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Operators on Some Vector-Valued Orlicz Sequence Spaces

Yılmaz YILMAZ, M. Kemal ÖZDEMİR, İhsan SOLAK and Murat CANDAN İnönü University, Faculty of Science, Department of Mathematics, 44280 Malatya, TURKEY

e-mail: yyilmaz@inonu.edu.tr

Abstract

In this note, we give some sequences of operators which have the same function with a basis for some vector-valued Orlicz sequence spaces. Also, we characterize the space Β

(

hM(X),Y

)

of continuous operators from hM(X) into Y where M is an Orlicz function, X, Y are Banach spaces and hM(X) is the space of all X-valued sequences x=(xk) such that

<

=1 k

xk

M ρ for all ρ>0.

Exactly, we obtain that each TΒ

(

hM(X),Y

)

is equivalent, under certain conditions, to any sequence

=(Ak)k=1

A of operators AkΒ(X,Y).

2000 Mathematics Subject Classification: 46A45, 47A05, 47A67, 46A20.

Keywords: Representations of functionals, Orlicz sequence spaces, Operator spaces.

Bazı Vektör-Değerli Orlicz Dizi Uzayları Üzerindeki Operatörler

Özet

Bu çalışmada, bazı vektör-değerli Orlicz dizi uzayları için bir baz ile aynı işleve sahip olan bir operatör dizisi tanımladık. Ayrıca, bundan faydalanarak hM(X) uzayından Y uzayına sürekli operatörlerin Β

(

hM(X),Y

)

uzayını karakterize ettik. Burada M bir Orlicz fonksiyonu, X, Y Banach uzayları ve hM(X),

<

k=1

xk

M ρ her ρ>0 için

olacak şekildeki tüm X-değerli x=(xk) dizilerinin uzayıdır. Aslında, tam olarak, bazı şartlar altında, her bir TΒ

(

hM(X),Y

)

operatörünün AkΒ(X,Y) operatörlerinin bir A=(Ak)k=1 dizisine denk olduğu sonucuna ulaştık.

2000 Matematik Konu Sınıflaması: 46A45, 47A05, 47A67, 46A20.

Anahtar Kelimeler: Fonksiyonellerin temsilleri, Orlicz dizi uzayları, Operator uzayları.

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1. Introduction

After J. Lindenstrauss & L. Tzafriri [1] introduced the Orlicz sequence space l , many M variations of these spaces investigated and some of them generalized to vector-valued case. For example, a general case of vector-valued Orlicz sequence spaces is given by D. Ghosh & P. D.

Srivastava [2]. In many respect, the properties of operators on some Orlicz sequence spaces (or Orlicz function spaces) are more important than structural investigation of the spaces.

Especially, dealing with some integral equation and introducing an existence theorem to the equation, operators on some Orlicz spaces which has been given with respect to the equation plays a crucial role. Some related results with applications of these spaces presented in some cites such as [3,4]. Further, in the operator theory and operator algebras, characterizations of operators are useful in giving an example or a counterexample to some structural problems. In connection with the vector-valued sequence space theory, the operators from X -valued Orlicz sequence space hM( X) (or lM(X)), where X is a Banach space, into another Banach space Y are our main interest. J. Lindenstrauss & L. Tzafriri, [5], characterized the functionals defined on lM by finding its continuous duals. Dealing with continuous duals of an abstract topological vector space X or representation of operators defined from X into another topological vector space Y , bases of the space X have an important role. This illustrated in cite [6] for some well- known scalar sequence spaces in detail. But, finding a basis for vector-valued sequence spaces is not possible, in general. In this note, we give some sequences of operators which have the same function with a basis for some vector-valued Orlicz sequence spaces. Also, by using this notion, the representation of continuous operators from hM( X) into another Banach space Y are presented.

