[proofplan] We convert the problem for $L^m$-valued $(0,q)$-forms into the standard Bochner--Kodaira--Nakano estimate on $(n,q)$-forms by tensoring with the inverse canonical bundle. The curvature of $K_X^{-1}\otimes L^m$ is the sum of a fixed curvature term and $m$ times the positive curvature of $L$, so for $m$ large the positive part dominates uniformly on all $(n,q)$-forms with $q\ge 1$. This gives a coercive estimate for the $\bar\partial$-Laplacian. The Hilbert-space existence theorem for $\bar\partial$ then produces the minimal solution, and elliptic regularity upgrades it to a smooth form. [/proofplan]

[step:Replace $(0,q)$-forms with $(n,q)$-forms valued in $K_X^{-1}\otimes L^m$]Let $K_X := \Lambda^{n,0}T^*X$ denote the canonical bundle, equipped with the Hermitian metric induced by $\omega$. Define \begin{align*} E_m &:= K_X^{-1}\otimes L^m \end{align*} with the product Hermitian metric $g_m$ induced by the dual metric on $K_X^{-1}$ and by $h^m$ on $L^m$. There is a natural bundle isomorphism \begin{align*} \mathcal{I}_m:\Lambda^{0,q}T^*X\otimes L^m &\longrightarrow \Lambda^{n,q}T^*X\otimes E_m \\ \alpha &\longmapsto \alpha \otimes \operatorname{id}_{K_X}, \end{align*} where $\operatorname{id}_{K_X}$ denotes the canonical identity endomorphism of $K_X$ under the trace-contraction identification \begin{align*} K_X\otimes K_X^{-1}\cong \operatorname{End}(K_X). \end{align*} Equivalently, this is the tensor contraction identification \begin{align*} \Lambda^{0,q}T^*X\otimes L^m \cong \Lambda^{n,q}T^*X\otimes K_X^{-1}\otimes L^m. \end{align*} Because the Hermitian metrics on both sides are induced from the same Kähler metric $\omega$ and the same line bundle metric $h^m$, $\mathcal{I}_m$ is an isometry for the corresponding $L^2$ norms. The map $\mathcal{I}_m$ is holomorphic in the coefficient bundle and commutes with $\bar\partial$. Therefore, if $f$ is a smooth $L^m$-valued $(0,q)$-form with $\bar\partial f=0$, then \begin{align*} F_m := \mathcal{I}_m f \end{align*} is a smooth $E_m$-valued $(n,q)$-form with $\bar\partial F_m=0$ and \begin{align*} \|F_m\|_{L^2(X,\omega,g_m)}=\|f\|_{L^2(X,\omega,h^m)}. \end{align*}[/step]

[guided]The Bochner--Kodaira--Nakano curvature term has the strongest positivity on $(n,q)$-forms. Our theorem is stated for $(0,q)$-forms, so we first rewrite the same objects in top holomorphic degree. Let $K_X := \Lambda^{n,0}T^*X$ be the canonical bundle. Since \begin{align*} \Lambda^{n,q}T^*X = \Lambda^{n,0}T^*X\otimes \Lambda^{0,q}T^*X = K_X\otimes \Lambda^{0,q}T^*X, \end{align*} tensoring by $K_X^{-1}$ gives the bundle identification \begin{align*} \Lambda^{0,q}T^*X\otimes L^m \cong \Lambda^{n,q}T^*X\otimes K_X^{-1}\otimes L^m. \end{align*} We therefore define \begin{align*} E_m := K_X^{-1}\otimes L^m. \end{align*} The isomorphism \begin{align*} \mathcal{I}_m:\Lambda^{0,q}T^*X\otimes L^m &\longrightarrow \Lambda^{n,q}T^*X\otimes E_m \\ \alpha &\longmapsto \alpha \otimes \operatorname{id}_{K_X} \end{align*} is the tensor-contraction identification obtained by viewing $\operatorname{id}_{K_X}$ as the canonical identity endomorphism under \begin{align*} K_X\otimes K_X^{-1}\cong \operatorname{End}(K_X). \end{align*} The metrics on both sides are induced from $\omega$ and $h^m$, so this