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Homework Help: Uniform Convergence

  1. Sep 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Alright, here is the problem. Given a compact metric space [tex]X[/tex], and a sequence of functions fn which are continuous and [tex]f_{n}:X->R[/tex] (reals), also [tex]f_n->f[/tex] (where f is an arbitrary function [tex]f:X->R[/tex]). Also, given any convergent sequence in [tex]X[/tex] [tex]x_{n}->x[/tex], [tex]f_{n}(x_{n})->f(x)[/tex]. The problem is to show that fn converges uniformly to f.


    3. The attempt at a solution
    Alright, I can prove this relatively easily if I can prove that f (the limit function) is continuous. However, I don't know if this is possible, does anyone see a way to do this? Only little hints if you see a way.
     
    Last edited: Sep 4, 2010
  2. jcsd
  3. Sep 5, 2010 #2
    Nevermind continuity, I don't think the continuity of f directly matters in this situation. However, maybe a start on the problem would be to assume that f is not uniformly continuous, and then, picking an epsilon e>0, for each n, pick one x in X such that |fn(x)-f(x)|>=e, and so create a sequence of points in X, which then must have a convergent subsequence due to the compactness of X. From there on, I am stumped, even if this is the correct way to solve the problem.
     
  4. Sep 5, 2010 #3

    Office_Shredder

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    I would think proof by contradiction is a good way to go. If the fn do not converge to f uniformly, then for all [tex]\epsilon[/tex], there exists an infinite subsequence of functions [tex]f_{n_k}[/tex] such that for each such function, there is a point [tex]x_{n_k}[/tex] with [tex]|f(x_{n_k})-f_{n_k}(x_{n_k})|>\epsilon[/tex]

    Relabel these sequences as [tex]f_k[/tex] and [tex]x_k[/tex] for convenience. [tex]x_k[/tex] has a convergent subsequence which we will label [tex]y_k[/tex]. Now let [tex]f_k[/tex] just be the functions corresponding to the points [tex]y_k[/tex]

    What can you say about [tex]|f(y_k)-f_k(y_k)|[/tex]
     
  5. Sep 5, 2010 #4
    Haha, took a walk and that's exactly what came to me. I think I had a proof of a similar theorem stuck in my head and wanted to follow that technique.
     
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