[proofplan] The estimate is first proved for $(0,q)$-forms that are smooth up to $\overline{\Omega}$ and satisfy the $\bar\partial^*$-boundary condition. For such forms the weighted Morrey–Kohn–Hörmander identity expresses $\|\bar\partial u\|_\phi^2 + \|\bar\partial^*_\phi u\|_\phi^2$ as the sum of a complex-Hessian interior term, a nonnegative gradient term, and a boundary integral of the Levi form. Pseudoconvexity forces the boundary integral to be nonnegative because the $\bar\partial^*$-boundary condition places the coefficient vectors in the complex tangent space, so it may be discarded. Choosing the weight $\phi = t|z|^2$ makes the Hessian term equal to $tq\|u\|_\phi^2$ via a combinatorial identity, yielding a weighted lower bound; comparing the weighted and unweighted adjoints and choosing $t$ so the lower-order error is absorbed converts this into the unweighted estimate with an explicit constant. A graph-norm density theorem extends the estimate from smooth forms to the full domain $\operatorname{Dom}(\bar\partial) \cap \operatorname{Dom}(\bar\partial^*)$. [/proofplan]

[step:Set up the weighted formalism and record the Morrey–Kohn–Hörmander identity] Throughout, $dV := d\mathcal{L}^{2n}$ is Lebesgue measure on $\mathbb{C}^n \cong \mathbb{R}^{2n}$ and $dS := d\mathcal{H}^{2n-1}$ is the induced surface measure on $b\Omega := \partial\Omega$. Fix a defining function $\rho \in C^\infty(V)$ on an open neighbourhood $V \supset \overline{\Omega}$ with $\Omega \cap V = \{\rho < 0\}$ and $\nabla\rho \neq 0$ on $b\Omega$. We write the Wirtinger derivatives $\partial_{z_j} = \tfrac12(\partial_{x_j} - i\partial_{y_j})$ and $\partial_{\bar z_j} = \tfrac12(\partial_{x_j} + i\partial_{y_j})$, and abbreviate $\rho_{z_j} := \partial_{z_j}\rho$ and $\rho_{j\bar k} := \partial_{z_j}\partial_{\bar z_k}\rho$. A $(0,q)$-form is written \begin{align*} u &= \sideset{}{'}\sum_{|J| = q} u_J \, d\bar z_J, \qquad d\bar z_J = d\bar z_{j_1} \wedge \cdots \wedge d\bar z_{j_q}, \end{align*} where the primed sum runs over strictly increasing multi-indices $J = (j_1 < \cdots < j_q)$ and each coefficient is a map $u_J : \Omega \to \mathbb{C}$. We extend the coefficients to arbitrary $q$-tuples by total antisymmetry: $u_{j_1\cdots j_q} := \operatorname{sgn}(\sigma)\, u_{\sigma(j_1) \cdots \sigma(j_q)}$ for the permutation $\sigma$ that sorts the indices increasingly, and $u_{j_1\cdots j_q} := 0$ if two indices coincide. For an increasing $(q-1)$-tuple $K$ and an index $j \in \{1,\dots,n\}$, $u_{jK}$ denotes the value at the $q$-tuple $(j, k_1, \dots, k_{q-1})$. The pointwise norm is $|u|^2 = \sideset{}{'}\sum_{|J|=q} |u_J|^2$. For $\phi \in C^\infty(\overline{\Omega})$ we use the weighted inner product and norm \begin{align*} (f, g)_\phi &= \int_\Omega \langle f, g\rangle\, e^{-\phi}\, dV, \qquad \|f\|_\phi^2 = (f,f)_\phi, \qquad \langle f, g\rangle = \sideset{}{'}\sum_{|J|=q} f_J\,\overline{g_J}, \end{align*} and write $\bar\partial^*_\phi$ for the Hilbert-space adjoint of $\bar\partial$ with respect to $(\cdot,\cdot)_\phi$. The