Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.1. pp. 27-41 , 2010 Applied Mathematics
The Double Sequences N. Subramanian
Department of Mathematics, SASTRA University, Tanjore-613 401, India e-mail: nsm aths@ yaho o.com
Received Date: October 8, 2008 Accepted Date: May 3, 2010
Abstract. Let 2denote the space of all double entire sequence. Let 2denote
the space of all double analytic sequences, let 2denote the space of all double
sequences. This paper is devoted to a study of the general properties of 2:
Key words: Double di¤erence sequence spaces; entire sequence; analytic se-quence; gai sequences; dual.
2000 Mathematics Subject Classi…cation: 40A05; 40C05; 40D05. 1. Introduction
Throughout w; and denote the classes of all, gai and analytic scalar valued single sequences, respectively.
We write w2 for the set of all complex sequences (x
mn); where m; n 2 N; the
set of positive integers. Then, w2 is a linear space under the coordinate wise
addition and scalar multiplication.
Some initial works on double sequence spaces is found in Bromwich[4]. Later on, they were investigated by Hardy[8], Moricz[12], Moricz and Rhoades[13], Basarir and Solankan[2], Tripathy[20], Colak and Turkmenoglu[6], Turkmenoglu[22], and many others.
Let us de…ne the following sets of double sequences:
Mu(t) := n (xmn) 2 w2: supm;n2Njxmnjtmn< 1 o ; Cp(t) := n
(xmn) 2 w2: p limm;n!1jxmn jtmn= 1 for some 2 C
o ; C0p(t) := n (xmn) 2 w2: p limm;n!1jxmnjtmn = 1 o ; Lu(t) := n (xmn) 2 w2:P1m=1 P1 n=1jxmnjtmn< 1 o ; Cbp(t) := Cp(t)TMu(t) andC0bp(t) = C0p(t)TMu(t) ;
where t = (tmn) is the sequence of strictly positive reals tmn for all m; n 2 N
and p limm;n!1 denotes the limit in the Pringsheim’s sense. In the case
tmn= 1 for all m; n 2 N; Mu(t) ; Cp(t) ; C0p(t) ; Lu(t) ; Cbp(t) and C0bp(t) reduce
to the sets Mu; Cp; C0p; Lu; Cbp and C0bp; respectively. Now, we may summarize
the knowledge given in some document related to the double sequence spaces. G•okhan and Colak [27,28] have proved that Mu(t) and Cp(t) ; Cbp(t) are
com-plete paranormed spaces of double sequences and gave the ; ; duals of the spaces Mu(t) and Cbp(t) : Quite recently, in her PhD thesis, Zelter [29] has
essentially studied both the theory of topological double sequence spaces and the theory of summability of double sequences. Mursaleen and Edely [30] have recently introduced the statistical convergence and Cauchy for double sequences and given the relation between statistical convergent and strongly Cesaro sum-mable double sequences. Nextly, Mursaleen [31] and Mursaleen and Edely [32] have de…ned the almost strong regularity of matrices for double sequences and applied these matrices to establish a core theorem and introduced the M core for double sequences and determined those four dimensional matrices transform-ing every bounded double sequences x = (xjk) into one whose core is a subset of
the M core of x: More recently, Altay and Basar [33] have de…ned the spaces BS; BS (t) ; CSp; CSbp; CSr and BV of double sequences consisting of all double
series whose sequence of partial sums are in the spaces Mu; Mu(t) ; Cp; Cbp; Cr
and Lu; respectively, and also examined some properties of those sequence spaces
and determined the duals of the spaces BS; BV; CSbp and the (#) duals
of the spaces CSbp and CSr of double series. Quite recently Basar and Sever
[34] have introduced the Banach space Lq of double sequences corresponding
to the well-known space `q of single sequences and examined some properties
of the space Lq: Quite recently Subramanian and Misra [35] have studied the
space 2
M(p; q; u) of double sequences and gave some inclusion relations.
We need the following inequality in the sequel of the paper. For a; b; 0 and 0 < p < 1; we have
(1) (a + b)p ap+ bp
The double series P1m;n=1xmn is called convergent if and only if the double
sequence (smn) is convergent, where smn=Pm;ni;j=1xij(m; n 2 N) (see[1]).
