Domains of Holomorphy, Plurisubharmonicity and Pseudoconvexity

If $a\in E$ and $f\in \mathcal{O}(\mathrm{\Omega })$, then $\mathrm{\mid }f(a)\mathrm{\mid }\le {\sup }_E\mathrm{\mid }f\mathrm{\mid }$ by definition of the supremum. Hence [L1] gives $a\in {\widehat{E}}_{\mathrm{\Omega }}$, so $E\subseteq {\widehat{E}}_{\mathrm{\Omega }}$.

For compact $K$, [L1] gives ${\widehat{K}}_{\mathrm{\Omega }}={\bigcap }_{f\in \mathcal{O}(\mathrm{\Omega })}\{a\in \mathrm{\Omega }:\mathrm{\mid }f(a)\mathrm{\mid }\le {\sup }_K\mathrm{\mid }f\mathrm{\mid }\}\mathrm{.}$ Each set in the intersection is closed in $\mathrm{\Omega }$ because $f$ is continuous, so ${\widehat{K}}_{\mathrm{\Omega }}$ is closed in $\mathrm{\Omega }$. The coordinate functions $z\mapsto z_j$ are holomorphic on $\mathrm{\Omega }$, so [L1] also gives $\mathrm{\mid }{a}_j\mathrm{\mid }\le {\sup }_{z\in K}\mathrm{\mid }{z}_j\mathrm{\mid }$ for every $a\in {\widehat{K}}_{\mathrm{\Omega }}$ and every coordinate $j$. Thus ${\widehat{K}}_{\mathrm{\Omega }}$ is coordinate-bounded.

Step 1.1 applied to $E$ gives ${\widehat{E}}_{\mathrm{\Omega }}\subseteq {\widehat{{\widehat{E}}_{\mathrm{\Omega }}}}_{\mathrm{\Omega }}$. For the reverse inclusion, let $a\in {\widehat{{\widehat{E}}_{\mathrm{\Omega }}}}_{\mathrm{\Omega }}$. Then [L1] gives $\mathrm{\mid }f(a)\mathrm{\mid }\le {\sup }_{{\widehat{E}}_{\mathrm{\Omega }}}\mathrm{\mid }f\mathrm{\mid }$ for every $f\in \mathcal{O}(\mathrm{\Omega })$, while the definition of ${\widehat{E}}_{\mathrm{\Omega }}$ itself gives ${\sup }_{{\widehat{E}}_{\mathrm{\Omega }}}\mathrm{\mid }f\mathrm{\mid }\le {\sup }_E\mathrm{\mid }f\mathrm{\mid }$. So $\mathrm{\mid }f(a)\mathrm{\mid }\le {\sup }_E\mathrm{\mid }f\mathrm{\mid }$ for every holomorphic $f$, and another use of [L1] shows $a\in {\widehat{E}}_{\mathrm{\Omega }}$.

By [L3], every closed polydisc ${\bar{\mathrm{\Delta }}}_r(a)$ with $a\in K$ lies in $\mathrm{\Omega }$. Since $K$ is compact, the union $K_r$ is bounded and closed in ${\mathbb{C}}^m$, hence compact, and it lies in $\mathrm{\Omega }$. Thus $K_r⋐\mathrm{\Omega }$.

Fix $a\in K$. Because ${\bar{\mathrm{\Delta }}}_r(a)\subseteq \mathrm{\Omega }$, the function $f$ is holomorphic on ${\mathrm{\Delta }}_r(a)$, so [L2] gives $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }f(a)\mathrm{\mid }\le \alpha !r^{-({\alpha }_1+\cdots +{\alpha }_m)}{\sup }_{z\in {\bar{\mathrm{\Delta }}}_r(a)}\mathrm{\mid }f(z)\mathrm{\mid }\le \alpha !r^{-({\alpha }_1+\cdots +{\alpha }_m)}{\sup }_{z\in K_r}\mathrm{\mid }f(z)\mathrm{\mid }\mathrm{.}$ Therefore the holomorphic function ${\mathrm{\partial }}^{\alpha }f$ is bounded on $K$ by the displayed constant.

The function ${\mathrm{\partial }}^{\alpha }f$ is holomorphic on $\mathrm{\Omega }$, so [L1] propagates the step-1.2 bound from $K$ to ${\widehat{K}}_{\mathrm{\Omega }}$. This is exactly the stated estimate.

By [L4], one has $K\subseteq {\widehat{K}}_{\mathrm{\Omega }}$, so ${\delta }_{\mathrm{\Omega }}({\widehat{K}}_{\mathrm{\Omega }})\le {\delta }_{\mathrm{\Omega }}(K)$. It remains to prove the reverse inequality.

Fix $w\in {\widehat{K}}_{\mathrm{\Omega }}$ and a number $s$ with $0<s<{\delta }_{\mathrm{\Omega }}(K)$. For every holomorphic $f$ on $\mathrm{\Omega }$, [L2] gives uniform bounds on all derivatives of $f$ at $w$ of the form $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }f(w)\mathrm{\mid }\le C_s\alpha !s^{-({\alpha }_1+\cdots +{\alpha }_m)}$ for a constant $C_s$ depending on $f$, $K$, and $s$ but not on $\alpha$. By [L3], the Taylor series of $f$ at $w$ therefore converges on the full polydisc ${\mathrm{\Delta }}_s(w)$ and defines a holomorphic function there that agrees with $f$ on some smaller polydisc already contained in $\mathrm{\Omega }$.

If ${\mathrm{\Delta }}_s(w)$ were not contained in $\mathrm{\Omega }$, then the smaller overlap from step 1.2 and the larger domain ${\mathrm{\Delta }}_s(w)$ would extend every holomorphic function on $\mathrm{\Omega }$, contradicting [L1]. Hence ${\mathrm{\Delta }}_s(w)\subseteq \mathrm{\Omega }$, so ${\delta }_{\mathrm{\Omega }}(w)\ge s$. Since this holds for every $w\in {\widehat{K}}_{\mathrm{\Omega }}$ and every $s<{\delta }_{\mathrm{\Omega }}(K)$, one gets ${\delta }_{\mathrm{\Omega }}({\widehat{K}}_{\mathrm{\Omega }})\ge {\delta }_{\mathrm{\Omega }}(K)$. Together with step 1.1, this proves the equality.

Property 1 implies property 2 by [L1].

Assume property 3. If $\mathrm{\Omega }={\mathbb{C}}^m$, property 1 holds directly from [L3]. Otherwise fix $p\in \mathrm{\partial }\mathrm{\Omega }$ and a connected open set $C\subseteq \mathrm{\Omega }$ with $p\in \bar{C}$. Let $(D_n)$ be an increasing compact exhaustion of $\mathrm{\Omega }$. Construct increasing holomorphically convex compact sets $K_n$ and points $p_n\to p$ recursively. Start with $K_1:={\widehat{D_1}}_{\mathrm{\Omega }}$. Once $K_n⋐\mathrm{\Omega }$ is chosen, choose $p_n\in C\setminus K_n$ with $\mathrm{\mid }{p}_n-p\mathrm{\mid }<1\mathrm{/}n$, and put $K_{n+1}:={\widehat{K_n\cup D_{n+1}\cup \{p_n\}}}_{\mathrm{\Omega }}\mathrm{.}$ Property 3 and idempotence of hulls make each $K_n$ compactly contained in $\mathrm{\Omega }$, and $p_n\in K_j$ whenever $j>n$.

Since $p_n\mathrm{\notin }{\widehat{K}}_n{}_{\mathrm{\Omega }}=K_n$, choose $g_n\in \mathcal{O}(\mathrm{\Omega })$ with $\mathrm{\mid }{g}_n(p_n)\mathrm{\mid }>{\sup }_{K_n}\mathrm{\mid }{g}_n\mathrm{\mid }$. Taking a sufficiently large positive power makes the ratio of these two quantities arbitrarily large; scaling that power then gives $f_n\in \mathcal{O}(\mathrm{\Omega })$ such that ${\sup }_{K_n}\mathrm{\mid }{f}_n\mathrm{\mid }<2^{-n}\text{and}\mathrm{\mid }{f}_n(p_n)\mathrm{\mid }>n+1+{\sum }_{j<n}\mathrm{\mid }{f}_j(p_n)\mathrm{\mid }\mathrm{.}$ Every compact subset of $\mathrm{\Omega }$ lies in some $K_N$, so ${\sum }_n{f}_n$ converges uniformly on compact subsets to a holomorphic function $f$. Moreover, $p_n\in K_j$ for $j>n$, so ${\sum }_{j>n}\mathrm{\mid }{f}_j(p_n)\mathrm{\mid }<{\sum }_{j>n}{2}^{-j}<1.$ The reverse triangle inequality and the second displayed bound give $\mathrm{\mid }f(p_n)\mathrm{\mid }>n$.

Assume property 2, and let $K⋐\mathrm{\Omega }$ be compact. By [L2], the hull ${\widehat{K}}_{\mathrm{\Omega }}$ is closed in $\mathrm{\Omega }$ and coordinate-bounded. Property 2 gives ${\delta }_{\mathrm{\Omega }}({\widehat{K}}_{\mathrm{\Omega }})={\delta }_{\mathrm{\Omega }}(K)>0$, so every point of ${\widehat{K}}_{\mathrm{\Omega }}$ carries a positive-radius polydisc contained in $\mathrm{\Omega }$. Hence every Euclidean limit point of ${\widehat{K}}_{\mathrm{\Omega }}$ still lies in $\mathrm{\Omega }$, and the closedness from [L2] makes ${\widehat{K}}_{\mathrm{\Omega }}$ closed in ${\mathbb{C}}^m$. Being closed and bounded in finite-dimensional Euclidean space, it is compact; the positive boundary-radius lower bound keeps it away from $\mathrm{\partial }\mathrm{\Omega }$. Thus ${\widehat{K}}_{\mathrm{\Omega }}⋐\mathrm{\Omega }$, so property 3 holds.

Suppose toward a contradiction that domains $U_1,U_2$ witness failure of property 1 as in [L3]. Choose $a\in U_1$ and $b\in U_2\setminus \mathrm{\Omega }$. By [L4], join them by a path in $U_2$, and let $p$ be its first point on $\mathrm{\partial }\mathrm{\Omega }$. The part of the path before $p$ lies in the connected component $C$ of $\mathrm{\Omega }\cap U_2$ containing $U_1$, so $p\in \bar{C}$. Apply steps 1.2 and 2.1 with this $p$ and $C$, obtaining $f\in \mathcal{O}(\mathrm{\Omega })$ and $p_n\in C$ with $p_n\to p$ and $\mathrm{\mid }f(p_n)\mathrm{\mid }>n$. By the assumed simultaneous-extension property, $f$ has an extension $F\in \mathcal{O}(U_2)$ agreeing with $f$ on $U_1$. The identity theorem [L5] gives $F=f$ on $C$, but continuity makes $F$ bounded near $p\in U_2$, contradicting $\mathrm{\mid }F(p_n)\mathrm{\mid }>n$. Thus property 3 implies property 1, and the three properties are equivalent.

View ${\mathbb{C}}^m$ as ${\mathbb{R}}^{2m}$ by writing $z_j=x_j+i{y}_j$. Since $K$ is compact, it is closed, so [L1] gives real numbers ${\alpha }_1,\dots ,{\alpha }_m,{\beta }_1,\dots ,{\beta }_m$ and a real number $\beta$ such that ${\sum }_{j=1}^m{\alpha }_j{x}_j+{\sum }_{j=1}^m{\beta }_j{y}_j\le \beta <{\sum }_{j=1}^m{\alpha }_j\mathrm{Re}{p}_j+{\sum }_{j=1}^m{\beta }_j\mathrm{Im}{p}_j$ for every $z=(z_1,\dots ,z_m)\in K$.

