New insights into linear maps which are anti-derivable at zero

Abstract

Let $A$ be a Banach algebra admitting a bounded approximate unit and satisfying property $\mathbb{B}$. Suppose $T:A\to X$ is a continuous linear map, where $X$ is an essential Banach $A$-bimodule. We prove that the following statements are equivalent: $(i)$ $T$ is anti-derivable at zero (i.e., $ab=0$ in $A$ $\Rightarrow T(b)\cdot a+b\cdot T(a)=0$); $(ii)$ There exist an element $\hat{I}¾\in X^{\ast \ast }$ and a linear map (actually a bounded Jordan derivation) $d:A\to X$ satisfying $\hat{I}¾\cdot a=a\cdot \hat{I}¾\in X$, $T(a)=d(a)+\hat{I}¾\cdot a$, and $d(b)\cdot a+b\cdot d(a)=-2\hat{I}¾\cdot (ba),$ for all $a,b\in A$ with $ab=0$. Assuming that $A$ is a C${}^{\ast }$-algebra we show that a bounded linear mapping $T:A\to X$ is anti-derivable at zero if, and only if, there exist an element $\hat{I}\cdot \in X^{\ast \ast }$ and an anti-derivation $d:A\to X$ satisfying $\hat{I}\cdot \cdot a=a\cdot \hat{I}\cdot \in X$, $\hat{I}\cdot \cdot [a,b]=0$ {\rm(}i.e., $L_{\hat{I}}\cdot :A\to A$, $L_{\hat{I}}\cdot (a)=\hat{I}\cdot \cdot a$ vanishes on commutators{\rm)}, and $T(a)=d(a)+\hat{I}\cdot \cdot a$, for all $a,b\in A$. The results are also applied for some special operator algebras.