5.4 Local potentials for closed (1,1)-currents618FE2108D26

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. $\square$

Proof. Fix a point and a coordinate chart around it, identified with an open set $\Omega \subset \mathbb C^ n$, and write $T$ as above with trace $\mu =\sum _ jt_{j\bar j}$, a positive Radon measure on $\Omega $. Choose a test function $\chi $ on $\Omega $ with $0\leq \chi \leq 1$ that equals one on a neighborhood $W$ of the point (lemma 2.5.1), and let $E_{2n}$ be the fundamental solution of proposition 3.1.13. By proposition 3.4.11, the function $u_0=4E_{2n}*(\chi \mu )$ is locally integrable and subharmonic, and $\Delta u_0=4\chi \mu $; in particular $\Delta u_0=4\mu $ on $W$. The current $R=T-\mathrm{dd}^{\mathrm{c}}u_0$ on $W$ is real and closed, and its coefficients $r_{j\bar k}=t_{j\bar k}-(u_0)_{j\bar k}$ have trace $\sum _ lr_{l\bar l}=\mu -\Delta u_0/4=0$, by (5.2.9). Using both relations (5.4.1) for $R$, we find

By Weyl's lemma (theorem 3.1.11) every coefficient $r_{j\bar k}$ is smooth, so $R$ is a smooth real closed $(1,1)$-form on $W$. By lemma 5.4.2, $R=\mathrm{dd}^{\mathrm{c}}v$ on a coordinate neighborhood $U\subset W$ of the point, with $v$ smooth and real. Then $u=u_0+v$ satisfies $\mathrm{dd}^{\mathrm{c}}u=T$ on $U$. Moreover $\Delta v=4\sum _ lr_{l\bar l}=0$, so $v$ is harmonic. Since the sum of a subharmonic function and a real harmonic function is subharmonic (see the discussion following theorem 3.4.5), $u$ is a locally integrable subharmonic function. By the uniqueness in theorem 3.4.12, it is the canonical subharmonic representative of its class.

Now let $u$ and $\widetilde u$ be canonical local potentials of $T$ on the same coordinate neighborhood $U$. Their classes in $L^1_{\mathrm{loc}}(U)$ differ by a real distribution with vanishing complex Hessian, hence with vanishing Laplacian. By Weyl's lemma, this difference is a smooth real harmonic function $h$, and $\mathrm{dd}^{\mathrm{c}}h=0$. Hence, as above, $u+h$ is a locally integrable subharmonic function, and it agrees almost everywhere with $\widetilde u$. By the uniqueness in theorem 3.4.12, $\widetilde u=u+h$ at every point of $U$, including the points where $u=-\infty $. $\square$

Proof. We first work near a point. Fix $w_0\in D$, a ball $B\Subset \Omega $ centered at $F(w_0)$, and a connected open neighborhood $W$ of $w_0$ with $F(W)\subset B$. The mollifications $u_\varepsilon =u*\rho _\varepsilon $, with the radial mollifier $\rho _\varepsilon $ of section 2.5, are defined on $B$ for small $\varepsilon $, and by (3.4.13) they decrease pointwise to $u$ as $\varepsilon \downarrow 0$. Their complex Hessians are positive semidefinite at every point: for $\xi \in \mathbb C^ n$, the function $\sum _{j,k}(u_\varepsilon )_{j\bar k}\xi _ j\bar\xi _ k$ is the mollification of the positive measure $\sum _{j,k}u_{j\bar k}\xi _ j\bar\xi _ k$ (matrix criterion), hence nonnegative by proposition 2.5.3. Since $F$ is holomorphic, the chain rule gives

so the complex Hessian of $u_\varepsilon \circ F$ is positive semidefinite, and in particular its Laplacian is nonnegative. A smooth function with nonnegative Laplacian is subharmonic, because by proposition 3.1.5 its spherical means increase with the radius and tend to its value at the center. Thus, along a sequence $\varepsilon _ k\downarrow 0$, the smooth subharmonic functions $u_{\varepsilon _ k}\circ F$ decrease to $u\circ F$ on $W$. By theorem 3.4.9, $u\circ F$ is subharmonic on $W$, and it is either identically $-\infty $ there or locally integrable, in which case $u_{\varepsilon _ k}\circ F\to u\circ F$ in $L^1_{\mathrm{loc}}(W)$. In the second case the currents $\mathrm{dd}^{\mathrm{c}}(u_{\varepsilon _ k}\circ F)$, which are positive by the matrix criterion, converge weakly to $\mathrm{dd}^{\mathrm{c}}(u\circ F)$, and positivity passes to the limit; so $\mathrm{dd}^{\mathrm{c}}(u\circ F)\geq 0$ on $W$.

It remains to pass from these neighborhoods to the connected components of $D$. Let $D_0$ be a component and let $A$ be the set of points of $D_0$ having a neighborhood on which $u\circ F\equiv -\infty $. The set $A$ is open. It is also closed in $D_0$: if $w_0\in D_0$ lies in the closure of $A$, the connected neighborhood $W$ of $w_0$ constructed above meets $A$, so $u\circ F=-\infty $ on a nonempty open subset of $W$; since a locally integrable function is finite almost everywhere, the alternative on $W$ forces $u\circ F\equiv -\infty $ on $W$, and $w_0\in A$. Hence either $A=D_0$, or $A=\varnothing $; in the latter case $u\circ F$ is locally integrable and subharmonic with positive complex Hessian near every point of $D_0$, hence on $D_0$. $\square$