R E S E A R C H
Open Access
The binomial sequence spaces of
nonabsolute type
Mustafa Cemil Bi¸sgin
**Correspondence:
[email protected] Faculty Of Arts And Sciences, Department Of Mathematics, Recep Tayyip Erdo ˘gan University, Zihni Derin Campus, Rize, 53100, Turkey
Abstract
In this paper, we introduce the binomial sequence spacesbr0,sandbr,s
c of nonabsolute
type which include the spacesc0andc, respectively. Also, we prove that the spaces
br0,sandbr,s
c are linearly isomorphic to the spacesc0andc, in turn, and we investigate some inclusion relations. Moreover, we obtain the Schauder bases of those spaces and determine their
α
-,β
-, andγ
-duals. Finally, we characterize some matrix classes related to those spaces.MSC: 40C05; 40H05; 46B45
Keywords: matrix transformations; matrix domain; Schauder basis;
α
-,β
- andγ
-duals; matrix classes1 The basic definitions and notations
Letwbe the set of all real (or complex) valued sequences. Thenwbecomes a vector space under point-wise addition and scalar multiplication. A sequence space is a vector sub-space ofw. We use the notations of∞,c,c, andp for the spaces of all bounded, null,
convergent, and absolutelyp-summable sequences, respectively, where ≤p<∞. A Banach sequence space is called aBK-space provided each of the mapspn:X−→C
defined bypn(x) =xnis continuous for alln∈N[]. By taking into account the definition
above, one can say that the sequence spaces∞,c, andcareBK-spaces with their usual sup-norm defined byx∞=supk∈N|xk|andpis aBK-space with itsp-norm defined by
xp= ∞
k= |xk|p
p ,
wherep∈[,∞).
LetA= (ank) be an infinite matrix with complex entries andx∈w, then theA-transform
ofxis defined by
(Ax)n=
∞
k=
ankxk (.)
and is assumed to be convergent for alln∈N[]. For brevity in the notation, we henceforth prefer that the summation without limits runs from to∞.
LetXandYbe two arbitrary sequence spaces andA= (ank) be an infinite matrix. Then
the domain ofAis denoted byXAand defined by
XA=
x= (xk)∈w:Ax∈X
, (.)
which is also a sequence space and the class of all matricesAsuch thatX⊂YAis denoted
by (X:Y). Moreover,A= (ank) is called conservative ifc⊂cA. Furthermore,A= (ank)
is calledm-multiplicative, iflimn→∞(Ax)n=mlimn→∞xnfor allx∈cand the class of all
m-multiplicative matrices is denoted by (X:Y)m. Specially,A= (ank) is called regular, if
A= (ank) is -multiplicative.
The spaces of all bounded and convergent series are denoted bybsandcsand are defined by aid of the matrix domain of the summation matrixS= (snk) such thatbs= (∞)Sand
cs=cS, respectively, whereS= (snk) is defined by
snk=
⎧ ⎨ ⎩
, ≤k≤n,
, k>n,
for allk,n∈N. A matrixA= (ank) is said to be a triangle ifank= fork>nandann= for
alln,k∈N. Furthermore, a triangle matrix uniquely has an inverse, which is also a triangle matrix.
The theory of matrix transformations is of great importance in the summability which was obtained by Cesàro, Borel, Riesz and others. Therefore, many authors have defined new sequence spaces by using this theory. For example, (∞)Nq andcNq in [],XpandX∞ in [],c˜andc˜ in [],ar andarc in []. Moreover, many authors have constructed new
sequence spaces by using especially the Euler matrix. For instance,erander
cin [],erpand
er
∞ in [] and [],er(),erc(), ander∞() in [],er((m)),erc((m)) ander∞((m)) in [],
er
(B(m)),erc(B(m)), ander∞(B(m)) in [],er(,p),erc(,p), ander∞(,p) in [],er(u,p) and er
c(u,p) in [].
In this paper, we introduce the binomial sequence spaces br,sandbr,s
c of nonabsolute
type which include the spacescandc, respectively. Also, we prove that the spacesbr,sand br,s
c are linearly isomorphic to the spacescandc, in turn and investigate some inclusion
relations. Moreover, we obtain the Schauder basis of those spaces and determine theirα-,
β-, andγ-duals. Finally, we characterize some matrix classes related to those spaces.
