[proofplan] We work in the Hilbert complex formed by the maximal $L^2$ Dolbeault operators. The strictly pseudoconvex subelliptic estimate gives compactness of the form-domain inclusion and a coercive estimate after removing the harmonic space. The Hilbert-space closed range criterion then gives closed range for the adjacent $\bar\partial$ maps, while the closed positive quadratic form associated with $\Box_q$ gives the self-adjoint Neumann Laplacian. Finally, the coercive estimate on $(\ker\Box_q)^\perp$ gives a bounded inverse there, which is the $\bar\partial$-Neumann operator. [/proofplan]

[step:Set up the Hilbert complex and the Neumann quadratic form]For $0\le k\le n$, write \begin{align*} H_k=L^2(\Omega;\Lambda^{0,k}T^*\Omega). \end{align*} The maximal $L^2$ Dolbeault operator \begin{align*} \bar\partial_k:\operatorname{Dom}(\bar\partial_k)\subset H_k\to H_{k+1} \end{align*} is closed and densely defined, and its Hilbert adjoint is \begin{align*} \bar\partial_k^*:\operatorname{Dom}(\bar\partial_k^*)\subset H_{k+1}\to H_k. \end{align*} Because $\bar\partial_{k+1}\bar\partial_k=0$ in the sense of distributions, these operators form a Hilbert complex. Define the form domain in degree $q$ by \begin{align*} \mathcal D_q:=\operatorname{Dom}(\bar\partial_q)\cap\operatorname{Dom}(\bar\partial_{q-1}^*)\subset H_q. \end{align*} Define the quadratic form \begin{align*} Q_q:\mathcal D_q\times\mathcal D_q\to\mathbb C \end{align*} by \begin{align*} Q_q(u,v)=(\bar\partial_q u,\bar\partial_q v)_{H_{q+1}}+(\bar\partial_{q-1}^*u,\bar\partial_{q-1}^*v)_{H_{q-1}}. \end{align*} Since $\bar\partial_q$ and $\bar\partial_{q-1}^*$ are closed, $\mathcal D_q$ is complete for the graph norm \begin{align*} \|u\|_{\mathcal D_q}^2:=\|u\|_{H_q}^2+Q_q(u,u). \end{align*} Thus $Q_q$ is a closed non-negative Hermitian form on $H_q$. By the representation theorem for closed non-negative Hermitian forms, there is a unique non-negative self-adjoint operator \begin{align*} \Box_q:\operatorname{Dom}(\Box_q)\subset H_q\to H_q \end{align*} such that \begin{align*} (\Box_q u,v)_{H_q}=Q_q(u,v) \end{align*} for every $u\in\operatorname{Dom}(\Box_q)$ and every $v\in\mathcal D_q$. This operator is precisely the standard $\bar\partial$-Neumann Laplacian \begin{align*} \Box_q=\bar\partial_{q-1}\bar\partial_{q-1}^*+\bar\partial_q^*\bar\partial_q \end{align*} with its Neumann domain.[/step]

