# Infinitely differentiable functions: how to prove that $e^\frac{1}{x^2-1}$ has derivative of any order?

Let $f:\mathbb{R}\to\mathbb{R}$ be a function given by

$$f(x)=\left\{\begin{matrix} \exp\left(\frac{1}{x^2-1}\right) & \text{if }|x|<1\\ 0 & \text{if }|x|\geq 1 \end{matrix}\right.$$

I would like to prove that $f\in C^\infty$, that is, $f\in C^k$ for all $k\in \mathbb{N}$. I think that it can be done by induction on $k$. If $|x|>1$, the problem is trivial. On other points, the base case is the simplest and the only that I'm be able to do. Can someone help me?

Thanks.

• Hint: we only really have to worry about a very small number of distinct points. Except at these points, show that every derivative of $f$ is a rational function multiplied by $f$ (using induction), and use that to verify differentiability by definition for those problematic points only. – Jonathan Y. Aug 26 '13 at 1:46
• @JonathanY. Is enough know that "every derivative of $f$ is a rational function multiplied by $f$" to verify differentiability at $-1$ and $1$? Or we need something more about the derivatives of $f$? – Pedro Aug 26 '13 at 1:53
• Further hint: exponentials and polynomials are in different growth classes. – anon Aug 26 '13 at 2:35
• I would do the derivative at $|x|=1$ manually (limit definition), then you can write $f'(x)$ again as a two-part function, and then do it again (manually). What you cannot do is take the derivative for $|x|<1$ and take limits of that expression (because there are examples $x^2 \sin (1/x)$ where this gives you the wrong answer, when $f$ is not continuously differentiable) – Evan Aug 26 '13 at 2:53
• @Pedro You should consider accepting one of the answers below if they satisfy you. I think Peter's answer is fabulous. – Vishal Gupta Aug 31 '13 at 2:05

Do it for $$f(x)=\begin{cases}\exp\left(-\frac 1 x\right)&x>0\\ 0&x\leq 0\end{cases}$$

Note that everywhere but in the origin, $$f$$ is infinitely differentiable. Moreover, for $$x>0$$

\eqalign{ f'\left( x \right) &= \frac{1}{{{x^2}}}f\left( x \right) \cr f''\left( x \right) &= \left( {\frac{1}{{{x^4}}} - \frac{2}{{{x^3}}}} \right)f\left( x \right) \cr f'''\left( x \right)&= \left( {\frac{1}{{{x^6}}} - \frac{6}{{{x^5}}} + \frac{6}{{{x^4}}}} \right)f\left( x \right)\cr &\&c \cr}

You can thus prove inductively that for $$x>0$$, $$f^{(k)}(x)=P_{2k}(x^{-1})f(x)$$ where $$P_{2k}$$ is a polynomial of degree $$2k$$.

As $$x\to 0^+$$ this amounts to looking at $$\lim_{x\to +\infty}P(x)\exp(-x)=0$$ for any polynomial $$P$$.

So, for any $$k$$, the limit as $$x\to 0$$ of the derivative is $$0$$. Now we use a slightly underrated theorem

Theorem (Spivak) Suppose $$f$$ is continuous at $$x=a$$, that $$f'(x)$$ exists for all $$x$$ in a neighborhood of $$a$$. Suppose moreover that $$\lim_{x\to a}f'(x)$$ exists. Then $$f'(a)$$ exists and $$f'(a)=\lim_{x\to a}f'(x)$$

Proof By definition, $$f'(a)=\lim_{h\to 0 }\frac{f(a+h)-f(a)}h$$

Consider $$h>0$$. For $$h$$ sufficiently small, $$f$$ will be continuous over $$[a,a+h]$$, and differentiable over $$(a,a+h)$$. Thus, by Lagrange, we can find $$a<\alpha_h such that $$\frac{f(a+h)-f(a)}h=f'(\alpha_h)$$

As $$h\to 0^+$$; $$\alpha_h\to a$$, and since the limit exists, $$f'(a)^+=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}h=\lim_{h\to 0^+}f'(\alpha_h)=\lim_{x\to a}f'(x)$$ The case $$h<0$$ is analogous. $$\blacktriangle$$.

The above lets you conclude that indeed $$f^{(k)}(0)=0$$ for all $$k$$, whence $$f$$ is $$C^k$$ for any $$k$$. Now, note your function is $$g(x)=f(1-x^2)$$

• Is this a typo at "this amounts to looking at..." Did you mean to say $exp(-\frac{1}{x})$ in the expression? – Tyler Hilton Sep 23 '13 at 4:51

Taking derivatives you happen to have $p(x)e^{\frac{1}{x^2 - 1}}$, where $p(x)$ is a rational polynomial function that blows up at $x = \pm 1$. What you need know is to prove the continuity of these derivatives. This, as anon suggests, turns out to be a very simple problem if you use the fact that the exponential function is way faster than every rational polynomial function. Thus, $\lim_{x \to \pm 1}p(x)e^{\frac{1}{x^2 - 1}} = 0$ $\forall p$. This means that even for $x = \pm 1$, the critical cases where the two branches need to match, every derivative of the function is continuous (in other words $\lim_{x \to \pm 1^-} f^k(x) = \lim_{x \to \pm 1^+} f^k(x)$, and this is exactly the continuity condition you were looking for).

I hope it helps :D

• You will need to calculate the limit manually at $\pm 1$. – Vishal Gupta Aug 31 '13 at 2:04