[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*}
[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]
[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]
