[proofplan]
The proof uses [Kohn's basic estimate](/theorems/3687) to obtain coercivity of the $\bar{\partial}$-Neumann quadratic form in positive degree. The closed form representation theorem then produces a bounded inverse $N_q$ for the self-adjoint operator $\Box_q$. For $\bar{\partial}$-closed data $f$, the commutation of $\bar{\partial}$ with the Neumann Laplacian forces $\bar{\partial}N_qf$ to be harmonic, and the basic estimate eliminates harmonic forms. The canonical solution is therefore $\bar{\partial}^*N_qf$, and its minimality follows from Hilbert-space orthogonality to $\ker \bar{\partial}$.
[/proofplan]
[step:Use Kohn's basic estimate to invert the Neumann Laplacian]
For each integer $q$ with $1\le q\le n$, define the [Hilbert space](/page/Hilbert%20Space)
\begin{align*}
H_q:=L^2_{(0,q)}(\Omega).
\end{align*}
Let
\begin{align*}
Q_q:\operatorname{Dom}(Q_q)\times \operatorname{Dom}(Q_q)\to \mathbb{C}
\end{align*}
be the closed non-negative sesquilinear form
\begin{align*}
Q_q[\alpha,\beta]
:=
(\bar{\partial}_q\alpha,\bar{\partial}_q\beta)_{H_{q+1}}
+
(\bar{\partial}_{q-1}^*\alpha,\bar{\partial}_{q-1}^*\beta)_{H_{q-1}},
\end{align*}
with form domain
\begin{align*}
\operatorname{Dom}(Q_q)
=
\operatorname{Dom}(\bar{\partial}_q)\cap \operatorname{Dom}(\bar{\partial}_{q-1}^*).
\end{align*}
Here $\bar{\partial}_n=0$ when $q=n$, and when $q=n$ the term involving $H_{q+1}$ is interpreted as $0$.
We use [Kohn's basic estimate](/theorems/3687) in the following precise form. Since $\Omega\subset\mathbb{C}^n$ is bounded, has $C^\infty$ boundary, and is pseudoconvex in the Levi sense, for every positive degree $1\le q\le n$ the $\bar{\partial}$-Neumann form has no harmonic obstruction and there is a constant $C_q>0$ such that every $\alpha\in \operatorname{Dom}(Q_q)$ satisfies
\begin{align*}
\|\alpha\|_{H_q}^2
\le
C_q\left(
\|\bar{\partial}_q\alpha\|_{H_{q+1}}^2
+
\|\bar{\partial}_{q-1}^*\alpha\|_{H_{q-1}}^2
\right)
=
C_q Q_q[\alpha,
\alpha].
\end{align*}
This is the cited external result: Kohn basic estimate for bounded smooth pseudoconvex domains in positive bidegree, in the coercive form without projecting away harmonic forms.
The closed form representation theorem applied to the closed non-negative form $Q_q$ gives a non-negative self-adjoint operator associated to $Q_q$. We define $\operatorname{Dom}(\Box_q)$ to be this represented operator domain:
\begin{align*}
\operatorname{Dom}(\Box_q)
:=
\left\{
\alpha\in \operatorname{Dom}(Q_q):
\text{ there exists }g\in H_q\text{ such that }Q_q[\alpha,\beta]=(g,\beta)_{H_q}\text{ for all }\beta\in\operatorname{Dom}(Q_q)
\right\},
\end{align*}
and for such $\alpha$ we set $\Box_q\alpha:=g$. In the $\bar{\partial}$-Neumann setting this represented operator is precisely
\begin{align*}
\Box_q=\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*+\bar{\partial}_q^*\bar{\partial}_q,
\end{align*}
with the boundary conditions encoded by the operator domain above. This is the cited external result: closed form representation theorem for the $\bar{\partial}$-Neumann form.
