[proofplan] Work in the standard data of Skoda's division setting: a pseudoconvex [open set](/page/Open%20Set) $\Omega\subset\mathbb{C}^m$, holomorphic maps $g=(g_1,\dots,g_p):\Omega\to\mathbb{C}^p$ and $f:\Omega\to\mathbb{C}$, a plurisubharmonic weight $\psi:\Omega\to[-\infty,\infty)$, the integer $q=\min(m,p-1)$, and a real parameter $\alpha>1$. The proof is one pointwise inequality fed into one functional inequality. Differentiating $v^0=\overline{g}f/|g|^2$ shows that $\bar\partial v^0$ is assembled from the vectors $w_k$ obtained by projecting $\overline{\partial_{z_k}g}$ off the direction $\overline{g}$; the Gram matrix of the $w_k$ equals $|g|^2$ times the complex Hessian of $\log|g|^2$, and the $w_k$ span a space of dimension at most $q$ because they are orthogonal to $\overline{g}$. A rank-$q$ trace inequality therefore bounds the pointwise pairing $\langle a,\bar\partial v^0\rangle$ by the curvature density of $\log|g|^2$ tested against $a$. Integrating, applying the Cauchy–Schwarz inequality, and inserting the curvature density into the Morrey–Kohn–Hörmander basic estimate produces the bound with constant $1/\alpha$; since $\alpha>1$, this is dominated by $\alpha/(\alpha-1)$, which is the asserted estimate. [/proofplan]

[step:Differentiate $v^0$ and express $\bar\partial v^0$ through vectors orthogonal to $\overline{g}$]We fix notation. Write $Z(g)=\{x\in\Omega:g(x)=0\}$ and let \begin{align*} s:\Omega\setminus Z(g)&\to(0,\infty)\\ x&\mapsto |g(x)|^2=\sum_{l=1}^p|g_l(x)|^2, \end{align*} a smooth positive function off $Z(g)$. On $K=\mathbb{C}^p$ we use the standard Hermitian inner product $\langle u,w\rangle=\sum_{l=1}^p u_l\overline{w_l}$, linear in the first argument, with $|u|^2=\langle u,u\rangle$. A $K$-valued $(0,1)$-form is written $a=\sum_{k=1}^m a_{\bar k}\,d\bar z_k$ with coefficient maps $a_{\bar k}:\Omega\setminus Z(g)\to\mathbb{C}^p$; in the Euclidean Kähler metric the pointwise inner product of two such forms is $\langle a,b\rangle=\sum_{k=1}^m\langle a_{\bar k},b_{\bar k}\rangle$ and $(a,b)_\Phi=\int_\Omega\langle a,b\rangle\,e^{-\Phi}\,d\mathcal{L}^{2m}$, where $\Phi=\alpha q\log s+\psi$. The minimal solution is the map \begin{align*} v^0:\Omega\setminus Z(g)&\to\mathbb{C}^p\\ x&\mapsto\frac{f(x)}{s(x)}\,\overline{g(x)}, \end{align*} with components $v^0_l=\overline{g_l}\,f/s$. Since $f$ and the $g_l$ are holomorphic, $\partial_{\bar z_k}f=0$ and $\partial_{\bar z_k}g_l=0$, whereas $\partial_{\bar z_k}\overline{g_l}=\overline{\partial_{z_k}g_l}$ and $\partial_{\bar z_k}s=\sum_{l'=1}^p g_{l'}\,\overline{\partial_{z_k}g_{l'}}$. Hence the coefficient of $d\bar z_k$ in $\bar\partial v^0$ has $l$-th component \begin{align*} (\bar\partial v^0)_{\bar k,l}=\partial_{\bar z_k}\!\Big(\frac{\overline{g_l}\,f}{s}\Big)=f\,\partial_{\bar z_k}\!