Holomorphic closure of chiral superfields ($\bar{D}f(\Phi )=0$)

A nonsingular holomorphic function of chiral superfields remains chiral because ${\bar{D}}_{\dot{\alpha }}f(\Phi )={\sum }_i{f}_i(\Phi ){\bar{D}}_{\dot{\alpha }}{\Phi }_i=0$. Dependence on conjugate superfields generally destroys the constraint. This explains the holomorphic form of a superpotential.