A smooth psh exhaustion gives Hartogs pseudoconvexity on bounded domains

For each fixed nonzero $\xi \in {\mathbb{C}}^n$ define $d_{\xi }(z)=\sup \{r>0:z+\{t:\mathrm{\mid }t\mathrm{\mid }<r\}\xi \subset \mathrm{\Omega }\},u_{\xi }=-\log d_{\xi }\mathrm{.}$ These radii are positive and finite because $\mathrm{\Omega }$ is open and bounded. If $r<d_{\xi }(z)$, the closed directional disc of radius $r$ is compact in $\mathrm{\Omega }$, and sufficiently small translations stay in $\mathrm{\Omega }$. Thus $d_{\xi }$ is lower semicontinuous, so $u_{\xi }$ is upper semicontinuous.

Fix an affine base disc $z(\zeta )=a+\zeta v$ with $z(\bar{\mathbb{D}})\subset \mathrm{\Omega }$, and a continuous real harmonic majorant $h$ on the closed unit disc with $h\ge u_{\xi }\circ z$ on its boundary. For $0<s<1$ the function $h_s(\zeta )=h(s\zeta )$ is harmonic on a disc of radius greater than $1$, and $h_s\to h$ uniformly on the closed unit disc. Given $\eta >0$, choose $s$ so that $\mathrm{\mid }{h}_s-h\mathrm{\mid }<\eta$ there; by [F3] choose a holomorphic $H$ on a disc of radius greater than $1$ with $\mathrm{Re}H=h_s+\eta$. Then $\mathrm{\Phi }(\zeta ,t)=z(\zeta )+t{e}^{-H(\zeta )}\xi$ is holomorphic near the closed base disc, and for $\mathrm{\mid }\zeta \mathrm{\mid }=1$, $\mathrm{\mid }t\mathrm{\mid }<1$, its value belongs to $\mathrm{\Omega }$, since $e^{-\mathrm{Re}H}\le d_{\xi }(z(\zeta ))$. Compactness of the base disc also puts all its images in $\mathrm{\Omega }$ for sufficiently small $\mathrm{\mid }t\mathrm{\mid }$.

Let $r_{\ast }$ be the supremum of radii $r\le 1$ for which $\mathrm{\Phi }(\bar{\mathbb{D}}\times \{t:\mathrm{\mid }t\mathrm{\mid }<r\})\subset \mathrm{\Omega }$. Suppose $r_{\ast }<1$, and fix $r_{\ast }<b<1$. The image of $\mathrm{\partial }\mathbb{D}\times \{\mathrm{\mid }t\mathrm{\mid }\le b\}$ is a compact subset of $\mathrm{\Omega }$ by step 2.1. Let $M$ be the maximum of $S$ on this image. For each $\mathrm{\mid }t\mathrm{\mid }<r_{\ast }$, [F1] makes $S\circ \mathrm{\Phi }(\cdot ,t)$ subharmonic, continuous on the closed base disc; its boundary values are at most $M$, so [F4] bounds it everywhere by $M$. All these images therefore lie in the fixed compact sublevel $K=\{S\le M\}\subset \mathrm{\Omega }$. By continuity their limits with $\mathrm{\mid }t\mathrm{\mid }\le r_{\ast }$ also lie in $K$. Uniform continuity on a slightly larger compact product then increases the admissible radius beyond $r_{\ast }$, contradicting its definition. Thus $r_{\ast }=1$, and $d_{\xi }(z(\zeta ))\ge e^{-h_s(\zeta )-\eta }$ throughout the base disc.

Step 3.1 gives $u_{\xi }\circ z\le h_s+\eta$. Choose $s\to 1$ and $\eta \to 0$ with the stated uniform error, to conclude $u_{\xi }\circ z\le h$. The affine-disc normalization covers every closed disc in every complex line in $\mathrm{\Omega }$. Hence [F2], together with the upper semicontinuity of step 1.1, makes each $u_{\xi }$ plurisubharmonic. The dilation of the majorant in step 2.1 ensures that $H$ and $\mathrm{\Phi }$ are defined past the base boundary; no boundary continuity of an arbitrary harmonic conjugate is assumed.

Write $\mathrm{\parallel }\xi {\mathrm{\parallel }}_{\mathrm{\infty }}={\max }_{j<n}\mathrm{\mid }{\xi }_j\mathrm{\mid }$. A sup-norm polydisc of radius $r$ consists exactly of all directional discs of radius $r$ with $\mathrm{\parallel }\xi {\mathrm{\parallel }}_{\mathrm{\infty }}=1$, so ${\delta }_{\mathrm{\Omega }}(z)={\inf }_{\mathrm{\parallel }\xi {\mathrm{\parallel }}_{\mathrm{\infty }}=1}{d}_{\xi }(z),u(z):=-\log {\delta }_{\mathrm{\Omega }}(z)={\sup }_{\mathrm{\parallel }\xi {\mathrm{\parallel }}_{\mathrm{\infty }}=1}{u}_{\xi }(z)\mathrm{.}$ By [F5], ${\delta }_{\mathrm{\Omega }}$ is a positive continuous distance function on $\mathrm{\Omega }$, so $u$ is continuous. On any compactly contained affine circle, each $u_{\xi }$ satisfies its submean inequality and is at most $u$ on the circle. Therefore $u_{\xi }$ at the center is at most the circle average of $u$; taking the supremum gives that same bound for $u$ at the center. Its continuity and these submean inequalities make it psh by [F1] and [F4]. This proves the exact Hartogs convention of [F5].