[proofplan] The proof is the Hilbert-space adjoint identity applied twice. The domain of $\Box_q$ ensures that both composed terms $\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*u$ and $\bar{\partial}_q^*\bar{\partial}_q u$ are defined, and that the two adjoint pairings are legitimate. Expanding $\Box_q u$ by definition and moving each operator to its adjoint gives exactly the two squared $L^2$ norms. [/proofplan]

[step:Expand the $\bar{\partial}$-Neumann Laplacian on its domain] Fix $u\in\operatorname{Dom}(\Box_q)$. By the definition of $\operatorname{Dom}(\Box_q)$, one has \begin{align*} u\in\operatorname{Dom}(\bar{\partial}_q)\cap\operatorname{Dom}(\bar{\partial}_{q-1}^*) \end{align*} and \begin{align*} \bar{\partial}_q u\in\operatorname{Dom}(\bar{\partial}_q^*),\qquad \bar{\partial}_{q-1}^*u\in\operatorname{Dom}(\bar{\partial}_{q-1}). \end{align*} Therefore both summands in the definition of $\Box_q u$ belong to $H_q$, and \begin{align*} (\Box_q u,u)_{L^2}=(\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*u,u)_{L^2}+(\bar{\partial}_q^*\bar{\partial}_q u,u)_{L^2}. \end{align*} [/step]

[step:Move each operator across the inner product by the adjoint relation]Apply the Hilbert-space adjoint relation to the operator \begin{align*} \bar{\partial}_{q-1}:\operatorname{Dom}(\bar{\partial}_{q-1})\subset H_{q-1}\to H_q. \end{align*} Since $\bar{\partial}_{q-1}^*u\in\operatorname{Dom}(\bar{\partial}_{q-1})$ and $u\in\operatorname{Dom}(\bar{\partial}_{q-1}^*)$, we obtain \begin{align*} (\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*u,u)_{L^2}=(\bar{\partial}_{q-1}^*u,\bar{\partial}_{q-1}^*u)_{L^2}. \end{align*} Thus \begin{align*} (\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*u,u)_{L^2}=\|\bar{\partial}_{q-1}^*u\|_{L^2}^2. \end{align*} Apply the Hilbert-space adjoint relation to the operator \begin{align*} \bar{\partial}_q:\operatorname{Dom}(\bar{\partial}_q)\subset H_q\to H_{q+1}. \end{align*} Since $u\in\operatorname{Dom}(\bar{\partial}_q)$ and $\bar{\partial}_q u\in\operatorname{Dom}(\bar{\partial}_q^*)$, we obtain \begin{align*} (\bar{\partial}_q^*\bar{\partial}_q u,u)_{L^2}=(\bar{\partial}_q u,\bar{\partial}_q u)_{L^2}. \end{align*} Thus \begin{align*} (\bar{\partial}_q^*\bar{\partial}_q u,u)_{L^2}=\|\bar{\partial}_q u\|_{L^2}^2. \end{align*}[/step]

[guided]The only point requiring care is the domain bookkeeping for unbounded operators. For a densely defined operator $A:\operatorname{Dom}(A)\subset H\to K$ with Hilbert-space adjoint $A^*:\operatorname{Dom}(A^*)\subset K\to H$, the adjoint relation says that \begin{align*} (Av,w)_K=(v,A^*w)_H \end{align*} whenever $v\in\operatorname{Dom}(A)$ and $w\in\operatorname{Dom}(A^*)$. First use this relation with \begin{align*} A=\bar{\partial}_{q-1}:\operatorname{Dom}(\bar{\partial}_{q-1})\subset H_{q-1}\to H_q. \end{align*} Here the vector playing the role of $v$ is $\bar{\partial}_{q-1}^*u$, and the vector playing the role of $w$ is $u$. The definition of $\operatorname{Dom}(\Box_q)$ gives both required domain conditions: \begin{align*} \bar{\partial}_{q-1}^*u\in\operatorname{Dom}(\bar{\partial}_{q-1}) \end{align*} and \begin{align*} u\in\operatorname{Dom}(\bar{\partial}_{q-1}^*). \end{align*} Therefore the adjoint relation applies and gives \begin{align*} (\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*u,u)_{L^2}=(\bar{\partial}_{q-1}^*u,\bar{\partial}_{q-1}^*u)_{L^2}. \end{align*} By the definition of the Hilbert norm induced by the $L^2$ [inner product](/page/Inner%20Product), \begin{align*} (\bar{\partial}_{q-1}^*u,\bar{\partial}_{q-1}^*u)_{L^2}=\|\bar{\partial}_{q-1}^*u\|_{L^2}^2. \end{align*} Now use the same adjoint relation with \begin{align*} A=\bar{\partial}_q:\operatorname{Dom}(\bar{\partial}_q)\subset H_q\to H_{q+1}. \end{align*} This time the vector playing the role of $v$ is $u$, and the vector playing the role of $w$ is $\bar{\partial}_q u$. The definition of $\operatorname{Dom}(\Box_q)$ gives \begin{align*} u\in\operatorname{Dom}(\bar{\partial}_q) \end{align*} and \begin{align*} \bar{\partial}_q u\in\operatorname{Dom}(\bar{\partial}_q^*). \end{align*} Hence \begin{align*} (\bar{\partial}_q u,\bar{\partial}_q u)_{L^2}=(u,\bar{\partial}_q^*\bar{\partial}_q u)_{L^2}. \end{align*} Since the inner product is linear in the first argument and conjugate-linear in the second, this identity is equivalently the standard energy pairing \begin{align*} (\bar{\partial}_q^*\bar{\partial}_q u,u)_{L^2}=(\bar{\partial}_q u,\bar{\partial}_q u)_{L^2}. \end{align*} Thus \begin{align*} (\bar{\partial}_q^*\bar{\partial}_q u,u)_{L^2}=\|\bar{\partial}_q u\|_{L^2}^2. \end{align*} The domain assumptions in $\operatorname{Dom}(\Box_q)$ are precisely what prevent any hidden boundary term or undefined pairing from appearing.[/guided]

[step:Add the two adjoint identities] Substituting the two identities from the previous step into the expansion of $(\Box_q u,u)_{L^2}$ gives \begin{align*} (\Box_q u,u)_{L^2}=\|\bar{\partial}_{q-1}^*u\|_{L^2}^2+\|\bar{\partial}_q u\|_{L^2}^2. \end{align*} Reordering the two non-negative summands yields \begin{align*} (\Box_q u,u)_{L^2}=\|\bar{\partial}_q u\|_{L^2}^2+\|\bar{\partial}_{q-1}^*u\|_{L^2}^2. \end{align*} This is the asserted identity. [/step]