Define $a_j:={\alpha }_j-i{\beta }_j$ and $L(z):={\sum }_{j=1}^m{a}_j{z}_j$. Then $L$ is complex-linear and $\mathrm{Re}L(z)={\sum }_{j=1}^m{\alpha }_j\mathrm{Re}{z}_j+{\sum }_{j=1}^m{\beta }_j\mathrm{Im}{z}_j\mathrm{.}$ Substituting this identity into step 1.1 gives the stated strict separation.

Let $p\in \mathrm{\Omega }\setminus \mathrm{conv}(K)$. Since $\mathrm{conv}(K)$ is compact and convex, [L1] gives a complex-linear functional $L$ such that $\mathrm{Re}L(z)\le \beta <\mathrm{Re}L(p)$ for every $z\in \mathrm{conv}(K)$, hence in particular for every $z\in K$. The holomorphic function $h(z):=\exp (L(z))$ then satisfies $\mathrm{\mid }h(z)\mathrm{\mid }\le e^{\beta }<\mathrm{\mid }h(p)\mathrm{\mid }$ on $K$. By [L2], this excludes $p$ from ${\widehat{K}}_{\mathrm{\Omega }}$.

Step 1.1 proves ${\widehat{K}}_{\mathrm{\Omega }}\subseteq \mathrm{conv}(K)$. Because $\mathrm{\Omega }$ is convex, [L3] gives $\mathrm{conv}(K)\subseteq \mathrm{\Omega }$. In finite-dimensional Euclidean space the convex hull of a compact set is compact, so ${\widehat{K}}_{\mathrm{\Omega }}$ is contained in a compact subset of $\mathrm{\Omega }$. Therefore ${\widehat{K}}_{\mathrm{\Omega }}⋐\mathrm{\Omega }$, and $\mathrm{\Omega }$ is holomorphically convex.

By [L1], the convex domain $\mathrm{\Omega }$ is holomorphically convex.

Applying [L2] to step 1.1 shows that $\mathrm{\Omega }$ is a domain of holomorphy.

The two parametrizations describe the same affine line because $a+(\alpha \mu +\beta )v=(a+\beta v)+\mu (\alpha v)$. Thus the second restriction is just $(u_{a,v})\circ \varphi$ on the corresponding one-variable domain.

A nonconstant affine map sends discs to discs and is biholomorphic onto its image, so the upper-semicontinuity and disc-submean inequalities of [L2] are preserved under composition with $\varphi$ and with ${\varphi }^{-1}$. Therefore $u_{a,v}$ is subharmonic or identically $-\mathrm{\infty }$ on one component exactly when $(u_{a,v})\circ \varphi$ is subharmonic or identically $-\mathrm{\infty }$ on the corresponding component. By [L1], the line-test definition is independent of the chosen affine parametrization.

Fix $a\in \mathrm{\Omega }$ and $v\in {\mathbb{C}}^m$, and define ${\varphi }_{a,v}(\lambda )=a+\lambda v$ on a small disc about $0$ whose image lies in $\mathrm{\Omega }$. By the chain rule, $\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }\lambda \mathrm{\partial }\bar{\lambda }}(u\circ {\varphi }_{a,v})(0)={\sum }_{j=1}^m{\sum }_{k=1}^m\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{z}_j\mathrm{\partial }{\bar{z}}_k}(a)v_j\overline{v_k}={\mathcal{L}}_u(a;v)\mathrm{.}$ Since $\mathrm{\Delta }=4{\mathrm{\partial }}_{\lambda }{\mathrm{\partial }}_{\bar{\lambda }}$ in one complex variable, [L3] says that $u\circ {\varphi }_{a,v}$ is subharmonic exactly when ${\mathcal{L}}_u(a;v)\ge 0$.

