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Math Help - continuous, sequence, integral

  1. #1
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    continuous, sequence, integral

    Suppose that f: [0,1] \rightarrow \mathbb{R} is continuous. For each n \in \mathbb{N}, define g_n : [0,1] \rightarrow \mathbb{R} by g_n(x)=f(x^n) for all  x \in [0,1]. Prove that the sequence (\int^1_0 g_n(x)) converges to f(0).
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    This question really confuses me. I don't see how this sequence converges to f(0). A few helpful hints on this one would be appreciated. Thanks in advance.
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  2. #2
    Moo
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    Hello,

    Firstly, you have to prove that \lim_{n \to \infty} \int_0^1 g_n(x) ~dx=\int_0^1 \lim_{n \to \infty} g_n(x) ~dx, by using the dominated convergence theorem for example (f is continuous over a closed set, --> it is bounded)

    Secondly, you have to prove that g_n(x) \xrightarrow[n \to \infty]{} f(0)
    In order to show that, write the epsilon definition of this limit : x^n \xrightarrow[n \to \infty]{} 0, when x \in [0,1) (you can "miss" the case x=1, because it's a neglectible case when dealing with integrals... dunno if you know how to deal with it)
    Then use the epsilon-delta method on the continuity of f, and you're done.
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    Quote Originally Posted by Moo View Post
    Hello,

    Firstly, you have to prove that \lim_{n \to \infty} \int_0^1 g_n(x) ~dx=\int_0^1 \lim_{n \to \infty} g_n(x) ~dx, by using the dominated convergence theorem for example (f is continuous over a closed set, --> it is bounded)

    Secondly, you have to prove that g_n(x) \xrightarrow[n \to \infty]{} f(0)
    In order to show that, write the epsilon definition of this limit : x^n \xrightarrow[n \to \infty]{} 0, when x \in [0,1) (you can "miss" the case x=1, because it's a neglectible case when dealing with integrals... dunno if you know how to deal with it)
    Then use the epsilon-delta method on the continuity of f, and you're done.
    What happens for the case x=1? Can we overlook the case x=1 or does this case also satisfy the requirement that (\int^1_0 g_n(x)) converges to f(0)?
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  4. #4
    Moo
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    Quote Originally Posted by maursvolta41 View Post
    What happens for the case x=1? Can we overlook the case x=1 or does this case also satisfy the requirement that (\int^1_0 g_n(x)) converges to f(0)?
    It does satisfy the requirement. Because when dealing with an integral, a single point can be omitted (in the case of a continuous function)


    If you know some measure theory, you can say that {1} is a set of Lebesgue measure equal to 0. So x^n converges almost everywhere to 0.
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