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Math Help - Some problems concerning uniform convergence

  1. #1
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    Some problems concerning uniform convergence

    Hi, I have some problems with a couple of exercises concerning uniform convergence, and am hoping for some helpful hints.

    Problem 1: Let (X, d_X) be a metric space, f, f_n : X -> C (complex numbers) for n = 1,2,... and f_n converges uniformly to f on X. If each f_n is continuous at x in X, and if x_n -> x in X, prove that lim n->infinity f_n(x_n) = f(x).

    Problem 2: Suppose (X, d_X) is a metric space, that f_n : X -> C is continuous for each n, and that {f_n} converges pointwise to f on X. If there exists a sequence {x_n} in X such that x_n -> x in X but f_n(x_n) does not converge to f(x), show that {f_n} does not converge uniformly to f on X.

    I'm not sure how hard these really are, but they're so full of information that I can't figure out where to begin (plus, I'm in a hurry, exam's on Friday .
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  2. #2
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    OK, I think I've solved the first one:

    Fix \epsilon >0. Now choose N_1 such that n \geq N_1 implies | f_n (x) - f(x) | \leq \frac{\epsilon}{2}. (Possible since f_n \to f uniformly.) Next, choose \delta > 0 such that |p-x| < \delta implies | f_n (p) - f_n (x) | < \frac{\epsilon}{2}. (Possible since f_n is continuous at x.) Now, choose N_2 such that n \geq N_2 implies |x_n-x| < \delta. (Possible since x_n \to x.) Let N be the maximum of { N_1 , N_2 }. Then n \geq N implies |x_n-x| < \delta which again implies |f_n (x_n) - f_n (x) | < \frac{\epsilon}{2}, but it also implies | f_n (x) - f(x) | \leq \frac{\epsilon}{2}. But then | f_n (x_n) - f(x) | \leq | f_n (x_n) - f_n (x) | + | f_n (x) - f(x) | < \epsilon, so f_n (x_n) \to f(x).

    After some consideration it seems to me that the second problem is a direct consequence of the first, because if we assume that {f_n} does converge uniformly to f we get a contradiction. Is this correct?
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  3. #3
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    Quote Originally Posted by claves View Post
    OK, I think I've solved the first one:

    Fix \epsilon >0. Now choose N_1 such that n \geq N_1 implies | f_n (x) - f(x) | \leq \frac{\epsilon}{2}. (Possible since f_n \to f uniformly.) Next, choose \delta > 0 such that |p-x| < \delta implies | f_n (p) - f_n (x) | < \frac{\epsilon}{2}. (Possible since f_n is continuous at x.) Now, choose N_2 such that n \geq N_2 implies |x_n-x| < \delta. (Possible since x_n \to x.) Let N be the maximum of { N_1 , N_2 }. Then n \geq N implies |x_n-x| < \delta which again implies |f_n (x_n) - f_n (x) | < \frac{\epsilon}{2}, but it also implies | f_n (x) - f(x) | \leq \frac{\epsilon}{2}. But then | f_n (x_n) - f(x) | \leq | f_n (x_n) - f_n (x) | + | f_n (x) - f(x) | < \epsilon, so f_n (x_n) \to f(x).

    After some consideration it seems to me that the second problem is a direct consequence of the first, because if we assume that {f_n} does converge uniformly to f we get a contradiction. Is this correct?
    Just one thing: How do you claim that this \delta works for all f_n with n>N (notice that it depends on your choice of n)?

    Your answer to the second one is correct though
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  4. #4
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    Quote Originally Posted by Jose27 View Post
    Just one thing: How do you claim that this \delta works for all f_n with n>N (notice that it depends on your choice of n)?
    Thanks for pointing that out. I've tried to find a solution, but I am struggling with it, so any ideas would be greatly appreciated.
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  5. #5
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    New try:

    First observe that since each f_n is continuous at x, f is continuous at x.

    Fix \epsilon >0. Now choose N_1 such that n \geq N_1 implies | f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}. (Possible since f_n \to f uniformly.) Next, choose \delta > 0 such that |p-x| < \delta implies | f (p) - f (x) | < \frac{\epsilon}{2}. (Possible since f is continuous at x.) Now, choose N_2 such that n \geq N_2 implies |x_n-x| < \delta. (Possible since x_n \to x.) Let N be the maximum of \{ N_1 , N_2 \}. Then n \geq N implies |x_n-x| < \delta which again implies |f (x_n) - f (x) | < \frac{\epsilon}{2}, but it also implies | f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}. But then | f_n (x_n) - f(x) | \leq | f_n (x_n) - f (x_n) | + | f (x_n) - f(x) | < \epsilon, so f_n (x_n) \to f(x).

    This is me, btw: \Rightarrow
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  6. #6
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    Quote Originally Posted by claves View Post
    New try:

    First observe that since each f_n is continuous at x, f is continuous at x.

    Fix \epsilon >0. Now choose N_1 such that n \geq N_1 implies | f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}. (Possible since f_n \to f uniformly.) Next, choose \delta > 0 such that |p-x| < \delta implies | f (p) - f (x) | < \frac{\epsilon}{2}. (Possible since f is continuous at x.) Now, choose N_2 such that n \geq N_2 implies |x_n-x| < \delta. (Possible since x_n \to x.) Let N be the maximum of \{ N_1 , N_2 \}. Then n \geq N implies |x_n-x| < \delta which again implies |f (x_n) - f (x) | < \frac{\epsilon}{2}, but it also implies | f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}. But then | f_n (x_n) - f(x) | \leq | f_n (x_n) - f (x_n) | + | f (x_n) - f(x) | < \epsilon, so f_n (x_n) \to f(x).

    This is me, btw: \Rightarrow


    Now, if you're interested you could try an prove the converse, namely:

    If for every sequence x_n\rightarrow x we have f_n(x_n)\rightarrow f(x) then f_n\rightarrow f uniformly on compact subsets of X
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