How to prove that $\ f$ is uniformly continuous here?

Question

Suppose $f: (0, 1) \to \mathbb{R}$ is differentiable with $|f'(x)| \le 1$. w.t.s that $f$ uniformly continuous

How can this be done?

Here was my attempt:

I used differentiability with a triangle inequality and found that $\delta = \epsilon$ had a potential of working, but I got lost in the details and felt it was not rigorous.

Edit (with respect to comments):

let $\epsilon > 0$, for $x \in (0, 1)$, we have

$\exists \delta > 0$ s.t

$|y - x| < \delta$ and $y \in (0, 1)$ implies

$\frac{f(y) - f(x)}{y - x} - f'(x) < \epsilon$

Notice that since the derivative is bounded, $f$ is Lipschitz continuous, and Lipchitz continuity implies uniform continuity. — unconscious statistician, Dec 12 '23 at 09:11
If you mean that we can get a value that matches our differentiation quotient, how does that help us? @CalvinKhor — user129393192, Dec 12 '23 at 23:03
I'm curious to know what you found not rigorous in your proof attempt. — SuhailSherif, Dec 12 '23 at 23:27
I "let" the epsilon for the differentiability one "go to 0" and found that we had what we wanted bound by delta @SuhailSherif (just the top of the differentiation quotient) and so claimed separately that delta doesn't depend on x (just epsilon), so we had uniform continuity, but I am not sure if I took it separately, if it would still hold, due to my letting of epsilon go to 0 ... — user129393192, Dec 12 '23 at 23:40
@user129393192 that's a good start for intuition. Now if only we could make that work when epsilon didn't tend to 0 in the differentiation. That is, if you know that the absolute value of the derivative is always at most 1, what can you say about the difference between f(x) and f(x+h), where h is not gong to 0? — SuhailSherif, Dec 12 '23 at 23:53
I do not have the $|f(x + h)|$ definition. I was using the fact that the derivative quotient $- f'(x)|$ goes to 0. @SuhailSherif — user129393192, Dec 13 '23 at 05:17
@user129393192 that's just mean value theorem. Put the denominator on the other side and use the given information on $f'$. then try your epsilon delta again — Calvin Khor, Dec 13 '23 at 10:56
@CalvinKhor i don't see how to put the denom on the other side. I added some detail to the post — user129393192, Dec 13 '23 at 20:02

yoyo · Answer 1 · 2023-12-13T22:43:50.920

1

[Probably repeating what the comments link to.] The idea is that "bounded derivative" means that the function doesn't "expand" too much, which is sort of the idea behind uniform continuity (by way of Lipschitz continuity).

More precisely, for $f:(0,1)\to\mathbb{R}$ as you describe, we have $|f(y)-f(x)|=|f'(z)(y-x)|\leq M|y-x|$ for some $z\in(x,y)$ by the mean value theorem, where $M\geq \sup\{|f'(x)| : 0<x<1\}$. In converting this to some kind of "$\epsilon-\delta$" argument, if $|y-x|<\delta$ then $|f(y)-f(x)|< M\delta:=\epsilon$. Or if you want to flip parameter dependence around, given $\epsilon$, take $\delta<\epsilon/M$.

edited Dec 13 '23 at 22:43

answered Dec 13 '23 at 20:28

yoyo

9,943

【yes, precisely】 – Calvin Khor Dec 13 '23 at 21:54
So we take arbitrary $x$, $y$ and use MVT. This assumes that $x < y$ correct? And since it is an open interval, we cover everybody in it, since by way of the $z(x, y)$, we can get the above inequality for any $x$, $y$? Is this understanding correct? – user129393192 Dec 14 '23 at 01:51

score 0 · Answer 2 · answered Dec 13 '23 at 20:25

To prove that a function is uniformly continuous (assuming that you want to do it directly), you need to say, for each positive number $ \epsilon $, which positive number $ \delta $ to use (and then check that this works). You said that you want to use $ \delta = \epsilon $, and I agree that this should work. So to make it rigorous, you just need to prove that, whenever $ x $ and $ y $ are numbers in $ ( 0 , 1 ) $, if $ \lvert y - x \rvert < \delta $, then $ \lvert f ( y ) - f ( x ) \rvert < \epsilon $. Since you've chosen $ \delta = \epsilon $, you can change this to: if $ \lvert y - x \rvert < \epsilon $, then $ \lvert f ( y ) - f ( x ) \rvert < \epsilon $.

You do not want to go back to the definition of the derivative at each point and say something like what you added at the end: for each $ \epsilon $ and each $ x $, there is a $ \delta $ such that [etc]. Here, different values of $ x $ could give you different values of $ \delta $, and this is only useful to prove that $ f $ is continuous, not uniformly continuous. (You could start with a definition of uniform differentiability, since $ f $ is uniformly differentiable, but you'd have to prove that first, which seems like an unnecessary detour to me.)

Still, you need to bring in the derivative of $ f $ somehow. And the comment by @CalvinKhor tells you what to use: the Mean-Value Theorem. Since $ f ' $ always exists and $ \lvert f ' \rvert \leq 1 $, you know that any difference quotient satisfies the same inequality: $$ \Biggl \lvert \frac { f ( y ) - f ( x ) } { y - x } \Biggr \rvert \leq 1 \text . $$ (This is because the difference quotient is equal to $ f ' ( z ) $ for some $ z $, by the MVT.) From this fact, you can prove that if $ \lvert y - x \rvert < \epsilon $, then $ \lvert f ( y ) - f ( x ) \rvert < \epsilon $.

How to prove that $\ f$ is uniformly continuous here?

2 Answers2