A locally bounded punctured slice has a holomorphic parameter extension

Fix $z^{\mathrm{\prime }}\in U$. The slice $w\mapsto f(z^{\mathrm{\prime }},w)$ is holomorphic on the punctured disc $0<\mathrm{\mid }w\mathrm{\mid }<R$ and bounded on $0<\mathrm{\mid }w\mathrm{\mid }<\eta$. Therefore [L1] gives a holomorphic extension $g_{z^{\mathrm{\prime }}}$ to the full disc $\mathrm{\mid }w\mathrm{\mid }<R$.

Fix one coordinate of $z^{\mathrm{\prime }}$ and hold the others fixed. For $\mathrm{\mid }\zeta \mathrm{\mid }=\eta$ the integrand $(z^{\mathrm{\prime }},\zeta )\mapsto f(z^{\mathrm{\prime }},\zeta )\mathrm{/}\zeta$ is jointly continuous in that parameter and $\zeta$, and for fixed $\zeta$ it is holomorphic in the chosen coordinate. So [L2] makes $F$ separately holomorphic on $U$. The same circle formula gives local bounds on compact subsets of $U$, so $F$ is locally bounded there.

Applying [L3] to the holomorphic function $g_{z^{\mathrm{\prime }}}$ on the circle $\mathrm{\mid }\zeta \mathrm{\mid }=\eta$ gives $g_{z^{\mathrm{\prime }}}(0)=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=\eta }\frac{f(z^{\mathrm{\prime }},\zeta )}{\zeta }d\zeta =F(z^{\mathrm{\prime }})\mathrm{.}$ So $F(z^{\mathrm{\prime }})$ is exactly the removable value of the slice at $w=0$.

Step 2.1 identifies $F(z^{\mathrm{\prime }})$ with the extension value at $w=0$ for every $z^{\mathrm{\prime }}$. Step 1.2 makes $F$ separately holomorphic and locally bounded, so [L4] upgrades it to a holomorphic function on $U$.