unweighted case $\phi \equiv 0$ recovers $\|\cdot\| = \|\cdot\|_{L^2}$ and $\bar\partial^* = \bar\partial^*_0$. Let \begin{align*} \mathcal{D}^{(0,q)} := C^\infty_{(0,q)}(\overline{\Omega}) \cap \operatorname{Dom}(\bar\partial^*), \end{align*} the space of $(0,q)$-forms with coefficients smooth up to the boundary that lie in $\operatorname{Dom}(\bar\partial^*)$. For such forms the $\bar\partial^*$-boundary condition takes the explicit form (citing a result not yet in the wiki: the boundary characterization of $\operatorname{Dom}(\bar\partial^*)$) \begin{align*} \sum_{j=1}^n \rho_{z_j}(p)\, u_{jK}(p) = 0 \qquad \text{for all } p \in b\Omega \text{ and all increasing } K \text{ with } |K| = q-1, \tag{BC} \end{align*} and, moreover, $\bar\partial^*_\phi u = \vartheta_\phi u$ on $\mathcal{D}^{(0,q)}$, where $\vartheta_\phi$ is the formal weighted adjoint; the same condition (BC) characterizes $\operatorname{Dom}(\bar\partial^*_\phi)$ for every smooth $\phi$, since multiplying by the smooth nonvanishing factor $e^{-\phi}$ does not alter the leading boundary symbol. We record the **Morrey–Kohn–Hörmander identity** (citing a result not yet in the wiki: the Morrey–Kohn–Hörmander identity): for every $\phi \in C^\infty(\overline{\Omega})$ and every $u \in \mathcal{D}^{(0,q)}$, \begin{align*} \|\bar\partial u\|_\phi^2 + \|\bar\partial^*_\phi u\|_\phi^2 &= \sideset{}{'}\sum_{|K|=q-1}\sum_{j,k=1}^n \int_\Omega \phi_{j\bar k}\, u_{jK}\,\overline{u_{kK}}\, e^{-\phi}\, dV \\ &\quad + \sideset{}{'}\sum_{|J|=q}\sum_{k=1}^n \int_\Omega \big|\partial_{\bar z_k} u_J\big|^2\, e^{-\phi}\, dV \\ &\quad + \sideset{}{'}\sum_{|K|=q-1}\sum_{j,k=1}^n \int_{b\Omega} \rho_{j\bar k}\, u_{jK}\,\overline{u_{kK}}\, e^{-\phi}\, dS, \tag{MKH} \end{align*} where $\phi_{j\bar k} = \partial_{z_j}\partial_{\bar z_k}\phi$. All three terms on the right carry a $+$ sign. [/step]

[step:Discard the boundary integral using pseudoconvexity]Fix $\phi \in C^\infty(\overline{\Omega})$ and $u \in \mathcal{D}^{(0,q)}$. We show the boundary integral in (MKH) is nonnegative and may be dropped, along with the manifestly nonnegative gradient term. Fix $p \in b\Omega$ and an increasing $(q-1)$-tuple $K$, and set $a^{(K)}(p) := (u_{1K}(p), \dots, u_{nK}(p)) \in \mathbb{C}^n$. Denote by $\mathcal{L}_\rho(p)$ the Levi form of $\rho$ at $p$, the sesquilinear form on $\mathbb{C}^n$ defined by \begin{align*} \mathcal{L}_\rho(p)(a,b) := \sum_{j,k=1}^n \rho_{j\bar k}(p)\, a_j\,\overline{b_k}, \qquad a,b \in \mathbb{C}^n. \end{align*} The boundary condition (BC) states $\sum_{j=1}^n \rho_{z_j}(p)\, a^{(K)}_j(p) = 0$, i.e. $a^{(K)}(p) \in \ker\big(\partial\rho|_p\big) = T^{1,0}_p(b\Omega)$, the complex tangent space. Since $\Omega$ is pseudoconvex, [The Levi Condition Is a Boundary Test](/theorems/3392) gives that the Levi form is positive semidefinite on the complex tangent space, so \begin{align*} \sum_{j,k=1}^n \rho_{j\bar k}(p)\, a^{(K)}_j(p)\,\overline{a^{(K)}_k(p)} = \mathcal{L}_\rho(p)\big(a^{(K)}(p), a^{(K)}(p)\big) \ge 0. \end{align*} Summing over the finitely