IS S N 1 3 0 3 –5 9 9 1
and
ON SOME NEW DOUBLE SEQUENCE SPACES OF INVARIANT MEANS DEFINED BY ORLICZ FUNCTIONS
and VAKEEL A. KHAN AND SABIHA TABASSUM
Abstract. The sequence space BV was introduced and studied by Mur-saleen[14]. In this paper we extend BV to 2BV (p; r; s) and study some
properties and inclusion relations on this space.
1. Introduction
Let l1; and c denote the Banach spaces of bounded and convergent sequences x = (xi), with complex terms, respectively, normed by kxk1 = sup
i jxij, where
i 2 N: Let be an injection of the set of positive integers N into itself having no …nite orbits that is to say, if and only if, for all i = 0; j = 0; j(i) 6= i and T be the
operator de…ned on l1 by (T (xi)1i=1) = (x (i))1i=1:
A continuous linear functional on l1 is said to be an invariant mean or -mean if and only if
(1) (x) 0; when the sequence x = (xi) has xi 0 for all i;
(2) (e) = 1; where e = f1; 1; 1; :::::::g and (3) (x (i)) = (x) for all x 2 l1:
If x = (xi) write T x = (T xi) = (x (i)): It can be shown that
V = x = (xi) : 1
X
m=1
tm;i(x) = L uniformly in i; L = lim x (1)
where m o; i > 0:
Received by the editors Agu. 01, 2011, Accepted: Dec. 26, 2011. 2000 Mathematics Subject Classi…cation. 46E30, 46E40, 46B20.
Key words and phrases. Invariant means, double sequence spaces, Orlicz Function.
c 2 0 1 1 A n ka ra U n ive rsity
1
tm;i(x) =
xi+ x (i)+ :::: + x m(i)
m + 1
and t 1;i= 0 (2)
. Where m(i) denote the mth iterate of (i) at i: In the case is the translation mapping, (i) = i + 1 is often called a Banach limit and V ; the set of bounded sequences of all whose invariant means are equal, is the set of almost convergent sequence. Subsequently invariant means have been studied by Ahmad and Mur-saleen[1], Mursaleen[12,13], Raimi[15] and many others.
The concept of paranorm is closely related to linear metric spaces. It is gen-eralization of that of absolute value. Let X be a linear space. A Paranorm is a function g : X ! R which satis…es the following axioms: for any x; y; x0 2 X,
; 02 C,
(i) g( ) = 0; (ii) g(x) = g( x);
(iii) g(x + y) g(x) + g(y)
(iv) the scalar multiplication is continuous, that is ! 0, x ! x0 imply
x ! 0x0:
Any function g which satis…es all the condition (i)-(iv) together with the con-dition
(v) g(x) = 0 if only if x = ,
is called a Total Paranorm on X and the pair (X; g) is called Total paranormed space. It is well known that the metric of any linear metric space is given by some total paranorm (cf.[18],Theorm 10.42,p183])
An Orlicz Function is a function M : [0; 1) ! [0; 1) which is continuous, nondecreasing and convex with M (0) = 0, M (x) > 0 for x > 0 and M (x) ! 1, as x ! 1. If convexity of M is replaced by M(x + y) M (x) + M (y) then it is called Modulus function.
An Orlicz function M satis…es the 2 condition (M 2 2 for short ) if there
exist constant k 2 and u0> 0 such that
M (2u) KM (u) whenever juj u0.
1The second author is supported by Maulana Azad National Fellowship under the University Grants Commision of India.
An Orlicz function M can always be represented in the integral form M (x) =
x
R
0
q(t)dt; where q known as the kernel of M; is right di¤erentiable for t 0; q(t) > 0 for t > 0; qis non-decreasing and q(t) ! 1 as t ! 1:
Note that an Orlicz function satis…es the inequality M ( x) M (x) for all with 0 < < 1; since M is convex and M (0) = 0:
W.Orlicz used the idea of Orlicz function to construct the space (LM): Lin-desstrauss and Tzafriri [9] used the idea of Orlicz sequence space;
lM := x 2 w : 1
X
k=1
M jxkj < 1; for some > 0 which is Banach space with the norm the norm
kxkM = inf > 0 : 1
X
k=1
M jxkj 1 :
The space lM is closely related to the space lp, which is an Orlicz sequence space
with M (x) = xp for 1 p < 1:
Orlicz functons have been studied by V.A.Khan[3,5,6,7,8] and many others. Throughout a double sequence is denoted by x = (xij):A double sequence is a
double in…nite array of elements xij 2 R for all i; j 2 N: Let2l1 and2c denote the
Banach spaces of bounded and convergent double sequence x = (xi;j) respectively.