The coercive estimate gives surjectivity and bounded invertibility by the closed range form criterion, whose hypotheses are: the form $Q_q$ is densely defined, closed, non-negative, and satisfies $\|\alpha\|_{H_q}^2\le C_qQ_q[\alpha,\alpha]$ on its domain. Density and closedness are part of the definition of the $\bar{\partial}$-Neumann form above, and the displayed Kohn estimate is exactly the coercivity hypothesis. Equivalently, applying Lax-Milgram to the [Hilbert space](/page/Hilbert%20Space) $\operatorname{Dom}(Q_q)$ with inner product $Q_q[\cdot,\cdot]$ gives, for each $f\in H_q$, a unique $w\in\operatorname{Dom}(Q_q)$ such that
\begin{align*}
Q_q[w,\beta]=(f,\beta)_{H_q}
\end{align*}
for every $\beta\in\operatorname{Dom}(Q_q)$; by the definition of the represented operator, $w\in\operatorname{Dom}(\Box_q)$ and $\Box_qw=f$. The same coercive estimate also gives
\begin{align*}
\|w\|_{H_q}^2\le C_qQ_q[w,w]=C_q(f,w)_{H_q}\le C_q\|f\|_{H_q}\|w\|_{H_q},
\end{align*}
so $\|w\|_{H_q}\le C_q\|f\|_{H_q}$. Thus the inverse is bounded on all of $H_q$. Consequently there is a linear operator
\begin{align*}
N_q:H_q\to \operatorname{Dom}(\Box_q)
\end{align*}
such that $N_q:H_q\to H_q$ is bounded and
\begin{align*}
\Box_qN_q f=f
\end{align*}
for every $f\in H_q$.
[guided]
The first task is to turn the analytic estimate into an operator inverse. For degree $q$, we work in the [Hilbert space](/page/Hilbert%20Space)
\begin{align*}
H_q:=L^2_{(0,q)}(\Omega).
\end{align*}
The relevant energy form is
\begin{align*}
Q_q[\alpha,\beta]
:=
(\bar{\partial}_q\alpha,\bar{\partial}_q\beta)_{H_{q+1}}
+
(\bar{\partial}_{q-1}^*\alpha,\bar{\partial}_{q-1}^*\beta)_{H_{q-1}},
\end{align*}
defined on
\begin{align*}
\operatorname{Dom}(Q_q)
=
\operatorname{Dom}(\bar{\partial}_q)\cap \operatorname{Dom}(\bar{\partial}_{q-1}^*).
\end{align*}
This form measures exactly the two pieces appearing in the $\bar{\partial}$-Neumann Laplacian.
[Kohn's basic estimate](/theorems/3687) says the following precise statement. Because $\Omega$ is bounded, pseudoconvex in the Levi sense, and has $C^\infty$ boundary, and because $q$ is in the positive degree range $1\le q\le n$, there exists a constant $C_q>0$ such that
\begin{align*}
\|\alpha\|_{H_q}^2
\le
C_q\left(
\|\bar{\partial}_q\alpha\|_{H_{q+1}}^2
+
\|\bar{\partial}_{q-1}^*\alpha\|_{H_{q-1}}^2
\right)
=
C_qQ_q[\alpha,\alpha]
\end{align*}
for every $\alpha\in \operatorname{Dom}(Q_q)$. This version has no projection onto the orthogonal complement of harmonic forms; the estimate itself rules out harmonic forms in positive degree. The result being cited here is an external theorem not yet in the wiki: Kohn basic estimate for bounded smooth pseudoconvex domains in positive bidegree.
The closed form representation theorem associates to the closed non-negative form $Q_q$ a self-adjoint non-negative operator. Its domain is
\begin{align*}
\operatorname{Dom}(\Box_q)
:=
\left\{
\alpha\in \operatorname{Dom}(Q_q):
\text{ there exists }g\in H_q\text{ such that }Q_q[\alpha,\beta]=(g,\beta)_{H_q}\text{ for all }\beta\in\operatorname{Dom}(Q_q)
\right\},
\end{align*}
and $\Box_q\alpha:=g$. In this setting the represented operator is exactly
\begin{align*}
\Box_q=\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*+\bar{\partial}_q^*\bar{\partial}_q,
\end{align*}
with the $\bar{\partial}$-Neumann boundary conditions included in this operator domain. This identifies the abstract form operator with the concrete $\bar{\partial}$-Neumann Laplacian. The result being cited here is an external theorem not yet in the wiki: closed form representation theorem for the $\bar{\partial}$-Neumann form.
