Department of Mathematics

Colorado State University

Math 317: Homework 6

Due: Friday, March 15, 2019
  1. (19.4)

    1. Prove that if is uniformly continuous on a bounded set , then is a bounded function on . (Hint: Try proof by contradiction.)

      Proof. Assume that is uniformly continuous on but suppose to the contrary that is unbounded on . Then for all , there exists as number so that . Use these to define the sequence .

      is bounded, so by the Bolzano-Weierstrauss theorem, there exists a convergent, hence Cauchy, subsequence . As is assumed to be uniformly continuous, this means that is itself a Cauchy, hence convergent, sequence. Say it converges to the number .

      But for all , we have that . This implies that the sequence cannot possibly converge to , a contradiction. Hence must be bounded on .

    2. Explain why (a) gives a proof that is not uniformly continuous on .

      is unbounded on , so it cannot be uniformly continuous.

  2. (23.1, 23.2) For each of the following power series, find the radius of convergence and determine the exact interval of convergence.

    1. The interval of convergence is .

    2. The interval of convergence is .

    3. The interval of convergence is .

    4. Notice that the sequence of terms is if for some , or otherwise (if the series index starts at 0 instead of 1, then , but this does not change the tail of the sequence). Hence the limit superior of the roots of terms is going to be .

      The interval of convergence is .

  3. (24.4) For , let .

    1. Find , the pointwise limit of .

    The pointwise limit is the function

    To see this, take to be in any of these cases and evaluate the limit .

    1. Determine whether uniformly on .

    It cannot, as the pointwise limit is not continuous on , while each of the functions are.

    1. Determine whether uniformly on .

    It cannot, as the pointwise limit is not continuous on , while each of the functions are.