A sequence x = (xmn)is said to be double analytic if supmnjxmnj1=m+n < 1:
The vector space of all double analytic sequences will be denoted by 2.A
sequence x = (xmn) is called double entire sequence if jxmnj1=m+n ! 0 as
m; n ! 1: The double entire sequences will be denoted by 2. A sequence
x = (xmn) is called double gai sequence if ((m + n)! jxmnj)1=m+n ! 0 as
m; n ! 1: The double gai sequences will be denoted by 2. Let denote the set of all …nite sequences.
Consider a double sequence x = (xij): The (m; n)th section x[m;n] of the
se-quence is de…ned by x[m;n]=Pm;n
the double sequence whose only non zero term is a 1 in the (i; j)thplace for each i; j 2 N:
An FK-space(or a metric space)X is said to have AK property if (=mn) is a
Schauder basis for X. Or equivalently x[m;n]! x.
An FDK-space is a double sequence space endowed with a complete metriz-able; locally convex topology under which the coordinate mappings x = (xk) !
(xmn)(m; n 2 N) are also continuous.
If X is a sequence space, we give the following de…nitions: (i)X0 is the continuous dual of X;
(ii)X = a = (amn) :P1m;n=1jamnxmnj < 1; for each x 2 X ;
(iii)X = a = (amn) :P1m;n=1amnxmnis convegent; f or each x 2 X ;
(iv)X =na = (amn) : supmn 1 PM;Nm;n=1amnxmn < 1; for each x 2 X
o ; (v)Let X beanF K space ; then Xf =nf (=
mn) : f 2 X 0o ; (vi)X =na = (amn) : supmnjamnxmnj1=m+n< 1 o ;
X :X ; X are called (orK •othe T oeplitz)dual of X; (or generalized K •othe T oeplitz) dual of X; dual of X; dual of X respectively:X is de…ned by Gupta and Kamptan [24]. It is clear that X X and X X ; but X X does not hold, since the sequence of partial sums of a double convergent series need not to be bounded.
The notion of di¤erence sequence spaces (for single sequences) was introduced by Kizmaz [36] as follows
Z ( ) = fx = (xk) 2 w : ( xk) 2 Zg
for Z = c; c0 and `1; where xk = xk xk+1 for all k 2 N: Here w; c; c0 and
`1 denote the classes of all, convergent,null and bounded sclar valued single sequences respectively. The above spaces are Banach spaces normed by
kxk = jx1j + supk 1j xkj
Later on the notion was further investigated by many others. We now introduce the following di¤erence double sequence spaces de…ned by
Z ( ) = x = (xmn) 2 w2: ( xmn) 2 Z
where Z = 2; 2and x
mn= (xmn xmn+1) (xm+1n xm+1n+1) = xmn
2. Lemma
As in single sequences (see [23, Theorem 7.27]). Let X be an FK-space : Then
(i)X Xf; (ii)If X has AK, X = Xf; (iii)If X has AD, X = X :
3. De…nitions and Preliminaries
Let w2 denote the set of all complex double sequences. A sequence x = (xmn)
is said to be double analytic if supmnjxmnj1=m+n < 1: The vector space of
all prime sense double analytic sequences will be denoted by 2: A sequence
x = (xmn) is called prime sense double entire sequence if jxmnj1=m+n ! 0 as
m; n ! 1: The double entire sequences will be denoted by 2: The space 2
and 2 is a metric space with the metric (2) d(x; y) = supmn
n
jxmn ymnj1=m+n: m; n : 1; 2; 3; :::
o
forall x = fxmng and y = fymngin 2:
A sequence x = (xmn) is called prime sense double gai sequence if
((m + n)! jxmnj)1=m+n ! 0 as m; n ! 1: The double gai sequences will be
denoted by 2: The space 2is a metric space with the metric (3) d(x; y) = supe mn
