On Certain Topological Properties of Double Paranormed Null Sequence Space

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On Certain Topological Properties of Double Paranormed Null Sequence Space

Narayan Prasad Pahari

Central Department of Mathematics Tribhuvan University, Kirtipur, Kathmandu, Nepal

Abstract The aim of this paper is to introduce and study a

new class c ((S , || . || ),

(amn) in various sequence spaces c0, c, , p depending upon the mode of m and n tending to infinity lead to several

0 mn

mn , u ) of double sequences with

u

their terms in a normed space S as a generalization of the familiar sequence space c0 . We investigate the condition in terms of and so that a class is contained in or equal to another class of same kind and thereby derive the conditions of their equality. We further explore some of the preliminary results that characterize the linear topological structures of

spaces, see Maddox [12].

A paranormed space (S, G) is a linear space S with zero element together with a function G : S R+ (called a paranorm on S) which satisfies the following axioms:

the space c0

((Smn

, || . ||mn

),

, u ) when topologized it with

PN1: G () = 0; PN2: G () = G () for all S;

suitable natural paranorm .

Keywords: Sequence space, Double sequence, Paranormed space, GK-space.

I.INTRODUCTION

We begin with recalling some notations and basic definitions that are used in this paper.

s

Let S be a normed space over C , the field of complex numbers. Let (S) denotes the linear space of all sequences = (sk ) with sk S , k 1 with usual coordinate wise operations .We shall denote (C) by . Any subspace S of is then called a sequence space. A normed space valued sequence space or a generalized sequence space is a linear space of sequences with their terms in a normed space. Several workers like Kamthan and Gupta [6], Khan [7], Kolk [8], Köthe [9], Maddox [11], Malkowski and Rakocevic [13], Pahari [16,17,18], Ruckle

  1. etc. have introduced and studied some properties of vector and scalar valued single sequence spaces, when sequences are taken from a Banach space.

    The theory of single sequence spaces has also been extended to the spaces of double sequences and studied by several workers. Boos Leiger [3], Gupta and Kamthan [4], Milovidov and Povolotzki [14], Morics [15], Rao [20] and many others have made their significant contributions and enriched the theories in this direction. In the recent years, Savas [22], Subramanian et al [23] and many others have

    PN3: G ( + ) G () + G ( ) for all , S; and

    PN4: Scalar multiplication is continuous.

    Note that the continuity of scalar multiplication is equivalent to

    1. if G (n) 0 and n as n , then

      G (n n) 0 as n , and

    2. if n 0 as n and be any element in S, then

G (n ) 0, see Wilansky [24].

A paranorm is called total if G () = 0 implies =

The concept of paranorm is closely related to linear metric space; see Wilansky [24] and its studies on sequence spaces were initiated by Maddox [10] and many others. Basariv and Altundag [1], Bhardwaj and Bala [2], Khan [7], Parasar and Choudhary [19], and many others further studied various types of paranormed sequence spaces .

Concerning Kproperty of scalar sequence spaces, see Kamthan and Gupta [6], GAKspace have been defined for vector valued sequence spaces and also they are defined for Banach space valued function space, see Gupta and Patterson [5]. We now introduce the following definition for double sequence spaces:

Let V(Smn) be a class of sequences { s = (smn), smn Smn ; m, n 1}.The topological sequence space (V(Smn), ) equipped with the linear topology is said to be a GK-space if the

introduced and studied various types of double sequence spaces using orlicz function.

The notion of convergence of a single sequence (an) leads to various notions of convergence for a double sequence (amn) by using many senses. The double sequence

map Pij : V(Smn) Sij, defined by P

for each i, j 1.

ij (s) = sij

is continuous

1

k 1/ umn z

m = m(k) n = n(k) k 1 and

  1. THE CLASS c0 ((Smn , || . ||mn), , u ) OF DOUBLE

    SEQUENCES

    smn

    = mn mn

    otherwise.