\Big(\frac{\overline{g_l}}{s}\Big)=\frac{f}{s}\Big(\overline{\partial_{z_k}g_l}-\frac{\overline{g_l}\,\partial_{\bar z_k}s}{s}\Big). \end{align*} Define, for $k=1,\dots,m$, the vector $w_k\in\mathbb{C}^p$ by \begin{align*} w_{k,l}:=\overline{\partial_{z_k}g_l}-\frac{\partial_{\bar z_k}s}{s}\,\overline{g_l},\qquad l=1,\dots,p, \end{align*} so that \begin{align*} (\bar\partial v^0)_{\bar k}=\frac{f}{s}\,w_k . \end{align*} Each $w_k$ is orthogonal to $\overline{g}$: using $\overline{(\overline g)_l}=g_l$, \begin{align*} \langle w_k,\overline{g}\rangle=\sum_{l=1}^p w_{k,l}\,g_l=\sum_{l=1}^p g_l\,\overline{\partial_{z_k}g_l}-\frac{\partial_{\bar z_k}s}{s}\sum_{l=1}^p|g_l|^2=\partial_{\bar z_k}s-\frac{\partial_{\bar z_k}s}{s}\,s=0 . \end{align*} Hence every $w_k$ lies in the orthogonal complement $(\mathbb{C}\,\overline{g})^\perp\subset\mathbb{C}^p$, a subspace of complex dimension $p-1$. In particular $\dim\operatorname{span}\{w_1,\dots,w_m\}\le\min(m,p-1)=q$.[/step]

[guided]The quantity to control, $(a,\bar\partial v^0)_\Phi$, involves $\bar\partial v^0$, so we compute it explicitly. Why is $\bar\partial v^0\neq0$? The vector $v^0=\overline{g}f/s$ solves the division problem $g\cdot v^0=\sum_l g_l\overline{g_l}f/s=f$ at every point of $\Omega\setminus Z(g)$, but it is not holomorphic: it carries the antiholomorphic factor $\overline{g}$. Its failure to be holomorphic is precisely the obstruction $\bar\partial v^0$. Differentiating, holomorphy of $f$ and $g$ gives $\partial_{\bar z_k}f=0$ and $\partial_{\bar z_k}g_l=0$, so the only surviving antiholomorphic derivatives are $\partial_{\bar z_k}\overline{g_l}=\overline{\partial_{z_k}g_l}$ and $\partial_{\bar z_k}s=\sum_{l'} g_{l'}\overline{\partial_{z_k}g_{l'}}$. Thus \begin{align*} (\bar\partial v^0)_{\bar k,l}=f\,\partial_{\bar z_k}\!\Big(\frac{\overline{g_l}}{s}\Big)=\frac{f}{s}\Big(\overline{\partial_{z_k}g_l}-\frac{\overline{g_l}\,\partial_{\bar z_k}s}{s}\Big), \end{align*} and we package the bracket as the vector $w_k$ with $w_{k,l}=\overline{\partial_{z_k}g_l}-\tfrac{\partial_{\bar z_k}s}{s}\overline{g_l}$, giving $(\bar\partial v^0)_{\bar k}=\tfrac{f}{s}w_k$. What is $w_k$ geometrically? The term $\tfrac{\partial_{\bar z_k}s}{s}\overline{g}$ is exactly the [orthogonal projection](/theorems/437) of $\overline{\partial_{z_k}g}$ onto the line $\mathbb{C}\overline{g}$: indeed the projection is $\tfrac{\langle\overline{\partial_{z_k}g},\overline{g}\rangle}{|\overline{g}|^2}\overline{g}$, and $\langle\overline{\partial_{z_k}g},\overline{g}\rangle=\sum_l\overline{\partial_{z_k}g_l}\,g_l=\partial_{\bar z_k}s$ while $|\overline{g}|^2=s$. So $w_k$ is $\overline{\partial_{z_k}g}$ with its $\overline{g}$-component removed, i.e. its [orthogonal projection](/theorems/437) onto $(\mathbb{C}\overline{g})^\perp$. This explains the orthogonality