Proof.
We first observe that the complex Hessian $\mathcal H(q)(x)$ of an upper test function $q$ for $u$ at $x$ is positive semidefinite. Indeed, let $\xi \in \mathbb C^ n\setminus \{ 0\} $, and let $r>0$ be so small that the closed disc $\{ x+\zeta \xi :|\zeta |\le r\} $ lies in the neighborhood of $x$ on which $q\ge u$. The function $\zeta \mapsto u(x+\zeta \xi )$ is subharmonic near this disc and finite at $\zeta =0$, so theorem 3.4.5(2) gives
\[ q(x)=u(x)\le \frac1{2\pi }\int _0^{2\pi }u(x+r\mathrm e^{\mathrm i\theta }\xi )\, \mathrm d\theta \le \frac1{2\pi }\int _0^{2\pi }q(x+r\mathrm e^{\mathrm i\theta }\xi )\, \mathrm d\theta . \]
Write $q_ j=\partial _{z_ j}q$ and $q_{jk}=\partial _{z_ j}\partial _{z_ k}q$. Since $q$ is real and of class $C^2$, Taylor's formula gives
\[ q(x+h)=q(x)+2\operatorname{Re}\sum _ jq_ j(x)h_ j+\operatorname{Re}\sum _{j,k}q_{jk}(x)h_ jh_ k+\sum _{j,k}q_{j\bar k}(x)h_ j\bar h_ k+o(|h|^2). \]
For $h=r\mathrm e^{\mathrm i\theta }\xi $ the terms containing $\mathrm e^{\mathrm i\theta }$ and $\mathrm e^{2\mathrm i\theta }$ have mean zero over $\theta \in [0,2\pi ]$, so the right-hand side of the preceding display equals $q(x)+r^2\sum _{j,k}q_{j\bar k}(x)\xi _ j\bar\xi _ k+o(r^2)$. Dividing by $r^2$ and letting $r\downarrow 0$, we obtain $\sum _{j,k}q_{j\bar k}(x)\xi _ j\bar\xi _ k\ge 0$.
(1) implies (2). Assume (1), let $q$ be an upper test function for $u$ at $x$, and suppose that $\det \mathcal H(q)(x)<f(x)$. Since $\mathcal H(q)(x)$ is positive semidefinite, we can choose $\delta >0$ so small that $q_\delta =q+\delta |z-x|^2$ satisfies $\mathcal H(q_\delta )(x)=\mathcal H(q)(x)+\delta I>0$ and still $\det \mathcal H(q_\delta )(x)<f(x)$. Since $q$ is of class $C^2$ and $f$ is continuous, there is $r>0$ such that $\overline{B(x,r)}\subset \Omega $, $q\ge u$ on $\overline{B(x,r)}$, the complex Hessian $\mathcal H(q_\delta )$ is positive definite on $\overline{B(x,r)}$, and $\det \mathcal H(q_\delta )<f$ on $\overline{B(x,r)}$; in particular, $q_\delta $ is plurisubharmonic on $B(x,r)$ by proposition 5.9.2. Put $b=q_\delta -\delta r^2/2$. Then $b=q+\delta r^2/2>u$ on the sphere $\partial B(x,r)$, while $b(x)=u(x)-\delta r^2/2<u(x)$.
The Borel set $S=\{ z\in B(x,r):b(z)<u(z)\} $ has positive Lebesgue measure. Otherwise $\max \{ u,b\} =b$ almost everywhere on $B(x,r)$; since both functions are plurisubharmonic, the uniqueness assertion of theorem 5.9.1 gives $\max \{ u,b\} =b$ everywhere, contradicting $b(x)<u(x)$. The function $b-u$ is lower semicontinuous and positive on $\partial B(x,r)$, so $\varliminf _{z\to \zeta }(b(z)-u(z))>0$ for every $\zeta \in \partial B(x,r)$. The comparison principle theorem 7.3.11, applied on $B(x,r)$ with the bounded plurisubharmonic functions $b$ and $u$ in place of $u$ and $v$, therefore gives
\[ \int _ S f\eta ^ n\le \int _ S(\mathrm{dd}^{\mathrm{c}}u)^ n\le \int _ S(\mathrm{dd}^{\mathrm{c}}b)^ n=\int _ S\det \mathcal H(q_\delta )\, \eta ^ n. \]
Here $(\mathrm{dd}^{\mathrm{c}}b)^ n=\det \mathcal H(q_\delta )\, \eta ^ n$ on $B(x,r)$ because $b$ is a plurisubharmonic $C^2$ function there (see the beginning of this section), and the integrals are finite because $(\mathrm{dd}^{\mathrm{c}}u)^ n$ is a Radon measure on $\Omega $ and $\overline{B(x,r)}$ is a compact subset of $\Omega $. Since $\det \mathcal H(q_\delta )<f$ on $S$ and $S$ has positive measure, the right side is strictly smaller than the left side, a contradiction. Thus (2) holds.
(2) implies (3). Let $q$ be an upper test function for $u$ at $x$, and let $B$ be positive definite with $\det B=1$. Since $\mathcal H(q)(x)$ is positive semidefinite, the infimum formula of lemma 7.8.1 and (2) give
\[ L_ Bq(x)=\operatorname {tr}\bigl (B\mathcal H(q)(x)\bigr )\ge n\bigl (\det \mathcal H(q)(x)\bigr )^{1/n}\ge nf(x)^{1/n}. \]
Since $u$ is real-valued and upper semicontinuous, lemma 7.8.2, applied to $L_ B$ and the continuous function $nf^{1/n}$, gives (3).
(3) implies (1). The mollifications $u_\varepsilon =u*\rho _\varepsilon $ of section 2.5 are smooth plurisubharmonic functions on $\Omega _\varepsilon $ which decrease to $u$ as $\varepsilon \downarrow 0$ (proposition 5.9.2). Fix $B$ as in (3). The distribution $L_ Bu-nf^{1/n}$ is positive, so its mollification is a nonnegative function on $\Omega _\varepsilon $ (proposition 2.5.3(3)). Since $L_ B$ has constant coefficients, proposition 2.5.3(1) identifies the mollification of $L_ Bu$ with $L_ Bu_\varepsilon $, and we obtain the pointwise inequality
\[ \operatorname {tr}\bigl (B\mathcal H(u_\varepsilon )\bigr )\ge n\, (f^{1/n}*\rho _\varepsilon )\quad \text{on }\Omega _\varepsilon \qquad (B>0,\ \det B=1). \]
Since $\mathcal H(u_\varepsilon )$ is positive semidefinite (proposition 5.9.2), taking the infimum over $B$ and using lemma 7.8.1, we obtain $\det \mathcal H(u_\varepsilon )\ge (f^{1/n}*\rho _\varepsilon )^ n$, which is the last assertion of the proposition. By the formula recalled at the beginning of this section, it reads
\[ (\mathrm{dd}^{\mathrm{c}}u_\varepsilon )^ n\ge (f^{1/n}*\rho _\varepsilon )^ n\eta ^ n\quad \text{on }\Omega _\varepsilon . \]
As $\varepsilon \downarrow 0$, the density on the right converges locally uniformly to $f$ by lemma 2.5.1(1), and $(\mathrm{dd}^{\mathrm{c}}u_\varepsilon )^ n\to (\mathrm{dd}^{\mathrm{c}}u)^ n$ weakly by theorem 7.3.7, applied along any sequence $\varepsilon _ j\downarrow 0$. Testing against nonnegative functions in $C_ c(\Omega )$ gives (1).
$\square$
Pome
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