    Let u = (umn) and v = (vmn) be any double sequences of

    Then we see that s c0 ((Smn , || . ||mn), , u ) . Since

    u 1 v

    m(k)n(k) k mn

    strictly positive real numbers and = (mn) and = (mn) be

    double sequences of non-zero complex numbers. Let (S , ||

    || m(k) n(k) s m(k) n(k) || m(k)n(k) = , k 1 and |mn smn| mn =

    mn

    . ||mn) ,m, n 1 be normed space over the field C of complex

    , otherwise. But for each k 1,

    numbers with zero element .We now introduce and study

    the following class of Banach space valued double sequences:

    ||

    m(k) n(k)

    s m(k) n(k)

    um(k)n(k) vm(k) n(k) um(k) n(k)

    || = k /

    m(k)n(k)

    > k 1/k > e1/2

    c ((S

    , || . ||

    ),

    shows that s c0 ((Smn , || . ||mn), , u ) , a

    0 mn

    mn , u ) = {s = (smn): smn Smn, m, n 1, and

    contradiction. This cmpletes the proof.

    || mn smn ||umn 0 as m + n }.

    mn Theorem 3.2: For any = (mn) , c0 ((Smn , || . ||mn), , v

    Further, by u = (umn)

    , we mean sup umn

    < We

    ) c0 ((Smn , || . ||mn),

    , u ) if and only if

    denote A() = max(1, ||) and the zero element of this class

    by = ( ) for all m,n .

    lim sup

    m + n

    vmn < .

    u

    mn

    mn

  2. SOME CONTAINMENT RELATIONS

    Proof:

    s

    Let the condition hold, and = (smn) c0 ((Smn , || .

    In this section we investigate the conditions in terms of u

    || ),

    and , so that a class c ((S , || . || ),

    mn , v ) . Then there exists a constant L > 0 such

    0 mn

    mn , u ) is contained in

    that vmn < L umn for all sufficiently large values of m,n.

    ||

    or equal to another class of same kind and thereby derive the conditions of their equality.

    Further || mn

    smn

    umn mn

    0 as m + n together

    Theorem 3.1: For any = (mn), c0 ((Smn , || . ||mn), , )

    u

    with

    || s ||umn (|| s

    ||vmn)1/Lfor all sufficiently large

    lim inf v

    mn mn mn

    mn mn mn

    c0 ((Smn , || . ||mn), , v ) if and only if

    mn >

    m + n

    0.

    umn

    values of m, n implies that s c0 ((Smn , || . ||mn), , u

    ) and hence

    Proof: For the sufficiency of the condition, suppose that

    c ((S

    , || . ||

    ),

    0 mn

    mn , v ) c0 ((Smn , || . ||mn), , u .

    the condition hold, and s = (smn) c0 ((Smn , || . ||mn), , u ) . Then there exists a constant K > 0 such that vmn > K umn for

    Conversely let the inclusion hold but

    lim sup

    m + n

    all

    sufficiently large values of m,n. Further || mn smn ||umn < 1

    vmn

    umn

    = Then there exists subsequences (m(k)) of

    mn

    for all sufficiently large values of m,n and so

    || s ||umn || s ||umn )K for all sufficiently large

    (m) and (n(k)) of (n) respectively for each k 1;

    v m(k) n(k) < k m(k) n(k).

    mn mn mn

    mn mn mn

    Thus for zmn

    Smn

    with ||z

    mn||mn

    = 1 the sequence

    values of m,n which implies that s c0 ((Smn , || . ||mn), , v

    ). Hence

    s = (smn) defined as

    c ((S

    , || . ||

    ),

    1

    smn = mn k

    1/vmn z m = m(k) n = n(k) k 1 and

    mn

    0 mn

    mn , u ) c0 ((Smn , || . ||mn), , v ).

    otherwise.