Why does coercivity produce an inverse on all of $H_q$? The closed range form criterion applies to a densely defined closed non-negative form whose form norm controls the ambient Hilbert norm. Those hypotheses hold here: $Q_q$ is the closed $\bar{\partial}$-Neumann form, and Kohn's estimate gives $\|\alpha\|_{H_q}^2\le C_qQ_q[\alpha,\alpha]$. Equivalently, for each $f\in H_q$, the map
\begin{align*}
\beta\mapsto (f,\beta)_{H_q}
\end{align*}
is bounded on the [Hilbert space](/page/Hilbert%20Space) $\operatorname{Dom}(Q_q)$ equipped with the inner product $Q_q[\cdot,\cdot]$, because
\begin{align*}
|(f,\beta)_{H_q}|
\le
\|f\|_{H_q}\|\beta\|_{H_q}
\le
C_q^{1/2}\|f\|_{H_q}Q_q[\beta,\beta]^{1/2}.
\end{align*}
The [Riesz representation theorem](/theorems/221) on this form [Hilbert space](/page/Hilbert%20Space) gives a unique $w\in\operatorname{Dom}(Q_q)$ such that
\begin{align*}
Q_q[w,\beta]=(f,\beta)_{H_q}
\end{align*}
for every $\beta\in\operatorname{Dom}(Q_q)$. By the represented-operator domain definition, this means $w\in\operatorname{Dom}(\Box_q)$ and $\Box_qw=f$. Finally,
\begin{align*}
\|w\|_{H_q}^2
\le
C_qQ_q[w,w]
=
C_q(f,w)_{H_q}
\le
C_q\|f\|_{H_q}\|w\|_{H_q},
\end{align*}
so $\|w\|_{H_q}\le C_q\|f\|_{H_q}$. Hence the inverse
\begin{align*}
N_q:H_q\to \operatorname{Dom}(\Box_q)
\end{align*}
exists, is bounded as an operator $H_q\to H_q$, and satisfies $\Box_qN_qf=f$ for every $f\in H_q$.
[/guided]
[/step]
[step:Show that $\bar{\partial}N_qf$ vanishes for closed data]
Let $f\in H_q$ satisfy $f\in \operatorname{Dom}(\bar{\partial}_q)$ and $\bar{\partial}_q f=0$. Define
\begin{align*}
w:=N_qf\in \operatorname{Dom}(\Box_q)\subset H_q.
\end{align*}
Then
\begin{align*}
\Box_qw=f.
\end{align*}
If $q=n$, then $\bar{\partial}_nw=0$ by degree reasons. Assume $q<n$. Since $w\in\operatorname{Dom}(\Box_q)$ and $\Box_qw=f\in\operatorname{Dom}(\bar{\partial}_q)$, we invoke the following precise commutation theorem for the $\bar{\partial}$-Neumann Laplacian on bounded $C^\infty$ pseudoconvex domains: if $a\in\operatorname{Dom}(\Box_q)$, $q<n$, and $\Box_qa\in\operatorname{Dom}(\bar{\partial}_q)$, then $\bar{\partial}_qa\in\operatorname{Dom}(\Box_{q+1})$ and
\begin{align*}
\Box_{q+1}(\bar{\partial}_qa)=\bar{\partial}_q(\Box_qa).
\end{align*}
The theorem includes the $\bar{\partial}$-Neumann boundary/domain conditions in the conclusion, so it supplies the non-trivial assertion that $\bar{\partial}_qw\in\operatorname{Dom}(\Box_{q+1})$. Applying it to $a=w$ gives
\begin{align*}
\Box_{q+1}(\bar{\partial}_qw)
=
\bar{\partial}_q(\Box_qw)
=
\bar{\partial}_q f
=
0.
