Linear maps which are anti-derivable at zero

Abstract

Let $T:A\to X$ be a bounded linear operator, where $A$ is a C${}^{\ast }$-algebra, and $X$ denotes an essential Banach $A$-bimodule. We prove that the following statements are equivalent: $(a)$ $T$ is anti-derivable at zero (i.e. $ab=0$ in $A$ implies $T(b)a+bT(a)=0$); $(b)$ There exist an anti-derivation $d:A\to X^{\ast \ast }$ and an element $\xi \in X^{\ast \ast }$ satisfying $\xi a=a\xi ,$ $\xi [a,b]=0,$ $T(ab)=bT(a)+T(b)a-b\xi a,$ and $T(a)=d(a)+\xi a,$ for all $a,b\in A$. We also prove a similar equivalence when $X$ is replaced with $A^{\ast \ast }$. This provides a complete characterization of those bounded linear maps from $A$ into $X$ or into $A^{\ast \ast }$ which are anti-derivable at zero. We also present a complete characterization of those continuous linear operators which are ${}^{\ast }$-anti-derivable at zero.