Let T:A→X\documentclass[12pt]{minimal}
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\begin{document}$$T:A\rightarrow X$$\end{document} be a bounded linear operator, where A is a C∗\documentclass[12pt]{minimal}
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\begin{document}$$\hbox {C}^*$$\end{document}-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\documentclass[12pt]{minimal}
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\begin{document}$$ab =0$$\end{document} in A implies T(b)a+bT(a)=0\documentclass[12pt]{minimal}
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\begin{document}$$T(b) a + b T(a)=0$$\end{document});(b)There exist an anti-derivation d:A→X∗∗\documentclass[12pt]{minimal}
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\begin{document}$$d:A\rightarrow X^{**}$$\end{document} and an element ξ∈X∗∗\documentclass[12pt]{minimal}
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\begin{document}$$\xi \in X^{**}$$\end{document} satisfying ξa=aξ,\documentclass[12pt]{minimal}
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\begin{document}$$\xi a = a \xi ,$$\end{document}ξ[a,b]=0,\documentclass[12pt]{minimal}
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\begin{document}$$\xi [a,b]=0,$$\end{document}T(ab)=bT(a)+T(b)a-bξa,\documentclass[12pt]{minimal}
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\begin{document}$$T(a b) = b T(a) + T(b) a - b \xi a,$$\end{document} and T(a)=d(a)+ξa,\documentclass[12pt]{minimal}
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\begin{document}$$T(a) = d(a) + \xi a,$$\end{document} for all a,b∈A\documentclass[12pt]{minimal}
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\begin{document}$$a,b\in A$$\end{document}.