\end{align*}
This is the cited external result not yet in the wiki: commutation of $\bar{\partial}$ with the $\bar{\partial}$-Neumann Laplacian, including its operator-domain conclusion.
Thus $\bar{\partial}_qw\in \ker \Box_{q+1}$. Applying the basic estimate in degree $q+1$ gives $\ker \Box_{q+1}=\{0\}$, and therefore
\begin{align*}
\bar{\partial}_qw=0.
\end{align*}
[guided]
We now use the hypothesis that the data $f$ is $\bar{\partial}$-closed. Define
\begin{align*}
w:=N_qf.
\end{align*}
By the definition of $N_q$, this element belongs to $\operatorname{Dom}(\Box_q)$ and satisfies
\begin{align*}
\Box_qw=f.
\end{align*}
The goal is to prove that the second term in the Neumann Laplacian,
\begin{align*}
\bar{\partial}_q^*\bar{\partial}_qw,
\end{align*}
does not contribute. For that we show first that $\bar{\partial}_qw=0$.
If $q=n$, then $\bar{\partial}_n$ maps $(0,n)$-forms to $(0,n+1)$-forms, and there are no non-zero $(0,n+1)$-forms on $\Omega\subset\mathbb{C}^n$. Hence $\bar{\partial}_nw=0$.
Assume now that $q<n$. The commutation theorem for the $\bar{\partial}$-Neumann Laplacian states the full unbounded-operator assertion needed here: if $a\in\operatorname{Dom}(\Box_q)$ and $\Box_qa\in\operatorname{Dom}(\bar{\partial}_q)$, then $\bar{\partial}_qa\in\operatorname{Dom}(\Box_{q+1})$ and
\begin{align*}
\Box_{q+1}(\bar{\partial}_qa)=\bar{\partial}_q(\Box_qa).
\end{align*}
The important point is that the theorem includes the $\bar{\partial}$-Neumann boundary conditions in the conclusion; it is not merely a formal algebraic commutation of differential expressions. These hypotheses hold here because $w=N_qf\in\operatorname{Dom}(\Box_q)$ and $\Box_qw=f\in\operatorname{Dom}(\bar{\partial}_q)$. The result being cited here is an external theorem not yet in the wiki: commutation of $\bar{\partial}$ with the $\bar{\partial}$-Neumann Laplacian, including its operator-domain conclusion. Applying this identity to $w$ gives
\begin{align*}
\Box_{q+1}(\bar{\partial}_qw)
=
\bar{\partial}_q(\Box_qw)
=
\bar{\partial}_q f.
\end{align*}
Since $f$ is $\bar{\partial}$-closed, $\bar{\partial}_q f=0$, so
\begin{align*}
\Box_{q+1}(\bar{\partial}_qw)=0.
\end{align*}
Therefore $\bar{\partial}_qw$ is harmonic in degree $q+1$.
[Kohn's basic estimate](/theorems/3687) in degree $q+1$ eliminates harmonic forms. Indeed, if $\gamma\in\ker\Box_{q+1}$, then
\begin{align*}
0
=
(\Box_{q+1}\gamma,\gamma)_{H_{q+1}}
=
Q_{q+1}[\gamma,\gamma],
\end{align*}
and the coercive estimate gives $\|\gamma\|_{H_{q+1}}=0$. Applying this to $\gamma=\bar{\partial}_qw$ yields
\begin{align*}
\bar{\partial}_qw=0.
\end{align*}
[/guided]
[/step]
[step:Reduce the Neumann equation to a $\bar{\partial}$ equation]
Define
\begin{align*}
u:=\bar{\partial}_{q-1}^*w=\bar{\partial}_{q-1}^*N_qf\in H_{q-1}.
