Lemma 6.5.20822DB927561

Proof. The idea is that, if $u$ is the minimal solution in $L^2(\Omega ,\mathrm e^{-p})$, then $u\mathrm e^\psi $ is the minimal solution of a twisted equation in $L^2(\Omega ,\mathrm e^{-p-\psi })$, to which we apply Hörmander's estimate. Put $m=\sup _\Omega |\psi |<\infty $. Since $\mathrm e^{-m}\le \mathrm e^{\pm \psi }\le \mathrm e^ m$, the spaces $L^2(\Omega ,\mathrm e^{-p})$ and $L^2(\Omega ,\mathrm e^{-p-\psi })$ consist of the same functions, with equivalent norms, and so contain the same holomorphic functions. By the discussion preceding theorem 6.4.7, applied with the weight $p-\psi $, which is bounded above on compact sets, $\alpha \in L^2_{\mathrm{loc},(0,1)}(\Omega )$; so $\bar\partial \alpha $ is defined. Since $\mathrm e^{-p-\psi }\le \mathrm e^{2m}\mathrm e^{\psi -p}$, we have $\int _\Omega |\alpha |_{A^{-1}}^2\mathrm e^{-p-\psi }\, \mathrm dV\le \mathrm e^{2m}E<\infty $. Hence theorem 6.4.9, applied with the plurisubharmonic function $p$ and the smooth function $q=\psi $, whose complex Hessian matrix $A$ is positive semidefinite, shows that the equation $\bar\partial u=\alpha $ has a solution in $L^2(\Omega ,\mathrm e^{-p-\psi })=L^2(\Omega ,\mathrm e^{-p})$. Let $u$ be the minimal solution in $L^2(\Omega ,\mathrm e^{-p})$, given by lemma 6.4.10(2).

The function $v=u\mathrm e^\psi $ belongs to $L^2(\Omega ,\mathrm e^{-p-\psi })$ and solves

For every holomorphic $g$ in $L^2(\Omega ,\mathrm e^{-p-\psi })$ we have $\int v\bar g\, \mathrm e^{-p-\psi }\, \mathrm dV=\int u\bar g\, \mathrm e^{-p}\, \mathrm dV=0$, so $v$ is the minimal solution of $\bar\partial v=\beta $ in $L^2(\Omega ,\mathrm e^{-p-\psi })$, by lemma 6.4.10(2).

We now estimate $v$ by theorem 6.4.9, applied again with $p$ and $q=\psi $. The form $\beta =\bar\partial v$ is $\bar\partial $-closed, and it lies in $L^2_{\mathrm{loc},(0,1)}(\Omega )$ because $\alpha $ and $u$ are locally square integrable and $\psi $ is smooth. Since $E<\infty $ and $h<1$, the dual energies of $\alpha $ and $\bar\partial \psi $ are finite almost everywhere, and at such points both vectors lie in the subspace on which the dual energy is a Hermitian quadratic form. The triangle inequality $|\alpha +u\, \bar\partial \psi |_{A^{-1}}\le |\alpha |_{A^{-1}}+|u|h^{1/2}$ therefore gives

So theorem 6.4.9 yields a solution of $\bar\partial w=\beta $ whose squared norm in $L^2(\Omega ,\mathrm e^{-p-\psi })$ is at most this integral, and the minimal solution $v$ has no larger norm. As $\int |v|^2\mathrm e^{-p-\psi }\, \mathrm dV=\int |u|^2\mathrm e^{\psi -p}\, \mathrm dV$, we obtain

At almost every point, let $\langle \cdot ,\cdot \rangle _{A^{-1}}$ be the Hermitian inner product associated with the dual energy on the subspace where it is finite; by the polarization identity $\langle x,y\rangle _{A^{-1}}=\frac14\sum _{k=0}^3\mathrm i^ k|x+\mathrm i^ ky|_{A^{-1}}^2$, the function $\langle \alpha ,\bar\partial \psi \rangle _{A^{-1}}$ is measurable. We expand

the second inequality being the Cauchy–Schwarz inequality. The function $h|u|^2\mathrm e^{\psi -p}$ is integrable, since $h<1$ almost everywhere and $u\in L^2(\Omega ,\mathrm e^{-p})$; moving its integral to the left, we find that $J=\int (1-h)|u|^2\mathrm e^{\psi -p}\, \mathrm dV$, which is finite, satisfies

If $a=0$, then $h=0$ almost everywhere on $\{ \alpha \ne 0\} $, the last integral vanishes, and $J\le E$, as claimed. Let $a>0$. The last integrand vanishes outside $\{ \alpha \ne 0\} $, where $h\le a$ and hence $h/(1-h)\le a/(1-a)$. For $t>1$, the inequality $2xy\le t^{-1}x^2+ty^2$ with $x=(1-h)^{1/2}|u|$ and $y=(h/(1-h))^{1/2}|\alpha |_{A^{-1}}$ gives there

Integrating, we obtain

Since $J$ is finite, this gives $J\le \frac t{t-1}\bigl (1+\frac{ta}{1-a}\bigr )E$. Choose $t=1+a^{-1/2}$. Then $1+ta/(1-a)=(1+\sqrt a)/(1-a)=1/(1-\sqrt a)$ and $t/(t-1)=1+\sqrt a$, so $J\le (1+\sqrt a)E/(1-\sqrt a)$. $\square$