If $u$ is plurisubharmonic, then every line restriction is subharmonic by [L1], so step 1.1 gives ${\mathcal{L}}_u(a;v)\ge 0$ for every $a$ and $v$. Conversely, if the Levi form is semipositive everywhere, then step 1.1 and [L3] show that every affine-line restriction is subharmonic. Applying [L1] again, $u$ is plurisubharmonic on $\mathrm{\Omega }$.

A decreasing limit of upper semicontinuous functions is upper semicontinuous, so $u$ is upper semicontinuous on $\mathrm{\Omega }$. If $u\equiv -\mathrm{\infty }$ on some connected component, the first alternative of the statement holds and there is nothing further to prove.

Fix $a\in \mathrm{\Omega }$ and $v\mathrm{\ne }0$. On every connected component of the line domain $\{\lambda :a+\lambda v\in \mathrm{\Omega }\}$, each restriction $\lambda \mapsto u_n(a+\lambda v)$ is subharmonic or identically $-\mathrm{\infty }$ by [L1]. Therefore [L2] makes the limit restriction $\lambda \mapsto u(a+\lambda v)$ subharmonic or identically $-\mathrm{\infty }$ there. Since the componentwise $-\mathrm{\infty }$ alternative was excluded in step 1.1, [L1] now shows that $u$ is plurisubharmonic on $\mathrm{\Omega }$.

For $a\in \mathrm{\Omega }$ and $v\in {\mathbb{C}}^m$, the chain rule [L2] gives ${\mathcal{L}}_{u\circ F}(a;v)={\mathcal{L}}_u(F(a);DF(a)v)\mathrm{.}$

Because $u$ is $C^2$ and plurisubharmonic, [L1] makes the right-hand side of step 1.1 nonnegative for every $a$ and $v$. Applying [L1] again, $u\circ F$ is plurisubharmonic on $\mathrm{\Omega }$.

Restrict every function in parts 1 and 2 to an affine complex line in $\mathrm{\Omega }$. By [L1], each restriction is subharmonic or identically $-\mathrm{\infty }$ on every connected component of the line domain, and [L2] shows that nonnegative finite sums and finite maxima preserve subharmonicity there. Another use of [L1] therefore gives parts 1 and 2.

For part 3, fix an affine complex line. Its restriction $v$ of $u$ is a real-valued subharmonic function by [L1]. The cited one-variable source result for convex nondecreasing compositions of subharmonic functions makes $\varphi \circ v$ subharmonic on each connected component. Because upper semicontinuity is preserved by nondecreasing composition, [L1] shows that $\varphi \circ u$ is plurisubharmonic.

The local upper bounds imply that $v$ never takes the value $+\mathrm{\infty }$. By the added hypothesis, $v$ is upper semicontinuous. Because $\mathcal{F}$ is nonempty and each member is not identically $-\mathrm{\infty }$ on a component, the same is true of $v$.

Fix an affine complex line in $\mathrm{\Omega }$ and one of the connected components of its line domain. Restricting the family $\mathcal{F}$ there, [L1] gives a nonempty family of subharmonic functions that is still locally bounded above. Its restricted supremum is exactly the restriction of $v$, and that restriction is upper semicontinuous because $v$ is. Therefore [L2] makes the restriction of $v$ subharmonic on that component. Thus the line test from [L1] is satisfied.

Step 1.1 gives the upper-semicontinuity and nontriviality conditions, and step 1.2 gives the line test. By [L1], $v$ is plurisubharmonic on $\mathrm{\Omega }$.

Choose a small Euclidean ball $B(a,r)\subseteq \mathrm{\Omega }$. Fix $z\in B(a,r)$, and restrict $u$ to the affine complex line through $a$ and $z$. By [L1], that restriction is subharmonic on the connected line-domain component containing the segment from $a$ to $z$, and the global bound makes its value at $a$ a finite maximum. Hence [L2] makes it constant on that component, so $u(z)=u(a)$. Thus $u$ is constant on $B(a,r)$.