many increasing $K$ and integrating against the nonnegative density $e^{-\phi}\,dS$ shows the boundary integral in (MKH) is $\ge 0$. The gradient term is a sum of integrals of $|\partial_{\bar z_k} u_J|^2 \ge 0$, hence $\ge 0$ as well. Discarding both nonnegative terms in (MKH) yields \begin{align*} \|\bar\partial u\|_\phi^2 + \|\bar\partial^*_\phi u\|_\phi^2 \;\ge\; \sideset{}{'}\sum_{|K|=q-1}\sum_{j,k=1}^n \int_\Omega \phi_{j\bar k}\, u_{jK}\,\overline{u_{kK}}\, e^{-\phi}\, dV. \tag{$\dagger$} \end{align*}[/step]

[guided]Our goal is to extract a lower bound on $\|u\|_\phi^2$ from (MKH), so we want to keep the interior Hessian term (which, for the weight chosen in the next step, will reproduce $\|u\|_\phi^2$) and throw away everything else — but only after checking that what we throw away has the right sign. There are two terms to dispose of. The gradient term $\sideset{}{'}\sum_{|J|=q}\sum_k \int_\Omega |\partial_{\bar z_k} u_J|^2 e^{-\phi}\,dV$ is an integral of squared moduli, hence nonnegative with no further argument; dropping it can only decrease the right-hand side, preserving an inequality of the form $\text{LHS} \ge (\text{Hessian term})$. The boundary integral is the subtle one — its integrand $\rho_{j\bar k} u_{jK}\overline{u_{kK}}$ has no definite sign in general, and this is exactly where both pseudoconvexity and the boundary condition are consumed. Why should it be nonnegative here? Fix a boundary point $p$ and a multi-index $K$, and collect the coefficients into the vector $a^{(K)}(p) = (u_{1K}(p), \dots, u_{nK}(p))$. The quantity $\sum_{j,k} \rho_{j\bar k}(p)\, a^{(K)}_j\, \overline{a^{(K)}_k}$ is precisely the Levi form $\mathcal{L}_\rho(p)$ evaluated on $a^{(K)}(p)$. Now, the Levi form is **not** positive semidefinite on all of $\mathbb{C}^n$; pseudoconvexity only guarantees positivity on the complex tangent space $T^{1,0}_p(b\Omega) = \ker(\partial\rho|_p) = \{\xi : \sum_j \rho_{z_j}(p)\xi_j = 0\}$, by [The Levi Condition Is a Boundary Test](/theorems/3392). So we must check that $a^{(K)}(p)$ actually lands in that subspace. This is exactly what the boundary condition (BC) provides: membership $u \in \operatorname{Dom}(\bar\partial^*)$ forces $\sum_j \rho_{z_j}(p)\, u_{jK}(p) = 0$, i.e. $a^{(K)}(p) \in T^{1,0}_p(b\Omega)$. Without (BC) the coefficient vector could have a normal component and the Levi form could be negative — this is precisely why the basic estimate requires $u \in \operatorname{Dom}(\bar\partial^*)$ and not merely $u \in \operatorname{Dom}(\bar\partial)$. Hence $\mathcal{L}_\rho(p)(a^{(K)}(p), a^{(K)}(p)) \ge 0$ pointwise on $b\Omega$. Summing over the finitely many increasing $(q-1)$-tuples $K$ and integrating against $e^{-\phi}\,dS \ge 0$ keeps the sign, so the entire boundary integral is $\ge 0$. Dropping both the gradient and the boundary terms from the equality (MKH) leaves the inequality ($\dagger$).[/guided]

