What is this derivative called: $$ D_\gamma f = \lim_{dt\to 0} \frac{f(\gamma(dt))-f(\gamma(0))}{||\gamma(dt)-\gamma(0)||} $$ It is not the same as the directional derivative or the Gateux derivative on wikipedia. Perhaps it is not as general since it uses a vector norm. (or perhaps it is but I don't see it)

  • $\begingroup$ Is there some reason you are interested in this derivative? That limit will very rarely exist... $\endgroup$ – Eric Wofsey Jun 29 '18 at 6:37
  • 1
    $\begingroup$ Are there hypothesis on $\gamma$ ? $\endgroup$ – nicomezi Jun 29 '18 at 6:38
  • $\begingroup$ (Notice, for instance, that if $\gamma(t)=t$, then $D_\gamma f$ will not exist for any ordinarily differentiable function $f:\mathbb{R}\to\mathbb{R}$ such that $f'(0)\neq 0$.) $\endgroup$ – Eric Wofsey Jun 29 '18 at 6:40
  • $\begingroup$ @EricWofsey: Can you explain? For me, $\gamma(t) = t$ results in $D_\gamma f = \lim_{h\searrow0}\frac{f(h) - f(0)}{h - 0} = f'(0)$. $\endgroup$ – gerw Jun 29 '18 at 6:50
  • $\begingroup$ Notice the norm at the denominator. @gerw $\endgroup$ – nicomezi Jun 29 '18 at 6:52

In the case that $\gamma$ is differentiable at $0$, this is very similar to the definition of the Hadamard derivative. There, the existence of $$\frac{f(\bar x + t_n \, h_n) - f(\bar x)}{t_n}$$ is required for all sequences $t_n \searrow 0$ and $h_n \to h$, see https://en.wikipedia.org/wiki/Hadamard_derivative.


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.