# A New Difference Sequence Set of Orderα and Its Geometrical Properties

Authors: Mikail Et; Vatan Karakaya
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We introduce a new class of sequences named asmαΔr,ϕ,p and, for this space, we study some inclusion relations, topological properties, and geometrical properties such as order continuous, the Fatou property, and the Banach-Saks property of type p.


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## Body

## 1. Introduction, Definitions, and Preliminaries

Byw we denote the space of all complex (or real) sequences. If x∈w, then we simply write x=(xk) instead of x=(xk)k=0∞. We will write l∞, c, and c0 for the sequence spaces of all bounded, convergent, and null sequences, respectively. Also by l1 and lp we denote the spaces of all absolutely summable and p-absolutely summable series, respectively.The notion of difference sequence spaces was generalized by Et and Çolak [1] such as X(Δr)={x=(xk):Δrx∈X}, for X=l∞, c, and c0. They showed that these sequence spaces are BK-spaces with the norm
(1)∥x∥Δ=∑i=1r‍|xi|+∥Δrx∥∞,
where r∈N, Δ0x=x, Δx=(xk-xk+1), Δrx=(Δrxk)=(Δr-1xk-Δr-1xk+1), and Δrxk=∑v=0r‍(-1)v(rv)xk+v. Recently difference sequences and related concepts have been studied in ([2–13]) and by many others.LetE be a sequence space. Then E is called(i)
solid (ornormal) if (αkxk)∈E for all sequences (αk) of scalars with |αk|≤1 for all k∈N, whenever (xk)∈E,(ii)
symmetric if (xk)∈E implies (xπ(k))∈E, where π is a permutation of N,(iii)
monotoneprovided E contains the canaonical preimages of all its step spaces,(iv)
sequence algebra if x·y∈E, whenever x,y∈E.It is well known that ifE is normal then E is monotone.Throughout this paperφs denotes the class of all subsets of N; those do not contain more than s elements. Let (ϕn) be a nondecreasing sequence of positive numbers such that nϕn+1≤(n+1)ϕn for all n∈N. The class of all sequences (ϕn) is denoted by Φ. The sequence space m(ϕ) was introduced by Sargent [14] and he studied some of its properties and obtained some relations with the space lp. Later on it was investigated by Tripathy and Sen [15] and Tripathy and Mahanta [16].Let us recall that a sequence{v(i)}i=1∞ in a Banach space X is called Schauderbasis of X (or basis for short) if for each x∈X there exists a unique sequence {λ(i)}i=1∞ of scalars such that x=∑i=1∞‍λ(i)v(i); that is, lim⁡n→∞∑i=1n‍λ(i)v(i)=x.A sequence spaceX with a linear topology is called a K-space if each of the projection maps Pi:X→C defined by Pi(x)=x(i) for x=(x(i))i=1∞∈X is continuous for each natural i. AFréchet space is a complete metric linear space and the metric is generated by an F-norm and a Fréchet space which is a K-space is called an FK-space; that is, a K-space X is called an FK-space if X is a complete linear metric space. In other words, X is an FK-space if X is a Fréchet space with continuous coordinate projections. All the sequence spaces mentioned above are FK spaces except the space c00.AnFK-space X which contains the space c00 is said to have the propertyAK if for every sequence {x(i)}∈X, x=∑i=1∞‍x(i)e(i) where e(i)=(0,0,…,1(ith-place),0,0,…).A Banach spaceX is said to be aKöthe sequence space (see [17, 18]) if X is a subspace of w such that(i)
ifx∈w,y∈X and |x(i)|≤|y(i)| for all i∈N, then x∈X and ∥x∥≤∥y∥;(ii)
there exists an elementx∈X such that x(i)>0 for all i∈N.We say thatx∈X is order continuous if for any sequence (xn) in X such that xn(i)≤|x(i)| for each i∈N and xn(i)→0 (n→∞) we have that ∥xn∥→0 holds.A Köthe sequence spaceX is said to be order continuous if all sequences in X are order continuous. It is easy to see that x∈X is order continuous if and only if ∥(0,0,…,0,x(n+1),x(n+2),…)∥→0 as n→∞.A Köthe sequence spaceX is said to have the Fatouproperty if, for any real sequence x and any {xn} in X such that xn↑x coordinatewisely and sup⁡n∥xn∥<∞, we have that x∈X and ∥xn∥→∥x∥.A Banach spaceX is said to have the Banach-Saksproperty if every bounded sequence {xn} in X admits a subsequence {zn} such that the sequence {tk(z)} is convergent in X with respect to the norm, where
(2)tk(z)=1k(z1+z2+⋯+zk),∀k∈N.Some of recent works on geometric properties of sequence space can be found in the following list ([19–21]).


