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Math Help - Pointwise Convergent & Monotone

  1. #1
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    Pointwise Convergent & Monotone

    Given f_n(x) \rightarrow F pointwise on a domain D, where each f_n(x) is monotone (either increasing or decreasing) on D, show that F is monotone on D .

    so what I have is:

    assume that f_n(x) is monotone increasing then:

    f_{n+1}(x) \geq f_x(n) therefore  f_{n+1}(x) -f_x(n) \geq 0

    by definition (the one that I am using) a function is pointwise convergent if and only if \lim_{n\rightarrow \infty} f_n(x) = F(x) and if and only if given an \varepsilon >0 \ \exists \ n_0 (dependent on both \varepsilon and x) \in \ \mathbb{N} such that |f_n(x)-F(x)|<\varepsilon

    so based on that if I let:

    \lim_{n\rightarrow \infty} f_n(x) = F(x) and \lim_{n\rightarrow \infty} f_{n+1}(x) = F'(x)

    therefore \lim_{n\rightarrow \infty} f_{n+1}(x) = F'(x) \geq \lim_{n\rightarrow \infty} f_n(x) = F(x) so F'(x) -F(x) \geq 0

    hence as f_n(x) increase so does F(x)

    is this correct?
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  2. #2
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    Quote Originally Posted by lllll View Post
    Given f_n(x) \rightarrow F pointwise on a domain D, where each f_n(x) is monotone (either increasing or decreasing) on D, show that F is monotone on D .

    so what I have is:

    assume that f_n(x) is monotone increasing then:

    f_{n+1}(x) \geq f_x(n) therefore  f_{n+1}(x) -f_x(n) \geq 0
    No. The function f_n(x) is increasing in x, not n. The definition should be that if y>x then f(y)>f(x).

    by definition (the one that I am using) a function is pointwise convergent if and only if \lim_{n\rightarrow \infty} f_n(x) = F(x) and if and only if given an \varepsilon >0 \ \exists \ n_0 (dependent on both \varepsilon and x) \in \ \mathbb{N} such that |f_n(x)-F(x)|<\varepsilon

    so based on that if I let:

    \lim_{n\rightarrow \infty} f_n(x) = F(x) (correct) and \lim_{n\rightarrow \infty} f_{n+1}(x) = F'(x) No. This should be \color{red}\lim_{n\rightarrow \infty} f_n(y) = F(y)
    ... from which you should be able to deduce that F(y) > F(x).
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