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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!

2 Answers 2

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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!2011-03-13
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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.

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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    @Jose Not sure this qualifies as a *detour* if one keeps in mind the general case of an interval $[a,b]$... but nevermind.2011-03-13