Extension to analytic function What does it mean that a function $g$ "extends to an analytic function"? In the particular context, $g$ is the Fourier transform of a function $f$ which satisfies a growth condition. Does it mean the same thing as "analytic continuation"? 
 A: The term "analytic continuation" is usually only used when one has a holomorphic function $f$ on an open set $U$ in $\mathbb{C}$ (or, more generally, in a complex manifold), and continues it across the boundary of $U$, so there is an open connected set $V$ with $V\cap U\neq\varnothing$ and $V \not\subset U$, and a holomorphic $g$ defined on $V$ such that $g \equiv f$ on a connected component of $V\cap U$. $g$ is then called an analytic continuation of $f$. Note that $g$ need not coincide with $f$ on all of $V\cap U$, and it may not be possible to have a holomorphic function $h$ defined on a proper open superset of $U$ whose restriction to $U$ is $f$. An example is $\log$, where $U$ is the slit plane $\mathbb{C}\setminus \{x \in \mathbb{R} : x \leqslant 0\}$, where we can analytically continue $\log$ across the branch-cut, but there is no extension to a holomorphic function on a strictly larger domain. (There is no fundamental reason why the term should not be used in the analogous real-analytic setting, but I have not come across that yet.)
The Fourier transform of $f$ is a priori only defined on $\mathbb{R}^n$, and that it "extends to an analytic function" means that there is a holomorphic function $G$ defined on an open set $U\subset \mathbb{C}^n$ containing $\mathbb{R}^n$ whose restriction to $\mathbb{R}^n$ is $g$. (That is the case if and only if $g$ is real-analytic.)
An extension of a function is defined on a larger set than the original, an analytic continuation need not be defined on the whole domain of the original (but to be a true continuation, its domain must include points not in the domain of the original).