[guided]The first point is that all operators here are unbounded operators, so their domains are part of the data. For $0\le k\le n$, the [Hilbert space](/page/Hilbert%20Space) in degree $k$ is \begin{align*} H_k=L^2(\Omega;\Lambda^{0,k}T^*\Omega). \end{align*} The operator \begin{align*} \bar\partial_k:\operatorname{Dom}(\bar\partial_k)\subset H_k\to H_{k+1} \end{align*} is the maximal $L^2$ realization: a form $u\in H_k$ lies in $\operatorname{Dom}(\bar\partial_k)$ exactly when its distributional $\bar\partial u$ is represented by an element of $H_{k+1}$. This realization is closed and densely defined. Its Hilbert-space adjoint is the closed densely defined operator \begin{align*} \bar\partial_k^*:\operatorname{Dom}(\bar\partial_k^*)\subset H_{k+1}\to H_k. \end{align*} In degree $q$, the natural energy space is not all of $H_q$ but the common domain \begin{align*} \mathcal D_q:=\operatorname{Dom}(\bar\partial_q)\cap\operatorname{Dom}(\bar\partial_{q-1}^*). \end{align*} On this domain define \begin{align*} Q_q:\mathcal D_q\times\mathcal D_q\to\mathbb C \end{align*} by \begin{align*} Q_q(u,v)=(\bar\partial_q u,\bar\partial_q v)_{H_{q+1}}+(\bar\partial_{q-1}^*u,\bar\partial_{q-1}^*v)_{H_{q-1}}. \end{align*} This is the energy form for the $\bar\partial$-Neumann problem. The graph norm associated with the two closed operators is \begin{align*} \|u\|_{\mathcal D_q}^2:=\|u\|_{H_q}^2+Q_q(u,u). \end{align*} Because $\bar\partial_q$ and $\bar\partial_{q-1}^*$ are closed, a [Cauchy sequence](/page/Cauchy%20Sequence) in this norm converges in $H_q$, its $\bar\partial_q$ images converge in $H_{q+1}$, and its $\bar\partial_{q-1}^*$ images converge in $H_{q-1}$. Closedness of the two operators then puts the limit back in both domains. Hence $\mathcal D_q$ is complete for $\|\cdot\|_{\mathcal D_q}$, and $Q_q$ is a closed non-negative Hermitian form. The representation theorem for closed non-negative Hermitian forms now applies to $Q_q$. It produces a unique non-negative self-adjoint operator \begin{align*} \Box_q:\operatorname{Dom}(\Box_q)\subset H_q\to H_q \end{align*} such that \begin{align*} (\Box_q u,v)_{H_q}=Q_q(u,v) \end{align*} for every $u\in\operatorname{Dom}(\Box_q)$ and every $v\in\mathcal D_q$. The identity from [citetheorem:9220] identifies this represented operator with the usual $\bar\partial$-Neumann Laplacian \begin{align*} \Box_q=\bar\partial_{q-1}\bar\partial_{q-1}^*+\bar\partial_q^*\bar\partial_q \end{align*} on its standard Neumann domain.[/guided]

[step:Use the subelliptic estimate to obtain compactness and finite-dimensional harmonic space] Let \begin{align*} \mathcal H_q:=\ker\Box_q. \end{align*} By the energy identity [citetheorem:9220], \begin{align*} \mathcal H_q=\{u\in\mathcal D_q:\bar\partial_q u=0\text{ and }\bar\partial_{q-1}^*u=0\}. \end{align*} We use the standard subelliptic estimate for smoothly bounded strictly pseudoconvex domains: there exist constants $\varepsilon>0$ and $C_{\mathrm{sub}}>0$ such that every $u\in\mathcal D_q$ satisfies \begin{align*} \|u\|_{H^\varepsilon(\Omega)}\le C_{\mathrm{sub}}\bigl(Q_q(u,u)^{1/2}+\|u\|_{H_q}\bigr). \end{align*} Here $H^\varepsilon(\Omega;\Lambda^{0,q}T^*\Omega)$ is the [Sobolev space](/page/Sobolev%20Space) of order $\varepsilon$ of $(0,q)$-forms. Since $\Omega$ is bounded with smooth boundary, the Rellich [compactness theorem](/theorems/2748) gives a compact inclusion \begin{align*} H^\varepsilon(\Omega;\Lambda^{0,q}T^*\Omega)\hookrightarrow H_q. \end{align*} Therefore the inclusion \begin{align*} \mathcal D_q\hookrightarrow H_q \end{align*} is compact when $\mathcal D_q$ is equipped with $\|\cdot\|_{\mathcal D_q}$. The unit ball of $\mathcal H_q$ is bounded in $\mathcal D_q$, because $Q_q(u,u)=0$ on $\mathcal H_q$. Its image in $H_q$ is compact. A Hilbert space has compact closed unit ball only when it is finite-dimensional, so $\mathcal H_q$ is finite-dimensional. [/step]

