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

• kingstrick
You need to explicitly show how the uniform continuity on these two intervals implies uniform continuity on [0,\infty). This can be done using the triangle inequality, for example.
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,∞).

## 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.

kingstrick said:
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?

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

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.

kingstrick said:
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.

jgens said:
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?

kingstrick said:
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 $[0,\infty)$?

jgens said:
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?

kingstrick said:
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 said:
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

kingstrick said:
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.

## What does it mean for a function to be continuous on an interval?

When a function is continuous on an interval, it means that it has no breaks or gaps in its graph over that interval. This means that the function can be drawn without lifting the pen from the paper, and there are no sudden jumps or holes in the graph.

## What is uniform continuity?

Uniform continuity is a stronger form of continuity that requires a function to have a consistent rate of change over its entire domain. This means that even if the function is not continuous at a specific point, the difference in output values for points close to each other will not be too large.

## What is the relationship between uniform continuity and continuity on a subset of an interval?

If a function is continuous on a subset of an interval, it means that it is continuous at every point within that subset. This also implies that the function is uniformly continuous on that subset, as it must have a consistent rate of change to be continuous at every point.

## How is uniform continuity different from local continuity?

Uniform continuity is a global property of a function, meaning it applies to the entire domain. Local continuity, on the other hand, only applies to a specific point or neighborhood of points. A function can be locally continuous at a point but not uniformly continuous on its entire domain.

## Why is the statement "If a function is continuous on an interval and uniformly continuous on a subset of that interval, then it is uniformly continuous on the entire interval" important?

This statement is important because it allows us to extend the property of uniform continuity from a subset of an interval to the entire interval. This simplifies the analysis of a function's continuity, as we only need to consider one interval instead of multiple subsets within that interval.

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