F is cont. on an interval I and unif. conti. on a subset of I then F is unif cont.

[kingstrick]

Homework Statement

Show that if F is continuous on [0,∞) and uniformaly continuous on [a,∞) for some positive constant , then f is uniformaly continuous on [0,∞).

Homework Equations

The Attempt at a Solution

Let I:= [0,∞) and A:=[a,∞) where a ≠0 and I is contained in ℝ and A is contained in I. Also assume that F is continuous on I but uniformally continuous on A. Lastly, assume that F is not uniformally continuous on [0,∞). Since by assumption F is unif. Cont. on A, the complement of A in I is where the uniform continuity must fail. Since F is continuous on I, and I is an interval, then there exist c,d in I and a k in ℝ where f(c) < k < f(d). Then there exist a b in I between c and d where f(b) = k. Since c and d are arbitrary values in I, then any value between c and d can be found. Lastly since 0 is contained in the interval it too must be continuous at that point (by assumption). Therefore f is continuous at every point in I. Thus F is uniformaly continuous on [0,∞).

**My proofs are always long winded like this**
If at all possible, could somebody first, help me if i solved this problem wrong, and if i am right, show me a more concise way to write this problem up.

 

[jgens]

Since F is continuous on I, and I is an interval, then there exist c,d in I and a k in ℝ where f(c) < k < f(d). Then there exist a b in I between c and d where f(b) = k. Since c and d are arbitrary values in I, then any value between c and d can be found. Lastly since 0 is contained in the interval it too must be continuous at that point (by assumption). Therefore f is continuous at every point in I. Thus F is uniformaly continuous on [0,∞).

This part of the argument does not show anything. All that you do here is restate that f is continuous on [0,\infty) (which you assumed) and say that continuous functions satisfy the Intermediate Value Theorem.

**My proofs are always long winded like this**
If at all possible, could somebody first, help me if i solved this problem wrong, and if i am right, show me a more concise way to write this problem up.

By assumption you know that f is uniformly continuous on [a,\infty) and continuous on [0,\infty). Since [0,a] is compact and f is continuous on [0,a], it follows that f is uniformly continuous on [0,a]. Now use the fact that f is uniformly continuous on [0,a] and uniformly continuous on [a,\infty) to complete the proof.

 

[kingstrick]

i was told that i needed to prove compactness. how do i do that?

 

[HallsofIvy]

What definition of "compactness" are you using?

 

[kingstrick]

We have not learned compactness yet. so I have to prove it if I'm going to use it

 

[jgens]

If you have not learned compactness and results related to compactness, then prove this theorem: If 0 &lt; a and f is continuous on [0,a], then f is uniformly continuous on [0,a].

This theorem is usually proved in a basic calculus course, so you might already have the result.

 

[kingstrick]

Is this better?

Let I:= [0,∞) and A:=[a,∞) where a > 0 and I is contained in ℝ and A is contained in I. Also assume that F is continuous on I but uniformally continuous on A. Since f is continuous on A, it must be continuous at every point in A. Therefore f is continuous at the point a. This means that f is continuous on the closed bounded interval [0,a]. By the uniform continuity theorem, since f is continuous on [0,a], and [0,a] is closed and bounded, then f is uniformaly continuous on [0,a]. Therefore f is uniformaly continuous on the intervals [0,a] and [a,∞]. Thus f is uniformaly continuous on I.

 

[jgens]

Is this better?

It is better, yes. But the proof is not complete and can be streamlined considerably.

Let I:= [0,∞) and A:=[a,∞) where a > 0 and I is contained in ℝ and A is contained in I. Also assume that F is continuous on I but uniformally continuous on A. Since f is continuous on A, it must be continuous at every point in A. Therefore f is continuous at the point a. This means that f is continuous on the closed bounded interval [0,a].

What you have above is correct, but unnecessarily wordy. Try something like: "Assume that f is continuous on [0,\infty) and uniformly continuous [a,\infty) for some 0 &lt; a. Since f is continuous on [0,\infty), it follows that f is continuous on [0,a]."

By the uniform continuity theorem, since f is continuous on [0,a], and [0,a] is closed and bounded, then f is uniformaly continuous on [0,a].

This is certainly true. Just make sure you can assume the result for this proof.

Therefore f is uniformaly continuous on the intervals [0,a] and [a,∞]. Thus f is uniformaly continuous on I.