    For the necessity of the condition suppose the inclusion

    holds, but lim inf

    vmn

    = 0. Then there exist

    is in c0 ((Smn , || . ||mn), , u ) but for each k 1,

    m + n

    umn

    || m(k) n(k) s m(k) n(k) ||um(k)n(k)

    = kvm(k) n(k)/um(k) n(k)

    subsequences (m(k)) of (m) and (n(k)) of (n) respectively such that

    k vm(k) n(k) < u m(k) n(k), k 1.

    Now taking zmn Smn with ||zmn||mn = 1, we define a

    m(k)n(k)

    > k 1/k

    > e1/2.

    sequence s = (smn) by

    This shows that s c0 ((Smn ,||. ||mn), , u ), a contradiction. This completes the proof.

    Theorem 3.3: For any = (mn),c0 ((Smn , || . ||mn), c0 ((Smn , || . ||mn),

    ) = c0 ((Smn , || . ||mn),

    , u which shows that s

    c0 ((Smn , || . ||mn),

    , u ) but s

    0 <

    lim inf

    m + n

    vmn

    umn

    , v ) if and only if

    lim sup

    m + n

    vmn < .

    u

    mn

    , u ), a contradiction. This

    completes the proof.

    u u

    Theorem 3.5: For any = ( umn), c0 ((Smn , || . ||mn), , )

    Theorem 3.4: For any u

    = (umn), c0 ((Smn , || . ||mn), , u )

    c0 ((Smn , || . ||mn), , u )

    c0 ((Smn , || . ||mn),

    lim sup

    mn umn

    , u )

    if and only if

    m + n

    .

    lim inf

    mn umn mn

    Proof:

    if and only if

    m + n mn

    > 0.

    Proof:

    For the sufficiency of the condition, suppose that

    Let the condition hold, and s = (smn) c0 ((Smn , || . ||mn),

    lim inf

    umn

    , u ) . Then there exists 0 < L < such that

    mn

    > 0 and s = (smn)

    | |umn < L|

    |umn, for all sufficiently large values of

    mn

    m + n mn

    mn mn

    c ((S , || . || ),

    m,n and so || mn smn ||umn

    || mn smn ||umn

    mn

    0 mn

    mn , u ) . Then there exists a constant

    umn

    K > 0 such that K < mn

    for sufficiently large

    implies that s c0 ((Smn , || . ||mn), , u ) . This shows

    mn

    values of m, n. Thus

    mn

    mn

    K || mn smn ||umn < || mn smn ||umn large

    for all sufficiently

    that c0 ((Smn , || . ||mn), , u ) c0 ((Smn , || . ||mn), , u )

    .

    Conversely let

    values of m, n and so || s

    ||umn 0 implies that

    c0 ((Smn , || . ||mn), , u ) c0 ((Smn , || . ||mn), , u )

    mn mn mn

    but

    lim sup

    mn

    umn

    = .Then there exist

    mn

    ||mn smn ||umn 0 and hence s c0 ((Smn , || . ||mn), , u ).

    This proves that

    m + n mn

    subsequences (m(k)) of (m) and n(k)) of (n) respectively such that for each k 1

    c0 ((Smn , || . ||mn), , u ) c0 ((Smn , || . ||mn), , u ) .

    | m(k) n(k)

    |um(k)n(k) > k |

    m(k) n(k)

    |um(k)n(k).

    For the necessity, suppose that

    Now taking z

    S , such that || z

    || = 1, we

    lim inf

    mn

    umn

    =0. Then there exist

    mn mn

    mn mn

    m + n mn

    define the sequence s = (smn) by

    m(k) n(k) |

    subsequences (m(k)) of (m) and (n(k)) of (n) respectively such that for each k 1, k | m(k) n(k)

    1 1/ umn

    k

    smn = mn

    otherwise.

    zmn m = m(k) n = n(k) k 1 and

    |

    um(k)n(k)

    < | um(k)n(k).

