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Duals of generalized Orlicz Hilbert sequence spaces and matrix transformations

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Abstract

In this paper, wedefine the sequence spaces hc (?(m)v,M,u,p), h0 (?(m) v,M,u,p) and h?(?(m)v,M,u,p) resulting from the infinite Hilbert matrix and the Musielak-Orlicz function. We give some topological properties and inclusion relations of these newly created spaces. We also identified ??, ?? and ??duals of the spaces. Finally, we tried to characterize some matrix transformations between these spaces.
Filomat 37:27 (2023), 9089–9102
https://doi.org/10.2298/FIL2327089B
Published by Faculty of Sciences and Mathematics,
University of Niˇ
s, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Duals of generalized Orlicz Hilbert sequence spaces and matrix
transformations
Damla Barlaka, C¸ gdem A. Bektas¸b
aDepartment of Statistics , Dicle University, Diyarbakır, Turkey
bDepartment of Mathematics, Fırat University, Elazı˘g, Turkey
Abstract. In this paper, we define the sequence spaces hc(m)
v,M,u,p,h0(m)
v,M,u,pand h(m)
v,M,u,p
resulting from the infinite Hilbert matrix and the Musielak-Orlicz function. We give some topological prop-
erties and inclusion relations of these newly created spaces. We also identified α, βand γduals of the
spaces. Finally, we tried to characterize some matrix transformations between these spaces.
1. Introduction and Preliminaries:
Hilbert defined the Hilbert matrix in 1894. The Hilbert matrix played both several branches of mathe-
matics and computational sciences.The n×nmatrix H=hi,j=1
i+j1i,jNis a Hilbert matrix [1]. We
consider the infinite Hilbert matrix Has follows:
H=
1 1/2 1/3 1/4...
1/2 1/3 1/4... .
1/3 1/4... . .
1/4... . . .
....
...
. .
and it can be showed in integral form as follows:
H=(hi,j)=Z1
0
xi+j2dx
The inverse of Hilbert matrix H1is defined by
H1=h1
i,j=(1)i+ji+j1 n+i1
nj! n+j1
ni! i+j1
i1!2
2020 Mathematics Subject Classification. 40F05; 46A45; 15B05; 40C05.
Keywords. Orlicz function, dierence operator, Hilbert matrix, matrix transformations, α,γduals.
Received: 28 December 2022; Accepted: 05 June 2023
Communicated by Dragan S. Djordjevi´
c
Email addresses: damla.barlak@dicle.edu.tr (Damla Barlak), cbektas@firat.edu.tr (C¸ i ˘
gdem A. Bektas¸)
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9090
for all i,jN.
Let us we denote the space of all real or complex sequence with w.We write the sequence spaces of
all convergent, null and bounded sequences by c,c0and l, respectively. Also we will denote the space
of all bounded, convergent and absolutely convergent series with bs,cs,and l1respectively. Let A=(ank )
be an infinite matrix of real or complex numbers and X,Ybe subsets of w.We write An(x)=Pkankxkand
Ax =An(x)for n,kN. For a sequence space X, the matrix domain of an infinite matrix Ais defined by
XA={x=(xk)w:Ax X}
which is also a sequence space. We denote with (X,Y)the class of all matrices Asuch that A:XY.
A matrix A=(ank)is called a triangle if ank =0 for k>nand ann ,0 for all nN. For the triangle
matrices A,B and a sequence x,A(Bx)=(AB)xholds. We remark that the triangle matrix A uniquely has an
inverse matrix A1=Band the matrix Bis also triangle.
Let Xbe a normed sequence space. A sequence (bn)in Xis called a Schauder basis for Xif for every
xXthere is a unique sequence (αn)of scalars such that
lim
n
x
n
X
k=0
αkbk
=0.
