Krdytkyu

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

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

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

$f_n to f$ pointwisely in $mathbb{R} iff limlimits_{ntoinfty}f_n(x)=f(x)$ $forall x in mathbb{R}$.

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

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

$implies limlimits_{n to infty}sup{|f_n(x)-f(x)|:xinmathbb{R}} = 0$.

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.

Since $fto 1$ and $fto 0$ as $xto infty$ and $xto -infty$ 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. $nge N$ 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 $[-M,M]$, $|f_n(x)-f(x)|<2epsilon$ 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 $nge N_i$ implies $|f_n(t_i)-f(t_i)|<epsilon$. Fix $T = max{N_i, N}$ and pick $xin A_i$ and $nge T$. It follows that $f_n(t_{i+1})-f_n(x)le f_n(t_{i+1})-f_n(t_i)le (f(t_{i+1})+epsilon)-(f(t_i)-epsilon)le 3epsilon$. By the Triangle Ineq.
$$begin{align}|f_n(x)-f(x)|&le |f_n(x)-f_n(t_{i+1})|+|f_n(t_{i+1})-f(t_{i+1})| + |f(t_{i+1})-f(t_i)| \ &le 5epsilon end{align}$$
But $A_i$ cover $[-M,M]$, so this holds for all $x in [-M,M]$. Thus, $nge T$ forces $|f_n(x)-f(x)|le 5epsilon$ and the convergence is uniform.

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 $pminfty$ in order to have the uniform convergence also on the complement of $[-R,R]$.