R E S E A R C H
Open Access
Cesàro summable difference sequence space
Vinod K Bhardwaj
1*and Sandeep Gupta
2*Correspondence:
vinodk_bhj@rediffmail.com 1Department of Mathematics,
Kurukshetra University, Kurukshetra, 136119, India
Full list of author information is available at the end of the article
Abstract
The difference sequence spacesc0(
),c(
) and
∞(
) were introduced by Kizmaz (Can. Math. Bull. 24:169-176, 1981). In this paper, we introduce the Cesáro summable difference sequence spaceC1(
) which strictly includes the spacesc0(
) andc(
)
but overlaps with
∞(
). It is shown that the newly introduced spaceC1(
) turns out
to be an inseparable BK space which does not possess any of the following: AK property, monotonicity, normality and perfectness. The Köthe-Toeplitz duals ofC1(
)
are computed and as an application, the matrix classes (C1(
),
∞), (C1(
),c;P) and
(C1(
),c0) are also characterized.
MSC: 40C05; 40A05; 46A45
Keywords: sequence space; BK space; Schauder basis; Köthe-Toeplitz duals; matrix map
1 Notations and definitions
Byswe shall denote the linear space of all complex sequences overC(the field of complex numbers).∞,candcdenote the spaces of all bounded, convergent and null sequences x= (xk) with complex terms, respectively, normed byx∞=supk|xk|.
Throughout this paper, unless otherwise specified, we writekfor∞k= andlimnfor
limn→∞.
The definitions given below may be conveniently found in [–].
A complete metric linear space is called a Frèchet space. LetXbe a linear subspace ofs such thatXis a Frèchet space with continuous coordinate projections. Then we say that Xis an FK space. If the metric of an FK space is given by a complete norm, then we say thatXis a BK space.
We say that an FK spaceXhas AK, or has the AK property, if (ek), the sequence of unit
vectors, is a Schauder basis forX. A sequence spaceXis called
(i) normal (or solid) ify= (yk)∈Xwhenever|yk| ≤ |xk|,k≥, for somex= (xk)∈X, (ii) monotone if it contains the canonical preimages of all its stepspaces,
(iii) sequence algebra ifxy= (xkyk)∈Xwheneverx= (xk),y= (yk)∈X,
(iv) convergence free when, ifx= (xk)is inXand ifyk= wheneverxk= , then
y= (yk)is inX.
The idea of dual sequence spaces was introduced by Köthe and Toeplitz [] whose main results concernedα-duals; theα-dual ofX⊂sbeing defined as
Xα=
a= (ak)∈s:
k
|akxk|<∞for allx= (xk)∈X
.
In the same paper [], they also introduced another kind of dual, namely, theβ-dual (see [] also, where it is called theg-dual by Chillingworth) defined as
Xβ=
a= (ak)∈s:
k
akxkconverges for allx= (xk)∈X
.
Obviously,φ⊂Xα⊂Xβ, whereφis the well-known sequence space of finitely non-zero
scalar sequences. Also, ifX⊂Y, thenYη⊂Xηforη=α,β. For any sequence spaceX, we
denote (Xδ)ηbyXδη, whereδ,η=αorβ. It is clear thatX⊂Xηη, whereη=αorβ.
For a sequence spaceX, ifX=XααthenXis called a Köthe space or a perfect sequence
space.
A sequence spacex= (xk) of complex numbers is said to be (C, ) summable (or Cesàro
summable of order ) to∈Ciflimkσk=, whereσk=k k
i=xi. ByCwe shall denote the linear space of all (C, ) summable sequences of complex numbers overC,i.e.,
C=
x= (xk)∈s:
k
k
i= xi
∈c .
It is easy to see thatCis a BK space normed by
x=sup
k
k
k
i= xi
, x= (xk)∈C.
The notion of difference sequence space was introduced by Kizmaz [] in as follows:
X() =x= (xk)∈s: (xk)∈X
forX=∞,c,c; wherexk=xk–xk+ for allk∈N(the set of natural numbers). For a detailed account of difference sequence spaces, one may refer to [–] where many more references can be found.
