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Here L-superfield means a left-chiral superfield: ${\bar{D}}_{\dot{\alpha }}\Phi =0$, and its complex conjugate is an antichiral superfield, $D_{\alpha }{\Phi }^{\ast }=0$. Interpret the two barred derivatives as the contracted scalar ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. The key point is chiral projection by squared supercovariant derivatives. Since the two components of $\bar{D}$ anticommute, any product of three barred derivatives repeats a component and is zero. Thusfor every superfield $X$, even when $X$ is antichiral. In particular, ${\bar{D}}^2{\Phi }^{\ast }$ is chiral.

The right side is chiral too. Because $\Phi$ is even, the graded derivation rule gives ${\bar{D}}_{\dot{\alpha }}({\Phi }^2)=({\bar{D}}_{\dot{\alpha }}\Phi )\Phi +\Phi ({\bar{D}}_{\dot{\alpha }}\Phi )=0$. Constant $m$ and $\lambda$ do not contribute derivatives. Consequentlyso applying the chiral constraint to the equation imposes no additional incompatible condition. This is the holomorphic closure of chiral superfields; a generic nonholomorphic right side would not satisfy it.

Both sides have zero Grassmann parity and are Lorentz scalars: the two dotted derivative indices are contracted with the invariant epsilon tensor, and $\Phi$ is a scalar superfield. Moreover, each covariant derivative anticommutes with every supercharge, so its squared contraction commutes with the supercharges. Hence the equation transforms covariantly under supersymmetry. It is the chiral field equation of motion of the Wess–Zumino model with superpotential $\mathcal{W}(\Phi )=m{\Phi }^2/2+\lambda {\Phi }^3/3$, for which ${\mathcal{W}}^{\prime }=m\Phi +\lambda {\Phi }^2$, using the normalization in the PDF. With mass dimensions $[\Phi ]=1$, $[\bar{D}]=1/2$, $[m]=1$ and $[\lambda ]=0$, each side has dimension two. No component field equations of motion are needed for this consistency argument.

A four-dimensional $\mathcal{N}=1$ chiral superfield is a complex scalar superfield satisfyingThe supersymmetric covariant derivatives anticommute with the supercharges, so the constraint is preserved by supersymmetry. Use left Grassmann derivatives and the printed convention ${\bar{D}}_{\dot{\alpha }}=-{\bar{\partial }}_{\dot{\alpha }}-i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\partial }_{\mu }$. The conjugate antichiral superfield satisfies $D_{\alpha }{\Phi }^{†}=0$.

Treating derivatives with respect to the Grassmann variables as graded derivatives and using $\{{\partial }_{\alpha },{\theta }^{\beta }\}={\delta }_{\alpha }{}^{\beta }$ givesThe terms in $\{{\mathcal{D}}_{\alpha },{\mathcal{D}}_{\beta }\}$ and $\{{\overline{\mathcal{D}}}_{\dot{\alpha }},{\overline{\mathcal{D}}}_{\dot{\beta }}\}$ cancel pairwise, so both vanish.

A chiral superfield obeys ${\overline{\mathcal{D}}}_{\dot{\alpha }}\Phi =0$, while an antichiral superfield obeys ${\mathcal{D}}_{\alpha }{\Phi }^{†}=0$. Since ${\overline{\mathcal{D}}}_{\dot{\alpha }}{y}^{\mu }=0$ for $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$, the general chiral expansion isIn the original coordinates this iswith signs following the conventions of the question.

A holomorphic function $W(\Phi )$ is chiral. Its highest component transforms into a spacetime divergence, so the F-term actionis supersymmetric even though it integrates over only chiral half of superspace.

The non-renormalization theorem says that perturbative loop corrections are full-superspace D-terms and cannot generate a new local superpotential. Holomorphy and spurion symmetries therefore preserveThe Kähler potential is renormalized, however. If its kinetic term is $Z{\Phi }^{†}\Phi$, canonical normalization ${\Phi }_c=Z^{1/2}\Phi$ givesup to scheme and scale conventions. Thus superpotential parameters are holomorphic invariants while physical masses and couplings still run through wave-function renormalization.