identification preserves pointwise norms and hence preserves $L^2$ norms after integration against the Kähler volume measure $dV_\omega(x)$: \begin{align*} \|\mathcal{I}_m\alpha\|_{L^2(X,\omega,g_m)}^2 = \int_X |\mathcal{I}_m\alpha(x)|_{\omega,g_m}^2\,dV_\omega(x) = \int_X |\alpha(x)|_{\omega,h^m}^2\,dV_\omega(x) = \|\alpha\|_{L^2(X,\omega,h^m)}^2. \end{align*} Because the identification uses holomorphic tensor factors, it commutes with $\bar\partial$. Thus $\bar\partial f=0$ implies $\bar\partial F_m=0$ for $F_m:=\mathcal{I}_m f$.[/guided]

[step:Show that the curvature term grows at least linearly in $m$]Let $\Theta_h(L)$ denote the Chern curvature of $(L,h)$. Since $(L,h)$ is positive, the real $(1,1)$-form $i\Theta_h(L)$ is positive. Because $X$ is compact, there exists a constant $\lambda>0$ such that \begin{align*} i\Theta_h(L)(\xi,\bar{\xi})\ge \lambda\,\omega(\xi,\bar{\xi}) \end{align*} for every point $x\in X$ and every tangent vector $\xi\in T_x^{1,0}X$. Let $\Theta_{g_m}(E_m)$ denote the Chern curvature of $(E_m,g_m)$. By additivity of Chern curvature under tensor products of Hermitian line bundles, \begin{align*} i\Theta_{g_m}(E_m) = i\Theta(K_X^{-1})+m\,i\Theta_h(L). \end{align*} The fixed curvature form $i\Theta(K_X^{-1})$ is smooth on compact $X$, so there exists $B>0$ such that, as Hermitian forms, \begin{align*} i\Theta(K_X^{-1})\ge -B\,\omega. \end{align*} Hence \begin{align*} i\Theta_{g_m}(E_m)\ge (m\lambda-B)\omega. \end{align*} For an $E_m$-valued $(n,q)$-form $\alpha$, the commutator $[i\omega,\Lambda_\omega]$ acts as multiplication by $q$. Therefore \begin{align*} \bigl([i\Theta_{g_m}(E_m),\Lambda_\omega]\alpha,\alpha\bigr)_{\omega,g_m} \ge q(m\lambda-B)|\alpha|_{\omega,g_m}^2. \end{align*} Choose $m_0\in\mathbb{N}$ so large that $m\lambda-B\ge \frac{\lambda}{2}m$ for all $m\ge m_0$. Then, for every $m\ge m_0$, \begin{align*} \bigl([i\Theta_{g_m}(E_m),\Lambda_\omega]\alpha,\alpha\bigr)_{\omega,g_m} \ge \frac{\lambda}{2}m|\alpha|_{\omega,g_m}^2 \end{align*} because $q\ge 1$.[/step]

[guided]The curvature of $E_m=K_X^{-1}\otimes L^m$ has one part that grows with $m$ and one part that is fixed. The growing part comes from $L^m$. Let $\Theta_h(L)$ be the Chern curvature of $L$. Positivity of $(L,h)$ means that $i\Theta_h(L)$ is a positive Hermitian $(1,1)$-form. Since $X$ is compact and the smallest eigenvalue of $i\Theta_h(L)$ relative to $\omega$ varies continuously, there is a uniform constant $\lambda>0$ such that \begin{align*} i\Theta_h(L)(\xi,\bar{\xi})\ge \lambda\,\omega(\xi,\bar{\xi}) \end{align*} for all $x\in X$ and all $\xi\in T_x^{1,0}X$. Curvature is additive under tensor products of Hermitian line bundles. Thus for \begin{align*} E_m=K_X^{-1}\otimes L^m \end{align*} we have \begin{align*} i\Theta_{g_m}(E_m) = i\Theta(K_X^{-1})+m\,i\Theta_h(L). \end{align*} The curvature $i\Theta(K_X^{-1})$ is independent of $m$. Since it is smooth on compact $X$, its eigenvalues relative to $\omega$ are bounded below. Hence there exists $B>0$ such that \begin{align*} i\Theta(K_X^{-1})\ge -B\,\omega. \end{align*} Combining this with the lower bound for $i\Theta_h(L)$ gives \begin{align*} i\Theta_{g_m}(E_m) \ge (m\lambda-B)\omega. \end{align*} Now we use the special feature of top holomorphic