Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   Calculus & Beyond Homework (http://www.physicsforums.com/forumdisplay.php?f=156)
-   -   bounded continous implies uniformly continuous (http://www.physicsforums.com/showthread.php?t=581810)

alanlu Feb27-12 07:40 AM

bounded continous implies uniformly continuous
 
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.

micromass Feb27-12 08:32 AM

Re: bounded continous implies uniformly continuous
 
So you know a characterization of uniform continuity using sequences??
That is: f is uniform continuous iff for all equivalent sequences [itex](x_n)_n[/itex] and [itex](y_n)_n[/itex] holds that [itex](f(x_n))_n[/itex] and [itex](f(y_n))_n[/itex] are also equivalent.

alanlu Feb27-12 04:39 PM

Re: bounded continous implies uniformly continuous
 
Was not aware of this characterization until now. My text does not mention this, so there must be another way.

fauboca Feb27-12 05:42 PM

Re: bounded continous implies uniformly continuous
 
Quote:

Quote by alanlu (Post 3787762)
Was not aware of this characterization until now. My text does not mention this, so there must be another way.

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

Ansatz7 Feb27-12 07:12 PM

Re: bounded continous implies uniformly continuous
 
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)?

jbunniii Feb27-12 08:08 PM

Re: bounded continous implies uniformly continuous
 
Quote:

Quote by Ansatz7 (Post 3787934)
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)?

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

[tex]\sup_{x \in K} f(x) = \max_{x \in K} f(x)[/tex]
and
[tex]\inf_{x \in K} f(x) = \min_{x \in K} f(x)[/tex]

How do you propose to use this fact?

jbunniii Feb27-12 08:09 PM

Re: bounded continous implies uniformly continuous
 
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?

Ansatz7 Feb27-12 08:39 PM

Re: bounded continous implies uniformly continuous
 
Quote:

Quote by jbunniii (Post 3788034)
This is true for a continuous function f defined on a compact set K:

[tex]\sup_{x \in K} f(x) = \max_{x \in K} f(x)[/tex]
and
[tex]\inf_{x \in K} f(x) = \min_{x \in K} f(x)[/tex]

How do you propose to use this fact?

Choose ε > 0 and for each x[itex]\in[/itex][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!)

jbunniii Feb27-12 11:17 PM

Re: bounded continous implies uniformly continuous
 
Quote:

Quote by Ansatz7 (Post 3788070)
Choose ε > 0 and for each x[itex]\in[/itex][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 = sup δ(x) on the interval. Then for any x on the interval |x - y| < δ0 implies...?

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

Dick Feb27-12 11:36 PM

Re: bounded continous implies uniformly continuous
 
Quote:

Quote by jbunniii (Post 3788316)
I don't see that it implies much of anything. What am I missing?

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?

Ansatz7 Feb27-12 11:52 PM

Re: bounded continous implies uniformly continuous
 
Quote:

Quote by jbunniii (Post 3788316)
I don't see that it implies much of anything. What am I missing?

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.

jbunniii Feb27-12 11:59 PM

Re: bounded continous implies uniformly continuous
 
Quote:

Quote by Dick (Post 3788343)
Well, ansatz7 did have disclaimer at the end. But I'm also not sure why you need the max and min. Pick a finite subcover 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?

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:

Quote:

A(d) = { u in [a, b] : x, y in [a, u], x - y < d implies f(x) - f(y) < e }

jbunniii Feb28-12 12:01 AM

Re: bounded continous implies uniformly continuous
 
Quote:

Quote by Ansatz7 (Post 3788363)
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.

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.

Ansatz7 Feb28-12 12:04 AM

Re: bounded continous implies uniformly continuous
 
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.

Quote:

Quote by Ansatz7 (Post 3787934)
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)?


alanlu Feb28-12 07:28 AM

Re: bounded continous implies uniformly continuous
 
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.


All times are GMT -5. The time now is 02:32 AM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums