3
$\begingroup$

I would like to know how to calculate the derivative of a non-analytic smooth function?

Suppose $f:\mathbb R\rightarrow \mathbb R$ is in $\mathcal C^\infty\backslash \mathcal C^\omega$ and in particular has no Taylor series expansion at $x$. The right (left) derivative at point $x$ is:

\begin{equation} f'(x)=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\Big(f(x+\epsilon)-f(x)\Big) \end{equation}

The limit, due to smoothness, exists even though we can't expand, but there is no way to find the limit by means of a calculation or is there?

thanks a lot!

$\endgroup$
1
$\begingroup$

It's not clear to me what your asking but maybe this will help. The classic example of nonanalytic smooth function is $f : \mathbb{R} \to \mathbb{R}$ where $f(x) = e^{\frac{-1}{x}}$ on $(0, \infty)$ and $f(x) =0$ on $(-\infty, 0]$. In particular the function isn't analytic at $0$ but we are still able to calculate derivatives.

$\endgroup$
  • $\begingroup$ Thanks! You're right. It's not quite clear. My problem is this: Suppose I have two vector fields $X$ and $Y$ and the Lie derivative $\mathcal L_{X}Y=\frac{d}{dt}[(\Phi^X_{-t})_*Y_{\Phi^X_t(x)}]$, then the class of $\Phi^X$ is determined by $X$, but suppose it's smooth but not analytic. The standard result that $\mathcal L_XY=[X,Y]$ somehow rests on the idea of Taylor expanding $\Phi^X_t$, which in the nonanalytic case doesnt seem to work... $\endgroup$ – Marlo Apr 13 '15 at 7:44
  • $\begingroup$ But in your example what would be the right derivative of $f$ at zero? And how is it defined? $\endgroup$ – Marlo Jun 18 '16 at 19:48

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.