if $\{f_n\}$ converges point wise to $f$ on $[a,b]$. $f_n$ is nondecreasing for each n. Prove that $f$ is nondecresing.

Question

if $\{f_n\}$ converges point wise to $f$ on $[a,b]$. $f_n$ is nondecreasing for each n. Prove that $f$ is nondecresing.

My attempt:

Suppose for the sake of contradiction that $\exists x_1,x_2\in[a,b]$ such that $x_1<x_2$ but $f(x_1)>f(x_2)$.

Since $f_n(x_1)\to f(x_1)$ as $n\to\infty$, then: $$\forall \epsilon>0, \exists N_1\in\mathbb{N}\text{ such that }|f(x_1)-f_n(x_1)|<\epsilon, \forall n>N_1$$

Also, Since $f_n(x_2)\to f(x_2)$ as $n\to\infty$, then: $$\forall \epsilon>0, \exists N_2\in\mathbb{N}\text{ such that }|f(x_2)-f_n(x_2)|<\epsilon, \forall n>N_1$$

choose $\epsilon=\frac{D}{3}=\frac{f(x_1)-f(x_2)}{3}>0$, then \begin{align} f_n(x_2)-f_n(x_1)&=f_n(x_2)-f(x_2)+f(x_2)-f(x_1)+f(x_1)-f_n(x_1)\\ &\leq |f_n(x_2)-f(x_2)|+|f(x_2)-f(x_1)|+|f(x_1)-f_n(x_1)|\\ &\leq \dfrac{2D}{3}+|f(x_2)-f(x_1)|\\ &\leq \dfrac{2D}{3}+D\\ \end{align}

I could not get a contradiction.

Answer 1

Use $\epsilon = \frac{f(x_2) - f(x_1)}{2}>0$. Then we have:

$$f_n(x_2) < \epsilon + f(x_2) = \frac{f(x_1) + f(x_2)}{2} = f(x_1) - \epsilon < f_n(x_1)$$ which is a contradiction, as $f_n$ is non decreasing.

Answer 2

You assumed that $f(x_1)-f(x_2) =: \epsilon >0$. You have that eventually $f_n(x_i)$ is within $\frac{\epsilon}{2}$ of $f(x_i)$ for $i=1,2$. You also have that $f_n(x_1) < f_n(x_2)$.

For the contradiction we simply have the following:

$$f(x_1) \leq f_n(x_1) + \frac{\epsilon}{2} < f_n(x_2) + \frac{\epsilon}{2} \leq f(x_2) + \epsilon = f(x_1)$$.