# What is the precise definition of 'uniformly differentiable'?

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?

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Where did you see this term? Can you provide a context? – Jesse Madnick Nov 19 '12 at 2:25
@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.. – Katlus Nov 19 '12 at 2:36

Let $f:[a,b] \to \mathbb{R}$. Differetiability 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 $\lim_{h \to 0} |\frac{f(x+h) - f(x) - hf'(x)}{h}| = 0$. 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 $|\frac{f(x+h) - f(x) - hf'(x)}{h}| < \epsilon$. 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.
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. – Betty Mock Sep 9 '13 at 1:18