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Quick Proof rigor check.

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f be a real uniformly continuous function on the bounbed set E in R. Prove that f is bounded on E.

    2. Relevant equations

    3. The attempt at a solution

    Since f is uniformly continuous, [tex]\left|f(x)-f(y)\right|< \epsilon[/tex] if [tex]\left|x-y\right|< \delta[/tex]. Since E is bounded, there exists some maximal distance between x and y; let M be the upper limit and P be the lower limit of E. Then

    [tex]\left|x-y\right|< \\left|M-P\right|<[/tex]. If M, P are not in E, then define two sequences {x_n} and {y_n} such that lim n->infinity {x_n}=M and lim n->infinity {y_n} = P.

    Then the definition of uniform continuity implies that if we let [tex] \delta = abs(M-P)[/tex], so that abs(x-y) < delta for all x, y then abs(f(x)-f(y))< epsilon for all f(x), f(y), with epsilon equal to lim n->infinity {f(x_n)}=M and lim n->infinity {f(y_n)}

    I feel fairly solid about this, but I'm finding the difference between absolute continuity and continuity a bit confusing and would like a rigor/correctness check.

  2. jcsd
  3. Feb 23, 2009 #2
    Here's where your proof goes wrong:

    You can't just pick a delta like this. You can only pick an epsilon and then prove existence of delta using the definition of uniform continuity. Here's how I would prove it:

    Choose arbitrary epsilon > 0. Then there exists delta>0 such that as long as |x-y|<delta, |f(x)-f(y)|<epsilon. Choose N so large that N delta > |M-P|. Now, choose arbitrary x0 in E. Then we must have f(x0) - epsilon N < f(x) < f(x0) + epsilon N for all x in E (check this). This proves the theorem.
  4. Feb 23, 2009 #3
    Hm, ok. So that is the same idea I think; or at least that makes complete sense and I was just being silly.

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