Proof.
The idea is to modify the weight far out in $\Omega $ so that lemma 6.4.2 applies and the error term in (6.4.5) is absorbed by additional convexity, and then to let the modification recede to the boundary.
Step 1: auxiliary functions. The matrix $H_\varphi $ is positive semidefinite because $\varphi $ is plurisubharmonic, and by the discussion preceding the theorem, applied with $A=H_\varphi $, the finiteness of $M$ implies that $\alpha \in L^2_{\mathrm{loc},(0,1)}(\Omega )$. By theorem 6.3.4, $\Omega $ has a nonnegative smooth strictly plurisubharmonic exhaustion function $s$. Fix an integer $t\ge 1$. Let $\kappa _0$ be the function fixed at the beginning of this section, put $c_\kappa =\sup |\kappa _0'|$, and for integers $\nu \ge 1$ put $\chi _\nu =\kappa _0(s-t-1-\nu )$. Then $0\le \chi _\nu \le 1$, and $\chi _\nu \in C^\infty _ c(\Omega )$, because $\{ s\le t+2+\nu \} $ is compact; $\chi _\nu =1$ on $\{ s\le t+1+\nu \} $, hence on a neighborhood of any given compact set for large $\nu $; and $\bar\partial \chi _\nu =\kappa _0'(s-t-1-\nu )\, \bar\partial s$ vanishes on $\{ s<t+2\} $ and satisfies $|\bar\partial \chi _\nu |^2\le c_\kappa ^2|\bar\partial s|^2$. We put
\[ \psi =\bigl (1-\kappa _0(s-t-1)\bigr )\log \bigl (1+c_\kappa ^2|\bar\partial s|^2\bigr ). \]
Then $\psi $ is smooth and nonnegative; $\psi $ and $\bar\partial \psi $ vanish on $\{ s\le t+1\} $, since $1-\kappa _0(x-t-1)$ and its derivative vanish for $x\le t+1$; and $|\bar\partial \chi _\nu |^2\le \mathrm e^\psi $ for every $\nu $, because $\mathrm e^\psi =1+c_\kappa ^2|\bar\partial s|^2$ on $\{ s\ge t+2\} $ and $\bar\partial \chi _\nu =0$ on $\{ s<t+2\} $.
Next we choose a smooth convex nondecreasing function $\gamma \colon \mathbb R\to [0,\infty )$ with $\gamma =0$ on $(-\infty ,t]$ such that
D557EE261608
6.4.8
\begin{equation} \gamma (s)\ge 2\psi ,\qquad H_{\gamma (s)}\ge (1+t)|\bar\partial \psi |^2 I, \qquad \int _\Omega |\alpha |^2\mathrm e^{-\varphi -\gamma (s)+\psi }\, \mathrm dV<\infty . \end{equation}
For integers $l\ge 1$ the sets $K_ l=\{ t+l\le s\le t+l+1\} $ are compact. The smallest eigenvalue of $H_ s$ is a positive continuous function on $\Omega $ (see the discussion following proposition 5.9.2), so its minimum $\lambda _ l$ over $K_ l$ is positive (if $K_ l=\varnothing $, put $\lambda _ l=1$ and $m_ l=P_ l=I_ l=0$), and the numbers $m_ l=(1+t)\max _{K_ l}|\bar\partial \psi |^2$, $P_ l=2\max _{K_ l}\psi $ and $I_ l=\int _{K_ l}|\alpha |^2\mathrm e^{-\varphi +\psi }\, \mathrm dV$ are finite, since $\alpha \in L^2_{\mathrm{loc},(0,1)}(\Omega )$. Choose a smooth nondecreasing function $g\colon \mathbb R\to [0,\infty )$ that vanishes on $(-\infty ,t]$ and satisfies, for every $l\ge 1$,
\[ g\ge m_ l/\lambda _ l\ \text{on }[t+l,t+l+1],\qquad \int _{t+l-1}^{t+l}g(x)\, \mathrm dx\ge \max \bigl \{ P_ l,\log ^+(2^ lI_ l)\bigr \} . \]
For instance, let $\kappa _1(x)=1-\kappa _0(2x)$, a smooth nondecreasing function with $\kappa _1=0$ on $(-\infty ,0]$ and $\kappa _1=1$ on $[1/2,\infty )$, and put $g(x)=\sum _{l\ge 1}c_ l\kappa _1(x-t-l+1)$ with $c_ l=\max \{ m_ l/\lambda _ l,2P_ l,2\log ^+(2^ lI_ l)\} $. The sum is locally finite, since only the terms with $l<x-t+1$ are nonzero at $x$, and all its terms vanish for $x\le t$. The $l$-th term equals $c_ l$ on $[t+l-1/2,\infty )$, so $g\ge c_ l$ on $[t+l,t+l+1]$ and $\int _{t+l-1}^{t+l}g(x)\, \mathrm dx\ge c_ l/2$.
