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Let $ f(x)= \begin{cases} \exp\left(\frac{-1}{1-|x|^2}\right), &\text{ if } |x| < 1, \\ 0, &\text{ if } |x|\geq 1. \end{cases} $ Prove that $f$ is infinitely differentiable everywhere. ($x$ belongs to $\mathbb{R}^n$ for fixed $n$.)

Well, this is obvious for $|x|>1$ and easy enough for the first derivative at $|x|=1$, but I can't seem to use the definition of the Gateaux derivative to show it for $|x|<1$. Any advice would be appreciated.

(This is not homework.)

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    My comments at http://math.stackexchange.com/questions/59842 for the one-variable case at the origin might be of help.2011-09-16

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We can show by induction that $\partial_{\alpha}f(x)=\begin{cases} \frac{P_{\alpha}(x)}{(1-|x|^2)^{2|\alpha|}}\exp\left(\frac 1{|x|^2-1}\right)&\mbox{ if }|x|<1,\\\ 0&\mbox{ otherwise}, \end{cases}$ where $\alpha\in\mathbb N^n$ and $P_{\alpha}$ is a polynomial. It's true for $\alpha=0$, and if $\alpha=e_k$ and $|x|<1$, $\partial_{e_k}f(x)=-\exp\left(\frac 1{|x|^2-1}\right)\frac{2x_k}{(|x|^2-1)^2},$ which shows that $f$ is also differentiable at $|x|=1$ and $P_{e_k}(x)=-2x_k$. If we assume that the property is true for $|\alpha|\leq p$ and $|\alpha|=p+1$ then let $k$ such that $\alpha_k\neq 0$, and put \alpha'=\alpha-e_k. Then |\alpha'|=p and for $|x|<1$ we have \begin{align*} \partial_{\alpha}P(x)&=\frac{\partial_{e_k}P_{\alpha'}(x)}{(1-|x|^2)^{2|\alpha'|}}\exp\left(\frac 1{|x|^2-1}\right)+\frac{P_{\alpha'}(x)(-2|\alpha'|-1)2x_k}{(1-|x|^2)^{2|\alpha'|+1}}\exp\left(\frac 1{|x|^2-1}\right)\\ &+\exp\left(\frac 1{|x|^2-1}\right)\frac{P_{\alpha'}(x)}{(1-|x|^2)^{2|\alpha'|}}\frac{2x_k}{(1-|x|^2)^2}\\ &=\exp\left(\frac 1{|x|^2-1}\right)\frac 1{(1-|x|^2)^{2|\alpha|}}\Big(\partial_{e_k}P_{\alpha'}(x)(1-|x|^2)^2\\ &- 2(1-|x|^2)P_{\alpha'}(x)x_k+2x_kP_{\alpha'}(x) \Big)\\ &=\exp\left(\frac 1{|x|^2-1}\right)\frac 1{(1-|x|^2)^{2|\alpha|}}\left(\partial_{e_k}P_{\alpha'}(x)(1-|x|^2)^2+2|x|^2x_kP_{\alpha'}(x)\right). \end{align*} So we got the induction formula P_{\alpha'+e_k}(x)=\partial_{e_k}P_{\alpha'}(x)(1-|x|^2)^2+2|x|^2x_kP_{\alpha'}(x), and $\partial_{\alpha}f(x)=0$ if $x=1$.