# Bounded continous implies uniformly continuous

1. Feb 27, 2012

### alanlu

I'm trying to show that continuous f : [a, b] -> R implies f uniformly continuous.

f continuous if for all e > 0, x in [a, b], there exists d > 0 such that for all y in [a, b], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e.

f uniformly continuous if for all e > 0, there exists d > 0 such that for all x and y in [a, b], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e.

I constructed
A(d) = { u in [a, b] : x, y in [a, u], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e }
A = Ud > 0 A(d)

And I think I need to show sup A = b and b is in A, but I'm stuck.

2. Feb 27, 2012

### micromass

Staff Emeritus
So you know a characterization of uniform continuity using sequences??
That is: f is uniform continuous iff for all equivalent sequences $(x_n)_n$ and $(y_n)_n$ holds that $(f(x_n))_n$ and $(f(y_n))_n$ are also equivalent.

3. Feb 27, 2012

### alanlu

Was not aware of this characterization until now. My text does not mention this, so there must be another way.

4. Feb 27, 2012

### fauboca

Isn't this by definition. f is continuous on a compact set so that it is uniformly continuous.

5. Feb 27, 2012

### Ansatz7

Isn't it true that on a compact set (which [a,b] is), the sup is equivalent to the maximum (and similarly, the inf is the same as a minimum)?

6. Feb 27, 2012

### jbunniii

This is true for a continuous function f defined on a compact set K:

$$\sup_{x \in K} f(x) = \max_{x \in K} f(x)$$
and
$$\inf_{x \in K} f(x) = \min_{x \in K} f(x)$$

How do you propose to use this fact?

7. Feb 27, 2012

### jbunniii

P.S. It would help to know what definition of "compact" you are using. There are several definitions that are equivalent on the real line.

Does your definition involve open covers and finite subcovers, or convergent subsequences, or "closed and bounded", or what?

8. Feb 27, 2012

### Ansatz7

Choose ε > 0 and for each x$\in$[a, b], choose δ(x) such that |x - y| < δ => |f(x) - f(y)| < ε (obviously this choice of δ(x) isn't unique, but just pick one for each x). Let δ0 = inf δ(x) (I had incorrectly written sup instead of inf before) on the interval. Then for any x on the interval |x - y| < δ0 implies...?

(Disclaimer: I don't really know much math, so people who actually know this stuff should correct me if I'm speaking nonsense!)

Last edited: Feb 27, 2012
9. Feb 27, 2012

### jbunniii

I don't see that it implies much of anything. What am I missing?

Last edited: Feb 27, 2012
10. Feb 27, 2012

### Dick

Well, ansatz7 did have a disclaimer at the end that it might be garbage, which it is. But I'm also not sure why you need the max and min. Pick a finite subcover (assuming finite subcover is the intended definition of compact) of the delta neighborhoods and pick the min of those deltas. So if |x-y|<delta then shouldn't x and y be in the same delta neighborhood or at worst in overlapping delta neighborhoods? Isn't that the vague picture hint?

Last edited: Feb 27, 2012
11. Feb 27, 2012

### Ansatz7

What can you then say about |f(x) - f(y)| for any x and y that satisfy |x - y| < δ0? I never did analysis formally, but I think this is valid.

EDIT: Sorry, I very stupidly wrote sup when I meant inf in my post above, so obviously it made no sense - way too tired to be useful. I'll go back and edit now.

12. Feb 27, 2012

### jbunniii

Pretty much. Some of the deltas in the argument will have to be delta/2 to make it work.

Also, I confused Ansatz7 with the original poster (alanlu), who made no mention of compactness of [a,b] and perhaps doesn't have the appropriate machinery (Heine-Borel) available. That may have been what he was getting at with this Spivak-style construction:

13. Feb 28, 2012

### jbunniii

Right, inf would make more sense. However, the inf of infinitely many delta(x) could be zero. This is where the compactness is necessary: to reduce the infinite cover to a finite cover, so "inf" becomes "min" and is strictly positive.

14. Feb 28, 2012

### Ansatz7

Right, I know that this only works because [a, b] is compact, as I stated in my first post in the thread. I believe it was you who asked how I would use this fact, which is where everything else came from.

15. Feb 28, 2012

### alanlu

Ah thanks! Actually, I did arrive at inf { d(x) }, but I wasn't sure how to turn that into something that is guaranteed to be > 0.