Q:

A question about real analytic functions 5

Hi everyone: Suppose that $f$ is a real analytic function on a domain $D$ of $\mathbb{R}^{m}$, $(m\geq2)$, which means at each point $x_{0}\in D$, $f(x)$ is equal to the sum of its Taylor's series for all $|x-x_{0}|<r$, and some $r>0$. Now, suppose that the above Taylor's series of $f(x)$ at $x_{0}$ is convergent for $|x-x_{0}|<R$  (and $\lbrace x:|x-x_{0}|<R\rbrace\subset D$) for some $R>r$; can we conclude that this Taylor's series equals $f(x)$ for $|x-x_{0}|<R$?

A:

The key uniqueness result is that if $g$ is analytic at $x_0 \in D$ and
there is a sequence of points $x_k \to x_0$ such that $g(x_k) = 0$, then
the Taylor expansion of $g$ about $x_0$ is zero.
Suppose $f$ is analytic at $x_0 \in D$ and the Taylor series for
$f$ has a radius of convergence $\rho \ge R$, and $B(x_0,R) \subset D$.
Then the Taylor series representation
is valid on $B(x_0,R)$.
To see this, let $\phi$ be the function defined by the Taylor series of $f$
at $x_0$ and let $g= f-\phi$ and note that $g$ is analytic. Suppose that
$g$ is not zero on $B(x_0,R)$ and let
$\delta = \inf \{ |x-x_0|| g (x) \neq 0 \} $. Then $\delta < R$, $\delta \ge r$ and we can find
some $y$ with $|y| = \delta$ and $y_k \to y$ such that $g(y_k) \neq 0$.
Since we have $g(y) = 0$, and can find a sequence of points $w_k \to y$ (with
$|w_k| < \delta$) such that $g(w_k) = 0$, the uniqueness result
shows that $g$ has a Taylor expansion that is zero in a neighbourhood
of $y$, which contradicts the definition of the $y_k$. Hence $g$ is
zero on $B(x_0,R)$ and so $f=\phi$ on $B(x_0,R)$.   

