[proofplan] The proof is the standard $L^2$ method for $\bar{\partial}$. First, for a smooth metric and compactly supported forms, the Bochner-Kodaira-Nakano identity gives a coercive lower bound for the $\bar{\partial}$-Laplacian on $L$-valued $(0,1)$-forms; the curvature of $L$ contributes $q$, while the Ricci curvature can lose at most $K$. Completeness of $X$ supplies cutoff functions with gradients tending to zero, so the coercive estimate extends from compactly supported forms to the Hilbert-space graph domain. The Hilbert-space range theorem then solves $\bar{\partial}u=f$ with the stated norm bound. Finally, singular metrics are handled under the stated regularization hypothesis by monotone smooth approximation, uniform estimates, weak compactness, and lower semicontinuity. [/proofplan] [step:Establish the coercive estimate for smooth compactly supported forms] First assume that $h$ is smooth and satisfies $i\Theta_h(L)\ge q\omega$. Let \begin{align*} \bar{\partial}_0:\operatorname{Dom}(\bar{\partial}_0)\subset L^2_{(0,0)}(X,L;h,\omega)&\to L^2_{(0,1)}(X,L;h,\omega),\\ \bar{\partial}_1:\operatorname{Dom}(\bar{\partial}_1)\subset L^2_{(0,1)}(X,L;h,\omega)&\to L^2_{(0,2)}(X,L;h,\omega) \end{align*} denote the maximal closed extensions of the Dolbeault operator, meaning the distributional $\bar{\partial}$ operators whose distributional images lie in the indicated $L^2$ spaces. Let $\bar{\partial}_0^*$ denote the Hilbert-space adjoint of $\bar{\partial}_0$. Let $C_c^\infty(X,(T^*X)_{0,1}\otimes L)$ denote the space of smooth compactly supported $L$-valued $(0,1)$-forms, where $(T^*X)_{0,1}$ denotes the anti-holomorphic cotangent bundle. Let $\alpha\in C_c^\infty(X,(T^*X)_{0,1}\otimes L)$. The [Bochner-Kodaira-Nakano identity](/page/Bochner-Kodaira-Nakano%20Identity) for $L$-valued $(0,1)$-forms, with the Ricci term included in the curvature of $(T^*X)_{0,1}$ with the displayed sign convention, gives \begin{align*} \|\bar{\partial}_0^*\alpha\|_{L^2(h,\omega)}^2 + \|\bar{\partial}_1\alpha\|_{L^2(h,\omega)}^2 = \|\nabla^{0,1}\alpha\|_{L^2(h,\omega)}^2 + \int_X \left\langle \bigl(i\Theta_h(L)+\operatorname{Ric}(\omega)\bigr)\alpha(x), \alpha(x) \right\rangle_{\omega,h} \,dV_\omega(x), \end{align*} where $\nabla_{0,1}$ is the $(0,1)$-part of the Chern connection on $(T^*X)_{0,1}\otimes L$. This is the cited Bochner-Kodaira-Nakano identity for line bundles in the convention where the curvature of the coefficient bundle $L\otimes (T^*X)_{0,1}$ contributes $i\Theta_h(L)+\operatorname{Ric}(\omega)$ on $(0,1)$-forms. More explicitly, if $(e_1,\dots,e_n)$ is a local unitary frame for $T^{1,0}X$ and $\alpha=\sum_{j=1}^n \alpha_j\,\overline{e_j}^*\otimes s$ in a local holomorphic frame $s$ of $L$, then a real $(1,1)$-form $A\ge c\omega$ acts on $(0,1)$-forms by \begin{align*} \left\langle A\alpha,\alpha\right\rangle_{\omega,h} \ge c\sum_{j=1}^n |\alpha_j|_h^2 = c|\alpha|_{\omega,h}^2. \end{align*} Thus the convention-dependent curvature term reduces to the displayed pointwise operator lower