Suppose that $\{f_n\}$ is a sequence of nondecreasing functions which map the unit interval into itself. Suppose that $$\lim_{n\rightarrow \infty} f_n(x)=f(x)$$ pointwise and that $f$ is a continuous function. Prove that $f_n(x) \rightarrow f(x)$ uniformly as $n \rightarrow \infty$, $0\leq x\leq1$. Note that the functions $f_n$ are not necessarily continuous.

This is one of the preliminary exam from UC Berkeley, the solution goes like this:

Because $f$ is continuous on $[0,1]$, which is compact, it is then uniformly continuous. Hence there exists $\delta >0$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$.

We then partition the interval with $x_0=0, \cdots ,x_m=1$ such that the distance $x_{i}-x_{i-1}$ is less than $\delta$.

Note that since there are only finite number of $x_m$, there is $N\in \mathbb{N}$ such that if $n\geq N$ then $|f_n(x_i)-f(x_i)|<\epsilon$ where $i=0,\cdots, m$

Now if $x\in[0,1]$, then $x\in[x_{i-1},x_i]$ for some $i\in\{1, \cdots m\}$.

My question is how to use the nondecreasingness to arrived at this inequality, for $n\geq N$


Can someone please help, I have been staring at the inequality for about a day now. Thanks.


BenjaLim has already dealt with the first half of the inequality. For the second half, note

$$f_n(x)-f(x_{i-1})\leq f_n(x_i)-f(x_{i-1})= (f_n(x_i)-f(x_i))+(f(x_i)-f(x_{i-1}))<\epsilon+\epsilon=2\epsilon.$$

  • $\begingroup$ Thanks bobobinks and BenjaLim !!! $\endgroup$ – KWO Jun 29 '12 at 1:28
  • $\begingroup$ @bobobinks +1 I did not see that. $\endgroup$ – user38268 Jun 29 '12 at 1:30
  • $\begingroup$ Thanks everyone!!! I did not see that too. How to "declare" that the problem has been solved? $\endgroup$ – KWO Jun 29 '12 at 1:31
  • $\begingroup$ Thanks all, I am logging off. Bye $\endgroup$ – KWO Jun 29 '12 at 1:37
  • $\begingroup$ @KWO I have shown you how to finish off the problem, please see my edit below. $\endgroup$ – user38268 Jun 29 '12 at 1:43

For the first part of the inequality we know that there is a natural number $N$ such that

$$ n\geq N \implies |f_n(x_{i-1}) - f(x_{i-1})| < \epsilon$$

Opening up the absolutely value signs, this means in particular that for $n \geq N$,

$$-\epsilon + f(x_{i-1}) < f_n(x_{i-1}).$$

Now suppose that $x \in [x_{i-1},x_i]$. Then this means that $x_{i-1} \leq x$ and because each $f_n$ is non-decreasing, for all $f_n$ such that $n$ is sufficiently large we have that

$$f_n(x_{i-1}) \leq f_n(x)$$

and so using the first inequality I obtained, we have that

$$f(x_{i-1}) - \epsilon < f_n(x).$$

Edit: Thanks to bobokinks we can now finish the problem. From you inequality we get that there is $N \in \Bbb{N}$ such that for all $n \geq N$, we have

$$|f_n(x) - f(x_{i-1})| < 2\epsilon.$$

I was an idiot before and proved nonsense. Here is the correct version: We have $$\begin{eqnarray*} |f_n(x) - f(x)| &<& |f_n(x) - f(x_{i-1})| + |f(x_{i-1}) - f(x)| \\ &<& 2\epsilon + \epsilon\\ &=& 3\epsilon. \end{eqnarray*}$$

The latter quantity being less than $\epsilon$ comes from uniform continuity of $f$ on $[0,1]$ and the fact that we chose each $[x_{i-1},x_i]$ to be of width less than $\delta$. Since $\epsilon > 0$ was arbitrary, we have proven that for any $x \in [0,1]$, as long as $n \geq N$ we have $f_n \longrightarrow f$ uniformly.

  • $\begingroup$ I got the first part too, is the second part that trouble me. $\endgroup$ – KWO Jun 29 '12 at 1:00
  • 1
    $\begingroup$ @KWO You did not state which part you did not get, so I just thought that you did not get any part of the inequality at all. $\endgroup$ – user38268 Jun 29 '12 at 1:02
  • $\begingroup$ any thought about the second? $\endgroup$ – KWO Jun 29 '12 at 1:02
  • $\begingroup$ @KWO Btw I think your inequality only holds when $n \geq N$. $\endgroup$ – user38268 Jun 29 '12 at 1:02
  • $\begingroup$ @KWO I am working on that now. $\endgroup$ – user38268 Jun 29 '12 at 1:03

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