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Available Online: http://saspjournals.com/sjpms 65

Scholars Journal of Physics, Mathematics and Statistics ISSN 2393-8056 (Print)

Abbreviated Key Title: Sch. J. Phys. Math. Stat. ISSN 2393-8064 (Online)

©Scholars Academic and Scientific Publishers (SAS Publishers)

(An International Publisher for Academic and Scientific Resources)

Arens Regularity of Bilinear Mapping and Reflexivity

Abotaleb Sheikhali, Nader Kanzi

E-mail address: Abotaleb.sheikhali.20@gmail.com, Nad.kanzi@gmail.com

Depatment of Mathematics, Payame Noor University (PNU), Tehran, Iran

*Corresponding author

Abotaleb Sheikhali

Article History

Received: 20.12.2017

Accepted: 20.01.2018

Published: 30.01.2018

DOI:

10.21276/sjpms.2018.5.1.3

Abstract: Let X, Y and Z be normed spaces. In this article we give a tool to

investigate Arens regularity of a bounded bilinear map f : X Y Z. Also, under

some assumptions on and , we give some new results to determine

reflexivity of the spaces.

Keywords: Arens regular, bilinear map, topological center, factor, second dual.

2010 Mathematics Subject Classification. 46H20, 46H25

INTRODUCTION AND PRELIMINARIES

Arens showed in [1] that a bounded bilinear map f : X Y Z on normed

spaces, has two natural different extensions , from into .

When these extensions are equal, f is said to be Arens regular. Throughout the article,

we identify a normed space with its canonical image in the second dual.

We denote by the topological dual of a normed space X. We write for

and so on. Let X, Y and Z be normed spaces and f : X × Y Z be a bounded bilinear

mapping. The natural extensions of fare as following:

(i) , give by where

, , ( is said the adjoint of ).

(ii) , give by

where , , .

(iii) , give by

where , , .

Let now : Y X Z be the flip of f defined by (y, x) = f (x, y), for every x X and y Y . Then is a

bounded bilinear map and it may extends as above to : . In general, the mapping :

is not equal to . When these extensions are equal, then f is Arens regular.

One may define similarly the mappings : and : and the higher rank

adjoints. Consider the nets and converge to and in the topologies,

respectively, then

and

So Arens regularity of is equivalent to the following

If the limits exit for each . The map is the unique extension of such that

is continuous for each and is continuous for

each . The left topological center of is defined by

Since is the unique extension of such that the map is

continuous for each , we can set

.

The right topological center of f may therefore be defined as

Again since the map is continuous for each , we have

.

Abotaleb Sheikhali & Nader Kanzi.; Sch. J. Phys. Math. Stat., 2018; Vol-5; Issue-1 (Jan-Feb); pp-65-68

Available Online: http://saspjournals.com/sjpms 66

A bounded bilinear mapping is Arens regular if and only if or equivalently . It is

clear that . If then the map is said to be left strongly irregular. Also and if

then the map is said to be right strongly irregular. A bounded bilinear mapping f : X Y Z is said to factor if

it is onto.

Investigate Arens regularity of bounded bilinear maps

S. Mohammadzadeh and Vishki H.R proved in [6] acriterion concerning to the regularity of a bounded bilinear

map. They showed that is Arens regular if and only if In the section we provide the same

conditions of Arens regularity. First, we give a similar lemma to the

Lemma 2.1. For a bounded bilinear map from into the following statements are equivalent:

(i) is Arens regular;

(ii) ;

(iii) .

Proof. If (i) hold then is Arens regular. Therefor . For every and we

have

Therefore .

(ii)

(iii) Let and we have

(iii)

(i) Let For every and we have

.

It follows that is Arens regular and this completes the proof

Theorem 2.2. Bounded bilinear map from into is Arens regular if and only if

Proof. Let and be arbitrary. If is Arens regular Then Therefore

.