## 2. Inclusion and Topological Properties of the Spacemα(Δr,ϕ,p)

In this section we introduce a new class of sequences and establish some inclusion relations. Also we show that this space is not perfect and normal.Letr be a fixed positive integer, α∈(0,1] any real number, and p a positive real number such that 1≤p<∞. Now we define the sequence space mα(Δr,ϕ,p) as
(3)mα(Δr,ϕ,p)={x∈w:sup⁡s≥1,σ∈φs1ϕsα∑n∈σ‍|Δrxn|p<∞}.
In the case p=1, we will write mα(Δr,ϕ) instead of mα(Δr,ϕ,p) and in the special case m=0 and p=1 we will write mα(ϕ) instead of mα(Δr,ϕ,p).The proof of each of the following results is straightforward, so we choose to state these results without proof.Theorem 1.
Letϕ∈Φ, α∈(0,1], and let p be a positive real number such that 1≤p<∞. Then the sequence space mα(Δr,ϕ,p) is a BK-space normed by
(4)∥x∥=∑i=1r‍|xi|+sup⁡s≥1,σ∈φs1ϕsα(∑n∈σ‍|Δrxn|p)1/p.Theorem 2.
Letϕ∈Φ, α∈(0,1], and let p be a positive real number such that 1≤p<∞; then mα(Δr,ϕ)⊂mα(Δr,ϕ,p).Theorem 3.
Letα and β be fixed real numbers such that 0<α≤β≤1 and p a positive real number such that 1≤p<∞; then mα(Δr,ϕ,p)⊂mβ(Δr,ϕ,p).Theorem 4.
Letα and β be fixed real numbers such that 0<α≤β≤1 and p a positive real number such that 1≤p<∞. For any two sequences (ϕs) and (ψs) of real numbers such that ϕ,ψ∈Φ. Then mα(Δr,ϕ,p)⊂mβ(Δr,ψ,p) if and only if sup⁡s≥1(ϕsα/ψsβ)<∞.Proof.
Letx∈mα(Δr,ϕ,p) and sup⁡s≥1(ϕsα/ψsβ)<∞. Then
(5)sup⁡s≥1,σ∈φs1ϕsα∑n∈σ‍|Δrxn|p<∞
and there exists a positive number K such that ϕsα≤Kψsβ and so that 1/ψsβ≤K/ϕsα for all s. Therefore for all s we have
(6)1ψsβ∑n∈σ‍|Δrxn|p≤Kϕsα∑n∈σ‍|Δrxn|p.
Now taking supremum over s≥1 and σ∈φs we get
(7)sup⁡s≥1,σ∈φs1ψsβ∑n∈σ‍|Δrxn|p≤sup⁡s≥1,σ∈φsKϕsα∑n∈σ‍|Δrxn|p
and so x∈mβ(Δr,ψ,p).
Conversely letmα(Δr,ϕ,p)⊂mβ(Δr,ψ,p) and suppose that sup⁡s≥1(ϕsα/ψsβ)=∞. Then there exists an increasing sequence (si) of naturals numbers such that lim⁡i(ϕsiα/ψsiβ)=∞. Let B∈R+, where R+ is the set of positive real numbers; then there exists i0∈N such that ϕsiα/ψsiβ>B for all si≥i0. Hence ϕsiα>Bψsiβ and so 1/ψsiβ>B/ϕsiα. Then we can write
