A note on Köthe Toeplitz duals of certain sequence spaces and their matrix transformations

(1)

Internat.

.

Math. & Math. Sci.

VOL. 18 NO. 4 (1995) 681-688

A

NOTE ON

K(THE-TOEPLITZ

DUALS OF

CERTAIN

SEQUENCE SPACES AND

THEIR MATRIX

TRANSFORMATIONS

681

B. CHOUDHARYandS.K. MISHRA

DepartmentofMathematics Indian InstituteofTechnology Hauz Khas, NewDelhi- 110016

India

(In thememoryofLateProfessorB. Kuttner)

(Received

November 5, 1992 and inrevisedform

September

22,

1993)

ABSTRACT.

In

thispaperwedefine thesequence spaces

S’oo(p),

Sc(p)

and

Sc0(p)

and determine

theK6the-Toeplitzdualsof

Sf(p).

Wealso obtainnecessaryand sufficientconditions foramatrix

A

tomap S’(R)(p) tof(R)and investigatesome related

problems.

KEY

WORDSAND PHRASES.

Sequence

spaces, K6the-Toeplitzduals,Matrixtransformations.

1992AMS SUBJECT CLASSIFICATION CODES. 40H05

INTRODUCTION.

If

{Pk}

is a sequenceofstrictly positivereal numbers, then

.(p) {x: sup

Xk

k

c(p) --{x"

xk-

pk Oforsome

t};

Co(p)

{x"

Ix

k

p _0}.

Fordetaileddiscussion onthesespaceswe refer

[1,4,5,6,7,8].

Recently Kizmaz

[3]

definedthefollowingsequence spaces:

If

Ax

(Xk-Xk/l),

then

.(A)

{x

{x#:

Ax e

.

};

c(A) ={x

={x:ZXxec};

Co(A)

={x

={x):

Axec

o}.

Thesespacesarc Banachspaceswith norm

x

1

Xl[

+ Ax

Furthermore, since

(R)(A)

is a Banachspace with continuous co-ordinates

(that

is,

Ix-x

| 0

implies Xk

Xkl

0 for each keN,as n oo

),

it is a

BK-space.

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682 B. CHOUDHARY AND S. K. MISHRA

X a

(ak)

k[

_ak

Xk[

< for each x e

X };

k=l

X a (a

k)

akx

_kisconvergent for each x X

};

k=l

X

and

X

’

arecalledthe

a-(or

K6the-Toeplitz)and

fl-(or

generalizedK6the-Toeplitz),dualspaces

of

X

respectively.

We

now definesome newsequence spaces. If

Ax

_Xk Xk.1,wedefine

s.(p) {x

(x}.

axe

_.(p)

};

Sc(p) {x

xk}"

axe

c(p)

};

Sco(p)

{x

{x}"

ax

e

Co(p)

We observe thatif x

k(for

all ke

N)

then xe

Se(p)

butx

e(R)(p).

PROPERTIES

(i)

se(R)(p)

and

Sc(p)

are

paranormed

spaces with the

paranorm

g

(X)

supk

Axk

Pk/M

where

M

max

(1,

sup

Pk)

if andonlyif 0 < infpk < sup_Pk <

"

(ii)

If p

{Pk}

is aboundedsequence, then

Sco(p)

is aparanormedspacewiththe

paranorm

g(x) sup

lAxkl

pu/M

k

The

proof

of thesepropertiesare similar totheproof givenin

[6, Th.1].

2. DUALS

THEOREM

1._{Let Pk >} 0 foreveryk. Then

(s,.(p))"

I"1

Y

(Y,):

N’"

Y, <

N-I n=l m=l

PROOF.

Weneed toprovethat

(se(R)(p))

isthesetofallsequences ysuchthat,forevery

positive integer

N,

N/p

y,I

< oo.

n=l m=l

Ifx

se,(p),

then

by

definition,

]AXnl

"<

N1/P"

So

lAX

nl

p" _is

bounded,sothat,forsome

N,

m=l

lAX

nip"

<_ _N thus

(2.1)

(by therelation

x,

A

xv)

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KTHE-TOEPLITZ

DUALS OF CERTAIN SEQUENCE SPACES 683

Thus, if

holds,then

(2.2)

Hence,

(2.2)

isasufficient condition fory

Conversely, if

N

isgivenwecan define x

so

that(2.2)

isnecessaryforytobein(Se(p))".