[step:Derive coercivity on the orthogonal complement of harmonic forms] We claim that there exists a constant $C_q>0$ such that every \begin{align*} u\in\mathcal D_q\cap\mathcal H_q^\perp \end{align*} satisfies \begin{align*} \|u\|_{H_q}^2\le C_q Q_q(u,u). \end{align*} Suppose not. Then for each $j\in\mathbb N$ there exists \begin{align*} u_j\in\mathcal D_q\cap\mathcal H_q^\perp \end{align*} such that \begin{align*} \|u_j\|_{H_q}=1 \end{align*} and \begin{align*} Q_q(u_j,u_j)\le \frac{1}{j}. \end{align*} The sequence $(u_j)_{j=1}^{\infty}$ is bounded in $\mathcal D_q$, so compactness of the inclusion $\mathcal D_q\hookrightarrow H_q$ gives a subsequence, still denoted $(u_j)_{j=1}^{\infty}$, and an element $u\in H_q$ such that \begin{align*} u_j\to u \end{align*} in $H_q$. Since $\bar\partial_q u_j\to 0$ in $H_{q+1}$ and $\bar\partial_{q-1}^*u_j\to 0$ in $H_{q-1}$, closedness of $\bar\partial_q$ and $\bar\partial_{q-1}^*$ gives $u\in\mathcal D_q$, \begin{align*} \bar\partial_q u=0, \end{align*} and \begin{align*} \bar\partial_{q-1}^*u=0. \end{align*} Thus $u\in\mathcal H_q$. Since $\mathcal H_q^\perp$ is closed in $H_q$ and each $u_j\in\mathcal H_q^\perp$, also $u\in\mathcal H_q^\perp$. Hence $u=0$. But $u_j\to u$ in $H_q$ and $\|u_j\|_{H_q}=1$, so $\|u\|_{H_q}=1$, a contradiction. The coercive estimate follows. [/step]

[step:Apply the Hilbert complex closed range criterion to the adjacent Dolbeault maps] We use the following standard Hilbert-space closed range criterion: if $A:\operatorname{Dom}(A)\subset X\to Y$ is a closed densely defined operator between Hilbert spaces, then $\operatorname{Range}(A)$ is closed if and only if there is a constant $C_A>0$ such that \begin{align*} \|x\|_X\le C_A\|Ax\|_Y \end{align*} for every $x\in\operatorname{Dom}(A)\cap\ker(A)^\perp$. Apply this criterion to the Hilbert complex in degree $q$. The coercive estimate \begin{align*} \|u\|_{H_q}^2\le C_q\bigl(\|\bar\partial_q u\|_{H_{q+1}}^2+\|\bar\partial_{q-1}^*u\|_{H_{q-1}}^2\bigr) \end{align*} on $\mathcal D_q\cap\mathcal H_q^\perp$ implies, by the standard complex-level form of the criterion, that $\operatorname{Range}(\bar\partial_{q-1})$ and $\operatorname{Range}(\bar\partial_q^*)$ are closed. Taking Hilbert adjoints, closed range of $\bar\partial_q^*$ is equivalent to closed range of $\bar\partial_q$. Therefore both \begin{align*} \bar\partial_{q-1}:H_{q-1}\supset\operatorname{Dom}(\bar\partial_{q-1})\to H_q \end{align*} and \begin{align*} \bar\partial_q:H_q\supset\operatorname{Dom}(\bar\partial_q)\to H_{q+1} \end{align*} have closed range. If $q=n$, then $H_{n+1}=\{0\}$ and $\bar\partial_n=0$, so its range is $\{0\}$ and is closed. [/step]

[step:Invert the Neumann Laplacian on the orthogonal complement of its kernel]For every $u\in\operatorname{Dom}(\Box_q)$, the form representation and [citetheorem:9220] give \begin{align*} (\Box_q u,u)_{H_q}=Q_q(u,u). \end{align*} If additionally $u\in\operatorname{Dom}(\Box_q)\cap\mathcal H_q^\perp$, the coercive estimate gives \begin{align*} \|u\|_{H_q}^2\le C_q(\Box_q u,u)_{H_q}. \end{align*} Applying the [Cauchy-Schwarz inequality](/theorems/432) in $H_q$ to the [inner product](/page/Inner%20Product) on the right gives \begin{align*} \|u\|_{H_q}^2\le C_q\|\Box_q u\|_{H_q}\|u\|_{H_q}. \end{align*} If $u\ne 0$, division by $\|u\|_{H_q}$ yields \begin{align*} \|u\|_{H_q}\le C_q\|\Box_q u\|_{H_q}. \end{align*} The same inequality is immediate when $u=0$. Thus $\Box_q$ is bounded below on $\operatorname{Dom}(\Box_q)\cap\mathcal H_q^\perp$. Since $\Box_q$ is self-adjoint, this bounded-below estimate implies that \begin{align*} \operatorname{Range}(\Box_q)=\mathcal H_q^\perp \end{align*} and that the inverse \begin{align*} \Box_q^{-1}:\mathcal H_q^\perp\to\mathcal H_q^\perp \end{align*} is bounded with operator norm at most $C_q$. Define \begin{align*} N_q:H_q\to H_q \end{align*} by \begin{align*} N_q f=\Box_q^{-1}f \end{align*} for $f\in\mathcal H_q^\perp$, and \begin{align*} N_q f=0 \end{align*} for $f\in\mathcal H_q$. Since $H_q=\mathcal H_q\oplus\mathcal H_q^\perp$, this defines a bounded operator on all of $H_q$.[/step]

