Lemma 6.3.3 (Richberg)52930869627B

Proof. Step 1: local approximations. Let $(K_ m)_{m\ge 1}$ be a compact exhaustion of $\Omega $ (section 2.2), so that $K_ m\subset \operatorname{Int}K_{m+1}$ and $\bigcup _ m\operatorname{Int}K_ m=\Omega $, and put $K_0=K_{-1}=\varnothing $. For each $m\ge 1$, the compact set $K_ m\setminus \operatorname{Int}K_{m-1}$ is covered by finitely many balls $B(c,r/3)$ with $c\in K_ m\setminus \operatorname{Int}K_{m-1}$ and $B(c,r)\Subset \operatorname{Int}K_{m+1}\setminus K_{m-2}$. Collecting these balls for all $m$, we obtain a sequence of balls $B_ i=B(c_ i,r_ i)\Subset \Omega $ whose inner balls $B(c_ i,r_ i/3)$ cover $\Omega $. The family $(B_ i)$ is locally finite: a compact subset of $\Omega $ lies in some $K_ m$, and a ball chosen at a stage $m'\ge m+2$ lies in the complement of $K_{m'-2}\supset K_ m$, so the compact set meets only the finitely many balls chosen at the stages $m'\le m+1$.

The function $u=(u-a|z|^2)+a|z|^2$ is continuous and plurisubharmonic by proposition 5.9.5. Let $\rho _ h$, $h>0$, be the radial mollifiers of section 2.5. By proposition 5.9.2(2), we have $u*\rho _ h\ge u$, and $(u-a|z|^2)*\rho _ h=u*\rho _ h-a|z|^2*\rho _ h$ is smooth and plurisubharmonic, hence has positive semidefinite complex Hessian by proposition 5.9.2(1); by lemma 2.5.1, $u*\rho _ h\to u$ locally uniformly as $h\to 0$. Since $\rho _ h$ is radial, $\int y\rho _ h(y)\, \mathrm dV(y)=0$, and expanding $|z-y|^2=|z|^2-2\operatorname{Re}\langle z,y\rangle +|y|^2$ gives $|z|^2*\rho _ h=|z|^2+\int |y|^2\rho _ h(y)\, \mathrm dV(y)$, whose complex Hessian matrix is $I$. Hence the complex Hessian matrix of $u*\rho _ h$ is at least $aI$.

For each $i$ choose $\delta _ i$ with $0<\delta _ i<a/2$ and $\delta _ ir_ i^2<\min _{\overline B_ i}\varepsilon $, put $\eta _ i=\delta _ ir_ i^2/6$, and choose $h_ i>0$ so small that $u*\rho _{h_ i}$ is defined near $\overline B_ i$ and satisfies $0\le u*\rho _{h_ i}-u<\eta _ i$ there. Define

The quadratic term is at least $\delta _ ir_ i^2/3=2\eta _ i$ on the inner ball $B(c_ i,r_ i/3)$, is less than $-2\eta _ i$ where $|z-c_ i|>\sqrt7\, r_ i/3$, and is at most $4\delta _ ir_ i^2/9$ everywhere. Therefore

the last because $2\eta _ i+4\delta _ ir_ i^2/9=7\delta _ ir_ i^2/9<\varepsilon $ on $B_ i$. The complex Hessian matrix of $u_ i$ is at least $(a-\delta _ i)I\ge (a/2)I$.

Step 2: a regularized maximum. Fix an even smooth function $\theta \ge 0$ on $\mathbb R$ with support in $(-1,1)$ and integral one, for instance the mollifier $\rho _{1/2}$ of section 2.5 in dimension $d=1$. For an integer $p\ge 1$, positive numbers $\eta _1,\ldots ,\eta _ p$ and $t=(t_1,\ldots ,t_ p)\in \mathbb R^ p$, define

The substitution $y_ i=\eta _ ix_ i$ and the evenness of $\theta $ show that $M_\eta $ is the convolution of the continuous function $t\mapsto \max _ it_ i$ with the smooth compactly supported function $y\mapsto \prod _ i\eta _ i^{-1}\theta (y_ i/\eta _ i)$, so $M_\eta $ is smooth by differentiation under the integral sign (section 2.5). It has the following properties.

1. $M_\eta $ is convex and nondecreasing in each variable, since for fixed $x$ the function $t\mapsto \max _ i(t_ i+\eta _ ix_ i)$ is.

