Linear maps characterized by special products on standard operator algebras

TL;DR

This paper characterizes linear maps on standard operator algebras that satisfy a specific zero-product condition, providing a deeper understanding of their structure in the context of Banach spaces.

Contribution

It offers a new characterization of linear maps based on special product conditions in standard operator algebras, expanding the theoretical framework.

Findings

Identifies conditions under which linear maps satisfy a zero-product relation

Provides a characterization of such maps in standard operator algebras

Enhances understanding of algebraic structures in Banach space operators

Abstract

Let A be a unital standard algebra on a complex Banach space X with dimX >1. We characterize the linear maps D; T : A --> B(X) satisfying aT(b) + D(a)b= 0 whenever a,b in A are such that ab = 0.

Equations

$$ab=0⟹a\tau (b)+\delta (a)b=0,(a,b\in \mathcal{A}).$$

$$AB=0⟹A\tau (B)+\delta (A)B=0,(A,B\in \mathcal{A}).$$

$$\delta (A)=AS-RA,\tau (A)=AT-SA$$

$$AB=0⟹A\delta (B)+\delta (A)B=0,(A,B\in \mathcal{A}).$$

$$\delta (A)=AS-RA$$

$$A\tau (X)+\delta (A)X=AX\tau (I)+\delta (AX).$$

$$AP\tau (Q)+\delta (AP)Q=0.$$

$$AP\tau (I)-AP\tau (P)+\delta (AP)-\delta (AP)P=0.$$

$$AP\tau (I)+\delta (AP)=AP\tau (P)+\delta (AP)P.$$

$$AQ\tau (P)+\delta (AQ)P=0.$$

$$A\tau (P)-AP\tau (P)+\delta (A)P-\delta (AP)P=0.$$

$$A\tau (P)+\delta (A)P=AP\tau (P)+\delta (AP)P.$$

$$A\tau (P)+\delta (A)P=AP\tau (I)+\delta (AP).$$

$$A\tau (X)+\delta (A)X=AX\tau (I)+\delta (AX)$$

$$\tau (XA)+\delta (I)XA=X\tau (A)+\delta (X)A.$$

$$P\tau (QA)+\delta (P)AQ=0$$

$$Q\tau (PA)+\delta (Q)PA=0.$$

$$P\tau (A)+\delta (P)A=P\tau (PA)+\delta (P)PA$$

$$\tau (PA)+\delta (I)PA=P\tau (PA)+\delta (P)PA.$$

$$\tau (PA)+\delta (I)PA=P\tau (A)+\delta (P)A.$$

$$\tau (XA)+\delta (I)XA=X\tau (A)+\delta (X)A$$

$$\delta (AB)=A\delta (B)+\delta (A)B-A\delta (I)B.$$

$$\delta (X)=\tau (X)-X\tau (I)+\delta (I)X,$$

$$\delta (AX)=\tau (AX)-AX\tau (I)+\delta (I)AX$$

$$\tau (AX)=A\tau (X)+\delta (A)X-\delta (I)AX$$

$$\tau (ABX)=AB\tau (X)+\delta (AB)X-\delta (I)ABX,$$

$$\tau (ABX)$$

$$=AB\tau (X)+A\delta (B)X-A\delta (I)BX+\delta (A)BX-\delta (I)ABX$$

$$\delta (AB)X=A\delta (B)X+\delta (A)BX-A\delta (I)BX$$

$$\delta (AB)=A\delta (B)+\delta (A)B-A\delta (I)B$$

$$\tau (AB)=A\tau (B)+\tau (A)B-A\tau (I)B.$$

Full text

MSC(2010): 47L10, 47B49, 47B47.

Keywords: Standard operator algebra, linear map, zero product.

