[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 $f\in H_q$.[/step]

[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: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*}[/step]

[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: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 $\bar{\partial}_{q-1}u=f$.[/step]

[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]