2. Prerequisites

For some Banach space X , B denotes the unit sphere of X , i.e. X BX =

{

xX: x 1

}

. Specially, we use B instead of M h (X)

BM in the context. Furthermore, X denotes the * continuous dual of X , and for an operator T from X to another Banach space Y , we denote the adjoint operator of T by T such that * (T*f)(x)= f(T(x)) for all x∈ . X

We recall, [7,1], that an Orlicz function is a function M:[0,∞)→[0,∞) which is continuous, non-decreasing and convex with M(0)=0, 0M(u)> for all u>0 and M(u)→∞ as u→∞. An Orlicz function M can always be represented in the following integral form:

, ) ( )

(u 0 p t dt M =

u

where p , known as the kernel of M , is right-differentiable for t≥0, 0p(0)= , 0p(t)> for

>0

t , p is non-decreasing and p(t)→∞ as t→∞.

Consider the kernel p(t) associated with Orlicz function M(u), and let

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} ) ( : sup{

)

(s t p t s

q = ≤ .

Then q possesses the same properties as the function p . Suppose now

ds s q v

N( )=

0v ( ) .

Then N is an Orlicz function. The functions M and N are called mutually complementary Orlicz functions, and they satisfy the Young inequality,

) ( ) (u N v M

uv≤ + for u,v≥0.

An Orlicz function M is said to satisfy the ∆ -condition for small u or at 0 if for each 2 k >0 there exist Rk >0 and uk >0 such that M(ku)≤RkM(u), for all u∈(0,uk] [8].

The space l consists all sequences M (x of scalars such that k)

⎟⎟<

⎜⎜ ⎞

=1 k

xk

M ρ for some ρ>0,

and it becomes a Banach space which is called an Orlicz sequence space with the Luxemburg norm

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟⎠≤

⎜⎜⎝

> ⎛

=

=

1 :

0 inf

) 1

( k

k M

M x

x ρ ρ .

The space l is closely related to the space M l which is an Orlicz sequence space with p

up

u

M( )= , (1≤ p<∞).

Another definition of l , [8], is given by the complementary function to M as follows: M

⎭⎬

⎩⎨

⎧ = ∈ ∈

=

= N

k k k k

M x x w x y y l

l ~

all for converges, :

) (

1

,

where N is the complementary function to M , and ~lN

is the collection of all x in w with

( )

<

k=1N xk . Clearly, ~l ⊆N lN

, and l is normed by the Orlicz norm M

( )

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ≤

=

∑ ∑

=

=

1 :

sup

1

1 k

k k

k

M xky N y

x .

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It was shown that these two norms on l are equivalent. M

An important closed subspace of l is M hM, and introduced by Y. Gribanov as follows:

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟⎠<∞ >

⎜⎜⎝

∈ ⎛

=

=

=

0 all for , :

) (

1

ρ ρ

k

k k

M

M x w x x

h .

Immediately, we can introduce the vector-valued extension of the spaces l and M h for any M

Banach space X . Therefore,

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟<∞ >

⎜⎜⎝

∈ ⎛

=

=

0 some for , :

) ( )

(

1

ρ ρ

k

k M

M x X

s x

l X ,

where )s(X is the space of all X -valued sequences and ⋅ is the norm of X . )lM(X is a Banach space with the Luxemburg norm

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟≤

⎜⎜

> ⎛

=

=

1 :

0 inf

) 1

( k

k M

M x

x ρ ρ ,

and it coincides with l whenever M X =Χ, the set of complex numbers. Further, define the closed subspace hM(X) of lM(X) by x=(xk)∈hM(X) iff

⎟⎟<

⎜⎜⎝

=1 k

xk

M ρ for all ρ>0.

If M satisfies the ∆ -condition then 2 hM(X)=lM(X).

3. Some Results on Vector-valued Orlicz Sequence Spaces

Let we begin this section with introducing another definition of lM(X) by the complementary function N to M .

⎭⎬

⎩⎨

⎧ ∈ = ∈

=

=

)

~ ( ) ( all for converges, )

( : ) ( )

( *

1

X f

f x

f X s x

X k N

k k k

M l

l ,

where )~ ( *

N X

l is the class of all sequences f =(fk) such that

=

( )

<∞

1

k N fk .