bound. The hypotheses $i\Theta_h(L)\ge q\omega$ and $\operatorname{Ric}(\omega)\ge -K\omega$ imply, pointwise on $X$, \begin{align*} \left\langle \bigl(i\Theta_h(L)+\operatorname{Ric}(\omega)\bigr)\alpha(x), \alpha(x) \right\rangle_{\omega,h} \ge (q-K)|\alpha(x)|_{\omega,h}^2. \end{align*} Since $\|\nabla^{0,1}\alpha\|_{L^2(h,\omega)}^2\ge 0$, integrating the pointwise lower bound gives \begin{align*} \|\bar{\partial}_0^*\alpha\|_{L^2(h,\omega)}^2 + \|\bar{\partial}_1\alpha\|_{L^2(h,\omega)}^2 \ge (q-K)\|\alpha\|_{L^2(h,\omega)}^2. \end{align*} [guided] We first prove the estimate in the setting where all integrations by parts are justified directly: $h$ is smooth, $i\Theta_h(L)\ge q\omega$, and the form $\alpha$ has compact support. Define the closed Hilbert-space operators \begin{align*} \bar{\partial}_0:\operatorname{Dom}(\bar{\partial}_0)\subset L^2_{(0,0)}(X,L;h,\omega)&\to L^2_{(0,1)}(X,L;h,\omega),\\ \bar{\partial}_1:\operatorname{Dom}(\bar{\partial}_1)\subset L^2_{(0,1)}(X,L;h,\omega)&\to L^2_{(0,2)}(X,L;h,\omega), \end{align*} where both are maximal distributional extensions of the Dolbeault operator with $L^2$ output. The adjoint $\bar{\partial}_0^*$ is the Hilbert-space adjoint with respect to the $L^2$ inner products induced by $h$ and $\omega$. For $\alpha\in C_c^\infty(X,(T^*X)_{0,1}\otimes L)$, the [Bochner-Kodaira-Nakano identity](/page/Bochner-Kodaira-Nakano%20Identity) for $L$-valued $(0,1)$-forms states, in the sign convention where $(T^*X)_{0,1}$ contributes $\operatorname{Ric}(\omega)$, that \begin{align*} \|\bar{\partial}_0^*\alpha\|_{L^2(h,\omega)}^2 + \|\bar{\partial}_1\alpha\|_{L^2(h,\omega)}^2 = \|\nabla^{0,1}\alpha\|_{L^2(h,\omega)}^2 + \int_X \left\langle \bigl(i\Theta_h(L)+\operatorname{Ric}(\omega)\bigr)\alpha(x), \alpha(x) \right\rangle_{\omega,h} \,dV_\omega(x). \end{align*} This is the standard Bochner-Kodaira-Nakano identity for line bundles in this convention. To see how the curvature hypothesis enters, choose a local unitary frame $(e_1,\dots,e_n)$ for $T^{1,0}X$ and write $\alpha=\sum_{j=1}^n \alpha_j\,\overline{e_j}^*\otimes s$ in a local holomorphic frame $s$ of $L$. If a real $(1,1)$-form $A$ satisfies $A\ge c\omega$, then its induced action on $(0,1)$-forms satisfies \begin{align*} \left\langle A\alpha,\alpha\right\rangle_{\omega,h} \ge c\sum_{j=1}^n |\alpha_j|_h^2 = c|\alpha|_{\omega,h}^2. \end{align*} This explains the sign convention: the coefficient-bundle curvature term appearing in the identity is exactly the operator induced by $i\Theta_h(L)+\operatorname{Ric}(\omega)$ on $(0,1)$-forms. The term $\|\nabla^{0,1}\alpha\|_{L^2(h,\omega)}^2$ is nonnegative, so the entire lower bound comes from this curvature term. Now use the two curvature assumptions. The inequality $i\Theta_h(L)\ge q\omega$ means that the curvature of $L$ contributes at least $q|\alpha(x)|_{\omega,h}^2$ at each point. The Ricci lower bound $\operatorname{Ric}(\omega)\ge -K\omega$ means that the Ricci contribution can decrease this by at most $K|\alpha(x)|_{\omega,h}^2$. Therefore \begin{align*} \left\langle \bigl(i\Theta_h(L)+\operatorname{Ric}(\omega)\bigr)\alpha(x), \alpha(x) \right\rangle_{\omega,h} \ge (q-K)|\alpha(x)|_{\omega,h}^2. \end{align*} Substituting this pointwise inequality into the Bochner identity and integrating with respect to $dV_\omega(x)$, we obtain \begin{align*} \|\bar{\partial}_0^*\alpha\|_{L^2(h,\omega)}^2 + \|\bar{\partial}_1\alpha\|_{L^2(h,\omega)}^2 \ge (q-K)\|\alpha\|_{L^2(h,\omega)}^2. \end{align*} This is the key coercive estimate. The strict inequality $q>K$ is exactly what makes the right-hand side positive. [/guided] [/step] [step:Extend the coercive estimate to the full graph domain using completeness] Let $\alpha\in\operatorname{Dom}(\bar{\partial}_1)\cap\operatorname{Dom}(\bar{\partial}_0^*)$. Since $(X,\omega)$ is complete, the [Gaffney cutoff lemma](/page/Gaffney%20Cutoff%20Lemma) applies to the complete Riemannian metric underlying $\omega$ and gives a sequence of smooth cutoff functions \begin{align*} \chi_m:X&\to[0,1] \end{align*} such that $\chi_m\in C_c^\infty(X)$, $\chi_m\to 1$ pointwise on $X$, and \begin{align*} \sup_{x\in X}|\nabla\chi_m(x)|_\omega\to 0. \end{align*} This is the cited Gaffney cutoff lemma on complete Riemannian manifolds. Define \begin{align*} \alpha_m:X&\to \Lambda^{0,1}T^*X\otimes L,\\ x&\mapsto \chi_m(x)\alpha(x). \end{align*} The [Gaffney graph-norm density theorem](/page/Gaffney%20Graph-Norm%20Density%20Theorem) on complete Hermitian manifolds applies because $X$ is complete and the operators are the maximal $L^2$ Dolbeault operators. Thus there is a sequence $(\beta_{m,k})_{k=1}^\infty$ in $C_c^\infty(X,(T^*X)_{0,1}\otimes L)$ such that $\beta_{m,k}\to\alpha_m$ in the graph norm of $\operatorname{Dom}(\bar{\partial}_1)\cap\operatorname{Dom}(\bar{\partial}_0^*)$ as $k\to\infty$. Applying the compact-support estimate to $\beta_{m,k}$ and passing to the limit in this graph norm gives the same estimate for $\alpha_m$. The product rule gives \begin{align*} \bar{\partial}_1(\chi_m\alpha) &= \chi_m\bar{\partial}_1\alpha + \bar{\partial}\chi_m\wedge \alpha, \\ \bar{\partial}_0^*(\chi_m\alpha) &= \chi_m\bar{\partial}_0^*\alpha - (\partial\chi_m)\lrcorner \alpha. \end{align*} The error terms satisfy \begin{align*} \|\bar{\partial}\chi_m\wedge \alpha\|_{L^2(h,\omega)} + \|(\partial\chi_m)\lrcorner \alpha\|_{L^2(h,\omega)} \le 2\sup_{x\in X}|\nabla\chi_m(x)|_\omega\,\|\alpha\|_{L^2(h,\omega)}. \end{align*} Hence $\bar{\partial}_1(\chi_m\alpha)\to \bar{\partial}_1\alpha$, $\bar{\partial}_0^*(\chi_m\alpha)\to \bar{\partial}_0^*\alpha$, and $\chi_m\alpha\to\alpha$ in $L^2(h,\omega)$. Passing to the limit in the compact-support estimate gives \begin{align*} \|\bar{\partial}_0^*\alpha\|_{L^2(h,\omega)}^2 + \|\bar{\partial}_1\alpha\|_{L^2(h,\omega)}^2 \ge (q-K)\|\alpha\|_{L^2(h,\omega)}^2. \end{align*} [guided] The compact-support estimate is not yet enough because the Hilbert-space solvability argument uses the full domains of $\bar{\partial}_1$ and $\bar{\partial}_0^*$. Completeness is the hypothesis that