Prove to me that uniform continuity on [0,a] and [a,\infty) implies uniform continuity on [0,\infty). This is the missing part of your argument.

 

[kingstrick]

It is better, yes. But the proof is not complete and can be streamlined considerably.



What you have above is correct, but unnecessarily wordy. Try something like: "Assume that f is continuous on [0,\infty) and uniformly continuous [a,\infty) for some 0 &lt; a. Since f is continuous on [0,\infty), it follows that f is continuous on [0,a]."



This is certainly true. Just make sure you can assume the result for this proof.



Prove to me that uniform continuity on [0,a] and [a,\infty) implies uniform continuity on [0,\infty). This is the missing part of your argument.

would arguing [o,a+1] is unif cont and [a-1,∞] is unif cont work?

 

[jgens]

would arguing [o,a+1] is unif cont and [a-1,∞] is unif cont work?

If you argued that, I would still say your proof is incomplete. And while showing that f is uniformly continuous on [0,a+1] is easy with the work you already have, showing that f is uniformly continuous on [a-1,\infty) is not necessarily possible. For example, what happens if a = 2^{-1} and f is discontinuous on (-\infty,0)?

To complete the proof, note the following: Fix 0 &lt; \varepsilon. Since f is uniformly continuous on [0,a] there exists 0 &lt; \delta_1 such that |f(x_1)-f(x_2)| &lt; \varepsilon whenever x_1,x_2 \in [0,a] satisfy |x_1-x_2| &lt; \delta_1. Since f is uniformly continuous on [a,\infty) there exists 0 &lt; \delta_2 such that |f(x_1)-f(x_2)| &lt; \varepsilon whenever x_1,x_2 \in [a,\infty) satisfy |x_1-x_2| &lt; \delta_2. Now can you need to find a 0 &lt; \delta which will make f satisfy uniform continuity on [0,\infty)?

 

[kingstrick]

If you argued that, I would still say your proof is incomplete. And while showing that f is uniformly continuous on [0,a+1] is easy with the work you already have, showing that f is uniformly continuous on [a-1,\infty) is not necessarily possible. For example, what happens if a = 2^{-1} and f is discontinuous on (-\infty,0)?

To complete the proof, note the following: Fix 0 &lt; \varepsilon. Since f is uniformly continuous on [0,a] there exists 0 &lt; \delta_1 such that |f(x_1)-f(x_2)| &lt; \varepsilon whenever x_1,x_2 \in [0,a] satisfy |x_1-x_2| &lt; \delta_1. Since f is uniformly continuous on [a,\infty) there exists 0 &lt; \delta_2 such that |f(x_1)-f(x_2)| &lt; \varepsilon whenever x_1,x_2 \in [a,\infty) satisfy |x_1-x_2| &lt; \delta_2. Now can you need to find a 0 &lt; \delta which will make f satisfy uniform continuity on [0,\infty)?

if i make ε/2 = delta then when i combine the two statements to make [0,∞) i would get ε, right?

 

[jgens]

if i make ε/2 = delta then when i combine the two statements to make [0,∞) i would get ε, right?

Remember that in this case \varepsilon is fixed. You need to find a \delta which will work for the fixed \varepsilon.

 

[kingstrick]

if i make ε/2 = delta then when i combine the two statements to make [0,∞) i would get ε, right?

Can i make δ=max(δ₁,δ₂) or add the δ's to signify union. I am really confused with picking a delta independent of ε

 

[jgens]

Can i make δ=max(δ₁,δ₂) or add the δ's to signify union. I am really confused with picking a delta independent of ε

Taking the maximum will not work. A good thing to remember with limit type problems is that you typically want to make things smaller. So given the choice of δ₁ and δ₂, you want to end up with the smaller of the two.

And you are not picking δ independent of ε. We picked δ₁ and δ₂ based on ε, and now we are picking a δ that will work for ε on [0,∞). I think a big part of your confusion is that you are thinking of δ as a function of ε. I am going to discourage you from thinking this way. While it is true that δ is related to ε (in particular, once ε is fixed there is a limited set of values from which δ can be chosen), it is not true that there is only one δ that works for a given ε. So if we have fixed 0 < ε and found 0 < δ that works with ε, then for each n in N it follows that δ_n = n^-1δ also works with ε.