Uniform Continuity Homework: Show h is Uniformly Continuous on [0, ∞)

Click For Summary
To show that a function h is uniformly continuous on [0, ∞) given its continuity on [0, ∞) and uniform continuity on [a, ∞) for some positive constant a, one can approach the problem using the epsilon-delta definition of continuity. The discussion highlights the importance of establishing uniform continuity on the interval [0, a] first, as any continuous function on a closed interval is uniformly continuous. Once uniform continuity is confirmed for both [0, a] and [a, ∞), the next step involves combining the deltas from these intervals to create a single delta that works for all of [0, ∞). The conversation emphasizes that uniform continuity requires a consistent delta across the entire interval, which cannot be achieved by simply taking the minimum of deltas from separate intervals. This methodical approach leads to a comprehensive proof of uniform continuity on the entire interval.
rainwyz0706
Messages
34
Reaction score
0

Homework Statement


Show that if h is continuous on [0, ∞) and uniformly continuous on [a, ∞),
for some positive constant a, then h is uniformly continuous on [0, ∞).



Homework Equations





The Attempt at a Solution


I'm thinking of using the epsilon-delta definition of continuity and look at two cases: 1. x>a, 0<y<a. 2. 0<x,y<a. but I'm not exactly sure if that works. Could anyone please give me some hints? Any help is appreciated!
 
Physics news on Phys.org
Consider the following questions:

1) Is h uniformly continuous on [0,a]?
2) If h is uniformly continuous on [0,a] and on [a,\infty) does it follow that h is uniformly continuous on [0,\infty)?
 
Let me think about this a bit in abstract terms.
If the limit for x going to a of f(x) = f(a) for every x in some interval I, we say that f is continuous on I. The definition of limit then says that: for every a in I, and for every epsilon, we can find a delta (depending on epsilon and a), such that ... "
Uniform continuity is a bit stronger. It says that for all points a we can choose the same delta. So the statement is "for every epsilon, we can find a delta (depending on epsilon), such that ... for all a in I simultaneously".

In general, continuity does not imply uniform continuity. You cannot simply take the minimum of all the deltas in your interval, for example, because in general this does not exist. Technically, you should be talking about an infimum, but this might be zero and that is not allowed.

Now you find yourself in the situation, where you know that for some infinite interval there is a strictly positive delta which works for all the points in that interval. This suggests to me a proof in two steps:
* find a strictly positive delta on the finite closed interval [0, a] (basically proving that any continuous function on [a, b] is uniformly continuous)
* combine these two deltas into a single one which is > 0 and works for all of [0, ∞)
 
CompuChip said:
* combine these two deltas into a single one which is > 0 and works for all of [0, ∞)

Yup. And the new overall δ would be the:

a. Max
b. Min
c. sum
d. difference
e. other

of the two δ's you have?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 40 ·
2
Replies
40
Views
4K
Replies
22
Views
2K
Replies
26
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
1K
Replies
3
Views
4K
Replies
7
Views
1K