# Sequence of monotone functions converging to a continuous limit, is the convergence uniform?

I'm reading some extreme value theory and in particular regular variation in Resnick's 1987 book Extreme Values, Regular Variation, and Point Processes, and several times he has claimed uniform convergence of a sequence of functions because "monotone functions are converging pointwise to a continuous limit". I am finding this reasoning a little dubious.

EDIT: Thanks to some comments below I realized I was confused, I really want each $f_n$ to be a monotone function on $[a,b]$, so it isn't quite like Dini's theorem. Formally I am wondering:

Let $f_n:[a,b]\rightarrow\mathbb{R}$ be a sequence of non-decreasing functions converging pointwise to the continuous function $f$. Is the convergence uniform?

My thoughts:

I have been thinking that since the set of discontinuities of each function $f_n$ is at most countable, the set of discontinuities of the entire sequence is at most countable...something something? Any thoughts would be greatly appreciated!

• @DavidMitra Isn't the limit then $f(0) = 1$, $f(x)=0$ otherwise and therefore not continuous?
– mlk
Jun 14, 2014 at 17:41
• @DavidMitra But then $f_n$ is not monotone, which was one of the requirements.
– mlk
Jun 14, 2014 at 17:43
• @mlk $f_n$ is monotone, fix $x$ and increase $n$. Jun 14, 2014 at 17:44
• Ahh, sorry, the "sequence of non-decreasing functions" was a bit misleading, I thought you wanted a theorem for functions monotonic in $x$, not in $n$. (Which I think, I have seen somewhere, but cannot remember the name of)
– mlk
Jun 14, 2014 at 17:48
• @DavidMitra Can anything be done if we add the hypothesis that each $f_n$ is a monotone function? Jun 14, 2014 at 17:48

Let $f:=\lim f_n$, and let $\varepsilon >0$ be given. We have to show that if $n$ is large enough, then $$\vert f_n(t)-f(t)\vert\leq\varepsilon\quad \hbox{for all t\in[a,b]}\, .$$

Since $f$ is assumed to be continuous, it is uniformly continuous on the compact interval $[a,b]$. So we may find a subdivision $a=t_0<t_1\dots <t_K=b$ of $[a,b]$ such that the oscillation of $f$ on each interval $[t_i,t_{i+1}]$ is less than $\varepsilon/2$.

Since $f_n(t_i)\to f(t_i)$ as $n\to\infty$ for $i=0,\dots ,K$, one can find $N$ such that if $n\geq N$, then $$\vert f_n(t_i)-f(t_i)\vert\leq\varepsilon/2\quad\hbox{for i=0,\dots K}\, .$$

Let us check that $\vert f_n(t)-f(t)\vert\leq \varepsilon$ for every $n\geq N$ and all $t\in [a,b]$.

Fix $n\geq N$, and take any $t\in [a,b]$. One can choose $i$ such that $t\in [t_i,t_{i+1}]$. Since the functions $f_n$ are non-decreasing we have $$f_n(t_i)\leq f_n(t)\leq f_n(t_{i+1})\, .$$ Since $\vert f_n(t_i)-f(t_i)\vert$ and $\vert f_n(t_{i+1})-f(t_{i+1})\vert$ are not greater than $\varepsilon/2$, it follows that $$f(t_i)-\varepsilon/2\leq f_n(t)\leq f(t_{i+1})+\varepsilon/2\, .$$ Moreover, since the oscillation of $f$ on $[t_i,t_{i+1}]$ is less than $\varepsilon/2$ and since $t\in [t_i,t_{i+1}]$, we also have $f(t_i)\geq f(t)-\varepsilon/2$ and $f(t_{i+1})\leq f(t)+\varepsilon/2$. Altogether, this gives $$f(t)-2\varepsilon/2\leq f_n(t)\leq f(t)+2\varepsilon/2\, ,$$ which concludes the proof.

