0
$\begingroup$

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 $|x-x_{0}|0$. Now, suppose that the above Taylor's series of $f(x)$ at $x_{0}$ is convergent for $|x-x_{0}|r$; can we conclude that this Taylor's series equals $f(x)$ for $|x-x_{0}|

  • 0
    We know that the two series coincide on $B(x_{0},R)$, because they already coincide on the smaller ball, $B(x_{0},r)$. The question is weather the sum of the series equals $f$ on $B(x_{0},R)$? Another way of formulating the same problem is this: Can we say that the sum of the series $$\sum\frac{1}{n!}f^{(n)}(x_{0})(x-x_{0})$$ equals $f(x)$ for all $x\in B(x_{0},R)$, if we know that $f$ is real analytic on $D$ and $B(x_{0},R)\subset D$? – 2017-02-08

1 Answers 1

1

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 $B(x_0,R)$.