2 The binomial sequence spaces of nonabsolute type
In this chapter, we introduce the binomial sequence spacesbr,s andbr,s
c of nonabsolute
type and prove that the spacesbr,sandbr,s
c are linearly isomorphic to the spacescandc, respectively. Moreover, we deal with an inclusion relation related to those spaces.
Letr,s∈Randr+s= . Then the binomial matrixBr,s= (br,s
nk) is defined by
brnk,s= ⎧ ⎨ ⎩
(s+r)n
n
k
sn–krk, ≤k≤n,
, k>n,
(i) supn∈Nnk=|(s+r)n n
k
sn–krk|<∞,
(ii) limn→∞(s+r)n n
k
sn–krk= ,
(iii) limn→∞nk= (s+r)n
n
k
sn–krk= .
Therefore, the binomial matrixBr,s= (brnk,s) is regular forsr> . Unless stated otherwise, we assume thatsr> .
Here, we would like to emphasize that if we taker+s= , we obtain the Euler matrix Er= (er
nk). So the binomial matrixBr,s= (b r,s
nk) generalizes the Euler matrixEr= (ernk).
By considering the definition of the binomial matrixBr,s= (br,s
nk), we define the binomial
sequence spacesbr,sandbr,s
c as follows:
br,s=
x= (xk)∈w: lim n→∞
(s+r)n
n
k=
n k
sn–krkxk=
and
brc,s=
x= (xk)∈w: lim n→∞
(s+r)n
n
k=
n k
sn–krkxkexists
.
The sequence spacesbr,sandbr,s
c can be redefined by using the notion of (.) as follows:
br,s= (c)Br,s and br,s
c =cBr,s. (.)
It is clear thatbr,s⊂br,s
c . Letx= (xk) be an arbitrary sequence. Then theBr,s-transform of
x= (xk) is defined by
Br,sxk=yk=
(s+r)k
k
j=
k j
sk–jrjxj (.)
for allk∈N.
Now, we want to start with the following theorem related to the theory ofBK-spaces, which is of great importance in the characterization of matrix transformations between sequence spaces.
Theorem . The binomial sequence spaces br,s and br,s
c are BK -spaces with their
sup-norms defined by
xbr,s=xbrc,s=B
r,sx
∞=sup
n∈N
Br,sxn.
Proof The sequence spacescandcareBK-spaces according to theirsup-norms. More-over, the binomial matrixBr,s= (br,s
nk) is a triangle matrix and (.) holds. By combining
these three facts and Theorem .. of Wilansky [], we deduce that the binomial se-quence spacesbr,sandbr,s
c areBK-spaces. This completes the proof.
Let|x|= (|xk|) for allk∈N. Because ofxbr,s=|x|br,sandxbcr,s=|x|brc,sfor at least one sequence in the binomial sequence spacesbr,sandbrc,s,b
r,s
andbrc,sare sequence spaces
Theorem . The binomial sequence spaces br,sand br,s
c are linearly isomorphic to the
sequence spaces cand c,respectively.
Proof Because a repetition of similar statements is redundant, the proof of the theorem is given for only the spacebr,s
c . For this purpose, we should show the existence of a linear
bijection between the spacesbrc,sandc. Let us consider the transformationLdefined by L:brc,s−→c,L(x) =Br,sx. Then it is obvious that, for allx∈brc,s,L(x) =Br,sx∈c. Moreover, it is clear thatLis a linear transformation andx= wheneverL(x) = . Because of this,L is injective.
Lety= (yk)∈cbe given. We define a sequencex= (xk) by
xk=
rk
k
j=
k j
(–s)k–j(s+r)jyj
for allk∈N. Then we obtain
Br,sxn= (s+r)n
n
k=
n k
sn–krkxk
=
(s+r)n n
k=
n k
sn–k
k
j=
k j
(–s)k–j(s+r)jyj
=yn
for alln∈N. So,Br,sx=yand sincey∈c, we conclude thatBr,sx∈c. Hence, we deduce
thatx∈br,s
c andL(x) =y. On account of thisLis surjective.
