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On topological properties of spaces obtained by the double band matrix

  • Suzan Zeren EMAIL logo and Çiğdem Bektaş
Published/Copyright: September 22, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

Let λ denote any one of the spaces ℓ∞ and ℓp and λ(Ť) be the domain of the band matrix Ť. We study ℓp(Ť) for 1 ≤ p ≤ ∞ and give some inclusions and its topological properties. Also, we define the alpha−, beta− and gamma− duals of the space ℓp(Ť). Finally, we give some matrix mappings.

Keywords: Sequence spaces; Matrix transformations; α−, β− and γ− duals
MSC 2010: 46A45; 40C05; 11B39

1 Introduction

Let ω, ℓ∞, c, c0 and ℓp denote the sequence spaces of all real or complex valued, bounded, convergent, null and absolutely p− summable sequences, respectively, where 1 ≤ p < ∞. Besides, let bs and cs denote the spaces of all bounded and convergent series, respectively. bvp is the space consisting of all sequences (xk) such that (xk − xk+1) ∈ ℓp. We assume that 1p+1q=1 for p, q > 1.

A sequence space X is called a K− space if pn : X → ℂ defined by pn(x) = xn is continuous for all n ∈ ℕ. A K− space X is called an FK− space if X is a complete linear metric space. If FK− space is normable then it is called a BK− space.

ℓ∞, c and c0 have the same norm given by

∥x∥ℓ∞=supk|xk|

for all k ∈ ℕ and these spaces are BK− spaces. ℓp is a BK− space with

∥x∥ℓp=∑k|xk|p

for 0 < p < 1 and

∥x∥ℓp=(∑k|xk|p)1/p

for 1 ≤ p < ∞.

Let A = (ank) be an infinite matrix and X, Y be sequence spaces. If Ax = (An(x)) exists for all x ∈ X and lies in Y, then A defines a matrix mapping from X into Y, where

An(x)=∑kankxk(n∈N) (1)

provided the series on the right side of (1) converges for each n ∈ ℕ.

For a sequence space X, the matrix domain of an infinite matrix A in X is defined by

XA={x∈ω:Ax∈X}. (2)

Also, (2) is a sequence space.

2 The sequence space ℓp(Ť)

We introduce the sequence space ℓp(Ť) where 1 ≤ p ≤ ∞ by using the band matrix Ť. Then, we analyze some topological properties of ℓp(Ť) and give some inclusion relations related to this space.

Let us define ℓp(Ť) and ℓ∞(Ť) as follows:

ℓp(Tˇ)=x=(xn)∈ω:∑n|rtnxn+s1tnxn−1|p<∞(1≤p<∞),ℓ∞(Tˇ)=x=(xn)∈ω:supnrtnxn+s1tnxn−1<∞,

where Ť = (tnk) is in ℓp and ℓ∞, respectively, also, is the double band matrix defined as follows:

tnk=rtn,k=ns1tn,k=n−10,k>n or 0≤k<n−1

for all k, n ∈ ℕ where r, s ∈ ℝ∖ {0} and t = (tn) ∈ c∖ c0 with tn > 0 for all n ∈ ℕ. If we rewrite ℓp(Ť) and ℓ∞(Ť) by using (2), we have

ℓp(Tˇ)=(ℓp)Tˇ(1≤p<∞) and ℓ∞(Tˇ)=(ℓ∞)Tˇ.

We have the inverse matrix Tˇ−1=(tnk−1) as follows:

tnk−1=1r(−sr)n−ktk∏j=kn1tj2,0≤k≤n0,k>n

for all k, n ∈ ℕ.

Additionally, we will frequently use y = (yn) by the Ť-transform of a sequence x = (xn), i.e.,

yn=Tˇn(x)=rt0x0,n=0rtnxn+s1tnxn−1,n≥1 (3)

for all n ∈ ℕ.

