Lemma 6.14.287D4C2281D7C

Proof. In dimension zero there is nothing to prove, so let $n\geq 1$. Choose concentric balls $B\Subset B'\Subset B''$ such that $\theta $ is defined on a neighborhood of $\overline{B''}$. By the last assertion of lemma 5.4.2, applied to the closed real $2$-form $\theta $ on the ball $B''$, there is a smooth real one-form $\alpha $ on $B''$ with $\mathrm d\alpha =\theta $. Since $\theta $ has type $(1,1)$, the $(0,2)$-part $\bar\partial \alpha ^{0,1}$ of $\mathrm d\alpha $ vanishes. The ball $B'$ is pseudoconvex by corollary 6.3.5, and $\alpha ^{0,1}$ is bounded on $B'$. Hence theorem 6.4.9, applied on $B'$ with $p=0$ and $q=|z|^2$, whose complex Hessian is the identity, gives a function $f\in L^2(B')$ with $\bar\partial f=\alpha ^{0,1}$.

The function $f$ is smooth. Indeed, let $y\in B'$ and choose polydiscs $P'\Subset P\Subset B'$ centered at $y$. By lemma 5.4.2, applied to the smooth $\bar\partial $-closed form $\alpha ^{0,1}$ on $P$, there is a smooth function $f_0$ on $P'$ with $\bar\partial f_0=\alpha ^{0,1}$. Then $\bar\partial (f-f_0)=0$ on $P'$ in the sense of distributions, so $\Delta (f-f_0)=4\sum _ j\partial _{z_ j}\partial _{\bar z_ j}(f-f_0)=0$ there, and by theorem 3.1.11 the locally integrable function $f-f_0$ agrees almost everywhere on $P'$ with a smooth function. Thus $f$ agrees almost everywhere near each point with a smooth function; two such smooth functions coincide on the overlap of their domains, being continuous and equal almost everywhere, so we may take $f$ smooth on $B'$. Since $\alpha $ is real, $\alpha ^{1,0}=\overline{\alpha ^{0,1}}=\partial \bar f$, and the $(1,1)$-part of $\mathrm d\alpha =\theta $ gives

Thus $h=4\pi \operatorname{Im}f$ has the required property on $B\subset B'$.

For the last assertion, let $h$ be real and pluriharmonic on a coordinate ball. Then $\partial h$ is a $(1,0)$-form with holomorphic coefficients, since $\bar\partial \partial h=0$, and it is closed. By the last assertion of lemma 5.4.2, applied to the closed one-form $2\partial h$ on this ball, there is a smooth function $F$ on the whole ball with $\mathrm dF=2\partial h$. As $\mathrm dF$ has type $(1,0)$, $F$ is holomorphic, and $\mathrm d(\operatorname{Re}F)=\partial h+\bar\partial h=\mathrm dh$. Hence $h-\operatorname{Re}F$ is constant. $\square$