Lemma 6.4.10961105705CF9

Proof. (1) The real and imaginary parts of $h$, given by $\operatorname{Re}\langle h,\chi \rangle $ and $\operatorname{Im}\langle h,\chi \rangle $ for real test functions $\chi $, are real distributions. Since derivatives of distributions commute, (5.2.9) shows that the Laplacian of $h$ is $4\sum _ j\partial _{z_ j}\partial _{\bar z_ j}h=0$. As the Laplacian has real coefficients, the real and imaginary parts of $h$ are harmonic distributions. By Weyl's lemma (theorem 3.1.11) they are given by smooth functions, and hence $h$ is given by a smooth function $g$. By integration by parts, the classical derivatives $\partial _{\bar z_ j}g$ represent the distributions $\partial _{\bar z_ j}h=0$, so they vanish almost everywhere by proposition 2.5.3(4), and hence everywhere, being continuous functions. Thus $g$ is holomorphic.

(2) Let $L\subset \Omega $ be compact, and let $m_ L$ be a real upper bound for $\varphi $ on $L$. For $u\in L^2(\Omega ,\mathrm e^{-\varphi })$ we have $\int _ L|u|^2\, \mathrm dV\le \mathrm e^{m_ L}\int _ L|u|^2\mathrm e^{-\varphi }\, \mathrm dV$. Hence $L^2(\Omega ,\mathrm e^{-\varphi })\subset L^2_{\mathrm{loc}}(\Omega )$, so that $\bar\partial u$ is defined in the sense of distributions, and convergence in $L^2(\Omega ,\mathrm e^{-\varphi })$ implies convergence in $L^2_{\mathrm{loc}}(\Omega )$, hence in the sense of distributions. The space $L^2(\Omega ,\mathrm e^{-\varphi })$ is a Hilbert space, by the completeness argument at the beginning of section 6.2, applied to the positive measure $\mathrm e^{-\varphi }\, \mathrm dV$. The set $\mathcal K$ is a linear subspace. It is closed: if $g_ j\in \mathcal K$ converge to $g$ in $L^2(\Omega ,\mathrm e^{-\varphi })$, then $g_ j\to g$ in the sense of distributions, so $\bar\partial g=\lim _ j\bar\partial g_ j=0$, and by (1) $g$ agrees almost everywhere with a holomorphic function, that is, $g\in \mathcal K$.

Let $u_0$ be a solution of $\bar\partial u=\alpha $ in $L^2(\Omega ,\mathrm e^{-\varphi })$. By (1), an element $u$ of this space is a solution if and only if $\bar\partial (u-u_0)=0$, that is, $u-u_0\in \mathcal K$. Thus the solutions form the nonempty closed convex set $u_0+\mathcal K$, which by proposition 6.2.1 contains a unique element $u_{\min }$ of smallest norm. If a solution $u$ is orthogonal to $\mathcal K$, then $\| u+k\| ^2=\| u\| ^2+\| k\| ^2\geq \| u\| ^2$ for every $k\in \mathcal K$, so $u$ has smallest norm, that is, $u=u_{\min }$. Conversely, if $P$ is the orthogonal projection onto $\mathcal K$ (proposition 6.2.1), then $u_{\min }-Pu_{\min }$ is a solution orthogonal to $\mathcal K$, hence equal to $u_{\min }$; so $u_{\min }$ is orthogonal to $\mathcal K$. $\square$