Doube sequence spaces have been studied by Moricz and Rhoads[11], E.Savas and R.F.Patterson[16], V.A.Khan[4] and many others.
Let be an injection having no …nite orbits and T be the operator de…ned on
2l1 by
T ((xi;j)1i;j=1) = (x (i;j))1i;j
The idea of -convergence for double sequences has recently been introduced in [2] and further studied by Mursaleen and Mohiuddine [12].
For double sequences,
2V = x = (xi;j) : 1 X m=1 1 X n=1
tmnpq(x) = L uniformly in p; q; L = lim x see[16]
(3) tmnpq(x) = 1 (m + 1)(n + 1) 1 X i=1 1 X j=1 x i(p); j(q); p; q = 0; 1; 2::: (4)
t0;0;p;q(x) = xpq; t 1;0;p;q(x) = xp 1;q(x); t0; 1;p;q(x)
= xp;q 1; t 1; 1;p;q(x) = xp 1;q 1;
and x i(p); j(q)= 0 for all i or j or both negative.
A double sequence space E is said to be solid if ( i;jxi;j) 2 E, whenever
(xi;j) 2 E, for all double sequences ( i;j) of scalars with j i;jj 1; for all i; j 2 N:
Let
K = f(ni; kj) : i; j 2 N; n1< n2< n3< :::: and k1< k2< k3< :::g N N
and E be a double sequence space. A K-step space of E is a sequence space
E
K= f( i;jxi;j) : (xi;j) 2 Eg:
A canonical pre-image of a sequence (xni;kj) 2 E is a sequence (bn;k) 2 E de…ned
as follows:
bnk=
(
ank if (n; k) 2 K;
0 otherwise :
A canonical pre-image of step space EK is a set of canonical pre-images of all
elements in EK:
A double sequence space E is said to be monotone if it contains the canonical pre-images of all its step spaces.
A double sequence space E is said to be symmetric if (xi;j) 2 E implies
(x (i); (j)) 2 E; where is a permutation of N:
2. Main Results
Lemma 1A sequence space E is solid implies E is monotone. Mursaleen[14] de…ned the sequence space
BV = fx 2 l1:
X
m
j m;i(x)j < 1; uniformly in ig; (5)
where m;i(x) = tm;i(x) tm 1;i(x)
assuming that tm;i(x) = 0 for m = 1
m;n(x) = 8 < : 1 m(m+1) m P n=1 n[xn(i) xn 1(i)] (m 1) xi (m = 0): (6) We de…ne 2BV = fx 22l1: X m;n j mnpq(x)j < 1; uniformly in p and qg; (7) where mnpq(x) = 8 > > > < > > > : 1 m(m+1)n(n+1) m P i=1 n P j=1 ij[x i(p); j(q) x i 1(p); j(q) x i(p); j 1(q)+ x i 1(p); j 1(q)] (m; n 1) xij m or n or both zero : (see[12]) (8)
Let M be an Orlicz function, p = (pi) be any sequence of strictly positive
real numbers and r 0. V.A.Khan[5] de…ned the following sequence space: BV (M; p; r) = x = (xi) : 1 X m=1 1 mr M j m;i(x)j pi < 1; uniformly in i and for some > 0 :
Let p = (pij) be any double sequence of strictly positive real numbers and r; s 0.
