6.3 Pseudoconvexity and boundary distance86B639827902

Proof. Fix $z_0$ and $a$, and put $g(\theta )=v(z_0+\mathrm e^{\mathrm i\theta }a)$, an upper semicontinuous function on the circle. By lemma 3.4.4 there are continuous real functions $g_ m$ decreasing to $g$ whose integrals decrease to the integral of $g$, which may be $-\infty $. Fix $m$ and $\delta >0$. We approximate $g_ m$ uniformly by real trigonometric polynomials, using the Fejér kernel

The two expressions agree because the product $\sum _{k,l=0}^{N-1}\mathrm e^{\mathrm i(k-l)\theta }$ contains exactly $N-|h|$ pairs $(k,l)$ with $k-l=h$, for each integer $h$ with $|h|<N$. By the first expression $F_ N\ge 0$, and summing the geometric series gives $F_ N(\theta )=\sin ^2(N\theta /2)/(N\sin ^2(\theta /2))\le 1/(N\sin ^2(\theta /2))$ for $0<|\theta |\le \pi $; by the second, $F_ N$ has integral one with respect to the normalized measure $\mathrm d\theta /(2\pi )$ on $[-\pi ,\pi ]$.

Let $q_ N=g_ m*F_ N$ be the convolution with respect to $\mathrm d\theta /(2\pi )$, that is, $q_ N(\theta )=\frac1{2\pi }\int _{-\pi }^{\pi }g_ m(\theta -\tau )F_ N(\tau )\, \mathrm d\tau $. For $0<\delta '\le \pi $ let $\omega (\delta ')$ be the largest oscillation of $g_ m$ over arcs of length at most $\delta '$, which tends to zero with $\delta '$ by uniform continuity. Splitting the integral at $|\tau |=\delta '$, we obtain

so $q_ N\to g_ m$ uniformly as $N\to \infty $. Inserting the second expression for $F_ N$ and substituting $\sigma =\theta -\tau $, we see that $q_ N(\theta )=\sum _{|k|<N}c_ k\mathrm e^{\mathrm ik\theta }$ with $c_ k=(1-|k|/N)\frac1{2\pi }\int _{-\pi }^{\pi }g_ m(\sigma )\mathrm e^{-\mathrm ik\sigma }\, \mathrm d\sigma $. Since $g_ m$ is real, $c_{-k}=\bar c_ k$, so $q_ N(\theta )=\operatorname{Re}P(\mathrm e^{\mathrm i\theta })$ for the polynomial $P(t)=c_0+2\sum _{0<k<N}c_ kt^ k$, and $\operatorname{Re}P(0)=c_0$ is the mean of $g_ m$. Choose $N$ with $|q_ N-g_ m|<\delta $. Then $\operatorname{Re}(P(t)+\delta )\ge g_ m\ge g$ on the circle, and the hypothesis gives

Letting $\delta \to 0$ and then $m\to \infty $, we obtain the asserted inequality. $\square$

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 $v(z)\le \max _{i\in I(z)}(u_ i(z)+\eta _ i)<u(z)+\varepsilon (z)$. $\square$

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 $s=|z|^2$. $\square$

Proof. Let $\Omega \subset \mathbb C^ n$ and $\Omega '\subset \mathbb C^ m$ be pseudoconvex, with nonnegative continuous plurisubharmonic exhaustion functions $\rho $ and $\rho '$. The function $(z,w)\mapsto \rho (z)+\rho '(w)$ is continuous and plurisubharmonic on $\Omega \times \Omega '$, by proposition 5.9.3 applied to the two projections and by proposition 5.9.5. Its sublevel set $\{ \le c\} $ is a closed subset of the compact set $\{ \rho \le c\} \times \{ \rho '\le c\} $, so $\Omega \times \Omega '$ is pseudoconvex.

Next let $s$ be as in the statement and $c\in \mathbb R$. The function $x\mapsto 1/(c-x)$ is convex and increasing on $(-\infty ,c)$, so the complex Hessian matrix of $1/(c-s)$ on the open set $\{ s<c\} $, namely $(c-s)^{-2}(s_{j\bar k})+2(c-s)^{-3}\bigl (\partial _{z_ j}s\, \overline{\partial _{z_ k}s}\bigr )$ by the chain rule, is positive semidefinite. Hence $|z|^2+1/(c-s)$ is a smooth plurisubharmonic function on $\{ s<c\} $, by proposition 5.9.2(1). For $C\le 0$ its sublevel set $\{ \le C\} $ is empty. For $C>0$ this sublevel set is closed in $\{ s<c\} $ and contained in the set $\{ s\le c-1/C\} $, which is a compact subset of $\{ s<c\} $; hence it is compact, and $\{ s<c\} $ is pseudoconvex. Moreover $\{ s<c\} $ is contained in the compact set $\{ s\le c\} $, so it is relatively compact in $\Omega $.

Finally, if $\Omega $ is pseudoconvex, theorem 6.3.4 provides a smooth plurisubharmonic exhaustion function $s$, and the sets $\Omega _ j=\{ s<j\} $, $j\ge 1$, are pseudoconvex, relatively compact in $\Omega $, increasing, and have union $\Omega $. (Some of them may be empty or disconnected; definition 6.3.1 allows arbitrary open sets.) $\square$

Proof. We construct a function $\psi $ with the properties required in theorem 5.12.1; the corollary then follows from that theorem. If $\Omega =\mathbb C^{m+k}$, let $\psi (z,w)=|z|^2+|\operatorname{Re}w|^2$. Otherwise let $d=\operatorname {dist}(\cdot ,\mathbb C^{m+k}\setminus \Omega )$ and put

Imaginary translations are isometries of $\mathbb C^{m+k}$ that preserve $\Omega $ and its complement, so $d$ and $\psi $ are invariant under them. The function $\psi $ is plurisubharmonic by proposition 5.9.5: $-\log d$ is plurisubharmonic by theorem 6.3.4, and $|z|^2+|\operatorname{Re}w|^2$ is plurisubharmonic by proposition 5.9.2(1), since its complex Hessian matrix is the diagonal matrix with entries $1$ in the $z$-directions and $\frac12$ in the $w$-directions, by the identity $(\operatorname{Re}w_ l)^2=\tfrac 12|w_ l|^2+\tfrac 12\operatorname{Re}(w_ l^2)$ for $1\le l\le k$. Finally, let $c\in \mathbb R$. Since $\psi $ is continuous, the set $\{ (z,t)\in \Omega _{\mathbb R}:\psi (z,t)\le c\} $ is closed in $\Omega _{\mathbb R}$. It is contained in $\{ |z|^2+|t|^2\le c\} $ and, if $\Omega \ne \mathbb C^{m+k}$, in the set $\{ d\ge \mathrm e^{-c}\} $, which is closed in $\mathbb C^{m+k}$ and contained in $\Omega $; it is therefore a closed and bounded subset of $\mathbb C^ m\times \mathbb R^ k$, hence compact. $\square$