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Thread: Pointwise Convergent & Monotone

  1. #1
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    Pointwise Convergent & Monotone

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

    so what I have is:

    assume that f_n(x) is monotone increasing then:

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

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

    so based on that if I let:

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

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

    hence as $\displaystyle f_n(x)$ increase so does $\displaystyle F(x)$

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

    so what I have is:

    assume that f_n(x) is monotone increasing then:

    $\displaystyle f_{n+1}(x) \geq f_x(n) $ therefore $\displaystyle f_{n+1}(x) -f_x(n) \geq 0$
    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 $\displaystyle \lim_{n\rightarrow \infty} f_n(x) = F(x)$ and if and only if given an $\displaystyle \varepsilon >0 \ \exists \ n_0$ (dependent on both $\displaystyle \varepsilon $ and $\displaystyle x$) $\displaystyle \in \ \mathbb{N}$ such that $\displaystyle |f_n(x)-F(x)|<\varepsilon$

    so based on that if I let:

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