Put $S:=\{z\in \mathrm{\Omega }:u(z)=u(a)\}$. It is nonempty. Every point of $S$ is another point where the finite global maximum is attained, so the argument of step 1.1 makes $S$ open. Since $u\le u(a)$ everywhere and upper semicontinuity makes $\{u\ge u(a)\}$ closed, $S=\{u\ge u(a)\}$ is closed. Connectedness of $\mathrm{\Omega }$ gives $S=\mathrm{\Omega }$, so $u$ is constant.

Fix an affine complex line in $\mathrm{\Omega }$. The restriction of $f$ to that line is a one-variable holomorphic function, and by the componentwise hypothesis it is not identically zero on the connected component under consideration. Therefore [L2] makes the restriction of $u=\log \mathrm{\mid }f\mathrm{\mid }$ subharmonic or identically $-\mathrm{\infty }$ there.

The function $u$ is upper semicontinuous because it is a logarithm of a continuous modulus away from the zero set and has value $-\mathrm{\infty }$ on the zero set. Step 1.1 is exactly the line test from [L1], so $u$ is plurisubharmonic on $\mathrm{\Omega }$.

Write $b(z)=-\log {\delta }_{\mathrm{\Omega }}(z)$, with $b\equiv 0$ when $\mathrm{\Omega }={\mathbb{C}}^m$ as specified in [L1], and assume $b$ is plurisubharmonic. The function $q(z)={\sum }_{j<m}\mathrm{\mid }{z}_j{\mathrm{\mid }}^2$ is plurisubharmonic by [L3]. For a proper domain, the distance function ${\delta }_{\mathrm{\Omega }}$ is positive and continuous on $\mathrm{\Omega }$ by [L5]: the triangle inequality makes distance to its nonempty complement $1$-Lipschitz. Thus $b$ is continuous in that case, and the whole-space convention is continuous as well. Therefore [L2] makes $u(z):=\max \{b(z),q(z)\}$ a continuous plurisubharmonic function.

For $c<0$, the sublevel $K_c:=\{z\in \mathrm{\Omega }:u(z)\le c\}$ is empty since $q\ge 0$. For $c\ge 0$, it is closed in $\mathrm{\Omega }$ by continuity and lies in $\{q\le c\}$, so it is bounded. If $\mathrm{\Omega }={\mathbb{C}}^m$, it is closed in the whole space and hence compact by [L4]. For a proper domain, every $z\in K_c$ also satisfies ${\delta }_{\mathrm{\Omega }}(z)\ge e^{-c}$. If a sequence from $K_c$ converges in ${\mathbb{C}}^m$ to $a$, the $1$-Lipschitz distance-to-complement function gives distance at least $e^{-c}$ at $a$, so $a\in \mathrm{\Omega }$; continuity of $u$ then puts $a\in K_c$. Thus $K_c$ is closed in the whole space, bounded and contained in $\mathrm{\Omega }$, hence compact by [L4]. All sublevels are compact, so [L1] gives the promised continuous plurisubharmonic exhaustion.

Because $\rho$ and $\tilde{\rho }$ vanish on the same $C^2$ hypersurface, have nonzero differentials there, and define the same negative side, one has $\tilde{\rho }=h\rho$ near $p$ for a positive $C^1$ function $h$.

Let $v$ be a complex tangent vector at $p$, so $\mathrm{\partial }\rho (p)\cdot v=0$. The second-order expansion of $h\rho$ at $p$ uses only first derivatives of $h$ because $\rho (p)=0$; every mixed product term contains $\mathrm{\partial }\rho (p)\cdot v$ or its conjugate and therefore vanishes on $v$. Consequently ${\mathcal{L}}_{\tilde{\rho }}(p;v)=h(p){\mathcal{L}}_{\rho }(p;v)\mathrm{.}$ Since $h(p)>0$, the two Levi forms have the same sign on complex tangent vectors.