[proofplan] The identity is the compact-support energy identity for $\square=\bar{\partial}\bar{\partial}^*+\bar{\partial}^*\bar{\partial}$. Because $q\ge 1$, the form $\bar{\partial}^*u$ has bidegree $(p,q-1)$, so the formal adjoint formula applies to the first term. For the second term, the same formula is applied to $u$ and $\bar{\partial}u$ when $q<n$; when $q=n$, $\bar{\partial}u=0$ by bidegree. Linearity in the first argument and conjugate symmetry then convert the two terms into the two squared norms. [/proofplan] [step:Expand the complex Laplacian on compactly supported forms] Let $u\in\Omega_c^{p,q}(\Omega)$. Since $q\ge 1$, the formal adjoint formula gives \begin{align*} \bar{\partial}^*u\in\Omega_c^{p,q-1}(\Omega). \end{align*} Also \begin{align*} \bar{\partial}u\in\Omega_c^{p,q+1}(\Omega) \end{align*} when $q<n$, while $\bar{\partial}u=0$ when $q=n$ because there are no $(p,n+1)$-forms. By the definition of $\square$, \begin{align*} \square u=\bar{\partial}\bar{\partial}^*u+\bar{\partial}^*\bar{\partial}u. \end{align*} Linearity of the $L^2$ inner product in the first argument gives \begin{align*} (\square u,u)_{L^2} = (\bar{\partial}\bar{\partial}^*u,u)_{L^2} + (\bar{\partial}^*\bar{\partial}u,u)_{L^2}. \end{align*} [/step] [step:Convert the $\bar{\partial}\bar{\partial}^*$ term into a norm] We apply the formal adjoint identity to \begin{align*} v=\bar{\partial}^*u\in\Omega_c^{p,q-1}(\Omega) \end{align*} and to the second argument $u\in\Omega_c^{p,q}(\Omega)$. The bidegrees match the hypotheses of the integration-by-parts formula: $v$ has bidegree $(p,q-1)$, $u$ has bidegree $(p,q)$, and both forms are smooth with compact support in $\Omega$. Therefore \begin{align*} (\bar{\partial}\bar{\partial}^*u,u)_{L^2} &=(\bar{\partial}^*u,\bar{\partial}^*u)_{L^2}\\ &=\|\bar{\partial}^*u\|_{L^2}^2. \end{align*} [/step] [step:Convert the $\bar{\partial}^*\bar{\partial}$ term into a norm] If $q=n$, then $\bar{\partial}u=0$, and hence \begin{align*} (\bar{\partial}^*\bar{\partial}u,u)_{L^2} =0 =\|\bar{\partial}u\|_{L^2}^2. \end{align*} Assume now that $q<n$. Then $\bar{\partial}u\in\Omega_c^{p,q+1}(\Omega)$. Applying the formal adjoint identity with first argument $u\in\Omega_c^{p,q}(\Omega)$ and second argument $\bar{\partial}u\in\Omega_c^{p,q+1}(\Omega)$ gives \begin{align*} (\bar{\partial}u,\bar{\partial}u)_{L^2} =(u,\bar{\partial}^*\bar{\partial}u)_{L^2}. \end{align*} Using conjugate symmetry of the Hermitian inner product, \begin{align*} (\bar{\partial}^*\bar{\partial}u,u)_{L^2} &=\overline{(u,\bar{\partial}^*\bar{\partial}u)_{L^2}}\\ &=\overline{(\bar{\partial}u,\bar{\partial}u)_{L^2}}\\ &=\|\bar{\partial}u\|_{L^2}^2. \end{align*} Thus the same identity holds for every $1\le q\le n$. [/step] [step:Combine the two norm identities] Substituting the two identities into the expansion of $(\square u,u)_{L^2}$ gives \begin{align*} (\square u,u)_{L^2} &= \|\bar{\partial}^*u\|_{L^2}^2 + \|\bar{\partial}u\|_{L^2}^2\\ &= \|\bar{\partial}u\|_{L^2}^2 + \|\bar{\partial}^*u\|_{L^2}^2. \end{align*} This proves the asserted energy identity. [/step]