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Lemma 1. For each x∈lM(X),

( )

<

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ≤

=

∑ ∑

=

=

1 :

) ( sup

1

1 k

k k

k

M fk x N f

x

defines a norm on lM(X). This norm is said to be Orlicz norm on lM(X).

Lemma 2. On lM(X), the norms ⋅ M and ⋅ (M) are equivalent, and x (M)x M ≤2 x (M). Proofs of these lemmas easily can be given in a similar way followed in cite [8, Theorem 8.9], by using the inequality

=

=

1 1

) (

k

k k k

k

k x f x

f ,

and the fact that x∈lM(X) iff

( )

xk k=1lM.

In general, for hM(X), a Schauder basis isn’t known. But it can be easily verified that hM(X) is separable whenever X is (see [2]). As an extension of the classical case, lM(X) may not be separable even if X is separable. Generally, the separability of Orlicz sequence spaces depends on whether M satisfies the ∆ -condition. 2

Now, for hM(X), let we give a theorem which has the same function with a basis.

Theorem 3. For k =1,2,K, let Ik:XhM(X) and Pk :hM(X)→X be defined by

⎟⎠

⎜ ⎞

=⎛

= 0,0,K,0, ,0,K )

( k k th position

k u u e u

I and Pk(x)=xk,

respectively. Then, for each xhM(X)

=

=

1

) )(

(

k

k

k P x

I

x o ,

that is ( )( ) 0

)

1 (

= M

n

k Ik Pk x

x o as n→∞.

Proof.

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. 1 :

0 inf

) , , , 0 , , 0 , 0 ( )

)(

(

1

) 2 (

1 )

1 (

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟⎠≤

⎜⎜⎝

> ⎛

=

=

+

= + +

=

n k

k n M n M

n k

k k

M x x x x

P I x

ρ ρ

K K

o

Since

k=1M

( )

xρk < for every ρ>0 we can find some positive integer m=m(ρ) such that

( )

1

1

k=m+ xk

M ρ . This shows that

0 1 :

0 inf

1

⎪⎭→

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟≤

⎜⎜ ⎞

>

+ k=n

xk

M ρ

ρ as n→∞.

Lemma 4. Let M be an Orlicz function. The sets

( )

⎭⎬⎫

⎩⎨

⎧ ∈ ≤

=

Λ

=

1 :

) (

1 1

k

xk

M X

s x

and

{

( ): ( ) 1

}

2 = ∈ ≤

Λ x s X x M

are identical.

Proof. Let x∈Λ1, this means

k=1M

( )

xρk 1 for ρ=1. Hence, x (M) ≤1, i.e., x∈Λ2. Conversely, let x∈Λ2, that is

1 1 :

0 inf

1

⎪⎭≤

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟≤

⎜⎜⎝

>

= k

xk

M ρ

ρ .

This means

k=1M

( )

xρk 1 for some ρ≤1. Therefore

k=1M

( )

xk 1 since M is non- decreasing.

4. Characterizations of Operators

Our main result is the following theorem which states the representation of operators from hM(X) into another Banach space Y .

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Theorem 5. Let X , Y be Banach spaces and M , N be mutually complementary Orlicz functions. Then, Β(hM(X),Y) is equivalent by the mapping T →(ToIk) to the Banach space

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ = ∈ = <∞

= =

k k N

B k f

N A A s X Y A A f

E

Y

1

* ) ( sup :

)) , ( ( ) (

*

Β ,

where each I is defined as in Theorem 3. By an equivalence, we mean a one to one, onto, k linear isometry.

Proof. It can be easily verified that E is a Banach space with the norm A . Let N )

), ( (h X Y

T∈Β M and say Ak =ToIk for each k . This implies (Ak oPk)(x) =0 so that 0

) ( k =

k x

A for each k . Since each xhM(X) has the representation =

=

1( )( )

k Ik Pk x

x o we

can write

=

=

=

=

1 1

) ( )

)(

(

k

k k k

k

k x A x

I T

Tx o .

