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

## 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,∞).

## 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.

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jgens
Gold Member

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.

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

HallsofIvy
Homework Helper

What definition of "compactness" are you using?

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

jgens
Gold Member

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.

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
Gold Member

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.

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
Gold Member

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 $[0,\infty)$?

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
Gold Member

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 $\varepsilon$.

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(δ12) or add the δ's to signify union. I am really confused with picking a delta independent of ε

jgens
Gold Member

Can i make δ=max(δ12) 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 δn = n-1δ also works with ε.