AB-space is a complete normed space. A topological sequence space in which all coordinate functionals
πk, πk(x)=xk,are continuous is called a K-space. A BK-space is defined as a K-space which is also a B-space,
that is, a BK-space is a Banach space with continuous coordinates. For example, the space lp(1 p<) is
BK-space with xp=P
k=0|xk|p1
pand c,c0and lare BK-space with x=sup
k
|xk|.
Kızmaz [2] was firstly introduced the concept of the dierence operator in the sequence spaces. Further
Et and C¸ olak [3] generalized the idea of dierence sequence spaces of Kızmaz.Besides this topic was studied
by many authors ([4], [5]).Now, the dierence matrix = (δnk)defined by
δnk =((1)nk,(n1kn)
0,(0<n1 or n>k).
The dierence operator order mis defined (m):ww,(1)xk=(xkxk1)and (m)x=(1)xk(m1)xk
for m2.
The triangle matrix (m)=δ(m)
nk defined by
δ(m)
nk =
(1)nk m
nk!,(max{0,nm} kn)
0,(0k<max{0,nm}or n>k)
for all k,nNand for any fixed mN.
Let v=(vk)be any fixed sequence of nonzero complex numbers. We define operators (m):wwby
mN,(0)
vxk=vkxk,vxk=(vkxkvk1xk1),(m)
vxk= (m1)
vxk(m1)
vxk1and so that
(m)
vxk=
m
X
i=0
(1)i m
i!vkixki
Polat [6] and Kirisci and Polat [7] have defined some new sequence spaces using Hilbert matrix. Let hc,h0
and hbe convergent Hilbert , null convergent Hilbert and bounded Hilbert sequence spaces, respectively.
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9091
An Orlicz function is a function M: [0,)[0,) which is continuous, non-decreasing and convex
such that M(0)=0,M(x)>0 for x>0 and M(x) as x .Lindenstrauss and Tzafriri [8] used the
idea of Orlicz function to construct the following sequence space
M=
xw:
X
k=1
M |xk|
ρ!<,for some ρ > 0
lMis a Banach space with the norm
x=inf
ρ > 0 :
X
k=1
M |xk|
ρ!1
which is called an Orlicz sequence space. A sequence M=(Mk)of Orlicz functions is called the Musielak-
Orlicz function (see [9], [10]). For more details on sequence spaces, see ([11], [12], [13])and the references
there in.
LetXbe a linear metric space. A function p:XRis called a paranorm, if
(P1)p(x)0 for xX,
(P2)p(x)=p(x)for all xX,
(P3)px+yp(x)+pyfor all x,yX,
(P4)If (λn)is a sequence of scalars with λnλas n and (xn)is a sequence of vectors with
p(xnx)0 as n ,then p(λnxnλx)0 as n .
A paranorm pfor which p(x)=0 implies x=0 is called total paranorm and the pair X,pis called a
total paranormed space. It is well known that the metric of any linear metric space is given by some total
paranorm (see ([9], Theorem 10.4.2, page 183)).
2. Main Results
In this section we define the sequence spaces hc(m)
v,M,u,p,h0(m)
v,M,u,pand h(m)
v,M,u,pand
give some relations between them. These sequence spaces are linear and BK-spaces. We prove that the new
Hilbert sequence spaces hc(m)
v,M,u,p,h0(m)
v,M,u,pand h(m)
v,M,u,pare isometrically isomorphic
to the space c,c0and lrespectively.
Definition 2.1. Let M=(Mk)be a sequence of Orlicz functions, v=(vk)be any fixed sequence of non-zero
complex numbers. Also, let p=pkand u=(uk)be the bounded sequence and sequence of positive real
numbers, respectively and H=hi,jbe an infinite Hilbert matrix. In the present paper we have defined the
following sequence spaces:
hc(m)
v,M,u,p=
x=(xk)w: lim
n→∞
n
X
k=1
1
n+k1
Mk
uk(m)
vxk
ρ
pk
exists,for some ρ > 0
,
h0(m)
v,M,u,p=
x=(xk)w: lim
n→∞
n
X
k=1
1
n+k1
Mk
uk(m)
vxk
ρ
pk
=0,for some ρ > 0
,
h(m)
v,M,u,p=
x=(xk)w: sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk
ρ
pk
<,for some ρ > 0
.
