# Understanding Arnold's definition of “differentiable”

I'm looking at Arnold's Mathematical Methods of Classical Mechanics at the beginning of chapter 3 p 55 which defines when a functional is differentiable. Slightly paraphrasing and skipping a couple of details:

A functional $\Phi$ is said to be differentiable if $\Phi(\gamma+h)-\Phi(\gamma)=F+R$

where $F(\gamma,h)$ is linear in $h$ and $R\sim O(h^2)$ in the sense that for $|h|<\epsilon$ and $\left|\frac{dh}{dt}\right|<\epsilon$ then $|R|<C\epsilon^2$.

I've been telling myself to think of $\gamma$ as a curve, $h$ as a slight variation of the curve, and $F$ as the differential or "principal variation" of the functional, and $R$ the "error."

This is my first time seeing this definition of "differentiable" and also of $O(h^2)$, and I have a pair of questions about the latter.

1. In the 50 pages preceding, I can't seem to find out what $|\cdot |$ means here. Is $|h|$ total variation of the curve $h$, or something like that?

2. How can I make sense of the constraints $|h|<\epsilon$ and $\left|\frac{dh}{dt}\right|<\epsilon$ controlling $R$? The best I've come up with is "if $h$ does not go up and down too much and the speed doesn't go up and down too much, then $R$ will be under control." It might be helpful to have a prototypical example of a situation where $h$ slow but too wavy, and a situation where $h$ is not wavy but the speed varies wildly. Good heuristics are welcome too.

• $|h|$ is $\|h\|_{\infty}$ on $[t_0,t_1]$ for the next theorem. The definition is left ambiguous to adjust it to each case. see, for example, footnote 26. Think of it as a template of many definitions, parametrized by the spaces of curves chosen, the norms chosen, etc. – user577471 Jul 26 '18 at 20:35
• @HGLandcaster ....each case? well... I would think it is quite unusual if the definition of $O(h^2)$ is variable based on the problem presented. Are you sure it's not just the sup norm everywhere? – rschwieb Jul 26 '18 at 20:40
• It might be enough for all what follows, I didn't continue reading much more. – user577471 Jul 26 '18 at 20:43
• It is not unusual for $O(h^2)$ and differentials to depend on the norms. It just happens that in finite dimensions they are equivalent. – user577471 Jul 26 '18 at 20:47
• @HGLandcaster ok, a bit of a surprise, but good to know. Thanks! – rschwieb Jul 26 '18 at 22:59