[proofplan]
The proof is a coercivity argument. We take an arbitrary $\bar\partial_E$-harmonic $E$-valued $(n,q)$-form and apply the Bochner-Kodaira-Nakano identity with the same curvature-commutator convention as in the hypothesis. Harmonicity makes the Dolbeault Laplacian term vanish, while the curvature lower bound forces the $L^2$ norm of the form to vanish. The Dolbeault-Hodge identification for holomorphic vector bundles then converts the absence of [harmonic representatives](/theorems/2747) into the vanishing of $H^q(X,K_X\otimes E)$.
[/proofplan]
[step:Apply the Bochner-Kodaira-Nakano identity to a harmonic representative]
Let
\begin{align*}
\alpha\in A^{n,q}(X,E)
\end{align*}
be a $\bar\partial_E$-harmonic form. Thus
\begin{align*}
\bar\partial_E\alpha=0
\end{align*}
and
\begin{align*}
\bar\partial_E^*\alpha=0,
\end{align*}
where $\bar\partial_E^*$ denotes the $L^2$-adjoint of $\bar\partial_E$ with respect to $\omega$ and $h$.
Let $\nabla'$ denote the $(1,0)$-part of the Chern connection on $(E,h)$, and let $(\nabla')^*$ denote its $L^2$-adjoint. The hypotheses for the Bochner-Kodaira-Nakano identity are satisfied because $(X,\omega)$ is compact Kähler, $(E,h)$ is a Hermitian holomorphic vector bundle, and $\alpha$ is a smooth $E$-valued form. By that identity, with curvature convention matching the commutator $[i\Theta(E,h),\Lambda]$ in the hypothesis, we have
\begin{align*}
\|\bar\partial_E\alpha\|_{L^2}^2+\|\bar\partial_E^*\alpha\|_{L^2}^2
=
\|\nabla'\alpha\|_{L^2}^2+\|(\nabla')^*\alpha\|_{L^2}^2+\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}.
\end{align*}
Since $\alpha$ is harmonic, the left-hand side is zero. Hence
\begin{align*}
0
=
\|\nabla'\alpha\|_{L^2}^2+\|(\nabla')^*\alpha\|_{L^2}^2+\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}.
\end{align*}
[guided]
We begin with an arbitrary harmonic representative because the theorem asserts that every such representative must vanish. The word harmonic here means harmonic for the Dolbeault Laplacian
\begin{align*}
\Delta_{\bar\partial,E}=\bar\partial_E\bar\partial_E^*+\bar\partial_E^*\bar\partial_E.
\end{align*}
Equivalently, on a compact Hermitian manifold, harmonicity gives the two first-order equations
\begin{align*}
\bar\partial_E\alpha=0
\end{align*}
and
\begin{align*}
\bar\partial_E^*\alpha=0.
\end{align*}
The identity we need is the Bochner-Kodaira-Nakano identity with the sign convention in which the curvature term is exactly
\begin{align*}
\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}.
\end{align*}
Here $\nabla'$ is the $(1,0)$-part of the Chern connection on $(E,h)$, and $(\nabla')^*$ is its $L^2$-adjoint. The hypotheses of the Bochner-Kodaira-Nakano identity are met: $(X,\omega)$ is compact Kähler, $(E,h)$ is a Hermitian holomorphic vector bundle, and $\alpha$ is a smooth $E$-valued $(n,q)$-form. Applied to $\alpha$, the identity gives
\begin{align*}
\|\bar\partial_E\alpha\|_{L^2}^2+\|\bar\partial_E^*\alpha\|_{L^2}^2
=
\|\nabla'\alpha\|_{L^2}^2+\|(\nabla')^*\alpha\|_{L^2}^2+\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}.
\end{align*}
The reason this identity is useful is that harmonicity annihilates the left-hand side. Indeed, both summands on the left are squared $L^2$ norms of zero forms, so
\begin{align*}
\|\bar\partial_E\alpha\|_{L^2}^2+\|\bar\partial_E^*\alpha\|_{L^2}^2=0.
\end{align*}
Therefore the identity becomes
\begin{align*}
0
=
\|\nabla'\alpha\|_{L^2}^2+\|(\nabla')^*\alpha\|_{L^2}^2+\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}.
\end{align*}
This is the central point: the analytic equation for harmonic forms has been converted into a sum of two nonnegative energy terms plus the curvature term appearing in the hypothesis.
[/guided]
[/step]
[step:Use the curvature lower bound to force the harmonic form to vanish]
The squared norms
\begin{align*}
\|\nabla'\alpha\|_{L^2}^2
\end{align*}
and
\begin{align*}
\|(\nabla')^*\alpha\|_{L^2}^2
\end{align*}
are nonnegative. Therefore the preceding identity implies
\begin{align*}
0\ge \bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}.
\end{align*}
The assumed lower bound applies to this smooth $E$-valued $(n,q)$-form $\alpha$, so
\begin{align*}
\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}\ge c\|\alpha\|_{L^2}^2.
\end{align*}
Combining the two inequalities gives
\begin{align*}
0\ge c\|\alpha\|_{L^2}^2.
\end{align*}
Since $c>0$ and $\|\alpha\|_{L^2}^2\ge 0$, it follows that
\begin{align*}
\|\alpha\|_{L^2}^2=0.
\end{align*}
Thus $\alpha=0$ as an $L^2$ form. Since $\alpha$ is smooth, it vanishes identically on $X$.
[/step]
[step:Identify Dolbeault cohomology with harmonic representatives]
By definition of the canonical bundle, $K_X=\Omega_X^n$. By [citetheorem:9104], applied with $p=n$, the natural map from $\bar\partial_E$-harmonic $E$-valued $(n,q)$-forms to Dolbeault cohomology is an isomorphism:
\begin{align*}
\ker\left(\Delta_{\bar\partial,E}:A^{n,q}(X,E)\to A^{n,q}(X,E)\right)\cong H^q(X,K_X\otimes E).
\end{align*}
The hypotheses of that theorem are satisfied because $X$ is compact Kähler and $E\to X$ is a holomorphic Hermitian vector bundle. Since every element of the harmonic space on the left is zero by the preceding step, the [vector space](/page/Vector%20Space) on the right is zero. Hence
\begin{align*}
H^q(X,K_X\otimes E)=0.
\end{align*}
[/step]
