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Homework Statement
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.
Homework 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
The Attempt at a Solution
(only if):
Assume fn is uniformly convergent. Then
\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} Thus
\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 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 \left| f_n (x) - f(x) \right| < \epsilon \forall x \in E, \forall m,n \geq N . 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 \left| f_n (x) - f_m(x) \right| < \epsilon \forall x \in E, \forall m,n \geq N. Let f(x) be an accumulation point of \{f_n(x) : f \in E\} and, let n*>= N where f_{n*} \in (f(x) - \epsilon, f(x) + \epsilon) thus \left| f_{n*} (x) - f(x) \right| < \frac{\epsilon}{2}. Therefore
\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
Way number two should work, but gets ugly. I don't think way number one works.
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