Moreover, we have for allx∈br,s c
L(x)
∞=Br,sx∞=xbrc,s.
SoLis norm preserving. Consequently,Lis a linear bijection. Then we obtain the fact that the spacesbr,s
c andcare linearly isomorphic, that is,brc,s∼=c. This completes the proof.
Theorem . The inclusions er
⊂b
r,s
and erc⊂bcr,sstrictly hold,where erand ercare Euler
sequence spaces of nonabsolute type.
Proof Ifr+s= , we obtainEr=Br,s. So, the inclusioner
⊂br,sholds. Assume that <r< ands= . Now, we define a sequencex= (xk) such thatxk= (–r)kfor allk∈N. Then it is
obvious thatx= ((–r)k) /∈candErx= ((– –r)k) /∈c. On the other hand,Br,sx= ((+r)k)∈ c. As a consequence,x= (xk)∈br,s\er.
This shows that the inclusioner
⊂br,sstrictly holds. Another part of the theorem can be proved in a similar way. This completes the proof.
Theorem . The inclusions c⊂br,sand c⊂brc,sstrictly hold.But the sequence spaces
br,sand∞do not include each other.
Proof If we consider regularity of the binomial matrixBr,s, we can easily conclude that Br,sx∈cwheneverx∈c. This means thatx∈br,s
for allx∈c, namelyc⊂b
r,s
we define a sequenceu= (uk) such thatuk= (–)kfor allk∈N. Then we obtainBr,sx=
((s–r s+r)
k)∈c
. As a consequence,uis inbr,sbut not inc. So, the inclusionc⊂br,sis strict. By using a similar way, one can show that the inclusionc⊂br,s
c is strict.
To prove the second part of the theorem, we consider the sequencese= (, , , . . .) and v= (vk) defined byvk= (–sr)k for allk∈N, where|sr|> . Then we obtainBr,se=eand
Br,sv= (, , , . . .). Hence,eis in
∞but not inbr,sandvis inbr,sbut not in∞. This shows that the sequence spacesbr,sand∞overlap but these spaces do not include each other.
This completes the proof.
Definition .(see []) An infinite matrixA= (ank) is called coregular, if A= (ank) is
conservative andχ(A) =limn→∞kank–klimn→∞ank= .
By taking into account the regularity of the binomial matrix Br,s= (brnk,s), we obtain
χ(Br,s) = = . So, the binomial matrixBr,s= (br,s
nk) is coregular.
Definition .(see []) LetA= (ank) be an infinite matrix with bounded columns. Then
Ais defined to be of typeMiftA= impliest= for everyt∈.
Definition .(see []) For a conservative triangleA= (ank),c⊂cA. Its closurec¯incAis
called the perfect part ofcA. Ifcis dense,A= (ank) is called perfect.
Now we give the following two theorems, which are needed.
Theorem .(see []) A regular triangle A= (ank)is of type M if there exists a(zi)∈∞
with zi=zj(i=j)and(zi)∞< such that
∀i∈N,∃(xki)k∈∞, ∀n∈N:z
n i =
n
k= ankxki.
Theorem .(see []) A coregular triangle is perfect if and only if it is of type M.
Theorem . Each regular binomial matrix Br,s= (brnk,s)is perfect. Proof We know that the regular binomial matrixBr,s= (br,s
nk) is coregular. So, for the proof,
we should show thatBr,s= (br,s
nk) is of typeM.
LetDr,s= (dr,s
nk) be the inverse ofBr,s= (b r,s
nk). Then we have for everyz∈C
uk(z) = k
v=
drkv,szv= rk
k
v=
k v
(–s)k–v(s+r)vzv
= rk
(s+r)z–sk.
One can easily verify thatsupk∈Nsup{|uk(z)|:|(s+r)z–s|<|r|} ≤ and that n
k=
brnk,suk(z) = n
k= brnk,s
k
v=
for all z∈C. Therefore, if we choose a sequence (zi) withzi=zj(i=j),(zi)∞< and
|(s+r)zi–s|<|r|(i∈N), thenxki=uk(zi) (i,k∈N) fit Theorem .. Thus, the regular binomial matrixBr,s= (brnk,s) is perfect. This completes the proof.
