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

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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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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