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Problem about uniformly continuous

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Let I be the interval I=[0,infinity). Let f: I to R be uniformly continuous. Show there exist positive constants A and B such that |f(x)|<=Ax+B for all x that belongs to I.


    2. The attempt at a solution
    Proof by contradiction.
     
  2. jcsd
  3. Sep 20, 2008 #2

    CompuChip

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    Proof by contradiction:
    Let I be the interval I = [0, infinity). Let f: I to R be uniformly continuous.

    Suppose that for all positive A and B, there is an x = x(A, B) in I such that |f(x)| > Ax + B.

    All you have to derive a contradiction with is the uniform continuity of f. Can you write this out in terms of f(x) (i.e. the definition)?
     
  4. Mar 3, 2011 #3
    Can you complete the answer please ??
    I am interesting to know how we can prove this .
     
  5. Mar 3, 2011 #4

    CompuChip

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    I didn't solve it myself at the time, but some quick scribbles show that if you use the definition and the reverse triangle inequality it should work out.
    How far did you get?
     
  6. Mar 3, 2011 #5

    jtbell

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    Staff: Mentor

    We don't provide complete answers here, only hints. If you're trying to do this problem yourself, please start a new thread instead of hijacking an old one.
     
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