 Solution

A positive holomorphic line bundle $F$ admits a Hermitian metric whose Chern connection curvature satisfies that $i{F}_F$ is a positive real (1, 1)-form. In a local holomorphic local frame with squared length $e^{-\phi }$, the local formula for the Chern connection on a line bundle gives $F_F=\partial \bar{\partial }\phi$, so positivity means $i\partial \bar{\partial }\phi$ is positive definite. Its closedness makes $\omega =i{F}_F$ a Kähler form; use this form to define the operators below.

Let $n={\dim }_{\mathbb{C}}X>0$, choose a Hermitian metric on $E$, and equip $G_m=E\otimes F^{-m}$ with the tensor-product metric. The curvature of a tensor product connection givesOn $G_m$-valued zero-forms, the Lefschetz commutator is $[L,\Lambda ]=-n\mathrm{id}$. Thus the Bochner-Kodaira-Nakano identity givesThis last operator is a fixed smooth self-adjoint bundle endomorphism. Compactness supplies a finite $C$ such that $⟨R_{E}v,v⟩\ge -C\Vert v{\Vert }^2$ at every point. For a holomorphic section $s$ of $G_m$, $D^{\prime \prime }s=0$ and $D^{\prime \prime \ast }s=0$ by degree, so integrating the identity givesChoose an integer $m_0$ with $m_{0}n>C$. Then $\boxed{H^0(X,E\otimes F^{-m})=0(m\ge m_0)}$. The threshold depends on the fixed bundle $E$, as the curvature bound makes explicit. Positive complex dimension is necessary: on a zero-dimensional manifold positivity is vacuous and a nonzero fibre has nonzero sections for every twist.