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Uniformly continuous and bounded

  1. Apr 28, 2006 #1
    Let [tex]f[/tex] be a real uniformly continuous function on the bounded
    set [tex]A[/tex] in [tex]\mathbb{R}^1[/tex]. Prove that [tex]f[/tex] is bounded
    on [tex]A[/tex].

    Since f is uniformly continuous, take [tex]\epsilon = m, \exists \delta > 0[/tex]
    such that
    [tex]|f(x)-f(p)| < \epsilon [/tex]
    whenever [tex]|x-p|<\delta[/tex] and [tex]x,p \in A[/tex]
    Now we have
    [tex]|f(x)| < m + |f(p)| [/tex]

    Obviously i should show [tex]b + |f(p)|[/tex] is bounded, but no idea how.
    Could someone help me? thx
     
    Last edited: Apr 28, 2006
  2. jcsd
  3. Apr 28, 2006 #2

    AKG

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    Science Advisor
    Homework Helper

    What is b? And why would you want to show that b + |f(p)| is bounded? You mean bounded as a function of p? Well that's no easier than showing directly that f is bounded, so this doesn't seem to be a worthwhile approach.

    The idea is simple. Fix [itex]\epsilon > 0[/itex]. There exists [itex]\delta > 0[/itex] such that, well, you know. Since A is bounded, you can choose a FINITE number of points x1, ..., xn in A such that the intervals of radius [itex]\delta[/itex] about the xi cover A. The total variation of f on these intervals is at most [itex]2\epsilon[/itex], so the total variation of f over all of A is at most [itex]2n\epsilon[/itex].

    Oh, and if this was homework, it should have been in the homework section.
     
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