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I already know that the function $$ f(x) = \begin{cases} \exp(- \frac{1}{x^2}), \quad x > 0 \\ 0 , \quad x \leq 0 \end{cases} $$

is infinitely differentiable throughout $\mathbb R$. The only real problem, of course, lies in showing that $f^{(k)} (0) = 0$ for any positive integer $k$. What I have not been able to deduce is that

$$ \phi(x) = \begin{cases} \exp(- \frac{1}{1 - x^2}), \quad |x| < 1 \\ 0 , \quad |x| \geq 1 \end{cases} $$

is also infinitely differentiable throughout $\mathbb R$, using the previous function. The problem now is finding out what happens at $x = 1,-1$. Does the substitution $\zeta ^2 = 1 - x^2$ work, or is there another way to prove this?

Thank you all!

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up vote 5 down vote accepted

This is much easier if you first show that

$g(x) = \begin{cases} e^{-\frac{1}{x}} & \text{for } x \gt 0 \\\\ 0 & \text{for } x \leq 0 \end{cases}$

is smooth. Then it's easy to see that $f(x) = g(x^2)$ and $\phi(x) = g(1-x^2)$, i.e. $f$ and $\phi$ are compositions of smooth functions.

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That's much better indeed, thank you! – Jose L. Lykón Mar 13 '11 at 1:59

Hint 1: For every positive $c$ and every real $a$, the function $f_{a,c}$ defined by $f_{a,c}(x)=\exp(-c/(x-a))$ if $x>a$ and $f_{a,c}(x)=0$ if $x\le a$ is infinitely differentiable on the real line.

Hint 2: use $f_{a,c}$ to build a function infinitely differentiable on the real line, zero for $x\ge a$ and positive for $x<a$.

Hint 3 : use the decomposition of $1/(1-x^2)$ as a linear combination of $1/(1\pm x)$ to write $\phi$ as a product of two infinitely differentiable functions.

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This is a good detour as well, thanks a bunch. – Jose L. Lykón Mar 13 '11 at 2:02
@Jose Not sure this qualifies as a detour if one keeps in mind the general case of an interval $[a,b]$... but nevermind. – Did Mar 13 '11 at 8:22

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