degree. Let $\Lambda_\omega$ denote the adjoint, with respect to the Kähler metric $\omega$, of exterior multiplication by $\omega$. On $E_m$-valued $(n,q)$-forms, the Lefschetz commutator $[i\omega,\Lambda_\omega]$ acts by the scalar $q$. Therefore, if $\alpha$ is an $E_m$-valued $(n,q)$-form, then \begin{align*} \bigl([i\Theta_{g_m}(E_m),\Lambda_\omega]\alpha,\alpha\bigr)_{\omega,g_m} \ge q(m\lambda-B)|\alpha|_{\omega,g_m}^2. \end{align*} The hypothesis $q\ge 1$ is used exactly here: it gives a strictly positive scalar on top-degree forms. Choose $m_0\in\mathbb{N}$ such that $m\lambda-B\ge \frac{\lambda}{2}m$ whenever $m\ge m_0$. Then \begin{align*} \bigl([i\Theta_{g_m}(E_m),\Lambda_\omega]\alpha,\alpha\bigr)_{\omega,g_m} \ge \frac{\lambda}{2}m|\alpha|_{\omega,g_m}^2 \end{align*} for all $m\ge m_0$ and all $E_m$-valued $(n,q)$-forms $\alpha$.[/guided]

[step:Apply the Bochner--Kodaira--Nakano identity to obtain a coercive estimate]Let \begin{align*} A_m:C^\infty\bigl(X;\Lambda^{n,q}T^*X\otimes E_m\bigr)&\to C^\infty\bigl(X;\Lambda^{n,q+1}T^*X\otimes E_m\bigr),\\ \alpha&\mapsto \bar\partial\alpha \end{align*} and let $A_m^*$ denote its $L^2(X,\omega,g_m)$ adjoint on $E_m$-valued forms. The Bochner--Kodaira--Nakano identity for compact Kähler manifolds, applied to the Hermitian holomorphic line bundle $(E_m,g_m)$, gives for every smooth $E_m$-valued $(n,q)$-form $\alpha$: \begin{align*} \|A_m\alpha\|_{L^2(X,\omega,g_m)}^2 + \|A_m^*\alpha\|_{L^2(X,\omega,g_m)}^2 \ge \int_X \bigl([i\Theta_{g_m}(E_m),\Lambda_\omega]\alpha(x),\alpha(x)\bigr)_{\omega,g_m} \,dV_\omega(x). \end{align*} Here we are citing a result not yet in the wiki: Bochner--Kodaira--Nakano identity. Using the curvature lower bound from the previous step, for every $m\ge m_0$ and every smooth $E_m$-valued $(n,q)$-form $\alpha$, \begin{align*} \|A_m\alpha\|_{L^2(X,\omega,g_m)}^2 + \|A_m^*\alpha\|_{L^2(X,\omega,g_m)}^2 \ge \frac{\lambda}{2}m \|\alpha\|_{L^2(X,\omega,g_m)}^2. \end{align*}[/step]

[guided]We now convert curvature positivity into an analytic inequality. Define the operator \begin{align*} A_m:C^\infty\bigl(X;\Lambda^{n,q}T^*X\otimes E_m\bigr)&\to C^\infty\bigl(X;\Lambda^{n,q+1}T^*X\otimes E_m\bigr),\\ \alpha&\mapsto \bar\partial\alpha. \end{align*} Let $A_m^*$ be its adjoint with respect to the $L^2$ inner product induced by $\omega$ and $g_m$. The Bochner--Kodaira--Nakano identity says that, on a compact Kähler manifold and for a Hermitian holomorphic vector bundle, the $\bar\partial$ energy controls the curvature commutator. In the present line-bundle case it gives \begin{align*} \|A_m\alpha\|_{L^2(X,\omega,g_m)}^2 + \|A_m^*\alpha\|_{L^2(X,\omega,g_m)}^2 \ge \int_X \bigl([i\Theta_{g_m}(E_m),\Lambda_\omega]\alpha(x),\alpha(x)\bigr)_{\omega,g_m} \,dV_\omega(x). \end{align*} Here we are citing a result not yet in the wiki: Bochner--Kodaira--Nakano identity. The previous step proved the pointwise estimate \begin{align*} \bigl([i\Theta_{g_m}(E_m),\Lambda_\omega]\alpha(x),\alpha(x)\bigr)_{\omega,g_m} \ge \frac{\lambda}{2}m|\alpha(x)|_{\omega,g_m}^2 \end{align*} for every $x\in X$. Integrating this pointwise inequality with respect to the Kähler volume measure $dV_\omega(x)$ gives \begin{align*} \int_X \bigl([i\Theta_{g_m}(E_m),\Lambda_\omega]\alpha(x),\alpha(x)\bigr)_{\omega,g_m} \,dV_\omega(x) \ge \frac{\lambda}{2}m \int_X|\alpha(x)|_{\omega,g_m}^2\,dV_\omega(x). \end{align*} Therefore \begin{align*} \|A_m\alpha\|_{L^2(X,\omega,g_m)}^2 + \|A_m^*\alpha\|_{L^2(X,\omega,g_m)}^2 \ge \frac{\lambda}{2}m \|\alpha\|_{L^2(X,\omega,g_m)}^2. \end{align*} This is the coercive estimate: the $\bar\partial$ complex has a spectral gap in degree $q$ of size at least \begin{align*} \frac{\lambda}{2}m. \end{align*}[/guided]

