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`Chinese Journal of MathematicsVolume 2014 (2014), Article ID 698621, 3 pageshttp://dx.doi.org/10.1155/2014/698621`
Research Article

## A Characterization of Jordan Homomorphism on Banach Algebras

Department of Mathematics, Ayatollah Borujerdi University, Borujerd, Iran

Received 20 October 2013; Accepted 17 December 2013; Published 29 January 2014

Academic Editors: C.-J. Zhao and Q. Zhu

Copyright © 2014 Abbas Zivari-Kazempour. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let and be Banach algebras and let be a Jordan homomorphism. We show that, under special hypotheses, is ring homomorphism. Some related results are given as well.

#### 1. Introduction

Let and be Banach algebras and let be a linear map. Then is called Jordan homomorphism if or equivalently, for all . Moreover, if is multiplicative, that is, then is called ring homomorphism. It is obvious that ring homomorphisms are Jordan, but the converse is false, in general. In fact, the converse is true under certain conditions. For example, each Jordan homomorphism from a commutative Banach algebra into is a ring homomorphism.

In [1], Żelazko proved that each Jordan homomorphism of Banach algebra into a semisimple commutative Banach algebra is ring homomorphism (see also Theorem 1.1 of [2]).

In [3], Eshaghi Gordji studied the concept of -Jordan homomorphisms for complex algebras. The linear map is said to be -Jordan homomorphism if, for all , . A 2-Jordan homomorphism is a Jordan homomorphism, in the usual sense.

He proved that each 3-Jordan homomorphism between commutative algebras and is 3-ring homomorphism [3]. Moreover, he showed that Zelazko’s theorem is valid for 3-Jordan homomorphism.

Throughout the paper, all Banach algebras are assumed to be over the complex field .

The proof of the following lemma is contained in [4].

Lemma 1. Let and be Banach algebras and let be a Jordan homomorphism. Then for all ,

#### 2. Jordan Homomorphism

In general, the kernel of an -Jordan homomorphism may not be an ideal. For example, let be the algebra of all matrices having on and below the diagonal. In this algebra product of any elements is equal to , so any linear map from into itself is a 3-Jordan homomorphism but its kernel does not need to be an ideal (see also [3], for the case ).

For Jordan homomorphism, we have the following.

Proposition 2. Let be a Jordan homomorphism from Banach algebra into a semiprime commutative Banach algebra . Then is an ideal of .

Proof. Let and be arbitrary. Since is Jordan and , we have It follows that . Put , and then Hence, is a central nilpotent element of , but semiprime algebra contains not of central nilpotent non-zero elements; therefore, . Thus, . Similarly, .

Theorem 3. Let be a Jordan homomorphism of Banach algebra into a semiprime commutative Banach algebra . If there exist linear functional on such that , then is ring homomorphism.

Proof. Let , be arbitrary elements of . Then the result is clear, if or is zero. We first assume that . Then , and so by the above proposition , it follows that Therefore, is ring homomorphism. Now suppose that , and both of them be nonzero. Take , , , and . Then , so Hence, , and the above argument shows that which implies that This completes the proof.

For certain calculations, we use the following notation which is called the Lie product of and :

Theorem 4. Let be a Jordan homomorphism from a commutative Banach algebra , with its range dense into a semiprime Banach algebra . Then is ring homomorphism.

Proof. By Lemma 6.3.2(e) of [4], we have for all . Since is commutative, , so The continuity of Lie product and density of the range of imply that is a central nilpotent element of and so it must be zero. Therefore, Thus, if , then we conclude that Therefore, is ring homomorphism.

As a consequence, we deduce the following result.

Corollary 5. Let be a Jordan homomorphism from a Banach algebra , with its range dense into a semiprime Banach algebra . If is commutative, then so is .

Theorem 6. Let be a Jordan homomorphism of Banach algebra into a unital Banach algebra with unit element . Then, for all with , .

Proof. Suppose there exist with and . Take . Then , so Thus, by Lemma 1 we have Since is Jordan and , (15) implies that and so , by (16). Therefore, It follows that , which is a contradiction. This completes the proof.

Let be a Banach algebra. It is well known that, on the second dual space of , there are two multiplications, called the first and second Arens products which make into a Banach algebra (see [5, 6]). By definition, the first Arens product on is induced by the left -module structure on . That is, for each , , and , we have Similarly, the second Arens product on is defined by considering as a right -module.

An element is called mixed unit if it is a right unit for and a left unit for . It is well known that an element is a mixed unit if and only if it is a weak* cluster point of some bounded approximate identity in [7].

Theorem 7. Let and be Banach algebras and let be a Jordan homomorphism. Then is a Jordan homomorphism.

Proof. Let and ; then we have It follows that and thus for all , Therefore, . Now we have Therefore, , as required.

The proof of the next result is the same as that of the above theorem.

Theorem 8. Let and be Banach algebras and let be a 3-Jordan homomorphism. Then is a 3-Jordan homomorphism.

Corollary 9. Suppose is a 3-Jordan homomorphism between Banach algebras. If is a commutative and semisimple, then is 3-ring homomorphism.

Proof. It follows of above Theorem and Theorem 2.5 of [3].

Proposition 10. Let be a 3-Jordan homomorphism between commutative Banach algebras. If has a BAI, then defined by is a homomorphism.

Proof. This follows of Theorem 2.2 of [3] and Theorem 3.2 of [8].
We recall that an element of -algebra is said to be partial isometry if .

Theorem 11. Let and be -algebras and let be a Jordan homomorphism preserving the involution. If is weakly norm continuous, then .

Proof. By Lemma 1, for all we have Thus, preserves the partial isometries. Since every partial isometry has norm , hence for partial isometries and positive scalers with , we conclude that By assumption is weakly norm continuous, so it follows that .

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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