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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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$|x|^2$ is just $x\cdot x$, so for $|x|<1$ the function is just $e^{1/(x\cdot x-1)}$ which is nice and $C^\infty$ by virtue of being composed from such functions. So the only real work you have to do is at $|x|=1$. –  Henning Makholm Sep 16 '11 at 2:28
    
This makes sense, thank you very much. –  somesaymaths Sep 16 '11 at 3:56
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If $f$ is smooth then the radially symmetric function $f(r)$ is smooth except perhaps at $r=0$, so your situation reduces to looking at $e^{-1/(1-r^2)}$ when $r=1$. I think the proof at that point involves splitting into left and right limits in the derivative definition, showing what form the derivatives on the left come in and then using the fact that $P(y)e^{-y}\to0$ as $y\to\infty$ regardless of how fast the rational function $P$ grows. –  anon Sep 16 '11 at 9:08
    
My comments at math.stackexchange.com/questions/59842 for the one-variable case at the origin might be of help. –  Dave L. Renfro Sep 16 '11 at 16:36

1 Answer 1

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 $x=1$.

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