Lemma 7.7.20E5A3E632224

Proof. Let $H_\varepsilon =H*\rho _\varepsilon $. Fix $a_0\in U$ and $r>0$ with $\overline{B(a_0,4r)}\subset U$, put $K=\overline{B(a_0,3r)}$, and let $M_0=\sup _ K|H|$, which is finite because $H$ is locally bounded. For $0<\varepsilon <r$, the regularization $H_\varepsilon $ is smooth and plurisubharmonic on a neighborhood of $\overline{B(a_0,2r)}$ (proposition 5.9.2), and $|H_\varepsilon |\le M_0$ on $\overline{B(a_0,2r)}$, since its values there are averages of values of $H$ on $K$. For $x\in \overline{B(a_0,2r)}$ and $|h|\le h_ K$,

by (7.7.3), because $x-y\in K$ for $|y|\le \varepsilon $. Replacing $h$ by $th$ with $0<t\le 1$, dividing by $t^2$ and letting $t\to 0$, we get $D^2H_\varepsilon (x)[h,h]\le 2C_ K|h|^2$ for $|h|\le h_ K$, and hence, by homogeneity, for all $h\in \mathbb C^ n=\mathbb R^{2n}$. On the other hand, the complex Hessian of the smooth plurisubharmonic function $H_\varepsilon $ is positive semidefinite (proposition 5.9.2), and $D^2H_\varepsilon [h,h]+D^2H_\varepsilon [\mathrm ih,\mathrm ih]=4\sum _{j,k}(H_\varepsilon )_{j\bar k}h_ j\bar h_ k\ge 0$; hence $D^2H_\varepsilon (x)[h,h]\ge -2C_ K|h|^2$ as well. Thus the second derivatives of the $H_\varepsilon $ are bounded on $\overline{B(a_0,2r)}$ by a constant $\Lambda _0$ independent of $\varepsilon $. For $x\in \overline{B(a_0,r)}$ and a unit vector $e$, Taylor's formula

gives $|\partial _ eH_\varepsilon (x)|\le 2M_0/r+\Lambda _0r=\Lambda $. By the mean value theorem on the convex set $B(a_0,r)$, $|H_\varepsilon (x)-H_\varepsilon (y)|\le \Lambda |x-y|$ for $x,y\in B(a_0,r)$. Since $H_\varepsilon \to H$ pointwise as $\varepsilon \downarrow 0$, by (3.4.13), the same inequality holds for $H$. Hence $H$ is continuous on $B(a_0,r)$, and, as $a_0$ was arbitrary, on $U$.

Moreover, $H_\varepsilon \to H$ in $L^1$ on compact subsets of $U$, by dominated convergence, and hence in the sense of distributions; and every ball $B_0\Subset U$ is covered by finitely many balls $B(a_0,2r)$ as above, so the second derivatives of the $H_\varepsilon $ are bounded on $B_0$ uniformly in small $\varepsilon $. By the converse part of proposition 2.8.5, the distributional real Hessian of $H$ has locally bounded measurable coefficients $A_{ab}$; hence $H\in C^{1,1}_{\mathrm{loc}}(U)$, and almost every $x\in U$ satisfies the second-order expansion of the statement.

The functions $H_{j\bar k}$ are bounded measurable functions on compact subsets of $U$. By proposition 2.5.3, the complex Hessian of $H_\varepsilon $ has the coefficients $(H_{j\bar k})_\varepsilon $; these are bounded on compact sets uniformly in small $\varepsilon $, and they converge to $H_{j\bar k}$ at every Lebesgue point of $H_{j\bar k}$, hence almost everywhere (theorem 2.8.3). In particular, the matrix $(H_{j\bar k})$ is positive semidefinite almost everywhere, as an almost everywhere limit of the positive semidefinite complex Hessians of the smooth plurisubharmonic functions $H_\varepsilon $. Since $H$ is continuous, $H_\varepsilon \to H$ locally uniformly (lemma 2.5.1(1)), so $(\mathrm{dd}^{\mathrm{c}}H_\varepsilon )^ n\to (\mathrm{dd}^{\mathrm{c}}H)^ n$ weakly on every ball $B_0\Subset U$ as $\varepsilon \downarrow 0$ along a sequence, by proposition 7.2.6 applied on $B_0$. By (7.2.4), $(\mathrm{dd}^{\mathrm{c}}H_\varepsilon )^ n=n!\pi ^{-n}\det ((H_{j\bar k})_\varepsilon )\, \mathrm dV_{2n}$, and the determinants converge almost everywhere and are bounded on compact sets uniformly in small $\varepsilon $. Hence, for every test function $\chi $ supported in such a ball $B_0$, dominated convergence gives

By lemma 2.5.2, every test function on $U$ is a finite sum of test functions supported in balls $B_0\Subset U$, so this identity holds for all test functions on $U$, and the two Radon measures coincide by proposition 2.5.3(4). $\square$