    Then, || m(k) n(k) s m(k) n(k) ||um(k)n(k) =

    1, k 1 and

    Now for zmn Smn with ||zmn||mn = 1, define the

    sequence s = (smn) by

    || mn

    smn

    m(k)n(k) k

    mn

    ||umn = 0, otherwise

    1

    =

    k 1/ u

    mn zmn m = m(k) n = n(k) k 1 and

    smn

    mn

    m(k)n(k)

    otherwise.

    shows that s c0 ((Smn , || . ||mn), , u ). But on the other hand

    Now, || m(k) n(k) s m(k) n(k) ||um(k)n(k) =

    1, k 1

    || m(k) n(k) s m(k) n(k) ||um(k)n(k)

    > 1, for all k = 1

    m(k)n(k) k

    mn

    and || mn smn ||umn = 0, otherwise. But for each k 1,

    || m(k) n(k) s m(k) n(k) ||um(k) n(k)

    implies that s c0 ((Smn ,||.||mn), , u ) ,a contradiction. This completes the proof.

    On combining Theorems 3.4 and 3.5 we get :

    = ||

    1

    m(k)n(k)

    k 1/ um(k)n(k) u

    z ||um(k) n(k)

    Theorem 3.6: For any u = (umn), c0 ((Smn , || . ||mn), , u )

    m(k) n(k)

    m(k)n(k)

    m(k) n(k)

    m(k) n(k)

    m(k)n(k)

    = c0 ((Smn

    , || . ||mn), , ) if and only if

    u

    m(k) n(k)um(k) n(k) 1

    lim inf

    umn

    lim sup

    umn

    0 <

    =

    m(k) n(k)

    . k > 1

    mn

    m + n mn

    mn

    m + n mn

    < .

  3. LINEAR TOPOLOGICAL STRUCTURES

Then for each m = m(k) n = n(k) k 1, we have

|| mn smn ||umn

= || m(k) n(k) s m(k) n(k) ||um(k)n(k)

OF c0 ((Smn , || . ||mn), , u )

In this section, we shall investigate some results that

mn

1

= k, k 1

m(k)n(k)

characterize the linear topological structure of

and || mn smn ||umn = 0, otherwise,

c ((S

, || . || ), mn

0 mn

mn , u ) when topologized it with suitable

natural paranorm . As far as the linear space structure of

u

c0 ((Smn , || . ||mn), , ) over the field C of complex numbers is concerned, we throughout take the

shows that s c0 ((Smn , || . ||mn), , u ) .But on the otherhand, for each m = m(k) n = n(k) k 1and for scalar = 2 we have

umn

um(k) n(k)

coordinatewise operations i.e., for s

scalar ,

= (smn), t

= (tmn) and

|| mn ( smn) ||mn = || m(k) n(k) 2 s m(k) n(k) ||m(k)n(k)

2k

s + t = (smn + tmn) and s = (smn)

u

and we see below that is necessary and sufficient

> k 1

s u

showing that c0 ((Smn , || . ||mn), , ), a

condition for linearity of c0 ((Smn , || . ||mn), , ).

u

contradiction. This completes the proof. Hence c0

((Smn , || . ||mn),

if and only if u =

Theorem 4.1: c0 ((Smn , || . ||mn), , u ) forms a linear space over the set of complex number C

= (u

, u ) is a linear space if and only if u

mn) .

(umn) .

Proof:

Consider u

,we define

= (umn) and s c0 ((Smn , || . ||mn), , u )

For the sufficiency of the condition, assume that

umn

G, u(s) = supmn|| mn smn ||mn (4.1)

u = (umn) and s = (smn), t = (tmn) c0 ((Smn , || .

||mn),

mn mn

umn

Theorem 4.2 : Let u = (umn) and Smn be a normed

, u ), m, n 1.So that ||

s ||mn 0

mn

and ||mn tmn ||umn 0 as m + n . Then we have

mn mn mn

|| mn (smn + tmn) ||umn || mn smn ||umn + || mn tmn ||umn

0 , as m + n .

space for each m,n 1. Then (c0 ((Smn , || . ||m), , u ) , G,u)

forms

a total paranormed space.

