16
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jonathan Y. Aug 26 '13 at 1:46
  • $\begingroup$ @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 $f$? $\endgroup$ – Pedro Aug 26 '13 at 1:53
  • $\begingroup$ Further hint: exponentials and polynomials are in different growth classes. $\endgroup$ – anon Aug 26 '13 at 2:35
  • $\begingroup$ 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) $\endgroup$ – Evan Aug 26 '13 at 2:53
  • $\begingroup$ @Pedro You should consider accepting one of the answers below if they satisfy you. I think Peter's answer is fabulous. $\endgroup$ – Vishal Gupta Aug 31 '13 at 2:05
15
$\begingroup$

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 $a<\alpha_h<a+h$ 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 $$g(x)=f(1-x^2)$$

$\endgroup$
  • 2
    $\begingroup$ Is this a typo at "this amounts to looking at..." Did you mean to say $exp(-\frac{1}{x})$ in the expression? $\endgroup$ – Tyler Hilton Sep 23 '13 at 4:51
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.