4
$\begingroup$

Since I'm not familiar with manifold concept, let's restrict ourselves to functions with real domain.

Let $A\subset \mathbb{R}$ and $f:A\rightarrow \mathbb{R}^k$.

What is '$f$ is uniformly differentiable on $A$' referring to?

  • 0
    Where did you see this term? Can you provide a context?2012-11-19
  • 0
    @Jesse Rudin PMA p.115, since i thought that it could be defined even if $f'$ is not continuous, i posted this. If this term is not generally used, let me know..2012-11-19
  • 0
    It is also present in Estep "Practical Analysis in One Variable" Chapter 32.2018-05-30

1 Answers 1

12

Let $f:[a,b] \to \mathbb{R}$. Differentiability means that the limit $$ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ (with the obvious modifications for $x = a,b$) exists, in which case we denote the limit as $f'(x)$. This definition can be rephrased as saying that there is a function $f':[a,b] \to \mathbb{R}$ which satisfies $$ \lim_{h \to 0} \left |\frac{f(x+h) - f(x) - hf'(x)}{h} \right| = 0. $$
The uniformity here means that we can approximate uniformly in $x$.

More precisely, given an $\epsilon > 0$ we may find a $\delta > 0$ so that whenever $0 < |h| < \delta$, then $$ \left|\frac{f(x+h) - f(x) - hf'(x)}{h}\right| < \epsilon. $$ It's easy to show that a differentible function is uniformly differentiable if and only if it's differentiable with a continuous derivative. I believe this is what Rudin has you prove.

Outside of Rudin's book, I don't know if I've ever heard the term "uniformly differentiable" used exactly, and a quick Google search seems to suggest that the term is primarily connected with that problem.

  • 3
    Just to clarify -- whenever a $\delta$ is chosen, people never say that $\delta$ depends on x. The "uniformity" of whatever kind (continuity, convergence, etc) always means that $\delta$ does not depend on x.2013-09-09
  • 0
    What about vector functions as Rudin asked?2013-12-30
  • 0
    @anonymous After 5 years, this page becomes the 1st suggestion from a search engine. Besides, it appears that Bartle's Introduction to Real Analysis also introduces the term "uniformly differentiable" in one its exercises.2017-02-01
  • 0
    A proof that a differentiable function is uniformly differentiable iff it is differentiable with a continuous derivative can be [found here](https://math.stackexchange.com/questions/2556707/a-function-is-uniformly-differentiable-if-its-derivative-is-uniformly-continuous/).2018-07-18