3 The Schauder basis and
α
-,β
-,γ
- and continuous dualsIn this chapter, we construct the Schauder basis and designate theα-,β-,γ-, and contin-uous duals of the binomial sequence spacesbr,sandbr,s
c .
Given a normed space (X, · X), a set{xk:xk∈X,k∈N}is called a Schauder basis for
Xif for allx∈Xthere exist unique scalarsλk,k∈N, such thatx=
kλkxk;i.e.,
x–
n
k=
λkxk
X
−→
asn→ ∞[].
Theorem . Let μk={Br,sx}k for all k∈Nand l=limk→∞μk. We define a sequence
g(k)(r,s) ={g(k)
n (r,s)}n∈Nas follows:
gn(k)(r,s) = ⎧ ⎨ ⎩
, ≤n<k,
rn n
k
(–s)n–k(s+r)k, n≥k,
for all fixed k∈N.Then the following statements hold. (a) The sequence{g(k)(r,s)}
k∈Nis a Schauder basis for the binomial sequence spacebr,s,
and everyx∈br,shas a unique representation of the form
x=
k
μkg(k)(r,s).
(b) The set{e,g()(r,s),g()(r,s), . . .}is a Schauder basis for the binomial sequence space br,s
c ,and anyx∈brc,shas a unique representation of the form
x=le+
k
[μk–l]g(k)(r,s).
Proof (a) Obviously we have
Br,sg(k)(r,s) =e(k)∈c, k∈N,
where e(k) is a sequence with in thekth place and zeros elsewhere. So, the inclusion {g(k)(r,s)} ⊂br,s
holds.
Given a sequencex= (xk)∈br,sandm∈N, we define
x[m]=
m
k=
μkg(k)(r,s).
Then, if we apply the binomial matrixBr,s= (br,s
nk) tox[m], we have
Br,sx[m]=
m
k=
μkBr,sg(k)(r,s) = m
k=
and
Br,sx–x[m]n= ⎧ ⎨ ⎩
, ≤n≤m,
(Br,sx)
n, n>m,
for allm,n∈N.
Now, let > be arbitrarily given. Then we may choose am∈Nsuch that
Br,sxm<
for allm≥m. Therefore,
x–x[m]br,s
=supn≥m
Br,sxn< sup
n≥m
Br,sxn≤ <
for allm≥m. This shows that
x=
k
μkg(k)(r,s).
To complete the proof of part (a), we should show the uniqueness of this representation. We assume that there exists a representation
x=
k
λkg(k)(r,s).
Due to the transformation,Ldefined in the proof of Theorem . is continuous; we can write
Br,sxn=
k
λk
Br,sg(k)(r,s)n=
k
λke(nk)=λn
for alln∈N, which contradicts the fact that (Br,sx)
n=μnfor alln∈N. Hence, everyx∈br,s has a unique representation, as desired.
(b) We know that{g(k)(r,s)} ⊂br,sandBr,se=e∈c. So, the inclusion{e,g(k)(r,s)} ⊂brc,s trivially holds.
For a given arbitrary sequencex= (xk)∈brc,s, we define a sequencey= (yk) such that
y=x–le, wherel=limk→∞μk. Then it is obvious thaty= (yk)∈br,s. By considering the part (a), one can say thaty= (yk) has a unique representation. This implies thatx= (xk)
has a unique representation, as desired in part (b). This completes the proof.
By taking into account the results of Theorems . and ., we give the following result.
Corollary . The binomial sequence spaces br,sand br,s
c are separable.
LetXandYbe two arbitrary sequence spaces. The multiplier space ofXandYis sym-bolized withM(X,Y) and defined by
M(X,Y) =y= (yk)∈w:xy= (xkyk)∈Yfor allx= (xk)∈X
By considering the definition ofM(X,Y), theα-,β-, andγ-duals of a sequence space X are defined by
Xα=M(X,), Xβ=M(X,cs) and Xγ=M(X,bs),
respectively.
ByX∗, we denote the space of all bounded linear functionals onX.X∗is called the con-tinuous dual of a normed spaceX.
