Lemma 7.3.5C04BF828EE91

Proof. Let $\rho _\varepsilon $ be the mollifier of section 2.5 on $\mathbb {R}^2$; it is a nonnegative even radial probability density supported in the disc of radius $\varepsilon $. Put $M_\varepsilon =\max *\rho _\varepsilon $, that is, $M_\varepsilon (x)=\int \max (x-y)\rho _\varepsilon (y)\, \mathrm dy$, where $\max (s,t)=\max \{ s,t\} $; it is smooth, since $\rho _\varepsilon $ is (section 2.5). Since $\rho _\varepsilon \ge 0$ has integral $1$, and $x\mapsto \max (x-y)$ is convex and increasing in each variable for each $y$, the function $M_\varepsilon $ is convex and increasing in each variable. Since $\rho _\varepsilon $ is even, $\int y_ i\rho _\varepsilon (y)\, \mathrm dy=0$ for $i=1,2$; integrating $\max (s-y_1,t-y_2)\ge s-y_1$ and $\max (s-y_1,t-y_2)\ge t-y_2$ against $\rho _\varepsilon $ therefore gives $M_\varepsilon (s,t)\ge s$ and $M_\varepsilon (s,t)\ge t$, that is, $M_\varepsilon \ge \max $. Since $|\max (x-y)-\max (x)|\le |y|\le \varepsilon $ on the support of $\rho _\varepsilon $, we have $M_\varepsilon \le \max +\varepsilon $. If $s-t>2\varepsilon $ and $|y|\le \varepsilon $, then $(s-y_1)-(t-y_2)\ge s-t-2\varepsilon >0$, so $M_\varepsilon (s,t)=\int (s-y_1)\rho _\varepsilon (y)\, \mathrm dy=s$; similarly $M_\varepsilon (s,t)=t$ if $t-s>2\varepsilon $.

It remains to show that $M_\varepsilon $ increases with $\varepsilon $. For fixed $x\in \mathbb R^2$, the function $g(\varepsilon )=\int \max (x-\varepsilon y)\rho _1(y)\, \mathrm{d}y$, defined for all real $\varepsilon $, equals $M_\varepsilon (x)$ for $\varepsilon >0$ and $\max (x)$ for $\varepsilon =0$; it is convex, since $\varepsilon \mapsto \max (x-\varepsilon y)$ is convex, and even, since $\rho _1$ is even. For $0\le \varepsilon _1<\varepsilon _2$, writing $\varepsilon _1=\lambda (-\varepsilon _2)+(1-\lambda )\varepsilon _2$ with $\lambda \in (0,1]$, we get $g(\varepsilon _1)\le \lambda g(-\varepsilon _2)+(1-\lambda )g(\varepsilon _2)=g(\varepsilon _2)$. This proves the monotonicity, and together with $\max \le M_\varepsilon \le \max +\varepsilon $ this gives $M_\varepsilon \downarrow \max $ as $\varepsilon \downarrow 0$.

Finally, let $a$ and $b$ be smooth and plurisubharmonic, and write $(a_1,a_2)=(a,b)$. Denoting by $\partial _ kM_\varepsilon $ and $\partial _ k\partial _ lM_\varepsilon $ the partial derivatives evaluated at $(a,b)$, the chain rule gives

The first sum is positive because $\partial _ kM_\varepsilon \ge 0$, $M_\varepsilon $ being increasing in each variable, and the forms $\mathrm i\partial \bar\partial a_ k$ are positive, since the smooth plurisubharmonic functions $a_ k$ have positive semidefinite complex Hessians (proposition 5.9.2). For the second, note that the real Hessian $H=(\partial _ k\partial _ lM_\varepsilon )$ is positive semidefinite: for $v\in \mathbb R^2$, the function $t\mapsto M_\varepsilon (x+tv)$ is convex, so its second derivative $v^ tHv$ is nonnegative. Writing $H=\sum _ m\lambda _ mv_ mv_ m^ t$ with $\lambda _ m\ge 0$ and real orthonormal vectors $v_ m$, and putting $\theta _ m=\sum _ k(v_ m)_ k\partial a_ k$, we get $\mathrm i\sum _{k,l}\partial _ k\partial _ lM_\varepsilon \, \partial a_ k\wedge \bar\partial a_ l=\sum _ m\lambda _ m\mathrm i\theta _ m\wedge \bar\theta _ m$, because $\bar\partial a_ l=\overline{\partial a_ l}$ for real $a_ l$; this is a positive $(1,1)$-form. Hence the complex Hessian of $M_\varepsilon (a,b)$ is positive semidefinite, and $M_\varepsilon (a,b)$ is plurisubharmonic by proposition 5.9.2. $\square$