11.31.9 Continuous potentials and pluriharmonic traces951E64D10409

Proof. Set $v=a-b$. The identity $\mathrm{dd}^{\mathrm{c}}(v^2)=2\mathrm{d}v\wedge \mathrm{d}^ cv+2v\mathrm{dd}^{\mathrm{c}}v$, integrated against $G$, and the equation $\mathrm{dd}^{\mathrm{c}}G=0$ give

The measures $(\mathrm{dd}^{\mathrm{c}}a+C\omega )\wedge G$ and $(\mathrm{dd}^{\mathrm{c}}b+C\omega )\wedge G$ are positive, and each has mass $CM$, because the terms $\mathrm{dd}^{\mathrm{c}}a\wedge G$ and $\mathrm{dd}^{\mathrm{c}}b\wedge G$ integrate to zero. Their difference $\mathrm{dd}^{\mathrm{c}}v\wedge G$ therefore has total variation at most $2CM$, which proves the upper bound. The lower bound holds because $\mathrm{d}v\wedge \mathrm{d}^ cv=(\mathrm{i}/2\pi )\partial v\wedge \bar\partial v$ is a positive $(1,1)$-form.

For the second estimate, work in a local unitary frame for $\omega $. Positivity of $G$ means that its coefficients form a positive semidefinite matrix of measures. These entries are absolutely continuous with respect to the trace measure $\mu $: the diagonal entries are dominated by the trace, and the off-diagonal entries by the Cauchy–Schwarz inequality for positive semidefinite matrices. The resulting matrix of densities has trace bounded by a constant depending only on the normalization. Pairing $\mathrm{d}v\wedge G$ or $\mathrm{d}^ cv\wedge G$ with a test one-form $\eta $, and applying the matrix Cauchy–Schwarz inequality and then the integral Cauchy–Schwarz inequality, we obtain

A finite cover by such frames and a partition of unity give the global estimate. Taking the supremum over smooth test forms of norm at most one yields the total variation bound. When $M=0$, positivity gives $G=0$, and both estimates hold trivially. $\square$

Proof. We first define the gradient currents $\mathrm{d}u\wedge G$ and $\mathrm{d}^ cu\wedge G$ as limits, then define the product by an integration-by-parts formula, and finally prove joint continuity.

Step 1: gradient currents. Choose smooth approximants $a_ k\to u$ as in lemma 5.42.5. There is a common constant $C>0$ such that $\mathrm{dd}^{\mathrm{c}}a_ k\geq -C\omega $. Write $M=\int _ X\omega \wedge G$. Since $a_ k$ converges uniformly, lemma 11.31.10 shows that $\mathrm{d}a_ k\wedge G$ and $\mathrm{d}^ ca_ k\wedge G$ are Cauchy sequences in total variation. Denote their limits, which are currents of order zero, by $\mathrm{d}u\wedge G$ and $\mathrm{d}^ cu\wedge G$. Applying the same lemma to two uniformly convergent smooth sequences with a common lower Hessian bound shows that the limits do not depend on the sequence. Uniform convergence also gives $a_ kG\to uG$ in total variation.

Step 2: the product. Define

in the sense of distributions. For smooth $u$, the Leibniz rule and $\mathrm{dd}^{\mathrm{c}}G=0$ show that this agrees with the usual product. Consequently $(\alpha +\mathrm{dd}^{\mathrm{c}}a_ k)\wedge G$ converges in distributions to the right-hand side of (11.31.12). With $\varepsilon _ k\to 0$ as in lemma 5.42.5, the measures $(\alpha +\mathrm{dd}^{\mathrm{c}}a_ k+\varepsilon _ k\omega )\wedge G$ are positive, with masses $\int _ X\alpha \wedge G+\varepsilon _ kM$. Their distributional limit is therefore a positive measure, and testing against the constant function one gives its mass. The uniform mass bound upgrades distributional convergence to weak convergence of measures. If $G$ is smooth, the uniform convergence of $a_ k$ also identifies the limit with the product of the current $\alpha +\mathrm{dd}^{\mathrm{c}}u$ and the smooth form $G$.

Step 3: independence of the decomposition. Adding a smooth function $h$ to $u$ adds $\mathrm{d}h\wedge G$ and $\mathrm{d}^ ch\wedge G$ to the gradient currents, so replacing $(\alpha ,u)$ by $(\alpha -\mathrm{dd}^{\mathrm{c}}h,u+h)$ leaves the product unchanged. If two continuous decompositions define the same $(1,1)$-current, the difference of their potentials has smooth $\mathrm{dd}^{\mathrm{c}}$; taking its trace in a coordinate chart and applying elliptic regularity shows that this difference is smooth. Thus the product depends only on the current $\alpha +\mathrm{dd}^{\mathrm{c}}u$ and on $G$.

Step 4: joint continuity. For $u$ and for each $u_ j$, keep only those smooth approximants from lemma 5.42.5 whose curvature errors are at most one. Choosing $C$ with $\alpha \leq (C-1)\omega $, all these approximants satisfy the same lower Hessian bound $-C\omega $. The numbers $M_ j=\int _ X\omega \wedge G_ j$ are bounded, by weak convergence. Fix a smooth approximant $a_ k$ of $u$, and put $\delta _{j,k}=\| u_ j-a_ k\| _\infty $. Applying lemma 11.31.10 to $a_ k$ and to smooth approximants of $u_ j$, and passing to the gradient limits, we find that the total variations of the differences $(\mathrm{d}u_ j-\mathrm{d}a_ k)\wedge G_ j$ and $(\mathrm{d}^ cu_ j-\mathrm{d}^ ca_ k)\wedge G_ j$ are at most $C'M_ j\sqrt{\delta _{j,k}}$, where $C'$ is independent of $j$ and $k$.

For a smooth real test function $\phi $, formula (11.31.12) now gives

where $C_\phi $ depends on the first two derivatives of $\phi $, on $\omega $ and on $C$. The same estimate holds with $u_ j,G_ j$ replaced by $u,G$. For fixed $k$, the products $(\alpha +\mathrm{dd}^{\mathrm{c}}a_ k)\wedge G_ j$ of a fixed smooth form with $G_ j$ converge weakly to $(\alpha +\mathrm{dd}^{\mathrm{c}}a_ k)\wedge G$. Let first $j\to \infty $ and then $k\to \infty $; since $u_ j\to u$ uniformly, the error tends to zero. This proves convergence in distributions, and positivity together with the bounded masses $\int _ X\alpha \wedge G_ j$ upgrades it to weak convergence of measures. $\square$