[step:Choose the weight $\phi = t|z|^2$ and evaluate the Hessian term by a combinatorial identity]For a parameter $t > 0$ take the weight \begin{align*} \phi : \overline{\Omega} \to \mathbb{R}, \qquad \phi(z) = t|z|^2 = t\sum_{l=1}^n z_l \bar z_l, \end{align*} which is smooth on $\overline{\Omega}$. Its complex Hessian is $\phi_{j\bar k} = \partial_{z_j}\partial_{\bar z_k}\phi = t\,\delta_{jk}$, so the right-hand side of ($\dagger$) becomes \begin{align*} \sideset{}{'}\sum_{|K|=q-1}\sum_{j,k=1}^n \int_\Omega t\,\delta_{jk}\, u_{jK}\,\overline{u_{kK}}\, e^{-\phi}\, dV = t \int_\Omega \Bigg(\sideset{}{'}\sum_{|K|=q-1}\sum_{j=1}^n |u_{jK}|^2\Bigg) e^{-\phi}\, dV. \end{align*} We claim the combinatorial identity \begin{align*} \sideset{}{'}\sum_{|K|=q-1}\sum_{j=1}^n |u_{jK}|^2 = q\,|u|^2 \qquad \text{pointwise on } \Omega. \tag{$\ast$} \end{align*} Indeed, fix an increasing $J$ with $|J| = q$. By antisymmetry $u_{jK} = 0$ unless $j \notin K$ and $\{j\} \cup K = J$ as sets; the pairs $(j, K)$ contributing to $|u_J|^2$ are exactly those with $j \in J$ and $K = J \setminus \{j\}$ (reordered increasingly), of which there are precisely $q$, and for each $|u_{jK}|^2 = |u_J|^2$ since reordering only multiplies the coefficient by a sign. Summing over $J$ gives ($\ast$). Substituting ($\ast$) and using $|u|^2 = \langle u, u\rangle$ together with the definition of $\|\cdot\|_\phi$, the right-hand side of ($\dagger$) equals $t q \|u\|_\phi^2$. Hence \begin{align*} \|\bar\partial u\|_\phi^2 + \|\bar\partial^*_\phi u\|_\phi^2 \;\ge\; t q\, \|u\|_\phi^2. \tag{$\star$} \end{align*}[/step]

[guided]We have reduced matters to producing a positive multiple of $\|u\|_\phi^2$ out of the interior Hessian term $\sideset{}{'}\sum_{K}\sum_{j,k}\int_\Omega \phi_{j\bar k} u_{jK}\overline{u_{kK}} e^{-\phi}\,dV$. The whole point of introducing a weight is that we get to **design** the Hessian $\phi_{j\bar k}$. The simplest strictly plurisubharmonic choice is $\phi(z) = t|z|^2$: a direct computation gives $\partial_{z_j}\partial_{\bar z_k}\big(t\sum_l z_l\bar z_l\big) = t\,\delta_{jk}$, i.e. the Hessian is $t$ times the identity matrix. This collapses the double sum over $j,k$ to a single diagonal sum, turning the Hessian term into $t\int_\Omega \big(\sideset{}{'}\sum_K\sum_j |u_{jK}|^2\big)e^{-\phi}\,dV$. The remaining question is how the quantity $\sideset{}{'}\sum_K\sum_j |u_{jK}|^2$ relates to the pointwise norm $|u|^2 = \sideset{}{'}\sum_J |u_J|^2$. It is **not** equal to $|u|^2$; there is a multiplicity factor. To see it, ask: for a fixed increasing $J$ with $|J|=q$, in how many ways does $|u_J|^2$ appear in $\sideset{}{'}\sum_K\sum_j|u_{jK}|^2$? A nonzero term $u_{jK}$ requires $j \notin K$ (else antisymmetry kills it) and $\{j\}\cup K = J$. Since $K$ is increasing of length $q-1$, it is determined by which element $j \in J$ we remove, and there are exactly $q$ choices of $j$. Reordering $(j,K)$ into increasing order only contributes a sign, so $|u_{jK}|^2 = |u_J|^2$ for each of these $q$ pairs. Therefore each $|u_J|^2$ is counted $q$ times, giving the identity ($\ast$): $\sideset{}{'}\sum_K\sum_j|u_{jK}|^2 = q|u|^2$. This factor of $q$ is the reason the constant in the basic