Let $S$ be the set of all rational numbers on $[0,1]$, $P=\{ \frac{1}{p}: p\geq 2\}$, $T=\{0,1\}$, and $S\backslash(P\cup T)=\{ r_1,r_2,\dots\}$.
Let $f_m:[0,1]\to [0,1]$, for all $m\in\mathbb{N}$, and $f:[0,1]\to [0,1]$ are functions that defined by
\begin{align*}
f_m(x)=\begin{cases}
\frac{1}{n+1},\quad x=r_n,n\leq m,\\
0,\quad x=0,\\
1,\quad others.
\end{cases}
\end{align*}
and
\begin{align*}
f(x)=\begin{cases}
\frac{1}{n+1},\quad x=r_n,n=1,2,3,\dots,\\
1,\quad x=0,\\
0,\quad others.
\end{cases}
\end{align*}
I want to check whether the sequence $(f_m^2)$ converges uniformly to $f^2$ or not. Where $f_m^2=f_m\circ f_m$ and $f^2=f\circ f$.
I try to prove that using the proposition from "An Introduction to Modern Analysis" by Vicente Montesinos, et.al.
And this is what I did: Take any $x\in [0,1]$.
- For $x=0$, I have $|f_m^2(x)-f^2(x)|=|0-0|=0$.
- For $x\in\{r_1,r_2,\dots,r_m\}$, I have $|f_m^2(x)-f^2(x)|=|0-0|=0$.
- For $x\in\{r_{m+1},r_{m+2},\dots\}$, I have $|f_m^2(x)-f^2(x)|=|1-0|=1$.
- For the others, I have $|f_m^2(x)-f^2(x)|=|1-1|=0$.
Then, $\sigma_m=\sup\{|f_m^2(x)-f^2(x)|:x\in[0,1]\}=\sup\{0,1\}=1$, such that $\lim\limits_{m\to\infty} \sigma_m=1\neq 0$. Thus, $(f_m^2)$ does not converge uniformly to $f^2$.
Is that correct?
Since the $f_m^2$ and $f^2$ are iterated functions, so I am not sure whether I can use Proposition 466 or not.
Thanks for any help.