1. Introduction

One of the interesting issues in mathematics is the determination of the structure of linear (additive) mappings on algebras (rings) that act through zero products in the same way as certain mappings, such as homomorphisms, derivations, centralizers, etc. For instance, see [1, 3, 4, 5, 8, 9, 11, 12, 13, 14, 15, 7, 16, 17, 20, 21, 22, 23] and references therein. Among these issues, one can point out the problem of characterizing a linear (additive) map $\delta$ from an algebra (ring) $\mathcal{A}$ into an $\mathcal{A}$-bimodule $\mathcal{M}$, which satisfy

[TABLE]

where $\delta:\mathcal{A}\rightarrow\mathcal{M}$ and $\tau:\mathcal{A}\rightarrow\mathcal{M}$ are linear (additive) maps. The condition (1) has also been studied by some authors and the mappings $\delta$ and $\tau$ have been characterized on different algebras (rings) (see [10, 19]). In [2], the authors consider linear maps $\delta,\tau:\mathcal{A}\rightarrow\mathcal{M}$ satisfying (1) and prove that if the unital algebra $\mathcal{A}$ is generated by idempotents, then $\delta$ and $\tau$ are of the form $\delta(a)=d(a)+\delta(1)a$ and $\tau(a)=d(a)+a\tau(1)$ ($a\in\mathcal{A}$), where $d:\mathcal{A}\rightarrow\mathcal{M}$ is a derivation. Also, characterizations of the maps $\delta$ and $\tau$ are given if $\mathcal{A}$ is assumed to be a triangular algebra under some constraints on the bimodule $\mathcal{M}$. In this paper, we describe the linear mappings of the standard operator algebras in a Banach space that satisfy (1).

2. The main results

Throughout this paper, all algebras and vector spaces will be over the complex field $\mathbb{C}$. Let $\mathcal{X}$ be a Banach space. We denote by $B(\mathcal{X})$ the algebra of all bounded linear operators on $\mathcal{X}$, and $F(\mathcal{X})$ denotes the algebra of all finite rank operators in $B(\mathcal{X})$. Recall that a standard operator algebra is any subalgebra of $B(\mathcal{X})$ which contains $F(\mathcal{X})$. We shall denote the identity matrix of $B(\mathcal{X})$ by $I$. In Theorem 2.1 of this article we characterize the linear maps $\delta,\tau:\mathcal{A}\rightarrow B(\mathcal{X})$ satisfying (1), where $\mathcal{A}$ is a unital standard operator algebra.

Theorem 2.1**.**

Let $\mathcal{X}$ be a Banach space, $dim\mathcal{X}\geq 2$, and let $\mathcal{A}\subseteq B(\mathcal{X})$ be a unital standard operator algebra. Suppose that $\delta$ and $\tau$ be linear maps from $\mathcal{A}$ into $B(\mathcal{X})$ satisfying

[TABLE]

Then there exist $R,S,T\in B(\mathcal{X})$ such that

[TABLE]

for all $A\in\mathcal{A}$.

From Theorem 2.1, one gets the following corollary, which is already proved in [18, Theorem 6]. So it can be said that Theorem 2.1 is a generalization of [18, Theorem 6].

Corollary 2.2**.**

Let $\mathcal{X}$ be a Banach space, $dim\mathcal{X}\geq 2$, and let $\mathcal{A}\subseteq B(\mathcal{X})$ be a unital standard operator algebra. Assume that $\delta:\mathcal{A}\to B(\mathcal{X})$ is a linear map satisfying

[TABLE]

Then there exist $R,S\in B(\mathcal{X})$ such that

[TABLE]

for all $A\in\mathcal{A}$ and $R-S\in Z(B(\mathcal{X}))$.

3. Proof of Theorem 2.1

We proof Theorem 2.1 through the following lemmas.

Lemma 3.1**.**

For all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$, we have

[TABLE]

Proof.

Let $P\in\mathcal{A}$ be an idempotent operator of rank one. Set $Q=I-P$. Then for all $A\in\mathcal{A}$, we obtain $APQ=0$. So by assumption we have

[TABLE]

Therefore,

[TABLE]

Hence

[TABLE]

Since $AQP=0$ ($A\in\mathcal{A}$), it follows that

[TABLE]

So

[TABLE]

Consequently

[TABLE]

By comparing (2) and (3), we obtain

[TABLE]

By [4, Lemma 1.1], every element $X\in F(\mathcal{X})$ is a linear combination of rank-one idempotents, and so

[TABLE]

for all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$. ∎

Lemma 3.2**.**

For all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$, we have

[TABLE]

Proof.

Let $P\in\mathcal{A}$ be a rank-one idempotent operator, and $Q=I-P$. So $PQA=0$ and $QPA=0$ for all $A\in\mathcal{A}$. By assumption we have

[TABLE]

and

[TABLE]

From these equations we have the followings, respectively.

[TABLE]

and

[TABLE]

Comparing these equations, we get

[TABLE]

Now, by [4, Lemma 1.1] we have

[TABLE]

for all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$. ∎

Lemma 3.3**.**

For all $A,B\in\mathcal{A}$, we have

[TABLE]

Proof.

Taking $A=I$ in Lemma 3.1, we find that

[TABLE]

for all $X\in F(\mathcal{X})$. Since $F(\mathcal{X})$ is an ideal in $\mathcal{A}$, it follows from (4) that

[TABLE]

for all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$. From this equation and Lemma 3.1, we obtain

[TABLE]

for all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$. From (5), we have

[TABLE]

for all $A,B\in\mathcal{A}$ and $X\in F(\mathcal{X})$. On the other hand,

[TABLE]

for all $A,B\in\mathcal{A}$ and $X\in F(\mathcal{X})$. By comparing (6) and (3), we see that

[TABLE]

for all $A,B\in\mathcal{A}$ and $X\in F(\mathcal{X})$. Since $F(\mathcal{X})$ is an essential ideal in primitive algebra $B(\mathcal{X})$, it follows that

[TABLE]

for all $A,B\in\mathcal{A}$. ∎

Lemma 3.4**.**

For all $A,B\in\mathcal{A}$, we have

[TABLE]

Proof.

From Lemma 3.2 and (4) we conclude that

[TABLE]

for all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$. Now, by using (3) for all $A,B\in\mathcal{A}$ and $X\in F(\mathcal{X})$, we calculate $\tau(XAB)$ in two ways and we obtain the followings.

[TABLE]

and

[TABLE]

Comparing these equations, we get

[TABLE]

for all $A,B\in\mathcal{A}$ and $X\in F(\mathcal{X})$. Since $F(\mathcal{X})$ is an essential ideal in $B(\mathcal{X})$, it follows that

[TABLE]

for all $A,B\in\mathcal{A}$ and $X\in F(\mathcal{X})$. ∎

Lemma 3.5**.**

For all $A\in\mathcal{A}$, we have

[TABLE]

Proof.

It follows from (4) and Lemma 3.3 that

[TABLE]

for all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$. On the other hand, according to the Lemma 3.4, for all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$, we have

[TABLE]

By comparing these equations, we find that

[TABLE]

for all $A\in\mathcal{A}$ and $X\in F(\mathcal{X})$. Since $F(\mathcal{X})$ is an essential ideal in $B(\mathcal{X})$, it follows that

[TABLE]

for all $A\in\mathcal{A}$. ∎

Now, by considering the obtained results we are ready to prove Theorem 2.1.

Proof of Theorem 2.1: Define the linear map $\Delta:\mathcal{A}\to B(\mathcal{X})$ by $\Delta(A)=\delta(A)-\delta(I)A$. It follows from Lemma 3.3 that

[TABLE]

So $\Delta$ is a derivation and according to [6, Theorem 2.5.14] there exists $S\in\mathcal{B}(\mathcal{X})$ such that $\Delta(A)=AS-SA$ for all $A\in\mathcal{A}$. Set $R=S-\delta(I)$. From the definition of $\Delta$ we conclude that $\delta(A)=AS-RA$ for all $A\in\mathcal{A}$. Also, by Lemma 3.5, we have $\Delta(A)=\tau(A)-A\tau(I)$ for all $A\in\mathcal{A}$. Set $T=S+\tau(I)$. Hence $\tau(A)=AT-SA$ for all $A\in\mathcal{A}$. The proof of theorem is complete.