computation \begin{align*} \langle w_k,\overline{g}\rangle=\sum_{l} g_l\overline{\partial_{z_k}g_l}-\frac{\partial_{\bar z_k}s}{s}\sum_l|g_l|^2=\partial_{\bar z_k}s-\partial_{\bar z_k}s=0 . \end{align*} The consequence we will exploit: the $m$ vectors $w_1,\dots,w_m$ all live in the $(p-1)$-dimensional space $(\mathbb{C}\overline{g})^\perp$, so they span at most $\min(m,p-1)=q$ dimensions. This is the origin of the rank $q$ in the statement — no separately authored lemma is needed.[/guided]

[step:Identify the complex Hessian of $\log|g|^2$ with the Gram matrix of the $w_k$]Let $\theta:=\log s$ and write $\theta_{j\bar k}:=\partial_{z_j}\partial_{\bar z_k}\theta$ for its complex Hessian, $1\le j,k\le m$. From $\partial_{\bar z_k}\theta=\partial_{\bar z_k}s/s$, \begin{align*} \theta_{j\bar k}=\partial_{z_j}\!\Big(\frac{\partial_{\bar z_k}s}{s}\Big)=\frac{\partial_{z_j}\partial_{\bar z_k}s}{s}-\frac{(\partial_{z_j}s)(\partial_{\bar z_k}s)}{s^2}, \end{align*} where $\partial_{z_j}\partial_{\bar z_k}s=\sum_{l=1}^p\partial_{z_j}g_l\,\overline{\partial_{z_k}g_l}$ and $\partial_{z_j}s=\sum_{l=1}^p\overline{g_l}\,\partial_{z_j}g_l$. We claim the Gram identity \begin{align*} \langle w_k,w_j\rangle=s\,\theta_{j\bar k},\qquad 1\le j,k\le m . \end{align*} Indeed, since $\overline{w_{j,l}}=\partial_{z_j}g_l-\dfrac{\partial_{z_j}s}{s}\,g_l$ (using $\overline{\partial_{\bar z_j}s}=\partial_{z_j}s$), \begin{align*} \langle w_k,w_j\rangle&=\sum_{l=1}^p\Big(\overline{\partial_{z_k}g_l}-\frac{\partial_{\bar z_k}s}{s}\overline{g_l}\Big)\Big(\partial_{z_j}g_l-\frac{\partial_{z_j}s}{s}g_l\Big)\\ &=\sum_{l}\overline{\partial_{z_k}g_l}\,\partial_{z_j}g_l-\frac{\partial_{z_j}s}{s}\sum_l\overline{\partial_{z_k}g_l}\,g_l-\frac{\partial_{\bar z_k}s}{s}\sum_l\overline{g_l}\,\partial_{z_j}g_l+\frac{\partial_{z_j}s\,\partial_{\bar z_k}s}{s^2}\sum_l|g_l|^2\\ &=\partial_{z_j}\partial_{\bar z_k}s-\frac{\partial_{z_j}s\,\partial_{\bar z_k}s}{s}-\frac{\partial_{\bar z_k}s\,\partial_{z_j}s}{s}+\frac{\partial_{z_j}s\,\partial_{\bar z_k}s}{s}\\ &=\partial_{z_j}\partial_{\bar z_k}s-\frac{\partial_{z_j}s\,\partial_{\bar z_k}s}{s}=s\,\theta_{j\bar k}, \end{align*} where we used $\sum_l\overline{\partial_{z_k}g_l}\,g_l=\partial_{\bar z_k}s$ and $\sum_l\overline{g_l}\,\partial_{z_j}g_l=\partial_{z_j}s$. Consequently the Hermitian matrix $(\theta_{j\bar k})_{j,k=1}^m=\frac1s\big(\langle w_k,w_j\rangle\big)_{j,k}$ is the Gram matrix of $w_1,\dots,w_m$ divided by $s$; it is positive semidefinite, and \begin{align*} \operatorname{rank}\big(\theta_{j\bar k}\big)_{j,k=1}^m=\dim\operatorname{span}\{w_1,\dots,w_m\}\le q . \end{align*} This recovers, in coordinates, the semipositivity of $i\partial\bar\partial\log|g|^2$, consistent with the [Log-Modulus of a Holomorphic Function is PSH](/theorems/3405).[/step]

