I am self-learning Real-Analysis. I'd like a clue or hint on how to proceed with this exercise problem from Stephen Abbott's Understanding Analysis without revealing the entire proof.

[Abbott, 6.2.10] This exercise and the next explore partial converses of the Continuous Limit Theorem (Theorem 6.2.6). Assume that $f_n \to f$ pointwise on $[a,b]$ and the limit function $f$ is continuous on $[a,b]$. If each $f_n$ is increasing (but not necessarily continuous), show that $f_n \to f$ uniformly.

Proof Attempt.

I tried a direct proof, but I am not exactly sure how to use the fact that each $f_n$ is increasing.

I tried a proof with contradiction.

We are given that, $f_n \to f$ pointwise on $[a,b]$ and the limit function $f$ is continuous on $[a,b]$. Further each $f_n$ is increasing. We proceed by contradiction.

Assume that $f_n$ does not converge uniformly to $f$. Thus,

\begin{align*} (\exists \epsilon_0 > 0), (\forall N \in \mathbf{N}) : (\exists n_0 \geq N),(\exists x_0 \in [a,b]) : |f_n(x_0) - f(x_0)| \geq \epsilon_0 \end{align*}

Since, $f$ is continuous on the compact set $[a,b]$, it is uniformly continuous on $[a,b]$. Now, I am not sure, how to proceed.

Any hints would be super-helpful guys.

  • 3
    $\begingroup$ Yes this one is rather known as “false Dini theorem”. It is in fact a result due to Pólya. See the French Wikipedia page about this. $\endgroup$
    – nejimban
    Jan 14, 2022 at 16:36
  • 1
    $\begingroup$ Asked and answered many times on this site — for example, here. $\endgroup$
    – RRL
    Jan 14, 2022 at 16:57
  • 1
    $\begingroup$ But the OP sates clearly that he/she want hints, not the full solution so it's no surprise that he/she is not googling it $\endgroup$ Jan 15, 2022 at 0:07

1 Answer 1


Hints: justify and apply the following claims:

$1).\ f$ is increasing.

$2).\ $ There is a partition $\{a,x_1,\cdots x_{n-2},b\}$ such that $f(x_i)-f(x_{i-1})<\epsilon;\ 0\le i\le n.$

$3).\ $ if $x\in [a,b]$ then $x_i\le x\le x_{i-1}$ for some $0\le i\le n$ and $f_n(x_{i-1})-f(x_{i-1})-\epsilon\le f_n(x)-f(x)\le f_n(x_i)-f(x_i)+\epsilon.$

$4).\ $ There is an integer $N$ such that $|f_n(x_i)-f(x_i)|<\epsilon$ for each $0\le i\le n$ and $n>N.$


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