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    Real analysis (Uniform continuous)

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    Let {fn} be a sequence of continuous functions, all defined on [a,b]. Suppose {fn(a)} is a diverging sequence of real numbers. Prove that {fn} does not converge uniformly on (a,b]. (Notice a is not included in this interval.)

    Assume {fn} converges unifromly and then use the diverging sequence of real numbers as a contradiction to f being continuous in the end.

    © BrainMass Inc. brainmass.com December 24, 2021, 10:26 pm ad1c9bdddf
    https://brainmass.com/math/real-analysis/real-analysis-uniform-continuous-475453

    SOLUTION This solution is FREE courtesy of BrainMass!

    Proof:
    Since is uniformly convergent on the interval , assume the limit function is , then for any , we can find some , such that for all , we have for all . Then for all , we have

    Since are continuous function on , then we can find some , such that for any , we have and . Thus we have

    So is a Cauchy sequence and thus must be convergent. However, the condition says that is a divergent sequence. We get a contradiction.
    Therefore, is not uniformly convergent on the interval .

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:26 pm ad1c9bdddf>
    https://brainmass.com/math/real-analysis/real-analysis-uniform-continuous-475453

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