[proofplan] The proof is the standard Bochner-Kodaira curvature-gap argument. Positivity of the Hermitian holomorphic line bundle gives a uniform positive lower bound for its curvature on the compact manifold. Tensor powers multiply this curvature lower bound by the tensor exponent, and the curvature commutator on $L^m$-valued $(n,q)$-forms inherits a positive lower bound proportional to $m q$. The Bochner-Kodaira-Nakano identity converts this pointwise lower bound into an $L^2$ estimate, and the remaining tensor-power factor is discarded because the estimate is only needed for positive tensor powers. [/proofplan]

[step:Extract a uniform positive curvature lower bound from positivity of $L$] Let $dV_\omega$ denote the Riemannian volume measure induced by $\omega$. The $L^2$ [inner product](/page/Inner%20Product) on $A^{n,q}(X,L^m)$ is denoted by $(\cdot,\cdot)_{L^2}$, and its norm is $\|\cdot\|_{L^2}$. Define $\Theta(L,h)$ to be the curvature form of the Chern connection of the Hermitian holomorphic line bundle $(L,h)$. By the curvature positivity hypothesis in the theorem statement, the real $(1,1)$-form $i\Theta(L,h)$ is positive. For each point $x\in X$, let $\mu_{\min}(x)$ denote the smallest eigenvalue of $i\Theta(L,h)_x$ with respect to $\omega_x$. The function $\mu_{\min}:X\to \mathbb R$ is continuous and positive. Since $X$ is compact, it has a positive minimum. Define \begin{align*} c:=\min_{x\in X}\mu_{\min}(x)>0. \end{align*} Thus, as Hermitian forms on $T_xX$ for every $x\in X$, \begin{align*} i\Theta(L,h)_x\ge c\,\omega_x. \end{align*} [/step]

[step:Compute the curvature lower bound for $L^m$ on $(n,q)$-forms]For an integer $m\ge 1$, let $h^m$ be the induced Hermitian metric on $L^m$. Let $L_\omega$ denote the Lefschetz operator given by wedging with $\omega$, and let $\Lambda$ denote its formal adjoint with respect to the pointwise Hermitian inner product induced by $\omega$ and $h^m$. The Chern connection on $L^m$ is the $m$-fold tensor power connection, so its curvature is \begin{align*} \Theta(L^m,h^m)=m\Theta(L,h). \end{align*} Fix $x\in X$. Choose a unitary coframe $dz_1,\dots,dz_n$ for $T_x^*X$ with respect to $\omega_x$ that diagonalizes $i\Theta(L,h)_x$: \begin{align*} i\Theta(L,h)_x=\sum_{j=1}^n \lambda_j\,i\,dz_j\wedge d\bar z_j. \end{align*} The lower bound from the previous step gives $\lambda_j\ge c$ for every $j\in\{1,\dots,n\}$. If $q>n$, then $A^{n,q}(X,L^m)=0$, so the desired estimate is vacuous. Assume now that $1\le q\le n$. By [citetheorem:9100], the curvature commutator $[i\Theta(L^m,h^m),\Lambda]$ acts diagonally on the standard components of an $L^m$-valued $(n,q)$-form. Since the holomorphic index set has size $n$, the eigenvalue on a component with antiholomorphic index set $J\subset\{1,\dots,n\}$ and $|J|=q$ is \begin{align*} m\sum_{j\in J}\lambda_j. \end{align*} Because each $\lambda_j\ge c$ and $|J|=q$, \begin{align*} m\sum_{j\in J}\lambda_j\ge mqc. \end{align*} Therefore, pointwise on $X$, \begin{align*} \left\langle [i\Theta(L^m,h^m),\Lambda]\alpha,\alpha\right\rangle\ge mqc\,|\alpha|^2. \end{align*} Integrating with respect to $dV_\omega$ gives \begin{align*} \left([i\Theta(L^m,h^m),\Lambda]\alpha,\alpha\right)_{L^2}\ge mqc\,\|\alpha\|_{L^2}^2. \end{align*}[/step]

