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Use both Kähler identities from the preceding parts, so that ${\partial }^{\ast }=i[\Lambda ,\bar{\partial }]$ and ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. For odd operators $a,b$, write $\{a,b\}=ab+ba$. Direct expansion givesSince $\partial \bar{\partial }+\bar{\partial }\partial =0$, we obtainTogether with part (b), this proves the full Kähler Laplacian identity

Now take the exact form $\eta$. Being $\bar{\partial }$-exact, it is orthogonal to the harmonic space, so $H\eta =0$, and it is also $\bar{\partial }$-closed. Choose $\alpha =G\eta$ for the Dolbeault Green operator of the Dolbeault Laplacian. Commutation gives $\bar{\partial }\alpha =G\bar{\partial }\eta =0$. ThereforeSuppose also $\partial \eta =0$, and put $v={\bar{\partial }}^{\ast }\alpha$, $u=\partial v$. The mixed identities and $({\bar{\partial }}^{\ast }{)}^2=0$ giveThus $u=\partial {\bar{\partial }}^{\ast }\alpha$ is harmonic for the Dolbeault Laplacian. By the Kähler Laplacian identity it is also $\partial$-harmonic and hence ${\partial }^{\ast }u=0$. But it is $\partial$-exact, soConsequently $\boxed{\partial {\bar{\partial }}^{\ast }\alpha =0}$: the form $v$ is $\partial$-closed.

For every harmonic form $h$ of the same bidegree as $v$, $⟨v,h⟩=⟨\alpha ,\bar{\partial }h⟩=0$. Harmonic spaces for the two Dolbeault Laplacians coincide, so $v$ has no $\partial$-harmonic component. Apply the $\partial$ version of the Dolbeault Hodge decomposition. Since $v$ is $\partial$-closed, it has no ${\partial }^{\ast }$-coexact component either. Thus $v=\partial \varphi$, with $\varphi \in A^{p-1,q-1}$. We conclude the requested ddbar lemma:For example $\varphi ={\partial }^{\ast }Gv$ is a choice, since the same Green operator serves ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. If $p=0$ or $q=0$, the relevant negative-bidegree space is zero and the argument forces $\eta =0$. Compactness, the absence of boundary and the Kähler condition are essential to this global harmonic argument.

The Hermitian metric induces pointwise Hermitian inner products on the exterior powers of the complex cotangent bundle, with different bidegrees orthogonal, and a positive volume form $d{V}_g$. Thus the global $L^2$ inner product on smooth differential forms of type $(p,q)$ isLet ${\bar{\partial }}^{\ast }$ be the formal adjoint of the Dolbeault operator, defined by integration by parts, and let ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. This Dolbeault Laplacian preserves bidegree, is elliptic and nonnegative, and hasOn a compact Hermitian manifold without boundary, the Dolbeault Hodge decomposition on a compact Hermitian manifold says that the smooth harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$ is finite-dimensional andMore precisely, there are the orthogonal harmonic projection $H$ and a Dolbeault Green operator $G$ taking smooth forms to smooth forms, with $G=0$ on harmonics and ${\Delta }_{\bar{\partial }}G=G{\Delta }_{\bar{\partial }}=I-H$. Elliptic regularity ensures these are statements about smooth forms, not only $L^2$ completions. The Green operator commutes with $\bar{\partial }$ and ${\bar{\partial }}^{\ast }$, since the Laplacian does and its inverse is unique on the orthogonal complement of the kernel.

Expanding $\alpha =H\alpha +{\Delta }_{\bar{\partial }}G\alpha$ givesThe three summands are orthogonal: harmonic forms are killed by both adjoints, and $⟨\bar{\partial }u,{\bar{\partial }}^{\ast }v⟩=⟨{\bar{\partial }}^{2}u,v⟩=0$. If $\alpha$ is $\bar{\partial }$-closed, its coexact component is orthogonal to every closed form; taking its inner product with $\alpha$ shows that component has norm zero. Conversely the first two components are closed. Thus $\boxed{\ker \bar{\partial }={\mathcal{H}}_{\bar{\partial }}^{p,q}\oplus \bar{\partial }{A}^{p,q-1}}$, with out-of-range form spaces taken to be zero.