Prove $f_n \to f$ uniformly on $\mathbb{R}$

Question

Denote by $D$ the set of all continuous, increasing functions $f: \mathbb{R} \to [0, \infty)$ such that $\lim \limits_{x \to - \infty} f(x) = 0 $ and $\lim \limits_{x \to + \infty} f(x) = 1 $. If $f_n,f \in D$ for all $n \geq 1$ and $f_n \to f$ pointwisely on $\mathbb{R}$, prove that $f_n \to f$ uniformly on $\mathbb{R}$.

My take on it:

NTS: $f_n\to f$ uniformly on $\mathbb{R}$ if $M_n = \sup|f_n(x) - f(x)|$ exists for all sufficiently large n, and $\lim\limits_{n \to \infty} M_n = 0, x \in \mathbb{R}$.

$\lim\limits_{x \to \infty} f(x) = 1$ $\iff$ $\forall \epsilon > 0$ given $\exists M \gt0$ such that $|f(x) - 1| \lt \epsilon$,

$\lim\limits_{x \to - \infty}f(x) = 0 \iff \forall \epsilon \gt 0$ given $\exists M \gt 0$ such that $|f(x)| \lt \epsilon$

$f_n \to f$ pointwisely in $\mathbb{R} \iff \lim\limits_{n\to\infty}f_n(x)=f(x)$ $\forall x \in \mathbb{R}$.

$\forall \epsilon \gt 0, \exists M \gt0$ such that $|f_n(x) - f(x)|\lt\epsilon$ $\forall n \geq M, \forall x \in \mathbb{R}$

$\implies \lim\limits_{n \to \infty}|f_n(x)-f(x)|=0$

$\implies \lim\limits_{n \to \infty}\sup{|f_n(x)-f(x)|:x\in\mathbb{R}} = 0$.

If someone can please help me and let me know if I am going in the right direction and how to proceed forward.

Thanks in advance.

Answer 1

No, the last step does not work: pointwise convergence does not imply uniform convergence. You are not using the increasing property of such functions.

Hint. Use the second theorem given HERE on the compact set $[-R,R]$ (it is a variant of Dini's Theorem) and use the assumption about the limits at $\pm\infty$ in order to have the uniform convergence also on the complement of $[-R,R]$.

Answer 2

Since $f\to 1$ and $f\to 0$ as $x\to \infty$ and $x\to -\infty$ respectively, for fixed $\epsilon>0$ there exists $M$ s.t. $|x|\ge M$ implies either $f(x)>1-\epsilon$ or $f(x)<\epsilon$. By the convergence of $f_n$, there exists $N$ s.t. $n\ge N$ implies $f_n(M)>1-\epsilon$ and $f_n(-M)<\epsilon$. But each $f_n$ is increasing, so $|x|>M$ implies either $f_n(x)>1-\epsilon$ or $f_n(x)<\epsilon$. Therefore, for $x$ outside $[-M,M]$, $|f_n(x)-f(x)|<2\epsilon$ by the Triangle Inequality.

By the continuity of $f$ on the compact $[-M,M]$, there exists $\delta>0$ s.t. $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. Partition $[-M,M]$ into intervals $A_i=[t_i,t_{i+1})$ of length shorter than $\delta$. For each $t_i$ there exists $N_i$ where $n\ge N_i$ implies $|f_n(t_i)-f(t_i)|<\epsilon$. Fix $T = \max\{N_i, N\}$ and pick $x\in A_i$ and $n\ge T$. It follows that $f_n(t_{i+1})-f_n(x)\le f_n(t_{i+1})-f_n(t_i)\le (f(t_{i+1})+\epsilon)-(f(t_i)-\epsilon)\le 3\epsilon$. By the Triangle Ineq. $$\begin{align}|f_n(x)-f(x)|&\le |f_n(x)-f_n(t_{i+1})|+|f_n(t_{i+1})-f(t_{i+1})| + |f(t_{i+1})-f(t_i)| \\ &\le 5\epsilon \end{align}$$ But $A_i$ cover $[-M,M]$, so this holds for all $x \in [-M,M]$. Thus, $n\ge T$ forces $|f_n(x)-f(x)|\le 5\epsilon$ and the convergence is uniform.