2. $\max _ it_ i\le M_\eta (t)\le \max _ i(t_ i+\eta _ i)$. The upper bound holds because $|x_ i|<1$ on the support of the integrand. For the lower bound, integrate $\max _ j(t_ j+\eta _ jx_ j)\ge t_ i+\eta _ ix_ i$ and use $\int x_ i\theta (x_ i)\, \mathrm dx_ i=0$.

3. $M_\eta (t_1+c,\ldots ,t_ p+c)=M_\eta (t)+c$ for $c\in \mathbb R$. Differentiating in $c$, together with the first property, gives $\partial M_\eta /\partial t_ i\ge 0$ and $\sum _ i\partial M_\eta /\partial t_ i=1$.

4. If $t_ i+\eta _ i\le \max _{j\ne i}(t_ j-\eta _ j)$ for some $i$, then deleting the variable $t_ i$ and the parameter $\eta _ i$ does not change the value of $M_\eta $. Indeed, for $|x_ j|<1$ we have $t_ i+\eta _ ix_ i<t_ i+\eta _ i\le \max _{j\ne i}(t_ j-\eta _ j)<\max _{j\ne i}(t_ j+\eta _ jx_ j)$, so the $i$-th term never realizes the maximum, and integrating out $x_ i$ removes it.

For smooth real functions $w_1,\ldots ,w_ p$, the chain rule gives

The real Hessian matrix $G=(\partial ^2M_\eta /\partial t_ i\partial t_ l)$ is positive semidefinite, because the second derivative of the convex function $s\mapsto M_\eta (t+s\tau )$ is nonnegative for every $\tau \in \mathbb R^ p$. Hence the matrix given by the second sum is positive semidefinite: for $b\in \mathbb C^ n$ its quadratic form equals $\sum _{i,l}G_{il}\zeta _ i\bar\zeta _ l$ with $\zeta _ i=\sum _ j\partial _{z_ j}w_ i\, b_ j$, and writing $\zeta =\xi +\mathrm i\xi '$ with $\xi ,\xi '\in \mathbb R^ p$, this is $\xi ^{\mathsf T}G\xi +\xi '^{\mathsf T}G\xi '\ge 0$ by the symmetry of $G$. Therefore, by the third property, if each $w_ i$ has complex Hessian matrix at least $(a/2)I$, so does $M_\eta (w_1,\ldots ,w_ p)$.

Step 3: gluing. For $z\in \Omega $ let $I(z)=\{ i:z\in B_ i\} $, a finite nonempty set, and define $v(z)=M_\eta \bigl ((u_ i(z))_{i\in I(z)}\bigr )$ with the parameters $(\eta _ i)_{i\in I(z)}$. We show that $v$ is smooth near a fixed point $z_0$. Let $I_0=\{ i:z_0\in B_ i\} $ and $I_1=\{ i:z_0\in \partial B_ i\} $, finite sets by local finiteness, and choose $j$ with $z_0\in B(c_ j,r_ j/3)$. There is a neighborhood $W$ of $z_0$ that is contained in $B(c_ j,r_ j/3)$ and in $B_ i$ for $i\in I_0$, that meets $\overline B_ i$ only for $i\in I_0\cup I_1$, and whose intersection with $B_ i$, for $i\in I_1$, lies in the shell $\sqrt7\, r_ i/3<|z-c_ i|<r_ i$. For $z\in W$ we have $I_0\subset I(z)\subset I_0\cup I_1$, $j\in I_0$, and every $i\in I(z)\cap I_1$ satisfies

By the fourth property of $M_\eta $, applied to these indices one at a time, $v=M_\eta \bigl ((u_ i)_{i\in I_0}\bigr )$ on $W$, which is smooth. Thus $v$ is smooth on $\Omega $, and by Step 2 its complex Hessian matrix is at least $(a/2)I$; hence $v-\tfrac a2|z|^2$ is smooth with positive semidefinite complex Hessian, and so it is plurisubharmonic by proposition 5.9.2(1). Finally, every $z$ lies in some inner ball $B(c_ j,r_ j/3)$, and the second property of $M_\eta $ gives $v(z)\ge u_ j(z)>u(z)$ and $v(z)\le \max _{i\in I(z)}(u_ i(z)+\eta _ i)<u(z)+\varepsilon (z)$. $\square$