We can obtain the following statements according to the special cases of t = (tn), r, s :

  1. If tn = 1 for all n ∈ ℕ, r = 1 and s = −1, then ℓp(Ť) = bvp.

  2. If r = 1 and s = −1, then ℓp(Ť) = ℓp(T) and ℓ∞(Ť) = ℓ∞(T). (see [1])

  3. If tn=fnfn+1 for all n∈N, then ℓp(Tˇ)=ℓp(F^(r,s)) and ℓ∞(Tˇ)=ℓ∞(F^(r,s)). (see [2])

  4. If tn=fnfn+1 for all n∈N,r=1 and s=−1 then ℓp(Tˇ)=ℓp(F^) and ℓ∞(Tˇ)=ℓ∞(F^). (see [3])

  5. If tn = e for all n ∈ ℕ, r = 1 and s = −1, then Ť is the difference matrix Δ. (see [4])

  6. If tn = 1 for all n ∈ ℕ, then ℓp(Tˇ)=ℓ^p and ℓ∞(Tˇ)=ℓ^∞. (see [5])

By using a new double band matrix, many sequence spaces have recently been defined by several authors, see for instance [6-9].

We now may begin the following theorem.

Theorem 2.1

ℓp(Ť) is a Banach space with the norm ∥x∥ℓp(Tˇ)=∥Tˇx∥ℓp, i. e.,

∥x∥ℓp(Tˇ)=(∑n|Tˇn(x)|p)1/p,1≤p<∞supn|Tˇn(x)|,p=∞.

Theorem 2.2

It is obvious that ℓp(Ť) is a BK- space for 1 ≤ p ≤ ∞.

Proof

The proof is easy. □

Theorem 2.3

The sequence space ℓp(Ť) is linearly isomorphic to the space ℓp, i. e., ℓp(Ť) ≅ ℓp for 1 ≤ p ≤ ∞.

Proof

By using (3), we define transformation Ť : ℓp(Ť) → ℓp. The linearity and injectivity of Ť is clear.

Let y = (yn) ∈ ℓp and define the sequence x = (xn) as follows:

xn=1r∑κ=0n[(−sr)n−k(∏j=kn1tj2)tkyk](n,k∈N) (4)

Then, by using (3) and (4), we obtain

Tˇn(x)=rtnxn+s1tnxn−1=rtn1r∑k=0n[(−sr)n−k(∏j=kn1tj2)tkyk]+s1tn1r∑k=0n−1[(−sr)n−k−1(∏j=kn−11tj2)tkyk]=tn{tn−1yn+∑k=0n−1[(−sr)n−k(∏j=kn1tj2)tkyk]}−1tn∑k=0n−1[(−sr)n−k(∏j=kn−11tj2)tkyk]=yn

for all n ∈ ℕ. This shows that Ťx = y. Since y ∈ ℓp, we obtain Ťx ∈ ℓp and hence x ∈ ℓp(Ť) . Thus Ť is surjective.

Morever for any x ∈ ℓp(Ť), we have

∥y∥ℓp=∥Tˇx∥ℓp=∥x∥ℓp(Tˇ)

which means that Ť preserves the norm in the case of 1 ≤ p ≤ ∞. Hence Ť is an isometry. Consequently, Ť is a linear bijection which shows that the spaces ℓp(Ť) and ℓp are linearly isomorphic. □

Lemma 2.4

([10]). A product ∏n(1+an) with positive terms an is convergent if and only if the series ∑nan converges.

Theorem 2.5

The inclusion ℓp(Ť) ⊂ ℓq(Ť) is strictly satisfied for 1 ≤ p > q < ∞.

Proof

Let 1 ≤ p < q < ∞. Then, it follows by the inclusion ℓp ⊂ ℓq that the inclusion ℓp(Ť) ⊂ ℓq(Ť) holds. Further, because of the inclusion ℓp ⊂ ℓq is strict, there is a sequence x ∈ ℓq ∖ ℓp. Let y = (yn) be as follows:

yn=1r∑k=0n[(−sr)n−k(∏j=kn1tj2)tkxk](n∈N).

