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