2 Motivation and introduction
During the last years, a large amount of work has been carried out by many mathemati-cians regarding various generalizations of difference sequence spaces of Kizmaz. Keeping aside some exceptions (see, for instance, [, ]), in most of these works, the underlying spaces have remained the same,i.e.,∞,candc. In the present work, we take the
oppor-tunity to introduce a difference sequence space with underlying space asC. We observe that
(i) Cc()as((–)k)∈Cbut((–)k) /∈c(), (ii) c()Cas(k)∈c()but(k) /∈C, and
(iii) c⊂c()∩C.
Thus the sequence spacesCandc() overlap but do not contain each other. Similarly,C and∞also overlap without containing each other as is clear from the fact thatC∞, ∞Candc⊂C∩∞. Note that the sequence ((–)k–
√
k) is (C, ) summable but not bounded, whereas the sequencex= (xk) given byx= ,x= and
xk= ⎧ ⎨ ⎩
, if i–<k≤(i–) (i= , , . . .);
is bounded but not (C, ) summable. This has motivated the authors to look for a new sequence space which properly includes the spacesC,c() and∞.
We now introduce a sequence spaceC(), Cesàro summable difference sequence space, as follows:
C() =
x= (xk)∈s: (xk)∈C
.
The overall picture regarding inclusions among the already existing spaces∞,c,c,C,
∞(),c(),c() and the newly introduced spaceC() is as shown below:
C ⊂ C()
∪ ∪
c ⊂ c ⊂ ∞
∩ ∩ ∩
c() ⊂ c() ⊂ ∞()
∩
C()
In this paper we show thatC() strictly includes the spacesc() andc() but overlaps with∞(). It is shown that the newly introduced spaceC() is an inseparable BK space which does not possess any of the following: AK property, monotonicity, normality and perfectness. The Köthe-Toeplitz duals ofC() are computed, and as an application, the matrix classes (C(),∞), (C(),c;P) and (C(),c) are also characterized.
3 Inclusion theorems and topological properties ofC1(
)
We begin with elementary inclusion theorems justifying thatC() is much wider than ∞,Candc().
Theorem . ∞⊂C(),the inclusion being strict.
Proof Letx= (xk)∈∞. Then there existsM> such that|x–xk+| ≤Mfor allk≥, and so kki=xi→ ask→ ∞. For strict inclusion, observe that (k)∈C() but (k) /∈∞.
Theorem . C⊂C(),the inclusion being strict.
Proof Forx= (xk)∈C, we havelimkkxk= , and sok k
i=xi→ ask→ ∞. Inclusion
is strict in view of the example cited in Theorem ..
Theorem . c()⊂C(),the inclusion being strict.
Proof Inclusion is obvious sincec⊂C. To see that the inclusion is strict, consider the
sequencex= (xk) = (, , , , , , . . .).
Proof SinceXY, there is a sequencex= (xk)∈Xsuch thatx∈/Y. Consider the sequence
y= (yk) = (, –x, –x–x, –x–x–x, . . .). Theny∈X() buty∈/Y().
Remark . We have already observed thatC∞ and∞C, so by Remark ., it follows that neitherC()⊆∞() nor∞()⊆C(). Also, we havec()⊂C()∩ ∞(). In view of this and Theorem ., we can say thatC() strictly includesc() and hencec() but overlaps with∞().
We now study the linear topological structure ofC().
Theorem . C()is a BK space normed by
x=|x|+sup k
k
k
i=
xi
, x= (xk)∈C().
The proof is a routine verification by using ‘standard’ techniques and hence is omitted.
Theorem . C()is not separable.
Proof LetAbe the set of all sequencesxa,xb, . . . , where
xa= (k+a)k= ( +a, +a, . . .), xb= (k+b)k= ( +b, +b, . . .), . . .
with|a–b|>;a,b∈R. Clearly,A⊂C() andAis uncountable. LetDbe any dense set inC().
Define a mapf:A→Das follows:
Letxa∈A⊂C(). AsDis dense inC(), so there exists somezxa∈Dsuch thatxa–
zxa<.
We setf(xa) =zxa.
Forxa,xb∈A, we have
xa–xb =( +a) – ( +b)+sup k
k
k
i=
(xa–xb)i
≥ |a–b|
> .
