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Uniform Continuity and Supremum

  1. Dec 14, 2011 #1
    thanks!!!!
     
    Last edited: Dec 14, 2011
  2. jcsd
  3. Dec 14, 2011 #2

    Dick

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    What kinds of functions have you tried and why do you think they aren't working?
     
  4. Dec 14, 2011 #3

    LCKurtz

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    Hint: That's a difference quotient.
     
  5. Dec 14, 2011 #4

    micromass

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    Further hint: find a function whose derivative is infinite/doesn't exist...
     
  6. Dec 14, 2011 #5
    Does x*sin(1/x) work since its derivative is undefined at x=0 which is in [-1,1]?
     
  7. Dec 14, 2011 #6

    micromass

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    Well,

    1) is it uniform continuous??
    2) Do the difference quotients go to infinity??

    Just saying that the derivative is undefined isn't really enough...

    PS There is a much easier example
     
  8. Dec 14, 2011 #7
    I thought that any continuous function on a closed and bounded interval is also uniformly continuous. And in the case of x*sin(1/x) the derivative appears to go to infinity at x=0.
     
  9. Dec 14, 2011 #8

    micromass

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    Correct.

    Not really. Rather, the derivative does not exist (since it oscillates too much). Nevertheless the supremum you mention does indeed go to infinity. (you might want to give a further proof if it is not clear)
     
  10. Dec 14, 2011 #9
    I feel silly for making things more complicated than necessary. What was the easier example you had in mind? I was thinking square root x would work if the interval was [0,1].
     
  11. Dec 14, 2011 #10

    micromass

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    Don't feel silly. Your example is very elegant.

    The square root is indeed the one I had in mind. You just need to modify it a bit.
     
  12. Dec 14, 2011 #11
    So it would just be sqrt(x+1) for the [-1,1] interval as another solution?
     
  13. Dec 14, 2011 #12

    micromass

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    That should do it.
     
  14. Dec 14, 2011 #13
    thanks
     
    Last edited: Dec 14, 2011
  15. Dec 14, 2011 #14

    Dick

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    It's not even true. There are functions that differentiable at x=0, that aren't even differentiable anywhere else.
     
  16. Dec 14, 2011 #15
    Sorry, there's a condition also that f: R-->R and must be differentiable on R
     
  17. Dec 14, 2011 #16

    Dick

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    Then maybe use the mean value theorem?
     
  18. Dec 15, 2011 #17

    I like Serena

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    Just out of curiosity, renjean and chairbear, what is the reason you deleted your questions?
     
  19. Dec 15, 2011 #18

    micromass

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    It's because they are cheating. Please report these kind of things.
     
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