n
((m + n)! jxmn ymnj)1=m+n: m; n : 1; 2; 3; :::
o
forall x = fxmng and y = fymngin 2:
De…ne the sets:
2 M = n x 2 w2: M jxmnj1=m+n ! 1 (m; n ! 1) for some > 0o 2 M = n x 2 w2: supmn M jxmnj 1=m+n < 1 for some > 0o The space 2M is a metric space with the metric
d (x; y) = infn > 0 : supmn M jxmn ymnj
1=m+n
1o and the space 2
M is a metric space with the metric
d (x; y) = infn > 0 : supmn M jxmn ymnj 1=m+n : m; n = 1; 2; 3; o 4. Main Results 4.1. Proposition 2 2
Proof: Let x 2 2:
Then we have ((m + n)! jxmnj)1=m+n! 0 as m; n ! 1:
Here, we get jxmnj1=m+n ! 0 as m; n ! 1: Thus we have x 2 2 and so 2 2:
4.2. Proposition
2
M 6= 2:
Proof: Let y = (ymn) be an arbitrary point in 2M : If y is not in 2; then
for each natural number p, we can …nd an index mpnp such that
(4) M ympnp 1=mp+np! > p; (p = 1; 2; 3; ) De…ne x = fxmng by (5) M xmn = ( 1
pm+n; for (m; n) = (mp; np) for somep 2 N
0; otherwise Then x is in 2M; but for in…nitely mn;
(6) M jymnxmnj > 1:
Consider the sequence z = fzmng ; where M z11 = M x11 s with
(7) s = 1 X m=1 1 X n=1 M xmn ; M zmn = M xmn : Then z is a point of 2 M: Also, P P M zmn = 0: Hence, z is in 2 M; but, by
(6),P PM zmnymn does not converge:
(8) )X Xxmnymndiverges:
Thus, the sequence y would not be in 2
M : This contradiction proves that
(9) 2M 2:
If we now choose M = id; where id is the identity and y1n = x1n = 1 and
ymn= xmn= 0 (m > 1) for all n; then obviously x 2 2M and y 2 2; but
(10) 1 X m=1 1 X n=1 xmnymn= 1: Hence; y =2 2M
From (9) and (10), we are granted 2M 6= 2: 4.3. Proposition
The dual space of 2is 2
Proof: First, we observe that 2 2; by Proposition 4.1. Theorefore 2 2 : But 2 6= 2; by Proposition 4.2. Hence
(11) 2 2
Next we show that 2 2: Let y = (y
mn) 2 2 : Consider f (x) = P1 m=1 P1 n=1xmnymnwith x = (xmn) 2 2 x = [(=mn =mn+1) (=m+1n =m+1n+1)] = 0 B B B B B B B B B B @ 0; 0; :::0; 0; ::: 0 0; 0; :::0; 0; ::: 0 : : : 0; 0; :::(m+n)!1 ; (m+n)!1 ; ::: 0 0; 0; :::0; 0; ::: 0 1 C C C C C C C C C C A 0 B B B B B B B B B B @ 0; 0; :::0; 0; ::: 0 0; 0; :::0; 0; ::: 0 : : : 0; 0; :::(m+n)!1 ; (m+n)!1 ; ::: 0 0; 0; :::0; 0; ::: 0 1 C C C C C C C C C C A n ((m + n)! jxmnj)1=m+n o = 0 B B B B B B B B B B B B B B @ 0; 0; :::0; 0; ::: 0 0; 0; :::0; 0; ::: 0 : : : 0; 0; :::(m+n)!1 ; (m+n)!1 ; ::: 0 0; 0; :::(m+n)!1 ; (m+n)!1 ; ::: 0 0; 0; :::0; 0; ::: 0 1 C C C C C C C C C C C C C C A :
Hence converges to zero.
Therefore [(=mn =mn+1) (=m+1n =m+1n+1)] 2 2:
Hence d ((=mn =mn+1) (=m+1n =m+1n+1) ; 0) = 1: But
each m; n: Thus (ymn) is a double bounded sequence and hence an analytic
se-quence. In other words y 2 2: But y = (y
mn) is arbitrary in 2 : Therefore
(12) 2 2
From (11) and (12) we get 2 = 2:
4.4. Proposition
2 has AK
Proof: Let x = (xmn) 2 2 and take the [m; n]th sectional sequence of
x: We have d x; x[r;s] = supmn
n
((m + n)! jxmnj)1=m+n: m r; n s
o ! 0 as [r; s] ! 1: Therefore x[r;s]! x in 2 as r; s ! 1: Thus 2has AK.
4.5. Proposition
2 is solid
Proof: Let jxmnj jymnj and let y = (ymn) 2 2: We have
((m + n)! jxmnj)1=m+n ((m + n)! jymnj)1=m+n: But ((m + n)! jymnj)1=m+n 2 2; because y 2 2: That is ((m + n)! jy
mnj)1=m+n! 0 ) ((m + n)! jxmnj)1=m+n!
