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Math Help - prove uniform continuity given a periodic and continuous function

  1. #1
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    prove uniform continuity given a periodic and continuous function

    I'm stuck on how to do this proof.

    Quote Originally Posted by Gaugan. Introduction To Analysis, 5th ed, p105
    A function f:R\rightarrow R is periodic iff there is a real number h\neq0 such that f(x+h)=f(x) for all x\in R. Prove that if f:R\rightarrow R is periodic and continuous, then f is uniformly continuous.
    I thought I might start like this...

    Choose \epsilon>0. By definition of continuity, for each z_{\alpha}\in R, there exists \delta_{\alpha}>0 such that if |z-z_{\alpha}|<\delta_{\alpha} and z\in R, then |f(z)-f(z_{\alpha})|<\epsilon.

    That way, if I can somehow show that \inf\{\delta_{\alpha}\}>0, then I could just finish thusly:

    Choose \delta=\inf\{\delta_{\alpha}\}. Then for each x\in R and y\in\{z_{\alpha}\}=R, if |x-y|<\delta and x\in R, then |f(x)-f(y)|<\epsilon.

    But how do I show that \inf\{\delta_{\alpha}\}>0? Or, if that's impossible, then perhaps instead of setting \delta_{\alpha}>0, I could set it to \delta_{\alpha}\in(0,h/4), and then somehow show a relationship between |x-y| and |z-z_0|. But I wouldn't know how to do that, either.

    Please advise.
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  2. #2
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    Do you know that if a function is continuous on a closed interval then it is informally continuous on that interval?
    If h>0 in the definition of periodic, then is the function continuous on [0,h]?
    What can you do with that?
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  3. #3
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    Quote Originally Posted by Plato View Post
    Do you know that if a function is continuous on a closed interval then it is informally continuous on that interval?
    If h>0 in the definition of periodic, then is the function continuous on [0,h]?
    What can you do with that?
    Well that would be a simple matter.

    Choose \alpha_n,\beta_n\in R with \alpha_n<\beta_n and \cup_{n=1}^{\infty}[\alpha_n,\beta_n]=R \; \Rightarrow \; [\alpha_n,\beta_n] is closed and bounded* \; \Rightarrow \; [\alpha_n,\beta_n] is compact and f is continuous over [\alpha_n,\beta_n] \; \Rightarrow \; f is uniformly continuous over R.

    EDIT: Or am I missing something, here? The textbook obviously wants me to incorporate h somehow into my proof, which I have not done.



    *- Does "bounded" here mean that f 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?
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  4. #4
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    Is it true that on any interval [a,b] such that b-a<\delta <h the graph of f is identical to its graph an equivalent sub-interval of [0,2h]?
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  5. #5
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    Quote Originally Posted by Plato View Post
    Is it true that on any interval [a,b] such that b-a<\delta <h the graph of f is identical to its graph an equivalent sub-interval of [0,2h]?
    Of course. But I do not see how this helps. In fact, I do not see what if anything is wrong with the proof from my last post.

    By the way, thanks for the help so far. I really appreciate it.
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  6. #6
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    Quote Originally Posted by hatsoff View Post
    Of course. But I do not see how this helps. In fact, I do not see what if anything is wrong with the proof from my last post.
    By the way, thanks for the help so far. I really appreciate it.
    Well think about it.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Well think about it.
    Choose n\in \mathbb{Z}. Then z\in[0,h]\Rightarrow(z+hn)\in[hn,hn+h] with f(z)=f(z+hn). Since f is uniformly continuous over every interval [hn,hn+h] and \cup_{n\in \mathbb{Z}}[hn,hn+h]=R, f is uniformly continuous over R.

    But I still don't see why h is significant. Why not any real number?
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  8. #8
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    Quote Originally Posted by hatsoff View Post
    But I still don't see why h is significant. Why not any real number?
    This is my last comment on this topic!
    Any periodic function in essence copies the graph on intervals over and over again.
    Thus if it is uniformly continuous of an interval what does that mean?
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  9. #9
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    Quote Originally Posted by Plato View Post
    This is my last comment on this topic!
    Any periodic function in essence copies the graph on intervals over and over again.
    Thus if it is uniformly continuous of an interval what does that mean?
    How about this...

    For any x,y\in R with |x-y|<|h|, then \exists\;a,b\in[0,|h|] with f(x)=f(a) and f(y)=f(b), and |x-y|=|a-b|. Since f is continuous over the compact interval [0,|h|], then \exists\;\delta\in(0,|h|) so that |x-y|<\delta\;\Rightarrow\;|a-b|<\delta\;\Rightarrow\;|f(a)-f(b)|<\epsilon\;\Rightarrow\;|f(x)-f(y)|<\epsilon. So f 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:

    Quote Originally Posted by hatsoff View Post
    Choose \alpha_n,\beta_n\in R with \alpha_n<\beta_n and \cup_{n=1}^{\infty}[\alpha_n,\beta_n]=R \; \Rightarrow \; [\alpha_n,\beta_n] is closed and bounded \; \Rightarrow \; [\alpha_n,\beta_n] is compact and f is continuous over [\alpha_n,\beta_n] \; \Rightarrow \; f is uniformly continuous over R.
    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.
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