Let us give some properties to use in the next lemma:
sup
K∈F
n
k∈K
ank
p<∞, (.)
sup
n∈N
k
|ank|<∞, (.)
lim
n→∞ank=αk for eachk∈N, (.)
lim
n→∞
k
ank=α, (.)
whereFis the collection of all finite subsets ofNand ≤p<∞.
Lemma .(see []) Let A= (ank)be an infinite matrix.Then the following statements
hold:
(i) A= (ank)∈(c:) = (c:)⇔(.)holds withp= ;
(ii) A= (ank)∈(c:c)⇔(.)and(.)hold;
(iii) A= (ank)∈(c:c)⇔(.), (.)and(.)hold;
(iv) A= (ank)∈(c:∞) = (c:∞)⇔(.)holds;
(v) A= (ank)∈(c:p)⇔(.)holds with≤p<∞.
Theorem . Theα-dual of the binomial sequence spaces br,sand br,s c is
vr,s=
a= (ak)∈w:sup K∈F
n
k∈K
n k
(–s)n–kr–n(r+s)kan
<∞
.
Proof Let us consider the sequencex= (xn) that is defined in the proof of Theorem ..
Then, for givena= (an)∈w, we obtain
anxn= n
k=
n k
(–s)n–kr–n(r+s)kanyk=
Ur,syn
for alln∈N. By considering the equality above, we deduce thatax= (anxn)∈whenever x= (xk)∈br,sorbrc,sif and only ifUr,sy∈whenevery= (yk)∈corc. This shows that a= (an)∈ {br,s}α={bcr,s}α if and only if Ur,s∈(c:) = (c:). If we combine this and Lemma .(i), we obtain
a= (an)∈
br,sα=brc,sα ⇔ sup
K∈F
n
k∈K
n k
(–s)n–kr–n(r+s)kan
<∞.
Therefore,{br,s}α={br,s c }α=v
r,s
Theorem . Define the sets vr,s,vr,s,and vr,sby
vr,s=
a= (ak)∈w:sup n∈N
n k= n
j=k
j k
(–s)j–kr–j(r+s)kaj
<∞
,
vr,s=
a= (ak)∈w:
∞
j=k
j k
(–s)j–kr–j(r+s)kajexists for each k∈N
,
vr,s=
a= (ak)∈w: lim n→∞ n k= n
j=k
j k
(–s)j–kr–j(r+s)kajexists
.
Then the following statements hold. (I) {br,s}β=vr,s
∩vr,s, (II) {br,s
c }β=vr,s∩vr,s∩vr,s, (III) {br,s}γ ={br,s
c }γ=v r,s
.
Proof Leta= (an)∈w. If we bear in mind the sequencex= (xk) that is defined in proof of
Theorem ., we have
n
k= akxk=
n k= rk k j= k j
(–s)k–j(r+s)jyj
ak = n k= n
j=k
j k
(–s)j–kr–j(r+s)kaj
yk
=Hr,syn
for alln∈N, where the matrixHr,s= (hr,s
nk) is defined by
hrnk,s= ⎧ ⎨ ⎩
n j=k
j k
(–s)j–kr–j(r+s)ka
j, ≤k≤n,
, k>n,
for allk,n∈N. Then:
(I)ax= (akxk)∈cswheneverx= (xk)∈br,sif and only ifHr,sy∈cwhenevery= (yk)∈
c. This shows thata= (ak)∈ {br,s}β if and only ifHr,s∈(c:c). If we combine this and Lemma .(ii), we conclude that
sup
n∈N
n
k=
hrnk,s<∞, (.)
lim
n→∞h
r,s
nk exists for eachk∈N. (.)
These results show that{br,s}β=vr,s
∩v
r,s
.
(II) In a similar way, we obtaina= (ak)∈ {brc,s}βif and only ifHr,s∈(c:c). If we combine
this and Lemma .(iii), we deduce that (.), (.) hold and
lim
n→∞ n
k=
hrnk,s exists, (.)
which shows that{br,s c }β=v
r,s
∩v
r,s
∩v
r,s
(III)ax= (akxk)∈bswheneverx= (xk)∈br,sorbrc,sif and only ifHr,sy∈∞ whenever
y= (yk)∈corc. This means thata= (ak)∈ {br,s}γ={brc,s}γ if and only ifHr,s∈(c:∞) = (c:∞). By combining this and Lemma .(iv), we deduce that (.) holds. Hence,{br,s}γ= {br,s
c }γ=vr,s. This completes the proof.