[step:Use the coercive estimate to solve $\bar\partial U_m=F_m$ with the sharp $m^{-1}$ bound]For each integer $r\ge 0$, let $H_{m,r}$ be the [Hilbert space](/page/Hilbert%20Space) completion of smooth $E_m$-valued $(n,r)$-forms in the norm $\|\cdot\|_{L^2(X,\omega,g_m)}$. Define the closed densely defined operator \begin{align*} D_{m,r}:\operatorname{Dom}(D_{m,r})\subset H_{m,r}&\to H_{m,r+1} \\ \alpha&\mapsto \bar\partial\alpha \end{align*} as the maximal closed extension of $\bar\partial$ from smooth forms, and let $D_{m,r}^*$ denote its Hilbert-space adjoint. The estimate from the previous step extends by density to every \begin{align*} \alpha\in \operatorname{Dom}(D_{m,q})\cap \operatorname{Dom}(D_{m,q-1}^*) \end{align*} and gives \begin{align*} \|D_{m,q}\alpha\|_{L^2(X,\omega,g_m)}^2 + \|D_{m,q-1}^*\alpha\|_{L^2(X,\omega,g_m)}^2 \ge \frac{\lambda}{2}m\|\alpha\|_{L^2(X,\omega,g_m)}^2. \end{align*} Set \begin{align*} c_m:=\frac{\lambda}{2}m. \end{align*} We now use the standard closed-range estimate for Hilbert complexes (citing a result not yet in the wiki: closed-range minimal-solution theorem for Hilbert complexes). In the following form, it says that for a closed Hilbert complex \begin{align*} H_{m,q-1}\xrightarrow{D_{m,q-1}}H_{m,q}\xrightarrow{D_{m,q}}H_{m,q+1}, \end{align*} if the estimate \begin{align*} c_m\|\alpha\|_{L^2(X,\omega,g_m)}^2 \le \|D_{m,q}\alpha\|_{L^2(X,\omega,g_m)}^2 + \|D_{m,q-1}^*\alpha\|_{L^2(X,\omega,g_m)}^2 \end{align*} holds for all $\alpha\in \operatorname{Dom}(D_{m,q})\cap\operatorname{Dom}(D_{m,q-1}^*)$, then $\operatorname{Ran}D_{m,q-1}$ is closed, $\ker D_{m,q}=\operatorname{Ran}D_{m,q-1}$, and every $F\in\ker D_{m,q}$ has a unique minimal solution \begin{align*} U\in \operatorname{Dom}(D_{m,q-1})\cap(\ker D_{m,q-1})^\perp \end{align*} such that \begin{align*} D_{m,q-1}U=F \end{align*} and \begin{align*} \|U\|_{L^2(X,\omega,g_m)}^2\le c_m^{-1}\|F\|_{L^2(X,\omega,g_m)}^2. \end{align*} In the Dolbeault complex, the same theorem gives the weak normal condition \begin{align*} D_{m,q-2}^*U=0 \end{align*} for the minimal solution, with this term omitted when $q=1$. This quoted Hilbert-complex theorem packages the domain statements, closed-range conclusion, normal equation, and minimal-norm estimate; no additional differentiation of a weak Laplacian equation is being used here. Since $F_m$ is smooth and $\bar\partial F_m=0$, we have $F_m\in\ker D_{m,q}$. Applying the closed-range theorem gives an $E_m$-valued $(n,q-1)$-form \begin{align*} U_m\in\operatorname{Dom}(D_{m,q-1}) \end{align*} such that \begin{align*} \bar\partial U_m=D_{m,q-1}U_m=F_m \end{align*} and \begin{align*} \|U_m\|_{L^2(X,\omega,g_m)}^2 \le \frac{2}{\lambda m}\|F_m\|_{L^2(X,\omega,g_m)}^2. \end{align*}[/step]

