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 $0 < a$ 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 $0 < a$. 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 $0 < a$. 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 $0 < \varepsilon$. Since $f$ is uniformly continuous on $[0,a]$ there exists $0 < \delta_1$ such that $|f(x_1)-f(x_2)| < \varepsilon$ whenever $x_1,x_2 \in [0,a]$ satisfy $|x_1-x_2| < \delta_1$. Since $f$ is uniformly continuous on $[a,\infty)$ there exists $0 < \delta_2$ such that $|f(x_1)-f(x_2)| < \varepsilon$ whenever $x_1,x_2 \in [a,\infty)$ satisfy $|x_1-x_2| < \delta_2$. Now can you need to find a $0 < \delta$ 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 $0 < \varepsilon$. Since $f$ is uniformly continuous on $[0,a]$ there exists $0 < \delta_1$ such that $|f(x_1)-f(x_2)| < \varepsilon$ whenever $x_1,x_2 \in [0,a]$ satisfy $|x_1-x_2| < \delta_1$. Since $f$ is uniformly continuous on $[a,\infty)$ there exists $0 < \delta_2$ such that $|f(x_1)-f(x_2)| < \varepsilon$ whenever $x_1,x_2 \in [a,\infty)$ satisfy $|x_1-x_2| < \delta_2$. Now can you need to find a $0 < \delta$ 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.