Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $-\infty \le a < c < b \le ∞$ and $f:(a,b)\to \mathbb{R}$ is continuous on $(a,b)$.

(a) If $f$ is uniformly continuous on both $(a,c)$ and $(c,b)$, prove that $f$ is uniformly continuous on $(a,b)$.

(b) Give an example to show that the conclusion in part (a) may be false if $f$ is not continuous on $(a,b)$.

share|improve this question

closed as off-topic by John, Claude Leibovici, Davide Giraudo, Michael Hoppe, egreg Feb 16 at 10:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John, Claude Leibovici, Davide Giraudo, Michael Hoppe
If this question can be reworded to fit the rules in the help center, please edit the question.

Welcome to math.SE: since you are new, we want to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many users find the use of the imperative ("Prove", "Show", etc) to be rude when asking for help. Please consider rewriting your post. –  user17762 May 16 '12 at 21:56
The problem with the way this question is phrased is that it looks as if "doit" is passing on to us a question written by someone other than "doit", so it's not clear whether the poster even understands the question. –  Michael Hardy May 16 '12 at 22:01
@MichaelHardy I actually did not look at the questions posted by "doit" earlier. Now looking at the previous questions posted by "doit", Arturo has already left comment on a previous question of the user here (math.stackexchange.com/questions/144461/…) –  user17762 May 16 '12 at 22:04
The imperative seems to be connected to the username... –  Michael Greinecker May 16 '12 at 22:06
possible duplicate of [Uniform continuity on $[a,b]$ and $ [b,c]$ $\implies$ uniform continuity on $[a,c]$.](math.stackexchange.com/questions/582200/…) –  Tucker Rapu Feb 16 at 9:21
add comment

1 Answer 1

Let $\epsilon\gt 0$. You know that there exists $\delta_0$ such that any two points that are $\delta_0$ close and both are in $(a,c)$ will have images that are $\epsilon$-close. And you know that there is a $\delta_1$ such that any two points that are $\delta_1$ close and both are in $(c,d)$ will have images that are $\epsilon$-close. Taking the minimum of $\delta_1$ and $\delta_2$ will guarantee any two points that are in the same subinterval and are within $\delta$ of each other will have images that are $\epsilon$-close. The problem lies with the situation in which one point is in $(a,c)$ and the other point is in $(c,b)$.

Now, we know the function is continuous at $c$, so we also know there is a $\delta_2$ such that if a point is within $\delta_2$ of $c$, then its image is within $\epsilon$ of the image of $c$.

Now think about the triangle inequality, and try to put it all together...

(and don't repost the question yet again!)

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.