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

    After some consideration it seems to me that the second problem is a direct consequence of the first, because if we assume that $\displaystyle {f_n}$ does converge uniformly to $\displaystyle f$ 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 $\displaystyle \epsilon >0$. Now choose $\displaystyle N_1$ such that $\displaystyle n \geq N_1$ implies $\displaystyle | f_n (x) - f(x) | \leq \frac{\epsilon}{2}$. (Possible since $\displaystyle f_n \to f$ uniformly.) Next, choose $\displaystyle \delta > 0$ such that $\displaystyle |p-x| < \delta$ implies $\displaystyle | f_n (p) - f_n (x) | < \frac{\epsilon}{2}$. (Possible since $\displaystyle f_n$ is continuous at $\displaystyle x$.) Now, choose $\displaystyle N_2$ such that $\displaystyle n \geq N_2 $ implies $\displaystyle |x_n-x| < \delta$. (Possible since $\displaystyle x_n \to x$.) Let $\displaystyle N$ be the maximum of $\displaystyle { N_1 , N_2 }$. Then $\displaystyle n \geq N$ implies $\displaystyle |x_n-x| < \delta$ which again implies $\displaystyle |f_n (x_n) - f_n (x) | < \frac{\epsilon}{2}$, but it also implies $\displaystyle | f_n (x) - f(x) | \leq \frac{\epsilon}{2}$. But then $\displaystyle | f_n (x_n) - f(x) | \leq | f_n (x_n) - f_n (x) | + | f_n (x) - f(x) | < \epsilon$, so $\displaystyle f_n (x_n) \to f(x)$.

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

    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 $\displaystyle \delta$ works for all $\displaystyle f_n$ with $\displaystyle n>N$ (notice that it depends on your choice of $\displaystyle n$)?
    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 $\displaystyle f_n$ is continuous at $\displaystyle x$, $\displaystyle f$ is continuous at $\displaystyle x$.

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

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

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

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

    This is me, btw: $\displaystyle \Rightarrow$


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

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