Thus, we get

Tˇn(y)=xn

for all n ∈ ℕ which means that Ťy = x and since x ∈ ℓq ∖ ℓp, we have Ťy ∈ ℓq ∖ ℓp. This implies that the inclusion ℓp(Ť) ⊂ ℓq(Ť) is strict. □

Theorem 2.6

The inclusion ℓp(Ť) ⊂ ℓ∞(Ť) is strictly satisfied for 1 ≤ p < ∞.

Proof

Let x ∈ ℓp(Ť), then Ťx ∈ ℓp. Since ℓp ⊂ ℓ∞, Ťx ∈ ℓ∞. Hence x ∈ ℓ∞(Ť) which means that ℓp(Ť) ⊂ ℓ∞(Ť). If we define y = (yn) by

yn=1r∑k=0n[(−1)k(−sr)n−ktk(∏i=kn1ti2)](n∈N)

Hence, we obtain for every n ∈ ℕ that

Tˇn(y)=rtnyn+s1tnyn−1=(−1)n

which means that Ťy ∈ ℓ∞∖ ℓp and hence y ∈ ℓ∞(Ť)∖ ℓp(Ť). Hence, the inclusion ℓp(Ť) ⊂ ℓ∞(Ť) is strict. This concludes the proof. □

Lemma 2.7

The inclusion ℓp ⊂ ℓp(Ť) is strictly satisfied for 1 ≤ p ≤ ∞.

Proof

The proof is similar to [1]. □

Lemma 2.8

Neither of the spaces ℓ∞ and ℓp(Ť) includes the other one, where 1 ≤ p < ∞.

Proof

The proof is similar to [1]. □

3 The α-, β- and γ- duals of the space ℓp(Ť) and ℓ∞(Ť)

The α−, β− and γ− duals of sequence space λ are introduced as follows:

λα={a=(ak)∈ω:ax=(akxk)∈ℓ1 for all x=(xk)∈λ},λβ={a=(ak)∈ω:ax=(akxk)∈cs for all x=(xk)∈λ},λγ={a=(ak)∈ω:ax=(akxk)∈bs for all x=(xk)∈λ}.

In present section, we give the α−, β− and γ− duals of the space ℓp(Ť), where 1 ≤ p ≤ ∞.

Throughout the present section, let F be the collection of all nonempty and finite subsets of ℕ.

The following known results from [11, 12] are fundamental for our investigation.

supn∑k|ank|q<∞. (6)

limn→∞ank exists for all k∈N. (7)

supK∈F∑k|∑n∈Kank|q<∞. (8)

supn,k|ank|<∞. (9)

supk∑n|ank|<∞. (10)

limn→∞∑k|ank|=∑k|limn→∞ank|. (11)

Lemma 3.1

The necessary and sufficient conditions for A ∈ (λ : μ) when λ ∈ {ℓp, ℓ1, ℓ∞} and μ ∈ {ℓ∞, c, ℓ1} can be read from Table 1, where 1. (6). 2. (6) and (7). 3. (8). 4. (9). 5. (7) and (9). 6. (10). 7. (6) with q = 1.8. (7) and (11). 9. (8) with q = 1.

Table 1

The characterization of the class (λ1 : λ2) with λ1 ∈ {ℓp, ℓ1, ℓ∞} and λ2 ∈ {ℓ∞, c, ℓ1}

From To
ℓ∞ c ℓ1
ℓp 1. 2. 3.
ℓ1 4. 5. 6.
ℓ∞ 7. 8. 9.

