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!


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.

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  • $\begingroup$ 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

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