[proofplan] We regard the $L^2$ Dolbeault complex in degree $(p,r)$ as a closed Hilbert complex. The spectral-gap estimate gives a coercive graph-norm inequality on the part of the common graph domain orthogonal to the harmonic subspace, and this is exactly the subspace on which the closed range criterion tests the adjoint estimate for $T_{r-1}$. Once the range is closed, the standard [orthogonal decomposition](/theorems/436) of a [Hilbert space](/page/Hilbert%20Space) by a closed subspace identifies reduced cohomology with harmonic forms. Finally, if the coercive estimate also applies to harmonic candidates, a harmonic form has zero graph energy and therefore must vanish. [/proofplan] [step:Set up the closed Hilbert complex in the relevant degree] Let $H_r$ denote the [Hilbert space](/page/Hilbert%20Space) of $L$-valued $(p,r)$-forms that are square-integrable with respect to the Kähler metric $\omega$ and the Hermitian metric $h$, equipped with its $L^2$ inner product $(\cdot,\cdot)_{H_r}$ and norm $\|\cdot\|_{H_r}$. Define the closed densely defined operator \begin{align*} T_{r-1}: \operatorname{Dom}(T_{r-1}) \subset H_{r-1} &\to H_r \\ \beta &\mapsto \bar{\partial}\beta, \end{align*} and define \begin{align*} T_r: \operatorname{Dom}(T_r) \subset H_r &\to H_{r+1} \\ \alpha &\mapsto \bar{\partial}\alpha. \end{align*} By the [closed Hilbert complex property for maximal $L^2$ Dolbeault operators](/page/L2%20Dolbeault%20Complex), applied on the complete Kähler manifold $(X,\omega)$ with coefficients in the Hermitian holomorphic line bundle $(L,h)$, the maximal $L^2$ extensions of $\bar{\partial}$ form a closed Hilbert complex. Hence $T_rT_{r-1}=0$, each $T_j$ is closed and densely defined, and each $T_j$ has Hilbert-space adjoint $T_j^*$. In degree $r$, the common graph domain is \begin{align*} V_r := \operatorname{Dom}(T_r) \cap \operatorname{Dom}(T_{r-1}^*) \subset H_r. \end{align*} Define the harmonic subspace in degree $(p,r)$ by \begin{align*} \mathcal{H}_r := \ker T_r \cap \ker T_{r-1}^* \subset H_r. \end{align*} On $V_r$, define the quadratic form \begin{align*} q_r: V_r \times V_r &\to \mathbb{C} \\ (\alpha,\gamma) &\mapsto (T_r\alpha,T_r\gamma)_{H_{r+1}} + (T_{r-1}^*\alpha,T_{r-1}^*\gamma)_{H_{r-1}}. \end{align*} The hypothesis says that \begin{align*} q_r(\alpha,\alpha) \geq c\|\alpha\|_{H_r}^2 \end{align*} for every $\alpha \in V_r \cap \mathcal{H}_r^\perp$. [guided] The operators in the theorem are unbounded operators on $L^2$ spaces, so the first task is to name the Hilbert spaces and domains precisely. Let $H_r$ be the [Hilbert space](/page/Hilbert%20Space) of square-integrable $L$-valued $(p,r)$-forms, with norm $\|\cdot\|_{H_r}$. The Dolbeault operator in the preceding degree is the closed densely defined map \begin{align*} T_{r-1}: \operatorname{Dom}(T_{r-1}) \subset H_{r-1} &\to H_r \\ \beta &\mapsto \bar{\partial}\beta, \end{align*} and the Dolbeault operator in the current degree is \begin{align*} T_r: \operatorname{Dom}(T_r) \subset H_r &\to H_{r+1} \\ \alpha &\mapsto \bar{\partial}\alpha. \end{align*} The completeness of the Kähler metric is the analytic hypothesis needed for the [closed Hilbert complex property for maximal $L^2$ Dolbeault operators](/page/L2%20Dolbeault%20Complex). The theorem applies here because $(X,\omega)$ is complete Kähler and $(L,h)$ is a Hermitian holomorphic line bundle, so the maximal $L^2$ extensions of $\bar{\partial}$ are closed densely