# How to prove that a function is class $C^{\infty}(\mathbb{R}).$

I would like to prove that this function (with the $a, b\in \mathbb{R}$ satisfying $a < b$),

$$f(x)=\begin{cases} \exp\left(-\frac{1}{x-a}+\frac{1}{x-b}\right), & \text{if },x\in (a,b)\,, \\ 0, & \text{if }x\notin (a,b)\text{} \end{cases}$$

is class $C^\infty (\mathbb{R}).$

But how?

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Of course, the problem is at $a$ and $b$. You can show by induction that the $d$-th derivative of $f$ on $(a,b)$ is of the form $R_d(x)\exp\left(-\frac 1{x-a}+\frac 1{x-b}\right)$, where $R_d$ is a rational fraction. –  Davide Giraudo Dec 31 '11 at 0:41
@DavideGiraudo The main obstacle I see here is that $R_d(x)$ has poles at $a$ and $b$. –  Alex Becker Dec 31 '11 at 1:01
Suppose $a=0$ and $b=1$, because the same reasoning applies in general. So we have $f(x)=\exp\left(\frac{1}{x(x-1)}\right)$ on $(0,1)$. Compositions of $C^\infty$ functions are $C^\infty$, so the only possible problems are at $0$ and $1$. Since $f(x)=f(1-x)$, it is enough to handle $0$.
Let's start by showing that $f'(0)=0$. The left-hand difference quotients at $0$ are $0$, so it is enough to show that $\lim\limits_{x\searrow 0}\frac{f(x)}{x}=0$. Note that $\frac{1}{x(x-1)}<-\frac{1}{x}$ for $x\in(0,1)$, so it is enough to show that $\lim\limits_{x\searrow 0}\frac{e^{-1/x}}{x}=0$. This is probably easiest to see by a change of variables, $t=1/x$, to yield $\lim\limits_{x\searrow 0}\frac{e^{-1/x}}{x}=\lim\limits_{t\to\infty}\frac{e^{-t}}{1/t}=\lim\limits_{t\to\infty}\frac{t}{e^t}=0$.
So far we know that $f'$ exists everywhere. Now as Davide Giraudo indicates in a comment, you can show by induction that in $(0,1)$, $f^{(k)}(x)=R_k(x)f(x)$ for some rational function $R_k$ having poles only at $0$ and $1$. I'll omit proof of this. Suppose that for some $k$ we know that $f^{(k)}(0)=0$. To show that $f^{(k+1)}(0)=0$, we need to show that $\lim\limits_{x\searrow 0}\frac{R_k(x)f(x)}{x}=0$. Note that $R_k(x)=\frac{g(x)}{x^{n}}$ for some integer $n$ and a function $g$ that is continuous at $0$. Again, since $f(x)<e^{-1/x}$, it suffices to show that $\frac{e^{-1/x}}{x^{n+1}}\to 0$ as $x\searrow 0$, and this is straightforward from the same change of variables $t=1/x$. (The base case $k=1$ wasn't really necessary to show since we are already given the base case $k=0$, but I though it might be helpful to start with the simplest case and make it more explicit.)
Again, since $f(x)=f(1-x)$, this also gives $f^{(k)}(1)=0$ for all $k$, and this shows (in sketch form) why $f$ is infinitely differentiable.