(8)1ψsiβ∑n∈σ‍|Δrxn|p>Bϕsiα∑n∈σ‍|Δrxn|p
for all si≥i0. Now taking supremum over si≥i0 and σ∈φs we get
(9)sup⁡si≥i0,σ∈φs1ψsiβ∑n∈σ‍|Δrxn|p>sup⁡si≥i0,σ∈φsBϕsiα∑n∈σ‍|Δrxn|p.
Since (9) holds for all B∈R+ (we may take the number B sufficiently large), we have
(10)sup⁡si≥i0,σ∈φs1ψsiβ∑n∈σ‍|Δrxn|p=∞
when x∈mα(Δr,ϕ,p) with
(11)0<sup⁡s≥1,σ∈φs1ϕsα∑n∈σ‍|Δrxn|p<∞.
Hence x∉mβ(Δr,ψ,p). This contradicts to mα(Δr,ϕ,p)⊂mβ(Δr,ψ,p). Hence sup⁡s≥1(ϕsα/ψsβ)<∞.The following results are derivable easily from Theorem4.Corollary 5.
Letα and β be fixed real numbers such that 0<α≤β≤1 and p a positive real number such that 1≤p<∞. For any two sequences (ϕs) and (ψs) of real numbers such that ϕ,ψ∈Φ. Then one has(i)
mα(Δr,ϕ,p)=mβ(Δr,ψ,p) if and only if 0<inf⁡s≥1(ϕsα/ψsβ)<sup⁡s≥1(ϕsα/ψsβ)<∞,(ii)
mα(Δr,ϕ,p)=mα(Δr,ψ,p) if and only if 0<inf⁡s≥1(ϕsα/ψsα)<sup⁡s≥1(ϕsα/ψsα)<∞,(iii)
mα(Δr,ϕ,p)=mβ(Δr,ϕ,p) if and only if 0<inf⁡s≥1(ϕsα/ϕsβ)<sup⁡s≥1(ϕsα/ϕsβ)<∞.Theorem 6.
Considermα(Δr-1,ϕ,p)⊂mα(Δr,ϕ,p) and the inclusion is strict.Proof.
It follows from Minkowski’s inequality. To show the inclusion is strict, letϕn=1 for all n∈N, α=1, p=1, and x=(kr-1); then x∈mα(Δr,ϕ,p)∖mα(Δr-1,ϕ,p).Theorem 7.
The sequence spacemα(ϕ) is solid and hence monotone, but the sequence space mα(Δr,ϕ,p) is neither solid nor symmetric and sequence algebra for m≥1.Proof.
Letx∈mα(ϕ) and y=(yn) be sequences such that |xn|≤|yn| for each n∈N. Then we get
(12)sup⁡s≥1,σ∈φs1ϕsα∑n∈σ‍|xn|≤sup⁡s≥1,σ∈φs1ϕsα∑n∈σ‍|yn|.
Hence mα(ϕ) is solid and hence monotone. To show the space mα(Δr,ϕ,p) is normal, let ϕn=1, for all n∈N, α=1, p=1, and x=(kr-1), then x∈mα(Δr,ϕ,p); but (αkxk)∉mα(Δr,ϕ,p) when αk=(-1)k for all k∈N. Hence mα(Δr,ϕ,p) is not solid. The other cases can be proved on considering similar examples.Theorem 8.
Consider(13)lp(Δr)⊂mα(Δr,ϕ,p)⊂l∞(Δr).Proof.
It is omitted.Theorem 9.
If0<p<q, then mα(Δr,ϕ,p)⊂mα(Δr,ϕ,q).Proof.
Proof follows from the following inequality:(14)(∑k=1n‍|xk|q)1/q<(∑k=1n‍|xk|q)1/q,(0<p<q).


