Do you know that if a function is continuous on a closed interval then it is informally continuous on that interval?
If in the definition of periodic, then is the function continuous on ?
What can you do with that?
I'm stuck on how to do this proof.
I thought I might start like this...Originally Posted by Gaugan. Introduction To Analysis, 5th ed, p105
Choose . By definition of continuity, for each , there exists such that if and , then .
That way, if I can somehow show that , then I could just finish thusly:
Choose . Then for each and , if and , then .
But how do I show that ? Or, if that's impossible, then perhaps instead of setting , I could set it to , and then somehow show a relationship between and . But I wouldn't know how to do that, either.
Please advise.
Well that would be a simple matter.
Choose with and is closed and bounded* is compact and is continuous over is uniformly continuous over .
EDIT: Or am I missing something, here? The textbook obviously wants me to incorporate somehow into my proof, which I have not done.
*- Does "bounded" here mean that is bounded over the interval, or that the interval itself is bounded? My textbook indicates the latter, but that doesn't make sense to me. Aren't all closed intervals bounded?
How about this...
For any with , then with and , and . Since is continuous over the compact interval , then so that . So is uniformly continuous.
The problem here is that I don't know what I need to fix with my previous attempted proof. That is to say, I do not see any problem with the following:
If someone could point out my error, then I could work on fixing it. However, I do not see any such error, which leaves me at a loss.