# Stuck on proving uniform convergence

I am preparing for my analysis finals tomorrow and I have been stuck over two hours trying to solve the following problem:

Let $f_1, f_2, \ldots$ be a convergent (pointwise) sequence of monotonically increasing functions defined on $[a,b] \to \mathbb R$. (I.e., $f_n(x) \leq f_n(y)$ if $x \leq y$). Let $f$ be the limit of the above mentioned sequence. Assume $f$ is continuous. Show that the above sequence is uniformly convergent.

I am not sure how to approach the above problem. I have been trying to show the above by showing that the window of values taken by $f_N$ (given by $f_n(b) - f_n(a)$) converges and messing around with triangle inequalities to get the required inequality ($f_n(p) - f(p) < \varepsilon$). But this I realized was wrong because even if the windows converge the functions themselves can be increasing at different rates within the interval thus the ($f_n(p) - f(p)$ need not shrink at a constant rate at all points). I feel that the fact $f_n$ is defined in a compact interval and therefore is uniformly continuous comes into the picture somehow, but I can't connect the dots. Any suggestions?

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– Alex Youcis Dec 15 '11 at 3:24

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Given an $\epsilon > 0$, at every point $x$, continuity of $f(x)$ gives you what? Now you have a collection of things, one for each $x$ in your domain...this defines something useful! What nice property does compactness give you? Why does that matter and how does it show uniform convergence instead of pointwise convergence? – tomcuchta Dec 15 '11 at 3:27

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@AlexYoucis No,it's not dini's thm.In dini's thm,the value taken by the function for succesive n's is inreasing.Here,it is the function that is increasing.(i.e $f_n(x) <= f_n(y)$ if $x$ < $y$ ,not $f_{n+1}(p) >= f_n(p)$ for all p) – Thiagarajan Dec 15 '11 at 3:28
Ahh, misread it. I'm sorry. – Alex Youcis Dec 15 '11 at 3:30

## 1 Answer

Since $f$ is uniformly continuous and increasing, given $\varepsilon>0$, there exist points $a=x_0<x_1<x_2<\cdots<x_m=b$ such that $f(x_k)-f(x_{k-1})<\varepsilon$ for each $k$. There exists $N$ such that $n\geq N$ implies $|f_n(x_k)-f(x_k)|<\varepsilon$ for each $k$. For $x\in[a,b]$, $x$ is in $[x_k,x_{k+1}]$ for some $k$, and if $n\geq N$, then

$$|f(x)-f_n(x)|\leq|f(x)-f(x_k)|+|f(x_k)-f_n(x_k)|+|f_n(x_k)-f_n(x)|.$$

Take that last term:

$\begin{align*} |f_n(x_k)-f_n(x)|&=f_n(x)-f_n(x_k)\\ &\leq f_n(x_{k+1})-f_n(x_k)\\ &=|f_n(x_{k+1})-f_n(x_k)|\\ &\leq|f_n(x_{k+1})-f(x_{k+1})|+|f(x_{k+1})-f(x_k)|+|f(x_k)-f_n(x_k)|. \end{align*}$

Now you should be able to see why this gives you $|f(x)-f_n(x)|<5\varepsilon$. (Note where increasing is used.) There are probably more elegant solutions.

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Thank you.The key idea which I didn't find was the fact the I can use uniform continuity to partition $[a,b]$ into finitely many sub-intervals.Thanks for clearing that up.I can sleep now.:) – Thiagarajan Dec 15 '11 at 3:57
Thiagarajan: Actually you can get away without thinking about uniform continuity, but instead the intermediate value theorem.  Since $f$ is increasing and continuous, $f([a,b])=[f(a),f(b)]$, and if you break up $[f(a),f(b)]$ with points $f(a)=y_0<y_1<y_2<\cdots<y_m=f(b)$ such that $y_{k+1}-y_k<\varepsilon$ for each $k$, by continuity there exists $x_k$ such that $f(x_k)=y_k$. – Jonas Meyer Dec 15 '11 at 4:01
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