Lemma 5.4.27815541A95AD

Proof. We solve the $\bar\partial $-equation in one variable by a Cauchy integral, then on a polydisc one variable at a time, and combine it with the Poincaré lemma.

Step 1: one variable. The function $K(z)=1/(\pi z)$ on $\mathbb C$ is locally integrable, and $\partial _{\bar z}K=\delta _0$ in the sense of distributions. Indeed, for a test function $\varphi $, dominated convergence gives $\langle \partial _{\bar z}K,\varphi \rangle =-\pi ^{-1}\lim _{\varepsilon \to 0}\int _{|z|>\varepsilon }(\partial _{\bar z}\varphi /z)\, \mathrm dV_2$. In polar coordinates $z=r\mathrm e^{\mathrm i\vartheta }$ we have $\partial _{\bar z}=\frac12\mathrm e^{\mathrm i\vartheta }(\partial _ r+\frac{\mathrm i}r\partial _\vartheta )$ and $\mathrm dV_2=r\, \mathrm dr\, \mathrm d\vartheta $. Hence, by Fubini's theorem,

because $\int _0^{2\pi }\partial _\vartheta \varphi \, \mathrm d\vartheta =0$ by periodicity and $\int _\varepsilon ^\infty \partial _ r\varphi \, \mathrm dr=-\varphi (\varepsilon \mathrm e^{\mathrm i\vartheta })$, $\varphi $ having compact support. As $\varepsilon \to 0$, the right-hand side tends to $-\pi \varphi (0)$, since $\varphi $ is continuous at $0$; so $\langle \partial _{\bar z}K,\varphi \rangle =\varphi (0)$. Consequently, for a smooth function $a$ with compact support in $\mathbb C$, the function

is smooth, since in the second expression all derivatives may be taken under the integral sign by dominated convergence, and $\partial _{\bar z}f(z)=\int K(w)(\partial _{\bar z}a)(z-w)\, \mathrm dV_2(w)=\langle \partial _{\bar w}K,a(z-\cdot )\rangle =a(z)$, because $\partial _{\bar w}[a(z-w)]=-(\partial _{\bar z}a)(z-w)$. More generally, let $a(\zeta ,s)$ be smooth on $\mathbb C\times S$, where $S$ is an open subset of some $\mathbb R^ N$, and suppose that $a$ vanishes for $\zeta $ outside a fixed compact set. Then $f(z,s)=\pi ^{-1}\int a(z-w,s)w^{-1}\, \mathrm dV_2(w)$ is smooth on $\mathbb C\times S$, and it is holomorphic in those complex parameters in which $a$ is holomorphic, again by differentiation under the integral sign: for $(z,s)$ in a compact set, the integrand vanishes for $w$ outside a fixed bounded set, and its derivatives are bounded by constants times the integrable function $|w|^{-1}$.

Step 2: the $\bar\partial $-equation on a polydisc. Write $\Delta =\Delta _1\times \cdots \times \Delta _ n$ and $\Delta '=\Delta '_1\times \cdots \times \Delta '_ n$, and write $g=\sum _ ja_ j\, \mathrm d\bar z_ j$. For each $j$ choose a smooth function $\chi _ j$ with compact support in $\Delta _ j$ that equals one on an open neighborhood $D_ j$ of $\overline{\Delta '_ j}$ (lemma 2.5.1), and put $P_ k=D_1\times \cdots \times D_ k\times \Delta _{k+1}\times \cdots \times \Delta _ n$. We construct smooth functions $f_1,\ldots ,f_ n$ such that, for $k=0,\ldots ,n$, the $\bar\partial $-closed form $g^{(k)}=g-\bar\partial (f_1+\cdots +f_ k)=\sum _ ja^{(k)}_ j\mathrm d\bar z_ j$ satisfies $a^{(k)}_ j=0$ on $P_ k$ for $j\leq k$. For $k=0$ there is nothing to prove. Let $k\geq 1$. Since $g^{(k-1)}$ is $\bar\partial $-closed, $\partial _{\bar z_ i}a^{(k-1)}_ j=\partial _{\bar z_ j}a^{(k-1)}_ i=0$ on $P_{k-1}$ for $i<k$; thus the coefficients of $g^{(k-1)}$ are holomorphic in $z_1,\ldots ,z_{k-1}$ there. Define

By Step 1, applied with the parameters $z_ i$, $i\neq k$, the function $f_ k$ is smooth on $\Delta $ and holomorphic in $z_1,\ldots ,z_{k-1}$ on $P_{k-1}$, and $\partial _{\bar z_ k}f_ k=\chi _ ka^{(k-1)}_ k=a^{(k-1)}_ k$ on $P_ k$. Hence $g^{(k)}=g^{(k-1)}-\bar\partial f_ k$ has the required property. For $k=n$ we obtain $g=\bar\partial (f_1+\cdots +f_ n)$ on $P_ n\supset \Delta '$.

Step 3: the Poincaré lemma and real closed $(1,1)$-forms. In this step we write points of $\mathbb C^ n=\mathbb R^{2n}$ as $x=(x_1,\ldots ,x_{2n})$, and we may assume that $R$ is defined on a ball $B$ centered at $0$. For a smooth $k$-form $\gamma $ on $B$, $k\geq 1$, define the $(k-1)$-form $H\gamma $ by

where $x$ is regarded as a tangent vector at $tx\in B$; the form $H\gamma $ is smooth, by differentiation under the integral sign. We claim that $\mathrm{d}H\gamma +H\mathrm{d}\gamma =\gamma $. Since both sides are linear in $\gamma $, it suffices to verify this for $\gamma =c\, \mathrm dx_ I$ with $|I|=k$. Let $\iota _ x$ denote contraction with $x$, and write $\partial _ l=\partial /\partial x_ l$. Then $H\gamma =F\, \iota _ x\mathrm dx_ I$ with $F(x)=\int _0^1t^{k-1}c(tx)\, \mathrm dt$, and $H\mathrm d\gamma =\sum _ l\partial _ lF\, \iota _ x(\mathrm dx_ l\wedge \mathrm dx_ I)$, because $\partial _ lF(x)=\int _0^1t^ k(\partial _ lc)(tx)\, \mathrm dt$. Using $\mathrm d(\iota _ x\mathrm dx_ I)=k\, \mathrm dx_ I$ and $\iota _ x(\mathrm dx_ l\wedge \mathrm dx_ I)=x_ l\, \mathrm dx_ I-\mathrm dx_ l\wedge \iota _ x\mathrm dx_ I$, we find

where we used that $t^ kc(tx)$ vanishes at $t=0$, since $k\geq 1$. In particular, a closed $k$-form $\gamma $ on $B$ satisfies $\gamma =\mathrm d(H\gamma )$, and $H\gamma $ is real if $\gamma $ is real; after a translation this proves the last assertion of the lemma for every ball. Applied to the closed real $2$-form $R$, this gives $R=\mathrm d\alpha $ with the real $1$-form $\alpha =HR$. Since $\alpha $ is real, $\alpha ^{1,0}=\overline{\alpha ^{0,1}}$. The $(0,2)$-component of $\mathrm d\alpha =R$ is $\bar\partial \alpha ^{0,1}=0$, so by Step 2, applied to a polydisc in $B$ centered at $0$, there is a smooth function $f$ with $\alpha ^{0,1}=\bar\partial f$ on a smaller polydisc; then $\alpha ^{1,0}=\partial \bar f$. Hence

and $v=4\pi \operatorname{Im}f$ has the required properties. $\square$