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

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.

• Remark: $D$ is the collection of functions that correspond to the distribution function (CDF) of continuous random variables. – Sean Roberson Nov 27 '18 at 19:52

## 2 Answers

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.

• Nice (full) answer (+1) – Robert Z Nov 28 '18 at 6:44

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 $$[-R,R]$$.

• The version of Dini's Theorem you link to requires $f_n$ to converge monotonically to $f$ but in this problem, that need not be the case. Is there a more general version of the theorem? – Guacho Perez Nov 27 '18 at 21:41
• @GuachoPerez Thanks for pointing out. I edited my answer. – Robert Z Nov 28 '18 at 6:42