# Some Matrix Transformations of Convex and Paranormed Sequence Spaces into the Spaces of Invariant Means

Authors: M. Mursaleen; S. A. Mohiuddine
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We determine the necessary and sufficient conditions to characterize the matrices which transform convex sequences and Maddox sequences intoVσ(θ) and Vσ∞(θ).


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## Body

## 1. Introduction and Preliminaries

Byw, we denote the space of all real-valued sequences x=(xk)k=1∞. Any vector subspace of w is called a sequence space. We write that ℓ∞, c, and c0 denote the sets of all bounded, convergent, and null sequences, respectively, and note that c⊂ℓ∞; also cs and ℓp are the set of all convergent and p-absolutely convergent series, respectively, where ℓp:={x∈w:∑k=0∞|xk|p<∞} for 1≤p<∞. In the theory of sequence spaces, a beautiful application of the well-known Hahn-Banach extension theorem gave rise to the concept of the Banach limit. That is, the lim functional defined on c can be extended to the whole ℓ∞, and this extended functional is known as the Banach limit [1]. In 1948, Lorentz [2] used this notion of a weak limit to define a new type of convergence, known as the almost convergence. Later on, Raimi [3] gave a slight generalization of almost convergence and named it the σ-convergence.A sequence spaceX 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 X is called an FK-space if X is a complete linear metric space. An FK-space whose topology is normable is called a BK-space. An FK-space X is said to have AK property if X⊃ϕ and (e(k)) is a basis for X, where (e(k)) is a sequence whose only nonzero term is a 1 in kth place for each k∈ℕ and ϕ=span{e(k)}, the set of all finitely nonzero sequences. If ϕ is dense in X, then X is called an AD-space, thus AK implies AD. For example, the spaces c0, cs, and ℓp are AK-spaces.LetX and Y be two sequence spaces, and let A=(ank)n;k=1∞ be an infinite matrix of real or complex numbers. We write Ax=(An(x)), An(x)=∑kankxk provided that the series on the right converges for each n. If x=(xk)∈X implies that Ax∈Y, then we say that A defines a matrix transformation from X into Y, and by (X,Y), we denote the class of such matrices.Letσ be a one-to-one mapping from the set of natural numbers into itself. A continuous linear functional φ on the space ℓ∞ is said to be an invariantmean or a σ-mean if and only if (i) φ(x)≥0 if x≥0 (i.e., xk≥0 for all k), (ii) φ(e)=1, where e=(1,1,1,…), and (iii) φ(x)=φ((xσ(k))) for all x∈ℓ∞.Throughout this paper, we consider the mappingσ which has no finite orbits, that is, σp(k)≠k for all integer k≥0 and p≥1, where σp(k) denotes the pth iterate of σ at k. Note that a σ-mean extends the limit functional on the space c in the sense that φ(x)=limx for all x∈c, (cf. [4]). Consequently, c⊂Vσ, the set of bounded sequences all of whose σ-means are equal. We say that a sequence x=(xk) is σ-convergent if and only if x∈Vσ. Using this concept, Schaefer [5] defined and characterized σ-conservative, σ-regular, and σ-coercive matrices. If σ is translation, then Vσ is reduced to the set f of almost convergent sequences [2].The notion ofσ-convergence for double sequences has been introduced in [6] and further studied in [7–9].Recently, the sequence spacesVσ(θ) and Vσ∞(θ) have been introduced in [10] which are related to the concept of σ-mean and the lacunary sequence θ=(kr).In this section, we establish the necessary and sufficient conditions on the matrixA=(ank)n,k=1∞ which transforms r-convex sequences in to the spaces Vσ(θ) and Vσ∞(θ).By a lacunary sequence, we mean an increasing integer sequenceθ=(kr) such that k0=0 and hr:=kr-kr-1→∞ as r→∞. Throughout this paper, the intervals determined by θ will be denoted by Ir:=(kr-1,kr], and the ratio kr/kr-1 will be abbreviated by qr.A bounded sequencex=(xk) of real numbers is said to be σ-lacunary convergent to a number L if and only if limr→∞(1/hr)∑j∈Irxσj(n)=L, uniformly in n, and let Vσ(θ) denote the set of all such sequences, that is,(1.1)Vσ(θ)={x∈l∞:lim⁡m→∞1hr∑j∈Irxσj(n)=Luniformlyinn}.
In this case, L is called the (σ,θ)-limit of x. We remark that(i)
ifσ(n)=n+1, then Vσ(θ) is reduced to the space f(θ) (cf. [11]),(ii)
c⊂Vσ(θ)⊂ℓ∞.A bounded sequencex=(xk) of real numbers is said to be σ-lacunary bounded if and only if supr,n|(1/hr)∑j∈Irxσj(n)|<∞, and let Vσ∞(θ) denote the set of all such sequences, that is,(1.2)Vσ∞(θ)={x∈l∞:sup⁡r,n|1hr∑j∈Irxσj(n)|<∞}.
We remark that c⊂Vσ(θ)⊂Vσ∞(θ)⊂ℓ∞ and the spaces Vσ∞(θ) and Vσ∞(θ) are BK spaces with the norm(1.3)‖x‖=sup⁡r,n|1hr∑j∈Irxσj(n)|.