If we take Mk(x)=xfor all kNand (vk)=(1,1, ...),Mk(x)=xfor all kN, we obtain that hc(m)
v,u,p,
h0(m)
v,u,p,h(m)
v,u,pand hc(m),u,p,h0(m),u,p,h(m),u,p, respectively. Also if (uk)=(1) and
pk=(1),for all kN,we obtain hc(m)
v,M,h0(m)
v,Mand h(m)
v,M.
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9092
We define the sequence y=ynwhich will be frequently used, as the H(m)
v-transform of a sequence as
follows:
yn=H(m)
vx(M,u,p)
=
n
X
k=1
1
n+k1
Mk
ukPn
i=k(1)ikm
ikxkvk
ρ
pk
(2.1)
for each k,m,nN.
We will use the following inequality throughout the paper. If 0 <pksup pk=H,D=max 1,2H1,
then
|ak+bk|pkD|ak|pk+|bk|pk(2.2)
for all kand ak,bkC.
Theorem 2.2. Let M=(Mk)be a sequence of Orlicz functions, v=(vk)be any fixed sequence of non-zero
complex numbers. Also, let p=pkand u=(uk)be the bounded sequence and sequence of positive real
numbers, respectively. Then hc(m)
v,M,u,p,h0(m)
v,M,u,pand h(m)
v,M,u,pare linear spaces over
the complex field C.
Proof. We shall prove the assertion for h(m)
v,M,u,ponly and others can be proved similarly. Let
x=(xk),y=ykh(m)
v,M,u,pand α, β C. Then there exist positive numbers ρ1and ρ2such that
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk
ρ1
pk
<,
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vyk
ρ2
pk
<,
for some ρ1, ρ2>0.Let ρ3=max 2|α|ρ1,2βρ2. Since M=(Mk)is a non-decreasing and convex, using
(2.2) inequality, we have
supnPn
k=1
1
n+k1
Mk
uk(m)
vαxk+βyk
ρ3
pk
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vαxk
ρ3
+uk(m)
vβyk
ρ3
pk
Dsup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk
ρ1
pk
+Dsup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vyk
ρ2
pk
<.
Thus, it becomes αx+βyh(m)
v,M,u,p. This proves that h(m)
v,M,u,pis linear space. Similarly, it
can be proved that hc(m)
v,M,u,pand h0(m)
v,M,u,pare linear spaces.
Theorem 2.3. Let M=(Mk)be a sequence of Orlicz functions, v=(vk)be any fixed sequence of non-zero
complex numbers. Also, let p=pkand u=(uk)be the bounded sequence and sequence of positive real
numbers, respectively. Then h(m)
v,M,u,pis a paranorm space with the following paranorm,
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9093
1(x)=inf
ρ
pk
G: supnPn
k=1
1
n+k1"Mk uk(m)
vxk
ρ!#pk!1
G
1,for some ρ > 0
,
where 0 <pksup pk=Hand G=max (1,H).
Proof. (i) Clearly 1(x)0 for x=(xk)h(m)
v,M,u,p.Since Mk(0)=0,we get 1(θ)=0.
(ii) 1(x)=1(x)is trivial.
(iii) Let x,yh(m)
v,M,u,p. Then there exist pozitive numbers ρ1, ρ2such that
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk
ρ1
pk
1
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vyk
ρ2
pk
1
Let ρ=ρ1+ρ2. Then by using Minkowski’s inequality, we have
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk+yk
ρ
pk
=sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk+yk
ρ1+ρ2
pk
ρ1
ρ1+ρ2
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk
ρ1
pk
+ρ2
ρ1+ρ2
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vyk
ρ2
pk
1
and thus,
1x+y=inf
ρ
pk
G:
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk+yk
ρ
pk
1
G
1,for some ρ > 0
inf
ρ1
pk
G:
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk+yk
ρ1
pk
1
G
1,for some ρ > 0
+inf
ρ2
pk
G:
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
vxk+yk
ρ2
pk
1
G
1,for some ρ > 0
.