Theorem . {br,s}∗and{br,s
c }∗are equivalent to.
Proof To avoid a repetition of similar statements, the proof of the theorem is given for only the binomial sequence spacebr,s
c . For the proof, the existence of a linear surjective
norm preserving mappingL:{brc,s}∗−→should be shown. Suppose thatf∈ {br,s
c }∗. Now from Theorem .(b) we know that{e,g()(r,s),g()(r,s), . . .}
is a basis forbr,s
c , and eachx∈brc,shas a unique representation of the form
x=le+
k
[μk–l]g(k)(r,s).
By the linearity and continuity off, we get
f(x) =lf(e) +
k
[μk–l]f
g(k)(r,s) (.)
for allx∈brc,s. Now define a sequencex= (xk)∈brc,ssuch thatxbrc,s= as follows:
xk=
⎧ ⎨ ⎩
rk k
j= k
j
(–s)k–j(r+s)jsgnf(g(j)(r,s)), ≤k≤n,
rk n
j= k
j
(–s)k–j(r+s)jsgnf(g(j)(r,s)), k>n,
for allk,n∈N. Hence,
f(x)=
n
k=
fg(k)(r,s)≤ f (.)
since|f(x)| ≤ f · xonbr,s
c . It follows from (.) that
k
fg(k)(r,s)=sup
n∈N
n
k=
fg(k)(r,s)≤ f.
Now we write (.) as
f(x) =al+
k
akμk,
where
a=f(e) –
k
fg(k)(r,s), ak=f
g(k)(r,s)
By taking into account|limk→∞(Br,sx)k| ≤ xbrc,s= , we get f(x)≤ xbr,s
c
|a|+
k
|ak|
whence
f(x)≤ |a|+
k
|ak| (.)
and|f(x)| ≤ f, so we define for alln≥,
xk=
⎧ ⎨ ⎩
rk k
j= k
j
(–s)k–j(r+s)jsgna
j, ≤k≤n,
rk n
j= k
j
(–s)k–j(r+s)jsgna j+rk
k j=n+
k j
(–s)k–j(r+s)jsgna, k>n.
Thenx∈br,s
c ,xbrc,s= ,limk→∞(B
r,sx)
k=sgnaand so
f(x)= |a|+
n
k=
|ak|+sgna
∞
k=n+ ak
≤ f.
We know thatlimn→∞∞k=n+ak= whenevera= (ak)∈. Then, if we pass to the limit asn→ ∞in the last inequality, we have
|a|+
k
|ak| ≤ f. (.)
By combining (.) and (.), we conclude that
f=|a|+
k
|ak|,
which is the norm on.
Let us define a transformationLsuch thatL:{br,s
c }∗−→,L(f) = (a,a,a, . . .). Then we
have
L(f)=|a|+|a|+|a|+· · ·=f
L(f)being thenorm. Therefore,Lis norm preserving. It is obvious thatLis surjective
and linear. This completes the proof.
4 Some matrix classes related to the binomial sequence spaces
In this chapter, we characterize some matrix classes related to the binomial sequence spaces br,sandbr,s
c . Now, we start with two lemmas which are required in the proof of
the *theorems.
Lemma .(see []) Each matrix map between BK -spaces is continuous.
For brevity of notations, here and in the following, we prefer to use
tnkr,s=
n
j=k
j k
(–s)j–kr–j(r+s)kanj
for alln,k∈N.
Theorem . Let A= (ank)be an infinite matrix with complex entries.Then the following
statements hold.
(I) A∈(brc,s:p)if and only if
sup
K∈F
n
k∈K
tnkr,s
p
<∞, (.)
tnkr,s exists for allk,n∈N, (.)
k
trnk,s converges for alln∈N, (.)
sup
m∈N
m
k=
m
j=k
j k
(–s)j–kr–j(r+s)kanj
<∞, n∈N, (.)
where≤p<∞. (II) A∈(br,s
c :∞)if and only if (.)and(.)hold,and
sup
n∈N
k
trnk,s<∞. (.)
