Cartan-Thullen boundary-radius theorem

By [L4], one has $K\subseteq {\widehat{K}}_{\mathrm{\Omega }}$, so ${\delta }_{\mathrm{\Omega }}({\widehat{K}}_{\mathrm{\Omega }})\le {\delta }_{\mathrm{\Omega }}(K)$. It remains to prove the reverse inequality.

Fix $w\in {\widehat{K}}_{\mathrm{\Omega }}$ and a number $s$ with $0<s<{\delta }_{\mathrm{\Omega }}(K)$. For every holomorphic $f$ on $\mathrm{\Omega }$, [L2] gives uniform bounds on all derivatives of $f$ at $w$ of the form $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }f(w)\mathrm{\mid }\le C_s\alpha !s^{-({\alpha }_1+\cdots +{\alpha }_m)}$ for a constant $C_s$ depending on $f$, $K$, and $s$ but not on $\alpha$. By [L3], the Taylor series of $f$ at $w$ therefore converges on the full polydisc ${\mathrm{\Delta }}_s(w)$ and defines a holomorphic function there that agrees with $f$ on some smaller polydisc already contained in $\mathrm{\Omega }$.

If ${\mathrm{\Delta }}_s(w)$ were not contained in $\mathrm{\Omega }$, then the smaller overlap from step 1.2 and the larger domain ${\mathrm{\Delta }}_s(w)$ would extend every holomorphic function on $\mathrm{\Omega }$, contradicting [L1]. Hence ${\mathrm{\Delta }}_s(w)\subseteq \mathrm{\Omega }$, so ${\delta }_{\mathrm{\Omega }}(w)\ge s$. Since this holds for every $w\in {\widehat{K}}_{\mathrm{\Omega }}$ and every $s<{\delta }_{\mathrm{\Omega }}(K)$, one gets ${\delta }_{\mathrm{\Omega }}({\widehat{K}}_{\mathrm{\Omega }})\ge {\delta }_{\mathrm{\Omega }}(K)$. Together with step 1.1, this proves the equality.