bridges this gap. Let $\alpha\in\operatorname{Dom}(\bar{\partial}_1)\cap\operatorname{Dom}(\bar{\partial}_0^*)$. Since $(X,\omega)$ is complete, the [Gaffney cutoff lemma](/page/Gaffney%20Cutoff%20Lemma) supplies smooth compactly supported functions \begin{align*} \chi_m:X&\to[0,1] \end{align*} with $\chi_m\to 1$ pointwise and \begin{align*} \sup_{x\in X}|\nabla\chi_m(x)|_\omega\to 0. \end{align*} This cutoff lemma is a consequence of completeness of the underlying Riemannian metric. Define the $L$-valued $(0,1)$-form \begin{align*} \alpha_m:X&\to \Lambda^{0,1}T^*X\otimes L,\\ x&\mapsto \chi_m(x)\alpha(x). \end{align*} The form $\alpha_m$ has compact support but need not be smooth. By the [Gaffney graph-norm density theorem](/page/Gaffney%20Graph-Norm%20Density%20Theorem) on complete Kähler manifolds, there is a sequence $(\beta_{m,k})_{k=1}^\infty$ of compactly supported smooth $L$-valued $(0,1)$-forms such that $\beta_{m,k}\to\alpha_m$ in the graph norm of $\operatorname{Dom}(\bar{\partial}_1)\cap\operatorname{Dom}(\bar{\partial}_0^*)$ for the maximal Dolbeault operators. We apply the compact-support estimate to each $\beta_{m,k}$, then let $k\to\infty$; graph-norm convergence gives convergence of the three terms in the estimate, so the estimate holds for $\alpha_m$. The only point to check is that the cutoff errors vanish. The product rule gives \begin{align*} \bar{\partial}_1(\chi_m\alpha) &= \chi_m\bar{\partial}_1\alpha + \bar{\partial}\chi_m\wedge \alpha, \\ \bar{\partial}_0^*(\chi_m\alpha) &= \chi_m\bar{\partial}_0^*\alpha - (\partial\chi_m)\lrcorner \alpha. \end{align*} The wedge product by $\bar{\partial}\chi_m$ and contraction by $\partial\chi_m$ are pointwise bounded by $|\nabla\chi_m|_\omega$. Therefore \begin{align*} \|\bar{\partial}\chi_m\wedge \alpha\|_{L^2(h,\omega)} + \|(\partial\chi_m)\lrcorner \alpha\|_{L^2(h,\omega)} \le 2\sup_{x\in X}|\nabla\chi_m(x)|_\omega\,\|\alpha\|_{L^2(h,\omega)}. \end{align*} The right-hand side tends to $0$. Also, since $0\le \chi_m\le 1$ and $\chi_m\to 1$ pointwise, dominated convergence with respect to the measure $dV_\omega(x)$ gives $\chi_m\alpha\to\alpha$ in $L^2(h,\omega)$. Thus \begin{align*} \bar{\partial}_1(\chi_m\alpha)&\to \bar{\partial}_1\alpha,\\ \bar{\partial}_0^*(\chi_m\alpha)&\to \bar{\partial}_0^*\alpha,\\ \chi_m\alpha&\to\alpha \end{align*} in their corresponding $L^2$ spaces. Passing to the limit in the compact-support coercive estimate yields \begin{align*} \|\bar{\partial}_0^*\alpha\|_{L^2(h,\omega)}^2 + \|\bar{\partial}_1\alpha\|_{L^2(h,\omega)}^2 \ge (q-K)\|\alpha\|_{L^2(h,\omega)}^2. \end{align*} [/guided] [/step] [step:Solve the smooth metric case by the Hilbert-space range theorem] Set \begin{align*} H_0&:=L^2_{(0,0)}(X,L;h,\omega),\\ H_1&:=L^2_{(0,1)}(X,L;h,\omega),\\ H_2&:=L^2_{(0,2)}(X,L;h,\omega), \end{align*} and let $T:=\bar{\partial}_0:H_0\to H_1$, $S:=\bar{\partial}_1:H_1\to H_2$. Then $S T=0$. The estimate proved above says that, for every $\alpha\in\operatorname{Dom}(S)\cap\operatorname{Dom}(T^*)$, \begin{align*} \|T^*\alpha\|_{H_0}^2+\|S\alpha\|_{H_2}^2 \ge (q-K)\|\alpha\|_{H_1}^2. \end{align*} By the [Hilbert-space $L^2$ solvability lemma for closed complexes](/page/Hilbert-Space%20L2%20Solvability%20Lemma), applied to the closed densely defined operators $T$ and $S$ with $S T=0$ and coercivity constant $q-K>0$, every $f\in H_1$ with $Sf=0$ lies in $\operatorname{Range}(T)$, and there exists $u\in\operatorname{Dom}(T)$ such that $Tu=f$ and \begin{align*} \|u\|_{H_0}^2\le \frac{1}{q-K}\|f\|_{H_1}^2. \end{align*} Since $Sf=0$ is exactly $\bar{\partial}f=0$, this gives $\bar{\partial}u=f$ and \begin{align*} \int_X |u(x)|_h^2\,dV_\omega(x) \le \frac{1}{q-K} \int_X |f(x)|_{\omega,h}^2\,dV_\omega(x) \end{align*} in the smooth metric case. [guided] We now convert the coercive estimate into existence of a solution. Define the Hilbert spaces \begin{align*} H_0&:=L^2_{(0,0)}(X,L;h,\omega),\\ H_1&:=L^2_{(0,1)}(X,L;h,\omega),\\ H_2&:=L^2_{(0,2)}(X,L;h,\omega), \end{align*} and define the closed densely defined operators \begin{align*} T&:=\bar{\partial}_0:H_0\to H_1,\\ S&:=\bar{\partial}_1:H_1\to H_2. \end{align*} The identity $\bar{\partial}^2=0$ gives $S T=0$. The estimate from the previous step says that every $\alpha\in\operatorname{Dom}(S)\cap\operatorname{Dom}(T^*)$ satisfies \begin{align*} \|T^*\alpha\|_{H_0}^2+\|S\alpha\|_{H_2}^2 \ge (q-K)\|\alpha\|_{H_1}^2. \end{align*} This is precisely the hypothesis of the [Hilbert-space $L^2$ solvability lemma for closed complexes](/page/Hilbert-Space%20L2%20Solvability%20Lemma). The lemma requires that $T$ and $S$ are closed densely defined operators, that $S T=0$, and that the above estimate holds with constant $c>0$ on $\operatorname{Dom}(S)\cap\operatorname{Dom}(T^*)$; these conditions have been verified with $c=q-K$. The lemma then gives that every $f\in\ker S$ belongs to $\operatorname{Range}(T)$, and one may choose $u\in\operatorname{Dom}(T)$ with $Tu=f$ and \begin{align*} \|u\|_{H_0}^2\le c^{-1}\|f\|_{H_1}^2. \end{align*} Since the theorem assumes $\bar{\partial}f=0$, we have $Sf=0$, so the lemma gives a section $u\in L^2_{(0,0)}(X,L;h,\omega)$ satisfying $Tu=f$, which means $\bar{\partial}u=f$ in the distributional sense. The norm estimate becomes \begin{align*} \int_X |u(x)|_h^2\,dV_\omega(x) \le \frac{1}{q-K} \int_X |f(x)|_{\omega,h}^2\,dV_\omega(x). \end{align*} This proves the theorem when $h$ is smooth. [/guided] [/step] [step:Pass from smooth metrics to singular metrics by approximation] Assume now that $h$ is singular. By the regularization hypothesis in the theorem statement, write $h=e^{-\varphi}$ and choose smooth metrics $h_j=e^{-\varphi_j}$ with $\varphi_j\downarrow\varphi$ pointwise. Thus $h_j\uparrow h$ pointwise, so for every local section $s$ of $L$ the functions $|s|_{h_j}^2$ increase pointwise to $|s|_h^2$. The same hypothesis gives constants $q_j>K$ with $q_j\to q$ and \begin{align*} i\Theta_{h_j}(L)\ge q_j\omega. \end{align*} Apply the smooth case to $(L,h_j)$. For each $j$, there exists an $L$-valued section $u_j$ such that $\bar{\partial}u_j=f$ and \begin{align*} \int_X |u_j(x)|_{h_j}^2\,dV_\omega(x) \le \frac{1}{q_j-K} \int_X |f(x)|_{\omega,h_j}^2\,dV_\omega(x). \end{align*} Because $h_j\uparrow h$, the pointwise norms $|f|_{\omega,h_j}^2$ increase pointwise to $|f|_{\omega,h}^2$. By the [monotone convergence theorem](/theorems/509) applied with respect to $dV_\omega(x)$, \begin{align*} \int_X |f(x)|_{\omega,h_j}^2\,dV_\omega(x) \to \int_X |f(x)|_{\omega,h}^2\,dV_\omega(x). \end{align*} For a smooth metric $h_m$, let $L^2_{\mathrm{loc}}(X,L;h_m,\omega)$ denote the space of measurable sections $v:X\to L$ such that $v|_K\in L^2(K,L;h_m,\omega)$ for every compact set $K\subset X$. Thus, for each fixed $m$, the tail $(u_j)_{j\ge m}$ is uniformly bounded in $L^2_{\mathrm{loc}}(X,L;h_m,\omega)$, because $h_m\le h_j$ pointwise for $j\ge m$. Choose a compact exhaustion $(K_\ell)_{\ell=1}^\infty$ of $X$ with $K_\ell\subset K_{\ell+1}^\circ$. For each pair $(m,\ell)\in\mathbb{N}\times\mathbb{N}$, weak compactness in the [Hilbert space](/page/Hilbert%20Space) $L^2(K_\ell,L;h_m,\omega)$ gives a subsequence of the tail that converges weakly on $K_\ell$ with respect to $h_m$. Enumerating the [countable set](/page/Countable%20Set) $\mathbb{N}\times\mathbb{N}$ and taking the usual diagonal subsequence, there is a subsequence, still denoted $(u_j)$, and an $L$-valued measurable section $u:X\to L$ such that $u_j\rightharpoonup u$ weakly in $L^2(K_\ell,L;h_m,\omega)$ for every $m$ and every $\ell$. Let $L^*\to X$ denote the holomorphic dual line bundle of $L$, and let $\psi\in C_c^\infty(X,\Lambda^{n,n-1}T^*X\otimes L^*)$ be a smooth compactly supported $L^*$-valued $(n,n-1)$-form. Its support lies in some $K_\ell$, so the above [weak convergence](/page/Weak%20Convergence) with any fixed smooth metric $h_m$ is enough to pass the distributional identity $\bar{\partial}u_j=f$ to the limit. Hence $\bar{\partial}u=f$. Lower semicontinuity of the $L^2$ norm with respect to [weak convergence](/page/Weak%20Convergence) gives, for each fixed approximating metric $h_m$ and each compact exhaustion set $K_\ell$, \begin{align*} \int_{K_\ell} |u(x)|_{h_m}^2\,dV_\omega(x) \le \liminf_{j\to\infty} \int_{K_\ell} |u_j(x)|_{h_m}^2\,dV_\omega(x) \le \liminf_{j\to\infty} \int_X |u_j(x)|_{h_j}^2\,dV_\omega(x) \le \frac{1}{q-K} \int_X |f(x)|_{\omega,h}^2\,dV_\omega(x). \end{align*} Here the middle inequality uses $K_\ell\subset X$ and $h_m\le h_j$ for $j\ge m$, and the last inequality uses $q_j\to q$ together with the convergence of $\int_X |f(x)|_{\omega,h_j}^2\,dV_\omega(x)$. Letting $\ell\to\infty$ and applying monotone convergence to the increasing sequence $\mathbb{1}_{K_\ell}|u|_{h_m}^2$ gives the same bound over $X$ for the fixed metric $h_m$. Then letting $m\to\infty$ and applying monotone convergence to the increasing sequence $|u|_{h_m}^2$ yields \begin{align*} \int_X |u(x)|_h^2\,dV_\omega(x) \le \frac{1}{q-K} \int_X |f(x)|_{\omega,h}^2\,dV_\omega(x). \end{align*} This proves the singular metric case and completes the proof. [/step]