Nowwe raise thefollowing question"

Is

it true that

(se(R)(p))

is the set ofsequences ysuchthat,forevery positive integer

N,

nN/P,

yl

< ?

(2.3)

In

otherwords,is it true that

Itdoesnotfollow at once from Theorem that thisconjectureisfalse, sinceitisnot obviousthat

theassertionthat

(2.2)

holds for allNis notequivalenttothe assertion that

(2.3)

holds for all

N.

Indeed, thereare some sequences

{Pn}

for whichthese assertions are equivalent.

However,

for

general

{Pn}

theyneednotbeequivalent. We give examplestoshow that

(A)

Itispossibletochoose

{pn}

such that thereis ay {y}for which

(2.3)

holds for all

N,

but

(2.2)

doesnot.Thus

(2.3)

is not

always

sufficient.

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It

ispossibletochoose

{Pn}

suchthat there is a

y

lotwhich

(2.2)

holdsforall n, but

(2.3)

doesnot.Thus

(2.3)

is not

always

necessary.

EXAMPLE

1. Take

Thentake

(1,2,3

k

P2k=l/k

!

1

Y2k-1

-Y2k

0

Since Y2k 0,itisonlythe oddtermswhichcontribute to

(2.2)

or

(2.3).

For

theseterms wetake

Pn

1and thus thesum onthe left of

(2.3)

is

2k-1

N

k

k;l

(4)

684 B. CHOUDHARY AND S. K. MISHRA N/

N"-

N

-.

m=l

Thus the sum on theleftof

(2.2)

isgreaterthan or

equal

to

EXAMPLE

2. Take

Nk-1

oifN > 1.

Then take

2rr2(n T,r=2,3,4

Yo

(otherwise)

/

In

the sums

(2.2),

(2.3)

allthetermsvanishexceptfor n 2

r,

r 2,3,4,...

So

weneedconsideronly

those terms. Ifn

T,

r>_ ₂then thereare

2r-(r-1)

termsin the sum

N/p"for which

Pm

1 so that

m=l

’

NIp,.

(T-(r-I))N

+ N

m=l p=2

ButNkp

p

N

SOthat for fixed

N

NlOg

=2 =2

p

O

(r

’+1)

o

(29.

Thus,for fixed

N,

and n 2 wehave

sothat

Nlh)" 0(2r)

E

N,+.

ly.l

=0 1

n=l =1

.

<

But

n=l r=2

=

rlN

,-2 r2

(since

N

k

rlos

N)

i.e. forN=3,4,5

(5)

KOTHE-TOEPLITZ

DUALS OF CERTAIN SEQUENCE SPACES 685 We now considerthesecond dual of(Se(p)) i.e.(Se(R)(p))

Is

it true that

Z: sup

Nl/V=

m=l

In

other words,is it truethat

(S’(R)(P))

isthesetofsequencesz

{z,,}

which aresuch that, for

someN

In

orderto seethatthisconjectureis true, weshall firstprovealemma.

LEMMA

1.

Suppose

that, for eachN,

{a,

ts}

is asequenceofpositive numbers, and that,

forfixed n,

a.

ts) is non-decreasinginN. Let

X

denote the setofsequences {y,} which are such that, for allN,

Then

X"

isthe set ofall

(z.}

such that, forsomeN

z.

0

(a

)

(2.6)

PROOF. The result that

(2.6)

is sufficientfor z

X

is trivial; for, if

(2.6)

holds for

someNthen since

(2.5)

holds for allNitholds for thatparticular N, whence

The result that

(2.6)

is necessary is not so obvious.

Suppose

it isfalse that there issomeNfor

which

(2.6)

holds.

Then, forevery

N,

Zll

_is

unbounded.

Hence,

we can determine anincreasing sequence

{n}of

positive integerssuch that

Nowdefiney {y.} by

Yn

ann

0

(n

ns,

N= 1,2,3

(6)

686 B. CHOUDHARY AND S. K. MISHRA Now given anyfixedNwehaveforall M _> N

ytl

M

anMtd)

M

anMtl

(since

a.

iN)_is_{non-decreasing}_for_fixed

n).

