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Homework Help: A theorem on uniform convergence

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Theorem to be proved:
    fn is uniformly convergent on E iff for every epsilon >0 there is an N, where n,m >= N then | fn(x) - fm(x) | < epsilon, for all x in E.

    2. Relevant equations
    Definition: fn is uniformly continuous on E if there is an f such that for every epsilon >0, there is an N such that for n >= N: | fn(x) - f(x) | < epsilon

    3. The attempt at a solution
    (only if):
    Assume fn is uniformly convergent. Then
    [tex]\forall x \in E, \forall n,m >= N, \left| f_n (x) - f(x) \right| < \frac{\epsilon}{2} and \left| f_m (x) - f(x) \right| < \frac{\epsilon}{2} [/tex] Thus
    [tex]\left| f_n (x) - f_m (x) \right| \leq \left| f_n (x) - f(x) + f(x) - f_m(x) \right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon[/tex] Great.

    Now here comes the part I wish I was doing better. I just feel like something is off. So I have two ways.

    Way number 1):
    Assume [tex] \left| f_n (x) - f(x) \right| < \epsilon \forall x \in E, \forall m,n \geq N[/tex] . Then fn is cauchy, and because fn is cauchy, fn is bounded on E. Then let M = max{fn(x)} x in E. Also because fn is cauchy it is convergent, and because fn <= M for all x in E, fn is convergent for all x in E. By the definition of a convergent sequence | fn(x) - f(x) | < epsilon.

    Way number 2):
    Assume [tex] \left| f_n (x) - f_m(x) \right| < \epsilon \forall x \in E, \forall m,n \geq N[/tex]. Let f(x) be an accumulation point of \{f_n(x) : f \in E\} and, let n*>= N where [tex] f_{n*} \in (f(x) - \epsilon, f(x) + \epsilon) thus \left| f_{n*} (x) - f(x) \right| < \frac{\epsilon}{2} [/tex]. Therefore

    [tex] \left| f_n (x) - f(x) \right| = \left| f_n (x) - f_{n*}(x) + f_{n*}(x) - f(x) \right| \leq \left| f_n (x) - f_{n*}(x) \right| + \left| f_{n*}(x) (x) - f(x) \right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \forall x \in E, \forall m,n \geq N[/tex]

    Way number two should work, but gets ugly. I don't think way number one works.
    Last edited: Dec 2, 2008
  2. jcsd
  3. Dec 2, 2008 #2
    Assume fn is Cauchy. Then fn is bounded (easy to show). Thus it contains a convergent subsequence, call it gn. Let g = lim gn. By fn Cauchy we have |fn - fm| < e/2 and gn convergent to g implies |gn - g| < e/2
    So |fn - gn + gn - g| < (or equal) |fn - gn| + |gn - g| < e
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