We de…ne the following double sequence spaces as:
2BV (M; p; r; s) = x = (xij) : 1 X m=1 1 X n=1 1 mrns M j mnpq(x)j pij < 1; uniformly in p; q and for some > 0 :
For M (x) = x, we get 2BV (p; r; s) = x = (xij) : 1 X m=1 1 X n=1 1 mrnsj mnpq(x)j pij < 1; uniformly in p; q : For pi;j = 1 for all i; j we get
2BV (M; r; s) = x = (xij) : 1 X m=1 1 X n=1 1 mrns M j mnpq(x)j < 1; uniformly in p; q and for some > 0 :
For r; s = 0, we get 2BV (M; p) = x = (xij) : 1 X m=1 1 X n=1 M j mnpq(x)j pij < 1; uniformly in p; q and for some > 0 :
For M (x) = x and r; s = 0, we get
2BV (p) = x = (xij) : 1 X m=1 1 X n=1 j mnpq(x)jpij < 1; uniformly in p; q :
For pi;j = 1 for all i; j and r; s = 0, we get 2BV (M ) = x = (xij) : 1 X m=1 1 X n=1 M j mnpq(x)j < 1; uniformly in p; q
and for some > 0 : For M (x) = x; pi;j= 1 and r; s = 0,we get
2BV = x = (xij) : 1 X m=1 1 X n=1 j mnpq(x)j < 1; uniformly in p; q :
Theorem 1The sequence space2BV (M; p; r; s) is a linear space over the …eld C
of complex numbers.
Proof Let x = (xi;j) and y = (yi;j) 22BV (M; p; r; s) and ; 2 C: Then there
exist positive numbers 1and 2 such that 1 X m=1 1 X n=1 1 mrns M j mnpq(x)j 1 pij < 1 and 1 X m=1 1 X n=1 1 mrns M j mnpq(y)j 2 pij < 1 uniformly in p and q and r; s 0
De…ne 3= max(2j j 1; 2j j 2): Since M is non decreasing and convex we have,
1 X m=1 1 X n=1 1 mrns M j mnpq(x) + mnpq(y)j 3 pij < 1 1 X m=1 1 X n=1 1 mrns M j mnpq(x)j 3 +j mnpq(y)j 3 pij < 1
1 X m=1 1 X n=1 1 mrns 1 2 M mnpq(x) 1 + M mnpq(y) 2 < 1
uniformly in p and q and r; s 0.
This proves that2BV (M; p; r; s) is a linear space over the …eld C of complex
num-bers.
Theorem 2 For any Orlicz function M and a bounded sequence p = (pi;j) of
strictly positive real numbers, 2BV (M; p; r; s) is a paranormed space with
para-norm g((xij)) = sup i jx i;1j+sup j jx 1;jj+inf pij H : 1 X m=1 1 X n=1 M j mnpq(x)j pij H1 1 uniformly in p and q where H = max(1; sup
i;j
pi;j):
Proof Clearly g(0) = 0; g( (xij)) = g((xi;j)):
Using Theorem[1], for = = 1; we get
g(x + y) g(x) + g(y):
For continuity of scalar multiplication let 6= 0 be any complex number. Then by de…nition we have
g( (xij)) = sup
i j xi;1j+supj j x1;jj+inf pij H : 1 X m=1 1 X n=1 M j mnpq( x)j pij H1 1 uniformly in p and q = sup
i j jjxi;1j+supj j jjx1;jj+inf (j jr) pij H : 1 X m=1 1 X n=1 M j mnpq(x)j r pij H1 1 uniformly in p and q
where 1r =j j = max(1; j jHg((xi;j))
and therefore g( (xij)) converges to zero when g((xij)) converges to zero in 2BV (M; p; r; s).
Now let x be …xed element in2BV (M; p; r; s): There exist > 0 such that
g((xij)) = sup
i jxi;1j+supj jx1;jj+inf pij H : 1 X m=1 1 X n=1 M j mnpq(x)j pij H1 1 uniformly in p and q Now g( (xij)) = sup i j xi;1j + supj j x1;jj + inf pijH : 1 P m=1 1 P n=1 1 mrns M j mnpq( x)j pij H1 1 uniformly in p and q ! 0 as ! 0:
This copmletes the proof.
