A holomorphic function on a Hartogs figure extends to the full bidisc

Fix $\rho$ with $r<\rho <1$ and define $F_{\rho }$ by the Cauchy integral in [L2]. Since $\mathrm{\mid }\zeta \mathrm{\mid }=\rho$ and $\mathrm{\mid }w\mathrm{\mid }<1$ place $(\zeta ,w)$ inside $H(r,s)$ by [L1], the integral is well defined. For fixed $z$ with $\mathrm{\mid }z\mathrm{\mid }<\rho$, [L2] applied to the one complex parameter $w$ makes $w\mapsto F_{\rho }(z,w)$ holomorphic on $\mathrm{\mid }w\mathrm{\mid }<1$. For fixed $w$, the same theorem applied to the one complex parameter $z$ makes $z\mapsto F_{\rho }(z,w)$ holomorphic on $\mathrm{\mid }z\mathrm{\mid }<\rho$. The ML estimate gives local bounds on compact subsets, so [L6] upgrades $F_{\rho }$ to a jointly holomorphic function on $\{\mathrm{\mid }z\mathrm{\mid }<\rho ,\mathrm{\mid }w\mathrm{\mid }<1\}$.

If $\mathrm{\mid }w\mathrm{\mid }<s$, then the slice $z\mapsto f(z,w)$ is holomorphic on $\mathrm{\mid }z\mathrm{\mid }<1$. Therefore [L3] gives $F_{\rho }(z,w)=f(z,w)$ whenever $\mathrm{\mid }z\mathrm{\mid }<\rho$ and $\mathrm{\mid }w\mathrm{\mid }<s$. So $F_{\rho }$ extends $f$ across the missing core over that open overlap.

If $r<{\rho }_1<{\rho }_2<1$, then both $F_{{\rho }_1}$ and $F_{{\rho }_2}$ are holomorphic on $\{\mathrm{\mid }z\mathrm{\mid }<{\rho }_1,\mathrm{\mid }w\mathrm{\mid }<1\}$ by step 1.1, and step 2.1 shows that they agree on the nonempty open subset $\{\mathrm{\mid }z\mathrm{\mid }<{\rho }_1,\mathrm{\mid }w\mathrm{\mid }<s\}$. Hence [L4] forces $F_{{\rho }_1}=F_{{\rho }_2}$ on the whole connected domain $\{\mathrm{\mid }z\mathrm{\mid }<{\rho }_1,\mathrm{\mid }w\mathrm{\mid }<1\}$.

For each point $(z,w)\in \widehat{H}(r,s)$ choose any $\rho$ with $\max \{r,\mathrm{\mid }z\mathrm{\mid }\}<\rho <1$, and set $F(z,w):=F_{\rho }(z,w)$. Step 3.1 shows that this does not depend on the chosen $\rho$, so $F$ is well defined and holomorphic locally, hence holomorphic on all of $\widehat{H}(r,s)$.

Step 2.1 gives $F=f$ on the open set $\{\mathrm{\mid }w\mathrm{\mid }<s\}\subseteq H(r,s)$, so $F$ is a holomorphic extension of $f$ to the bidisc. If $G$ is another such extension, then $F$ and $G$ are holomorphic on the connected bidisc and agree on the same nonempty open overlap with $f$; [L4] gives $F=G$. Thus the extension is unique.

The Laurent-coefficient view in [L5] is compatible with the integral construction above: the Cauchy kernel removes the vanished negative part and rebuilds the same holomorphic continuation.