Therefore, 1x+y1(x)+1y.
Finally, we prove that the scalar multiplication is continuous. Let λbe any complex number. By
definition,
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9094
1(λx)=inf
ρ
pk
G:
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
v(λxk)
ρ
pk
1
G
1,for some ρ > 0
=inf
(|λ|t)
pk
G:
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
v(xk)
t
pk
1
G
1,for some ρ > 0
where t=ρ
|λ|>0. Since |λ|pkmax (1,|λ|sup pk), we have
1(λx)max (1,|λ|sup pk).inf
(t)
pk
G:
sup
n
n
X
k=1
1
n+k1
Mk
uk(m)
v(xk)
t
pk
1
G
1,for some ρ > 0
.
So the fact that scalar multiplication is continuous is due to the above inequality. This completes the proof
of the theorem.
Theorem 2.4. Let M=(Mk)be a sequence of Orlicz functions, u=(uk)be a sequence of positive real
numbers and v=(vk)be any fixed sequence of non-zero complex numbers. If for each k, 0 pkqk<,p=
pkand q=qkare bounded sequences of positive real numbers, then h0(m)
v,M,u,ph0(m)
v,M,u,q.
Proof. Let xh0(m)
v,M,u,p.Then
n
X
k=1
1
n+k1
Mk
uk(m)
v(xk)
ρ
pk
0 as n
This means that
1
n+k1
Mk
uk(m)
v(xk)
ρ
pk
<1
for large enough kvalues. Since Mkis increasing and pkqk, we have as n
n
X
k=1
1
n+k1
Mk
uk(m)
v(xk)
ρ
qk
n
X
k=1
1
n+k1
Mk
uk(m)
v(xk)
ρ
pk
0.
Thus xh0(m),M,u,q. This completes the proof.
Theorem 2.5. Let M=(Mk)be a sequence of Orlicz functions, v=(vk)be any fixed sequence of non-zero
complex numbers and φ=lim
t→∞
Mk(t)
t>0. Then h0(m)
v,M,u,ph0(m)
v,u,p.
Proof. Let φ > 0 to prove that h0(m)
v,M,u,ph0(m)
v,u,p.From the definition of φ, Mk(t)φ(t),
for all t>0.Since φ > 0,we have t1
φMk(t)for all t>0.
Let x=(xk)h0(m)
v,M,u,p. Thus, we have
n
X
k=1
1
n+k1
uk(m)
v(xk)
ρ
pk
n
X
k=1
1
n+k1
Mk
uk(m)
v(xk)
ρ
pk
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9095
which means that x=(xk)h0(m)
v,u,p.This completes the proof.
Theorem 2.6. Let M=M
kand M′′ =M′′
kbe sequences of Orlicz functions and v=(vk)be any fixed
sequence of non-zero complex numbers, then
h0(m)
v,M,u,ph0(m)
v,M′′ ,u,ph0(m)
v,M+M′′ ,u,p.
Proof. Let x=(xk)h0(m)
v,M,u,ph0(m)
v,M′′ ,u,p. Therefore,
n
X
k=1
1
n+k1
M
k
uk(m)(xk)
ρ
pk
as n
n
X
k=1
1
n+k1
M′′
k
uk(m)
v(xk)
ρ
pk
as n .
Then, we have
Pn
k=1
1
n+k1
M
k+M′′
k
uk(m)
v(xk)
ρ
pk
K
n
X
k=1
1
n+k1
M
k
uk(m)
v(xk)
ρ
pk
+K
n
X
k=1
1
n+k1
M′′
k
uk(m)
v(xk)
ρ
pk
0 as n .
Thus,
n
X
k=1
1
n+k1
M
k+M′′
k
uk(m)
v(xk)
ρ
pk
0 as n .