[guided]The right-hand side of the target estimate carries the weight $\Phi=\alpha q\log s+\psi$, whose curvature is governed by the complex Hessian of $\log s$. We compute that Hessian and tie it to the very vectors $w_k$ that build $\bar\partial v^0$; this is the bridge that lets a single algebraic inequality (next step) connect the pairing with $\bar\partial v^0$ to the curvature. Differentiating $\theta=\log s$ twice, \begin{align*} \theta_{j\bar k}=\frac{\partial_{z_j}\partial_{\bar z_k}s}{s}-\frac{(\partial_{z_j}s)(\partial_{\bar z_k}s)}{s^2},\qquad \partial_{z_j}\partial_{\bar z_k}s=\sum_l\partial_{z_j}g_l\,\overline{\partial_{z_k}g_l}. \end{align*} Why expect this to be the Gram matrix of the $w_k$? Because $w_k$ was obtained from $\overline{\partial_{z_k}g}$ by removing its $\overline{g}$-component, and removing the rank-one $\overline{g}$-direction is exactly the second (subtracted) term of $\theta_{j\bar k}$. Carrying out the inner product, with $\overline{w_{j,l}}=\partial_{z_j}g_l-\tfrac{\partial_{z_j}s}{s}g_l$, the four resulting sums collapse via $\sum_l\overline{\partial_{z_k}g_l}g_l=\partial_{\bar z_k}s$, $\sum_l\overline{g_l}\partial_{z_j}g_l=\partial_{z_j}s$, $\sum_l|g_l|^2=s$ to \begin{align*} \langle w_k,w_j\rangle=\partial_{z_j}\partial_{\bar z_k}s-\frac{\partial_{z_j}s\,\partial_{\bar z_k}s}{s}=s\,\theta_{j\bar k}. \end{align*} Two facts follow that we use later. First, $(\theta_{j\bar k})$ is a Gram matrix (up to the positive factor $1/s$), hence positive semidefinite — so the curvature density tested against any $a$ is nonnegative. Second, its rank equals the dimension of $\operatorname{span}\{w_k\}$, which Step 1 bounded by $q$. The rank $q$ in the statement is therefore the rank of $i\partial\bar\partial\log|g|^2$, not an externally supplied parameter.[/guided]

[step:Bound the pointwise pairing $\langle a,\bar\partial v^0\rangle$ by the curvature density via a rank-$q$ trace inequality]Using $(\bar\partial v^0)_{\bar k}=\frac fs w_k$, \begin{align*} \langle a,\bar\partial v^0\rangle=\sum_{k=1}^m\Big\langle a_{\bar k},\frac{f}{s}\,w_k\Big\rangle=\frac{\overline f}{s}\sum_{k=1}^m\langle a_{\bar k},w_k\rangle=\frac{\overline f}{s}\,P,\qquad P:=\sum_{k=1}^m\langle a_{\bar k},w_k\rangle . \end{align*} Introduce the curvature density of $\theta$ tested against $a$, \begin{align*} H_\theta:=\sum_{j,k=1}^m\theta_{j\bar k}\,\langle a_{\bar j},a_{\bar k}\rangle\ \ge 0, \end{align*} nonnegative because $(\theta_{j\bar k})$ is positive semidefinite. The Gram identity of the previous step gives $\sum_{j,k}\langle w_k,w_j\rangle\langle a_{\bar j},a_{\bar k}\rangle=s\,H_\theta$. [claim:Rank trace inequality] Let $a_{\bar 1},\dots,a_{\bar m}$ and $w_1,\dots,w_m$ be vectors in $\mathbb{C}^p$, and let $r:=\dim\operatorname{span}\{w_1,\dots,w_m\}$. Then \begin{align*} \Big|\sum_{k=1}^m\langle