Theorem 3.2

Define the sets d^1(r,s),d^2(r,s),d^3(r,s),d^4(r,s),d^5(r,s)andd^6(r,s) as follows:

d^1(r,s)={a=(ak)∈ω:supK∈F∑k∑n∈K1r(−sr)n−k∏j=kn1tj2tkanq<∞},d^2(r,s)={a=(ak)∈ω:supk∑n1r(−sr)n−k∏j=kn1tj2tkan<∞},d^3(r,s)={a=(ak)∈ω:supn∑k∑i=kn1r(−sr)i−k∏j=ki1tj2tkaiq<∞},d^4(r,s)={a=(ak)∈ω:limn→∞∑i=kn1r(−sr)i−k∏j=ki1tj2tkajexistsforallk∈N},d^5(r,s)={a=(ak)∈ω:supn,k∑i=kn1r(−sr)i−k∏j=ki1tj2tkai<∞},d^6(r,s)=a=(ak)∈ω:limn→∞∑k∑i=kn1r(−sr)i−k∏j=ki1tj2tkai=∑klimn→∞∑i=kn1r(−sr)i−k∏j=ki1tj2tkai<∞.

Let 1 < p < ∞. Then,

  1. (ℓp(Tˇ))α=d^1(r,s),(ℓ1(Tˇ))α=d^2(r,s)and(ℓ∞(Tˇ))α=d^1(r,s)withq=1.

  2. (ℓp(Tˇ))β=d^3(r,s)∩d^4(r,s),(ℓ1(Tˇ))β=d^4(r,s)∩d^5(r,s)and(ℓ∞(Tˇ))β=d^4(r,s)∩d^6(r,s).

  3. (ℓp(Tˇ))γ=d^3(r,s),(ℓ1(Tˇ))γ=d^5(r,s)and(ℓ∞(Tˇ))γ=d^3(r,s)withq=1.

Proof

Let us prove it only for the space ℓp(Ť), where 1 < p < ∞. Let us take any a = (an) ∈ ω. We have with the notation (4) that

anxn=an1r∑k=0n(−sr)n−k∏j=kn1tj2tkyk=∑k=0n[1r(−sr)n−k∏j=kn1tj2tkan]yk=∑k=0nbnk(r,s)yk=(By)n,(n∈N)

where B = (bnk(r, s)) is defined by

bnk(r,s)=1r(−sr)n−k∏j=kn1tj2tkan,0≤k≤n0,k>n(k,n∈N).

Hence, we have that ax = (anxn) ∈ ℓ1 whenever x = (xn) ∈ ℓp(Ť) iff By ∈ ℓ1 whenever y = (yn) ∈ ℓp. From the Table 1, we have the desired result that

(ℓp(Tˇ))α=d^1(r,s).

Now, we obtain

∑k=0nakxk=∑k=0nak{∑i=0k[1r(−sr)k−i∏j=ik1tj2tiyi]}=∑k=0n∑i=kn{1r(−sr)i−k∏j=ki1tj2tk}aiyk=∑k=0n∑i=kncik(r,s)aiyk=∑k=0ndnk(r,s)yk=(Dy)n(n∈N) (12)

D = (dnk(r, s)) defined by

dnk(r,s)=∑i=kncik(r,s)ai,0≤k≤n0,k>n

where

cik(r,s)=1r(−sr)i−k∏j=ki1tj2tk,0≤k≤n0,k>n

for all k, n ∈ ℕ. Therefore, we have that ax = (anxn) ∈ cs whenever x = (xn) ∈ ℓp(Ť) iff Dy ∈ c whenever y = (yn) ∈ ℓp. From the Table 1, we have that

(ℓp(Tˇ))β=d^3(r,s)∩d^4(r,s).

As this, we deduce by (12) that ax = (anxn) ∈ bs whenever x = (xn) ∈ ℓp(Ť) iff Dy ∈ ℓ∞ whenever y = (yn) ∈ ℓp. Thus, we obtain from the Table 1 that

(ℓp(Tˇ))γ=d^3(r,s).

This concludes the proof. □

4 Certain matrix mappings on the space ℓp(Ť)

In the present section, we analyze the matrix mappings from the space ℓp(Ť) to the spaces ℓ∞, c, c0 and ℓ1, where 1 ≤ p ≤ ∞.

