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I'm confused about the following exercise in the textbook Vector Calculus:

Let $f:(0,1) \rightarrow \mathbb{R}$ be s.t. $t \mapsto \frac{1}{t}$.

Part a) Prove that $f$ is continuous at every point $p \in (0,1)$

Part b) Prove that $f$ is not uniformly continuous

For part a, the solution given is as follows:

Let $\delta = \min( \frac{1}{2}p, \frac{1}{2}\epsilon p^2)$. Then from $|t-p| < \delta$ it follows that $t>\frac{1}{2}p$ and also $$|f(t)-f(p)| = \frac{|t-p|}{|tp|} < \frac{ \frac{1}{2}\epsilon p^2} { \frac{1}{2}p^2} = \epsilon$$

For part b, this proof is given:

Take $\epsilon = \frac{1}{2}$. Let $\delta$ be arbitrary. Then choose $n$ s.t. $\delta > \frac{1}{n}$, and let $p = \frac{1}{n}$ and $t = \frac{1}{n+1}$. We get that $|t-p| < \delta$ but $f(t) - f(p) = 1 > \epsilon$.

Textbook defines only uniform continuity explicitly:

A function $f:D \rightarrow \mathbb{R}$ is uniformly continuous on $D$ if to each $\epsilon>0$ there corresponds $\delta>0$ such that $$|f(t) - f(s)| < \epsilon \text{ whenever } |t-s| < \delta$$ for $t,s \in D$.

Part a is asking us to prove that given a point, the function is continuous at that point. I.e. $\forall p \forall \epsilon \exists \delta$. But part b is asking us to prove that $\forall \epsilon \exists \delta \forall p$.

While I understand the logical difference, I am extremely confused as to the interpretation of these definitions. How can a function be continuous at every point, but not uniformly continuous? Does this have to do with uniformity of slopes?

Uniform continuity, as defined here, seems to require that all points are some function of $epsilon$ away from each other, regardless of the choice of point. In other words, uniform continuity is independent of the point.

How should I interpret this notion intuitively?

Mani
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    Imagine you run a sock store. If your sign says "Bring us your feet - we'll have a sock that fits them", that's continuity. If your sign says "We have one model of sock, and it'll fit anybody's feet", that's uniform continuity. – JonathanZ Sep 03 '23 at 19:24
  • @JonathanZ Great analogy for foot fetishists lol – blargoner Sep 03 '23 at 19:27
  • @JonathanZ Fun analogy, but I'm confused as to what this means geometrically. How can I tell if a function is or is not uniformly continuous from the graph? – Mani Sep 03 '23 at 19:28
  • If it was uniformly continuous then it would have a finite limit as $x \downarrow 0$. – copper.hat Sep 03 '23 at 19:31
  • Honestly, when it come to your average smooth function, "uniform continuity" is almost always equivalent to "derivative is bounded over the interval in question", or graphically "there's a limit to how steep the tangent lines get". There are lots of exceptions to this description, but I think it's a great way to build insight. – JonathanZ Sep 03 '23 at 19:32
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    It does have to do with slopes in a sense. F fails to be uniformly continuous because it has unbounded slope. Uniform continuity generalizes this idea to apply to functions which might not be differetiable, but still captures the idea that the function has a limit on how quickly its values can change. – eyeballfrog Sep 03 '23 at 19:34
  • @Johnathan So is it correct to say a function is uniformly continuous if its first derivative is bounded in the domain? – Mani Sep 03 '23 at 19:38
  • Suppose $U\subseteq\mathbb R$. To say that $f:U\to\mathbb R$ is continuous means that $$ \forall x\in U:\forall\varepsilon>0:\exists\delta>0:\forall y\in U:|x-y|<\delta\implies |f(x)-f(y)|<\varepsilon , . $$ Since $\delta$ appears after both $x$ and $\varepsilon$, it is allowed to depend on both of them. For $f$ to be uniformly continuous, there is a stricter condition imposed: $\delta$ may depend on $\varepsilon$, but on $x$. – Joe Sep 03 '23 at 19:38
  • Here is a another way to look at it. Pick $x>0$. Then $f({1 \over x+1}) = f({1 \over x}) +1$, but ${1 \over x+1}, {1 \over x}$ are arbitrarily close together. – copper.hat Sep 03 '23 at 19:42
  • @Shagester - Yes, differentiable everywhere and $|f'(x)| \lt M$ everywhere imply uniform continuity. It follows pretty easily from the mean value theorem. – JonathanZ Sep 03 '23 at 19:50
  • @eyeballfrog bounded slope is more precisely captured by the Lipschitz property. Think about $\sqrt{x}$ on $[0, 1]$. – ronno Sep 03 '23 at 20:07
  • @ronno That's true, but it's more that when functions aren't differentiable, the idea of "bounded slope" splits into multiple concepts based on how much you want to bound the variation. – eyeballfrog Sep 04 '23 at 13:03
  • To me bounded slope is uniform continuity and $\delta$ can be chosen to be linear wrt $\epsilon$. – ronno Sep 04 '23 at 13:09

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The geometric interpretation of uniform continuity, for real-valued functions on intervals of $\mathbb R$, if you like thinking about the graph of a function is:

  • For your given $\epsilon$ value, draw a "band" around your graph by plotting $f(x) + \epsilon$ and $f(x) - \epsilon$. You can also think of them as lower and upper guardrails, limiting the motion of your graph.

  • Now try to wiggle the original graph back and forth horizontally by $\delta$, which corresponds to graphs of $f(x-\delta_1)$ for $| \delta_1 | \lt \delta$.

If you can do the back-and-forth wiggling and keep the wiggled graph within the band you created, congratulations, you have found a $\delta$ that works for that $\epsilon$ in the definition of uniform continuity.

Unfortunately, for the standard non-uniformly continuous example, $1/x$ on $(0,1)$, we are so bad at visualizing what's happening near $0$ (we really want to draw the $\epsilon$-band transverse to the graph, not purely in the $y$-axis direction) that we can't see how this fails. So I almost end up falling back on the "bounded derivative" version, even though it's less general.

JonathanZ
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