7.14 Gradients against positive currents166A7FDE3C64

Proof. We approximate $v$ by smooth plurisubharmonic functions $v_ l$ and pass to the limit, first in the gradients (Step 1) and then in each assertion of the lemma (Steps 2–5). All assertions are local, so we fix a compact set $K\subset U$ and an open neighborhood $W\subset U$ of $K$ on which smooth plurisubharmonic functions $v_ l$ decrease to $v$. For the convergence assertion, $W$ and the $v_ l$ are given. Otherwise we take for $W$ a relatively compact open neighborhood of $K$ in $U$ and for $v_ l$ the radial mollifications $v*\rho _{\varepsilon _ l}$, where $\varepsilon _ l\downarrow 0$ and $\varepsilon _1$ is less than the distance from $\overline W$ to $\mathbb C^ n\setminus U$; they are smooth and plurisubharmonic on $W$ and decrease to $v$ (proposition 5.9.2). Choose a compact set $K_1\subset W$ with $K\subset \operatorname{Int}K_1$, and a function $\sigma \in C^\infty _ c(W)$ with $\sigma \geq 0$ which equals one on a neighborhood of $K_1$. Since $v\leq v_ l\leq v_1$, the $v_ l$ are uniformly bounded on $\operatorname{Supp}\sigma $.

Step 1. Convergence of the gradients. For a smooth real function $w$, we have $\mathrm{dd}^{\mathrm{c}}(w^2)=2\, \mathrm dw\wedge \mathrm d^{\mathrm c}w+2w\, \mathrm{dd}^{\mathrm{c}}w$, and $\mathrm{dd}^{\mathrm{c}}(w^2\Theta )=\mathrm{dd}^{\mathrm{c}}(w^2)\wedge \Theta $ because $\Theta $ is closed. Hence

Apply this identity to $w=v_ l-v_ m$ and integrate against $\sigma $:

Both terms on the right tend to zero as $l,m\to \infty $. For the first, $|v_ l-v_ m|\leq v_ N-v$ for $l,m\geq N$, and $v_ N-v\to 0$ pointwise and boundedly on $\operatorname{Supp}\sigma $; since $\mathrm{dd}^{\mathrm{c}}\sigma \wedge \Theta $ is a signed measure with total variation bounded by a multiple of $\mu $ on $\operatorname{Supp}\sigma $, dominated convergence applies. For the second, expand it into the four integrals $\int \sigma v_ p\, \mathrm{dd}^{\mathrm{c}}v_ q\wedge \Theta $ with $p,q\in \{ l,m\} $, taken with the signs $+,-,-,+$. Given indices $l_ k,m_ k\to \infty $, pass to a subsequence along which both are strictly increasing; then $v_{l_ k}$ and $v_{m_ k}$ decrease to $v$, and the weighted form of theorem 7.3.7 on $W$, with the base current $\Theta $, shows that each of the four integrals tends to $\int \sigma v\, \mathrm{dd}^{\mathrm{c}}v\wedge \Theta $. Their signed sum therefore tends to zero, and since this holds for a subsequence of every such pair of index sequences, the second term tends to zero as $l,m\to \infty $. Since $\sigma \geq 0$, $\sigma =1$ on $K_1$ and $\mathrm dw\wedge \mathrm d^{\mathrm c}w\wedge \Theta $ is the positive measure $|H^{1/2}\partial w|^2\mu $, by (7.14.1), the left side bounds $\| H^{1/2}\partial v_ l-H^{1/2}\partial v_ m\| _{L^2(K_1,\mu )}^2$. Consequently $H^{1/2}\partial v_ l$ is a Cauchy sequence in $L^2(K_1,\mu )$; let $\Gamma \in L^2(K_1,\mu )$ be its limit. In particular, $H^{1/2}\partial v_ l$ converges in $L^2(K,\mu )$, which is the convergence assertion.

Step 2. The first derivative. Let $\eta $ be a smooth real one-form with compact support in $\operatorname{Int}K_1$, with $(1,0)$-part $\sum _ ke_ k\, \mathrm dz_ k$. By (7.14.2),

since $H^{1/2}e$ is bounded and vanishes outside $K_1$, and $\mu (K_1)<\infty $. On the other hand, $\mathrm dv_ l\wedge \Theta =\mathrm d(v_ l\Theta )$ since $\Theta $ is closed, and $v_ l\Theta \to v\Theta $ weakly on $W$ by dominated convergence on the coefficient measures; hence $\mathrm d(v_ l\Theta )\to \mathrm d(v\Theta )$ weakly. Therefore the pairing of $\mathrm d(v\Theta )$ with $\eta $, written $\int \eta \wedge \mathrm d(v\Theta )$, equals the last integral. Write $\langle H^{1/2}e,\Gamma \rangle =\sum _ ke_ kG_ k$, where $G_ k=\sum _ j(H^{1/2})_{jk}\bar\Gamma _ j$ belongs to $L^2(K_1,\mu )\subset L^1(K_1,\mu )$, so that

Thus, on real test one-forms $\eta $ supported in $\operatorname{Int}K_1$, the current $\mathrm d(v\Theta )$ acts by integrating against $\mu $ the coefficients $e_ k$ and $\bar e_ k$ of $\eta $, multiplied by the functions $-2\pi \mathrm i\, G_ k$ and $2\pi \mathrm i\, \bar G_ k$, which are integrable on $K_1$. Every test one-form is a complex linear combination of two real ones, so on $\operatorname{Int}K_1$ the current $\mathrm d(v\Theta )$ has measure coefficients, which are absolutely continuous with respect to $\mu $ and finite on $K_1$. In particular, for a bounded Borel function $a$,