defined operators and satisfy the complex relation $T_rT_{r-1}=0$. The adjoint $T_{r-1}^*:\operatorname{Dom}(T_{r-1}^*)\subset H_r\to H_{r-1}$ is therefore defined in the Hilbert-space sense. The forms on which both terms in the estimate are meaningful form the common graph domain \begin{align*} V_r := \operatorname{Dom}(T_r) \cap \operatorname{Dom}(T_{r-1}^*) \subset H_r. \end{align*} The harmonic subspace in degree $(p,r)$ is \begin{align*} \mathcal{H}_r := \ker T_r \cap \ker T_{r-1}^* \subset H_r. \end{align*} On the common graph domain, the energy form is \begin{align*} q_r: V_r \times V_r &\to \mathbb{C} \\ (\alpha,\gamma) &\mapsto (T_r\alpha,T_r\gamma)_{H_{r+1}} + (T_{r-1}^*\alpha,T_{r-1}^*\gamma)_{H_{r-1}}. \end{align*} The assumed spectral gap is the coercive inequality \begin{align*} q_r(\alpha,\alpha) \geq c\|\alpha\|_{H_r}^2 \end{align*} for every $\alpha \in V_r \cap \mathcal{H}_r^\perp$. Thus the estimate is not asserted on all of $V_r$ at this stage; it is asserted exactly on the part of the graph domain orthogonal to harmonic forms. This is the point where the constant $c>0$ enters the proof. [/guided] [/step] [step:Derive the closed range estimate for $T_{r-1}$] We prove that $\operatorname{Range}(T_{r-1})$ is closed in $H_r$. Let \begin{align*} M_r := \overline{\operatorname{Range}(T_{r-1})}^{H_r}. \end{align*} Since $T_rT_{r-1}=0$ and $T_r$ is closed, $M_r \subseteq \ker T_r$. If $\alpha \in M_r \cap \operatorname{Dom}(T_{r-1}^*)$, then $\alpha \in V_r$ and $T_r\alpha=0$. Moreover $M_r \perp \ker T_{r-1}^*$: indeed, for $\eta \in \operatorname{Dom}(T_{r-1})$ and $\zeta \in \ker T_{r-1}^*$, \begin{align*} (T_{r-1}\eta,\zeta)_{H_r} = (\eta,T_{r-1}^*\zeta)_{H_{r-1}} = 0, \end{align*} and passing to the $H_r$-closure gives $M_r \subseteq (\ker T_{r-1}^*)^\perp$. Since $\mathcal{H}_r \subseteq \ker T_{r-1}^*$, we have $\alpha \perp \mathcal{H}_r$. Hence $\alpha \in V_r \cap \mathcal{H}_r^\perp$, and the spectral-gap estimate gives \begin{align*} c\|\alpha\|_{H_r}^2 &\leq \|T_r\alpha\|_{H_{r+1}}^2 + \|T_{r-1}^*\alpha\|_{H_{r-1}}^2 \\ &= \|T_{r-1}^*\alpha\|_{H_{r-1}}^2. \end{align*} Thus \begin{align*} \|\alpha\|_{H_r} \leq c^{-1/2}\|T_{r-1}^*\alpha\|_{H_{r-1}} \end{align*} for every $\alpha \in M_r \cap \operatorname{Dom}(T_{r-1}^*)$. Now apply the [Hilbert-space closed range criterion](/page/Closed%20Range%20Theorem) to the closed densely defined operator $T_{r-1}$. The criterion states that $\operatorname{Range}(T_{r-1})$ is closed if and only if there is a constant $C>0$ such that \begin{align*} \|\alpha\|_{H_r} \leq C\|T_{r-1}^*\alpha\|_{H_{r-1}} \end{align*} for all $\alpha \in \overline{\operatorname{Range}(T_{r-1})}^{H_r}\cap \operatorname{Dom}(T_{r-1}^*)$. By definition, this testing subspace is exactly $M_r \cap \operatorname{Dom}(T_{r-1}^*)$, and the estimate above verifies the criterion on that subspace with $C=c^{-1/2}$. Therefore $\operatorname{Range}(T_{r-1})$ is closed in degree $(p,r)$. [guided] We want to prove closedness of the image of $\bar{\partial}$ in degree $(p,r)$, that is, closedness of $\operatorname{Range}(T_{r-1}) \subset H_r$. The [Hilbert-space closed range criterion](/page/Closed%20Range%20Theorem) applies to closed densely defined Hilbert-space operators. We have already verified in the setup step that $T_{r-1}$ is closed and densely defined, so the criterion says that it is enough to estimate vectors in the closure of the range by the