[guided]The coercive estimate must be applied to a precise Hilbert complex, not to an ambiguous symbol indexed by form degree. For each integer $r\ge 0$, define $H_{m,r}$ to be the $L^2$ Hilbert space of $E_m$-valued $(n,r)$-forms, obtained by completing smooth forms in the norm induced by $\omega$ and $g_m$. Define \begin{align*} D_{m,r}:\operatorname{Dom}(D_{m,r})\subset H_{m,r}&\to H_{m,r+1} \\ \alpha&\mapsto \bar\partial\alpha \end{align*} to be the maximal closed extension of $\bar\partial$, and let $D_{m,r}^*$ be its Hilbert-space adjoint. These operators are densely defined because smooth forms are dense in each $H_{m,r}$. The previous step proved the estimate first for smooth $E_m$-valued $(n,q)$-forms. Since $D_{m,q}$ and $D_{m,q-1}^*$ are closed and smooth forms form a core for the Dolbeault Laplacian on the compact Kähler manifold $X$, the same estimate extends to every \begin{align*} \alpha\in \operatorname{Dom}(D_{m,q})\cap \operatorname{Dom}(D_{m,q-1}^*) \end{align*} as \begin{align*} \|D_{m,q}\alpha\|_{L^2(X,\omega,g_m)}^2 + \|D_{m,q-1}^*\alpha\|_{L^2(X,\omega,g_m)}^2 \ge \frac{\lambda}{2}m\|\alpha\|_{L^2(X,\omega,g_m)}^2. \end{align*} Set \begin{align*} c_m:=\frac{\lambda}{2}m. \end{align*} This estimate is the closed-range input for solving $\bar\partial$. More explicitly, on the Hilbert complex \begin{align*} H_{m,q-1}\xrightarrow{D_{m,q-1}}H_{m,q}\xrightarrow{D_{m,q}}H_{m,q+1}, \end{align*} we use the standard closed-range minimal-solution theorem for Hilbert complexes. It says that the displayed estimate implies \begin{align*} \ker D_{m,q}=\operatorname{Ran}D_{m,q-1} \end{align*} and that every $F\in\ker D_{m,q}$ has a unique solution \begin{align*} U\in \operatorname{Dom}(D_{m,q-1})\cap(\ker D_{m,q-1})^\perp \end{align*} with \begin{align*} D_{m,q-1}U=F, \qquad \|U\|_{L^2(X,\omega,g_m)}^2\le c_m^{-1}\|F\|_{L^2(X,\omega,g_m)}^2. \end{align*} For the minimal solution in the Dolbeault complex, the same theorem also gives \begin{align*} D_{m,q-2}^*U=0, \end{align*} with the term omitted when $q=1$. This is the advertised minimal-solution estimate. The form $F_m$ is smooth and satisfies $\bar\partial F_m=0$, so $F_m\in\operatorname{Dom}(D_{m,q})$ and $D_{m,q}F_m=0$. Therefore the Hilbert-space theorem gives an $E_m$-valued $(n,q-1)$-form \begin{align*} U_m\in\operatorname{Dom}(D_{m,q-1}) \end{align*} such that \begin{align*} \bar\partial U_m=D_{m,q-1}U_m=F_m \end{align*} and \begin{align*} \|U_m\|_{L^2(X,\omega,g_m)}^2 \le \frac{1}{c_m}\|F_m\|_{L^2(X,\omega,g_m)}^2 = \frac{2}{\lambda m}\|F_m\|_{L^2(X,\omega,g_m)}^2. \end{align*} The positivity estimate eliminates harmonic obstructions in degree $q$, and the operator $D_{m,q-1}$ is the correctly typed previous-degree $\bar\partial$ operator with the same coefficient bundle $E_m$.[/guided]

[step:Return to $(0,q-1)$-forms and apply elliptic regularity]Define \begin{align*} u:=\mathcal{I}_m^{-1}U_m. \end{align*} Then $u$ is an $L^m$-valued $(0,q-1)$-form and, because $\mathcal{I}_m$ commutes with $\bar\partial$, \begin{align*} \bar\partial u = \mathcal{I}_m^{-1}(\bar\partial U_m) = \mathcal{I}_m^{-1}F_m = f. \end{align*} We chose $U_m$ as the minimal $L^2$ solution supplied by the closed-range theorem. Thus it also satisfies the weak normal condition \begin{align*} D_{m,q-2}^*U_m=0 \end{align*} with this term omitted when $q=1$. Together with $D_{m,q-1}U_m=F_m$, this gives the elliptic system for a minimal $\bar\partial$ solution. Define the degree-$(q-1)$ Dolbeault Laplacian by \begin{align*} \Box_{m,q-1}:= D_{m,q-1}^*D_{m,q-1} + D_{m,q-2}D_{m,q-2}^*, \end{align*} with the second summand omitted when $q=1$. Equivalently, in the weak sense, \begin{align*} \Box_{m,q-1}U_m=D_{m,q-1}^*F_m. \end{align*} The right-hand side is smooth because $F_m$ is smooth and $D_{m,q-1}^*$ is a first-order differential operator with smooth coefficients. Elliptic regularity for the Dolbeault Laplacian on compact Hermitian manifolds (citing a result not yet in the wiki: elliptic regularity for the Dolbeault Laplacian) therefore implies that $U_m$ is smooth. Hence $u$ is smooth. Finally, since $\mathcal{I}_m$ is an isometry, \begin{align*} \|u\|_{L^2(X,\omega,h^m)}^2 = \|U_m\|_{L^2(X,\omega,g_m)}^2 \le \frac{2}{\lambda m} \|F_m\|_{L^2(X,\omega,g_m)}^2 = \frac{2}{\lambda m} \|f\|_{L^2(X,\omega,h^m)}^2. \end{align*} Taking \begin{align*} C:=\frac{2}{\lambda} \end{align*} proves the desired estimate. The integer $m_0$ depends only on the lower positivity constant $\lambda$ and the fixed curvature bound $B$ for $K_X^{-1}$, not on $m$, $q$, or $f$; the constant $C=2/\lambda$ is independent of $m$ and $f$.[/step]

[guided]We now convert the solution back to the form type in the theorem. Define \begin{align*} u:=\mathcal{I}_m^{-1}U_m. \end{align*} Since $U_m$ is an $E_m$-valued $(n,q-1)$-form and $\mathcal{I}_m$ identifies $L^m$-valued $(0,q-1)$-forms with $E_m$-valued $(n,q-1)$-forms, the form $u$ is an $L^m$-valued $(0,q-1)$-form. Because $\mathcal{I}_m$ commutes with $\bar\partial$, the equation for $U_m$ becomes the desired equation for $u$: \begin{align*} \bar\partial u = \mathcal{I}_m^{-1}(\bar\partial U_m) = \mathcal{I}_m^{-1}F_m = f. \end{align*} It remains to justify smoothness. The Hilbert-space theorem gives the minimal $L^2$ solution $U_m$, and minimality supplies the weak normal equation \begin{align*} D_{m,q-2}^*U_m=0 \end{align*} with the term omitted for $q=1$. Since $D_{m,q-1}U_m=F_m$, the pair of equations \begin{align*} D_{m,q-1}U_m &= F_m, & D_{m,q-2}^*U_m &= 0 \end{align*} is an elliptic first-order system for $U_m$. Let \begin{align*} \Box_{m,q-1}:= D_{m,q-1}^*D_{m,q-1} + D_{m,q-2}D_{m,q-2}^*, \end{align*} again omitting the second summand when $q=1$. Equivalently, \begin{align*} \Box_{m,q-1}U_m=D_{m,q-1}^*F_m \end{align*} in the weak sense. The right-hand side is smooth because $F_m$ is smooth. Elliptic regularity for the Dolbeault Laplacian on compact Hermitian manifolds gives that $U_m$ is smooth. Hence $u=\mathcal{I}_m^{-1}U_m$ is smooth. Finally, the isometry property of $\mathcal{I}_m$ transfers the estimate without changing the constant: \begin{align*} \|u\|_{L^2(X,\omega,h^m)}^2 = \|U_m\|_{L^2(X,\omega,g_m)}^2 \le \frac{2}{\lambda m} \|F_m\|_{L^2(X,\omega,g_m)}^2 = \frac{2}{\lambda m} \|f\|_{L^2(X,\omega,h^m)}^2. \end{align*} Thus the theorem holds with \begin{align*} C:=\frac{2}{\lambda}. \end{align*} The integer $m_0$ depends only on the positivity constant $\lambda$ and the fixed curvature bound $B$ for $K_X^{-1}$, not on $m$, $q$, or $f$; the constant $C$ is independent of $m$ and $f$.[/guided]