[proofplan] Equip $X$ with the Kähler metric determined by the positive curvature form $\omega_L=i\Theta_h(L)$. The Dolbeault resolution and Hodge theory identify the sheaf cohomology group $H^q(X,\Omega_X^p\otimes L)$ with the space of harmonic $L$-valued $(p,q)$-forms. The [Bochner–Kodaira–Nakano identity](/theorems/3859) then shows that every such harmonic form has non-negative energy bounded below by $(p+q-n)$ times its squared norm. When $p+q>n$, this forces the harmonic representative to vanish, hence the cohomology group is zero. [/proofplan] [step:Use positivity of $L$ to choose the curvature Kähler metric] Let $\Theta_h(L)$ denote the Chern curvature of the Hermitian holomorphic line bundle $(L,h)$, and define the real $(1,1)$-form \begin{align*} \omega := i\Theta_h(L). \end{align*} By the positivity hypothesis on $(L,h)$, the form $\omega$ is positive definite. Since Chern curvature is closed, $\omega$ is a Kähler form on $X$. Let $\mu_\omega$ denote the Riemannian volume measure induced by $\omega$. For integers $p,q$ with $0 \le p,q \le n$, let \begin{align*} A_{p,q}(X,L) \end{align*} denote the complex [vector space](/page/Vector%20Space) of smooth $L$-valued $(p,q)$-forms on $X$. The Hermitian metric $h$ on $L$ and the Kähler metric $\omega$ define a pointwise Hermitian inner product $(\cdot,\cdot)_{\omega,h}$ on $L$-valued forms and the global $L^2$ inner product \begin{align*} (\alpha,\beta)_{L^2} := \int_X (\alpha(x),\beta(x))_{\omega,h}\,d\mu_\omega(x), \qquad \alpha,\beta \in A^{p,q}(X,L). \end{align*} Write \begin{align*} \|\alpha\|_{L^2}^2 := \int_X |\alpha(x)|_{\omega,h}^2\,d\mu_\omega(x). \end{align*} [/step] [step:Identify sheaf cohomology with harmonic $L$-valued forms] Let \begin{align*} \bar{\partial}_L:A_{p,q}(X,L)\to A_{p,q+1}(X,L) \end{align*} be the Dolbeault operator induced by the holomorphic structure on $L$. Let \begin{align*} \bar{\partial}_L^*:A_{p,q+1}(X,L)\to A_{p,q}(X,L) \end{align*} be its Hilbert-space adjoint with respect to $(\cdot,\cdot)_{L^2}$. Define the Dolbeault Laplacian \begin{align*} \Box_{\bar{\partial},L,p,q}:A_{p,q}(X,L)\to A_{p,q}(X,L) \end{align*} by \begin{align*} \Box_{\bar{\partial},L,p,q}\alpha := \bar{\partial}_L\bar{\partial}_L^*\alpha+\bar{\partial}_L^*\bar{\partial}_L\alpha. \end{align*} Let \begin{align*} \mathcal{H}_{p,q}(X,L) := \{\alpha\in A_{p,q}(X,L):\Box_{\bar{\partial},L,p,q}\alpha=0\} \end{align*} be the space of harmonic $L$-valued $(p,q)$-forms. By the [Dolbeault theorem for holomorphic vector bundles](/theorems/???) and the [Dolbeault–Hodge theorem for holomorphic vector bundles on compact Kähler manifolds](/theorems/???), the natural map from [harmonic representatives](/theorems/2747) to Dolbeault cohomology gives an isomorphism \begin{align*} \mathcal{H}_{p,q}(X,L)\cong H^q(X,\Omega_X^p\otimes L). \end{align*} Therefore it suffices to prove that $\mathcal{H}_{p,q}(X,L)=\{0\}$ whenever $p+q>n$. [guided] The sheaf $\Omega_X^p\otimes L$ is resolved by the Dolbeault complex of smooth $L$-valued $(p,\bullet)$-forms: \begin{align*} 0\to \Omega_X^p\otimes L \to A^{p,0}(X,L) \xrightarrow{\bar{\partial}_L} A^{p,1}(X,L) \xrightarrow{\bar{\partial}_L} \cdots \xrightarrow{\bar{\partial}_L} A^{p,n}(X,L) \to 0. \end{align*} Thus the sheaf cohomology group $H^q(X,\Omega_X^p\otimes L)$ is computed by the Dolbeault cohomology of this complex. Since $X$ is compact and $\omega$ is Kähler, elliptic Hodge theory applies to the Dolbeault Laplacian \begin{align*} \Box_{\bar{\partial},L}^{p,q}\alpha = \bar{\partial}_L\bar{\partial}_L^*\alpha+\bar{\partial}_L^*\bar{\partial}_L\alpha. \end{align*} It gives a unique harmonic representative in each Dolbeault cohomology class, hence an isomorphism \begin{align*} \mathcal{H}_{p,q}(X,L)\cong H^q(X,\Omega_X^p\otimes L). \end{align*} Consequently, the vanishing theorem is reduced to an analytic statement: prove that every smooth $L$-valued $(p,q)$-form $\alpha$ satisfying \begin{align*} \Box_{\bar{\partial},L,p,q}\alpha=0 \end{align*} must be zero when $p+q>n$. [/guided] [/step] [step:Apply the Bochner–Kodaira–Nakano identity to a harmonic representative] Let $\alpha\in\mathcal{H}_{p,q}(X,L)$. Since $\alpha$ is harmonic, \begin{align*} 0 = (\Box_{\bar{\partial},L,p,q}\alpha,\alpha)_{L^2} = \|\bar{\partial}_L\alpha\|_{L^2}^2 + \|\bar{\partial}_L^*\alpha\|_{L^2}^2. \end{align*} Let \begin{align*} \partial_L^*:A_{p,q}(X,L)\to A_{p-1,q}(X,L) \end{align*} denote the adjoint of the $(1,0)$ part of the Chern connection on $L$. Let \begin{align*} \Lambda_\omega:A_{p,q}(X,L)\to A_{p-1,q-1}(X,L) \end{align*} denote contraction with the Kähler form $\omega$, adjoint to exterior multiplication by $\omega$. We use the commutator convention $[A,B]:=AB-BA$ for composable operators $A$ and $B$. The [Bochner–Kodaira–Nakano identity](/theorems/???) for a Hermitian holomorphic line bundle over a compact Kähler manifold gives \begin{align*} \|\bar{\partial}_L\alpha\|_{L^2}^2 + \|\bar{\partial}_L^*\alpha\|_{L^2}^2 = \|\partial_L^*\alpha\|_{L^2}^2 + \int_X \bigl(([i\Theta_h(L),\Lambda_\omega]\alpha)(x),\alpha(x)\bigr)_{\omega,h} \,d\mu_\omega(x). \end{align*} Because $i\Theta_h(L)=\omega$, the [Lefschetz commutator formula](/theorems/???) on $(p,q)$-forms, with the same convention $[A,B]=AB-BA$, gives \begin{align*} [i\Theta_h(L),\Lambda_\omega]\alpha = [\omega\wedge,\Lambda_\omega]\alpha = (p+q-n)\alpha. \end{align*} Substituting this into the identity yields \begin{align*} 0 = \|\partial_L^*\alpha\|_{L^2}^2 + (p+q-n)\|\alpha\|_{L^2}^2. \end{align*} [guided] The analytic input is the [Bochner–Kodaira–Nakano identity](/theorems/???). We use the commutator convention $[A,B]:=AB-BA$ for composable operators $A$ and $B$. In this setting the identity applies because $X$ is compact Kähler, $L$ is a Hermitian holomorphic line bundle, and $\alpha$ is a smooth $L$-valued $(p,q)$-form. The identity states that \begin{align*} \|\bar{\partial}_L\alpha\|_{L^2}^2 + \|\bar{\partial}_L^*\alpha\|_{L^2}^2 = \|\partial_L^*\alpha\|_{L^2}^2 + \int_X \bigl(([i\Theta_h(L),\Lambda_\omega]\alpha)(x),\alpha(x)\bigr)_{\omega,h} \,d\mu_\omega(x). \end{align*} Here the curvature term is the decisive one. Since we chose the Kähler form to be the curvature form itself, \begin{align*} i\Theta_h(L)=\omega. \end{align*} Exterior multiplication by $\omega$ raises total degree by $2$, while $\Lambda_\omega$ is its adjoint. The [Lefschetz commutator formula](/theorems/???) on forms of bidegree $(p,q)$, under the convention $[A,B]=AB-BA$, is \begin{align*} [\omega\wedge,\Lambda_\omega]\alpha=(p+q-n)\alpha. \end{align*} Therefore the curvature integral becomes \begin{align*} \int_X \bigl(([i\Theta_h(L),\Lambda_\omega]\alpha)(x),\alpha(x)\bigr)_{\omega,h} \,d\mu_\omega(x) = (p+q-n) \int_X |\alpha(x)|_{\omega,h}^2\,d\mu_\omega(x). \end{align*} That is, \begin{align*} \int_X \bigl(([i\Theta_h(L),\Lambda_\omega]\alpha)(x),\alpha(x)\bigr)_{\omega,h} \,d\mu_\omega(x) = (p+q-n)\|\alpha\|_{L^2}^2. \end{align*} Since $\alpha$ is harmonic, the left-hand side of the Bochner–Kodaira–Nakano identity is also \begin{align*} \|\bar{\partial}_L\alpha\|_{L^2}^2 + \|\bar{\partial}_L^*\alpha\|_{L^2}^2 = 0. \end{align*} Combining the two equalities gives \begin{align*} 0 = \|\partial_L^*\alpha\|_{L^2}^2 + (p+q-n)\|\alpha\|_{L^2}^2. \end{align*} [/guided] [/step] [step:Conclude that the harmonic representative vanishes when $p+q>n$] Assume $p+q>n$. Then $p+q-n$ is a positive integer. The identity \begin{align*} 0 = \|\partial_L^*\alpha\|_{L^2}^2 + (p+q-n)\|\alpha\|_{L^2}^2 \end{align*} is a sum of two non-negative [real numbers](/page/Real%20Numbers). Hence each term is zero, and in particular \begin{align*} \|\alpha\|_{L^2}^2=0. \end{align*} By the definition of the $L^2$ norm, \begin{align*} \int_X |\alpha(x)|_{\omega,h}^2\,d\mu_\omega(x)=0. \end{align*} Since $\alpha$ is smooth and $|\alpha|_{\omega,h}^2$ is a continuous non-negative function on $X$, this implies $\alpha=0$ on $X$. Therefore \begin{align*} \mathcal{H}_{p,q}(X,L)=\{0\}. \end{align*} Using the harmonic representative isomorphism from the previous step, \begin{align*} H^q(X,\Omega_X^p\otimes L)=0 \end{align*} for every $p,q$ with $p+q>n$. [/step] [step:Specialize to the canonical bundle] Set $p=n$. Since $K_X=\Omega_X^n$, the preceding vanishing gives \begin{align*} H^q(X,K_X\otimes L) = H^q(X,\Omega_X^n\otimes L) = 0 \end{align*} whenever $n+q>n$. This condition is equivalent to $q>0$, so \begin{align*} H^q(X,K_X\otimes L)=0 \end{align*} for every integer $q>0$. [/step]