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Can anyone explicitly verify that the function $ f(x) = \left\{ \begin{array}{lr} 0 & : x = 0 \\ e^{-1/x^{2}} & : x \neq 0 \end{array} \right. $

is infinitely differentiable on all of $\mathbb{R}$ and that $f^{(k)}(0) = 0$ for every $k$?

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I'm sure lots of people can ... –  Neal Mar 13 '12 at 23:57
    
This is related math.stackexchange.com/questions/47948/…. You already know that that function is smooth away from zero. Now you just have to prove inductively that you have the same property in zero. The consequence of the mean value theorem showed in the link is enough to do that. –  Beni Bogosel Mar 13 '12 at 23:57

1 Answer 1

For $x\neq 0$ you get: $$\begin{split} f^\prime (x) &= \frac{2}{x^3}\ f(x)\\ f^{\prime \prime} (x) &= 2\left( \frac{2}{x^6} - \frac{3}{x^4}\right)\ f(x)\\ f^{\prime \prime \prime} (x) &= 4\left( \frac{2}{x^9} - \frac{9}{x^7} +\frac{6}{x^5} \right)\ f(x) \end{split}$$ In the above equalities you can see a path, i.e.: $$\tag{1} f^{(n)} (x) = P_{3n}\left( \frac{1}{x}\right)\ f(x)$$ where $P_{3n}(t)$ is a polynomial of degree $3n$ in $t$.

Formula (1) can be proved by induction. You have three base case, hence you have only to prove the inductive step. So, assume (1) holds for $n$ and evaluate: $$\begin{split} f^{(n+1)} (x) &= \left( P_{3n}\left( \frac{1}{x}\right)\ f(x) \right)^\prime\\ &= -\frac{1}{x^2}\ \dot{P}_{3n} \left( \frac{1}{x}\right)\ f(x) + P_{3n} \left( \frac{1}{x}\right)\ f^\prime (x)\\ &= \left[ -\frac{1}{x^2}\ \dot{P}_{3n} \left( \frac{1}{x}\right) +\frac{2}{x^3}\ P_{3n} \left( \frac{1}{x}\right)\right]\ f(x)\\ &= \left[ -t^2\ \dot{P}_{3n}( t) +2t^3\ P_{3n}( t)\right]_{t=1/x}\ f(x) \end{split}$$ where the dot means derivative w.r.t. the dummy variable $t$; now the function $-t^2\ \dot{P}_{3n}( t) +2t^3\ P_{3n}( t)$ is a polynomial in $t$ of degree $3n+3=3(n+1)$, therefore: $$f^{(n+1)}(x) = P_{3(n+1)} \left( \frac{1}{x}\right)\ f(x)$$ as you wanted.

Formula (1) proves that $f\in C^\infty (\mathbb{R}\setminus \{0\})$.

Now, for each fixed $n$, you have: $$\lim_{x\to 0} f^{(n)}(x) = \lim_{x\to 0} P_{3n}\left( \frac{1}{x}\right)\ f(x) =0$$ for $f\in \text{o}(1/x^{3n})$ as $x\to 0$. Therefore, by an elementary consequence of Lagrange theorem, any derivative of your functions is differentiable also in $0$. Thus $f\in C^\infty (\mathbb{R})$.

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