Therefore, x=(xk)h0(m)
v,M+M′′ ,u,pand this completes the proof.
Theorem 2.7. Let M=M
kand M′′ =M′′
kbe sequences of Orlicz functions and v=(vk)be any fixed
sequence of non-zero complex numbers, then
h0(m)
v,M,u,ph0(m)
v,M M′′ ,u,p
Proof. Let x=(xk)h0(m)
v,M,u,p.Then we have
lim
n→∞
n
X
k=1
1
n+k1
M
k
uk(m)
v(xk)
ρ
pk
=0.
Let ε > 0 and choose δ > 0 with 0 <δ<1 such that Mk(t)< ε, for 0 tδ.
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9096
Write yk=1
n+k1
M
k
uk(m)
v(xk)
ρ
and consider
n
X
k=1Mkykpk=X
1Mkykpk+X
2Mkykpk
where the first summation is over ykδand the second summation is over yk> δ. Since Mkis continuos,
we have
X
1Mkykpk< εH(2.3)
and for yk> δ, we use the fact that
yk<yk
δ1+yk
δ.
From the definition, we have for yk> δ
Mkyk<2Mk(1)yk
δ.
Hence,
X
1Mkykpkmax 1,2Mk(1)δ1HX
1ykpk(2.4)
From the equation (2.3) and (2.4), we have
h0(m)
v,M,u,ph0(m)
v,M M′′ ,u,p.
Theorem 2.8. Hilbert sequence spaces hc(m)
v,M,u,p,h0(m)
v,M,u,pand h(m)
v,M,u,pare iso-
metrically isomorphic to the space c,c0and lrespectively, that is, hc(m)
v,M,u,pc,h0(m)
v,M,u,p
c0and h(m)
v,M,u,pl.
Proof. We’ll just do the proof for h0(m)
v,M,u,pc0for the others it can be done similarly. To
demonstrate the theorem, we must show that there is linear bijection between the space h0(m)
v,M,u,p
and c0. For this, we consider the transformation Tdefined by the notation (2.1), from h0(m)
v,M,u,pto c0
by xy=Tx. The linearity of Tis obvious. Moreover, when Tx =θit is trivial that x=θ=(0,0,0...)and
hence Tis injective. Next, let y=ync0and the sequence x=(xn)is defined as follows:
xn=v1
n
n
X
k=1
n
X
k=1 m+ni1
ik!h1
ik
yk
where h1
ik is defined by (2.1). Then,
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9097
lim
n→∞ H(m)
vxM,u,p
n=lim
n→∞
n
X
k=1
1
n+k1
Mk
uk(m)
v(xk)
ρ
pk
=lim
n→∞
n
X
k=1
1
n+k1
Mk
ukPm
i=0(1)im
ixkivki
ρ
pk
=lim
n→∞
n
X
k=1
1
n+k1
Mk
ukPn
i=k(1)ikm
ikxkvk
ρ
pk
=lim
n→∞yn=0.
Thus, xh0(m)
v,M,u,p. As a result, it is clear that Tis surjective. Since it is linear bijection,
h0(m)
v,M,u,pand c0are linear isomorphic. This completes the proof.
Remark 2.9. It is well known that the spaces c,c0and lare BK-spaces. Let us considering the fact
that (m)
vis a triangle, we can say that the Hilbert sequence spaces hc(m)
v,M,u,p,h0(m)
v,M,u,pand
h(m)
v,M,u,pare BK-spaces with the norm defined by
xM,u,p
=
H(m)
vx
M,u,p
=sup
n
n
X
k=1
1
n+k1
Mk
ukPm
i=0(1)im
ixkivki
ρ
pk
(2.5)
Corollary 2.10. Define the space d(k)=d(k)
n(m)
v,M,u,pnN
d(k)
n(m)
v,M,u,p=
Pn
k=1
Mk
ukPn
i=k
m+ni1
ni
h1
ik vk
ρ
pk
,nk
0,n<k
for every fixed kN.The following statements hold:
(i) The sequence d(k)
n(m)
v,M,u,pis a basis for the space h0(m)
v,M,u,pand every xh0(m)
v,M,u,p
has a unique representation of the form
x=X
kH(m)
vxM,u,p
kd(k)
(ii) The set nt,d(1),d(2), ...ois a basis or the space hc(m)
v,M,u,pand every xhc(m)
v,M,u,phas a
unique representation of the form
x=st +X
kH(m)
vxM,u,p
ksd(k)
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9098
where t=tn(m)
v,M,u,p=Pn
k=1
Mk
ukPn
i=k
m+ni1
ni
h1
ik vk
ρ
pk
for all kNand s=lim
k→∞ H(m)
vxM,u,p
k.
