Uniform Continuity and Continuity

Aug 2010
961
101
Let S be the set of points on an interval and let f be continuous on S. Then at every point x there is a neighborhood of x st

|f(x)-f(y)|<ε if |x-y|<δ(x).

The set of all δ(x) form an open cover of S. If S is compact, (closed and bounded, [a,b]), then there is a finite collection of the δ(x) which covers S. Then δ = min δi(x) and f is uniformly continuous. That establishes sufficiency. Necesssity established by uniform continuity implies continuity.

If S is not bounded, we have to choose a minimum from an infinite collection of δ(x). If you show that δ(x) < δ for all x, you haven’t determined that a min δ exists for all x and you haven’t proved uniform continuity.*

Conclusion:
1) f(x) continuous on S compact ([a,b]) iff f(x) uniformly continuous.
2) f(x) continuous on [a,∞) and lim f(x)=k as x→∞, does not imply uniform continuity.

*Given all the δ(x) < δ, is there a minimum one(s) among them- a subtle point. For example, if you can show for a given ε, 0 < δ(x) < α, what is the minimum value of δ(x)?

EDIT: There is an ambiguity in the definition of Uniform Continuity, depending on whether, given ε for the interval, you require:
3) A SPECIFIC MIN δ(ε), or
4) ANY δ LESS THAN δ(ε)

The conclusions 1) and 2) above assume definition 3).
 
Last edited:
Aug 2010
961
101
Conclusion 2) in OP is incorrect, and everything past 1), and the sentence previous to 1), should be discarded.

Revised Conclusion:

1’) f(x) continuous on S compact ([a,b]) iff f(x) uniformly continuous.
2’) f(x) continuous on [a,∞) and lim f(x)=k as x→∞, implies uniform continuity.

A proof of 2) is given by slipeternal in post #19 of
http://mathhelpforum.com/advanced-applied-math/226156-uniform-continuity-2.html

A clarified and expanded version is given below. If the expansion is part of the standard proof, then my apologies for posting it Peer Math Review, and this post will just be a clean-up of the previous one.

1) Let b be smallest x st |f(x) - k│ < ε/4, x>b. Then
|f(x) - f(y)│ < ε/2 for all │x - y│in [b,∞). Because lim f(x)=k as x →∞.

2) There is a δ such that |f(x) - f(y)│ < ε/2 for │x - y│< δ, x,y in[a,b]. Heine-Borel.

3) If x<b and y>b, |f(x) - f(y)│≤ |f(x) - f(b)│+ |f(y) - f(b)│< ε/2+ ε/2 = ε for │x - b │< δ and all y>b, or │x - y│< δ because │x - y│< δ → │x - b │< δ.

4) Suppose b’>b. Then (abbreviated)
a) |f(x) - f(y)│ < ε/2 for all │x - y│in [b’,∞)
b) |f(x) - f(y)│< ε/2 for all │x - y│in [b,b’].
c) |f(x) - f(y)│< ε/2 for │x - y│< δ all x,y in [a,b].
So
|f(x) - f(y)│< ε for │x - y│< δ for all x,y in [b,∞).

The last step is very important because it shows δ depends on a single b.

So continuity implies uniform continuity is consistent for the both conditions, on [a,b], and on [a,∞) with limf(x)=k.