a_{\bar k},w_k\rangle\Big|^2\le r\sum_{j,k=1}^m\langle w_k,w_j\rangle\,\langle a_{\bar j},a_{\bar k}\rangle . \end{align*} [/claim] [proof] Let $A,W\in\mathbb{C}^{p\times m}$ be the matrices whose $k$-th columns are $a_{\bar k}$ and $w_k$. Treating vectors as $p\times1$ matrices, $\langle a_{\bar k},w_k\rangle=w_k^{*}a_{\bar k}=(W^{*}A)_{kk}$, so \begin{align*} \sum_{k=1}^m\langle a_{\bar k},w_k\rangle=\operatorname{tr}(W^{*}A)=\operatorname{tr}(AW^{*})=\operatorname{tr}B,\qquad B:=AW^{*}\in\mathbb{C}^{p\times p}. \end{align*} Likewise $(W^{*}W)_{jk}=\langle w_k,w_j\rangle$ and $(A^{*}A)_{kj}=\langle a_{\bar j},a_{\bar k}\rangle$, so \begin{align*} \sum_{j,k=1}^m\langle w_k,w_j\rangle\langle a_{\bar j},a_{\bar k}\rangle=\operatorname{tr}(W^{*}WA^{*}A)=\operatorname{tr}\big((AW^{*})^{*}AW^{*}\big)=\|B\|_{\mathrm{HS}}^2, \end{align*} where $\|B\|_{\mathrm{HS}}^2=\operatorname{tr}(B^{*}B)$ is the squared Hilbert–Schmidt norm. Since $B=AW^{*}$, we have $\operatorname{rank}B\le\operatorname{rank}(W^{*})=\operatorname{rank}W=r$. Let $\Pi\in\mathbb{C}^{p\times p}$ be the [orthogonal projection](/theorems/437) onto $\operatorname{Range}(B)$; then $\Pi=\Pi^{*}$, $\Pi B=B$, and $\operatorname{tr}\Pi=\operatorname{rank}B\le r$. With the Hilbert–Schmidt inner product $\langle X,Y\rangle_{\mathrm{HS}}=\operatorname{tr}(Y^{*}X)$, \begin{align*} \operatorname{tr}B=\operatorname{tr}(\Pi B)=\langle B,\Pi\rangle_{\mathrm{HS}}, \end{align*} and the [Cauchy–Schwarz Inequality](/theorems/432) for $\langle\cdot,\cdot\rangle_{\mathrm{HS}}$ gives \begin{align*} |\operatorname{tr}B|=|\langle B,\Pi\rangle_{\mathrm{HS}}|\le\|B\|_{\mathrm{HS}}\,\|\Pi\|_{\mathrm{HS}}=\|B\|_{\mathrm{HS}}\sqrt{\operatorname{tr}\Pi}\le\sqrt{r}\,\|B\|_{\mathrm{HS}} . \end{align*} Squaring yields $|\operatorname{tr}B|^2\le r\,\|B\|_{\mathrm{HS}}^2$, which is the claim. [/proof] Applying the claim with $r=\dim\operatorname{span}\{w_k\}\le q$ and the identity $\sum_{j,k}\langle w_k,w_j\rangle\langle a_{\bar j},a_{\bar k}\rangle=s\,H_\theta$, \begin{align*} |P|^2\le r\,s\,H_\theta\le q\,s\,H_\theta . \end{align*} Therefore \begin{align*} |\langle a,\bar\partial v^0\rangle|^2=\frac{|f|^2}{s^2}\,|P|^2\le\frac{q\,|f|^2}{s}\,H_\theta . \end{align*}[/step]

[guided]We must compare the pairing $P=\sum_k\langle a_{\bar k},w_k\rangle$ with the curvature quantity $\sum_{j,k}\langle w_k,w_j\rangle\langle a_{\bar j},a_{\bar k}\rangle=s\,H_\theta$. A naive term-by-term Cauchy–Schwarz, $|P|\le\sum_k|a_{\bar k}||w_k|$, would cost a factor $m$. The point of Skoda's mechanism is that the $w_k$ span only $r\le q$ dimensions, so the honest cost is $r$, not $m$. To see this cleanly, encode the data as matrices $A,W\in\mathbb{C}^{p\times m}$ with columns $a_{\bar k},w_k$. Then $P=\operatorname{tr}(W^{*}A)=\operatorname{tr}(AW^{*})=\operatorname{tr}B$ with $B=AW^{*}$, and the curvature quantity is $\|B\|_{\mathrm{HS}}^2$. So the whole comparison is: how large can $|\operatorname{tr}B|$ be relative to $\|B\|_{\mathrm{HS}}$, given $\operatorname{rank}B\le r$? The trace only 'sees' $B$ through the projection $\Pi$ onto its range: $\operatorname{tr}B=\operatorname{tr}(\Pi B)=\langle B,\Pi\rangle_{\mathrm{HS}}$. Cauchy–Schwarz then bounds this by $\|B\|_{\mathrm{HS}}\|\Pi\|_{\mathrm{HS}}$, and $\|\Pi\|_{\mathrm{HS}}=\sqrt{\operatorname{tr}\Pi}=\sqrt{\operatorname{rank}B}\le\sqrt r$. This is the rank-$q$ improvement; equality in spirit is approached when $B$ is a scalar on its range. Feeding $\operatorname{rank}B\le\operatorname{rank}W=r\le q$ and $\|B\|_{\mathrm{HS}}^2=s\,H_\theta$ back in gives $|P|^2\le q\,s\,H_\theta$. Since $\langle a,\bar\partial v^0\rangle=\tfrac{\overline f}{s}P$, the pointwise pairing obeys $|\langle a,\bar\partial v^0\rangle|^2=\tfrac{|f|^2}{s^2}|P|^2\le\tfrac{q|f|^2}{s}H_\theta$. Note how the singular powers of $s$ are already tracking toward the weight: one power of $1/s$ remains, which will combine with $e^{-\Phi}=s^{-\alpha q}e^{-\psi}$ to produce the exponent $-2(\alpha q+1)$ in the statement.[/guided]

[step:Integrate, apply Cauchy–Schwarz, and bound the curvature density by the Morrey–Kohn–Hörmander estimate]Set \begin{align*} I:=\int_\Omega|f|^2\,|g|^{-2(\alpha q+1)}\,e^{-\psi}\,d\mathcal{L}^{2m}. \end{align*} If $I=\infty$ the right-hand side of the asserted estimate is infinite and the inequality holds automatically, so assume $I<\infty$. Taking absolute values and using the pointwise bound of the previous step (recall $H_\theta\ge0$), \begin{align*} |(a,\bar\partial v^0)_\Phi|\le\int_\Omega|\langle a,\bar\partial v^0\rangle|\,e^{-\Phi}\,d\mathcal{L}^{2m}\le\sqrt q\int_\Omega\frac{|f|}{\sqrt s}\,\sqrt{H_\theta}\,e^{-\Phi}\,d\mathcal{L}^{2m}. \end{align*} Apply the [Cauchy–Schwarz Inequality](/theorems/432) in $L^2(\Omega,\mathcal{L}^{2m})$ to the factors $\dfrac{|f|}{\sqrt s}\,e^{-\Phi/2}$ and $\sqrt{H_\theta}\,e^{-\Phi/2}$: \begin{align*} |(a,\bar\partial v^0)_\Phi|\le\sqrt q\Big(\int_\Omega\frac{|f|^2}{s}\,e^{-\Phi}\,d\mathcal{L}^{2m}\Big)^{1/2}\Big(\int_\Omega H_\theta\,e^{-\Phi}\,d\mathcal{L}^{2m}\Big)^{1/2}. \end{align*} Since $e^{-\Phi}=s^{-\alpha q}e^{-\psi}$, the first factor is \begin{align*} \int_\Omega\frac{|f|^2}{s}\,e^{-\Phi}\,d\mathcal{L}^{2m}=\int_\Omega|f|^2\,s^{-(\alpha q+1)}e^{-\psi}\,d\mathcal{L}^{2m}=\int_\Omega|f|^2|g|^{-2(\alpha q+1)}e^{-\psi}\,d\mathcal{L}^{2m}=I, \end{align*} so squaring gives \begin{align*} |(a,\bar\partial v^0)_\Phi|^2\le q\,I\int_\Omega H_\theta\,e^{-\Phi}\,d\mathcal{L}^{2m}. \end{align*} Now bound the curvature density. Because $\Phi=\alpha q\,\theta+\psi$, the complex Hessians add: $\Phi_{j\bar k}=\alpha q\,\theta_{j\bar