If $ f_n\to f $ pointwise, and $f_n $ are increasing continuous functions, and $ f $ is continuous, then $f_n\to f $ uniformly

Question

let $ f_{n}:[a,b]\to\mathbb{R} $ be a sequence of increasing and continuous functions (each function is continuous and increasing, it dosent mean that the sequence is increasing). And assume that exists $ f:[a,b]\to\mathbb{R} $ such that $ f $ is continuous and $ f_{n}\to f $ pointwise. I have to prove that the convergence is also uniformly.

I tried to assume by contradiction that there is no uniformly convergence, but I got stuck. Now I have a new intuitio, since $ f_n $ and $ f $ are continious in $ [a,b] $ they are uniformly continuous. so actually for any $ x $ I can write:

$ |f_{n}\left(x\right)-f\left(x\right)|\leq|f_{n}\left(x\right)-f_{n}\left(x+\delta\right)|+|f\left(x\right)-f\left(x+\delta\right)|+|f_{n}\left(x+\delta\right)-f_{n}\left(x+\delta\right)| $

Where $ \delta $ would fit to some $ \varepsilon $ from the unifromly continuous definition, where the first two terms can be small as I want, I can continue with it untill $ x+k\delta $ "is close enough" to $ b $ where $ k \in \mathbb{R} $. But again, Im not sure how to do it since $ \delta $ would have to depend on $ n $.

Any hints would help. Thanks in advance

Answer 1

Let $\varepsilon>0$ and $$ V_n=\{x\in [a,b] : |f_n(x)-f(x)|<\varepsilon \}, \quad n\in\mathbb N. $$ Clearly, the $V_n$'s are open relative to $[a,b]$ and $\bigcup_{n\in\mathbb N}V_n=[a,b]$, and hence the $V_n$'s are an open cover of $[a,b]$.

Since $[a,b]$ is compact, there is a finite sub-cover $$ [a,b]= V_{n_1}\cup\cdots\cup V_{n_k} $$ where $n_1<n_2<\cdots<n_k$. But since $f_n$ is increasing, then $m<n\Longrightarrow V_m\subset V_n$, and hence $$ [a,b]= V_{n_k}, $$ i.e., $$ V_{n_k}=\{x\in [a,b] : |f_{n_k}(x)-f(x)|<\varepsilon \}=[a,b] $$ and hence $$ |f_{n_k}(x)-f(x)|<\varepsilon, $$ for all $x\in [a,b]$, and since $f_n$ is an increasing sequence, then $$ 0 \le f(x)-f_{n}(x)<\varepsilon, \quad \text{for all $n\ge n_k$ and $x\in[a,b]$.} $$

Answer 2

The answer by @Yiorgos is the original proof. Here is another one I would like tu submit. As $(f_n)$ converges increasingly, $f \geqslant f_n$ for all $n$ and $|f(x) - f_n(x)| = f(x) - f_n(x)$. Let $M_n = \sup_{x\in [a,b]}f(x) - f_n(x) \geqslant 0$. The fact that $f$ is the uniform limit of $(f_n)$ is equivalent to the fact that $M_n$ converges to $0$.

By the increasing hypothesis of $(f_n)$, $M_n$ is a decreasing sequence. It is moreover positive, thus have a limit, say $m$. By continuity, there exists $x_n \in [a,b]$ such that $M_n = f(x_n)-f_n(x_n)$. By compacity of $[a,b]$, $(x_n)$ has a converging subsequence $(x_{\varphi(n)})$ with limit $x \in [a,b]$.

To conclude, notice that $M_{\varphi(n)}=f(x_{\varphi(n)}) - f_{\varphi(n)}(x_{\varphi(n)})$, and thus : \begin{align} m = \lim_{n\to\infty}M_{\varphi(n)} = \lim_{n\to \infty}f(x_{\varphi(n)}) - f_{\varphi(n)}(x_{\varphi(n)}) = f(x) - f(x) = 0 \end{align}