[guided]The coercive estimate now becomes an inverse estimate for $\Box_q$. Take \begin{align*} u\in\operatorname{Dom}(\Box_q)\cap\mathcal H_q^\perp. \end{align*} The form identity for the $\bar\partial$-Neumann Laplacian gives \begin{align*} (\Box_q u,u)_{H_q}=Q_q(u,u). \end{align*} Since $u$ is orthogonal to the harmonic space, the coercive estimate from the previous step applies: \begin{align*} \|u\|_{H_q}^2\le C_qQ_q(u,u). \end{align*} Substituting the energy identity gives \begin{align*} \|u\|_{H_q}^2\le C_q(\Box_q u,u)_{H_q}. \end{align*} Now apply the Cauchy-Schwarz inequality in the Hilbert space $H_q$: \begin{align*} |(\Box_q u,u)_{H_q}|\le \|\Box_q u\|_{H_q}\|u\|_{H_q}. \end{align*} Therefore \begin{align*} \|u\|_{H_q}^2\le C_q\|\Box_q u\|_{H_q}\|u\|_{H_q}. \end{align*} If $u\ne 0$, dividing by $\|u\|_{H_q}$ gives \begin{align*} \|u\|_{H_q}\le C_q\|\Box_q u\|_{H_q}. \end{align*} If $u=0$, the same inequality holds. This estimate says that $\Box_q$ cannot send a nonzero vector in $\mathcal H_q^\perp$ close to zero without the vector itself being close to zero. Since $\Box_q$ is self-adjoint and $\ker\Box_q=\mathcal H_q$, the Hilbert-space bounded inverse theorem for self-adjoint closed-range operators gives \begin{align*} \operatorname{Range}(\Box_q)=\mathcal H_q^\perp \end{align*} and a bounded inverse \begin{align*} \Box_q^{-1}:\mathcal H_q^\perp\to\mathcal H_q^\perp. \end{align*} The operator norm of this inverse is at most $C_q$. Finally define the Neumann operator \begin{align*} N_q:H_q\to H_q \end{align*} by inverting $\Box_q$ on $\mathcal H_q^\perp$ and setting it equal to zero on $\mathcal H_q$: \begin{align*} N_q f=\Box_q^{-1}f \end{align*} for $f\in\mathcal H_q^\perp$, and \begin{align*} N_q f=0 \end{align*} for $f\in\mathcal H_q$. The [orthogonal decomposition](/theorems/436) \begin{align*} H_q=\mathcal H_q\oplus\mathcal H_q^\perp \end{align*} makes this definition unambiguous. It also proves boundedness of $N_q$ on all of $H_q$, because it is bounded on $\mathcal H_q^\perp$ and zero on $\mathcal H_q$.[/guided]

[step:Conclude closed range and Neumann solvability] The preceding steps prove that $\bar\partial_{q-1}$ and $\bar\partial_q$ have closed range. They also prove that $\Box_q$ maps \begin{align*} \operatorname{Dom}(\Box_q)\cap\mathcal H_q^\perp \end{align*} bijectively onto $\mathcal H_q^\perp$ with bounded inverse. Hence $\operatorname{Range}(\Box_q)=\mathcal H_q^\perp$, so $\Box_q$ has closed range. The operator $N_q$ constructed above satisfies \begin{align*} N_q|_{\ker\Box_q}=0 \end{align*} and is the bounded inverse of $\Box_q$ on $(\ker\Box_q)^\perp$. This is precisely the claimed $\bar\partial$-Neumann solvability statement. [/step]