k}+\psi_{j\bar k}$. The weight $\psi$ is plurisubharmonic, so by the [Levi Form Criterion for Smooth PSH Functions](/theorems/3403) its complex Hessian $(\psi_{j\bar k})$ is positive semidefinite, whence $\sum_{j,k}\psi_{j\bar k}\langle a_{\bar j},a_{\bar k}\rangle\ge0$ pointwise. Writing $H_\Phi:=\sum_{j,k}\Phi_{j\bar k}\langle a_{\bar j},a_{\bar k}\rangle$, we obtain $H_\Phi\ge\alpha q\,H_\theta$, i.e. \begin{align*} H_\theta\le\frac{1}{\alpha q}\,H_\Phi . \end{align*} By the Morrey–Kohn–Hörmander basic estimate — the weighted curvature inequality established in the proof of the [$\bar\partial$-Solvability on Pseudoconvex Domains (Hörmander–Kohn)](/theorems/3493) theorem — applied componentwise to the compactly supported form $a$ (the scalar weight $e^{-\Phi}$ and the Hermitian metric on $\mathbb{C}^p$ act diagonally on the $p$ components), \begin{align*} \int_\Omega H_\Phi\,e^{-\Phi}\,d\mathcal{L}^{2m}\le\|\bar\partial a\|_\Phi^2+\|\bar\partial_\Phi^* a\|_\Phi^2 . \end{align*} The hypotheses are met: $\Omega$ is pseudoconvex, and $a\in C_c^\infty(\Omega\setminus Z(g);\Lambda^{0,1}\otimes\mathbb{C}^p)$ has compact support disjoint from both $Z(g)$ and $\partial\Omega$, so $\Phi=\alpha q\log s+\psi$ is smooth on a neighbourhood of $\operatorname{supp}a$ and the underlying integration-by-parts identity carries no boundary term; the nonnegative gradient contribution $\sum_{j,k}\int_\Omega|\partial_{\bar z_j}a_{\bar k}|^2\,e^{-\Phi}\,d\mathcal{L}^{2m}$ is discarded. The compact support of $a$ inside $\Omega\setminus Z(g)$ is what makes the singular weight harmless — no cutoff across $Z(g)$ is required for this a priori estimate. Combining the last two displays, \begin{align*} \int_\Omega H_\theta\,e^{-\Phi}\,d\mathcal{L}^{2m}\le\frac{1}{\alpha q}\big(\|\bar\partial a\|_\Phi^2+\|\bar\partial_\Phi^* a\|_\Phi^2\big). \end{align*}[/step]

[guided]We now turn the pointwise bound $|\langle a,\bar\partial v^0\rangle|^2\le\tfrac{q|f|^2}{s}H_\theta$ into the global estimate. Why split the integrand as $\big(\tfrac{|f|}{\sqrt s}e^{-\Phi/2}\big)\big(\sqrt{H_\theta}\,e^{-\Phi/2}\big)$ before Cauchy–Schwarz? Because the first factor squared integrates to exactly the weighted $f$-integral $I$ that appears in the statement, while the second factor squared is the curvature integral $\int H_\theta e^{-\Phi}$, which the basic estimate controls. Cauchy–Schwarz automatically optimises the split, so \begin{align*} |(a,\bar\partial v^0)_\Phi|^2\le q\,I\int_\Omega H_\theta e^{-\Phi}\,d\mathcal{L}^{2m},\qquad I=\int_\Omega\frac{|f|^2}{s}e^{-\Phi}\,d\mathcal{L}^{2m}=\int_\Omega|f|^2|g|^{-2(\alpha q+1)}e^{-\psi}\,d\mathcal{L}^{2m}, \end{align*} using $e^{-\Phi}=s^{-\alpha q}e^{-\psi}$. Next we must convert $\int H_\theta e^{-\Phi}$ into the Dirichlet-type quantity $\|\bar\partial