# Pointwise Convergent & Monotone

• Mar 21st 2009, 06:32 PM
lllll
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?
• Mar 22nd 2009, 01:03 PM
Opalg
Quote:

Originally Posted by lllll
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).