## 3. Geometrical Properties of the Spacemα(Δr,ϕ,p)

In this section, we study some geometrical properties of the spacemα(Δr,ϕ,p). Some of these geometrical properties are the order continuous, the Fatou property, and the Banach-Saks property of type p. Let us start with the following theorem.Theorem 10.
The spacemα(Δr,ϕ,p) is order continuous.Proof.
To prove this theorem, we have to show thatmα(Δr,ϕ,p) is an AK-space. It is easy to see that mα(Δr,ϕ,p) contains c00 which is the space of real sequences which have only a finite number of nonzero coordinates. By using definition of AK-properties, we have that x={x(i)}∈mα(Δr,ϕ,p) has a unique representation x=∑i=1∞‍x(i)e(i); that is, ∥x-x[j]∥mα(Δr,ϕ,p)=∥(0,0,…,x(j),x(j+1),…)∥mα(Δr,ϕ,p)→0 as j→∞, which means that mα(Δr,ϕ,p) has AK. Hence, since BK-space mα(Δr,ϕ,p) containing c00 has AK-property, the space mα(Δr,ϕ,p) is order continuous.Theorem 11.
The spacemα(Δr,ϕ,p) has the Fatou property.Proof.
Letx be any real sequence from (w)+ and {xn} any nondecreasing sequence of nonnegative elements from mα(Δr,ϕ,p) such that xn(i)→x(i) as n→∞ coordinatewisely and sup⁡n∥xn∥mα(Δr,ϕ,p)<∞.
Let us denoteτ=sup⁡n∥xn∥mα(Δr,ϕ,p). Then, since the supremum is homogeneous, we have
(15)1τ(∑i=1r‍|xn(i)|+sup⁡s≥1,σ∈φs1ϕsα(∑i∈σ‍|Δrxn(i)|p)1/p)=∑i=1r‍|xn(i)τ|+sup⁡s≥1,σ∈φs1ϕsα(∑n∈σ‍|Δrxn(i)τ|p)1/p≤∑i=1r‍|xn(i)∥xn∥mα(Δr,ϕ,p)|+sup⁡s≥1,σ∈φs1ϕsα(∑n∈σ‍|Δrxn(i)∥xn∥mα(Δr,ϕ,p)|p)1/p=1∥xn∥mα(Δr,ϕ,p)∥xn∥mα(Δr,ϕ,p)=1.
Moreover, by the assumptions that {xn} is nondecreasing and convergent to x coordinatewisely and by the Beppo-Levi theorem, we have
(16)1τlim⁡n→∞(∑i=1r‍|xn(i)|+sup⁡s≥1,σ∈φs1ϕsα(∑i∈σ‍|Δrxn(i)|p)1/p)=∑i=1r‍|x(i)τ|+sup⁡s≥1,σ∈φs1ϕsα(∑i∈σ‍|Δrx(i)τ|p)1/p=∥xs∥mα(Δr,ϕ,p))≤1,
whence
(17)∥x∥mα(Δr,ϕ,p)))≤τ=sup⁡n∥xn∥mα(Δr,ϕ,p))=lim⁡n→∞∥xn∥mα(Δr,ϕ,p))<∞.
Therefore, x∈mα(Δr,ϕ,p)). On the other hand, since 0≤xn≤x for any natural number n and the sequence {xn} is nondecreasing, we obtain that the sequence {∥xn∥mα(Δr,ϕ,p))} is bounded from above by ∥x∥mα(Δr,ϕ,p)). As a result, lim⁡n→∞∥xn∥mα(Δr,ϕ,p))≤∥x∥mα(Δr,ϕ,p)), which together with the opposite inequality proved already yields that ∥x∥mα(Δr,ϕ,p))=lim⁡n∥xn∥mα(Δr,ϕ,p)).Theorem 12.
The spacemα(Δr,ϕ,p)) has the Banach-Saks property of the type p.Proof.
It can be proved with standard technic.



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