Immediately each Ak∈Β(X,Y) since AkTIk = T . Now, let us define the mapping

N

M X Y E

h

Ψ:Β( ( ), ) , by Ψ(T)= A=(Ak)k=1; Ak =ToIk. 0

)

( =

Ψ T if and only if each T oIk =0 and so T =0 by the definition of each I , i.e. k Ψ is one to one. Also, for an arbitrary AEN, if we define the operator T by

=

=

1

) (

k

k k x A Tx

on hM(X) then, by using the Young inequality, we have

( )

1 . )

sup (

) ( ) sup

sup ( sup

) ( sup

) (

1

*

*

1

* 1

*

*

*

*

*

*

*

*

=

=

=

=

=

=

=

=

=

=

⎟⎟

⎜⎜

⎝ + ⎛

⎟⎟

⎜⎜

≤ ⎛

⎟⎟+

⎜⎜

≤ ⎛

⎟⎟⎠

⎜⎜ ⎞

= ⎛

n m k n k

m

k k k

k B

f

n m k

N k k B k

f n

m

k k k N

k B

f n

m k

k B k

f

n m k

k B k

f n

m k

k k

A M x f

A f N A

x f

A f M

A f N A

x f A

x A f x

A

Y Y

Y Y

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Since ))(Ak*f)k=1∈lN(Β(X,Y for each fY* and

) 1

( * 1

*

⎟⎟≤

⎜⎜

k =

k N k

k

f A

f N A

from the cite [8, Proposition 8.12], we have

) 0 sup (

1

*

*

*

⎟⎟→

⎜⎜

=

= n

m

k k k N

k B

f A f

f N A

Y

as m,n→∞.

Also

1 ⎟⎟→0

⎜⎜

= n m k

k

A

M x as m,n→∞

since xhM(X). This means the series

Ak(xk) is convergent, i.e., T is well-defined.

Further, that the mapping Ψ is onto, i.e., T∈Β(hM(X),Y) comes from the following equalities chain

( )

. )

( sup

4) Lemma (by , 1 :

) )(

( sup sup

) )(

( sup sup

) ( sup

sup

) ( sup sup

* 1

1 1

* 1

* 1

1

*

*

*

*

A f

A

x M x

f A

x f A

x A f

x A Tx

T

k N B k

f

k

k k

k B k

f

k

k B k

x B f

k

k B k

f B x

k

k B k

x B

x

Y Y Y M M Y

M M

=

=

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ≤

=

=

⎟⎟⎠

⎜⎜⎝

= ⎛

=

=

=

=

=

=

=

=

This shows, in the same time, Ψ is an isometry.

Example 6. Let X =Y =c0 and M , N be mutually complementary Orlicz functions.

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟<∞ >

⎜⎜

∈ ⎛

= 0

0) ( ): ,for every 0

( k

M

M x c

s x c

h ρ

ρ

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Let

. 1,2, , ) 2 ( 1 ) 2 ( 1 0 0

0 ) 2 ( 1 ) 2 ( 1 0

0 0

) 2 ( 1 ) 2 ( 1

0 0

0 1

2 2

2 2

2 2

2

K

O M

M M

M

L L L L

=

⎟⎟

⎟⎟

⎟⎟

⎜⎜

⎜⎜

⎜⎜

= k

k k

k k

k k

k Ak

) , (c0 c0

Β is equivalent to infinite matrix class (c0,c0) since c is a BK-spaces and has AK-0 property [9, p.218]. So, each A∈Β(c0,c0) since A =sup

ia =k2 <∞

nik n

k . Now, we assert

that the sequence A=(Ak) defines a continuous linear operator from lM(c0) into c by the 0 virtue of Theorem 5. To show this, we shall denote that

. )