## 2. Convex Sequence Spaces

Pati and Sinha [12] defined r-convex sequences as follows: a real sequence x=(xk)k=0∞ is said to be r-convex,r∈ℕ, if Δrxk≥0 for all k∈ℕ, where Δrxk is defined by(2.1)Δ0xk=xk,Δ1xk=xk-xk+1,Δrxk=Δ(Δr-1xk),r∈N.
The space of all bounded r-convex sequences with r≥2 is denoted by SCr, that is,(2.2)SCr:={x∈l∞:Δrxk≥0∀k∈N},SC1:={x∈l∞:xk-xk+1≥0}.
It is clear that SC1⊆c.It is well known that (Zygmund [13]) a bounded convex sequence (xk) is nonincreasing. It is easy to prove the identity Δ(r+s)xk=Δr(Δsxk),r,s≥0,which shows that SCr⊂SCr-1, when r≥2. Properties of bounded r-convex sequences have been investigated by Rath [14]. Note that SCr⊂v⊂c⊂ℓ∞. Recently, using the generalized difference operator Δr, Çolak and Et [15], and Et and Çolak [16] defined and studied the sequence spaces c0(Δr),c(Δr), and ℓ∞(Δr). In this section, we establish the necessary and sufficient conditions on the matrix A=(ank)n,k=1∞ which transforms r-convex sequences into the spaces Vσ(θ) and Vσ∞(θ).Write(2.3)t(n,k,m)=1hm∑j∈Imaσj(n),k,g(r)(n,k,m)=∑j=1k(r-1k-j)t(n,j,m)kr-1,λmn(x)=∑k=1∞g(n,k,m)xk,
where for our convenience, we use g(n,k,m) instead of g(r)(n,k,m) for r≥2 throughout the paper.Theorem 2.1.
A∈(SCr,Vσ(θ)) if and only if(i)
supi,p|∑k=p∞aik|<∞,(ii)
there exists a constantM such that for s,n=1,2,…,(2.4)sup⁡m∑k=s∞|g(n,k,m)|≤M,(iii)
limmg(n,k,m)=αk, uniformly in n, for each k∈ℕ,(iv)
limm∑kg(n,k,m)=α, uniformly in n.Proof.
In [17], a characterization of A∈(SCr,Fℬ) was given, where Fℬ, in the sense of [18], is the bounded domain of a sequence ℬ=(B(i)) of matrices B(i)=(brk(i)). Now, by the setting
(2.5)brk(i)={1hr0ifk=σj(i),j∈Ir,otherwise.(r,k,i∈N),
then Vσ(θ)=Fℬ, and the proof follows from Theorem 2.1 of [17].Theorem 2.2.
A∈(SCr,Vσ∞(θ)) if and only if the condition (i) of Theorem 2.1 holds and
(2.6)sup⁡n,m∑k|t(n,k,m)|<∞.Proof.
Sufficiency
Suppose that the conditions (i) and (2.6) hold and x=(xk)∈SCr⊂ℓ∞. Therefore, Ax is bounded, and we have
(2.7)|λmn(x)|≤∑k|g(n,k,m)xk|≤(∑k|g(n,k,m)|)(sup⁡k|xk|).
Taking the supremum over n,m on both sides and using (2.6), we get Ax∈Vσ∞(θ) for x∈SCr.
Necessity
LetA∈(SCr,Vσ∞(θ)). Condition (i) follows as in the proof of Theorem 2.1. Write qn(x)=supm|λmn(Ax)|. It is easy to see that qn is a continuous seminorm on SCr, since for x∈SCr⊂ℓ∞,
(2.8)|qn(x)|≤M‖x‖,M>0.
Suppose that (2.6) is not true, then there exists x∈SCr with supnqn(x)=∞. By the principle of condensation of singularities (cf. [19]), the set
(2.9){x∈SCr:sup⁡nqn(x)=∞}
is of second category in SCr and hence nonempty, that is, there is x∈SCr with supnqn(x)=∞. But this contradicts the fact that qn is pointwise bounded on SCr. Now, by the Banach-Steinhaus theorem, there is a constant M such that
(2.10)qn(x)≤M∥x∥.
Now, we define a sequence x=(xk) by
(2.11)xk={sgng(n,k,m)kforeachm,n(1≤k≤k0),0fork>k0.
Then, x∈SCr. Applying this sequence to (2.10), we get (2.6).
This completes the proof of the theorem.