Theterms in

(2.5)

forwhich n is notequalto

nM

forsome

M

are0; hence the contributionto

(2.5)

of thesetermswith n _> n_Nisless thanor

equal

to

M

Sincethereareonlyafinite number ofterms with n _{< nN}theseries

(2.6)

converges.This holds

forevery

N;

henceyeX.

Butwhen n nNwehave

ynzn

Hence

ynz

divergessothat z X* n=l

Theconjecture preceding Lemma 1nowfollows from the result for

($4(R) (p))a by

taking

MATRIX

TRANSFORMATIONS

In

this section we findnecessaryandsufficientconditionsfor

A

(S’(R)(p),

a(R)).

Weneed the

followinglemma.

LEMMA

2.

Let

_Pk> 0 foreveryk. Then

(Sl,.(p))

a

a#:

at

N

"

converges,

Rt

N/v’ <

k=l m=l k--I

Where

Rk

E

av"

v=k

PROOF.

Suppose

thatx

S’(R)(p).

Then there isaninteger

whereneN.

Since

N>max(l,

SUPk

IAxklPk)

suchthat

akXk

=

RkAxk-P-/I

Ax

k

k=l k=l _k=l

k=l _k=l

itfollows that

RkAX

kis

absolutely

convergent.

(7)

KOTHE-TOEPLITZ DUALS OF CERTAIN SEQUENCE SPACES Also, by Corollary 2

[3],

the convergenceof

k

implies that

limR/x

Nup. 0

m=l

687

Hence,

itfollows from

(3.1)

that

akx

kis convergent for each x e S.(p). k--I

Thisgivesa e

(se(p)).

Conversely, suppose that a e

(Se=(p))

,

then by definition, x S..(p).

axt

is convergent for each

k;l

Since e (1,1,1 S.(p) andx N

v---I v=l =I

areconvergent.

By

using Corollary2

[3]

wefind that

k

Thus, weobtainfrom

(3.1)

that theseries

Rit

Axit

convergesfor each x

se(R)(p).

kl

Notethatx

se(R)(p)

if and

only

if

Ax

e(R)(p).

Thisimpliesthat

R

{Rk}

(e=(p))

.

Itnowfollows from Theorem2

[4]

that

converges for allN > 1.

Wenowfindnecessaryand sufficientconditionsforamatrix

A

tomap

se(R)(p)

to

e**.

THEOREM2.Letpk > 0 foreveryk.Then

A

(se(R)(p),

e(R))

if andonlyif

(i)

slE

am,

N’a,-

l<oo,

N>l;

k:l =I

(ii)

NIP

any

v--It

< ),N > I.

PROOF. Wefirstprovethattheseconditions arenecessary.

Suppose

that

A

e

x-I

b longsto

se. p),

the condition

(i)

---=1

holds.

In

orderto seethat

(ii)

isnecessarywe assumethat forN > 1,

N

(8)

688 B. CHOUDHARY AND S. K. MISHRA

Thenitfollows fromTheorem 3

[4]

thatB

(e(p),e). Hence,

thereis asequencext

e(R)(p)

such

that

suplXkl

pk _and

bkx

k

,

O(1).

k _k=l

Wenowdefine thesequence y by

k

Yk

Xv(k

e N),

Yo

0. Theny e SQ**(p)

v=l

and

akY

k

bcx

k * O(1).

k=l k=l

This contradictsthat

A

e

(se(p),e(R)).

Thus,

(ii)

is necessary.

We

nowprovethesufficiency partof the theorem.

Suppose

that

(i)

and

(ii)

ofthe theorem

hold. Then

A,

(.se(p))

for eachn eN.

Hence

A(x)

ax

_k convergesfor each n Nandfor each x

se(p).

Followingtheargument

k=l

usedin Lemma2,we findthat ifxe

se(p)

such that sup

IAxl

< N, then

k

k=l k=l v=k

Thisprovesthat

Ax

e(R).

Hence,

the theoremisproved.

ACKNOWLEDGEMENT. The authors thank the refree for

helpful

suggestions.

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,

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H.,

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(2) (1981),

169-176.

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C.G.,

and Maddox,

I.J.,

Matrix transformations between some classes of

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I.J.,

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I.J.,

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I.J., Spaces

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