Now
zxa–xb≥ xa–xb–xa–zxa
> –
=
and already we havexb–zxb<
, thereforezxa=zxb. Hencef is one-to-one. Asf(A)⊂D
soDis uncountable. Thus,C() has no countable dense set.
The result follows from the fact that if a normed space has a Schauder basis, then it is separable.
Corollary . C()does not have the AK property.
Theorem . C()is not normal(solid)and hence neither perfect nor convergence free.
Proof Takingx= (xk) = (k– ) andy= (yk) = ((–)k(k– )), we see thatx∈C() buty∈/
C() although|yk| ≤ |xk|,k≥ and soC() is not normal. It is well known [] that every perfect space, and also every convergence free space, is normal and consequentlyC() is
neither perfect nor convergence free.
Theorem . C()is neither monotone nor a sequence algebra.
Proof Takex= (xk) = (k)∈C(). Considery= (yk) where
yk= ⎧ ⎨ ⎩
xk, ifkis odd;
, ifkis even
i.e.,y= (, , , , , . . .). Then (yk) = (, –, , –, , . . .) and so (yk) /∈C,i.e., (yk) /∈C()
and henceC() is not monotone. To see thatC() is not a sequence algebra, takex=
y= (k) and observe thatx,y∈C() butxy= (k) /∈C().
4 Köthe-Toeplitz duals ofC1(
)
In this section we compute the Köthe-Toeplitz duals ofC() and show thatC() is not perfect.
Theorem .
C()α=
a= (ak) :
k
k|ak|<∞
=D.
Proof Leta= (ak)∈D. For anyx= (xk)∈C(), we have (kki=xi)∈c,i.e., (k(x–
xk+))∈cand so there exists someM> such that|xk| ≤M(k– ) +xfork≥ and hence
supkk–|x
k|<∞, which implies that
k
|akxk|=
k
k|ak|
k–|xk|
<∞.
Thus,a= (ak)∈[C()]α.
Conversely, leta= (ak)∈[C()]α. Then
k|akxk|<∞for allx= (xk)∈C(). Taking xk=kfor allk≥, we havex= (xk)∈C() whence
kk|ak|<∞.
Remark . It is well known [, ] that [c()]α= [c()]α= [
∞()]α=D, so we
con-clude that [c()]α= [c()]α= [
Theorem .
C()αα=
a= (ak) :sup k
k–|ak|<∞
=D.
Proof Taking m= and X =c in [, Theorem .], we have [c()]αα ={a= (a k) :
supkk–|a
k|<∞}and the result follows in view of Remark ..
Corollary . C()is not perfect.
The proof follows at once when we observe that the sequence ((–)k(k– ))∈[C
()]αα but does not belong toC().
Theorem .
C()β=
a= (ak) :
k
k|ak|<∞
=D.
Proof Leta= (ak)∈Dandx= (xk)∈C(). Then (k k
i=xi)∈c. Forn∈N, we have n
k=
akxk= – n
k=
(k– )ak
k–
k–
i=
xi
+x
n
k= ak.
Obviously, (ak) and ((k– )ak)∈. We definey= (yk) byy= andyk=k–ki=–xifor
allk≥. Theny∈cand sincecα=, the series∞
k=(k– )ak(k– k–
i= xi) converges
absolutely.
Conversely, if a= (ak)∈[C()]β, then
kakxk converges for all x= (xk)∈C().
In particular, taking xk = for all k, we have
kak converges and so ∞
k=(k– )× ak(k–
k–
i=xi) converges for allx= (xk)∈C(). Sincex= (xk)∈C() if and only if
y= (
k k
i=xi)∈c, we have ((k– )ak)∈cα.
Corollary . [c()]α= [c()]α= [
∞()]α= [C()]α= [C()]β.
5 Matrix maps
Finally, we characterize certain matrix classes. For any complex infinite matrixA= (ank),
we shall writeAn= (ank)k∈Nfor the sequence in thenth row ofA. IfX,Yare any two sets of
sequences, we denote by (X,Y) the class of all those infinite matricesA= (ank) such that
the seriesAn(x) =
kankxk converges for allx= (xk)∈X(n= , , . . .) and the sequence
Ax= (Anx)n∈Nis in Y for allx∈X.
