Theorem 6.3.4EA195CC9592D

Proof. Steps 1 and 2 deduce the plurisubharmonicity of $-\log \operatorname {dist}(\cdot ,\mathbb C^ n\setminus \Omega )$ from a submean inequality for the distance to the boundary in a fixed direction; Step 3 uses this function to construct a continuous exhaustion function, which lemma 6.3.3 smooths.

Step 1: distances in a fixed direction. Let $\Omega $ be pseudoconvex, with a continuous plurisubharmonic exhaustion function $\rho $. For $z\in \Omega $ and $\xi \in \mathbb C^ n$ put

so that $\delta (z,0)=\infty $. The function $\delta $ is lower semicontinuous on $\Omega \times \mathbb C^ n$: if $r<\delta (z,\xi )$, then the compact disc $z+\xi \overline\Delta _ r$ lies in $\Omega $, hence so does $z'+\xi '\overline\Delta _ r$ for all $(z',\xi ')$ close to $(z,\xi )$, and $\delta (z',\xi ')\ge r$ for these points. Thus $V=-\log \delta $ is an upper semicontinuous function on $\Omega \times \mathbb C^ n$ with values in $[-\infty ,\infty )$. We verify the hypothesis of lemma 6.3.2 for $V$ on the open set $\Omega \times \mathbb C^ n\subset \mathbb C^{2n}$.

Let $(z_0+ta,\xi _0+tb)$, $|t|\le 1$, be a closed disc in $\Omega \times \mathbb C^ n$, and let $P$ be a polynomial with $V(z_0+ta,\xi _0+tb)\le \operatorname{Re}P(t)$ for $|t|=1$, that is, $\delta (z_0+ta,\xi _0+tb)\ge |\mathrm e^{-P(t)}|$. Consider the holomorphic map

We have $h(t,0)=z_0+ta\in \Omega $ for $|t|\le 1$; and $h(t,w)\in \Omega $ for $|t|=1$ and $|w|<1$, because then $|w\mathrm e^{-P(t)}|<\delta (z_0+ta,\xi _0+tb)$. Let $R$ be the supremum of the numbers $r\ge 0$ with $h(\overline\Delta \times \overline\Delta _ r)\subset \Omega $. Since $h(\overline\Delta \times \{ 0\} )$ is a compact subset of $\Omega $ and $h$ is continuous, $R>0$. We show that $R\ge 1$.

Suppose that $R<1$. Then $h(\partial \Delta \times \overline\Delta _ R)$ is a compact subset of $\Omega $; let $c$ be the maximum of $\rho $ on it. For $|w|<R$ we have $h(\overline\Delta \times \{ w\} )\subset \Omega $, because every radius smaller than an admissible one is admissible; since this set is compact and $h$ is continuous, there is $r'>1$ such that $t\mapsto h(t,w)$ maps the disc $\Delta _{r'}$ into $\Omega $, and hence into one connected component of $\Omega $. The function $t\mapsto \rho (h(t,w))$ is finite, hence subharmonic on $\Delta _{r'}$ by proposition 5.9.3, and it is at most $c$ on $\partial \Delta $. By upper semicontinuity, $\varlimsup _{t\to \zeta }(\rho (h(t,w))-c)\le 0$ for every $\zeta \in \partial \Delta $, and the maximum principle of proposition 3.4.7 on the disc $\Delta $ gives $\rho (h(t,w))\le c$ for $|t|<1$, hence for $|t|\le 1$. Thus $h(\overline\Delta \times \Delta _ R)$ lies in the compact set $\{ \rho \le c\} $, and so does its closure $h(\overline\Delta \times \overline\Delta _ R)$. This compact subset of $\Omega $ has positive distance from $\mathbb C^ n\setminus \Omega $, and the uniform continuity of $h$ on compact sets gives $h(\overline\Delta \times \overline\Delta _{R'})\subset \Omega $ for some $R'>R$, contradicting the definition of $R$.

Hence $R\ge 1$. In particular $h(0,w)=z_0+w\mathrm e^{-P(0)}\xi _0\in \Omega $ for $|w|<1$, which means that $\delta (z_0,\xi _0)\ge \mathrm e^{-\operatorname{Re}P(0)}$, that is, $V(z_0,\xi _0)\le \operatorname{Re}P(0)$. By lemma 6.3.2, $V$ satisfies the circle submean inequality on every closed complex disc in $\Omega \times \mathbb C^ n$.

Step 2: the boundary distance. Now let $\Omega \ne \mathbb C^ n$ and put $d(z)=\operatorname {dist}(z,\mathbb C^ n\setminus \Omega )$. For a unit vector $\xi $ we have $\delta (z,\xi )\ge d(z)$, because $z+\xi \Delta _{d(z)}$ lies in the ball $B(z,d(z))\subset \Omega $. Equality holds if $\xi $ points from $z$ to a nearest point of the closed set $\mathbb C^ n\setminus \Omega $, since that point, $z+d(z)\xi $, lies in $z+\xi \Delta _ r$ for every $r>d(z)$. Hence

For fixed $\xi $, the function $z\mapsto V(z,\xi )$ satisfies the circle submean inequality on every closed disc $\{ z_0+ta:|t|\le 1\} \subset \Omega $, by Step 1 with $b=0$, and it is bounded above by $-\log d$. Therefore

and taking the maximum over $\xi $ gives the submean inequality for $-\log d$. Since $d$ is positive and $1$-Lipschitz, $-\log d$ is continuous. Its restriction to each connected component of each affine complex-line section of $\Omega $ is therefore subharmonic (definition 3.4.2): at a point $z_0+t_0a$ of such a section, the circle means over $\{ z_0+(t_0+r\mathrm e^{\mathrm i\theta })a\} $ dominate the value for all small $r>0$. Hence $-\log d$, which is finite, is plurisubharmonic on each connected component of $\Omega $ by definition 5.8.1, that is, on $\Omega $.

Step 3: smooth exhaustion functions. Let $\Omega \ne \mathbb C^ n$ be an open set on which $-\log d$ is plurisubharmonic, and put

Then $u$ is continuous, $u-|z|^2$ is plurisubharmonic by proposition 5.9.5, and $u$ is an exhaustion function of $\Omega $: the set $\{ u\le c\} $ is contained in the ball $\{ |z|^2\le c\} $ and in $\{ d\ge \mathrm e^{-c}\} $, so it is a closed and bounded subset of $\mathbb C^ n$ contained in $\Omega $, hence compact. In particular, $\Omega $ is pseudoconvex, which proves the converse statement.

By lemma 6.3.3 with $a=1$ and $\varepsilon =1$, there is a smooth function $s$ with $u\le s\le u+1$ such that $s-|z|^2/2$ is plurisubharmonic. Then $s\ge u\ge 0$, and the complex Hessian matrix of $s$ is at least $I/2$ by proposition 5.9.2(1) applied to $s-|z|^2/2$, so that $s$ is strictly plurisubharmonic. Moreover, since $s\ge u$, the sublevel set $\{ s\le c\} $ is a closed subset of the compact set $\{ u\le c\} $. Thus $s$ is a nonnegative smooth strictly plurisubharmonic exhaustion function. If $\Omega \ne \mathbb C^ n$ is pseudoconvex, Step 2 shows that $-\log d$ is plurisubharmonic, so this construction applies; for $\Omega =\mathbb C^ n$ we take $s=|z|^2$. $\square$