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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?

Katlus
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2 Answers2

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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.

lcv
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anonymous
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    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 09 '13 at 01:18
  • What about vector functions as Rudin asked? – Sigur Dec 30 '13 at 14:38
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    @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. – Alex Vong Feb 01 '17 at 07:15
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    A proof that a differentiable function is uniformly differentiable iff it is differentiable with a continuous derivative can be found here. – Guillaume F. Jul 18 '18 at 00:28
  • The term does also appear in a book I'm reading, Interpolation and Approximation with Splines and Fractals, page 212. The addition here is that k-differentiability is imposed so I imagine that in fact $k$-uniform differentiability is equivalent to be in $C^k$ – user1868607 Mar 12 '19 at 13:24
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    The theorem that a differentiable function with a continuous derivative must be uniformly differentiable is correct as you stated it, for a function on $[a,b]$. But if the domain is not compact, then the correct theorem is that a differentiable function with a uniformly continuous derivative is uniformly differentiable (and vice versa). – Toby Bartels Feb 10 '20 at 15:36
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    Uniform differentiability is also important in constructive analysis, since it is better behaved there than mere differentiability, and the theorem that pointwise differentiability with a uniformly continuous derivative implies uniform differentiability doesn't hold constructively (although the converse does). – Toby Bartels Feb 10 '20 at 15:36
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    @TobyBartels Actually, I believe the implication (uniformly continuous derivatives implies uniformly differentiable) does hold true constructively. You may be assuming the absence of dependent choice – wlad Apr 07 '20 at 05:35
  • @ogogmad : Yes, I believe that it works with Dependent Choice; someone (possibly you) has just been posting to the constructive-news mailing list about this. So my comment really should say that uniform differentiability is important to some varieties of constructive mathematics. (Although even if it's equivalent to differentiability with a uniformly continuous derivative, it's still a key idea, so maybe just remove the second half of that comment.) – Toby Bartels Apr 07 '20 at 19:33
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As a matter of fact, the most well-known concepts of differentiability are always pointwise, like (where $f^{-1}(\mathbb R)$ is the domain of $f$)

$$\forall \varepsilon>0, \exists \delta>0, \forall x \in f^{-1}(\mathbb R): 0<|x-x_0|\le \delta \to \left|\frac{f(x)-f(x_0)}{x-x_0}-L\right| \le \varepsilon$$

or the Cauchy-convergence form

$$\forall x_0,\forall \varepsilon>0, \exists \delta>0, \forall x_1,x_2 : \left.\begin{aligned}0<|x_1-x_0|\le \delta\\ 0<|x_2-x_0|\le \delta\end{aligned}\right\}\to \left|\frac{f(x_2)-f(x_1)}{x_2-x_1}-g(x_0)\right| \le \varepsilon$$

However, just like the concept of pointwise continuity ($\forall x_0 \forall \varepsilon \exists \delta \forall x$) is different from uniform continuity ($\forall \varepsilon \exists \delta \forall x_0 \forall x$), the not-generally-used concept "uniform differentiability" can be defined by erasing $x_0$, as

$$\begin{aligned} \forall \varepsilon>0, \exists \delta > 0, \forall x_1, x_2 \in f^{-1}(\mathbb R), \\ (0<x_2-x_1\le\delta) \to \forall \xi\in (x_1, x_2)&: \\ \qquad \qquad\left|\frac{f(x_2)-f(x_1)}{x_2-x_1}-g(\xi)\right| &\le \varepsilon \end{aligned}$$

where I do not use the term $g(x_2)$, but any representers $g(\xi)$ instead, like what Riemann integral uses, to keep the symmetry of the formula.

A page of a Chinese textbook The Straightforward Calculus about uniform differentiability

In fact, it is really compatible with the Lagrange mean value theorem associated with the differentiable-preserving property of close subintervals, which can be written as

$$\forall [u,v] \subset f^{-1}(\mathbb R), \exists \xi \in [u,v]:\frac{f(v)-f(u)}{v-u}=g(\xi)$$

where a quantifier $\forall \xi$ is changed into $\exists \xi$.

Roseleaves Wu
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