Proof:

It can be easily verified that G,u defined by (4.1) satisfy following properties of paranormed space.

Hence s + t c0 ((Smn , || . ||mn), , u ).

PN1: G,u(s) 0 and G,u(s) = 0 if and only if s =

Also it is clear that for any scalar , c0 ((Smn , || . ||mn), ,

u ) , since

s

PN2: G,u(s + t ) G

,u

(s) + G

,u( t )

mn

|| mn smn ||umn

= ||umn || mn smn ||umn

mn

mn

A()|| mn smn ||umn

0 as m + n .

PN3: G,u(s) A() G,u(s) where C.

Throughout the proof of the theorem, G,u will be denoted by G. Here we prove the continuity of scalar

multiplication i.e., PN4. Further for continuity of scalar multiplication, it is sufficient to show that

Conversely if u

= (umn) then there exist subsequences

(k)

  1. if s

    in G and k as k

    (m(k)) of (m) and (n(k)) of (n) respectively such that

    u m(k) n(k)

    > k for each k 1.

  2. if k 0 as k

and s c0 ((Smn , || . ||mn),

, u )

Now taking zmn

sequence s = (smn) by

Smn

with ||z

mn||mn

= 1, we a define

then k in G as k .

s

Now (a) is easily proved if we suppose that |k| L

1 k 1/ umn z

m = m(k) n = n(k) k 1 and

for all k 1 and consider,

s = mn mn

(k) u u

mn

mn otherwise.

G(k s ) supm,n|k| mn sup m,n|| mn smn || mn

(k)

A(L) G (s ).

s u

Now let c0 ((Smn , || . ||mn), , ) , |k| 1 for all k

N and > 0. Then there exists K such that

  1. Maddox, I.J., Infinite matrices of operators; Lecture Notes in Mathematics 786, Springer- Verlag Berlin, Heidelberg, New York, ( 1980).

  2. Malkowski, E. and Rakocevic, V. , An introduction into the theory of

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    mn

    || mn smn ||umn

    and hence

    < , for all m + n K

  3. Milovidov, S.P. and Povolotzki, A.I. , Dual spaces for conditional Köthe spaces of double real

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    smn

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    mn

    mn

    Now choose N1 so that |k|umn || mn smn ||umn < , for all

    k N1, and 2 m + n K.

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    s)

    Thus G (k

    for all k max(N, N1) which proves

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    (b). Hence G forms a total paranorm on

    c0 ((Smn), mn

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  6. Pahari , N.P., On Certain Topological Structures of Paranormed

    , u, || . || ).

    Theorem 4.3 : Let u = (umn) and Smn be a normed space for each m, n 1. Then

    Orlicz Space (S ((X ,||.||),, , u ), F ) of Vector Valued Sequences , International Jour. of Mathematical Archive. , 4(11) ( 2013) 231- 241.

  7. Pahari , N.P., On certain topological structures of

    (c0 ((Smn , || . ||mn),

    Proof:

    , u ) , G

    ,u

    ) is a GK-space.

    normed space valued generalized Orlicz function space. International Jour. of Scientific and Research (IJSER) 2(1) ( 2014) 6166.

    For each m,n 1, the continuity of linear map

  8. Parashar, S.D. and Choudhary, B., Sequence spaces

    defined by Orlicz functions, Indian

    Pij : c0 ((Smn , || . ||mn), , ) Sij where P

    u

    ij (s) = sij

    J. Pure Appl. Maths., 25(4), (1994), 419428.

  9. Rao, K Chandrasekhra , Space of matrix operators;

    1 1/uij

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    and so

    Bull. Cal. Math. Soc. 80, (1988), 9195.

    c0 ((Smn

    , || . ||mn), , ) is a GK-space.

    u

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