\end{align*}
Since $w\in \operatorname{Dom}(\Box_q)$, the expression $\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*w$ is defined. Using $\Box_qw=f$ and $\bar{\partial}_qw=0$, we obtain
\begin{align*}
f
=
\Box_qw
=
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*w
+
\bar{\partial}_q^*\bar{\partial}_qw
=
\bar{\partial}_{q-1}u.
\end{align*}
Thus $u\in \operatorname{Dom}(\bar{\partial}_{q-1})$ and $\bar{\partial}_{q-1}u=f$.
[/step]
[step:Use orthogonality to prove minimality of the canonical solution]
Let $v\in \operatorname{Dom}(\bar{\partial}_{q-1})$ be any other solution of
\begin{align*}
\bar{\partial}_{q-1}v=f.
\end{align*}
Then
\begin{align*}
h:=v-u\in \operatorname{Dom}(\bar{\partial}_{q-1})
\end{align*}
satisfies
\begin{align*}
\bar{\partial}_{q-1}h=0.
\end{align*}
Because $u=\bar{\partial}_{q-1}^*w$ and $h\in\ker \bar{\partial}_{q-1}$, the definition of the Hilbert-space adjoint gives
\begin{align*}
(u,h)_{H_{q-1}}
=
(\bar{\partial}_{q-1}^*w,h)_{H_{q-1}}
=
(w,\bar{\partial}_{q-1}h)_{H_q}
=
0.
\end{align*}
Therefore $u$ is orthogonal to $h=v-u$, and the Pythagorean identity in the [Hilbert space](/page/Hilbert%20Space) $H_{q-1}$ gives
\begin{align*}
\|v\|_{H_{q-1}}^2
=
\|u+h\|_{H_{q-1}}^2
=
\|u\|_{H_{q-1}}^2+\|h\|_{H_{q-1}}^2
\ge
\|u\|_{H_{q-1}}^2.
\end{align*}
Thus $u=\bar{\partial}_{q-1}^*N_qf$ has minimal $L^2$ norm among all $L^2$ solutions of $\bar{\partial}_{q-1}u=f$.
[guided]
Take any other solution
\begin{align*}
v\in \operatorname{Dom}(\bar{\partial}_{q-1})
\end{align*}
with
\begin{align*}
\bar{\partial}_{q-1}v=f.
\end{align*}
The difference between two solutions is
\begin{align*}
h:=v-u.
\end{align*}
Since both $v$ and $u$ solve the same equation, linearity of $\bar{\partial}_{q-1}$ gives
\begin{align*}
\bar{\partial}_{q-1}h
=
\bar{\partial}_{q-1}v-\bar{\partial}_{q-1}u
=
f-f
=
0.
\end{align*}
Thus $h\in\ker\bar{\partial}_{q-1}$.
The canonical solution has the special form
\begin{align*}
u=\bar{\partial}_{q-1}^*w.
\end{align*}
This places $u$ in the range of the adjoint. Elements in the range of an adjoint are orthogonal to the kernel of the original operator. Here this is a direct computation using the definition of $\bar{\partial}_{q-1}^*$:
\begin{align*}
(u,h)_{H_{q-1}}
=
(\bar{\partial}_{q-1}^*w,h)_{H_{q-1}}
=
(w,\bar{\partial}_{q-1}h)_{H_q}
=
(w,0)_{H_q}
=
0.
\end{align*}
So the decomposition
\begin{align*}
v=u+h
\end{align*}
is orthogonal in $H_{q-1}=L^2_{(0,q-1)}(\Omega)$. The Pythagorean identity gives
\begin{align*}
\|v\|_{H_{q-1}}^2
=
\|u+h\|_{H_{q-1}}^2
=
\|u\|_{H_{q-1}}^2+\|h\|_{H_{q-1}}^2
\ge
\|u\|_{H_{q-1}}^2.
\end{align*}
Hence no other $L^2$ solution can have smaller norm. This proves the minimality of $\bar{\partial}_{q-1}^*N_qf$ and completes the theorem.
[/guided]
[/step]