Theorem 3Suppose that 0 < pij qij< 1 for each m 2 N and r; s 0: Then
(i) 2BV (M; p) 2BV (M; q):
(ii) 2BV (M ) 2BV (M; r; s):
Proof(i)Suppose x 22BV (M; p): This implies that
M j mnpq(x)j
pij
1
for su¢ ciently large values m; n say m m0; n n0for some …xed m0; n02 N:
Since M is non decreasing, we have
1 P m=m0 1 P n=n0 M j mnpq(x)j qij P1 m=m0 1 P n=n0 M j mnpq(x)j pij < 1: uniformy in p; q. Hence x 22BV (M; q):
The second proof is trivial.
Corollary 1 If 0 pij 1 for each i and j, then 2BV (M; p) 2BV (M ): If
0 pij 1 for all i; j then2BV (M ) 2BV (M; p):
Theorem 4The sequence space2BV (M; p; r; s) is solid.
Proof Let x 22BV (M; p; r; s): This implies P1 m=1 1 P n=1 1 mrns M j mnpq(x)j pij < 1:
Let ( ij) be sequence of scalars such that j ijj 1 for all i; j 2 N: Then the
result follows from the following inequality
1 X m=1 1 X n=1 1 mrns M j ij mnpq(x)j pij X1 m=1 1 X n=1 1 mrns M j ij mnpq(x)j pij < 1: Hence x 22BV (M; p; r; s); for all sequences of scalars ( ij) with j ijj 1 for all
i; j 2 N whenever x 22BV (M; p; r; s):
From Theorem[4] and Lemma we have:
Corollary 2The sequence space2BV (M; p; r; s) is monotone.
Theorem 5Let M1; M2be Orlicz functions satisfying 2-condition and r; r1; r2;
s; s1; s2 0: Then we have
(i) if r; s > 1 then2BV (M; p; r; s) 2BV (M M1; p; r; s),
(ii) 2BV (M1; p; r; s) \2BV (M2; p; r) 2BV (M1+ M2; p; r; s);
(iii) if r1 r2and s1 s2 then2BV (M; p; r1; s1) 2BV (M; p; r2; s2):
Proof(i)Since M is continuous at 0 from right, for > 0, there exists 0 < < 1 such that 0 c implies M (c) < : If we de…ne
I1= m 2 N : M1 j mnpq(x)j for some > 0 : I2= m 2 N : M1 j mnpq(x)j > for some > 0 : then, when M1 j mnpq (x)j > we get M M1 j mnpq(x)j 2 M (1) M1 j mnpq(x)j
Hence for x 22BV (M; p; r; s) and r; s > 1 1 X m=1 1 X n=1 1 mrns M M1 j mnpq(x)j pij = X m2I1 X n2I1 1 mrns M M1 j mnpq(x)j pij + X m2I2 X n2I2 1 mrns M M1 j mnpq(x)j pij X m2I1 X n2I1 1 mrns[ ] pij + X m2I2 X n2I2 1 mrns 2 M (1) M1 j mnpq(x)j pij max( h; H) 1 X m=1 1 X n=1 1 mrns+ max 2 M (1) h 2M (1) H
(where 0 < h = inf pij pij H = sup i;j
pij< 1:)
(ii)The proof follows from the following inequality
1 mrns (M1+ M2) j mnpq (x)j pij C mrns M1 j mnpq (x)j pij +mCrns M2 j mnpq (x)j pij :
(iii)The proof is trivial.
Corollary 3Let M be an Orlicz function satisfying 2-condition. Then we have.
(i) if r; s > 1 then2BV (p; r; s) 2BV (M; p; r; s),
(ii) 2BV (M; p) 2BV (M; p; r; s);
(iii) 2BV (M ) 2BV (M; r; s):
Acknowledgments. The authors would like to record their gratitude to the reviewer for his careful reading and making some useful corrections which improved the presentation of the paper.
ÖZET: BV dizi uzay¬, Mursaleen taraf¬ndan tan¬mlanm¬¸s ve in-celenmi¸stir. Bu makalede ise BV uzay¬, 2BV (p; r; s) uzay¬na
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Current address : Department of Mathematics, A.M.U. Aligarh-202002 INDIA E-mail address : vakhan@math.com, sabihatabassum@math.com,