(

sup * 1

*0

<

= =

k k N

B f

f A A

c

Using the unit vector bases of c and 0 c0*=l1, we can easily show that

. 1,2, , ) 2 ( 1 0 0

0

) 2 ( 1 ) 2 ( 1 0 0

0 ) 2 ( 1 ) 2 ( 1 0

0 0

) 2 ( 1 1

2 2 2

2 2

2 2

* K

O M

M M

M

L L L L

=

⎟⎟

⎟⎟

⎟⎟

⎜⎜

⎜⎜

⎜⎜

= k

k k k

k k

k k

Ak

Hence

2 1

* 1

; 2 )

( k

a y a

y y f

Ak = k = kn n= kn = n+ n+

for each f∈l1. Further, for each

) 1

, ,

(a1 a2 K Bl

f = ∈ and for every ρ>0,

⎟⎟⎠

⎜⎜ ⎞

= ⎛

⎟⎟⎠

⎜⎜ ⎞

≤ ⎛

⎟⎟⎠

⎜⎜ ⎞

≤ ⎛

⎟⎟

⎜⎜

=

=

=

ρ π ρ

ρ ρ

1 6 1 1

1

2 1 2

1 2 1

1

1

*

k N N

k f f N

N A

k k k

k

since

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1 .

1 2 2 2 1

1 1 1

*

=

+

= +

=

n

n n k

k f

k k

a y a

f A

If we say

6 , : 1

0

inf 2

⎭⎬

⎩⎨

⎧ ⎟⎟⎠≤

⎜⎜ ⎞

> ⎛

= ρ N ρ π

K

then A f K

k N

k =

) 1 (

* )

( for each

l1

B

f ∈ . We get

<

A f = K

k N B k

f

2 )

(

sup * 1

l1

Since

. 2 )

( 2 )

( * 1 ( )

*f 1 A f K

Ak k= Nk k= N

Remark 7. lN(Β(X,Y))⊆EN. This follows from the fact that

* .

*f A f A f

Akk ⋅ = k

Indeed, if (Ak)∈lN(Β(X,Y)) and fBY*, we can write that

,

1 1

*

=

= ⎟⎟⎠

⎜⎜⎝

≤ ⎛

⎟⎟

⎜⎜

k

k k

k A

f N N A

ρ ρ

whence

) 1 ( )

1 (

* ) ( )

(Akf k= NAk k= N so that

) 1 (

* ) 1 2( )

(Akf k= NAk k= N from Lemma 2. This implies

. )

( 2 )

(

sup * 1 1 ( )

*

<

=

=

k k N k k N

B f

A f

A

Y

Example 8. The inclusion relation in Remark 7 may be strict. Let X =l1, Y =hM, and M , N be mutually complementary Orlicz functions such that M(1)=1. Define Ak: l1hM by

⎟⎟⎠

⎜⎜ ⎞

=⎛

= k k 0,0,K,0,kth positionk ,0,K

kx x e x

A

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for x=(xn)∈l1. Since h is equivalent to M* l by the cite [5, p.148], for some N fBhM*

=

n n ny x x

f( )

where y=(yn)∈lN and y (N) = f . Hence, (Ak*f)(x)= f(Akx)=xkyk so that

* .

k

kf y

A =

This implies

, 1

1 ( )

*

⎟≤

⎜⎜

k= N

k

y f N A

and so

. 1 sup

1 ( )

*

*

⎟≤

⎜⎜

k= N

k B

f y

f N A

hM

This means (Ak)∈EN. On the other hand, Ak =1 for each k . Let us show this assertion. For

hM

B x

,

) 1 ) 1 ( (since ,

1 :

0

) inf

( )

(

x

M x

M x e

x x

A

k

k k M

M k k

=

=

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ⎟⎟⎠≤

⎜⎜⎝

> ⎛

=

= ρ ρ

that is, Akxx(M) ≤1. Also, for x= , ek Akx(M) =1, whence the assertion proved. Therefore, ))

, ( ( )

(Ak ∉l ΒN l1 hM since

=1 (1 )=∞

k N ρ for every ρ>0.

