- Consider the function \(f\) defined on R by
\[ f (x) = \left\{ \begin{array}{c l} 0 & \text {if} x \leq 0, \\ e ^ {- 1 / x ^ {2}} & \text {if} x > 0. \end{array} \right. \]
Prove that \(f\) is indefinitely diferentiable on \(\mathbb { R } ,\) and that \(f ^ { ( n ) } ( 0 ) = 0\) for all \(n \geq 1\) Conclude that \(f\) does not have a converging power series expansion \(\scriptstyle \sum _ { n = 0 } ^ { \infty } a _ { n } x ^ { n }\) for x near the origin.
Solution
For \(x>0\), set \(t=1/x\). We claim that for every \(n\ge0\) there is a polynomial \(P_n\) such that \[ f^{(n)}(x)=P_n(1/x)e^{-1/x^2}. \] This is clear for \(n=0\), with \(P_0=1\). If it holds for \(n\), differentiation gives \[ f^{(n+1)}(x)=\left[-x^{-2}P_n'(1/x)+2x^{-3}P_n(1/x)\right]e^{-1/x^2}, \] which has the same form.
For every integer \(m\ge0\), \[ \lim_{x\downarrow0}x^{-m}e^{-1/x^2}=0. \] Indeed, with \(t=1/x\), this is \(t^m e^{-t^2}\to0\) as \(t\to\infty\). Hence for every \(n\), \[ \lim_{x\downarrow0}f^{(n)}(x)=0. \tag{1} \]
We now prove inductively that \(f\) is \(C^\infty\) at \(0\) and \(f^{(n)}(0)=0\) for every \(n\ge0\). For \(n=0\), continuity follows from (1). Suppose \(f^{(n)}\) has been extended continuously to \(0\) with value \(0\). Since it vanishes identically on \((-\infty,0]\), its derivative from the left at \(0\) is \(0\); from the right, \[ \frac{f^{(n)}(h)-f^{(n)}(0)}{h} =\frac{P_n(1/h)e^{-1/h^2}}{h}\longrightarrow0 \] by the same exponential-decay estimate. Thus \(f^{(n+1)}(0)=0\), and (1) shows continuity of the next derivative at \(0\). Therefore \(f\in C^\infty(\mathbb R)\) and all derivatives at \(0\) vanish.
If \(f\) were represented near \(0\) by a convergent power series \(\sum_{n\ge0}a_nx^n\), then necessarily \[ a_n=\frac{f^{(n)}(0)}{n!}=0 \] for all \(n\). The series would therefore be identically zero near \(0\), contradicting \(f(x)=e^{-1/x^2}>0\) for \(x>0\). Hence no convergent power-series expansion about \(0\) represents \(f\) on any neighborhood of \(0\).