## 3. Maddox Sequence Spaces

A linear topological spaceX over the real field ℝ is said to be a paranormed space if there is a subadditive function g:X→ℝ such that g(θ)=0,g(x)=g(-x), and scalar multiplication is continuous, that is, |αn-α|→0 and g(xn-x)→0 imply g(αnxn-αx)→0 for all x,xn in X and α,αn in ℝ, where θ is the zero vector in the linear space X. Assume here and after that x=(xk) is a sequence such that xk≠0 for all k∈ℕ. Let p=(pk)k=0∞ be a bounded sequence of positive real numbers with supkpk=H and M=max{1,H}. The sequence spaces(3.1)c0(p):={x∈ω:lim⁡k|xk|pk=0},c(p):={x∈ω:x-le∈c0(p)forsomel∈C},l∞(p):={x∈ω:sup⁡k|xk|pk<∞},l(p):={x∈ω:∑k=0∞|xk|pk<∞},
were defined and studied by Et and Çolak [16] and Pati and Sinha [12]. If pk=p(k=0,1,…) for some constant p>0, then these sets reduce to c0,c,l∞, and lp, respectively. Note that c0(p) is a linear metric space paranormed by
(3.2)g(x)=sup⁡k|xk|pk/M
where M=max(1,suppk). l∞(p) and c(p) fail to be linear metric spaces because the continuity of scalar multiplication does not hold for them, but these two turn out to be linear metric spaces if and only if infkpk>0. l(p) is linear metric space paranormed by h1(x)=(∑k|xk|pk)1/M. All these spaces are complete in their respective topologies; however, these are not normed spaces in general (cf. [20]).In this section, we characterize the matrix classes(l(p),Vσ(θ)) and (l∞(p),Vσ(θ)).LetAx be defined, then, for all r,n, we write(3.3)τr(Ax)=∑k=1∞t(n,k,r)xk,
where(3.4)t(n,k,r)=1hr∑j∈Ira(σj(n),k),
and a(n,k) denotes the element ank of the matrix A.Theorem 3.1.
A∈(ℓ(p),Vσ(θ)) if and only if there exists B>1 such that for every n,(i)
(3.5)sup⁡r∑k|t(n,k,r)|qkB-qk<∞,(1<pk<∞),(pk-1+qk-1=1);sup⁡r,k|t(n,k,r)|pk<∞,(0<pk≤1),(ii)
a(k)={ank}n=1∞∈Vσ(θ) for each k, that is, limrt(n,k,r)=uk uniformly in n.In this case, the (σ,θ)-limit of Ax is ∑kukxk.Proof.
Necessity
We consider the case1<pk<∞. Let A∈(ℓ(p),Vσ(θ)). Since ek∈ℓ(p), the condition (ii) holds. Put frn(x)=τrn(Ax), since τrn(Ax) exists for each r and x∈l(p), therefore {frn(x)}r is a sequence of continuous real functionals on l(p) and further supr|frn(x)|<∞ on l(p). Now condition (i) follows by arguing with uniform boundedness principle. The case 0<pk≤1 can be proved similarly.
Sufficiency
Suppose that the conditions(i) and (ii) hold and x∈ℓ(p). Now for every m≥1, we have
(3.6)∑k=1m|t(n,k,r)|qkB-qk≤sup⁡r∑k|t(n,k,r)|qkB-qk.
Therefore,
(3.7)∑k|uk|qkB-qk=lim⁡⁡mlim⁡r∑k=1m|t(n,k,r)|qkB-qk≤sup⁡r∑k|t(n,k,r)|qkB-qk<∞.
Thus, the series ∑kt(n,k,r)xk and ∑kukxk converge for each r and x∈ℓ(p). For a given ε>0 and x∈ℓ(p), choose k0 such that
(3.8)(∑k=k0+1∞|xk|pk)1/H<ε,
where H=supkpk. Since (ii) holds, therefore there exists r0 such that
(3.9)|∑k=1k0(t(n,k,r)-uk)|<ε,
for every r>r0. Hence, by the condition (ii), it follows that
