My question is motivated by the following: if $f\_1, f\_2: \mathbb{R}^m \rightarrow \mathbb{R}$ are real analytic functions and they agree on an open set $V$ then we actually have $f\_1 \equiv f\_2$. But how do real analytic functions depend on their values on $V$? More precisely is the following true:


Suppose $f\_\lambda:\mathbb{R}^m \rightarrow \mathbb{R}$ is a family of analytic functions indexed by $\lambda \in (0,1) \subset \mathbb{R}$. Let $V \subset \mathbb{R}^m$ be an open set and assume that the map $(\lambda,y) \rightarrow f\_\lambda(y)$ is analytic on $(0,1) \times V$. Then my question is: is the map analytic on $(0,1) \times \mathbb{R}^m$? 


I think this should be straightforward, but I am unable to come up with a rigorous argument or a counter example. 



Dependence of an analytic function on its values in an open set
Counterexample:
$$(\lambda,x)\mapsto \frac{(\lambda-1/2)^3}{x^2+(\lambda-1/2)^2}$$
It is analytic on $(0,1)\times V$ where $V=(1,2)$ but not analytic on $(0,1)\times R$.
The singularity is at $(1/2,0)$.