Proof Given a sequencex= (xk)∈brc,s, we suppose that the conditions (.)-(.) hold.
Then, by taking into account Theorem .(II), we conclude that{ank}k∈N∈ {brc,s}β for all
n∈N. Thus, theA-transform ofxexists. Let us consider a matrixUr,s= (ur,s
nk) defined by
urnk,s=trnk,sfor alln,k∈N. SinceUr,s= (ur,s
nk) satisfies Lemma .(v), we deduce thatUr,s=
(urnk,s)∈(c:p).
Now, we consider the following equality:
m
k= ankxk=
m
k=
m
j=k
j k
(–s)j–kr–j(r+s)kanjyk (.)
for alln,m∈N. If we take the limit (.) side by side asm→ ∞, we obtain
k
ankxk=
k
trnk,syk. (.)
By taking thep-norm (.) side by side, we have
Axp=U
r,sy
p<∞.
Conversely, suppose thatA∈(br,s
c :p). It is known thatbrc,sandpareBK-spaces. If we
combine this fact and Lemma ., we deduce that there is a constantM> such that
Axp≤Mxbrc,s (.)
for allx∈br,s c .
Now, we define a sequencex= (xk) such thatx=k∈Kg(k)(r,s) for all fixedk∈N, where
g(k)(r,s) ={g(nk)(r,s)}n∈NandK∈F. Since inequality (.) holds for allx∈brc,s, we have
Axp=
n
k∈K
trnk,s
pp
≤Mxbrc,s=M.
Therefore (.) holds.
According to the assumption,Acan be applied to the binomial sequence spacebr,s c . So,
it is trivial that the conditions (.)-(.) hold. This completes the proof of part (I). If we take Lemma .(iv) instead of Lemma .(v), then part (II) can be proved in a
similar way.
Theorem . Let A= (ank)be an infinite matrix with complex entries.Then,A∈(brc,s:c)
if and only if (.), (.),and(.)hold,and
lim
n→∞t r,s
nk=αk for all k∈N, (.)
lim
n→∞
k
tnkr,s=α. (.)
Proof Suppose thatAsatisfies the conditions (.), (.), (.), (.), and (.). Given an arbitrary sequencex= (xk)∈brc,swithlimk→∞xk=l, thenAxexists. SinceBr,s= (brnk,s) is
regular andy= (yk) is connected with the sequencex= (xk) by equation (.), we obtain
y= (yk)∈csuch thatlimk→∞yk=l.
By considering the conditions (.) and (.), we have
k
j=
|αj| ≤sup n∈N
j
tnjr,s<∞
for allk∈N. This shows us that (αk)∈. Bearing in mind the condition (.), we have
k
ankxk=
k
trnk,s(yk–l) +l
k
trnk,s, n∈N. (.)
By taking into account the conditions (.), (.), and (.), if we take the limit (.) side by side asn→ ∞, we write
lim
n→∞(Ax)n=
k
αk(yk–l) +lα, (.)
On the contrary, suppose thatA∈(br,s
c :c). It is well known that every convergent
se-quence is also bounded, namelyc⊂∞. By combining this fact and Theorem .(II), we
deduce that the necessity of the conditions (.), (.), and (.) holds. SinceAxexists and belongs tocfor allx∈brc,s, if we takeg(k)(r,s) ={gn(k)(r,s)}n∈N instead of an arbitrary
sequencex= (xk), we deduce thatAg(k)(r,s) ={tnkr,s}n∈N∈cfor allk∈N. This shows us that
the necessity of (.) holds.
Moreover, if we takex=ein (.), we obtainAx={ktnkr,s}n∈N∈c. The last result is the
necessity of (.). This completes the proof.
Corollary . Let A= (ank)be an infinite matrix with complex entries.Then A∈(brc,s:c)m
if and only if the conditions(.), (.),and(.)hold,and the conditions(.)and(.) hold withαk= for all k∈Nandα=m,in turn.