(3.10)|∑k=k0+1∞(t(n,k,r)-uk)|
is arbitrary small, and we have
(3.11)lim⁡r∑kt(n,k,r)xk=∑kukxk,
uniformly in n.
This completes the proof of the theorem.Theorem 3.2.
A∈(ℓ∞(p),Vσ(θ)) if and only if there exists N>1 such that(i)
Mn=supr∑k|t(n,k,r)|N1/pk<∞ for every n,(ii)
a(k)={ank}n=1∞∈Vσ(θ) for each k, that is, limrt(n,k,r)=uk uniformly in n,(iii)
limr∑k|t(n,k,r)-uk|=0 uniformly in n.In this case, the (σ,θ)-limit of Ax is ∑kukxk.Proof.
Necessity
LetA∈(ℓ∞(p),Vσ(θ)), then A∈(ℓ∞,Vσ(θ)), and the conditions (ii) and (iii) follow from Theorem 3 of Schaefer [5]. Now on the contrary, suppose that (i) does not hold, then there exists N>1 such that Mn=∞. Therefore, by Theorem 3 of Schaefer [5], the matrix
(3.12)B=(bnk)=(ankN1/pk)∉(l∞,Vσ(θ)),
that is, there exists x∈ℓ∞ such that Bx∉Vσ(θ). Now, let y=(N1/pkxk), then y∈ℓ∞(p) and Bx=Ay∉Vσ(θ), which contradicts that A∈(ℓ∞(p),Vσ(θ)). Therefore, (i) must hold.
Sufficiency
Suppose that the conditions hold andx∈ℓ∞(p), then for every n,
(3.13)|∑kt(n,k,r)xk|≤(sup⁡k|xk|pk)(sup⁡r∑k|t(n,k,r)|N1/pk)<∞.
Therefore Ax is defined. Now arguing as in Theorem 3.1, we get Ax∈Vσ(θ), and the series ∑kt(n,k,r)xk and ∑kukxk converge for x∈ℓ∞(p). Hence, by the condition (iii), we get
(3.14)lim⁡r∑kt(n,k,r)xk=∑kukxk,
uniformly in n.
This completes the proof of the theorem.Theorem 3.3.
Let1<pk<supkpk=H<∞ for every k, then A∈(ℓ(p),Vσ∞(θ)) if and only if there exists an integer N>1 such that(3.15)sup⁡r,n∑k|t(n,k,r)|qkN-qk<∞.Proof.
Sufficiency
Let (3.15) hold and that x∈ℓ(p) using the following inequality (see [21]):
(3.16)|ab|≤C(|a|qC-q+|b|p),
for C>0 and a,b, are two complex numbers (q-1+p-1=1), we have
(3.17)|τr(Ax)|=∑k|t(n,k,r)xk|≤∑kN[|t(n,k,r)|qkN-qk+|xk|pk],
where qk-1+pk-1=1. Taking the supremum over r,n on both sides and using (3.15), we get Ax∈Vσ∞(θ)) for x∈ℓ(p), that is, A∈(ℓ(p),Vσ∞(θ)).
Necessity
LetA∈(ℓ(p),Vσ∞(θ)). Write qn(x)=supr|τr(Ax)|. It is easy to see that for n≥0, qn is a continuous seminorm on ℓ(p), and (qn) is pointwise bounded on ℓ(p). Suppose that (3.15) is not true, then there exists x∈ℓ(p) with supnqn(x)=∞. By the principle of condensation of singularities [19], the set
(3.18){x∈l(p):sup⁡nqn(x)=∞}
is of second category in ℓ(p) and hence nonempty, that is, there is x∈ℓ(p) with supnqn(x)=∞. But this contradicts the fact that (qn) is pointwise bounded on ℓ(p). Now, by the Banach-Steinhaus theorem, there is constant M such that
(3.19)qn(x)≤Mg(x).
Now, we define a sequence x=(xk) by
(3.20)xk={δM/pk(sgn⁡t(n,k,r))|t(n,k,r)|qk-1S-1N-qk/pk,1≤k≤k0,0,fork>k0,
where 0<δ<1 and
(3.21)S=∑k=1k0|t(n,k,r)|qkN-qk.
Then it is easy to see that x∈ℓ(p) and g(x)≤δ. Applying this sequence to (3.19), we get the condition (3.15).
This completes the proof of the theorem.



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*Source: 612671-2012-05-17.xml*