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What does it mean that a function is uniformly continuous? (Geometrically)

I have this definition but i dont understand the definition geometrically. Can someone help me to understand this?

Let $D\subseteq\mathbb{R}$ and $f:D\rightarrow\mathbb{R}$, $f$ is uniformly continuous if $\forall\epsilon>0$ exists $\delta>0$ such that $\forall x,y\in D$ and $\text{|x-y|<}\delta$ then $|f(x)-f(y)|<\epsilon$

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  • It means, in some way, that the oscillation of the function is bounded. Check this. – Masacroso Mar 01 '17 at 23:46
  • @Masacroso I don't think what you say is always the correct intuition. For example, what is the "oscillation" of $f(x) = x^2$ for $x \in \mathbb{R}$? – Tom Mar 01 '17 at 23:48
  • Bounded first derivative $\implies$ uniformly continuous. These conditions aren't the same though, as this is uniformly continuous but not bounded first derivative (additionally, I don't believe that uniform continuity implies $f\in C^1$, but I don't explicitly know that it doesn't). – Mark Schultz-Wu Mar 01 '17 at 23:48
  • @Tom this is the reason what I said "in some way". The oscillation of $f(x)=x^2$, for a distance $\epsilon>0$, is clearly not bounded. – Masacroso Mar 01 '17 at 23:49
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    If a function is continuous but not uniformly continuous then somewhere along the domain, it becomes arbitrarily steep. The image two points that are nearby in the domain are not nearby. – Doug M Mar 01 '17 at 23:52
  • I personally think that this is one of those instances where "geometric intuition" is overrated. For instance, a basic fact is that a function $f:[0,1]\to \Bbb R$ is continuous if and only if it is uniformly continuous. I think any "geometric intuition" one could give you about "steepleness" and "slopes" would miserably fail in predicting this. –  Mar 01 '17 at 23:56
  • @G.Sassatelli Can you construct a continuous function $f : [0,1] \to \mathbb{R}$ that has a slope which tends to $\infty$ anywhere on the domain? If not, then wouldn't that "predict" the result based on that intuition? – Tom Mar 02 '17 at 00:01
  • @Tom Cantor's function is always there, nice and UC on $[0,1]$, with those seemingly very ripid slopes in strategic palces. –  Mar 02 '17 at 00:05
  • @G.Sassatelli But can you talk about slope anywhere except where it's zero? Don't get me wrong, there are many bizarre looking functions out there which defy "geometric intuition," but my point is that for those that we can picture, should we eschew using them to gain some heuristic? – Tom Mar 02 '17 at 00:11
  • @Tom Perhaps it's better to have a handbook of obstructions to uniform continuity (examples of what is not UC), rather than an imprecise interpretation of a kind of continuity which often coincides with ordinary continuity. That being said, it's just my opinion (which marks a personal limit). –  Mar 02 '17 at 00:15
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    By the way, @tom, there are indeed UC differentiable functions $f:\Bbb R\to \Bbb R$ such that $f'([0,\varepsilon])=\Bbb R$ for any $\varepsilon>0$. Consider $x^2\sin\frac1{x^2}$. –  Mar 02 '17 at 00:35
  • @G.Sassatelli Very good example. Of course, $f'$ is not defined at $0$, but you're right, the slope does tend towards infinity as $x \to 0$. I like your idea of creating a counter example booklet. – Tom Mar 02 '17 at 01:30
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    @Tom If you set $f(0)=0$, then $f'(0)=0$. –  Mar 02 '17 at 01:34
  • @G.Sassatelli Yes, of course. It was implicitly clear that $f(0) =0$ (for continuity). I should have said "continuous" rather than "defined".. my bad! – Tom Mar 02 '17 at 01:57
  • @G.Sassatelli Also: $f(x) = \sqrt{x}$ for $x \in [0,1]$. – Tom Mar 02 '17 at 02:19

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