[guided]The key point is that tensoring by $L^m$ multiplies the positive curvature by $m$, and on $(n,q)$-forms this positivity appears at least $q$ times. Let $\mu_{\min}:X\to\mathbb R$ denote the function assigning to each point $x\in X$ the smallest eigenvalue of the Hermitian form $i\Theta(L,h)_x$ with respect to $\omega_x$. Since $i\Theta(L,h)$ and $\omega$ are smooth Hermitian forms, $\mu_{\min}$ is continuous. Since $i\Theta(L,h)$ is positive, $\mu_{\min}(x)>0$ for every $x\in X$. Compactness of $X$ gives a positive minimum, and we denote it by \begin{align*} c:=\min_{x\in X}\mu_{\min}(x)>0. \end{align*} Therefore $i\Theta(L,h)_x\ge c\,\omega_x$ as Hermitian forms on $T_xX$ for every $x\in X$. Let $h^m$ denote the Hermitian metric induced on $L^m$. Let $L_\omega$ denote the Lefschetz operator given by wedging with $\omega$, and let $\Lambda$ denote its formal adjoint with respect to the pointwise Hermitian inner product induced by $\omega$ and $h^m$. The Chern connection on a tensor power is the [tensor product](/page/Tensor%20Product) connection, so the curvature of $L^m$ is \begin{align*} \Theta(L^m,h^m)=m\Theta(L,h). \end{align*} Fix a point $x\in X$. Since $\omega_x$ is Hermitian, we may choose a unitary coframe $dz_1,\dots,dz_n$ for $T_x^*X$ that diagonalizes the Hermitian form $i\Theta(L,h)_x$. Thus \begin{align*} i\Theta(L,h)_x=\sum_{j=1}^n \lambda_j\,i\,dz_j\wedge d\bar z_j. \end{align*} The inequality $i\Theta(L,h)_x\ge c\omega_x$ means in this unitary coframe that $\lambda_j\ge c$ for each index $j$. If $q>n$, then there are no nonzero $(n,q)$-forms, so the estimate holds without further argument. Assume $1\le q\le n$. By [citetheorem:9100], the curvature commutator for a line bundle is diagonal on the standard components of an $L^m$-valued $(p,q)$-form. We apply it with $p=n$. Because the holomorphic index set then contains every index in $\{1,\dots,n\}$, the only contribution is the sum over the antiholomorphic index set $J$, where $|J|=q$. For $L^m$, the eigenvalue on that component is \begin{align*} m\sum_{j\in J}\lambda_j. \end{align*} Since every $\lambda_j$ is at least $c$ and the set $J$ has exactly $q$ elements, \begin{align*} m\sum_{j\in J}\lambda_j\ge mqc. \end{align*} Thus every diagonal component of the commutator is bounded below by $mqc$. Therefore, for every smooth $L^m$-valued $(n,q)$-form $\alpha$, \begin{align*} \left\langle [i\Theta(L^m,h^m),\Lambda]\alpha,\alpha\right\rangle\ge mqc\,|\alpha|^2 \end{align*} at each point of $X$. Integrating this pointwise inequality over $X$ with respect to the volume measure $dV_\omega$ gives \begin{align*} \left([i\Theta(L^m,h^m),\Lambda]\alpha,\alpha\right)_{L^2}\ge mqc\,\|\alpha\|_{L^2}^2. \end{align*}[/guided]

[step:Apply the Bochner-Kodaira-Nakano identity and discard nonnegative terms] Let $\nabla_m$ denote the Chern connection on $L^m$, and let $\nabla_m'$ denote its $(1,0)$ part as an operator on $L^m$-valued forms. Let $(\nabla_m')^*$ denote the formal adjoint of $\nabla_m'$ with respect to the $L^2$ inner products induced by $\omega$ and $h^m$. We use the Bochner-Kodaira-Nakano identity in the coercive form supplied by [citetheorem:9101] for the Hermitian holomorphic line bundle $(L^m,h^m)$ over the compact Kähler manifold $(X,\omega)$. For every smooth $\alpha\in A^{n,q}(X,L^m)$, \begin{align*} \|\bar\partial_{L^m}\alpha\|_{L^2}^2+\|\bar\partial_{L^m}^*\alpha\|_{L^2}^2=\|\nabla_m'\alpha\|_{L^2}^2+\|(\nabla_m')^*\alpha\|_{L^2}^2+\left([i\Theta(L^m,h^m),\Lambda]\alpha,\alpha\right)_{L^2}. \end{align*} The first two terms on the right are squared $L^2$ norms, hence nonnegative. Using the curvature lower bound from the previous step, \begin{align*} \|\bar\partial_{L^m}\alpha\|_{L^2}^2+\|\bar\partial_{L^m}^*\alpha\|_{L^2}^2\ge mqc\,\|\alpha\|_{L^2}^2. \end{align*} [/step]

[step:Choose the constants and conclude the estimate] Set \begin{align*} m_0:=1, \qquad C_q:=\frac{1}{qc}. \end{align*} For every integer $m\ge m_0$, we have $m\ge 1$, so the estimate from the previous step implies \begin{align*} qc\,\|\alpha\|_{L^2}^2\le mqc\,\|\alpha\|_{L^2}^2\le \|\bar\partial_{L^m}\alpha\|_{L^2}^2+\|\bar\partial_{L^m}^*\alpha\|_{L^2}^2. \end{align*} Multiplying by $C_q=(qc)^{-1}$ gives \begin{align*} \|\alpha\|_{L^2}^2\le C_q\left(\|\bar\partial_{L^m}\alpha\|_{L^2}^2+\|\bar\partial_{L^m}^*\alpha\|_{L^2}^2\right). \end{align*} This is the desired estimate for all $m\ge m_0$ and all smooth $L^m$-valued $(n,q)$-forms $\alpha$. [/step]