Step 3. The gradient measure. Since $H^{1/2}\partial v_ l\to \Gamma $ in $L^2(K_1,\mu )$, the densities $|H^{1/2}\partial v_ l|^2$ converge to $|\Gamma |^2$ in $L^1(K_1,\mu )$, so the measures $\mathrm dv_ l\wedge \mathrm d^{\mathrm c}v_ l\wedge \Theta =|H^{1/2}\partial v_ l|^2\mu $ converge weakly on $\operatorname{Int}K_1$ to the positive measure $|\Gamma |^2\mu $. On the other hand, by the identity of Step 1,

weakly on $W$: the first term converges by dominated convergence on the coefficient measures of $v_ l^2\Theta $ and the weak continuity of $\mathrm{dd}^{\mathrm{c}}$, and the second by theorem 7.3.7. Hence the right side of (7.14.4) is the positive measure $|\Gamma |^2\mu $ on $\operatorname{Int}K_1$. Since $K$ was arbitrary and $K\subset \operatorname{Int}K_1$, the current $E_\Theta (v)$, which does not involve the approximations, is a positive measure on $U$.

Step 4. The Cauchy–Schwarz inequality. Let $\eta $, $\tau $ and $a$ be as in the statement, and apply Steps 1–3 with $K=\operatorname{Supp}\tau $; then $\operatorname{Supp}\eta \subset K\subset \operatorname{Int}K_1$. For a smooth real function $w$ on $W$, (7.14.2) and the Cauchy–Schwarz inequality give

Since $|H^{1/2}e|^2\leq |e|^2/\pi $, we have $16\pi ^2|H^{1/2}e|^2\leq C_\eta $ with $C_\eta =16\pi \sup |e|^2$, which depends only on $\eta $. The second integral is at most $\int \tau \, \mathrm dw\wedge \mathrm d^{\mathrm c}w\wedge \Theta $, because $\tau \geq 0$ and $\tau =1$ on $\operatorname{Supp}\eta $. Apply this to $w=v_ l$ and let $l\to \infty $. The left side converges to $|\int a\eta \wedge \mathrm dv\wedge \Theta |^2$ by Step 2, since $a\, \operatorname{Im}\langle H^{1/2}e,H^{1/2}\partial v_ l\rangle \to a\, \operatorname{Im}\langle H^{1/2}e,\Gamma \rangle $ in $L^1(K_1,\mu )$, and $\int \tau \, \mathrm dv_ l\wedge \mathrm d^{\mathrm c}v_ l\wedge \Theta \to \int \tau E_\Theta (v)$ by Step 3, since $\tau $ is continuous with compact support in $\operatorname{Int}K_1$. This proves (7.14.6).

Step 5. The chain rule. Both sides of (7.14.5) are complex-linear in $f$, so we may assume that $f$ is real. Let first $f$ be smooth. The functions $v_ l$ and $v$ take values in a compact interval $J$ on $\operatorname{Supp}\sigma $. Since $f(v_ l)\to f(v)$ boundedly, $f(v_ l)\Theta \to f(v)\Theta $ weakly on $W$ by dominated convergence, and hence $\mathrm d(f(v_ l)\Theta )\to \mathrm d(f(v)\Theta )$. By the chain rule for smooth functions and the closedness of $\Theta $, $\mathrm d(f(v_ l)\Theta )=f'(v_ l)\, \mathrm dv_ l\wedge \Theta $. Moreover $f'(v_ l)H^{1/2}\partial v_ l\to f'(v)\Gamma $ in $L^2(K_1,\mu )$: the difference is

where the first term tends to zero because $f'$ is bounded on $J$, and the second by dominated convergence. Now let $\eta $ be as in Step 2. Identity (7.14.2), multiplied by the continuous function $f'(v_ l)$, and Step 2 with $a=f'(v)$ give

Since the left side also tends to $\int \eta \wedge \mathrm d(f(v)\Theta )$, the currents $\mathrm d(f(v)\Theta )$ and $f'(v)\, \mathrm dv\wedge \Theta $ agree on $\operatorname{Int}K_1$. As $K$ was arbitrary, this proves (7.14.5) for smooth $f$.

Now let $f\in C^1(\mathbb R)$, and let $f_ m=f*\rho _{1/m}$ be its mollifications on $\mathbb R$. By proposition 2.5.3 and lemma 2.5.1, $f_ m\to f$ and $f_ m'=f'*\rho _{1/m}\to f'$ uniformly on compact subsets of $\mathbb R$. Since $v$ is locally bounded, $f_ m(v)\to f(v)$ and $f_ m'(v)\to f'(v)$ uniformly on compact subsets of $U$. As $\Theta $ and $\mathrm dv\wedge \Theta $ are currents of order zero, it follows that $f_ m(v)\Theta \to f(v)\Theta $ and $f_ m'(v)\, \mathrm dv\wedge \Theta \to f'(v)\, \mathrm dv\wedge \Theta $ weakly. Since $\mathrm d$ is weakly continuous, the identity $\mathrm d(f_ m(v)\Theta )=f_ m'(v)\, \mathrm dv\wedge \Theta $ passes to the limit. $\square$