0 as m; n ! 1: Therefore x = (xmn) 2 2: This completes the proof.
4.6. Proposition dual of 2 is 2
Proof: Let y 2 dual of 2: Then jx
mnymnj Mm+n for some constant
M > 0 and for each x 2 2: Therefore jy
mnj Mm+n for each m; n by taking
x = =mn= 0 B B B B B B B B B B @ 0; 0; :::0; 0; ::: 0 0; 0; :::0; 0; ::: 0 : : : 0; 0; :::(m+n)!1 ; 0; ::: 0 0; 0; :::0; 0; ::: 0 1 C C C C C C C C C C A :
This shows that y 2 2: Then
On the other hand, let y 2 2: Let > 0 be given. Then jymnj < Mm+nfor each
m; n and for some constant M > 0: But x 2 2: Hence ((m + n)! jxmnj) < m+n
for each m; n and for each > 0: i.e jxmnj <
m+n (m+n)!: Hence jxmnymnj = jxmnj jymnj < m+n (m+n)!M m+n= ( M )m+n (m+n)! ) y 2 2 (14) 2 2
From (13) and (14) we get 2 = 2:
4.7. Proposition
Let 2 denote the dual space of 2: Then we have 2 = 2:
Proof: We recall that
x = =mn= 0 B B B B B B B B B B @ 0; 0; :::0; 0; ::: 0 0; 0; :::0; 0; ::: 0 : : : 0; 0; :::(m+n)!1 ; 0; ::: 0 0; 0; :::0; 0; ::: 0 1 C C C C C C C C C C A : with 1 (m+n)! in the (m; n) th
position and zero other wise, with x = =mn; n ((m + n)! jxmnj)1=m+n o = 0 B B B B B B B B B @ 01=2; 0; :::0; 0; ::: 01=1+n : : : 01=m+1; 0; ::: (m+n)!(m+n)! 1=m+n ; 0; ::: 01=m+n+1 01=m+2; 0; :::0; 0; ::: 01=m+n+2 1 C C C C C C C C C A : = 0 B B B B B B B B B B @ 0; 0; :::0; 0; ::: 0 0; 0; :::0; 0; ::: 0 : : : 0; 0; :::11=m+n; 0; ::: 0 0; 0; :::0; 0; ::: 0 1 C C C C C C C C C C A :
which is a double sequence. Hence =mn2 2: Let us take
f (x) =P1m=1P1n=1xmnymn with x 2 2 and f 2 2 : Take x = (xmn) =
=mn2 2: Then
jymnj kfk d (=mn; 0) < 1 for each m; n
Thus (ymn) is a bounded sequence and hence an double analytic sequence. In
other words y 2 2: Therefore 2 = 2: 4.8. Proposition
2 = 2
Proof: Step 1:Let (xmn) 2 2and let (ymn) 2 2: Then we get jymnj1=m+n
M for some constant M > 0: Also (xmn) 2 2 ) ((m + n)! jxmnj)1=m+n = 2M1 ) jxmnj 2m+nMm+n1 (m+n)!: Hence P1 m=1 P1 n=1jxmnymnj P1m=1 P1 n=1jxmnj jymnj <P1m=1P1n=1 1 2m+nMm+n1 Mm+n(m+n)!1 <P1m=1P1n=12m+n1 (m+n)!1 < 1:
Therefore, we get that (xmn) 2 2 and so we have
(15) 2 2
Step 2: Let (xmn) 2 2 : This says that
(16) ) 1 X m=1 1 X n=1 jxmnymnj < 1 for each (ymn) 2 2
Assume that (xmn) =2 2; then there exists a sequence of positive integers
(mp+ np) strictly increasing such that
xmp+np > 1 2mp+np 1 (m + n)!; (p = 1; 2; 3; ) Take
ymp;np = 2 mp+np(m + n)! (p = 1; 2; 3; ) and ymn= 0 otherwise Then (ymn) 2 2: But P1 m=1 P1 n=1jxmnymnj =P P1p=1 xmpnpympnp > 1 + 1 + 1 + :
We know that the in…nite series 1+1+1+ diverges. HenceP1m=1P1n=1jxmnymnj
diverges. This contradicts (16). Hence (xmn) 2 2: Therefore
(17) 2 2
From (15) and (17) we get 2 = 2:
4.9. De…nition
Let p = (pmn) is a double sequece of positive real numbers. Then 2(p) =nx = (x
