This is Spivak's Calculus on Manifolds ex. 2-25, he says

Define $f:\mathbb{R}\to \mathbb{R}$ by $f(x) = \left\lbrace \begin{array}{l} e^{-x^{-2}} &\text{ if } x \neq 0\\ 0 &\text{ otherwise } \\ \end{array} \right.$. Show that $f$ is $C^{\infty}$ anf $f^{(i)}(0)=0\; \forall i$.

I haven't found a way to get a general solution. I only managed to prove:

$(1)$ $f$ is continuous at $0$.

This follows from $f(0)=0$ and $\lim_{x\to0} f(x)=\lim_{x\to 0}e^{-x^{-2}}=\lim_{x\to 0}\displaystyle\frac{1}{e^{1/x^2}}=0$ since $\displaystyle\frac{1}{x^2}\to+\infty$ if $x\to 0$ then $e^{1/x^2}\to+\infty$ if $x\to 0$, which means that $\displaystyle\frac{1}{e^{1/x^2}}\to 0$ if $x\to 0$ .


I got this by calculating the limit $f'(0)=\lim_{h\to 0}\displaystyle\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\displaystyle\frac{f(h)}{h}=\lim_{h\to 0}\displaystyle\frac{e^{-h^{-2}}}{h}=\lim_{h\to 0}\displaystyle\frac{1/h}{e^{h^{-2}}}$.

The last one follows from L'Hôpital: $\lim_{h\to 0}\displaystyle\frac{\frac{d}{dh}(1/h)}{\frac{d}{dh}(e^{h^{-2}})}=\lim_{h\to 0}\displaystyle\frac{(-1/h^2)}{(e^{h^{-2}})(-2h^{-3})}=\lim_{h\to 0}\displaystyle\frac{h}{2e^{h^{-2}}}=\displaystyle\frac{1}{2}\left(\lim_{h\to 0}\displaystyle\frac{1}{e^{h^{-2}}}\right)\left(\lim_{h\to 0} h\right) = 0$.

$(3)$ $f''(0)=0$

I found that $\displaystyle\frac{df}{dx} = \left\lbrace \begin{array}{l} 2x^{-3}e^{-x^{-2}} &\text{ if } x \neq 0\\ 0 &\text{ otherwise } \\ \end{array} \right.$, then I can get $f''(0)$ computing the limit $\lim_{h\to 0}\displaystyle\frac{f'(0+h)-f'(0)}{h} = \lim_{h\to 0}\displaystyle\frac{f'(h)}{h} = \lim_{h\to 0}\displaystyle\frac{2e^{-h^{-2}}}{h^4}= 2\lim_{h\to 0}\displaystyle\frac{1/h^4}{e^{h^{-2}}}=-2\lim_{h\to 0}\displaystyle\frac{-4h^{-5}}{-2h^{-3}e^{h^{-2}}}\\=4\lim_{h\to 0}\displaystyle\frac{1}{h^2e^{h^{-2}}}=4\lim_{h\to 0}\displaystyle\frac{h^{-2}}{e^{h^{-2}}})=-4\lim_{h\to 0}\displaystyle\frac{-2h^{-3}}{e^{h^{-2}}(-2h^{-3})}=-4\lim_{h\to 0}\displaystyle\frac{1}{e^{h^{-2}}}=0$, after using L'Hôpital twice.

But still I don't have a clue to prove that $f^{(i)}(0)=0$ for every $i$. One of the ideas I managed was to find a general form for $f^{(i)}$ but after I got $f'$ doesn't seem possible to do so considering that I would have to apply the product rule each time.

  • 3
    $\begingroup$ You can use a semi-explicit form. For $x\neq 0$, you can write $$f^{(n)}(x) = P_n\left(\frac{1}{x}\right)e^{-x^{-2}}$$ with a polynomial $P_n$. Assuming as induction hypothesis that $f^{(n)}(0) = 0$, you get $$f^{(n+1)}(0) = \lim_{x\to 0} \frac{1}{x}P_n\left(\frac{1}{x}\right)e^{-x^{-2}}.$$ $\endgroup$ – Daniel Fischer Jul 15 '14 at 15:49
  • $\begingroup$ math.stackexchange.com/questions/476195/… $\endgroup$ – lemon Jul 15 '14 at 15:49
  • $\begingroup$ You can conclude Daniel Fischer's comment by showing that $\lim_{x\rightarrow 0}\left(\frac{a}{x^n}\right)e^{-x^2}=0$ for any $n$ $\endgroup$ – Robert Wolfe Jul 15 '14 at 15:51

Prove by induction that for $x\neq0:\ $ $f^{(i)}(x)=R_i(x)e^{-x^{-2}}$, where $R_i$ is a certain rational function. Then show that $\lim_{x\to0}f^{(i)}(x)=0$.


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