Corollary 2.11. The Hilbert sequence spaces h0(m)
v,M,u,pand hc(m)
v,M,u,pare separable.
3. Characterizations of Matrix Transformation and α,βand γduals
Let A=(ank)be an infinite matrix of complex numbers, Xand Ybe subsets of the sequence space w.
Let x=(xk)and y=ykbe two sequences. Thus, we can write xy =xkyk,x1Y={aw:ax Y}
and M(X,Y)=xX1Y={aw:ax Y,for all xX}for the multiplier space of Xand Y. In the special
cases of Y={l1,cs,bs},we write xα=x1l1,xβ=x1cs,xγ=x1bs and Xα=M(X,l1),Xβ=M(X,cs),
Xγ=M(X,bs)for the αdual, βdual, γdual of X.By An=(ank)we denote the sequence in the nth row of
Aand write An(x)=P
k=1ankxknNand A(x)=(An(x)) ,provided Anxβfor all n.
We shall begin with the lemmas due to Stieglitz ve Tietz [15] which will be used in the computation of
the βand γduals of the Hilbert sequence spaces.
Lemma 3.1. [16] Let X,Ybe any two sequence spaces. A(X:YT)if and only if TA (X:Y),where A
is an infinite matrix and Tis a triangle matrix.
Lemma 3.2. (i) Let An=(ank)be an infinite matrix. Then A(c0:l)if and only if
sup
nX
k
|ank|<(3.1)
(ii) A(c0:c)if and only if (3.1) holds with
lim
nank exists for all k.(3.2)
(iii) A(c0:bs)if and only if
sup
nX
k
m
X
n=0
ank
<.(3.3)
(iv) A(c0:cs)if and only if (3.3) holds with
X
k
ank convergent for all k.(3.4)
Lemma 3.3. (i) Let An=(ank )be an infinite matrix. Then A(c:c)if and only if (3.1) and (3.2) hold with
lim
n→∞ X
k
ank exists.
(ii) A(l:c)if and only if (3.2) holds with
lim
nX
k
|ank|=X
klim
n→∞ ank .(3.5)
Lemma 3.4.(i) Let An=(ank)be an infinite matrix. Then A(c:cs)if and only if (3.3), (3,4) hold and
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9099
X
nX
k
ank convergent. (3.6)
(ii) A(l:cs)if and only if (3.2) holds and
lim
mX
k
X
n=m
ank
=0.(3.7)
Lemma 3.5. [17] Let U=(unk )be an infinite matrix of complex numbers for all n,kN. Let BU=(bnk )
be defined via a sequence a=(ak)wand inverse of the triangle matrix U=(unk)by
bnk =
n
X
j=k
ajujk
for all n,kN.Then,
Xα
U=na=(ak)w:BU(X:l1)o,
Xβ
U=na=(ak)w:BU(X:c)o
Xγ
U=na=(ak)w:BU(X:l)o.