adjoint. In the present notation, the required estimate is \begin{align*} \|\alpha\|_{H_r} \leq C\|T_{r-1}^*\alpha\|_{H_{r-1}} \end{align*} for every $\alpha \in \overline{\operatorname{Range}(T_{r-1})}^{H_r}\cap \operatorname{Dom}(T_{r-1}^*)$. Define the closed subspace \begin{align*} M_r := \overline{\operatorname{Range}(T_{r-1})}^{H_r}. \end{align*} Because $T_rT_{r-1}=0$, every actual boundary lies in $\ker T_r$. Since $T_r$ is closed, its kernel is a closed subspace of $H_r$, so taking the $H_r$-closure gives \begin{align*} M_r \subseteq \ker T_r. \end{align*} Now take $\alpha \in M_r \cap \operatorname{Dom}(T_{r-1}^*)$. The inclusion $M_r \subseteq \ker T_r$ gives $T_r\alpha=0$, and the domain assumption gives $\alpha \in \operatorname{Dom}(T_{r-1}^*)$. Hence $\alpha \in V_r$. We still must verify the orthogonality hypothesis in the spectral-gap estimate. Let $\eta \in \operatorname{Dom}(T_{r-1})$ and let $\zeta \in \ker T_{r-1}^*$. By the definition of the Hilbert-space adjoint, \begin{align*} (T_{r-1}\eta,\zeta)_{H_r} = (\eta,T_{r-1}^*\zeta)_{H_{r-1}} = 0. \end{align*} Thus $\operatorname{Range}(T_{r-1}) \perp \ker T_{r-1}^*$, and taking the $H_r$-closure gives \begin{align*} M_r \subseteq (\ker T_{r-1}^*)^\perp. \end{align*} Since $\mathcal{H}_r \subseteq \ker T_{r-1}^*$, every element of $M_r$ is orthogonal to $\mathcal{H}_r$. Therefore $\alpha \in V_r \cap \mathcal{H}_r^\perp$, so the spectral-gap estimate applies: \begin{align*} c\|\alpha\|_{H_r}^2 &\leq \|T_r\alpha\|_{H_{r+1}}^2 + \|T_{r-1}^*\alpha\|_{H_{r-1}}^2 \\ &= \|T_{r-1}^*\alpha\|_{H_{r-1}}^2. \end{align*} Taking square roots gives \begin{align*} \|\alpha\|_{H_r} \leq c^{-1/2}\|T_{r-1}^*\alpha\|_{H_{r-1}}. \end{align*} This is precisely the closed range criterion with $C=c^{-1/2}$ on the testing subspace $M_r \cap \operatorname{Dom}(T_{r-1}^*) = \overline{\operatorname{Range}(T_{r-1})}^{H_r}\cap \operatorname{Dom}(T_{r-1}^*)$. Therefore $\operatorname{Range}(T_{r-1})$ in degree $(p,r)$ is closed. [/guided] [/step] [step:Identify reduced cohomology with harmonic forms] Define the harmonic subspace in degree $(p,r)$ by \begin{align*} \mathcal{H}_r := \ker T_r \cap \ker T_{r-1}^* \subset H_r. \end{align*} Since $\operatorname{Range}(T_{r-1})$ is closed, the [orthogonal decomposition theorem for Hilbert spaces](/page/Orthogonal%20Decomposition) gives \begin{align*} H_r = \operatorname{Range}(T_{r-1}) \oplus \ker T_{r-1}^*. \end{align*} Intersecting this decomposition with $\ker T_r$ and using $\operatorname{Range}(T_{r-1}) \subseteq \ker T_r$ yields \begin{align*} \ker T_r = \operatorname{Range}(T_{r-1}) \oplus \mathcal{H}_r. \end{align*} The reduced $L^2$ Dolbeault cohomology in degree $(p,r)$ is \begin{align*} \overline{H}^{p,r}_{(2)}(X,L) := \ker T_r / \overline{\operatorname{Range}(T_{r-1})}^{H_r}. \end{align*} Because the range is closed, this is \begin{align*} \overline{H}^{p,r}_{(2)}(X,L) = \ker T_r / \operatorname{Range}(T_{r-1}). \end{align*} The direct-sum decomposition therefore assigns to each reduced cohomology class a unique representative in $\mathcal{H}_r$. [guided] After closed range is known, the cohomology statement is an [orthogonal projection](/theorems/437) argument. Define the space of harmonic $L^2$ forms in degree $(p,r)$ by \begin{align*} \mathcal{H}_r := \ker T_r \cap \ker T_{r-1}^* \subset H_r. \end{align*} Here $\ker T_r$ is the space of $\bar{\partial}$-closed forms and $\ker T_{r-1}^*$ is the space