Put $\gamma (x)=\int _{-\infty }^ xg(y)\, \mathrm dy$. Then $\gamma $ is smooth, nondecreasing and convex, because $\gamma '=g$ is nondecreasing, and $\gamma =0$ on $(-\infty ,t]$. Since $\Omega $ is the union of $\{ s\le t+1\} $ and the sets $K_ l$, we may check the first two conditions in (6.4.8) on each of these sets. On $\{ s\le t+1\} $ they hold because $\psi $ and $\bar\partial \psi $ vanish there while $\gamma \ge 0$ and $H_{\gamma (s)}\ge 0$. On $K_ l$ we have $\gamma (s)\ge \gamma (t+l)\ge P_ l\ge 2\psi $, and the chain rule
\[ H_{\gamma (s)}=\gamma '(s)H_ s+\gamma ''(s)\bigl (\partial _{z_ j}s\, \overline{\partial _{z_ k}s}\bigr )_{j,k}\ge \gamma '(s)H_ s \]
gives $H_{\gamma (s)}\ge (m_ l/\lambda _ l)\lambda _ lI\ge (1+t)|\bar\partial \psi |^2I$. Finally, the integral in (6.4.8) over the compact set $\{ s\le t+1\} $ is finite, and its integral over $K_ l$ is at most $\mathrm e^{-\gamma (t+l)}I_ l\le 2^{-l}$, because $\gamma (s)\ge \gamma (t+l)\ge \log ^+(2^ lI_ l)$ there; summing over $l$, we obtain the third condition.
Step 2: the estimate. Put $\varphi _0=\varphi +\gamma (s)$ and $\varphi _ k=\varphi _0+(k-3)\psi $ for $k=1,2,3$, and let $T$ and $S$ be the corresponding operators. By the last condition in (6.4.8), $\alpha \in \mathcal H_1=L^2_{(0,1)}(\Omega ,\mathrm e^{-\varphi _2})$, and $\alpha \in \ker S$. Let $f$ be a smooth compactly supported $(0,1)$-form. We apply (6.4.5) with $\varphi _0$ in place of $\varphi $ and with the integer $t$ fixed in Step 1. Since $H_{\varphi _0}=H_\varphi +H_{\gamma (s)}$, the second condition in (6.4.8) absorbs the last term of (6.4.5), and we obtain
\[ \int H_\varphi [f]\, \mathrm e^{-\varphi _0}\, \mathrm dV \le (1+t^{-1})\| T^*f\| _{\mathcal H_0}^2+\| Sf\| _{\mathcal H_2}^2. \]
Moreover $\varphi -2\varphi _2=-\varphi _0-\gamma (s)+2\psi \le -\varphi _0$ by the first condition in (6.4.8). Since $M<\infty $, the dual energy of $\alpha $ is finite almost everywhere, and the inequality $|\sum _ j\alpha _ j\bar f_ j|^2\le |\alpha |_{H_\varphi ^{-1}}^2H_\varphi [f]$ together with the Cauchy–Schwarz inequality gives
\[ \begin{split} |\langle \alpha ,f\rangle _{\mathcal H_1}|^2 & \le \int |\alpha |_{H_\varphi ^{-1}}^2\mathrm e^{-\varphi }\, \mathrm dV\cdot \int H_\varphi [f]\, \mathrm e^{\varphi -2\varphi _2}\, \mathrm dV \le M\int H_\varphi [f]\, \mathrm e^{-\varphi _0}\, \mathrm dV\\ & \le M(1+t^{-1})\bigl (\| T^*f\| _{\mathcal H_0}^2+\| Sf\| _{\mathcal H_2}^2\bigr ). \end{split} \]
Both sides are continuous for the graph norm of lemma 6.4.2, whose hypotheses hold by Step 1; hence the inequality extends to all $f\in \operatorname{Dom}T^*\cap \operatorname{Dom}S$. By theorem 6.2.4 there is $u_ t\in \mathcal H_0$ with $\bar\partial u_ t=\alpha $ and $\| u_ t\| _{\mathcal H_0}^2\le M(1+t^{-1})$. On $\{ s<t\} $ both $\gamma (s)$ and $\psi $ vanish, so $\varphi _1=\varphi $ there and
\[ \int _{\{ s<t\} }|u_ t|^2\mathrm e^{-\varphi }\, \mathrm dV\le M(1+t^{-1}). \]
Step 3: passage to the limit. For each integer $l\ge 1$ the functions $u_ t$ with $t\ge l$ have squared norms at most $2M$ in $L^2(\{ s<l\} ,\mathrm e^{-\varphi })$. By proposition 6.2.2 and a diagonal argument, there is a sequence $t_ i\to \infty $ such that $u_{t_ i}$ converges weakly in $L^2(\{ s<l\} ,\mathrm e^{-\varphi })$ for every $l$. The limits on different sublevel sets agree on their intersections, because restriction from $\{ s<l+1\} $ to $\{ s<l\} $ is a bounded linear map, and they define a function $u$ on $\Omega $. Weak convergence in these spaces implies convergence in the sense of distributions: a test function $\chi $ has support in some set $\{ s<l\} $, and $\int u_{t_ i}\chi \, \mathrm dV=\langle u_{t_ i},\bar\chi \mathrm e^\varphi \rangle _\varphi $ with $\bar\chi \mathrm e^\varphi \in L^2(\{ s<l\} ,\mathrm e^{-\varphi })$. Hence $\bar\partial u=\alpha $. By the lower semicontinuity of the norm in proposition 6.2.2, $\int _{\{ s<l\} }|u|^2\mathrm e^{-\varphi }\, \mathrm dV\le \varliminf _ iM(1+t_ i^{-1})=M$ for every $l$, and the monotone convergence theorem gives $\int _\Omega |u|^2\mathrm e^{-\varphi }\, \mathrm dV\le M$.
$\square$
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