Theorem 3.6. The α, βand γ-duals of the Hilbert sequence spaces defined as
hh0(m)
v,M,u,piα={a=(ak)w:W(c0:l1)},
hhc(m)
v,M,u,piα={a=(ak)w:W(c:l1)},
hh(m)
v,M,u,piα={a=(ak)w:W(l:l1)},
hh0(m)
v,M,u,piβ={a=(ak)w:W(c0:c)},
hhc(m)
v,M,u,piβ={a=(ak)w:W(c:c)},
hh(m)
v,M,u,piβ={a=(ak)w:W(l:c)},
hh0(m)
v,M,u,piγ={a=(ak)w:W(c0:l)},
hhc(m)
v,M,u,piγ={a=(ak)w:W(c:l)},
hh(m)
v,M,u,piγ={a=(ak)w:W(l:l)}.
Proof. We shall only compute the α, βand γ-duals of h0(m)
v,M,u,psequence space. Let h1
nis
defined by (2.1). Let us take any a=(ak)w. We define the matrix W=(wnk)by
wnk =
n
X
k=1
Mk
ukPn
i=k m+ni1
ni!h1
ik anv1
n
ρ
pk
.
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9100
Consider the equation
n
X
k=1
akxk=
n
X
k=1akv1
k(vkxk)
=
n
X
k=1
Mk
ukPk
i=1(Pk
j=i m+kj1
kj!h1
ij )akv1
kyi
ρ
pk
=
n
X
k=1
Mk
ukPk
i=1(Pk
j=i m+kj1
kj!h1
ij aiv1
i)yk
ρ
pk
=Wyn.(3.8)
Using (3.8), we have ax =(akxk)cs or bs whenever x=(xk)h0(m)
v,M,u,pif and only if Wy l1,cor l
whenever y=ykc0.Then, from Lemma 3.1 and Lemma 3.5, we obtain that a=(ak)hh0(m)
v,M,u,piα,
a=(ak)hh0(m)
v,M,u,piβor a=(ak)hh0(m)
v,M,u,piγif and only if W(c0:l1),W(c0:c)or
W(c0:l),which is required result.
Therefore, the α, βand γ-duals of Hilbert sequence spaces will be helpful in the characterization of
matrix transformations. Let Xand Ybe arbitrary subsets of w. We will show that the characterization of
the classes (X:YT)ve (XT:Y)can be reduced to (X,Y), where Tis a triangle. Since if the sequence spaces
h0(m)
v,M,u,pand c0are linearly isomorphic, then the equivalence class xh0(m)
v,M,u,pyc0
holds. So using Lemma 3.1 and 3.5, we get the desired result.
Theorem 3.7. Let us consider the infinite matrices A=(ank )and B=(bnk ). These matrices get associated
with each other by the relations:
bnk =
n
X
k=1
Mk
ukP
j=k m+nj1
nj!h1
jk anj v1
k
ρ
pk
(3.9)
for all k,m,nN.Then the following statements are true:
(i) Ah0(m)
v,M,u,p:Yif and only if {ank}kNhh0(m)
v,M,u,piβfor all nNand B(c0,Y),where
Ybe any sequence space;
(ii) Ahc(m)
v,M,u,p:Yif and only if {ank}kNhhc(m)
v,M,u,piβfor all nNand B(c,Y),
where Ybe any sequence space;
(iii) Ah(m)
v,M,u,p:Yif and only if {ank}kNhh(m)
v,M,u,piβfor all nNand B(l,Y),
where Ybe any sequence space.
Proof. We suppose that the relation in (3.9) holds between A=(ank)and B=(bnk).The spaces
h0(m)
v,M,u,pand c0are linearly isomorphic. Let Ah0(m)
v,M,u,p:Yand y=ykc0.Then
BH(m)
vexists and (ank)hh0(m)
v,M,u,piβfor all kN, it means that (bnk)c0for all k,nN. Hence, By
exists for each yc0.Thus, if we take m in the equality,
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9101
m
X
k=1
ankxk=
m
X
k=1
Mk
uk"Pk
i=1Pk
j=i m+kj1
kj!h1
ij #ank v1
k
ρ
pk
=X
k
bnk yk
for all m,nNwhich conclude that B(c0,Y).On the contrary, let (ank)kNhh0(m)
v,M,u,piβfor each
kNand B(c0,Y)and x=(xk)h0(m)
v,M,u,p. Then it is clear that Ax exists. Thus, we attain from the
following equality for all nN
X
k
bnk yk=X
k
ankxk
as m that Ax =By and it is easy to show that Ah0(m)
v,M,u,p:Y. This completes the proof.