of forms orthogonal to all $\bar{\partial}$-boundaries. Since $\operatorname{Range}(T_{r-1})$ is closed, the [orthogonal decomposition theorem for Hilbert spaces](/page/Orthogonal%20Decomposition) applies to the closed subspace $\operatorname{Range}(T_{r-1})\subset H_r$ and gives \begin{align*} H_r = \operatorname{Range}(T_{r-1}) \oplus \operatorname{Range}(T_{r-1})^\perp. \end{align*} By the definition of the Hilbert-space adjoint, \begin{align*} \operatorname{Range}(T_{r-1})^\perp = \ker T_{r-1}^*. \end{align*} Therefore \begin{align*} H_r = \operatorname{Range}(T_{r-1}) \oplus \ker T_{r-1}^*. \end{align*} Now restrict this decomposition to closed forms. Since $T_rT_{r-1}=0$, every boundary belongs to $\ker T_r$. Hence intersecting with $\ker T_r$ gives \begin{align*} \ker T_r = \operatorname{Range}(T_{r-1}) \oplus (\ker T_r \cap \ker T_{r-1}^*) = \operatorname{Range}(T_{r-1}) \oplus \mathcal{H}_r. \end{align*} The reduced cohomology is defined by \begin{align*} \overline{H}^{p,r}_{(2)}(X,L) := \ker T_r / \overline{\operatorname{Range}(T_{r-1})}^{H_r}. \end{align*} Because the range has just been proved closed, the closure can be removed: \begin{align*} \overline{H}^{p,r}_{(2)}(X,L) = \ker T_r / \operatorname{Range}(T_{r-1}). \end{align*} The direct-sum decomposition says exactly that every class has one and only one representative lying in $\mathcal{H}_r$. Existence comes from projecting a closed form onto the second summand, and uniqueness follows because the intersection of the two summands is $\{0\}$. [/guided] [/step] [step:Use the spectral gap on harmonic candidates to prove vanishing] Assume that the same estimate also holds for every $\alpha\in V_r$ satisfying $T_r\alpha=0$ and $T_{r-1}^*\alpha=0$. Let $\alpha \in \mathcal{H}_r$. Then $\alpha \in V_r$, $T_r\alpha=0$, and $T_{r-1}^*\alpha=0$. Applying the additional estimate to this harmonic candidate gives \begin{align*} c\|\alpha\|_{H_r}^2 &\leq \|T_r\alpha\|_{H_{r+1}}^2 + \|T_{r-1}^*\alpha\|_{H_{r-1}}^2 \\ &= 0. \end{align*} Since $c>0$, this implies $\|\alpha\|_{H_r}=0$, hence $\alpha=0$ in $H_r$. Thus $\mathcal{H}_r=\{0\}$. By the harmonic representative identification above, \begin{align*} \overline{H}^{p,r}_{(2)}(X,L)=0. \end{align*} [guided] The closed range argument only used the spectral gap on $V_r \cap \mathcal{H}_r^\perp$. To prove vanishing of cohomology, we now assume the additional hypothesis that the same coercive estimate is valid for every $\alpha\in V_r$ satisfying $T_r\alpha=0$ and $T_{r-1}^*\alpha=0$. Take any $\alpha \in \mathcal{H}_r$. By the definition \begin{align*} \mathcal{H}_r = \ker T_r \cap \ker T_{r-1}^*, \end{align*} we have $\alpha \in \operatorname{Dom}(T_r) \cap \operatorname{Dom}(T_{r-1}^*) = V_r$, together with \begin{align*} T_r\alpha = 0, \qquad T_{r-1}^*\alpha = 0. \end{align*} Applying the additional estimate to this $\alpha$ gives \begin{align*} c\|\alpha\|_{H_r}^2 &\leq \|T_r\alpha\|_{H_{r+1}}^2 + \|T_{r-1}^*\alpha\|_{H_{r-1}}^2 \\ &= 0. \end{align*} Because $c>0$, the last inequality forces $\|\alpha\|_{H_r}=0$. In a [Hilbert space](/page/Hilbert%20Space), norm zero means equality to the zero vector, so $\alpha=0$ in $H_r$. Therefore $\mathcal{H}_r=\{0\}$. The previous step identified reduced cohomology with the harmonic subspace. Since the harmonic subspace is zero, the reduced cohomology group is zero: \begin{align*} \overline{H}^{p,r}_{(2)}(X,L)=0. \end{align*} [/guided] [/step]