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Reading Haaser-Sullivan's Real Analysis

  1. Jul 26, 2010 #1
    Hi peeps!

    I was reading Haaser-Sullivan's Real Analysis and came across a problem for which I have a doubt. A part of it is stated like this : " For all x in the closed interval [a,b] in R, |g'(x)|<=1 '' (g(x) is, of course, a real-valued function of a real variable and that's all we know about it). Does that mean that for all x in [a,b], g'(x) is defined or that for all x in [a,b] such that g'(x) is defined, |g(x)|<=1?

    Thanks in advance!
     
  2. jcsd
  3. Jul 27, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Derivative!

    I would interpret it as saying that g'(x) is defined and between -1 and 1, for all x in [a,b].
     
  4. Jul 27, 2010 #3
    Re: Derivative!

    That's what I thought, thanks! I haven't touched the formal definition of differentiability, but in Calculus I learned that a function was differentiable on [a,b] iff its derivative exists on [a,b]. So the condition stated above is enough to show differentiability on [a,b] and thus, continuity and contraction?
     
  5. Jul 27, 2010 #4

    Mark44

    Staff: Mentor

    Re: Derivative!

    I think that all we can say is that since |g'(x)| is defined at every x in the interval, then |g(x)| is continuous on the same interval, but that g(x) is not necessarily continuous.

    What do you mean by "contraction?" Are you saying that |g(x)| <= x?
     
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