Theorem 4.1

Let λ be an arbitrary subset of ω and 1 ≤ p ≤ ∞. Then, we have A = (ank) ∈ (ℓp(Ť), λ) if and only if

E(m)=(enk(m))∈(ℓp,c)foralln∈N,E=(enk)∈(ℓp,λ),

where

enk(m)=∑i=km1r(−sr)i−k∏j=ki1tj2tkani,0≤k≤m0,k>m

and

enk=∑i=k∞1r(−sr)i−k∏j=ki1tj2tkani

for every k, m, n ∈ ℕ.

Proof

Suppose that A = (ank) ∈ (ℓp(Ť), λ) and x = (xk) ∈ ℓp(Ť). Then, we have from (4)

∑k=0mankxk=∑k=0mank∑i=0k1r(−sr)k−i∏j=ik1tj2tiyi=∑k=0m∑i=km1r(−sr)i−k∏j=ki1tj2tkaniyk=∑k=0menk(m)yk=En(m)(y)

for every m, n ∈ ℕ. Since Ax exists and Ax ∈ c, E(m) ∈ (ℓp, c). Letting m →∞ in the last equality, we have Ax = Ey. Then, we have that E ∈ (ℓp, λ).

Conversely, assume that the conditions hold and take an arbitrary x ∈ ℓp(Ť). Then, we obtain that (enk)keℕ ∈ ℓpβ from our assumptation and Theorem 3.2. Since An=(ank)keN∈(ℓp(Tˇ))β for every n ∈ ℕ, Ax exists. Then, we have from the above equality as m → ∞ that Ax = Ey and hence A∈(ℓp(Tˇ),λ).

Now, we list the following conditions:

supn∑k=0n|a~nk|q<∞ (13)

limn→∞a~nk exists for all k∈N (14)

supn,k|a~nk|<∞ (15)

limn→∞∑k=0n|a~nk|=∑k=0n|limn→∞a~nk|<∞ (16)

limn→∞a~nk=0 for all k∈N (17)

limn→∞∑k|a~nk|=0 (18)

where

a~nk=∑i=kn1r(−sr)i−k∏j=ki1tj2tkani. □

Theorem 4.2

Let 1 < p < ∞. Then we have

  1. A = (ank) ∈ (ℓp(Tˇ),ℓ∞) ⇔(13), (14) and (6) hold.

  2. A = (ank) ∈ (ℓp(Tˇ),c) ⇔ (13), (14), (6) and (7) hold.

  3. A = (ank) ∈ (ℓp(Tˇ),c0) ⇔ (13), (14), (17) and (6) hold.

  4. A = (ank) ∈ (ℓp(Tˇ),ℓ1) ⇔(13), (14) and (8) hold.

Theorem 4.3

  1. (i) A = (ank) ∈ (ℓ1(Tˇ),ℓ∞) ⇔(14), (15) and (9) hold.

  2. A = (ank) ∈ (ℓ1(Tˇ),c) ⇔(14), (15), (7) and (9) hold.

  3. A = (ank) ∈ (ℓ1(Tˇ),c0) ⇔(14), (15), (17) and (9) hold.

  4. A = (ank) ∈ (ℓ1(Tˇ),ℓ1) ⇔(14), (15) and (10) hold.

Theorem 4.4

  1. A = (ank) ∈ (ℓ∞(Tˇ),ℓ∞) ⇔(14), (16) and (6) with q = 1 hold.

  2. A = (ank) ∈ (ℓ∞(Tˇ),c) ⇔(14), (16), (7) and (11) hold.

  3. A = (ank) ∈ (ℓ∞(Tˇ),c0) ⇔(14), (16) and (18) hold.

  4. A = (ank) ∈ (ℓ∞(Tˇ),ℓ1) ⇔(14), (16) and (8) with q = 1 hold.

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Received: 2017-2-27
Accepted: 2017-6-30
Published Online: 2017-9-22

© 2017 Zeren and Bektaş

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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https://doi.org/10.1515/math-2017-0095
Keywords for this article
Sequence spaces; Matrix transformations; α−, β− and γ− duals
Creative Commons
BY-NC-ND 3.0

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