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Let $\Omega $ be a domain in $\mathbb{C}^n$. Consider the algebra $A(\Omega)=\mathcal{C}(\overline\Omega)\cap\mathcal{O}(\Omega).$ Let $\{F_\nu\}$ be a sequence in $A(\Omega)$ such that $F_\nu(\overline{\Omega})\subseteq\overline{\mathbb{D}}$ for every $\nu\in\mathbb{Z}_+.$ Now by Montel's Theorem we know that there is a subsequence of this sequence which converges uniformly on compact sets of $\Omega.$ Let $F$ be the limit function of this subsequence then we know this is holomorphic. The question here is if this F belongs to $A(\Omega)$ or not? I mean if we can extend $F$ continuously to the boundary ?

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In general no, even if $n=1$. Take for example $\Omega$ as the unit disc, let $F \in H^\infty \setminus A$ with $|F| < 1$ and let $F_\nu(z) = F( r_n z)$ where $r_n \searrow 1$.

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  • $\begingroup$ Thanks. So in general it is not true. The sequence of hol. functions that I got has this particular property that each member of the family maps a given fixed point on the boundary to the boundary of unit disc. Does it make any difference? $\endgroup$ – Abelvikram Jun 14 '13 at 3:01

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