estimate improves as $q$ grows, and it is also why the argument requires $q \ge 1$: for $q = 0$ there are no indices to contract and the Hessian term vanishes identically, so no positive lower bound on $\|u\|_\phi^2$ is produced — consistent with the basic estimate being false for functions ($q=0$). Plugging ($\ast$) back in, the Hessian term is exactly $tq\int_\Omega |u|^2 e^{-\phi}\,dV = tq\|u\|_\phi^2$, and ($\dagger$) becomes the weighted lower bound ($\star$).[/guided]

[step:Compare the weighted and unweighted adjoints and absorb the lower-order term]The inequality ($\star$) involves the weighted adjoint $\bar\partial^*_\phi$, whereas the theorem concerns the unweighted $\bar\partial^*$. We relate them. On $\mathcal{D}^{(0,q)}$ the formal adjoints satisfy $\vartheta_\phi u = e^{\phi}\,\vartheta\big(e^{-\phi} u\big)$, where $\vartheta u = -\sideset{}{'}\sum_{|K|=q-1}\big(\sum_{j=1}^n \partial_{z_j} u_{jK}\big)\,d\bar z_K$ is the unweighted formal adjoint. Computing, \begin{align*} \vartheta_\phi u = e^{\phi}\Bigg(-\sideset{}{'}\sum_{|K|=q-1}\sum_{j=1}^n \partial_{z_j}\big(e^{-\phi} u_{jK}\big)\Bigg)d\bar z_K = \vartheta u + \sideset{}{'}\sum_{|K|=q-1}\Bigg(\sum_{j=1}^n \phi_{z_j}\, u_{jK}\Bigg) d\bar z_K, \end{align*} since $\partial_{z_j}\big(e^{-\phi}u_{jK}\big) = e^{-\phi}\big(\partial_{z_j}u_{jK} - \phi_{z_j}u_{jK}\big)$. Using $\bar\partial^*_\phi u = \vartheta_\phi u$ and $\bar\partial^* u = \vartheta u$ on $\mathcal{D}^{(0,q)}$, we obtain $\bar\partial^*_\phi u = \bar\partial^* u + \beta u$ with \begin{align*} \beta u := \sideset{}{'}\sum_{|K|=q-1}\Bigg(\sum_{j=1}^n \phi_{z_j}\, u_{jK}\Bigg) d\bar z_K, \qquad \phi_{z_j} = t\,\bar z_j. \end{align*} Applying the [Cauchy–Schwarz inequality](/theorems/432) to the inner sum, for each $K$, \begin{align*} \Bigg|\sum_{j=1}^n \phi_{z_j}\, u_{jK}\Bigg|^2 \le \Bigg(\sum_{j=1}^n |\phi_{z_j}|^2\Bigg)\Bigg(\sum_{j=1}^n |u_{jK}|^2\Bigg) = t^2 |z|^2 \sum_{j=1}^n |u_{jK}|^2, \end{align*} because $\sum_j |\phi_{z_j}|^2 = t^2\sum_j |z_j|^2 = t^2|z|^2$. Summing over $K$ and invoking ($\ast$), \begin{align*} |\beta u|^2 \le t^2 |z|^2 \sideset{}{'}\sum_{|K|=q-1}\sum_{j=1}^n |u_{jK}|^2 = t^2 |z|^2\, q\, |u|^2 \le t^2 R^2 q\,|u|^2, \end{align*} where $R := \sup_{z \in \overline{\Omega}}|z| < \infty$ since $\Omega$ is bounded. Hence $\|\beta u\|_\phi \le t R \sqrt{q}\,\|u\|_\phi$, and by the triangle inequality together with $(a+b)^2 \le 2a^2 + 2b^2$, \begin{align*} \|\bar\partial^*_\phi u\|_\phi^2 \le \big(\|\bar\partial^* u\|_\phi + tR\sqrt{q}\,\|u\|_\phi\big)^2 \le 2\|\bar\partial^* u\|_\phi^2 + 2t^2 R^2 q\,\|u\|_\phi^2. \end{align*} Inserting this bound for $\|\bar\partial^*_\phi u\|_\phi^2$ into ($\star$) while keeping the $\|\bar\partial u\|_\phi^2$ term unchanged gives \begin{align*} t q\,\|u\|_\phi^2 \le \|\bar\partial u\|_\phi^2 + 2\|\bar\partial^* u\|_\phi^2 + 2t^2 R^2 q\,\|u\|_\phi^2. \end{align*} Now fix the value $t = \tfrac{1}{4R^2}$. Then $2t^2 R^2 q = \tfrac{q}{8R^2} = \tfrac{tq}{2}$, so the last term absorbs half of the left-hand side: \begin{align*} \frac{tq}{2}\,\|u\|_\phi^2 \le \|\bar\partial u\|_\phi^2 + 2\|\bar\partial^* u\|_\phi^2. \tag{$\sharp$} \end{align*} With this $t$, the weight obeys $0 \le \phi = t|z|^2 \le tR^2 = \tfrac14$ on $\overline{\Omega}$, hence $e^{-1/4} \le e^{-\phi} \le 1$, giving for every form $v$ \begin{align*} e^{-1/4}\|v\|^2 \le \|v\|_\phi^2 \le \|v\|^2. \end{align*} Applying the lower bound to $u$ and the upper bounds to $\bar\partial u, \bar\partial^* u$ in ($\sharp$), and using $\tfrac{tq}{2} = \tfrac{q}{8R^2}$, \begin{align*} \frac{q\, e^{-1/4}}{8R^2}\,\|u\|^2 \le \frac{tq}{2}\,\|u\|_\phi^2 \le \|\bar\partial u\|_\phi^2 + 2\|\bar\partial^* u\|_\phi^2 \le \|\bar\partial u\|^2 + 2\|\bar\partial^* u\|^2 \le 2\big(\|\bar\partial u\|^2 + \|\bar\partial^* u\|^2\big). \end{align*} Therefore, for all $u \in \mathcal{D}^{(0,q)}$, \begin{align*} \|u\|^2 \le \frac{16\, e^{1/4}\, R^2}{q}\big(\|\bar\partial u\|^2 + \|\bar\partial^* u\|^2\big), \tag{BE} \end{align*} which is the asserted estimate with $C_q = \dfrac{16\, e^{1/4}\, R^2}{q}$.[/step]

[guided]We hold the weighted bound ($\star$): $\|\bar\partial u\|_\phi^2 + \|\bar\partial^*_\phi u\|_\phi^2 \ge tq\|u\|_\phi^2$. The obstruction to reading off the theorem is that $\bar\partial^*_\phi \neq \bar\partial^*$: the weighted adjoint and the genuine $L^2$-adjoint are different first-order operators, so a weighted estimate is not automatically an unweighted one. We must quantify the discrepancy. The relation between formal adjoints is $\vartheta_\phi = e^\phi \vartheta(e^{-\phi}\,\cdot)$, which simply records that the weighted adjoint is conjugated by the weight $e^{-\phi}$. Carrying out the differentiation, the only new contribution comes from differentiating $e^{-\phi}$, producing a zeroth-order term in $u$: \begin{align*} \bar\partial^*_\phi u - \bar\partial^* u = \beta u = \sideset{}{'}\sum_{|K|=q-1}\Big(\sum_j \phi_{z_j} u_{jK}\Big) d\bar z_K. \end{align*} Crucially, $\beta u$ is of order zero: no derivatives of $u$ appear, only the multiplier $\phi_{z_j} = t\bar z_j$. This is what makes absorption possible. How large is it? By [Cauchy–Schwarz](/theorems/432) applied index-by-index and the combinatorial identity ($\ast$) again, $|\beta u| \le t|z|\sqrt q\,|u| \le tR\sqrt q\,|u|$, where $R = \sup_{\overline\Omega}|z|$ is finite because $\Omega$ is bounded — this is the only place boundedness of the domain enters. Integrating, $\|\beta u\|_\phi \le tR\sqrt q\,\|u\|_\phi$. Now substitute $\bar\partial^*_\phi u = \bar\partial^* u + \beta u$ into ($\star$). The elementary inequality $(a+b)^2 \le 2a^2 + 2b^2$ separates the genuine adjoint from the error, costing only a factor $2$: \begin{align*} tq\|u\|_\phi^2 \le \|\bar\partial u\|_\phi^2 + 2\|\bar\partial^* u\|_\phi^2 + 2t^2 R^2 q\,\|u\|_\phi^2. \end{align*} Here is the decisive scaling observation. The main term on the left grows **linearly** in $t$ (it is $tq\|u\|_\phi^2$), while the error term grows **quadratically** in $t$ (it is $2t^2R^2q\|u\|_\phi^2$). Therefore — contrary to what one might guess — we must take $t$ **small**, not large, so that the linear main term dominates the quadratic error. (Taking $t$ large would make the error swamp the main term and destroy the estimate.) The cleanest choice equalizes the error with half the main term: solving $2t^2R^2q = \tfrac12 tq$ gives $t = \tfrac{1}{4R^2}$. With this choice the error absorbs exactly half the left side, leaving ($\sharp$): $\tfrac{tq}{2}\|u\|_\phi^2 \le \|\bar\partial u\|_\phi^2 + 2\|\bar\partial^* u\|_\phi^2$. Finally we remove the weight. Because $\phi$ ranges only over $[0, \tfrac14]$ on $\overline\Omega$ for this $t$, the weight $e^{-\phi}$ is pinched between the positive constants $e^{-1/4}$ and $1$; hence weighted and unweighted $L^2$ norms are equivalent, with controllable constants. Bounding $\|u\|_\phi^2 \ge e^{-1/4}\|u\|^2$ from below and $\|\bar\partial u\|_\phi^2, \|\bar\partial^* u\|_\phi^2 \le \|\bar\partial u\|^2, \|\bar\partial^* u\|^2$ from above turns ($\sharp$) into the purely unweighted estimate (BE), and tracking the constants yields the explicit value $C_q = 16 e^{1/4} R^2 / q$. The strategy throughout was the standard one: introduce a weight to manufacture a positive zeroth-order term, then exploit boundedness of the domain to discard the weight at the cost of a finite multiplicative constant.[/guided]

[step:Extend the estimate to the full form domain by graph-norm density] It remains to pass from $\mathcal{D}^{(0,q)}$ to all of $\operatorname{Dom}(\bar\partial) \cap \operatorname{Dom}(\bar\partial^*)$. Equip the latter with the graph norm \begin{align*} \|u\|_{\mathrm{gr}} := \big(\|u\|^2 + \|\bar\partial u\|^2 + \|\bar\partial^* u\|^2\big)^{1/2}. \end{align*} Because $\Omega$ has $C^\infty$ boundary, $\mathcal{D}^{(0,q)} = C^\infty_{(0,q)}(\overline{\Omega}) \cap \operatorname{Dom}(\bar\partial^*)$ is dense in $\operatorname{Dom}(\bar\partial) \cap \operatorname{Dom}(\bar\partial^*)$ for the graph norm (citing a result not yet in the wiki: the Friedrichs–Hörmander graph-norm density theorem for $\bar\partial \oplus \bar\partial^*$). Let $u \in \operatorname{Dom}(\bar\partial) \cap \operatorname{Dom}(\bar\partial^*)$. By density choose a sequence $(u_m)_{m \ge 1} \subset \mathcal{D}^{(0,q)}$ with $u_m \to u$, $\bar\partial u_m \to \bar\partial u$, and $\bar\partial^* u_m \to \bar\partial^* u$, all in $L^2$. The estimate (BE) holds for each $u_m$: \begin{align*} \|u_m\|^2 \le \frac{16\, e^{1/4}\, R^2}{q}\big(\|\bar\partial u_m\|^2 + \|\bar\partial^* u_m\|^2\big). \end{align*} Each of the three $L^2$ norms is continuous with respect to $L^2$ convergence, so letting $m \to \infty$ gives \begin{align*} \|u\|^2 \le \frac{16\, e^{1/4}\, R^2}{q}\big(\|\bar\partial u\|^2 + \|\bar\partial^* u\|^2\big). \end{align*} Since $u \in \operatorname{Dom}(\bar\partial) \cap \operatorname{Dom}(\bar\partial^*)$ was arbitrary and $1 \le q \le n$ was fixed, this establishes Kohn's basic estimate with $C_q = \dfrac{16\, e^{1/4}\, R^2}{q}$, completing the proof. As $q \ge 1$, one may also take the $q$-independent constant $C_q = 16\, e^{1/4}\, R^2$. [/step]