Hartogs figures give local extension across polydisc shells

By translating by $a$ and scaling each coordinate by ${\rho }_j^{-1}$, we may reduce to the case $a=0$ and ${\rho }_j=1$ for all $j$. Then the shell is exactly $H(r,s)\times {\mathrm{\Delta }}_1^{m-2}$, with $H(r,s)$ in the $(z_1,z_m)$ variables and the middle variables passive.

Fix $\rho$ with $r<\rho <1$. On the domain $D_{\rho }:=\{z:\mathrm{\mid }{z}_1\mathrm{\mid }<\rho ,\mathrm{\mid }{z}_j\mathrm{\mid }<1\text{ for }2\le j\le m\}$ define [construct] $F_{\rho }(z):=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=\rho }\frac{f(\zeta ,z_2,\dots ,z_m)}{\zeta -z_1}d\zeta \mathrm{.}$ Because $\mathrm{\mid }\zeta \mathrm{\mid }=\rho >r$ and $\mathrm{\mid }{z}_j\mathrm{\mid }<1$ for $j\ge 2$, every point $(\zeta ,z_2,\dots ,z_m)$ lies in the shell from step 1.1, so the integral is well defined.

Fix all variables except one coordinate of $z\in D_{\rho }$. For the $z_1$ variable, the integrand in step 2.1 is jointly continuous on the contour times $\{\mathrm{\mid }{z}_1\mathrm{\mid }<\rho \}$ and holomorphic in $z_1$, so [L1] makes $z_1\mapsto F_{\rho }(z)$ holomorphic. For any coordinate $z_k$ with $2\le k\le m$, the denominator is constant and the slice $z_k\mapsto f(\zeta ,z_2,\dots ,z_m)$ is holomorphic on the unit disc because $f$ is holomorphic on the shell. Another use of [L1] makes $z_k\mapsto F_{\rho }(z)$ holomorphic. Therefore $F_{\rho }$ is separately holomorphic on $D_{\rho }$.

If $\mathrm{\mid }{z}_m\mathrm{\mid }<s$, then for fixed $z_2,\dots ,z_{m-1}$ the slice $z_1\mapsto f(z_1,z_2,\dots ,z_m)$ is holomorphic on the full unit disc. Applying [L2] on the circle $\mathrm{\mid }\zeta \mathrm{\mid }=\rho$ gives $F_{\rho }(z)=f(z)(\mathrm{\mid }{z}_1\mathrm{\mid }<\rho ,\mathrm{\mid }{z}_m\mathrm{\mid }<s,\mathrm{\mid }{z}_j\mathrm{\mid }<1\text{ for }2\le j\le m-1)\mathrm{.}$ So $F_{\rho }$ agrees with $f$ on a nonempty open subset of the shell.

Let $K\subseteq D_{\rho }$ be compact. Choose $\delta >0$ so that $\mathrm{\mid }{z}_1\mathrm{\mid }\le \rho -\delta$ on $K$. The set $\{(\zeta ,z_2,\dots ,z_m):\mathrm{\mid }\zeta \mathrm{\mid }=\rho ,z\in K\}$ is a compact subset of the shell, so $\mathrm{\mid }f\mathrm{\mid }$ has a finite bound $M_K$ there. The contour in step 2.1 has length $2\pi \rho$, and $\mathrm{\mid }\zeta -z_1\mathrm{\mid }\ge \delta$ on that contour for $z\in K$, hence $\mathrm{\mid }{F}_{\rho }(z)\mathrm{\mid }\le \frac{1}{2\pi }{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=\rho }\frac{M_K}{\mathrm{\mid }\zeta -z_1\mathrm{\mid }}\mathrm{\mid }d\zeta \mathrm{\mid }\le \frac{\rho M_K}{\delta }(z\in K)\mathrm{.}$ Thus $F_{\rho }$ is locally bounded on $D_{\rho }$, so [L4] upgrades step 3.1 to joint holomorphicity on $D_{\rho }$.

If $r<{\rho }_1<{\rho }_2<1$, then both $F_{{\rho }_1}$ and $F_{{\rho }_2}$ are holomorphic on $D_{{\rho }_1}$ by step 4.1, and step 3.2 shows that they agree on the nonempty open subset of $D_{{\rho }_1}$ where $\mathrm{\mid }{z}_m\mathrm{\mid }<s$. Therefore [L3] gives $F_{{\rho }_1}=F_{{\rho }_2}$ on all of $D_{{\rho }_1}$.

For each point $z$ of the full polydisc from step 1.1, choose any $\rho$ with $\max \{r,\mathrm{\mid }{z}_1\mathrm{\mid }\}<\rho <1$ and set $F(z):=F_{\rho }(z)$. Step 5.1 makes this definition independent of $\rho$, and step 4.1 shows that $F$ is holomorphic near each point. For fixed $\rho$, the open set $S(0,1;r,s)\cap D_{\rho }$ is connected: its central part $\mathrm{\mid }{z}_m\mathrm{\mid }<s$ meets its annular part $r<\mathrm{\mid }{z}_1\mathrm{\mid }<\rho$, and each part is connected. On this set $F_{\rho }$ and $f$ are holomorphic and agree on the nonempty central part by step 3.2, so [L3] gives $F_{\rho }=f$ throughout it. Every shell point lies in one such $D_{\rho }$, hence $F$ extends $f$ on the entire shell. If $G$ is another holomorphic extension to the full polydisc, then $F$ and $G$ agree with $f$ on the same nonempty open subset where $\mathrm{\mid }{z}_m\mathrm{\mid }<s$, so [L3] forces $F=G$. Thus the extension is unique.