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Math Help - Uniform convergence

  1. #1
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    Uniform convergence

    Suppose that f:[0,1] \rightarrow \mathbb{R} is continuous and that {f_n} is such that  f_n \rightarrow f pointwise.

    Show that  f_n \rightarrow f uniformly if and only if for every sequence  {x_n} such that  x_n \rightarrow x , we have that  f_n(x_n) \rightarrow f(x) .



    My work: I've finished the forward implication. This is my work on the reverse. We'll prove the contrapositive.

    Suppose that  f_n does not converge to f uniformly. So, there is some  \epsilon > 0 so that for any  N \in \mathbb{Z} , there is an  n \ge N and an  x_n so that  |f(x_n) - f_n(x_n)| \ge 2\epsilon .

    So,  {x_n} is a bounded sequence of reals, and so has a convergent subsequence, say {x_{n'}} such that  x_{n'} \rightarrow x \in [0,1]

    Since f is continuous at x, we can find an N \in \mathbb{Z} such that for any n > N,  |f(x) - f(x_n)| < \epsilon

    But then, if n' > N, we have that
     2\epsilon \le |f(x_{n'}) - f_{n'}(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + |f(x) - f(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + \epsilon

    or

     \epsilon \le |f_{n'}(x_{n'}) - f(x)|

    and so  f_{n'}(x_{n'}) does not converge to f(x).

    Something about this argument feels incredibly wrong, and I was wondering if anyone could point out where I went awry.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Math Major View Post
    Suppose that f:[0,1] \rightarrow \mathbb{R} is continuous and that {f_n} is such that  f_n \rightarrow f pointwise.

    Show that  f_n \rightarrow f uniformly if and only if for every sequence  {x_n} such that  x_n \rightarrow x , we have that  f_n(x_n) \rightarrow f(x) .



    My work: I've finished the forward implication. This is my work on the reverse. We'll prove the contrapositive.

    Suppose that  f_n does not converge to f uniformly. So, there is some  \epsilon > 0 so that for any  N \in \mathbb{Z} , there is an  n \ge N and an  x_n so that  |f(x_n) - f_n(x_n)| \ge 2\epsilon .

    So,  {x_n} is a bounded sequence of reals, and so has a convergent subsequence, say {x_{n'}} such that  x_{n'} \rightarrow x \in [0,1]

    Since f is continuous at x, we can find an N \in \mathbb{Z} such that for any n > N,  |f(x) - f(x_n)| < \epsilon

    But then, if n' > N, we have that
     2\epsilon \le |f(x_{n'}) - f_{n'}(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + |f(x) - f(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + \epsilon

    or

     \epsilon \le |f_{n'}(x_{n'}) - f(x)|

    and so  f_{n'}(x_{n'}) does not converge to f(x).

    Something about this argument feels incredibly wrong, and I was wondering if anyone could point out where I went awry.
    I don't see where you have produced an actual sequence. You start with some point x_n, and go from there. You need to produce a full-fledged sequence.
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  3. #3
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    For any  N \in \mathbb{Z} , there is an  n > N and an  x_n so that  |f(x_n) - f_n(x_n)| \ge 2\epsilon , which is where I get the sequence from. Although I guess it would've been more clear if I had labeled the x_n's as x_N?
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