In the case Y is finite dimensional, Dvoretzky-Rogers theorem assert that the inclusion relation (7) is an equivalence. To show this let us recall that m is the space of all scalar 0 sequences taking on only finitely many values, that is m0=span{ζ}, where ζ is the set of all sequences of zeros and ones.

Theorem 9. ))EN =lN(Β(X,Y when Y is finite dimensional.

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Proof. Suppose that (Ak)∈EN and x=(xk)∈hM(X). For each v=(vk)∈m0 we have )

( )

(v x h X

vx= k kM . Indeed, for each ρ>0

,

1 1

1

⎟⎟<

⎜⎜ ⎞

≤ ⎛

⎟⎟⎠

⎜⎜ ⎞

⎛ ⋅

⎟⎟≤

⎜⎜ ⎞

∑ ∑

=

=

= k

k k

k k k

k

k x

x M M v

x M v

λ ρ ρ

ρ

where λ=maxvk . Further,

k

k k k v x A ( )

converges for all vm0, since the operator =

=

1 ( )

k Ak xk

Tx well-defined on hM(X) by the virtue of Theorem 5. Now let (k be a strictly increasing sequence of natural numbers, and i) define )b=(bk by

if . , 0

if , 1

⎩⎨

= =

i i

k k k

k b k

Then, obviously bm0, and so

=

i k k k

k k

k b x Ai x i

A ( ) ( )

is convergent. This implies that the series

k k k x

A ( ) is subseries convergent, so it is unconditionally convergent. Hence, from Dvoretzky-Rogers theorem it is absolutely convergent since Y is finite dimensional. So, we have that

<

k

k k x A ( )

for each x=(xk)∈hM(X). Now, we can find some ykBX such that

k k

k A y

A ≤2

for eachk , since each Ak∈Β(X,Y). Further, define the sequence z=(zk) such that z=ukyk for each (uk)∈hM. Obviously zhM(X) and so

(13)

. 2

) ( 2 2

<

=

=

k k k k

k k k k

k k k k

k k

z A

y u A

y A u u

A

This shows that the real sequence

(

Ak

)

k=1hMα. But, from the cite [8], α-dual of h is M equivalent to l . Hence, N

⎟⎟<

⎜⎜ ⎞

k=1

Ak

N ρ

for some ρ>0. This means A=(Ak)∈lN(Β(X,Y)).

Corollary 10. Let M , N be mutually complementary Orlicz functions. Then )

( )]

(

[hM X *=lN X* .

This is a direct consequence of Theorem 5 and Theorem 9 for Y =Χ, the scalar field of )

(X

hM and X . References

1. J. Lindenstrauss and L. Tzafriri, On Orlicz sequence space, Israel J. Math. 10, 379–390, 1971.

2. D. Ghosh and P. D. Srivastava, On some vector valued sequence spaces using Orlicz function, Glas. Mat. Ser. III 34, 2, 819–826, 1999.

3. M. M. Rao and Z. D. Ren, Applications of Orlicz Spaces, Marcel Dekker, 2002.

4. M. M. Rao and Z. D. Ren, Theory of Orlicz Spaces, Marcel Dekker, 1991.

5. J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces I. Sequence spaces, Springer- Verlag, Berlin-Heidelberg-New York, 1977.

6. I. Singer, Bases in Banach Spaces 1, Springer-Verlag, Berlin, Heidelberg, New York, 1970.

7. M. A. Krasnosel’skii and Y. B. Rutickii, Convex Functions and Orlicz Spaces, Noordhoff Ltd., Groningen, Netherlands, 1961.

8. P. K. Kamthan and M. Gupta, Sequence Spaces and Series, Marcel Dekker, Inc., New York and Basel, 1981.

9. I. J. Maddox, Elements of Functional Analysis, Second Edition, Cambridge University Press, Cambridge, 1988.

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