Theorem 3.8. Let us assume that components of the infinite matrices A=(ank )and E=(enk)are
connected with the following relation
enk =
n
X
k=1
n
X
j=k
1
n+j1
Mk
ukPn
j=k(1)jkm
jkajkv1
k
ρ
pk
(3.10)
for all m,nNand Xbe any given sequence space. Then the following statements are true:
(i) A=(ank)X:h0(m)
v,M,u,pif and only if E(X:c0);
(ii) A=(ank)X:hc(m)
v,M,u,pif and only if E(X:c);
(iii) A=(ank)X:h(m)
v,M,u,pif and only if E(X:l).
Proof. Let us suppose that z=(zk)X. Using the relation (3.10), we have
m
X
k=1
enkzk=
m
X
k=1
n
X
k=1
n
X
j=k
1
n+j1
Mk
ukhPn
j=k(1)jkm
jkajkv1
kizk
ρ
pk
(3.11)
for all m,nN.Then, for m equation (3.11) gives us that (Ez)n=nH(m)
v(Az)on. Thus, we can obtain
that Az h0(m)
v,M,u,pif and only if Ez c0. This completes the proof.
Now, we give some conditions:
lim
kank =0 for all n,(3.12)
lim
n→∞ X
k
|ank|=0,(3.13)
lim
n→∞ X
kank an,k+1=0,(3.14)
sup
nX
kank an,k+1<,(3.15)
D. Barlak, C¸. A. Bektas¸ /Filomat 37:27 (2023), 9089–9102 9102
lim
kank an,k+1exists for all k,(3.16)
lim
n→∞ X
kank an,k+1=X
klim
n→∞ ank an,k+1,(3.17)
sup
nlim
n→∞ank<,(3.18)
Corollary 3.9. Let A=(ank )be an infinite matrix and X=h0(m)
v,M,u,p,Y=hc(m)
v,M,u,pand
Z=h(m)
v,M,u,p.Then, the following statements hold:
(a)A=(ank)(X,l)if and only if (3.1) holds with bnk instead of ank;
(b)A=(ank)(X,bs)if and only if (3.3) holds with bnk instead of ank;
(c)A=(ank)(Y,cs)if and only if(3.3), (3.4) and (3.6) hold with bnk instead of ank;
(d)A=(ank)(Z,c)if and only if (3.2) and (3.5) hold with bnk instead of ank;
(e)A=(ank)(Z,cs)if and only if (3.7) holds with bnk instead of ank;
fA=(ank)(X,c)if and only if (3.1) and (3.2) hold with bnk instead of ank;
1A=(ank)(X,cs)if and only if (3.3) and (3.4) holds with bnk instead of ank.
Corollary 3.10. Let A=(ank)be an infinite matrix and X=h0(m)
v,M,u,p,Y=hc(m)
v,M,u,pand
Z=h(m)
v,M,u,p.Then, the following statements hold:
(a)A=(ank)(l,X)if and only if (3.13) holds with enk instead of ank ;
(b)A=(ank)(bs,X)if and only if (3.12) and (3.14) hold with enk instead of ank ;
(c)A=(ank)(bs,Y)if and only if (3.12), (3.16) and (3.17) hold with enk instead of ank ;
(d)A=(ank)(cs,Y)if and only if (3.15) and (3.2) hold with enk instead of ank ;
(e)A=(ank)(bs,Z)if and only if (3.12) and (3.15) hold with enk instead of ank ;
fA=(ank)(cs,Z)if and only if (3.15) and (3.18) hold with enk instead of ank ;
1A=(ank)(cs,X)if and only if (3.2) and (3.15) hold with enk instead of ank .
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