(I-Pr)(Ax)‖l∞=sup⁡n>r‖An‖n(ϕ)*
for all r∈ℕ. This and (3.12) together imply (3.11). This completes the proof of the theorem.As an immediate consequence of Theorem3.2 and (2.5), we have the following.Corollary 3.3.
(a) IfA∈(n(ϕ),c0), then
(3.23)LAiscompactifflim⁡r→∞(sup⁡n>r‖An‖n(ϕ)*)=0.
(b) IfA∈(n(ϕ),c), then
(3.24)LAiscompactifflim⁡r→∞(sup⁡n≥r‖An-α‖n(ϕ)*)=0,
where α=(αk)  with αk=limn→∞ankfor all k∈ℕ.
(c) IfA∈(n(ϕ),ℓ∞), then
(3.25)LAiscompactiflim⁡r→∞(sup⁡n>r‖An‖n(ϕ)*)=0.Remark 3.4.
It is worth mentioning that the condition in (3.25) is only a sufficient condition for the operator LA to be compact, where A∈(n(ϕ),ℓ∞). In the following example, we show that it is possible for LA to be compact while limr→∞(supn>r∥An∥n(ϕ)*)≠0.
Choose a fixedm∈ℕ such that xm′≠0 for some x′=(xk′)∈Sn(ϕ). Now, we define the matrix A=(ank) by anm=1 and ank=0 for all k≠m(n∈ℕ). Then, we have Ax=xme for all x=(xk)∈w, hence A∈(w,ℓ∞)⊂(n(ϕ),ℓ∞). Also, since LA is of finite rank, LA is compact. On the other hand, we have An=e(m) and hence ∥An∥n(ϕ)*=supx∈Sn(ϕ)|xm| for all n∈ℕ by (1.3). This implies that
(3.26)lim⁡r→∞(sup⁡n>r‖An‖n(ϕ)*)=sypx∈Sn(ϕ)⁡|xm|≥|xm′|>0.



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