• f(ti)−ε/2≤fn(t)≤fn(ti+1)+ε/2. this line is wrong, as the right side of the inequality should be symmetric and just be the limit function. That gives the desired end form Nov 29, 2017 at 5:04
• @NikPronko: thanks for pointing out the typo. It should be OK now. Dec 9, 2017 at 10:59
• @CharlieTian Thanks, for the same reason. Dec 9, 2017 at 11:00
• I have tried to include definitions of pointwise, montone, continuity and uniform continuity but to no avail. Can you explain how did you come up with the partition technique? Is this the first question you pull this off or are there precedents for this established pattern? Jun 22, 2021 at 3:01

Definition. Let $$X$$ be a set and let $$Y$$ be a metric space. A sequence of mappings $$u_n:X \to Y$$ converges uniformly to $$u:X \to Y$$ if $$\lim_{n \to \infty} \sup_{x \in X} d(u_n(x),u(x)) = 0.$$

Proof. Assume $$(u_n)$$ does not converge uniformly to $$u$$. There exists $$\varepsilon > 0$$ and a sequence $$(x_n) \subset [a,b]$$ such that for all $$n$$, $$\varepsilon \le |u_n(x_n) - u(x_n)|.$$ Since $$[a,b]$$ is compact, $$(x_n)$$ contains a convergent subsequence. Assume, without loss of generality, that $$(x_n)$$ converges to $$x_0 \in [a,b]$$. By the Heine-Cantor theorem, $$u:[a,b] \to \mathbb{R}$$ is uniformly continuous. Thus, there exists $$\delta > 0$$ such that $$$$x,y \in (x_0-\delta,x_0+\delta) \implies |u(x) - u(y)| < \frac{\varepsilon}{2}.\tag{1}\label{1}$$$$ Since $$(x_n)$$ converges to $$x_0 \in (x_0 - \delta, x_0 + \delta)$$, there exists $$N$$ such that if $$n \ge N$$, then $$x_n \in (x_0-\delta,x_0+\delta)$$. Since $$(u_n)$$ converges to $$u$$ pointwise, there exists $$K \ge N$$ such that if $$k \ge K$$, then $$$$|u_k(x_0+\delta) - u(x_0+\delta)| < \frac{\varepsilon}{2} \tag{2}\label{2}$$$$ and $$$$|u_k(x_0-\delta) - u(x_0-\delta)| < \frac{\varepsilon}{2}. \tag{3}\label{3}$$$$ Fix $$k \ge K$$. We can deduce from $$\eqref{2}$$ and $$\eqref{3}$$ that $$u_k(x_0+\delta) < u(x_0 + \delta) + \frac{\varepsilon}{2}$$ and $$u(x_0-\delta) - \frac{\varepsilon}{2} < u_k(x_0-\delta).$$ Since $$u_k$$ is nondecreasing and $$x_0 - \delta < x_k < x_0 + \delta$$, $$u(x_0-\delta) - \frac{\varepsilon}{2} < u_k(x_0-\delta) \le u_k(x_k) \le u_k(x_0+\delta) < u(x_0 + \delta) + \frac{\varepsilon}{2}.$$ Subtract by $$u(x_k)$$ to get $$$$u(x_0-\delta) - u(x_k) - \frac{\varepsilon}{2} < u_k(x_k) - u(x_k) < u(x_0 + \delta) - u(x_k) + \frac{\varepsilon}{2}.\tag{4}\label{4}$$$$ By $$\eqref{1}$$, we know that $$-\frac{\varepsilon}{2} \le u(x_0-\delta) - u(x_k)$$ and $$u(x_0 + \delta) - u(x_k) \le \frac{\varepsilon}{2}.$$ Combining these deductions with $$\eqref{4}$$, we obtain $$-\varepsilon=-\frac{\varepsilon}{2} -\frac{\varepsilon}{2} < u_k(x_k) - u(x_k) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$ whence $$|u_k(x_k) - u(x_k)| < \varepsilon.$$ By assumption, $$\varepsilon \le |u_k(x_k) - u(x_k)|$$, so we get the contradiction $$\varepsilon < \varepsilon$$.