a\|_\Phi^2+\|\bar\partial_\Phi^* a\|_\Phi^2$. The bridge is the Morrey–Kohn–Hörmander basic estimate, the same curvature inequality that powers $\bar\partial$-solvability on pseudoconvex domains. It states, for the chosen weight, $\int H_\Phi e^{-\Phi}\le\|\bar\partial a\|_\Phi^2+\|\bar\partial_\Phi^* a\|_\Phi^2$ where $H_\Phi=\sum_{j,k}\Phi_{j\bar k}\langle a_{\bar j},a_{\bar k}\rangle$ uses the full Hessian of $\Phi$. We verify its hypotheses carefully. (i) $\Omega$ is pseudoconvex. (ii) $a$ is smooth with compact support contained in $\Omega\setminus Z(g)$; hence $\Phi$ is smooth on a neighbourhood of $\operatorname{supp}a$, and the integration-by-parts identity behind the estimate has no boundary contribution and no singularity from $Z(g)$. (iii) The identity also contains the nonnegative term $\sum_{j,k}\int|\partial_{\bar z_j}a_{\bar k}|^2 e^{-\Phi}\ge0$, which we drop to get the inequality. (iv) Because the weight and metric are diagonal in the $p$ components of $\mathbb{C}^p$, the scalar estimate applies to each component and sums. Finally we connect $H_\theta$ to $H_\Phi$. Since $\Phi=\alpha q\,\theta+\psi$ and Hessians are additive, $H_\Phi=\alpha q\,H_\theta+\sum_{j,k}\psi_{j\bar k}\langle a_{\bar j},a_{\bar k}\rangle$. The hypothesis that $\psi$ is plurisubharmonic enters here: by the Levi form criterion its Hessian is positive semidefinite, so the $\psi$-term is $\ge0$ and may be dropped, leaving $H_\theta\le\tfrac1{\alpha q}H_\Phi$. (What would fail without $\psi$ psh? The $\psi$-term could be negative and we could no longer bound $H_\theta$ by $H_\Phi$ alone.) Chaining the three facts gives $\int H_\theta e^{-\Phi}\le\tfrac1{\alpha q}(\|\bar\partial a\|_\Phi^2+\|\bar\partial_\Phi^* a\|_\Phi^2)$.[/guided]

[step:Combine the estimates and absorb the constant using $\alpha>1$] Substituting the curvature bound of the previous step into $|(a,\bar\partial v^0)_\Phi|^2\le q\,I\int_\Omega H_\theta\,e^{-\Phi}\,d\mathcal{L}^{2m}$, \begin{align*} |(a,\bar\partial v^0)_\Phi|^2\le q\,I\cdot\frac{1}{\alpha q}\big(\|\bar\partial a\|_\Phi^2+\|\bar\partial_\Phi^* a\|_\Phi^2\big)=\frac{1}{\alpha}\,I\,\big(\|\bar\partial a\|_\Phi^2+\|\bar\partial_\Phi^* a\|_\Phi^2\big). \end{align*} For $\alpha>1$ the constant satisfies $\tfrac1\alpha\le\tfrac{\alpha}{\alpha-1}$: this is equivalent to $\alpha-1\le\alpha^2$, i.e. $\alpha^2-\alpha+1\ge0$, which holds for every real $\alpha$, while $\alpha>1$ ensures $\tfrac{\alpha}{\alpha-1}>0$. Therefore \begin{align*} \left|(a,\bar\partial v^0)_\Phi\right|^2\le\frac{\alpha}{\alpha-1}\Big(\int_\Omega|f|^2|g|^{-2(\alpha q+1)}e^{-\psi}\,d\mathcal{L}^{2m}\Big)\big(\|\bar\partial a\|_\Phi^2+\|\bar\partial_\Phi^* a\|_\Phi^2\big), \end{align*} which is exactly the asserted estimate. $\blacksquare$ [/step]

## Formalized Name

Skoda's Weighted $L^2$ Estimate