Domains of holomorphy are Hartogs pseudoconvex

If $\mathrm{\Omega }={\mathbb{C}}^m$, it is Hartogs pseudoconvex by the whole-space convention in [L2]. Assume henceforth that $\mathrm{\Omega }$ is proper. Fix an affine map $\lambda \mapsto z_0+\lambda w_0$ from the closed unit disc into $\mathrm{\Omega }$, and put $u(\lambda ):=-\log {\delta }_{\mathrm{\Omega }}(z_0+\lambda w_0)\mathrm{.}$ By [L3], $u$ is continuous on the closed disc. The trigonometric-polynomial algebra on the unit circle is unital, separates points through the coordinate function, and is self-adjoint because $\bar{z}=z^{-1}$ on the circle. Given $\epsilon >0$, [L4] therefore gives a complex trigonometric polynomial within $\epsilon$ of the real function $u$; taking its real part gives a real trigonometric polynomial $q$ with $\mathrm{\mid }q-u\mathrm{\mid }<\epsilon$. Write $q(e^{it})=\mathrm{Re}p(e^{it})$ for a holomorphic polynomial $p$, and replace $p$ by $p+\epsilon$. Then $u(\lambda )<\mathrm{Re}p(\lambda )\le u(\lambda )+2\epsilon (\mathrm{\mid }\lambda \mathrm{\mid }=1)\mathrm{.}$

For $\eta \in {\mathbb{C}}^m$ with ${\max }_j\mathrm{\mid }{\eta }_j\mathrm{\mid }<1$ and $t\in [0,1]$, define ${\mathrm{\Phi }}_t^{\eta }(\lambda ):=z_0+\lambda w_0+t{e}^{-p(\lambda )}\eta \mathrm{.}$ On $\mathrm{\mid }\lambda \mathrm{\mid }=1$, the perturbation has sup norm strictly less than ${\delta }_{\mathrm{\Omega }}(z_0+\lambda w_0)$, so each boundary circle ${\mathrm{\Phi }}_t^{\eta }(\mathrm{\partial }\mathbb{D})$ lies in $\mathrm{\Omega }$. The initial disc at $t=0$ also lies in $\mathrm{\Omega }$, so [L1] applied to the family $t\mapsto {\mathrm{\Phi }}_t^{\eta }$ gives ${\mathrm{\Phi }}_1^{\eta }(\mathbb{D})\subseteq \mathrm{\Omega }$. Since this holds for every $\eta$ in the unit polydisc, the whole equal-radius polydisc of radius $e^{-\mathrm{Re}p(\lambda )}$ around $z_0+\lambda w_0$ lies in $\mathrm{\Omega }$ for every $\mathrm{\mid }\lambda \mathrm{\mid }<1$. Thus $-\log {\delta }_{\mathrm{\Omega }}(z_0+\lambda w_0)\le \mathrm{Re}p(\lambda )(\mathrm{\mid }\lambda \mathrm{\mid }<1)\mathrm{.}$

Evaluating step 2.1 at $0$ and averaging the boundary values of the real part of the polynomial $p$ gives $u(0)\le \mathrm{Re}p(0)=\frac{1}{2\pi }{\int }_0^{2\pi }\mathrm{Re}p(e^{it})dt\le \frac{1}{2\pi }{\int }_0^{2\pi }u(e^{it})dt+2\epsilon \mathrm{.}$ Letting $\epsilon \downarrow 0$ gives the submean inequality. Because the affine closed disc was arbitrary and $u$ is continuous, [L5] makes every affine-line restriction subharmonic. By [L2], $-\log {\delta }_{\mathrm{\Omega }}$ is plurisubharmonic, so $\mathrm{\Omega }$ is Hartogs pseudoconvex.