# Analytic extension from a closed analytic subset

I have the following question: Let $\Omega \subset \mathbb{R}^{n}$ be an open set and consider $X \subset \Omega$ an analytic subset. By this I mean that there exists analytic functions $f_{1},...,f_{k}$ defined on $\Omega$ such that $X=\{ x\in \Omega |f_{1}(x)=...=f_{k}(x)=0 \}$. Let $f:X\rightarrow \mathbb{R}$ be a analytic function on $X$ and assume that there exists a open neighbourhood $U$ of $X$ in $\Omega$ and a analytic function $g:U\rightarrow \mathbb{R}$ that extends $f$. Is it then true that there exists a analytic function $h:\Omega \rightarrow \mathbb{R}$ that extends $g$, in particular $f$?

I am reading the paper "A Note on the Extension of Analytic Functions off Real Analytic Subsets" by G. Nardelli and A. Tancredi. Here is the link: http://www.mat.ucm.es/serv/revmat/vol9-1/vol9-1d.pdf . There, in section 2 Theorem 1 it is mentioned that there exists such an $h$ that extends $f$. But, my question is, does this $h$ also extend $g$?

I am hoping for a lot of answers.

Cheers Juno

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No, since $g$ might "blow up" near the boundary of $U$ in $\Omega$. Let $n = 1$, $\Omega = \mathbf{R}$, $X = \{0\}$, $f = 0$, $U = (-1,1)$, and $g$ an analytic function on $U$ vanishing at 0 but unbounded near $1$ and/or $-1$. –  Marguax Oct 7 '13 at 17:26