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Check for what $n\in\mathbb{N}\cup \left\{+\infty\right\}$ $f\in C^n(\mathbb{R})$ (that is $f$ is $n$ times differentiable, I wasn't sure that this designation is common), if:

a) $f(x)=|x|^m, \ m\in\mathbb{R}$

b) $f(x)= \begin{cases} e^{-1/x}, \ x>0 \\ 0, \ x\le 0 \end{cases} $

I wasn't given in school any convenient theorem to check such things but I really want to learn it. How can I check it?

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1 Answer 1

I give some hints, and add more details if needed.

a) The problem is at $0$, since if $x\neq 0$ $f$ is infinitely derivable in a neighborhood of $x$. We try to compute the first derivative at $0$. We compute $\lim_{h\to 0}\frac{f(h)-f(0)}h=\lim_{h\to 0}\frac{|h|^m}h$. We observe that the limit doens't exist if $m-1\leq 0$. What is the limit when $m>1$? I guess this should help you to see the pattern.

b) Show by induction that if $x>0$ and $d$ is an integer then $f^{(d)}(x)=P_d\left(\frac 1x\right)e^{1/x}$, where $P_d$ is a polynomial.

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why in a) limit doesn't exist if $m\le 1$? why and why only if? –  xan Apr 12 '12 at 17:55
If $m-1< 0$, for $h>0$ $\frac{|h|^m}h=h^{m-1}$ converges to $+\infty$, and to $-\infty$ if $h\to 0^-$. If $m=1$, the right-limit is $1$ and the left-limit is $-1$. –  Davide Giraudo Apr 12 '12 at 18:01
ok, the way I see it: $f\in C^{0}(\mathbb{R})$. Is it correct? –  xan Apr 12 '12 at 18:11
It depend on $m$, for $m<0$ it's not well defined at $0$, and cannot be extended continuously. –  Davide Giraudo Apr 12 '12 at 18:17

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