Lemma .(see []) Let A= (ank)be an infinite matrix with complex entries.Then A∈
(br,s
∞:c)if and only if (.)and(.)hold,and
lim
n→∞
k
tr,s nk=
k
lim
n→∞t
r,s
nk, (.)
lim
m→∞
k
m
j=k
j k
(–s)j–kr–j(r+s)kanj
=
k
tnkr,s, n∈N. (.)
Theorem . (brc,s:c)m∩(br∞,s:c) =∅.
Proof We suppose that (br,s
c :c)m∩(b∞r,s:c)=∅. Then there is at least a matrixA= (ank)
such that the conditions of Corollary . and Lemma . hold forA= (ank). If we consider
the conditions (.) and (.), we conclude that
lim
n→∞
k
trnk,s= .
This result contradicts the condition (.). So, the classes (brc,s:c)mand (br∞,s:c) are
dis-joint. This last step completes the proof of the theorem.
Now, by using Lemma ., we can give some more results.
Corollary . Given an infinite matrix A= (ank)with complex entries,we define a matrix
Eu,v= (eu,v
nk)as follows:
eunk,v= (u+v)n
n
j=
n j
vn–jujajk
for all n,k∈N,where u,v∈Rand uv> .Then the necessary and sufficient conditions in order that A belongs to any of the classes(br,s
c :bu∞,v), (bcr,s:bup,v), (bcr,s:buc,v)and(bcr,s:buc,v)m
are obtained by taking Eu,v= (eu,v
nk)instead of A= (ank)in the required ones in Theorems
Corollary . Given an infinite matrix A= (ank)with complex entries,we define a matrix
C= (cnk)as follows:
cnk=
n+
n
j= ajk
for all n,k∈N.Then the necessary and sufficient conditions in order that A belongs to any of the classes(br,s
c :X∞), (bcr,s:Xp), (brc,s:c),˜ and(brc,s:c)˜mare obtained by taking C= (cnk)
instead of A= (ank)in the required ones in Theorems., .and Corollary.,where Xp,
X∞,andc are defined in˜ []and[],respectively.
Corollary . Given an infinite matrix A= (ank)with complex entries, we define two
matrices C= (cnk)and E= (enk)as follows:
cnk=ank–an+,k and enk=ank–an–,k
for all n,k∈N.Then the necessary and sufficient conditions in order that A belongs to any of the classes(br,s
c :∞()), (brc,s:c()), (bcr,s:p()),and(brc,s:c())mare obtained by taking
C= (cnk)or E= (enk)instead of A= (ank)in the required ones in Theorems., .and
Corollary.,where∞()and c()are defined in[]andp()is defined in[].
Corollary . Given an infinite matrix A= (ank)with complex entries,we define a matrix
E= (enk)as follows:
enk=
n+
n
j=
+tja jk
for all n,k∈N,where <t< .Then the necessary and sufficient conditions in order that A belongs to any of the classes(br,s
c :at∞), (brc,s:atp), (bcr,s:atc),and(bcr,s:atc)mare obtained by
taking E= (enk)instead of A= (ank)in the required ones in Theorems., .and
Corol-lary.,where at
∞,atp,and atcare defined in[]and[],respectively.
Corollary . Given an infinite matrix A= (ank)with complex entries,we define a matrix
E= (enk)as follows:
enk= n
j= ajk
for all n,k∈N.Then the necessary and sufficient conditions in order that A belongs to any of the classes(br,s
c :bs), (brc,s:cs),and(bcr,s:cs)mare obtained by taking E= (enk)instead of
A= (ank)in the required ones in Theorems., .and Corollary..
5 Conclusion
By considering the definition of the binomial matrixBr,s= (br,s
nk), we deduce thatBr,s= (b r,s nk)
means of orderr. Moreover, the binomial matrixBr,s= (br,s
nk) is not a special case of the
weighed mean matrices. So, this paper filled up a gap in the existent literature.
Competing interests
The authors declare that they have no competing interests.
Author’s contributions
The authors read and approved the final manuscript.
Acknowledgements
We would like to express our thanks to the anonymous reviewers for their valuable comments.
Received: 13 August 2016 Accepted: 22 November 2016
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