mn) : ((m + n)! jxmnj)1=m+n pmn
! 0 as m; n ! 1o
suppose that pmnis a constant for each m; n then 2(p) = 2:
4.10. Proposition Let 0 pmn qmnand let
n qmn pmn o be bounded. Then 2(q) 2(p) Proof: Let (18) x 2 2(q) Therefore we have (19) ((m + n)! jxmnj)1=m+n qmn ! 0 as m; n ! 1: Let tmn= ((m + n)! jxmnj)1=m+n qmn and mn= pmn=qmn: Since pmn qmn;
we have 0 mn 1: Take 0 < < mn: De…ne
(20) umn= ( tmn; if (tmn 1) 0; if (tmn< 1) ; vmn= ( 0; if (tmn 1) tmn; if (tmn< 1)
tmn= umn+ vmn; tmnmn = umnmn+ vmnmn: Now it follows that u mn mn umn tmn; vmnmn umn: Since tmnmn = umnmn + vmnmn; then tmnmn tmn+ vmn: ((m + n)! jxmnj)1=m+n qmn mn ((m + n)! jxmnj)1=m+n qmn ) ((m + n)! jxmnj)1=m+n qmn pmn=qmn ((m + n)! jxmnj)1=m+n qmn ) ((m + n)! jxmnj)1=m+n qmn pmn ((m + n)! jxmnj)1=m+n qmn : But ((m + n)! jxmnj)1=m+n qmn ! 0 as m; n ! 1: (by (14)). Therefore ((m + n)! jxmnj)1=m+n pmn ! 0 as m; n ! 1: Hence (21) x 2 2(p)
From (18) and (21) we get 2(q) 2(p) :
4.11. Proposition
(a)Let 0 < inf pmn pmn 1: Then 2(p) 2
(b)Let 1 pmn suppmn< 1: Then 2 2(p) :
Proof: (a)Let x 2 2(p)
(22) ((m + n)! jxmnj)1=m+n pmn ! 0 as m; n ! 1 Since 0 < inf pmn pmn 1; (23) ((m + n)! jxmnj)1=m+n ((m + n)! jxmnj)1=m+n pmn
From (22) and (23) it follows that
(24) x 2 2
Thus 2(p) 2: This completes the proof.
Proof (b): Let pmn 1 for each mn and suppmn< 1 and let x 2 2:
(25) ((m + n)! jxmnj)1=m+n ! 0 as m; n ! 1
Since 1 pmn suppmn< 1; we have
(26) ((m + n)! jxmnj)1=m+n pmn
((m + n)! jxmnj)1=m+n pmn
! 0 as m; n ! 1 (by using (25)). Therefore x 2
2(p) :
4.12. Proposition
Let 0 < pmn qmn< 1 for each m; n: Then x 2 2(p) x 2 2(q) :
Proof: Let x 2 2(p)
(27) ((m + n)! jxmnj)1=m+n pmn
! 0 as m; n ! 1
This implies that ((m + n)! jxmnj)1=m+n 1 for su¢ ciently large m; n: We
get (28) ((m + n)! jxmnj)1=m+n qmn ((m + n)! jxmnj)1=m+n pmn ) ((m + n)! jxmnj)1=m+n qmn
! 0 as m; n ! 1 (by using (27)). Since x 2
2(q) ; hence x 2 2(p) 2(q) :
4.13. Proposition
2 = 2for = ; ; ; f
Proof: Step 1: 2 has AK by Proposition (4.4). Hence by Lemma (2) (ii), we
get 2 = 2 f: But 2 = 2: Hence
(29) 2 = 2
Step 2: Since AK implies AD, hence by Lemma 2(iii) we get 2 = 2 :
Therefore
(30) 2 = 2
Step 3: 2is normal by Proposition (4.5). Hence, [24, Proposition 2.7], we get
(31) 2 = 2 = 2
From (29),(30) and (31), we have 2 = 2 = 2 = 2 f = 2:
5. Ackowledgement
I wish to thank the referees for their several remarks and valuable suggestions that improved the presentation of the paper and also I wish to thank Prof. Dr. Ahmet Sinan Cevik,Department of Mathematics, Faculty of Science, Selcuk
University,Campus, 42075 Konya - Turkey, for his valuable moral support in connection with my paper presentation.
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