# 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 '14 at 17:41
• @DavidMitra But then $f_n$ is not monotone, which was one of the requirements. – mlk Jun 14 '14 at 17:43
• @mlk $f_n$ is monotone, fix $x$ and increase $n$. – BlueBuck Jun 14 '14 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 '14 at 17:48
• @DavidMitra Can anything be done if we add the hypothesis that each $f_n$ is a monotone function? – BlueBuck Jun 14 '14 at 17:48

## 1 Answer

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 – Charlie Tian Nov 29 '17 at 5:04
• @NikPronko: thanks for pointing out the typo. It should be OK now. – Etienne Dec 9 '17 at 10:59
• @CharlieTian Thanks, for the same reason. – Etienne Dec 9 '17 at 11:00