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In a sense, I did just ask this question. However, since the question is about how to write a beautiful looking proof and my proof is entirely rewritten, it seems like it should be a new question. What I'm looking for here are (aside from any actual mathematical problems in the proof of course) is advice on how to better structure what I'm writing or to better communicate the ideas. Anything from overall structure, word choice, what letters I give my variables, anything at all. Anyhow, the question from Spivak:

Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by

$$f(x)=\begin{cases} e^{-x^{-2}}& x\ne 0\\ 0 & x=0\\ \end{cases}$$

Show that $f$ is a $C^\infty$ function, and $f^{(i)}(0)=0$ for all $i$.

Now, here is my proof.

First, we show that $f\in C^\infty$ for $x\ne 0$ by induction.

Let $-g(x)=\frac{1}{x^2}$. Then we can write $f$ as $f(x)=e^{g(x)}$. Let $P={p:\mathbb{R}\rightarrow\mathbb{R} : p(x)=\sum \frac{a_n}{x^n},a_n\in\mathbb{R}}$. Now we can show by induction that $f^{(n)}(x)=e^{g(x)}p_n(x)$ for some $p_n \in P$.

Let $p_0=\frac{1}{x^0}=1$. Then $p\in P$ and $f(x)=e^{g(x)}p_0(x)$.

Assume $p_n\in P$ such that $p_n=\sum_{i=0}^\infty\frac{a_i}{x^i}$

\begin{align*} (e^{g(x)}p_n(x)) &= e^{g(x)}g'(x) p_n(x) + e^{g(x)} p_n'(x)\\ &= e^{g(x)}(g'(x) p_n(x) + p_n'(x))\\ &= e^{g(x)}(\frac{2}{x^3} p_n(x) + p_n'(x))\\ &= e^{g(x)}(\frac{2}{x^3} \sum \frac{a_i}{x^i} + \sum \frac{-ia_i}{x^{i+1}})\\ &= e^{g(x)}(\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}})\\ \end{align*}

Since $\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}}\in P$, it follows that $p_n\in P \Rightarrow p_{n+1}\in P$. Thus by induction, $f^{(n)}=e^{g(x)}p_n(x)$ where $p_n\in P$ and $f\in C^\infty$ for $x\ne 0$.

Now we show by induction that $f^{(n)}(0)=0$ for all $n$.

First, we prove that for all $n \ge -1$,

$$m(n)=\lim_{h\rightarrow 0} \frac{1}{h^n e^{\frac{1}{h^2}}}=0$$

For $n=-1$ we have $\lim_{h\rightarrow 0}\frac{1}{h^{-1} e^{\frac{1}{h^2}}}=\lim_{h\rightarrow 0}\frac{h}{e^{\frac{1}{h^2}}}=0$

For $n=0$ we have $\lim_{h\rightarrow 0}\frac{1}{h^{0} e^{\frac{1}{h^2}}}=\lim_{h\rightarrow 0}\frac{1}{e^{\frac{1}{h^2}}}=0$

Now assume that $m(n-1)=0$ and $m(n-2)=0$.

\begin{align*} m(n) &= \lim_{h\rightarrow 0} \frac{1}{h^n e^{\frac{1}{h^2}}}\\ &= \lim_{h\rightarrow 0} \frac{h^{-n}}{e^{\frac{1}{h^2}}}\\ &= \lim_{h\rightarrow 0} \frac{-nh^{-n-1}}{e^{\frac{1}{h^2}} (-2 h^{-3})}\tag{L'H\^{o}pital's rule}\\ &= \frac{n}{2} \lim_{h\rightarrow 0} \frac{h^{-n-1+3}}{e^{\frac{1}{h^2}}}\\ &= \frac{n}{2} \lim_{h\rightarrow 0} \frac{1}{h^{n-2}e^{\frac{1}{h^2}}}\\ &= \frac{n}{2} m(n-2)=0\\ \end{align*}

Thus by induction, $m(n)=0$.


$$f^{(n)}(x)=e^{g(x)}(\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}})=\sum 2a_i m(i+3)+\sum-im(i+1)$$

it follows that $f^{(n)}=0$ for all $n$.

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1 Answer 1

I think the first part of your proof is a little clumsy. For $x\neq 0$ you don't need to use induction: note that what you have for $x\neq 0$ is a composite of $C^\infty$ functions.

When you write $(e^{g(x)}p_n(x))'=e^{g(x)}g′(x)p_n(x)+e^{g(x)}p_n′(x)$ you are utilizing the chain rule. Thus, you are assuming that the composite $(e^{g(x)}p_n(x))$ is differentiable. Of course this function is differentiable, but this is what you want to prove: that $f^{(n)}$ is differentiable for $x\neq 0$. The point is that you are proving even more: you're giving the general form of the derivatives.

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At this stage in the textbook, we haven't proved anything about whether composites of $C^\infty$ functions are differentiable, so I dont't have that option. As to assuming whether the composite is infinitely differentiable, I thought I showed that as long as $p_n$ was a polynomial of $x^{-1}$ that it was differentiable explicitly by showing the form of the derivative. How would you more clearly show that result? –  Unkz Sep 17 '10 at 1:56
You don't need it for $x \neq 0$, but I guess you do need it for the $x=0$ part, which is the harder part of the question. –  Aryabhata Sep 17 '10 at 6:30
I can't think about a more clear solution, but look that all you need is the chain rule: "the composite of differentiable functions is differentiable". Then, "applying it n times", you can prove the assertion without making reference to the explicit form of the function f. –  Ronaldo Sep 21 '10 at 0:09

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