Domains of Holomorphy, Plurisubharmonicity and Pseudoconvexity — Examples

By [L1], the bidisc ${\mathbb{D}}^2$ is a convex domain in ${\mathbb{C}}^2$.

Applying [L2] to step 1.1 shows that ${\mathbb{D}}^2$ is holomorphically convex.

For $\rho (z)=\mathrm{\mid }{z}_1{\mathrm{\mid }}^2+\cdots +\mathrm{\mid }{z}_m{\mathrm{\mid }}^2-1$, one has $\frac{{\mathrm{\partial }}^2\rho }{\mathrm{\partial }{z}_j\mathrm{\partial }{\bar{z}}_k}={\delta }_{jk},$ so [L2] gives ${\mathcal{L}}_{\rho }(z;v)=\mathrm{\mid }{v}_1{\mathrm{\mid }}^2+\cdots +\mathrm{\mid }{v}_m{\mathrm{\mid }}^2\ge 0$ for every $z$ and every $v\in {\mathbb{C}}^m$.

In particular the Levi form is nonnegative on every complex tangent vector at every boundary point of the unit ball. By [L1], the unit ball is Levi pseudoconvex.

The half-space $H$ is convex because if $\mathrm{Re}{z}_1>0$ and $\mathrm{Re}{w}_1>0$, then $\mathrm{Re}((1-t)z_1+t{w}_1)>0$ for every $t\in [0,1]$.

Applying [L1] to step 1.1 shows that $H$ is a domain of holomorphy.

Let $a$ satisfy $\mathrm{\mid }a\mathrm{\mid }\le 1$, and let $f$ be entire. Then $f$ is holomorphic on the unit disc and continuous on its closure, so [L2] gives $\mathrm{\mid }f(a)\mathrm{\mid }\le {\sup }_{\mathrm{\mid }z\mathrm{\mid }=1}\mathrm{\mid }f(z)\mathrm{\mid }$. By [L1], this shows $a\in {\widehat{E}}_{\mathbb{C}}$. Hence the closed unit disc lies in the hull.

If $\mathrm{\mid }a\mathrm{\mid }>1$, take the entire function $f(z)=z$. Then $\mathrm{\mid }f(a)\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }>1={\sup }_{z\in E}\mathrm{\mid }f(z)\mathrm{\mid }$, so [L1] excludes $a$ from the hull. Therefore no point outside the closed unit disc lies in ${\widehat{E}}_{\mathbb{C}}$. Together with step 1.1, this identifies the hull exactly.

Let $P=\{(z_1,z_2):\mathrm{\mid }{z}_1\mathrm{\mid }\le r_1,\mathrm{\mid }{z}_2\mathrm{\mid }\le r_2\}$. If $a\in P$ and $f$ is holomorphic on ${\mathbb{D}}^2$, then $f$ is continuous on the closed polydisc $P$, and the distinguished boundary of $P$ is exactly $T$. Therefore [L2] gives $\mathrm{\mid }f(a)\mathrm{\mid }\le {\sup }_T\mathrm{\mid }f\mathrm{\mid }$. By [L1], every point of $P$ lies in ${\widehat{T}}_{{\mathbb{D}}^2}$.

If $a=(a_1,a_2)$ lies in ${\mathbb{D}}^2\setminus P$, then either $\mathrm{\mid }{a}_1\mathrm{\mid }>r_1$ or $\mathrm{\mid }{a}_2\mathrm{\mid }>r_2$. In the first case the holomorphic coordinate function $f(z_1,z_2)=z_1$ satisfies $\mathrm{\mid }f(a)\mathrm{\mid }>{\sup }_T\mathrm{\mid }f\mathrm{\mid }$, and in the second case $f(z_1,z_2)=z_2$ does the same. Hence [L1] excludes every point outside $P$ from the hull. Together with step 1.1, this identifies ${\widehat{T}}_{{\mathbb{D}}^2}$ exactly.

The equal-radius polydisc about $z$ stays in the half-space exactly while its first-coordinate radius is smaller than $\mathrm{Re}{z}_1$, so ${\delta }_H(z)=\mathrm{Re}{z}_1$. The function $u(z)=-\log (\mathrm{Re}{z}_1)$ is $C^2$ on $H$ and satisfies $\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{z}_1\mathrm{\partial }{\bar{z}}_1}(z)=\frac{1}{4(\mathrm{Re}{z}_1{)}^2},\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{z}_j\mathrm{\partial }{\bar{z}}_k}(z)=0\text{ for }(j,k)\mathrm{\ne }(1,1)\mathrm{.}$

Hence the Levi form of $u$ is ${\mathcal{L}}_u(z;v)=\frac{\mathrm{\mid }{v}_1{\mathrm{\mid }}^2}{4(\mathrm{Re}{z}_1{)}^2}\ge 0.$ By [L2], $u=-\log {\delta }_H$ is plurisubharmonic on $H$.

Every holomorphic function on $\mathrm{\Omega }$ extends to the full bidisc by [L1]. Thus the whole bidisc is a common larger domain across the missing origin for every holomorphic function on $\mathrm{\Omega }$.

The extension statement in step 1.1 contradicts the domain-of-holomorphy condition in [L2], so $\mathrm{\Omega }$ is not a domain of holomorphy. Applying the equivalence in [L2], $\mathrm{\Omega }$ is not holomorphically convex either. This refutes the statement.

The function $f(z_1,z_2)=1\mathrm{/}(z_1{z}_2-1)$ is holomorphic on $U$. Let $p$ be a point of the removed hypersurface $z_1{z}_2=1$. If $f$ extended holomorphically to a neighborhood $V$ of $p$, then $(z_1{z}_2-1)F$ would be holomorphic on $V$ and equal to $1$ on the nonempty open set $V\cap U$. By [L2], it would equal $1$ on all of $V$, impossible at $p$ where $z_1{z}_2-1=0$. So the same function is singular at every boundary point of $U$, and [L1] makes $U$ a domain of holomorphy.

The points $(2,0)$ and $(0,2)$ lie in $U$, but their midpoint $(1,1)$ lies on the removed hypersurface and therefore is not in $U$. Hence $U$ is not convex. This refutes the statement.

The domain $\mathrm{\Omega }$ is a domain in ${\mathbb{C}}^2$, so it is one instance of the claimed class of domains in ${\mathbb{C}}^n$.

By [L1], this instance fails the expected several-variable convexity/domain-of-holomorphy package. In particular, the statement "every domain in ${\mathbb{C}}^n$ is a domain of holomorphy" is already false in dimension $2$.

The function $1\mathrm{/}{z}_1$ is holomorphic on $U_1$. If it extended across a point of the missing hyperplane $\{z_1=0\}$, then $z_{1}F$ would be holomorphic there and equal to $1$ on a nonempty open subset of the extension domain, so [L2] would force $z_{1}F\equiv 1$, impossible where $z_1=0$. The same argument with $1\mathrm{/}{z}_2$ shows that $U_2$ is also a domain of holomorphy.

The union $U_1\cup U_2$ is exactly ${\mathbb{C}}^2\setminus \{(0,0)\}$. By [L3], every holomorphic function on this punctured space extends across the origin, so [L1] shows that $U_1\cup U_2$ is not a domain of holomorphy. Thus the statement is false.