Conversely, suppose and let and be two nets that are converge to and

in the-topologies, respectivety. Then

Therefore is Arens regular and this completes the proof

Corollary 2.3. For a bounded bilinear map f : X Y Z, the following statements are equivalent:

(i) ;

(ii) and are Arens regular;

(iii)

Proof. The implication (i) (ii) follows from the fact that . Now Theorem

2.2 implies the Arens regularity of , or equivalenty From which

Abotaleb Sheikhali & Nader Kanzi.; Sch. J. Phys. Math. Stat., 2018; Vol-5; Issue-1 (Jan-Feb); pp-65-68

Available Online: http://saspjournals.com/sjpms 67

Therefore the Arens regularity of follows again by Theorem 2.2. Thus is Arens regular.

(ii) (iii) If is Arens regular. Then

Now if is Arens regular. Then we have

The equalities (2-1) and (2-2) together establish the assertion.

(iii) (i) First we show that . For every and

As lies in thus and the proof

Theorem 2.4. Let and A be normed spaces and is a bounded bilinear map. If

factor and is Arens regular. Then is Arens regular.

Proof. Let factor. Thus for every there exists and such that .

Suppose that and , and be bounded nets −converging to and

respectively. For every we have

It follows that is Arens regular

As an cosequnce of this theorem we have the following result:

Corollary 2.5. Let and be normed spaces and is a bounded bilinear map. If

factor and is Arens regular. Then is Arens regular.

Arens regularity and reflexivity

In this section, we show that with which assumptions left strongly irregular property is equivalent to the right

strongly irregular property.

Theorem 3.1. For a bounded bilinear map f : X Y Z,

Abotaleb Sheikhali & Nader Kanzi.; Sch. J. Phys. Math. Stat., 2018; Vol-5; Issue-1 (Jan-Feb); pp-65-68

Available Online: http://saspjournals.com/sjpms 68

(i) If factor then both and are Arens regular if and only if is reﬂexive.

(ii) If factor then both and are Arens regular if and only if is reﬂexive.

Proof. We only give a proof for (ii), A similar proof applies for (i). Let and are Arens regular by Corollary 2.3

. On the other hand factors, So . Therefore

Conversely, using is obvious

As an immediate consequnce of Theorem 3.1 and , we have the next Corollary.

Corollary 3.2. If one of the two following statement is assumed:

(i) and are Arens regular and factor;

(ii) and are Arens regular and factor;

Then every adjoint map and every ﬂip map of is Arens regular.

Corollary 3.3. Let and are Arens regular and factor. Then is left strongly irregular if

and only it is right strongly irregular.

Proof. The follows by applying Theorem 3.1 and

If is reﬂexive. Then obviously bounded bilinear map from into is Arens regular. But from Arens regularity

does not imply the reﬂexivity of . The next Theorem, we use the Theorem 2.2 and show that if .

Then is reﬂexive.

Theorem 3.4. Let bounded bilinear map from into is Arens regular and let is a Banach space. If

for some . Then is reﬂexive.

Proof. Let deﬁne by for every . Obviously for

every . We have

Therefore = for every . Now Theorem 2.2 implies that . Since

thus is onto. Therefore from into is onto. Let so there exists

such that = = Thus is reﬂexive

REFERENCES

1. Arens A, The adjoint of a bilinear operation, Proc. Amer. Math. Soc, 2 (1951), 839-848.

2. Barootkoob S, Mohammadzadeh S and Vishki H.R, Topological centers of certain Banach module action, bulletin of

the Iranian Mathematical Society, Vol. 35 No. 2 (2009), 25-36.

3. Dales H.G, Banach algebras and automatic continuity, London Math. Soc. Monographs 24 (Clarendon Press,

Oxford,2000)

4. Dales H.G, Rodrigues-Palacios A, and Velasco M.V, The second transpose of a derivation, J. London Math. Soc.

64(2) (2001), 707-721.

5. Eshaghi Gordji M and Filali M, Arens regularity of module actions, Studia Math. 181 (3) (2007), 237-254.

6. Mohammadzadeh S and Vishki H.R, Arens regularity of module actions and the second adjoint of a derivation, Bull

Austral. Mat. Soc. 77 (2008), 465-476.

7. Ulger A, Weakly compact bilinear forms and Arens regularity, Proc. Amer. Math. Soc. 101 (1978), 697-701.

8. Sheikhali A, Sheikhali A, Akhlaghi N, Arens regularity of Banach module actions and the strongly irregular

property, J. Math. Computer Sci, 13 (2014), 41-46.