The following theorem is well known.
Theorem . [, p.]Let X and Y be BK spaces and suppose that A= (ank)is an infinite
matrix such that(kankxk)n∈N∈Y for each x∈X,i.e.,A∈(X,Y),then A:X→Y is a
bounded linear operator.
Proof Suppose thatsupn∞k=(k– )|ank|<∞andx= (xk)∈C(). Proceeding as in
The-orem ., we have∞k=|ank k–
i= xi|<∞.
Form∈N,
m
k=
ankxk= – m k= ank k– i= xi +x m k= ank,
which yields the absolute convergence ofkankxkfor eachn∈N, and finally we have
k
ankxk ≤
sup
k≥
k–
k– i= xi sup n ∞ k=
(k– )|ank|
+xsup
n
k
(k– )|ank|
for alln∈N.
Conversely, by Theorem ., we have
k
ankxk
=An(x)≤sup n
An(x)=Ax∞≤ Ax (.)
for eachn∈Nandx= (xk)∈C().
Choose anyn∈Nand anyr∈Nand define
xk= ⎧ ⎨ ⎩
(k– )sgnank, if <k≤r;
, otherwise.
Thenx= (xk)∈c⊂C() withx= . Inserting this value ofx= (xk) in (.) , we have
r
k=
(k– )|ank| ≤ A. (.)
Lettingr→ ∞and noting that (.) holds for everyn∈N, we are through. Remark . Ifx= (xk)∈C(), then there exists some∈Csuch thatlimkkki=xi=.
We shall calltheC() limit of the sequence (xk) and by (C(),c;P) we shall denote that
subset of (C(),c) for whichC() limits are preserved.
Theorem . A∈(C(),c;P)if and only if
(i) supn∞k=(k– )|ank|<∞, (ii) limn
k(k– )ank= –, (iii) limnank= for eachk, (iv) limn
kank= .
Proof Let the conditions (i)-(iv) hold and suppose thatx= (xk)∈C() with limkk× k
i=xi=. It is implicit in (i) that, for eachn∈N,
k(k– )|ank|converges. It follows
that∞k=(k– )ank(k– k–
i= xi) converges, whence
k
ankxk= – ∞
k=
(k– )ank
k–
k–
i=
xi–
–
k
(k– )ank+x
k
ank. (.)
Letk=k k
i=xi–,H=supk|k|andM=supn
Then, for anyp∈N, we have
∞
k=
(k– )ank
k–
k–
i=
xi–
≤H
p
k=
(k– )|ank|+Msup k>p
|k–|
and hence
lim sup
n
∞
k=
(k– )ank
k–
k–
i=
xi–
≤Msupk>p
|k–|.
Lettingp→ ∞, we have∞k=(k– )ank(k– k–
i= xi–)→ asn→ ∞. Making use of
this and also of (ii) and (iv) in (.), we get the result. Conversely, letA∈(C(),c;P). Then (
kankxk)n∈N∈cfor allx= (xk)∈C(). By the same argument as in Theorem ., we have supn∞k=(k– )|ank|<∞. Takingx=ek∈
C(), we get (ank)n∈N ∈c with limnank = for eachk. Also, forx= (k– ), we have
(k(k– )ank)n∈N∈cwithlimn
k(k– )ank= –, and finallyx= (, , , . . .)∈C() yields
limn
kank= .
Theorem . A∈(C(),c)if and only if
(i) supn∞k=(k– )|ank|<∞, (ii) limnk(k– )ank= , (iii) limnank= for eachk, (iv) limn
kank= .
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
VKB and SG contributed equally. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Kurukshetra University, Kurukshetra, 136119, India.2Department of Mathematics, Arya PG
College, Panipat, 132103, India.
Acknowledgements
The authors are grateful to the referee for his/her valuable comments and suggestions, which have improved the presentation of the paper.
Received: 15 March 2013 Accepted: 17 June 2013 Published: 5 July 2013 References
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doi:10.1186/1029-242X-2013-315
