Reading this, this and this Q&A's I've understood that a uniformly continuous differentiable function on $\mathbb R$ need not have a bounded derivative. There have been some attempts at giving some intuition for uniform continuity, but I still haven't found any precise formulation in terms of already familiar things like a limit or a derivative.

So, can uniform continuity of a differentiable function be formulated using the old familiar things like a limit or derivative or anything of the similarly "basic" type — instead of the usual $\varepsilon-\delta$ definition?

E.g. for continuity of a function $f:\mathbb R\to\mathbb R$ we can use the $\varepsilon-\delta$ definition

$$\forall x\in\mathbb R\;\forall\varepsilon>0\,\exists\delta(\varepsilon)>0:\forall x'\in\mathbb R:(|x'-x|<\delta\implies |f(x')-f(x)|<\varepsilon),$$

which can be rewritten in terms of a limit:

$$\forall x\in\mathbb R\;\lim_{x'\to x}f(x')=f(x).$$

Can similar thing be done for uniform continuity?

  • $\begingroup$ And how would you define a limit or derivative without $\varepsilon - \delta$? $\endgroup$ – Sten Feb 5 '15 at 15:28
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    $\begingroup$ @Sten consider them already defined. I.e. I want to, using these black boxes, formulate a criterion of uniform continuity. $\endgroup$ – Ruslan Feb 5 '15 at 15:39
  • $\begingroup$ OK that makes sense. $\endgroup$ – Sten Feb 5 '15 at 15:45

Uniform continuity of a function $f\colon\mathbb R\to\mathbb R$ is defined as follows:

$$\forall\varepsilon>0\,\exists\delta>0:\forall x,x'\in\mathbb R:\left(|x-x'|<\delta\implies|f(x)-f(x')|<\varepsilon\right).\tag1$$

Denoting $x-x'=2a$, we can equivalently say:

$$\forall\varepsilon>0\,\exists\delta>0:\forall x\in\mathbb R\,\forall a\in(-\delta,\delta):|f(x)-f(x+2a)|<\varepsilon.\tag2$$ This is the same as $$\forall\varepsilon>0\,\exists\delta>0:\forall a\in(-\delta,\delta):\sup_{x\in\mathbb R}|f(x)-f(x+2a)|<\varepsilon,\tag3$$ or, collapsing further, $$\underset{a\to0\vphantom{\mathbb R}}{\lim\vphantom{p}}\sup_{x\in\mathbb R}|f(x)-f(x+a)|=0.\tag4$$

Now $(4)$ is the final answer to the OP.


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