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

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Homework Help Overview

The discussion revolves around the concept of uniform continuity for a function h defined on the interval [0, ∞). The original poster seeks to demonstrate that if h is continuous on [0, ∞) and uniformly continuous on [a, ∞) for some positive constant a, then h must also be uniformly continuous on the entire interval [0, ∞).

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the epsilon-delta definition of continuity and consider different cases for the intervals involved. Questions are raised about the uniform continuity of h on the interval [0, a] and whether uniform continuity on both [0, a] and [a, ∞) implies uniform continuity on [0, ∞).

Discussion Status

Some participants have offered abstract reasoning regarding the definitions of continuity and uniform continuity, suggesting a two-step approach to the proof. There is an ongoing exploration of how to combine deltas from different intervals to establish uniform continuity across the entire domain.

Contextual Notes

There is a focus on the implications of continuity versus uniform continuity, with participants questioning the assumptions made about the behavior of the function h on the interval [0, a].

rainwyz0706
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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!
 
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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?
 

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